Question

Arrange the following in increasing order of their basic strength:
(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH_2, C₆H₅CH₂NH_2.

Answer 1

(i) Basic strength:- Basic strength of an amine is determined by availability of electron from nitrogen.

Basic strength increasing order:-

(C₂​H₅​)₂​NH > C₂​H₅​NH₂​ > NH_3​ > C₆​H_5​CH₂​NH₂​ > C₆​H₅​NH₂

(ii) Basic strength:- Amine basic strength is determined by availability of electron [lone pair] from nitrogen.

Increasing order of basic strength:-

(C_2​H₅​)₂​NH > (C₂​H₅​)₃​N > C₂​H₅​NH₂​ > C_6​H₅​NH_2​

(C₂​H₅​)₃​N is less basic than (C₂​H₅​)₂​NH because of steric hindrance, electrons are less available from (C₂​H₅​)₃​N than (C₂​H_5​)₂​N

(iii) C₆​H₅​NH_2​ < C₂​H₅​CH₂​NH2​ < (CH₃​)₃​N < CH₃​NH₂​ < (CH₃​)₂​NH

Answer 2

(i) Considering the inductive effect of alkyl groups, NH₃, C₂H₅NH_2, and (C₂H₅)₂NH can be arranged in the increasing order of their basic strengths as:

Again, C₆H₅NH₂ has proton acceptability less than NH3. Thus, we have:

Due to the −I effect of C₆H₅ group, the electron density on the N-atom in C₆H₅CH₂NH₂ is  lower than that on the N-atom in C₂H₅NH₂, 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,
C₂H₅NH₂, (C₂ H₅)₂NH_2, and their basic strengths as follows:

Again, due to the −R effect of C6H5 group, the electron density on the N atom in C₆H₅ NH₂ is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C₆H₅NH₂ is lower than that of C₂H₅NH₂. 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, CH₃NH₂, (CH₃)₂NH, and (CH₃)₃N can be arranged in the increasing order of their basic strengths as

In C₆H₅NH₂, 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 C₆H₅CH₂NH₂, 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
C₆H₅CH₂NH₂ than in C₆H₅NH₂ i.e., C₆H₅CH₂NH₂ is more basic than C₆H₅NH₂. Again, due to the −I effect of C6H5 group, the electron density on the N−atom in
C₆H₅CH₂NH₂ is lower than that on the N−atom in (CH₃)₃N. Therefore, (CH₃)₃N is more basic

than C₆H₅CH₂NH₂. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.