Question

An optically active alkyl bromide \(C_4H_9Br\) reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at \(333\,K\) to form compound [D]. The functional group in compound D is confirmed by :

• A. Haloform test
• B. Lucas test
• C. Silver mirror test
• D. Benedict test

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to trace a multi-step reaction sequence starting from an optically active alkyl bromide and identify the final product 'D' and its characteristic test.
Step 2: Detailed Explanation:
1. Starting Material: \(C_{4}H_{9}Br\) is optically active. The only optically active isomer of bromobutane is 2-bromobutane (\(CH_{3}-CHBr-CH_{2}CH_{3}\)).
2. Step 1 (\(C_{4}H_{9}Br \xrightarrow{\text{alc. KOH}} [A]\)): Elimination reaction. Major product (Saytzeff) is But-2-ene (\(CH_{3}-CH=CH-CH_{3}\)).
3. Step 2 (\([A] \xrightarrow{Br_{2}} [B]\)): Addition of bromine. [B] is 2,3-dibromobutane (\(CH_{3}-CHBr-CHBr-CH_{3}\)).
4. Step 3 (\([B] \xrightarrow{\text{alc. KOH, } NaNH_{2}} [C]\)): Double dehydrohalogenation. [C] is But-2-yne (\(CH_{3}-C \equiv C-CH_{3}\)).
5. Step 4 (\([C] \xrightarrow{Hg^{2+}/H_{2}SO_{4}} [D]\)): Hydration of alkyne.
\[ CH_{3}-C \equiv C-CH_{3} + H_{2}O \xrightarrow{Hg^{2+}/H^{+}} [CH_{3}-C(OH)=CH-CH_{3}] \xrightarrow{\text{taut.}} CH_{3}-CO-CH_{2}CH_{3} \]
Product [D] is Butan-2-one (Ethyl methyl ketone).

6. Confirmation of Functional Group: Butan-2-one is a methyl ketone (contains \(CH_{3}CO-\) group). Methyl ketones give a positive Haloform (Iodoform) test when treated with \(X_{2}/NaOH\).
It will not give Silver mirror (Tollens) or Benedict tests as those are for aldehydes. Lucas test is for alcohols.
Step 3: Final Answer:
Compound D is butan-2-one, confirmed by the Haloform test.