WBCHSE Class 12 Chemistry Unit 10 Haloalkanes And Haloarenes Notes

Class 12 Chemistry Unit 10 Haloalkanes And Haloarenes Haloalkanes And Haloarenes Introduction

Replacement of H-atom (s) of an aliphatic or aromatic hydrocarbon with halogen atom(s) results in the formation of compounds known as alkyl halide (haloalkane) or aryl halide (haloarene). Such halogen-substituted hydrocarbons are collectively called halohydrocarbons.

• Haloalkanes contain halogen atom(s) bonded to sp³-hybridised C-atom of an alkyl group whereas haloarenes contain halogen atom(s) bonded to sp²-hybridised carbon atom(s) of an aryl group.
• The difference in hybridisation of the C-atom in the C—X bond is responsible for the different characteristics of the two classes of halohydrocarbons. Nature produces a large number of organic compounds containing halogens, some of which are used in medicines.
• For example, the chlorine-containing antibiotic chloramphenicol, produced by soil microorganisms, is a very effective medicine used for the treatment of typhoid. Halohydrocarbons find wide applications in agriculture, industry and in our daily life.
• For example, chloroform and carbon tetrachloride are used as solvents in most of the organic reactions. DDT and BHC are two important chlorine-containing insecticides. Freons or chlorofluorocarbons like CF₂CL₂, CFCl₃ etc., are used as refrigerants.
• Some synthetic halogen compounds are used as medicines. For example, chloroquine is used for the treatment of malaria, diclofenac sodium is used as an analgesic, and halothane (CF₃CHClBr) is used as an anaesthetic (mixed with a very low concentration of O₂ or O₂-N₂O gas mixture), etc. Halohydrocarbons are also used as starting compounds in the synthesis of a variety of other important organic compounds.

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Unit 10 Haloalkanes And Haloarenes Classification Of Haloalkanes And  Haloarenes

Haloalkanes and haloarenes are broadly classified into two types depending on the number of halogen atoms present as well as on the nature of the C- X bond.

Classification based on the number of halogen atoms:

Halohydrocarbons may be classified as mono-, di-, tri-, or tetra- depending on whether they contain one, two, three or four halogen atoms respectively in their molecules.

Example: CH₃Cl: Chloromethane or methyl chloride (mono haloalkane)

CH₂Cl₂: Dichloromethane or methylene chloride (haloalkane)

CHCl₃: Trichloromethane or chloroform (trihaloakane)

Classification Based On The Nature Of C—X Bond:

Monohalocompounds are further classified into three types based on the type of hybridisation of the carbon atom bonded to the halogen atom.

Compounds Containing C_{sp^{3 —}}X Bonds: 

This class is further divided into four types:

1. Alkyl halides or haloalkanes (R—X): In alkyl halides, the halogen atom (X) is attached to an alkyl group (R). They form a homologous series represented by the general formula C_nH_2n+1 X. They are further classified as primary, secondary or tertiary halides, according to the nature of the carbon atom to which the halogen atom is bonded.

If the halogen atom is linked to a primary (1°) carbon atom, the alkyl halide is called a primary (1°) alkyl halide; if the halogen atom is linked to a secondary (2°) carbon atom, the alkyl halide is called a secondary (2°) alkyl halide and if the halogen atom is linked to a tertiary (3°) carbon atom, the alkyl halide is called a tertiary (3°) alkyl halide.

2. Allylic halides: In these compounds, the halogen-containing sp³ – hybridised carbon is attached to a double-bonded carbon (an allylic carbon). Therefore, these monohalo compounds are characterised by a

group. Their general formula is C_nH_2n-1X, where X = F, CL, Br or I and n = 1, 2, 3 … Allylic halides are further classified as 1°, 2° or 3°.

Allylic halides Example:

3. Benzylic halides: In these compounds, the halogen-containing sp³-hybridised carbon atom is linked to an aromatic carbon (benzylic carbon). These are further classified as 1°, 2° or 3°.

Benzylic halides Example:

1. C₆H₅CH₂Cl  → Benzyl chloride

(1° benzylic halide)

2. C₆H₅CHClCH₃ → α-phenyl ethyl chloride                          

(2° benzylic halide)

3.C₆H₅C(CH₃)2CI → α-methyl-α-phenyl ethyl chloride

(3° benzylic halide)

4. Propargyl halides: In these halides, halogen is attached to an sp³ -hybridised carbon atom which is attached to a triple-bonded carbon. These may be further classified as 1°, 2° or 3°.

Propargyl halides Example:

Compounds containing C_sp²—X bond: These are of the following types:

1. Vinylic halides: In these compounds, the halogen atom is bonded to a _sp²-hybridised carbon atom of a carbon-carbon double bond.

Vinylic halides Example:

2. Aryl halides: In these compounds, the halogen atom is bonded to an _sp²  hybridised carbon atom of the aromatic ring.

Aryl halides Example:

Compounds containing C_sp — X bond: In these compounds, the halogen atom is bonded to an sp- hybridised carbon atom of a carbon-carbon triple bond.

Example:

H-C ≡ C-Br → Bromoethyne

CH₃C ≡ C – Cl → 1-chloroprene

Unit 10 Haloalkanes And Haloarenes Nomenclature Of Haloalkanes And Haloarenes

Nomenclature of haloalkanes or alkyl halides:

1. In the IUPAC system, the alkyl halides are named as halosubstituted hydrocarbons, i.e., as haloalkanes. The numbering of the parent straight chain alkane is done in such a way that the carbon bearing the halogen atom gets the lowest possible number.

Nomenclature of haloalkanes or alkyl halides Example: 

2. In the common system, the dihalogen derivatives of alkanes containing two identical halogen atoms attached to the same carbon atom are called alkylidene dihalides.

They are also called geminal dihalides or simply gem-dihalides. In the IUPAC system, they are called dihaloalkanes. The position (or the locant) of the halogen is repeated twice and then prefixed to the name of the dihaloalkane.

Nomenclature of haloalkanes or alkyl halides Example:

CH₃CHBr₂→ IUPAC: 1,1-dibromoethane

Common: Ethylidene dibromide

CH₃CCl₂CH₃→  2,2-dichloro propane

Isopropylidene dichloride

3. In the common system, the dihalogen derivatives of alkanes in which the two identical halogen atoms are present on the adjacent carbon atoms of the carbon chain are called alkylene dihalides.

They are also called vicinal dihalides or simply vie-dihalides. In the IUPAC system, these are called dihaloalkanes. The locants of the two halogen atoms are prefixed to the name of dihaloalkane.

Example:

ClCH₂CH₂Cl → IUPAC: 1,2-dichloroethane

Common: Ethylene dichloride

CH₃CHBrCH₂Br → 1,2- dibromo propane

Propylene dibromide

Nomenclature of haloarenes or aryl halides:

1. In the common as well as in the IUPAC system, aryl halides are called haloarenes.

2. To indicate the relative positions of the two halogen atoms in dihalogen derivatives, the prefixes o-, m- and p- are used in the common system but numerical prefixes 1,2-, 1,3- and 1,4- respectively are used in the IUPAC system.

Nature Of The C—X Bond:

1. The halogen atom is relatively more electronegative than the carbon atom. Therefore, the C— X bond is a polar covalent bond in which the carbon atom bears a partial +ve charge and halogen atom bears a partial negative charge.

2. The size of the halogen atom increases down the group in the periodic table. Thus, the C —X bond length increases and the bond enthalpy decreases from C —F to C—I.

3. Electronegativity decreases from F to I. Thus, the polarity of the C—X bond and hence, the dipole moment of the haloalkane is expected to decrease. However, the dipole moment of CH₃F is slightly lower than that of CH₃Cl.

• This can be explained by the relation μ = e × d. F-atom has a much smaller size and much higher partial negative charge than Cl-atom. Hence, the C — F bond is much shorter than the C — Cl bond.
• As a result, the product of charge (e) and the distance between the two charge centres or the bond length (d), is., dipole moment (μ) of CH₃F turns out to be slightly lower than that of CH₃Cl.

Some physical properties of halomethanes (CH₃—X):

The dipole moment of ethyl chloride is greater than that of vinyl chloride:

• Vinyl chloride is a resonance hybrid. Due to resonance a π -moment (μ_π) is developed in the molecule and it acts in opposition to the strong σ -moment (μ_σ) arising due to the high electronegativity of chlorine.
• Thus in vinyl chloride, the cr -moment is neutralised partially by the n -moment Accordingly, the resultant moment has a low value and it acts in the direction of the σ -moment.
• In ethyl chloride, however, there is no n -moment and it is only the strong cr -moment involving the highly polar C—Cl bond reinforced by the +1 effect of the — CH₃ group. So dipole moment of ethyl chloride is greater than that of vinyl chloride.

Class 12 Chemistry Unit 10 Haloalkanes And HaloarenesPreparation Of Haloalkanes

Preparation of haloalkanes from alcohols Haloalkanes are best prepared by the replacement of the hydroxyl ( — OH) group of an alcohol by a halogen atom.

By the action of halogen acids:

• Alcohols can be converted into haloalkanes by reacting them with halogen acids (HX).
• R—OH (alcohol) + HX → R—X (haloalkane) + H₂O
• For a given alcohol, the order of reactivity of the halogen acids is HI > HBr > HCl for S_N2 reactions. For a given halogen acid, the order of reactivity of alcohols is: 3° > 2° > 1° alcohol.

Explanation: It is a nucleophilic substitution reaction.

The conjugate acid of alcohol, R—OH₂, obtained by accepting a proton from the acid, acts as the substrate and is converted into an alkyl halide (RX) by S_N1 or S_N2 reaction. Halide ion acts as the nucleophile and water (weaker base) acts as a leaving group.

It is an S_N2 reaction for 1° alcohol and methanol:

• Nucleophilicity, i.e., a tendency to donate an electron pair to the C-atom, decreases as:
• I^– > Br^– > Cl^– .Hence, the reactivity of the halogen acids decreases in the order: HI > HBr > HC1.

It is an S_N1 reaction for 2° and 3° alcohol:

A 3° carbocation is more stable than 2° carbonation.  Thus, 3° alcohol is more reactive than 2° alcohol. Order of reactivity of alcohols in S_N1 reaction: 3° > 2° > 1°.

Preparation of chloroalkanes or alkyl chlorides: 1° and 2° chloroalkanes are prepared by reacting an alcohol with gaseous HCl in the presence of anhydrous ZnCl₂. This is known as Grove’s process.

Preparation of chloroalkanes or alkyl chlorides Examples:

Function of anhydrous ZnCl₂: Anhydrous ZnCl₂ is a good Lewis acid. It forms a complex with the alcohol by coordinating with an unshared pair of electrons on the O-atom, providing a better-leaving group for the reaction. As a result, the S_N2 reaction occurs easily.

• Reaction with 3° alcohol to yield tertiary haloalkane proceeds by simply shaking it with concentrated HCl at room temperature.
• (CH₃)₃C-OH + HCl(conc.) → (CH₃)₃C—Cl + H₂O
• 2-methyl propane-2-ol (3°) → 2-chloro-2-methylpropane

Preparation of bromoalkanes or alkyl bromides:

1. Bromo alkanes are prepared by refluxing a suitable alcohol with HBr (48%) in the presence of a small amount of cone. H₂SO₄.

Preparation of bromoalkanes or alkyl bromides Example:

2. HBr is generated in situ (during the reaction) by heating sodium bromide with concentrated H₂SO₄.

Preparation of bromoalkanes or alkyl bromides Example: 

Preparation of iodoalkanes or alkyl iodides:

1. Iodoalkanes are prepared by refluxing a suitable alcohol with HI (57%).

Preparation of iodoalkanes or alkyl iodides Example:

2. HI is also generated by heating KI with 95% H₃PO₄.

Example:

• 2° and 3° alcohols are normally not used to prepare bromoalkanes, as cone H₂SO₄ dehydrates the alcohols to alkenes.
• Phosphoric acid (H₃PO₄) is used, instead of H₂SO₄, in the preparation of HI because the latter (a strong oxidising agent) oxidises HI to I₂ :
• 2HI + H₂SO₄—>2H₂O + SO₂ +I₂
• Haloalkanes cannot be prepared by reacting alcohols with NaCl, NaBr or Nal because, though the halide ion, X^–(mainly Br^– and I^–), is a good nucleophile, it cannot displace the OH^– ion, which is a strong base and a poor leaving group:

By the action of phosphorus halides:

1. Chloroalkanes or alkyl chlorides are prepared by treating alcohols with phosphorus pentachloride (PCl₅) or phosphorus trichloride (PCl₃).

By the action of phosphorus halides Example:

2. Bromoalkanes and iodoalkanes can be prepared by the action of phosphorus tribromide (PBr₃) and phosphorus triiodide (PI₃) respectively on suitable alcohols. PBr₃ and PI₃ are not very stable compounds. Therefore, these are normally prepared by the action of red phosphorus with bromine and iodine respectively.

By the action of phosphorus halides Example:

By the action of thionyl chloride: 

Chloroalkanes can be prepared by refluxing alcohols with thionyl chloride (SOCl₂) in the presence of pyridine.

By the action of phosphorus halides Example: 

The thionyl chloride method for the preparation of chloroalkanes is preferred to other methods because the gaseous by-products (SO₂ and HCl) escape leaving behind chloroalkanes in an almost pure state.

Preparation of haloalkanes from hydrocarbons:

Haloalkanes are also prepared from hydrocarbons (alkanes, alkenes and alkynes).

Preparation of haloalkanes from alkanes:

Haloalkanes (chloroalkanes and bromoalkanes) are prepared by allowing alkanes to react with halogens (Cl₂ or Br₂) in the presence of heat (250-400°C) or light. In this free radical substitution reaction, a mixture of mono-, di-, and polyhaloalkanes is obtained.

Preparation of haloalkanes from alkanes Example: A mixture of 4 substituted compounds is obtained when methane is allowed to react with chlorine in the presence of diffused sunlight at ordinary temperature.

• Methyl chloride is obtained by using an excess of methane. However, this method is not suitable for laboratory preparation because the boiling points of these haloalkanes are quite close and cannot be easily separated. In industry, a mixture of haloalkanes can be separated by using fractionating columns.
• If an alkane contains two or more non-equivalent H- atoms then all the isomeric monohalo derivatives are obtained.
• The reactivity of different types of hydrogens (1 °, 2 ° or 3 ) are different. Thus, different amounts of monohalo derivatives are obtained. The reaction occurs through the formation of free radicals.
• The order of stability follows the sequence: 3° > 2° > l°. Therefore, the reactivity of various types of hydrogen in the halogenation reactions is in the order: 3°H> 2°H> 1°H.

Preparation of haloalkanes from alkanes Example: Propane reacts with chlorine producing 2-chloropropane (major product) and butane reacts with bromine producing 2-bromobutane (major product).

• Bromine is less reactive than chlorine towards alkanes but the selectivity of bromine is greater than that of chlorine. The relative rate of substitution of 3°, 2° and 1° hydrogen atoms by chlorine is 5: 3.8 :1 at 298K and by bromine is 1600: 82: 1 at 400K.
• In industry, free radical chlorination is used for the preparation of alkyl chlorides from alkanes (or cycloalkanes) containing equivalent H-atoms. Free radical bromination is used only for the preparation of 3° alkyl bromide. free radicals are stabilised by resonance.

Therefore, benzylic and allylic H -atoms are highly reactive and so, the haloalkanes are the only products obtained.

The reversible reaction of alkane with iodine occurs at a very slow rate. HI (a strong reducing agent) reduces alkyl iodide to alkane. The reaction is usually carried out in the presence of oxidising agents like HIO₃, HNO₃, HgO etc., which destroy HI as it is formed and thus drive the reaction in the forward direction.

= \(\mathrm{CH}_4 \text { (methane) }+\mathrm{I}_2 \rightleftharpoons \mathrm{CH}_3 \mathrm{I}+\mathrm{HI}\)

= \(5 \mathrm{HI}+\mathrm{HIO}_3 \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}\)

F₂ is a strong oxidising agent. Fluorination of alkanes occurs explosively through the cleavage of C — C bonds. The reaction is highly exothermic. However, the reaction may be carried out smoothly by diluting F₂ with inert gases like N₂ or Ar.

The order of reactivity of halogens towards alkane is:

F₂ > CI₂ > Br₂ > I₂.

Fluorination, chlorination and bromination of alkanes are exothermic while iodination is endothermic.

Preparation of haloalkanes from alkenes: By addition of halogen acids: Haloalkanes can be prepared by allowing halogen acids, HX, to react with alkenes. The order of reactivity of halogen acids is: HI > HBr > HCl > HF. Symmetrical alkenes produce only one haloalkane while unsymmetrical alkenes produce a mixture of two haloalkanes. In this mixture, the major product is one which is obtained according to Markownikoff’s rule.

Preparation of haloalkanes from alkenes Example:

Markownikoff’s Addition:

• In the presence of organic peroxides like benzoyl peroxide (PhCOOOPh), di-tert-butyl peroxide (Me₃COOCMe₃), etc., the reaction of HBr (but not of HF, HC1 or HI) with unsymmetrical alkenes takes place contrary to the Markownikoff’s rule, i.e., anti Markownikoff addition occurs.
• The negative part of the attacking reagent (Br) becomes attached to the carbon carrying more hydrogen atoms while the positive part (H) becomes attached to the carbon containing a lesser number of hydrogen atoms. This is known as the Peroxide effect or Kharasch effect.

Anti-Markonnikoff addition:

By allylic halogenation: When alkenes (except ethylene) are heated with Cl₂ or Br₂ at a high temperature of about 773K, the allylic hydrogen (the hydrogen attached to carbon next to the double bond) is substituted with the halogen atom forming allyl halides.

N-Bromosuccinimide (NBS) is an important reagent of allylic bromination, e.g., when propene, is treated with NBS in the presence of light and peroxide (initiator), bromination occurs at the methyl group giving a good yield of 3 -bromoprop-1-ene.

Sulphuryl chloride (SO₂Cl₂) is an important reagent for allylic chlorination, e.g., when propene, is heated with sulphuryl chloride in the presence of light and a trace amount of peroxide, allylic chlorination of propene occurs to yield 3-chloroprop-1-ene.

Preparation Of Haloalkanes By Other Methods:

Preparation Of Haloalkanes By Halogen Exchange

Finkelstein reaction: This reaction is particularly useful for preparing iodoalkanes. Iodoalkanes are obtained by heating the corresponding chloroalkanes or bromoalkanes with sodium iodide in acetone.

Finkelstein reaction Example:

It is an S_N2 reaction. Hence, the iodoalkane produced has a higher yield, if the preparation begins with a primary (1°) alkyl halide.

Swarts reaction: Alkyl fluorides are prepared by heating chloroalkanes or bromoalkanes with inorganic fluorides like AsF₃ SbF₃, CoF₂, AgF and Hg₂F₂.

Swarts reaction Example:

From Silver Salts Of Carboxylic Acids: Hunsdiecker Reaction: Bromoalkanes are prepared by refluxing the silver salt of a carboxylic acid with bromine in carbon tetrachloride. The reaction is known as the Borodine Hunsdiecker reaction or simply the Hunsdiecker reaction. Chloroalkane is also prepared by this method but the yield is low.

From Silver Salts Of Carboxylic Acids Example: 

• Alkyl bromide produced from carboxylic acid contains one C-atom less than the carboxylic acid. So, this reaction is used to reduce the length of the carbon chain.
• The amount of alkyl halide obtained, depends on the alkyl group of the carboxylic acid, in the order:
• primary > secondary > tertiary.
• Alkyl iodides are not prepared by this method because the reaction between the silver salt of a carboxylic acid and iodine (2:1) forms an ester instead of an iodoalkane. This reaction is called Birnbaum-Simonini reaction.

Physical Properties Of Haloalkanes

1. Physical state: Methyl fluoride, methyl chloride, methyl bromide, ethyl fluoride and ethyl chloride are gases at ordinary temperatures. Methyl iodide and alkyl halides up to C₁₈ are liquids at room temperature. The higher alkyl halides are solids at normal temperatures.
2. Colour: Alkyl halides are colourless in pure state, but bromides and iodides develop colour when exposed to light.
3. Smell: They are normally odourless, but some low-boiling liquid halides like chloroform, have a sweet smell.
4. Solubility: Alkyl halides are soluble in organic solvents like alcohol, ether, benzene etc., but sparingly soluble in water.

Alkyl halides are polar in nature. Therefore, they are expected to be soluble in the polar solvent, water. However, they are practically insoluble In water because they cannot form hydrogen bonds with water.

• In dissolution, energy must be supplied to overcome the intermolecular forces already existing. For a haloalkane to dissolve in water, energy is required to overcome the van der Waals forces and dipole-dipole attractions between the haloalkane molecules and to break the hydrogen bonds between water molecules.
• This energy is not released during bond formation between the haloalkane and water molecules. Hence, haloalkanes are slightly soluble in water.

5. Melting and boiling points: In alkyl halides, the molecules are held together by weak van der Waals forces and weak dipole-dipole interactions. So, they have relatively low melting and boiling points.

• Due to greater polarity as well as higher molecular mass when compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole attractions and van der Waals forces) are stronger in haloalkanes. So, they have higher boiling points when compared to alkanes containing the same number of carbon atoms.

1. The boiling points of haloalkanes increase in the order: RF < RCl < RBr < RI. With an increase in the size and mass of the halogen, the magnitude of van der Waals forces of attraction increases while the dipole-dipole attraction decreases.

The effect of increasing van der Waals forces of attraction is greater than the effect of decreasing dipole-dipole attractions. Therefore, the boiling point increases from RF to RI.

2. For alkyl halides containing the same halogen atom, the boiling point increases with an increase in the size of the alkyl group. This is because with an increase in the size of the alkyl group, the magnitude of the van der Waals forces of attraction increases.

3. For isomeric alkyl halides, the boiling points decrease with increase in branching. This is because, with increase in branching, the surface area of the alkyl halide decreases. Consequently, the magnitude of van der Waals forces of attraction decreases.

4. The boiling points of chloro, bromo and iodo compounds increase with an increase in the number of halogen atoms. This is because an accumulation of Cl atoms on CH₄ increases the molecular weight and size, which increases the van der Waals forces of attraction.

Polyfluoroalkanes usually have low boiling points. For example, hexafluoroethane (C₂F₆) boils at -79°C, even though its molecular weight (138) is close to that of decane (b.p. 174°C).

Fluorine is highly electronegative. Hence, the electrons of poly fluoroalkanes are tightly held and their molecules do not get easily polarised. Therefore, van der Waals forces of attraction between them are much weaker, leading to their low boiling points.

6. Density: Alkyl fluorides and chlorides are lighter than water while bromides, iodides and polychloro derivatives are heavier (denser) than water.

1. The density of alkyl halides decreases with an increase in the number of carbon atoms (i.e., the size of the R group).

2. The density increases with increase in the number of halogen atoms.

3. The density increases with increase in the atomic mass of the halogen atom, RI>RBr>RCI

7. Stability: C—X bond strength decreases in the order: C—F > C—Cl > C—Br > C—I. Therefore, the stability of haloalkanes having the same alkyl group decreases in the order: R—F > R—Cl > R —Br > R—I. Alkyl iodides turn violet when exposed to light. This is because the weaker C—I bond decomposes easily in the presence of light to give iodine (I₂):

= \(2 \mathrm{R}-\mathrm{I} \stackrel{h \nu}{\longrightarrow} \mathrm{R}-\mathrm{R}+\mathrm{I}_2\)

Due to higher stability and easy availability, organic chlorides are the most commonly used solvents in industry, rather than bromides and iodides.

8. Inflammability: Inflammability of organic halogens is less than that of the corresponding hydrocarbons, which decreases with an increase in halogen atoms.

• Since halo-hydrocarbons are solvents for fats and oils and do not catch fire easily, polychloro compounds like tri- and tetrachloroethylene are extensively used as solvents for dry cleaning.
• Carbon tetrachloride or tetrachloromethane (CCl₄) is used as a fire extinguisher in the name of ‘pyrene’.

Unit 10 Haloalkanes And Haloarenes Nucleophilic Substitution Reactions Of Haloalkanes Or Alkyl Halides

Nucleophilic substitution reactions:

In the haloalkanes, a carbon atom is bonded to a more electronegative halogen atom (F, Cl, Br or I). As a consequence, the carbon atom acquires a partial positive charge and the halogen atom acquires a partial negative charge.

• Due to the presence of a small positive charge on carbon, a haloalkane molecule is very susceptible to attack by a nucleophilic reagent (i.e., a reagent possessing a negative charge or an unshared pair of electrons).
• Thus, when a nucleophile stronger than the halide ion (X^–) attacks the positively polarised C- atom of an alkyl halide, the halogen atom with its bonding pair of electrons is displaced and a new bond between the carbon atom and the incoming nucleophile is formed. These reactions are called nucleophilic substitution reactions.

Leaving group: In a nucleophilic substitution reaction, the atom or group which departs with its bonding pair of electrons is called the leaving group. Halide ions (except F~ ) are good leaving groups.

Better the leaving group faster is the nucleophilic substitution reaction. Among the alkyl halides, I^– is the best-leaving group and F^– is the worst.

The leaving ability of groups (fugacity) increases with decreasing basicity. Since the basic order of halides is

→ \(\mathrm{F}^{\ominus}>\mathrm{Cl}^{\ominus}>\mathrm{Br}^{\ominus}>
\mathrm{I}^{\ominus}\), their leaving ability increases in the order of

→ \(\mathrm{F}^{\ominus}<\mathrm{Cl}^{\ominus}<\mathrm{Br}^{\ominus}<\mathrm{I}^{\ominus}\).

• Also, to be effective a leaving group should be linked to the central carbon atom with a relatively weak bond. The C X bond strength increases in the order of C —I < C —Br < C —Cl < C —F. Therefore the order of leaving ability is as mentioned above. Therefore, iodoalkanes undergo nucleophilic substitution reactions at the fastest rate, while fluoroalkanes are practically unreactive under ordinary conditions.
• The reactivity of haloalkanes is due to the presence of a polar C—X bond. In methyl halides, C—X bond polarity follows the sequence: CH₃ —Cl(1.94D) > CH₃ —Br(1.79D) > CH₃—I(1.64D).
• Thus order of reactivity of the haloalkanes is expected to be: R—Cl > R—Br > R — I. But the observed reactivity sequence is found to be exactly opposite to this order, i.e., R—I > R—Br > R—Cl.
• The observed sequence of reactivity is in accordance with the leaving group ability of the halide ion (i.e.,\(\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}\) ).

Some important nucleophilic substitution reactions of alkyl halides or haloalkanes:

= \(\mathrm{Nu}^{\ominus}+\mathrm{R}-\mathrm{X} \rightarrow \mathrm{R}-\mathrm{Nu}+\mathbf{X}^{\ominus}\left(\mathrm{R}=\mathrm{Me} / 1^{\circ} / 2^{\circ}, \mathrm{X}=\mathrm{Cl} / \mathrm{Br} / \mathrm{I}\right)\)

Mechanism of nucleophilic substitution reactions: Nucleophilic substitution reactions are represented as S_N (Substitution Nucleophilic). Depending on the number of species (molecule or ion) involved in the rate-determining step of the reaction, the S_N reactions are classified as S_N1 and S_N2.

• The number of particles taking part in the rate-determining step (the slowest step of the reaction) is called the molecularity of the reaction.
• The molecularity of the reaction is 1 when only one particle participates in the rate-determining step and it is 2 when two particles participate in the rate-determining step.
• If the molecularity is 1, the mechanism of the substitution reaction is called S_N1 (Substitution Nucleophilic Unimolecular) and if the molecularity is 2, the mechanism of the reaction is called S_N2 (Substitution Nucleophilic Bimolecular).

S_N2 Reaction (Substitution Nucleophilic Bimolecular):

S_N2 Reaction Definition: The reactions taking place by S_N2 mechanism i.e., the reaction in which the r.d. step is bimolecular, are called Sn2 reactions.

Characteristics: The expulsion of the leaving group from the substrate (bond breaking) and the attachment of the nucleophile with the substrate (bond formation) takes place simultaneously, in a single step. Hence, it is the rate-determining step (r.d.s) or rate-limiting step of the reaction. o The rate of an S_N2 reaction depends on the

S_N2 Reaction Example: Methyl chloride undergoes hydrolysis in aqueous solution of KOH to produce methyl alcohol. This reaction occurs by the S_N2 mechanism.

S_N2 Reaction Reaction: The nucleophile (OH^–) attacks the carbon atom of CH₃CI from the side exactly opposite to that of the C — Cl bond (backside attack). This is because, in this mode of attack, the electrostatic force of repulsion between the negatively charged nucleophile and the negatively charged leaving group is minimal as the angular distance between them is maximum.

• The backside attack is also sterically more feasible. As the electron pair of OH^– ion approaches the carbon atom, the C —Cl bonding electrons get closer to the Cl atom. Thus, a state is reached when the C — Cl bond is neither completely cleaved nor the C — OH bond formed completely.
• It is called the transition state, in which the carbon is half-bonded to both hydroxyl and chlorine and full-bonded to the three hydrogen atoms. In the transition state, the negative charge of OH^– is partially reduced while Cl^– atom acquires a partial negative charge and also, the tetrahedral (sp³) carbon of CH₃CI becomes trigonal (sp²).
• The carbon and the three attached hydrogen atoms are coplanar (every bond angle is 120°) and the plane is perpendicular to the line in which the nucleophile OH^–, carbon and the leaving group (Cl^–) exist.
• Once the transition state is formed, the OH^δ- group further approaches the carbon atom to form a 100% covalent bond while the Cl^– atom is eliminated as Cl^– by taking full possession of the CCl bonding electrons. In the substituted compound CH₃OH, the carbon is once again tetrahedral (sp³). The energy required to break the C — Cl bond is supplied by the formation of the C — O bond.

Rate of the reaction is controlled by the concentration of both the nucleophile (OH^–) and the substrate (CH₃Cl), i.e., these two particles are involved in the transition state of the rate-determining step (the only step here). Therefore, it is an S_N2 reaction.

∴ Rate of the reaction = \(k\left[\mathrm{CH}_3 \mathrm{Cl}\right]\left[\mathrm{OH}^{-}\right]\)

The orbital picture of the transition state clearly shows why a backside attack always occurs in an S_N2 reaction.

Orbital picture of S_N2 transition state: In the transition state, the carbon atom which undergoes nucleophilic attack is sp² hybridised. The p-orbital of the carbon atom is placed perpendicular to the plane in which the three sp² orbitals exist. One lobe of the p- orbital overlaps with the leaving group L^– and the other lobe overlaps with the nucleophile Nu^–. Hence, a frontside attack does not take place. For a frontside attack to occur, the nucleophile and the leaving group should overlap with a single lobe of the p-orbital, which is impossible.

S_N1 reaction (Substitution Nucleophilic Unimolecular):

S_N1 reaction Definition: The reactions taking place by S_N1 mechanism i.e., the reactions in which the r.d. step is unimolecular, are called S_N1 reactions.

S_N1 reaction Characteristics: These reactions occur in two steps. In the first step, the leaving group (normally an anion) is detached from the substrate to form a stable carbocation intermediate.

• This step is slow and hence, is the rate-determining step of the reaction. In the second step, the nucleophile (an anion or a neutral molecule) combines with the carbocation to form the substituted product through the formation of a covalent bond.
• The rate of an S_N1 reaction depends only on the concentration of the substrate, i.e., the rate of the reaction oc [substrate]. The S_N1 reactions in which the solvent acts as the nucleophile are called solvolysis.

