A)
X Y Z \[C{{H}_{3}}CH=C{{H}_{2}}\] \[C{{H}_{3}}CH\left( Br \right)C{{H}_{2}}Br\] \[C{{H}_{3}}CH\left( OH \right)C{{H}_{3}}\]
B)
\[C{{H}_{3}}CH=C{{H}_{2}}\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\] \[C{{H}_{3}}C{{H}_{2}}OC{{H}_{2}}C{{H}_{3}}\]
C)
\[C{{H}_{3}}CH=C{{H}_{2}}\] \[C{{H}_{3}}CH\left( Br \right)C{{H}_{3}}\] \[C{{H}_{3}}C{{H}_{2}}OC{{H}_{2}}C{{H}_{3}}\]
D)
\[C{{H}_{3}}CH=C{{H}_{2}}\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OC{{H}_{3}}\]
Correct Answer: D
Solution :
\[C{{H}_{3}}CHBrC{{H}_{3}}\xrightarrow{alc.\,KOH}C{{H}_{3}}CH=C{{H}_{2}}\] \[\downarrow \]HBr, peroxide \[C{{H}_{3}}\underset{Z}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}OC{{H}_{3}}\xleftarrow{C{{H}_{3}}ONa}C{{H}_{3}}\underset{Y}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}Br\]You need to login to perform this action.
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