Step 1: Understanding the Concept:
Phenol reacts with formaldehyde (\(\text{HCHO}\)) in the presence of acid in an electrophilic substitution (Lederer-Manasse type) reaction. The \(-\text{OH}\) group activates the ring and directs the new group to the ortho and para positions.
Step 2: Key Formula or Approach:
The electrophile is the protonated formaldehyde, \(\text{CH}_2=\text{OH}^+\). It attacks the ring and introduces a \(-\text{CH}_2\text{OH}\) group.
Step 3: Detailed Explanation:
The first product is o-hydroxybenzyl alcohol, a benzene ring with \(-\text{OH}\) and \(-\text{CH}_2\text{OH}\) on neighbouring carbons (the p-isomer forms too). This is option D.
Option A (benzoic acid) and option C (benzaldehyde) need oxidation, which is not present here. Option B (benzyl alcohol) has no \(-\text{OH}\) on the ring, so the phenol group would have to be lost, which does not happen.
Final Answer:
Phenol and formaldehyde with acid give o-hydroxybenzyl alcohol, option (D).
\[ \boxed{\text{o-Hydroxybenzyl alcohol}} \]