S_N1 reaction Example: Hydrolysis of tert-butyl bromide by aqueous KOH solution leads to the formation of tert-butyl alcohol. The reaction takes place by the S_N1 mechanism,

Reaction Mechanism:

• In the first step, the C— Br bond of ferf-butyl bromide dissociates to form ferf-butyl cation (intermediate) and bromide ion (Br^–). This step is endothermic and hence, takes place slowly. Therefore, it is the rate-determining step of the reaction. The energy required for breaking the C Br bond is available from the energy released due to solvation of these two ions (solvation energy).
• In the second step, the nucleophile (OH^– ion) attacks the carbocation to form tert butyl alcohol, the substitution product. Due to the formation of a bond, this step is exothermic and hence, occurs readily.
• 

Rate of the reaction depends only on the concentration of the substrate (CH₃)₃CBr, i.e., only the (CH₃)₃CBr molecule is involved in the transition state of the rate-determining step (the first step). Therefore, it is an S_N1 reaction.

∴  Rate of the reaction = K [(CH₃)₃CBr]

Factors affecting the rates of S_N1 and S_N2 reactions:

Structure of the substrate R—X:

1. In an S_N2 reaction, the attack by the nucleophile on the or-carbon (i.e., the carbon bearing the halogen) occurs from the backside (the side exactly opposite to that of the leaving group). Therefore, the presence of bulky substituents on or near the a-carbon atom tends to hinder the approach of the nucleophile towards the a-carbon due to steric hindrance and so, the reaction occurs with difficulty.
2. Evidently, the greater the number of alkyl groups, i.e., the greater the steric hindrance, the slower the reaction. Of all the alkyl halides, methyl halides [e.g., CH₃Cl), are the most reactive in S_N2 reaction. This is because they experience the least steric hindrance due to the presence of three small H atoms on the or-carbon.
3. In 1° alkyl halides (for example., CH₃CH₂Cl) there is one alkyl group attached to a -carbon atom. Hence, due to greater steric hindrance, they are less reactive than methyl halides. In 2° alkyl halides (e.g., Me₂CHCl), there are two alkyl groups attached to the a -carbon atom.
4. Hence, due to greater steric hindrance, they are less reactive than 1° alkyl halides. Due to the presence of three alkyl groups on the a -carbon atom, the steric hindrance experienced by the 3° alkyl halides (for example., Me₃CCl) is the maximum and so, they are unreactive in S_N2 reactions.
5. Thus, the S_N2 reactivity of various alkyl halides follows the order: methyl halides > primary (1°) alkyl halides > secondary (2°) alkyl halides >> tertiary (3°) alkyl halides.
6. Steric hindrance and relative S_N2 reactivity of methyl, ethyl, isopropyl and tert-butyl halides are shown below:

• A carbocation is formed in the rate-determining step of an S_N1 reaction. Therefore, the reactivity of the substrate depends on the stability of the resulting carbocation. The greater the stability of the carbocation, the easier its formation and hence, the faster the rate of the reaction, i.e., higher the reactivity of the substrate.
• The stability of carbocations decreases in the order: of 3° carbocation >2° carbocation > 1° carbocation > CH₃. Therefore, the reactivity of the alkyl halides towards the S_N1 reaction decreases in the same order: 3° > 2° > 1° > methyl halides. Therefore the reactivity of various types of alkyl halides towards S_N2 and S_N1 reactions follows the reverse order:

Allyl and benzyl carbocations are stabilised by resonance. Therefore, allyl and benzyl halides are more reactive than other simple primary halides, towards S_N1 reaction.

• From the above discussion, it may be concluded that methyl and primary alkyl halides undergo substitution predominantly by the S_N2 mechanism, whereas tertiary halides undergo substitution by the S_N1 mechanism.

Secondary, primary allylic and primary benzylic halides may react either by S_N1 or S_N2 mechanism or by both mechanisms without much preference, depending on the condition of the reaction (i.e., nature of the nucleophile, nature of the solvent, etc.)

Nature of the solvent:

• Polar protic solvents like CH₃OH, C₂H₅OH, H₂O, etc. are suitable for S_N1 reaction because in such solvents both the carbocation and the halide ion are stabilised by solvation.
• Molecules of the protic solvents, solvate a halide ion by forming hydrogen bonds with it. Molecules of the protic solvent, solvate a carbocation by orienting their negative ends around the cation and by donating the unshared electron pairs to the vacant orbitals of the cation.

Aprotic polar solvents, such as dimethyl sulphoxide, Me₂S=O(DMSO); N, N-dimethylformamide,(CH₃)₂NCHO (DMF); acetone (CH₃COCH₃), etc. are suitable for S_N2 reactions as they preferentially solvate cations. They do not solvate anions as they are unable to form H-bonds. Thus, the anions remain relatively free and highly reactive as nucleophiles.

Concentration and reactivity of the nucleophile:

• The nucleophile does not participate in the rate-determining step of an S_N1 reaction. The rates of S_N1 reactions are unaffected both by the concentration and the nature of the nucleophile. The rates of S_N2 reactions, however, depend on both the concentration and the nature of the nucleophile.
• This is because the nucleophile is involved in the one-step reaction. A strong nucleophile is one that reacts rapidly with a given substrate. In a group of nucleophiles containing identical nucleophilic atoms, nucleophilicity parallels basicity, i.e., a stronger base is a stronger nucleophile.

Oxygen compounds exhibit the given order of nucleophilic reactivity:

RO^– > HO^– >> RCOO^– > ROH > H₂O

• However, when the attacking atoms are elements belonging to the same group of the periodic table, the nucleophilicity increases down the group because the attacking atom becomes progressively less electronegative.
• The larger the size, the more polarisable the nucleophile. For instance, the order of nucleophilicity of the halide ions is: I^– > Br^– > Cl^– > F^– (in protic solvent). The nucleophilicity decreases with an increase in electronegativity from left to right, for example: CH₃^– > NH₂^– > OH^– > F^–.

Nature of the leaving group:

• The rates of both S_N1 and S_N2 reactions are influenced by the nature of the leaving group because, in both reactions, the expulsion of the leaving group from the substrate occurs in the rate-determining step.
• The best-leaving groups are the ions or molecules that are the weakest bases. For instance, in the following series, the leaving-group ability {fugacity) increases with decreasing basicity:
• NH^–₂<CH₃O^–<HO^–<PhO^–<NH^–₃<CN^–-<F^–<Cl^–<Br^–<I^–
• The leaving group ability of the halide ions also depends on the strength of the C—X bond. C—I bond dissociation enthalpy (234 kJ.mol^-1) is the lowest and C—F bond dissociation enthalpy (452 kJ.mol^-1) is the highest. Therefore, iodide is the best while fluoride is the poorest leaving group.

Comparison Between S_N2 And S_N1 Reaction:

Possible mechanistic courses depending on the structure of the substrate:

Ambident nucleophile:

The nucleophiles containing more than one (generally two) nucleophilic centres i.e., the nucleophiles capable of reacting at more than one site, are called ambident nucleophiles.

Example: Cyanide ion, CN^– (nucleophilic centres: C and N) and nitrite ion, NO^–₂  (nucleophilic centres: N and O ) are two ambident nucleophiles.

In an S_N1 reaction, the more electronegative atom (N atom of CN^– ion and O atom of NO^– ion) is involved in a nucleophilic attack on the positively charged carbon atom of the carbocation. On the other hand, in an S_N2 reaction, the less electronegative and larger (more polarisable) atom (C of CN- and N of NO^–₂ ) is involved in nucleophilic attack on the positively polarised carbon atom of the substrate.

= \(\mathrm{R}-\mathrm{Br}+\mathrm{K}^{+} \mathrm{CN}^{-} \stackrel{\mathrm{S}_{\mathrm{N}^2}}{\longrightarrow} \mathrm{R}-\mathrm{CN} \text { (alkyl cyanide) }+\mathrm{KBr}\)

= \(\mathrm{R}-\mathrm{Br}+\mathrm{AgCN} \stackrel{\mathrm{S}_{\mathrm{N}} 1}{\longrightarrow} \mathrm{R}-\mathrm{NC} \text { (alkyl isocyanide) }+\mathrm{AgBr}\)

Stereochemistry of nucleophilic substitution reactions: In an S_N2 reaction, a complete inversion of the configuration of the substrate occurs while in an S_N1 reaction racemisation occurs.

• To understand the stereochemistry of S_N1 and S_N2 reactions, some basic principles of stereochemistry and terms such as plane polarised light, optical activity, chirality, enantiomers, diastereoisomers, retention, inversion, racemisation etc., are discussed below:

Plane polarised light:

• Ordinary light consists of electromagnetic waves of various wavelengths (A = 4000-7000A). On the other hand, monochromatic light consists of waves of single wavelength.
• It can be obtained either by passing visible light through a prism or by using a source emitting light of a single wavelength e.g., sodium vapour lamp emits yellow light (A = 5893A).
• Both visible and monochromatic light consist of waves which vibrate in all the planes perpendicular to the line of propagation of the light.
• These are unpolarised light. If such a beam of light is passed through a Nicol prism (developed by William Nicol, a Scottish physicist, from a particular crystalline form of CaCO₃ known as calcite), the light coming out of the prism vibrates in only one plane.
• All other planes are eliminated. A beam of light which vibrates only in a single plane is called plane polarised light.

Optical Activity:

Optical Activity Definition: The compounds which can rotate the plane of I polarisation of a plane-polarised light when such a light is passed through them or their solution are called optically active compounds and the phenomenon is called optical activity.

The angle through which the plane of polarised light rotates, i.e., the angle of rotation (a) can be measured with an instrument known as a polarimeter.

Stereoisomers that rotate the plane of polarisation of a plane-polarised light are said to be optically active isomers or optical isomers and the phenomenon is called optical isomerism.

Optical isomers may be divided into two types on the basis of the rotation of the plane of polarised light. These are:

Dextrorotatory compounds: The optical isomers which rotate the plane of polarisation in a clockwise direction. (Latin ‘dexter1 meaning right). They are represented by d or (+) sign.

Laevorotatory compounds: The optical isomers which rotate the plane of polarisation in a counterclockwise direction (Latin ‘laevus’ meaning left). They are represented by the l or (-) sign.

Symmetric and asymmetric molecules:

Stereocentre: The four valencies of an sp³-hybridised carbon atom are directed towards the corners of a regular tetrahedron and if all the atoms or groups attached to a carbon atom are different, such a carbon is called asymmetric carbon or stereogenic centre or stereocentre.

Asymmetric molecule: If a molecule contains no element of symmetry, such as plane of symmetry, centre of symmetry and alternating axis of symmetry, then it is called an asymmetric molecule. Every asymmetric molecule is optically active.

Symmetric molecule: If a molecule contains at least any one of the elements of symmetry, such as the plane of symmetry centre of symmetry or alternating axis of symmetry, then it is called a symmetric molecule. Every symmetric molecule is optically inactive.

Characteristics Of Symmetric And Asymmetric Molecules: A symmetric molecule or object and its mirror image are identical in all respects and are superimposable on each other.

Characteristics Of Symmetric And Asymmetric Molecules Example: Ethanol (CH₃CH₂OH) and 2-chloropropane (CH₃CHClCH₃) are examples of symmetric molecules. Most of the objects we see are symmetric.

• A ball, a book, a table, a chair, a cup, and solid models of the alphabets H, O, A etc., are examples of symmetric objects. These objects and their mirror images are identical in all respects and are superimposable on each other. All these substances have a plane of symmetry.
• An asymmetric molecule or object and its mirror image are non-superimposable to each other.

Example: Lactic acid (CH₃CHOHCOOH) and 1,2- dichloropropane (CH₃CHClCH₂Cl) are examples of asymmetric molecules. Left and right hands left and right shoes, the solid models of the alphabets P, Q and R, etc. are examples of asymmetric objects.

The left hand is the mirror image of the right hand. If we put our left hand on our right hand and vice-versa, they do not coincide, i.e., they are non-superimposable on each other.

Chirality

Chirality Definition: Objects that are non-superimposable on their mirror images are called chiral objects and the property is called chirality. Objects that are superimposable in their mirror images are called achiral objects. Optical activity is an important physical property of chiral objects.

Test for chirality: Chiralilty can be determined by the non-superimposability of an object and its mirror image. For this, models of an organic molecule and its mirror image are constructed.

• If these two models are superimposable (i.e., if the two models can be placed on each other in such a way that atoms of one structure coincide with the other), then it is symmetric or achiral. If they are non-superimposable, then it is asymmetric or chiral.

Test for chirality Example: The achirality of 2-bromopropane (CH₃CHBrCH₃) can be determined by its superimposability.

• Three-dimensional structures of 2-bromopropane (A) and its mirror image (B) are drawn. To check the superimposability of B over A, B is rotated through 180° in such a way that the C—Br bond in the new structure C projects in the same direction as in the structure A, making A and C, superimposable.

• The presence of asymmetric C-atom in 2-bromobutane (CH₃CHBrCH₂CH₃) makes it a chiral molecule. This f can be determined by its non-superimposability.
• Three-dimensional structures of 2-bromobutane (D) and its mirror image (E) are drawn. To check the non-superimposability of E 180° in such a way that the C — Br bond in the new structure F projects in the same direction as in structure D.
• Now F is superimposed over D and it is found that D and F are non-superimposable irrespective of the way it is turned or twisted without breaking the bonds.

From the above discussion, it may be concluded that the presence of a single asymmetric carbon atom makes the molecule chiral and hence optically active. Therefore, an asymmetric carbon is also called a chiral carbon. In chiral molecules, the asymmetric carbon or the chiral carbon is denoted by an asterisk (*).

Symmetry elements:

Molecular chirality arises due to the absence of the three elements of symmetry:

1. Plane of symmetry
2. Centre of symmetry

Alternating axis of symmetry (S_n): A molecule is said to have an alternating axis of symmetry of n fold (or order) if rotation of the molecule about the axis by 360° In followed by reflection through a plane, perpendicular to this axis produces an indistinguishable structure with the original.

An alternating axis of symmetry is designated as S_n. It is also called rotation-reflection symmetry, triflic acid, for example, has an S₂ axis. The molecule is, therefore, achiral and optically inactive.

Asymmetric carbon atom and optical activity: Molecules with only one asymmetric carbon atom are asymmetric and optically active and those with more than one asymmetric carbon atom may be symmetric or asymmetric, (may be active or inactive).

• For example, (+)-tartaric acid containing two asymmetric carbon atoms is optically active while meso-tartaric acid is optically inactive.
• However, the presence of an asymmetric carbon in a compound is not an essential condition for exhibiting optical activity. For example, some substituted alkenes (abC=C=Cab) having different terminal substituents are optically active, even though they contain no asymmetric carbon atom. This is because these molecules are chiral as a whole.

Example:

• Therefore, the presence of an asymmetric carbon or a chiral carbon is not the only condition for a molecule to show optical activity.
• A molecule is optically active provided the entire molecule is chiral. From the above discussion, it can be stated that a compound will be optically active if any one of the following conditions is fulfilled: the compound and its mirror image are non-superimposable on each other, the compound contains only one asymmetric carbon atom and none of the elements of symmetry, such as plane of symmetry, centre of symmetry and alternating axis of symmetry should be present in the compound.
• E. Eliel has mentioned a molecule which does not contain any element of symmetry, yet is optically active. He further said that according to this principle, there is a possibility of the existence of such other molecules.
• So, a necessary and sufficient condition for a compound to be optically active is the non-superimposability of the molecule on its mirror image i.e., chirality.

Meso-compounds: Stereoisomers which are achiral and optically inactive inspite of the presence of multiple chiral centres are called meso-compounds, e.g., meso-tartaric acid is a meso-compound.

Enantiomers and diastereoisomers:

Enantiomers: A pair of stereoisomers which are non-superimposable mirror images of each other are called enantiomers or enantiomorphs.

Enantiomers Example: Lactic acid or 2-hydroxypropanoic acid (CH₃CHOHCOOH) has 2 optically active isomers which are enantiomers of each other. The three-dimensional representations (flying-wedge projection formula) and Fischer projection formula of lactic acid are shown below:

• A pair of enantiomers have identical physical properties, such as melting point, boiling point, solubility, density, etc. They differ only in their interaction with the plane-polarised light. One enantiomer rotates the light to the right (dextrorotatory) and the other an equal magnitude to the left (laevorotatory).
• They have identical chemical properties except their reactions with chiral reagents. Such a pair of optical isomers is also called an optical antipode.

Diastereoisomers: A pair of stereoisomers that are not enantiomers, are called diastereoisomers or diastereomers.

Diastereoisomers Example: Optically active (+) -tartaric acid and optically inactive meso-tartaric acid represent a pair of diastereoisomers.

Similarly, cts-but-2-ene and trans-but-2-ene represent a pair of diastereoisomers.

Diastereoisomers have different physical properties like melting point, boiling point, solubility, dipole moment, specific rotation etc. They also differ in chemical properties. They react at different rates with both chiral and achiral reagents.

Racemic Mixture Or Racemic Modification And Racemisation:

Racemic Mixture: An equimolecular mixture of a pair of enantiomers is known as racemic mixture or racemic modification.

• The specific rotations of two enantiomers are equal in magnitude but opposite in direction. Therefore, the positive rotation caused by the molecules of one enantiomer is cancelled by the negative rotation caused by the same number of molecules of the other enantiomer. As a consequence, racemic modification is always optically inactive.
• The (+) – rotation of one enantiomer is compensated by the (-)- rotation of the other. So, it is said that a racemic modification is optically inactive due to external compensation. A racemic modification is represented by prefixing dl- or (±) before the name of the compound.

Racemic Mixture Example: Racemic lactic acid CH₃CHOHCOOH is written as (±) -CH₃CHOHCOOH or dl-CH₃CHOHCOOH.

Racemisation: The process of producing racemic modification starting from either of the pure enantiomers is called racemisation.

This may be carried out by applying heat or by allowing the pure enantiomer to react with a chemical reagent. Racemisation is a thermodynamically favourable process because it possesses entropy of mixing, ΔS. ΔS is a positive quantity. Therefore, in the expression ΔG = ΔH – TΔS, ΔG is negative (assuming ΔH is constant).

Retention Of Configuration:

If the spatial arrangement of bonds with respect to an asymmetric carbon in a chiral molecule remains the same before and after the reaction, retention of configuration occurs.

Explanation: In the transformation of the substrate XCabc into the product YCabc, if the hypothetical path, traced by the substituents a, b, c (a→b→c), remains the same (clockwise as shown in the following figures) in both the substrate and the product, the reaction is said to have taken place with retention of configuration. XCabc and YCabc have the same relative configuration.

In general, during a reaction, if none of the bonds to the chiral centre or stereocentre is broken, the product has the same relative configuration of groups around the stereocentre as that of the substrate. Such a reaction is said to proceed with retention of configuration.

Retention Of Configuration Example: When 2-methylbutan-l-ol is heated with concentrated HCl, 1-chloro-2-methylbutane is obtained. In this reaction, no bond to the asymmetric carbon is broken. Therefore, the reaction must proceed with the retention of the configuration.

Inversion Of Configuration:

If the spatial arrangement of the bonds with respect to an asymmetric carbon in a chiral compound suffers a change after the reaction, the compound is said to have undergone inversion of configuration.

Inversion Of Configuration Explanation: In the transformation of the substrate XCabc into the product YCabc, if the hypothetical path traced by the substituents a, b, c(a→b→c) changes (clockwise in XCabc and anticlockwise in YCabc), the substrate is said to have undergone the reaction with inversion of its configuration.

Inversion Of Configuration Example: When 2-chlorobutane is subjected to alkaline hydrolysis under SN2 conditions, the product 2-butanol is obtained with an inversion of configuration.

Inversion, retention and racemisation: There are three outcomes of a chemical reaction involving bond cleavage and bond formation at an asymmetric carbon atom. Let us, for example, consider the substitution of the group by Y in the following reaction.

• If (1) is the only product obtained, the reaction is said to proceed with the retention of configuration. If (2) is the only product obtained, the reaction is said to proceed with the inversion of configuration.
• If, however, a 1: 1 mixture of (1) and (2) is obtained, the process is called racemisation. The product is optically inactive because one isomer rotates the plane of polarisation of a plane-polarised light in a direction opposite to that of the other isomer but to the same extent.

Stereochemistry Of S_N2 Reactions: In S_N2 reactions, the attack of the nucleophile occurs from the side directly opposite to the leaving group. The backside attack causes the arrangement of the three nonreacting groups on the central carbon to be turned inside out like an umbrella in a high wind.

As a result, S_N2 reactions of optically active alkyl halides are always accompanied by an inversion of configuration at the chiral centre.

Stereochemistry Of S_N2 Reactions Example: When (-)-2-bromooctane is heated with NaOH solution, (+)-2-octanol is formed with the —OH group occupying the position opposite to what bromine had occupied.

The reactant and the product of the S_N2 reactions are usually not enantiomers (except when R — I is treated with I^–). So, they may have the same or opposite signs of optical rotation.

Stereochemistry Of S_N1 Reactions: S_N1 reactions are accompanied by racemisation which means if the alkyl halide is chiral (optically active), the product is a racemic mixture.

• The intermediate carbocations formed in the first step (slow and r.d. step) of SN1 reactions are sp²-hybridized and planar (achiral).
• Therefore, the attack of the nucleophile on it, takes place from either side (front and rear) giving a 1:1 mixture of the two enantiomers. In one enantiomer, retention of configuration while in the other, inversion of configuration occurs. The resulting (racemic) mixture is optically inactive.

Example:  Hydrolysis of [+] or (-)-3-bromo-3-methylhexane in aqueous acetone results in the formation of (±)-3- methylhexan-3 -ol

D, L-system of nomenclature: Relative configuration: Stereoisomers differing in configuration must be properly designated for their stereochemical identities.

• The oldest system of nomenclature of optical isomers having chiral centres is D, L-system, introduced by Emil Fischer.
• The glyceraldehyde [HOCH₂CH(OH)CHO] molecule was chosen by Fischer as the standard for defining configurations of molecules with chiral centres.
• Fischer arbitrarily labelled the (+)-enantiomer as D-enantiomer and (-)-enantiomer as L-enantiomer and drew the projection formulas of D-and L-glyceraldehyde as follows:

• The enantiomer in which H is on the left-hand side of the vertical line and OH on the right-hand side was designated by ‘D’ and its mirror image (H on the right-hand side and OH on the left-hand side) by ‘L’.
• A molecule containing a single chiral centre may be designated as D or L by imagining a resemblance between the ligands on its chiral centre and those in glyceraldehyde. The enantiomer having the ‘same’ groups in the same place as D-glyceraldehyde is designated as ‘D’ and the one having the ‘same’ groups in the same place as L-  glyceraldehyde is designated as ‘L’.
• Asymmetric molecules with the general formula R — CHX—R’ are always written by the Fischer projection with R — C—R; (the longest chain) as the vertical chain with C-l carbon (according to IUPAC) at the top. If X is on the right-hand side of the horizontal line, the designation used is D and when X is on the left hand side, it is designated as L (X is taken to be a negative group).

• Any stereoisomer which is obtained from or converted into D-glyceraldehyde belongs to D-series. Similarly, stereoisomer that is obtained from or converted into L-glyceraldehyde belongs to L-series.
• For instance, the following sequence of reactions, in which there is no rupture of a bond to the chiral carbon atom, establishes that the configuration of (-) -lactic acid is the same as that of D-(+) – glyceraldehyde.

Thus, D, L-representation of configuration is a relative method, that indicates the relative configuration of a molecule. It does not indicate the absolute configuration, i.e., the actual arrangements of groups in space around chiral carbon.

D, L-symbols have no relation with the sign of rotation of the optically active compound. A D-compound may be (+) or (-). Similarly, an L-compound may be (+) or (-): In the case of compounds containing more than one chiral carbon atom, the relative configuration of each of the chiral carbon is represented by D, and L-symbols separately. In the given isomer of 2,3-dihydroxybutanoic acid, the — OH group at C-2 is on the right-hand side and the —OH group at C- 3 is on the left-hand side of the horizontal line and so its name dihydroxybutanoic acid.

R, S-system of nomenclature: Absolute configuration: An important drawback of the D, L -system for specification of configuration around a chiral carbon is its suitability for molecules which are not related with a particular glyceraldehyde, To overcome this problem R. S. Cahn, Sir Christopher Ingold and V.

• Prelog evolved a new unambiguous system of assigning absolute configuration to chiral molecules. This system is named as CIP (Cahn, Ingold, Prelog) system or R, S – system. Such specifying configurations are independent of any reference compound. Hence, the system is also termed as ‘Absolute configuration’ assignment.
• R, S -symbols have no relation with the sign of rotation of the optically active compound. A f+) -compound may be R or S. Similarly, a (-) -compound may be R or S.

Conventions for assigning R, S-designation: For assigning R, S-designation to any chiral centre, the conventions to be followed are: Four different atoms or groups attached to a chiral centre are identified and to each of these substituents, a priority symbol 1, 2, 3, 4 or a, b, c, d based on sequence rules (discussed later) is assigned.

• The priority decreases as 1 > 2 > 3 > 4 or a>b>c>d. The chiral centre is then viewed from the opposite side of the lowest priority group (4 or d) and a hypothetical path is traced from the first priority group, through the second, to the third (1→2→3 or a→b→c).
• If the path describes a clockwise motion, then the stereocentre has an R configuration and if the path describes a counterclockwise motion, then the stereocentre has an S configuration. [R and S are from the Latin words Rectus and Sinister meaning right and left respectively.]
• The designation is written in italics within parentheses followed by a hyphen before the name of the compound. Each of the chiral centres containing more than one chiral centre is designated by R, and S -symbols in the same way. In the Fischer projection of a chiral

• carbon, if the lowest priority group occupies a position in the vertical line (upper or lower) and the hypothetical path on moving from a to b to c (i.e., a→b→c or 1→2→3) describes a clockwise motion, then the chiral centre is said to have ‘ R’ configuration. If this path traces a counterclockwise motion, then the chiral centre is said to have an ‘S’ configuration.
• However, if the lowest priority group occupies a position in the horizontal line (left or right) in the Fischer projection and the path a→b→c (or 1→2→3 ) describes a clockwise motion then the chiral centre is said to have ‘S’ configuration; if this path traces a counter-clockwise motion, the configuration of the chiral centre is R.

Sequence rules for the determination of the priority of ligands (i.e., atoms or groups attached to the chiral centre) according to CIP convention:

• If all four atoms or groups attached to the chiral centre are different, then each group is assigned a priority on the basis of the atomic number of the atom that is directly attached to the chiral centre.
• The group with the highest atomic number is given the highest priority, a; the group with next lower atomic number is given the next lower priority, b; and so on.

Example: In 1 -bromo-1 -chloroethane (CH₃ — C H ClB r), the chiral centre is bonded to CH₃, H, Cl and Br. The atomic numbers of atoms directly attached to the chiral centre show that the priority order should be Br > Cl > CH₃ > H.

• If one of the four groups is replaced by a lone pair of electrons (as In carbanions, amines and other 9uch compounds having pyramidal geometry) then the lone pair has the lowest priority.
• When a priority cannot be assigned on the basis of the atomic number of the atoms that are directly bonded to the chiral centre, then the next set of atoms in the unassigned groups is examined. This process of exploration is continued until a decision can be made. We assign a priority to the first point of difference.

Example: In 2-chlorobutane (CH₃CHClCH₂CH₃; C* = chiral carbon), four ligands attached to the chiral centre are — CH₃, —H, —Cl, — CH₂CH₃.

• Out of these Cl is the highest (a) and H is the lowest (d) priority group. Decisions regarding priority between — CH₃ and —CH₂CH₃ groups cannot be made based on the atom directly attached to (chiral carbon), because the carbon atom of each group.
• However, when we examine the methyl (— CH₃) group, we find that the next set of atoms consists of three hydrogen atoms (H, H, H). But in the ethyl group ( —CH₂CH₃), the next set of atoms consists of one carbon atom and two hydrogen atoms (C, H, H). Since carbon has a higher atomic number than hydrogen, therefore, we assign the ethyl group ( —CH₂CH₃) the higher priority i.e., —CH₂CH₃ > — CH₃.

• When isotopic atoms are attached directly to the chiral centre, then the atom with a higher mass number gets the higher priority. Thus, deuterium (D) gets priority over ordinary hydrogen (H).
• Similarly, — ¹⁴CH₃ > —¹²CH₃; —¹⁸OH>— ¹⁶OH; —¹⁵NH₂> — ¹⁴NH₂ etc.

Example: In 1-chloro-l-deuterioethane, the priority sequence of different groups attached to the chiral centre is: Cl > CH₃ > D > H.

On the basis of rule (D we have the following priority sequence involving several pairs of groups.

→ \(-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3>-\mathrm{CH}_2 \mathrm{CH}_2-\mathrm{H} ;\)

→ \(-\mathrm{CH}_2 \mathrm{OH}>-\mathrm{CH}_2 \mathrm{NH}_2\)

→ \(-\mathrm{CH}_2 \mathrm{CHFBr}>-\mathrm{CH}_2 \mathrm{CHFCl}\)

• (Decision is reached at the bold atoms)
• Rule (3) must not be used until rule (2) is completely exhausted. Thus, —CH₂CH₂CH₃ > —CD₂CH₃ because propyl > ethyl (rule 2); but — CH₂CD₂CH₃> —CH₂CH₂CH₃ (rule 3).
• When a ligand bifurcates, we must proceed along the branch providing the highest priority until a difference is encountered.
• The decision must be made at the earliest opportunity and once made cannot be changed from consideration of substituents further along the chain. Thus we have,

For the purpose of assigning priority to groups containing multiple bonds (e.g.,—CH =0,—CH=CH₂, —C = N etc.), both atoms involved in such bonding are treated as if they were duplicated or triplicated as the case may be. The duplicated or triplicated atoms are placed within parentheses and are made up to ligancy four with phantom atoms (0) having atomic number zero. Phantom atoms (0) have lower priorities than H. Thus an aldehyde group ( —CH=O) can be treated as if it were

Omitting the phantom atoms, it can be represented as,

Example:

• The first atom (i.e., the atom attached to the chiral centre) of —CO₂H, —CHO, —C ≡ CH and — CH=CH₂ are connected respectively to (O, O, O), (O, O, H), (C, C, C) and (C, C, H). Therefore, the priority order is: —CO₂H > —CHO > —C = CH > —CH=CH₂.
• When two groups attached to the chiral centre have the same structure but different configurations then the group with (R) -configuration gets priority over the group with (S) -configuration. Similarly, Z> E.

