Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Long Question And Answers

Question 1. Depict the bonding in the following compounds In terms of atomic orbitals involved and predict all the bond angles:

1. CD₃CH=CH₂
2. CH₃OCH₃

Answer:

1. The 3 carbon atoms are sp³ , sp² and sp² -hybridised respectively. Therefore the bond angles about these carbons are 109.5°, 120° and 120° respectively corresponding to tetrahedral and trigonal planar geometries.

2. The carbon and oxygen atoms are all sp³ -hybridized. So, the bond angles are nearly 109.5° corresponding to tetrahedral geometry.

Class 11 Organic Chemistry

Question 2. Mention the number of primary (1°), secondary (2°) and tertiary (3°) hydrogen atoms in the following

Answer:

Hydrogen Atoms Given in the table:

Question 3. How many alkyl groups can be derived from the alkane, (CH₃)₂ CHCH₂ CH(CH₃)₂ and why? Write their IUPAC names
Answer:

Since this hydrocarbon molecule contains 3 types of nonequivalent hydrogen atoms, the removal of these hydrogen atoms gives 3 different alkyl groups.

These are as follows:

Question 4. Write the IUPAC names of the following compounds

Class 11 Organic Chemistry

Answer:

1. 3-ethyl-4-methylhept-5-en-2-one
2. 3,3,5-trlmethylhex-1 -en-2-ol.
3. l-bromo-4-metlvylheptan-3-on«.
4. l-etliyl-4-methylcyclohexane.
5. Cyclohexylcyclohexnne.
6. 1,3-dlcyclopropylpropanc.
7. 2-metliyl-2-cyclopropylpropnne.
8. N -ethyl- N -methylpropan-2-nmine.
9. 4-hydroxy-4-methylpontan-2-one.
10. 3-methylpent-l-ene.

Question 5. Arrange the given carbocations In order to increase stability and explain two-order

Answer:

The order of increasing stability of these carbocation Is :

Being an aromatic one [(4 n + 2)n -electron system, where n = 1], the carbocation (I) is the most stable. The carbocation (II) is effectively resonance stabilized. So, its stability is greater than that of (ill) and (IV) (which are not resonance-stabilized) but less than that of (I).

The carbocation (III) is stabilized by +1 and the hyperconjugation effect of the methyl group and its stability is less than (II). The carbocation (IV) is destabilized by the stronger -I effect of the — CF₃ group, so it is the least stable one.

Class 11 Organic Chemistry

Question 6. Terf-Butyl chloride (Me₃CCl) does not participate in D+ S_N2 reaction—explain with reasons
Answer:.

Due to severe steric hindrance caused by three methyl groups, the backside attack on the central carbon by the nucleophile becomes completely inhibited and it is for this reason, that terf-butyl chloride does not participate in the S_N2 reaction

Question 7. Write the resonance structures of CH₂=CH —CHO and compare their stabilities.
Answer:

The compound is a resonance hybrid of three structures:

The stability order of these structures is: I > II > III.

The uncharged structure (I) is the most stable one. The charged structure (II) is moderately stable because the more electronegative oxygen atom bears the negative charge and the less electronegative carbon atom bears the positive charge.

Also, the octet of carbon is not filled up. The charged structure (III) is the least stable because the more electronegative O-atom bears the positive charge and the less electronegative C-atom bears the negative charge. Also, the octet of the O-atom is not filled up

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Question 8. Which is more stable and why: (CH₃)₃C, (CD₃)₃C
Answer:

Since D is more electron-releasing than H, — CD₃ is more electron-releasing than — CH₃. So, (CD₃)₃C⁺ is expected to be more stable than (CH₃)₃C⁺. But actually, the is true and this can be explained in terms of hyperconjugation. Since the C—H bond is weaker than the C— D bond the hyper conjugative stability of (CH₃)₃C⁺ is greater than thatof (CD₃)₃C⁺

Class 11 Organic Chemistry

Question 9.

1. How many stereoisomers of formula, CH₃ would be possible if methane was a pyramid with a rectangular base? Draw them.
2. How many stereoisomers of formula, CH₂YZ would be possible if methane was a pyramid with a square base? Draw them.
3. What is the relationship (diastereoisomers, enantiomers, conformational isomers, homomers i.e., identical structures or constitutional isomers) between the members of given pairs of structures?

Answer:

1. Two stereoisomers Mirror (enantiomers) are Q possible if methane H / was a pyramid with a rectangular base.

2. Three stereoisomers (I, II, and III) are possible if methane was a pyramid with a square base. (I and II) are enantiomers. (I and III) and (II and III) are two pairs of diastereoisomers.

Class 11 Organic Chemistry

3.

1. . Diastereoisomers
2. Homomers
3. Geo¬ metrical isomers
4. Constitutional isomers
5. Conformational isomers
6. Enantiomers.

Question 10. Although Quorine is more electronegative than iQfi chlorine, fluorobenzene has a lower dipole moment (p = 1.63D ) than chlorobenzene (μ = 1.75D ).
Ans.

Both fluorine in fluorobenzene and chlorine in structure) chlorobenzene withdraws electrons from the ring by -I effect and donate electrons to the ring by the + R effect. Because of the smaller size of fluorine, the +R effect involving orbitals of similar sizes (2p of both F and C) is much stronger. So, the moment due to the stronger -I effect of fluorine is considerably neutralized by the moment due to +R effect.

Hence, fluorobenzene possesses a net dipole moment which is relatively low (1.63D). On the other hand, because of the larger size of chlorine, the + R effect involving orbitals of dissimilar sizes (3p of Cl and 2p of C) is much weaker than the -I effect which is somewhat lower due to lower electronegativity of chlorine. So, the moment due to +R effect is much smaller than the moment due to -I effect. Hence, chlorobenzene possesses a relatively high net dipole moment (1.75D

Class 11 Organic Chemistry

Question 11. The negatively charged carbon atom in the structure

1. Is sp² -hybridized while the negatively charged carbon atom in
2. Is sp³ -hybridised —Explain.

Answer:

The negatively charged carbon atom of a resonance-stabilized carbanion is sp² -hybridized. The carbanion (I) is resonance stabilized. So the negative carbon atom Is sp².
hybridized

On the other hand, the carbanion (II) Is not resonance stabilized because a double bond cannot be formed at the bridgehead position of small bicycle systems (Hrodt’s rule). Hence, the negatively charged carbon of the carbanion (II) is sp³ -hybridized

Question 12. Mention the state of hybridisation of the starred (*) carbon atoms in each of the following compounds.

Answer:

1. sp²
2. sp
3. sp²
4. sp
5. sp
6. sp³

Question 13. How many σ and π -bonds are present in each of the following molecules?

1. CH₃-C≡-CH= CH₂

2. CH₂=CH-CH=C=CHCH₃ 

3. 

Answer:

1. σ -bonds =10, π -bonds =3
2. σ -bonds = 13 , π -bonds = 3
3. σ -bonds =10, π -bonds = 3

Class 11 Organic Chemistry

Question 14.  Which atoms in each of the following molecules remain in the same plane and why?

1.  CH₃CH= CH₃
2.  C₆H₅C≡ CCH₃
3. CH₃CH=C=C=CHCH₃
4. CH₃COCH₂CH₃

Answer:

sp²-carbon atoms and the atoms attached to them lie in one plane.

sp² -carbon atoms and the atoms attached to them lie in one plane. Also, sp carbon atoms & the atoms attached to them lie not only in one plane but also in one line.

Lie in one plane

Lie in one plane

Question 15. Mention the number of primary (1°), secondary (2°), tertiary (3°), and quaternary (4°) C -atoms present in the given molecules: °
Answer:

Answer:

Question 16. Write down the IUPAC name of a hydrocarbon having a 4° C-atom with molecular formula, C₆H₁₄ . How many monochrome derivatives of this hydrocarbon is possible? Write their structures
Answer:

The hydrocarbon corresponding to the molecularformula C6H14 and containing one tertiary carbon atom is CH₃C(CH₃)₂CH₂CH₃ (C-2 is a quaternary carbon atom). Its IUPAC name is 2,2-dimethylbutane.

Since the alkane contains three types of non-equivalent hydrogen atoms, three monobromo derivatives of the alkane are possible. These are: (CH₃)₃CCHBrCH₃ and (CH₃)₃CCH₂CH₂Br

Question 17. Racemic tartaric acid and meso-tartaric add are both optically inactive—why?
Answer:

Racemic tartaric add is an equimolar mixture of (+) and (-)-tartaric adds. In racemic tartaric add, therefore, the rotatory power of (+) enantiomer is neutralised by the rotatory power of the (-) enantiomer (external compensation) and for this reason, the racemic tartaric add is optically inactive.

On the other hand, the meso-tartaric add is optically inactive because it has a plane of symmetry and it is superimposable on its mirror image. In fact, in this case, the optical rotation of one half of the molecule is exactly canceled by the optical rotation of the other half (internal compensation).

Class 11 Organic Chemistry

Question 18. How many isomers of butene are possible? What type 1 of isomerism do they exhibit?
Ans.

Three structural isomers of butene are possible: CH₃CH₂CH=CH₂ (But-l-ene), CH₃CH=CHCH₃ (But-2- ene), and (CH₃)₂C=CH₂ (2-methylpropene). Again, but-2- ene exists as two geometrical isomers (diastereoisomers):

Hence, there are in total 4 isomers of butene: but-1-ene,  cis but-2-ene, irans-but-2-ene, 2-methylpropene

Question 19. Give examples of

1. An optically inactive compound containing an asymmetric carbon atom
2. An optically active compound containing no asymmetric carbon.

Answer:

1. Meso-tartaric acid containing two asymmetric carbon atoms is optically inactive because it has a plane of symmetry, l.e., tire molecule is superimposable on its mirror Image

2. Penta-2,3-diene (an abC=C=Cab type of allene) is optically active because it is not superimposable on its mirror image.

Question 20. Name a compound having two similar asymmetric carbon atoms and give its structure. What type of isomerism docs it exhibit? Draw Fischer projection formulas of these isomers and comment on their optical activity. How are they related to each other?
Answer:

Tartaric acid has two similar asymmetric carbon atoms (HOOC — *CHOH —*CHOH —COOH). The compound exhibits optical isomerism.

Fisher projection formulas of its isomers are as follows:

Class 11 Organic Chemistry

The relations among the isomers are as follows: I and II are enantiomers; I and III are diastereoisomers and II and III are diastereoisomers

Question 21.

1. Give the structure and IUPAC name of an optically active alkane having the lowest molecular mass. Is there another alkane of the same molecular mass that is also optically active?
2. Give an example of a compound that exhibits both optical & geometrical isomerism

Answer:

Such a compound must contain an asymmetric carbon atom which will remain attached to a H-atom and three different alkyl groups (smaller size).

So, the optically active alkane having the lowest molecular mass’ is, 3-methylhexane [CH₃CH₂  *CH(CH₃)CH₂CH₂CH₃ ]. Another optically active alkane with the same molecular mass is 2,3-dimethyl pentane [CH₃*CH₂CH(CH₃)CH(CH₃)₂] which is a chain isomer of the first one.

Pent-3-en-2-oI [CH₃*CH(OH)CH=CHCH₃] exhibits both geometrical and optical isomerism because the compound contains an asymmetric carbon atom and each of the doubly bonded carbon atoms is attached to two different groups.

Question 22. The following two isomers may be called diastereoisomers but not enantiomers —why? Explain why these are optically inactive

Answer:

The given pair of isomers have the same structure but different configurations. They are neither superimposable nor bear mirror image relationships with each other. So they are related to each other as a pair of diastereoisomers and not as enantiomers. Each of these two isomers has a plane of symmetry, i.e., each of them is superimposable on its mirror image and so, these are optically inactive.

Question 23. p nitrophenol is more acidic than phenol.
Answer:

In p-nitrophenol, the electron-attracting — NO₂ group by its stronger -R effect and relatively weaker -I effect makes the oxygen atom relatively more positively polarised compared to the oxygen atom of phenol. As a result, the O — H bond in nitrophenol dissociates more easily to give H⁺ ions. For this reason, p-nitrophenol is more acidic than phenol

Class 11 Organic Chemistry
Answer:

In p-nitroaniiine, the electron-attracting — NO₂ group by its stronger -R effect and relatively weaker -I effect makes the nitrogen atom of the — NH₂ group relatively more positive compared to the nitrogen atom of aniline. As a result, the availability of the unshared pair of electrons on nitrogen atom in p-nitroaniline is highly reduced as compared to the unshared electron pair on nitrogen in aniline. For this reason, p-nitroaniline behaves as a weaker base compared to aniline.

Question 24. Dipole moment of vinyl chloride CH₂=CHCI) is less than the dipole moment of ethyl chloride (CH₃CH₂Cl) —explain.
Answer:

In vinyl chloride, the moment caused by the -I effect of Cl-atom (μσ) is partially neutralized by the moment caused by its +R effect (pn). As a result, the value of net moment of vinyl chloride decreases and it is lower than that of ethyl chloride in which only the stronger -I effect of chlorine operates.

Question 25. Arrange the following ions in order of increasing basicity and explain the order

1. CH₃ ^–CH(I)
2. CH ≡^–C(II)
3. CH₂ =^–CH(III)

Answer:

The order  of increasing basicity of the given ions is:

⇒ \(\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}(\mathrm{II})<\mathrm{CH}_2=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}(\mathrm{III})<\mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2(\mathrm{I})\)

C -atoms bearing the negative charge in carbanions (I), (II) & (III) are sp³, sp and sp² -hybridized respectively. Percentages of s -character of these three hybrid orbitals are 25%, 50% and 33% respectively.

As the s -character of hybrid orbital increases, C -atoms bearing the negative charge in carbanions (I), (II) & (III) are sp³, sp and sp² -hybridized respectively. Percentages of s -character of these three hybrid orbitals are 25%, 50% and 33% respectively. As the s -character of hybrid orbital increases,

Question 26. Give example:

1. A non-nucleophiiic anion
2. A planar carbocation
3. An aromatic carbocation
4. An aromatic carbanion
5. A reagent which acts as a source of carbanion
6. A reaction that does not proceed through intermediate
7. An aprotic polar solvent
8. An ambident nucleophile
9. A neutral electrophile
10. A group which stabilizes a carbocation
11. A group which stabilizes a carbanion
12. An alkyl group which does not supply electrons to a double bond by hyperconjugation
13. A carbocation that can be stored for years.

Answer:

1. BF^–₄

2. Benzyl cation

3. Cyclopropenvl cation

4. Cyclopentadienyl anion

5. Grignard reagents \(\left(\mathrm{R}^{\delta-}-\mathrm{M}^{\delta+} \mathrm{gX}\right)\)

6. S_N2 reaction

7. Dimethyl formamide [DMP, Me₂NCHO]

8. ^–CN \((: \stackrel{\ominus}{\mathrm{C}} \equiv \mathrm{N}: \longleftrightarrow: C=\stackrel{\ominus}{\mathrm{N}}:)\)

[Nucleophiles having two or more available sites for nucleophilic attack are called ambident nucleophiles)

9. Dichlorocarbene (:CCl₂)

10. \(-\ddot{O}:\mathrm{CH}_3\)

11. —NO₂

12. —C(CH₃)₃

13. Triphenylmethylfluoroborate Ph₃⁺CBF₄^–

Question 27. Explain the given basicity order in aqueous medium: )₂NH(2°) > CH₃NH₂(1°) > (CH₃)₃N(3°)
Answer:

The basic strength of amines in the aqueous medium depends on two factors:

Increased electron density on the N-atom makes an amine more basic.

So considering the +1 effect of different numbers of alkyl groups on the N -atom, the basic strength of amines should follow the order:

Again, basicity of an amine increases as stabilization of the conjugate acid, through solvation, increases. The conjugate acid of primary amine attains maximum stability through intermolecular H -bond formation with three molecules of water, while the conjugate base of tertiary amine attains minimum stabilisation through such H-bond formation with only one molecule of water.

Class 11 Organic Chemistry

Thus on the basis of stability of the conjugate acids, the basic strength of amines should follow the order: CH₃—NH₂ > (CH₃)₂NH > (CH₃)₃N As a consequence of these two opposite orders of basicity practically we find the given sequence of basicity in aqueous medium: (CH₂)₂NH(2°) > CH₃NH₂(1°) > (CH₃)₃N(3°)

Question 28. Which of the two: 2NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?
Answer:

O₃ NCH₂CH₃O^– is expected to be more stable than CH₃CH₂O^–. —NO₂ group by strong-I effect disperses the negative charge on O-atom in   ion and stabilises it. On the other hand, the CH₃CH₂— group by its +1 effect tends to intensify the negative charge on oxygen atom in and hence destabilizes it.

Question 29. CH₃Cl undergoes hydrolysis more easily than C₆H₅Cl. Explain.
Answer:

Unshared electron-pair on chlorine atom in chloro-benzene becomes involved in resonance interaction with the n -electrons of a benzene ring. As a result, the C — Cl bond assumes some double bond character. Thus, the C — Cl bond becomes much stronger and so the displacement of chlorine atom from the ring becomes difficult, i.e., the compound does not undergo hydrolysis easily.

On the other hand, the C —Cl bend in CH₃— Cl gets no opportunity to assume a double bond character. So r undergoes hydrolysis readily under ordinary conditions.

Question 30. Benzyl chloride participates in S_N1 eaction even though it is a primary (1°) substrate. Explain.
Answer:

Carbocation produced from benzyl chloride in first step (rate-determining step) of S_N1 reaction is stabilized by resonance. Thus, benzyl chloride participates in S_N1 reaction even though it is a primary (1 °) alkyl halide.

Class 11 Organic Chemistry

Question 31. Bond Dissociation enthalpy of C₆H₅CH₂ – H bond is much less than CH₂-H bond Explain
Answer:

Benzyl radical (C₆H5₅C^•H₂) produced by homolytic cleavage of C_sp³ —H bond of toluene is considerably stabilised by resonance. But, the stability of methyl radical (^•CH₃, obtained by homoiytic fission of the C—H bond of methane Is not stabilized by any factor and in fact, it is very much unstable. Thus, the C—H bond dissociation enthalpy of toluene is much less than the C —H bond dissociation enthalpy of methane. ,

Question 32. N,N, 2,6-Tetramethylaniline is more basic than N- dimethylaniline. Explain.
Answer:

Because of steric interaction involving two ortho-methyl groups and two methyl groups attached to nitrogen, the unshared electron-pair on N is not involved in resonance interaction with n -electrons (steric inhibition of resonance) o: ring. To avoid steric strain, the — NMe₂ group rotates about C — N bond axis and thereby loses coplanarity with the ring. As a result, nitrogen can easily donate its unshared electron pair to a proton.

On the other bund, no such steric inhibition occurs in N- dimethylaniline  because the two ortho H-atoms are relatively much smaller in size. The unshared electron-pair on N-atom become involved in resonance interaction with the ring and therefore, is not fully available for taking up a proton. This explains why N, N ,2, O-tetramethylaniline is more basic than N, N -dimethylaniline.

Class 11 Organic Chemistry

Question 33. Chloroform is more acidic than fluoroform. Explain.
Answer:

CF^–₃  the conjugate base of fluoroform (CHF₃ ), is stabilized by -I effect of 3 F-atoms. But CCl^–₃, the conjugate base of chloroform (CHCl₃ ), is relatively more stabilized by somewhat weaker -I effect of 3 Cl-atoms along with rf-orbital resonance (Cl has vacant rf-orbital). So chloroform is more acidic than fluoroform.

Question 34. How can you separate benzoic acid and nitrobenzene from their mixture by the technique of extraction using an appropriate chemical reagent?
Answer:

The mixture is shaken with a dilute sodium bicarbonate solution when benzoic acid gets converted to sodium benzoate and dissolves in water leaving nitrobenzene behind. The mixture is extracted with ether or chloroform when nitrobenzene goes into the organic layer. After separating the organic layer, it is distilled to get nitrobenzene. The aqueous layer is acidified with dilute HC1 when benzoic acid gets precipitated. It is obtained by filtration.

C₆H₅COOH + NaHCO₃→C₆H₅COONa(Sodium benzoate (Soluble)) + CO₂ + H₂O

Question 35.

1. Which atoms in a toluene molecule always remain in the same plane and why?
2. Which atoms in a propyne molecule remain in a straight line and why?

Answer:

1. An sp² -a hybridized carbon atom and the atoms directly attached to it always remain in the same plane. Therefore, in toluene ), all the atoms except 3 H-atoms of methyl group {i.e, seven C – and five H -atoms) remain in the same plane.

2. An sp -hybridized carbon atom and the atoms directly attached to it remain in a straight line. Therefore, in the propyne molecule (CH₃—C = CH), all the atoms except the 3 hydrogen atoms of the methyl group remain in the same straight line.

Question 36. Write the state of hybridization of C -atoms in the following compounds and predict the shape of each of the molecules :

1. H₂C=O
2. CH₂Cl
3. HC = N
4. CH₂=C=CH₂
5. CH₂=C=C=CH₂

Answer:

1. sp² -hybridised C-atom, trigonal planar;
2. sp² hybridised C-atom, tetrahedral;
3. sp -hybridised C-atom, linear;
4. sp² , sp and sp² -hybridised C -atoms respectively, elongated tetrahedron;
5. sp², sp , sp and sp² -hybridised C atom respectively, planar

Question 37.

1. Expand each of the following condensed formulas into their complete structural formulas:

1. HOCH₂CH₂NH₂
2. CH₃CH=CHCOCH₃
3. CH₃C ≡ CCH₂COOH

2. Write bond-line formulas of the following two compounds

1. CH₃CH₂CH₂CH₂CHBrCH₂CHO
2. (C₂H₅)₂CHCH₂OH

Answer:

1. Condensed Formulas:

2. Bond Line Formulas:

Question 38.

1. Expand each of the following bond-line formulas to show all the atoms including C and H.

2. How many  σ and π -bonds are present in

1. CH₂=CH—CN and
2. CH₂=C=CHCH₃

Answer:

1. Bond-line formulas:

Class 11 Organic Chemistry

2. σ and π -bonds:

1. σ _{C -H} = 3: σ _{C -C} = 2: π _{C= C} = 1: σ _{C – N} = 1

π _{C= N}= 2 i.e., total σ -bond = 6 and total  π -bond = 3

2.  1. σ _{C -H} = 6: σ _{C -C} = 3: π _{C= C} = 2:

i.e total σ – bond = 9 and total π – bond = 2

Question 39. Which of the given compounds may exist as two or more isomeric forms? Give the structures and names of the possible isomers.

1. CHBr₃
2. C₂H₂Cl₄
3. C₃H₈
4. C₂H₅F
5. C₂H₄Br₂
6. C₆H₄Cl₂

Answer:

1. No isomer is possible

2. Two isomers are possible : ClCH₂CCl₃ (1,1,1,2-tetrachloroethane), Cl₂CHCHCl₂ (1, 2,2-tetrachloroethane)

3. No isomer is possible

4. No isomer is possible;

5. Two isomers are possible : BrCH₂CH₂Br (1, 2- dibromoethane), CH₃CHBr₂ (1,1-dibromoethane)

6. Three isomers are possible:

Question 40. Write the structures and IUPAC names of the H—C—C=C-C—C-OH compounds with the molecular formula, C₄H₈O₂. H (in) H
Answer:

The molecular formula, C₄H₈O₂conforms to the general formula of monocarboxylic acids and esters. Therefore the following structures of monocarboxylic acids and esters can be written with the given formula

Question 41. Write the structures and IUPAC names of the compounds with molecular formula, C₄H₁₀.
Answer:

The molecular formula, C₄H₁₀ conforms to the general formula of monohydric alcohols and ethers. Therefore, the structures of the following monohydric alcohols and ethers can be written with the given molecular formula

Class 11 Organic Chemistry

Question 42. Which of the following compounds will exhibit tautomerism and which do not? Give reasons.

1. CH₃COCH₃
2. C₆H₅COC₆H₅
3.  C₆H₅COCH₃
4. C₆H₅CHO
5. Me₃CCOCMe₃

Answer:

Tautomerism is possible for those carbonyl compounds which contain at least one α-H atom (the H-atom attached to a carbon atom adjacent to the C= O group) Therefore, compounds 1  and 3 containing a-H atom exhibit tautomerism while compounds 2, 4 and 5 containing no α -H atom do not exhibit tautomerism

Question 43. Designate the following pairs as metamers, chain isomers, position isomers, functional isomers, and stereoisomers. Also, indicate which are not isomers at all
Answer:

1. (CH₃)₂CHC(CH₃)₃, (CH₃)₄C

2.  CH₃CH₂CH₂OH, CH₃OCH₂CH₃ CH₃

3.

4. (CH₃)CHCOCH₃, (CH₃)₂CHCH₂CHO COOH

5. CH₃OCH₂CH₂CH₃, CH₃CH₂OCH₂CH₃

6.

Answer:

1. The molecular formulas of these two compounds are not identical. Thus, these two are not isomers
2. Functional isomers
3. Position isomers
4. Functional isomers
5. Metamers
6. Stereoisomers (geometrical isomers)

Question 44. Which of the following compounds will exhibit geometrical isomerism and why?

Answer:

1. One of the doubly bonded carbon atoms is attached to two identical atoms (Cl). Therefore, the compound will not exhibit geometrical isomerism.
2. Each doubly bonded C-atom is attached to two different groups (C-2 is attached to CH3 and H, while C-3 is attached with Cl and C₂H₅ ). So, it will exhibit geometrical isomerism.
3. The compound will not exhibit geometrical isomerism because each of the two terminal doubly bonded carbon is attached to two identical atoms (H).
4. The compound will exhibit geometrical isomerism because each of the two ring carbons is attached to two different groups (H and CH₃).
5. The compound will exhibit geometrical isomerism because each of the two ring carbons is attached to two different groups (H and CH₃ ).
6. The compound will not exhibit geometrical isomerism because one of the two doubly bonded carbon atoms is attached to two identical groups (ring moiety

Question 45. Which of the following compounds are optically active and why?

Class 11 Organic Chemistry

Answer:

1. The compound contains one asymmetric C-atom (CH₃CHBrCH₂CH₃) . So, it is not superimposable on its mirror image and hence, it is optically active.
2. The molecule has a plane of symmetry and it is superimposable on its mirror image. Therefore, it is optically inactive.
3. The molecule is not superimposable on its mirror image. So, it is optically active
4. The molecule is not superimposable on its mirror image. So, it is optically active.
5. This planar compound is superimposable on Its mirror image. So, it is optically inactive.

Question 46. Which type of stereoisomerism is exhibited by the compound, CH₃CH=CH —CH=CHC₂H₅? How many stereoisomers are possible? Draw the structures and designate them as E/Z
Answer:

The compound exhibits geometrical isomerism because the groups attached to each of the terminal doubly bonded carbon atoms are different. The number of geometrical isomers = 2ⁿ (n = number of double bonds) =2² = 4. These are as follows:

Question 47. Name a compound having; two dissimilar asymmetric carbon atoms and write Its structure. What type of isomerism does It exhibit? Draw Fischer projection formulas of the Isomers and comment on their optical activity. How are (lie Isomers related to each other? m Explain the orders of acidity of carboxylic acids:
Answer:

Compound containing two dissimilar asymmetric carbon atoms  \(\left(\mathrm{CH}_3 \stackrel{*}{\mathrm{C}} \mathrm{HOH} \stackrel{*}{\mathrm{C}} \mathrm{HBr} \mathrm{CH}_3\right)\).

The compound may have 2ⁿ(n = no. of dissimilar asymmetric carbon atom)2² = 4 possible stereoisomers. projection formulas of these Isomers are as follows

Class 11 Organic Chemistry

The relations among the isomers are as follows: (I, II) and (III, IV) are two pair of enantiomers whereas (I, III), (I, IV), (II, III), and (II, IV) are four pairs of diastereomer.

Question 48. Arrange cis-but-2-ene, trans-but-2-ene, and but-I-ene H3C in increasing order of their stability and give reason.
Answer:

But-2 -ene (CH₃CH=CHCH₃) contains sixa-H atoms while but-l-ene (CH₃CH₂CH=CH₂) contains only two o-H -atoms capable of participating in hyperconjugation. Therefore, because of more effective hyperconjugation, but-2- ene is thermodynamically more stable than but-l-ene.

Again, due to steric interaction between two methyl groups on the same side of the double bond in cis-but-2-ene, it is relatively less stable than the trans-isomer in which no such steric interaction operates between the methyl groups situated on the opposite sides of the double bond.

Question 49. CH₃Cl is unreactive towards S_N1 reaction—why?
Answer:

Stability of the carbocation obtained in the first step (rate¬determining step) of an S_N1 reaction determines whether the reaction will lake place or not. Since methyl cation [⁺ CH₃ ] is a very unstable one, methyl chloride is unreactive toward S_N1 reaction

Question 50. Explain the orders of acidity of carboxylic acids

1. Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
2. CH₃CI₂COOH>(CH₃)₂CHCOOH >(CH₃)₃CCOOH

Answer:

1. -I effect explains this order of acid strength. As the number of halogen atoms on the a -carbon decreases, the overall -I effect decreases and as a consequence, the acid strength decreases

2. + 1 effect explains the given order of acid strength. As the number of methyl groups attached to the a -carbon atom increases, the overall +1 effect increases and consequently, the acid strength decreases

Question 51. Explain why an organic liquid vaporizes below its boiling point when it undergoes steam distillation.
Answer:

In steam distillation, sum of the vapour pressures of water and organic liquid become equal to the atmospheric pressure. This means that both of them distill at a pressure much lower than the atmospheric pressure, i.e., both of them will vapourise at a temperature that is less than their normal boiling points.

Question 52. 

1. Write the state of hybridization of C -atoms mentioned In each of the following compounds:

1. C-4 of Pcnt-l-cn-4-yne
2. C-l of I’ropanoic acid
3. C-3 of Penta-2,3-dienc,
4. C-3 of Pcntan-3-one and
5. C. -3 of 3,3-dietliylpcntane

2. Which atoms of each of the following molecules/ions always remain in the same plane?

1. CH₃CH = CHCH₃

2. C₆H₅C ≡ C—CN

3. C₆H₅CH₃

4. CH₂=C=C=CH₂

5. CH₃COCH₃

6. CH₃CONH₂

7. Cl₃ C —CH=CH—^–CH₂

8. (CD₃)₃C⁺

^{11. –}CH₂COCH₂CH₃

13. (CH₃)₂⁺CH — NH₂

Answer:

1. If four valencies ofcarbon atom are satisfied by four single bonds, then it is sp³ -hybridised. Iffour valencies are satisfied . by one double bond and two single bonds, then it is sp² -hybridised. Iffour valencies are satisfied by one triple bond and one single bond or by two double bonds, then it is sp -hybridised.

Class 11 Organic Chemistry

2. An sp² -hybridized C -atom and the atoms directly attached to it remain in the same plane. Again, an sp -sp-hybridized C atom along with die atoms with which it is directly attached remain in a straight line. Therefore, in a molecule containing bond sp² -and sp -hybridized carbons, all the atoms remain in the same plane (except 1, 2-dienes).

A negatively charged C -atom (or a heteroatom containing lone pairs of electrons such as N, O, etc.) adjacent to a double bond is sp² – hybridized and tire atoms attached to that carbon or heteroatom remain in the plane of the system containing the double bond.

Therefore, the atoms in the given molecules/ions that remain in the same plane are as follows:

The boiling point of it pure organic liquid is 70°C. There arc two samples of tills liquid having boiling ranges:

1. 76-78°C and
2. 69-78°C respectively.

Question 53. 

1. The electronic configuration of C-atom is: ls²2s²2p², yet its valency is four —why?
2. The four C —H bonds of methane molecule are equivalent — explain with reasons.

Answer:

During a chemical reaction, the carbon atoms gain energy and promote one of the two electrons of 2s -orbital to the higher 2p_z -orbital. Thus in the excited state, the electronic configuration of carbon becomes  ls²2s¹2p¹x2p¹y2p¹z.

At this condition, the valence shell of the C-atom contains four unpaired electrons. Thus, the C -atom can form four covalent bonds using four unpaired electrons. This explains why carbon having electronic configuration, ls²2s²2p² is tetravalent.

The equivalency of four C—H bonds in methane (CH4) be explained by the concept of hybridization of orbitals. In the excited state, the four valence orbitals of carbon, i.e., one 2s and three 2p orbitals possessing slightly different energies mix up and result in the formation of four equivalent sp3 -hybrid orbitals. These hybrid orbitals overlap with the four Is -orbitals of four H -atoms to form four C—H bonds which are also equivalent (same bond length and bond strength)

Class 11 Organic Chemistry

Question 54. 

1. Arrange sp, sp² & sp³ -orbitals in increasing order of:

1. Bond length
2. Bond angle
3. Bond energy
4. Size of orbitals and
5. S -character.

2. Organic compounds are usually water-insoluble. Why?

3. Write the structure of the smallest hydrocarbon having the empirical formula C₂H. What is the shape of the molecule?

4. Draw the p -p-orbitals involved in forming n -n-bonds in the molecule, CH₂=C =CH_2, and predict whether the molecule is planar or not.
Answer:

1.

1. sp—sp<sp²—sp²<sp³—sp³
2. sp³ < sp² < sp
3. sp³—sp³ < sp²—sp² < sp—sp
4. sp < sp² < sp³
5. sp³ < sp² < sp

2. Organic compounds are covalent. Thus they do not get ionised. Moreover, they are usually less polar or non-polar compounds and hence do not dissolve in highly polar solvent, water.

3. The compound is (C₂H)₂ or C₄H₂ and its structure is HC = C —C = CH. The shape of the molecule is linear because all the C -atoms are sp -hybridized.

The molecule is non-planar because the two planes containing one C-atom and two H-atoms are perpendicular to each other

Class 11 Organic Chemistry

Question 55. Give the IUPAC names of the following compounds:

1. CH₃CHClCHBrCH₃

2. CH₃CHFCOCH₂CH₃

3.  (CH₃)₂CHCH₂OH

4. CH₃COOCH(CH₃)₂

5. CH₃CHBrCH(CH₃)COOII

6. CH₃CHOHCH₂CHO

7.

8.

9. HC ≡ CCH(CH₃)CH=CH₂

10. CH₃OCH(CH₃)CH₂CH₃

11. CH₃CHClCH₂CONH₂

12. BrCH₂CBr₂(CH₂)₃CHCl₂

13.

14. CH₃COCH₂COCH₃

Answer:

1. 2-bromo-3-chlorobutane
2. 2-fluoropentan-3-one
3. 2-methylpropan-l-ol
4. Isopropyl ethanoate
5. 3-bromo-2-methyl butanoic acid
6. 3-hydroxy butanal
7. 2,3,5-trimethyl-4-propylheptane (the chain containing a maximum number of substituents is considered as the principal chain).
8. 3-ethyl-2, 4, 5-trimethyl heptane
9. 3-methylpent-l-en-4-yne
10. 2-methoxybutane
11. 3-chloro-butanamide
12. 5,5,6-tribromo-l,l-dichlorohexane
13. 5-sec-butyl-4-isopropyl decane or 4-(l-methyl ethyl)-5-  (1-methyl propyl) decane
14. Pentane-2,4-dione

Class 11 Organic Chemistry

Question 56. Write structures of the following: 

1. Hept-5-en-l-yne
2. 1-bromo-2-ethoxyethane
3. 3-chloropropanoyl bromide
4. 1-chloroprocaine-2-amine
5. 4-iodo-3- nitro butanal
6. 3-phenyl prop-2-enoic acid
7. Ethanoic methanoic anhydride
8. 2-carbomoylpropanoic acid
9. Pentane-2,4-dione
10. 5-formyl-3-oxo pentanoic acid
11. Ferf-butyl alcohol
12. But-2-ene-1,4-dioic acid
13. Trimethylacetic acid
14. Diethylbutane-1,4-dioate
15. 3-(carboxymethyl) pentanoic acid
16. 1,3-dimethyl cyclo hex-l-ene

Answer:

 Question 57. Draw resonance structures of the following compounds.

1. C₆H₅NO₂
2. CH₃CH=CHCHO
3. C₆H₅CHO
4. C₆H₅CH₂
5. CH₃CH=CHCH₂

Class 11 Organic Chemistry

Question 58. 

1.  A mixture of ether and water can be separated by simple distillation.
2. Water present in rectified spirit can be removed by azeotropic distillation.
3. Benzoic add can be extracted from its aqueous solution using benzene.
4. Sugar containing NaCl as an impurity can be purified by crystallization using ethanol but not water

Answer:

1. There is a considerable difference between the boiling points of ether and water Hence, at the boiling point of more volatile ether, the vapors almost exclusively consist of ether and at the boiling point of less volatile water, the vapors almost entirely consist of water Thus, these can be separated by simple distillation.
2. A mixture of water and rectified spirit form an azeotropic mixture, i.e., the constituents of this mixture cannot be separated into their components by fractional distillation. So, azeotropic distillation is required to remove the water-rectified spirit
3. Benzene is immiscible ’with water but benzoic add is highly soluble in benzene,. Hence, benzoic add can be extracted from its aqueous solution using benzene.
4. Sugar is soluble in hot ethanol whereas common salt remains insoluble. Thus, impure sugar can be purified by crystallization. However, purification is not possible using water as a solvent because both components become readily soluble in water

Question 59. In the Lassa goe’s test, NH₂OH.HCl responds to the test for the element chlorine but not for the element nitrogen, explain
Answer:

As there is no carbon (C) atom in NH₂OH.HCl, NaCN is not produced in the first step by the reaction between sodium (Na) and C

Hence, the formation of sodium ferrocyanide and ferric ferrocyanide (prussian blue) in the subsequent step does not take place. Therefore NH₂OH HCl does not respond to the Lassaigne’s test for nitrogen.

But chlorine (Cl) present in NH₂OH-HCl combines with Na metal to form soluble NaCl which reacts with AgNO₃ in the subsequent step to produce a white precipitate of AgCl which is soluble in ammonium hydroxide

Na + Cl → NaCl ; NaCl + AgNO₃→AgCl ↓(white) + NaNO₃

AgCl + 2NH₄OH → [Ag(NH₃)₂]Cl (water soluble) + 2H₂O

Question 60. What are hybridization states of each C- atom in the compounds:

1. CH₂=C= O
2. CH₃CH=CH₂
3. (CH₃)₂CO
4. CH₂=CHCN
5. C₆H₆

Answer:

Class 11 Organic Chemistry

Question 61. Indicate the σ and π bonds in the following Heptan-4-one molecules:

1. C₆H₆
2. C₆H₁₂
3. CH₂CI₂,
4. CH₂=C=CH₂
5. CH₃NO₂
6. HCONHCH₃

Answer:

Question 62. Write bond-line formulas for:

1. Isopropyl alcohol,
2. 2,3-dimethyl butanal
3. Heptan-4-one

Answer:

Question 63.  Give the IUPAC names of the following compounds:

Class 11 Organic Chemistry

Answer:

1. Propylbenzene
2. 3-methylpentanenitrile,
3. 2,5-dimethyl heptane
4. 3-bromo:3-chloroheptane,
5. 3-chloro-propanal
6. 2,2-dichloroethanol.

Question 64. Which of the following represents the correct IUPAC CHO name for the compounds concerned?

1. 2,2-dimetliyIpentane or 2-dimethyl pentane
2. 2,4,7-trimethylolethane or 2,5,7-trimethylolethane
3. 2-chloro-4-methyl pentane or 4-chloro-2-methyl pentane
4. But-3-yne-l-ol or But-4-ol-l-yne.

