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1. |
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. | \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\underset{C{{H}_{2}}C{{H}_{3}}}{\overset{{}}{\mathop{\underset{|}{\overset{{}}{\mathop{C}}}\,}}}\,=C{{H}_{2}}\] |
Answer» C. \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=CHC{{H}_{3}}\] | |