Question

Find the product for
$CH_{3}CH_{2}-O-CH_{2}-CH_{2}-O-CH_{2}-C_{6}H_{5}+HI$
(excess)

• A. $HO-CH_{2}CH_{2}OH, C_{6}H_{5}CH_{2}-I, CH_{3}CH_{2}-I$
• B. $C_{6}H_{5}CH_{2}-OH, CH_{3}CH_{2}-I, I-CH_{2}CH_{2}-OH$
• C. $I-CH_{2}CH_{2}-I, C_{6}H_{5}CH_{2}-I, CH_{3}CH_{2}-OH$
• D. $HO-CH_{2}CH_{2}-OH, C_{6}H_{5}CH_{2}-I, CH_{3}CH_{2}-OH$

Answer 1

Answer: (A)

Presence of excess of $HI$ favours $S_{N}1$ mechanism.

So, formation of products is controlled by the stability of the carbocation resulting in the cleavage of $C - O$ bond in protonated ether. Thus the product for given equation are

$C_6H_5CH_2I, CH_3CH_2I, HOCH_2 - CH_2OH$