Answer:

1. 2,2-dimethyl pentane (two alkyl groups are on the same carbon and hence the locant is repeated twice).
2. 2,4,7-trimethyloctane (since 2,4,7 locant set is lower than the set 2,5,7).
3. 2-chloro-4-methylpentane (alphabetical order of substituents is maintained).
4. But-3-yne-l-ol (using lower locant for the principal functional group)

Question 65. Draw formulas for the first 5 members of each homologous series beginning with the given compounds:

1. HCOOH
2. CH₃COCH₃
3. H-CH=CH₂

Answer:

1. HCOOH , CH₃COOH, CH₃CH₂COOH,CH₃CH₂CH₂COOH,CH₃CH₂CH₂CH₂COOH

2. CH₃COCH₃ , CH₃COCH₂CH₃, CH₃COCH₂CH₂CH₃ ,CH₃CH₂COCH₂CH₃, CH₃COCH₂CH₂CH₂CH₃

3. CH=CH₂ , CH₃CH=CH₂ , CH₃CH₂CH=CH₂ , CH₃CH₂CH₂CH=CH₂, CH₃CH₂CH₂CH₂CH=CH₂

Question 66. Give condensed and bond line structural formulas i and identify the functional group(s) present, if any, I for:

1. 2,2,4-trimethylpentane
2. 2-hydroxyl, 2,3-propane tricarboxylic acid
3. Hexanediol

Answer:

Question 67. Identify the functional groups in given compounds:

Answer:

1.  —OH (phenolic hydroxyl), —CHO (aldehyde), — OMe (methoxy).

2. — NH₂ [1° amino (aromatic)], —CO₂—CH₂— (ester), -N(C₂H₅)₂ (3° amino)

3.  —CH=CH— (ethylenic double bond), — NO₂ (nitro)

Question 68. 0.4 g of an organic compound containing N was Kjeldahlised and NH₃ obtained was passed into 50 mL (N/2) H₂SO₄ solution. The volume of the acid solution was increased to 150 mL by adding distilled water. 20 mL of this acid solution required 31 mL (N₂O) NaOH for complete neutralization. Calculate the percentage of N.
Answer:

20 mL of (partially neutralized) diluted acid solution

= 31mL \(\frac{1}{20}\) NaOH solution.

= 15.5

Strength of(partially neutralised) diluted acid solution

= \(31 \times \frac{1}{20} \times \frac{1}{20}(\mathrm{~N})=\frac{31}{400}(\mathrm{~N})\)

Amount of H₂SO₄ present in 150 mL (partially Amount of H2S04 present in 150 mL (partially

= \(\frac{31 \times 150}{400 \times 1000}\)

Now, 50mLof \(\frac{1}{2}(\mathrm{~N}) \mathrm{H}_2 \mathrm{SO}_4\) solution contains = \(\frac{1 \times 50}{2 \times 1000}\) g-equiv. H₂SO₄

NH₃ produced by decomposition 0.4 g of the organic compound = \(\left(\frac{50}{2000}-\frac{31 \times 150}{400000}\right)\)

= 0.013375 g-equivalent

= 0.013375 X×17 g

Now, 0.013375 x 17 g NH3 = \(\frac{14}{17} \times 0.013375\)

% of nitrogen in the organic compound

= \(\frac{14 \times 0.013375}{0.4} \times 100\)

= 46.81

Class 11 Organic Chemistry

Question 69. Expand each of the following condensed formulas into their complete structural formul as:

1. CH₃CH₂COCH₂CH₂CI
2. CH₃CH= CH(CH₂)₄CH₃
3.  BrCH₂CH₂C=CCH₂CH₃

Answer:

Question 70. Write down the condensed structural formula and bond¬ line structural formula for each of the following molecules:

1. ICH₂CH₂CH₂CH₂CH(CH₃)CH₂

2.

Answer:

Question 71. Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen:

Answer:

.

Question 72. Explain why alkyl groups act as electron donors when attached to a n system.
Answer:

Due to hyperconjugation (cr, n conjugation), alkyl groups act as electron donors when attached to a n -system. This is shown in the case of propane—

Question 73. Draw the resonance structures for the following compounds. Show the electron shift using curvedarrow notation:

1. C₆H₅OH
2. C₆H₅NO₂
3. CH₃CH=CHCHO
4. C₆H₅— CHO
5. C₆H₅—CH₂
6. CH₃CH=CHCH₂

Answer:

Question 74. Identify the reagents underlined in the following ey=o + H2O equations as nucleophiles or electrophiles

Answer:

1. Nucleophile OH^–
2. Nucleophile (^–CN)
3. Electrophile (CH₃⁺CO)

Question 75. Classify the following reactions in one of the reaction type studied in this unit.

Answer:

1. Nucleophilic substitution
2. Electrophilic addition
3. P -Elimination
4. Nucleophilic substitution & rearrangement

Question 76. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Class 11 Organic Chemistry

Answer:

1. Structural isomers (position isomers as well as metamers)
2. Geometrical isomers
3. Resonance contributors

Question 77. For the given bond cleavages, use curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocadon and carbanion.

Answer:

Question 78. Write down the IUPAC names of the alkyl groups having the molecular formula, C₄H₆.
Answer:

Four alkyl groups are possible. These are

Question 79. Give the structural difference O aldehyde C & ketonic groups.
Answer:

An aldehyde group (—CHO) is a carbonyl group in which one valency of the carbonyl carbon is satisfied by a H-atom, and the other valency is satisfied by another atom or an alkyl group. On the other hand, the keto group is also a carbonyl group in which two valencies of the carbonyl carbon are satisfied by two alkyl groups

Question 80. Both formic acid (HCOOH) and acetic acid (CH₃COOH) contain the same functional group, yet there are some differences in their chemical properties—explain.
Answer:

The structural formula of formic acid is such that it can be said to contain  said to contain a  as well as

So it exhibits the properties of both —CHO and

—COOH groups. But acetic acid contains only

—COOH group and hence it exhibits the properties of compounds containing only carboxyl group

Class 11 Organic Chemistry

Question 81. Label the primary (1°), secondary (2°), tertiary (3°), and quaternary (4°) carbon atoms in the following compound:

Answer:

Question 82. Write down the IUPAC and common names of each of the given compounds:

1. CH₃CH= CH₂
2. CH₃C=CCH₃
3. CH₃CHOHCH₃
4. CH₃OCH₂CH₂CH₃
5. CH₃CH₂CHO
6. CH₃COC₂H₅
7. C₂H₅COOH
8. C₂H₂COCl
9. CH₃CONH₂
10. CH₃CO₂C₂H₅
11. CH₃CH₂NH₂
12. CH₃NHCH₂CH₃
13. (CH₃)₂NCH₂CH₃
14. CH₃CH₂CN

Answer:

1. Propene; Propylene
2. But-2-yen; Dimethylacetylene
3. Propan-2-ol; Isopropyl alcohol
4. 1-methoxypropane; Methyl /t-propyl ether
5. Propanal; Propionaldehyde
6. Butan-2-one; Ethyl methyl ketone
7. Propanoic acid; Propionic acid
8. Propanoyl chloride; Propionyl chloride
9. Propanamide; Propionamide
10. Ethyl ethanoate; Ethyl acetate
11. Ethanamine; Ethylamine
12. Methylethanamine; Ethylmethylamine
13. N, N-dimethylethanolamine; Ethyldimethylamine
14. Propanenitrile; Ethyl cyanide

Question 83. Write down the structures of the following compounds:

1. 2-Iodopropane
2. Hex-3-yne
3. Pent-l-ene
4. 2,2-Dichloropropane
5. 1, l, 1, 2-Tetrachloroethane
6. Propan-2-ol
7. Propane-1,3-diol
8. Butane-1,2,3-triol.
9. 2-Methoxypropane
10. 2-Methylpentanoic acid
11. 2,2-Dimethylbutanal
12. Pentan-3-one 
13. Butanoyl chloride
14. Aceticformic anhydride
15. Ethylmethanoate
16. N-Methylmethanamine
17. N-Ethyl-N-methylhexanamine
18. Butanenitrile.

Answer:

1. CH₃CHICH₃
2. CH₃CH₂C=CCH₂CH₃
3. CH₃CH₂CH₂CH=CH₂
4. CH₃CCl₂CH₃
5. CH₂Cl CCl₃,
6. CH₃CH(OH)CH₃
7. HOCH₂CH₂CH₂OH
8. CH₃CH(OH)CH(OH)CH₂OH
9. CH₃CH(OCH₃)CH₃
10. CH₃CH₂CH₂CH(CH₃)COOH
11. CH₃CH₂C(CH₃)₂CHO,
12. CH₃CH₂COCH₂CH₃
13. CH₃CH₃CH₂COCl
14. CH₃COOCHO
15. HCOOCH₂CH₃
16. CH₃NHCH₃
17.  CH₃CH₂N(CH₃)CH₂CH₃
18. CH₃CH₂CH₂CN

Class 11 Organic Chemistry

Question 84. Write down the IUPAC names of the following compounds:

Answer:

1. 3-ethyl-5-methyl heptane,
2. 2,2-dimethylbutane
3. 2,2,4-trimethylpentane,
4. 3-ethyl-2,2,4-trimethylpentane
5. 4-(1,1-dimethyl ethyl)heptane
6. 3,4-diethylhexane
7. 6-ethyl-2-methyl-5-(1,1-dimethyl ethyl)octane

Question 85. What is wrong with the following names? Draw the structures they represent and write their correct names.

1. 1,1-dimethyl hexane
2. 3-methyl-5-methyl heptane
3. 4, A-dimethyl-3-ethyl pentane
4. 3, 4,7-trimethylolethane
5. 3,3-diethyl-2,A, Atrimethylpentane

Answer:

1. (CH₃)₂CHCH₂CH₂CH₂CH₂CH₃ – 2-methylheptane
2. CH₃CH₂CH₂C(CH₃)₂CH₂CH₂CH₂CH₃ – 4,4-dimethyl octane
3. CH₃CH₂CH(CH₃)CH₂CH(CH₂CH₃)₂–  3-ethyl-5-methyl heptane
4. CH₃C(CH₃)₂CH(CH₂CH₃)₂ –  3-ethyl-2, 2-dimethyl pentane
5. CH₃CH₂CH(CH₃)CH(CH₃)CH₂CH₂CH(CH₃)₂ –  2,5,6-trimethylolethane
6. (CH₃)₂CHC(CH₂CH₃)₂C(CH₃)₃ –  3,3-diethyl-2,2,4-trimethylpentane

Question 86. Give the IUPAC name of the following alkane containing complex substituents:

Answer:  3-ethyl-7,7-fels(2,4-dimethylhexyl)-5,9,11-trimethyltridecane

Question 87. Write the IUPAC names of the following compounds:

Answer:

1. 3-ethyl pent-1-ene,
2. 3-ethyl hex-1-en-5-yne
3. 2-ethyl-3,3-dimethyl but-1-ene
4. Pent-3-en-1-yne
5. 3-methyihexa-1,5-diene
6.  3-isobutylhept-1-en-4-yne
7. 3-propylhept-l-ene,
8. 3-methyl-4-methylidenehept-1-en-6-yne
9. Hexa- 1,3-dien-5-yne,
10. 5-methylhepta-1,2,6-triene

Question 88. Write down the structures of the following compounds

1. Pent-3-en-l-yne
2. 3-methylpenta-l, 4-diyne
3. 3-(2-methylpropyl)hept- 1-en-4-yne
4. 3-ethylpenta-l,3-diene
5. 5-ethynylhepta-l,3,6-triene
6. 4-ethyl-4-methylhex-l-yne

Answer:

Class 11 Organic Chemistry

Question 89. Write down the IUPAC names of the following compounds:

Answer:

Question 90. Write down the structures of the following:

1. 2-methyl butanol chloride
2. 5-chloro-3-ethylpentan-2-one
3. Diethyl butane-1, A-dioate
4. Methyl-2-methyl prop- 2-en-l-oate 
5. 3-phenyl prop-2-enoic acid
6. Propane- 1,2,3-tricarboxamide.

Answer:

Question 91. Give the IUPAC names of the following compounds:

1. CH₃COCH₂COOC₂H₅
2. H₂NCH₂CH₂CH₂COOH
3. CH₃CH(CN)CH₂COCH₃

Answer:

Question 92. Write down the structures of the following compounds:

1. 3-formylpentanoic acid
2. 3-hydroxyl-oxopentanal
3. 2, 3-dihydroxy butane dioic acid
4. 3-hydroxy cyclohexanone
5. 3-hydroxy-3-methyl butane-2-one

Answer: 

Question 93. Write the structures of the following compounds:

1. 2-chloro-2-methylbutan-l-ol
2. 4-amino-2-ethylpent-2-enal
3. Hex-A-yn-2-one
4. 1-bromo-3-chloracyclohex-1-ene
5. But-2-ene-l, 4-dioic acid
6. 4-nitropent-l-yne Ethyl 3-methoxy – 4-nitrobutanoate

Answer:

Question 94. What type of structural isomerism is exhibited by the following pairs of isomers?

1. CH₃CHCOOH and CH₃COOCH₃

2. CH₃ —C≡C — CH₃ and CH₃CH₂C≡CH

3. CH₂= CHOH and CH₃CHO

4. CH₂ = CH(CH₂)₃CH₃ and C₆H₆

6. CH₃CH₂CH₂OH and (CH₃)₂CHOH

Answer:

1. Functional group isomerism
2. Position isomerism
3. Tautomerism (special case of functional group isomerism)
4. Ring-chain isomerism
5. Position isomerism
6. Position isomerism

Question 95. Which two of the following compounds are

1. Position isomers
2. Tautomers
3. Ring-chain isomers
4. Metamers
5. Chain isomers and
6. Functional isomers

Answer:

1. Position isomers: (h) and (k),
2. Tautomers: (a) and (f)
3. Ring-chain isomers: (e) and (j)
4. Metamers : (c) and (g)
5. Chain isomers: (b) and (i)
6. Functional isomers: (d) and (l)

Question 96. Identify the optically active and optically inactive compounds:

1. CH₃CHOHC₂H₅

2. CH₃CH₂OH

3. C₂HgCHBrCH(CH₃)₂

4.

5. CH₃CH=CHC₂H₅

Answer:

(1), (3), and (4) will be optically active as each of these molecules contain one asymmetric center.

But (2) & (5) are optically inactive as they do not have a symmetric center

Question 97. Which of the following will exhibit geometrical or cis-trans isomerism and which of them will not? Give reasons.

1. CH₃CH=CBr₂

2. BrCH=CHCH₂CH₃

3. CH₂=CH —CH=CH₂

5. CH₂=CHCH=CHCH=CH₂

Answer:

(2), (5), (6), (7), and (8) exhibit geometrical isomerism.

But (1), (3), and (4) do not exhibit geometrical isomerism.

Question 98. Draw the Fischer projectionformulas ofall stereoisomers CH₃CHBrCHClCOOH. Mention how they are related to each other
Answer:

Enantiomers: 1 and 2, 3 and 4

Diastereomers: 1 and 3; 1 and 4; 2 and 3; 2 and 4

Question 99. Write down the structure and IUPAC names of two isomeric optically active alkanes having lowest molecular mass.
Answer:

Question 100. Which of the following compounds are meso-compounds and which are not? Give reasons

Answer:

(2) and (3) are meso compounds (they are optically inactive due to the presence of center of symmetry). (1) is optically active. It contains two asymmetric centers and it is not superimposable on its mirror image.

Question 101. Arrange in order of decreasing basic strength and show(XI) Pari-n reasons: CH₃—CH=NH,  CH₃—C=N, CH₃ — NH₂

Ongoing from 

in CH₃NH₂, CH₃CH=NH and CH₃C=N, the unshared electron pairs are in sp³ , sp² and sp -orbitals respectively. As the s -character of the hybrid orbital (containing lone pair) of N -atom increases, the electrons are drawn closer to the nitrogen nucleus and hence electron donating ability of the amino nitrogen decreases causing a decrease in basicity.

Thus basic strength decreases in the sequence:

Question 102. Arrange in order of increasing acidity and give reasons: CH₃CH₂OH, (CH₃)₃ COH, CH₃OH, (CH₃)₂CHOH
Answer:

Since alkyl groups have +1 effect, there will be an increased electron displacement towards the oxygen atom on going from primary to secondary to tertiary alcohol. This may be represented (qualitatively) as follows:

Question 103. Arrange the following anions in order of increasing stability and give reasons: CH₂=^–CH, CH₃^–CH₂, CH = C^–
Answer:

The greater the -ve charge on the oxygen atom, the closer is the displacement of the covalent pair in the O —H bond to the hydrogen atom, hence separation of a proton becomes increasingly difficult. Thus the acid strengths of alcohols will be in the order:

(CH₃)₃COH < (CH₃)₂CHOH < CH₃CH₂OH < CH₃OH

Ongoing from \(\mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2→\mathrm{CH}_2=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}→\mathrm{HC} \equiv \stackrel{\ominus}{\mathrm{C}}\) it is seen that the unshared electron pairs of the carbanion carbons are in sp³, sp² and sp -hybrid orbitals respectively. As the s -s-character of the hybrid orbitals increases, the electrons are drawn closer to the nucleus of the carbanion carbon and hence ability to hold the electron pair increases, causing successive increase in stability. Stability order

⇒ \(\mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2<\mathrm{CH}_2=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}<\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}\)

Question 104. Which of the following pairs do not represent two resonance structures and why?

Answer:

The two structures which differ in the positions of atoms are not resonance structures. Thus the pair of structures given in (1), (2), and (5) do not represent resonance structures

Question 105. In between CH₃COOH and CH₃COO^–, which one is more resonance stabilized and why?
Answer:

Both CH₃COOH and CH₃COO^– are resonance hybrids of two canonical forms. But one of the resonance structures of CH₃COOH involves separation of charge, while none of the resonance structures of CH₃COO^– involves any separation of charge (also these structures are equivalent). Hence CH₃COO^– is more resonance stabilized compared to CH₃COOH

Question 106. Which N -atom of guanidine is more basic and why
Answer:

Protonation on tire doubly bonded N -atoms produce a cation (conjugate acid) which is stabilized by resonance involving three equivalent canonical forms. On the other hand protonation on either of the singly bonded N-atoms produces a cation (conjugate acid) which is not stabilised by resonance. Thus the doubly bonded N-atom of guanidine is more basic.

Question 107. Which of the two N atoms of the following compound undergoes protonation and why?
Answer:

N-atom ofthe ring ‘A’ undergoes protonation because the resultant cation (conjugate acid) is stabilized by resonance. The n-atom of the ring ‘B’ does not undergo protonation because in that case the resulting cation will not be stabilized by resonance.

Question 108. Which resonance structure in each of the following compounds contributes more towards the hybrid and why?
Answer:

• The 1st structure contributes more because both C and N have octet of electrons in their valence shells.
• The 1st structure contributes more because it involves no separation of charge
• The 2nd structure contributes more as the -ve charge is on the more electronegative O-atom.
•  The 1st structure contributes more because there is separation of charge. The 2nd structure involves the separation of charge; also the +ve charges, on the adjacent atoms, repel each other.
• The 1st structure contributes more compared to the 2nd structure. The 2nd structure is highly unstable as it contains a negative N-atom.

Question 109. Which of the following compounds can be represented as a resonance hybrid and which of them can the? Give reasons.

1.  CH₃CH₂OH,

2.CH₃CONH₂

3. CH₃CH=CHCH₂NH₂

4. H₂N—CH=CH— NO₂

5. 

Answer:

Structures and (3) can not be represented as resonance hybrids because lone pairs on the O-atom or N-atom can not undergo delocalization. However, structures (2), (4) and (5) can be represented as resonance hybrids.

Question 110.  Why are the three carbon-oxygen bonds In carbonate (CO₃^-2) ion equal in length?

This is so because CO₃^-2 ion is a resonance hybrid of three equivalent canonical forms.

Question 111. Which one between phenol and cyclohexanol is more acidic and why?
Answer:

Phenol is a stronger acid than cyclohexanol.

It can be explained as follows:

1. Due to resonance, the O-atom of OH group acquires a +ve charge and so the release of the proton is facilitated

2. When phenol ionises, the formed phenoxide ion is also a resonance hybrid, but it is more stabilized by resonance than a unionized phenol molecule because of the spreading of a negative charge only. In the unionised molecule, resonance involves the separation of charge

Such effects are not possible in case of cyclohexanol and hence proton release is not facilitated.

Question 112. Arrange the following ions in order of increasing stability and give your reasons
Answer:

Question 113. Which one between 2-methylbut-2-ene and 2-methylbut-1- ene has higher heat of hydrogenation and why?
Answer:

In structure III, +ve charge on the carbon is involved in I delocalization with the benzene ring

In structure (I), +ve charge on the carbon is involved in delocalization not only with the benzene ring but also with N-atom of the \(\ddot{\mathrm{N}}\)Me₂ group, thereby making this structure more stable than (III).

Structure (II) is least stable because +ve charge on the carbon can not be involved in delocalization with the aromatic ring (steric inhibition of resonance)

Question 114. Arrange the following ions in order of increasing stability and give your reasons
Answer:

2-Methylbut-2-ene contains nine hyperconjugable α-H -atoms, so this molecule is involved in effective hyperconjugation. As a result this molecule gains extra stability and it has relatively lower heat of hydrogenation. On the other hand, 2-methylbut-1-ene contains only five hypercoagulable α-H -atoms and so the effect of hyperconjugation stabilizing this molecule is relatively less. Thus it has a relatively higher heat of hydrogenation.

Question 115. The C—C bond in acetaldehyde (CH₃CHO) is shorter than that in ethane while the C— C bond in trifluoro acetaldehyde (CF₃CHO) is essentially the same as that in ethane. Explain
Answer:

Acetaldehyde molecule contains three α-H -atoms. These H-atoms are involved in hyperconjugation with the double bond of the carbonyl group. So C—C bonds in acetaldehyde have partial double bond character. In ethane the C—C bond has pure single bond character.

Thus C — C bond in acetaldehyde is shorter than that in ethane. Trifluoroacetaldehyde does not have a-H -atoms. So hyper-conjugation is not possible in CF₃CHO

Thus, the C — C bond in CF₃CHO is essentially the same as that in ethane.

Question 116. Arrange the following isomeric alkenes in order of increasing stability and give your reasons:

1. (CH₃)₂C=C(CH₃)₂ [I]
2. CH₂=CHCH₂CH₂CH₃ [II]
3. CH₃CH=CHCH(CH₃)₂[ III]
4. CH₃CH =C(CH₃)CH₂CH₃[IV)

Answer:

The stability of an alkene is determined by the number of hyperconjugative structures, which in turn is dictated by the number of α-H -atoms (concerning the olefinic carbons) present in the molecule. The greater the number of hyper conjugative structures, the higher is the stability of the alkene Now the number of ar-H -atoms in the alkene I, II, III, and IV are 12, 2, 4 and 8 respectively. Thus stability of the alkanes follows the sequence: II < III < IV < I .

Question 117. Which one of the following two conformations of butane is more stable and why?

Answer:

Eclipsed conformation I, in which methyl groups on two adjacent carbons are just opposite to each other. In this conformation steric strain and bond opposition strain are maximum, hence this conformation is most unstable.

Anti-conformation II, in which methyl groups are as far apart as possible, is most stable due to minimum repulsion between methyl groups. Note that, there is no bond opposition strain in this conformation.

Question 118. Which of the 2 geometric isomers I of Me₃CCH=CHCMe₃ has a higher heat of combustion and why?
Answer:

cis-isomer is less stable because of a very large steric hindrance between two bulky t-butyl groups lying on the same side of the double bond. On the other hand, transisomer is more stable because the bulky f-butyl groups are on the opposite sides of the double bond. Thus cis isomer has a higher heat of hydrogenation

Question 119. Which one between C₆H₅CH₃ and CH₄ has lower Csp³—H bond dissociation enthalpy and why?
Answer:

Bond dissociation enthalpy of C₆H₅CH₂—H is less than (II) that of H₃C — H as  C₆H₅CH₂ is more stable (stabilized by resonance) than that of ^•CH₃ (which has no resonance stabilization).

Question 120. Arrange the following carbocations in order of increasing stability and explain the order:

Answer:

In cation in,(III) +ve charge is not delocalized due to steric inhibition of resonance. However, +ve charge is delocalized in both the cations I and But the extent of delocalization of +ve charge is higher in H due to the additional effect involving the methoxy group

Question 121. Arrange the following carbanions in order of increasing stability and explain the order

Answer:

The stability of carbanions increases as the extent of delocalization of the -ve charge increases. Carbanion I is most stable as the -ve charge is delocalised not only by the benzene ring but also by the -NO₂ group. Carbanion HI is moderately stable as the -ve charge is delocalised only by the benzene ring. Carbanion U is least stable because of the +R effect of the -OMe group (although the -ve charge is delocalized by the benzene ring). Thus the sequence of stabilityis II < III < I

Question 122. Classify the following species as electrophile or nucleophile and explain your choice:

Answer:

Nucleophile : CH₃C^– , CHgCOO^– , CH₂=CH₂

Electrophile : Cl+ , BF₃ , (CH₃)₃C⁺ , R— X

CH₃O^–– and CH₃COO^– are negatively charged species having available unshared electron pairs on O-atom. So these are nucleophiles. CH₂=CH₂ can also act as a nucleophile as it contains loosely bound ir -electrons.

In the species Cl⁺ , BF₃ and Me₃C⁺ , there are electron deficiency (having sex tet of electrons) on the valence shells of Cl, B, and C-atom respectively. So these are electrophiles. In the alkyl halides (R—X) there is electron deficiency on the a -carbon due to strong -I effect of halogen atom. So RX can act as electrophile.

Question 123. Formulate the following as a two-step reaction and designate the nucleophile and electrophile in each step:  CH₂= CH₂ + Br₂ → BrCH₂CH₂Br
Answer:

Question 124. CN^– and NO₂^– are called ambient nucleophiles. Explain
Answer:

Nucleophiles that have more than one (generally two) suitable atoms through which they can attack the substrate called ambident nucleophiles. Each of the ^–CN ions and NO₂^– contain two atoms through which they can be involved in nucleophilic attack (these atoms are indicated by arrows). So these are ambident nucleophiles.

Question 125. Mention the type of each of the following reactions
Answer:

Answer:

1. Nucleophilic substitution (S_N2)
2. Electrophilic addition
3. Free-radical substitution
4. Elimination reaction (E2)
5. Rearrangement reaction

Question 126. Calculate the double bond equivalent (DBE) of each of the given compounds:

1. C₁₃H₉BrS
2. C₁₂H₁₆N₂O₄

Answer:

Double bond equivalent (DBE) of C₁₃H₉BrS

= \(\frac{13(4-2)+9(1-2)+1(1-2)+1(2-2)}{2}+1\)

= 9

DBE of  C₁₂H₁₆N₂O₄

= \(\frac{12(4-2)+16(1-2)+2(3-2)+4(2-2)}{2}+1\)

= 6

Question 127. Calculate the double bond equivalent (DBE) of a compound having molecular formula, C₅H₈. On catalytic hydrogenation, the compound consumes 1 mol of hydrogen. Write the structures of all the possible isomers of the compound
Answer:

DBE of C₅H₈ \(=\frac{5(4-2)+8(1-2)}{2}+1\) = 2

On hydrogenation, it consumes 1 mol of H₂. So it contains one double bond and one ring.

Thus possible structures of the compounds are:

Question 128. Arrange the following carbocations in order of increasing stability and explain the order:
Answer:

Sequence of stability: (4) > (2) > (3) > (1) Carbocation (4) is a primary carbocation, but it is most stable due to resonance.

Question 129. Which of the carbocations is the most stable?

1. CH₃CH₂⁺CH₂,
2. CH₂=CH⁺CH₂,
3. C₆H₅ ⁺CH₂
4. All are equally stable.

Answer:

(1), (2), and (3) are all primary carbocations. Cations (1), (2), and (3) have 0, 2, and 4 resonance structures respectively. So carbocation (3) is most stable

Question 130.  Which one between the two CH₃CO^– and CH₃ CH₃COC^– HCOCH₃ is more stable and why?

Negative charge CH₃COC^– H₂ is involved in delocalization with the n electrons of only one carbonyl group. On the other hand, the carbonation-carbon of on the carbanion-carbon of charge e -ve on n CH₃COC^–HCOCH₃ is involved in delocalization with the π -electrons of two carbonyl groups. Thus the second carbanion is more stable

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Short Question And Answers

Question 1. Why is impure glycerol purified by distillation under reduced pressure?
Answer:

The boiling point of glycerol is 563K under normal pressure and it undergoes decomposition below this temperature. Thus, simple distillation cannot be used for its purification. Under a reduced pressure of 12 mm Hg, the boiling point of glycerol is reduced and then at 453 K it can be distilled without getting decomposed.

Question 2. Why is it necessary to use acetic acid and not sulphuric acid for the acidification of sodium extract for testing sulfur by lead acetate test?
Answer:

To detect the presence of sulfur, sodium extract is acidified with acetic acid because lead acetate being soluble does not interfere with the test. If sulphuric acid is used, lead acetate will react with sulphuric acid to form a white precipitate of lead sulfate which will interfere with the test.

Pb(CH₃COO)₂  ( Lead acetate )+ H₂SO₄ →PbSO₄  (Lead sulphate)↓(White) + 2CH₃COOH

Question 3. The presence of N in hydroxylamine hydrochloride cannot be detected by Lassaigne’s test—why?
Answer:

When hydroxylamine hydrochloride (NH₂OH.HCl) is fused with metallic sodium, NaCN is not obtained as the compound contains no carbon. Thus, the presence of nitrogen in this compound cannot be detected by Lassaigne’s test.

Class 11 Organic Chemistry

Question 4. How can it be possible to detect the presence of nitrogen in hydrazine hydrochloride?
Answer:

During fusion of the compound with metallic sodium if some starch or charcoal is added, carbon of starch or charcoal combines with nitrogen of the compound to form NaCN which will further indicate the presence of nitrogen in the compound a proton. On the other hand, no such sterlc inhibition occurs in N, N-dimethylaniline because the two orlp H-atoms are relatively much smaller in size. The unshared electron-pair on N-atom becomes involved In resonance interaction with the ring and therefore, is not fully available for taking up a proton. This explains why N,N ,2,6-tetramethylsilane is more basic than N,N -dimethylaniline

Question 5. Chloroform is more acidic than fluoroform. Explain.
Answer:

CF₃^–, tire conjugate base of fluoroform (CHF₃^–), is stabilised by -I effect of 3 F-atoms. But CCl^–₃ the conjugate base of chloroform (CHCl₃), is relatively more stabilized by the somewhat weaker -I effect of 3 Cl-atoms along with d-orbital resonance (Cl has vacant d-orbital). So chloroform is more acidic than fluoroform

Question 6. How can you separate benzoic acid and nitrobenzene from their mixture by the technique of extraction using an appropriate chemical reagent?
Answer:

The mixture is shaken with a dilute sodium bicarbonate solution when benzoic acid gets converted to sodium benzoate and dissolves in water leaving nitrobenzene behind. The mixture is extracted with ether or chloroform when nitrobenzene goes into the organic layer. After separating the organic layer, it is distilled to get nitrobenzene. The aqueous layer is acidified with dilute HCl when benzoic acid gets precipitated.It is obtained by filtration

Question 7. Why is impure glycerol purified by distillation under reduced pressure?
Answer:

The boiling point of glycerol is 563K under normal pressure and it undergoes decomposition below this temperature. Thus, simple distillation cannot be used for its purification. Under a reduced pressure of 12 mm Hg, the boiling point of glycerol is reduced and then at 453 Kit can be distilled without getting decomposed

Question 8. Why is it necessary to use acetic acid and not sulphuric acid for the acidification of sodium extract for testing sulphur by lead acetate test?
Answer:

To detect the presence of sulfur, sodium extract is acidified with acetic acid because lead acetate being soluble does not interfere with the test. If sulphuric acid is used, lead acetate will react with sulphuric acid to form a white precipitate of lead sulphate which will interfere with the test.

Class 11 Organic Chemistry

Question 9. The presence of N in hydroxylamine hydrochloride cannot be detected by Lassaigne’s test—why? p
Answer:

When hydroxylamine hydrochloride (NH₂OH HCl) is fused with metallic sodium, NaCN is not obtained as the compound contains no carbon. Thus, the presence of nitrogen in this compound cannot be detected by Lassaigne’s tes

Question 10. How can it be possible to detect the presence of nitrogen in hydrazine hydrochloride?
Answer:

During fusion of the compound with metafile sodium if some starch or charcoal is added, carbon of starch or charcoal combines with nitrogen of the compound to form NaCN which will further indicate the presence of nitrogen in the compound.

Question 11. Can you separate two liquids A (b.p. 413 K) and B (b.p. 403 K) present in a mixture by simple distillation?
Answer:

The two components cannot be separated by simple distillation. This is because the vapors of both liquids will be formed simultaneously and will condense together in the receiver. The separation of these two liquids can be done by fractional distillation.

Question 12. Will CCl₄ give a white precipitate of AgCl on heating it with silver nitrate solution? Give a reason for your
Answer.

When CCl₄ is heated with AgNO₃ solution, a white precipitate of AgCl will not be formed. This is because CCl₄ being a covalent compound with a strong C— Cl bond does not ionize to give Cl” ions required for the formation of the precipitate of AgCl

Question 13. Is it possible to distinguish between phenylhydrazine hydrochloride and hydrazine hydrochloride by Lassaigne’s test? Give reason.
Answer:

Lassaigne’s test can be used to distinguish between phenylhydrazine hydrochloride (C₆H₅NHNH₂ HCl) and hydrazine hydrochloride (NH₂NH₂ HCI). This is because the former containing carbon and nitrogen produces NaCN when fused with metallic sodium while the latter containing no carbon does not produce NaCN when fused with sodium

Question 14. Define R_f value. What is called descending paper chromatography
Answer:

If the solvent is placed at the top and the upper end of the chromatography paper is dipped in it, then the solvent moves downwards. This is called descending paper chromatography.

Class 11 Organic Chemistry

Question 15. Tendency of carbon to exhibit catenation is much higher than that of Si and S—why
Ans.

The C— C bond dissociation enthalpy (348.6 kj. mol^-1) is much higher than that of Si—Si (228.4 kj -mol-1) and S—S (224.2 kj- mol^-1) bond dissociation enthalpies and since the formation of C—C bond is thermodynamically much favorable, the tendency of carbon to exhibit catenation is much higher than that of silicon and sulfur

Question 16. Melting and boiling points of organic compounds are usually very low— Why?
Answer:

Covalent organic compounds usually exist as single molecules. The attractive forces operating among these less polar or non-polar molecules are very low. As a result of this, the melting and boiling points of these compounds are usually very low

Question 17. Write the IUPAC name of the compound,    mentioning the secondary prefix, primary prefix, a word root, primary suffix & secondary suffix respectively
Answer:

In the compound    , the secondary prefix: is bromo (at C-3); the primary prefix: is cyclo; the word root: is pent; the primary suffix: is ane (e is to be omitted), and the secondary suffix: is ol (at C-l). Therefore, the IUPAC name of the compound is

Class 11 Organic Chemistry

Question 18. Which one of them is more pure and why?
Answer:

The first sample (boiling range: 76-78°C) Is more pure because its boiling range is shorter

Question 19. The wind is on an azeotropic mixture? Give example.
Answer:

An azeotropic mixture is a mixture of two or more liquids having a constant boiling point. The most familiar example of an azeotropic mixture is a mixture of ethanol and water in the ratio of 95.6: 4.4. It boils at a temperature of 78.5°C.

Question 20. A mixture contains two organic solids, A and B. The solubilities of A and B in water are 12 g per 100 mL and 3 g per 100 mL respectively. How will you separate A and B from this mixture?
Answer:

The two components can be separated by fractional crystallization. When the saturated hot solution of this mixture is allowed to cool, the less soluble compound, B crystallizes out first leaving the more soluble compound, A in the mother liquor. The mother liquor is then concentrated and allowed to cool when the compound A crystallizes out.

Question 21. What is seeding?
Answer:

Seeding is a process of inducing crystallization by adding a crystal of the pure substance into its saturated solution.

Class 11 Organic Chemistry

Question 22. Suggest methods for the separation of the components in each of the following mixtures:

1. A mixture of liquid A (b.p. 366 K) and liquid B (b.p. 355.5 K).
2. A mixture of liquid C (b.p. 360 K) and liquid D (b.p. 420 K).

Answer:

1. The two liquids, A and B can be separated by fractional distillation because the boiling points of them differ by just 10.5 K.
2. The two liquids, C and D can be separated by simple distillation because the boiling points of them differ widely by 60 K.

Question 23. A mixture contains three amino acids. How can they be Identified?
Answer:

When the mixture is subjected to separation by paper chromatography, three spots on the paper become visible at different heights from the starting line by placing the paper under UV light. Then they can be identified by determining their R_f values and comparing these values with the R_f val of the pure compounds.

Question 24. The R_f values of X and Fin a mixture determined by TLC method in a solvent mixture are 0.75 and 0.52 respectively. If the mixture is separated by column chromatography using the same solvent mixture as the mobile phase, which of the two components, will elute first and why?
Answer:

The higher R_f value of X (0.75) indicates that it is less strongly adsorbed as compared to compound Y having a lower R_f value (0.52). Therefore, if the mixture is separated by column chromatography using the same solvent mixture as the mobile phase, X will be eluted first.

Question 25. Why is an organic compound fused with sodium for
Answer:

When an organic compound is fused with metallic sodium, these elements present in the compound are converted into water-soluble sodium salts (NaCN, NaX, and Na₂S). The presence of cyanide ion (CN^–), halide ion (X^–), and sulfide ion (S^2-) in the solution can then be 40 What is seeding? confirmed by using suitable reagents

Question 26.  What type of fission of a covalent bond produces free radicals? Give an example with proper sign.
Answer;

Homolytic fission of covalent bonds produces free radicals.

Question 27.

1. Write down the IUPAC name of the following compound

Question 28. Draw the structure of the following compound: 3,4-dimethyl pentanoic acid
Answer:

1. 2-bromo-2-chloroethanol.

2.

Question 29. Draw the canonicals of CH₃COOH and CH₃COO^–. In which case resonance is more important?
Answer:

Class 11 Organic Chemistry

Equivalent structures (more stable) Resonance is more important for CH₃COO^– as it involves two equivalent resonating structures and the negative charge is always on the electronegative O-atom.

Question 30. Write the principle of estimation of carbon and hydrogen in an organic compound.
Answer:

A known amount of dry and pure organic compound is heated with cupric oxide(CuO) in a hard glass test tube. As a result, the carbon (C) and hydrogen (H) present in the compound are oxidized to carbon dioxide (CO₂) and water (H₂O ) respectively

Knowing the amounts of CO₂ and H₂O formed in the reaction it is possible to calculate the percentages of C and H present in the compound

Question 31.

1. Write down the IUPAC name of the following compound: CH₃COCH₂CH – Cl – COCH₃ sodium nit
2. Write down the structural formula of the following compound: Hex-1 -en-4-yne

Answer:

Question 32.

1. Arrange the following radicals in increasing order of-I effect: I, Br, Cl, F’
2. Write the structural formula of the following compound: 5-amino pent-3-enoic acid

Answer:

1. I <Br<Cl<P

2.   \(\mathrm{HOO} \stackrel{1}{\mathrm{C}}-\stackrel{2}{\mathrm{C}} \mathrm{H}_2-\stackrel{3}{\mathrm{C}} \mathrm{H}=\stackrel{4}{\mathrm{C}} \mathrm{H}-\stackrel{5}{\mathrm{C}} \mathrm{H}_2-\mathrm{NH}_2\)

Question 33.

