Question

Which of the following set(s) of reagents convert benzoic acid (\(X\)) to primary amine (\(Y\)) with one carbon atom more than that in \(X\)? \[ \begin{aligned} \text{I. }&\mathrm{NH_3/\Delta;\ Br_2,\ OH^-} \text{II. }&\mathrm{LiAlH_4;\ H_2O} \text{III. }&\mathrm{LiAlH_4,\ H_2O;\ NaBr/H_2SO_4;\ KCN;\ H_2/Ni} \end{aligned} \] The correct answer is

• A. I, II only
• B. II, III only
• C. III only
• D. II only

Hint:

To increase the carbon chain by one and obtain a primary amine: \[ \boxed{ \mathrm{R{-}X} \xrightarrow{\mathrm{KCN}} \mathrm{R{-}CN} \xrightarrow{\mathrm{H_2/Ni}} \mathrm{R{-}CH_2NH_2} } \] The cyanide ion contributes one additional carbon atom.

Answer 1

The Correct Option is C

Step 1: Requirement of the reaction. Benzoic acid contains \[ 7 \] carbon atoms. The required primary amine must contain \[ 8 \] carbon atoms, i.e., one carbon more than benzoic acid.

Step 2: Check Set-I. \[ \mathrm{C_6H_5COOH} \xrightarrow{\mathrm{NH_3,\Delta}} \mathrm{C_6H_5CONH_2} \xrightarrow{\mathrm{Br_2/OH^-}} \mathrm{C_6H_5NH_2} \] The Hofmann bromamide reaction removes one carbon atom. Thus, aniline has only \[ 6 \] carbon atoms. Hence, \[ \boxed{\text{Set-I is incorrect}.} \]

Step 3: Check Set-II. \[ \mathrm{C_6H_5COOH} \xrightarrow{\mathrm{LiAlH_4}} \mathrm{C_6H_5CH_2OH} \] Only benzyl alcohol is formed; no amine is obtained. Hence, \[ \boxed{\text{Set-II is incorrect}.} \]

Step 4: Check Set-III. \[ \mathrm{C_6H_5COOH} \xrightarrow{\mathrm{LiAlH_4}} \mathrm{C_6H_5CH_2OH} \] \[ \mathrm{C_6H_5CH_2OH} \xrightarrow{\mathrm{NaBr/H_2SO_4}} \mathrm{C_6H_5CH_2Br} \] \[ \mathrm{C_6H_5CH_2Br} \xrightarrow{\mathrm{KCN}} \mathrm{C_6H_5CH_2CN} \] \[ \mathrm{C_6H_5CH_2CN} \xrightarrow{\mathrm{H_2/Ni}} \mathrm{C_6H_5CH_2CH_2NH_2} \] The nitrile introduces one additional carbon atom, and reduction gives a primary amine. Hence, \[ \boxed{\text{Set-III is correct}.} \] Therefore, \[ \boxed{\text{III only}} \] is the correct answer. Thus, \[ \boxed{(C)} \] is the correct answer.