In the following sequence of reactions:
`CH_(3)-Br overset(KCN)rarr A overset(H_(3)overset(+)(O))rarr B overset(LiAlH_(4))rarr C` the end product (C ) is :

Options

A. Methane
B. Acetaldehyde
C. Ethyl alcohol
D. acetone

Explanation

To solve the given sequence of reactions, let's break it down step by step.

### Step 1: Reaction of Methyl Bromide with KCN
The first reactant is methyl bromide (CH₃Br). When it reacts with potassium cyanide (KCN), a nucleophilic substitution reaction occurs. The cyanide ion (CN⁻) acts as a nucleophile and attacks the carbon atom bonded to the bromine atom, displacing the bromine atom.

**Reaction:**
\[ \text{CH}_3\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CN} + \text{KBr} \]

**Product A:**
The product of this reaction is methyl cyanide (CH₃CN), also known as acetonitrile.

### Step 2: Hydrolysis of Methyl Cyanide
Next, we perform hydrolysis on methyl cyanide (CH₃CN) using water (H₂O) in the presence of an acid (H⁺). During hydrolysis, the nitrile group (CN) is converted into a carboxylic acid.

**Reaction:**
\[ \text{CH}_3\text{CN} + 2 \text{H}_2\text{O} \overset{H^+}{\rightarrow} \text{CH}_3\text{COOH} + \text{NH}_3 \]

**Product B:**
The product of this hydrolysis is acetic acid (CH₃COOH).

### Step 3: Reduction of Acetic Acid with Lithium Aluminum Hydride
In the final step, we reduce acetic acid (CH₃COOH) using lithium aluminum hydride (LiAlH₄), which is a strong reducing agent. This reduction converts the carboxylic acid into an alcohol.

**Reaction:**
\[ \text{CH}_3\text{COOH} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{LiAlO}_2 + \text{H}_2 \]

**Product C:**
The final product of this reaction is ethanol (CH₃CH₂OH).

### Conclusion
The end product (C) of the given sequence of reactions is **ethanol (C₂H₅OH)**.

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