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(A) The reactant is $4-hydroxybenzaldehyde$,which contains both an aldehyde group $(-CHO)$ and a phenolic hydroxyl group $(-OH)$.When $PhMgBr$ (a Grig

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(A) The reactant is $4-hydroxybenzaldehyde$,which contains both an aldehyde group $(-CHO)$ and a phenolic hydroxyl group $(-OH)$.When $PhMgBr$ (a Grignard reagent) is added,it acts as a strong base and reacts with the acidic phenolic $-OH$ group first.$4-hydroxybenzaldehyde + PhMgBr \rightarrow 4-formylphenoxide magnesium bromide + Benzene$The $Ph^-$ from $PhMgBr$ abstracts the acidic proton from the phenolic $-OH$ group to form benzene $(C_6H_6)$ and the magnesium salt of the phenol.Upon workup with $aq. NH_4Cl$,the magnesium salt is protonated back to the original $4-hydroxybenzaldehyde$.Therefore,the final product $P$ is $4-hydroxybenzaldehyde$,which contains $1$ hydroxyl group.

The product of the following reaction is $P$.
$4-hydroxybenzaldehyde + PhMgBr (1 \text{ equiv.}) \xrightarrow{aq. NH_4Cl} P$
The number of hydroxyl groups present in the product $P$ is . . . . . . .

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The product of the following reaction is $P$. $4-hydroxybenzaldehyde + PhMgBr (1 \text{ equiv.}) \xrightarrow{aq. NH_4Cl} P$The number of hydroxyl groups present in the product $P$ is . . . . . . .