A) Ethylene bromide
B) Ethanol
C) Ethyl bromide
D) Ethylidene bromide
Correct Answer: C
Solution :
\[CH\equiv CH\underset{{{H}_{2}}S{{O}_{4}}}{\mathop{\xrightarrow{{{H}_{2}}O/H{{g}^{2+}}}}}\,{{H}_{3}}C-\underset{(\times )}{\overset{\overset{O}{\mathop{||}}\,}{\mathop{C}}}\,-H\]\[C{{H}_{3}}-\underset{(y)}{\mathop{C{{H}_{2}}}}\,-OH\xrightarrow{{{P}_{\text{4}}}\text{/}B{{r}_{2}}}\underset{(z)}{\mathop{{{C}_{2}}{{H}_{5}}Br}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