Example:

Priority sequence of groups:

—CH=CHCH₃ > — CH=CHCH₃ > — CH₃—H

A list of common substituents arranged in increasing order of priority according to the CIP sequence rules

Lone pair of electrons (:) (lowest) < H <D <—CH₃

⇒ \(<-\mathrm{C}_2 \mathrm{H}_5<-\mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2<-\mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}<-\mathrm{CH}_2 \mathrm{C}_6 \mathrm{H}_5\)

⇒ \(<-\mathrm{CH}\left(\mathrm{CH}_3\right)_2<-\mathrm{CH}=\mathrm{CH}_2<-\mathrm{C}\left(\mathrm{CH}_3\right)_3<-\mathrm{C} \equiv \mathrm{CH}\)

⇒ \(<-\mathrm{C}_6 \mathrm{H}_5<-\mathrm{CH}_2 \mathrm{OH}<-\mathrm{CHO}<-\mathrm{COR}<-\mathrm{CONH}_2\)

⇒ \(<-\mathrm{CO}_2 \mathrm{H}<-\mathrm{CO}_2 \mathrm{R}<-\mathrm{COX}<-\mathrm{CX}_3<-\mathrm{NH}_2<-\mathrm{NHCH}_3\)

⇒ \(<-\mathrm{N}\left(\mathrm{CH}_3\right)_2<-\mathrm{NO}<-\mathrm{NO}_2<-\mathrm{OH}<-\mathrm{OCH}_3\)

⇒ \(<-\mathrm{OC}_6 \mathrm{H}_5<-\mathrm{OCOR}<-\mathrm{F}<-\mathrm{SH}<-\mathrm{SR}<-\mathrm{SOR}\)

⇒ \(<-\mathrm{SO}_2 \mathrm{R}<-\mathrm{SO}_3 \mathrm{H}<-\mathrm{Cl}<-\mathrm{Br}<-\mathrm{I} \text { (highest). }\)

Example: Priority of groups attached to asymmetric C-atom of 1-bromo-l-chloroethane: —Br > —Cl > —CH₃ > H. In the given enantiomer of the compound, a→b→c traces counterclockwise path and thus, the configuration is S.

The priority of four groups attached to the asymmetric C- atom of alanine is: —NH₂ > —COOH > —CH₃ > —H. In the following enantiomer of alanine, a→b→c traces a clockwise path and thus, the configuration is R.

The sequence of priority of the four groups attached to the asymmetric carbon atom of 2-butanol is: — OH > —CH₂CH₃ > —CH₃ > —H. In the following enantiomer of 2-butanol, the lowest priority group (i.e., H) falls on the vertical line and a—>c traces a clockwise path. Therefore, the configuration of the stereocentre is R.

Priority of four groups attached to asymmetric C-atom of glyceraldehyde: —OH > —CHO > CH₂OH > H. In the following enantiomer of glyceraldehyde, the lowest priority group (i.e., H) falls on the horizontal line and →b→c traces a clockwise path. Therefore, the configuration of the stereocentre is S.

In an S_N2 reaction of a chiral substrate, if the priority of the leaving group in the substrate and that of the entering group (nucleophile) in the substitution product are the same, then an R-substrate is converted into an S-product and vice versa. However, if the priorities are different, the configurational designation may or may not remain the same.

Example:

Unit 10 Haloalkanes And HaloarenesChemical Reactions Of Haloalkanes

Due to the presence of a polar C^δ+—X^δ- bond, haloalkanes are highly reactive. They are used in the synthesis of a large variety of useful organic compounds. Reactions of haloalkanes may be divided into the following five categories: Nucleophilic substitution reactions, Elimination reactions,  Reactions with metals, Reductions and Rearrangements.

Nucleophilic substitution reactions:

Substitution by hydroxyl( —OH) group: formation of alcohol

When haloalkanes are boiled with aq. solution of alkali (NaOH or KOH) or with silver oxide suspended in water (Ag₂O/H₂O or ‘AgOH’), they undergo hydrolysis to form alcohols.

Substitution by alkoxy (—OR) group: formation of ethers:

When haloalkanes are heated with an alcoholic solution of sodium or potassium alkoxide, ethers are obtained.

Substitution by cyano (—CN) group: formation of alkyl cyanides (alkane nitriles):

When haloalkanes are heated with ethanolic solution of potassium cyanide, alkyl cyanides are obtained.

• Alkyl cyanides are important in the preparation of acid amides, carboxylic acids and primary amines. Acid amides (RCONH₂) are obtained when alkyl cyanides are partially hydrolysed with a cone.
• HCl or cone. H₂SO₄ or treated with alkaline hydrogen peroxide. Carboxylic acids (RCOOH) are obtained when they are completely hydrolysed with dilute mineral acids or alkalies.
• Primary amines are obtained when they are reduced with sodium/ alcohol (Mendius reduction) or catalytically with H₂/Ni or with lithium aluminium hydride (LiAlH₄).

The reaction of alkyl halides with potassium cyanide (KCN) offers a convenient method for increasing the length of the carbon chain by one carbon at a time, i.e., for converting a homologue into its next higher homologue.

Substitution by isocyanide (-NC) group: Formation of alkyl isocyanides

When haloalkanes are heated with aq. ethanolic solution of silver cyanide (AgCN) and alkyl isocyanides is obtained.

Haloalkanes (mainly 1° or 2° bromides or iodides) react with sodium or potassium nitrite to form nitroalkanes.

Substitution by nitro (-NO₂) group: Formation of nitroalkanes

Haloalkanes (mainly 1° are 2° bromides or iodides) react with sodium or potassium nitrite to form nitroalkanes.

Silver nitrate (AgNO₂) gives nitro compounds only when RX is a primary bromide or iodide.

Substitution by nitrite (—ONO) group: formation of alkyl nitrites

Haloalkanes (2° or 3° alkyl halides) react with aqueous- alcoholic solution of silver nitrite to form alkyl nitrites.

Substitution by carboxylate (—OCOR) group: formation of esters

When haloalkanes are heated with ethanolic solution of silver salt of a fatty acid, esters are obtained.

Substitution by amino (—NH₂) group: formation of the amines: When haloalkanes are heated with ethanolic solution of ammonia at 373K in a sealed tube, a mixture of 1°, 2° and 3° amines along with quarternary ammonium salts is obtained. The reaction is known as Hoffmann ammonolysis.

Formation Of The Amines Example:

Substitution by thiol (—SH) group: Formation of thioalcohol: When haloalkanes are heated with aq. ethanolic solution of sodium or potassium hydrosulphide, thioalcohols are formed.

Example:

Substitution by alkylthio (— SR)group: Formation of thioethers

When haloalkanes are heated with aqueous ethanolic solution of sodium or potassium mercaptide (NaSR or KSR) or sodium (Na₂S), thioethers are obtained.

Formation Of Thioethers Examples:

Substitution by alkynyl (—C≡CR) group: Formation Of Higher Alkynes

When treated with sodium alkynide, alkyl halides from higher alkynes.

Formation Of Higher Alkynes Example:

This substitution reaction is used for converting lower alkyne into a higher alkyne.

Substitution of chlorine or bromine by iodine:

Chloroalkanes or bromoalkanes react with Nal dissolved in methanol or acetone to form iodoalkanes. This reaction is known as Finkelstein reaction.

Substitution by hydrogen: Reduction

Haloalkanes (1° or 2°) react with LiAlH₄ (hydride ion donor) to form alkanes.

Reduction Example:

Elimination Reactions (dehydrohalogenation):

• When a haloalkane is heated with a concentrated alcoholic solution of potassium hydroxide, a hydrogen atom is eliminated from the β -carbon (i.e., the carbon adjacent to the halogenated carbon) and a halogen atom is eliminated from the u-carbon {i.e., the halogenated carbon.) Therefore, an alkene is formed due to the elimination of a molecule of hydrogen halide.
• Since these reactions involve the loss of β – hydrogen atoms, they are known as -elimination reactions. These are also called 1,2-elimination reactions as it involves the removal of adjacent atoms. The most common mechanism for dehydrohalogenation is the E2 mechanism.

Elimination Reactions Example:

Orientation Of Elimination:

If the halogen atom is present on the terminal carbon atom of the alkyl halide, dehydrohalogenation can take place only in one direction to yield only the terminal alkene. For example, when 1-chlorobutane is heated with alcoholic KOH, only but-l-ene is obtained.

Saytzeff Elimination: If, however, the halogen atom in an alkyl halide is present on any carbon atom within the chain {i.e., not on a terminal carbon), the alkyl halide undergoes dehydrohalogenation in two or more different directions depending on the number of /?-hydrogen atoms available.

• When this occurs, one of the products usually predominates, and that becomes the major product. The Russian chemist Alexander Saytzeff in 1875 formulated a rule which can predict the major product to be obtained. This rule is called the Saytzeff rule.
• ln dehydrohalogenation reactions of alkyl halides (R—X, X = Cl, Br, I), the more highly substituted alkene, i.e., the alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms is obtained as the major product. This mode of elimination is called Saytzeff elimination.

Saytzeff Elimination Example: When 2-bromobutane is heated with alcoholic KOH, but-2-ene is obtained as the major product and but-l-ene is obtained as the minor product.

• If the more substituted alkene (the Saytzeff product) is capable of exhibiting cis-trans isomerism, the more stable trans-isomer is always formed as the major product.
• For example, in the above reaction, 71% of /raws-but-2-ene and 10% of ds-but-2-ene are obtained. Due to steric interaction between the two methyl groups on the same side of the double bond, the as-isomer is relatively less stable than the transisomer and so it is obtained as the minor product.

Cause of predominance of the more substituted alkene in saytzeff elimination:

• The base-promoted dehydrohalogenation of an alkyl halide is an ideal E2 elimination in which both the C—H and C —L bonds are broken and the carbon-carbon double bond is formed at the same time. Hence, the transition state of the reaction possesses considerable double bond character.
• Highly substituted alkene is more stable due to the hyperconjugative effect. For the same reason, a highly substituted transition state having a double bond character is more stable. The product obtained through this stable transition state is formed readily.
• The rate of dehydrohalogenation of an alkyl halide producing a more stable alkene is higher than that producing a less stable one. The difference in the rate of reaction is also due to the increase in the number of β – hydrogen atoms. With an increase in the number of β-H- atoms, the chances of elimination increases.

Saytzeff Elimination Example: The rates of dehydrochlorination of chloroethane (producing an unsubstituted alkene), 2-chloropropane (producing a monosubstituted alkene) and 2-chIoro-2- methylpropane (producing a disubstituted alkene) follow the order: CH₃CH₂C1 < CH₃CHClCH₃ < (CH₃)₃CCl. However, the ease of dehydrohalogenation of alkyl halides having the same alkyl group but different halogen atoms increases in the order: R—Cl < R—Br < R—I and this is because the leaving group ability increases from Cl^– to I^–.

• There are two important mechanisms for β-elimination reactions: The E2 mechanism and the El mechanism. The reaction of alcoholic KOH with bromoethane proceeds through the E2 mechanism. An E₂ reaction is a single-step process.
• The base OH- receives a proton from the β -carbon, forming H₂O (a by-product), the electron pair of the β C—H bond forms the new n bond and the leaving group Br~ departs with the bonding electron pair of the C— Br bond.
• Two bonds are broken (C— H and C — Br) and two new bonds are formed (H— OH and the n bond) in a single step. The transition state, therefore, contains four partial bonds with the negative charge distributed over the base and the leaving group.
• The carbon-carbon single bond acquires a partial double bond character. Since the rate-determining step (in this case, it is the only step) involves the substrate as well as the base, the mechanism is designated as E₂ (elimination bimolecular).
• The rate of the reaction is proportional to the molar concentrations of both the substrate and the base, i.e., the reaction rate = k[CH₃CH₂Br][OH^–]. Expulsions of H and Br take place from opposite sides, i.e. a trans-elimination occurs.

β-elimination may also take place by the E1cB mechanism. It has been discussed along with the EI mechanism.

Hofmann Elimination:

Hofmann Rule: In an E₂ reaction, if the substrate contains positively charged groups like \(-\stackrel{\oplus}{\mathrm{N}} R_3 \text { or }-\stackrel{\oplus}{\mathrm{S}} \mathrm{R}_2\), i.e., if the leaving group is neutral like \(\ddot{\mathrm{N}} \mathrm{R}_3 \text { or } \ddot{\mathrm{S}} \mathrm{R}_2\) and if the substrate is an alkyl fluoride, i.e., if the leaving group is a very bad one like, the less substituted alkene is obtained as the major product. This type of elimination reaction is called Hofmann elimination.

• The alkyl fluorides containing very poor leaving group (F^–) undergo Hofmann elimination when treated with a base. Due to the presence of a very poor leaving group, the breaking of the C — H bond starts well before the breaking of the C — L bond.
• A strong electron-withdrawing (-1) leaving group causes the development of partial positive charges on all the neighbouring carbon atoms. This renders the β -hydrogens acidic so that they are easily abstracted by a base.
• Thus, the transition state possesses little alkene character but considerable carbanion character. Any factor that stabilises a carbanion also stabilises the transition state. Since the order of carbanion stability is primary > secondary > tertiary, the abstraction of β -hydrogen by a base takes place preferentially from the methyl group (primary carbanion). As a result, the reaction produces a predominance of the less substituted alkene.

Hofmann Rule Example: In 2 -fluorobutane, due to the +1 effect of the CH₃ group, the β-H atom on C-3 is less acidic than the β-H atom on C-l. Hence, the β-H on C-l undergoes abstraction by the base to form a less substituted alkene, but-l-ene as the major product through the more stable carbanion-like transition state.

Eliminatiorvversus substitution reaction:

When an alkyl halide having β-H atoms is reacted with a base or a nucleophile, it may undergo two competing reactions: substitution (S_N1 or S_N2) and elimination (El and E2 ). The following factors determine the preferred mode of reaction: The nature of the alkyl halide, the strength and size of the base or nucleophile, and the conditions of the reaction (nature of the solvent and temperature).

1. Tertiary or 3° alkyl halides: 3° alkyl halides (R₃CX) undergo reaction by other mechanisms except S_N2.

With Strong Bases: They undergo elimination reaction by the E2 mechanism. S_N2 reaction does not take place due to steric hindrance.

With Strong Base Example:

With weak nucleophiles or bases: A mixture of S_N1 and El products is obtained.

Example:

2. Primary Or 1° Alkyl Halides: Primary alkyl halides (RCH₂X) take part in S_N2 and E2 reactions. With strong nucleophiles or bases: In a more polar solvent and at a relatively low temperature, a substitution reaction by the S_N2 mechanism takes place.

Primary Or 1° Alkyl Halides Example:

In a less polar solvent (e.g., alcohol) and at a much higher temperature, an E2 reaction occurs in the presence of a strong base.

Example:

With A Strong And Bulkier Base: An elimination reaction by the E2 mechanism occurs because the bulkier base cannot act as a nucleophile.

With A Strong And Bulkier Base Example:

3. Secondary Or 2° Alkyl Halides: Secondary alkyl halides (R₂CHX) react by all the mechanisms.

With Strong Bases And Nucleophiles: Forms a mixture of S_N2 and E2 reaction products.

Secondary Or 2° Alkyl Halides Example:

With Strong And Sterically Hindered Bases: An elimination reaction by the E2 mechanism occurs because the bulkier base cannot act as a nucleophile.

With Strong And Sterically Hindered Bases Example:

With Weak Nucleophiles Or Bases: Forms a mixture of S_N1 and El products.

With Weak Nucleophiles Or Bases Example:

Reaction with metals: Most of the organic chlorides, bromides and iodides react with certain metals like Mg, Li, Pb etc., to form compounds containing carbon-metal bonds. Such compounds are called organometallic compounds.

Reaction with magnesium:

1. An important class of organometallic compounds obtained by the reaction of alkyl halides with metallic magnesium in dry ether is alkyl magnesium halides. These organomagnesium compounds were discovered by Victor Grignard in 1900 and are called Grignard reagents.
2. Grignard reagents are represented by the general formula RMgX, where R= alkyl or aryl group (e.g., methyl, ethyl, phenyl, etc.) and X = Cl, Br, I.
3. The C^δ-—Mg^δ+ bond is a highly polar covalent bond because carbon is more electronegative than magnesium and attracts the bonding electrons towards itself. On the other hand, the bond between magnesium and halogen is ionic in nature.
4. In the formation of Grignard reagents, the reactivity of the alkyl halide containing the same alkyl group follows the order: R—I > R—Br > R—Cl.

Preparation: In the laboratory, Grignard prepared by reacting clean and dry metallic magnesium with dry alkyl or aryl halides in anhydrous alcohol-free ether. The resulting Grignard reagent remains dissolved in ether. It is this ethereal solution of Grignard reagent which is used in various chemical reactions.

Example:

Due to much higher polarity of the C — Mg bond, the Grignard reagent serves as a good source of carbanions <R^–) and so they are very reactive. They react with any compound containing active or acidic hydrogen (e.g., H₂O, ROH, H₂S, RSH, NH₃, 1° amine, 2° amine, terminal alkyne, etc.) to form hydrocarbons. Therefore, the alkyl halides are easily converted into alkanes through the formation of Grignard reagents.

Due to such reactivity, the apparatus and the reagents used in the preparation of Grignard reagents must be completely dry and free from alcohol. Also, the reaction atmosphere must be free of O₂ and CO₂ because RMgX reacts with them and gets destroyed.

Grignard reagents are very useful for the preparation of different classes of important organic compounds like alcohols, aldehydes, ketones, carboxylic acids, etc.

Synthetic uses of Grignard reagents:

Reaction with lithium:

Alkyl or aryl halides react with Li in a hydrocarbon solvent in an inert atmosphere to yield alkyl-lithiums or aryl-lithiums.

Example:

Alkyl-lithiums are covalent colourless liquids or readily fusible solids. Carbon-lithium bond is a polar covalent bond (^δ-C — Li^δ+) as lithium is less electronegative than carbon. The organolithium compounds behave both as nucleophiles and bases. Organolithium compounds are usually more reactive and less susceptible to steric hindrance than Grignard reagents.

Reaction with lead: The most important organometallic compound of lead is tetraethyllead or TEL [Pb(C₂H₅)₄]. It has been used to improve antiknock rating of gasoline. However, due to air pollution, this is not used nowadays. TEL may be prepared by the reaction of lead-sodium alloy with ethyl chloride.

Reaction With Sodium: Alkyl halides react with sodium in dry ether to form alkanes containing double the number of carbon atoms present in the alkyl halide. The reaction is known as the Wurtz reaction. It is an excellent method for the preparation of symmetrical alkanes.

Reaction With Sodium Example:

Corey-House Synthesis: It is the synthesis of alkane from alkyl halides (R’X) and Gilman reagent, R₂CuLi in the presence of dry ether. Symmetrical as well as unsymmetrical alkanes are synthesised by this reaction.

Corey-House Synthesis Example:

 Reduction Of Haloalkanes: Haloalkanes can be reduced to alkanes by a variety of reagents. This reduction is called hydrogenolysis.

With H₂ in the presence of Ni, Pd or Pt as catalyst:

 Reduction Of Haloalkanes Example:

With HI In The Presence Of Red Phosphorus:

 Reduction Of Haloalkanes Example:

With Zn or Sn and HC1 or Zn-Cu couple and alcohol:

Example:

Rearrangement of haloalkanes:

When an alkyl halide is heated at 573K or at a lower temperature in the presence of anhydrous aluminium chloride as a catalyst, it undergoes rearrangement to form an isomeric alkyl halide.

Reaction Mechanism:

Preparations Of Haloarenes: Although haloalkanes are easily prepared by the replacement of the —OH group by a halogen atom, haloarenes or aryl halides cannot be prepared from phenols. Because replacement of the phenolic —OH group by halogen atom is not possible. In fact, the carbon-oxygen bond in phenol has considerable double bond character due to the delocalisation of unshared electron pairs on oxygen with the ring π-electrons. Hence, cleavage of carbon-oxygen bonds in phenol is difficult under ordinary conditions.

From Aromatic Hydrocarbons:

By Halogenation Of The Aromatic Ring:

Chlorination And Bromination: Chloroarenes and bromoarenes are prepared by the reaction of aromatic hydrocarbons with chlorine or bromine in the presence of Lewis acids like anhydrous ferric or aluminium halide at a low temperature (310-320K) in the absence of sunlight.

Chlorination And Bromination Example:

Important information regarding chlorination and bromination: When an excess of Br₂ or Cl₂ is used, a second halogen atom is introduced in the ring mainly at o- and p-positions since halogens are o, p-directing.

Toluene undergoes chlorination in the presence of anhydrous FeCl₃ to form a mixture of o^– and p^–chlorotoluenes, since the —CH₃ group is o, p-directing.

• Similarly, bromination of toluene with Br2 in the presence of anhydrous FeBr3 produces a mixture of obromotoluene (minor) and p-bromotoluene (major). Halogenation of arenes is an electrophilic aromatic substitution reaction.
• The Lewis acid acts as a halogen carrier in this reaction. Fe is used instead of FeX₃ because, under the reaction conditions, Fe reacts with to form FeX₃. Example: Benzene reacts with Cl₂ or Br₂ in the presence of Fe to form chlorobenzene or bromobenzene.

Iodination: Iodoarenes are not prepared by direct iodination of arenes because iodine is least reactive among halogens, the reaction is reversible and the HI formed reduces the aryl iodide back to arene while itself is oxidised to iodine.

However, when the reaction is carried out in the presence of oxidising agents like nitric acid, iodic acid (HIO₃), mercuric iodide, etc., the HI produced is either oxidised to iodine or is eliminated as mercuric iodide. As a result, the reaction proceeds in the forward direction to produce iodobenzene.

⇒ \(2 \mathrm{HI}+2 \mathrm{HNO}_3 \rightarrow 2 \mathrm{NO}_2+\mathrm{I}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(5 \mathrm{HI}+\mathrm{HIO}_3 \rightarrow 3 \mathrm{H}_2 \mathrm{O}+3 \mathrm{I}_2 ; 2 \mathrm{Hg}+\mathrm{HgO} \rightarrow \mathrm{HgI}_2+\mathrm{H}_2 \mathrm{O}\)

• The reaction also takes place by the substitution reaction involving the electrophile I (iodonium ion) obtained by the oxidation of I₂ by the oxidising agent.
• Fluoroarenes or aryl fluorides are not prepared by direct fluorination of arenes since the reaction is violent and cannot be controlled easily.

Raschig process: Chlorobenzene is manufactured by this process. It involves passing a mixture of benzene vapour, air and hydrogen chloride over a hot copper chloride catalyst.

By-side chain halogenation (benzylic halogenation):

Direct halogenation: Side chain substituted aryl halides or aralkyl halides are prepared by direct halogenation of a suitable arena under appropriate conditions.

Example: When Cl₂ is passed through boiling toluene in the presence of sunlight, chlorophenylmethane (benzyl chloride) is obtained.

When an excess of Cl₂ gas is passed, the benzyl chloride formed reacts further to give benzal chloride at first and then benzotrichloride.

Benzylic Halogenation: When the side chain of the arena is larger than a methyl group, halogenation occurs preferentially at the benzylic carbon (i.e., the carbon directly attached to the ring). This process is called benzylic halogenation. It occurs because the benzylic radical (C₆H₅CH —CH₃) expected to be formed during this reaction is stabilised by resonance.

Halogenation By Excess Cl₂: When ethylbenzene, is allowed to react with excess Cl₂ in the presence of sunlight, 1,1-dichloro-l-phenylethane through the formation of 1-chloro-l-phenylethane is obtained.

Halogenation By Excess Cl₂ Example:

Halogenation by sulphuryl chloride: Benzylic chlorination can also be carried out by treating an arene with sulphuryl chloride (SO₂Cl₂) at 475K in the presence of light and traces of peroxide.

Halogenation by NBS: Benzylic bromination can be carried out by treating an arene with NBS (N-bromosuccinimide) in the presence of light and traces of peroxide.

From Diazonium Salts:

Preparation Of Chloroarenes And Bromoarenes:

Sandmeyer Reaction: Chloroarenes (aryl chlorides) and bromoarenes (aryl bromides) are prepared by treating freshly prepared diazonium salt with cuprous chloride (CuCl) dissolved in HCl or cuprous bromide (CuBr)dissolved in HBr, respectively.

Sandmeyer Reaction Example:

Benzene diazonium chloride can be easily prepared by diazotisation of aniline (C₆H₅NH₂) with nitrous acid (NaNO₂/HCl) at low temperatures (0-5°C).

Gattermann Reaction: If copper powder in the presence of halogen acid (HCl or HBr) is used instead of cuprous halide dissolved in the corresponding halogen acid (CuCl/ HCl or CuBr/HBr), the yield of chlorobenzene or bromobenzene increases. Such a modified form of the Sandmeyer reaction is known as the Gattermann reaction.

Gattermann Reaction Example:

Preparation Of Iodoarenes (aryl iodides):

l Iodoarenes (aryl iodides) can be prepared by simply warming the diazonium salt solution with an aqueous solution of KI. It is, however, the best method for the preparation of iodoarenes.

Preparation Of Iodoarenes Example:

Preparation Of Fluoro Arenes (aryl fluorides): When fluoroboric acid (HBF₄) is added to a solution of diazonium salt, insoluble diazonium fluoroborate is precipitated. This salt is separated and dried. When the dry salt is heated, fluoroarene or aryl fluoride is obtained. This reaction is called the Balz-Schiemann reaction.

Preparation Of Fluoro Arenes Example:

From The Silver Salt Of Aromatic Acids: Aryl bromides (like alkyl bromides) are prepared by refluxing the silver salt of aromatic acids with bromine dissolved in carbon tetrachloride. This reaction is called the Hunsdiecker reaction.

From The Silver Salt Of Aromatic Acids Example:

Physical Properties Of Haloarenes

1. Physical state and colour: Haloarenes are generally colourless liquids or crystalline solids.

2. Density: Haloarenes are heavier than water. Their densities increase from fluorobenzene to iodobenzene.

3. Solubility: These are insoluble in water because they do not form hydrogen bonds with water. However, they are quite soluble in organic solvents like benzene, ether, chloroform, carbon tetrachloride, etc.

4. Boiling and melting points: Boiling points of halo¬ arenes are low due to the absence of hydrogen bonding. The boiling points of halobenzene increase with the increase in the size of the halogen atom from fluorine (F) to iodine (I).

• The boiling points of isomeric dihalobenzenes are nearly the same but the melting points of the p-isomer is higher than that of the o -and m -isomer. It is due to the fact that the molecules of the p -isomer are more symmetrical and so they are closely packed in the crystal lattice.
• As a consequence, the intermolecular forces of attraction increases leading to higher melting points as compared to the other isomers. For the same reason, the solubility of the para-isomer is less than that of ortho- and meta-isomer in a particular solvent.

Dipole moments: Fluorobenzene (μ = 1.63D) , chlorobenzene (p = 1.75D) and bromobenzene (μ = 1.72D) have dipole moments lower than corresponding halomethanes. But, iodobenzene has higher dipole moment (μ = 1.71D) than that of CH₃I (μ = 1.636D).

• In a halobenzene molecule, the halogen atom withdraws electrons from the ring by its -I effect and donates electrons to the ring by the +R effect. Thus, there operates two moments acting in opposite directions, one from the halogen atom towards the ring (μ_R) and the other from the ring towards the halogen atom (μ_l). Since μ_l > μ_R, halobenzenes possess a net dipole moment.
• With the increase in the size of the halogen atom, strengths of both μ_R and μ_l decrease but the resonance effect decreases more rapidly than the inductive effect. As a result, the net moment increases from fluorobenzene to chlorobenzene but remains almost the same in chlorobenzene, bromobenzene and iodobenzene.
• The dipole moments of isomeric dihalobenzenes decrease as the angle between the two halogen atoms increases, i.e., their dipole moments decrease in the order: o-dihalobenzene > m – dihalobenzene > p-dihalobenzene.

Chemical Properties Of Haloarenes

Aryl and vinyl halides possess extremely low reactivity as compared to alkyl halides, towards nucleophilic substitution reactions. In fact, aryl and vinyl halides remain inert under the same conditions in which alkyl halides easily undergo nucleophilic substitution.

Example: Attempts to convert aryl or vinyl halides into phenol (or alcohols), ethers, amines, or nitriles by treatment with the usual nucleophilic reagents are also unsuccessful. However, aryl and vinyl halides cannot be used in place of alkyl halides in the Friedel-Crafts reaction.

Causes of extremely low reactivity of aryl and vinyl halides:

Resonance effect: In an aryl halide molecule (e.g., chlorobenzene), an unshared pair of electrons on the halogen atom is involved in resonance interaction with the ring; π-electrons. A similar resonance interaction occurs in vinyl halide (e.g., vinyl chloride). As a result, the C—X bond acquires some double bond character.

• On the other hand, in an alkyl halide molecule (e.g., CH₃—Cl), the Cl atom is attached to the carbon atom by a pure single bond. Consequently, the C—X bond in aryl or vinyl halides is much stronger than in alkyl halides and hence cannot be cleaved easily. So, they are inert towards S_N1 or S_N2 reactions.
• Aryl or vinyl halides are stabilised by resonance but alkyl halides are not. Due to this, the activation energy (E_act) for the displacement of the halogen atom from aryl or vinyl halides is much greater than that from alkyl halides. Thus, aryl or vinyl halides are reasonably less reactive than alkyl halides towards nucleophilic substitution reaction.

The hybridisation state of the C -atom in the C—X bond: In alkyl halides, the C atom in the C—X bond is sp³ – hybridised while in aryl or vinyl halides, it is sp² – hybridised. Since an sp² -hybrid orbital having greater character is smaller than an sp³ -hybrid orbital having lesser s-character, the C—X bond in aryl or vinyl halides (obtained by sp²-p overlap) is shorter and hence stronger than the C —X bond in alkyl halides (obtained by sp³-p overlap). Thus, aryl or vinyl halides do not take part in S_N1 or S_N2 reactions involving cleavage of the C —X bond.

Presence of electron-rich π -cloud: Since the benzene ring in aryl halides and a double bond in vinyl halides are electron-rich systems, therefore, they repel the electron-rich nucleophile and as a result, S_N2 reactions involving backside attack do not take place.

Instability of aryl or vinyl cations: Aryl or vinyl cations expected to be formed as a result of self-ionisation is not at all stabilised by resonance because the sp² -hybrid orbital of +ve carbon is perpendicular to the p-orbitals of the phenyl ring or the p-orbital of the sp-hybridised positive carbon is perpendicular to the p-orbitals of the vinyl group. Thus, these cations are not formed and hence, aryl and vinyl halides do not undergo S_N1 reactions.

Important Reactions Of Haloarenes:

Nucleophilic Substitution Reactions:

Substitution By Hydroxyl Group: Due to very low reactivity, aryl halides do not undergo nucleophilic substitution reaction to yield phenol even when refluxed with NaOH solution for a few days. However, the reaction occurs under drastic conditions.

Substitution By Hydroxyl Group Example: When chlorobenzene is heated with NaOH solution (6-8%) at 350°C under 300 atm. pressure, sodium phenoxide is obtained which on acidification produces phenol (Dow process). This reaction is the basis of the Dow process for the manufacture of phenol. The nucleophilic substitution occurs through the formation of a very reactive neutral intermediate known as benzyne.

However, if there is an electron-attracting group like —N02 at the positions ortho or para to the Cl atom, the nucleophilic substitution reaction occurs easily. Example: When p-chloronitrobenzene is heated with 15% NaOH solution at 170°C (much less than the temperature required by chlorobenzene), p-nitrophenol is obtained.

The electron-attracting — NO₂ group reduces the electron density of the benzene ring by its -R effect and thereby activates the halogen towards nucleophilic displacement. It also helps to stabilise the intermediate carbanion by effective resonance interaction.

Reaction Mechanism: The reaction proceeds through the following steps:

First Step: Nucleophilic attack by OH^– on the ring carbon-bearing chlorine to form a resonance-stabilised carbanion intermediate.

Second Step: Loss of Cle from the intermediate carbanion to yield the substitution product.

