A)
B) Alcohol
C) Acetylene
D) Benzaldehyde
Correct Answer: C
Solution :
\[C{{H}_{3}}-COOH\xrightarrow{LiAl{{H}_{4}}}C{{H}_{3}}-C{{H}_{2}}-OH\underset{443\,K}{\mathop{\xrightarrow{{{H}^{+}}}}}\,\] \[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{B{{r}_{2}}}\underset{\underset{Br\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,-\underset{\underset{Br\,\,}{\mathop{|\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,\underset{KOH}{\mathop{\xrightarrow{alc.}}}\,\] \[\underset{\text{Acetylene}}{\mathop{CH\equiv CH}}\,+2KBr+2{{H}_{2}}O\]
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