NCERT Solutions for Class 12 Chemistry Chapter 13 - Amines

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1. Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

(i) (CH₃)₂ CHNH₂

(ii) CH₃(CH₂)2NH₂

(iii) CH₃NHCH(CH₃)₂

(iv) (CH₃)3CNH₂

(v) C₆H₅NHCH₃

(vi) (CH₃CH₂)2NCH₃

(vii) m-BrC₆H₄NH₂

Ans. (i) Propan-2-amine (1° amine)

(ii) Propan-1-amine (1° amine)

(iii) N-methylpropan-2-amine (2° amine)

(iv) 2-Methylpropan-2-amine (1° amine)

(v) N-Methylbenzamine or N-methylaniline (2° amine)

(vi) N-Ethyl-N-methylethanamine (3° amine)

(vii) 3-Bromobenzenamine or 3-bromoaniline (1° amine)

(viii) N, N-dimethylethanamine (3° amine)

2. Give one chemical test to distinguish between the following pairs of compounds:

(i) Methylamine and dimethylamine

(ii) Secondary and tertiary amines

(iii) Ethylamine and aniline

(iv) Aniline and benzylamine

(v) Aniline and N-Methylaniline.

Ans. (i) Methylamine being a 1° amine, can be distinguished by the carbylamine test from dimethylamine.

Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide give foul-smelling isocyanides or carbylamines. Methylamine (which is an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.

$$\underset{\text{Methylamine (1°)}}{\text{CH}_{3}-\text{NH}_{2}} + \text{CHCl}_{3} + 3\text{KOH}\xrightarrow{\Delta}\underset{\text{Methylisocyanide(Foul smell)}}{\text{CH}_{3}-\text{NC}} + 3\text{KCl + 3H}_{2}\text{O}\\\text{(CH}_{3})_{2}\text{NH} + \text{CHCl}_{3} + 3\text{KOH}\xrightarrow{\Delta}\text{No reaction} $$

(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl).

Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N-diethylamine reacts with Hinsberg’s reagent to form N, N-diethyl-benzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent.

(iii) Ethylamine and aniline can be distinguished using the Azo-dye test. A dye is obtained when aromatic amines react with HNO₂(NaNO₂ + dil. HCl) at 0-5°C to form benzene diazonium chloride followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red or orange in colour.

Aliphatic amines such as ethylamine give a brisk effervescence due to the evolution of N₂ gas under similar conditions.

$$\text{CH}_{3}\text{CH}_{2}-\text{NH}_{2} + \text{HONO}\xrightarrow{\text{0-5\degree\text{C}}}\text{C}_{2}\text{H}_{5}\text{OH} + \text{N}_{2}\uparrow{} + \text{H}_{2}\text{O}$$

(iv) Aniline and benzylamine can be distinguished with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

On the other hand, aniline reacts with HNO₂ at low temperature to form stable diazonium salt. Thus, nitrogen gas is not evolved. Also it gives azo-dye test.

$$\underset{\text{Aniline}}{\text{C}_{6}\text{H}_{5}\text{NH}_{2}}\xrightarrow[\text{273-278\space K}]{\text{NaNO}_{2} + \text{HCl}}{\text{C}_{6}\text{H}_{5}\text{N}_{2}^{\normalsize+}\text{Cl}^{\normalsize- }} + \text{NaCl}+ \text{2H}_{2}\text{O}$$

(v) Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul smelling isocyanides or carbylamines. Aniline, C₆H₅NH₂ is a primary aromatic amine and N-methylaniline, C₆H₅NH(CH₃) is a secondary amine. Hence, aniline will give positive carbylamine test while N-methylaniline will not.

3. Account for the following

(i) pK_b of aniline is more than that of methylamine

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) Although amino group is o and p–directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Ans. (i) pK_b of aniline is more than that of methylamine: Higher the pKb value lower is the basic strength.

In aniline due to resonance, the electrons on the N-atom are delocalised over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.

On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pK_b of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not, Ethylamine when added to water forms intermolecular H-bonds with water. Hence, it is soluble in water.

