1   Classify the following amines as primary, secondary and tertiary:$\\$ (i)$\\$ (ii)$\\$ (iii) $( C_2 H _5 ) _2 CHNH _2$$\\$ (iv) $( C _2 H _5 )_ 2 NH$

Solution :

(i)$ 1^o$$\\$ (ii)$ 3^o$$\\$ (iii) $1^o$$\\$(iv)$ 2^o$

2   (i) Write the structures of different isomeric amines corresponding to the molecular formula, . $C _4 H _{11} N$$\\$ (ii) Write $IUPAC$ names of all the isomers.$\\$ (iii) What type of isomerism is exhibited by different pairs of amines?

Solution :

(i), (ii) The structures and their $IUPAC$ names of different isomeric amines corresponding to the molecular formula, $C _4 H _{11} N$ are given below:$\\$ (a)$ CH_ 3 -CH _2 -CH _2 -CH _2 -NH _2$ Butanamine $(1 ^0 )$$\\$ (iii) Position isomers : (a) and (b); (e) and (g)$\\$ Chain isomers: (a) and (c); (a) and (d); (b)and(c); (b)and(d)$\\$ Metamers: (e) and (f). (f) and (g)$\\$ Functional isomers: All $1 ^0$ amines are functional isomers of $2 ^0$ and $3 ^0$ amines and vice-versa

(i), (ii) The structures and their $IUPAC$ names of different isomeric amines corresponding to the molecular formula, $C _4 H _{11} N$ are given below:$\\$ (a)$ CH_ 3 -CH _2 -CH _2 -CH _2 -NH _2$ Butanamine $(1 ^0 )$$\\$ (iii) Position isomers : (a) and (b); (e) and (g)$\\$ Chain isomers: (a) and (c); (a) and (d); (b)and(c); (b)and(d)$\\$ Metamers: (e) and (f). (f) and (g)$\\$ Functional isomers: All $1 ^0$ amines are functional isomers of $2 ^0$ and $3 ^0$ amines and vice-versa

3   How will you convert:$\\$ (i)Benzene into aniline$\\$ (ii)Benzene into N,N-dimethylaniline$\\$ (iii) $Cl -$$\\$ $( CH _2 )_ 4 - Cl$ into Hexane $-1, 6-$ diamine

Solution :

4   Arrange the following in increasing order of their basic strength:$\\$ (i) $C _2 H _5 NH _2 , C_ 6 H _5 NH _2 , NH _3 , C_ 6 H_ 5 CH_ 2 NH_ 2$ and $(C_ 2 H_ 5 )_ 2 NH$$\\$ (ii)$ C_ 2 H_ 5 NH _2 , (C_ 2 H _5 )_ 2 NH, (C_ 2 H _5 )_ 3 N, C_ 6 H _5 NH _2$$\\$ (iii)$ CH _3 NH_ 2 , (CH _3 ) _2 NH, (CH _3 ) _3 N, C_ 6 H_ 5 NH _2 , C_ 6 H_ 5 CH _2 NH_ 2$

Solution :

(i) Considering the inductive effect of alkyl groups, $NH_ 3 , C_ 2 H_ 5 NH _2$ , and $(C _2 H_ 5 )_ 2 NH$ can be arranged in the increasing order of their basic strengths as:$\\$ $NH_3 < C_2H_5NH_2 < (C_2H_5)_2 NH$$\\$ Again, $C _6 H_ 5 NH_ 2$ has proton acceptability less than $NH_ 3 $. Thus, we have:$\\$ $C_6H_5NH_2 < NH_3 < C_2H_5NH_2 < (C_2H_5)_2NH$$\\$ Due to the -I effect of $C_ 6 H _5$ group, the electron density on the N-atom in $C_ 6 H_ 5 CH _2 NH_ 2$ is lower than that on the N-atom in $C _2 H_ 5 NH_ 2$ , but more than that in $NH _3$ . Therefore, the given compounds can be arranged in the order of their basic strengths as:$\\$