The mechanism discussed above is also known as the S_NAr mechanism. Similarly, the alkaline hydrolysis of nitrophenol occurs and the mechanism of the reaction may be shown as follows:

• The intermediate carbanion obtained by a similar nucleophilic attack by OH^– ion on m-chloronitrobenzene is not stabilised by the -R effect of the — NO₂ group because this group is not in proper conjugation with the negative charge. The carbanion is stabilised by resonance involving only the double bonds of the ring.
• Therefore, the intermediate carbanion obtained from chloronitrobenzene is less stable than the carbanion obtained from o-and p-chloronitrobenzenes. Thus, an — NO₂ group meta to the Cl atom cannot activate the ring towards nucleophilic attack and hence mchloronitrobenzene reacts at an extremely slower rate than the other two isomers.

Although the — NO₂ group cannot stabilise the intermediate carbanion obtained from m-chloronitrobenzene by resonance, it can stabilise the carbanion by the -1 effect and in fact, this carbanion is relatively more stable than that expected to be formed from chlorobenzene. For this reason, mchloronitrobenzene is more reactive than chlorobenzene towards nucleophilic substitution reactions.

Aromatic Nucleophilic Substitution (SNAr) Reactions: The mechanism of aromatic nucleophilic substitution is known as the S_NAr (Substitution Nucleophilic Aromatic) mechanism. With the increase in the number of electron-attracting (-R) groups at ortho- and para-positions with respect to the halogen atom, the reactivity of haloarene progressively increases. As a result, the reaction can be carried out under less drastic or rather normal conditions.

Aromatic Nucleophilic Substitution (SNAr) Reactions Example: When two — NO₂ groups are present, the reaction can be carried out in a basic medium at 95°C. But when three — NO₂ groups are present, it can be carried out in the presence of water (neutral nucleophile) at only 50°C.

Substitution by the amino (-NH₂) group: When chlorobenzene is heated at 202°C with aqueous ammonia under 60 atm pressure in the presence of a cuprous oxide catalyst), aniline is obtained.

Substitution by the cyano (—CN) group: When bromobenzene is heated with cuprous cyanide in the presence of pyridine or DMF, cyanobenzene is obtained.

Electrophilic Substitution Reactions: Aryl halides undergo usual electrophilic substitution reactions like nitration, halogenation, sulphonation, FriedelCrafts reactions, etc. of the benzene ring.

Characteristics: The behaviour of halogen atoms is quite abnormal towards electrophilic aromatic Substitution. They are ortho-, and para-directing, even though they are deactivating in nature. Halogen atoms withdraw electrons from the ring by the -I effect but donate electrons to the ring by the +R effect.

• As the ^–I effect is stronger than the +R COCH₃ chloroacetophenone (major) effect, halogens by their net electron withdrawal, deactivate the ring, i.e., the electrophilic substitution reactions of halobenzenes occur at a rate slower than benzene.
• In fact, a halogen atom by its ^–I effect withdraws electrons from all positions of the ring i.e., it deactivates all the positions and by the +R effect it donates electrons to only o-and p-positions, i.e., it makes the deactivation less for the ortho-and para-positions than for the metaposition.
• As a result, o-and p-substitution products are predominant. Due to steric crowding at the orthoposition, the para-isomer is formed as the major product.

Halogenation:

Nitration:

Sulphonation:

Friedel-Crafts reaction: Alkylation, Acylation.

Reaction With Metals:

Reaction with sodium: The reaction of sodium with aryl halides may take place in two ways such as:

Wurtz-Fittig reaction: When an aryl halide is treated with an ethereal solution of an alkyl halide in the presence of sodium, an alkyl benzene is obtained. The reaction is known as the Wurtz-Fittig reaction.

An important application of the Wurtz-Fittig reaction is that a linear side chain can be easily introduced Into the benzene ring by this reaction (this can’t be done by Friedel-Crafts reaction).

Fitting reaction: When only aryl halides are treated with sodium in dry ether, diaryls are produced. This reaction is known as the Fittig reaction.

Reaction with copper (Ullmann reaction): When an aryl iodide is heated with Cu-powder in a sealed tube, a diaryl is obtained. The reaction is called the Ullmann reaction.

Reaction With Magnesium: A Grignard reagent is obtained when an ethereal solution of aryl bromide or iodide is treated with metallic magnesium. Aryl chlorides form Grignard reagents only when the reaction is carried out in dry tetrahydrofuran (THF) as a solvent.

Reaction With Magnesium Example:

Reaction With Lithium: Aryl lithiums are obtained when aryl bromides or iodides are treated with Li in dry ether.

Reaction With Lithium Example:

In principle, organic halides contain different types of halogens:

• Ionic halogen: Halogens present in benzene diazonium halides and quaternary ammonium halides (C₆H₅ and N₂X and R₄NX, where X = Cl, Br or I).
• Labile halogen: Halogens present in alkyl, allyl and benzyl halides (R—X, CH₂=CH—CH₂X and C₆H₅CH2X, where X = Cl, Br or I).
• Inert halogen: Halogens present in aryl and vinyl halides (C₆H₅X and CH₂=CH—X, X = Cl, Br or I). These three types of halogens can be distinguished AgNO₃ test.
• AgNO₃ test in warm conditions: Organic halide is boiled with an aqueous solution of KOH. The reaction mixture is then cooled and acidified with dil. HNO₃.
• AgNO₃ solution is then added to the solution. The appearance of a white or yellow precipitate indicates the presence of labile hydrogen in the compound.
• AgNO₃ test in cold condition: AgNO₃ solution is added to a cold aqueous solution of organic halide acidified with dil. HNO₃ . The formation of a white or yellow precipitate indicates the presence of ionic halogen in the compound.
• A negative AgNO₃ test suggests the presence of inert halogen in the compound.
• Nuclear substituted and side chain substituted aryl halides can be distinguished by oxidation followed by Lassaigne’s test If the oxidation product of the aryl halide contains halogen (detected by Lassaigne’s test), it must be a nuclear-substituted aryl halide. However, if the oxidation product contains no halogen, it is a side chain substituted aryl halide.

Class 12 Chemistry Unit 10 Haloalkanes And Haloarenes Polyhalogen Compounds

Carbon compounds containing more than one halogen atom, are usually called polyhalogen compounds. Many of these find applications in industry and agriculture. A few polyhalogen compounds and their physiological effects are discussed below.

Dichloromethane (methylene chloride, CH₂CI₂):

Uses of dichloromethane: It is extensively used as a solvent for paint remover and for cleansing metal surfaces. It is also used as a process solvent in the manufacture of drugs. It is used as a propellent in aerosols owing to its high volatility (low b.p.).

Physiological effects:

• Methylene chloride affects the human central nervous system (CNS). Exposure to lower levels of methylene chloride in air can impair hearing and vision to some extent. At higher levels, it can cause dizziness, nausea, tingling and numbness in the fingers and toes.
• In humans, direct contact of methylene chloride with the skin causes intense burning and mild redness of the skin because it dissolves some fatty tissues. When inhaled, it is metabolised by the body to carbon monoxide which may lead to carbon monoxide poisoning.

Trichloromethane (chloroform, CHCI₃):

Properties Of Trichloromethane: In the presence of sunlight, chloroform undergoes slow oxidation by air to form phosgene (carbonyl chloride), which is an extremely poisonous gas.

⇒ \(2 \mathrm{CHCl}_3+\mathrm{O}_2 \stackrel{h v}{\longrightarrow} 2 \mathrm{COCl}_2 \text { (Phosgene) }+2 \mathrm{HCl}\)

Therefore, chloroform is always stored in dark-coloured air-tight bottles in order to cut off light and avoid any contact with air. Oxidation of chloroform is further retarded by the addition of a small amount (0.6-1%) of pure ethanol which acts as an antioxidant. Even if phosgene is formed in trace amounts, it reacts with ethanol to give non-poisonous diethyl carbonate.

When chloroform is heated with Ag-powder, acetylene is produced:

Carbylamine reaction: In the presence of alcoholic potassium hydroxide, chloroform reacts with primary amine (alkyl or aryl) to form alkyl or aryl isocyanide with a foul smell. This reaction is also known as Hoffman’s isocyanide test.

⇒ \(\mathrm{RNH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \longrightarrow \mathrm{RNC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}\)

Formation of chloropicrin: Chloroform is treated with nitric acid to form chloropicrin which is used as a broad-spectrum antibiotic, and fungicide. It is commonly known as tear gas.

Formation of chlorine: In the presence of solid KOH or NaOH, chloroform reacts with acetone vigorously to form chlorine. It is highly toxic in nature.

Uses of trichloromethane:

In industry, chloroform is used as an important solvent for oils, fats, waxes, alkaloids, iodine, rubbers, resins, etc. It is mainly used for the manufacture of the freon refrigerant R-22.

Physiological effects:

• Chloroform was once used as a general anaesthetic in surgery. However, because of its high toxicity, it has now been replaced by less toxic and relatively safer anaesthetic agents such as ether. When the vapours of chloroform are inhaled, the central nervous system is depressed.
• It has been observed that breathing about 900 parts of chloroform per million (900 ppm) for a very short time causes dizziness, fatigue and headache. Chronic chloroform exposure may cause damage to the liver (where it is metabolised to phosgene), heart and kidneys. Some people develop sores when their skin comes in contact with chloroform.

Triiodomethane (iodoform, CHI₃):

Properties of triiodomethane:

Iodoform is a yellow crystalline solid with a characteristic unpleasant odour. It is insoluble in water but soluble in organic solvents like alcohol, ether, etc.

When iodoform is heated with silver powder, acetylene is produced:

Uses of triiodomethane: Iodoform was used earlier as an antiseptic for dressing wounds. In fact, when it comes in contact with the skin, it decomposes and slowly liberates iodine which accounts for its antiseptic properties. However, because of its unpleasant smell, it has now been replaced by other formulations containing iodine.

Tetrachloromethane (carbon tetrachloride, CCl₄):

Preparation of tetrachloromethane: Carbon tetrachloride is manufactured by chlorination of methane in the presence of light.

Uses of tetrachloromethane: It is largely used in the manufacture of refrigerants and propellants for aerosol cans and some pharmaceuticals. It is used as a raw material in the synthesis of chlorofluorocarbons (freons) and other chemicals.

• It is widely used as an industrial solvent for fats, oils, waxes, resins and lacquers.
• It was widely used (until the mid-1960s) as a fire extinguisher (its vapours are non-inflammable) under the name pyrene, as a dry-cleaning fluid, as a degreasing agent and as a stain remover in homes.

Physiological Effects: Exposure to carbon tetrachloride vapours causes liver cancer in humans and permanent damage to nerve cells.

• The most common effects are dizziness, lightheadedness, nausea and vomiting. In severe cases, these effects can lead to stupor, unconsciousness or even death.
• Exposure to CCl₄ vapours can make the heart beat irregular or may even stop it When its vapours come in contact with eyes, it causes irritation.

Environmental Effects: When tetrachloromethane is released into the air, it rises to the upper atmosphere and depletes the ozone layer. Depletion of the ozone layer leads to increased exposure to ultraviolet rays which may cause health hazards in humans such as skin cancer, eye diseases and disruption of the immune system.

• Chlorofluorocarbons (CFCs) Or Freons: The chlorofluoro derivatives of methane and ethane in which all the H-atoms are replaced by halogen atoms are collectively called freons.
• Chlorofluorocarbons (CFCs) Or Freons Example: Trichlorofluoroethane, CFCl₃ (called Freon-11), dichlorodifluoromethane, CF₂Cl₂(called Freon-12), etc.
• Properties Of Freons: Freons are extremely stable, low boiling inflammable, non-toxic, non-corrosive and highly unreactive compounds. They can be easily liquefied by applying pressure at room temperature.
• Uses Of Freons: Out of various freons, Freon-12 is the most common refrigerant that is used in refrigerators and air conditioners. It is prepared from carbon tetrachloride by Swarts reaction.

Freon-12 is also used as a propellant in aerosols and foams to spray out deodorants, cleansers, shaving creams, hair sprays and insecticides.

Environmental effects: CFCs slowly rise to the stratosphere where sunlight catalyses their decomposition. This decomposition process contributes to the destruction of the ozone layer which shields the earth’s surface from harmful ultraviolet radiation.

P, p’-dichlorodiphenyltrichloroethane (DDT):

DDT, the first chlorine-containing organic insecticide, was originally prepared in the year 1873. Its insecticidal properties were discovered in 1939 by the Swiss scientist Paul Muller. He was awarded the Nobel Prize in Physiology and medicine for this discovery. The IUPAC name of DDT is 11,1-trichloro-2,2- bis (p-chlorophenyl) ethane.

Preparation of DDT: DDT is prepared by heating chlorobenzene (2 mol) with chloral or trichloroacetaldehyde (1 mol) in the presence of a cone. H₂SO₄ . It is, in fact, an electrophilic aromatic substitution reaction.

• Uses of DOT: DDT is a cheap but powerful insecticide which is widely used to kill mosquitoes and other insects. It is an effective insecticide for sugar cane and fodder crops. It is particularly effective against Anopheles sp. mosquitoes which spread malaria. In fact, DDT has saved millions of lives worldwide by eliminating malaria.
• Toxic effects of DDT: DDT is harmful to man and animals. It is a highly stable, fat-soluble compound. It enters the human body mainly through food. Since it is not completely biodegradable, i.e., it is not rapidly metabolised in the body, it gets deposited and stored in fatty tissues. This accumulation of DDT affects the reproductive system, nervous system, kidneys and liver. So, the use of DDT is banned in many countries.
• Environmental effects: As DDT is a very stable compound, it remains intact in the soil for years. DDT used in agriculture goes to water bodies through rainwater and causes water pollution as it is highly toxic to aquatic organisms. In water, it gets absorbed by the fish and enters the food chain when contaminated fish is consumed. Similarly, in the soil, it gets absorbed by the plants and enters the food chain for animal and human consumption.

Unit 10 Haloalkanes And Haloarenes Very Short Question And Answer

Question 1. Give examples of 2° benzylic halide and vinylic halide.
Answer: 2° benzylic halide: C₆H₅CHClCH₃ (1-chloro-1-phenylethane); vinylic halide: CH₂=CHCl (Chloroethene).

Question 2. Which one among the methyl halides (CH₃X, X = F, Cl, Br, I) possesses the highest dipole moment?
Answer: Methyl chloride (CH₃Cl).

Question 3. Which one of the following reagents is the most acceptable for preparing chloroalkanes from alcohols? SOCl₂, HCl, PCl₅, PCl₃
Answer: Thionyl chloride (SOCl₂).

Question 4. Name a suitable reagent for allylic bromination.
Answer: N-bromosuccinimide or NBS

Question 5. Arrange the following compounds in order of decreasing boiling point:

1. CH₃CH₂CHBrCH₃
2. CH₃CH₂CH₂CH₂Br
3. (CH₃)₃CBr

Answer:  Order of decreasing boiling point: 2 > 1 > 3.

Question 6. Arrange o-, m- and p-dichlorobenzene in order of increasing boiling point.
Answer: M-ClC₆H₄Cl < p-ClC₆H₄Cl < o-ClC₆H₄Cl

Question 7. Arrange bromomethane, bromoform, chloromethane and dibromomethane in order of increasing boiling point.
Answer: Chloromethane < bromomethane < dibromomethane < bromoform.

Question 8. Arrange methyl halides (CH₃X, X = F, Cl, Br, I) in the order of increasing S_N2 reactivity.
Answer: CH₃— F < CH₃—Cl < CH₃ — Br < CH₃ — I .

Question 9. Arrange the following ions/molecules in the order of decreasing nucleophilicity in ethanol: 1. CH₃COO-, OH^–, CH₃O^–, H₂O 2. F^–, Br^–, I^–, Cl^–.
Answer:

1. CH₃O^– > OH^– > CH₃COO^– > H₂O
2. I^– > Br^– > Cl^– > F^–

Question 10. Which of the following compounds cannot be easily prepared by free radical chlorination? (CH₃)₃CCH₂CI, CH₃CH₂CH₂CH₂CH₂CI,
Answer: CH₃(CH₂)₃CH₂Cl (all hydrogens of the corresponding alkane are not equivalent).

Question 11. Which one out of CN^–, NO^–₂, CH₃COO^– and! CH3COC^–HCO₂C₂H₅ is not an ambident: nucleophile?
Answer:  CH₃COO^–.

Question 12. Which one out of fluorobenzene and chlorobenzene is more polar?
Answer: Chlorobenzene

Question 13. Which one out of (CH₃)₃CCH₂Br ,(CH₃)₃CBr, CH₃CH₂Br and CH₂=CHBr is not unreactive towards S_N2 reaction?
Answer: CH₃CH₂Br.

Question 14. Which one out of CH₂=CH—CH₂Cl, (CH₃)₃CC1, C₆H₅CH₂Cl, CH₂=CH—Cl and CH₃OCH₂Cl is unreactive towards S_N1 reaction?
Answer:  CH₂=CH—Cl (backside attack is not possible).

Question 15. Identify the compounds which are unreactive towards both S_N1 and S_N2 reactions:

Answer: 1,4 and 5

Question 16.  Which one out of OH^–, OCH^–₂, I^– and F^– is a good nucleophile as well as a good leaving group?
Answer: I^–

Question 17. Arrange the following compounds in order of increasing tendency of alkaline hydrolysis: 1-chloro- 2,4,6-trinitrobenzene (1), l-chloro-4-nitrobenzene (2), 1-chloro-3-nitrobenzene (3), 1-chloro-2,4- dinitrobenzene (4), chlorobenzene (5).
Answer: 5<3<2<4< 1.

Question 18. Which one of the following molecules is chiral?
 

Answer: CH₃CHClCH₂CH₃ (it contains only one asymmetric carbon).

Question 19. Write the name of the compound with R/Sdesignation obtained when (R) 2-bromo octane is subjected to alkaline hydrolysis (S_N2).
Answer: (S)-2-octanol

Question 20. Arrange the following compounds in increasing E2 reactivity (dehydrobromination): (CH₃)₂CHBr(1), (CH₃)₃CBr(2), CH₃CH₂Br(3)
Answer: 3<1<2.

Question 21. Which one of the following compounds does not react with RMgX to form RH?

1. (CH₃)₂NH
2. CH₃CH₂OH
3. H₂O
4. (CH₃)₃N
5. CH₃C≡CH

Answer: (CH₃)₃N

Question 22. 1,1,1-trichloro-2,2-bis (4-chlorophenyl) ethane is an insecticide. Write down the name by which it is popularly known.
Answer: DDT.

Question 23. Which monochloro derivative of m-xylene gives a white precipitate when treated with AgNO₃ solution?
Answer: The monochloro derivatives of xylene are:

The fourth compound produces a white precipitate when treated with AgNO₃ solution and this is because the Cl atom is attached to a benzylic carbon.

Question 24. Arrange the following compounds in order of increasing S_N1 reactivity:

Answer: The order of increasing S_N1 reactivity is: 2 < 3 < 1.

Question 25. Which of the following objects are achiral: Screw, shoe, ball, chair, hand, ear, nose, a spool of thread.
Answer: Ball, chair, and nose.

Question 26. Which monochloro derivative of pentane is chiral?
Answer: CH₃CHClCH₂CH₂CH₃ (2-chloropentane).

Question 27. Which of the following halides are inert towards S_N2 reaction?

CH₂=CHBr, (CH₃)₃CCl, (CH₃)₃CCH₂Cl , , CH₂=CHCH₂Br

Answer: The first four halides are unreactive toward the S_N2 reaction.

Question 28. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{OH} \stackrel{\mathrm{HBr}}{\longrightarrow} A+B \text {; Identify } A \text { and } B \text {. }\)
Answer: 

Question 29. Arrange the following groups in order of decreasing priority according to the sequence rules: —CH₃, —COOH, —H, —C₆H₅, —F
Answer:  — F > —OH > —COOH > —C₆H₅ > —CH₃ > — H .

Question 30. Arrange the following halides in order of increasing E2 reactivity (dehydrochlorination): CH₃CH₂Cl, (CH₃)₂CHCl, (CH₃)₃CCI
Answer: 

The order of increasing E2 reactivity (dehydrochlorination) is: CH₃CH₂CI < (CH₃)₂CHCl < (CH₃)₃CCl

Question 31. Identify the hydrocarbon (C₅H₁₀) which on photochemical chlorination gives only one monochloro derivative.
Answers: Cyclopentane 

Question 32. Which methyl halide has the highest density?
Answers: Methyl iodide (CH₃I)

Question 33. Arrange in the order of increasing boiling point:

1. CH₃CH₂CH₂CH₂Br
2. (CH₃)₃CBr
3. (CH₃)₂CHCH₂Br

Answers: The increasing order of b.p. 2 < 3 < 1

Question 34. Which methyl halide molecule has the highest dipole moment?
Answers: Methyl chloride (CH₃Cl)

Question 35. Mention the type of mechanism involved in chlorination of CH₄.
Answers: Free radical substitution

Question 36. Arrange CH₃F, CH₃I, CH₃Br, and CH₃Cl in the order of increasing S_N2 reactivity.
Answers:

The order of increasing S_N2 reactivity: CH₃F < CH₃Cl < CH₃Br < CH₃I

Question 37. Arrange the compounds in the order of increasing S_N1 reactivity:

1. ClCH₂CH=CHCH₂CH₃
2. CH₃CCl=CHCH₂CH₃
3. CH₃CH=CHCH₂CH₂Cl

Answers: Order of increasing S_N1 reactivity: 2 < 3 < 1

Question 38. Which of the given compounds remain inert in S_N2 reactions?

1. CH₂=CHBr,
2. (CH₃) CCl,
3. (CH₃)₃CCH₂Cl,
4. 
5. CH₂=CH —CH₂Br

Answers: All except the last one

Question 39. What type of compounds are formed when alkyl halides react with Mg in dry ether?
Answers: Organometallic compound; 

Question 40. Give an example of a haloalkane that does not give a Wurtz reaction.
Answers: Tert-butyl chloride (Me₃CCl)

Question 41.CH₃CH=CHCH₂OH \(\stackrel{\mathrm{HBr}}{\longrightarrow}\) A + B; Identify A and B.
Answers: A⇒CH₃CHBrCH=CH₂(3-bromobut-l-ene), B ⇒ CH₃CH=CHCH₂Br (l-bromobut-2-ene)

Question 42. Give an example of nucleophilic catalysis.
Answers: The reaction of primary alkyl chloride with sodium acetate is speeded up in the presence of Nal

Question 43. Explain why alkyl halides are not generally prepared in the laboratory by free radical halogenation.
Answers:  Isomeric monohalo derivatives are obtained and polyhalogenation may also take place

Question 44. Which substitution reactions occur with inversion of configuration?
Answers: S_N2

Question 45. Which substitution reactions occur with inversion as well as retention of configuration?
Answers: S_N1

Question 46. Which of the following molecules are chiral i.e., optically active?

1. CH₃CHBrC₂H5
2. CH₃CH₂OH
3. CH₃Cl
4. CH₃CH=C=CHCH₃

Answers: 1 and 4

Question 47. Predict the major products expected to be formed when isopropyl chloride is treated with (CH₃)₃CO^– & OH^– separately.
Answers: Propene CH₂=CHCH₃ and isopropyl alcohol (Me₂CHOH) respectively

Question 48. What change in optical activity has been observed when KI is added to optically active solution of (+)-2-iodooctane in acetone?
Answers: The solution becomes optically inactive due to racemisation.

Question 49. Mention the one-step reactions among S_N1, S_N2, E1 and E2.
Answers: S_N2 and E2

Question 50. Arrange the compounds in the order of increasing E2 (dehydro-chlorination) reactivity:

CH₃CH₂Cl, (CH₃)₂CHCl, (CH₃)₃CCI

Answers: The increasing order of reactivity is: CH₃CH₂Cl < (CH₃)₂CHCl < (CH₃)₃CCl

Question 51.CH₃CH₂CHClCH =CH₂ is optically active. Explain why it becomes optically inactive when reduced with H₂/Pt.
Answers:

On reduction, the optically active compound is converted into CH₃CH₂CHClCH₂CH₃ and the compound becomes achiral

Question 52. Write the structure of (R)-CH₃CHBrC₂H₅.
Answers:

To write the structure of the given compound, Br→C₂H₅→CH₃ should be arranged in a clockwise manner to get R-configuration.

Question 53. What do you mean by a chiral molecule?
Answers: The compound which cannot be superimposed on its mirror image

Question 54. (-) -Lactic acid (0.5 mol) is mixed with (+)-lactic acid (1 mol). Predict whether the mixture will be optically active or not.
Answers: Remains optically active

Question 55. Mention the ratio of l-chloro-2-methylpropane and 2-chloro-2-methylpropane obtained on photochemical chlorination of 2-methylpropane.
Answers: They will be obtained in 5: 9 ratio

Question 56. Mention the state of hybridisation of the central carbon atom in an S_N2 transition state.
Answers: sp²

Question 57. Give an example of a compound which reacts with RMgX to give another Grignard reagent.
Answers: Propyne (CH₃C = CH)

Question 58. In which direction does an equimolar mixture of two enantiomers rotate the plane of polarised light?
Answers: The mixture does not rotate plane-polarized light

Question 59. How many asymmetric carbon atoms are there in tartaric acid?
Answers: Two

Question 60. Mention the nucleophilic involved when ethyl chloride reacts with LiAlH₄ to yield ethane.
Answers: H^–

Question 61. Explain which CN^– is called an ambident nucleophile.
Answers: Because CN^– can act as a nucleophile both through carbon and nitrogen

Question 62. Write down the IUPAC name of freon-12.
Answers: Dichlorodifluoromethane

Question 63. Predict the product obtained when Cl₂ gas is passed through boiling toluene in the presence of sunlight.
Answers: C₆H₅CH₂Cl (benzyl chloride)

Question 64. Mention the product expected to be formed when benzene diazonium tetrafluoroborate is heated.
Answers: Fluorobenzene (C₆H₅F)

Question 65. Mention the test by which you can distinguish between C₆H₅CH₂Cl and o-chlorotoluene.
Answers: AgNO₃ test

Question 66. Arrange the following compounds in the order of increasing reactivity towards nucleophilic substitution reaction:

1. 4-chloronitrobenzene
2. Chlorobenzene
3. 1-chloro- 2,4,6-trinitrobenzene
4. l-chloro-2,4-dinitrobenzene.

Answers: The order of increasing reactivity: 2 < 1 < 4 < 3

Question 67. Which halobenzene has the lowest dipole moment?
Answers: Iodobenzene (C₆H₅I)

Question 68. Which of the isomeric dichlorobenzenes has the highest melting point?
Answers: P-Dichlorobenzene

Question 69. Arrange in the order of increasing S_N1 reactivity:

1. C₆H₅CH₂Cl
2. (C₆H₅)3CCl
3. C₆H₅CH₂CH₂Cl

Answers: Order of decreasing S_N1 reactivity: 2 > 1 > 3

Question 70. In which position of chlorobenzene the electron density is relatively low?
Answers: Meta-position

Question 71. Which halide is the most reactive in the S_NAr reaction of

Answers:

Question 72. C₆H₅Br + \(\mathrm{Mg} \stackrel{\text { ether }}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{D}_2 \mathrm{O}}{\longrightarrow} \mathrm{B}\) Write the names and structures of A and B.
Answers:

A⇒C₆H₅MgBr (phenylmagnesium bromide), B⇒C₆H₅D (deuterobenzene)

Question 73. Predict the type of reaction involved in the preparation of DDT from chlorobenzene.
Answers: Electrophilic substitution reaction

Question 74. Which poisonous gas is formed when chloroform is exposed to air and light?
Answers: Phosgene (COCl₂)

Question 75. What is pyrene? Mention the value of its dipole moment.
Answers: Carbon tetrachloride (CCl₄), its dipole moment (μ) = OD

Question 76.Write the IUPAC name of Freon-12.
Answers: Dichlorodifluoromethane

Question 77. Which one out of the Wurtz-Fittig and Friedel-Crafts reactions is more acceptable for the preparation of propylbenzene?
Answers: Wurtz-Fittig reaction because in this case there is no possibility of the formation of an isomeric compound.

Question 78. Which halogen present in hexachlorobenzene and benzene hexachloride is inert?
Answers: Hexachlorobenzene as it is an aryl halide

Question 79. What type of reaction is involved in benzylic halogenation?
Answers: Free radical substitution reaction

Question 80. What type of compound is phenyl lithium?
Answers: An organometallic compound

Question 81. Give the product obtained when PhMgBr is treated with D₂O.
Answers: C₆H₅D (deuteriobenzene)

Question 82. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\mathrm{Br}_2 / h v}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{NaCN}}{\longrightarrow} \mathrm{B}\). Identify A and B and explain their formations.
Answers:

A ⇒ C₆H₅CHBrCH₃ (because benzyl free-radical is resonance-stabilized), B ⇒ C₆H₅CH(CN)CH₃ (because CN^– is a stronger nucleophile than Br^–).

Unit 10 Haloalkanes And Haloarenes Short Question And Answers

Question 1. Which one of the isomeric dichlorobenzenes has the highest melting point and why?
Answer:

The melting point of p-dichlorobenzene is the highest. The most symmetrical molecules of the p-isomer pack closely in the crystal lattice. As a result, the intermolecular forces of attraction are much stronger. Therefore, the p-isomer melts at a temperature higher than that of the o-and m-isomers.

Question 2. How can halohydrocarbons be classified based on the type of hybridisation of carbon bonded to the halogen atom? Give example.
Answer:

Based on the type of hybridisation of carbon to which the halogen is attached,

Halohydrocarbons are of the following three types:

1. C _sp3—X (Example, CH₃—Cl),

2. C_sp2-X (Example,).

3. C_sp2-X (Example, HC≡CCl).

Question 3. Alkyl halides are insoluble in water, even though they are polar—why?
Answer:

• Polar molecules of haloalkanes (R^δ+—X^δ-) are held together by dipole-dipole attractions. On the contrary, water molecules remain associated through intermolecular hydrogen bonding. However, alkyl halide molecules are unable to form hydrogen bonds with water molecules. So, alkyl halides are insoluble in water.
• This can be interpreted alternatively as follows. For a haloalkane to dissolve in water, energy is required to overcome the dipole-dipole attractions of haloalkanes and to break the hydrogen bonds of water molecules.
• Such an amount of energy is not released when dipole-dipole attractions are set up between haloalkane and water molecules. Therefore, alkyl halides are practically insoluble in water.

Question 4. Write the structures of two isomeric lowest molecular-weight alkanes which are chiral.
Answer:

Question 5. Explain why phosphoric acid and not sulphuric acid are used with KI during the preparation of alkyl iodides from alcohols.
Answer:

KI is expected to give HI on reacting with H₂SO₄ and it is HI which is expected to convert alcohols (ROH) to alkyl iodides (R—I). However, H₂SO₄ is a strong oxidising agent. It oxidises HI to I₂ which cannot react with alcohols.

⇒ \(\mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{KHSO}_4+2 \mathrm{HI}\)

⇒ \(2 \mathrm{HI}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{I}_2+\mathrm{SO}_2\)

To overcome this problem, the non-oxidising acid H₃PO₄ is used instead of H₂SO₄.

⇒ \(\mathrm{ROH}+\mathrm{KI}+\mathrm{H}_3 \mathrm{PO}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{R}-\mathrm{I}+\mathrm{KH}_2 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O}\)

Question 6. CH₃Cl undergoes hydrolysis much more easily as compared to C₆H₅Cl. Explain with reasons.
Answer:

In chlorobenzene, C₆H₅Cl, an unshared pair of elect-turns on the Cl-atom, is involved in resonance with the ring electrons. As a result, the C—Cl bond acquires some double bond character. Thus, the C— Cl bond becomes much stronger and the expulsion of the Cl-atom from the ring cannot be affected easily. So, it is not easy to hydrolyse chlorobenzene, as it is carried out under extreme conditions (high temperature and pressure).

