A compound of molecular formula C$$_8$$H$$_8$$O$$_2$$ reacts with acetophenone to form a single cross-aldol product in the presence of base. The same compound on reaction with concentrated NaOH forms benzyl alcohol as one of the products. The structure of the compound is:

We are told that the unknown organic compound has the molecular formula $$C_8H_8O_2$$. We also know two important pieces of reactivity:

(i) It reacts with acetophenone $$\left(C_6H_5COCH_3\right)$$ in the presence of base to give a single cross-aldol (more precisely, Claisen-Schmidt) condensation product.

(ii) When treated with concentrated $$NaOH$$ it produces benzyl alcohol $$\left(C_6H_5CH_2OH\right)$$ among the products.

Let us analyse each clue step by step and compare with the four given structures.

Step 1 Checking the molecular formula of every option

First we write the molecular formula for every candidate compound, counting all carbon, hydrogen and oxygen atoms.

A. Methoxybenzaldehyde (p-$$CH_3O\!-\!C_6H_4\!-\!CHO$$) Ring $$C_6H_4$$ gives $$C_6H_4$$     Side groups: $$CH_3O$$ contributes $$C_1H_3O_1$$, and $$CHO$$ contributes $$C_1H_1O_1$$.     Total: $$C_{6+1+1}=C_8,\;H_{4+3+1}=H_8,\;O_{1+1}=O_2.$$     So option A exactly matches $$C_8H_8O_2.$$

B. Methyl benzoate $$\left(C_6H_5COOCH_3\right)$$     Ring $$C_6H_5$$ plus $$COOCH_3$$ (which is $$C_2H_3O_2$$).     Total: $$C_{6+2}=C_8,\;H_{5+3}=H_8,\;O_2.$$     This also fits $$C_8H_8O_2.$$

C. p-Methylbenzoic acid (p-$$CH_3\!-\!C_6H_4\!-\!COOH$$)     Ring $$C_6H_4$$, plus $$CH_3$$ ( $$C_1H_3$$ ), plus $$COOH$$ ( $$C_1H_1O_2$$ ).     Total: $$C_{6+1+1}=C_8,\;H_{4+3+1}=H_8,\;O_2.$$     Again it fits the formula.

D. p-Hydroxy methyl acetophenone (HO-$$C_6H_4\!-\!COCH_3$$)     Ring $$C_6H_4$$, plus $$OH$$ ( $$O_1H_1$$ ), plus $$COCH_3$$ ( $$C_2H_3O_1$$ ).     Total: $$C_{6+2}=C_8,\;H_{4+1+3}=H_8,\;O_{1+1}=O_2.$$     This also matches $$C_8H_8O_2.$$

Thus the molecular formula alone keeps all four options in the race, so we must rely on the given reactions.

Step 2 Interpreting the reaction with acetophenone under basic conditions

The phrase “reacts with acetophenone to form a single cross-aldol product in the presence of base” is the textual description of a Claisen-Schmidt condensation. The general fact we must recall is:

In an aldol/Claisen-Schmidt reaction, the likelihood of getting a single product is highest when one carbonyl component has no α-hydrogens. Such a carbonyl compound can act only as an electrophile (acceptor) and never as an enolate donor.

Among ordinary carbonyl groups, an aromatic aldehyde $$Ar-CHO$$ possesses no hydrogens at the α-position because the carbonyl carbon is directly attached to the aromatic ring, not to a carbon bearing hydrogens. Hence compounds of the type $$Ar-CHO$$ satisfy the “no α-H” condition and always give a single cross-aldol product with a ketone such as acetophenone.

We now examine which options contain an aldehyde group and lack α-hydrogens.

A. Methoxybenzaldehyde - contains $$\;CHO$$ attached to an aromatic ring. No α-H present.

B. Methyl benzoate - is an ester. Esters do not undergo the ordinary aldol mechanism with acetophenone.

C. p-Methylbenzoic acid - is a carboxylic acid, not an aldehyde, and therefore not eligible for aldol condensation.

D. p-Hydroxy methyl acetophenone - is a ketone with a $$CH_3$$ group next to the carbonyl; it does have α-hydrogens, so multiple self- and cross-aldol products would be possible. This contradicts the “single product” statement.

Therefore, only option A is consistent with the first observation.

Step 3 Interpreting the reaction with concentrated NaOH giving benzyl alcohol

Now we recall another well-known reaction:

The Cannizzaro reaction states that an aldehyde devoid of α-hydrogens, when treated with strong base (concentrated $$NaOH$$), undergoes self-redox to give one molecule reduced to an alcohol and another oxidised to a carboxylate salt.

Symbolically, for an aldehyde $$Ar-CHO$$ with no α-H,

$$2\,Ar-CHO \;\xrightarrow[\;]{\;conc.\;NaOH\;} Ar-CH_2OH \;+\; Ar-COO^-Na^+$$

One of the products is always the corresponding benzyl alcohol $$Ar-CH_2OH$$. This is exactly the second clue in the question.

Which of the options can undergo a Cannizzaro reaction?

A. Methoxybenzaldehyde - yes, it is an aldehyde lacking α-H; Cannizzaro feasible.

B. Methyl benzoate - ester; Cannizzaro not possible.

C. p-Methylbenzoic acid - carboxylic acid; Cannizzaro not possible.

D. p-Hydroxy methyl acetophenone - a ketone with α-H; Cannizzaro does not occur.

Again, only option A satisfies the second observation.

Step 4 Final decision

Both experimental facts - exclusive cross-aldol product formation with acetophenone and the Cannizzaro reaction giving benzyl alcohol - point unambiguously to an aromatic aldehyde that lacks α-hydrogens. Among the given choices, that description fits only p-methoxybenzaldehyde.

Hence, the correct answer is Option A.