Identify the product D in the following series of reaction
`CH_(3)COOH overset(LiAlH_4) rarr A underset(443K)overset(H^+) rarr B overset(Br_2)rarr C overset(alc.)underset(KOH) rarr D`

Options

A. Methane
B. Alcohol
C. Acetylene
D. Benzaldehyde

Explanation

To identify the product D in the given series of reactions, let's break down each step systematically.

### Step 1: Reduction of Acetic Acid (CH₃COOH)
- **Reaction**: CH₃COOH + LiAlH₄ → A
- **Explanation**: Lithium Aluminium Hydride (LiAlH₄) is a strong reducing agent. It reduces carboxylic acids to primary alcohols. Therefore, acetic acid (CH₃COOH) will be reduced to ethanol (CH₃CH₂OH).
- **Product A**: Ethanol (CH₃CH₂OH)

### Step 2: Dehydration of Ethanol
- **Reaction**: A (Ethanol) → B (at 443 K with H⁺)
- **Explanation**: When ethanol is heated in the presence of an acid (H⁺), it undergoes dehydration to form an alkene. The major product at this temperature is ethylene (C₂H₄).
- **Product B**: Ethylene (C₂H₄)

### Step 3: Bromination of Ethylene
- **Reaction**: B (C₂H₄) + Br₂ → C
- **Explanation**: Ethylene reacts with bromine (Br₂) in an electrophilic addition reaction, resulting in the formation of 1,2-dibromoethane (C₂H₄Br₂).
- **Product C**: 1,2-Dibromoethane (C₂H₄Br₂)

### Step 4: Elimination Reaction
- **Reaction**: C (1,2-Dibromoethane) + alc. KOH → D
- **Explanation**: When 1,2-dibromoethane is treated with alcoholic potassium hydroxide (alc. KOH), it undergoes elimination (dehydrohalogenation) to form an alkyne. The elimination of bromine and hydrogen bromide results in the formation of acetylene (C₂H₂).
- **Product D**: Acetylene (C₂H₂)

### Final Answer:
The product D in the series of reactions is **Acetylene (C₂H₂)**.

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