Question

Identify the product ' B ' in the following sequence of reactions.
Methyl magnesium bromide $\xrightarrow{\text{CdCl}_2} \text{A} \xrightarrow{\text{CH}_3\text{COCl}} \text{B}$}

• A. Dimethyl cadmium
• B. Propanone
• C. Butanone
• D. Propanal

Hint:

Dialkyl cadmium reagents are less reactive than Grignard reagents and are used to prepare ketones without further reduction to alcohols.

Answer 1

The Correct Option is B

Step 1: Concept Grignard reagents react with anhydrous cadmium chloride to form dialkyl cadmium, which further reacts with acyl halides to form ketones.

Step 2: Meaning Methyl magnesium bromide ($CH_3MgBr$) reacts with $CdCl_2$ to form dimethyl cadmium ($(CH_3)_2Cd$) as product 'A'.

Step 3: Analysis Product 'A' ($(CH_3)_2Cd$) reacts with acetyl chloride ($CH_3COCl$) to replace the chlorine atom with a methyl group.

Step 4: Conclusion The reaction $2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$ yields acetone, which is also known as propanone. Final Answer: (B)