1. Whyis(CH₃)₃C⁺ more stable than CH₃CH⁺₂?
2. Indicate the electrophilic centre of the following compounds: CH₃CHO, CH₃CN

Answer:

1. Due to the +1 effect of three electron-donating CH₃ groups, (CH₃)₃C⁺ is more stable than CH₃⁺CH₂ (which contains only one CH₃ group attached to C⁺).  Furthermore (CH₃)₃ C⁺ is stabilized by 9 hyper conjugative structures, while CH₃CH₂ is stabilized by only three hyper conjugative structures

2.

Question 34. Name IUPAC name of the following:

Answer:

1. Propan-1,2,3-trio
2. 3,3-dichlorobutanoic acid

Class 11 Organic Chemistry

Question 35. Explain the order of basicity of the following compounds:

1. CH₃—CH₂— NH₂
2.  CH₃ —CH=N H
3. CH₃—CH₂—CN

Answer: CH₃— CH₂—CN < CH₃—CH=N-H < CH₃—CH₂— NH₂

Question 36. A compound having molecular formula CgH18 can form only one monobromo derivative. Draw the structure of the compound.
Answer:

Since, the compound forms only one monobromo derivative, all hydrogen atoms are equivalent. Electrophilic center CH₃—C— H.

Thus the compound will be

Question 37. Is 2-hydroxypropanoic acid optically active? Explain.
Answer:

2-hydroxy propanoic acid \(\stackrel{3}{\mathrm{C}} \mathrm{H}_3-\stackrel{2 *}{\mathrm{C}} \mathrm{H}(\mathrm{OH})-\stackrel{1}{\mathrm{C}} \mathrm{OOH}\)1 is optically active as there is one chiral center situated at C-2

Question 38. Write the IUPAC names of the compound CH₂=CHCH₂CH₂C=CH & CH₃CH=CHCH₂C=CH

Question 39. Which of the two : O₂NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?_
Answer:

  ion is more stable than the effect of the — NOz group causes the dispersal of ~ve charge on the O-atom. On the other hand, —CH₂CH₃ group has +1 effect which tends to intensify the -ve charge on the O -atom leading to the destabilization of the ion.

Question 40. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulfur, and halogens.
Answer:.

Nitrogen, sulfur, and halogen atoms, present in organic Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulfur, and halogens. Ans. Nitrogen, sulfur, and halogen atoms, are present in organic

Question 41. Name a suitable technique to separate the components from a mixture of calcium sulfate & camphor.
Answer:

Camphor is sublimable but CaSO₄ is not Therefore sublimation of the mixture gives camphor on the inner surface of the funnel while CaSO₄ is left in the china dish

Question 42. Will CCl₄ give a white precipitate of AgCl on heating it with silver nitrate? Give a reason for your answer.
Answer:

CCl₄ is an covalent compound. Thus, it does not ionize to give Cl- ions. Hence AgNO₃ does not react with CCl₄ even under hot conditions to form a white precipitate of AgCl

Question 43. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:

CO₂ is an acidic oxide. Thus, it reacts with the strong base KOH to form K₂CO₃ (salt). 2KOH + CO₂ →K₂ CO₃ + H₂ O

So, during the estimation of carbon, an increase in the mass of the U-tube containing KOH solution is produced from the organic compound. Thus % of carbon in the organic compound can be estimated by using the equation:

⇒ \(\% \text { of carbon }=\frac{12}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of organic compound }} \times 100\)

Question 44. Why is it necessary to use acetic acid and not sulphuric acid for the acidification of sodium extract for testing sulfur by lead acetate test?
Answer:

For testing sulfur, the sodium extract is acidified with acetic acid because lead acetate remains soluble in an acetic acid medium and hence does not interfere with the test. If H₂SO₄ is used, lead acetate itself will react with H₂SO₄ to form a white precipitate of lead sulfate.

Pb(OCOCH₃) + H₂SO₄→PbSO₄ ↓ (white ppt)+ 2CH₃COOH

Question 45. Explain why chlorine but not nitrogen in hydroxylamine hydrochloride (NH₂OH-HCl) can be detected by Lassaigne’s test
Answer:

Hydroxylamine hydrochloride (NH₂OH-HCl) contains nitrogen but does not contain carbon as an element. So, on fusion with metallic sodium, it cannot form sodium cyanide. Cyanide ion is essential to produce Prussian blue. Thus nitrogen cannot be detected by Lassaigne’s test. However hydroxylamine hydrochloride contains chlorine as an element and so on fusion with Na, it forms NaCl. Thus, chlorine can be detected by Lassaigne’s test.

Question 46. Differentiate between the principles of Dumas method & Kjeldahl’s method.
Answer:

In Dumas’s method, the nitrogenous organic compound is decomposed to produce gaseous nitrogen. The measured volume of N₂ is used to calculate the % of nitrogen in the given compound. In Kjeldahl’s method, the nitrogenous organic compound is decomposed to produce

(NH₄)₂ SO₄ which is further decomposed to give NH₃. The amount of NH3 so formed is used to calculate the % of nitrogen in the compound.

Class 11 Organic Chemistry

Question 47. 0.495 g of organic compound on combustion gave 0.99 g of CO₂ and 0.405 g of water. Calculate the percentages of carbon and hydrogen in the compound
Answer:

Amount of C in die compound = \(\frac{12}{44} \times 0.99\) 0.99 = 0.27

Amount of H in die compound = \(\frac{2}{18} \times 0.405\) x 0.405 = 0.045

v % of C in the compound = \(\frac{12}{44} \times 0.99 \times \frac{100}{0.495}\) = 54.54

% of H in the compound = \(\frac{2}{18} \times 0.405 \times \frac{100}{0.495}\) = 9.09

Question 48. 0.50 g of an organic compound when analysed by Dumas method produced 62.0 mL of nitrogen at STP. Determine the percentage of nitrogen in the compound.
Answer:

Mass of 62.0 mL (STP) of N₂ = \(\frac{28 \times 62.0}{22400}\)

-. % of N in the compound = \(=\frac{28 \times 62.0}{22400} \times \frac{100}{0.50}\)

= 15.5

Question 49. Is it possible to distinguish between hydrazine and phenylhydrazine by Lassaigne’s test? Give your reason.
Answer:

For the detection of nitrogen in an organic compound by Lassaigne’s test, the compound must contain both C and N, to permit the formation of NaCN.

Phenylhydrazine contains both C and N. So it gives a positive test for nitrogen in Lassaigne’s test. On the other hand, hydrazine contains nitrogen but does not contain carbon so it gives a negative test for nitrogen. Thus the two compounds can be distinguished by Lassaigne’s test

Question 50. Give an example of a ketone that does not exhibit tautomerism.
Answer:

2,2,4,4-tetramethylpentan-3-one

Question 51. Arrange in the order of increasing enol content and give
CH₃COCH₃ ,  , CH₃ COCH₂COCH₃
Answer:

Question 52. Write the structure and the IUPAC name of the alkane having the lowest molecular mass and which on bromination produces three monobromo derivatives.
Answer:

Pentane (CH₃CH₂CH₂CH₂CH₃) is the desired hydrocarbon with the lowest molecular mass which contains three types of non-equivalent H-atoms.

Question 53. How many types of non-equivalent H -atoms are there in each of the following compounds

Answer:

(1) 2 types (2) 3 types (3) 4 types (4) 6 types (5) 5 types (6) 7 types (7) 2 types.

Class 11 Organic Chemistry

Question 54.  Write the structure and the IUPAC name of an alkane (C₁₈H₃₆) which on bromination produces only 1 monobromo derivative.
Answer:

Question 55. Mention the type of substitution reactions in which the attacking reagents are NO₂⁺, OH^-, or ^•Cl
Answer:

Attacking reagent NO₂⁺ : Electrophilic substitution

Attacking reagent OH^– : Nucleophilic substitution.

Attacking reagent Cl^• : Free-radical substitution

Question 56. Suggest a method to purify:

1. Iodine containing traces of common salt,
2. Kerosene containing a little of water and
3. Camphor contains little benzoic acid.

Answer:

1. On sublimation, I₂ sublimes leaving behind NaCl. Alternatively, I₂ can be extracted with CCl₄ and the extract on evaporation gives I₂.
2. Kerosene and water are immiscible liquids having different densities. So they can be separated by using a separating funnel.
3. On boiling with water, benzoic acid dissolves but camphor remains insoluble, which can be separated by filtration (under hot conditions).

Question 57.  Suggested method for the separation of each of the following mixtures:

1. A mixture of liquidA (b.p. 365 K) and liquid B (b.p. 356K)
2. A mixture of liquid C (b.p. 395 K) and liquid D (b.p. 360 K)

Answer:

1. By fractional distillation
2. By ordinary distillation.

Question 58. The Rj values of two compounds, X and Y in a mixture determined by TLC are 0.66 and 0.41 respectively. If the mixture is separated by column chromatography using the same solvent mixture as the mobile phase, which one of the two compounds will be eluted first and why?
Answer:

Compound X (Rj- = 0.66) will be eluted first. The component having a higher R_f value is adsorbed less strongly by the stationary phase (adsorbent) and hence it is eluted first.

Question 59. Give an example of each of

1. Adsorption chromatography and
2. Partition chromatography

Answer:

1. Thin layer chromatography
2. Paper chromatography

Question 60. Is it possible to get pure benzoic acid from a sample containing impurities of naphthalene through the process of crystallization using benzene asa solvent? Give reason.
Answer:

It is not possible to purify impure benzoic acid by recrystallization using benzene as a solvent because both naphthalene and benzoic acid are quite soluble in benzene. Purification is however possible if hot water be used as the solvent because benzoic acid is soluble in hot water but naphthalene is not

Class 11 Organic Chemistry

Question 61. Write down the bond-line structural formulas

1. 2-methylbutane
2. 3,3 -dimethyl hexane
3. 2 -bromooctane and
4. Chlorocyclopentane.

Answer:

Question 62. Arrange in increasing order of strength and give reasons: CH₂=CHCOOH, HC=CCOOH, CH₃CH₂COOH
Answer:

Electronegativity of the hybridized carbon atoms increases in the sequence C_sp²<C_sp²<C_sp

The strength of carboxylic acid increases as the electronegativity of a carbon increases. This sequence of acid strength

⇒ \(\stackrel{\beta}{\mathrm{C}} \mathrm{H}_3 \stackrel{\alpha}{\mathrm{C}} \mathrm{H}_2 \mathrm{COOH}<\stackrel{\beta}{\mathrm{C}} \mathrm{H}_2=\stackrel{\alpha}{\mathrm{C}} \mathrm{HCOOH}<\stackrel{\beta}{\mathrm{C}} \equiv \stackrel{\alpha}{\mathrm{C}} \mathrm{COOH}\)

Question 63.  Arrange the following free radicals in order of increasing stability and explain the order

The stability of alkyl free radicals increases as the number of a -hydrogens increases. This is so because the extent of delocalization of the unpaired electron of any free radical increases with an increase in the number of or-H atoms. Since the number of or-H atoms in (1), (2), and (3) are 1, 5 and 0 respectively, the sequence of stability is given by: (3)< (1)< (2).

Question 64. Designate the species as electrophile or nucleophile obtained on heterolytic cleavage of C— C bond in ethane.
Answer: 

Ethane undergoes heterolytic bond fisssion to give a carbocation (methyl cation, ⁺CH₃) and a carbanion (methyl anion, ^–CH₃ ). Methyl cation is an electrophile, while methyl anion is a nucleophile

Class 11 Organic Chemistry

Question 65. Although BF₄^– is an anion, it is not a nucleophile—why?
Answer:

In BF₄^– ion the central boron atom is negatively charged but it does not have any unshared electron-pair to act as a nucleophile

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Multiple Choice Questions

Question 1. The ease of dehydrohalogenation of an alkyl halide with alcoholic KOH is—

1. 3° <2° < 1°
2. 3° < 2° > 1°
3. 3° > 2° > 1°
4. 3° > 2° < 1°

Answer: 2. 3° < 2° > 1°

Question 2. Which will exhibit optical isomerism

Answer: 2

Question 3. Which of the following is sec-butyl phenyl vinyl methane—

Answer: 3

Question 4. The correct states of hybridisation of C₂ and C₃ in compound H₃C —CH=C=CH —CH₃

1. sp², sp³
2. sp², sp
3. sp², sp³
4. sp, sp

Answer: 2. sp², sp

Question 5. Under identical conditions, the S_N1 reaction will most efficiently with—

1. Tert-butyl chloride
2. 2-methyl-l-chloropropane
3. 2-chloroquine
4. Vinyl Chloride

Answer: 1. Tert-butyl chloride

Question 6. Which one of the following characteristics belongs to an electrophile—

1. It is any species having electron deficiency which reacts at an electronrich C-centre
2. It is any species having electron enrichment, that reacts at an electron deficientC-centre
3. It is cationic in nature
4. It is anionic in nature

Answer:  1. It is any species having electron deficiency which reacts at an electronrich C-centre

Class 11 Organic Chemistry

Question 7. The most stable enol tautomer of MeCOCH₂CO₂Et 

1. CH₂=C(OH)CH₂CO₂Et
2. MeC(OH)—CHCO₂Et
3. MeCOCH=C(OH)OEt
4. CH₂=C(OH)CH=C(OH)OEt

Answer: 2. MeC(OH)—CHCO₂Et

Question 8. Order of stability of the carbocations

1. Ph₂C⁺CH₂Me

2. PhCH₂CH₂C⁺HPh

3. PhCH⁺CHMe

4. Ph₂C(Me)CH₂  is

1. 4 > 3 > 1 > 3
2. 1 > 2 > 3 > 4
3. 2 > 1 > 4 > 3
4. 1>4>3>2

Answer: 2. 1 > 2 > 3 > 4

Question 9. MeCH₂CH=CH₂ is stable than Me₂C=CH₂ because—

1. Inductive effect of group.
2. Resonance effect of Me-group.
3. Hyperconjugative effect of Me-group.
4. Resonance and inductive effects of Me-group.

Answer: 3. Hyperconjugative effect of Me-group.

Question 10. (+) and (-)-Lactic acid has the same molecular formula, C₃H₆O₃. They are related as

1. Structure isomers
2. Geometric isomers
3. Optical isomers
4. Homomers

Answer: 3. Optical isomers

Question 11. Which of the following statements is correct for 2-butene—

1. The C₁ —C₂ bond is an sp³ —sp³ σ -bond
2. The C₂— C₃ bond is an sp³—sp² σ -bond
3. The C₁—C₂ bond is an sp³—sp³ σ-bond
4. The C₁— C₂ bond is an sp²—sp² σ -bond

Answer: 3. The C₁—C₂ bond is an sp³—sp³ σ-bond

Class 11 Organic Chemistry

Question 12. Basicity of aniline is less than methyl amine, because—

1. Hyperconjugation effect of Me-group in MeNH₂
2. Resonance effect of the phenyl group in aniline
3. Molar mass of methylamine is less than that of aniline
4. Resonance effect of Me-group in MeNH₂

Answer: 2. Resonance effect of the phenyl group in aniline

Question 13. Tautomerism is exhibited by—

Answer: 1, 2 and 3

Question 14. Amongst the following, the one which can exist in free state as a stable compound is—

1. C₇H₉O
2. C₈H₁₂O
3. C₆H₁₁O
4. C₁₀H₁₇O₂

Answer:  2. C₈H₁₂O

Question 15. The correct pair of compounds which gives blue colouration/ precipitate and white precipitate, respectively, when their Lassaigne’s test is separately done is—

1. NH₂NH₂ HCl and C1CH₂COOH

2. NH₂CSNH₂ and PhCH₂Cl

3. NH₂CH₂COOH and NH₂CONH₂

Answer: 4.

Question 16. The IUPAC name of the compound X is—

1. 4-cyano-4-methyl-2-oxopentane
2. 2-cyano-2-methyl-4-oxopentane
3. 2,2,-dimethyl-4-oxopentanenitrile
4. 4-cyano-4-methyl-2-pentanone

Answer: 3. 2,2,-dimethyl-4-oxopentanenitrile

Class 11 Organic Chemistry

Question 17. The optically active molecule is

Answer: 3

Question 18. (+) -2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amount of SbCl5, due to the formation of—

1. Carbanion
2. Free-radical
3. Carbene
4. Carbocation

Answer: 4. Carbocation

Question 19. The order of decreasing ease abstraction of hydrogen atoms in the following Hb molecule is—

1. H_a>H_b>H_c
2. H_a>H_c>H_b
3. H_b>H_a>H_c
4. H_c > H_b>H_a

Answer: 2. H_a>H_c>H_b

Question 20.The most likely protonation site in the given molecule is

1. C-1
2. C-2
3. C-3
4. C-6

Answer: 1. C-1

Question 21. The 4-th higher homologue of ethane is—

1. Butane
2. Pentane
3. Hexane
4. Heptane

Answer: 3. Hexane

Question 22. Among the following structures the one which is not a resonating structure of others is—

Answer: 4.

Question 23. The correct order of decreasing length of the bond as indicated by the arrowin the following structures is—

Answer: 3

Question Question 24. IUPAC name ofthe molecule, is

1. 5,6-dimethylhept-2-ene
2. 2,3-dimethylhept-5-ene
3. 5,6-dimethylhept-3-ene
4. 5-isopropylhex-2-ene

Answer:  1. 5,6-dimethylhept-2-ene

Class 11 Organic Chemistry

Question 25. The correct statementregarding the given compound is—

1. All three compounds are chiral
2. Only 1 and 2 are chiral
3. 1 and 3 are diastereomers
4. Only 1 and 3 are chiral

Answer: 4. Only 1 and 3 are chiral

Question 26. In Lassaigne’s test for the detection of nitrogen in an organic compound, the appearance of blue coloured

1. Ferric ferricyanide
2. Ferrous ferricyanide
3. Ferric ferrocyanide
4. Ferrous ferrocyanide

Answer: 3. Ferric ferrocyanide

Question 27. The reaction of methyl trichloroacetate (Cl₃CCO₂Me) with sodium methoxide (NaOMe) generates

1. Carbocation
2. Carbene
3. Carbanion
4. Carbon radical

Answer: 2. Carbene

Question 28. In a mixture, two ommtlomors are found to bo present in tho amount of 0f*% mu) 15% respectively. The enantiomeric excess (e,o) Is—

1. 85%
2. 15%
3. 70%
4. 60%

Answer: 3. 70%

Question 29. In the following compound, the number it ‘sp’ hybridised carbon is CH₂=C=CH-CH-C≡CH –

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Question 30. Which of the following statements Is/are correct

Answer: 2 and 4

Question 31. The correct order of add strengths of benzoic acid (X), hydroxybenzoic acid (Y) and p-nitrobenzoic acid (Z) is—

1. Y> Z > X
2. Z > Y > X
3. Z > X > Y
4. Y > X > Z

Answer:  3. Z > X > Y

Question 32. in the IUPAC system, PhCH₂CH₂CO₂H is named as—

1. 3-phenylpropanoid acid
2. Benzoyl acetic acid
3. Carboxyethyl benzene
4. 2-phenylpropanoid acid

Answer: 1. 3-phenylpropanoid acid

Class 11 Organic Chemistry

Question 33. The major product(s) obtained in the reaction Is/are

Answer: 1 and 4

Question 34. The possible product (s) to be obtained from the reaction of cyclobutyl amine with HNOz is/are—

Answer:  1 and 3

Question 35.  Identify the compound that exhibits tautomerism—

1. 2-pentanone
2. 2-butene
3. Lactic acid
4. Phenol

Answer:  2. 2-butene

Question 36. How many chiral compounds are possible on monochlorination of 2-methylbutane—

1. 2
2. 4
3. 6
4. 8

Answer: 1. 2

Question 37. Identify the compound that exhibits tautomerism—

1. 2-pentanone
2. 2-butene
3. Lactic acid
4. Phenol

Answer: 1. 2-pentanone

Question 38. The order of stability of the following carbocations is—

1. 3>1>2
2. 3>2>1
3. 2>3>1
4. 1>2>3

Answer: 1. 3>1>2

Class 11 Organic Chemistry

Question 39. Arrange the compounds in order of decreasing acidity—

Answer: 4.

Question 40. A solution of (-)-l-chloro-l-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl₅, due to the formation of—

1. Free radical
2. Carbanion
3. Carbene
4. Carbocation

Answer:  4. Carbocation

Question 41. In S_N2 reactions, the correct order of reactivity for the compounds: CH₃Cl, CH₃CH₂Cl, (CH₃)₂CHCl and (CH₃)₃CCl is _____

1. (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl > (CH₃)₃CC;
2. CHgCl > (CH₃)₂CHCl > CH₃CH₂CI > (CH₃)₃CCI
3. CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl
4. CH₃CH₂Cl > CH₃CI > (CH₃)₂CHCl > (CH₃)₃CCl

Answer: 3. CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl

Question 42. For the estimation of nitrogen, 1.4 g of an organic compounejÿwas digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10

1. 5%
2. 6%
3. 10%
4. 3%

Answer: 3. 10%

Question 43. In Carius method of estimation of halogens, 250 g of an organic compound gave 141 g AgBr. Percentage of Br in the compound(Ag = 108, Br = 80)

1. 48
2. 60
3. 24
4. 36

Answer: 3. 24

Question 44. Which of the following compounds will exhibit

1. 2-phenyl-l-butene
2. 1,1 – diphenyl 1 – propane
3. 1-phenyl-2-buten
4. 3 – phenyl – 1 – butene

Answer: 3. 1-phenyl-2-buten

Question 45. The increasing order of S_N1 reactivity of the following compounds is—

1. 1<3<2
2. 2<3<1
3. 3<2<1
4. 2<1<3

Answer:  4. 2<1<3

Question 46. The resonance stability is minimum for the compound—

Answer: 2.

Question 47. Which of the following compounds will be suitable for Kjeldahl’s method of nitrogen estimation—

Answer: 4

Question 48. The increasing order of basicity of the following compounds is—

Answer: 1

Question 49 . Consider the reactions:

Answer: 3

Question 50. Which undergoes nucleophilic substitution most easily—

Answer: 1

Question 51. IUPAC name ofthe compound,

1. Trans-2-chloro-3-iodo-2-pentene
2. Cis-3-iodo-4-chloro-3-pentane
3. Trans-3-iodo-4-chloro-3-pentene
4. Cis-2-chloro-3-iodo-2-pentene

Answer: 1. Trans-2-chloro-3-iodo-2-pentene

Question 52. Considering the state of hybridisation of C-atoms, which one among the following is linear

1. CH₃—C₂—CH₂—CH₃
2. CH₃—CH=CH—CH₃
3. CH₃-C = C-CH₃
4. CH₂=CHCH₂C= C

Answer: 3. CH₃-C = C-CH₃

Question 53. Which is a nucleophilic substitution reaction—

Answer: 4

Question 54. Which is most reactive towards an electrophilic reagent—

Answer: 3.

Question 55. The correct order of increasing bond length of C—H , C — O, C—C and C=C is _

1. C—H < C—O < C—C < C=C
2. C—H<C=C<C—O<C—C
3. C—C < C=C < C—O < C—H
4. C—O<C—H<C—C<C=C

Answer: 2. C—H<C=C<C—O<C—C

Question 56. RCHO + NH₂NH₂→RCH=N—NH₂ What sort of reactions it—

1. Electrophilic addition-eliminationreaction
2. Free radical addition-eliminationreaction
3. Electrophilic substitution-elimination reaction
4. Nucleophilic addition-eliminationreaction

Answer: 4. Nucleophilic addition-eliminationreaction

Question 57. Which of the following acids do not exhibit optical

1. Maleic acid
2. Ammino acids
3. Lactic acid
4. Tartaric acid

Answer: 1. Maleic acid

Question 58. The correct order of decreasing acid strength of trichloroacetic acid (I), trifluoroacetic acid (n), acetic acid (in) and formic acid (IV) is—

1. 2>1> 4>3
2. 2>4>3 >1
3. 1> 2>3>4
4. 1>3>2>4

Answer: 1. 2>1> 4>3

Question 59. Which nomenclatureisnot according to IUPAC system

Answer: 3

Question 60. Structure of the compoimd whose IlIPAC name is 3-ethyl- 2-hydroxy-4-methylhex-3-en-5-ynoic acid is

Answer: 4

Question 61. Structure of isobutyl group an organic compound is —

Answer: 2

Question 62. The order of stability of the following tautomeric forms is—

1. 2>3>1
2. 1>2>3
3. 3>2>1
4. 2>1>3

Answer: 3. 3>2>1

Question 63. Which of the following compounds will undergo racemisation when solution of KOH hydrolyses—

1. 1 and 2
2. 2 and 4
3. 3 and 4
4. 1 and 4

Answer: Nonn of  these

Question 64. Most reactive towards nucleophilic addition reaction is—

Answer:  4

Question 65. In the Kjeldahl’s metliod for estimation ofnitrogenpresent in a soil sample, ammonia evolved from 0.75 g of sample neutralised 10 mL of 1M H₂SO₄ The percentage of nitrogenin the soilis—

1. 37.33
2. 45.33
3. 35.33
4. 43.33

Answer: 1. 37.33

Question 66. The number of structural isomers possible from the molecular formula C₃H₉N is

1. 4
2. 5
3. 2
4. 3

Answer:  1. 4

Question 67.  In an S_N1 reaction on chiral centres there is__________

1. 100% racemisation
2. Inversion more tlian retention leading to partial racemisation
3. 100% retention
4. 100%Inversion

Answer:  2. Inversion more tlian retention leading to partial racemisation

Question 68. Which of the following statements is not correct for a nucleophile—

1. Nucleophile is a Lewis acid
2. Ammonia is a nucleophile
3. Nucleophiles attack low electrons density sites
4. Nucleophiles are not electron seeking

Answer: 1, 2,3 and 4

Question 69. Two possible stereo-structures of CH3CHOH-COOH, which are optically active, are called—

1. Diastereomers
2. Atropisomers
3. Enantiomers
4. Mesotners

Answer: 3. Enantiomers

Question 70. In which of tire following molecules, all the atoms are copluuar—

Answer: 4.

Question 71. The correct order of add strengtlis of the given carboxylic acid is—

Answer: 3

Question 72. Which among the given molecules can exhibit tautomersim

Answer: 4

Question 73. The correct order of add strengtlis of the given carboxylic acid is—

Answer: 2

Question 74. The pair of electrons in the given carbanion is presentin which ofthe following orbitals 

1. sp
2. 2p
3. sp
4. sp

Answer: 1. sp

Question 75. The correct statement about the basicity of aryl amines is—

1. Aryi amines are in general more basic than alkyl amines because the N-atom in aryl amines is sp hybridised
2. Aryi amines are in general less basic than alkyl amines because the unshared pair of electrons on nitrogen in aryi amines undergoes effective delocalisation with the ring π-electrons
3. Aryi amines are in general more basic than alkyl amines because the unshared pair of electrons on nitrogen in aryi amines does not undergo delocalisation with the ring – -electrons
4. Aryi amines are more basic than alkyi amines due to

Answer: 2. Aryi amines are in general less basic than alkyl amines because the unshared pair of electrons on nitrogen in aryi amines undergoes effective delocalisation with the ring π-electrons

Question 76.  Which one of the following statements for the given
reactions is correct—

1. Is a substitution reaction but (2) and (3) are addition reactions.
2. (1) and (2) are elimination reactions, but (3) is an addition reaction.
3. (1) is an elimination reaction, (2) is a substitutionreaction and (3) is an addition reaction.
4. (1) is an elimination reaction, but (2) and (3) are substimtion reactions.

Answer:  3. (1) is an elimination reaction, (2) is a substitutionreaction and (3) is an addition reaction.

Question 77. The IUPAC name ofthe compound 

1. 5-formylhex-2-en-3-one
2. 5-methyM-oxohex-2-en-5-al
3. 3-keto-2-methylhex-5-enal
4. 3-keto-2-methylhex-4-enal

Answer:  4. 3-keto-2-methylhex-4-enal

Question 78. Which one is the most acidic compound—

Answer: 3

Question 79. The most suitable method of separation of 1 : 1 mixture of ortho andpara- nitrophenols is—

1. Chromatography
2. Crystallisation
3. Steam distillation
4. Sublimation

Answer:  3. Steam distillation

Question 80. The correct statement regarding electrophile is—

1. Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
2. Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
3. Electrophiles can be either neutral or positively charged species and can form a bond by accepting a pair ofelectrons from a nucleophile
4. Electrophile is a negatively charged species and can form a bond accepting a pair of electrons from a nucleophile

Answer: 3.  Electrophiles can be either neutral or positively charged species and can form a bond by accepting a pair ofelectrons from a nucleophile

Question 81. Which of the following is correct with respect to -I effect of the substituents (R = alkyl) —

1. —NR₂ > —OR > —F
2. —NH₂ < —OR < —F
3. —NH₂ > —OR > —F
4. —NR₂ < —OR < —F

Answer: 2. —NH₂ < —OR < —F

Question 82. Which ofthe following carbocations is expected to be most stable—

Answer: 3.

Question 83. 3. Which of the following molecules represents the order of hybridisation sp², sp², sp, sp fromleft to right atoms—

1. CH₃ —CH=CH —CH₃
2. HC=C —C=CH
3. CH₃=CH—CH=CH₂
4. CH₂=CH-C=CH

Answer: 4. CH₂=CH-C=CH

Question 84. S_N2 reaction readily occurs in—

Answer: 1.

Question 85. The correct decreasing order of pK_a is

1. 2>4>1>3
2. 4>2>3>1
3. 3>2>4>1
4. 4>1>2>3

Answer:  1. 2>4>1>3

Question 86.  The correct decreasing order of pK_b is—

1. 1>2>3>4
2. 3>4>2>1
3. 2>3>4>1
4. 4>2>1>3

Answer: 4. 4>2>1>3

Question 87. Find the dihydroxycyclopentane—

1. 1
2. 2
3. 3
4. 4

Answer: 3. 3

Question 88. Decreasing order ofnucleophilicity is—

1. OH^–>NH^–>CH₃O^–>RNH₂
2. NH^–>OH^–>CH₃O^–>RNH₂
3. NH^–>CH₃O^–>OH^–>RNH₂
4. CH₃O^–>NH^–>OH^–>RNH₂

Answer:  3. NH^–>CH₃O^–>OH^–>RNH₂

Question 89. pK_a increases in benzoic acid when substituent “x” is bonded at para-position, then “x” is—

1. —COOH
2. —NO₂
3. —CN
4. —OCH₃

Answer: 4. —OCH₃

Question 90. The IUPAC name of the given compound is (CH₃)₃CCH₂C(CH₃)₃

1. 2,3,4,4-tetramethylpentane
2. 1,2,2,4-tetramethylpentene
3. 2,2,4,4-tetramethylpentane
4. 3,3-dimethylpentane

Answer: 3. 2,2,4,4-tetramethylpentane

Question 91. The purity ofan organic compound is determined by—

1. Chromatography
2. Crystallisation
3. Melting orboilingpoint
4.  Both (1) and (3)

Answer: 4.  Both (1) and (3)

Question 92. Lassaigne’s test for the detection ofnitrogen fails in—

1. H₂N—CO—NHNH₂ .HCI
2. NH₂—NH₂HCI
3. C₆H₅—NH —NH₂ . HCl
4.  C₆H₅CONH₂

Answer: 2. NH₂—NH₂HCI

Question 93. Among the following, the achiral amino acid is—

1. 2-ethylalanine
2. 2-methylglycine
3. 2-hydroxymethylserine
4. Tryptophan

Answer: 3. 2-hydroxymethylserine

Question 94. Arrange the following nucleophiles in the decreasing order of nucleophilicity—

Answer:  4

Question 95. Which ofthe following is an electrophile—

1. CCl₂
2. CH₃^–
3. H₂O
4. NH₃

Answer: 1. CCl₂

Question 96. Give IUPAC name ofthe following compound—

1. 5-hydroxy cyclo hex-3-en-1-one
2. 3-hydroxy cyclo hex-5-en-1-one
3. 8-hydroxy cyclo hex-3-en-1-one
4. 7-hydroxy cyclo hex-5-en-1-one

Answer:  1. 5-hydroxy cyclo hex-3-en-1-one

Question 97. Which of the following is the correct order of acidic strength of the following compounds

1. 1 > 2> 3
2. 2 > 3> 1
3. 1 > 3> 2
4. 3 > 2> 1

Answer:  2. 2 > 3> 1

Question 98. IUPAC name ofthe given compound is

1. 2-methoxy-4-bromonitrobenzene
2. 3-bromo-6-nitro-l-methoxybenzene
3. 3 -methoxy-4-nitrobromobenzene
4. 5-bromo-2-nitro-l-methoxybenzene

Answer: 4. 5-bromo-2-nitro-l-methoxybenzene

Question 99. The number of <r- and n -bonds in pent-l-en-4-yne molecule is respectively—

1. 8 and 2
2. 10 and 3
3. 6 and 4
4. 7 and 2

Answer: 2. 10 and 3

Question 100. The hybrid orbitals involved in the formation of the C₂— C₃ bond in the following compound, CH₂=CH — CH₂—CH₂—C≡CH are—

1. sp-sp²
2. sp-sp³
3. sp²-sp³
4. sp³-sp³

Answer: 3.  sp²-sp³

Question 101. The increasing order of electronegativity of the carbon atoms C-2, C-3 and C-4 in the compound CH₃ — C=C — CH₂ — CH=CH₂ is—

1. C-3 < C-2 < C-4
2. C-4 < C-3 < C-2
3. C-2 < C-4 < C-3
4. C-3 < C-4 < C-2

Answer: 1.  C-3 < C-2 < C-4

Question 102. CH₃CONH₂→CH₃CN; In this conversion, the change in hybridisation state of the carbon atom of the functional group is—

1. sp³-sp
2. sp³-sp
3. sp-sp³
4. sp³-sp³

Answer: 2. sp³-sp

Question 103. The correct shapes of CCl₄ and CCl₂=C=C=CCl₂ molecules are respectively—

1. Linear and tetrahedral
2. Planar and pyramidal
3. Tetrahedral and planar
4. Tetrahedral and linear

Answer: 3. Tetrahedral and planar

Question 104. The number of C and H-atoms that lie in the same plane in a toluene (C₆H₅CH₃) molecule is respectively—

1. 7 and 5
2. 6 and 5
3. 7 and 3
4. 6 and 3

Answer: 1.  7 and 5

Question 105. The number of primary, secondary, tertiary and quaternary carbon atoms in 2,2,4-trimethylpentane is respectively—

1. 5,1,1 and 1
2. 1,1,1 and 5
3. 4,1,1 and 2
4. 1,5,1 and 1

Answer: 1. 5,1,1 and 1

Question 106. The number of primary, secondary and tertiary H-atoms in molecule is respectively—

1. 14,9 and 2
2. 15,9 and 1
3. 15,8 and 2
4. 15,8 and 1

Answer: 4.  15,8 and 1

Question 107. In which of the following molecules does all the atoms lie on the same straight line—

1. HC = C—C = CH
2. HC = C—CH₃
3. HC≡CN
4. C₃O₂

Answer: 2. HC = C—CH₃

Question 108. Hybridisation states of C-2, C-3, C-5 and C-6 in the com¬pound, (CH₃)₃CCH =CHCHC=CH are respectively—

1. sp, sp³, sp² and sp³
2. sp³, sp², sp² and sp
3. sp, sp², sp² and sp³
4. sp, sp², sp³ and sp²

Answer: 1.sp, sp³, sp² and sp³

Question 109. IUPAC name of the compound,

1. 1,1-diethyl-2,2-dimethyl pentane
2. 4,4-dimethyl-5,5-diethyl pentane
3. 5,5-diethyl-4,4-dimethyl heptane
4. 3-ethyl-4,4-dimethylheptane

Answer: 4.  3-ethyl-4,4-dimethylheptane

Question 110. IUPAC name of the compound,

1. 4-propyl-3-methylhex-5-en-2-one
2. 3-propyl-5-methylhex-l-en-5-one
3. 3-methyl-4-propylhex-5-en-2-one
4. 3-methyl-4-vinylheptan-2-one

Answer:  3.  3-methyl-4-propyl hex-5-en-2-one

Question 111. Bond lengths of C—H, C—O, C—C and C=C follow the sequence—

1. C—H < C — O < C—C < C=C
2. C—H < C=C < C—O< C — C
3. C—C<C=C<C—O<C—H
4. C—O<C—H<C—C<C=C

Answer: 2. C—H < C=C < C—O < C — C

Question 1112. IUPAC name of the compound

 

1. 1-chloro-2-nitro-4-methylbenzene
2. 1-chloro-4-methyl-2-nitrobenzene
3. 2-chloro-1-nitro-5-methylbenzene
4. m-nitro-p-chlorotoluene

Answer:  2. 1-chloro-4-methyl-2-nitrobenzene

Question 113. The correct IUPAC name of the alkyl group is—

1. 2-ethyl-3-sec-butyl propyl
2.  2,4-diethyl pentyl
3. 2-etliyl-4-methylhexyl
4. 2-methyl-4-ethylhexyl

Answer: 3.  2-etliyl-4-methylhexyl

Question 114. The hybridization states of the carbon atom of the amido and cyano group are respectively—

1. sp³ and sp²
2. sp² and sp
3. sp and sp²
4. sp³ and sp

Answer: 2.  sp² and sp

Question 115. Which of the following pair of compounds are isomers—

1. CH₃CH₂OH, CH₃OCH₃

2. CH₃OC₃H₇, C₂H₅OC₂H₅ .

3. CH₃CH₂CHO, CH₃COCH₃

4.

Answer: 4.

Question 116. Two aliphatic compounds will not be considered isomers if they are—

1. Aldehyde and ketone
2. Ether and alcohol
3. Ether and aldehyde
4. Carboxylic acid and ester

Answer: 3.  Ether and aldehyde

Question 117. The number of organic compounds with molecular formula C₄H₁₀ are—

1. 7
2. 5
3. 6
4. 8

Answer: 1. 7

Question 118. Only a monosubstituted compound is obtained when an alkane reacts with chlorine in presence of UV light. The alkane is—

1. Propane
2. Pentane
3. Butane
4. Cyclohexane

Answer: 4.  Cyclohexane

Question 119. Two enantiomers rotate the plane of polarisation of plane polarised light

1. In different directions but keeping the angle same
2. In the same direction but with different angles
3. In the same direction and in the same angle
4. In different directions with different angles

Answer: 1.  In different directions but keeping the angle same

Question 120. Which of the following is an optically active compound—

1. CH₃CHClCH=:CH₂
2. CH₃CHCl₂
3. Meso-tartaric acid
4. CH₃CH=C=CH₂

Answer: 1.  CH₃CHClCH=:CH₂

Question 121. Which of the following compounds exhibit both geometrical and optical isomerism—

1. CH₃CHClCH=C(CH₃)₂
2. CH₃CH=CH — CHBrCH₂CH₃
3. CH₂=C=CH—CH=CHCH₃
4. CH₃CH₂CH=CH₂

Answer: 2.  CH₃CH=CH — CHBrCH₂CH₃

Question 122. Which of the following compounds does not exhibit tautomerism—

1. CH₃CH₂N=O
2. CH₃NO₂
3. CH₃COCH₃
4. (CH₃)₃CCOC₆H₅

Answer: 4.  (CH₃)₃CCOC₆H₅

Question 123. The enol content in which of the following compounds is maximum—

1. CH₃COCH₂COCH₃

3. CH₃COCH₃

4. CH₃CHO

Answer: 1. CH₃COCH₂COCH₃

Question 124. The optically active alkane of the lowest molecular mass which is also chiral is—

1. 3-methyl hexane
2. 2, 3-dimethyl pentane
3. 2-methyl hexane
4. 2,5-dimethyl hexane

Answer: 1.  3-methylhexanec

Question 125. Which of the following compounds possesses a center of symmetry—

1. Trans-1,3-dimethylcyclobutane
2. Cis-1,3-dimethylcyclobutane
3. Trans-1-ethyl-3-methyl cyclobutane
4. Cis-l-ethyl-3-methyl cyclobutane

Answer: 1.  Trans-1,3-dimethylcyclobutane

Question 126. The compounds ds-2-butene and frans-2-butene can be differentiated by

1. The number of products obtained due to their chlorination
2. The number of products obtained due to their bromination
3. Their reaction with H₂ in presence of nickel catalyst
4. Their respective boiling points

Answer:  4. Their respective boiling points

Question 127. Which of the following is optically active—

Answer: 4.

