Amines

Chemistry

NCERT

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)$(C _2 H _5 )_ 3 N+HCl \to$