`CH_(3)COOHoverset(LiAlH_(4))(rarr)(A), (A)+CH_(3)COOHoverset(H_(3)O^(+))(rarr)(B)+H_(2)O` . In the above reaction 'A' and 'B' respectively are `:`

Options

A. `CH_(3)COOC_(2)H_(5),C_(2)H_(5)OH`
B. `CH_(3)CHO,C_(2)H_(5)OH`
C. `C_(2)H_(5)OH,CH_(3)CHO`
D. `C_(2)H_(5)OH,CH_(3)COOC_(2)H_(5)`

Explanation

To solve the given reaction sequence, we will analyze each step carefully.

### Step 1: Identify the Reactants
The first reactant is acetic acid (CH₃COOH) and it reacts with lithium aluminium hydride (LiAlH₄).

### Step 2: Reduction of Acetic Acid
Lithium aluminium hydride (LiAlH₄) is a strong reducing agent. When acetic acid is treated with LiAlH₄, it undergoes reduction to form ethanol (C₂H₅OH or CH₃CH₂OH).

**Reaction:**
\[ \text{CH}_3\text{COOH} + \text{LiAlH}_4 \rightarrow \text{C}_2\text{H}_5\text{OH} \]
Thus, we identify:
- A = C₂H₅OH (Ethanol)

### Step 3: Reaction of Ethanol with Acetic Acid
Next, the product A (ethanol) reacts with acetic acid (CH₃COOH) in the presence of hydronium ion (H₃O⁺). This reaction is an esterification reaction.

**Reaction:**
\[ \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \overset{\text{H}_3\text{O}^+}{\rightarrow} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]
Thus, we identify:
- B = CH₃COOC₂H₅ (Ethyl acetate)

### Final Answer:
- A = Ethanol (C₂H₅OH)
- B = Ethyl acetate (CH₃COOC₂H₅)