The reaction of benzyl chloride with sodium cyanide followed by reduction with hydrogen in the presence of nickel gives

Options

A. `beta`-phenylethylamine
B. N-isobutylaniline
C. benzylamine
D. aniline

Answer

To solve the question regarding the reaction of benzyl chloride with sodium cyanide followed by reduction with hydrogen in the presence of nickel, we can break it down into the following steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Benzyl chloride (C6H5CH2Cl)
- Sodium cyanide (NaCN)

### Step 2: Nucleophilic Substitution Reaction
When benzyl chloride reacts with sodium cyanide, a nucleophilic substitution occurs. The cyanide ion (CN⁻) acts as a nucleophile and attacks the carbon atom bonded to the chlorine atom in benzyl chloride. The chlorine atom is a good leaving group and leaves as chloride ion (Cl⁻).

**Reaction:**
\[ \text{C}_6\text{H}_5\text{CH}_2\text{Cl} + \text{NaCN} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{CN} + \text{NaCl} \]

**Product:**
- The product formed is benzyl cyanide (C6H5CH2CN).

### Step 3: Reduction of Benzyl Cyanide
The next step involves the reduction of benzyl cyanide (C6H5CH2CN) using hydrogen gas (H2) in the presence of nickel (Ni) as a catalyst. This reduction converts the nitrile group (CN) into an amine group (NH2).

**Reaction:**
\[ \text{C}_6\text{H}_5\text{CH}_2\text{CN} + 2\text{H}_2 \xrightarrow{\text{Ni}} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \]

**Product:**
- The final product is benzylamine (C6H5CH2NH2).

### Final Answer
The reaction of benzyl chloride with sodium cyanide followed by reduction with hydrogen in the presence of nickel gives **benzylamine (C6H5CH2NH2)**.

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