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(A) The starting material is $1,3$-benzodioxole. Step $(i)$: Treatment with excess $HI$ causes the cleavage of the cyclic acetal (dioxole ring). The $

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(A) The starting material is $1,3$-benzodioxole. Step $(i)$: Treatment with excess $HI$ causes the cleavage of the cyclic acetal (dioxole ring). The $C-O$ bonds are broken,resulting in the formation of a catechol derivative ($1,2$-dihydroxybenzene) and formaldehyde $(CH_2O)$,but in this specific fused system,the ring opening leads to $2$-hydroxybenzyl alcohol (salicyl alcohol). Step $(ii)$: The Reimer-Tiemann reaction $(CHCl_3, NaOH)$ introduces a formyl group $(-CHO)$ at the ortho position relative to the phenolic $-OH$ group. Thus,the product is $2$-hydroxy$-3-$(hydroxymethyl)benzaldehyde.

List-$I$	List-$II$
$A$. Benzaldehyde	$i$. Phenolphthalein
$B$. Phthalic anhydride	$ii$. Benzoin condensation
$C$. Phenyl benzoate	$iii$. Oil of wintergreen
$D$. Methyl salicylate	$iv$. Fries rearrangement

Predict the major product of the following reaction:
$1,3$-benzodioxole $\xrightarrow[(ii) CHCl_3, NaOH]{(i) HI \text{ (excess)}}$ ?

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Predict the major product of the following reaction:$1,3$-benzodioxole $\xrightarrow[(ii) CHCl_3, NaOH]{(i) HI \text{ (excess)}}$ ?