Question
Isobutyl magnesium bromide with dry ether and ethyl alcohol gives :
A.
B.
C.
D.
Answer :
Solution :
\[{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{2}}MgBr\xrightarrow{{{C}_{2}}{{H}_{5}}OH}{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{3}}\]
\[{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{2}}MgBr\xrightarrow{{{C}_{2}}{{H}_{5}}OH}{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{3}}\]