Ethyl alcohol cannot be used as a solvent for $C_{2} H_{5} M g I$ because:

C H_{3} M g I

reacts with alcohol giving in methane.

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the reaction between them is explosive in nature.

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C H_{3} M g I

is converted to

C_{2} H_{5} M g I

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alcohol is immisicible with

C H_{3} M g I

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Solution

The correct option is B $C H_{3} M g I$ reacts with alcohol giving in methane.
Ethyl alcohol cannot be used as a solvent for $C H_{3} M g I$ because of $C H_{3} M g I$
reacts with alcohol giving in methane.

$C H_{3} M g I + C_{2} H_{5} - O H \to C H_{4} + C_{2} H_{5} - O - M g I$

$H$ from ethanol reacts with $C H_{3}^{-}$ group from $C H_{3} M g I$ to form methane.

So, the correct option is $A$

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