A) formaldehyde
B) acetaldehyde
C) acetone
D) None of these
Correct Answer: B
Solution :
\[\underset{(A)}{\mathop{C{{H}_{3}}CHO}}\,+HCN\xrightarrow[{}]{{}}\underset{(B)}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ CN \end{smallmatrix}}{\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{CH}}}\,}}\,\xrightarrow[{}]{{{H}_{2}}O}\] \[\underset{(C)}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ COOH \end{smallmatrix}}{\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{{{C}^{*}}}}}\,}}\,-H\] Compound [C] contains asymmetric carbon atom hence it shows optical isomerism. It also gives iodoform test due to the presence of \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,\] group.
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