Question 128. Which of the following is non-superimposable on its mirror image—

Answer: 3

Question 129. The number of structural isomers that are formed when one H-atom of diphenylmethane  is substituted by chlorine is—

1. 4
2. 6
3. 8
4. 7

Answer: 1. 4

Question 103. The number of isomers formed by a compound whose molecular formula is C₂BrClFI is—

1. 3
2. 4
3. 5
4. 6 
   

Answer: 4. 6

Question 131.  In this reaction, the hydroxy acid obtained is—

1. (+)-enantiomer
2. (-)-enantiomer
3. 50% (+) and 50% (-)- enantiomer
4. 20% (+) and 80% (-)-enantiomer

Answer: 3.  50% (+) and 50% (-)- enantiomer

Question 132. Which of the following compounds will produce the most stable carbocation in the presence of an acid—

1. (CH₃)₂CHCH₂OH
2. CH₂=CH—CH₂OH
3. (CH₃)₂CHOH
4. (CH₃)₃COH

Answer: 2. CH₂=CH—CH₂OH

Question 133. The correct order of stability of the given carbanions:

1. 1>2>3
2. 2>1>3
3. 3>2>1
4. 2>3>1

Answer: 1. 1>2>3

Question 134. The most stable carbocation is—

Answer: 4.

Question 135. The most stable carbanion is—

Answer:  2

Question 136. Carbocation which does not undergo rearrangement is-

1. (CH₃)₂CH⁺CH₂
2. (CH₃)₂CH⁺CHCH₃
3.  (CH₃)₃⁺C
4. (CH₃)₃C⁺CH₂

Answer: 3.  (CH₃)₃⁺C

Question 137. Which of the following carbocations is quite stable and can even be stored in the laboratory as a salt— in which hyperconjugation does not occur—

1. (CH₃)₂ ⁺CH
2. (C₆H₅)₃⁺C
3.  CH₂=CH-⁺CH₂
4. ⁺CH₂CH₂C₆H₅

Answer: 2. (C₆H₅)₃⁺C

Question 138. Which of the given alicyclic compounds is most active—

1. C₆H₅C(CH₃)₃
2. C₆H₅CH₃
3. (CH₃)₂C=CH₂
4. CD₃CH=CH₂

Answer: 1. C₆H₅C(CH₃)₃

Question 139.  Which of the given resonance structures is most stable—

Answer: 4

Question 140. Which ofthe given alicyclic compounds is most active—

Answer: 2.

Question 141.

1. ^•CH₃
2. CH₃^•CH₂
3. CH₂=CH^•CH₂
4. C₆H₅^•CH₂

The correct order of stability of these free radicals is—

1. 1>2>3>4
2. 3>2>1>4
3. 4>3>2>1
4. 4>7>2>3

Answer: 3. 4>3>2>1

Question 142. The structures of carbocation and carbanion are respectively—

1. Linear and planar
2. Trigonal planar and trigonal pyramidal
3. Tetrahedral and trigonal planar
4. Trigonal pyramidal and tetrahedral

Answer: 2.  Trigonal planar and trigonal pyramidal

Question 143. Correct order of stability of the given three carbanions is—

Answer: 3

Question 144. Which of the following alkenes is most stable—

1. (CH₃)₂C=C(CH₃)₂
2. (CH₃)₂C=CHCH₂CH₃
3. CH₃CH₂CH=CHCH₂CH₃
4. CH₃CH₂CH₂CH₂CH=CH₂

Answer: 1.  (CH₃)₂C=C(CH₃)₂

Question 145. In which of the following compounds, the extent of resonance between the benzene ring and halogen atom is maximum—

Answer: 1

Question 146. The compound whose basicity is maximum in gaseous and aqueous medium is—

1. NH₃
2. CH₃NH₂
3. (CH₃)₂NH
4. (CH₃)₃N

Answer: 3. (CH₃)₂NH

Question 147. (+)-1-Chloro-1-phenylethane undergoes racemisation in presence of the small amount of SbCl₅.The intermediate formed in this process is—

1. A carbene
2. A carbocation
3. A carbanion
4. A free radical

Answer: 2.  A carbocation

Question 148. In which of the following compounds, presence of nitrogen cannot be detected by Lassaigne’s test—

1. NH₂NH₂.HCI
2. C₆H₅NHNH₂.HCI
3. PhN≡NPh
4. NH₂CONH₂

Answer: 1.  NH₂NH₂.HCI

Question 149. Which of the following compounds is responsible for the formation of Prussian blue during the detection of nitrogen by Lassaigne’s test—

1. Na₄[Fe(CN)₆]
2. Fe₄[Fe(CN)₆]
3. Fe₂[Fe(CN)₆]
4. Fe₃[Fe(CN)6]₂

Answer: 2.  Fe₄[Fe(CN)₆]

Question 150. The process by which essential oils can be extracted from flowers is—

1. Distillation
2. Crystallization
3. Vacuum distillation
4. Steam distillation

Answer: 4.  Steam distillation

Question 151. The process that is suitable for detecting two different types of ink in any handwritten ancient document is—

1. Column chromatography
2. Solvent extraction
3. Distillation
4. Thin layer chromatography

Answer: 4.  Thin layer chromatography

Question1 52. Detection of which of the following functional groups is required to confirm the presence of nitrogen in the corresponding compound—

1. Amido
2. Carboxyl
3. Carbonyl
4. Alkoxycarbonyl

Answer: 1.  Amido

Question 153. Which of the following compounds does not exhibit geometrical isomerism—

Answer: 3.

Question 154. The number of geometrical isomers of the compound, CH₃CH=CH — CHCH₃ is_____

1. 3
2. 2
3. 4
4. 5

Answer: 1.  3

Question 155. Types of non-equivalent H-atoms in CH₃CH(OH)CH(OH)CH₃

1. 2
2. 4
3. 3
4. 6

Answer: 3. 3

Question 156. The number of compounds formed on monobromination

  is_________ 

1. 3
2. 2
3. 5
4. 4

Answer: 4.  4

Question 157.  Types of non-equivalent H-atoms in 

1. 2
2. 4
3. 3
4. 6

Answer: 2. 4

Question 158. The double bond equivalent of C₈H₉CIO is—

1. 4
2. 6
3. 3
4. 2

Answer: 1.  4

Question 159. The correct order of basicity of

1. CH₃NH₂
2. (CH₃)₂NH and
3. (CH₃)₃N

in chlorobenzene is—

1. 1<3< 2
2. 2 <3< 1
3. 1< 2< 3
4. 2< 1< 3

Answer: 3. 1< 2< 3

Question 160. The correct order of stability of 

1. (CH₃)₂⁺CH

2. CH₃⁺CHOCH₃

3. ClCH₂⁺CHCH₃ is—

1. 1> 2 > 3
2. 2 > 3 > 1
3. 2 > 1> 3
4. 3 > 2 > 1

Answer: 3.  2 > 1> 3

Question 161. The correct order of stability of these carbanions is—

1. CH₃^–CH —CO —CH₃

2. CH₃CH₂-CO- ^–CH₂ and

3.  ^–CH₂CH₂— CO — CH₃ 

1. 3<1<2
2. 1<3<2
3. 2<3<1
4. 3<2<1

Answer: 1. 3<1<2

Question 162. In Kjeldahl’s method, CuSO₄ is used to—

1. Catalyze the reaction
2. Oxidise the reaction
3. Reduce the reaction
4. Increase boiling point

Answer: 1. Catalyse the reaction

Question 163. The number of optically active isomers among five probable alcohols of molecular formula C₄H₁₀O is—

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 164. Which compound gives most unstable enol—

Answer: 1.

Question 165. If 3.4% sulfur is present in insulin, then the minimum molecular mass of insulin will be—

1. 350
2. 470
3. 560
4. 940

Answer: 4. 940

Question 166. The number of enantiomer pairs formed on monochlorination of 2,3-dimethylbutane is—

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 167. There are three different asymmetric carbon atoms in a compound. The number of possible stereoisomers of this compound is—

1. 8
2. 3
3. 9
4. 6

Answer: 1. 8

Question 168. In which of the following compounds, the nucleophilic character of N-atom is maximum—

Answer: 1

Question 169. Which of the following resonance structures is incorrect 

1. x>z>y
2. z>x>y
3. x>y>z
4. z>x>y

Answer: 3. x>y>z

Question 170. Number of electrons in the p-orbital of methyl cation is—

1. 2
2. 3
3. 4
4. None of these

Answer: 4. None of these

Question 171. Which compound can exhibit geometrical isomerism—

1. Acetone-oxime
2. Isobutene
3. Acetophenone-oxime
4. Benzophenone-oxime

Answer: 3. Acetophenone-oxime

Question 172. In the given compound, hydrogen atom linked to which carbon atom is most acidic in nature—

1. C_α
2. C_β
3. C_γ
4. C_δ

Answer:   4. C_γ

Question 173. In which oi the following molecules, all the constituent carbon atoms have the same state of hybridisation—

1. HC≡C —C≡N
2. CH₃—C≡C —CH₂CH₃
3. CH₂=C=C=CH₂
4. CH₂=CH —CHO

Answer: 4.  CH₂=CH —CHO

Question 174. In which of the following molecules, all the atoms lie in the same plane—

1. CH₂=C=CH₂
2. CCI₂=C=C=CH₂
3. C₆H₅C=CH
4. CH₂=CH—C=CH

Answer: 2 and 3

Question 175. Which of the following express a homologous series—

1. Methanol, ethanol, 1-propanol
2. 1-hexene, 2-hexene, 3-hexene
3. Formic acid, acetic acid, propionic acid
4. Methane, methanol, methanal

Answer: 1 and 3

Question 176. Which of the following statements are incorrect—

1. Heat of hydrogenation of CH₃CH₂CH=CH₂ is less than that of (CH₃)₂C =CH
2. CCl₃ is more stable than CF₃
3. Bond lengths of three carbon-oxygen bonds in carbonate (CO^2-₃) ion are not equal
4. Free radicals are paramagnetic

Answer: 1 and 3

Question 177.  Which of the following orders of stability are correct—

1. CH₃⁺CHOCH₃ > CH₃⁺CHCH2OCH₃
2. F⁺CH₂>FCH₂⁺CH₂
3. PhCH₂⁺CH₂>Ph⁺CH₂
4. FCH₂COO^–< CH₃COO^–

Answer:   a and 2

Class 11 Organic Chemistry

Question 178. Which process is not represented correctly—

Answer: 1 and 3

Question 179. Which ofthe following sets represent only electrophiles ______

1. BF₃, NH₃, H₂O
2. AICI₃, SO₃, ⁺NO₂
3. ⁺NO₂, ⁺CH₃, CH₃⁺CO
4. C₂H₅^–, ^•C₂H₅, C₂H⁺₅

Answer:  2 and 3

Question 180. Delocalisation in hyperconjugation occurs—

1. In case of rr -bond electrons of C — H bond of any alkyl group directly linked to a double bond (f)
2. In case of cr -bond electrons of C — H bond of any alkyl group directly linked to a positive carbon atom
3. In case of a-electrons of C=C
4. In case of lone pair of electrons

Answer:  1 and 2

Class 11 Organic Chemistry

Question 181. Which of the following statements are incorrect—

1. Sodium extract is first boiled with dilute HC1 during detection of halogens by Lassaigne’s test
2. If in an organic compound, both nitrogen and sulphur are present, then blood-red colouration is observed during detection of nitrogen by Lassaigne’s test
3. Organic compounds which dissociate at their melting points are purified by vacuum distillation
4. In paper chromatography, the stationary phase is solid while the mobile phase is liquid

Answer: 1 and 4

Question 182. Which of the following compounds give a red colouration in Lassaigne’s test during detection of nitrogen—

Answer:  3 and 4

Question 183. Which of the following mixtures is responsible for blue colouration in Lassaigne’s test during detection of N—

1. NH₂NH₂ + charcoal
2. NH₄Cl + NaNO₃
3. C₆H₅COOH + NaNO₃
4. NH₂NH₂ + NH₄Cl

Answer: 1 and 3

Question 184. In which of the following structures, the arrangement of the four groups/atoms is different from that of  _________ 

Answer:   1 and 3

Class 11 Organic Chemistry

Question 185. Which ofthe following are optically active compounds ___________

Answer:  2 and 3

Question 186. In which of the following compounds, the number of hyperconjugable hydrogen atoms are same—

Answer:  2 and 4

Question 187. For which of the following compounds, the number of compounds formed on monobromination are same—

Answer:   1 and 2.

Class 11 Organic Chemistry

Question 188. The carbocations which attain stabilityby resonance are___

Answer: 1,3,and 4

Question 189. Mixtures that can be separated by simple distillation are

1. A mixture of ether and toluene
2. A mixture of hexane and toluene
3. A mixture of benzene and chloroform
4. A mixture of 95% ethanol and 5% water

Answer:  1 and 3

Question 190. Compounds that can be purified by steam distillation are—

1. Acetic acid
2. o-nitrophenol
3. Ethanol
4. Nitrobenzene

Answer:  2 and 4

Question 191. Which of the following statements are incorrect—

1. Quantitative estimation of nitrogen in any compound can be done by kjeldahl’s method
2. Quantitative estimation of sulphur in organic compounds can be done by Dumas method
3. Quantitative estimation of halogens in organic compounds can be done by Carius method
4. In Liebig’s method of detecting carbon and hydrogen in organic compounds carbon is converted into carbon dioxide while hydrogen is converted into water

Answer: 1 and 3

Question 192. In which of the following processes, any organic liquid vapourises below its boiling point—

1. Vacuum distillation
2. Distillation
3. Steam distillation
4. Sublimation

Answer: 1 and 3

Class 11 Organic Chemistry

Question 193. Which exhibit optical and geometrical isomerism—

1. CH₃CHCICH=CH₂
2. CH₃-CHCI —CH=CH₂
3. CH₃CH=CH— CH(CH₃)₂
4. CH₃CH=CH—CH=C=CHCH₃

Answer: 2 and 4

Question 194. Which of the following are correct statements—

1. (CF₃)₃C⁺ is more stable than (CH₃)₃C⁺ ,
2. Na⁺does not act as an electrophile
3. Ph₃C⁺ can be stored in the form of Ph₃C⁺BF^–₄ salt
4. CH₃CH₂O^– is less stable than O₂NCH₃CH₂O^–

Answer:  2 and 3

Question 195.  The nucleophilic reagents are—

1. OH^–
2. :NH₃
3. CCl₂
4. CN^–

Answer: 1,2 and 4

Question 196. The electrophilic reagents are—

1. ⁺NO₂
2. Cl⁺
3. H₂O
4. SO₃

Answer: 1, 2, 4

Class 11 Organic Chemistry

Question 197. Which of the following statements regarding Lassaigne’s test are correct—

1. N, S and halogens are detected by converting them into their corresponding inorganic salts
2. This test is done to detect N, S and hydrogen
3. Organic compounds are fused with sodium metal
4. Different halogens can be distinguished

Answer:  1,3 and 4

Question 198. Which of the following exhibit keto-enol tautomerism—

1. C₆H5COC₆H₅
2. C₆H₅COCH=CH₂
3. C₆H₅COCH₂COCH₃
4. CH₃COCH₂COCH₃

Answer:  2. C₆H₅COCH=CH₂

Question 199. Which of the following do not exhibit optical activity—

1. 3-methyl-l-pentene
2. 2-methyl-2-pentene
3. 4-methyI-l-pentene
4. 3-methyl-2-pentene

Answer: 2,3, and 4

Question 200. The correct statements are—

1. Racemic mixture is an equimolecular mixture of a pair of enantiomers
2. Configuration of a molecule means stable three¬dimensional arrangement of the groups attached to a specific atom of the molecule
3. Melting & boiling points of 2 enantiomers are different
4. A molecule may be optically active or inactive if more than one asymmetric carbon is present in the molecule

Answer: 1,2 and 4

Question 201. Which of the following are planar—

1. Ferf-butyl radical
2. Ferf-butyl carbocation
3. Ferf-butyl carbanion
4. Allyl carbanion

Answer: 1,2 and 4

Class 11 Organic Chemistry

Question 202. Which can act as both electrophile and nucleophile—

1. CH₃OH
2. CH₃Cl
3. CH₃CN
4. HCHO

Answer: 3 and 4

Question 203. Which of the following can act neither as an electrophile nor as a nucleophile

1. H₃O⁺
2. R₄N⁺
3. CN^–
4. SO₃

Answer: 1 and 2

Question 204. Which of the following conditions favour E2 reaction—

1. A strong base of high concentration
2. A solvent of low polarity
3. 3° alkyl halide as the substrate
4. Alkyl iodide

Answer: 1,2,3,4

Question 205. Compounds that will not exhibit geometrical isomerism—

Answer:  1,2,3,

Class 11 Organic Chemistry

Question 206. In which ofthe following, a plane ofsymmetryis present—

Answer: 1,2,4

Question 207. Three stereoisomers of CH₂YZ are possible if tire structure of methane be—

1. Reactangular planar
2. Square planar
3. square pyramidal
4. Octahedral

Answer: 1 and 2

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Very Short Question And Answers

Question 1. Which property of carbon is responsible for forming straight chains, branched chains, or rings?
Answer: Catenation property of carbon.

Question 2. Wind are the reason fort the existence of a large number of organic compounds.
Answer:

Reasons are: 

1. Catenation property of carbon
2. Its tendency to combine with other non-metals and
3. Phenomenon of isomerism exhibited by carbon compounds.

Class 11 Organic Chemistry

Question 3. Find the number of <r -and ;r -boud(s) in the molecule: CH₃CH₂CH==CH—C = 
Answer: Number of σ -bonds = 13 and number of π-bonds = 3

Question 4. Predict the state of hybridization of the starred carbon atoms: 

1. H*C ≡ CCH₃

2. 

Answer:

1. sp
2. sp².

Question 5. What is the shape of the molecule: C₆H₆CN?
Answer: Planar.

Question 6. Give the shape of the molecule: HC ≡ C — C ≡CCl
Answer: Linear

Question 7. What is the state of hybridization of a carbon atom linked to two other atoms by two double bonds?
Answer: Sp -hybridization.

Question 8.  Arrange the following in order of increasing carbon-carbon bond length: ethane, ethylene, and acetylene.
Answer:

Acetylene (HC ≡ CH) < Ethylene (CH₂=CH₂) < Ethane (CH₃—CH₃).

Class 11 Organic Chemistry

Question 9. What will be the shape of a hydrocarbon molecule containing two sp² – & one sp³ -hybridized C-atom?
Answer: Three-dimensional.

Question 10.  Arrange in order of increasing bond dissociation enthalpy:  C_sp— C_sp  —C_sp³ – C_sp³ —, C_sp² —C_sp² 
Answer:

C_sp³ — C_sp³ < C_sp² — C_sp²  < C_sp— C_sp

Question  11. Arrange the starred C-atoms in the following compound in order of increasing s -character of their hybridization states:
¹C*H₃—²CH=³C*H—⁴CH=⁵*C=⁶CH—⁷CH₂—⁸CH₃
Answer:

C-1 <C-3<C-5.

Question  12. Which is the correct bond-line structural formula of CH₂=CH—C = CCH₂CH₃ ?

Answer:

No. 2 s the correct bond-line structural formula.

Question 13. Write the names of an alicyclic compound and a heterocyclic compound.
Answer:

Cyclohexane and pyridine, respectively.

Class 11 Organic Chemistry

Question 14. Give one example of each benzenoid and non-benzenoid aromatic compound.
Answer: Toluene  and azulene

Question 15. Write down the IUPAC name of the compound represented by the swastika sign.
Answer: 3,3-diethyl pentane

.

Question 16. Which one is the correct name of an alkyne containing five carbon atoms? Pent-2-yne or Pent-3-yne
Answer: Pent-2-yne.

Question 17. Mention the name of the alkyl group that may be obtained by removal of one 2° H-atom from propane.
Answer:  Isopropyl.

Class 11 Organic Chemistry

Question 18. How many alkyl groups are expected to be obtained from CH₃CH₂CH₂CH(CH₃)CH₂CH₃ by the removal of different non-equivalent H -atoms?
Answer: Seven alkyl groups because there are seven types of non-equivalent H -atoms.

Question 19. Which of the following has no existence? 

1. 1° H – atom
2. 3° C -atom
3. 2° H -atom
4. 4°H -atom

Answer: 4°H-atom has no existence

Class 11 Organic Chemistry

Question 20. How many 4°C-atoms are there in 2,2.3,3- tetramethylbutane?
Answer:. The number of 4°C- atoms is 2.

Question 21. How many 3°H-atoms are there in 4-ethyl-2- methylhexan e?
Answer: The number of 3° H-atoms is 2.

Question 22. Write the bond-line structural formula of an alkane with five carbon atoms which has only primary hydrogen atoms.
Answer: Bond-line structure of the alkane:

Question 23. Give examples of two terminal functional groups.
Answer: . —COOH, — CHO

Class 11 Organic Chemistry

Question 24. How many univalent groups are expected to be obtained from toluene?
Answer: 4 univalent groups can be expected

Question 25. What are the primary suffixes used to write IUPAC names of CH₃CH₃, CH₂=CH_2, and HC ≡CH?
Answer: ‘anej ‘ene’ and ‘yne’ respectively.

Question 26. Give an example of a saturated hydrocarbon that can be represented by the general formula, C_nH_2n.
Answer: Cyclopentane

Question 27. How many 7r -bonds are there in 3-methylidene-l, 4- pentadiene?
Answer: The number of n -bonds is 3.

Question 28. Write the names of the alkyl group(s) that may be obtained from (CH₃)₄C.
Answer: Neopentyl or 2,2-dimethylpropyl.

Class 11 Organic Chemistry

Question 29. Write the IUPAC name of a hydrocarbon containing one sp, two sp² & two sp³ -hybridized C-atoms
Answer: Penta-2,3-diene or 3-methyIbuta-l,2-diene.

Question 30. How many alkyl groups are possible having the molecular formula, C₄H₉?
Answer: 4 different alkyl groups are possible.

Question 31. Which type of isomerism is exhibited by n-pentane and neopentane?
Answer: Chain isomerism.

Question 32. Write down the structure and the IUPAC name of the tautomer of butanal
Answer: CH₃CH₂CH=CH— OH (But-l-en-l-ol).

Question 33. How many structural isomers will be obtained by the displacement of two H-atoms of propane by two Cl- atoms? Write their structures.
Answer: Question 65. A compFour (ClCH₂CH₂CH,Cl, ClCH₂CHClCH₃ , Cl₂CHCH₂CH₃ and CH₃CCl₂CH₃).

Question 34. Write structures and names of two compounds which are position isomers as well as metamers
Answer:

Question  35. How are the two compounds, CH₂=CHCH₂CH₃ and related to each other?
Answer: These two compounds are ring-chain isomers.

Question 36. Which two of the following are geometrical isomers?

Answer: 2 and 4 are geometrical isomers.

Question 37. Which of the following compounds do not exhibit geometrical isomerism

1. 

2. PhCH=CHPh

3. Me2C=NOH

4. CH₂=CH—CH=CH —CH=CH₂

Answer:  1 and 3 do not exhibit geometrical isomerism

Class 11 Organic Chemistry

Question 38. Which of the given compounds are optically active?

Answer: 3 and 4 are optically active compounds

Question 39. Mention the type of the following reaction: Me₃CCH₂OH + HBr Me₂CBrCH₂CH₃ + H₂O
Answer: It is a substitution and rearrangement reaction.

Question 40. How many types of non-equivalent H-atoms are present in the given compound?

Answer: Two types of non-equivalent H-atoms are present

Question 41. In which of the given compounds, all the H-atoms are; equivalent?

Answer: 1 and 3

Question 42. Calculate the double bond equivalent (DBE) of the compound having the molecular formula, C₆H₈ Is the compound aromatic?
Answer: DBE = 3 ; The compound is not aromatic

Question 43. How many monobrow derivatives are possible for each of Ortho, meta, and para-xylene?
Answer: Three, four, and two respectively

Question 44. Arrange the following groups in order of decreasing strength of -I effect groups in order of decreasing — ⁺NR₃, —NO₂, —F, —CN.
Answer:  — ⁺NR₃ >NO₂>—CN >—F

Question 45. Arrange the following free radicals in the decreasing order of their stability:

Answer: 5 >4>1>2>3

Question 46. In which C—C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least?
Answer: C-2 —C-3 bond.

Question 47. Arrange the following carbocations in increasing order of stability

Answer: 4<2<1<3

Question 48. Arrange the following compounds in increasing order of number of hypercoagulable hydrogen atoms:

Answer: 6<2<3<5 <4<1

Question 49. Arrange the following compounds in order of increasing bond dissociation enthalpy:

1. CH₃ — H
2. (CH₃)₂CH-H
3. (CH₃)₃C—H
4. CH₃CH₂ —H

Answer: 3<2<4<1

Question 50. Arrange the following in increasing order of stability:

Answer: 4<2<3<1< 5

Question 51. How can aniline be purified?
Answer: By steam distillation as well as vacuum distillation.

Question 52. How can glycerol be purified?
Answer: By distillation under reduced pressure.

Class 11 Organic Chemistry

Question 53. Suggest a method to separate a mixture of o-hydroxybenzaldehyde and p-hydroxybenzaldehyde.
Answer: Steam distillation.

Question 54. How will you separate a mixture of two solid pounds of different solubilities in the same solvent?
Answer: By fractional crystallization.

Question  55. An organic liquid decomposes below its boiling point. How can it be purified?
Answer: By distillation under reduced pressure.

Question 56. Which technique can be used to purify iodine-containing traces of common salt?
Answer: Sublimation.

Question 57. Suggest a method for the purification of a liquid containing non-volatile Impurities.
Answer: Simple distillation

Question 58. How can aniline (b.p. 184°C) be separated from petroleum ether (b.p. 40-60°C)?
Answer: By simple distillation

Question 59. Out of water and benzene, which can be used to purify benzoic acid containing naphthalene by fractional crystallization?
Answer: Water (because both are soluble in benzene).

Question 60. Give an example of a chromatographic technique in which both the mobile and stationary phases are liquids
Answer: Paper chromatography.

Question 61. Mention two distillation processes in which organic liquids boil at temperatures below their respective boiling points. 
Answer: Distillation under reduced pressure & steam distillation.

Question 62. Explain why the Lassaigne’s extract should not be prepared by using tap water.
Answer: Because tap water contains chloride ion (Cl )

Question 63. Give an example of a compound that does not contain halogen but gives Beilstein’s test
Answer:  Pyridine (C₆H₅N)

Question 64. In Carius’s method for estimation of phosphorus, the precipitate of which compound is finally obtained?
Answer: Ammonium phosphomolybdate [(NH₄)₃PO₄-12MoO₃] or magnesium pyrophosphate (Mg₂P₂O₇).

Question 65. Give an example of a nitrogenous organic compound to which Kjeldahl’s method for the estimation of nitrogen is not applicable.
Answer: Azobenzene (C₆H₅— N=N—C₆H₅)

Question 66. What is the C — C=C bond angle value in a benzene (C₆H₆) molecule?
Answer: 120°

Question 67.  What is the H — C = C bond angle value in an acetylene (C₂H₂) molecule?
Answer: 180°

Question 68.  Mention the state of hybridization of C and N-atoms in
Answer: sp, sp²

Question 69. Mention the state of hybridization of the carbon atoms present in the molecule, CH₃CH =C =CHCH₂CH₃.
Answer: sp, sp, sp, sp, sp, sp

Question 70.  Give the name of a simple organic molecule that has a cylindrical n -electron cloud.
Answer: Acetylene

Question 71.  Give an example of a molecule in which all atoms lie in the same plane.
Answer: CH₂=CH₂

Question 72.  Give an example of a molecule in which all the atoms lie in a straight line.
Answer: HC = CH

Question 73. Calculate the number of cr and n -bonds in the molecule, CH3CH=CH—C = C —CHO.
Answer: σ -bond:12, π -bonds

Question 74. What are the possible values of n if CH₂=(C)_n—CH₂ is a planar molecule?
Answer: n = 0,2,4, — etc

Question 75. What are the possible values of n if CH₂— (C)n_n=CH₂ is a non-planar molecule?
Answer: n = 1, 3, 5, etc.

Question 76. Write the structure of a hydrocarbon molecule that contains one 4° carbon atom.
Answer:  (CH₃)₄C;

Question 77. Give an example of a compound that contains primary (1°), secondary (2°), and tertiary (3°) H -atoms.
Answer: CH₃CH₂CH(CH₃)₂

Question 78. How many alkyl groups can be derived from the alkane, CH₃(CH₂)₃CH₃?
Answer: Three

Question 79.  Write the group prefix used for the — COOH group.
Answer: ‘oic acid’;

Question 80.  Write the IUPAC name:
Answer: (3-ethyl-l-methyl)-pentyl -CH(CH₃)CH₂C(C₂H₅)CH₂CH₃

Question 81.  Write the IUPAC name: CH₂=CH—CH=CH — C = CH
Answer:  Hexa-l,3-dien-5-yne

Question 82.  Write the structure and name of an alkane having five C- atoms which on bromination gives only one monobromo derivative.
Answer:  Neopentane [(CH₃)₄C];

Question 83. Write structures of two compounds which are metamers as well as position isomers.
Answer: CH₃CO(CH₂)₂CH₃, C₂H₅COC₂H₅

Question 84. Give an example of a carbonyl compound in which tautomerism does not take place.
Answer: Benzaldehyde (C₆H₅CHO)

Question 85.  Give an example of a 3° free radical containing six hyperconjugable hydrogens.
Answer: (CH₃CH₂)₃C

Question 86.Which is the most stable carbocation having the formula,
Answer: Terf-butyl cation, Me₃C

Question 87. Which has greater resonance stabilization—PhNH₂ or PhNH₃?
Answer: PhNH₂

Question 88. Write the name of a cyclic compound that is isomeric with but-l-ene.
Answer: Cyclobutane

Question 89.  Write the names of two non-polar solvents that are commonly employed for crystallization.
Answer: Benzene and carbon tetrachloride

Question 90.  What type of furnace is used in the Carius method for the estimation of halogens?
Answer: Bomb furnace

Question 91. Mention the type of chromatography in which both the mobile and stationary phases are liquid.
Answer: Paper chromatography

Question 92. With the help of which type of distillation process glycerol can be purified?
Answer: Distillation under reduced pressure

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Fill In The Blanks

Question 1. When four alkyl groups are attached to a carbon atom, that particular C-atom is called _______________
Answer: 4°

Question 2. The shape of the molecule containing only sp² -hybridized carbon atoms is _______________
Answer: Planar

Question 3. The C-2 atom ofpropa-1,2-dieneis ________hybridised.
Answer: sp

Question 4. The shape of the molecule containing only sp-hybridized carbon atoms is_______________
Answer: Linear

Question 5. The successive members of a homologous series differ by _______________ mass units.
Answer: 14

Question 6. The molecule, HC = C—CH —CH — CH₃ contains _______________ a -bonds.
Answer: 10

Question 7. The molecule contains_______________ 2°H -atoms.
Answer: 4

Question 8. The general formula of dihydric alcohol is _______________
Answer: C_nH_2n(OH)₂

Question 9. The compound, 5-(l,2-dimethylpropyl)-6-ethyldecane contains _______________ 3° carbon atoms.
Answer: 4

Question 10. Stereoisomers have _______________ atom-to-atom bonding sequence or connectivity.
Answer: Same

Question 11. The number of isomers of a benzenoid aromatic compound having molecular formula, C₇H₈O is _______________
Answer: 5

Question 12. Ethoxyethane and 2-methoxy propane are related as _______________
Answer: 12 metamers

Question 13. The amount of negative charge present on each O-atom of carbonate ion is _______________
Answer: =-2/3

Question 14. The homolytic fission of a covalent bond requires _______________ energy than that required by its heterolytic c fission.
Answer: Less

Question 15. _______________ involves delocalization of σ -electrons of C — H bond of an alkyl group directly attached to an unsaturated system or to an atom with a vacant or singly p-orbital.
Answer: Hyperconjugation

Question 16. In paper chromatography, both the stationary and mobile phases are _______________
Answer: Liquid

Question 17. An impure sample of benzoic acid containing a little sodium chloride can be purified by _______________
Answer: Sublimation

Question 18. In steam distillation, the organic liquid boils at a _______________ temperature than its normal boiling point.
Answer: Lower

Question 19. In Cariu’s method of estimation, chlorine present in an organic compound is converted into _______________.
Answer: AgCl

Question 20. _______________ distillation is used to remove water from rectified spirit.
Answer: Azeotropic

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Warm-Up Question And Answers

Question 1. Why are the four C —Cl bonds in CCI, equivalent?
Answer:

In the formation of a CCI₄ molecule, a carbon atom with sp³ -hybridization (containing four equivalent sp³ – hybrid orbitals) uses its hybrid orbitals to form four C — Cl bonds with four Cl -atoms. So these C— Cl bonds are all equivalent.

Question 2. Which atoms in each of the following molecules lie in the same line and why?
Answer:

1. SP -carbon atoms and the atoms attached to them lie in the same line 

2.

Question 3. A π-bond is weaker and more reactive than a σ-bond. sp -carbon atoms and the atoms attached to them lie in
Answer: 

End-on overlap gives rise to σ -bonds, and lateral overlap gives rise to n -bonds. Die lateral overlap in a π-bond cannot be as effective as the overlap in a σ bond. Hence, a σ -bond is always stronger than a π -bond.

Question 4. What is the shape of each of the given compounds?

1. H₂C=O
2. CH₃CI
3. HCN

Answer:

1. Planar trigonal
2. Tetrahedral
3. Linear

Question 5. Arrange C_sp—C_sp, C_sp² —C_sp², and C_sp³ —C_sp³ σ -bonds in order of increasing bond length and explain the order.
Answer: C_sp—H < C_sp²— H < C_sp³ —H. For an explanation, see bond lengths

Question 6. Arrange C_sp—C_sp, C_sp² —C_sp², and C_sp³ —C_sp³ σtrbonds in order of increasing bond dissociation enthalpy and explain the order
Answer:

C_sp³—C_sp³, C_sp² —C_sp², and C_sp —C_sp  – For explanation see bond strength

Question 7. Which is the correct bond-line structural formula of CH₃CH₂C ≡ CCH₂CH₃: 
Answer:

Question 8. Identify the saturated compounds:

1. CH₃CH₂CH=O
2. C₂H₅OH

Answer:

1. CH₃CH₂CHO and
2. CH₃CH₂OH

Question 9. Write down the structure of an alkane that contains only primary (1°) carbon atoms and primary (1°) hydrogen atoms.

⇒ \(\mathrm{H}_3 \stackrel{\mathrm{l}^{\circ}}{\mathrm{C}}-\stackrel{1^{\circ}}{\mathrm{C}} \mathrm{H}_3\) – Ethane

Question 10. Give examples of the following:

1. A mixed ether
2. A tertiary alcohol,
3. An aromatic aldehyde
4. A mixed anhydride and (v)a secondary amine

Answer:

1. Ethyl methyl ether (CH₃CH₂OCH₃)
2. Tert-butyl alcohol [(CH₃)₃COH]
3. Benzaldehyde (C₆H₅CHO)
4. Dimethylamine [(CH₃)₂NH]

Question 11. Arrange the following functional groups in order of preference as the principal functional groups:
Answer:  —CONH₂,—NH₂,—CHO, —CN, —COOH, —O

Question 12. Write the structures and IUPAC names of two metamers having molecular formula, C₅H₁₀O.
Answer: CH₃COCH₂CH₂CH₃ (but-2-one) and CH₃CH₂COCH₂CH₃ (pentan-3-one)

Question 13. Arrange the following atoms or groups in increasing order of -l effect: — L, —Br, —Cl, —F
Answer: —I < —Br < —Cl < — F.

Question 14. Arrange in decreasing order of their strength and give Newtons; CH₃CH₂COOH > Me₂CHCOOH > Me₃CCOOH
Answer:

The strength of carboxylic acid decreases as the electron-releasing effect of the alkyl group attached to the —COOH group increases. Thus, acid strength decreases in the order: CH₃CH₂COOH > Me₂CHCOOH > Me₃CCOOH

Question 15. Why is BU₃N more basic than BuNH₂, in the C₆H₅Cl medium?
Answer:

Chlorobenzene is an aphotic solvent. In such solvents, the basic strength of the amine increases as the number of electron-donating alkyl groups on the amino nitrogen increases.

Question 16. Explain the following observation
Answer:

Resonance is inhibited due to steric hindrance. So, electrophilic substitution at p -position does not occur.

Question 17. Label the following carbonations as 1°, 2° or 3°:

 
Answer:

(1)2° (2) 3° (3) 2° (4) 1°

Question 18. Which one of the two carb anions is less stable and why?
Answer: The second one is less stable as it is antiaromatic in nature 4π -electron system).

Question 19.  What are the shapes of the free radicals ^•CH₃, ^•CF₃ and why?
Answer:

CF₃: Pyramidal (sp³ -hybridised C-atom),

CH₃ : Planar (sp² -hybridised C-atom)

Question 20. Why does: CCl₂ act as an electrophile?
Answer: Dichlorocarbene is an electron-deficient molecule. There is only a sextet of elections in the valence shell of the carbon atom of this molecule (: CCl₂). So this molecule acts as an electrophile (to fulfill the octet of carbon)

Question 21. Which one out of the S_N1 and S_N2 reactions is more susceptible to steric effect and why?
Answer: The S_N2 reaction is susceptible to steric effect because, in the transition state, the carbon atom undergoing nucleophilic attack is attached to five atoms or groups.

Question 22. Which of the following reactions do not involve an intermediate and why?— S_N1, S_N2, E1, E2
Answer: S_N2 and E2 reactions are one-step processes and hence intermediate is not involved in such reactions

Question 23. Explain the reason for the fusion of an organic compound with metallic sodium in Lasagne’s test
Answer: The purpose of the fusion of an organic compound with metallic sodium is to convert nitrogen, sulfur, and halogen present in the organic compound to water-soluble sodium cyanide, sodium sulphide, and sodium halide respectively

Question 24. How will you purify a sample of benzoic acid that contains traces of common salt? 
Answer: By sublimation. Benzoic acid sublimes by, leaving behind NaCl.

Question 25. Explain why glycerol cannot be purified by simple distillation. Mention a method that can be useful.
Answer: Glycerol cannot be purified by simple distillation because it decomposes at its boiling point. It can however be purified by distillation under reduced pressure.

Question 26. How do you separate a mixture of o-nitro phenol and p-nitro phenol?
Answer: o -and p -nitro phenol can be separated from a mixture by steam distillation.

Note: o-nitro phenol is steam volatile

Question 27. In the fusion test of organic compounds, the nitrogen of an organic compound is converted to—sodium nitrate, sodium nitrify,  sodium amide, and sodium cyanide.
Answer: Sodium cyanide

Question 28. Why is Lasagne’s extract not prepared with tap water?
Answer: Tap water often contains chloride ions, which will interfere in the test for halogen.

