A) ethane
B) propene
C) but-1-ene
D) but-2-ene
Correct Answer: D
Solution :
The aldehyde with molecular mass \[44\,\,u\] is \[C{{H}_{3}}CHO\] (acetaldehyde). Therefore, the symmetrical alkene is but-2-ene. \[\underset{but-2-ene}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,\xrightarrow[(ii)\,\,Zn/{{H}_{2}}O]{(i)\,\,{{O}_{3}}}\underset{acetaldehyde}{\mathop{2C{{H}_{3}}CHO}}\,\]
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