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Question

(B) The reaction of $3,4$-methylenedioxybenzaldehyde (piperonal) with nitromethane $(CH_3NO_2)$ in the presence of a base $(NaOH)$ is a Henry reaction

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(B) The reaction of $3,4$-methylenedioxybenzaldehyde (piperonal) with nitromethane $(CH_3NO_2)$ in the presence of a base $(NaOH)$ is a Henry reaction (nitro-aldol condensation). This reaction produces a $\beta$-hydroxy nitro compound as an intermediate,which then undergoes dehydration to form the nitroalkene,$(A)$,which is $3,4$-methylenedioxy-$\beta$-nitrostyrene. The structure of $(A)$ is a benzene ring with a methylenedioxy group at the $3,4$-positions and a $-CH=CH-NO_2$ group at the $1$-position. When $(A)$ is treated with $HCl$ and heated,the methylenedioxy group $(-O-CH_2-O-)$ undergoes hydrolysis to yield the corresponding catechol derivative (a benzene ring with two $-OH$ groups at the $3,4$-positions) and the side chain $-CH=CH-NO_2$ remains intact. Therefore,the final product $(B)$ is $3,4$-dihydroxy-$\beta$-nitrostyrene.

The product $(B)$ is:

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