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Two isomeric compounds ‘A’ & ‘B’ have the formula C₃H₆Cl₂ with Aq. KOH solution ‘A’ gives propionaldehyde and ‘B’ gives acetone. Then A & B are

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${CH}_{3} - {CCl}_{2} - {CH}_{3} and {CH}_{3} - {CH}_{2} - {CHCl}_{2}$

${CH}_{3} - CHCl - {CHCl}_{2} and {CH}_{3} - {CH}_{2} - {CHCl}_{2}$

${CH}_{3} - {CH}_{2} - {CHCl}_{2} and {CH}_{3} - {CCl}_{2} - {CH}_{3}$

${CH}_{3} - CHCl - {CHCl}_{2} and {CH}_{3} - {CCl}_{2} - {CH}_{3}$

answer is C.

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Terminal gem dihalides gives aldelydes & internal gem dihalides gives ketones on alkaline hydrolysis

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