A) \[{{(C{{H}_{3}})}_{2}}COHC{{H}_{2}}C{{H}_{3}}\]
B) \[C{{H}_{3}}C{{H}_{2}}CH\left( C{{H}_{3}} \right)-O-C{{H}_{2}}C{{H}_{3}}\]
C) \[C{{H}_{3}}{{\left( C{{H}_{2}} \right)}_{3}}-O-C{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}{{\left( C{{H}_{2}} \right)}_{4}}-O-C{{H}_{3}}\]
Correct Answer: C
Solution :
[c] \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}\xrightarrow{HBr/{{H}_{2}}{{O}_{2}}}\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{{{C}_{2}}{{H}_{5}}ONa}\] \[C{{H}_{3}}{{(C{{H}_{2}})}_{3}}O{{C}_{2}}{{H}_{5}}\] In presence of \[{{H}_{2}}{{O}_{2}},HBr\] adds in anti-Markovnikov's way (peroxide effect).
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