A) ethene
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. \[\underset{\text{but}-2-\text{ene}}{\mathop{C{{H}_{3}}CH}}\,=CHC{{H}_{3}}\xrightarrow[(ii)\,Zn/{{H}_{2}}O]{(i)\,{{O}_{3}}}\underset{\text{acetaldehyde}}{\mathop{2C{{H}_{3}}CHO}}\,\]
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