question_answer

The major product of the following reaction is: \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\underset{Br}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[{{C}_{2}}{{H}_{5}}OH]{{{C}_{2}}{{H}_{5}}ONa}\]

A) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\underset{O{{C}_{2}}{{H}_{5}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{2}}-C{{H}_{3}}\]

B) \[{{C}_{6}}{{H}_{5}}CH=\underset{C{{H}_{3}}}{\overset{{}}{\mathop{\underset{|}{\overset{{}}{\mathop{C}}}\,}}}\,-C{{H}_{2}}-C{{H}_{3}}\]

C) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=CHC{{H}_{3}}\]

D) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\underset{C{{H}_{2}}C{{H}_{3}}}{\overset{{}}{\mathop{\underset{|}{\overset{{}}{\mathop{C}}}\,}}}\,=C{{H}_{2}}\]

Correct Answer: B

Solution :

[b]