NCERT Solutions for Class 12 Chemistry Chapter 9 Amines

Page No. 262

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(i)

Answer :

The nitrogen atom is directly connected with only one carbon atom, so it is a primary aromatic amine.

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(ii)

Answer :

In this compound, the nitrogen atom is directly connected to 3 carbon atoms. So, it is a tertiary aromatic amine

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(iii) $(C_2{H}_5{)}_{2}CHN{H}_2$

Answer :

$(C_2{H}_5{)}_{2}CHN{H}_2$

Here, the N atom is directly connected to only one carbon atom. So it is a primary aliphatic amine.

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(iv) $(C_2{H}_5{)}_{2}NH$

Answer :

$(C_2{H}_5{)}_{2}NH$

In structure, we can see that N is directly connected to 2 carbon atoms. So, it is a secondary amine.

Question 9.2 (i) Write structures of different isomeric amines corresponding to the molecular formula,

$C_4{H}_{11}N$

Answer :

different isomeric amines corresponding to the molecular formula, $C_4{H}_{11}N$

(i)${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2(v){\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NHCH}}_3$

(ii) (vi)

(iii) (vii)

(iv) (viii)

Question 9.2 (ii) Write IUPAC names of all the isomers.

Answer :

IUPAC name of all the isomers-

$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}N{H}_2$	Butanamine
	Butan-2-amine
	2-Methylpropanamine
	N-Methylpropan-2-amine
	2-Methylpropan-2-amine
	N,N-Dimethylethanamine
${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NHCH}}_3$	N-Methylpropanamine
	N-Ethylethanamine

Question 9.2 (iii) What type of isomerism is exhibited by different pairs of amines?

Answer :

$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}N{H}_2$	Butanamine (Chain isomerism + position isomerism)
	Butan-2-amine (chain isomerism + position isomerism)
	2-Methylpropanamine (chain isomerism)
	N-Methylpropan-2-amine (position isomerism + metamerism)
	2-Methylpropan-2-amine (chain isomerism)
	N,N-Dimethylethanamine
$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}NHC{H}_3$	N-Methylpropanamine (Position isomerism)
	N-Ethylethanamine (no isomerism)

Question 9.3 (i) How will you convert

Benzene into aniline

Answer :

Nitration of benzene gives nitrobenzene. And after that, reduce the nitro group by catalytic hydrogenation.

Question 9.3 (ii) How will you convert

Benzene into N, N-dimethylaniline

Answer :

Nitration of benzene gives nitrobenzene, and after catalytic hydrogenation of nitrobenzene, it gives aniline. Aniline reacts with two moles of chloromethane to form N, N-dimethylaniline.

Question 9.3(iii) How will you convert

$Cl-(C{H}_2{)}_4-Cl$ into hexan-1,6-diamine

Answer :

On reacting the given reactant with ethanolic sodium cyanide, the CN molecules replace both chlorine atoms. And after that catalytic hydrogenation, we get our desired product. (reduce the CN to $-C{H}_{2}N{H}_2$ in both sides)

Question 9.4 (i) Arrange the following in increasing order of their basic strength:

$C_2{H}_{5}N{H}_2,$ $C_6{H}_{5}N{H}_2,$ $N{H}_3,$ $C_6{H}_{5}C{H}_{2}N{H}_2$ and $(C_2{H}_5{)}_{2}NH$

Answer :

Considering the inductive effect, solvation effect and steric hindrance of the alkyl group, which decides the basic strength of alkylamines. The order of basic strength in ethyl-substituted amine is-

$N{H}_3<C_2{H}_{5}N{H}_2<(C_2{H}_5{)}_{2}NH$

and order in the benzene-substituted ring-

$C_6{H}_{5}N{H}_2<C_6{H}_{5}C{H}_{2}N{H}_2$

Due to the -R effect of benzene $C_6{H}_{5}N{H}_2$ has less electron density than ammonia. So, the final order is -

$C_6{H}_{5}N{H}_2<N{H}_3<C_6{H}_{5}C{H}_{2}N{H}_2<C_2{H}_{5}N{H}_2<(C_2{H}_5{)}_{2}NH$

Question 9.4 (ii) Arrange the following in increasing order of their basic strength:

$C_2{H}_{5}N{H}_2,$ $(C_2{H}_5{)}_{2}NH,$ $(C_2{H}_5{)}_{3}N,$ $C_6{H}_{5}N{H}_2$

Answer :

$C_6{H}_{5}N{H}_2<C_2{H}_{5}N{H}_2<(C_2{H}_5{)}_{3}N<(C_2{H}_5{)}_{2}NH$

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group, the increasing order of basicity in ethyl as a substituted group is shown above.

