Identify A in the following chemical reaction.

The reaction begins with a crossed Cannizzaro reaction between $$\mathrm{4\text{-}Methoxybenzaldehyde}$$ and $$\mathrm{HCHO}$$ in presence of $$\mathrm{NaOH}$$.

Since neither aldehyde contains $$\mathrm{\alpha}$$-hydrogens, disproportionation occurs.

Formaldehyde is preferentially oxidised to formate, while the aromatic aldehyde is reduced to:

$$\mathrm{4\text{-}Methoxybenzyl\ Alcohol}$$

In the second step, $$\mathrm{NaH}$$ deprotonates the alcohol to form an alkoxide ion.

The alkoxide attacks $$\mathrm{CH_3CH_2Br}$$ through an $$\mathrm{S_N2}$$ mechanism in DMF, forming:

$$\mathrm{4\text{-}Methoxybenzyl\ Ethyl\ Ether}$$

In the final step, concentrated $$\mathrm{HI}$$ and heat cleave both ether linkages.

In the aryl methyl ether, iodide attacks the methyl group, forming a phenol group $$\mathrm{(-OH)}$$.

In the benzylic ether, iodide attacks the benzylic carbon, replacing the ethoxy group with:

$$\mathrm{(-CH_2I)}$$

The final major product contains a phenol group and an iodomethyl group at para positions on the benzene ring.

Correct Option: $$\mathrm{A}$$