Question 29. Write down the formula of Prussian blue.
Answer: Fe₄[Fe(CN)₆]₃

Class 11 Organic Chemistry

Question 30. Why do diazonium salts not respond to Lasagne’s test?
Answer: Under hot conditions, diazonium salts decompose to liberate N₂ gas, and hence they do not respond to Lasagne’s test for nitrogen.

Question 31. Beilstein test cannot be considered as a confirmatory test for the presence of halogen in a compound—why?
Answer: Many halogen-free compounds,

For example – Certain derivatives of  Pyridine and quinoline, purines, acid amides, urea, thiourea, cyano compounds, etc. Give this test, presumably owing to the formation of volatile copper cyanides

Question 32. What is the role of CuSO₄ and K₂ SO₄ used in Kjeldahl’s method for the estimation of nitrogen?
Answer: Potassium sulfate increases the boiling point of H₂ SO₄ and thus ensures a complete reaction, while copper sulfate catalyzes the reaction.

Question 33. Which method is used to estimate N in foodstuffs
Answer: Kjeldahl’s method is largely used for the estimation of nitrogen in foodstuffs, drugs, and fertilizers

Question 34. For which compounds, Kjeldahl’s method is not applicable for the estimation of nitrogen?
Answer: Kjeldahl’sand Techniques method does not apply to compounds] containing nitrogen in the ring

For example –  Pyridine, quinoline, etc.) and compounds containing nitrogen directly linked to an oxygen atom, NO_2, or another nitrogen atom i.e., azo (— N=N— ) compounds.

Question 35. The weight of which compound is finally taken in the Carius method for the estimation of phosphorus
Answer: In this method, the amount of ammonium phosphomolybdic formed is weighed in the final step

Class 11 Chemistry Hydrocarbons Long Questions And Answers

Class 11 Hydrocarbons Q&A

Question 1. How can an eclipsed conformation of ethane be converted into a staggered conformation?
Answer:

In an ethane molecule, if one carbon atom is kept fixed around the C—C bond axis and the other carbon atom is rotated at a minimum angle of 60°, then the eclipsed conformation is converted to the staggered conformation.

Question 2. Give examples of a chiral conformation and an achiral conformation of n-butane.
Answer:

Gauche-staggered conformation of n-butane is chiral because it cannot be superimposed on its mirror image.

However, the fully eclipsed conformation of n-butane is achiral as it can be superimposed on its mirror image.

Question 3. Arrange the following conformations of n-butane according to their increasing stability:

1. Gauche-staggered
2. Fully eclipsed
3. Eclipsed and
4. Ante-Mggered

Answer:

When 2-iodopropane is used as the alkyl halide in the Wurtz reaction, the alkane obtained is 2,3-dimethylbutane.

Question 4. Which of the following alkanes cannot be prepared by the Wurtz reaction in good yield?

1. (CH₃)₂CHCH₂CH(CH₃)₂
2. (CH₃)₂CHCH₂CH₂CH(CH₃)₂
3. (CH₃)₃CCH₂CH₂CH₂CH₃
4. CH₃CH₂C(CH3)₂CH₂CH₃
5. (CH₃)₃C-C(CH₃)₃

Class 11 Hydrocarbons Q&A

Answer:

(1), (3) and (4) are three unsymmetrical alkanes. So, these cannot be prepared by Wurtz reaction in good yield. Again, for preparing alkane (5), a 3° alkyl halide is required. So, in spite of being a symmetrical alkane, (5) cannot be prepared by the Wurtz reaction.

Question 5. How will you prepare methane and ethane starting from ethanoic acid?
Answer:

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 6. How many monochloro derivatives are obtained on chlorination of n-pentane, isopentane and neopentane? Write down their structures.
Answer:

There are three and four types of non-equivalent hydrogen atoms in n-pentane (CH₃CH₂CH₂CH₂CH₃) and isopentane [CH₃CH(CH₃)CH₂CH₃] respectively. Whereas, in neopentane [(CH₃)₄C], all H -atoms are equivalent.

Therefore, chlorination of n-pentane, isopentane and neopentane form three, four and one monochloride derivatives respectively.

Question 7. Chlorination of cyclohexane to prepare chlorocyclohexane is more practicable than the chlorination of methylcyclohexane to prepare l-chloro-l-methylcyclohexane— explain.
Answer:

There are five types of non-equivalent H -atoms in methylcyclohexane

When it undergoes chlorination, four other monochloride derivatives are formed along with 1-chloro-1-methylcyclohexane

As a result, the yield of the desired product is low’—’ and Low it is difficult to separate the product from the mixture. On the other hand, all H -atoms in cyclohexane are equivalent and thus, only chlorocyclohexane

Is formed as the product for this reason, Achlorination of cyclohexane to prepare chlorocyclohexane is more feasible than the chlorination of methylcyclohexane to prepare 1-chloro-l-methylcyclohexane.

Class 11 Hydrocarbons Q&A

Question 8. Write the IUPAC names of the following compounds:

Answer:

Question 9. Write the structural formula:

1. 3-(1- methyl ethyl) hex-2-ene; 
2. 4-ethyl- 2, 4- dimethyl hept-1- ene

Answer:

Question 10. Write the IUPAC names and structures of the alkenes having the molecular formula C₅H₁₀.
Answer:

IUPAC names and structures of the alkenes having the molecular formula C₅H₁₀ are as follows

Question 11. Write the mechanism of acid-catalysed dehydration of butyl alcohol.
Answer:

Dehydration of isopropyl alcohol in the presence of concentrated H₂SO₄ is an El reaction. The reaction occurs in three steps. The second step of the reaction is the slowest, i.e., it is the die rate-determining step of the reaction.

Step 1: Protonation of the alcohol.

Step 2: Elimination of water molecules and formation of carbocation

Step 3: Elimination of proton from carbocation

Question 12. Write the structures of A and B obtained in from given reactions
Answer:

Answer:

A is R —CHBr —CH₃ and B is RCHBrCH₂Br. The alkene, HBr and the formed alkyl bromide (A) are all colourless. So, the left reaction cannot be used to detect ethylenic unsaturation.

On the other hand, the alkene and the formed dibromoalkane (B) are colourless but bromine has a reddish-brown colour.

So, the right reaction can be used to detect ethylenic unsaturation because decolourisation of bromine takes place in this reaction.

Question 13. How can a double bond be created in a molecule of a compound which has a carbon-carbon single bond?
Answer:

A double bond is created in a molecule of a compound containing a carbon-carbon single bond by the can given method.

Question 14. Which reaction is used to detect ethylenic unsaturation and why? Write the structures and IUPAC names of the compounds expected to be obtainedin thegiven reactions:

⇒ \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HCl}→{\text { Peroxide }}\)

⇒ \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr}→{\text { Peroxide }}\)

Answer:

Class 11 Hydrocarbons Q&A

Question 15. Write the product of the given reaction. Explain its formation:

Answer:

Class 11 Hydrocarbons Q&A

Due to -R and -I -effect of — NO₂ group, carbocation [I] is less stable than carbocation [II]. So the reaction proceeds through carbocation [II] and the major product formed is

Question 16. State Markownikoffs rule. Explain with an example. How would you convert ethylene to acetylene? Identify the compound in the reaction—

Answer:

Markownikoffs rule and example: Conversion of ethylene to acetylene

Question 17. Ethane can be dried by passing through concentrated H2S04 but not ethylene—why?
Answer:

Ethane being a saturated hydrocarbon does not react with concentrated H₂SO₄

CH₃ —CH₃  (Ethane) + cone. H₂SO₄ . So, ethane can be dried by passing through concentrated H₂SO₄.

On the other hand, ethylene being an unsaturated hydrocarbon, when passed through concentrated H₂SO₄ gets absorbed by the acid and forms ethyl hydrogen sulphate. So, ethylene cannot be dried by passing through concentrated H₂SO₄.

Class 11 Hydrocarbons Q&A

Question 18. Identify the compounds A, B and C in the following reaction and write their names:

Answer:

Carbonyl compounds B and C contain three and two carbon atoms respectively. There are also three carbon atoms on one side of the double bond and two carbon atoms on the other side of the double bond in the alkene.

Therefore, two alkenes with molecular formula C₅H₁₀ are: 2-methylbut-2-ene [CH₃ —C(CH₃)=CHCH₃] and pent-2-ene (CH₃CH₂CH=CHCH₃) . If A (C₅H₁₀) is 2-methylbut-2-ene, thenB (C₃H₆O) &C(C₂H₄O) areacetone (CH₃COCH₃) and acetaldehyde (CH₃CHO) respectively.

If A is pent-2-ene, then B and C are propanal (CH₃CH₂CHO) and acetaldehyde (CH₃CHO) respectively.

Class 11 Hydrocarbons Q&A

Question 19. An alkene on ozonolysis produces propanone and propanal. The alkene is—

1. 2-methyl pent-2-ene
2. 3-methyl pent-2- ene
3. A-methyl pent-2-ene
4. Hex-3-ene.

Answer:

The products formed on ozonolysis are propanone and propanal. Therefore, the alkene can be determined as—

The alkene is (1) 2-methylpent-2-ene.

Question 20. How can cis- and trans-hydroxylation of cis-2-butene be carried out? Comment on the optical activity of theformed products.
Answer:

Osmium tetroxide adds to the double bond of cis-2- butene to form osmic ester which is hydrolysed by an aqueous ethanolic solution of sodium bisulphite. In this case, the two —OH groups get attached to the doubly bonded carbon atoms from the same side of the double bond and form 1,2-diol.

So, cis-hydroxylation takes place in case of this reaction. On the other hand, cis-2-butene reacts with peracids to form the corresponding epoxide. The resulting epoxide on hydrolysis with dilute acid or alkali yields 1,2-diol.

Epoxidation followed by hydrolysis causes the addition of two —OH groups from the opposite sides of the double bond. So, trans-hydroxylation takes place in case of this reaction.

Question 21. What product would you expect from the following reaction?

Class 11 Hydrocarbons Q&A

Answer:

An alkyne should be formed when a vicinal dihalide is refluxed with ethanolic KOH. However, in the given case, an alkyne docs not form because a six-membered ring system cannot accommodate a linear portion like C—C=C —C. So, the compound formed is 1,3-cyclohexadiene.

Question 22. Identifyhand B in the following reactions

Answer:

Question 23. Give IUPAC names of the following compounds:

1. CH₃CH₂-CH=C=CH₂

2.

3.  CH₂=CH-CH(CH₃)-(CH₃)C=CH₂

Answer:

IUPACname: 1,2-pentadiene.

IUPAC name: 1-methyl-1,4-cyclohexadiene.

IUPAC name: 2,3-dimethyl-1,4-pentadiene

Class 11 Hydrocarbons Q&A

Question 24. What are the two planar conformations of 1,3-butadiene? Which conformation is less stable and why?
Answer:

The two planar conformations of1,3-butadiene are—

Due to steric hindrance or strain, s-ds-conformation is less stable.

Question 25. Between 1,3- and 1,4-cyclohexadiene, which compound has a lower value of heat of hydrogenation and why?
Answer:

1,3-cyclohexadiene being a conjugated diene is more stable than 1,4-cyclohexadiene which is a nonconjugated diene. So, heat of hydrogenation of 1,3- cyclohexadiene has a lesser value than that of 1,4- cyclohexadiene.

Question 26. Calculate the double bond equivalent of benzene from its molecular formula.
Answer:

Double bond equivalent (DBE) of compound,

⇒ \(\mathrm{DBE}=\frac{\sum n(v-2)}{2}+1\)

Where, n is the number of different atoms present in the molecule and v is the valency of each atom. The molecular formula of benzene is C₆H₆.

So, DBE of benzene = \(\frac{6(4-2)+6(1-2)}{2}+1=4\)

Question 27. What knowledge about the carbon-carbon bond length in benzene may be obtained from valence bond theory?
Answer:

The benzene molecule is a resonance hybrid of Kekule structures (1) and (2) and the contribution of each hybrid structure is 50% i.e., equal.

The single bonds (C—C) and the double bonds (C—C) in structure (1) become double and single bonds respectively in structure (2). As the two equally stable resonance structures (1) and (2) contribute equally to the hybrid.

it may be said that, any two adjacent carbon atoms of a benzene molecule are linked by a bond intermediate between a single and a double bond. So, all the carbon-carbon bonds of benzene are equivalent and their lengths are equal (1.39A).

Again, the bond order of each bond is the same (1.5). So, it can be said that all carbon-carbon bonds are equal in length.

⇒ \(\text { Bond order }=\frac{\text { Double bond }+ \text { Single bond }}{2}=\frac{2+1}{2}=1.5\)

Class 11 Hydrocarbons Q&A

Question 28. Which of the following representations is correct and why?

Answer:

According to representation

(1), it seems that structures (1) and (2) have a separate existence. There is no separate existence of structures (1) and (2).

So, (1) and (2) cannot be related by ’. Thus, the representation (i) is incorrect. Since (1) and (2) are two resonance structures which have no separate existence. So, (1) and (2) can be related by’ •*-»ÿ ‘. Thus, the representation

(2) is correct.

Question 29. What is the basic difference between aromatic and anti¬ aromatic compounds?
Answer:

Monocyclic planar conjugated polyene systems containing (4n + 2) delocalised; π-electrons (n = 0, 1,2,3, are called aromatic compounds.

Monocyclic planar conjugated polyene systems containing 4n delocalised π- electrons (n = 1,2,3,…) are called antiaromatic compounds.

Question 30. Will cyclooctatetraene exhibit aromatic character? Explain.
Answer:

Since cyclooctatetraene does not contain (4n + 2)n electrons, it does not exhibit an aromatic character.

As cyclooctatetraene has 4n – electrons (n=2), it should be an antiaromatic compound. However, the ring of this compound is very large in size and so it does not exist in the unstable planar shape, rather it forms a tub shaped structure. As a result, conjugation is lost and so cyclooctatetraene is a non-aromatic compound.

Class 11 Hydrocarbons Q&A

Question 31. Using the theory of aromaticity, explain the finding that A and B are different compounds, but Cand D are identical.
Answer:

As A is an antiaromatic compound (4nπ -electron system, n = 1 ), it becomes unstable due to the delocalisation of π electrons. As delocalisation of π -electrons does not take place for A, B is not the resonance structure of A.

B is the structural isomer of A. So, A and B are two different compounds. Again, C is an aromatic compound [(4n + 2)π electron-system, n = 1 ] which attains stability due to the delocalisation of electrons. So, delocalisation of electrons takes place for C. D is the resonance structure of C, i.e., Cand D are same compound.

Question 32. Classify each of the given species aromatic, antiaromatic and nonaromatic

Class 11 Hydrocarbons Q&A

Answer:

1. Is an antiaromatic compound because the B atom contains a vacant p -orbital,
2. Is a non-aromatic compound because one carbon atom of the ring does not have a p -orbital.
3. Behaves as an aromatic compound with (4n + 2)π-electrons (n = 1) because of the vacant d -orbital and lone pair of electrons of the sulphur atom

Is an aromatic ion with (4n+2)π- electrons, n = 0

Question 33. Which is the smallest aromatics species?
Answer:

The smallest aromatic species is cyclopropenyl cation.

Question 34. Writestructuralformulae of isomeric nitrotoluenes.
Answer:

Structural formulae ofisomeric nitrotoluenes are

Question 35. Write structural formulae of isomeric dibromophenols.
Answer:

Structural formulae of isomeric dibromo phenols are

Class 11 Hydrocarbons Q&A

Question 36. More than three dibromobenzenes are not possible—explain.
Answer:

Considering the resonance structures of benzene, it is easy to understand that positions 1,2- and 1,6- are indistinguishable. Similarly, positions 1,3- and 1,5- are indistinguishable. Thus, in case of bromobenzene, only three isomers are possible which are as follows

Question 37. Write the IUPAC names of the given compounds

Class 11 Hydrocarbons Q&A

Answer:

1. 1,2-dihydroxybenzene
2. 1-phenylpropanoid-l-one
3. 2-hydroxybenzoic acid
4. Al-phenylethylamine
5. l-bromo-3-chlorobenzene
6. 3-phenylpropanoid acid
7. 2,4,6-trinitrotoluene
8. 4-hydroxy-3-methoxy benzaldehyde

Question 38. Write structures and IUPAC names:

1. Mesitylene
2. Styrene
3. Pyrogallol
4. Picric acid
5. Salicylaldehyde
6. Benzophenone
7. TNT
8. Phthalic acid
9. Anthranilic acid.

Answer:

Class 11 Hydrocarbons Q&A

Question 39. Classify the following groups as o-/p-or m- directing group and activating or deactivating group:

1. -NO₂
2. -Cl
3. -C₂H₅
4. -CP₃
5. OH
6. — NHCOCH₃
7. —NH₃
8. — O
9. —COCH₃

Answer:

1. — NO₂ (deactivating and m -directing),
2. —Cl (deactivating and o/p -directing),
3. —C₂H₅ (activating and o-lp- directing),
4. — CF₃ (deactivating and m -directing),
5. —OH (activating and o-lp- directing),
6. — NHCOCH₃ (activating and o-lp- directing),
7. — NH₃ (deactivating and m directing),
8. — Oe (activating and o-lp-directing
9. — COCH₃ (deactivating and m -directing).

Question 40. Explain each of the following observations:

1. Although —Cl is a deactivating group, it is o-lp-directly.
2. The —CH₃ group is an o-/p- directing group, even though the carbon atom contains no unshared pair of electrons.
3. The —OCH₃ group is an activating and o-/p directing group.
4. The — CCl₃ group is a m -directing group, even though the carbon atom is not bonded to a more electronegative atom bya double or triple bond.

Answer:

3. —OCH₃  group is an o-/p-directing group because an unshared pair of electrons on O-atom participate in resonance (+R -effect) and increase the electron density of the ring at ortho- and para-positions.

So, electrophiles (E) preferably enter the ortho- and para-positions. Due to an increase in electron density, the ring becomes more activated than the unsubstituted benzene towards an electrophilic substitution reaction. Thus, — OCH₃ is an activating group.

4. —CCl₃ is an electron-withdrawing group because of its -I effect which is attributed to the presence of three highly electronegative Cl -atoms. Consequently, it decreases the electron density of the benzene ring, especially at the ortho- and para-positions.

So, —CCl₃ is a deactivating group which makes the ring less reactive towards electrophilic substitution and substitution occurs preferably at meta-position.

Class 11 Hydrocarbons Q&A

Question 41. How will you prepare benzene from the given compounds?

1. C₆H₅COOH
2. C₆H₅CMe₃ 
3. C₆H₅CH₂Cl
4. C₃H₅Br

Answer:

Class 11 Hydrocarbons Q&A

Question 42. Write two processes to convert C₆H₆ into C₆H₅D.
Answer:

Question 43. Write the formed.

 

Answer:

Class 11 Hydrocarbons Q&A

Question 44. Why groups like — CHO, — NO₂, — B(OR) — PBr₃ and — SR₂ act as meta-directing groups?
Answer:

Since the carbon atom of the:

1. — CHO group is bonded to the oxygen atom by a double bond, the nitrogen atom of the
2. — NO₂ group is linked with the oxygen atom by a double bond, the boron atom of the
3. —B (OR)₃ group contains a vacant p -orbital and the phosphorus and sulphur atoms of the groups

—⁺ PBr₃ and — ⁺SR₂-, have vacant d-orbitals, all of these groups reduce the electron densities of ortho- and para- positions by their -R effect. Consequently, the electron density at the metaposition becomes relatively higher and the electrophile preferably enters the meta-position. Thus, these groups behave as meta-directing groups

Question 45. Arrange the compounds in increasing order of their rate of nitration and give reason: Benzene, Toluene, Nitrobenzene, Hexadeuterobenzene (C₆D₆).
Anwer:

Increasing order of rate of nitration of given compounds is: nitrobenzene < benzene = hexadeuterobenzene < toluene. The electron-attracting nitro (— NO₂) group decreases the electron density of the nitrobenzene ring and as a result, its nitration occurs at a rate slower than that of benzene.

On the other hand, the electron-repelling methyl (— CH₃) group increases the electron density of the toluene ring and as a result, its nitration proceeds at a rate faster than that of benzene. Benzene and hexadeuterobenzene (C₆D₆) undergo nitration at the same rate because in aromatic electrophilic substitution reaction, cleavage of C — H or C — D bond does not occur at the rate-determining step.

Class 11 Hydrocarbons Q&A

Question 46. A mixture of benzene and bromine solution remains unchanged for indefinite period oftime, but ifan iron nail is added to the solution, bromination of benzene occurs rapidly—explain.
Answer:

Benzene is an aromatic compound having no ethylenic unsaturation. So benzene does not participate in addition reaction with bromine. Again, substitution reaction of benzene does not take place with the poor electrophile bromine alone.

So a solution of bromine in benzene remains stable (f.e., unchanged) for indefinite period of time. However, when an iron nail is added to the solution, bromination of benzene occurs to yield bromobenzene because iron then acts as a halogen carrier. The red solution of bromine becomes colourless

2Fe + 3Br₂→2FeBr₃; Br₂ + FeBr₃ → Br⁺ FeBr^–

C₆H₆ + Br⁺ FeBr^–₄ →C₆H₅Br + HBr + FeBr₃

Question 47. Write the monosubstituted compounds formed in each of the following reactions and state whether each reaction is faster or slower than that of benzene.

1. Nitration of C₆H₅NHCOCH₃,
2. Bromination of C₆H₅CBr₃
3. Chlorination of C₆H₅CMe₃
4. Nitration of C₆H₅—C₆H₅
5. Nitration of C₆H₅—COOCH₃
6. Sulphonation of C₆H₅CHMe₂
7. Nitration of C₆H₅CN,
8. Bromination of C₆H₅I,
9. Nitration of C₆H₅-C₆H₄C₆H₅

Answer:

1. p-O₂NC₆H4NHCOCH₃ (for this compound nitration occurs faster than benzene).

2. m-BrC₆H₄CBr₃ (for this compound bromination occurs slower than benzene)

3. p-ClC₆H₄CMe₃ (for this compound chlorination occurs faster than benzene)

4. p-O₂NC₆H₄C₆Hg (for this compound nitration occurs faster than benzene)

5. m-O₂NC₆H₄COOMe (for this compound nitration occurs slower than benzene)

6. p-HSO₃C₆H₄CHMe₂ (for this compound sulphonation occurs faster than benzene)

7. m-O₂NC₆H₄CN (for this compound nitration occurs slower than benzene)

8. p- BrC₆H₄I (reaction occurs slower than benzene)

9.

(reaction occurs faster than benzene in the middle ring because it is attached to two activating — CgH5 groups on both sides.)

Class 11 Hydrocarbons Q&A

Question 48. Write three methods by which alkyl side chains can be introduced into the benzene ring
Answer:

The: methods by which alkyl side chains, can’ be introduced into the benzene ring are—

1. By using (a) CH₃CH₃X, AlCl₃,  CH₂=CH₂ HF and CH₃CH₂OH, BF₃ or concentrated H₂SO₄ in Friedel-Crafts alkylation reaction.

2. By acylation of benzene using CH₃COCI or (CH₃CO)₂O, AlCl₃ followed by . Clemmensen reduction  the formed ketone.

4. By reacting CH₃CH₂Br with Ph₂CuLi according to Corey-House synthesis.

Question 49. Classify the following groups based on their orientation and reactivity:

Answer:

1,4 and 5 are activating and ortho-/para- directing groups. 3 and 6 are deactivating and meta-directing groups. 2 is a deactivating group (due to — NO₂ ) and is ortho-/para directing >C=C<

Question 50. 1-butyne and 2-butyne are allowed to react separately with the reagents given below:

1. Na, liquid NH₃ ; 
2. H₂ (1 mole), Pd-BaSO₄, quinoline,
3. H₂SO₄,H₂O, H₂SO₄ ; 
4. H₂/Pt.

Which reagent(s) will produce the same product in both cases? Write the structures, of products formed in these cases
Answer:

Both 1-butyne & 2-butyne react separately with reagents 1 and 2 to produce 1-butene and 2-butene respectively

However, reagents(3 & 4 react with 1-butyne & 2-butyne separately to yield same products (2-butanone & butane)

Class 11 Hydrocarbons Q&A

 

Question 51. Ethyne reacts with dil. H2S04 in presence of Hg2+ salts to give acetaldehyde, but with HC1, under similar conditions, it gives vinyl chloride. Account for such observation.
Answer:

In the first step, ethyne reacts with Hg²⁺ in to form cyclic complex (I) . This is then attacked by more nuleophilic H₂O , 2— rather than, weakly nucleophilic SO₄, to form unstable vinyl alcohol which then tautomerism to give acetaldehyde

It HCl is used instead of H₂SO₄ then the complex (I) is attacked by more nucleophilic Cl-, rather than weakly nucleophilic H₂O, to give vinyl chloride

An alkane has a molecular mass of 72. Give structure of all the possible isomers along with their IUPAC OH names

Let the alkane be C_nH_2n+2. Its molecular

=12n + (2n + 2) = 14n + 2

.*. 14n + 2 = 72 , thus n – 5 and hence the alkane is C₅H₁₂

The isomers of the alkane C₅H₁₂ are—

Question 52. Find the number of structural and configurational isomers of a bromo compound C₅H₉Br formed by the addition of HBR to 2-pentyne.
Answer:

Addition of one molar proportion of HBr to CH₃—CH₂—C=C—CH₃ produces two structural isomers 1 and 2

Each of these structural isomers can exist as a pair of geometrical isomers (cis and trans) and hence there are four possible configurational isomers

Question 53. Identify the products P and Q in the following reaction:

Answer: In absence of light, the reaction occurs via polar mechanism

Question 54. Identify the product ‘T’ in the following reaction and the major product. account for its formation.

Answer:

The product‘T is iodobenzene.

Explanation: Since I is less electronegative than Cl, so I+ is the effective electrophile that takes part in the reaction

Question 55. Identify the major product obtained on; monobromination (Br₂/FeBr₃) of meta methyl anisole and account for its formation
Answer:

Both —CH₃ and —OCH₃ are o-/p-directing groups. Therefore, the possible positions of attack which are facilitated by these groups are indicated by arrows as shown below

Attack by the’electrophile (Br⁺) is disfavoured at C₂ because this position is most crowded. Again -I effect of — OCH₃ group does not favour attack at C₆ . So most favourable attack occurs at C₄ , thereby producing 4-bromo-3-methylanhole as

Question 56. The enthalpy of hydrogenation of cyclohexene is -119.5 kj. mol^-1 ,. If the resonance energy of benzene is 150.4 kj. mol^-1 , estimate its enthalpy of Br2/FeBr3 hydrogenation.
Answer:

Enthalpy of hydrogenation of cyclohexene

=-119.5 kj- mol^-1

So enthalpy of hydrogenation of hypothetical cyclohexatriene

= 3 × -119.5 kj- mol^-1,

In otherwards, the calculated (or theoretical) enthalpy of hydrogenation of benzene =-3 × -119.5 kj- mol^-1,

Let the actual (i.e., experimental) enthalpy of hydrogenation of benzene = × kj- mol^-1

Now, R. E. of benzene = calculated enthalpy of hydrogenation of benzene- actual enthalpy of hydrogenation of benzene

o,-150.4 = -3 X 119.5 —x

x = -3 ×119.5 + 150.4 = -208.1 kj. mol^-1

Question 57. How low will you prove:

1. Acidic character of acetylene.
2. Presence of terminal =CH2 group in 1-pentene.
3. Presence of acetylenic hydrogen in 1-butyne.
4. 2-butene is a symmetrical alkene. 1-butyne

Answer:

When acetylene is added to water and shaken, the resulting solution does turn blue litmus red. The following reactions in which the H-atoms of acetylene are replaced by metal atom prove the acidic character ofacetylene.

1. HBr (a) Acetylene (HC=CH) reacts with sodium in two steps to form monosodium acetylide (HC = CNa) and disodium acetylide (NaC = CNa) respectively and in each case, H is evolved.

2. When acetylene gas is passed through ammoniacal Cu₂Cl₂ or AgNO₃ solution, metallic acetylide is precipitated in each case

3. Ozonolysis of 1-pentene leads to the formation of formaldehyde (HCHO) as one of the products. This proves that there is a terminal =CH₂ group present in 1-pentene

When 1-butyne is treated with ammoniacal cuprous chloride, a red precipitate of cuprous 1-butynide is obtained. Again, when an aqueous solution of silver nitrate is added to the alcoholic solution of1-butyne, a white precipitate of silver 1-butynide is obtained

Ozonolysis of any symmetrical alkene results in the formation of only one carbonyl compound (2 moles). Since 2-butene, on ozonolysis, produces two moles of acetaldehyde (CH₃ CHO), it must be a symmetrical alkene. Its structure is: CH₃ CH=CH CH₃

Question 58. Identify the major product obtained in each of the following reactions and explain its formation

Answer:

1. 

Although the alkene is an unsymmetrical one, Markownikoff’s rule is not directly applicable here because there are same number of H-atoms attached to double-bonded carbons. Out of the 2 carbocations ( C₆H₅CHCH₂CH₃ & C₆H₅CH₂CHCH₃ ) obtained in the first step of the reaction, the first one (a benzylic carbocation) is more stable because it is stabilised by resonance involving the benzene ring. So, this carbocation is formed more easily and readily and in the second step, it combines with Br ion to give 1-bromo-l-phenylpropane as the major product

2. 

This reaction occurs according to Markownikoff’s rule. Chlorine is more electronegative than iodine. So, in I^δ+– Cl^δ- molecule, the I-atom with a partial positive charge combines first with the alkene as an electrophile. Between the two carbocations, (CH₃)₂⁺CCH₂I and  (CH₃)₂⁺CICH₂ formed in the first one being a 3° carbocation is relatively more stable. So, it is formed more easily and readily and in the  second step, it combines with Cl to give 2-chloro-liodo-2-methylpropane as the major product

3. 

Double bond is more reactive than triple bond towards electrophilic addition reactions. For this reason, bromine (1 mole) is added mainly to the double bond of the compound to produce 4, 5 -dibromopent-1-yne as the major product.

Question 59. Write the names and structures of the two alkenes (molecular formula: C₄H₆) which give the compound when added to HBr in the absence of organic peroxide but different compounds when added to HBr in the presence of peroxide
Answer:

When but-1-ene (CH₃CH₂CH=CH₂) and but-2-ene (CH₃CH=CHCH₃) react with HBr in absence of an organic peroxide, the same product is obtained.

However, in the presence of an organic peroxide, the addition of HBr to but-1-ene occurs contrary to Markownikoff’s rule and hence differences are obtained from these two cases

1.

2.

Question 60. Mention two reactions in which ethylene and benzene behave differently and two reactions in which they behave similarly.
Answer:

Benzene, unlike ethylene, fails to discharge the red colour of bromine in CCl₄ or the reddish-violet (purple) colour of potassium permanganate solution because, unlike ethylene the π -electron system of benzene possesses extraordinary
O₂ stabilisation.

So, in these two reactions ethylene and benzene behave differently.

In the following two reactions, ethylene and benzene behave similarly.

1. Both ethylene and benzene bums with sooty flame to produce CO₂ and H₂O

⇒ \(\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒  \(2 \mathrm{C}_6 \mathrm{H}_6+15 \mathrm{O}_2 \longrightarrow 12 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}\)

2. Both ethylene and benzene react with ozone to form ozonide (an additional

⇒ \(\mathrm{C}_2 \mathrm{H}_4+\mathrm{O}_3 \longrightarrow \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_3\) (Ethylene (ozonides)

⇒  \(\mathrm{C}_6 \mathrm{H}_6+3 \mathrm{O}_3 \longrightarrow \mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_9\) (Benzene tri ozonide)

Question 61. Distinguish between:

Answer:

1. Toluene (C₆H₅CH₃), on oxidation by alkaline KMnO₄ solution followed by acidification gives a white crystalline precipitate of benzoic acid (C₆H₅COOH). On the other hand, terf-butylbenzene does not undergo such an oxidation reaction by an alkaline KMnO₄ solution.

2. O -xylene and m -xylene, on oxidation by alkaline KMnO₄ solution, produce phthalic acid and isophthalic acid respectively. Phthalic acid, when heated, forms phthalic anhydride which responds to phthalein test On the other hand, isophthalic acid on heating does not produce any anhydride.

Question 62. Write the name & structure of the following compounds:

1. An unsaturated aliphatic hydrocarbon which forms monosodium salt.
2. An organic compound which causes depletion of the ozone layer.
3. An alkane which is used as a fuel for household cooking.
4. An alkyl bromide which reacts with alcoholic KOH to form only 1-butene.
5. An alkene which reacts with HBr in the presence or absence of peroxide to give the same product.
6. A compound containing iodine which, when heated with silver powder, produces acetylene.
7. An alkyl bromide (C₄H₉Br) does not participate in Wurtz reaction.
8. An alkene which on ozonolysis forms glyoxal and formaldehyde.

Answer:

1. Propyne (CH₃C= CH).
2. Dichlorodifluoromethane (CF₂Cl₂).
3. Butane(CH₃CH₂CH₂CH₃).
4. 1-bromobutane (CH₃CH₂CH₂CH₂Br).
5. 2-butene (CH₃CH=CHCH₃).
6. Iodoform (CHI₃).
7. Tert-butylbromide (Me₃CBr).
8. 1,3-butadiene (CH₂=CH—CH=CH₂)

Question 63. Identify A … G in the following reaction sequence

Answer:

Question 64. Write structures and names of the compounds A to Q in the following reaction sequences:

Answer:

Question 65. How will you distinguish between each of the given pairs of compounds by a single chemical test? 

1. Ethylene and acetylene
2. Ethane and acetylene
3. 1-butyne and 2-butyne
4. Ethane and ethylene.
5. Propene and propyne
6. 1-butene and 2-butene
7. 2-pentene and benzene
8. Benzene and cyclohexene.

Answer:

The distinction between two compounds should be written in a tabular form. A reagent which either causes a colour change, or evolution of a gas or the appearance of a precipitate should be selected for this purpose.

Question 66. How will you carry out the following transformations:

1. Acetylene → Acetone
2. Acetylene →Dldeuteroacetylene(C₂D₂)
3. Acetylene → Acetylenedicarboxyllc acid
4. 1-butyne → 2-butyne
5. Propene → 1-propanol  Propyne →  Propanal

Answer:

Question 67. Boiling points of three isomeric pentanes are 36.2°C, 28°C and 9.5°C respectively. Identify the compounds and give reason.
Answer:

Strength of van der Waals forces depends on the area of products, A and B obtained in the following reactions: contact between molecules. Area of contact between straight-chain n -pentane (CH₃CHCH₂CH₂CH₃) molecules is maximum. So, the extent of van der Waals’ attraction among its molecules is maximum. For’this reason, its boiling point is highest (36.2°C) . On the other hand, area of contact between spherical neopentane [(CH₃)₄C] molecules is minimum.

So, _ the extent of van der Waals’ attraction among its molecules is the minimum. For this reason, its boiling point is the lowest (9.5°C) . Again, the area of contact between isopentane [(CH₃)₂CHCH₂CH₃] molecules is intermediate between n-pentane and neopentane and so, its boiling point (28°C) is intermediate between the other two isomers.

Question 68. Write the structure and the name of the monobromoderivative which is obtained as the major product when n-butane reacts with bromine in the presence of light. Why is it produced in larger amount?
Answer:

n -butane reacts with bromine in the presence of light to give 2-bromobutane as the major product.

The reaction occurs through free radical mechanism. As 2° free radical (CH₃CH₂CHCH₃) is relatively more stable than 1° free radical (CH₃CH₂CH₂CH₂) , displacement of 2° H-atom occurs rapidly to give 2-bromobutane as major product

Question 68. Possible methods for preparation of 4-methyl-2- pentyne arc given. Which method in desirable & why?

Answer:

In both the methods given, the reaction in the second step (reddish-brown) proceeds through S_N2 pathway and it is known that an S_N2 reaction is very susceptible to steric effect So, the product will be obtained in good yield if in the second step, methyl or primary alkyl bromide is used. In the second step of the second method, methyl bromide (CH₃Br) has been used. Hence, the second method is desirable.

Question 70.  Three separate cylinders contain methane, ethylene acetylene respectively. How will you identify them?
Ana.

The three gases are first separately passed through the ammoniacal solution of cuprous chloride. The gas, which gives a red precipitate, is acetylene. The gases in the remaining two cylinders are separately passed through a solution of bromine in CCl₄. The gas, which decolourises the reddish-brown solution of bromine, is ethylene. Hence, the remaining gas in the other cylinder is methane.

Question 71. Give example alkene which on oxidation by acidic solution of KMn04 or on ozonolysis gives the same compound. Give reason.
Answer:

A terminal =CR₂ group of an alkene gets converted into a ketone when the alkene is heated with an acidic solution of KMnO₄ or subjected to ozonolysis. Hence, an example of such an alkene is 2,3-dimethyl but-2-ene

Question 72. Write the formulas and names of the alkenes which on hydrogenation form 2-methylpentane.
Answer:

The carbon skeleton of the probable alkenes is

As there are four different positions of the double bond in the given carbon skeleton, four alkenes are possible which form 2-methylpentane on hydrogenation. The probable alkenes are:

Question 73. Write two possible methods of preparing 2-methylpropane by Corey-House synthesis. Out of these two methods, which one is better and why?
Answer:

Two possible methods of preparing 2-methylpropane by Corey-House synthesis are as follows-

In Corey-House synthesis, the third step is an S_N2 reaction (sensitive to steric effect). So, this step is highly favourable for methyl or primary halides, less’ favourable for secondary alkyl halides arid1does not occur in case of tertiary alkyl halides. In methods (1) and (2), a secondary halide and methyl halide have been used respectively In ease of the third step. So, method

(1) Is better than the method

(2) for preparing 2-methyl propane by Corey-House synthesis.

Question 74. A or B A and B are the two geometrical isomers. Identify them.
Answer:

The alkene which gives only acetaldehyde on ozonolysis is 2-butene (CH₃– CH —CH CH₃).

CH₃CH=O+O=CHCI ⇒ CH₃CH=CHCH₃

So, A and B are the two geometrical isomers of 2-butene:

Question 75. Dlazomethane (CH₂N₂), on decomposition forms singlet methylene (: CH₂) which gets attached to different non-equivalent C—H bonds of alkanes to form various alkanes. Name the alkanes formed when pentane (CH₃CH₂CH₂CH₂CH₃) reacts with singlet methylene. Assuming methylene to be highly reactive and less selective, calculate the probable amounts of the formed alkanes.
Answer:

Three alkanes are formed when pentane reacts with singlet methylene because there are three non-equivalent C—H bonds in pentane molecules. So, the alkanes formed are:

As methylene is highly reactive and less selective, its insertion occurs a random fashion. So, the amounts of the formed compounds are calculated the basis of probitblllity factor and number ofequivalent C— H bonds. For example, Percentage of hexane,

(CH₃CH₂CH₂CH₂CH₂CH₃) = \(\frac{6}{12}\) × 100 = 50

Percentage of 2-methyl pentane

CH₃—CH – CH₃—CH₂CH₂CH₃  =\(\frac{6}{12}\) × 100 = 50

percentage of 3-methyl pentane = \(\frac{2}{12}\) × 100 = 16.7

Question 76. How will you prepare (CH₃)₂CD¹⁴ CH₃ from propane (CH₃CH₂CH₃) ?
Answer:

Question 77. How will you prepare \({ }^{14} \mathrm{CH}_3{ }^{14} \mathrm{CH}_2{ }^{14} \mathrm{CH}_3\) taking \({ }^{14} \mathrm{CH}_3 \mathrm{I}\) as the only source of carbon?
Answer:

Question 78. In the reaction of 2-pentene with HI, the two isomeric iodopentanes are produced in almost equal amounts —why?
Answer:

The two doubly bonded carbon atoms in 2-pentene are bonded to the same number (one) of H-atoms. So, the two isomeric iodopentanes are produced in nearly equal amounts.