Question 9.4 (iii) Arrange the following in increasing order of their basic strength:

(iii) $C{H}_{3}N{H}_2,$ $(C{H}_3{)}_{2}NH,$ $(C{H}_3{)}_{3}N,$ $C_6{H}_{5}N{H}_2,$ $C_6{H}_{5}C{H}_{2}N{H}_2.$

Answer :

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group, which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in the case of methyl-substituted amines is -

$C_6{H}_{5}N{H}_2<C_6{H}_{5}C{H}_{2}N{H}_2<(C{H}_3{)}_{3}N<C{H}_{3}N{H}_2<(C{H}_3{)}_{2}NH$

Question 9.5 (i) Complete the following acid-base reactions and name the products:

$C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2+HCl\to$

Answer :

The above-given reaction is an acid-base reaction. So salt is formed.

$C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2+HCl\to$ $C{H}_3(C{H}_2{)}_{2}N{H}_3^{+}C{l}^{-}$

Question 9.5 (ii) Complete the following acid-base reactions and name the products:

$(C_2{H}_5{)}_{3}N+HCl\to$

Answer :

$(C_2{H}_5{)}_{3}N+HCl\to$ $(C_2{H}_5{)}_3{N}^{+}HC{l}^{-}$

It is an acid-base reaction, so the N will accept H from $HCl$ and form a salt.

Question 9.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Answer :

The methyl iodide reacts with aniline to give N, N-dimethylaniline.

With the excess of methyl iodide in the presence of sodium carbonate solution ($N{a}_{2}C{O}_3$), N,N-dimethylaniline produces N,N,N-trimethyl anilinium carbonate.

Question 9.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Answer :

When aniline reacts with benzoyl chloride, HCl is produced as a by-product and N-phenylbenzamide is produced as a major product. The lone pair of the nitrogen atom attacks the acidic carbon of benzoyl chloride.

Question 9.8 Write structures of different isomers corresponding to the molecular formula, $C_3{H}_{9}N.$ Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Answer :

Structures of different isomers corresponding to the molecular formula $C_3{H}_{9}N$ and their IUPAC name-

(i)		Propan-1-amine
(ii)		Propan-2-amine
(iii)		N – Methylethanamine
(iv)		N,N-Dimethylmethanamine

The structures (i) and (ii) will liberate nitrogen gas ($N_2$) on treatment with nitrous acid.

Question 9.9 (i) Convert

3-Methylaniline into 3-nitrotoluene

Answer :

On diazotisation, reaction $N{H}_2$ will convert to $N_{2}Cl$ . And then reacts with fluoroboric acid, followed by $NaN{O}_2/Cu/\mathrm{\Delta }$ that converts $N_{2}Cl$ into the nitro group ($-N{O}_2$) and evolution of dinitrogen gas.

Question 9.9 (ii) Convert

Aniline into 1,3,5-tribromobenzene.

Answer :

Aniline on bromination in the presence of water gives 2,4,6 tribromoaniline, which on further reacting with nitrous acid converts $N{H}_2$ into $N_{2}Cl$. Now, with the help of $H_{3}P{O}_2/water$, it removes the $N_{2}Cl$ group from the benzene ring and we get our final product 1,3,5-tribromobenzene.

NCERT Solutions for Class 12 Chemistry Chapter 9 Amines (Exercise Questions)

The exercise questions are solved here with detailed explanations.

Question 9.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) $(C{H}_3{)}_{2}CHN{H}_2$

Answer :

The IUPAC name of the compound is Propan-2-amine

Since $N$ (nitrogen) is connected to only one carbon. Hence, it is a primary amine.

Question 9.1 (ii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$C{H}_3(C{H}_2{)}_{2}N{H}_2$

Answer :

The IUPAC name is Propan-1-amine

It is a primary amine (nitrogen connects with only one carbon atom)

Here is the structure of the compound $C{H}_3(C{H}_2{)}_{2}N{H}_2$ -

Question 9.1 (iii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$C{H}_{3}NHCH(C{H}_3{)}_2$

Answer :

$C{H}_{3}NHCH(C{H}_3{)}_2$

N-Methylpropan-2-amine

Since N atom is connected with two carbon atoms, it is a two-degree amine.

