Consider the reaction of HI with the followin | vedclass

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(C) Compound $I$ (tetrahydropyran) is a cyclic ether. Upon reaction with excess $HI$,the ether ring undergoes cleavage.In the first step,the protonate

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$I$ only

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(C) Compound $I$ (tetrahydropyran) is a cyclic ether. Upon reaction with excess $HI$,the ether ring undergoes cleavage.In the first step,the protonated ether is attacked by $I^-$ to form $HO-(CH_2)_5-I$.In the second step,the hydroxyl group is converted to an iodide by $HI$,resulting in the formation of $I-(CH_2)_5-I$ ($1$,$5$-diiodopentane).Compound $II$ contains a phenolic ether linkage (where the oxygen is attached to an aromatic ring). Cleavage of the $C(aryl)-O$ bond is extremely difficult due to the partial double bond character of the $C-O$ bond and the stability of the aromatic ring,so it does not form a di-iodide under standard conditions.

Consider the reaction of $HI$ with the following compounds:
$I$: Tetrahydropyran
$II$: Isochroman (or similar cyclic ether fused to benzene)
Which forms a di-iodide on reaction with $HI$ (excess)?

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Consider the reaction of $HI$ with the following compounds:$I$: Tetrahydropyran$II$: Isochroman (or similar cyclic ether fused to benzene)Which forms a di-iodide on reaction with $HI$ (excess)?