In the following sequence of reaction, the product C is
`CH_(3)-Br overset(KCN) rarr A overset(H_(2)O^(+))rarr B overset(LiAlH_(4))underset("Ether")rarr C`

Options

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

Answer

To determine the product C in the given sequence of reactions, we will follow each step carefully.

### Step 1: Reaction of CH₃Br with KCN
- **Starting Material:** CH₃Br (Methyl bromide)
- **Reagent:** KCN (Potassium cyanide)
- **Reaction:** The bromine atom (Br) in methyl bromide is replaced by a cyano group (CN) through a nucleophilic substitution reaction.

**Product A:** CH₃CN (Acetonitrile)

### Step 2: Hydrolysis of CH₃CN with H₃O⁺
- **Starting Material:** CH₃CN (Acetonitrile)
- **Reagent:** H₃O⁺ (Hydronium ion, which is a combination of H₂O and H⁺)
- **Reaction:** The acetonitrile undergoes hydrolysis, where the cyano group is converted into a carboxylic acid group. This is an oxidation process.

**Product B:** CH₃COOH (Acetic acid)

### Step 3: Reduction of CH₃COOH with LiAlH₄
- **Starting Material:** CH₃COOH (Acetic acid)
- **Reagent:** LiAlH₄ (Lithium aluminum hydride) in ether
- **Reaction:** Lithium aluminum hydride is a strong reducing agent that reduces the carboxylic acid to an alcohol. The carbonyl group (C=O) in acetic acid is reduced to a hydroxyl group (–OH).

**Product C:** CH₃CH₂OH (Ethanol)

### Final Answer
The product C formed in this sequence of reactions is **Ethanol (CH₃CH₂OH)**.

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