Question 9.1(iv) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$(C{H}_3{)}_{3}CN{H}_2$

Answer :

$(C{H}_3{)}_{3}CN{H}_2$

2-methyl-propan-2-amine

It is a one-degree amine

Question 9.1 (v) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$C_6{H}_{5}NHC{H}_3$

Answer :

$C_6{H}_{5}NHC{H}_3$

N-Methylbenzamine or N-methylaniline

Here N atom is connected to two C atoms. So, it is a secondary amine.

Question 9.1 (vi) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$(C{H}_{3}C{H}_2{)}_{2}NC{H}_3$

Answer:

$(C{H}_{3}C{H}_2{)}_{2}NC{H}_3$

N,N-Diethylmethylamine

N is connected with three carbon atoms so that it is a tertiary amine

Question 9.1 (vii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$m-Br{C}_6{H}_{4}N{H}_2$

Answer :

$m-Br{C}_6{H}_{4}N{H}_2$
3-bromoaniline or 3-bromobenzenamine

It is a primary amine

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine

Answer :

Carbylamine test

This test is used to distinguish between aliphatic and aromatic primary amines. Heating with chloroform $(CHC{l}_3)$ and potassium hydroxide ( $KOH$ ) gives isocyanides or carbylamines, which have a foul smell. Two-degree and 3-degree amines do not show this reaction.

Here, methylamine is primary and dimethylamine is a secondary amine.

The first one is 1-degree and the second one is 2-degree amine.

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(ii) Secondary and tertiary amines

Answer :

Secondary and tertiary amines can be distinguished by reacting them with Hinsberg's reagent, which is also called benzenesulphonyl chloride. $(C_6{H}_{5}S{O}_{2}Cl)$ In the case of primary amine, the product is soluble in alkali but not in the secondary amine case. And tertiary amine does not react with this reagent.

In the case of a secondary amine

Tertiary amine + benzenesulphonyl chloride $\to$ no reaction

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(iii) Ethylamine and aniline

Answer :

Ethylamine and aniline can be differentiated by the azo-dye test.

On reacting aniline with $(NaN{O}_2+dil.HCl)$ followed by 2-naphthol, gives a colored product. But when it is ethylamine, it gives brisk effervescence due to the evolution of $N_2$ gas under the same conditions.

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(iv) Aniline and benzylamine

Answer :

Benzylamine reacts with nitrous acid to form a diazonium salt, which is unstable, and it also gives alcohol with the evolution of nitrogen gas. On the other hand, aniline reacts with nitrous acid to form a stable diazonium salt.

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(v) Aniline and N-methylaniline.

Answer :

Aniline and N-methylaniline can both be distinguished by the carbylamine test. Aniline is the primary aromatic compound so it gives s positive test, but N-methyl aniline is secondary and does not give this test.

Structure of both compounds-

Question 9.3 Account for the following:

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

Answer:

Here is the structure of aniline (left) and methylamine

Due to the resonance in aniline, the electrons of the nitrogen atoms are delocalised over the benzene ring. So, because of that, the electron density at the nitrogen atom decreases and is less available for donation.

On the other hand, in the case of methylamine, the methyl ($C{H}_3$ ) group is an electron-donating group (+I effect) which increases the electron density at the N atom. Hence $p{K}_b$ value of aniline is higher than that of methylamine.

Question 9.3 Account for the following:

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

Answer:

$R=C_2{H}_5$

The extent of intermolecular hydrogen bonding in ethylamine is very high. Hence, it is soluble in water. But aniline does not undergo hydrogen bonding with water to a large extent due to the presence of a bulky hydrophobic group $(C_6{H}_5)$. Hence, it is insoluble in water.

Question 9.3 Account for the following:

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

Answer :

${\mathrm{H}}_3\mathrm{C}-{\mathrm{NH}}_2$ $-\mathrm{OH}$

Due to the +I effect of the methyl ($C{H}_3$) group, methylamine is more basic than water. So in water $C{H}_{3}N{H}_2$ produces hydroxide ion by accepting $H^{+}$ ions from water.

$C{H}_{3}N{H}_2+H_{2}O\to C{H}_3{N}^{+}{H}_3+O{H}^{-}$

Ferric chloride ($FeC{l}_3$) dissociates in water into- $FeC{l}_3\to F{e}^{3+}+3C{l}^{-}$

The hydroxide ion ($O{H}^{-}$) react with $F{e}^{3+}$ and form a precipitate of $Fe(OH{)}_3$ (reddish brown ppt)

$F{e}^{3+}+3O{H}^{-}\to Fe(OH{)}_3$

Question 9.3 Account for the following:

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

Answer :

Nitration is carried out in a strongly acidic medium, and aniline is protonated to form the anilinium ion, which is meta-directing. Because of this reason, aniline on nitration gives a substantial amount of meta-nitroaniline.

