A) ethyl alcohol \[+HBr\]
B) ethanol\[+B{{r}_{2}}\]
C) alcohol \[+HBr\]
D) None of the above
Correct Answer: D
Solution :
Commercially ethyl bromide is manufactured by the reaction of ethyl alcohol with 48%\[HBr\]in the presence of cone.\[{{H}_{2}}S{{O}_{4}}\]. \[\underset{ethyl\,alcohol}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,+HBr\xrightarrow[{}]{Conc.{{H}_{2}}S{{O}_{4}}}\underset{ethyl\text{ }bromide}{\mathop{{{C}_{2}}{{H}_{5}}Br}}\,+{{H}_{2}}O\]
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