When isobutyl magnesium bromide in dry ether is treated with abso…

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MCQ

When isobutyl magnesium bromide in dry ether is treated with absolute ethyl alcohol, the products formed are

A
$\mathop {\mathop {C{H_3} - CH - C{H_2}OH}\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,\,\,\,\,\,} $ and $C{H_3}C{H_2}MgBr$
B
$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ and $Mg(OH)Br$
✓
$\mathop {\mathop {C{H_3} - CH - C{H_3}}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,} $ and $C{H_3} - C{H_2}OMgBr$
D
$\mathop {\mathop {C{H_3} - CH - C{H_3}}\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}} ,C{H_2} = C{H_2}$ and $Mg(OH)Br$

✓

Answer

Correct option: C.

$\mathop {\mathop {C{H_3} - CH - C{H_3}}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,} $ and $C{H_3} - C{H_2}OMgBr$

c
$(c)$

$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - \boxed{Mg - Br + {C_2}{H_5}O}H} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_{3\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array} \to $

$\mathop {\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}} \\
| \\
{C{H_3}}
\end{array}}\limits_{Isobu\tan e} $ $+$ $\underset{\begin{smallmatrix}
Ethoxy\,magnesium \\
bromide
\end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{5}}OMgBr}}\,$