A) ethylene bromide
B) ethanol
C) ethyl bromide
D) ethylidene bromide
Correct Answer: C
Solution :
The reaction is as follows \[CH\equiv CH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{{{H}_{2}}O/H{{g}^{2+}}}C{{H}_{3}}-CHO\] \[\xrightarrow[(reduction)]{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{{{P}_{4}}/B{{r}_{2}}}\underset{Ethyl\,bromide}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,\]You need to login to perform this action.
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