Question

Identify 'A' in following reaction.
$$\mathrm{A} \xrightarrow{\text{Dimethyl cadmium}} \text{propanone} + \text{cadmium chloride}$$

• A. Ethyl chloride
• B. Ethylidene dichloride
• C. Ethanoyl chloride
• D.

Hint:

Organocadmium reagents are deliberately chosen in organic synthesis because they are less reactive than Grignard reagents. They readily attack highly reactive acid chlorides to form ketones, but they are too weak to attack the resulting ketone, preventing unwanted secondary alcohol formation.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to determine the identity of starting material 'A' that reacts with dimethyl cadmium to produce propanone (acetone) and a byproduct of cadmium chloride ($\mathrm{CdCl}_2$).

Step 2: Key Formula or Approach:
Dialkylcadmium reagents ($\mathrm{R}_2\mathrm{Cd}$) react cleanly with acyl chlorides (acid chlorides, $\mathrm{R'COCl}$) to synthesize ketones. The reaction follows this general stoichiometric path:
$$2\mathrm{R'COCl} + \mathrm{R}_2\mathrm{Cd} \rightarrow 2\mathrm{R'COR} + \mathrm{CdCl}_2$$

Step 3: Detailed Explanation:
Let's analyze the structural fragments of the given products:
The desired ketone product is propanone: $\mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}_3$.
The organometallic reagent is dimethyl cadmium: $(\mathrm{CH}_3)_2\mathrm{Cd}$, which supplies the methyl group ($-\mathrm{CH}_3$).
Matching this to our general reaction equation:
$$2\mathrm{R'COCl} + (\mathrm{CH}_3)_2\mathrm{Cd} \rightarrow 2\mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}_3 + \mathrm{CdCl}_2$$ By isolating the fragments, the $\mathrm{R'}$ acyl segment from starting material 'A' must be a two-carbon chain containing the carbonyl unit ($\mathrm{CH}_3-\mathrm{CO}-$).
Therefore, the starting material 'A' must be acetyl chloride, which is systematically named ethanoyl chloride ($\mathrm{CH}_3\mathrm{COCl}$).
$$2\mathrm{CH}_3\mathrm{COCl} + (\mathrm{CH}_3)_2\mathrm{Cd} \rightarrow 2\mathrm{CH}_3\mathrm{COCH}_3 + \mathrm{CdCl}_2$$

Step 4: Final Answer:
The compound 'A' is ethanoyl chloride, which corresponds to option (C).