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 :

N atom is directly connected with only one C 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 N atom directly connected with 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 N atom directly connected with only one C 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 clearly see that N is directly connected with 2 carbon atom. 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 ) $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}N{H}_2$ (v) $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}NHC{H}_3$

(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
$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}NHC{H}_3$	N-Methylpropanamine
	N-Ethylethanamine

Question 13.2(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 now 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 on reacting 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 atom. 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. 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 benzene substituted ring-

$C_6{H}_{5}NH2<C_6{H}_{5}C{H}_{2}N{H}_2$

Due to the -R effect of benzene $C_6{H}_{5}NH2$ has less electron density than ammonia. So, the final order is -

$C_6{H}_{5}NH2<N{H}_3<C_6{H}_{5}C{H}_{2}N{H}_2<C_2{H}_{5}N{H}_2<(C_2{H}_5{)}_{2}NH$

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 case of methyl substituted amines are -

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

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-phenyl benzamide is produced as a major product.
lone pair of N 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 structure (i) and (ii) will liberate nitrogen gas ( $N_2$ ) on treating 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 $NaN{O}_2/Cu/\mathrm{\Delta }$ convert $N_{2}Cl$ 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)

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 1-methyl-ethanamine.

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

structure of the compound $(C{H}_3{)}_{2}CHN{H}_2$

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 :

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-Methyl-2-methylethanamine

since N atom is connected with two C atom, So 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-Methylpropane-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 with two C atom. 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-Ethyl-N-methylethanamine

N is connected with three carbon atom 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 on heating with chloroform $(CHC{l}_3)$ and potassium hydroxide ( $KOH$ ) gives isocyanides or carbylamines which has foul smelling. two degree and 3-degree amines do not show this reaction.

Here methylamine is primary and dimethylamine is a secondary amine.

first one is 1-degree and second one is 2-degree

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 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 with $(NaN{O}_2+dil.HCl)$ and 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 diazonium salt, which is unstable and 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 both can be distinguished by 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 compound-

primary aromatic amine secondary aromatic amine
13.2 Give one chemical test to distinguish between the following pairs of compounds. (v) Aniline and N-methylaniline.
Edit Q

Question 9.3 Account for the following:

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

Answer :

Here is the structure of aniline and methylamine

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

On the other hand, In the case of methylamine, the methyl ( $C{H}_3$ ) group is electron donating group (+I effect) which increase the electron density at 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 oh 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 $(C_6{H}_6)$ . 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 : ${\mathrm{H}}_3\mathrm{C}-{\mathrm{NH}}_2$ $\mathrm{H}-\mathrm{OH}$

Due to +I effect of 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 strongly acidic medium, 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 fridal craft reaction, we use $AlC{l}_3$ which is acidic in nature and aniline is also a show basic nature. Thus there is an acid-base reaction between them and form a salt.

Due to the positive charge on the N-atom. the electrophilic substitution in 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 goes under resonance due to which the dispersal of positive charge is more and we know that higher is the resonance higher is the stability. Therefore diazonium salt of aromatic amines is more stable than those 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. So, therefore to obtain pure and only 1-degree amine 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 rest two compound. In ethylamine, there is only one electron donating group (+I effect) but in diethylamine, there is two electron donating group so the electron density is much higher than ethylamine.

In between $C_6{H}_{5}NHC{H}_3,$ and $C_6{H}_{5}N{H}_2$ they are the weak base because of delocalisation of lone pair electron 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 high electron density of electron at N atom so that it can donate an electron. On the other hand in aromatic amines the lone pair of electron delocalised in the benzene ring so the availability of electron is less.

$(C_2{H}_5{)}_{2}NH$ is more basic because of +I effect of the two ethyl group (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 more number of electron donating group.

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 from N atom, 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- $C_6{H}_{5}N{H}_2<C_6{H}_{5}NHC{H}_3<C_6{H}_{5}C{H}_{2}N{H}_2$

$C_6{H}_{5}NHC{H}_3,$ it is more basic than aniline due to the presence of one electron donating group which increases the electron density at N atom. And also 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 donation group. More is the electron donating group, higher is the +I effect and if the size of the donating group is large +I effect should also 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. More is the extent; higher is 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, Oxygen atom is more electronegative than N atom, so the strength of hydrogen bonding is higher in $C_2{H}_{5}OH,$ than $C_2{H}_{5}OH,$

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 :

More extensive is the hydrogen bonding, more 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. On increasing the molecular mass of amines decreases its solubility in water because of an 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 it with an excess of ammonia molecule to produce methanamide ( $C{H}_{3}CON{H}_2$ ) which by Hoffmann bromamide degradation reaction gives us desired product (methenamine)

$C{H}_{3}COOH+SOC{l}_2\to C{H}_{3}COCl\stackrel{N{H}_3(excess)}{\to }C{H}_{3}CON{H}_2$ $\stackrel{B{r}_2+KOH}{\to }C{H}_{3}N{H}_2$

Question 9.5 How will you convert:

(ii) Hexanenitrile into 1-aminopentane

Answer :

First, acid hydrolysis of hexanenitrile which converts the CN group into COOH group and then react with $SOC{l}_2$ which replace COOH by $COCl$ and then the excess of ammonia which replace Cl by $N{H}_2$ and then Hoffmann bromamide degradation reaction which gives us desired product.