On the other hand, the C—Cl bond in CH₃— Cl is unable to acquire any double bond character because similar electron delocalisation is not possible. Thus, it is easy to hydrolyse CH₃—Cl under normal conditions.

Question 7. An alkene reacts with HCl to yield only 1-chloro-1-methyl cyclohexane. Identify the alkene, and give a reaction.
Answer:

Question 8. Which one of the following compounds will undergo S_N1 reaction readily?
Answer:

The compound B undergoes S_N1 reaction readily because the resulting carbocation (PhCH₂) is stabilised by resonance. The compound A undergoes S_N1 reaction relatively slowly because the resulting 2° carbocation is not resonance stabilised.

Question 9. Allyl chloride undergoes hydrolysis at a rate faster than w-propyl chloride – why?
Answer:

Hydrolysis (solvolysis) is in general an S_N1 reaction. Allyl chloride undergoes easy hydrolysis because the resulting allyl cation (obtained on C— Cl bond cleavage) is stabilised considerably by resonance. On the other hand, n-propyl chloride, which is expected to produce a very unstable 1° carbocation, does not undergo hydrolysis by the S_N1 mechanism. It undergoes hydrolysis by the relatively slow S_N2 process.

Question 10.  In ethanol, A undergoes solvolysis (S_N1 ), whereas B does not—why.
Answer:

• The carbocation obtained from the C—Cl bond cleavage of A is a very stable aromatic carbocation [cyclopropenyl cation having (4n + 2)π- electrons, where n = 0], For this reason, A undergoes solvolysis (S_N1) at a faster rate.
• On the other hand, the carbocation expected to be obtained on the C — Cl bond cleavage of B is an unstable anti-aromatic carbocation [cyclopentadienyl cation, having 4n it- electrons, where n = 1 ]. Therefore, the compound B does not undergo solvolysis (S_N1 ) in ethanol.

Question 11. How will you get iodoethane from ethanol if only Nal is present in the laboratory?
Answer:  Idodoethane will be obtained if ethanol is allowed to react with Nal in the presence of a cone. H₂SO₄.

Question 12. Arrange the following compounds in order of increasing SN1 reactivity and explain the order:
Answer:

Answer: A better-leaving group causes a faster S_N1 reaction. Since, the ability of halide ions as a leaving group increases in the order: Cl Bre< Ie, therefore, the S_N1 reactivity of the three tertiary substances (2, 3 and 4) increases in the order: 2 < 4 < 3. Being a secondary substrate compound I exhibit less S_N1 reactivity than the tertiary substrate 2 containing the same leaving group.

Hence, the S_N1 reactivity of the given halides increases in the order: 1 < 2 < 4 < 3.

Question 13. Which one of each of the following pairs does not undergo S_N2 reaction and why?

(CH₃)₃CCH₂CI, CH₃CH₂CI

Answer: Both compounds are expected to undergo S_N2 reaction because both of them are primary substrates. However, the first one is a neopentyl halide (three alkyl groups at the β -carbon) and thus, due to steric hindrance, it does not undergo S_N2 reaction. Being a primary substrate, the second alkyl halide CH₃CH₂Cl undergoes S_N2 reaction easily.

The second compound is a tertiary substrate and the backside attack on a -carbon is not at all possible because of its cage-like structure. So it is inert towards S_N2 reaction. On the other hand, being a secondary substrate the first compound undergoes S_N2 reaction.

Question 14. Which one of each pair undergoes Sÿl reaction readily and why?

1. CH₂=CHCH₂CH₂CI, CH₃CH=CHCH₂CI
2. CH₃OCH₂CI, CH₃OCH₂CH₂CI

The carbocation obtained from the C- Cl bond cleavage of the second compound is stabilised by resonance. Because of this, it undergoes S_N1 reaction readily. On the other hand, being a primary substrate the first compound does not undergo S_N1 reaction.

The primary carbocation obtained from the C- Cl bond cleavage of the first compound is considerably stabilised by resonance. Because of this, it undergoes S_N1 reaction readily.

The second compound is expected to produce a very unstable carbocation which cannot be stabilized by resonance ( —OCH₃ group is not directly attached to the a -carbon) and for this reason, it does not undergo S_N1 reaction.

Question 15. Identify products A and B and comment on their yields. CH₃CH₂CH=CHCH₃ + HCl→A + B

Since the double-bonded carbons of the alkene contain same number of hydrogen atoms, the intermediate 2° carbocations leading to the formation of A and B are nearly equally stable. For this reason, the compounds A and B are obtained at nearly equal rates and in nearly equal amounts.

Question16. A hydrocarbon (72g-mol_1) reacts with chlorine in CH₃ CH₂CI CH₃ in the presence of light to produce one monochloro and two di-chloro derivatives. Write down the structure of the hydrocarbon.
Answer:

The molecular formula of the hydrocarbon having a molecular mass of 72g.mol^–1 is C₃H₁₂. Since it produces one monochloro and two dichloro derivatives while reacting with C_l2 in the presence of light, all the H atoms of this hydrocarbon are equivalent. Therefore, the hydrocarbon is neopentane, (CH₃)4C.

Question 17. Predict the product expected to be obtained when HCl is added to isobutylene. Explain the mechanism.
Answer:

Isobutylene (Me₂C=CH₂) undergoes addition with HCl according to the Markownikoff’s rule and tert-butyl chloride is obtained as the only product. The other product (1-chloro-2- methylpropane) is not obtained because it is expected to be obtained through the formation of a very unstable 1° carbocation.

Question 18.

1. How CH₃MgI could be used to prepare methoxy ethane? (Write arrowhead equation only)
2. Identify the compounds which will respond to the iodoform test:

CH₃CH₂COCH₂CH2I, CH₃CH₂COCH₂I

CH₃CH₂COCH₃, CH₃COOCH₂CH₃

Answer:

Question 19. Indicate the reagents for the following transformation.

Answer: 

Question 20. A molecule has the following structure.

Which one is the IUPAC name—

1. (R) -2-chloro-2-methyl butane
2. (S) -2-chloro-3-methyl butane
3. (R) -2-methyl-3-Chlorobutane
4. (S) -2-methyl-3-Chlorobutane

Answer: 2. (S) -2-chloro-3-methyl butane

Question 21.

1. By what type of mechanism, the following compound undergoes alkaline hydrolysis?

Show the mechanism of the reaction.

2. Write the environmental effect of DDT

Question 22. Which of the following is an example of freon—

1. BrCH₂CH₂Cl
2. CCl₂F₂
3. CCl₂Br₂
4. ICH₂CH₂F

Answer: 2. CCl₂F₂

Question 23. What will happen when bromoethane reacts with an aqueous solution of sodium hydroxide? Write the mechanism of the reaction.
Answer:

When bromoethane reacts with an aqueous solution of sodium hydroxide,methanol is produced.

CH₃Br+NaOH → CH₃OH+NaBr

Question 24. Which of the following compounds most readily undergoes solvolysis by S_N1 mechanism—

1. CH₃I

3. CH₃CHCICH₃

4. (CH₃)₃CCl

Answer: 4. (CH₃)₃CCl

Question 25. 

1. Write the IUPAC name mentioning the R/S notation of the following compound:

2. Write the reagent in case of the first reaction and the organic product in case of the second reaction:

Answer:

1. (R)-2-Chlorobutane.

Question 26. For the compounds CH₃Cl, CH₃I, CH₃Br and which of the following is the correct order of C — halogen bond length—

Answer: 1

Question 27. In which of the following two compounds S_N2 reaction is faster? Give reason.

CH₃CH₂CH₂C1 and CH₃CH₂CH₂I

Answer: The S_N2 reaction of CH₃CH₂CH₂I is faster as iodine is a better-leaving group.

Question 28. Give the IUPAC name of the following compound.

Answer: IUPAC Name: 3-bromo-2-methyl prop-1-ene

Question 29.

1. Draw the structure of the major monohalo product in the following reaction:

2. Which halogen compound in each of the following pairs will react faster in the S_N2 reaction:

1. CH₃Br or CH₃I
2. (CH₃)₃C—Cl or CH₃—Cl

Answer:

 

2. I^– is a better-leaving group than Br^– and so CH₃I will undergo S_N2 reaction with OH^– at a relatively faster rate.

Due to less steric hindrance, CH₃Cl undergoes S_N2 reaction at a much faster rate than (CH₃)₃CCl (a 3° substrate).

Question 30. Identify the chiral molecule in the following pair

Answer:

Question 31. Name the following according to the IUPAC system:

Answer:

1. 2-bromotoluene;
2. 2,2- dimethylchloropropane

Question 32. Give the IUPAC names of the following compounds:

3. H₂C=CH₂—Cl

Answer: 

1. 2-bromobutane;
2. 1,3-dibromo benzene;
3. 3-chloropropene

Question 33. How are the following conversions carried out?

1. Benzyl chloride to Benzyl alcohol.
2. Ethyl magnesium chloride to Propan-l-ol

Answer:

Question 34. Out of which is more reactive towards S_N1 reaction and why?
Answer:

is more reactive towards S_N1 reaction as it is a secondary halide and the formed 2° carbocation is very stable. On the other hand, is a primary halide and the 1° carbocation formed which is not so stable.

Question 35. Give reasons— The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
Answer:

In chlorobenzene, the C-atom of the C—Cl bond is sp² – hybridised, thus it is more electronegative than the corresponding sp³ -hybridised carbon of chlorocyclohexane. Consequently, the attraction of bonding electron pair away from carbon by chlorine is less pronounced in chlorobenzene than in cyclohexyl chloride and so, the molecular dipole moment is smaller in chlorobenzene.

Further, chlorobenzene is a resonance hybrid of the given resonating structures and so, there exists a n -moment. Thus n -moment acts in opposition to that of the strong cr -moment arising out of the highly polar C The resultant dipole moment of chlorobenzene—- Clisbond. Thus smaller than that of cyclohexyl chloride (it has no resonance structure and so, no n-moment).

Question 36. Which of the following reactions is S_N1 type?

Answer: Reaction (2) is S_N1 type, as it is accompanied by retention of configuration.

Question 37. How do you convert:

1. Chlorobenzene to toluene.
2. But-I-ene to But-2-ene.
3. Ethanol to ethyl iodide.

Or, What happens when:

1. n-butyl chloride is treated with alcoholic KOH.
2. 2-Chloropropane is treated with sodium in the presence of dry ether.

Answer:

Or,

Question 38. Following compounds are given to you: 2-bromopentane, 2-bromo-2-methyl butane, 1-bromopentane

1. Write the compound which Is most reactive towards S_N2 reaction.
2. Write the compound which is optically active.
3. Write the compound which is most reactive towards -the elimination reaction.

Answer:

1. 1-bromopentane is a primary halide, hence, due to less steric hindrance it undergoes S_N2 reaction faster.
2. 2-bromopentane is optically active as C-2 is asymmetric carbon.
3. 2-bromo-2-methylbutane is most reactive towards the P-elimination reaction. This is because tertiary alkyl halides on dehydrohalogenation form the most substituted alkene.

Question 39. Write the structure of 3-bromo-2-methyiprop-l-ene.
Answer:

Question 40. Out of which is an example of allylic halide?
Answer:

Question 41. Out of which is an example of vinylic halide?
Answer: 

Question 42. Out of which is an example of a benzylic halide?
Answer:

Question 43. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why?
Answer:

In chlorobenzene C₆H₅Cl, an unshared pair of electrons on the Cl-atom, is involved in resonance with the ring n electrons. As a result, the C—Cl bond acquires some double bond character. Thus, the C —Cl bond becomes much stronger and the expulsion of the Cl-atom from the ring cannot be affected easily. So, it is not easy to hydrolyse chlorobenzene.

On the other hand, the C—Cl bond in benzyl chloride (Ph— CH₂—Cl) is unable to acquire any double bond character because similar electron delocalisation is not possible. Thus, it is easy to hydrolyse benzyl chloride.

Question 44.

1. Identify the chiral molecule in the following pair:

2. Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.

3. Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-l methylcyclohexane with alcoholic KOH.

Answer:

Question 45. Draw the structures of all the eight structural isomers that have the molecular formula CgHuBr. Name each isomer according to the IUPAC system and classify them as primary, secondary, or tertiary bromide.
Answer:

Question 46. Write IUPAC names of the following:

Answer:

1. 4-bromopent-2-ene,
2. 3-bromo-2-methylbut-l-ene
3. 4-bromo-3-methyl pent-2-ene
4. 1-bromo-2-methylbut-2-ene
5. 1-bromobut-2-ene
6. 3-bromo-2-methylpropene.

Question 47. Write structures of the following compounds:

1. 2-chloro-3-methyl pentane
2. l,4-dibromo but-2-ene
3. 1-chloro-4-methylcyclohexane
4. 4-fert-butyl-3-iodoheptane
5. 1-bromo-4-sec-butyl-2-methylbenzene

Answer:

Question 48. Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH3)2CHCH2CH3.
Answer:

The given hydrocarbon contains four different types of non-equivalent H-atoms. So it will form four different monochloro derivatives. These are —

Question 49. Write the products of the following reactions:

Answer:

Question 50. Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:

One method of converting an alcohol to alkyl iodide is to treat the alcohol with KI in the presence of an acid catalyst.

⇒ \(\mathrm{R}-\mathrm{OH}+\mathrm{KI} \underset{\Delta}{\stackrel{\mathrm{H}^{\oplus}}{\longrightarrow}} \mathrm{R}-\mathrm{I}+\mathrm{H}_2 \mathrm{O}+\mathrm{K}^{+}\)

For simplicity, we may consider that KI reacts with H⁺ to form HI, which then reacts with alcohol to form alkyl iodide.

⇒ \(\mathrm{KI}+\mathrm{H}^{+} \rightarrow \mathrm{K}^{+}+\mathrm{HI}\)

⇒ \(\mathrm{R}-\mathrm{OH}+\mathrm{HI} \rightarrow \mathrm{R}-\mathrm{I}+\mathrm{H}_2 \mathrm{O}\)

If the acid under consideration is an oxidising acid, such as conc.H₂SO₄, then during the reaction it will oxidise HI to I₂ (iodine) and as a result, the alcohol will not be converted to alkyl iodide.

⇒ \(2 \mathrm{HI}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{I}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2\)

However, a non-oxidising acid, such as H₃PO₄ can be used in the reaction.

⇒ \(\mathrm{ROH}+\mathrm{KI} \underset{\left(\mathrm{H}^{+}\right)}{\stackrel{\mathrm{H}_3 \mathrm{PO}_4}{\longrightarrow}} \mathrm{R}-\mathrm{I}+\mathrm{H}_2 \mathrm{O}+\mathrm{K}^{+}\)

Question 51. Write structures of different dihalogen derivatives of propane.
Answer:

Four isomeric halogen derivatives of propane are:

Question 52. Among the isomeric alkanes of molecular formula C₅Hi₂, identify the one that on photochemical chlorination yields —

1. A single monochloride.
2. Three isomeric monochlorides.
3. Four isomeric monochlorides.

Answer:

All the H-atoms are equivalent. So, it forms a single monochloride derivative.

It has three different types of equivalent hydrogens designated as a, b, and c. So, it will form three different monochloro derivatives.

It has four different types of equivalent hydrogens designated as a, b, c, and d. So, it will form four isomeric monochloro derivatives.

Question 53. Draw the structures of major monohalo products in each of the following reactions:

Question 54. Arrange each set of compounds in order of increasing boiling points.

1. Bromomethane, Bromoform, Chloromethane, Dibromomethane.
2. 1-chloropropane, Isopropyl chloride, 1-chlorobutane.

1. For alkyl halides containing the same alkyl group, the boiling point increases with an increasing molecular mass of the halogen. So, the boiling point of bromethane is greater than chloromethane. But, for bromo derivatives of methane, boiling point increases as the number of Bratoms in the molecule increases. Thus, boiling point increases in the sequence: Chloromethane (CH₃Cl) < Bromomethane (CH₃Br) < Dibromomethane (CH₂Br₂) < Bromoform (CHBr₃).

2. For alkyl halides containing the same halogen, the boiling point increases as the size of the alkyl group increases. So, the boiling point of 1-chlorobutane is greater than that of 1-chloropropane and isopropyl chloride (CH₃— CHCICH₃). But for isomeric alkyl halides, the boiling point decreases as branching increases. So, the boiling point of 1-chloropropane is greater than that of isopropyl chloride. Thus boiling point increases in the sequence: Isopropyl chloride <l-chloropropane < I-chlorobutane.

Question 55. Haloalkanes react with KCN to form alkyl cyanides as the main product while AgCN forms isocyanides as the chief product. Explain.
Answer:

The cyanide ion (CN^–) is an ambident nucleophile because it can attack electrophilic centres through two different nucleophilic centres (C and N).

In the presence of AgCN, the reaction proceeds by the SN1 mechanism because the Ag⁺ ion coordinates with the halogen atom and promotes the formation of a carbocation by precipitating silver halide. Attack on the carbocation then takes place through the more electronegative and more dense nitrogen atom (the more positive charge centre is Attacked by the more negative charge centre). As a result, alkyl cyanide (R—NC) is obtained predominantly.

In the presence of KCN, on the other hand, the reaction proceeds by the S_N2 mechanism because unlike silver ion K+ ion cannot promote ionisation of the carbon-halogen (C—X) bond. In this case, the cyanide ion attacks R—X through the more polarisable (larger in size and less electronegative) and more nucleophilic carbon atom (a less positive charge centre is attacked by a less negative one). As a result, an alkyl cyanide (R—CN) is obtained predominantly.

Question 56. In the following pairs of halogen compounds, which would undergo S_N2 reaction faster?

1. Cyclohexyl methyl chloride is a 1° alkyl chloride, while cyclohexyl chloride is a 2° alkyl chloride and hence, the former undergoes S_N2 reaction at a faster rate.
2. 1^– is a better-leaving group (because of its large size and low electronegativity) than Cl- ion (small size, high electronegativity). So, n-butyl iodide undergoes S_N2 reaction at a faster rate than n-butyl chloride.

Question 57. Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions:

The four isomeric bromobutanes,

1. C₆H₅CH₂Br,
2. C₆H₅CH(C₆H₅)Br,
3. C₆H₅CH(CH₃)Br,
4. C₆H₅C(CH₃)(C₆H₅)Br

Answer:

1. The four isomeric bromobutanes are as follows:

In the S_N1 pathway, the reactivity of the alkyl halides increases as the stability of the corresponding carbocations increases. Since the stability of the carbocations increases in the order:

CH₃CH₂CH₂CH₂(I°)<(CH₃)₂CHCH₂(1°) <CH₃CH₂CHCH₃(2°)<(CH₃)₃C(3°). [Remember that the electron-donating +I effect of (CH₃)₂CH — is greater than that of CH₃CH₂CH₂—, the reactivity increases in the order (1) < (2) < (3) < (4).

In the S_N2 pathway, the reactivity of the alkyl halides decreases as the steric hindrance around the electrophilic carbon (i.e., α-carbon) increases. Thus the reactivity follows the sequence: (1) > (2) > (3) > (4).

2. The order of stability of the carbocations derived from the given alkyl bromides is: C₆H₅C⁺H₂ < C₆H₅C⁺H(CH₃) < C₆H₅C⁺H(C6H5)< C₆H₅C⁺(CH₃)(C₆H₅).

So In S_N1 pathway reactivity is: C₆H₅CH₂Br< C₆H₅CH(CH₃)Br < C₆H₅CH(C₆H₅)Br < C₆H₅C(CH₃)(C₆H₅)Br.

In the S_N2 pathway, the reactivity of the alkyl halides decreases as the steric hindrance around the electrophilic carbon (i.e., α-carbon) increases. Since the C₆H₅ group is larger than the CH₃ group, the reactivity follows the sequence:

C₆H₅C(CH₃)(C₆H₅)Br <C₆H₅CH(C₆H₅)Br < C₆H₅CH(CH₃)Br< C₆H₅CH₂Br.

Question 58. Identify chiral and achiral molecules in each of the following pair of compounds.
Answer:

1. In structure (a), the central carbon is attached to four different groups (CH₃, Br, OH and H) and hence, it represents a chiral molecule. In structure (b), the central carbon is attached to two identical groups (i.e., two Br-atoms) and hence, this molecule is achiral.
2. In structure (a), central carbon is attached to four different groups (CH₃, CH₂CH₂CH₃, OH and H), hence, it represents a chiral molecule. In structure (b), central carbon is attached to two identical groups (two CH₂CH₃ groups) and hence, it represents an achiral molecule.
3. In structure (a), C_-2 is attached to four different groups (CH₃, CH₂CH₃, H and Br ) hence, it represents a chiral molecule. In structure (b), no carbon is attached to four different groups and so, it represents an achiral molecule.

Question 59. Although chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Why?
Answer:

This can be explained by considering the relative stabilities of the σ-complexes formed during ortho-, para- and mete-substitution.

The σ-complexes for ortho – and para-substitution are stabilised by four resonating structures, while the σ-complex for meta-substitution is stabilised by only three resonating structures. In other words, the σ-complexes for o- and p-substitution are more stable than that for the meta substitution. Hence, chlorobenzene undergoes electrophilic substitution preferably at the ortho and para positions.

However, Cl-atom withdraws electrons from the ring by the -1 effect and at the same time it increases electron density in the ring by the +R effect. Since the -I effect of Cl is stronger than that of its +R effect, the ring system becomes deactivated. Hence, chlorobenzene undergoes electrophilic substitution at a much slower rate than unsubstituted benzene.

Question 60. Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism? Explain your answer.

Answer:

1. Out of the two isomeric alkyl bromides, the a -carbon of 1° -halide is less hindered and more positively polarised. Since the electron donating +1 effect of one ethyl and one methyl group is greater than that of a single n-butyl group, 1-bromobutane will react more rapidly by the S_N2 mechanism.
2. Out of the two isomeric alkyl bromides, the a -carbon of 2° alkyl bromide is less hindered and more positively polarised. Since the overall electron donating +1 effect of three methyl groups is greater than the overall effect of one ethyl and methyl group, 2-bromobutane reacts more rapidly by the S_N2 mechanism.
3. Out of the two isomeric 1°-alkyl bromides, the former has a methyl group on the δ-carbon while the latter has a methyl group on the β-carbon. As a result, the α-carbon of the latter compound suffers greater steric hindrance than that of the former compound. So, 3-methyl-1- bromobutane reacts more rapidly by S_N2 mechanism.

Question 61. In the following pairs of halogen compounds, which compound undergoes a faster S_N1 reaction?

Answer:

The reactivity of alkyl halides in the S_N1 pathway increases as the stability of carbocation derived from an alkyl halide increases.

3° carbocations are more stable than 2° carbocations, therefore, (CH₃)₃CCl undergoes S_N1 reaction at a much faster rate than (C₂H₅)₂CHCl.

2° carbocations are more stable than 1° carbocations. Therefore, CH₃(CH₂)₄CHClCH₃ undergoes S_N1 reaction at a much faster rate than CH₃(CH₂)₄CH₂Cl

Question 62. Identify A, B, C, D, E, R and R1 in the following:

Answer:

Cyclohexylmagnesium bromide Cyclohexane

Question 63. Name the following halides according to the IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

1. (CH₃)₂CHCH(Cl)CH₃
2. CH₃CH₂CH(CH₃)CH(C₂H₅)C1
3. CH₃CH₂C(CH₃)₂CH₂I
4. (CH₃)₃CCH₂CH(Br)C₆H₅
5. CH₃CH(CH₃)CH(Br)CH₃
6. CH₃C(C₂H5)₂CH₂Br
7. CH₃C(Cl)(C₂H₅)CH₂CH₃
8. CH₃CH=C(Cl)CH₂CH(CH₃)₂
9. CH₃CH=CHC(Br)(CH₃)₂
10. p-CIC₆H₄CH₂CH(CH₃)₂
11. m-ClCH₂C₆H₄CH₂C(CH₃)₃
12. o-Br-C₆H₄CH(CH₃)CH₂CH₃

Answer:

1. 2-chloro-3-methyl butane (2° alkyl halide)
2. 3-chloro-4-methyl hexane (2° alkyl halide)
3. 1-iodo-2,2-dimethylbutane (l°alkyl halide)
4. 1-bromo-3,3-dimethyl-1-phenyl butane (2° benzylic halide)
5. 2-bromo-3-methyl butane (2° alkyl halide)
6. 1-bromo-2-ethyl-2-methyl butane (1° alkyl halide)
7. 3-chloro-3-methyl pentane (3°alkyl halide)
8. 3-chloro-5-methyl hex-2-ene (vinylic halide)
9. 4-bromo-4-methyl pent-2-ene (allylic halide)
10. 1-chloro-4-(2-methylpropyl) benzene (aryl halide)

1-chloromethyl-3-(2,2-dimethylpropyl) benzene (1° benzylic halide)

1 -bromo-2-(1 -methylpropyl) benzene (aryl halide)

Question 64. Give the IUPAC names of the following compounds:

1. CH₃CH(Cl)CH(Br)CH₃
2. CHF₂CBrClF
3. CICH₂C = CCH₂Br
4. (CCl₃)₃CCl
5. CH₃C(p-ClC6H4)₂CH(Br)CH₃
6. (CH₃)₃CCH=CIC₆H₄I-p

Answer:

1. 2-bromo-3-chlorobutane

2. 1-bromo-1-chloro-1,2,2-trifluoroethane

3. 1-bromo-4-chlorobut-2-yne

Question 65. Write the structures of the following organic halogen compounds.

1. 2-chloro-3-methyl pentane
2. p -bromochlorobenzene
3. l-chloro-4-ethylcyclohexane
4. 2-(2-chlorophenyl)-1-Idooctane
5. 2-bromobutane
6. 4-tert-butyl-3-iodoheptane
7. 1-bromo-4-sec-butyl-2-methylbenzene
8. l,4-dibromo but-2-ene

Answer:

Question 66. Which one of the following has the highest dipole moment:

1. CH₂Cl₂
2. CHCl₃
3. CCl₄

Answer:

The three-dimensional structures of the three compounds along with the directions of different bond moments are shown below:

Due to the symmetrical structure, the dipole moment of CCl₄ is zero. In compound

The resultant of two C —Cl bond moments is reinforced by two C — H bond moments. But in compound

The resultant of two C—Cl bond moments is reinforced by one C— H bond moment but opposed by a third C— Cl bond moment. This shows that the dipole moment of CH₂Cl₂ is greater than that of CHCl₃. Thus, CH₂Cl₂ has the highest dipole moment.

Question 67. A hydrocarbon C₅H₁₀ does not react with chlorine in the dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:

The molecular formula of the hydrocarbon is C₅H₁₀ (i.e., C_nH_2n type). So it is either an alkene or a cycloalkane. But the compound does not react with Cl₂ in the dark. So it is not an alkene but must be a cycloalkane. Since it reacts with Cl₂ in bright sunlight to give a single monochloro derivative (C₅H₉Cl), all the ten H-atoms of the hydrocarbon must be equivalent. Thus, the given hydrocarbon is cyclopentane.

Question 68. Write the isomers of the compound having a formula C₄H₉Br.
Answer: 

D.B.E. for C₄H₉Br = \((4+1)-\frac{9+1}{2}=0\)

Thus it is a saturated acyclic compound, i.e., an alkyl halide. The possible isomers are

Question 69. Write the equations for the preparation of 1-iodo-butane from—

1. 1-butanol
   
2. 1-chlorobutane
3. but-l-ene.

Answer:

Question 70. What are ambident nucleophiles? Explain with an example.
Answer:

Nucleophiles which have more than one, generally two suitable atoms through which they can attack the substrate are called ambident nucleophiles. An example is the cyanide ion (CN^–). It can attack the substrate either through carbon to form cyanide or through nitrogen to form isocyanide.

Question 71. Which compound in each of the following pairs will react faster in the S_N2 reaction with OH-?

1. CH₃Br or CH₃I
2. (CH₃)₃CC1 or CH₃Cl

I^– is a better-leaving group than Br^–. So CH₃I reacts faster with OH^– ion in the S_N2 pathway.

S_N2 reactivity of alkyl halides increases as the hindrance around the a -carbon (i.e., the carbon undergoing nucleophilic attack) decreases and also the +ve polarisation of the α -carbon increases. Both these factors are favourable for CH₃Cl and hence, it reacts faster in S_N2 reaction with OH^–.

Question 72. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

1. 1-bromo-1-methylcyclohexane
2. 2-chloro-2-methyl butane
3. 2,2,3-trimethyl-3-bromopentane.

Answer:

1. In a base-catalysed dehydrohalogenation reaction, the halogen atom is lost from the α-carbon and a hydrogen atom is lost from the β-carbon atom. The given substrate contains two types of β-hydrogens and hence, two different alkenes are formed in the reaction.

Product ‘A’ is more substituted alkene (Saytzeff product) and hence, it is formed in a predominant amount.

2. The substrate contains two different types of β-hydrogens and hence, two different products (alkenes) are formed. The more substituted alkene I (i.e., the Saytzeff product) is formed in the predominant amount.

3. The substrate contains two different types of β-hydrogens and hence, in principle, two different alkenes (1 and 2) are produced. However, the more substituted alkene I (i.e., Saytzeff product) is formed as the major product.

Question 73. How will you bring about the following conversions?

1. Ethanol to but-l-yne
2. Ethane to bromoethene
3. Propene to 1-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1- ene to but-2-ene
9. 1-chlorobutane to n-octane
10. Benzene to biphenyl.

Answer:

Question 74. Explain why Alkyl halides, though polar, are immiscible with water.
Answer:

Alkyl halides are polar compounds and their molecules are held together by relatively weaker attractive forces involving dipole-dipole interaction. On the other hand, molecules of water (H₂O) are held together by very strong attractive forces involving intermolecular H-bonding. Since molecules of alkyl halides cannot form H-bonds with H₂O molecules, alkyl halides are immiscible in water.

Question 75. Write the structure of the major organic product in each of the following reactions:

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{NaI} \underset{\text { heat }}{\stackrel{\text {acetone  }}{\longrightarrow}}\)

⇒ \(\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}+\mathrm{KOH} \underset{\text { heat }}{\stackrel{\text { ethanol }}{\longrightarrow}}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{NaOH} \stackrel{\text { water }}{\longrightarrow}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{KCN} \stackrel{\text { aq. ethanol }}{\longrightarrow}\)

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{ONa}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl} \longrightarrow\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}+\mathrm{SOCl}_2 \longrightarrow\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr} \stackrel{\text { peroxide }}{\longrightarrow}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_3\right)_2+\mathrm{HBr} \longrightarrow\)

Question 76. Write the mechanism of the following reaction:

⇒ \(n \mathrm{BuBr}+\mathrm{KCN} \stackrel{\mathrm{EtOH}-\mathrm{H}_2 \mathrm{O}}{\longrightarrow} n \mathrm{BuCN}\)

Answer:

KCN ionises in solution to form cyanide ion , which is an ambident nucleophile.

It attacks the carbon atom of C — Brbond of the substrate preferably through the C-atom rather than the N-atom, because carbon has a more polarisable electron cloud owing to its larger size and lower electronegativity.

Thus n-butyl cyanide is the predominant product. Since the C —C bond is stronger than the C — N bond, it favours the formation of this product.

Question 77. Arrange in order of reactivity towards S_N2 reaction:

1. 2-bromo-2-methyl butane, 1-bromopentane, 2- bromo pentane
2. l-bromo-3-methyl butane, 2- bromo-2-methyl butane, 2-bromo-3-methyIbutane
3. 1-bromobutane, l-bromo-2,2-dimethyl propane, 1-bromo-2-methylbutane, 1 -bromo-3-methylbutane.

