When $H I$ is added in excess to the unsymmetrical ether, it yields:

P h C H_{2} O H

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P h I

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H O C H_{2} C H_{2} P h - N O_{2}

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I C H_{2} C H_{2} P h N O_{2}

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Solution

The correct option is D $I C H_{2} C H_{2} P h N O_{2}$
When HI is added in excess to the unsymmetrical ether, it results in the formation of phenol and $I C H_{2} C H_{2} P h N O_{2}$
$P h - O - C H_{2} C H_{2} P h N O_{2} H I - - \to I C H_{2} C H_{2} P h N O_{2} + P h O H$
Iodine is attached to aliphatic carbon atom and -OH group is attached to aromatic C atom.

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