A) ketone
B) ethanol
C) acetaldehyde
D) propionaldehyde
Correct Answer: C
Solution :
\[CH=CH+{{H}_{2}}O\xrightarrow[{}]{HgS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}}\] \[\underset{CHOH}{\overset{C{{H}_{2}}}{\mathop{||}}}\,\xrightarrow[{}]{900\,K}\underset{X}{\mathop{C{{H}_{3}}CHO}}\,\] Thus X is identified as acetaldehydeYou need to login to perform this action.
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