But aniline does not undergo H-bonding with water to a very large extent due to the presence of a bulky hydrophobic –C₆H₅ group. Hence, aniline is insoluble in water.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide: Methylamine is more basic than water due to the +I effect of the CH₃ group. Therefore, in water, methylamine produces OH^– ions by accepting H⁺ ions from water.

$$\text{CH}_{3}-\text{NH}_{2} + \text{H - OH}\xrightarrow{}\text{CH}_{3}-\text{NH}^{\normalsize+}_{3} + \text{OH}^{\normalsize-}$$

Then, OH^– ion reacts with Fe³⁺ ion, from dissociated FeCl₃ to form a precipitate of hydrated ferric oxide.

$$\text{2Fe}^{3+} + 6\text{OH}^{\normalsize-}\xrightarrow{}\underset{\text{Hydrated ferric oxide}}{\text{Fe}_{2}\text{O}_{3}.3\text{H}_{2}\text{O}}$$

(iv) Although amino group is o, p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline: Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).

For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Craft’s reaction : A Friedel-Craft’s reaction is carried out in the presence of AlCl₃ but AlCl₃ is a Lewis acid while aniline is a strong base. Thus, aniline reacts with AlCl₃ to form a salt.

Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo the Friedel-Craft’s reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines:

The diazonium ion undergoes resonance as shown below:

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines: Gabriel phthamide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

4. Arrange the following:

(i) In decreasing order of pK_b values:

C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)2NH and C₆H₅NH₂

(ii) In increasing order of basic strength:

C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂ NH and CH₃NH₂.

(iii) In increasing order of basic strength:

(а) Aniline, p-nitroaniline and p-toluidine

(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂

(iv) In decreasing order of basic strength in gas phase:

C₂H₅NH₂, (C₂H₅)2NH, (C₂H₅)3N and NH₃

(v) In increasing order of boiling point:

C₂H₅OH, (CH₃)2NH, C₂H₅NH₂

(vi) In increasing order of solubility in water:

C₆H₅NH₂, (C₂H₅)2NH, C₂H₅NH₂

Ans. (i) Higher pK_b value, lower is the basic strength. In C₂H₅NH₂, only one –C₂H₅ group is present while in (C₂H₅)2NH, two –C₂H₅ groups are present. Thus, the +I effect is more in (C₂H₅)2NH than in C₂H₅NH₂. Therefore, the electron density over the N-atom is more in (C₂H₅)2NH than in C₂H₅NH₂. Hence, (C₂H₅)2NH is more basic than C₂H₅NH₂.

C₆H₅NHCH₃ is more basic than C₆H₅NH₂ and less basic than (C₂H₅)2NH and C₂H₅NH₂ due to the delocalization of the lone pair in the former two. Hence, the order of increasing basicity of the given compounds is as follows:

C₆H₅NH₂ < C₆H₅NHCH₃ < C₂H₅NH₂ < (C₂H₅)2NH

Therefore, decreasing pK_b values are:

C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)2NH

(ii) C₆H₅N(CH₃)₂ is more basic than C6H5NH2 due to presence of +I effect of two –CH₃ groups in C₆H₅N(CH₃)₂. Further, CH₃NH₂ contains one –CH₃ group while (C₂H₅)2NH contains two –C₂H₅ groups. Thus, (C₂H₅)2NH is more basic than C₂H₅NH₂.

Now, C₆H₅N(CH₃)₂ is less basic than CH₃NH₂ because of the –I effect of –C₆H₅ group.

Hence, the decreasing order of the basic strengths of the given compounds is as follows:

(C₂H₅)₂NH > CH₃NH₂ > C₆H₅N(CH₃)₂ > C₆H₅NH₂

In p-toluidine, the presence of electron-donating –CH₃ group increases the electron density on the N-atom. Thus, p-toluidine is more basic than aniline.

On the other hand, the presence of electron withdrawing –NO₂ group decreases the electron density over the N-atom in p-nitroaniline. Thus, p-nitroaniline is less basic than aniline. Hence, the increasing order of the basic strengths of the given compounds is as follows:

p-Nitroaniline < Aniline < p-Toluidine.