Answer:

The given bromoalkanes are isomeric compounds. S_N2 reactivity of these alkyl bromides is determined by the hindrance at C-atom, undergoing nucleophilic attack; the greater the steric hindrance, the slower will be the reaction.

Thus, from the structures of the substrates we predict the following sequence of increasing reactivity: A < C < B.

2. S_N2 reactivity decreases as 1° > 2° > 3° (because steric hindrance at the a -carbon increases in this sequence).

Thus, the sequence of S_N2 reactivity is A > C > B.

S_N2 reactivity of 1° alkyl halides decreases as the alkyl substituents on the carbon chain get closer to the α-carbon.

Thus S_N2 reactivity of unsubstituted 1° alkyl halide, A, is maximum. The reactivity of ϒ-substituted 1° alkyl halide, D, is somewhat less than that of A.

The reactivity of -substituted alkyl halides B and C is less than that of D. Out of B and C, the former is less reactive because its a -carbon becomes more crowded by the presence of two methyl groups on the β-carbon. Thus S_N2 reactivity decreases in the sequence: A > D > C > B.

Question 78. Out of C₆H₅CH₂Cland C₆H₅CHClC₆H₅, which is! more easily hydrolysed by aqueous KOH?
Answer:

In an aqueous solution, hydrolysis of phenyl-substituted methyl halides will occur preferably by the S_N1 pathway. Thus the reactivity of the given compounds will be determined by the stability of the corresponding carbocations. As (C₆H₅)₂C⁺H is more stable than C₆H₅C⁺H₂, C₆H₅ClC₆H₅ undergoes hydrolysis at a much faster rate than C₆H₅CH₂Cl.

More stable carbonation (greater no. of resonance structure)

Less stable carbonation (lesser no.of resonance structure)

Question 79. p -dichlorobenzene has higher m.p. and solubility than those of o -and m -isomers. Discuss.

Ans. The para-isomer, with a more symmetrical structure, can be packed more closely in the crystal lattice and thus has stronger intermolecular forces of attraction than those of the ortho- and meta-isomers.

Crystal lattice must be broken for the melting or dissolution of a compound. Thus, a large amount of energy is needed to melt or dissolve p -dichlorobenzene than ortho- & meta- dichlorobenzenes. Thus m.p. of para- isomer is greater and its solubility is lower than ortho- and meta- isomers.

Question 80. How the following conversions can be carried out?

1. Propene to propane-1-ol Ethanol to but-1-yne
2. 1-bromopropane to 2-bromopropane
3. Toluene to benzyl alcohol
4. Benzene to 4-bromonitrobenzene
5. Ethanol to propane nitrile
6. Benzyl alcohol to 2-phenyl ethanoic acid
7. Aniline to chlorobenzene
8. 2-chlorobutane to 3, 4-dimethyl hexane
9. 2-methyl-1-propene to 2-chloro-2- methylpropane
10. Ethyl chloride to propanoic acid
11. But-l-ene to n -butyliodide
12. 2-chloropropane to 1-propanol
13. Isopropyl alcohol to iodoform
14. Chlorobenzene to p -nitrophenol
15. 2-bromopropane to 1-bromopropane
16. Chloroethane to butane
17. Benzene to diphenyl
18. tert-butyl bromide to isobutyl bromide
19. Aniline to phenyl isocyanide

Answer:

Question  81. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:

In an aqueous solution, OH^– ion (obtained from KOH) acts as a strong nucleophile and causes nucleophilic substitution of alkyl chloride to give alcohol. In an alcoholic medium, alkoxide ion (RO^–) is formed to some extent and this acts as a strong base to cause an elimination reaction, thereby producing alkene as the major product.

Question 82. Primary alkyl halide C₄H₉Br (.A) reacted with alcoholic KOH to give compound (B). Compound (B) is reacted with HBr to give (C) which is an isomer of (A). When (A) is reacted with sodium metal it gives compound (D), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (A) and write the equations for all the reactions.
Answer:

There are two 1° alkyl halides with the molecular formula C₄H₉Br. These are:

So compound (A) is either n-butyl bromide or isobutyl bromide, n-butyl bromide reacts with metallic sodium (Wurtz reaction) to give n-octane.

So the structure of compound (A) will be such that it will not form n-octane on reaction with Na-metal. This shows that compound (A) is isobutyl bromide. Its reaction with Na-metal can be formulated as follows, thereby showing that compound (D) is 2,5 dimethylhexane.

The reaction of compound (A) with alcoholic KOH can be formulated as follows:

This shows that compound (B) is isobutene. Finally, the reaction of compound (B) with HBr can be formulated as follows, thereby indicating that the compound (C) is tert-butyl bromide [which is an isomer of compound (A)].

Question 83. What happens when—

1. N -butyl chloride is treated with alcoholic KOH,
2. Bromobenzene is treated with, Mg in the presence of dry ether,
3. Chlorobenzene is subjected to hydrolysis,
4. Ethyl chloride is treated with aqueous KOH,
5. Methyl bromide is treated with sodium in the presence of dry ether,
6. Methyl chloride is treated with KCN.

Answer:

Question 84. Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does the preparation of aryl iodides require the presence of an oxidising agent?
Answer:

The iodination of hydrocarbon (RH or ArH) is a reversible reaction. To shift the equilibrium to the forward direction, HI formed during the reaction should be removed by using an oxidising agent, e.g., HNO₃ or HIO₃.

Question 85. Out of o-and p-dibromobenzene which one has a higher melting point and why?
Answer:

Due to its symmetrical structure, the para-isomer fits in the crystal lattice better than the ortho-isomer. So p -dibromobenzene has higher m.p. than its ortho-isomer.

Question 86. Which of the compounds will react faster in the S_N1 reaction with the OH^– ion? CH₃ —CH₂—Cl or C₆H₅-CH₂-CI
Answer:

In the rate-determining step of the S_N1 reaction a carbocation intermediate is formed. Since C₆H₅C⁺H₂is more stable than CH₃C⁺H₂, therefore C₆H₅CH₂Cl will react faster than CH₃CH₂Cl in the S_N1 pathway.

Question 87. Why does iodoform have appreciable antiseptic properties?
Answer:

The antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

Question 88. Haloarenes and haloalkanes are less reactive than haloalkenes. Explain.
Answer:

Due to resonance, C—Xbond in haloarenes and haloalkenes have a partial double bond character, but it is a pure single bond in haloalkanes. Hence, the C—X bonds of haloarenes and haloalkenes are stronger than that of haloalkanes. Accordingly, haloarenes and haloalkenes are less reactive than haloalkanes.

Question 89. Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.
Answer:

The role of Lewis acid catalysts (FeCl₃, FeBr₃, etc.) in the preparation of aryl chlorides or bromides from arenes is to generate the electrophiles Cl⁺ or Br⁺ from the corresponding halogens. These electrophiles then react with arenes.

Question 90. Which of the following compounds 1 and 2 will not react with a mixture of NaBr and H₂SO₄? Explain why?

1. CH₃CH₂CH₂OH
2. 

Answer: 

Bromine produced by the interaction of NaBr and H₂SO₄ reacts with phenol but not with propanol.

2NaBr + 3H₂SO₄→2NaHSO₄ + Br₂ + SO₂ + 2H₂O

CH₃CH₂CH₂OH + Br₂→No reaction

Question 91. Which of the products will be the major product in the reaction given below? Explain.

Answer:

This addition reaction proceeds via the formation of a carbocation intermediate. Since 2° carbocation CH₃C⁺H — CH₃ is more stable than 1 ° carbocation CH₃CH₂CH₂, therefore, the major product is CH₃—CHI—CH₃.

Question 92. Why is the solubility of haloalkanes in water very low?
Answer:

For a haloalkane to dissolve in water, energy is required to overcome the attractive forces between the haloalkane molecules and break the H-bonds between H₂O molecules. Less energy is released when new attractions are set up between the haloalkane and the water molecules as these are not as strong as the original H-bonds in water. So the solubility of haloalkanes in water is very low.

Question 93. Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho-, para- directing or meta-directing.

Answer: 

The given molecule is a resonance hybrid of the following structures:

These resonance structures indicate that there are partial —charges in the ortho and para positions of the ring system of the actual molecule. Hence, the functional group present in the molecule is ortho para directing in electrophilic substitution.

Question 94. Classify the following compounds as primary, secondary, and tertiary halides.

1. 1-bromobut-2-ene
2. 4-bromopent-2-ene
3. 2-bromo-2-methylpropane

Answer:

Question 95. Compound ‘A’ with molecular formula C₄H₉Bris treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of the compound and KOH both.

1. Write down the structural formula of both compounds ‘A’ and ‘B’.
2. Out of these two compounds, which one will be converted to the product with an inverted configuration

Answer:

The rate of reaction of compound A (MF: C₄HaBr) with aq. KOH depends upon the concentration of the compound ‘A’ only. So the reaction occurs by the S_N1 pathway. This shows that the compound ‘A’ is f-butyl bromide[(CH₃)₃CBr].

Now, compound ‘B’ is an optically active isomer of compound ‘A’. So compound B must be 2-bromobutane. The rate of reaction of compound B with aq. KOH depends on the concentration of both substrate and nucleophile, so the reaction proceeds by S_N2 mechanism, causing inversion of configuration.

Question 96. Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, C₇H₈ is treated with Cl₃ in the presence of FeCl₃.
Answer:

From the high C: H ratio of compound A(MF: C-Hg), it appears that it is an aromatic compound and it must be toluene (C₆H₅CH₃). Since the —CH₃ group is ortho/para directing therefore, compound ‘A’ undergoes electrophilic substitution on reaction with Cl₂ in the presence of FeCl₃, to form a mixture of o-and p-chlorotoluenes.

Question 97. Identify the products A and B in the following reaction:

CH₃—CH₂—CH==CH—CH₃ + HCl → A + B

Answer:

Carbocation 2 (five hyperconjugative structures) is somewhat more stable than carbocation 1 (four hyperconjugative structures) and so the yield of product (B) is slightly greater than that of A.

Question 98. Which of the following compounds will have the highest melting point and why?

Answer:

Compound 2 has the most symmetrical structure because in this compound both the CH₃ groups and the Cl-atoms are para to each other. Due to such symmetry, compound II fits into crystal lattice better than other isomers (i.e. 1 and 2) and hence, it has the highest melting point.

Question 99. Write down the structure and IUPAC name for neo-pentyl bromide.
Answer:

Question 100. A hydrocarbon of molecular mass 72g-mol^-1 gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.
Answer:

Let the hydrocarbon be C_nH2_n+2. Its molecular mass = 12n + (2n + 2) = 14n + 2

∴ 14n + 2 = 72, or, n = 5 .

So the hydrocarbon is C₅H₁₂. It forms a single monochloro derivative and so all the twelve H-atoms are equivalent. Thus it must be neopentane, (CH₃)₄C. This structure is in accordance with the fact that the compound forms only one monochloride derivative and two dichloro derivatives.

Question 101. Name the alkene which will yield 1-chloro-1- methylcyclohexane by its reaction with HCl. Write the reactions involved.
Answer:

There are two possible structures for the alkene under consideration. These are A and B.

Question 102. Which of the following haloalkanes reacts with aqueous KOH most easily? Explain by giving a reason.

1. 1-bromobutane
2. 2-bromobutane
3. 2-bromo-2-methylpropane
4. 2-chlorobutane

Answer:

2-bromo-2-methylpropane is a 3° alkyl halide, which will react most readily with aqueous KOH because of its ability to form a stable 3° carbocation.

Question 103. Why can aryl halides not be prepared by the reaction of phenol with HCl in the presence of ZnCl₂?
Answer:

In phenol, the lone pair on the O-atom of the —OH group is involved in resonance with the benzene ring and so, the C —O bond has a partial double bond character. Thus, breaking of this bond leading to the formation of chlorobenzene is rather difficult.

Question 104. Which of the following compounds would undergo SN1 reaction faster and why?

Answer:

Compound ‘B’ undergoes S_N1 reaction faster than compound ‘A’, because the carbocation derived from the former is more stable than that derived from the latter.

Question 105. Allyl chloride is hydrolysed more readily than n-propyl chloride. Why?
Answer:

Allyl chloride shows high reactivity because of its ability to undergo hydrolysis via the S_N1 pathway invoking a more stable carbocation (stabilised by resonance) intermediate. On the other hand, carbocation derived from n-propyl chloride is not stabilised by resonance. Thus, allyl chloride is hydrolysed more readily than n-propyl chloride.

Question 106. Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?
Answer:

Grignard reagents are very susceptible to reaction with compounds containing active hydrogen. Thus they react very rapidly with H₂O (moisture) to give the corresponding alkane.

R —MgX + H-OH → R-H + Mg(X)OH Alkane

Question 107. How do polar solvents help in the first step in S_N1 mechanism?
Answer:

In the first step i.e., rate-determining step (r.d.s.) of the S_N1 reaction, the substrate undergoes ionisation to form a carbocation intermediate Such ionisation process is favoured in the presence of a polar solvent (i.e, solvent having high dielectric constant). Furthermore, especially the protic polar solvents stabilise the carbocation. thereby favouring the S_N1 mechanism.

Question 108. Write a test to detect the presence of a double bond in a molecule.
Answer:

Detection of double bond: 1. When Br₂ in CCl₄ is added j to the solution of an unsaturated compound in CCl₄, the orange colour of Br₂ disappears due to the formation of a colourless dibromo derivative.

2. Baeyer’s test; When a cold dilute alkaline solution of KMnO₄ is added to the aqueous solution or suspension of an unsaturated compound, the pink colour of KMnO₄ solution is discharged due to the formation of the; corresponding 1,2-diol.

Question 109. Diphenyls are a potential threat to the environment. How are these produced from aryl halides?
Answer:

By Fittig reaction: Diphenyls are produced by treating haloarenes with sodium metal in dry ether.

By Ullmann biaryl synthesis: By heating an iodoarene with Cu-powder in a sealed tube.

Question 110. What are the IUPAC names of the insecticide DDT and benzene hexachloride? Why is their use banned in India and other countries?
Answer:

DDT: 2,2-bis(4-chlorophenyl)-1,1,1-trichloroethane. Benzene hexachloride: 1,2,3,4,5,6-hexachlorocycIo- hexane. The use of these insecticides has been banned because they are not biodegradable. They enter the food chains of animals and get deposited and stored in their fatty tissues over a period of time which ultimately affects their reproductive systems.

Question 111. Elimination reactions (especially β-elimination) are as common as the nucleophilic substitution reaction in the case of alkyl halides. Specify the reagents used in both cases.
Answer:

Both elimination (especially β-elimination) and nucleophilic substitution reactions occur simultaneously and compete with each other. However, one reaction can be made to be predominated over the other by a proper choice of reagents and reaction conditions. Thus the use of strong and bulkier bases and high temperatures favor elimination reactions, whereas weaker and smaller bases and lower temperatures favour substitution reactions.

Question 112. How will you obtain monobromobenzene from aniline?

Answer:

Question 113. Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution:

Answer:

The introduction of electron-withdrawing —I and —R groups (e.g., NO₂) in the ortho and para positions of halobenzenes causes an increase in reactivity towards nucleophilic substitution because the stability of anionic intermediate increases by the presence of such groups. The higher the number of such groups at ortho and para positions with respect to the halogen atom, the higher the reactivity. Thus the reactivity of the given compounds follows the sequence; 3 > 2 > 1.

Question 114. Tert-butyl bromide reacts with aq. NaOH by the S_N1 mechanism while n-butyl bromide reacts by the S_N2 mechanism. Why?
Answer:

The reaction by the S_N1 mechanism is favoured by the ability of the substrate to form a stable carbocation intermediate. Since tert-butyl bromide is capable of forming a stable carbocation [(CH₃)₃ C⁺], therefore, it reacts with XaOH by the S_N1 mechanism. On the other hand, a reaction by the S_N2 mechanism is favoured in the absence of steric hindrance around a -carbon of the substrate. Thus, n-butyl bromide (a primary alkyl halide) reacts with NaOH by S_N2 mechanism.

Question 115. Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.
Answer:

This electrophilic addition reaction proceeds preferably via the formation of a more stable 3° carbocation, thereby forming tert-butyl chloride as the major product.

Question 116. How can you obtain iodoethane from ethanol when no other iodine-containing reagent except Nal is available in the laboratory?
Answer:

Question 117. 1-butanol reacts with NaBr in the presence of a cone. H₂SO₄ to form 1-bromobutane. However, the reaction does not take place in the absence of H₂SO₄. Explain.
Answers:

OH^– is a very poor leaving group, so it cannot be displaced by Bre; in the presence of H₂SO₄, the —OH group becomes converted into — OH₂ which is a much better leaving group hence, substitution occurs.

Question 118. In free radical chlorination, the tendency of substitution of various types of hydrogens follows the order: 3° > 2° > 1°. Explain.
Answers: Since the order of stability of the free radicals is 3° > 2° > 1°

Question 119. On free radical chlorination, 1-butene produces mainly 3-chlorobut-1-ene—why?
Answers: The allyl free radical (CH₂=CH — CHCH₃) is relatively more stable

Question 120. Why n-butane undergoes chlorination at 25 °C in the presence of light to form 72% 2-chlorobutanol and 28% 1-chlorobutane?
Answers:

1-chlorobutane/2-chlorobutanol =(no. of 1°H -atoms ÷ no. 2°H -atoms) × (reactivity of 1°H ÷ reactivity of 2°H) = (6/4) × (1/3.8) = (6/15.2)= 28%/72%

Question 121. Arrange in the order of increasing S_N1 reactivity and explain:

Answers: 1 < 2 < 4 < 3, because the order of leaving ability of the leaving groups is: I^– > Br^– > Cl^– and 3° substrate is more reactive than 1° substrate

Question 122. Arrange in the order of increasing S_N2 reactivity and explain the order:

CH₃CI, CH₃Br, CH₃CH₂Cl, (CH₃)₂CHCl

Answers: (CH₃)₂CHCl < CH₃CH₂Cl < CH₃Cl < CH₃Br

Question 123. Solvolysis of (A) in ethanol takes place readily while the solvolysis of (B) does not virtually take place—why?
Answers: Because the carbocation obtained from A i.e., is aromatic and very stable but the carbocation obtained from B, i.e.,, is anti-aromatic and very unstable

Question 124. Which one of each pair will not undergo S_N2 reaction and why?

1. (CH₃)₃CCH₂Cl, CH₃CH₂Cl

Answers:

1. Due to severe steric hindrance caused by 3 methyl groups at the p-position, the first one is, in fact, inactive;
2. Due to its cage-like structure, a backside attack on the second chloride is not possible so it is unreactive

Question 125. After standing in aqueous acid, (R)-2-butanol is found to have lost its optical activity. Explain this observation.
Answers: An achiral carbocation is formed which undergoes attack by H₂O from both sides equally well to form (±) -2-octanol and so, the solution turns optically inactive

Question 126. Which one of each pair of compounds will undergo S_N1 reaction readily and why?

1. CH₂CH —CH₂CH₂Cl,
2. CH₃CH=CHCH₂Cl
3. CH₃OCH₂Cl, CH₃OCH₂CH₂Cl

Answers:

1. The second one as it forms resonance-stabilised allyl cation (CH₃CH =CHCH₂);
2. First, it forms a resonance-stabilised carbocation

Unit 10 Haloalkanes And Haloarenes Long Questions And Answer

Question 1. Explain the following order of dipole moments of chloromethanes: CH₃Cl > CH₂Cl₂ > CHCl₃ > CCl₄.
Answer:

• The CCl₄ molecule has a highly symmetrical tetrahedral configuration. So, the resultant of three C—Cl bond moments cancels the fourth C—Cl bond moment. Hence, the molecule possesses no net dipole moment = OD). The dipole moment of CH₂Cl₂ in which two C — Cl bonds make an angle of nearly 109°28/ is expected to be larger than that of CH₃Cl because the resultant of two C—Cl bond moments must be greater than one C—Cl bond moment.
• Also, the dipole moment of CH₃Cl is expected to be equal to that of CHCl₃ because the moment of a — CCl₃ group is equal to that of a C—Cl bond and the moment of a — CH₃ group is equal to that of a C —H bond. Thus, the expected order of dipole moment is: CH₂Cl₂ > CH₃Cl ≈ CHCl₃ > CCl₄
• This order, however, does not agree with the experimental dipole moment values. This anomaly may be explained by considering the moment acting in the opposite direction induced by each C—Cl dipole in the other. Since there is only one C—Cl bond in CH₃Cl, the opposing induced moment is absent in it. So, it possesses the largest dipole moment.
• In CH₂Cl₂, there are two C—Cl bonds. Thus, there is an opposing induced moment that partly cancels the resultant moment of the two C—Cl bonds. So, its dipole moment is smaller than CH₃Cl. In CHCl₃, there are three C —Cl bonds.
• So, the magnitude of the opposing induced moment is relatively large which considerably reduces the resultant of the three C—Cl bond moments. Therefore, its dipole moment is smaller than that of CH₂Cl₂ and much smaller than CH₃Cl. So, the actual order of dipole moments is CH₃Cl > CH₂Cl₂ > CHCI₃ > CCl₄.

Question 2. Assign R/S Designation to the chiral centres:

Answer:

1. S
2. R
3. R
4. S

Question 3. Which one of the following will react with the AgNO₃ solution to form a white precipitate:

3. CH₂=CH-Cl

Answer:

Question 4. Label the following pairs of structures as enantiomers, diastereoisomers or homomers:

Answer:

1. Enantiomers
2. Homomers (i.e., identical structures)
3. Diastereoisomers.

Question 5. Discuss the nature of the C—X bond in the haloarenes.
Answer:

In haloarenes, the sp²-hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C—Xbond more tightly than sp³-hybridised carbon in haloalkane with a lesser s-character.

Thus, the C—X bond of haloarenes is less polar than that of haloalkanes.

The polarity of the C—Xbond of haloarenes is further decreased due to the involvement of a lone pair of electrons of the halogen atom in delocalisation with the n-electron system of the aromatic ring.

This is supported by the observation that the dipole moment of CH₃Cl(1.83 D) is greater than that of chlorobenzene (1.69 D).

Question 6. Which one of the following methods is suitable for the preparation of fert-butyl ethyl ether and why?

Answer:

• The S_N2 reaction that occurs between an alkoxide and an alkyl halide (Williamson synthesis) is very susceptible to steric hindrance. Since a tertiary alkyl chloride is used in the method, due to severe steric hindrance at the reacting centre, the reaction does not take place.
• Hence, this method is not suitable for the preparation of fert-butyl ethyl ether. On the other hand, a primary alkyl chloride used in this method undergoes SN2 reaction smoothly due to much less steric hindrance at the reacting centre. Therefore, this method is suitable for the preparation of tert-butyl ethyl ether.

Question 7. If (+)-2-methyl-I-butanol rotates the plane of polarised light towards the right-hand side (in the clockwise direction) and the angle of rotation is 5.90°, indicate the angle of rotation and the direction towards which the plane of polarised light will rotate by (-) -2-methyl-1-butanol.
Answer:

5.90° towards the left (in the counterclockwise direction).

Question 8. Identify the alkene A obtained in the following reaction and predict whether it is a cis- or a trans-isomer:

Answer:

It is an E2 reaction in which trans-elimination occurs almost throughout the reaction. Therefore, in this reaction, H and Cl must be eliminated from the opposite sides and as a result, tram-1,2-diphenyl propane is obtained. The stereochemical course of the reaction may be shown as follows:

Question 9. Arrange the following bromides in the order of increasing S_N1 reactivity and example the order:
Answer:

The higher the stability of the carbocation obtained on dissociation of an alkyl halide, greater is the reactivity of the alkyl halide towards S_N1 reaction. Dissociation of (1) leads to formation of a stable aromatic ion [(4n + 2)π electron, n = 1].

Dissociation of (2) leads to formation of a relatively less stable carbocation (resonance stabilised by the double bond).

Dissociation of (3) is expected to form a very unstable antiaromatic carbocation (4nn electron, n = 1). Hence, the dissociation does not actually take place in this case.

Therefore, the order of increasing SN1 reactivity of the bromides is: 3 < 2 < 1.

Question 10. Which one of the following two methods is appropriate for the preparation of neopentyl chloride (Me₃CCH₂Cl) and why?

⇒ \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{OH} \stackrel{\mathrm{HCl}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\left(\mathrm{CH}_3\right)_4 \mathrm{C} \stackrel{\mathrm{Cl}_2 / h v}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{Cl}+\mathrm{HCl}\)

Answer:

All the H-atoms of neopentane (Me₄C) are equivalent and the intermediate radical does not undergo rearrangement.

So, the second method leads to the formation of neopentyl chloride as the only product. This method is, therefore, appropriate for the preparation of neopentyl chloride.

On the other hand in the first reaction, the intermediate carbocation is highly unstable and readily rearranges to a more stable 3° carbocation. So, 2-chloro-2-methylbutane is obtained almost exclusively instead of neopentyl chloride. Therefore, the first method is not appropriate for the preparation of neopentyl chloride.

Question 11. Identify the isomeric alkanes (molecular formula = C₅H₁₂) which produce

1. A single monochloride,
2. Three isomeric monochlorides and four isomeric monochlorides on photochemical chlorination.
   

Answer:

The alkane which produces only one monochloride must have equivalent H-atoms. Therefore, the alkane is neopentane or 2,2-dimethylpropane [(CH₃)₄C].

The alkane which produces three monochlorides must have three types of non-equivalent hydrogen atoms and hence, the alkane is pentane . Substitution of H-atoms labeled a, b, and c lead to the formation of three monochlorides.

The alkane which produces 4 monochlorides must have four types of non-equivalent hydrogens, and hence, the alkane is 2-methylbutane . Substitution of H-atoms labelled a, b, c, and d lead to the formation of 4 monochlorides.

Question 12. Which one of the following is a meso-compound?

Answer: Compound 2

Question 13. Arrange the following compounds in order of increasing SN2 reactivity and explain: 

1. 2-bromo-2-methybutane, 1-bromopentane, 2-bromopentane;
2. Ethyl bromide,isobutyl bromide, n-propyl bromide, neopentyl bromide

Answer:

The S_N2 reactivity of alkyl halides depends on the steric hindrance at the reacting centre. The tendency of nucleophilic attack increases with a decrease in steric crowding.

The steric hindrance in S_N2 transition state progressively increases from primary (1°) halide to tertiary (3°) halide. Thus, the S_N2 reactivity of these compounds follows the order: 1-bromopentane > 2-bromopentane > 2-bromo-2-methylbutane.

The S_N2 attack on the halogenated carbon is progressively hindered with increase in the number of alkyl groups at the β-carbon atom.

There are no alkyl groups at the _N-carbon atom of ethyl bromide while n-propyl bromide contains one, isobutyl bromide contains two and neopentyl bromide contains three alkyl groups at the _N-carbon atom. Therefore, the S_N2 reactivity of these bromides decreases in the order: ethyl bromide > n-propyl bromide > isopropyl bromide > neopentyl bromide.

Question 14. Which one of the pairs undergoes hydrolysis by aqueous KOH readily and why?

1. C₆H₅CH₂Cl and C₆H₅CHClC₆Hg
2. (CH₃)₃CCl and (CD₃)₃CCl

Answer:

1. Both the halides undergo alkaline hydrolysis by S_N1 mechanism because the carbocations obtained on their ionisation are stabilised by resonance.

Water is a polar solvent, it also creates an appropriate environment for their easy ionisation. The higher the stability of the carbocation, higher is the V reactivity of the halide.

The carbocation (C₆H₅C⁺HC₆H₅) obtained from C₆H₅CHClC₆H₅ is stabilised by resonance involving two phenyl groups while the carbocation (C₆H₅C⁺H₂) obtained from C₆H₅C⁺H₂Cl is stabilised by resonance involving only one phenyl group. Hence, the carbonation C₆H₅C⁺HC₆H₅ is relatively more stable than the carbocation C₆H₅C⁺H₂. Thus, hydrolysis of C₆H₅CHClC₆H₅ by aqueous KOH is much faster than that of C₆H₅CH₂Cl.

2. These 3° alkyl halides undergo alkaline hydrolysis by S_N1 mechanism because the 3° carbocation obtained on their ionisation are stabilised by inductive as well as hyperconjugative effect.

Also, water which is highly polar and a protic solvent, makes the ionisation favourable by stabilising the resulting ions. The high dielectric constant of water also favours ionisation by decreasing the electrostatic force between the incipient ions.

Since D is more electron-releasing than H, therefore, —CD₃ is expected to be a better electron-donating (+1) group than —CH₃. Thus, (CD₃)₃C⁺ is expected to be more stable than (CH₃)₃C⁺. But actually (CH₃)₃C⁺ Is relatively more stable than (CD₃)₃C⁺. This is because the hyperconjugative stabilisation of (CH₃)₃C⁺ involving a relatively weaker C—H bond is much greater than the hyperconjugative stabilisation (CD₃)₃C⁺ involving relatively stronger C—D bond. Hence, due to formation of more stable carbocation, the hydrolysis of (CH₃)₃CCl (solvolysis) is much faster than that of (CD₃)₃CCl.

Question 15. Explain why the following general reaction takes place readily in the presence of a small amount of Nal.

R — Cl + R’ —ONa→ROR’ + NaCl

Answer:

Although R’O^– is a good nucleophile, the given S_N2 reaction occurs at a slower rate as Cl is not a good leaving 0 group. In presence of I^– , the reaction actually takes place in two steps:

I^– is a stronger nucleophile (more polarisable) than R’O^–. It readily displaces Cl^– from RCl. I^– is also a good leaving group (weaker base and weaker C—I bond). So, it is
readily displaced by RO^– and once again reacts in the first step. Thus, in the presence of I^– as a catalyst, the overall reaction occurs readily (This is an example of a nucleophilic catalysis).

Question 16. An alkyl chloride (RCl) reacts with aqueous KOH to form an alcohol predominantly while it reacts with alcoholic KOH to form an alkene predominantly. Explain these observations.
Answer:

• A stronger base and a less polar solvent favours elimination (E2) more than substitution (S_N2). In such a condition, an alkene is formed predominantly. Alcohol is a much less polar solvent than water and the base OEt^– (formed by the reaction between EtOH and OH^–) present in alcoholic medium is stronger than OH^–.
• So, alkyl chloride reacts with alcoholic KOH to form an alkene predominantly by an E2 reaction. Since OH^– is a weaker base than OEt^– and water is more polar than alcohol, an alkyl chloride reacts with aqueous alkali to form an alcohol, predominandy by an S_N2 reaction.

Question 17. Predict the products (A and B) in the following reactions and designate them as R or S. Mention the stereochemical course (inversion or retention of configuration) involved in each case.

Answer: 1. The reaction occurs as follows:

In this reaction, none of the bonds to the asymmetric carbon is broken. Therefore, the product has the same relative configuration of the groups around the asymmetric carbon as that of the reactant.

Hence, the reaction proceeds with the retention of configuration. The path, 1→2→3 describes a clockwise motion and so the stereocentre is said to have R configuration. Therefore, the compound A is (R)-1-chloro-2-methylbutane.

2. The reaction occurs as follows:

It is an S_N2 reaction and so it proceeds with inversion of configuration at the asymmetric centre. The path, 1→2→3 describes a counterclockwise motion and so, the stereocentre is said to have S configuration. Therefore, the compound B is (S)-2-octanol.

