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JEE CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

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Analyzing +4 | -1
C4H8Cl2(X)→ aq. Acetone C4H8O(Y)→ΔNaOH,I2CHI3+\mathrm{C}_4 \mathrm{H}_8 \mathrm{Cl}_2(\mathrm{X}) \xrightarrow{\text { aq. Acetone }} \mathrm{C}_4 \mathrm{H}_8 \mathrm{O}(\mathrm{Y}) \xrightarrow[\Delta]{\mathrm{NaOH}, \mathrm{I}_2} \mathrm{CHI}_3+ Acid Salt (Z)(\mathrm{Z}). Compound X\mathrm{X} can be:
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Correct option: C
CH3−C(Cl2)−CH2−CH3\mathrm{CH}_3-\mathrm{C}\left(\mathrm{Cl}_2\right)-\mathrm{CH}_2-\mathrm{CH}_3
The correct answer is C (CH3−C(Cl2)−CH2−CH3\mathrm{CH}_3-\mathrm{C}(\mathrm{Cl}_2)-\mathrm{CH}_2-\mathrm{CH}_3) because it forms a methyl ketone upon reaction with aqueous acetone, which then undergoes the iodoform reaction.

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