(b) C₆H₅NHCH₃ is more basic than C₆H₅NH₂ due to the presence of electron-donating –CH₃ group in C6H5NHCH3. Again, in C₆H₅NHCH₃, –C₆H₅ group is directly attached to the N-atom. However, it is not so in C₆H₅CH₂NH₂. Thus, in C₆H₅NHCH₃, the – I effect of –C6H5 group decreases the electron density over the N-atom. Therefore, C₆H₅CH₂NH₂ is more basic than C₆H₅NHCH₃. Hence, the increasing order of the basic strengths of the given compounds is as follows:

C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂

(iv) In the gas phase, there is a no solvation effect. Hence, the basic strength only depends upon the + I effect. Also, greater the number of alkyl groups, the higher is the + I effect. Hence, decreasing basic strength follows order:

(C₂H₅)3N > (C₂H₅)2NH > C₂H₅NH₂ > NH₃

(v) The boiling point of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. (CH₃)₂NH contains only one H-atom whereas C₂H₅NH₂ contains two H-atoms.

Hence, the boiling point of C₂H₅NH₂ is higher than that of (CH₃)₂NH. Further, O is more electronegative than N. Thus, C₂H₅OH forms stronger H-bonds than C₂H₅NH₂. As a result, the boiling point of C₂H₅OH is higher than that of C₂H₅NH₂ and (CH₃)₂NH. Therefore, compounds can be arranged in the increasing order of their boiling points as follows:

(CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(vi) The more extensive the H-bonding, the higher is the solubility. C₂H₅NH₂ contains two H-atoms whereas (C₂H₅)2NH contains only one H-atom. Thus, C₂H₅NH₂ undergoes more extensive H-bonding than (C₂H₅)2NH. Aniline has a bulky hydrophobic part. Hence, less solubility of C6H5NH2 than C2H5NH2 and (C₂H₅)2NH. Hence, the increasing order of their solubility in water is as follows:

C₆H₅NH₂ < (C₂H₅)2NH < C₂H₅NH₂

5. How will you convert:

(i) Ethanoic acid into methanamine

(ii) Hexanenitrile into 1-aminopentane

(iii) Methanol to ethanoic acid.

(iv) Ethanamine into methanamine

(v) Ethanoic acid into propanoic acid

(vi) Methanamine into ethanamine

(vii) Nitromethane into dimethylamine

(viii) Propanoic acid into ethanoic acid.

$$\textbf{Ans.}\space\text{(i) CH}_{3}\text{COOH}\xrightarrow[\text{-SO}_{2}-\text{HCl}]{\text{SOCl}_{2}}\text{CH}_{3}\text{COCl}\xrightarrow[\text{NH}_{4}\text{Cl}]{\text{NH}_{3}(\text{excess})}\text{CH}_{3}\text{CONH}_{2}\xrightarrow{\text{Br}_{2}/\text{NaOH}}\underset{\text{Methanamine}}{\text{CH}_{3}\text{NH}_{2}}$$

$$\text{(iii)}\space\underset{\text{Methanol}}{\text{CH}_{3}\text{OH}}\xrightarrow[-\text{POCl}_{3}]{\text{PCl}_{5}}\text{CH}_{3}\text{Cl}\xrightarrow{\text{KCH(alc)}}\text{CH}_{3}\text{CN}\xrightarrow{\text{H}_{3}\text{O}^{\normalsize+}}\underset{\text{Ethanoic acid}}{\text{CH}_{3}\text{COOH}}$$

