$C_{6}H_{5}CH_{2}OH\xrightarrow{HBr} A\xrightarrow{KCN} B \xrightarrow[ ext{Partial hydrolysis}]{H_{2}O/H^{+}} C \xrightarrow{NaOBr} D$
D' in the above reaction is :

Question

$C_{6}H_{5}CH_{2}OH\xrightarrow{HBr} A\xrightarrow{KCN} B \xrightarrow[	ext{Partial hydrolysis}]{H_{2}O/H^{+}} C \xrightarrow{NaOBr} D$
D' in the above reaction is :

Accepted Answer

To understand why Option 2, "Phenyl methanamine," is the correct answer, let's analyze the reaction sequence step by step:

1. **First Step: Formation of A**
   - Starting with $C_{6}H_{5}CH_{2}OH$ (benzyl alcohol), it reacts with HBr. This reaction involves the substitution of the hydroxyl group (-OH) with a bromine atom, forming benzyl bromide ($C_{6}H_{5}CH_{2}Br$). This is a typical nucleophilic substitution reaction where the alcohol is converted to an alkyl halide.

2. **Second Step: Formation of B**
   - Benzyl bromide ($C_{6}H_{5}CH_{2}Br$) then reacts with KCN. This reaction is another nucleophilic substitution where the bromine atom is replaced by a cyano group (-CN), resulting in the formation of benzyl cyanide ($C_{6}H_{5}CH_{2}CN$).

3. **Third Step: Formation of C**
   - Benzyl cyanide ($C_{6}H_{5}CH_{2}CN$) undergoes partial hydrolysis in the presence of water and an acid catalyst (H⁺). Partial hydrolysis of a nitrile typically converts it into an amide. Therefore, benzyl cyanide is converted to benzyl amide ($C_{6}H_{5}CH_{2}CONH_{2}$).

4. **Fourth Step: Formation of D**
   - Finally, benzyl amide ($C_{6}H_{5}CH_{2}CONH_{2}$) reacts with NaOBr (sodium hypobromite) in the Hofmann rearrangement reaction. This reaction involves the conversion of an amide to an amine with one fewer carbon atom. Therefore, benzyl amide is converted to phenyl methanamine ($C_{6}H_{5}CH_{2}NH_{2}$).

Thus, the final product D is phenyl methanamine, which corresponds to Option 2.

**Explanation of Incorrect Options:**

1. **Option 1: Benzoic acid**
   - Benzoic acid ($C_{6}H_{5}COOH$) would be formed if the nitrile group underwent complete hydrolysis, converting it to a carboxylic acid. However, the reaction specifies partial hydrolysis, leading to an amide, not a carboxylic acid.

3. **Option 3: Phenyl ethanoic acid**
   - Phenyl ethanoic acid ($C_{6}H_{5}CH_{2}COOH$) would require the conversion of the nitrile to a carboxylic acid, followed by a rearrangement or reduction. The given reaction sequence does not support this transformation.

4. **Option 4: Phenyl ethanamine**
   - Phenyl ethanamine ($C_{6}H_{5}CH_{2}CH_{2}NH_{2}$) would require an additional carbon atom in the chain, which is not introduced in the reaction sequence. The Hofmann rearrangement reduces the carbon count by one, not increases it.

In conclusion, the reaction sequence and transformations clearly support that the final product D is phenyl methanamine, making Option 2 the correct answer.