Question 9.3 Account for the following:

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

Answer :

In the Friedel-Crafts reaction, we use $AlC{l}_3$, which is a Lewis acid, and aniline has basic nature. Thus, there is an acid-base reaction between them and they form a salt.

Due to the positive charge on the N-atom. The electrophilic substitution in the benzene ring is deactivated.

Question 9.3 Account for the following:

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

Answer :

In diazonium salt, the structure undergoes resonance due to which the dispersal of positive charge is more and we know that the higher the resonance, the higher is the stability. Therefore, the diazonium salt of aromatic amines is more stable than that of aliphatic amines.

Question 9.3 Account for the following:

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

Answer :

Gabriel synthesis is used for primary amines. Secondary and tertiary amines are not formed by this method. Therefore, to obtain pure and only 1-degree amine, the Gabriel phthalimide reaction is preferred.

Question 9.4 Arrange the following:

(i) In decreasing order of the pKb values:

$C_2{H}_{5}N{H}_2,$ $C_6{H}_{5}NHC{H}_3,$ $(C_2{H}_5{)}_{2}NH$ and $C_6{H}_{5}N{H}_2$

Answer :

$C_2{H}_{5}N{H}_2$ and $(C_2{H}_5{)}_{2}NH$ are aliphatic amines, so they are more basic than the other two compounds. In ethylamine, there is only one electron-donating group (+I effect), but in diethylamine, there are two electron-donating groups, so the electron density is much higher than in ethylamine.

Between $C_6{H}_{5}NHC{H}_3$ and $C_6{H}_{5}N{H}_2$, they are weak bases because of delocalisation of lone pair electrons in the benzene ring. Aniline is less basic than N-methylaniline( $C_6{H}_{5}NHC{H}_3,$ ) due to the absence of an electron-donating group. Thus, the decreasing order of pKb values is-

$C_6{H}_{5}N{H}_2>C_6{H}_{5}NHC{H}_3>C_2{H}_{5}N{H}_2>(C_2{H}_5{)}_{2}NH$

Question 9.4 Arrange the following:

(ii) In increasing order of basic strength:

$C_6{H}_{5}N{H}_2,$ $C_6{H}_{5}N(C{H}_3{)}_2,$ $(C_2{H}_5{)}_{2}NH$ and $C{H}_{3}NH$

Answer :

The increasing order of basic strength-

$C_6{H}_{5}N{H}_2<C_6{H}_{5}N(C_2{H}_5{)}_2<C{H}_{3}N{H}_2<(C_2{H}_5{)}_{2}NH$

Aliphatic amines are more basic than aromatic amines due to the presence of a high electron density of electron at the nitrogen atom so that it can donate an electron. On the other hand, in aromatic amines, the lone pair of electrons is delocalised in the benzene ring, so the availability of electrons is less.

$(C_2{H}_5{)}_{2}NH$ is more basic because of the +I effect of the two ethyl groups (so electron density is highest in this compound), while methylamine has one ethyl group. In aromatic amines, diethylaniline is more basic due to the presence of a greater number of electron-donating groups.

Question 9.4 Arrange the following:

(iii) In increasing order of basic strength:

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

Answer :

Structure of the following compounds-

Aniline, p-nitroaniline, p-toluidine

In p-toluidine, the methyl group increases the electron density at the N atom (+I effect of methyl), so it is more basic than aniline. In p-nitroaniline, the nitro group is an electron-withdrawing group, so it decreases the electron density at nitrogen, so it is less basic than aniline.

Question 9.4 Arrange the following:

(iii) In increasing order of basic strength:

(b) $C_6{H}_{5}N{H}_2,$ $C_6{H}_{5}NHC{H}_3,$ $C_6{H}_{5}C{H}_{2}N{H}_2.$

Answer :

The increasing order of basic strength- $C_6{H}_{5}N{H}_2<C_6{H}_{5}NHC{H}_3<C_6{H}_{5}C{H}_{2}N{H}_2$

$C_6{H}_{5}NHC{H}_3$ is more basic than aniline due to the presence of one electron-donating group, which increases the electron density at nitrogen atom. But in this compound N atom is directly attached with the benzene ring, so due to -R effect $C_6{H}_{5}NHC{H}_3$ is less basic than $C_6{H}_{5}C{H}_{2}N{H}_2.$

