When ethylamine reacts with methyl magnesium bromide, the product is

C H_{3} C H_{3}

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C H_{4}

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C H_{3} C H_{2} C H_{3}

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C H_{3} C H_{2} C H_{2} C H_{3}

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Solution

The correct option is B $C H_{4}$
$C H_{2} C H_{2} N H_{2} + C H_{3} M g B r \to C H_{4} + M g (N H C_{2} H_{5}) B r$

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