Question

Identify substrate 'A' in the following reaction. 2nA Dimethyl cadmium → 2n Propanone + n Cadmium chloride}

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

Hint:

Dialkyl cadmium reagents are used to synthesize ketones from acyl chlorides.

Answer 1

The Correct Option is A

Concept:
Dialkyl cadmium reagents react with acid chlorides to give ketones. General reaction: \[ 2RCOCl + (R')_2Cd \rightarrow 2RCOR' + CdCl_2 \] This reaction is used to prepare ketones without further addition.
Step 1: Given reagent and product
Reagent: \[ (CH_3)_2Cd \] Product formed: \[ \text{Propanone } (CH_3COCH_3) \]
Step 2: Structure of propanone
Propanone has formula: \[ CH_3COCH_3 \] It is a ketone containing:

• Acyl part = $CH_3CO-$
• Alkyl added from cadmium reagent = $CH_3$

Step 3: Identify acid chloride A
Since methyl group comes from $(CH_3)_2Cd$, the required acid chloride must be: \[ CH_3COCl \] which is ethanoyl chloride (acetyl chloride).
Step 4: Write balanced reaction
\[ 2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2 \] Products are:

• Propanone
• Cadmium chloride

This matches the given reaction.
Step 5: Final Answer
Therefore, compound $A$ is: \[ \boxed{CH_3COCl} \] Quick Tip:
Acid chloride + dialkyl cadmium $\Rightarrow$ ketone only (no tertiary alcohol formation).