Question 9.4 Arrange the following:

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

$C_2{H}_{5}N{H}_2,$ $(C_2{H}_5{)}_{2}NH,$ $(C_2{H}_5{)}_{3}N,$ and $N{H}_3$

Answer :

Decreasing order of basic strength in the gas phase-

$(C_2{H}_5{)}_{3}N>(C_2{H}_5{)}_{2}NH>C_2{H}_{5}N{H}_2>N{H}_3$

In the gas phase, there is no solvation effect. So, the basicity directly depends on the number of electron-donating groups. More is the electron electron-donating groups, the higher the +I effect and if the size of the donating group is large +I effect should also be high.

Question 9.4 Arrange the following:

(v) In increasing order of boiling point:

$C_2{H}_{5}OH,$ $(C{H}_3{)}_{2}NH,$ $C_2{H}_{5}N{H}_2$

Answer :

The boiling point of the compounds depends on the extent of hydrogen bonding of the compound. The more the extent, the higher the boiling point. $C_2{H}_{5}N{H}_2$, it contains two H atoms, so it has a higher boiling point than $(C{H}_3{)}_{2}NH$. On the other hand, the Oxygen atom is more electronegative than the N atom, so the strength of hydrogen bonding is higher in $C_2{H}_{5}OH$ than in $C_2{H}_{5}N{H}_2$.

Thus, the increasing order of boiling point is-

$(C_2{H}_5{)}_{2}NH<C_2{H}_{5}N{H}_2<C_2{H}_{5}OH$

Question 9.4 Arrange the following:

(vi) In increasing order of solubility in water:

$C_6{H}_{5}N{H}_2,$ $(C_2{H}_5{)}_{2}NH,$ $C_2{H}_{5}N{H}_2.$

Answer :

The more extensive the hydrogen bonding, the greater is the solubility in water. $C_2{H}_{5}N{H}_2$ contains two H atom while $(C_2{H}_5{)}_{2}NH$ contains only one. So, the $C_2{H}_{5}N{H}_2$ undergoes extensive hydrogen bonding. Hence its solubility is more than $(C_2{H}_5{)}_{2}NH$ in water. An increase in the molecular mass of amines decreases their solubility in water because of the increase in bulky hydrophobic groups.

Thus, the increasing order of solubility in water is-

$C_6{H}_{5}N{H}_{2<}(C_2{H}_5)NH<C_2{H}_{5}N{H}_2$

Question 9.5 How will you convert:

(i) Ethanoic acid into methanamine

Answer :

Ethanoic acid on reacting with $SOC{l}_2$ replaces the OH by $Cl$ and then reacts with an excess of ammonia molecule to produce methanamide ($C{H}_{3}CON{H}_2$), which, by the Hoffmann bromamide degradation reaction, gives us desired product (methenamine)

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{SOCl}}_2\to {\mathrm{CH}}_3\mathrm{COCl}\stackrel{{\mathrm{NH}}_3\text{ (excess) }}{\to }{\mathrm{CH}}_3{\mathrm{CONH}}_2\stackrel{{\mathrm{Br}}_2+\mathrm{KOH}}{\to }{\mathrm{CH}}_3{\mathrm{NH}}_2$

Question 9.5 How will you convert:

(ii) Hexanenitrile into 1-aminopentane

Answer :

First, hexanenitrile undergoes acid hydrolysis, converting the nitrile group (-CN) into a carboxylic acid $(-\mathrm{COOH})$. This carboxylic acid is then treated with thionyl chloride $\left({\mathrm{SOCl}}_2\right)$ to form the corresponding acid chloride $(-\mathrm{COCl})$. Reacting this with excess ammonia $\left({\mathrm{NH}}_3\right)$ replaces the chlorine atom with an amine group, yielding a primary amide ($-{\mathrm{CONH}}_2$).
Finally, the Hofmann bromamide degradation is carried out, which removes one carbon atom and produces the desired primary amine as the final product.

Question 9.5 How will you convert:

(iii) Methanol to ethanoic acid

Answer :

When methanol reacts with phosphorus pentachloride, OH^- is reduced by Cl^- to form chloromethane, which on reacting with ethanolic sodium cyanide, gives $C{H}_{3}CN$ and followed by acidic hydrolysis, it produces ethanoic acid ($C{H}_{3}COOH$)

Question 9.5 How will you convert:

(iv) Ethanamine into methanamine

Answer :

Ethanamine reacts with nitrous acid to form an azo compound, which further reacts with water to form ethanol and this on oxidation gives ethanoic acid. After treating with an excess of ammonia, the ethanoic acid becomes ethanamide, which further reacts with bromine and a strong base (Hoffmann bromamide degradation reaction), to form methanamine.

