WBCHSE Class 11 Chemistry Hydrocarbons Question And Answers And Multiple Choice Questions

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