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${({CH}_{3})}_{2} CO \to_{(HCl)}^{NaCN} A \to_{Δ}^{H_{3} O^{+}} B$
In the above sequence A and B are

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${({CH}_{3})}_{2} C (OH) CN, {({CH}_{3})}_{2} C (OH) COOH$

${({CH}_{3})}_{2} C (OH) CN, {({CH}_{3})}_{2} C {(OH)}_{2}$

${CH}_{3} CH (OH) CN, {({CH}_{3})}_{2} CHCOOH$

${({CH}_{3})}_{2} C (OH) CN, {({CH}_{3})}_{2} C = O$

answer is A.

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Detailed Solution

Product A - Carbonyl group on treatment with cyanohydrin in acidic condition gives cyanohydrin derivative.

Product B - Cyanohydrin derivative produced on acidic hydrolysis gives Propionic acid.

${(C H_{3})}_{2} C (O H) C N \to_{▵}^{H_{3} O^{+}} {(C H_{3})}_{2} C (O H) C O O H$

Hence, the correct option is option (A) ${(C H_{3})}_{2} C (O H) C N, {(C H_{3})}_{2} C (O H) C O O H$

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