The product P of the following reaction is
`CH_(3)-COOH overset(LiAlH_(4))to overset(SOCl_(2))to overset(Koverset(14)(C)N)to overset(H_(3)O^(o+)) to P`

Answer

To determine the product P from the given reaction sequence, we will analyze each step carefully.

### Step-by-Step Solution:

1. **First Reaction: Reduction of Acetic Acid**
- Starting with acetic acid (CH₃COOH), we treat it with lithium aluminium hydride (LiAlH₄), which is a strong reducing agent.
- **Reaction:**
\[ \text{CH}_3\text{COOH} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O} \]
- **Product:** Ethanol (CH₃CH₂OH).

2. **Second Reaction: Conversion of Ethanol to Ethyl Chloride**
- Next, we take the ethanol (CH₃CH₂OH) and react it with thionyl chloride (SOCl₂).
- **Reaction:**
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{SOCl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{SO}_2 + \text{HCl} \]
- **Product:** Ethyl chloride (CH₃CH₂Cl).

3. **Third Reaction: Nucleophilic Substitution with Potassium Cyanide**
- The ethyl chloride (CH₃CH₂Cl) is then reacted with potassium cyanide (KCN).
- **Reaction:**
\[ \text{CH}_3\text{CH}_2\text{Cl} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{KCl} \]
- **Product:** Ethyl cyanide (CH₃CH₂CN).

4. **Fourth Reaction: Hydrolysis of Ethyl Cyanide**
- Finally, we hydrolyze the ethyl cyanide (CH₃CH₂CN) in the presence of hydronium ions (H₃O⁺).
- **Reaction:**
\[ \text{CH}_3\text{CH}_2\text{CN} + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH} \]
- **Product:** Ethanoic acid (CH₃CH₂COOH).

### Final Product:
The final product P is **Ethanoic acid (CH₃CH₂COOH)**.

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