The end product (B) in the following sequence of reactions is `CH_(3)Cl overset(KCN)rarr A overset(H_(2)O //H^(+))rarr B`

Options

A. `CH_(3)COOH`
B. HCOOH
C. `CH_(3)NH_(2)`
D. `CH_(3)COCH_(3)`

Explanation

To solve the question, we need to analyze the sequence of reactions starting from the reactant CH₃Cl and ending with the product B.

### Step-by-Step Solution:

1. **Identify the Starting Compound:**
The starting compound is CH₃Cl (methyl chloride).

2. **First Reaction with KCN:**
When CH₃Cl reacts with KCN, the cyanide ion (CN⁻) acts as a nucleophile. The mechanism involved here is nucleophilic substitution (S_N2). The nucleophile attacks the carbon atom bonded to chlorine, resulting in the displacement of the chlorine atom.
- **Reaction:**
\[ CH₃Cl + KCN \rightarrow CH₃CN + KCl \]
- **Product A:** Methyl cyanide (CH₃CN).

3. **Second Reaction with H₂O and H⁺:**
The product A (CH₃CN) undergoes hydrolysis in the presence of water (H₂O) and an acid (H⁺). During hydrolysis, the cyanide group (CN) is converted into a carboxylic acid group (COOH).
- **Reaction:**
\[ CH₃CN + H₂O \overset{H^+}{\rightarrow} CH₃COOH \]
- **Product B:** Acetic acid (CH₃COOH).

4. **Final Answer:**
The end product B is CH₃COOH (acetic acid).

### Summary:
The sequence of reactions starting from CH₃Cl leads to the formation of CH₃COOH after the nucleophilic substitution with KCN and subsequent hydrolysis.