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: Understanding the Concept:
This sequence involves the preparation of a ketone from an acid chloride using an organocadmium reagent.
Organocadmium compounds are less reactive than Grignard reagents and selectively react with acid chlorides to stop at the ketone stage.
Step 2: Key Formula or Approach:
The reactions follow this general scheme:
1. $2\text{RMgX} + \text{CdCl}_2 \rightarrow \text{R}_2\text{Cd} + 2\text{Mg(X)Cl}$
2. $\text{R}_2\text{Cd} + 2\text{R}'\text{COCl} \rightarrow 2\text{R}'\text{COR} + \text{CdCl}_2$
Step 3: Detailed Explanation:
In the first step, methyl magnesium bromide ($\text{CH}_3\text{MgBr}$) reacts with cadmium chloride to form dimethyl cadmium (A):
\[ 2\text{CH}_3\text{MgBr} + \text{CdCl}_2 \rightarrow (\text{CH}_3)_2\text{Cd} + 2\text{Mg(Br)Cl} \]
In the second step, dimethyl cadmium (A) reacts with acetyl chloride ($\text{CH}_3\text{COCl}$) to form propanone (B):
\[ (\text{CH}_3)_2\text{Cd} + 2\text{CH}_3\text{COCl} \rightarrow 2\text{CH}_3\text{COCH}_3 + \text{CdCl}_2 \]
The chemical name of $\text{CH}_3\text{COCH}_3$ is propanone (acetone).
Step 4: Final Answer:
The product ' B ' is Propanone.