`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 question, we need to follow the steps of the reactions involving toluene, N-bromosuccinimide (NBS), and the cyanide ion (CN⁻).

### Step-by-Step Solution:

1. **Identify the Reactants**:
We start with toluene (C₇H₈), which has the structure:
\[ \text{Ph-CH}_3 \]
where "Ph" represents the phenyl group (benzene ring).

2. **Reaction with NBS**:
Toluene reacts with N-bromosuccinimide (NBS) to undergo allylic bromination. In this case, the hydrogen atom on the carbon adjacent to the benzene ring (the benzylic position) is replaced by a bromine atom. The reaction can be represented as:
\[ \text{Ph-CH}_3 \xrightarrow{\text{NBS}} \text{Ph-CH}_2Br \]
This gives us our product (1):
\[ \text{Product (1): Ph-CH}_2Br \]

3. **Reaction with Cyanide Ion (CN⁻)**:
The next step involves the reaction of product (1) with the cyanide ion (CN⁻). The bromine atom is a good leaving group, and the carbon attached to bromine becomes electrophilic due to the electronegativity of bromine. The cyanide ion acts as a nucleophile and attacks this carbon, leading to the substitution of bromine with the cyanide group. The reaction can be represented as:
\[ \text{Ph-CH}_2Br + \text{CN}^- \rightarrow \text{Ph-CH}_2CN + \text{Br}^- \]
This gives us our final product (2):
\[ \text{Product (2): Ph-CH}_2CN \]

### Final Answer:
The product (2) in the sequence is:
\[ \text{Ph-CH}_2CN \]