`Ph CH_(3) overset( NBS ) to (1) overset( CN^(-))to (2) ` the product (2) in the sequence is :

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, we will analyze the reactions involving phenyl with a methyl group (Ph-CH₃) first with N-bromo-succinimide (NBS) and then with the cyanide ion (CN⁻).

### Step 1: Reaction of Ph-CH₃ with NBS
1. **Identify the starting compound**: The starting compound is toluene (Ph-CH₃).
2. **Understand the role of NBS**: NBS is used for allylic bromination, which involves the formation of a free radical.
3. **Mechanism of reaction**:
- When toluene reacts with NBS, a bromine radical is generated.
- The bromine radical abstracts an allylic hydrogen (the hydrogen on the carbon adjacent to the methyl group).
- This forms a free radical on the carbon adjacent to the methyl group, which can stabilize through resonance.
4. **Formation of Product 1**: The bromine radical then adds to the carbon where the free radical is formed, resulting in the product:
- **Product 1**: Benzyl bromide (Ph-CH₂Br).

### Step 2: Reaction of Product 1 with CN⁻
1. **Identify Product 1**: The product from the first reaction is benzyl bromide (Ph-CH₂Br).
2. **Understand the role of CN⁻**: The cyanide ion (CN⁻) is a good nucleophile.
3. **Mechanism of reaction**:
- The cyanide ion will attack the carbon atom that is bonded to the bromine in benzyl bromide.
- The bromine atom is a good leaving group and will leave as bromide ion (Br⁻).
4. **Formation of Product 2**: The nucleophilic attack by CN⁻ leads to the formation of:
- **Product 2**: Benzyl cyanide (Ph-CH₂CN).

### Final Products
- **Product 1**: Benzyl bromide (Ph-CH₂Br)
- **Product 2**: Benzyl cyanide (Ph-CH₂CN)

### Summary
The final product (2) in the sequence is benzyl cyanide (Ph-CH₂CN).

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