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

Concept: The sequence involves elimination, halogenation, double elimination and hydration of an alkyne. Step 1: {Identify compound A} Optically active \(C_4H_9Br\) is \(2\)-bromobutane. With alcoholic KOH: \[ 2\text{-bromobutane} \rightarrow but\text{-}2\text{-ene} \] Thus \[ A = \text{but-2-ene} \] Step 2: {Reaction with bromine} \[ \text{but-2-ene} + Br_2 \rightarrow 2,3\text{-dibromobutane} \] Thus \[ B = 2,3\text{-dibromobutane} \] Step 3: {Double elimination} With alcoholic KOH and sodamide: \[ 2,3\text{-dibromobutane} \rightarrow but\text{-}2\text{-yne} \] Thus \[ C = \text{but-2-yne} \] Step 4: {Hydration of alkyne} In presence of \(HgSO_4/H_2SO_4\): \[ but\text{-}2\text{-yne} \rightarrow butan\text{-}2\text{-one} \] Thus \[ D = \text{butan-2-one} \] This contains a methyl ketone group. Step 5: {Test for methyl ketone} Methyl ketones give positive haloform test. Thus the functional group is confirmed by: \[ \text{Haloform test} \]