Treatment of benzene with CO/HCl in the presence of anhydrous `"AlCl"_(3)//"CuCl" ` followed by reaction with `Ac_(2)"O"//"NaOAc"` gives compound X as the major product. Compound X upon reaction with `Br_(2)//Na_(2)CO_(3)`, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with `H_(2)//Pd-C`, followed by `H_(3)PO_(4)` treatment gives Z as the major product.
The compound Y is

Options

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D.

Answer

To solve the question step by step, let's break down the reactions and identify the compounds formed at each stage.

### Step 1: Reaction of Benzene with CO and HCl
- **Starting Material**: Benzene (C₆H₆)
- **Reagents**: CO/HCl in the presence of anhydrous AlCl₃ and CuCl
- **Reaction Type**: Gatterman-Koch reaction
- **Product**: Benzaldehyde (C₆H₅CHO)

**Explanation**: In this reaction, benzene reacts with carbon monoxide and hydrochloric acid in the presence of a Lewis acid catalyst (AlCl₃ and CuCl) to form benzaldehyde.

### Step 2: Reaction of Benzaldehyde with Acetic Anhydride
- **Starting Material**: Benzaldehyde (C₆H₅CHO)
- **Reagents**: Acetic anhydride (Ac₂O) and sodium acetate (NaOAc)
- **Product**: Cinnamic acid (C₆H₅CH=CHCOOH)

**Explanation**: Benzaldehyde undergoes a reaction with acetic anhydride in the presence of sodium acetate, leading to the formation of cinnamic acid, which has the structure C₆H₅CH=CHCOOH.

### Step 3: Reaction of Cinnamic Acid with Br₂/Na₂CO₃
- **Starting Material**: Cinnamic acid (C₆H₅CH=CHCOOH)
- **Reagents**: Bromine (Br₂) and sodium carbonate (Na₂CO₃)
- **Reaction Type**: Electrophilic addition
- **Product**: 2-Bromocinnamic acid (C₆H₅CHBr-CH(COOH)₂)

**Explanation**: The double bond in cinnamic acid reacts with bromine, resulting in the addition of bromine across the double bond, forming 2-bromocinnamic acid.

### Step 4: Heating with Moist KOH
- **Starting Material**: 2-Bromocinnamic acid
- **Reagents**: Moist potassium hydroxide (KOH)
- **Product**: 2-Phenylprop-2-enoic acid (Y)

**Explanation**: The bromine atom is eliminated during the reaction with moist KOH, resulting in the formation of an alkyne, specifically 2-phenylprop-2-enoic acid (Y).

### Step 5: Reaction of Cinnamic Acid with H₂/Pd-C
- **Starting Material**: Cinnamic acid (C₆H₅CH=CHCOOH)
- **Reagents**: Hydrogen gas (H₂) and palladium on carbon (Pd-C)
- **Product**: 2-Phenylpropanoic acid (Z)

**Explanation**: Cinnamic acid is hydrogenated in the presence of Pd-C, resulting in the formation of 2-phenylpropanoic acid.

### Final Answer
The compound Y formed after the reactions is **2-Phenylprop-2-enoic acid**.

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