$$\underset{\text{Ethanoic acid}}{\text{CH}_{3}\text{COOH}}\xrightarrow{\text{LiAlH}_{4}}\text{CH}_{3}\text{CH}_{2}\text{OH}\xrightarrow[\text{Or} \text{PI}_{3}]{\text{P+I}_{2}}\text{CH}_{3}\text{CH}_{2}\text{I}\xrightarrow[\text{-Kl}]{\text{-KCl}}\text{CH}_{3}\text{CH}_{2}\text{CN}\xrightarrow{\text{H}^{\normalsize+}/\text{H}_{2}\text{O}}\underset{\text{Propanoic acid}}{\text{CH}_{3}\text{CH}_{2}\text{COOH}}\\\text{(vi)\space}\text{CH}_{3}\text{NH}_{2}\xrightarrow[\text{-N}_{2},-\text{H}_{2}\text{O}]{\text{HONO}}\text{CH}_{3}\text{OH}\xrightarrow[\text{Or PI}_{3}]{\text{P + I}_{2}}\text{CH}_{3}\text{I}\xrightarrow[\text{-KI}]{\text{KCl}}\text{CH}_{3}\text{CN}\xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH}}\text{CH}_{3}\text{CH}_{2}\text{NH}_{2}\\\text{(vii)}\space\underset{\text{Nitromethane}}{\text{CH}_{3}\text{NO}_{2}}\xrightarrow{\text{Sn/HCl}}\text{CH}_{3}\text{NH}_{2}\xrightarrow{\text{CHCl}_{3},\text{KOH},\Delta}\text{CH}_{3}\text{NC}\xrightarrow{\text{LiAIH}_{4}}\underset{\text{Dimethylamine}}{\text{CH}_{3}\text{NHCH}_{3}}$$

6. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Ans. Primary, secondary and tertiary amines can be identified and distinguished by Hinsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C₆H₅SO₂Cl). The three types of amines react differently with Hinsberg’s reagent. Therefore, they can be easily identified using Hinsberg’s reagent.

(i) Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali.

(ii) Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.

There is no H-atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali.

(iii) On the other hand, tertiary amines do not react with Hinsberg’s reagent at all.

7. Write short notes on the following:

(i) Carbylamine reaction

(ii) Diazotisation

(iii) Hofmann’s bromamide reaction

(iv) Coupling reaction

(v) Ammonolysis

(vi) Acetylation

(vii) Gabriel phthalimide synthesis

Ans. (i) Carbylamine reaction: Also called as isocyanide reaction which is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

$$\underset{\text{Primaryamine}}{\text{R - NH}_{2}} + \underset{\text{Chloroform}}{\text{CHCl}_{3}} + \underset{\text{Potassiumhydroxide}}{3\text{KOH}(\text{alc.})}\xrightarrow{\Delta}\underset{\text{Carbylamine}}{\text{R-NC}} + 3\text{KCl + 3H}_{2}\text{O}\\\text{For example,}\\\underset{\text{Methanamine}}{\text{CH}_{3} - \text{NH}_{2}} + \text{CHCl}_{3} + \text{3KOH(alc.)}\xrightarrow{\Delta}\underset{\text{Methyl carbylamine or methyl isocyanide}}{\text{CH}_{3}-\text{NC} + }\text{3KCl + 3H}_{2}\text{O}$$

(ii) Diazotisation: Aromatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.

For example, on treatment with NaNO₂ and HCl at 273−278 K, aniline produces benzenediazonium chloride, with NaCl and H₂O as by-products.

(iii) Hoffmann’s bromamide reaction: When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom. Carboxylic group leaves as metal carbonate by-product.

(iv) Coupling reaction: The reaction of joining two aromatic rings through the −N=N−bond is known as coupling reaction. Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form coloured azo compounds.

It can be observed that, the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution.

(v) Ammonolysis: When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (– NH₂) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

$$\text{R}-\text{N}^{\normalsize+}\text{H}_{3}\text{X}^{\normalsize-} + \text{NaOH}\xrightarrow{}\underset{\text{Amine}}{\text{R-NH}_{2} + \text{H}_{2}\text{O} + \text{NaX}}$$

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines and also a quaternary ammonium salt as shown.

$$\underset{(1)\degree}{\text{RNH}_{2}}\xrightarrow{\text{RX}}\underset{\text{(2\degree)}}{\text{R}_{2}\text{NH}}\xrightarrow{\text{RX}}\underset{\text{(3\degree)}}{\text{R}_{3}\text{N}}\xrightarrow{\text{RX}}\underset{\text{Quaternaryammonium salt}}{\text{R}_{4}\text{N}^{\normalsize+}\text{X}^{\normalsize-}}$$

(vi) Acetylation: Acetylation (or ethanoylation) is the process of introducing an acetyl group into a molecule.