Question 9.5 How will you convert:

(v) Ethanoic acid into propanoic acid

Answer :

Acetic acid, on reduction with lithium aluminium hydride followed by acid hydrolysis, gives ethanol, which on reacting with $PC{l}_5$ gives ethyl chloride. Ethyl chloride further reacts with ethanolic sodium cyanide to form propanenitrile, which on acidic hydrolysis gives the desired product.

Question 9.5 How will you convert:

(vi) Methanamine into ethanamine

Answer :

Methanamine on diazotisation gives methane diazonium salt, which on further hydrolysis forms methanol. Methanol on reacting with $PC{l}_5$, followed by ethanolic sodium cyanide ($NaCN$), produces ethane nitrile. This on reduction with $H_2/Pd$ gives ethanamine.

Question 9.5 How will you convert:

(vii) Nitromethane into dimethylamine

Answer :

Conversion of Nitromethane into dimethylamine is shown below:

${\mathrm{CH}}_3{\mathrm{NO}}_2\underset{\mathrm{H}}{\overset{\mathrm{Sn}/\mathrm{HCl}}{\to }}{\mathrm{CH}}_3{\mathrm{NH}}_2\stackrel{{\mathrm{CHCl}}_3/\mathrm{KOH}}{\to }{\mathrm{CH}}_3\mathrm{NC}$

Reduction of Isocyanide

$${\mathrm{CH}}_3\mathrm{NC}\underset{\mathrm{H}}{\overset{\mathrm{Na}/{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}{\to }}{\mathrm{CH}}_3{\mathrm{NHCH}}_3$$

Question 9.5 How will you convert:

(viii) Propanoic acid into ethanoic acid?

Answer :

Propanoic acid on reacting with an excess of ammonia, gives propanamide, which further reacts with bromine and potassium hydroxide (Hoffmann bromamide degradation reaction) to give ethanamine. On diazotisation, ethanamine is converted into an azo salt, which on treatment with water forms ethanol. Ethanol on oxidation, provides ethanoic acid.

Question 9.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Answer :

Primary, secondary, and tertiary amines are distinguished by Hinsberg's reagent test. In this test, the amines are allowed to react with benzene sulphonyl chloride (Hinsberg's reagent). All three types of amines give different products.

1.Primary amine reacts with benzene sulphonyl chloride $(C_6{H}_{5}S{O}_{2}Cl)$ gives N-ethylbenzene sulphonamide which is soluble in alkali.

2. When secondary amines react, it gives N,N-diethylbenzene sulphonamide, which is insoluble in alkali.

3. Tertiary amine does not react with Hinsberg's reagent.

Question 9.7 Write short notes on the following:

(i) Carbylamine reaction

Answer :

Carbylamine reaction-

Aliphatic and aromatic primary amines on reacting with chloroform and ethanolic KOH form isocyanides or carbylamine, which is a foul-smelling substance. Secondary and tertiary amines do not give this reaction. This reaction is known as the carbylamine reaction or isocyanide test. It is used to distinguish primary amines.

$R-N{H}_2+CHC{l}_3+KOH\underset{heat}{\overset{{}^{\mathrm{\Delta }}}{\to }}R-NC+3KCl+3H_{2}O$

Question 9.7 Write short notes on the following:

(ii) Diazotisation

Answer :

Aromatic primary amines react with nitrous acid ($NaN{O}_2+2HCl$) at low temperature to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotisation.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2+{\mathrm{NaNO}}_2+2\mathrm{HCl}\stackrel{273-278\mathrm{K}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}{\mathrm{Cl}}^{-}+\mathrm{NaCl}+2{\mathrm{H}}_2\mathrm{O}$

Question 9.7 Write short notes on the following:

(iii) Hofmann’s bromamide reaction

Answer :

Hofmann’s bromamide reaction-
When an amide is treated with bromine in aqueous solution or an ethanolic solution of sodium hydroxide, it produces primary amines with one less carbon atom than the parent compound. This reaction is known as the Hoffmann bromide degradation reaction.

R = alkyl group like $C{H}_3,C_2{H}_5....etc$

Question 9.7 Write short notes on the following:

(iv) Coupling reaction

Answer :

The reaction of joining diazonium salt to the aromatic ring (like phenol) through a $-N=N-$ bond is known as a coupling reaction. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt ($C_6{H}_5{N}_2^{+}C{l}^{-}$) to give p-hydroxyazobenzene (orange in colour).

