Question

The product Z of the given reaction sequence is: \(C_2H_4 \xrightarrow[(ii)\ KMnO_4/H^+]{(i)\ H_2O/H^+} X \xrightarrow{SOCl_2} Y \xrightarrow{(C_2H_5)_2Cd} Z\)

• A. An acid chloride
• B. A ketone
• C. An aldehyde
• D. An ester

Hint:

Dialkyl cadmium reagents stop at ketone stage, unlike Grignard reagents which further react to alcohols.

Answer 1

The Correct Option is B

Step 1: Track the sequence from ethene.
We start with $C_2H_4$ and pass it through hydration, oxidation, then $SOCl_2$, then diethylcadmium. Let us name each product on the way to Z.
Step 2: Hydrate ethene.
With water and acid, ethene adds water to give ethanol: $C_2H_4 + H_2O \rightarrow C_2H_5OH$.
Step 3: Oxidise to the acid (X).
Strong oxidation by $KMnO_4/H^+$ takes the primary alcohol all the way to a carboxylic acid: $C_2H_5OH \rightarrow CH_3COOH$. So $X = CH_3COOH$.
Step 4: Make the acid chloride (Y).
Thionyl chloride converts the carboxylic acid into its acid chloride: $CH_3COOH \xrightarrow{SOCl_2} CH_3COCl$. So $Y = CH_3COCl$.
Step 5: React with diethylcadmium.
Dialkyl cadmium reagents convert acid chlorides into ketones: $CH_3COCl + (C_2H_5)_2Cd \rightarrow CH_3COC_2H_5$. So $Z = CH_3COC_2H_5$.
Step 6: Identify the class of Z.
$CH_3COC_2H_5$ (butan-2-one) has a carbonyl between two carbon groups, so it is a ketone.
\[ \boxed{\text{A ketone}} \]