In the following reaction sequence, the compound J is an
interemediate
`Iunderset(CH_(3)COONa)overset((CH_(3)CO)_(2)O)(to)J underset({:((ii)SOCl_(2)),((iii)"anhyd".AlCl_(3)):})overset((i)H_(2),Pd//C)(to)K`
`J(C_(9)H_(8)O_(2))` gives effervsecence on the treament with` NaHCHO_(3)`
and positive Baeyr's test.
The compound K, is

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, let's break down the reaction sequence and identify the compounds involved.

### Step 1: Identify Compound I
The question states that compound I is a benzene ring attached to an aldehyde group. This can be represented as:
- **Compound I:** C₆H₅CHO (Benzaldehyde)

### Step 2: Reaction of Compound I to Form Compound J
Compound I reacts with sodium acetate (CH₃COONa) in the presence of acetic anhydride ((CH₃CO)₂O). This reaction is known as the Perkin reaction, which leads to the formation of a cinnamic acid derivative. The product J can be represented as:
- **Compound J:** C₉H₈O₂ (Cinnamic Acid: C₆H₅-CH=CH-COOH)

### Step 3: Properties of Compound J
The problem states that compound J gives effervescence upon treatment with sodium bicarbonate (NaHCO₃) and gives a positive Baeyer's test.
- The effervescence indicates the presence of a carboxylic acid group, which is consistent with the structure of cinnamic acid.
- A positive Baeyer's test indicates the presence of an alkene (C=C bond), which is also present in cinnamic acid.

### Step 4: Reaction of Compound J to Form Compound K
Next, compound J (Cinnamic Acid) undergoes hydrogenation in the presence of Pd/C (palladium on carbon) to reduce the double bond (C=C) to a single bond (C-C), resulting in:
- **Intermediate after hydrogenation:** C₆H₅-CH₂-COOH (Phenylpropanoic acid)

### Step 5: Reaction of Intermediate with SOCl₂
The intermediate then reacts with thionyl chloride (SOCl₂), which converts the carboxylic acid group (-COOH) into an acyl chloride (-COCl):
- **Intermediate after reaction with SOCl₂:** C₆H₅-CH₂-COCl

### Step 6: Friedel-Crafts Acylation
Finally, the acyl chloride undergoes Friedel-Crafts acylation in the presence of anhydrous AlCl₃, which introduces an acyl group onto the benzene ring:
- **Final Compound K:** C₉H₉Cl (Benzoyl chloride derivative)

### Conclusion
The final compound K is a product of Friedel-Crafts acylation, which is a benzene ring with an acyl group.

### Final Answer
- **Compound K:** C₉H₉Cl (Benzoyl chloride derivative)

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