`C_6H_6overset(CH_3COCl)underset(AlCl_3)rarroverset(Zn//Hg//HCl)rarr(P)`
In the reaction sequence product (P) will be

Options

A. `C_6H_5-CH_3`
B. `C_6H_5-CH_2-CH_3`
C. `C_6H_5-CH_2-CHO`
D. `C_6H_5-CH=CH_2`

Answer

To solve the question, we need to analyze the given reaction step by step.

### Step 1: Identify the Reactants
The reactants are benzene (C₆H₆) and acetyl chloride (CH₃COCl) in the presence of aluminum chloride (AlCl₃).

### Step 2: Friedel-Crafts Acylation
The reaction of benzene with acetyl chloride in the presence of aluminum chloride is a Friedel-Crafts acylation reaction. In this reaction, the acetyl chloride reacts with aluminum chloride to form an acylium ion (CH₃C^+=O).

**Reaction:**
\[ \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \]

### Step 3: Formation of the Product from Friedel-Crafts Acylation
The acylium ion (CH₃C^+=O) acts as an electrophile and attacks the benzene ring, resulting in the formation of acetophenone (C₆H₅COCH₃).

### Step 4: Reduction with Zinc Amalgam and HCl
The next step involves the reduction of the acetophenone product using zinc amalgam (Zn/Hg) and hydrochloric acid (HCl). This reduction converts the carbonyl group (C=O) of acetophenone into a methylene group (–CH₂–).

**Reaction:**
\[ \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{Zn/Hg/HCl}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \]

### Step 5: Final Product
The final product (P) obtained after the reduction is ethylbenzene (C₆H₅CH₂CH₃).

### Conclusion
Therefore, the product (P) of the reaction sequence is:
\[ \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \]

---