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution when treated with acid chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of – NH₂ or – NH group by the acetyl group, which in turn leads to the production of amides. To shift the equilibrium to the right hand side, the HCl formed during the reaction is removed as soon as it is formed. This reaction is carried out in the presence of a base (such as pyridine) which is stronger than the amine.

When amines react with benzoyl chloride, the reaction is also known as benzoylation. For example,

(vii) Gabriel phthalimide synthesis: Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

8. Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

(ii) Benzene to m-bromophenol

(iii) Benzoic acid to aniline

(iv) Aniline to 2, 4, 6-tribromofluorobenzene

(v) Benzyl chlor ide to 2-phenylethanamine

(vi) Chlorobenzene to p-Chloroaniline

(vii) Aniline to p-bromoaniIine

(viii) Benzamide to toluene

(ix) Aniline to benzyl alcohol.

9. Give the structures of A,B and C in the following reactions:

$$\text{(i)}\space\text{CH}_{3}\text{CH}_{2}\text{I}\xrightarrow{\text{NaCN}}\text{A}\xrightarrow[\text{Partial hydrolysis}]{\text{OH}^{\normalsize-}}\text{B}\xrightarrow{\text{NaOH + Br}_{2}}\text{C}\\\text{(ii)\space}\text{C}_{6}\text{H}_{5}\text{N}_{2}\text{Cl}\xrightarrow{\text{CuCN}}\text{A}\xrightarrow{\text{H}_{2}\text{O/H}^{\normalsize+}}\text{B}\xrightarrow{\text{NH}_{3}}\\\text{(iii)\space}\text{CH}_{3}\text{CH}_{2}\text{Br}\xrightarrow{\text{KCN}}\text{A}\xrightarrow{\text{LiAH}_{4}}\text{B}\xrightarrow[0\degree\text{C}]{\text{HNO}_{2}}\\\text{(iii)\space}\text{CH}_{3}\text{CH}_{2}\text{Br}\xrightarrow{\text{KCN}}\text{A}\xrightarrow{\text{LiAlH}_{4}}\text{B}\xrightarrow[\text{0}\degree\text{C}]{\text{HNO}_{2}}\text{C}\\\text{(iv)\space\text{C}}_{6}\text{H}_{5}\text{NO}_{2}\xrightarrow{\text{Fe/HCl}}\text{A}\xrightarrow[\text{273 K}]{\text{NaNO}_{2} + \text{HCl}}\text{B}\xrightarrow[\Delta]{\text{H}_{2}\text{O}/\text{H}^{\normalsize+}}\text{C}\\\text{(v)\space}\text{CH}_{3}\text{COOH}\xrightarrow[\Delta]{\text{NH}_{3}}\text{A}\xrightarrow{\text{NaOBr}}\text{B}\xrightarrow{\text{NaNO}_{2}/\text{HCl}}\text{C}\\\text{(vi)\space}\text{C}_{6}\text{H}_{5}\text{NO}_{2}\xrightarrow{\text{Fe/HCl}}\text{A}\xrightarrow[\text{273 K}]{\text{HNO}_{2}}\text{B}\xrightarrow{\text{C}_{6}\text{H}_{5}\text{OH}}\text{C}$$

