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

Concept: Basic strength depends upon the availability of the lone pair of electrons on the nitrogen atom. Greater availability of the lone pair results in greater basic strength. Electron-donating groups increase basicity, whereas electron-withdrawing effects and resonance decrease basicity.

Step 1: Comparing aniline and ammonia. In aniline, \[ C_6H_5NH_2 \] the lone pair on nitrogen participates in resonance with the benzene ring. Because of resonance, the lone pair becomes less available for protonation. Therefore, aniline is less basic than ammonia. \[ C_6H_5NH_2 \lt NH_3 \]

Step 2: Comparing benzylamine with ammonia. In benzylamine, \[ C_6H_5CH_2NH_2 \] the amino group is separated from the benzene ring by a \(CH_2\) group. Hence the lone pair does not participate significantly in resonance. Therefore benzylamine is more basic than ammonia. \[ NH_3 \lt C_6H_5CH_2NH_2 \]

Step 3: Comparing ethylamine and benzylamine. Ethyl group exerts a positive inductive effect \((+I)\). This increases electron density on nitrogen and enhances basicity. Hence, \[ C_6H_5CH_2NH_2 \lt C_2H_5NH_2 \]

Step 4: Comparing ethylamine and diethylamine. Secondary aliphatic amines generally show greater basic strength than primary amines due to stronger electron-releasing inductive effects. Thus, \[ C_2H_5NH_2 \lt (C_2H_5)_2NH \]

Step 5: Combining all comparisons into a single order. The overall increasing order becomes \[ C_6H_5NH_2 \lt NH_3 \lt C_6H_5CH_2NH_2 \lt C_2H_5NH_2 \lt (C_2H_5)_2NH \] This matches Option (A). \[ \boxed{\text{Option (A)}} \]