The two 2° carbocations (CH₃CH₂CH₂CHCH₃ and CH₃CH₂CHCH₂CH₃ ) obtained on the addition of proton at C-2 or C-3 are almost equally stable. So, the reaction proceeds through the two routes nearly at the same rate and consequently, the two isomeric iodopentanes are formed in nearly equal amounts.

Question 79. From the following two reactions, arrange HC = CH, NH₃ and H₂O in the increasing order of their acidic character.

1. HC = CH + NaNH₂→ HC = CNa + NH₃
2. HC = CNa + H₂O → HC = CH + NaOH

Answer:

In reaction no. , HC ≡ CH exhibits its acidic character and produces NH₃ from NaNH₂. So HC = CH is more acidic than NH₃.

On the other hand, in reaction (2), water exhibits its acidic character and produces HC = CH from HC = CNa. So, H₂O is more acidic than HC = CH. Thus, the increasing order of acidic character: NH₃ < HC=CH < H₂O

Class 11 Hydrocarbons Q&A

Question 80. Unlike acetylene, ethylene dissolves in concentrated sulphuric acid—why?
Answer:

In the first step of the reaction with concentrated H₂SO₄, ethylene forms an ethyl cation (CH₃C⁺H₂) and acetylene forms a vinyl cation (CH₂ =⁺CH) by accepting a H® ion.

Since vinyl cation is less stable than ethyl cation, in the case of acetylene, the first step (rate-determining step) of the reaction does not occur easily. Thus, unlike ethylene, acetylene fails to dissolve in concentrated H₂SO₄

Question 81. Write the structure of the product expected to be formed when CH₂=CH—CH₃(C = 14C) is subjected to free radical chlorination.
Answer:

Question 82. Identify the products obtained when ethylene gas is passed through bromine water in the presence of sodium chloride.
Answer:

Class 11 Hydrocarbons Q&A

Question 83. Which alkenes are formed on dehydrating the following alcohols in the presence of acid? Give the reaction mechanism.

Answer:

Class 11 Hydrocarbons Q&A

Question 84. The conjugated dienes are more reactive than alkenes which in turn are more reactive than alkynes towards electrophilic addition reactions —explain.
Answer:

The reactivity of alkenes, alkynes or conjugated dienes towards electrophilic addition reaction depends on the stability of the intermediate carbocation obtained in the rate-determining step by addition of the electrophile (E⁺).

Out of the three carbocations (la, Ila and IUa), (IlIa) is the most stable because it is stabilised by resonance. Again, out of (la) and (Ila), (IIla) is less stable because the positive charge in it is placed on a more electronegative sp2 -hybfiflis6d carbon atom.

Thus, the stabilities of these carbocations follow the order IIIa> la > Ila. Therefore, the order of activity of these compounds is: conjugated diene > alkene > alkyne.

Question 85. Calculate the resonance energy of 1,3-butadiene from the following data

Class 11 Hydrocarbons Q&A

Answer:

The heat liberated due to hydrogenation of one double bond = 30 kcal – mol^-1

The heat liberated due to hydrogenation of two double bonds = 30 × 2 = 60 kcal – mol^-1 .

Heat liberated due to hydrogenation of 1,3-butadiene (CH=CH—CH=CH₂) = 57 kcal -mol^-1 .

Therefore, resonance energy of 1,3-butadiene = 60- 57 = 3 kcal – mol^-1

Question 86. Dehydration of alcohols to alkene is carried out by treating with cone. H₂SO₄ but not with cone. HCl or HNO₃. Give reasons.
Answer:

Dehydration of alcohol proceeds via the formation of a carbocation intermediate. If HCl is used as the dehydrating agent then chloride ion (Cl^–), being a good nucleophile, attacks at carbonium ion carbon (Cl⁺) thereby producing alkyl chloride as the substitution product together with the alkene as the elimination product.

Class 11 Hydrocarbons Q&A

If cone. H₂SO₄ Is used as the reagent then H₂SO₄ ion derived from H₂SO₄ does not act as a nucleophile. Instead the carbocation loses a proton from the β -carbon atom to give alkene (R—CH=CH₂) as the elimination product.

If cone. HNO₃ is used as the reagent then it being a strong oxidising agent, brings about oxidation of the alcohol first to an aldehyde or a ketone and then to a carboxylic acid.

Class 11 Hydrocarbons Q&A

Question 87. How will you prepare ethylbenzene by using ethyne as the only organic substance and any other inorganic substance of your choice?
Answer:

Ethylbenzene (C₆H₅C₂H₅) may be prepared from ethyne (acetylene) through the following steps:

Class 11 Hydrocarbons Q&A

Question 88. Explain why bromination of benzene requires FeBr₃ as a catalyst, while bromination of anisole (C₆H₅OCH₃) does not require any catalyst.
Answer:

Since the benzene molecule is not so reactive,© for bromination it requires more reactive bromine cation (Br) or the complex Br— Br—FeBr₃ as the electrophile. Due to the presence of electron-donating (+R) methoxy (— OCH₃) group.

The anisole ring becomes much more reactive towards an electrophilic substitution reaction. When the non-polar bromine molecule comes in contact with the anisole ring, it 6+ 6 — becomes partially polarised (Br— Br)and its positive end (weak electrophile) undergoes easy attack by anisole. Therefore, due to the greater reactivity of anisole, its bromination requires no catalyst.

Question 89. Neither vinyl chloride (CH₂=CH—Cl) nor chlorobenzene (C₆H₅ —Cl) can be used as an alkylating agent in the Friedel-Crafts reaction—why?
Answer:

In vinyl chloride or chlorobenzene, the unshared pair of electrons on the Cl atom is involved in resonance interaction with the σ -electron system and as a result, the C—Cl bond in both cases acquires some double bond character.

The Lewis acid AlCl₃ is incapable of breaking such a strong C—Cl bond. Moreover, even if the C— Cl bond breaks, the carbocations produced would be unstable (due to a positive charge on an sp² – hybridised carbon atom).

Hence such a bond is very much reluctant to undergo cleavage. For this reason, vinyl chloride or chlorobenzene cannot be used as an alkylating agent in the Friedel-Crafts reaction.

Question 90. Two methods for the preparation of propylbenzene are given below-

Question 91. Which one of the two methods is more effective for the preparation of propylbenzene? Give reason.
Answer:

Method 2 is more effective for the preparation of propylbenzene. This is because in method 1, the alkylating agent containing a chain of three carbon atoms isomerises to give isopropyl benzene as the principal product.

Class 11 Hydrocarbons Q&A

Moreover, the alkyl group activates the benzene ring towards further substitution. So, there is a possibility of polyalkylation of benzene. However, although method 2 involves two steps, the desired propylbenzene is obtained as the only product in a higher yield. In the first step of the reaction.

The CH₃CH₂CO — group is introduced into the ring. Since the acyl group has no possibility of isomerisation, no other isomeric group can enter the ring.

Furthermore, the acyl group being an electron-attracting one deactivates the ring and consequently, polyacylation cannot take place. Hence in the first step, only propiophenone (C₆H₅COCH₂CH₃) is produced and in the second step, it is reduced by Clemmensen method to give only propylbenzene.

Question 92. Which ring (A or B) is in each of the following! compounds will undergo nitration readily and why?

Class 11 Hydrocarbons Q&A

Answer:

• Ring A is attached to the electron-donating or activating group — :O:COPh, whereas ring B is attached to the electron-attracting or deactivating — COOPh group. So, ring A is more reactive than ring B towards electrophilic substitution reaction. Hence, ring A will undergo nitration readily.
• Since a deactivating — N02 group is attached to ring A, it is relatively less reactive than ring B towards electrophilic substitution. Consequently, ring B undergoes nitration readily.
• Ring A is attached to an electron-donating or activating — CH₃ group while ring B is linked to an electron-attracting or deactivating — CF₃ group. So, ring A is relatively more reactive than ring B towards electrophilic substitution. Hence, ring A undergoes nitration at a faster rate.
• Ring B undergoes nitration readily. The reason is similar to that given in the case of compound (1).

Question 93. Write the names and structures of the compounds formed during the Friedel-Crafts reaction of benzene with

1. CH₂Cl₂
2. CHCl₃ and
3. CCl₄

Answer:

Class 11 Hydrocarbons Q&A

Question 94. Show the formation of the electrophile in each case:

1. Cl₃/AlCl₃
2. Br₂/Fe
3. Conc.HNO₃ + conc.H₂SO₄

Answer:

Class 11 Hydrocarbons Q&A

Question 95. How will you prepare the following compounds from benzene?

1. PhCH₂CH₂Ph,
2. PhCH₂CH₂CH₂Ph and
3. PhCH₂CH₂CH₂CH₂Ph

Answer:

Class 11 Hydrocarbons Q&A

Question 96. An optically active compound A (C₁₀H₄) gets oxidised to benzoic acid (C₆H₅COOH) by alkaline KMnO₄ However, compound B, which is an optically inactive isomer of A does not get oxidised by alkaline KMn04. Identify A and B.

Answer:

As A is oxidised to C₆H₅COOH, A is a substituted benzene which has only one side chain consisting of four carbon atoms. Again, as A is optically active, there must be an unsymmetric carbon atom present in the side chain.

So, the side chain is —CH(CH₃)CH₂CH₃ and A is sec-butylbenzene, C₆H₅CH(CH₃)CH₂CH₃. B, an isomer of A does not get oxidised by alkaline KMnO₄.

Thus, there is no benzylic hydrogen in the compound. So, the side chain is — C(CH₃)₃. The compound B is tert-butylbenzene, C₆H₅C(CH₃)₃

Question 97. Considering the stability of <r -complex, explain why — OCH₃ is o-lp- orienting while —NO₂ is mefa-orientlng.
Answer:

Electrophilic substitution reaction in anisole proceeds via the following reaction mechanism:

Class 11 Hydrocarbons Q&A

There is an extraordinarily stable (every atom has its octet fulfilled) resonance structure in both ortho- and para-a -complex, but there is no such resonance structure in the meta-σ- complex. So, ortho- and para-cr -complex are more stable than meta-cr complex. Consequently, electrophilic substitution proceeds easily and rapidly via ortho- and para-a -complex resulting in ortho and para- substituted compounds as major products. Thus, — OCH₃ is ortho-/para-orienting group.

Electrophilic substitution reaction in nitrobenzene proceeds via the following reaction mechanism:

Both ortho- and para- complexes are extraordinarily unstable resonance structures (due to the presence of a positive charge on two adjacent atoms). However, in meta-σ -complex there is no such resonance structure and so it is more stable than ortho- and para- σ -complex.

Thus, the reaction proceeds rapidly via the meta- σ -complex and the meta- substituted compound is obtained as the major product Thus, — NO is /nefa-orienting group.

Question 98.

Identify (A)-(F) in the following reaction

Class 11 Hydrocarbons Q&A

Answer:

Question 99.

1. What will be the major product when propyne is treated with aqueous H₂SO₄? Explain the equation.

2. An organic compound (A), C₇H₈O is insoluble in aqueous NaHCO₃ but soluble in NaOH. (A), on treatment with bromine water rapidly forms compound (B), C₇H₅OBr₂. Give structures of (A) & (B). What will be (A) if it does not dissolve in NaOH solution but shows the reaction given above?

Answer:

Propyne does not react with aqueous H₂SO₄ in the absence of Hg²⁺ ion. In the presence of an Hg²⁺ ion, propyne reacts with aqueous H₂SO₄ to give the unstable compound prop-2-enol (according to Markownikoff’s rule) which tautomerism to give acetone.

The problem is solved by assuming that the compound ‘B’ has the molecular formula C₇H₆OBr₂

Class 11 Hydrocarbons Q&A

Question 100.  Write the structural formula of the compounds A to F:

Answer:

Question 101. Both Br₂(g) and NO₂(g) are reddish-brown gaseous substances. How will you chemically distinguish between them?
Answer:

Class 11 Hydrocarbons Q&A

Question 102. Draw the structural formula of the compound from A toF.
Answer:

Question 103. Convert:

1. 2-propanol → 1-propanol
2. 2-butene→Ethane

Answer: 

Question 104. Write the IUPAC names of the following compounds:

Answer:

Class 11 Hydrocarbons Q&A

Question 105. For the Riven compounds write structural formulas and IUFPAC names for all possible isomers having the number of double or triple bonds as indicated:

1. C₄H₈ (one double bond)
2. C₅H₈ (one triple bond)

Answer:

Question 106. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

1. Pent-2-ene
2. 3,4-dimethyIhept-3-ene
3. 2-ethyl but-1-ene
4. I-phenyl but-1-ene

Answer:

Class 11 Hydrocarbons Q&A

Question 107. Explain why the following systems are not aromatic.

Answer:

1.

There are no p-orbitals on one of the H H CH₃H CH₃ carbon atoms forming the ring structure of this system and It is not a cyclic conjugated polyene containing (4n + 2)n -electrons. So, the system is not aromatic.

2.

There are no p-orbitals on one of the carbon atoms forming the ring structure of this system and it is not a cyclic conjugated polyene containing (4n + 2)n -electrons. So, the system is not aromatic.

3.

Cyclooctatetraene has a non-planar structure and there are 8π -electrons in it. So, cyclooctatetraene is a non-aromatic compound.

Question 108.  How will you convert benzene into

1. p-nitrobromobenzene
2. m-nltrochlorobenzene
3. p-nitrotoluene
4. Acetophenone

Answer:

Class 11 Hydrocarbons Q&A

Question 109. In H₃C—CH₂—C(CH₃)₂—CH₂CH(CH₃)₂, identify 1°, 2°, and 3° carbon atoms and give the number of H atoms bonded to each one of these.
Answer:

Number of H-atoms attached to 1° carbon atom = 15

Number of H-atoms attached to 2° carbon atom = 4

Number of H-atoms attached to 3° carbon atom = 1

Question 110. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?
Answer:

Class 11 Hydrocarbons Q&A

As the products A, B and C cannot be obtained from any one of the two Kekule structures, this confirms that o-xylene is a resonance hybrid of the two Kekule structures 1 and 2.

Question 111. Arrange benzene, n-hexane and ethynein decreasing order of acidic behaviour. Also give reason for this behaviour.
Answer:

The hybridisation state of carbon in the compounds benzene, n-hexane and ethyne are as follows—

The nucleus. Thus, the correct order of decreasing acidic behaviour is: ethyne > benzene > n-hexane.

Question 112. How would you convert the given compounds into benzene?

1. Ethyne
2. Ethene
3. Hexane

Answer:

Class 11 Hydrocarbons Q&A

Question 113. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Answer:

The structural formula of 2-methyl butane is—

The structures of different alkenes by putting double bonds at different positions along with satisfying the tetravalency of each carbon atom which gives 2-methyl butane on hydrogenation are as follows-

Question 114. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E⁺.

1. Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
2. Toluene, p-H₃C —C₆H₄—NO₂, P-O₂N—C₆H₄—NO₂

Answer:

The electron density of the benzene nucleus increases in the presence of an electron-donating group (activating group). Consequently, electrophiles can easily attack the benzene nucleus. On the other hand, the electron density of the benzene nucleus decreases in the presence of the electron-withdrawing group (deactivating group). This makes electrophilic substitution difficult for the benzene nucleus.

Therefore, the order of the different compounds according to their decreasing relative reactivity with an electrophile E+ is—

Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene

Toluene > p-CH₃C₆H₅NO₂ > p-O₂NC₆H₄NO₂

Class 11 Chemistry Hydrocarbons Short Question And Answers

Question 1. Propane may be obtained by the hydrolysis of n-propyl magnesium bromide and another alkyl magnesium bromide. Write the name and structure of that alkylmagnesium bromide.
Answer:

Propane can be obtained by hydrolysis of isopropyl magnesium bromide (Me₂CHMgBr)

Question 2. When a concentrated aqueous solution of sodium formate is used in Kolbe’s electrolysis method, no alkane is obtained—why?
Answer:

In Koibe’s electrolysis method, two— R groups of two RCOONa molecules combine to form the alkane, and R— R and two CO₂ molecules are obtained from two COONa groups. As, there is no alkyl group in the salt, sodium formate (HCOONa), no alkane is formed on electrolysis of its concentrated aqueous solution.

Class 11 Hydrocarbons Q&A

Question 3. Prepare 2,2-dimethylpropane by Corey-House synthesis.
Answer:

Question 4. How will you convert: \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHBr} \longrightarrow \mathrm{CH}_3 \mathrm{CHDCH}_3\)
Answer:

Question 5. Identify RI and R’l in the following reaction \(\mathrm{RI}+\mathrm{R}^{\prime} \mathrm{I} →{\mathrm{Na} / \text { ether }} \text { Butane + Propane + Ethane }\). What is the role of ether in this case?
Answer:

When the Wurtz reaction is carried out with a mixture of HI and R’l, two hydrocarbons with an even number of carbon atoms (R— R and R’—R’) and one hydrocarbon with an odd number of carbon atoms (R—R’) are formed. Among the formed alkanes, butane (CH₃CH₂CH₂CH₃) and ethane (CH₃CH₃) are respectively R— R and R’—R’ whereas, propane (CH₃CH₂CH₂) is R — R’.

So, it is evident that R is an alkyl group containing two carbon atoms, i.e., ethyl group (CH₃CH₂ — ) and Rr is a one carbon atom-containing alkyl group, i.e., methyl group (—CH₃). Therefore, RI is ethyl iodide (CH₃CH₂I) and R’l is methyl iodide (CH₃I). The role ofether in this case is that it acts as a solvent.

Question 6. How will you convert methyl bromide to (1) methane and (2) ethane in one step.
Answer:

1.

2.

Question 7. Arrange the following compounds in order of their increasing stability and explain the reason: 2-butene, propene, 2-methyl but-2-ene
Answer:

The order of increasing stability of the given alkenes is— propene < 2-butene < 2-methyl but-2-ene

The number of hypercoagulable hydrogen in propene, 2-butene and 2-methylbut-2-ene molecules are 3, 6 and 12 respectively. With an increase in several hypercoagulable hydrogens, stability due to the hyperconjugation effect of the molecules increases.

Class 11 Hydrocarbons Q&A

Question 8. Between the position isomer and chain isomer of but-1ene, which one exhibits geometrical isomerism and why?
Answer:

Position isomer of but-1-ene is but-2-ene (CH₃CH=CHCH₃) and chain isomer of but-1-ene is 2-methylpropene [(CH₃)₂C=CH₂] . In a molecule of but-2-ene, different groups are attached to the C-atom which is linked to the double bond.

So, but-2-ene exhibits geometrical isomerism. On the other hand, in a molecule of 2-methylpropene, two similar atoms (H-atom) are attached to the C-atom linked to the double bond. So, 2-methylpropene does not exhibit geometrical isomerism.

Question 9. Write the structures of the two alkenes obtained when 2-butanol is heated with excess ofconcentrated H₂SO₄. Which is obtained predominantly?
Answer:

Stability due to hyperconjugation is higher in the case of but-2-ene compared to that of but-1-ene. so, but-2-ene is obtained predominantly.

Question 10.

Answer:

Question 11. Identify the compounds obtained on heating (CH₃)₄NOHO
Answer:

Compounds formed: trimethyl amine and methyl alcohol.

Question 12. What is the major product formed in the reaction between CH₂=CH—NMe₃Ie and HI ? Write its structure.
Answer:

Question 13. What happens when a mixture of ethylene and O₂ gas is passed through a solution of PdCI₂ in the presence of CuCl₂  at high pressure and 50°C?
Answer:

When a mixture of ethylene and O₂  gas is passed through a PdCl₂  solution in the presence of CuCl₂  at high pressure and 50°C, acetaldehyde is formed as the product.

Class 11 Hydrocarbons Q&A

Question 14. Write the IUPAC names of the following compounds:

1. HC=C-CH(CH₃)₂
2. CH₃-C=C-C(CH₃)₃

Answer:

IUPAC name: 3 – Methylpent-1-yen

IUPAC name: 4,4-dimethylpent-2-yen

Question 15. C = C bond length is shorter than C and C —C —why?
Answer:

The σ -bond In C≡C is formed due to the overlapping of two small sp -hybridised orbitals. In C—C, the σ -bond is formed due to the overlapping of two bigger sp² -hybridised orbitals whereas in C — C.

the σ –bond is formed due to the overlapping of two even bigger sp³ -hybridised orbitals. Again, the multiplicity of the bond between two atoms increases, and the atoms come closer to each other leading to a decrease in bond length. So, bond length follows the order: C=C < C=C < C— C.

Question 16. Write structures and IUPAC names of the alkynes having molecular formula C₅H₈.

Class 11 Hydrocarbons Q&A
Answer:

Structures and IUPAC names of the alkynes having molecular formula C₅H_n are-

⇒ \(\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \mathrm{H}_2 \stackrel{2}{\mathrm{C}} \equiv \stackrel{1}{\mathrm{C}} \mathrm{H} \quad \text { (Pent-1-yne) }\)

⇒ \(\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \equiv \stackrel{2}{\mathrm{C}} \stackrel{1}{\mathrm{C}} \mathrm{H}_3 \quad \text { (Pent-2-yne) }\)

Question 17. What happens when the ethanolic solution of 1,1,2,2-tetrabromoethane is heated with zinc dust?
Answer:

When the ethanolic solution of 1,1,2,2-tetrabromoethane is heated with Zn dust, acetylene is formed as the product.

Question 18. What happens when the gas obtained by the action of water on calcium carbide is passed through an ammoniacal AgNO₃ Solution? Identify the solution through which acetylene gas is passed to form a red precipitate.
Answer:

Acetylene is formed by die action of water on calcium carbide. When acetylene gas is passed through ammoniacal silver nitrate solution, a white precipitate of silver acetylide (Ag₂C₂) is obtained.

Question 19. Which gas is used in carbide lamps or Hawker’s lamps? How does the gas produce a bright illuminating flame in the lamp?
Answer:

Acetylene gas is used in carbide lamps or Hawker’s lamps for producing bright illuminating flame. The percentage of carbon in acetylene is greater than in saturated hydrocarbons having the same number of carbon atoms.

As a result, incomplete combustion of acetylene gas takes place and the heated carbon particles thus formed produce Hluminating flame and bright light.

Question 20. What is Lindlar’s catalyst? Mention its use.
Answer:

Use of Lindlar’s catalyst:

This catalyst is used to add 1 molecule H₂ i.e., for partial hydrogenation of an alkyne. As cis-hydrogenation takes place in this case, a cis-alkene can be prepared from a non-terminal alkyne by using this catalyst.

Question 21. Which out of ethylene & acetylene is more acidic and why?
Answer:

The greater the s -the character of a hybridised carbon atom, the greater will be its electronegativity. The s -character of sp hybridised carbon atom of acetylene is greater than that of the sp² -hybridised carbon atom of ethylene.

so the electronegativity of the carbon atom of acetylene (CH=CH) is greater than the carbon atom of ethylene (CH₂=CH₂).

The tendency of H -atom attached to the more electro¬ negative carbon atom to be removed as a proton (H+) is relatively higher. So, the acidity of ethylene is less than that of acetylene, i.e., acetylene is more acidic than ethylene.

Question 22. How will you convert ethyne into ethanol?
Answer:

Question 23. Which out of ethyne and propyne is more acidic and why?
Answer:

In ethyne (HC=CH), there are two terminal hydrogen atoms whereas, in propyne (CH₃C=CH) there is only one. Apart from this, one electron-donating —CH₃ group is attached to the carbon atom present on the other side of the triple bond in the propyne molecule.

This decreases the acidity of the alkynyl hydrogen atom. So, ethyne is more acidic than propyne

Question 24. How will you distinguish between but-1-yne and but-2-yne?
Answer:

But-1-yne (CH₃CH₂C=CH) being a terminal alkyne reacts with a solution of ammoniacal stiver nitrate (AgNO₃) to form a white precipitate of silver 1-butynide.

However, but-2-yne (CH₃C=CCH₃) being a non-terminal alkyne does not undergo this type of reaction. So, but-1-yne and but-2-yne can be distinguished by observing the result of the above-mentioned test using ammoniacal stiver nitrate solution.

Class 11 Hydrocarbons Q&A

Question 25. How will you convert propylene into propylene?
Answer:

Question 26. What is the expected shape ofa benzene molecule in the absence of resonance?
Answer:

In the absence of resonance, benzene will be considered as the compound, 1,3,5-cyclohexatriene. The structure of this hypothetical compound is—

The structure will be such because each carbon-carbon bond length will be unequal. As, C=C is shorter than C—C, the structure of the molecule will appear as an irregular hexagon instead ofa regular one.

Question 27. Write structures of two aromatic ions in which there is a p-orbital containing 2 electrons and the other in which there is a vacant-orbital.
Answer:

Cyclopentadienyl anion has two electrons in one of its p -orbital whereas, cyclopropenyl cation  has a vacant p -orbital.

Question 28. Write the name and structural formula of the dibromobenzene which forms three mononitro compounds.
Answer:

Question 29. What happens when each of the following compounds is heated with acidified K₂Cr₂O₇ solution—
Answer:

Ethylbenzene:

Ethylbenzene gets oxidised to benzoic acid.

Question 30. Benzene exhibits a greater tendency towards substitution reactions but a lesser tendency towards addition reactions — Explain.
Answer:

Benzene attains stability by resonance. If benzene undergoes addition reactions, then it no longer participates in resonance. The aromaticity of benzene is no longer retained due to loss of conjugation.

Consequently, the extra stability of benzene is lost. So, benzene has a lower tendency to undergo an addition reaction. However, during the formation of the substitution product, the aromaticity and stability of benzene remain intact. So, benzene has a higher tendency towards substitution reactions.

Question 31. Benzene burns with a luminous sooty flame but methane bums with a non-luminous flame with no black smoke. Why?
Answer:

Due to the high percentage of carbon, elementary carbon is produced during the burning of benzene. As a result, black smoke is formed. The presence of hot carbon particles in the flame makes the flame luminous. In methane, the percentage of carbon is low so no elementary carbon is produced during the burning of methane. Thus, methane burns with a non-luminous flame with no black smoke.

Class 11 Hydrocarbons Q&A

Question 32. Name the electrophiles which participate in the following reactions: nitration, chlorination, Frieclel-Crafts alkylation, Friedel-Crafts acylation and sulphonation.
Answer:

Electrophiles which participate in nitration, chlorination, Friedel-Crafts alkylation, Friedel-Crafts acylation and sulphonation reactions are nitronium ion (⁺NO₂), positively charged chlorine ion (Cl⁺) or chlorine-iron (III) Chloride complex (Cl — Cl⁺— ^–FeCl₃), carbocation (R), B: acylium ion (R⁺CO) and sulphur trioxide (SO₃) respectively.

Question 33. Why iodobenzene cannot be prepared directly by combining benzene and iodine in the presence of iron filings? Why is this reaction possible in the presence of nitric acid?
Answer:

Iodobenzene cannot be directly prepared by combining benzene and iodine because the reaction is reversible. However, in the presence of nitric acid, this reaction becomes possible because nitric acid oxidises the hydrogen iodide as it is formed and so drives the reaction to the right.

Question 34. Write the structure of the compound 4-(1-isopropyl butyl)-3-propyl undecane.
Answer:

Question 35. Write the structure of 2,2,3-trimethylpentane and label the 1°, 2°, 3° and 4° carbon atoms.
Answer:

Question 36. Write the IUPAC names of two different optically active alkanes with lowest molecular mass.
Answer:

2 optically active alkanes with the lowest molecular masses-

Question 37. What is the state of hybridisation of the quaternary carbon atom present in the neopentane molecule? Write the IUPAC name and structure of the alkane formed by the combination of neopentyl and tert-butyl groups.
Answer:

The state of hybridisation of the quaternary carbon atom In a nooporitanu molecule is sp³

The alkane formed by the combination of neopentyl and terf-butyl groups Is—

IUPAC name: 2,2,4,4-tetramethylpentane

Question 38. How many chain isomers will be obtained on replacement of different H-atoms of n-pentane? Write their structures and IUPAC names.
Answer:

Three non-equivalent H-atoms are present in a pentane (CH₃CH₂CH₂CH₂CH₃) molecule. Thus, the replacement of these H-atoms by —CH₃ groups results in three isomeric alkanes. These are as follows —

CH₃CH₂CH₂CH₂CH₂CH₃ (hexane)

CH₃CH₂CH₂CH(CH₃)₂(2-methylpentane)

CH₃CH₂CH(CH₃)CH₂CH₃(3-methylpentane)

Class 11 Hydrocarbons Q&A

Question 39. Write the trivial and IUPAC names of the branched chain alkane with the lowest molecular mass.
Answer:

The trivial name of the branched chain alkane with the lowest molecular mass is isobutane and its IUPAC name is 2-methylpropane.

Question 40. Write the IUPAC name and structure of the alkane having formula C₈H₁₈ and containing a maximum number of methyl groups.
Answer:

The alkane of formula C₈H₁₈ containing the maximum number of methyl groups is

IUPAC name: 2,2,3,3-tetramethylbutane

Question 41. Benzene cannot be used as a solvent in ozonolysis of unsaturated hydrocarbons—why?
Answer:

Benzene itself reacts with ozone to form a triozonide. So, benzene cannot be used as a solvent in ozonolysis of unsaturated hydrocarbons

Question 42. Which one of the following three alkyl halides does not undergo Wurtz reaction and why?

1. CH₃CH₂Br
2. CH₃I
3. (CH₃)₃CB

Answer:

1. The second step of the Wurtz reaction is an S_N2 reaction.
2. An S_N2 reaction is very susceptible to steric effect so, a 3° alkyl halide does not take part in S_N2 reaction.
3. So, (CH₃)₃CBr being a 3° alkyl halide does not undergo Wurtz reaction

Question 43. Which compound is the strongest acid and why?

1. HC ≡CH
2. C₆H₆
3. C₂H₆
4. CH₃OH

Answer:

CH₃OH is the strongest acid because here the H-atom is bonded to oxygen which is more electronegative titan carbon irrespective of its state of hybridisation

Class 11 Hydrocarbons Q&A

Question 44. Give two equations for the preparation of propane from Grignard reagent.
Answer:

Question 45. How can CH₃D be prepared from CH₄ ?
Answer:

Question 46. The preparation of which of the following alkanes by Wurtz reaction is not practicable?

1. (CH₃)₃C-C(CH₃)₃
2. CH₃—CH(CH₃)—CH₂CH₃
3. (CH₃)₂CHCH₂CH₂CH(CH₃)₂

Answer:

1. Not practicable. Although it is a symmetrical alkane, prepared from a carboxylic acid? its preparation requires a f-alkyl halide, (CH₃)₃CX, which does not participate in Wurtz reaction.
2. Not practicable. It is an unsymmetrical alkane.
3. Practicable, because it is a symmetrical alkane

Question 47. When a concentrated aqueous solution of a mixture of sodium salts of two monocarboxylic acids is subjected to electrolysis, ethane, propane and butane are liberated at the anode. Write the structures and names of the two starting sodium salts
Answer:

Ethane may be produced from two CH₃COONa molecules, propane from 1 CH₃COONa molecule and one CH₃CH₂COONa molecule and butane (CH₂CH₂CH₂CH₃) from 2CH₃CH₂COONa molecules. Evidentlyethane, propane & butane may be obtained from the mixture of sodium ethanoate (CH₃COONa) & sodium propionate (CH₃CH₂COONa)

Question 48. How can an alkane having one carbon atom less be prepared from a carboxylic acid?
Answer:

Question 49. Arrange the following compounds in the increasing order of their acidic character.

1. H₂O, CH₂=CH₂, NH₃
2. HC≡CH, CH₃OH, CH₃CH₃

Answer:

1. Increasing order of acidic character: CH₂=CH₂ < NH₃ < H₂O [Because, the increasing order of electronegativity: C₂ < N < O ]

2. Increasing order of acidic character: CH₃CH₃ < HC = CH < CH₃OH [increasing order of electronegativity: c ,<C. <ol

Question 50. An alkane (molecular mass 72) produces only one monochloro derivative. Give IIJPAC name of the compound. Give reasons.
Ana.

As the alkane (C_nH_2n+2) on monochlorination produces only one monochloride derivative, so all of its hydrogens are equivalent. The molecular mass of the alkane is 72, i.e.,12 × n +{2n + 2) = 72, or n = 5. Hence, the alkane contains a carbon atoms, l.e., it is one of the isomeric pentanes. Ihe pentane in which all the H-atoms are equivalent is neopentane, (CH₃)₄ C. The IUPAC name of the compound is 2,2-dimethylpropane.

Question 51. 
Answer:

A = CH₃CHO (Acetaldehyde);

B= CHCHOH

(Ethyl alcohol); C = CHCl₃

D = HCOONa (Sodium formate)

Class 11 Hydrocarbons Q&A

Question 52. Acetylene does not react with NaOH or KOH, even though it possesses acidic character —why?
Answer:

Acetylene (HC = CH) is a weaker acid than water (H₂O) and OH- is a weak base than HC=C^–. As a result, weak acid HC=CH does not react with weak base OH^– to form strong acid H₂O and strong base HC≡C. So acetylene fails to react with NaOH or KOH

Question 53.  Write the products obtained when propyne ion (CH₃C = ^–C:) is allowed to react with

1. H₂O,
2. CH₃OH
3. H₃ (liquid)
4. 1-hexene and
5. Hexane.

Also mention, where no reaction occurs
Answer:

1. CH₃C = CH + OH^–
2. CH₃C = CH + CH₃O^–
3. No reaction
4. No reaction
5. No reaction.

(Propyne is a weaker acid than H₂O and CH₃OH but it is a stronger acid than, NH3 1-hexene and hexane.]

Question 54. Write the names of the products obtained in the following cases
Answer:

Answer:

1. Br CH₂CH₂ COOH (3-bromopropanoic acid).
2. C₆H₅ COCH₃ (Acetophenone).

Question 55. How can allyl chloride be prepared from1-propanol?
Answer:

Question 56. trans-pent-2-ene is polar but frans-but-2-ene is non-polar —why
Answer:

In a trans-but-2-ene molecule, the two C-—CH₃ bond moments, oriented in opposite directions cancel each other and so the net dipole moment in a trans-but-2-ene molecule is zero. On the other hand, in trans-pent-2-ene molecule, although the C-*-CH₃ and C—C₂H₅ bond moments act in opposite directions, they are not equal in magnitude and so they cannot cancel each other. Hence, the molecule possesses a net dipole moment

Class 11 Hydrocarbons Q&A

Question 57. Convert acetylene into but-2-one.
Answer:

Question 58. Ozonolysis of an alkene leads to the formation of an aldehyde and an isomeric ketone having molecular formula, C₃H₆O. Identify the alkene.
Answer:

The aldehyde and the ketone having molecular formula, C₃H₆O are CH₃CH₂CHO and CH₃COCH₃ respectively.

Therefore, the starting alkene is 2-methylpent-2-ene. (CH₃)₂C= O + O=CHCH₂CH₃-(CH₃)₂C =CHCH₂CH

Question 59. An alkene having molecular formula, C₄H₈ reacts with HBr to form a tertiary alkyl bromide. Identify the alkene and the alkyl bromide.
Answer:

Question 60. Carry out the following transformation (in two steps): Methyl acetylene →1-bromopropan
Answer:

Question 61. H How can butan-2-one be prepared from acetylene?
Answer:

Question 62. Write the IUPAC names of the products obtained when buta-1,3-diene reacts with bromine in1:1 molar ratio
Answer:

Class 11 Hydrocarbons Q&A

Question 63. Distinguish between buta-1,3-diene and but-1-yne?
Answer:

But-1-yne (CH₃CH₂C = CH) reacts with ammoniacal cuprous chloride solution to give a red precipitate. Buta-1,3- diene, however, does not respond to this test

Question 64. Prove that benzene is insoluble in water and is lighter than water.
Answer:

Benzene is taken in a separating flask and water is added to it. The mixture is shaken and then allowed to stand until two layers are separated. The upper layer contains benzene while water settles at the bottom. This observation proves that benzene is insoluble in water and is lighter than water.

Question 65. Why does benzene burn with a sooty flame?
Answer:

As the carbon content in benzene molecule (C₆H₆, C: H = 1 : 1 ) is relatively higher as compared to the saturated hydrocarbon, hexane (C₆H₁₄ C: H = 1:2.3), elementary carbon is formed during burning of benzene. So, benzene bums with a sooty flame

Question 66. Nitration of aniline with 98% H₂SO₄ and cone. [Anilinium sulphate (water soluble)] HNO₃ occurs very slowly and mainly metasubstitution occurs —why?
Answer:

In the presence of 98% H₂SO₄, the — NH₂ group of aniline takes up a proton (H) and is converted to an electronattracting (-1) and meta-directing — NH₃ group. Electron lectrophilic attack of —NO₂ group occurs very slowly. This accounts for the extremely slow rate of nitration of aniline and formation of mainly mefa-substituted product under strongl acidic conditions

Question 67. Activating groups are ortho-/para- directing, while; the deactivating groups are meta- directing—why?
An8.

The activating groups by exerting their electron-donating +1 and/or +R effects increase electron densities at the orthoand para- positions to a larger extent than the meta- position. So, the electrophiles (E+) enter preferably the ortho- and parapositions. On the other hand, the deactivating groups by exerting their electron withdrawing -I and I or -R effects decrease electron densities of meta-positions to a lesser extent than the ortho- and para- positions. So the electrophile enters preferably the relatively more electron-rich meta- position.

Question 68. How will you remove traces of aniline present in benzene?
Answer:

If benzene containing traces of aniline (basic in nature) is shaken with cone. H₂SO₄ then aniline dissolves in the acid forming a salt.- The acid layer is thus removed. In this way traces of aniline present in benzene can be removed

Question 69. How can traces of phenol present in a sample of benzene be removed?
Answer:

If the sample of benzene containing traces of phenol (acidic in nature) is shaken with 10% NaOH solution then phenol reacts with NaOH to form sodium phenoxide which dissolves in the aqueous layer. The aqueous layer is then removed. In this way traces of phenol present in a sample of benzene can be removed.

