Identification of Functional Group via Chemical Tests from Reaction Sequence

To determine the functional group present in compound [D] and the corresponding test used for its confirmation, we analyze the reaction sequence step-by-step:

1. Identification of Optically Active Alkyl Bromide ($\text{C}_4\text{H}_9\text{Br}$)

The isomers of alkyl bromide with the molecular formula $\text{C}_4\text{H}_9\text{Br}$ are:

• $1$-Bromobutane (achiral)
• $2$-Bromobutane (chiral, optically active)
• $1$-Bromo-$2$-methylpropane (achiral)
• $2$-Bromo-$2$-methylpropane (achiral)

Since the given reactant is optically active, it must be $2$-bromobutane: $\text{CH}_3-\text{C}^* \text{H(Br)}-\text{CH}_2-\text{CH}_3$

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2. Formation of Compound [A]

When $2$-bromobutane undergoes dehydrohalogenation with ethanolic $\text{KOH}$ ($E2$ reaction), according to Saytzeff's rule, the most substituted and stable alkene is formed as the major product: $\text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{alc. KOH}} \underset{\text{Compound [A] (But-2-ene)}}{\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3}$

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3. Formation of Compound [B]

Electrophilic addition of bromine ($\text{Br}_2$) to but-2-ene yields a vicinal dibromide: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 + \text{Br}_2 \rightarrow \underset{\text{Compound [B] (2,3-dibromobutane)}}{\text{CH}_3-\text{CH(Br)}-\text{CH(Br)}-\text{CH}_3}$

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4. Formation of Compound [C]

Reaction of $2,3$-dibromobutane with ethanolic $\text{KOH}$ followed by sodamide ($\text{NaNH}_2$) causes double dehydrohalogenation to yield an alkyne: $\text{CH}_3-\text{CH(Br)}-\text{CH(Br)}-\text{CH}_3 \xrightarrow[\text{2. } \text{NaNH}_2]{\text{1. alc. KOH}} \underset{\text{Compound [C] (But-2-yne / But-1-yne)}}{\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3}$

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5. Formation of Compound [D]

Addition of water (hydration) to the alkyne in the presence of $\text{HgSO}_4$ and dilute $\text{H}_2\text{SO}_4$ at $333\text{ K}$ (Kucherov's reaction) yields an enol intermediate that undergoes tautomerism to form a ketone:

$\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 + \text{H}_2\text{O} \xrightarrow[\text{dil. } \text{H}_2\text{SO}_4, 333\text{ K}]{\text{HgSO}_4} \left[\text{CH}_3-\text{C(OH)}=\text{CH}-\text{CH}_3\right] \xrightarrow{\text{Tautomerism}} \underset{\text{Compound [D] (Butan-2-one)}}{\text{CH}_3-\overset{\underset{||}{\text{O}}}{\text{C}}-\text{CH}_2-\text{CH}_3}$

(Note: Even if isomerization to terminal alkyne but-1-yne occurs due to $\text{NaNH}_2$, Markovnikov's addition of water yields the exact same product, butan-2-one).

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6. Confirmation Test for Compound [D]

Compound [D] is butan-2-one, which contains a methyl ketone group ($\text{CH}_3-\text{C}(=\text{O})-$).

• Methyl ketones give a positive Haloform test (e.g., reaction with $\text{I}_2/\text{NaOH}$ produces a yellow precipitate of iodoform, $\text{CHI}_3$).
• Lucas test is used to distinguish alcohols ($1^\circ, 2^\circ, 3^\circ$).
• Silver mirror test and Benedict test are specific to aldehydes and reducing sugars.

Therefore, the functional group in compound [D] is confirmed by the Haloform test.

Correct Answer: A (Haloform test)