Aromatic Compounds Containing Nitrogen - Result Question 12
####12. The increasing basicity order of the following compounds is
(A) $CH _3 CH _2 NH _2$
(B) $CH _3 CH _2 NH$
(C) $H _3 C-\stackrel{N}{\stackrel{C}{N}-} CH _3$
(a) (D) $<$ (C) $<$ (B) $<(A)$
(b) $(A)<(B)<(C)<(D)$
(c) $(A)<(B)<(D)<(C)$
(d) $(D)<(C)<(A)<(B)$
(2019 Main, 9 Jan II)
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Answer:
Correct Answer: 12. (d)
Solution:
- If we consider Lewis basicity (basicity in aprotic solvents or in vapour phase), the order of basicity will be.
$$ \begin{aligned} & D<A<B<C \\ & 2^{\circ}<1^{\circ} \\ & 2^{\circ} \quad 3^{\circ} \end{aligned} $$
But, this order does not match with the options given. So, it has been asked on basicity of the amines in aqueous solution. When no phase is given, then basicity of amine is considered in aqueous solution as they are liquids. In aqueous solution, basicity of $2^{\circ}$-amines (aliphatic) is maximum because, of the thermodynamic stability of its conjugate acid.
$$ \underset{(B)}{Et _2 NH}+H _2 O \rightleftharpoons \underset{\text { Conjugate acid }}{Et _2 \stackrel{\oplus}{N} H _2}+\stackrel{\ominus}{O} H $$
$Et _2 \stackrel{\oplus}{N} H _2$ is a sterically symmetric tetrahedral ion as it contains equal number (two) of bulkier Et-group and small size $H$-atoms. Here, two $H$-atoms give additional stability through hydrogen bonding with $H _2 O$ (solvent) molecules.
Aromatic amines $(D)$ are always weaker bases than aliphatic amines, because of the conjugation of $l p$ of electrons of $N$ $(+R$-effect) with the benzene ring.
So, the correct order is $(D)<(C)<(A)<(B)$.