Question 70. How will you distinguish between benzene and hexane by a simple test in the laboratory?
Answer:

Percentage of carbon in benzene is much higher than corresponding alkane, hexane. So benzene bums with the formation ofelementary carbon and as a result, a sooty flame is produced. During burning of hexane no such sooty flame is produced. Thus, by observing the nature of the flame produced, the two compounds can be easily distinguished

Class 11 Hydrocarbons Q&A

Question 71. Aniline does not undergo Friedel-Crafts reaction, though it contains an electron-donating group—why?
Answer:

Aniline (Lewis base) combines with AlCl₃ (Lewis acid) by donating the unshared electron pair on nitrogen, to form a complex. As a result, the N-atom of — NH₂ group acquires a positive charge and this, being converted into an electronattracting group, decreases the electron density of the ring to such an extent that the Friedel-Crafts reaction does not occur

Question 72.  How much trisubstituted benzene may be obtained from o-, m- and p-chlorotoluene and why?
Answer:

Each of o – and m -chlorotoluene will give 4 trisubstituted benzenes, while p -chlorotoluene will give two trisubstituted benzenes. This is because there are 4 types of non-equivalent hydrogens (or positions) in each of the ortho- and meta¬ isomers and two types of non-equivalent hydrogens (or positions) in the para-isomer

Question 73. C₅H₁₂ and C₈H₁₈ are two alkanes which form one monochloride each on reacting with chlorine. Write the names of the alkanes and structural formulas of the chlorides.
Answer:

As both the alkanes form one monochloride, it can be said that all H-atoms in these two alkanes are equivalent. Therefore, the alkane with molecular formula C₆H₁₂ is (CH₃)₄C and the alkane with molecular formula C₈H₁₈ is (CH₃)₃C-C(CH₃)₃

Question 74. How can you convert methyl acetylene to acetone?
Answer:

At 60-80°C temperature, when methyl acetylene or propyne is passed through a dilute (20%) solution of sulphuric acid containing 1% H₂SO₄ , it combines with one molecule of water to form the unstable compound, 2-propanol. Addition of water molecule to unsymmetrical alkyne (propyne in this case) through the Markownikoff’s rule. 2-propenol! rapidly tautomerises to acetone

Question 75. Mention two reactions of benzene to show its behaviour is different from that of the open chain unsaturated hydrocarbons.
Answer:

1. Benzene does not decolourise bromine-water.
2. Benzene does not react with halogen acids such as HCl, HBr etc

Question 76. Is it possible to isolate pure staggered ethane or pure eclipsed ethane at room temperature? Explain.
Answer:

The energy difference between eclipsed and staggered conformations of ethane is only 2.8kcal .mol^-1, which is easily achieved by collisions among the molecules at room temperature. So it is not possible to isolate either pure staggered or pure eclipsed form of ethane at room temperature

Question 77. Explain why rotation about carbon-carbon double bond is hindered?.
Answer:

Carbon-carbon double bond consists of a σ -and a π bond. The π-bond is formed by lateral overlap of unhybridised p -orbitals of two carbon atoms above and below the plane of the carbon atoms. Ifan attempt be made to rotate one of the Catoms of the double bond with respect to the other, the p orbitals will no longer overlap, thereby causing fission of the n bond. Since breaking of a π -bond requires considerable amount of energy, which is not available at room temperature, so rotation about the carbon-carbon double bond is hindered.

Question 78. Arrange the following carbanions in order of their Hg+ decreasing stability

Answer:

Due to greater s -character sp -hybridised carbon is more electronegative than sp³ -hybridised carbon and hence can accomodate the -ve charge more effectively. So 1 and 2 are more stable than 3. Again —CH group has electrondonating +1 effect, therefore it interacts with the -ve charge on carbanion carbon and hence destabilises 1 relative to 2. Thus, stability of the given carbanions decreases in the sequence: B> 1> 3

Question 79. Explain why ferf-butylbenzene cannot be oxidised to benzoic acid by treatment with alkaline KMnO₄
Answer:

Alkylbenzenes can be oxidised to benzoic acid provided that the side chain contains one benzylic {hydrogen atom. Since terf-butylbenzene does not contain any benzylic hydrogen, so the alkyl chain cannot be oxidised to — COOH group.

Question 80. Explain why HF forms hydrogen bonding with acetylene even though it is non-polar in nature
Answer:

Due to sp -hybridisation of carbon, the electrons of the C —H bond of acetylene are attracted considerably towards carbon. Consequendy, each carbon carries a partial negative charge and each hydrogen carris a partial positive charge. Owing to the presence of partial positive charge on hydrogen, acetylene forms H -bond with the F -atom of the HF molecule

⇒ \(\begin{array}{r}
\delta+\delta-\delta-\delta+\delta-\delta+ \\
\cdots \mathrm{H}-\mathrm{C} \equiv \stackrel{\delta}{\mathrm{C}}-\mathrm{H} \cdots \mathrm{F}-\mathrm{H}
\end{array}\)

Question 81. One mole ofa symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. Identify the alkene
Answer:

Let, the aldehyde be C_nH_2n+1CHO

Molecular mass of this aldehyde

=[12n + (2n +1) + (12 +1 + 16)]u

= (14n + 30)u

Thus, 14n + 30 = 44

n = 1

So the aldehyde is CH₃CHO.

Obviously, the alkene is CH₃CH=CH— CH

Question 82. The addition of HBr to 1 -butene gives a mixture of mechanisms: A and B as the main products together with a small amount of another compound C. Identify A, B and C
Answer:

Question 83. An alkene, C₆H₁₂, reacts with HBr in the absence as well as in the presence of peroxide to give the same product. Find its structure.
Answer:

Symmetrical alkenes react with HBr in the presence or absence of peroxide to give the same product. Hence the given alkene may have the structure (1) or (2)

Question 84. How will you prepare a sample of propane free from other alkanes using ethyl bromide, methyl bromide and diethyl ether as the organic compounds, together . The product‘T is iodobenzene. with other inorganic materials
Answer:

It can be prepared by Corey-House synthesis:

Question 85. The catalytic hydrogenation of which of the following is most exothermic?

Answer:

Least substituted alkene, having the lowest number of hyperconjugative structures, has the least thermodynamic stability (i.e., highest energy content) and so it has the highest heat of hydrogenation. Now, out of the given compounds, (C) is the least substituted alkene and so it has the highest heat of hydrogenation

Question 86. Identify A, B, C & D in the following reaction—

Answer:

1. H —C≡C —H
2.  H —C ≡ CNa
3. H—C ≡ C—CH₃
4. H₂C=CH—CH₃

Question 87. Identify A, B, C Si D in the following reaction

Answer:

1.  CH₂=CH₂
2. CH= CH
3. CH₃CHO
4. CH₃—CH₃

Question 88. What organic compound is obtained when

1. Ethyl iodide is subjected to react with Zn -Cu couple/aqueous ethanol and 
2. Iodoform is heated with Ag powder?

Answer:

1. CH₃CH₃
2. C₂H₂
   

Question 89. Write structures of A and B in the following reactions.

Question 90.  Which one of these two reactions can be used for the identification of ethylenic unsaturation? Why? ½ + ½+½ + 1
Answer:

R-CH(Br)—CH₃

R—CH(Br) —CH₂Br

Question 91.  A hydrocarbon (A) is obtained when 1,2-dibromoethane reacts with alcoholic KOH. (A) decolourises alkaline KMnO₄ solution. (A) contains acidic hydrogen. Identify (A) with reasons.
Answer:

 

Question 92. Writes structures of A, B and C 

Answer:

Question 93. How will you convert?

1. HC = CH→H₃C-CH₂-CH₃

Answer:

Question 94. An organic compound (A) composed of C and H contains 85.71 %C. It shows M⁺ at mlz = 42 in the mass spectrum. The compound reacts with HBr in the absence of peroxide to yield an organic compound (B) which is isomeric with the compound (C) obtained when the compound reacts with HBr in the presence of peroxide. Identify A, B and C.
Answer:

⇒ \(C: H=\frac{85.71}{12}: \frac{14.29}{1}=7.14: 14.29=1: 2\)

Emperical formula = CH₂, Molar formula = (CH₂)_n

According to the problem, n × (12 + 2) = 42.

n = 3. So, the actual molecular formula is C₃H₆. DBE = 1. So, probable structures of CH₆ are—

Structure A is accepted asit can undergo given reactions:

Question 95.

1. Write equations of all the steps of the reaction of methane with chlorine in the presence of diffused sunlight

2. Identify A and B

Answer:

⇒ \(\mathrm{A} \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OSO}_3 \mathrm{H}, \mathrm{B} \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Question 18. Convert benzene into aniline by using the following reagents in the correct order: alkaline KMnO₄ and then HCl NH₃, heat; Br₂/KOH; CH₃Cl /anhydrous AlCl₃.
Answer:

Question 96. Write structures of the organic products obtained in the following reactions :

Answer:

Question 97. Identify (X) and (Y) in the following reactions:

Answer:

Question 98. Identify(M) and (N)in the following reactions

Answer:

Question 99. Write the product of the following reaction:

Answer:

Question 100. Benzene on reaction with NOCl in presence of acid produces an organic compound (A). (A) on treatment with NaNH₂/liq.NH₃ furnishes another organic compound (15). (B) on treatment with NaNO₂/HBF₄ affords an organic compound (C) which on heating gives an organic compound (D). Identify(A), (B), (C) and (D).
Answer:

Question 101. Two different compounds produce only acetaldehyde on ozonolysis. Draw the structures of the two compounds
Answer:

Question 102. Write the name and the structural formula of the product obtained when hydrogen bromide reacts with propene in the presence of benzoyl peroxide.
Answer:

Question 103. Identify the compound in the following reaction

Answer:

Identify the compound Ain the following reaction:

Question 104.

1. Among benzene and toluene, which one will undergo nitration reaction easily and why?

2. Identify A, B, C and D

Answer:

If the electron density of the benzene ring increases, then the reactivity of the ring towards electrophilic substitution also increases. In toluene, the —CH₃ group increases the electron density of the ring and as a result, the reactivity of the ring also increases due to +1 and the hyperconjugation effect of the —CH₃ group. So, nitration occurs more easily for toluene than unsubstituted benzene.

Question 105. Write the name and structural formula of A in the following reaction.

Answer:

Question 106. What happens when What happens ether?
Answer:

Question 107. Two isomeric compounds A and B have the molecular formula C₃H.Br forms the same compound C on dehydrobromination. C on ozonolysis produces acetaldehyde and formaldehyde. Identify A, B and C
Answer:

⇒ \(A \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}\)

⇒ \(B \Rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)

⇒ \(C \Rightarrow \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2\)

Question 108. An alkene A on ozonolysis gives a mixture of ethanal & pentan-3-one. Write the structure & IUPAC name of A.
Answer:

Question 109. An alkene ‘A’ contains three C —C, eight C —H crbonds and one C — C π -bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of ‘A’.
Answer:

An aldehyde with molar mass 44u is ethanal (CH₃CH=:O). The formula of the alkene ‘A’ which on ozonolysis gives two moles of ethanol can be determined as follows—

⇒ \(\underset{\text { Ethanal }}{\mathrm{CH}_3 \mathrm{CH}}=\underset{\text { Ethanal }}{\mathrm{O}}+\underset{\text { But-2-ene (A) }}{\mathrm{O}}=\underset{\mathrm{CHCH}_3}{\mathrm{CH}_3-\mathrm{CH}}=\underset{\mathrm{CH}}{\mathrm{CH}}-\mathrm{CH}_3\)

There are three C — C cr -bonds, eight C —H tr -bonds and one C —C 7t -bond in but-2-ene.

Question 110. Propanal and pentan-3-one are the ozonolysis products of an alkene. What Is the structural formula of the alkene?
Answer:

Question 111. Write chemical equations of the combustion reaction of the following hydrocarbons: Butane Pantene Hexyne Toluene
Answer:

Question 112. Draw The Cis- and trans-structure of hex- 2 ene which isomer will have higher B>P and Why?
Answer:

The general formula of hex-2-ene is CH₃—CH₂—CH₂—CH=CH—CH₃. Structural formulas of ds-and trans-isomers of this compound are-

The ds-isomer being more polar than the trans-isomer has a higher value of dipole moment than that of the trans-isomer. Intermolecular dipole-dipole interaction in the case of cis-isomers is more than that in trans-isomers. So, the boiling point of the customer is higher.

Question 113. Why is benzene extraordinarily stable though it contains three double bonds?
Answer:

There are (4n + 2) delocalised 7r -electrons (n = 1) in the planar benzene molecule.

Consequently, it attains stability due to aromaticity. So, benzene is extraordinarily stable despite having three double bonds.

Question 114. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why
Answer:

Nitration of the benzene ring is an electrophilic substitution reaction. In this reaction, the presence of an activating group ( —CH₃) increases the reactivity of the benzene ring, whereas the presence of a deactivating group (—NO₂) decreases the reactivity of the benzene ring. Therefore, order of nitration is toluene > benzene > m-dinitrobenzene.

Question 115. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Answer:

Ethylation of benzene means the introduction of an ethyl group in the benzene ring. This reaction is carried out by Friedel-Crafts reaction of benzene with ethyl halide (chloride or bromide), ethene or ethanol. Lewis acids, other than anhydrous AlCl₃, that can be used in this reaction are anhydrous FeCl₃, SnCl₄, BF₃, HF etc.

Question 116. Why does an iline not participate in Friedel-Crafts reaction?
Answer:

Aniline reacts with AlCl₃ complexing \(\mathrm{C}_6 \mathrm{H}_5-\stackrel{\oplus}{\mathrm{N}} \mathrm{H}_2-\stackrel{\ominus}{\mathrm{AlCl}} \mathrm{Cl}_3\) which makes —NH₂ group electron-withdrawing nature. Consequently, the benzene ring becomes highly deactivated so aniline does not participate in the Friedel-Crafts reaction;

Question 117. The reaction of CH₂=CH—N(CH₃)₃I takes place contrary to Markownikoff’s rule—why?
Answer: 

The carbocation formed due to the addition of H+ to the carbon atom containing a higher number of hydrogen atoms becomes unstable because of the electron-attracting —NMe₃ group.

So, the reaction takes place contrary’ to Markownikoff’s rule;

Question 118. Mention the limitations of the Wurtzreaction.
Answer:

Limitations:

Tertiary alkyl halides do not respond to this reaction,

Methane cannot be prepared by this reaction and

Preparation of unsymmetrical alkanes cannot be done by this method;

Question 119. What product is formed when the given compound reacts with HBr and why?
Answer: The product obtained in

is because the carbocation formed in the first step

Is, resonance stabilised due to —OCH₃ group

Question 200. Two different compounds produce only acetaldehyde on ozonolysis. Draw the structures of the two compounds
Answer:

Question 201. Write the name and the structural formula of the product obtained when hydrogen bromide reacts with propene in the presence of benzoyl peroxide.
Answer:

Question 202. Identify the compound in the following reaction

Answer:

Identify the compound Ain the following reaction:

Question 203.

1. Among benzene and toluene, which one will undergo nitration reaction easily and why?

2. Identify A, B, C and D

Answer:

If the electron density of the benzene ring increases, then the reactivity of the ring towards electrophilic substitution also increases. In toluene, the —CH₃ group increases the electron density of the ring and as a result, the reactivity of the ring also increases due to +1 and the hyperconjugation effect of the —CH₃ group. So, nitration occurs more easily for toluene than unsubstituted benzene.

Hydrocarbons Multiple Choice Questions

Question 1. Which one of the following alkenes produces a tertiary alcohol on acid-catalysed hydration

1. CH₃—CH—CH=CH₂
2. CH₃—CH=CH—CH₃
3. (CH₃)₂C=CH₂
4. CH₃-CH=CH₂ 

Answer: 3. (CH₃)₂C=CH₂

Question 2. Only which one of the following compounds is obtained when excess of Cl₂ is passed through boiling toluene

Answer: 4

Question 3. The Friedel-Crafts reaction using MeCl and anhydrous AlCl₃ maybe carried out best with—

1. Benzene
2. Nitrobenzene
3. Toluene
4. Acetophenone

Answer: 3. Toluene

Due to its +1 and hyperconjugative effect, the —CH₃ group increases the electron density of the benzene ring. Thus, toluene becomes more susceptible towards electrophilic substitution reaction. So, Friedel-Crafts reaction using MeCl and anhydrous AlCl₃ is carried out best with toluene.

Question 4. Baeyer’s reagent is—

1. Alkaline potassium permanganate
2. Acidified potassium permanganate
3. Neutral potassium permanganate
4. Alkaline potassium manganate

Answer: 1. Alkaline potassium permanganate

Dilute aqueous solution of alkaline potassium permanganate (1-2%) is called Baeyer’s reagent

Question 5. 2-methylpropane monochloririation under photochemical condition gives—1

1. 2-chloro-2-methylpropane as major product
2. (1 : 1) mixture of l-chloro-2-methylpropane and 2-chloro-2-methylpropane
3. 1-chloro-2-methylpropane as major product
4. (1:9) mixture of l-chloro-2-methylpropane and 2- chloro-2-methylpropane

Answer: 3. 1-chloro-2-methylpropane as major product

The ratio of A and B in the mixture is 5: 9, though 3° H is more active than 1° H, but in this case number of 1°H is 9 times than that of 3°H.

Question 6. Treatment of with  NaNH₂/liq.NH₃ gives—

Answer: 4.

The given reaction proceeds through the formation of the intermediate,  via the benzyne mechanism

Question 7. The best method for the preparation of 2,2-dimethylbutane is via the reaction of—

1. Me₃CBr and MeCH₂Br in Na/ ether
2. (Me₃C)₂CuLi and MeCH₂Br
3. (MeCH₂)₂CuLi and Me₃CBr
4. Me₃CMgI and MeCH₂I

Answer: 2. (Me₃C)₂CuLi and MeCH₂Br

Corey-House synthesis is the best method for preparing 2,2-dimethylbutane. The corresponding chemical reaction is given as

Question 8. Reaction of benzene with Me₃CCOCl in the presence of anhydrous AlCl₃gives

Answer: 2

In this electrophilic substitution, removal of CO from the electrophile, acylium ion (Me₃CC⁺=O) results in a more stable tertiary butyl carbocation Me₃C⁺). Thus, C₆H₅CMe₃ is formed as the product.

Question 9. An optically active compound having molecular formula C₈H₁₆ on ozonolysis gives acetone (CH₃COCH₃) as one of the products. Structure of the compound is—

Answer: 2

Question 10. The reagents to carry out the following conversion are—

Answer: 4

Question 11.   Identify the correct method for the synthesis of the compound shown above from the following alternatives—

Answer: 2

Question 12. 1,4-dimethylbenzene on heating with anhydrous AlCl₃ and HCl produces—

1. 1,2-dimethylbenzene
2. 1,3-dimethylbenzene
3. 1,2,3-trimethylbenzene
4. Ethylbenzene

Answer: 21,3-dimethylbenzene

When 1,4-dimethylbenzene is heated with anhydrous AlCl₃ in the presence of HCl it undergoes isomerisation to

Question 13. The major product of the above reaction is—

Answer: 3

The major product in the given reaction can be determined by the following reaction mechanism

Question 14. Identify’X’ in the following sequence of reactions—

Answer: 2

Question 15. The major products obtained on ozonolysis of 2,3-dimethyl-1-butene followed by reduction with Zn and HO are—

1. Methanoic acid and 2-methyl-2-butanone
2. Methanal and 3-methyl-2-butanone
3. Methanal and 2,2-dimethyl-3-butanone
4. Methanoic acid and 2-methyl-3-butanone

Answer: 2. Methanal and 3-methyl-2-butanone

Question 16. Which one of the following compounds is not aromatic—

Answer: 2

Cyclooctatetraene is a tub-shaped compound and thus it is non-aromatic.

Question 17. An alkene on ozonolysis produces only one dicarbonyl compound. The alkene is—

Answer: 2

Question 18. The major product(s) obtained from the following reaction of1 mol of hexadeuteriobenzene is/are—

Answer: 1

Question 19. The isomerisation of 1-butyne to 2-butyne can be achieved by treatment with—

1. Hydrochloric acid
2. Ammoniacal silver nitrate
3. Ammoniacal cuprous chloride
4. Ethanolic potassium hydroxide

Answer: 4. Ethanolic potassium hydroxide

Question 20. The major product(s) obtained in the following reaction is (are)—

Answer: 1 and 4

Question 21. The number of possible organobromine compounds which can be obtained in the allylic bromination of 1- butene with N-bromo succinamide is—

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Question 22.

The species P, Q, R and S respectively are—

1. Ethene, ethyne, ethanal, ethane
2. Ethane, ethyne, ethanal, ethene
3. Ethene, ethyne, ethanal, ethanol
4. Ehyne, ethane, ethene, ethanal

Answer: 1. Ethene, ethyne, ethanal, ethane

Question 23. The number of alkenes which can produce 2-butanol by the successive treatment of

1. B₂H₆ in tetrahydrofuran solvent and

2. Alkaline H₂O₂solution is

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 24. CH₃C=CMgBr can be prepared by the reaction of—

1. CH₃ —C≡C—Br withMgBr₂
2. CH₃C≡CH with MgBr₂
3. CH₃C≡CH with KBr and Mg metal
4. HCH₃C≡CH with CHgMgBr

Answer: 4. HCH₃C≡CH with CHgMgBr

Question 25. Formaldehyde is one of the products obtained on ozonolysis of a compound. The presence of which of the following groups is proved by this observation—

1. Vinyl group
2. Isopropyl group CH₃
3. Acetylenic triple bond
4. Two ethylenic double bonds

Answer: 1. In ozonolysis, the vinyl group is converted into HCHO

Question 26. Which one of the following converts 2-hexyne into 2-hexene—

1. Li/NH₃
2. Pd/BaSO₄
3. LiAlH₄
4. Pt/H₂

Answer: 1. Li/NH₃

Question 27. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is—

1. 2-butene
2. Ethene
3. Acetylene
4. 2-butyne

Answer: 4. 2-butyne

Question 28. In the reaction   the product C is—

1. Acetyl chloride
2. Acetylene
3. Acetaldehyde
4. Ethylene

Answer: 4.  Ethylene

Question 29.  Which compound would give 5-keto-2-methyl hexanal OH upon ozonolysis?

Answer: 4

Question 30. The product (A) of the reaction given below CH₃ is—

Answer: 1

Question 31. The reaction of propene with HOCI(Cl₃ + H₂O) proceeds through the intermediate—

Answer: 1

Question 32. 2-chloro-2-methyl pentane in methanol yields on— reaction with sodium

1. Both 1 and 3
2. Only 2
3. Both 1 and 2
4. All of these

Answer: 4. All of these

Question 33. 3-methylpent-2-ene reacts with HBr in the presence of peroxide to yield an additional compound. How many three-dimensional isomers are possible for this addition compound— 

1. 2
2. 4
3. 6
4. 0

Answer: 2. 4

The resulting compound consists of two different asymmetric centres (shown by asterisks). Thus the number of three-dimensional isomers = 22 = 4

Question 34. Which one of the following compounds undergoes mononitration to yield a considerable amount of m product—

Answer: 1

During tire nitration of aniline, —NH₂ group gets protonated and form an anilinium ion ( —NH₃). The + —NH₃ group is deactivating and meta-orienting. Thus during the mono nitration of aniline, a significant amount of meta-product is obtained

Question 35. The major product of the following reaction is

Answer: 4

Due to the neopentyl type of structure, it cannot undergo S_N2 reaction

Question 36.  The trans-alkenes are formed by the reduction of alkynes with—

1. Na/liq. NH₃
2. H₂/Pd-C,BaSO₄
3. Sn/HCl
4. NaBH₄

Answer: 1. Na/liq. NH₃

Question 37.   The major products 1 and 3 are respectively

Answer: 3

Question 38. In the following reaction

Answer: 3.

Question 39. Which of the following reagents is used to distinguish between 1-butyne and 2-butyne—

1. HCl
2. OH
3. Br
4. NaNH₂

Answer: 4. NaNH₂

Question 40. Which one of the following compounds is most reactive towards electrophilic-nitration reaction—

1. Toluene
2. Benzene
3. Benzoic acid
4. nitrobenzene

Answer: 1. Toluene

Due to +1 and the hyperconjugation effect of —CH₃ group, toluene is the most reactive towards electrophilic nitration reaction

Question 41. Which of the following compounds will not undergo Friedel-Crafts reaction easily

1. Toluene
2. Cumene
3. xylene
4. Nitrobenzene

Answer: 4. Nitrobenzene

Groups like nitro ( —NO₂) withdraw electrons from the benzene ring and deactivate the ring to such an extent that it cannot be attacked by the relatively weak electrophile. So, nitrobenzene does not undergo FriedelCraft reaction

Question 42. The radical is aromatic because it has—

1. 6p-orbitals and 7 unpaired electrons
2. 6p-orbitals and 6 unpaired electrons
3. 7p- orbitals and 6 unpaired electrons
4. 7p-orbitals and 7 impaired electrons

Answer: 3. 7p- orbitals and 6 unpaired electrons

The given free radical is aromatic because it contains a benzene ring having 6n -electrons which remain delocalised. The seventh odd electron does not play any role in determining the aromaticity of the
free radical.

Question 43. Some meta-directing substituents in aromatic substitution are given. Which is most deactivating— 

1. COOH
2. —NO₂
3. —C=N
4. —SO₃H

Answer: 2. —NO₂

Due to -I and -R effect, —NOa is a highly deactivating and —CN, —SO₃H and moderately deactivating meta-orienting group

Question 44. Nitrobenzene, on reaction with cone. HNO₃/H₂SO₄ at 80-100°C forms one of the following products—

1. 1,4-dinitrobenzene
2. 1,2,4-trinitrobenzene
3. 1,2-dinitrobenzene
4. 1,3-dinitrobenzene

Answer: 4. 1,3-dinitrobenzene

Question 45. Identify the sequence of reactions

1. CH₃(CH₂)₃-O-CH₂CH₃
2. (CH₃)₂CH—O—CH₂CH₃
3. CH₃(CH₂)₄—O-CH₃
4. CH₃CH₂-CH(CH₃)-O-CH₂CH₃

Answer: 1. CH₃(CH₂)₃-O-CH₂CH₃

Question 46. Which of the following organic compounds has the same hybridisation as its combustion product (CO₂) —

1. Ethane
2. Ethyne
3. Ethene
4. Ethanolic

Answer: 2. Ethyne

C-atom in both ethyne and C02 is sp -hybridised

Question 47. The oxidation of benzene by V₂O₅in the presence of air produces—

1. Benzoic anhydride
2. Maleic anhydride
3. Benzoic acid
4. Benzaldehyde

Answer: 2. Maleic anhydride

Question 48. Which of the following is not the product of dehydration of

Answer: 2

 does not form because the intermediate carbonation i.e is highly stable, so it does not undergo rearrangement

Question 49. In the reaction with HCI, an alkane reacts by Markownikoff’s rule, to give a product 1-chloro-1 methylcyclohexane. The possible alkane is—

Answer: 3

Question 50. Because of the absence of torsional strain, staggered conformation is more stable than eclipsed conformation.

1. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is— the staggered conformation of etha HPgi*less stable than eclipsed conformation because staggered hasÿirsional strain
2. The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain
3. The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
4. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

Answer:  4. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

Question 51. Consider the nitration of benzene using a mixed cone. H₂SO₄ and HNO₃.If a large amount of KH₂SO₄ is added to the mixture, the rate of nitration will be—

1. Faster
2. Slower
3. Unchanged
4. Doubled

Answer: 2. Slower

⇒ \(\mathrm{HNO}_3+2 \mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{H}_3 \stackrel{\oplus}{\mathrm{O}}+\stackrel{\oplus}{\mathrm{N}} \mathrm{O}_2+2 \mathrm{HSO}_4^{\ominus}\)

Addition of KHSO₄ increases the concentration of

H₂SO^–₄ and due to the common ion effect, the production of

NO⁺₂ decreases which slow down the nitration process

Question 52. In the reaction

X and 7 are—

1. X = 1 -butyne; Y- 3 -hexyne
2. X = 2 -butyne; Y = 3 -hexyne
3. X = 2 -butyne; 7=2 -hexyne
4. X = 1 -butyne; 7=2 -hexyne

Answer: 1.  X = 1 -butyne; Y- 3 -hexyne

Question 53. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction—

1. CH₃ CH₂CH₂OH

2. CH₃ CH₂CH₂Br

3. CH₂=C=O

Answer: 2. CH₃ CH₂CH₂Br

Question 54. In the given reaction the product P is

Answer: 2

Question 55. The compound that will react most readily with gaseous bromine has the formula—

1. C₂H₂
2. C₄H₁₀
3. C₂H₄
4. C₃H₆

Answer: 4. C₃H₆

The reaction undergoes via a radical pathway. Among the given alkenes, propene can form the most stable radical intermediate and thus it undergoes the reaction faster than other alkenes.

Question 56. Which one is the correct order of acidity—

Answer: 1.

For a C-atom in the hybridised state, acidic character increases with increase of the s-character. Again the presence alkyl group reduces the acidic property due to its +1 effect. Hence the order of acidity of the given compounds:

Question 57. Nitration of aniline in a strong acidic medium also gives  -m-nitroaniline because

1. In acidic (strong) medium aniline is present as an anilinium ion
2. Inspite of substituents nitro group always goes to only m -position
3. In absence of substituents nitro group always goes to m -m-position
4. In electrophilic substitution reactions amino group is meta-directive

Answer: 1. In acidic (strong) medium aniline is present as an anilinium ion

Question 58. The compound C₇H₈ undergoes the following reactions 

The Product ‘C

 is___________

1. p -bromotoluene
2. m – bromotoluene
3. 3-bromo-2,4,6-trichlorotoluene
4. o -bromotoluene

Answer: 2. m – bromotoluene

Question 59. Hydrocarbon (d) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to a gaseous hydrocarbon containing less than four carbon atoms. A is—

1. CH₄
2. HC≡CH
3. CH₃ – CH₃
4. CH₂ = CH₂

Answer: 1.CH₄

Question 60. Identify the major products P, Q and R in the following sequence of reactions

 

Answer: 1

Question 61. When trans-butene is reacted with Br₂ the product formed is

3. Meso-compounds

4. Both 2 And 3

Answer: 4. Both 2 And 3

With trans-but-2-ene, the product of Br₂ addition is Br optically inactive due to the formation of symmetric me-so-compounds

Question 62. What is ‘A’in the following—

Answer: 3

Question 63.  Which of the following aromatic—

Answer: 4

Any planar cyclic system containing (4n + 2)n electrons and having a single cyclic n -electron cloud encompassing all the carbon atoms in the ring is aromatic

Question 64. Which of the following alkenes will give the same product by any method out of hydration, hydroboration-oxidation and oxymercuration-demarcation—

Answer: 2. CH₃CH=CHCH₃

CH₃CH=CHCH₃ is symmetrical and gives the same product by any of the given methods adopted

Question 65. Which of the following species is not aromatic

1. Benzene
2. Cyclooctatetraenyl dianion
3. Tropyliumion
4. Cyclopentadienyl cation

Answer: 4.  Cyclopentadienyl cation

On applying the Huckel’s rule, [(4n + 2)n -electron] system

Question 66. What will be compound A in the following reaction—

Answer: 1.

Question 67.

Answer: 3

Strong electron releasing group ( — OCH₃) generally predominates over the deactivating group ( —NO₂). Thus, o – and p – products will be formed. Due to steric hindrance ortho-product will be formed in lesser amount than para-product

Question 68. Which Brof the following compounds is aromatic in nature—

Answer: 1,3 and 4

1: Due to the presence of (4n + 2)πe in; it follows Huckel’s rule and therefore,it is aromatic. Due to the presence of an extra lone pair of electrons in , total electron comes out to be 4πe . Thus, it is antiaromatic.

3: In although it is cyclic and has conjugated Huckel’s (4n + 2)πe rule is not followed here and also ring is notplanar. Hence,it is non-aromatic.

4: It has 6πe in conjugation but not in the ring, hence it is non-aromatic

Question 69. In the given reaction

Answer: 2

Question 70. What is the decreasing order of boiling points for the following compounds

1. 1> 2 > 3
2. 2 >3> 1
3. 1 > 3 > 2
4. 3 > 2 > 1

Answer: 1.1> 2 > 3

Question 71. The correct order of decreasing boiling points of the following hydrocarbons is—

1. n-butane

2. 2-methylbutane

3. n-pentane

4. 2,2- dimethylpropane

1. (1) > (2) > (3) > (4)
2. (2) > (3) > (4) > (1)
3. (4) > (3) > (2) > (1)
4. (3) > (2) > (4) > (1)

Answer: 4. (3) > (2) > (4) > (1)

Question 72. Addition of HBr with 1-butane (CH₃CH₂CH=CH₂) forms a mixture consisting of 1,2 and 3

In the mixture—  

1. 1 and 2 exist as major products while 3 as minor product
2. 2 exists as major product and 3 as minor products
3. 2 exists as a minor product whileI and 3 as major products
4. 1 and II exist as minor products while 3 as majorproduct

Answer:  1and 2 exist as major products while 3 as minor product

Question 73. Which of the following alkenes does not exhibit geometrical isomerism—

1. 1 and 2 exist as major products while 3 as minor product
2. Exists as major product while and 3 as minor products
3. 2 exists as minor products while 3 as major products
4. 1 and 2 exist as minor products while 3 a major product

Answer:  4. 1 and 2 exist as minor products while 3 as major product

Question 74. When a mixture of concentrated aqueous solutions of sodium salts of two monocarboxylic acids is electrolysed, a mixture of ethane, propane and butane is obtained at the anode. The two acids are
—
CH₂COOH, CH₃CH₂COOH

1. CH₃COOH, HCOOH
2. CH₃CH₂COOH,CH₃CH₂CH₂COOH
3. (CH₃)₂CHCOOH, CHCOOH

Answer:  1. CH₂COOH, CH₃CH₂COOH

Question 75.  The products formed when CH₃I, CH₃CHO and CH₃CH₂COOH are respectively reduced by HI in presence of red phosphorus are—

1. CH₃CH₃, CH₃CH₃, CH₃CH₂CH₃
2. CH₄, CH₃CH₃, CH₃CH₂CH₃
3. CH₃CH₂CH₃, CH₄, CH₃CH₃
4. CH₄, CH₃CH₂CH₃, CH₃CH₂CH₂CH₃

Answer: 2. CH₄, CH₃CH₃, CH₃CH₂CH₃

Question 76. The decreasing order of boiling points of isomeric pentanes is—

1. n-pentane > isopentane > neopentane
2. Isopentane > n-pentane > neopentane
3. Neopentane > isopentane > n-pentane
4. n-pentane > neopentane > isopentane

Answer: 1. n-pentane > isopentane > neopentane

Question 77. The reactivity of different types of hydrogen during halogenation of alkanes follows the order—

1. 2°H > 1°H > 3°H
2. 1°H >2°H > 3°H
3. 2°H > 3°H > 1°H
4. 3°H > 2°H > 1°H

Answer: 4. 3°H > 2°H > 1°H

Question 78. The correct IUPAC name of the monochord derivative that forms as the major product during chlorination of 3-ethyl pentane is—

1. 1-chloro-3-ethyl pentane
2. 2-chloro-3-ethyl pentane
3. 3-chloro-3-ethyl pentane
4. 3-ethyl-3-chloroethane

Answer:  3. 3-chloro-3-ethyl pentane

Question 79. Which of the following symmetrical alkanes is not prepared by Wurtz reaction—

1. Ethane
2. Butane
3. 2,2,3,3-tetramethyl butane
4. 2,3-dimethylbutane

Answer: 3. 2,2,3,3-tetramethyl butane

Question 80. The octane numbers of 2,2,4-trimethylpentane isooctane) and n-pentane are respectively—

1. 50,50
2. 100,0
3. 0,100
4. 50,0

Answer: 2. 2. 100,0

Question 81. The increasing order of octane numberis

1. n-alkane < branched alkane < cycloalkane < aromatic hydrocarbon
2. Aromatic hydrocarbon < cycloalkane < branched alkane < n-alkane
3. Cycloalkane < branched alkane < aromatic hydrocarbon < n-alkane
4. Aromatic hydrocarbon < cycloalkane < n-alkane < branched alkane

Answer: 1. n-alkane < branched alkane < cycloalkane < aromatic hydrocarbon

Question 82. The amount of oxygen (in mole) required for the combustion of1 mol hydrocarbon (CÿHÿ,) is—

1. \(\left(x+\frac{y}{4}\right)\)
2. \(\left(y+\frac{x}{4}\right)\)
3. (x+y)
4. \(\left(x+\frac{y}{2}\right)\)

Answer: 1. \(\left(x+\frac{y}{4}\right)\)

Question 83. Which of the following compounds on reacting with Grignard reagent (RMgX) does not form an alkane—

1. CH₃C=CH
2. CH₃CH₂OCH₂CH
3. C₂H₅OH
4. H₂O

Answer:  2. CH₃CH₂OCH₂CH

Question 84. The change in the hybridisation state of carbon in ethane duringits combustion is—

1. sp³→sp
2. sp²→sp³
3. sp→sp³
4. sp→sp²

Answer: 1. sp³→sp

Question 85. The intermediate formed during chlorination of methane in diffused sunlight is

1. \(\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3\)
2. :CH₂
3. ^•CH₃
4. :C^–H₃

Answer:   3. ^•CH₃

Question 86. The product formed when 2 equivalent ofmetallic sodium reacts with l-bromo-3-chlorocyclobutane in ether medium is

Answer: 4

Question 87. Which of the following compounds are not formed when a mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction—

1. Butane
2. Propane
3. Pentane
4. Hexane

Answer: 2. Propane

Question 88. In Kolbe’s electrolytic process which of the following compounds does not lead to the formation of an alkane—

1. CH₃COONa
2. CH₃CH₂COONa
3. HCOONa
4. CH₃CH₂CH₂COOK

Answer: 3. HCOONa

Question 89. —I + Zn +I —R—>R —R + Znl2; The reaction is—

1. Frankland reaction
2. Grignard reaction
3. Wurtzreaction
4. Corey-House synthesis

Answer: 1. Frankland reaction

Question 90. Which of the following statements is not true—

1. Alkanes are non-polar compounds
2. Alkanes are insoluble in polar solvents like water
3. Among isomeric alkanes, n-alkane has the lowest boiling point
4. Higher alkanes (> C₁₇) are hard like wax

Answer: 3. Among isomeric alkanes, n-alkane has the lowest boiling point

Question 91. The reagent which is not used in the preparation of propene from 1-bromopropane is

1. Water/KOH
2. Ethanol/ C₂H₅O^–Na⁺
3. Ethanol/KOH
4. Tert-butylalcohol/(CH₃)₃CO^–K⁺

Answer: 1. Water/KOH

Question 92. The reducing agent which is not used to prepare RCH=CHR from RC=CR is—

1. Na/liq.NH₃
2. H₂/Lindlar’s catalyst
3. B₂H₆ /tetrahydrofuran
4. H₂/PtorPd

Answer: 4. H₂/PtorPd

Question 93. Decreasing order of stability of the given carbanions is—

1. CH₃—C≡^–C:

2. H —C=C^–:

3. CH₃—:C^–H₂

1. 1 > 2 > 3
2. 2 > 1> 3
3. 3 > 2 > 1
4. 3 > 1 > 2

Answer: 2. 2 > 1> 3

Question 94. Baeyer’s reagent which is used in the Baeyer’s test for detecting ethylenic unsaturation is—

1. An acidic solution of potassium permanganate
2. A dilute alkaline solution of potassium permanganate
3. An aqueous solution of bromine
4. A solution ofbrominein acetic acid

Answer: 2. A dilute alkaline solution of potassium permanganate

Question 95. Which of the following compounds on ozonolysis forms C02 along with other products—

1. CH₂=C=CHCH₃
2. CH₃CH=CH—CH=CH₂
3. CH₃CH=CH—CH=CH₂
4. CH₃CH=CH—CH=CH₂

Answer: 1. CH₂=C=CHCH₃

Question 96. Which of the following compounds on ozonolysis forms a carbonyl compound

1. CH₃CH=CH₂
2. (CH₃)₂C=CHCH₃
3. (CH₃)₂C=C(CH₃)₂
4. CH₂=CH —CH=CH₂

Answer:  3. (CH₃)₂C=C(CH₃)₂

Question 97. Alkene which forms CH₃COCH₂CH₂CH₂CH₂COCH₃ on ozonolysis is

Answer:  4.