Question 18. No correlation exists between the (R) and (S) designation and the direction of rotation of plane-polarised light. Justify the statement.
Answer:

The rotation of a compound, [(+) or (-)] is something that we measure in the laboratory with a polarimeter and it depends on how the molecule interacts with light. On the other hand, the (R) and (S) designation is our own artificial way of describing how the atoms or groups are arranged in space around a chiral centre, i.e., the configuration of a particular chirality centre.

Therefore, no necessary correlation exists between the (R) and (S) designation and the direction of rotation of plane-polarised light. Dextrorotatory or (+)- compounds may have R or S configuration. Similarly, levorotatory or (-)-compounds may have an R or S configuration.

Example:

Question 19. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with NaOEt in ethanol and identify the major product obtained in each case.

1.  3-bromo-2, 2, 3-trimethylpentane
2. 2-chloro-2-methylbutane
3. 1-bromo-l-methylcyclohexane

Answer:

According to the Saytzeff rule, alkyl halides undergo dehydrohalogenation to give more highly substituted alkene as the major product.

Thus, the halides (1), (2) and (3) undergo dehydrohalogenation in the presence of NaOEt/EtOH to yield 3,4,4-trimethylpent-2-ene(1), 2-methylbut-2-ene(3) and 1- methylcyclohexane (5) respectively as major products.

Question 20. How will you carry out the following transformations

1. Propene→propan-l-ol
2. Ethanol→but-2-yne
3. 1-bromopropane→2-bromopropane
4. Toluene→ benzyl alcohol
5. Benzyl alcohol→2-phenyl- ethanoic acid
6. Aniline→chlorobenzene
7. But-1- ene→n-butyl iodide
8. Propan-2-ol→acetylene
9. Chlorobenzene→p-nitrophenol
10. Benzene →biphenyl
11. Chlorobenzene → DDT
12. Ethanol→ but-l-yne
13. Propene→propyne
14. But-l-ene→ but-2-ene

Answer:

Question 21. (+)-2-iodobutane loses its optical activity when Nal is added to its acetone solution. Explain.
Answer:

(+)-2-iodobutane reacts with Nal in acetone to form (± ) – 2-iodobutane, which leads to racemisation. Thus the solution becomes optically inactive.

• Each S_N2 attack on (+)-2-iodobutane by I^– occurs with inversion of configuration resulting in the formation of (-) -2- iodobutane. As the concentration of the (-) -enantiomer in the reaction mixture increases, its tendency to react with l^– forming a (+)-enantiomer increases.
• At a certain time, the mixture contains equal amounts of (+) and (-)-2- iodobutane. Since both the enantiomers react with I^– at the same rate (because the activation energy is the same), their composition remains unchanged.
• As the rotation caused by the molecules of one enantiomer is exactly cancelled by equal and opposite rotation caused by the same number of molecules of the other enantiomer, the mixture is optically inactive.

Question 22. Identify the major product:

Answer:

Question 23. What is the main difference in using RS and RO as nucleophile in S_N2 reaction?
Answer:

RS^– is a weak base, while RO^– is a strong base. Therefore, when RS^– is used as a nucleophile in an S_N2 reaction, the elimination (E2) reaction is not a major problem. However, when RO^– is used as a nucleophile, an elimination reaction is a major problem.

Question 24. Arrange the halide ions in order of increasing reactivity towards CH₃Br(S_N2) in a gaseous medium and explain the order.
Answer:

Since the solvation effect is absent in the gaseous medium, the smaller anion (in which the negative charge is concentrated in a small volume) is more reactive and acts as a stronger nucleophile than the larger anion.

Therefore, the reactivity of halide ions towards CH₃Br in gaseous medium decreases in the order: F^– > Cl^– > Br^– > I^– (their size increases in the order: F^–< Cl^–< Br^–< I^–).

Question 25. S_N1 reactions are extremely uncommon in gas phase-why?
Answer:

In the rate-determining step (the 1st step) of an S_N1 reaction, a carbocation and an anion are obtained. The more the ions are solvated, the more they become stable and the easier is the cleavage of the C —L bonds, hence, faster is the S_N1 reaction. Solvent is absent in the gas phase and so, there is no question of stabilization of ions by solvation. For this reason, S_N1 reactions are extremely uncommon in the gas phase.

Question 26. Which one of the following compounds undergoes S_N1 reaction readily?

Answer:

Compound I undergoes S_N1 reaction readily, even though both of these two compounds axe tertiary substrates. This is because the carbocation obtained from I is stabilized by resonance, while the carbocation obtained from 2 is not.

Question 27. Explain why the rate of the following S_N2 reaction increases on increasing the polarity of the solvent:

⇒ \(\mathbf{R}_3 \mathbf{N}:+\mathbf{R}-\mathbf{X} \rightarrow \mathbf{R}_3 \stackrel{\oplus}{\mathbf{N}}-\mathbf{R} \mathbf{X}^{\ominus}\)

Answer:

Since the reactants are neutral, therefore, the transition state of this reaction is relatively more polar than the reactants.

When the polarity of the solvent increases, the transition state becomes relatively more stable as compared to the reactants. As a result, the activation energy of the reaction (E_a) decreases and the reaction rate increases.

Question 28. Explain the following observation:

Answer:

The substrate undergoes an S_N1 reaction readily because the C —Cl bond cleavage occurs readily in the presence of the Lewis acid Ag⁺ to produce a very stable aromatic carbocation (the cyclopropenyl cation) This cation then undergoes nucleophilic attack by CH₃COO^– at the three carbons to give 33.33% of I and 66.66% of II. This occurs because the three resonance structures are equivalent.

Question 29. Explain why CH₃Cl undergoes hydrolysis at a faster rate in the presence of Nal.

The hydrolysis of methyl chloride in aqueous medium takes place at a much slower rate because H₂O is a weak nucleophile (because it is a neutral nucleophile and the attacking oxygen atom is highly electronegative and less polarisable) and Cl^– is not a very good leaving group (because it is not a very weak base and the C—Cl bond is not very weak).

The iodide ion (I^–) catalyses the hydrolysis of CH₃Cl. Due to high polarisability and low solvation energy, I^– is a good nucleophile.

Again, for its weak basicity and low bond energy of the C—I bond, I^– is a good leaving group. For this dual ability to attack and depart readily, I^– can act as an effective catalyst in substitution reactions. In the presence of Nal, the hydrolysis of CH₃Cl takes place by the following two steps—

Each of these two reactions [(2) and (3)] takes place at a faster rate than the reaction (1) because I^– is a better nucleophile than H₂O and it is a better-leaving group than Cl^–.

As a result, the overall hydrolysis occurs at a faster rate in the presence of Nal. It is a case of nucleophilic catalysis.

Question 30. When dilute acid is added to optically active (R)-2- octanol and the mixture is allowed to stand for some time, the solution becomes optically inactive. Explain this observation.
Answer:

In the presence of acid, the —OH group of (R)-2-butanol undergoes protonation and then C—O bond dissociation occurs to form an achiral 2° carbocation (an S_N1 process).

The dissociation occurs smoothly because H20 is a very good leaving group. Nucleophilic attack on the carbocation by water then takes place from both sides with equal facility to yield equimolar amounts of (R) and (S)-2-butanol having opposite rotation.

It is for this reason, that optically active (R)-2-octanol becomes optically inactive due to the formation of a racemic mixture when allowed to stand with dilute acid.

Question 31. Some alkyl halides undergo substitution whereas some undergo elimination reaction on treatment with bases. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer:

Primary and secondary alkyl halides prefer to undergo substitution reaction by S_N2 mechanism. Thus n -butyl chloride and sec-butyl chloride react with aqueous NaOH to form the corresponding alcohols as the substitution product. On the other hand, tert- alkyl halides prefer to undergo elimination reaction (mainly by E2 mechanism). Thus, tert-butyl bromide reacts with ethanolic KOH to form isobutene as the predominant product.

Question 32. Some halogen-containing compounds are useful in daily life. Some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to a great extent. Name the class of these halocompounds. In your opinion, what should be done to minimise harmful effects of these compounds?
Answer:

A number of polyhalogen compounds, such as dichloromethane (CH₂Cl₂), chloroform (CHCl₃), iodoform (CHI₃), carbon tetrachloride (CCl₄), freons, and DDT have a variety of applications in our daily life. Freons are mostly used for aerosol propellants, refrigeration, and air conditioning purposes. Since freons (chlorofluoro compounds of methane and ethane) are highly stable compounds, they eventually go to the atmosphere from where they diffuse unchanged into the stratosphere. In the stratosphere, freons are able to initiate radical chain reactions that can upset the natural ozone balance causing depletion of ozone layer.

Thus more UV rays reach the earth which, in turn, damage the flora and fauna. To minimise the harmful effects of freons, hydrochlorofluorocarbons (HCFCs), such as CHClF₂ (called freon-22), CHCl₂F, and hydrofluorocarbons (HFCs), such as FCH₂CH₂F (HFC-152) may be used as substitutes for chloroflurocarbons. Due to the presence of C—H bonds, such compounds are generally destroyed at lower altitudes before they reach the stratosphere.

Question 33. Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides? How can we enhance the reactivity of aryl halides?
Answer:

In haloarenes, the sp²-hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C—Xbond more tightly than sp³ can be used for chlorination hybridised carbon in haloalkane with less s-character. Thus the C—X bond of haloarenes is less polar than that of haloalkanes. Polarity of C—Xbond of haloarenes is further decreased by the involvement of lone pair of electrons of halogen atom in delocalisation with the π- electron system of the aromatic ring.

Due to such delocalisation, the C —Xbond in haloarenes has partial double bond character, thereby making the C—Xbond stronger. In a nucleophilic substitution reaction, the approaching nucleophile is also repelled by the π-electron system of the aromatic ring. Thus aryl halides are less reactive towards nucleophilic substitution reaction than alkyl halides. The reactivity of aryl halides can be enhanced by the introduction of electron- withdrawing —R groups (e.g., —NO₂, —CN) at ortho or para positions with respect to the halogen atom. In the presence of such groups, the stability of anionic intermediate increases, and hence, the reaction rate increases.

Unit 10 Haloalkanes And Haloarenes Multiple Question And Answer

Question 1. By passing excess Cl₂(g) in boiling toluene, which one of the following compounds is exclusively formed—

Answer: 4.

Question 2. Under identical conditions, the S_N1 reaction will occur most efficiently with—

1. Tert-butyl chloride
2. 1-chlorobutane
3. 2-methyl-l-chloropropane
4. 2-chlorobutane

Answer: 1.

S_N1 reaction proceeds via the formation of a carbocation intermediate. Thus more stable the carbocation, the more reactive is the parent alkyl halide towards the S_N1 reaction. On removal of Cl^– ion tert-butyl chloride gives 3° carbocation which is stabilised by 9 hyperconjugative H atoms. Therefore, tert-butyl chloride is the most reactive. Hence, the S_N1 reaction will occur most efficiently with it.

Question 3. The best method for the preparation of 2,2-dimethyl butane is via the reaction of—

1. Me₃CBr and MeCH₂Brin Na ether
2. (Me₃C)₂2CuLiand MeCH₂Br
3. (MeCH₂)₂CuLiand Me₃CBr
4. Me₃CMgI and MeCH₂I

Answer: 2.

→ \(\left(\mathrm{Me}_3 \mathrm{C}\right)_2 \mathrm{CuLi}+\mathrm{MeCH}_2 \mathrm{Br} \stackrel{\text { dry ether }}{\longrightarrow} \mathrm{Me}_3 \mathrm{C}-\mathrm{CH}_2 \mathrm{Me}\)

Question 4. 2-methylpropane on monochlorination under photochemical condition gives—

1. 2-chloro-2-methylpropane as major product
2. 1:1 mixture of l-chloro-2-methylpropane and 2-chloro-2-methylpropane
3. l-chloro-2-methylpropane as a major product
4. 1:9 mixture of l-chloro-2-methylpropane and 2-chloro-2-methylpropane.

Answer: 1.

Thus, 1-chloro-2-methylpropane is a major product at normal temperatures under photochemical conditions.

Question 5. The compound that will have a permanent dipole moment among the following is—

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

Answer: 1.

The structures of the given compounds along with the direction of bond moments are given below:

CCl₄ and C₂Br₂ are symmetrical molecules. Hence in CCl₄, the resultant of two C —Cl bond moments cancels the resultant of the other two C—Cl bond moments. Again, in C₂Br₂, two C—Br bond moments cancel out each other.

In the case of frans-dichloroethene, two C —Cl bond moments are cancelled by each other. Again, two C— H bond moments are cancelled by each other. In CH₂Cl₂, the resultant of two C—Cl bond moments is reinforced by the resultant of two C—H bond moments. Therefore CH₂Cl₂ (dipole moment =1.62D) has a permanent dipole moment.

Question 6. The suitable reagent for nuclear iodination of aromatic compounds is—

1. kI/CH₃COCH₃
2. I₂/CH₃CN
3. kI/CH₃COOH
4. I₂/HNO₃

Answer: 4.

Iodobenzene cannot be prepared by direct iodination of the aromatic ring because iodine is least reactive among halogens, the reaction is reversible and the HI formed reduces iodobenzene to benzene and I₂.

But, on addition to HNO₃, HI produced is oxidised to I₂. As a result, the reaction proceeds in the forward direction to produce iodobenzene.

Question 7. Predict the product —

Answer: 3.

Br is directly attached with benzene ring, for which it cannot undergo substitution.

Question 8. The Major product of the reaction is—

Answer: 3.

In this reaction, 3° carbocation which is formed as an intermediate, is stabilised by resonance. The extent of stability of this tertiary carbocationic centre is higher than that of the primary carbocationic centre. Hence, Bre ion (nucleophile) attacks the 3° carbocationic centre predominantly. As a result, the product is the major product.

Question 9. Choose the correct statement(s) among the following—

Answer:  2,4

Question 10. Identify X in the following sequence of reactions—

Answer: 2.

Question 11. The major product(s) obtained from the following reaction of 1 mol of hexadeuterobenzene is/are—

Answer: 1.

On addition to water in the reaction medium, the precipitate of bromodeuterobenzene is separated out.

Question 12. The reaction sequence given below gives the product it.

The structure of the product R is—

Answer: 4.

Question 13. The number of possible organobromine compounds which can be obtained in the allylic bromination of 1- butene with N-bromosuccinimide is—

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

Answer: 4.

Question 14. Phenol is heated with a mixture of KBr and KBrOs. The major product is—

1. 3-bromophenol
2. 4-bromophenol
3. 2,4,6-tribromophenol
4. 2-bromophenol

Answer: 3.

Question 15. Which branched open-chain isomer of the hydrocarbon having molecular mass 72 u produces only one mono-substituted alkyl halide—

1. Pentane
2. Neopentane
3. Isohexane
4. Neohexane

Answer: 2.

Among the given four compounds, pentane has no branched open chain structure. In the case of neohexane, molecular mass exceeds 72u. All the H atoms present in neopentane are equivalent while all the H atoms present in isohexane are not equivalent. Therefore neopentane can produce only one mono-substituted alkyl halide.

Question 16. Compound (A) C₈H₉Br gives a white precipitate when warmed with alcoholic AgNO₃. Oxidation of A gives an acid (B) C₈H₆O₄, which is easily heated. The compound (A) is—

Answer: 4.

Question 17. In S_N2 reactions, the correct order of reactivity for the following compounds :

CH₃CI, CH₃CH₂Cl, (CH₃)₂CHCland(CH₃)₃CCl is—

1. (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl > (CH₃)₃CCl
2. CH₃Cl > (CH₃)₂CHCl > CH₃CH₂Cl > (CH₃)₃CCl
3. CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl
4. CH₃CH₂Cl > CH₃Cl > (CH₃)₂CHCl > (CH₃)₂CCl

Answer: 3.

Steric hindrance gets increased by increase in the number of alkyl groups on a -carbon. As a result, attack of nucleophile from backside to a -carbon in S_N2 reaction gets hampered. Thus, the order of increasing S_N2 reactivity is—

CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl

Question 18. The compound obtained when 1,1,1-trichloroethane is heated with silver dust is—

1. 2-butene
2. acetylene
3. ethene
4. 2-butyne

Answer: 4.

Question 19. The absolute configuration of is—

1. (2R, 3S)
2. (2S, 3R)
3. (2S, 3S)
4. (2R, 3R)

Answer: 2.

Question 20. The major product obtained in the following reaction is—

1. (+)C₆H₅CH(O^t Bu)CH₂C₆H₅
2. (-)C₆H₅CH(O^t Bu)CH₂C₆H₅
3. (±)C₆H₅CH(O^t Bu)CH₂C₆H₅
4. C₆H₅CH=CHC₆H₅

Answer: 4.

Question 21. The increasing order of reactivity of the following halides for S_N1 reaction is—

1. (3) < (2) < (1)
2. (2) < (1) < (3)
3. (1) < (3) < (2)
4. (2) < (3) < (1)

Answer: 2.

Question 22. The major product of the following reaction is—

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

Answer: 2.

A benzylic carbocation is stabilised by resonance with the n -electrons of benzene ring and the extent of resonance stabilisation is further increased by the presence of electron-donating —OCH₃ group in the para- position of a benzene ring. We know, 2° carbocation is more stable than 1° carbocation.

So the order of stability of die carbocations is benzylic carbocation > 2° carbocation > 1° carbocation. The feasibility of the S_N1 reaction is directed. proportional to the stability of carbocation. Hence, the increasing order of reactivity of the given halides for S_N1 reaction is (2) < (1) < (3).

Question 23. The increasing order of reactivity of the following halides for S_N1 Reaction is—

Answer:

Question 24. Consider the reactions: 

Mechanism of reactions (1) and (2) are respectively—

1. S_N1 and S_N2
2. S_N1 and S_N1
3. S_N2 and S_N2
4. S_N2 and S_N1

Answer: 3.

Bodi the reactions follow S_N2 mechanism. If they follow S_N1 mechanism, then the products obtained in both cases will be

Question 25. The IUPAC name of the compound is—

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

Answer: 1.

Question 26. Which of the following compounds will undergo racemisation when hydrolysed with a solution of KOH—

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

Answer: 4. 1 and 4

Question 27. What products are formed when the following compounds are treated with Br₂ in the presence of FeBr₃

Answer: 3.

Question 28. In S_N1 reaction on chiral centres, there is— 

1. 100% racemisation
2. Inversion more than retention leading to partial recemisation
3. 100% retention
4. 100% inversion

Answer: 2.

In S_N1 reaction, frontside attack by nucleophiles is not feasible till the leaving group is completely removed from the reactant. Thus, backside attack occurs to a greater extent. Hence, partial racemisation occurs.

Question 29. An alkene reacts with HCl in accordance with Markownikoff’s rule to give 1-chloro-1- methylcyclohexane. The alkene is—

Answer: 3.

Question 30. For the following reactions:

Which of the following statements is correct—

1. Is substitution, 2 and 3 are addition reactions
2. 1 and 2 are elimination reactions and 3 is addition reaction
3. 1 is elimination, 2 is substitution and 3 is addition reaction
4. 1 is elimination, 2 and 3 are substitution reactions

Answer: 3. 1 is elimination, 2 is substitution and 3 is addition reaction

Question 31. Consider the reaction:

This reaction will be the fastest in—

1. Water
2. Ethanol
3. Methanol
4. N, N’ -dimethylformamide (DMF)

Answer: 4.

For the given reaction, polar aprotic solvent (PMF) is to be used to make the reaction faster.

Question 32. Which of the following can be used as the halide component for Freidel-Crafts reaction—

1. Isopropyl chloride
2. Chlorobenzene
3. Bromobenzene
4. Chloroethene

Answer: 1. Isopropyl chloride

Question 33. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction—

1. H₃C—CH₂—CH₂Br
2. H₂C—CH₂—CH₂
3. H₃C—CH₂—CH₂OH
4. H₂O=C=O

Answer: 4.

Question 34. In the given reaction:

The product of P is 

Answer: 4.

Question 35. Identify A and predict the type of reaction-

Answer: 4.

Question 36. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. A is—

1. CH₄
2. CH≡CH
3. CH₃—CH₃
4. CH₂=CH₂

Answer: 1.

Question 37. The compound C₇H₈ undergoes the following reaction:

→ \(\mathrm{C}_7 \mathrm{H}_8 \stackrel{3 \mathrm{Cl}_2 / \Delta}{\longrightarrow} A \stackrel{\mathrm{Br}_2 / \mathrm{Fe}}{\longrightarrow} B \stackrel{\mathrm{Zn} / \mathrm{HCl}}{\longrightarrow} C\)

The product C is—

1. P-bromotoluene
2. M-bromotoluene
3. 3-bromo-2,4,6-trichlorotoluene
4. O-bromotoluene

Answer: 2.

Question 38. An aromatic compound C₇H₆C₂(A), gives AgCI on boiling with alcoholic Ag\02 solution and yields C₇H₇OCl on treatment with sodium hydroxide. A on oxidation gives monochlorobenzoic acid. The compound A is—

Answer: 1.

Question 39. Which of the following compounds are optically active—

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

Answer: 2. 2 and 3

Question 40. Which of the following is a chiral compound—

1. Hexane
2. N-butane
3. Methane
4. 2,3,4-trimethyl hexane

Answer: 4.

A chiral compound contains one or more chiral carbon atom(s). Chiral carbon atoms are usually bonded to four different atoms or groups of atoms.

Question 41. Which of the following compounds is not chiral—

1. 1-chloropentane
2. 2-chloropentane
3. l-chloro-2-methylpentane
4. 3-chloro-2-methylpentane

Answer: 1.

Here, no asymmetric carbon atom is present. Here it is not chiral.

Question 42. Which one of the following does not give a white precipitate with acidified silver nitrate solution—

1. C₆H₆Cl
2. CH₂=CH—Cl
3. CH₂=CH—CH₂—Cl
4. Both 1 and 2

Answer: 4.

An sp²-hybridized carbon atom is more electronegative than a sp²-hybridized carbon atom due to a greater 5-character. Therefore, the sp2 -hybridized carbon atom of the C—Cl bond in chlorobenzene has very little tendency to donate C —Cl bond pair electrons. Further due to the delocalization of a lone pair of electrons of Cl-atom over the benzene ring, the C —Cl bond in chlorobenzene gets a partial double bond character.

Hence removal of Cl^– ion from chlorobenzene is not feasible. Vinyl chloride (CH₂=CHCl) is unreactive in S_N2 reaction due to resonance.

In vinyl chloride, the lone pair of electrons on chlorine is in resonance with the C =C double bond. Hence, there is a partial double bond character in C—Cl bond. As a result, the removal of Cle is no feasible here.

Further, due to greater stabilisation of allylic carbocation intermediate by resonance, chlorine can easily be eliminated from CH₂=CH — CH₂—Cl, and gives a white precipitate with acidified AgNO₃ solution.

Question 43. (R)-2-Iodobutane is treated with Nal in acetone and allowed to stand for a long time. The product eventually formed is—

1. (R)-2-iodobutane
2. (S)-2-iodobutane
3. (± )-2-iodobutane
4. (± )-1,2-diiodobutane

Answer: 3.

(R)-2-iodobutane [(+)-2-iodobutane] reacts with Nal in acetone to form (±) -2-iodobutane, which leads to reacemisation.

Each S_N2 attack on (R)-2-iodobutane or (+)-2- iodobutane by Ie occurs with Inversion of configuration resulting in the formation of (S) -2-iodobutane or (-) -2- iodobutane.

As the concentration of the (-) -enantiomer in the reaction mixture increases, its tendency to react with Ie forming a (+)-enantiomer increases. At a certain time, the mixture contains equal amounts of (+) and (-) -2-iodobutane.

Since both the enantiomers react with I^– at the same rate (because the activation energy is the same), their composition remains unchanged.

Question 44. The correct order of reactivity in S_N1 reaction for the following compounds is—

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

Answer: 3.

S_N1 reaction proceeds via the formation of a carbocation intermediate. Hence, more stable the carbocation, more reactive is the parent alkyl halides towards S_N1 reaction. Compound (3) gives 2° carbocation which is stabilised by +M effect of — OMe group. Thus it is the most stable carbocation.

Compound (2) gives benzylic carbocation which is resonance stabilised. We know, 2° carbocation is more stable than 1° carbocation. Therefore, the correct order of reactivity of S_N1 reaction is 3 > 2 > 1 > 4.

Question 45. Major product of the given reaction is—

Answer: 3.

Due to steric hindrance, secondary alkyl halides prefer to undergo elimination reaction in the presence of OEt. Here OEt acts as a base. In this case, according to Saytzeff’s rule, a more stable alkene is formed in this reaction.

Question 46. The decreasing order of S_N2 reaction for the given compounds is—

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

Answer: 2.

In S_N2 reaction, the attack by the nucleophile on the α-carbon (i.e., the carbon bearing the halogen) occurs from the backside (the side exactly opposite to that of the leaving group).

Therefore, the presence of bulky substituents on or near the a -carbon atom, tends to hinder the approach of the nucleophile towards the a – carbon due to steric hindrance and so, the reaction occurs with difficulty.

Evidently, the greater the number of alkyl groups, i.e., greater the steric hindrance, slower the reaction. Since the steric hindrance follows the order: 4 > 3 > 1 > 2. So, rate of S_N2 reaction is 2 > 1 > 3 > 4.

Question 47. What is the product of the following reaction:

Answer: 2.

This is an S_N2 reaction that is always accompanied by iversion of configuaration.

Question 48. The order of reactivity of following alcohols with halogen acids is—

1. CH₃CH₂CH₂OH

2. CH₃CH₂CH(CH₃)OH

3. CH₃CH₂C(CH₃)₂OH

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

Answer: 2. 3 > 2 > 1

Explanation: The reactivity of alcohols with halogen acids decreases in the order: 3° > 2° > 1°.

Question 49. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature—

Answer: 4

Explanation: 3° alcohols are the most reactive and they react with cones. HCl at room temperature.

Question 50. Identify the compound Y in the following reaction.

Answer: 1.

Question  51. Toluene reacts with a halogen in the presence of iron (3) chloride giving ortho- and para- halo compounds. The reaction is—

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

Answer: 2. Electrophilic substitution reaction

Question 52. Which of the following is a halogen exchange reaction—

Answer: 1.

Hint: This is Finkelstein’s reaction.

Question 53. Which reagent will you use for the following reaction—

1. Cl₂/UV light
2. NaCl + H₂SO₄
3. Cl₂ gas in dark
4. Cl₂ gas in the presence of iron in dark

Answer: 1

Hint: Alkanes undergo free radical substitution with Cl₂ in presence of UV light.

Question 56. Arrange the following compounds in the increasing order of their densities—

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

Answer: 1. 1 < 2 < 3 < 4

Hint: Density increases with increasing molecular mass.

Question 57. Arrange the following compounds in increasing order of their boiling points—

1. (CH₃)₂CH—CH₂Br

2. CH₃CH₂CH₂CH₂Br

3. (CH₃)₃C-Br

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

Answer: 3. 3 < 1 < 2

Hint: As the branching decreases, intermolecular forces of attraction (van der Waals force) increase, and hence, boiling point increases.

Question 58. In which of the following molecules carbon atom marked with an asterisk (*) is asymmetric—

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

Answer: 2. 1,2,3

Hint: A carbon atom attached to four different groups is said to be asymmetric.

Question 59. Which of the following structures is enantiomeric with the molecule (A) given below:

Answer: 1.

Hint: An even number of interchanges of positions of any two groups on the asymmetric carbon retains the configuration, while odd number of such interchanges produces an enantiomer.

Question 60. Which of the following is an example of vic-dihalide—

1. Dichloromethane
2. 1, 2-dichloroethane
3. Ethylidene chloride
4. Allyl chloride

Answer: 2. 1,2-dichloromethane

Hint: In vicinal dihalides, halogen atoms are on the adjacent C-atoms.

Question 61. The position of —Br in the compound in CH₃CH =CHC(Br)(CH₃)₂ can be classified as—

1. Allyl
2. Aryl
3. Vinyl
4. Secondary

Answer: 1. Allyl

Hint: In the allylic halides, the halogen atom is on a carbon atom next to the doubly bonded carbon atom.

Question 62. Chlorobenzene is formed by reaction of chlorine with benzene in the presence of AlCl₃. Which of the following species attacks the benzene ring in this reaction—

1. Cl^–
2. Cl⁺
3. AlCl₃
4. [AlCl₄]^–

Answer: 2

Question 63. Ethylidene chloride is a/an—

1. Vic-dihalide
2. Gem-dihalide
3. Allylic halide
4. Vinylic halide

Answer: 2

Hint: Ethylidene chloride (CH₃CHCl₂) is a geminal dihalide.

Question 64. What is ‘A’ in the following reaction—

Answer: 3

Hint: Addition occurs in accordance with Markownikoff’s rule.

Question 65. A primary alkyl halide would prefer to undergo—

1. S_N1 reaction
2. S_N2 reaction
3. α-Elimination
4. Racemisation

Answer: 2

Hint: or -carbon of 1° alkyl halide is least crowded and so undergoes S_N2 reaction.

Question 66. Which of the following alkyl halides will undergo S_N1 reaction most readily—

1. (CH₃)₃C-F
2. (CH₃)₃C-Cl
3. (CH₃)₃C-Br
4. (CH₃)₃C-I

Answer: 4. (CH₃)₃C-I

Hint: Leaving group ability of halide ions is: I^–> Br^–> Cl^–> F^–.

Question 67. The correct IUPAC name for the compound CH₃-CH(C₂H₅)-CH₂-Br is

1. 1-bromo-2-ethylpropane
2. 1-bromo-2-ethyl-2-methylethane
3. 1-bromo-2-methylbutane
4. 2-methyl-1-bromobutane

Answer: 3. 1-bromo-2-methylbutane

Question 68. What should be the correct IUPAC name for diethylbromomethane—

1. 1-bromo-1,1-diethylmethane
2. 3-bromopentane
3. 1-bromo-1-ethylpropane
4. 1-bromopentane

Answer: 2. 3-bromopentane

Question 69. The reaction of toluene with chlorine in the presence of iron and in the absence of light yields—

Answer: 4

Hint: Cl is ortho/para-directing.

Question 70. Chloromethane on treatment with excess of ammonia yields mainly—

1. N, N-dimethylmethanamine
2. N-methylmethanamine (CH₃NH—CH₃)
3. Methenamine (CH₃NH₂)
4. Mixture containing all these in equal proportion

Answer: 3. Methenamine (CH₃NH₂)

Question 71. Molecules whose mirror image is non-superimposable over them are known as chiral. Which of the following molecules is chiral in nature—

1. 2-bromobutane
2. 1-bromobutane
3. 2-bromopropane
4. 2-bromopropan-2-ol

Answer: 1

Hint: 2-bromobutane (CH₃CHBrCH₂CH₃) contains asymmetric carbon.

Question 72. Reaction of C₆H₅CH₂Br with aqueous sodium hydroxide follows—

1. S_N1 mechanism
2. S_N2 mechanism
3. Any of the above two depending upon the temperature of reaction
4. Saytzeffrule

Answer: 1. S_N1 mechanism

Hint: S_N1 mechanism is preferred because of its ability to form highly stable carbocation (C₆H₅CH₂).

Question 73. Which of the carbon atoms present in the molecule given below are asymmetric—

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

Answer: 2

Hint: Carbon atoms marked ‘b’ and ‘c’ are attached to four different groups.

