A) \[C{{H}_{2}}=CHC{{H}_{2}}CH=CHC{{H}_{3}}\]
B) \[C{{H}_{2}}=CHCH=CHC{{H}_{2}}C{{H}_{3}}\]
C) \[C{{H}_{3}}CH=C=CHC{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}CH=CH-CH=CHC{{H}_{3}}\]
Correct Answer: D
Solution :
[d] \[C{{H}_{3}}-\underset{Br}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-\underset{Br}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\] \[\xrightarrow[E2]{KOH,\,\,C{{H}_{3}}OH\,\,\Delta }\] \[C{{H}_{3}}-CH=\underset{(Saytzeff\,\,product)}{\mathop{CH-CH=C{{H}_{2}}}}\,-C{{H}_{3}}\]You need to login to perform this action.
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