Question 9.5 How will you convert:

(iii) Methanol to ethanoic acid

Answer :

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

Question 9.5 How will you convert:

(iv) Ethanamine into methanamine

Answer :

React ethanamine with nitrous acid to form an azo compound, which further reacts with water to form ethanol, which on oxidising gives ethanoic acid. After treating with an excess of ammonia the ethanoic acid becomes ethanamide, which on further reacting with Bomine and a strong base (Hoffmann bromamide degradation reaction) to form methenamine.

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 on further reacting with ethanolic sodium cyanide to form propanenitrile, which on acidic hydrolysis gives the desired output.

Question 9.5 How will you convert:

(vi) Methanamine into ethanamine

Answer :

Methenamine on diazotisation gives methane diazonium salt, which on further hydrolysis form methanol. Methanol on reacting with $PC{l}_5$ followed by ethanolic sodium cyanide ( $NaCN$ ) produces Ethane nitrile, which on reduction with $H_2/Pd$ gives ethenamine.

Question 9.5 How will you convert:

(vii) Nitromethane into dimethylamine

Answer :

Conversion of Nitromethane into dimethylamine is shown below:

$$\begin{gathered} (Nitromethane)C{H}_{3}N{O}_2\underset{H}{\overset{Sn/HCL}{\to }}(Aminomethane)C{H}_{3}N{H}_2\stackrel{CHC{l}_3/KOH}{\to } \\ (Methylisocyanide)C{H}_{3}NC\underset{H}{\overset{Na/C_2{H}_{5}OH}{\to }}(Dimethylamine)C{H}_{3}NHC{H}_3 \end{gathered}$$

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 on further react with bromine and potassium hydroxide (Hoffmann bromamide degradation reaction) gives ethenamine. On diazotisation, ethenamine converted into azo salt which on treating 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 the Hinsberg's reagent test. In this test, the amines are allowing 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-diethyl benzene 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 foul smelling substance. Secondary and tertiary amine does not give this reaction. This reaction is known as carbylamine reaction or isocyanide test. It is used to distinguish primary amine.

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

Question 9.7 Write short notes on the following:

(ii) Diazotisation

Answer :

DIAZOTISATION-

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 salt is known as diazotisation.

$C_6{H}_{5}N{H}_2+NaN{O}_2+2HCl\stackrel{273-278K}{\to }{C}_6{H}_5{N}_2^{+}C{l}^{-}+NaCl+2H_{2}O$

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 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 :

Coupling reaction-
The reaction of joining of diazonium salt to the aromatic ring (like phenol) through $-N=N-$ bond is known ass 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) undergoes 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 ammonia molecule is known as ammonolysis.

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

• Above primary amine is 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 and 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 used for mainly aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide and produces potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives corresponding primary amine. Aromatic amines cannot be produced by this method due to the inability of aryl halides to do nucleophilic substitution reaction with 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 sandmeyers reaction ( $CuCN/KCN$ ) to benzonitrile and after acid hydrolysis benzonitrile becomes benzoic acid.

Question 9.8 Accomplish the following conversions:

(ii) Benzene to m-bromophenol

Answer :

First, We do nitration of benzene so that it can direct the bromine to meta position and then bromination of nitrobenzene after that reduce it with the help of $Sn/HCl$ followed by diazotisation reaction and then hydrolysis in warm water.

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 $N{H}_2$ and after that do Hoffmann bromide degradation reaction to remove -CO group.

Question 9.8 Accomplish the following conversions:

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

Answer :

First, we do bromination of aniline which gives 2,4,6-tribromoaniline after that do diazotisation reaction which converts $N{H}_2$ group into $N_{2}Cl$ and then react it with fluoroboric acid( $HB{F}_4+\mathrm{\Delta }$ ) gives us 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 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 reacting with $H_2/Pd$ , it 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 it gives N-phenylethanamide which on further reacting with $B{r}_2/C{H}_{3}COOH$ the $NHCOC{H}_3$ group direct the bromine to the para- position and 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 reduction 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 gives 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 ( $C_6{H}_{5}COOH$ ) and then reduction process with $LiAl{H}_4$ (lithium aluminium hydride) to get 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 $C{H}_{3}C{H}_{2}CN$ ( propionitrile) . 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) and the third reactant is Hoffmann bromide degradation reaction, So just remove the -CO group and it becomes 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 cynobenzene $(C_6{H}_{5}CN)$ . On acidic hydrolysis of cynobenzene give product (B) benzoic acid $(C_6{H}_{5}COOH)$ and benzoic acid react with ammonia (heat) it produces product (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, C are respectively $C{H}_{3}C{H}_{2}CN,C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$ and $C{H}_{3}C{H}_{2}C{H}_{2}OH$

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

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

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