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Question

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

A

$$CH_{3}-\underset{CH_{3}}{\underset{|}{CH}}-CH_{2}OH$$ and $$CH_{3}CH_{2}MgBr$$

B

$$CH_{3}-\underset{CH_{3}}{\underset{|}{CH}}-CH_{2}-CH_{2}-CH_{3}$$ and $$Mg(OH)Br$$

C

$$CH_{3}-\underset{CH_{3}}{\underset{|}{CH}}-CH_{3}$$ and $$CH_{3}-CH_{2}OMgBr$$

D

$$CH_{3}-\underset{CH_{3}}{\underset{|}{CH}}-CH_{3}$$, $$CH_{2}=CH_{2}$$ and $$Mg(OH)Br$$

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Solution

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Correct option is C. $$CH_{3}-\underset{CH_{3}}{\underset{|}{CH}}-CH_{3}$$ and $$CH_{3}-CH_{2}OMgBr$$

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