A) \[C{{H}_{2}}=C{{H}_{2}}~\]
B) \[{{C}_{2}}{{H}_{5}}OH\]
C) \[HCHO\]
D) acetaldehyde
Correct Answer: D
Solution :
\[CH\equiv CH+{{H}_{2}}O\xrightarrow[HgS{{O}_{4}}]{{{H}_{2}}S{{O}_{4}}}C{{H}_{2}}=\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H\] \[\xrightarrow{ketonization}\underset{\text{Acetaldehyde}}{\mathop{C{{H}_{3}}.CHO}}\,\]
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