Reactions-

Diazonium salt reacts with aniline to give p-aminoazobenzene (yellow colour)

Question 9.7 Write short notes on the following:

(v) Ammonolysis

Answer :

Ammonolysis-
When an alkyl or benzyl halide is going to react with the solution of ammonia(an ethanolic solution), it undergoes a nucleophilic substitution reaction in which the halogen atom is replaced by an amino ($-N{H}_2$) group. This process of cleavage of the carbon-halogen (C-X) bond by the ammonia molecule is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base, it gives a primary amine
$R{N}^{+}{H}_3{X}^{-}+NaOH\to R-N{H}_2+H_{2}O+NaX$

• Here, primary amine behaves as a nucleophile and can further react with an alkyl halide to form secondary and tertiary amines and finally quaternary ammonium salt.

Question 9.7 Write short notes on the following:

(vi) Acetylation

Answer :

Acetylation-
The introduction of an acetyl group ( $C{H}_{3}CO$ )in a molecule is known as acetylation.

Example-

Aliphatic and aromatic primary or secondary amines undergo acetylation reaction when treated with acid chlorides, anhydrides, or esters by nucleophilic substitution. Here Hydrogen atom of $N{H}_2$ or $NH$ is replaced by $C{H}_{3}CO$ the group.

Question 9.7 Write short notes on the following:

(vii) Gabriel phthalimide synthesis.

Answer :

Gabriel phthalimide synthesis.-
It is mainly used for aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide to produce the potassium salt of phthalimide. Upon heating with an alkyl halide and subsequent alkaline hydrolysis, this salt yields the corresponding primary amine. Aromatic amines cannot be produced by this method due to the inability of aryl halides to undergo nucleophilic substitution reaction with the anion produced by phthalimide.

Question 9.8 Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

Answer :

First, reduce nitrobenzene into aniline with the help of $H_2/Pd$. Now convert aniline to diazonium salt, followed by Sandmeyer's reaction ($CuCN/KCN$) to obtain benzonitrile, and after acid hydrolysis, this benzonitrile becomes benzoic acid.

Question 9.8 Accomplish the following conversions:

(ii) Benzene to m-bromophenol

Answer :

First, benzene is nitrated to form nitrobenzene, which is a meta-directing group and thus directs incoming substituents to the meta position. Next, bromination of nitrobenzene is carried out, introducing a bromine atom at the meta position relative to the nitro group.

The nitro group is then reduced to an amine using $\mathrm{Sn}/\mathrm{HCl}$ that forms m-bromoaniline. This is followed by diazotization (reaction with ${\mathrm{NaNO}}_2$ and HCl at $0-5^{\circ }\mathrm{C}$) to form the diazonium salt, which is finally subjected to hydrolysis with warm water that gives m-bromophenol as the final product.

Question 9.8 Accomplish the following conversions:

(iii) Benzoic acid to aniline

Answer :

On reacting benzoic acid with $SOC{l}_2$, we get benzene sulphonyl chloride and when it reacts with ammonia the chlorine is replaced by $N{H}_2$. Then the Hoffmann bromide degradation reaction is carried out to remove -CO group.

Question 9.8 Accomplish the following conversions:

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

Answer :

First, we will do the bromination of aniline which gives 2,4,6-tribromoaniline. After that, we will perform diazotisation reaction which converts $N{H}_2$ group into $N_{2}Cl$ and then we will react it with fluoroboric acid( $HB{F}_4+\mathrm{\Delta }$) to give us the desired product.

Question 9.8 Accomplish the following conversions:

(v) Benzyl chloride to 2-phenylethanamine

Answer :

On reacting benzyl chloride with the ethanolic sodium cyanide ($NaCN$), it replaces the Chlorine with the CN molecule. And then reduction with the help of $H_2/Ni$

Question 9.8 Accomplish the following conversions:

(vi) Chlorobenzene to $p-chloroaniline$

Answer :

On nitration of chlorobenzene, it gives p-nitrochlorobenzene, which is on further reaction with $H_2/Pd$, reduces $N{O}_2$ to $N{H}_2$

Question 9.8 Accomplish the following conversions:

(vii) Aniline to $p-bromoaniline$

Answer :

Aniline on reacting with the acetic anhydride in the presence of pyridine, gives N-phenylethanamide, which on further reaction with $B{r}_2/C{H}_{3}COOH$, directs the bromine to the para-position. After hydrolysis, we can recover the original $N{H}_2$ group at the same position

Question 9.8 Accomplish the following conversions:

(viii) Benzamide to toluene

Answer :

Use the Hoffmann bromamide degradation reaction so that it gives aniline and then do diazotisation reaction so that $N{H}_2$ group is replaced by $N_{2}Cl$ and then reduce it with the help of ($H_{2}O/H_{3}P{O}_2$) so that it reduced into the benzene ring. And then alkylation of benzene will give toluene as a final product.

