`CH_(3)COOH overset(LiAlH_(4))toA`
`A+CH_(3)COOH overset(H_(3)O^(+))to 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)`.

Answer

To solve the problem step-by-step, we will analyze the reactions given and identify the compounds 'A' and 'B'.

### Step 1: Identify the reaction of acetic acid with lithium aluminium hydride (LiAlH₄)
- The first part of the reaction is:
\[ \text{CH}_3\text{COOH} \overset{\text{LiAlH}_4}{\rightarrow} A \]
- Acetic acid (CH₃COOH) is a carboxylic acid. Lithium aluminium hydride (LiAlH₄) is a strong reducing agent that can reduce carboxylic acids to primary alcohols.
- Therefore, when acetic acid is treated with LiAlH₄, it will be reduced to ethanol (C₂H₅OH).

### Step 2: Determine the structure of compound A
- From the reduction of acetic acid, we find that:
\[ A = \text{C}_2\text{H}_5\text{OH} \quad (\text{Ethanol}) \]

### Step 3: Identify the reaction of compound A with acetic acid in the presence of H₃O⁺
- The second part of the reaction is:
\[ A + \text{CH}_3\text{COOH} \overset{\text{H}_3\text{O}^+}{\rightarrow} B + \text{H}_2\text{O} \]
- Here, ethanol (A) reacts with acetic acid in the presence of an acid catalyst (H₃O⁺). This reaction is known as esterification, where an alcohol reacts with a carboxylic acid to form an ester and water.

### Step 4: Determine the structure of compound B
- The reaction of ethanol with acetic acid will produce ethyl acetate (an ester):
\[ B = \text{C}_2\text{H}_5\text{OCOCH}_3 \quad (\text{Ethyl acetate}) \]

### Conclusion
- Therefore, the compounds 'A' and 'B' are:
- A = Ethanol (C₂H₅OH)
- B = Ethyl acetate (C₂H₅OCOCH₃)

### Final Answer
- A = C₂H₅OH (Ethanol)
- B = C₂H₅OCOCH₃ (Ethyl acetate)