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

C H_{3} - C H - C H_{2} O H a n d C H_{3} C H_{2} M g B r | C H_{3}

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C H_{3} - C H - C H_{2} - C H_{2} - C H_{3} a n d M g (O H) B r | C H_{3}

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C H_{3} - C H - C H_{3} a n d C H_{3} - C H_{2} O M g B r | C H_{3}

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C H_{3} - C H - C H_{3}, C H_{2} = C H_{2} a n d M g (O H) B r | C H_{3}

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Solution

The correct option is C $C H_{3} - C H - C H_{3} a n d C H_{3} - C H_{2} O M g B r | C H_{3}$

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