When $P h - O H$ reacts with $C_{2} H_{5} I$ with anhydrous $C_{2} H_{5} O H$ as the solvent in the presence of ethoxide ions, what is the product of this reaction?

P h - M e

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P h - O - M e

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P h - O - E t

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P h - E t

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Solution

The correct option is C $P h - O - E t$
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Similar questions

P h - O H

C_{2} H_{5} I

C_{2} H_{5} O H

P h - O H + C_{2} H_{5} I C_{2} H_{5} O N a - --------- \to anhyd. C_{2} H_{5} O H

P h - C O O - P h

P h - C O O H

C H_{3} M g I

Between pH value of 7 to 14, As pH increases:

P h - O - C H_{2} - P h

H I

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