JEE Advanced 2017 Paper 2 Chemistry: Paragraph 2

The 2,4-DNP test shows a carbonyl group, and the iodoform test shows a methyl ketone $\mathrm{CH_3CO-}$. So $\mathbf{X}$ is a methyl ketone, which eliminates 2-methylpentan-3-one. Among the remaining methyl ketones of formula $\mathrm{C_6H_{12}O}$ ($\mathrm{CH_3COC_4H_9}$) the candidates are hexan-2-one, 4-methylpentan-2-one, 3-methylpentan-2-one and 3,3-dimethylbutan-2-one. Optical activity requires a stereocentre. Only 3-methylpentan-2-one has a carbon bearing four different groups ($\mathrm{H}$, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$, $\mathrm{COCH_3}$). The others have no stereocentre. The haloform product $\mathbf{Y}$ is 2-methylbutanoic acid, $\mathrm{C_5H_{10}O_2}$, and Wolff-Kishner reduction gives 3-methylpentane, which is achiral, consistent with all the data.

The alcohol formed is 3-methylpentan-2-ol, $\mathrm{CH_3CH(OH)CH(CH_3)CH_2CH_3}$, which has two stereocentres (C2 and C3) and therefore four stereoisomers: $(2R,3R)$, $(2S,3S)$, $(2R,3S)$ and $(2S,3R)$. The C3 stereocentre of $(S)$-$\mathbf{X}$ is not touched by hydride, so it remains fixed. Hydride adds to either face of the planar carbonyl carbon, creating the new C2 stereocentre in both configurations. Since the two faces of the carbonyl are diastereotopic (because of the adjacent stereocentre) the two products form in unequal amounts, but both form. This gives exactly two stereoisomers, which are diastereomers of each other. Four would need both C3 configurations, which requires racemic $\mathbf{X}$, and one would need complete facial selectivity, which hydride reduction does not show.