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(B) The reaction given is the Reimer-Tiemann reaction,which involves the formylation of phenols using $CHCl_3$ and $NaOH$ followed by acidic workup.In

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$2-$hydroxy$-5-$methoxybenzaldehyde

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(B) The reaction given is the Reimer-Tiemann reaction,which involves the formylation of phenols using $CHCl_3$ and $NaOH$ followed by acidic workup.In the starting material,$4$-methoxyphenol,the $-OH$ group is a strong ortho/para directing group,and the $-OCH_3$ group is also an ortho/para directing group.The $-OH$ group is more strongly activating than the $-OCH_3$ group.Therefore,the formylation occurs at the position ortho to the $-OH$ group.Since the para position to the $-OH$ group is already occupied by the $-OCH_3$ group,the formylation occurs at the ortho position,resulting in $2$-hydroxy-$5$-methoxybenzaldehyde.

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