A)
B)
C)
D)
Correct Answer: A
Solution :
[a] \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}\xrightarrow[\text{Anti}\,\text{-}\,\text{Markownikoff }\!\!'\!\!\text{ s}\,\,\text{rule}]{\text{HBr/}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow[{{S}_{N}}2\,\,\text{reaction}]{{{C}_{2}}{{H}_{5}}ONa}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}O{{C}_{2}}{{H}_{5}}\]
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