Question 74. Which of the following compounds will give racemic mixture on nucleophilic substitution by OH^– ion—

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

Answer: 1

Hint: The compound (a) contains a chiral carbon attached to halogen, so it reacts with OH^– to form racemic alcohol.

26.

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

Answer: 3.

Hint: Rate of substitution of aryl halides increases in presence of electron-withdrawing — NO₂ group. Rate enhancement is maximum if it is in the ortho or para position.

27.

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

Answer: 4. 2 < 3 > 1

Rate of reaction of aryl halides decreases in presence of electron donating group. Decrease in rate is maximum if it is in the ortho or para position.

Question 75.

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

Answer: 4. 1 < 2 < 3

Hint: Reactivity towards nucleophilic substitution increases as the number of nitro groups in the ortho and para positions with respect to the halogen atom increases.

Question 76.

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

Answer: 3. 3 < 2 < 1

Hint: Reactivity towards nucleophilic substitution decreases as the number of electron-donating CH₃ groups in the ortho and para positions with respect to the halogen atom increases.

Question 77. Which is the correct increasing order of boiling points of the following compounds—

1-iodobutane, 1-bromobutane, 1-chlorobutane, butane

1. Butane < 1-chlorobutane < 1-bromobutane <1-iodobutane
2. 1-iodobutane < 1-bromobutane < 1-chlorobutane <butane
3. Butane < 1-iodobutane < 1-bromobutane <1-chlorobutane
4. Butane < 1-chlorobutane < 1-iodobutane <1-bromobutane

Answer: 1. Butane < 1-chlorobutane < 1-bromobutane <1-iodobutane

Hint: For alkyl halides containing the same alkyl group, boiling point increases as the atomic mass of the halogen increases.

Question 78. Which is the correct increasing order of boiling points of the following compounds—

1-bromoethane, 1-bromopropane, 1-bromobutane, bromobenzene

1. Bromobenzene < 1-bromobutane < 1-bromopropane <1-bromoethane
2. Bromobenzene < 1-bromoethane < 1-bromopropane < 1-bromobutane
3. 1-bromopropane < 1-bromobutane < 1-bromoethane < bromobenzene
4. 1-bromoethane < 1-bromopropane < 1-bromobutane < bromobenzene

Answer: 4. 1-bromoethane < 1-bromopropane < 1-bromobutane < bromobenzene

Hint: For alkyl or aryl halides containing the same halogen atom, boiling point increases as the mass of the hydrocarbon part increases.

Question 79. Consider the following reactions:

1. Which of the statements are correct about the above reaction—

1. 1 and 5 both are nucleophiles.
2. In 3 carbon atom is sp³-hybridised.
3. In 3 carbon atom is sp²-hybridised.
4. 1 and 5 both are electrophiles.

Answer:

1. 1 and 5 both are nucleophiles and

3. In 3 carbon atom is sp²-hybridised.

2. Which of the following statements are correct about this reaction—

1. The given reaction follows S_N2 mechanism.
2. 2 and 4 have opposite configuration.
3. 2 and 4 have same configuration.
4. The given reaction follows S_N1 mechanism.

Answer:

1. The given reaction follows S_N2 mechanism.

3. 2 and 4 have same configuration.

Hint: As shown in the mechanism it is one step reaction passing through a single transition state i.e., the reaction follows S_N2 mechanism. Neither (2) nor (4) contain any chiral carbon, so they have same configuration.

3. Which of the following statements are correct about the reaction intermediate—

1. Intermediate 3 is unstable because this carbon is attached to 5 atoms.
2. Intermediate 3 is unstable because carbon atom is sp²-hybridised.
3. Intermediate 3 is stable because carbon atom is sp²-hybridised.
4. Intermediate 3 is less stable than the reactant 2.

Answer:

1. Intermediate 3 is unstable because this carbon is attached to 5 atoms.

4. Intermediate 3 is less stable than the reactant 2.

Note: (3) is not an intermediate, it should properly be called a transition state.

Question 80. Basis of the following reaction:

Question 81. Haloalkanes contain halogen atom (s) attached to the sp³-hybridised carbon atom of an alkyl group. Identify haloalkane from the following compounds—

1. 2-bromopentane
2. Vinyl chloride (chloroethene)
3. 2-chloroacetophenone
4. Trichloromethane

Answer:

1. 2-bromopentane

4. Trichloromethane

Hint: 2-bromo-pentane (CH₃CHBrCH₂CH₂CH₃) and chloroform (CHCl₃) are alkyl halides.

Question 82. Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements—

1. Both the compounds form same product on treatment with alcoholic KOH.
2. Both the compounds form same product on treatment with aq. NaOH.
3. Both the compounds form same product on reduction.
4. Both the compounds are optically active.

Answer:

1. Both the compounds form same product on treatment with alcoholic KOH.

3. Both the compounds form same product on reduction.

Hint: Both ethylene chloride and ethylidene chloride form acetylene on treatment with alcoholic KOH. On reduction, they also give the same product, ethane.

Question 83. Which of the following compounds are gem-dihalides—

1. Ethylidene chloride
2. Ethylene dichloride
3. Methylene chloride
4. Benzyl chloride

Answer:

1. Ethylidene chloride

3. Methylene chloride

Hint: Ethylidene chloride (CH₃CHCl₂) and methylene chloride (CH₂Cl₂) are gem-dihalides because two halogen atoms are on the same carbon.

Question 84. Which of the following are secondary bromides—

1. (CH₃)₂CHBr
2. (CH₃)₃CCH₂Br
3. CH₃CH(Br)CH₂CH₃
4. (CH₃)₂CBrCH₂CH₃

Answer:

1. (CH₃)₂CHBr

2. CH₃CH(Br)CH₂CH₃

Hint: 2° bromides have the general structure, R—CHBr—R’.

Question 85. Which of the following compounds can be classified as aryl halides—

1. P-ClC₆H₄CH₂CH(CH₃)₂
2. P-CH₃CHCl(C₆H₄)CH₂CH₃
3. O-BrH₂C —C₆H4CH(CH3)CH₂CH₃
4. C₆H₅CI

Answer:

1. P-ClC₆H₄CH₂CH(CH₃)₂

4. C₆H₅CI

Hint: In aryl halides halogen atom is attached directly to aromatic ring.

Question 86. Alkyl halides are prepared from alcohols by treating with—

1. HCl + ZnCl₂
2. Red P + Br₂
3. H₂SO₄ + KI
4. All the above

Answer: 1,2.

1. HCl + ZnCl₂

2. Red P + Br₂

Hint: Alcohols do not form alkyl halides on treatment with KI in presence of H₂SO₄, because H₂SO₄ (oxidising agent) oxidises KI to I₂.

Question 87. Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of or.

1. CaF₂
2. CoF₂
3. Hg₂F₂
4. NaF

Answer: 2,3.

2. CoF₂

3. Hg₂F₂

Hint: Only transition metal fluorides such as CoF₂ and Hg₂F₂ react with alkyl chlorides or bromides to form alkyl fluorides.

Question 88. Which one of the following compounds is expected to be formed when chlorobenzene is treated with chloral in the presence of cone. H2SO₄ —

1. Gammexane
2. Hexachloroethane
3. Freon
4. DDT

Answer: 4. DDT

Question 89. Major product obtained in the following reaction is—

Answer: 4

Question 90. Which one of the following reagents can be used for chlorination of toluene leading to the formation of benzyl chloride (C₆H₅CH₂Cl)-

1. HOCl
2. SOCl₂
3. Cl₂
4. NaOCl

Answer: 4.Cl₂

Question 91. Which one of the following compounds is least willing to undergo nucleophilic substitution reaction—

1. (CH₃)₃C-C1
2. CH₂=CHCl
3. CH₃CH₂Cl
4. CH₂=CHCH₂Cl

Answer: 2. CH₂=CHCl

Question 92. The reagent required for carrying out the above conversion is—

1. Alcoholic KOH
2. Alcoholic KOH and then NaNH₂
3. Aqueous KOH and then NaNH₂
4. Zn/CH₃OH

Answer: 2. Alcoholic KOH and then NaNH₂

Question 93. Which one of the following reagents can be used to distinguish between chlorocyclohexane and chlorobenzene—

1. AgNO₃/C₂H₅OH
2. Ag(NH₃) OH
3. Na, HNO₃, AgNO₃
4. Br₂/CCl₄

Answer: 1. AgNO₃/C₂H₅OH

Question 94. Which one of the following compounds is expected to be formed when Cl₂ gas is passed through propene at 400°C —

1. PVC
2. Allyl chloride
3. Vinyl chloride
4. 1,2-dichloroethane

Answer: 2. Allyl chloride

Question 95. Chlorination of paraffins occurs by free radical mechanism in which the chain terminating reaction is—

1. Cl₂→2Cl
2. CH₃Cl + Cl→CH₂Cl + HCl
3. CH₃ + Cl2→CH₃Cl + Cl
4. Cl + Cl→Cl₂

Answer: 4. Cl + Cl→Cl₂

Question 96. Which one of the following compounds has the highest boiling point—

1. CH₃CH₂CH₂Cl
2. CH₃CH₂CH₂CH₂Cl
3. CH₃CH(CH₃)CH₂Cl
4. (CH₃)₃CCl

Answer: 2. CH₃CH₂CH₂CH₂Cl

Question 97. CH₃CH₂CHBrCH₃ + (CH₃)₃COK→ The major product in the above reaction is—

Answer: 4

Question 98. CH₃Br + Nu^–→CH₃ — Nu + B^–r [Nu^– (nucleophile) = PhO^–(1), AcO^–(2), HO^–(3), CH3O^–(4) The rate of the reaction by using various nucleophiles (1 to 4) decreases in the order—

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

Answer: 1. 4 > 3 > 1 > 2

Question 99. The existence of meso-isomer is possible in the case of—

1. 2-chlorobutane
2. 2,3-dichlorobutane
3. 2,3-dichloropentane
4. 2-hydroxypropanoic acid

Answer: 2. 2,3-dichlorobutane

Question 100. Which one of the following compounds will be obtained when ethyl formate is allowed to react with excess of CH₃MgI and the resulting compound is hydrolysed—

1. N-propyl alcohol
2. Isopropyl alcohol
3. Acetaldehyde
4. Acetone

Answer: 2. Isopropyl alcohol

Question 101. Which one of the following compounds will undergo only S_N2 reaction—

1. Benzyl chloride
2. Ethyl chloride
3. Chlorobenzene
4. Isopropyl chloride

Answer: 2. Ethyl chloride

Question 102. Which one of the following statements is correct regarding hydrolysis of tert-butyl bromide with aqueous NaOH—

1. The reaction occurs by S_N1 mechanism.
2. The intermediate in this reaction is a carbocation.
3. The rate of the reaction becomes doubled if the concentration of alkali is doubled.
4. The rate of the reaction becomes doubled if the concentration of tert-butyl bromide is doubled.

Answer: 3. The rate of the reaction becomes doubled if the concentration of alkali is doubled.

Question 103. Tetrahydrofuran reacts with excess of HI to form—

1. 1,4-diiodobutane
2. 1,4-butanediol
3. 2,5-diiodotetrahydrofuran
4. 4-iodo-1-butanol

Answer: 1. 1,4-diiodobutane

Question 104. Which one of the following reagents reacts with ethanol to form iodoform—

1. KI and aqueous KOH
2. I₂ and aqueous KOH
3. I₂ /aqueous KI
4. HI and HIO₃

Answer: 2. I₂ and aqueous KOH

Question 105. Which one of the following statements regarding benzyl chloride is not correct—

1. It produces a white precipitate with AgN03 solution.
2. It is an aromatic compound in which the side chain is substituted.
3. It takes part in the nucleophilic substitution reaction.
4. It is less reactive than vinyl chloride.

Answer: 4. It is less reactive than vinyl chloride.

Question 106. The compound obtained on chlorination of toluene in the presence of sunlight is hydrolysed with an aqueous solution of NaOH to give—

1. O-cresol
2. P-cresol
3. O-and p-cresol
4. 1,3,5-trihydroxy toluene

Answer: 3. O-and p-cresol

Question 107. The compound which is obtained on hydrolysis of 2,2-dichloropropane is—

1. Acetone
2. 2,2-propanediol
3. Isopropyl alcohol
4. Acetaldehyde

Answer: 1. Acetone

Question 108. Which one of the following methods is the best one for the preparation of alkyl halides—

1. ROH + SOCl₂→
2. ROH + PCl₅→
3. ROH + PCl₃→
4. \(\mathrm{ROH}+\mathrm{HCl} \stackrel{\mathrm{ZnCl}_2}{\longrightarrow}\)

Answer: 1. ROH + SOCl₂→

Question 109. In which of the following reactions, propanenitrile is obtained as the major product—

1. Ethylbromide + alcoholic KCN
2. Propyl bromide + alcoholic KCN
3. Propyl bromide + alcoholic AgCN
4. Ethyl bromide + alcoholic AgCN

Answer: 1. Ethylbromide + alcoholic KCN

Question 110. Which one of the following compounds cannot be hydrolysed with sodium hydroxide solution—

1. Vinyl chloride
2. Methyl chloride
3. Ethyl chloride
4. Isopropyl chloride

Answer: 1. Vinyl chloride

Question 111. The maximum number of CH₃I molecules which will react with CH₃NH₂ is—

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

Answer: 1. 3

Question 112. Dichlorination of propane leads to the formation of a mixture of different isomers. The number of isomers in the mixture is—

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

Answer: 3. 4

Question 113. Which has the lowest b.p.—

1. CH₃F
2. CH₃CI
3. CH₃Br
4. CH₃I

Answer: 1. CH₃F

Question 114. Which will react with AgNO₃ most readily—

1. CH₃CH=CHCl
2. CH₃CH₂CH₂Cl
3. CH₂=CHCH₂Cl
4. CH₃CH₂CHCl₂

Answer: 3. CH₂=CHCH₂Cl

Question 115. Which one of the following compounds does not respond to the haloform reaction—

1. CH₃CH₂CHOHCH₃
2. CH₃CH₂CH₂COCH₃
3. CH₃COCH₂COOC₂H₅
4. CH₃COC₆H₅

Answer: 3. CH₃COCH₂COOC₂H₅

Question 116. The compound expected to be formed (2 moles) in the above reaction is—

Answer: 2

Question 117. trans-2-phenyl-1-bromocyclopentane reacts with alcoholic KOH to form—

1. 4-phenylcyclopentene
2. 2-phenylcyclopentene
3. 1-phenylcyclopentene
4. 3-phenylcyclopentene

Answer: 4. 3-phenylcyclopentene

Question 118. In the solvolysis of which alkyl halide, the tendency of rearrangement is minimum—

1. Cis-1-chloro-2-methylcyclohexane
2. 2-chIoro-2-phenylpentane
3. 2-chloro-3-phenylpentane
4. 2-chloro-4-phenylpentane

Answer: 2. 2-chIoro-2-phenylpentane

Question 119. Which one of the following alkyl halides will react with sodium methoxide to form only one alkene—

1. 2-chloro-2-methylpentane
2. 3-chloro-2-methylpentane
3. 2-chloro-4-methylpentane
4. 3-chloro-3-ethylpentane

Answer: 4. 3-chloro-3-ethylpentane

Question 120. O-fluorotoluene reacts with sodium amide to form—

1. Only o-toluidine
2. Only m-toluidine
3. A mixture of o- and p-toluidine
4. A mixture of o- and m-toluidine

Answer: 4. A mixture of o- and m-toluidine

Question 121. Alkyl iodides react with NaCN to form alkyl cyanides plus a little amount of alkyl isocyanides. The reason for the formation two types of products is—

1. Ionic character of NaCN
2. Nucleophilic character of CN^–
3. Ambident character of CN^–
4. Electrophilic character of CN^–

Answer: 3. Ambident character of CN^–

Question 122. The S_N2 reaction of which of the following chlorine-containing compounds occurs with complete stereochemical inversion—

1. (C₂H₅)₂CHCI
2. (CH₃)₃CCl
3. (CH₃)₂CHCl
4. CH₃Cl

Answer: 4. CH₃Cl

Question 123. In which of the following compounds, the carbon- halogen bond is the weakest one—

1. Benzyl bromide
2. Bromobenzene
3. Vinyl bromide
4. Benzyl chloride

Answer: 1. Benzyl bromide

Question 124. Which one of the following compounds will react with bleaching powder to form trichloromethane—

1. Methanal
2. Phenol
3. Ethanol
4. Methanol

Answer: 3. Ethanol

Question 125. Ethyl bromide reacts with alcoholic AgN02 to form—

1. Ethene
2. Ethane
3. Ethyl nitrile
4. Nitroethane

Answer: 4. Nitroethane

Question 126. Chlorobenzene reacts with one of the following compounds to form aniline—

1. NH₃/Cu₂O
2. NH₃/H₂SO₄
3. NaNH₂, NH₃(l)
4. None of these

Answer: 3. NaNH₂, NH₃(l)

Question 127. When an ethereal solution of 1-bromo-3-chlorocyclobutane is treated with two equivalents of sodium, the compound expected to be formed is—

Answer: 4

Question 128.S_N1 reactivity of alkyl halides follows the order—

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

Answer: 2. 3°>2°>1°

Question 129. Which one is not favorable for S_N1 reaction—

1. 3° alkyl halide
2. Strong nucleophile
3. Polar solvent
4. Low concentration

Answer: 2. Strong nucleophile

Question 130. Which alkyl halide possesses highest tendency to undergo S_N2 reaction—

1. 2 -bromobutane
2. 1 -bromobutane
3. 2-bromo-2-methyipropane
4. 1-bromo-2-methylpropane

Answer: 2. 1 -bromobutane

Question 131. The order of S_N2 reactivity (reaction with (Kl/acetone) of the compounds P, Q, R, and S is—

1. P > Q > R > S
2. S > P > R > Q
3. P > R > Q > S
4. R > P > S > Q

Answer: 1. P > Q > R > S

Question 132. An alkyne containing sodium amide reacts with a bromoalkane to give 3-octyne. The formulas of the bromoalkane and the alkyne are respectively—

1. Br(CH₂)₄CH₃ and CH₃CH₂C = CH
2. Br(CH₂)₂CH₃ and CH₃(CH₂)₂C = CH
3. Br(CH₂)₄CH₃ and CH₃C = CH
4. Br(CH₂)₃CH₃ and CH₃CH₂C≡CH

Answer: 3. Br(CH₂)₄CH₃ and CH₃C = CH

Question 133. The IUPAC name of the compound

1. 4-bromo-2-cyanophenol
2. 2-bromo-5-hydroxybenzonitrile
3. 2-cyano-4-hydroxybromobenzene
4. 6-bromo-3-hydroxybenzonitrile

Answer: 1. 4-bromo-2-cyanophenol

Question 134. here the product P is—

1. Aq.NaOH
2. Alc.KOH
3. (C₂H₅)₃N
4. Both 2 and 3

Answer: 4

Question 135. here the product P is —

Answer: 2

Question 136. Which of the following reagents cannot substitute —OH group by Cl—

1. CH₃COCl
2. SOCl₂
3. PCl₃
4. PCl₅

Answer: 1. CH₃COCl

Question 137. Chlorobenzene gives aniline with—

1. NH₃/Cu₂O
2. NH₃/H₂SO₄
3. NaNH₂
4. None of these

Answer: 1 and 3

1. NH₃/Cu₂O

3. NaNH₂

Question 138. \(\mathrm{A} \stackrel{\mathrm{I}_2 / \mathrm{NaOH}}{\longrightarrow}\) iodoform + sodium succmate In this sequence, A can be—

1. Pentan-2-one
2. Acetophenone
3. 4-ketopentanoic acid
4. Hexan-2,5-dione

Answer: 3 and 4

3. 4-ketopentanoic acid

4. Hexan-2,5-dione

Question 139. Which of the following readily undergo nucleophilic substitution reaction by S_N1 mechanism in presence of butanol—

1. (CH₃)₃CBr
2. C₆H₅CH₂Br
3. BrCH₂CH=CH₂
4. (CH₃)₃CCH₂Br

Answer: All options are correct

Question 140. Product is/are—

Answer: 1 and 2 and 3

Question 141.

In this sequence, ‘A’ can be—

Answer: 2 and 3

Question 142. Ethyl bromide can be converted into diethyl ether by—

1. Reacting with sodium ethoxide
2. Heating with moist Ag₂O
3. Heating with dry Ag₂O
4. Reacting with ethylmagnesium bromide

Answer: 1 and 3

1. Reacting with sodium ethoxide

3. Heating with dry Ag₂O

Question 143. Which of the following are ambident nucleophiles—

1. NH₃
2. CN^–
3. NO₂^–
4. H₂O

Answer: 2 and 3

2. CN^–

3. NO₂^–

Question 144. Which of the following are S_N2 reactions—

Answer: 1 and 3

Question 145. Which of the following reagents or tests can’t be used to distinguish between allyl bromide and n-propyl bromide—

1. Br₂/CCl₄
2. KOH followed by acidifying with HN03 and adding aqueous AgNO₃
3. Lassaigne’s test
4. Alkaline KMnO₄

Answer: 2 and 3

Question146. The values of the dipole moment are the same for

Answer: 1 and 4

Question 147. Which of the following compounds reacts with Mg and then water to give toluene—

Answer: All options are correct

Question 148. Which of the following compounds undergo free radical halogenation to give a racemic mixture—

1. (CH₃)₃C
2. CH₃CH₂CH₂CH₃
3. CH₃CH₂CH(CH₃)2
4. CH₃CH₃

Answer: 2 and 3

Question 149. Which of the following reagents can be used for chlorination of toluene to yield benzyl chloride (C₆H₂CH₂Cl) —

1. SO₂Cl₂
2. SOCl₂
3. Cl₂
4. NaOCl

Answer: 1 and 2 and 3

1. SO₂Cl₂

2. SOCl₂

3. Cl₂

Question 150. In which of the following compounds X is bonded to an sp² carbon—

Answer: 1 and 4

Question 151. The compounds which produce yellow precipitate with I₂/NaOH are—

1. ICH₂COCH₂CH₃
2. CH₃COOCOCH₃
3. CH₃CONH₂
4. CH₃CH(OH)CH₂CH₃

Answer: 1 and 4

1. ICH₂COCH₂CH₃

4. CH₃CH(OH)CH₂CH₃

Question 152. Which of the following compounds react with water—

1. CHCl₃
2. CCl₃CHO
3. CCl₄
4. ClCH₂CH₂Cl

Answer: 2 and 4

2. CCl₃CHO

4. ClCH₂CH₂Cl

Question 153. The correct statements are—

1. Benzyl halides are more reactive than vinyl or aryl halides
2. Vinyl halides are more reactive than alkyl halides
3. Aryl halides are less reactive than alkyl halides
4. Aryl halides are more reactive than benzyl halides

Answer: 1 and 3

1. Benzyl halides are more reactive than vinyl or aryl halides

3. Aryl halides are less reactive than alkyl halides

Question 154. The reagents used for the preparation of DDT from chlorobenzene are—

1. Chloral(CCl₃CHO)
2. Cone. H₂SO₄
3. CH₃COCCl
4. ClCH₂COCH₂Cl

Answer: 1 and 2

1. Chloral(CCl₃CHO)

2. Cone. H₂SO₄

Question 155. \(‘ X^{\prime} \stackrel{\mathbf{I}_2 / \mathrm{NaOH}}{\longrightarrow}\) Iodoform + Sodium succinate ‘X’ is—

1. Pentan-2-one
2. Acetophenone
3. Hexane-2,5-dione
4. 4-ketopentanoic acid

Answer: 3 and 4

3. Hexane-2,5-dione

4. 4-ketopentanoic acid

Question 156. Which of the following compounds do not rotate the plane of polarised light in a particular direction?

1. 3-chloropentane
2. 2-bromooctane
3. (±) -lactic acid
4. N-ethyl-N-methylpropan-1-amine
5. Meso-tartaric acid 

Answer: 1, 2,3and 5

Question 157. Which of the following statements is correct about the mechanism of this reaction—

1. A carbocation will be formed as an intermediate in the reaction.
2. OH- will attack substrate 2 from one side and Cl- will leave it simultaneously from the other side.
3. An unstable intermediate will be formed in which OH- and Cl- will be attached by weak bonds.
4. Reaction proceeds through S_N 1 mechanism.

Answer:

A carbocation will be formed as an intermediate in the reaction.

Reaction proceeds through S_N 1 mechanism.

Hint: The reaction of 2° alkyl halide proceeds via S_N1 mechanism involving the generation of a 2° carbocation.

Question 158. Which of the following statements is correct about the kinetics of this reaction—

1. The rate of reaction depends on the concentration of only 2.
2. The rate of reaction depends on the concentration of both (a) and (b).
3. Molecularity of the reaction is one.
4. Molecularity of reaction is two.

Answer:

1. The rate of reaction depends on the concentration of only 2.

3. Molecularity of the reaction is one.

Hint: S_N1 reaction (as pointed out in the previous question) is unimolecular and its rate depends only on the concentration of the substrate.

Unit 10 Haloalkanes And Haloarenes Fill in the blanks

Question 1.The IUPAC name of BrCH₂CH₂CH₂Cl is _____
Answers: 1-Bromo-3-chloropropane

Question 2.C₆H₅C(CH₃)₂Cl is an example of ____ benzylic halide.
Answers: 3°

Question 3. Among the isomeric dichlorobenzenes the _____ isomer has the highest boiling point.
Answers: Para

Question 4. The Cl atom in chlorobenzene is _____ directing.
Answers: Ortho/para

Question 5. The bromoalkane obtained as the major product in free radical bromination of propene is ____
Answers: 2-Bromopropane

Question 6. CN^– is an ____ nucleophile.
Answers: Ambident

Question 7. The central carbon atom in the transition state of an S_N2 reaction is ____ hybridised.
Answers: sp²

Question 8. The compound obtained on alkaline hydrolysis of (S)-2-bromooctane is ____
Answers: R-2-octanol

Question 9. The S_N1 reactivity depends on the stability of the ____ obtained in the rate-determining step.
Answers: Carbocation

Question 10. The S_N2 reactivity ____ with increase in steric hindrance.
Answers: Decreases

Question 11. A chiral molecule is not ____ on its mirror image.
Answers: Superimposable

Question 12. CH₃Cl does not undergo ____ reaction while (CHq).CCl does not undergo ____ reaction.
Answers: sp² and sp²

Question 13. Tendency of nucleophilic substitution reaction ____ with increase in the no. of electron-attracting groups in halobenzene.
Answers: Increases

Question 14. The C—Cl bond in chlorobenzene is ____ in length than the C—Cl bond in benzyl chloride.
Answers: Shorter

Question 15. Chlorobenzene is converted into phenol through the formation of ____ (intermediate).
Answers: Benzyne

Question 16. Ethanol & propane react with I₂/NaOH to form ____
Answers: Iodoform

Question 17. Chlorobenzene reacts with ____ in the presence of cone. H₂SO₄ to form DDT.
Answers: Chloral

Question 18. Biphenyl is obtained when ____ is heated with Cu-powder in a sealed tube.
Answers: Iodobenzene

Question 19. Haloarenes are ____ reactive than haloalkanes towards nucleophilic substitution reactions.
Answers: Less

Question 20. Freon-12 (CF₂Cl₂) is a widely used as ____
Answers: Refrigerant

Unit 10 Haloalkanes And Haloarenes Matching Type

Question 1. 

Answer: 1. C; 2. D; 3. A; 4. B;

Question 2.

Answer: 1. C; 2. E; 3. A; 4. B; 5. D

Question 3.

Answer: 1. B; 2. D; 3. A; 4. C

Question 4.

Answer: 1. B; 2.D; 3. E; 4. A; 5. C

Question 5.

Answer: 1. A; 2. C; 3. B; 4. D

Unit 10 Haloalkanes And Haloarenes Assertion Reason Type Question And Answer

 In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.

(A) and (R) both are correct statements and (R) is correct explanation for (A).

(A) and (R) both are correct statements but (R) is not correct explanation for (A).

(A) is correct statement but (R) is wrong statement.

(A) and (R) both are incorrect statements.

(A) is wrong statement but (R) is correct statement.

Question 1. Assertion (A): Phosphorus chlorides (tri and penta) are preferred over thionyl chloride for the  preparation of alkyl chlorides from alcohols.

Reason (R): Phosphorus chlorides give pure alkyl halides.

Answer: 2. (A) and (R) both are correct statements but (R) is not correct explanation for (A).

Hint: Thionyl chloride is preferred over phosphorus chlorides for the preparation of alkyl chlorides from alcohols because SOCl2 gives pure alkyl chlorides.

Question 2. Assertion (A): Boiling points of alkyl halides decrease in the order: RI > RBr > RC1 > RF

Reason (R): The boiling points of alkyl chlorides, bromides, and iodides are considerably higher      than that of the hydrocarbon of comparable molecular mass.

Answer: 5. (A) is wrong statement but (R) is correct statement.

Hint: For alkyl halides containing the same alkyl group, the boiling points decrease as the atomic mass of the halogen decreases.

Question 3. Assertion (A): KCN reacts with methyl chloride to give methyl isocyanide.

Reason (R): CN^– is an ambident nucleophile.

Answer: 4. (A) is correct statement but (R) is wrong statement.

Hint: Correct assertion: KCN reacts with methyl chloride to give methyl cyanide as the predominant product.

Question 4. Assertion (A): Tert-butyl bromide undergoes Wurtz reaction to give 2,2,3,3-tetramethylbutane.

Reason (R): In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.

Answer: 5. (A) is wrong statement but (R) is correct statement.

Hint: Correct assertion—2,2,3,3-tetramethylbutane cannot be prepared by Wurtz reaction of fert-butyl

Question 5. Assertion (A): Presence of a nitro group at ortho-or porn-position increases reactivity of haloarenes towards nucleophilic substitution.

Reason (R): Nitro group, being an electron-withdrawing group decreases the electron density over the bvenzene ring.

Answer: 1. (A) and (R) both are correct statements and (R) is correct explanation for (A).

Hint: Reason is correct explanation of assertion.

Question 6. Assertion (A): In monohaloarenes, further electrophilic substitution occurs at ortho- and para- positions.

Reason (R): Halogen atom is a ring deactivator.

Answer: 5. (A) is wrong statement but (R) is correct statement.

Hint: Correct reason: Due to +R effect of halogen, negative charges appear at ortho and para positions of the ring.

Question 7. Assertion (A): Aryl iodides can be prepared by reaction of arenes with iodine in the presence of an oxidising agent.

Reason (R): Oxidising agent oxidises I₂ into HI.

Answer: 3. (A) is correct statement but (R) is wrong statement.

Hint: Correct reason: oxidising agent oxidises HI to

Question 8. Assertion (A): It is difficult to replace chlorine by —OH in chlorobenzene in comparison to that in chloroethane.

Reason (R): Chlorine-carbon (C —Cl) bond in chlorobenzene has a partial double bond character due to resonance.

Answer: 1. (A) and (R) both are correct statements and (R) is correct explanation for (A).

Hint: Reason is correct explanation of assertion.

Question 9. Assertion (A): Hydrolysis of (-)-2-bromooctane proceeds with inversion of configuration.

Reason (R): This reaction proceeds through the formation of a carbocation.

Answer: 3. (A) is correct statement but (R) is wrong statement.

Hint: Correct reason: Reaction proceeds by S_N2 mechanism.

Question 10. Assertion (A): Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene.

Reason (R): —NO₂ group is a m-directing group.

Answer: 4. (A) and (R) both are incorrect statements.

Hint: Correct assertion—chlorination of nitrobenzene leads to the formation of m -nitrochlorobenzene.