$$\text{(ii)\space}\underset{\text{Benzene diazonium chloride}}{\text{C}_{6}\text{H}_{5}\text{N}_{2}^{\normalsize+}\text{Cl}^{\normalsize-}}\xrightarrow{\text{CuCN}}\underset{\text{Cyanobenzene(A)}}{\text{C}_{6}\text{H}_{5}\text{CN}}\xrightarrow{\text{H}_{2}\text{O}/\text{H}^{\normalsize+}}\underset{\text{Benzoic acid (B)}}{\text{C}_{6}\text{H}_{5}\text{COOH}}\xrightarrow[\Delta]{\text{NH}_{3}}\underset{\text{Benzamide (C)}}{\text{C}_{6}\text{H}_{5}\text{CONH}_{2}}\\\text{(iii)\space}\underset{\text{Ethyl bromide}}{\text{CH}_{3}\text{CH}_{2}\text{Br}}\xrightarrow{\text{KCN}}\underset{\text{Ethylcyanide(A)}}{\text{CH}_{3}\text{CH}_{2}\text{CN}}\xrightarrow{\text{LiAlH}_{4}}\underset{\text{Propane-1-amine}}{\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{NH}_{2}}\xrightarrow[\text{0\degree C}]{\text{HNO}_{2}}\underset{\text{Propane-1-ol}}{\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{OH}}\\\text{(iv)\space}\underset{\text{Nitrobenzene}}{\text{C}_{6}\text{H}_{5}\text{NO}_{2}}\xrightarrow{\text{Fe/HCl}}\underset{\text{Aniline (A)}}{\text{C}_{6}\text{H}_{5}\text{NH}_{2}}\xrightarrow[\text{273 K}]{\text{NaNO}_{2} + \text{HCl}}\underset{\text{Benzene diazonium chloride(B)}}{\text{C}_{6}\text{H}_{5}-\text{N}_{2}^{\normalsize+}\text{Cl}^{\normalsize-}}\xrightarrow[\Delta]{\text{H}_{2}\text{O}/\text{H}^{\normalsize+}}\underset{\text{Phenol (C)}}{\text{C}_{6}\text{H}_{5}\text{OH}}\\\text{(v)}\underset{\text{Acetic acid}}{\text{CH}_{3}\text{COOH}}\xrightarrow[\Delta]{\text{NH}_{3}}\underset{\text{Ethanamide (A)}}{\text{CH}_{3}\text{CONH}_{2}}\xrightarrow{\text{NaOBr}}\underset{\text{Methanamine (B)}}{\text{CH}_{3}\text{NH}_{2}}\xrightarrow{\text{NaNO}_{2}/\text{HCl}}\underset{\text{Methanol}}{\text{CH}_{3}\text{OH}}$$

10. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.

Ans. It is given that compound `C’ having the molecular formula, C₆H₇N is formed by heating compound `B' with Br₂ and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound `B' is an amide and compound `C' is an amine. The only amine having the molecular formula, C₆H₇N is aniline, (C₆H₅NH₂).

Therefore, compound `B' (from which `C' is formed) must be benzamide, (C₆H₅CONH₂).

Further , benzamide is formed by heating compound `A' with aqueous ammonia.

Therefore, compound `A' must be benzoic acid.

The given reactions can be explained with the help of the following reaction:

11. Complete the following reactions:

$$\textbf{(i)}\space\textbf{C}_{6}\textbf{H}_{5}\textbf{NH}_{2} + \textbf{CHCl}_{3} + \textbf{alc. KOH}\xrightarrow{}\\\textbf{(ii)\space}\textbf{C}_{6}\textbf{H}_{5}\textbf{N}_{2}\textbf{Cl} + \text{H}_{3}\textbf{PO}_{2} + \textbf{H}_{2}\textbf{O}\xrightarrow{}\\\textbf{(iii)\space}\textbf{C}_{6}\textbf{H}_{5}\textbf{NH}_{2} + \textbf{H}_{2}\textbf{SO}_{4}(\textbf{conc})\xrightarrow{}\\\textbf{(iv)\space}\textbf{C}_{6}\textbf{H}_{5}\textbf{N}_{2}\textbf{Cl} + \textbf{C}_{2}\textbf{H}_{5}\textbf{OH}\xrightarrow{}\\\textbf{(v)\space}\textbf{C}_{6}\textbf{H}_{5}\textbf{NH}_{2} + \textbf{Br}_{2}\textbf{(aq)}\xrightarrow{}\\\textbf{(vi)\space}\textbf{C}_{6}\textbf{H}_{5}\textbf{NH}_{2} + \textbf{(CH})_{3}\textbf{(CO})_{2}\textbf{O}\xrightarrow{}\xrightarrow{}\\\textbf{(vii)\space C}_{6}\textbf{H}_{5}\textbf{N}_{2}\textbf{Cl}\xrightarrow[\textbf{(ii) NaNO}_{2}/\text{Cu},\Delta]{\textbf{(i)\space HBF}_{4}}$$

