Arrange the following amines in increasing order of basic strength in gaseous phase

(A) $(CH_3)_3N$
(B) $CH_3NH_2$
(C) $C_6H_5NH_2$
(D) $C_6H_5 CH_2NH_2$

Choose the correct answer from the options given below:

(C), (D), (B), (A)

The correct answer is Option (1) → (C), (D), (B), (A)

Correct Order (Increasing Basic Strength in Gaseous Phase):

$(C) < (D) < (B) < (A)$

that is

$C_{6}H_{5}NH_{2} < C_{6}H_{5}CH_{2}NH_{2} < CH_{3}NH_{2} < (CH_{3})_{3}N$

Complete statement:

In the gaseous phase, the basic strength of amines depends mainly on the $+I$ (electron-donating) effect of alkyl groups and the availability of the lone pair on nitrogen. Solvation effects are absent in gas phase.

Reasoning:

$(C)$ $C_{6}H_{5}NH_{2}$ (Aniline)

• Least basic. The lone pair on nitrogen is delocalized into the benzene ring due to resonance, so it is less available to accept a proton.

$(D)$ $C_{6}H_{5}CH_{2}NH_{2}$ (Benzylamine)

• More basic than aniline because the nitrogen lone pair is not directly involved in resonance with the ring. The benzene ring is separated by $CH_{2}$.

$(B)$ $CH_{3}NH_{2}$ (Methylamine)

• More basic due to the $+I$ effect of the methyl group, which pushes electron density toward nitrogen.

$(A)$ $(CH_{3})_{3}N$ (Trimethylamine)

• Most basic in the gaseous phase. Three methyl groups show strong $+I$ effect, increasing electron density on nitrogen and making the lone pair most available.

Hence, increasing order of basic strength:

Aniline < Benzylamine < Methylamine < Trimethylamine.