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Question: ‘D’ product $\xrightarrow[\text{(3) }AlCl_3]{\text{(1) }Ni/H_2\atop\text{(2) }SOCl_2}$ major product...

‘D’ product →(3) AlCl3(1) Ni/H2(2) SOCl2\xrightarrow[\text{(3) }AlCl_3]{\text{(1) }Ni/H_2\atop\text{(2) }SOCl_2} major product is?

A

1-indanone

B

1-indenone

C

2-indanone

D

Naphthalene

Answer

1-indanone

Explanation

Solution

The reaction sequence involves hydrogenation, reaction with thionyl chloride, and a Friedel-Crafts acylation.

  1. Hydrogenation (Ni/H2): This step likely reduces a double bond. If 'D' is 1-indenone, this step would reduce the double bond to form 1-indanone.

  2. SOCl2: Thionyl chloride is used to convert alcohols to alkyl chlorides or carboxylic acids to acyl chlorides.

  3. AlCl3: Aluminum chloride is a Lewis acid catalyst used in Friedel-Crafts reactions.

Given the options, if we start with 1-indenone, the first step (Ni/H2) would reduce the double bond to give 1-indanone. The subsequent steps with SOCl2 and AlCl3 are less likely to cause further significant changes to the structure, especially considering the provided options. Therefore, 1-indanone is the major product.