Question 9.8 Accomplish the following conversions:

(ix) Aniline to benzyl alcohol

Answer :

Aniline on treating with nitrous acid followed by sandmeyers reaction gives benzonitrile ( $C_6{H}_{5}C\equiv N$), which on acidic hydrolysis form benzoic acid ($C_6{H}_{5}COOH$). The benzoic acid is then reduced by $LiAl{H}_4$ (lithium aluminium hydride) to get the desired product.

Question 9.9 Give the structures of A, B and C in the following reactions:

(i) $C{H}_{3}C{H}_{2}I\stackrel{NaCN}{\to }A\underset{Partialhydrolysis}{\overset{O{H}^{-}}{\to }}B\stackrel{NaOH+B{r}_2}{\to }C$

Answer :

In the first reaction, the iodine is replaced by CN, so the structure of (A) is $C{H}_{3}C{H}_{2}CN$.

On partial hydrolysis, the CN converted in to $CON{H}_2$, so the structure of B is $C{H}_{3}C{H}_{2}CON{H}_2$ (propanamide). The third reaction is the Hoffmann bromide degradation reaction. So, the -CO group will be removed and it will become a primary aliphatic amine. Thus the structure of C is $C{H}_{3}C{H}_{2}N{H}_2$ (ethanamine).

Question 9.9 Give the structures of A, B and C in the following reactions:

(ii) $C_6{H}_5{N}_{2}Cl\stackrel{CuCN}{\to }A\stackrel{H_{2}O/H^{+}}{\to }B\underset{\mathrm{\Delta }}{\overset{N{H}_3}{\to }}C$

Answer :

The first part is the Sandmeyers reaction so CN replaces the $N_{2}Cl$ of the reactant and the product (A) is cyanobenzene $(C_6{H}_{5}CN)$. On acidic hydrolysis of cyanobenzene give product (B), benzoic acid $(C_6{H}_{5}COOH)$ and benzoic acid reacts with ammonia (heat) to produce

(C) benzamide $(C_6{H}_{5}CON{H}_2)$.

Question 9.9 Give the structures of A, B and C in the following reactions:

(iii) $C{H}_{3}C{H}_{2}Br\stackrel{KCN}{\to }A\stackrel{LiAl{H}_4}{\to }B\underset{0^{\circ }C}{\overset{HN{O}_2}{\to }}C$

Answer :

The structure of A,B and C is

$C{H}_{3}C{H}_{2}CN,C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$ and $C{H}_{3}C{H}_{2}C{H}_{2}OH$, respectively.

In the first reaction, the bromine is replaced by a cyanide molecule ($CN$) and after a reduction by $LiAl{H}_4$, the CN group becomes $C{H}_{2}N{H}_2$.

Question 9.9 Give the structures of A, B and C in the following reactions:

(iv) $C_6{H}_{5}N{O}_2\stackrel{Fe/HCl}{\to }A\underset{273K}{\overset{NaN{O}_2+HCl}{\to }}B\underset{H_{2}O/H^{+}}{\overset{\mathrm{\Delta }}{\to }}C$

Answer :

A- $C_6{H}_{5}N{H}_2$

B- $C_6{H}_5{N}_2^{+}C{l}^{-}$

C- $C_6{H}_{5}OH$

The first nitrobenzene reduced to aniline and then it reacts with $(NaN{O}_2/HCl)$ to form benzene diazonium salt and is hydrolysed to form phenol( $C_6{H}_{5}OH$).

Question 9.9 Give the structures of A, B and C in the following reactions:

(v) $C{H}_{3}COOH\underset{\mathrm{\Delta }}{\overset{N{H}_3}{\to }}A\stackrel{NaOBr}{\to }B\stackrel{NaN{O}_2/HCl}{\to }C$

Answer :

A- ethanamide ($C{H}_{3}CON{H}_2$)

B- methenamine ($C{H}_{3}N{H}_2$)

C- methanol ($C{H}_{3}OH$)