$$\textbf{Ans.}\space\text{(i)\space}\text{C}_{6}\text{H}_{5}\text{NH}_{2} + \text{CHCl}_{3} + 3\text{KOH}\xrightarrow{\Delta}\text{C}_{6}\text{H}_{5}\text{N}≡\text{C} + \text{3KCl + 3H}_{2}\text{O}\\\text{(ii)\space}2\text{C}_{6}\text{H}_{5}\text{N}_{2}\text{Cl} + \text{H}_{3}\text{PO}_{2} + \text{H}_{2}\text{O}\xrightarrow{\text{Cu}^{\normalsize+}}2\text{C}_{6}\text{H}_{6} + 2\text{N}_{2} + \text{H}_{3}\text{PO}_{3} + \text{2HCl}$$

$$\text{(iv)\space C}_{6}\text{H}_{5}\text{N}_{2}\text{Cl} + \text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Reduction}}\text{C}_{6}\text{H}_{6} + \text{CH}_{3}\text{CHO} + \text{N}_{2} + \text{HCl}$$

$$\text{(vi)\space}\underset{\text{Aniline}}{\text{C}_{6}\text{H}_{5}\text{NH}_{2}} + \underset{\text{Acetic anhydride}}{\text{CH}_{3}\text{CO} -\text{O}-\text{COCH}_{3}}\xrightarrow[\text{Or pyridine}]{\text{CH}_{3}\text{COOH}}\underset{\text{Acetanilide}}{\text{C}_{6}\text{H}_{5}\text{NHCOCH}_{3} + }\text{CH}_{3}\text{COOH}\\\text{(vii)\space}\underset{\text{Benzene diazonium chloride}}{\text{C}_{6}\text{H}_{5}\text{N}_{2}\text{Cl}}\xrightarrow[\text{-HCl}]{\text{HBF}_{4}}\underset{\text{Benzene dizonium fluoroborate}}{\text{C}_{6}\text{H}_{5}\text{N}_{2}^{\normalsize+}\text{BF}_{4}^{\normalsize-}}\xrightarrow[\Delta]{\text{NaNO}_{2}/\text{Cu}}\underset{\text{Nitrobenzene}}{\text{C}_{6}\text{H}_{5}\text{NO}_{2} + \text{NaF} + \text{NaBF}_{3}}$$

12. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Ans. Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (S_N2) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

13. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Ans. (i) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273–278 K) to form benzene diazonium salt. The reactions is known as diazotization reaction.

(ii) Primary aliphatic amines react with nitrous acid at low temperature to form primary alcohol and nitrogen gas accompanied by risk effervescence. For example:

$$\underset{\text{Ethylamine}}{\text{C}_{2}\text{H}_{5}\text{NH}_{2}} + \text{HONO}\xrightarrow{\text{273-278 K}}\underset{\text{Ethyl alcohol}}{\text{C}_{2}\text{H}_{5}\text{OH} + \text{N}_{2} + \text{H}_{2}\text{O}}$$

14. Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Ans. (i) Loss of proton from an amine gives an amide ion while loss of a proton from alcohol give an alkoxide ion.

R—NH₂ → R—NH^– + H⁺

R—O—H → R—O^– + H⁺

Since O is more electronegative than N, so it will attract positive species more strongly in comparison to N. Thus, RO^~ is more stable than RNH®. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.

(ii) Due to the presence
of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass.

(iii) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons

(a) Due to resonance in aniline and other aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus it is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia and aliphatic amines.

(b) Aromatic amines are more stable than corresponding protonated ion; Hence, they hag very less tendency to combine with a proton to form corresponding protonated ion, and thus they are less basic.