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, $C _2 H _5 NH_ 2 , (C_ 2 H_ 5 )_ 2 NH _2$ , and their basic strengths as follows:$\\$ Again, due to the -R effect of $C_ 6 H_ 5$ group, the electron density on the N atom in $C_ 6 H_ 5 NH_ 2$ is lower than that on the N atom in $C_ 2 H _5 NH_ 2$ . Therefore, the basicity of $C_ 6 H_ 5 NH_ 2$ is lower than that of $C_ 2 H_ 5 NH_ 2$ . Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:$\\$ (iii) Considering the inductive effect and the steric hindrance of alkyl groups,$ CH_ 3 NH_ 2 , (CH_ 3 )_ 2 NH,$ and $(CH_ 3 )_ 3 N$ can be arranged in the increasing order of their basic strengths as:$\\$ In $C_ 6 H_ 5 NH_ 2 , N$ is directly attached to the benzene ring. Thus, the lone pair of electrons on the N-atom is delocalized over the benzene ring. In $C _6 H _5 CH_ 2 NH_ 2 , N$ is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in $C_ 6 H_ 5 CH_ 2 NH_ 2$ than in $C_ 6 H_ 5 NH_ 2$ i.e.,$ C_ 6 H_ 5 CH_ 2 NH_ 2$ is more basic than $C _6 H_ 5 NH_ 2$ . Again, due to the -I effect of $C _6 H_ 5$ group, the electron density on the N-atom in $C_ 6 H_ 5 CH_ 2 NH _2$ is lower than that on the N-atom in $(CH _3 )_ 3 N$. Therefore, $(CH _3 )_ 3 N$ is more basic than $C _6 H_ 5 CH_ 2 NH_ 2$ . Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.$\\$

5   Complete the following acid-base reactions and name the products:$\\$ (i)$ CH_3 CH _2 CH _2 NH _2 + HCl \to$$\\$ (ii)$ (C _2 H _5 )_ 3 N+HCl \to $

Solution :

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

Solution :

Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of $Na_ 2 CO _3$ solution, N, N-dimethylaniline produces N, N, N-trimethylanilinium carbonate.

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

Solution :

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 $N _2$ gas on treatment with nitrous acid.

Solution :

In all, four structural isomers are possible. These are:

only $1^o$ amines react with $HNO_ 2$ to liberate $N _2$ gas

9   Convert:$\\$ (i)3-Methylaniline into 3-nitrotoluene$\\$ (ii)Aniline into 1,3,5- Tribromo benzene

Solution :

10   Write $IUPAC$ names of the following compounds and classify them into primary, secondary and tertiary amines.$\\$ (i) $(CH _3 )_ 2 CHNH _2$ $\\$ (ii) $CH _3 (CH _2 )_ 2 NH_ 2$ $\\$ (iii) $CH_ 3 NHCH (CH _3 )_ 2$ $\\$ (iv)$(CH_ 3 )_ 3 CNH$ $\\$ (v)$ C_ 6 H _5 NHCH$ $\\$ (vi)$ (CH _3 CH _2 ) _2 NCH_ 3$ $\\$ (vii) $m-BrC _6 H _4 NH_ 2$

Solution :

(i) 1-Methylethanamine ($1^ 0$ amine) $\\$ (ii) Propan-1-amine ($1^ 0$ amine) $\\$ (iii) N-Methyl-2-methylethanamine ($2^ 0$ amine) $\\$ (iv) 2-Methylpropan-2-amine ($1 ^0$ amine) $\\$ (v) N-Methylbenzamine or N-methylaniline ($2^ 0$ amine) $\\$ (vi) N-Ethyl-N-methylethanamine ($3^ 0$ amine) $\\$ (vii) 3-Bromobenzenamine or 3-bromoaniline ($1^ 0$ amine) $\\$

11   Why cannot vitamin C be stored in our body?

Solution :