The reaction sequence is as follows:Catechol | vedclass

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Question

(C) The reaction proceeds in two steps:Step $(i)$: $Catechol$ reacts with $CHCl_3$ and $KOH$ (Reimer-Tiemann reaction) to form $3,4-dihydroxybenzaldeh

Vedclass · Accepted Answer

$1,3-$Benzodioxole$-5-$carboxaldehyde

Vedclass · Answer

(C) The reaction proceeds in two steps:Step $(i)$: $Catechol$ reacts with $CHCl_3$ and $KOH$ (Reimer-Tiemann reaction) to form $3,4-dihydroxybenzaldehyde$ (protocatechualdehyde).Step $(ii)$: The resulting $3,4-dihydroxybenzaldehyde$ reacts with $CH_2I_2$ in the presence of $NaOH$. This is a Williamson ether synthesis type reaction where the two phenolic $-OH$ groups undergo cyclization with the methylene group $(CH_2)$ from $CH_2I_2$ to form a cyclic acetal/ether linkage.The final product $[X]$ is $1,3-benzodioxole-5-carboxaldehyde$ (also known as piperonal).

Column-$I$ (Reaction)	Column-$II$ (Product)
$(A)$ Dinitration of $4$-methylphenol	$(i)$ $2$-nitro-$4$-methylphenol + $2$-nitro-$5$-methylphenol
$(B)$ Mononitration of phenyl ethanoate	$(ii)$ $2,4$-dinitro-$5$-methylphenol
$(C)$ Dinitration of $3$-methylphenol	$(iii)$ $4$-nitrophenyl ethanoate
$(D)$ Mononitration of $3$-methylphenol	$(iv)$ $2,6$-dinitro-$4$-methylphenol

The reaction sequence is as follows:
$Catechol \xrightarrow{(i) CHCl_3 + KOH}{(ii) CH_2I_2 + NaOH} [X]$
Here $[X]$ is:

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The reaction sequence is as follows:$Catechol \xrightarrow{(i) CHCl_3 + KOH}{(ii) CH_2I_2 + NaOH} [X]$Here $[X]$ is: