Identify the product in the following reaction :

An aldehyde that possesses at least one $$\alpha$$-hydrogen undergoes the base-catalysed aldol reaction. In the given problem the reactant is ethanal (acetaldehyde) and the reagent is dilute aqueous $$NaOH$$ kept below about $$40^{\circ}\text{C}$$, so dehydration does not take place. Only the aldol addition is observed.

Step 1 - Enolate formation
$$CH_3CHO + OH^- \;\longrightarrow\; CH_2CHO^- + H_2O$$

Step 2 - Nucleophilic attack on a second carbonyl molecule
The enolate ion formed above attacks the carbonyl carbon of another acetaldehyde molecule:
$$CH_2CHO^- + CH_3CHO \;\longrightarrow\; CH_3CH(OH)CH_2CHO + OH^-$$

Step 3 - Protonation
The alkoxide produced in step 2 is protonated by water, regenerating the hydroxide ion used in step 1. The overall stoichiometry is therefore
$$2CH_3CHO \xrightarrow[\;dil.\;]{NaOH,\; 295\,\text{K}} CH_3CH(OH)CH_2CHO$$

The product is $$\beta$$-hydroxyaldehyde, specifically 3-hydroxybutanal (often called “aldol”). Among the options given, this structure corresponds to Option A.

Hence the correct answer is:
Option A which is: 3-hydroxybutanal (CH3CH(OH)CH2CHO).