What are the products obtained from the reaction given below?
$M e O E t H I - - \to ?$

M e I

and

E t O H

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M e O H

and

E t I

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M e I, M e O H, E t I

and

E t O H

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None of the above

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Solution

The correct option is B $M e I$ and $E t O H$
$M e O E t H I - - \to M e I + E t O H$
Due to $S N^{2}$ mechanism, iodide ion approaches $M e -$ as it provides less hindrance

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(C H_{3})_{3} - C - B r 80 % E t O H \to 20 % H_{2} O

M e - C \equiv C - M e N a / N H_{3} (l i q .) - ------- \to E t O H, - 33^{\circ} C X = d i l . a l k a l i n e K M n O_{4} - ----------- \to

N H_{3}

{C H}_{3} {C O}_{2} E t + M e M g l 2 e q u i v a l e n t (i i) H^{+} - ---- \to (i) E t h e r (A) + E t O H

(A)

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