Question

An aromatic compound A on treatment with ethanolic NaCN forms compound B which on reduction with $H_2/Ni$ gives a compound C of molecular formula $C_8H_{11}N$. The compounds A, B and C are

• A. chlorobenzene, phenylnitrile and 2-phenyl methanamine
• B. chlorobenzene, phenylnitrile and 2-phenyl ethanamine
• C. chlorobenzene, phenylethane nitrile and 2-phenyl methanamine
• D. chlorophenyl methane, phenylethane nitrile and 2-phenyl ethanamine
• E. dichlorobenzene, phenylnitrile and 2-phenyl methanamine

Hint:

$NaCN$ increases carbon chain by one carbon because cyanide carbon becomes part of product.

Answer 1

The Correct Option is D

Concept:
Alkyl halides react with $NaCN$ to form nitriles, and nitriles on catalytic hydrogenation give primary amines.
Step 1: Use final molecular formula.
Given final product: \[ C_8H_{11}N \] This corresponds to phenethylamine: \[ C_6H_5CH_2CH_2NH_2 \] which is 2-phenyl ethanamine.
Step 2: Find nitrile precursor B.
Reduction of nitrile: \[ R-CN \xrightarrow{H_2/Ni} R-CH_2NH_2 \] So precursor must be: \[ C_6H_5CH_2CN \] This is phenylethane nitrile.
Step 3: Find starting compound A.
Nitrile is formed from corresponding benzyl halide using ethanolic $NaCN$: \[ C_6H_5CH_2Cl \xrightarrow{NaCN} C_6H_5CH_2CN \] Thus A is chlorophenyl methane (benzyl chloride).
Step 4: Final matching.
\[ A=\text{chlorophenyl methane} \] \[ B=\text{phenylethane nitrile} \] \[ C=\text{2-phenyl ethanamine} \]
Hence, correct option is (D).