Question

Consider the following compounds : \[ C_2H_5NH_2,\quad (C_2H_5)_2NH,\quad C_6H_5CH_2NH_2,\quad NH_3,\quad C_6H_5NH_2 \] The correct increasing order of the above compounds on the basis of their basic strength is :

• A. \(C_6H_5NH_2 \lt NH_3 \lt C_6H_5CH_2NH_2 \lt C_2H_5NH_2 \lt (C_2H_5)_2NH\)
• B. \(NH_3 \lt C_6H_5CH_2NH_2 \lt C_6H_5NH_2 \lt C_2H_5NH_2 \lt (C_2H_5)_2NH\)
• C. \(C_6H_5CH_2NH_2 \lt (C_2H_5)_2NH \lt NH_3 \lt C_6H_5NH_2 \lt C_2H_5NH_2\)
• D. \(C_2H_5NH_2 \lt C_6H_5NH_2 \lt NH_3 \lt C_6H_5CH_2NH_2 \lt (C_2H_5)_2NH\)

Hint:

Aniline is less basic than ammonia because the nitrogen lone pair is involved in resonance. Aliphatic amines are generally more basic due to the electron-releasing \(+I\) effect of alkyl groups.

Answer 1

The Correct Option is A

Step 1: Place aniline and ammonia.
In aniline $C_6H_5NH_2$, the lone pair on $N$ delocalises into the $\pi$ system of the benzene ring through resonance, reducing its availability for protonation. So aniline is less basic than $NH_3$: $C_6H_5NH_2 < NH_3$.
Step 2: Place benzylamine and ethylamine.
In benzylamine $C_6H_5CH_2NH_2$, the $-CH_2-$ spacer interrupts conjugation, so the lone pair is not delocalised into the ring. Benzylamine is therefore more basic than $NH_3$. Ethylamine $C_2H_5NH_2$ has a stronger $+I$ (electron-donating) effect from the ethyl group than the benzylic carbon, making it more basic than benzylamine. Order so far: $C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2$.
Step 3: Place diethylamine at the top.
Diethylamine $(C_2H_5)_2NH$ bears two ethyl groups, each pushing electrons onto $N$ by the $+I$ effect, giving the highest electron density on nitrogen and the strongest basicity. Final order: \[ C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH \]
\[ \boxed{C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH} \]