Question 98. An alkene, on ozonolysis forms HCHO, CH₃COCHO and CH₃CHO. The alkene is—

1. CH₂=C(CH₃)—CH=CHCH₃
2. CH₂=C(CH₃) —CH₂ —CH=CH2
3.  CH₃—CH=CH—CH=CHCH₃
4. CH₃—CH=CH—CH₂—CH=CH₂

Answer: 1. CH₂=C(CH₃)—CH=CHCH₃

Question 99. CH₃COCOCH₃, CH₃COCHO and OHC—CHO are formed due to the ozonolysis of o-xylene. The ratio in which the compounds are formed—

1. 3:2:1
2. 2:3:1
3. 1:2:3
4. 3:1:2

Answer: 3. 1:2:3

Question 100. Which of the following reactions occurs following Markownikoff’s rule

Answer:  4

Question 101. For which of the following reaction, Markownikoff’s rule is applicable—

Answer:  3

Question 102. The intermediate which is formed in the first step of the reaction between CH₃CH=CH₂ and HBr is a

1. Carbanion
2. Carbocation
3. Free radical
4. Carbene

Answer: 2.

Question 103. Among all the HX compounds, only HBr reacts with CH₃CH=CH₂ in the presence ofperoxide according to antiMarkownikoff’s rule. This is because, in case of HBr, the third and fourth steps

1. CH₃—CH=CH₂ + Br→CH₃CHCH2Br and

2. CH₃CHCH₂Br + H —Br→CH₃CH₂CH₂Br + Br )

1. The third step is exothermic while the fourth step is endothermic
2. Both the steps are exothermic
3. The third step is endothermic while the fourth step is exothermic and the fourth steps
4. Both the steps are endothermic

Answer: 2. Both the steps are exothermic

Question 104. Which of the given compounds on reacting with HBr forms the same product in the presence and absence of peroxide—

Answer: 3

Question 105. The order of acidity of ethyne (1), ethane (2) and ethene (3) is

1.  2 >1 >3
2. 1 >3 >2
3. 1 >2 >3
4. 2 >3 >1

Answer: 3. 1 >2 >3

Question 106. The compound which does not form a red precipitate on reacting with ammoniacal cuprous chloride solution is—

Answer: 2

Question 107. The reagent that cannot be used to distinguish between ethylene and acetylene is—

1. Ammoniacal cuprous chloride
2. Br₂/H₂O
3. Dil. H₂SO₄/Hg²⁺
4. Ammoniacal silvernitrate solution

Answer:  2. Br₂/H₂O

Question 108. Two compounds, when subjected to ozonolysis separately,  CH₃COCH₂CH₃(2mol) The compounds are— each form and

1. Enantiomers
2. Diastereomers
3. Metamers
4. Tautomers

Answer:  1. Enantiomers

Question 109. The compound Y in the given reaction is –

Answer: 1

Question 110. An alkene (molecular formula : C₅H₁₀) on ozonolysis forms acetone as one of the products. The alkene is—

1. 2-methyl-1-butene
2. 3-methyl-1-butene
3. 2-methyl-2-butene
4. Cyclopentane

Answer: 2.  3-methyl-1-butene

Question 111. Which of the following compounds can be used to prepare both ethylene and acetylene—

1. CH₃CH₂OH
2. BrCH₂CH₂Br
3. CH ₃CH₂Br
4. BrCH₂CH₂OH

Answer: 2. BrCH₂CH₂Br

Question 112. Alkyl chloride on dehydrochlorination produces 2 alkenes (C₆H₁₂) which on ozonolysis form four compounds—

1. CH₃CHO
2. CH₃CH₂CHO,
3. CH₃COCH₃ and
4. (CH₃)₂CHCHO.

The alkenes are—

1. 4-methylpent-2-ene and 2-methylpent-2-ene
2. 2-methyl pent-2-ene and 2,3-dimethyl but-2-ene
3. 4-methylpent-2-en§ and hex-3-ene
4. 2-methylpent-2-ene and hex-3-ene

Answer: 1. 4-methyl pent-2-ene and 2-methyl pent-2-ene

Question 113. The compound that exhibits geometrical isomerism is—

1. C₂H₅Br
2. (CH)₂(COOH)₂
3. CH₃CHO
4. (CH₂)₂(COOH)₂

Answer: 2. (CH)₂(COOH)₂

Question 114.

1. X: cis-2-butene and Y: frans-2-butene
2. X: trans-2-butene and Y :cis-2-butene
3. X, Y both are cis-2-butene
4. X, Y both are trans-2-butene

Answer: 1. X: cis-2-butene and Y: frans-2-butene

Question 115. An alkene may be formed from a carbocation if—

1. One H- ion gets eliminated
2. One H+ ion gets added
3. One H+ ion gets eliminated
4. One H- ion gets added

Answer: 3. One H+ ion gets eliminated

Question 116. The number of moles of water produced when one mole acetylene undergoes complete combustion is—

1. 1 mol
2. 2 mol
3. 3 mol
4. 4 mol

Answer: 1.  1 mol

Question 117.  , A, B, C in the above reaction are respectively—

1. CH₃COCH₃, CH₃CHO, CO₂
2. CH₃COCOOH, CH₃COOH, CO₂
3. CH₃CH₂COOH, CH₃CHO, CO₂
4. CH₃COCH₃, CHgCOOH, CO₂

Answer: 4. CH₃COCH₃, CHgCOOH, CO₂

Question 118. The position of the double bond in an alkene can be determined by—

1. Hydrogenation
2. Ozonolysis
3. Hydroxylation
4. Hydroboration

Answer: 2. Ozonolysis

Question 119. Heavy water reacts with calcium carbide to form—

1. CaD₂
2. C₂D₂
3. Ca₂D₂O
4. CD₂

Answer: 2. C₂D₂

Question 120. The addition of HBr occurs most readily for—

Answer: 4.

Question 121.    In this reaction a is

Answer: 1.

Question 122. Which of the following compounds undergoes hydrolysis to form propyne-

1. Al₄C₃
2. Mg₂C₃
3. B₄C
4. La₄C₃

Answer: 2.Mg₂C₃

Question 123. Hydration of alkenes (except ethylene) in presence of acid produce—

1. 1° alcohols
2. 2° or 3° alcohols
3. Mixture of1° and 2° alcohols
4. Mixture of 2° and 3° alcohols

Answer: 2. 2° or 3° alcohols

Question 124. Cyclohexanone   react with witting reagents (Ph3P—CHR) to form—

Answer: 3

Question 125. When ethylene gas is passed through an aqueous solution of NaCl and Br₂ the compound whose formation is not possible is—

Answer: 3

Question 126. Here , X is

1. Cyclobutane
2. Cyclopropane
3. Cyclopentane
4. Cyclohexane

Answer: 2. Cyclopropane

Question 127. Which of the following alkynes cannot be converted into a terminal alkyne when heated with NaNH₂/ paraffin—

1. CHC₃=CCH₃
2. CH₃CH₂C=CCH₂CH₃
3. CH₃CH₂CH₂C=CCH₂CH₃
4. (CH₃)₂CHC=CCH(CH3)₂

Answer: 4. (CH₃)₂CHC=CCH(CH₃)₂

Question 128. Correct order of decreasing reactivity of the given com¬ pounds towards electrophilic substitution reaction is –

1. 3 > 1 > 2 > 4
2. 4 > 1 > 2 > 3
3. 1 > 2 > 3 > 4
4. 2 > 1 > 3 > 4

Answer: 1. 3 > 1 > 2 > 4

Question 129. Number of monochloride derivatives possible for diphenylmethane

1. 4
2. 8
3. 7
4. 18

Answer: 1. 4

Question 130. The compound which is formed in excess when Cl₂ reacts with toluene in presence of FeCl₃ is—

1. Benzyl chloride
2. o – and p -chlorotoluene
3. m -chlorotoluene
4. Benzoyl chloride

Answer: 2. o – and p -chlorotoluene

Question 131. The major product obtained when

Answer: 2

Question 132. Polysubstitution occurs for which of the following reactions—

Answer: 2

Question 133.  compound X is________________

Answer: 2.

Question 134. The compound which is most reactive in case of electrophilic attack is-

Answer: 1

Question 135. Benzene does not form additional compound because—

1. It has ring structure
2. Its double bond is verystrong
3. It has 6 equivalent h-atoms
4. Its aromatic stability is lost

Answer: 4. Its aromatic stability is lost

Question 136. In strong acidic and alkaline medium, p-aminophenol exists in (X) and (Y) forms respectively

Thus, in acidic and alkaline medium, electrophilic substitution occurs at-

1. a,c
2. a,d
3. b,c
4. b,d

Answer: 1. a,c

Question 137. In electrophilic substitution reaction of benzene—

1. The first step is exothermic but the second step is endothermic
2. The first step is endothermic but the second step is exothermic
3. Both the steps are exothermic
4. Both the steps are endothermic

Answer: 2.  The first step is endothermic but the second step is exothermic

Question 138. The method suitable for converting benzene into propylbenzene is 

Answer: 3

Question 139. An aromatic compound of molecular formula C₈H₁₀ reacts with a mixture of concentrated HN03 and concentrated H₂SO₄ to form a mononitro compound. The structural formula of C₈H₁₀ is

Answer: 4

Question 140. Nitrobenzene is prepared from benzene by using conc. HNO₃ and cone. H₂SO₄ . In the nitrating mixture, nitric acid acts as a/an—

1. Base
2. Acid
3. Reducing agent
4. Catalyst

Answer: 1.  Base

Question 141. On passing excess Cl₂(g) through boiling toluene, the only compound that forms is—

Answer: 4

Question 142. Which of the given participates in Friedel-Crafts reaction—

Answer: 4

Question 143. In which of the following compounds the ring on the left side undergoes electrophilic substitution reaction—

Answer: 4

Question 144. The increasing order of the rate of nitration reaction of the following compounds is- 

1. 2 < 3 < 1 < 4 < 5
2. 4 < 5 < 1 = 3 < 2
3. 4 < 1 = 3 < 5 < 2
4. 1 < 3 < 2 < 5 < 4

Answer: 2.  4 < 5 < 1 = 3 < 2

Question 145. The major product formed in the following reaction is-

Answer: 4.

Question 146.

Answer: 2

Question 147. The acidity of which of the following compounds is quite high compared to the rest of the given compounds—

Class 11 Hydrocarbons Q&A

Answer: 3

Question 148. Bromination takes place most rapidly in—

Answer: 2

Question 149. 

Answer: 1

Question 150. In case of trisubstituted benzene, if the substituents are different, then the number of isomers will be

1. 5
2. 8
3. 6
4. 10

Answer: 4. 10

Class 11 Hydrocarbons Q&A

Question 151. The chemical formula ofCetane is—

1. C₆H₁₂
2. (CH₃)₃C(CH₂)₁₁CH₃
3. CH₃(CH₂)₁₄CH₃
4. (C₂H₅)₄C

Answer: 3. CH₃(CH₂)₁₄CH₃

Question 152. Which of the following gets converted into an explosive when it is turned into liquid by applying high pressure—

1. Propane
2. n-butane
3. Isobutane
4. Acetylene

Answer: 4. Acetylene

Question 153. The product which is not obtained when ethylene reacts with K3 mixed with Br₂/H₂O is—

1. BrCHCH₂Br
2. BrCH₂CH₂OH
3. HOCH₂CH₂OH
4. BrCH₂CH₂I

Answer: 3. HOCH₂CH₂OH

Question 154. Which of the following does not form a sooty flame—

1. Toluene
2. Benzene
3. Mesitylene
4. Butane

Answer: 4. Butane

Question 155. Which of the following statements is incorrect—

1. Delocalisation of electrons occur between two n bonds in a propadiene molecule
2. Delocalisation of electrons occur between two n bonds in a molecule of 1, 3-butadiene
3. Cumulated polyenes with odd number of double bonds exhibit geometrical isomerism if their terminal groups are different
4. Cumulated polyenes with even number of double bonds exhibit optical isomerism if their terminal groups are different

Answer: 1.  Delocalisation of electrons occur between two n bonds in a propadiene molecule

Question 156. Which of the given is a benzenoid aromatic compound—

1. Anthracene
2. Pyrrole
3. Pyridine
4. Cyclopentadienyl anion

Answer: 1. Anthracene

Question 157. Gas used in Hawker’s lamp for emitting bright light is—

1. Acetylene
2. Ethylene
3. Methane
4. Propane

Answer: 1. Acetylene

Question 158. Benezene when subjected to ozonolysis (03 followed by Zn/H₂O ) forms—

Answer: 1

Question 159. BHC is a/an—

1. Fertiliser
2. Insecticide
3. Explosive
4. Solvent

Answer: 2.  Insecticide

Question 160. (CH₃)₃CMgCl reacts with D₂O to form—

1. (CH₃)₃CD
2. (CH₃)₃OD
3. (CD₃)₃CD
4. (CD₃)₃OD

Answer: 1.  (CH₃)₃CD

Question 161. Which two compounds undergo ozonolysis to produce CH₃CHO, CH₃COCHO and HCHO —

Class 11 Hydrocarbons Q&A

Answer: 2, 3

Question 162. Among the following oxidation reactions of methane, which two are controlled oxidation reactions —

Class 11 Hydrocarbons Q&A

Answer: 3,4

Question 163. Which of the following alkenes undergo ozonolysis to form a mixture of two ketones—

Answer: 3,4

Question 164. Which of the following compounds form the same product with HBr in presence and absence of peroxide—

1. Cyclohexene
2. But-2-ene
3. Hex-3-ene
4. 1-methylcyclohexene

Answer: 1,2,3

Class 11 Hydrocarbons Q&A

Question 165. The compounds which react with dilute H₂SO₄ in the presence of HgS04 to form methyl ketone are—

Answer: 2,3,4

Question 166. The compounds which only form glyoxal when subjected to ozonolysis are—

1. Ethene
2. Benzene
3. Toluene
4. Ethyne

Answer: 2,4

Question 167. In which of the following compounds, nitration take place at the para-position—

Class 11 Hydrocarbons Q&A

Answer: 2,4

Question 168. Which of the following groups are deactivating but ortho-J para-orienting—

1. —Cl
2. —CH=CH—COOH
3. —N=O
4. -CF₃

Answer: 1,2,3

Question 169. Which two of the following groups are used to block a definite position in the benzene ring-

1. -SO₃H
2. — CH₃
3. — CF₃
4. —CMe₃

Answer: 1,4

Question 170. Which of the following reactions do not occur—

Answer: 1,2

question 171. Which of the following reactions do not take place easily in the benzene ring—

1. Polyadenylation
2. Polynitration
3. Poly sulphonation
4. Polyalkylation

Answer: 1,2,3

Question 172. Polybrominated takes place in which of the given cases—

Class 11 Hydrocarbons Q&A

Answer: 2,3

Question 173. In which of the following reactions, toluene is obtained— When methanol reacts with PhMgBr

1. Na-salt of o -toluic acid is heated with sodalime
2. p-cresol is distilled in presence of Zn dust
3. Benzyl alcohol is heated in the presence of red
4. Phosphorous and concentrated HI

Answer: 2,3,4

Question 174. In which of the reactions, ferf-butylbenzene is formed—

Class 11 Hydrocarbons Q&A

Answer: 1,3,4

Question 175. Which on ozonolysis forms a mixture of two ketones—

Answer: 1,3,4

Question 176. Which of the following compounds undergo chlorination to produce a type of monochloroalkane—

Answer: 1,2,4

Question 177. Which undergoes nitration reaction faster than benzene-

1. C₆H₅CH₃
2. C₆H₅NHCOCH₃
3. C₆H₅COOH
4. C₆H₅CHO

Answer: 1,2

Question 178. Which undergoes nitration reaction slower than benzene—

1. C₆H₅CH=CHCOOH
2. C₆H₅CH=CH-NO₂
3. C₆H₅CMe₃
4. C₆H₅OCH₃

Answer: 1,2

Class 11 Hydrocarbons Q&A

Question 179. Which of the following cr -complexes are more stable than the cr -complex of benzene—

Answer: 1,4

Question 180. Which of the following compounds has 10 isomers—

1. CI₂C₆H₃NH₂
2. CH₃C6H4NO₂
3. BrClC6H₃CHO
4. O₂NC₆H₃BrCH₃

Answer: 3,4

Question 181. The compounds which get oxidised by alkaline KMn04 to form benzoic acid are —

1. Toluene
2. Ethylbenzene
3. Tert-butyl benzene
4. Benzyl chloride

Answer: 1,2,4

Question 182. Which of the following can be used to distinguish between ethylene and acetylene—

1. Bromine water
2. Ammoniacal Cu₂Cl₂
3. Ammoniacal AgNO solution
4. Dilute alkaline kmnO₄ solution

Answer: 2,3

Question 183. In which two compounds, homolytic cleavage of the C—Ha bond takes place most readily—

Answer: 1,2

Question 184. The compounds which do not participate in Friedel Crafts reaction are —

Class 11 Hydrocarbons Q&A

Answer: 1,2,4

Question 185. Which of the following cannot be used as an alkylating reagent in Friedel-Crafts reaction—

Answer: 2,4

Question 186. Which of the following facts are correct

Answer: 1,3

Question 187. The compounds which exist as liquids are—

1. C₅H₁₂
2. C₃H₈
3. C₂H₆
4. C₇H₁₆

Answer: 1,4

Question 188. Which of the given can be prepared by Wurtz reaction–

1. 2-methylpropane
2. 2,3-dimethyl butane
3. Hexane
4. All of them

Answer: 2,3

Question 189. Which of the following compounds do not produce acetylene on hydrolysis—

1. CaC₂
2. Al₄C₃
3. Be₂C
4. Zn(CH₄)₂

Answer: 2,3,4

Question 190. Markownikoff’s rule is applicable for which of the following reactions—

Answer: 1,2

Question 191. Which of the following options are correct with respect to Friedel-Crafts reaction —

1. Alkylation Reagent: CH₂=C₆H₅Cl
2. Solvent: C₆H₅NO₂, CS₂
3. Catalyst: AlCl₃ , H₂SO₄
4. All Of the Above

Answer: 2,3

Class 11 Hydrocarbons Q&A

Question 192. Lewisite and its antidote are—

1. Lewisite ClCH=CHAsC12
2. Antidote 1,1-dimercapto-l-propanol
3. Lewisite CH₂=CHAsCl₂
4. Antidote 2,3-dimercapto-l-propanol

Answer: 1,4

Question 193. Halogenation ofan alkene is a or an—

1. Substitution reaction
2. Elimination reaction
3. Addition reaction
4. Oxidation reaction

Answer: 1,4

Question 194. During detection of unsaturation in an unknown organic compound disappearance of the violet colour of dilute and cold KMn04 solution indicate—

1. Presence of ethylenic unsaturation in the compound
2. The presence of a group in the compound which gets easily oxidised by kmn04
3. Presence of only single covalent bond in the compound
4. All of the above are true

Answer: 1,2

Question 195. Which of the following options are correct—

1. Ortho- or para-orienting: — NR₂, —NHCOCH₃
2. Mete-orienting: —NO₃, —Cl
3. Ortho- or para-orienting: — CF₃, —SO₃H
4. Mete-orienting: —CHO, —COR

Answer: 1,4

Question 196. Which of the following statements are true for Kolbe’s electrolytic method—

1. It is an effective method for preparing symmetrical alkanes
2. Reduction of carboxylate ion occurs at the anode
3. Platinum electrodes are used in this method
4. Methane cannot be prepared by this method

Answer: 1,3,4

Question 197.   In this reaction, X and Y are-

1. X = CH₃COOH
2. X = HCOOH
3. F = CH₃COONa
4. Y = C₂H₅COONa

Answer: 1,3

Hydrocarbons Very Short Questions And Answers

Question 1. Which hydrocarbon is obtained on hydrolysis of Al₄C₃?
Answer: CH₄

Question 2. Name an alkane which cannot be prepared by the Wurtzreaction.
Answer: CH₄

Question 3. Which alkane is expected to be formed when ethyl magnesium bromide is allowed to react with water?
Answer: Ethane

Question 4. How many acyclic isomers of C₅H₁₂ are possible?
Answer: Isomer

Question 5. What is the main constituent of CNG ?
Answer: CH₄

Question 6. Which type of aliphatic hydrocarbon undergoes substitution reaction?
Answer: Saturated

Question 7. What is the name of the alkene obtained when an aqueous solution of potassium succinate is electrolysed?
Answer: Ethylene

Question 8. CH₃CH=CH₂HCl/peroxlde?
Answer: CH₃CHClCH

Question 9. Which alkene on ozonolysis yields only acetaldehyde?
Answer:  2-butene

Class 11 Hydrocarbons Q&A

Question 10. What is Baeyer’s reagent? What is its use?
Answer:  Alkaline KMnO₄ ,it is used to identify C=C and C=C;

Question 11. What is Lindlar’s catalyst?
Answer: Pd-CaCO₃/ (CH₃COO)₂Pb

Question 12. What is teflon?
Answer: Polytetrafluoroethylene

Question 13. 2-butanone and ethanal are obtained when an alkene containing five carbon atoms is subjected to ozonolysis. State the position of the double bond in the alkene.
Answer: Doublebondis at C-2 of the alkene containing five carbon atoms

Question 14. What is mustard gas?
Answer:  2,2′- dichloro-diethyl sulphide;

Question 15. Name a reagent which can be used to distinguish between 2-butyne and 1-butyne.
Answer: Ammoniacal Cu₂Cl₂

Question 16. Which alkyne is used in Hawker’s lamp?
Answer: HC=CH

Question 17. Mention the name of the compound obtained when acetylene reacts with arsenic chloride.
Answer: Lewisite

Question 18. What is the chemical name of Westron?
Answer: 1, 1,2,2- tetrachloroethane

Question 19. Mention one use ofWestrosol.
Answer: As an organic solvent

Question 20. What is obtained when acetylene is passed through a hot iron tube?
Answer: C₆H₆

Question 21. Give an example of an anti-knock compound.
Answer: Tetraethyl lead

Class 11 Hydrocarbons Q&A

Question 22. Which of the following cannot produce white precipitate by the action of ammoniacal AgNO₃—Acetylene, dimethyl acetylene, methyl acetylene, ethyl acetylene. one acts as the base
Answer: Dimethyl acetylene

Question 23.  Name the reagent which is used to carry out dihydroxylation of a double bond.
Answer: OsO₄ followed by hydrolysis;

Question 24. Which polymer is used to make carry bags? Name its monomer.
Answer: 5. Polyethylene or Polythene, ethylene;

Question 25. Which compound is formed as the major product when propyne reacts with 20% H₂SO₄ in the presence of 1% HgSO₄ at 80°C?
Answer: . Acetone (CH₃COCH₃)

Question 26. What is the state of hybridisation of each carbon atom in an aromatic ring?
Answer: sp²

Question 27. Name the compound obtained by ozonolysis of benzene.
Answer: Glyoxal

Question 28. Give an example of a group which increases the rate of aromatic electrophilic substitution reaction.
Answer: —NH₂

Question 29. Give an example of a group which decreases the rate of aromatic electrophilic substitution reaction.
Answer: —NO₂

Question 30. Name an ortho-/para-orienting group.
Answer: Methyl (-CH₃)

Question 31. Name a meta-orienting group.
Answer: Nitro ( —NO₂)

Question 32. Give an example of a reversible electrophilic substitution reaction.
Answer: Sulphonationreaction

Question 33. Which is the smallest aromatic molecule/ion?
Answer: cyclopropenyl cation

Question 34. What is the orientation of the deactivating halogen atoms?
Answer: Ortho-/para

Class 11 Hydrocarbons Q&A

Question 35. Which heterocyclic compound remains as an impurity in benzene obtained from fractional distillation of coal tar?
Answer: Thiophene

Question 36. Which reagent is used in Birch reduction?
Answer: Na / liquid NH₃, ethanol

Question 37. Which type of flame is observed during the combustion of benzene?
Answer: Sooty flame

Question 38. Nitration occurs at which position of the compound
Answer: Predominantly para position

Question 39. Which electrophile is involved in the desulphonation reaction
Answer:  Proton (H⁺);

Question 40. Give an example of a carcinogen
Answer: 1, 2-benzpyrene

Question 41. Between HNO₃ and H₂SO₄ which one acts as the base during formation ofN02 ion?
Answer: HNO₃

Question 42. Which is the rate-determining step in an aromatic electrophilic substitution reaction?
Answer: First step, i.e., formation of σ -complex;

Question 43. Which step in aromatic electrophilic substitution reaction is exothermic in nature?
Answer: Second step i.e., formation of substituted compound;

Question 44. Give an example of a neutral electrophile which participates in an electrophilic substitution reaction.
Answer: Sulphur trioxide;

Question 45. Give an example of a polynuclear hydrocarbon.
Answer: Anthracene

Question 46. Which compound other than anhydrous AlCl₃ can be used for the ethylation of benzene?
Answer: FeCl₃

Class 11 Hydrocarbons Q&A

Question 47. What is gamm exane? Mention its use.
Answer: BHC (Benzene hexachloride); as an insecticide

Question 48. What is the electrophile involved in nitration reaction?
Answer: NO₂

Question 49. Name the products obtained on pyrolysis of propane.
Answer: Propene, ethene, methane and H₂

Question 50. Which compound out of 1-butene, 1-butyne and 2-butyne is most acidic?
Answer: 1-butyne is the most acidic.

Question 51. Write the name of the compound obtained when n-heptane is subjected to aromatisation
Answer: Toluene (C₆H₅—CH₃).

Question 52. Which alkane cannot be prepared by Kolbe’s method?
Answer: Methane (CH₄).

Question 54. Write the names of the compounds obtained on ozonolysis of o-xylene.
Answer: Glyoxal, methyl glyoxal and dimethyl glyoxal

Question 55. What is lindane?
Answer: Benzene hexachloride (BHC), C₆H₆Cl₆.

Question 56. What is picric acid?
Answer:  2,4,6-trinitrophenol is known as picric acid

Question 57.  What is the name of the compound obtained when benzene is oxidised by air (02) in the presence of V205 catalyst heated at 500°C ?
Answer: Maleic anhydride

Question 58. Which group out of -NO, and -CgHg is an o-/p directing group and which one is a o-/p-directing group?
Answer: NO₂→m -directing; —C₆H₅→o-/p -directing

Question 59.  Which will undergo nitration at a faster rate: C₆H₆ or C₆H₅Cl?
Answer: C₆H₆ undergoes nitration at a comparatively faster rate

Question 60. Name a group which is o -/p -directing but is also a deactivating group.
Answer: Chloro(-Cl)

Question 61. A hydrocarbon on ozonolysis produces ethanal and methanal. 
Answer:  CH₃CH =CH₂

Question 62. Mention the product:
Answer: CH₃CHO

Class 11 Hydrocarbons Q&A

Question 63. Write the structure of an organic compound which reacts with water to yield methanal and hydrogen peroxide
Answer:

Question 64.  Benzene reacts with CH₃COCl in the presence of anhydrous AlCl₃ to form (an organic compound).
Answer: Acetophenone.

Question 65. Which reagent can be used for the following conversion?  HC≡CH→HC=CH₂
Answer: H₂,Pd-CaCO₃/Pb(OAc)₂ (Lindler’s catalyst)

Fill In The Blanks

Question 1. The formula of marsh gas is _______________
Answer: CH₄

Question 2.  _______________ are called paraffins.
Answer: Alkanes

Question 3. Beryllium carbide yields _______________
Answer: CH₄

Question 4. Dutch oil is _______________
Answer: 1,2-dichloroethane;

Question 5. Wurtz reaction is suitable for the preparation of _______________alkanes.
Answer: Symmetrical

Question 6. CHgCOCHg undergoes Clemmensen reduction to yield _______________
Answer: Propane

Question 7.___________ can be identified by Schryver’s colour test
Answer: CH

Question 8. Peroxide effect is applicable only for _______________
Answer: HBr

Question 9. _______________ is obtained when.a solution of sodium butanoate is electrolysed
Answer: n-hexane

Question 10. Isobutylmagnesium bromide reacts with water to form _______________
Answer: Isobutane

Class 11 Hydrocarbons Q&A

Question 11. _______________ on ozonolysis produces formaldehyde and acetaldehyde.
Answer: Propene

Question 12. _______________ is obtained as the major product when 2- butanol is dehydrated.
Answer: But 2 ene

Question Benzene is a polymer of _______________
Answer: Acetylene

Question 14. The simplest hydrocarbon which reacts with ammoniacal silver nitrate to produce a white precipitate is _________
Answer: Acetylene

Question 15. _______________ is obtained when 2-butyne is passed through a mixture of 20% H₂SO₄ and 1% H₂SO₄
Answer: 2 – butanone

Question 16. Hexamethylbenzene is the trimer of _______________
Answer: 2 butyne

Question 17. When a mixture of _________ and Ag – powder is heated ________________ is obtained as the product
Answer: Chloroform, acetylene

Question 18. The values of boiling and melting points of alkadienes are _ than the corresponding alkanes and alkenes containing same number of carbon atoms.
Answer: Higher

Question 19. Ozonolysis of acetylene forms _______________
Answer: Glyoxal

Question 20. Two molecules of HBr react with acetylene to form _______________
Answer: 1,1 dibromomoethane

Question 21. If an alkene forms only one type of carbonyl compound on ozonolysis, then it can be concluded that the alkene is _______________
Answer: Symmetrical

Question 22. C_xH_y _______________ xCO₂ + + y/2 H₂O Heat
Answer: \(\left(x+\frac{y}{4}\right) \mathrm{O}_2\)

Question 23. Number of isomeric tribromobenze is _______________
Answer: Three

Question 24. 1,3,5-trinitrobenzene is an_ compound. _______________
Answer: Explosive

Question 25. The resonance-stabilised carbocation formed in the first step of the electrophilic substitution reaction is called _______________
Answer: Sigma complex

Class 11 Hydrocarbons Q&A

Question 26. _______________ of benzene is carried out by using N⁺O₄ BF^–₄ salt.
Answer: Nitration

Question 27. When benzene is oxidised by atmospheric oxygen in the presence of V₂O₅ at high temperature, _______________ is obtained.
Answer: Maleic anhydride

Question 28. The product obtained due to Birch reduction of benzene when subjected to ozonolysis forms only _______________
Answer: Propanediol

Question 29. —COOH is a/an ______________ Group but COO is an __________ group
Answer: Deactivating, activating;

Question 30. NH₂ group _______________ electron density at ortho-/para positions of the rin
Answer: Increases;

Question 31.  —NO₂ group ______________ electron density at meta position of the ring
Answer: Decreases

Question 32. C —C bond lengths of benzene are _________________
Answer: Equivalent

Question 15. Fill in the blank __________(organic compound) is obtained when an aqueous solution of potassium succinate is electrolysed.
Answer: Ethylene

Class 11 Chemistry Warm Up Questions And Answers

Question 1. What are the chief constituents of LPG?
Answer: The chief constituents of LPG are n-butane and isobutane.

Question 2. Why do C—C bonds instead of C—H bonds of alkanes dissociate due to the effect of heat?
Answer: The bond energy of the C— C bond (ΔH = 83 kcal. mol^-1) is less than that of the C—H bond (ΔH= 99 kcal. mol^-1). So, the C—C bond dissociates more easily than the C—H bond.

Question 3. Write the IUPAC name of the straight-chain hydrocarbon consisting of 20 carbon atoms.
Answer: IUPAC’s name of the straight-chain hydrocarbon consisting of 20 carbon atoms is eicosane.

Question 4. Give the structures of the isomers of molecular formula C₅H₁₂–
Answer: CH₃CH₂CH₂CH₂CH₃ (n -pentane) CH₃CH(CH₃)CH,CH₃ (isopentane) and (CH₃)₄C (neopentane)

Question 5. Explain why dry ether is used in the Wurtzreaction.
Answer:

Dry ether is used because it is present in ether, then it may react with metallic sodium thereby rendering it ineffective

2Na + 2H₂O→2NaOH + H₂

Class 11 Hydrocarbons Q&A

Question 6. Predict whether Me3CBr will take part in Wurtz reaction or not
Answer:  Wurtz reaction proceeds through the S_N2 pathway. As tertiary alkyl halides do not participate in S_N2 reaction (due to steric effect), Me₃CBr does not participate in Wurtz reaction

Question 7. Explain why methane does not react with chlorine in the dark.
Answer:  The reaction does not take place because in the dark Cl —Cl bond does not dissociate to form Cl free radical;

Question 8. One molecule of a hydrocarbon produces one molecule each of acetone, methyl glyoxal and formaldehyde on ozonolysis.Identify the hydrocarbon.
Answer:

The hydrocarbon is 3, 4-dimethylpenta-l, 3-diene [CH₃—C(CH₃)=C(CH₃)—CH=CH₂] or, 2,4-dimethylpenta-1,3-diene [CH₂=C(CH₃)—CH=C(CH₃) —CH₃];

Question 9. Explain why 1-butyne reacts with ammoniacal silver nitrate to produce a white precipitate, but 2-butyne does not
Answer:

1-butyne (CH₃CH₂C=CH) being a terminal alkyne reacts with ammoniacal AgNO₃  solution to produce a, white precipitate but 2-butyne (CH₃C=CCH₃) being a non-terminal alkyne does not react with ammoniacal AgNO₃ solution;

Question 10. How will you detect the presence of acetylene in a gas mixture?
Answer:

If the gas mixture when passed through ammoniacal AgNO₃ solution or ammoniacal Cu₂Cl₂ solution forms a white or red precipitate, then the gas mixture contains acetylene

Question 11. Explain why the carbon-carbon bond in acetylene is shorter than the carbon-carbon bond in ethylene.
Answer:

cr -bond in acetylene (HC=CH) is formed due to the overlapping of two small sp-hybridised orbitals whereas in ethylene (H₂C=CH₂) itis formed by overlapping oftwo bigger sp² hybridised orbitals. So, the bond length of HC=CH <H₂C=CH₂;

Question 12. How will you distinguish between ethylene and acetylene?
Answer:

Acetylene reacts with ammoniacal AgNO₂ solution to form a white precipitate of silver acetylide (AgC=CAg) but ethylene does not give a similar reaction with ammoniacal AgNO₃ solution

Question 13. The population of which conformation increases with the rise in temperature?
Answer: The population of the less stable conformation Increases with the increase in temperature.

Class 11 Hydrocarbons Q&A

Question 14. What are the carbides which react with water to form methane commonly known as?
Answer: The carbides which react with water to form methane are commonly known as methanldes.

Question 15.

Question 16. Why are hydrocarbons insoluble in water but highly soluble in solvents like petroleum ether, benzene, carbon tetrachloride etc?
Answer:

An important principle regarding dissolution is ‘like dissolves like’. It means that polar molecules dissolve in polar solvents while non-polar molecules dissolve in nonpolar solvents. This dissolution process is thermodynamically favourable. Water is a highly polar solvent whereas, petroleum, ether, benzene, and carbon tetrachloride are non-polar solvents. As hydrocarbons are non-polar compounds, they are insoluble in water but soluble in petroleum ether, benzene and carbon tetrachloride.

Question 17. Why are the alkanes called paraffins?
Answer: Alkanes are called paraffin as their chemical reactivity is quite low (Latin: parum = little, affinis = affinity).

Question 18. What are the typical reactions of alkanes?
Answer: Typical reactions of alkanes are substitution reactions.

Question 19. Mention the type of mechanism through which halogenation of alkanes occurs.
Answer: Free-radical mechanism.

Question 20. What happens when methane is heated at 1000°C in the absence of air?
Answer:

Methane when heated at 1000°C in the absence of air, decomposes to form a fine powder of carbon which is known as carbon black:

⇒ \(\mathrm{CH}_4 →{1000^{\circ} \mathrm{C}} \mathrm{C}+2 \mathrm{H}_2 \uparrow\)

Question 21. What is the main constituent of natural gas which is used as a fuel?
Answer: The main constituent of natural gas which is used as a fuel is methane (90%).

Question 22. Why is light or heat essential for the chlorination of alkanes?
Answer:

Cl free radical is required for the initiation of the reaction between an alkane and chlorine, i.e., homolysis of the Cl—Cl bond is necessary. The energy required for this hemolysis is derived from light or heat. So, light or heat is essential for the chlorination of alkanes.

Class 11 Hydrocarbons Q&A

Question 23. Which gas is responsible for explosions in coal mines?
Answer: Methane is responsible for explosions in coal mines.

Question 24. Write the IUPAC name of freon – 113.
Answer: IUPAC name off neon-113, i.e., Cl₂FC— CClF₂ is 1,1,2-trichloro-1,2,2-trifluoroethane

Question 25. Which reaction helps locate the position of double bond in alkenes?
Answer:

The reaction which helps locate the position of double bond in alkenes is ozonolysis.

Question 26. An alkene (C₄Hg) reacts with HBr in the presence or in absence of peroxide to give the same compound. Identify the alkene.
Answer:

As tire given alkene (molecular formula: C₄HO) reacts with HBr to give the same product in the presence and absence of peroxide, the alkene is symmetrical. So, a symmetrical alkene with molecular formula C₄H₈ is but-2-ene (CH₃CH=CHCH₃).

Question 27. Calculate the number of sigma (or) and pi (n) bonds in methyl acetylene.
Answer:

In methyl acetylene (CH₃-C CH), there are 6 a-bonds and 2 bonds.

Question 28. Which of the following compounds will react with metallic sodium to produce H₂ gas?

1. C₂H₄
2. C₆H₆
3. C₂H₂
4. CH₃CH₂CH₃

Answer: 3. Acetylene (C₂H₂) reacts with metallic sodium to produce H2 gas:

⇒ \(\mathrm{HC} \equiv \mathrm{CH}+2 \mathrm{Na} \rightarrow \mathrm{NaC} \equiv \mathrm{CNa}+\mathrm{H}_2 \uparrow\)

Class 11 Hydrocarbons Q&A

Question 29. The C₂ — C₃ bond 1,3-butadiene possesses some double bond characteristics.
Answer:

The C₂—C₃ bond in 1,3-butadiene possesses some double bond character because of the delocalisation of n -electrons.

Question 30. An arena when oxidised forms 1,3-dicarboxylic acid. Write the numbers of side chains and their position in the arena.
Answer:

As the arena gets oxidised to a dicarboxylic acid, it has two side chains. It can be said that the two side chains are at 1,3- or meta-position of each other because a 1,3-dicarboxylic acid forms in the oxidation.

Question 32. Distinguish between benzene and toluene with the help of a chemical reaction.
Answer:

Toluene on oxidation by alkaline KMn04 and subsequent acidification produces shining white crystals of benzoic acid. Benzene, on the other hand, does not undergo oxidation with alkaline KMnO₄ to form any white precipitate.

Question 33. Between — NH₂ and —NO₂, which group facilitates nucleophilic substitution reaction in the benzene ring?
Answer:

The group which facilitates nucleophilic substitution reaction in the benzene ring is — NO₂ because it decreases the electron density of the benzene ring.

Question 34. Arrange in order of increasing reactivity towards electrophilic substitution: benzene, nitrobenzene, toluene, chlorobenzene.
Answer:

The order of increasing reactivity towards electrophilic substitution of the compounds is :

Nitrobenzene< Chloro¬ benzene < Benzene < Toluene.

Question 35. Name the halogen carrier in the chlorination of benzene.
Answer: The compound which acts as the halogen carrier in chlorination of benzene is either AlCl₃ or FeCl₃.

Question 36. Benzene undergoes de-sulphonation but not denitration. Why?
Answer:

Since sulphonation is a reversible reaction, benzene can undergo a desulphonation reaction. However, nitration is an irreversible reaction. So, benzene cannot undergo a nitration reaction.

Question 37. If the calculated and the experimental heats of combustion of benzene are 824.1 and 789.1 kcal mol^-1 respectively, then calculate the value of resonance energy of benzene.
Answer:

Resonance energy = calculated heat of combustion experimental heat of combustion = (824.1 – 789.1)kcal. mol^-1= 35 kcal .mol^-1