In the following reaction:

The best combination is:

We recall the general acid-catalysed acetal formation:

$$\text{Aldehyde}(RCHO) + 2\,\text{Alcohol}(R'OH) \xrightarrow{HCl} RCH(OR')_2 + H_2O$$

The mechanism involves these key steps:

1. Protonation of the carbonyl oxygen to give a more electrophilic $$RCH^+=O$$ species.
2. Nucleophilic attack of the alcohol oxygen on the carbonyl carbon to form a hemiacetal.
3. Proton transfers followed by loss of water and a second alcohol attack to give the acetal.

The slow, rate-determining step is the nucleophilic attack of the alcohol on the protonated carbonyl carbon. Hence:

• The electrophile should be as reactive and as unhindered as possible.
• The nucleophile (alcohol) should be as unhindered and as nucleophilic as possible.

Reactivity order of carbonyl compounds toward nucleophilic addition is

$$\text{HCHO} \; \lt \; CH_3CHO \; \lt \; R_2C=O$$

but when we talk of ease of attack, “<” actually reads “less hindered & more reactive → reacts faster,” so

Formaldehyde (HCHO) > Acetaldehyde ($$\text{CH}_3\text{CHO}$$) > Ketones

Among alcohols, steric bulk decreases nucleophilicity:

MeOH (1°) > EtOH (1° larger) > 2° alcohols > t-BuOH (3°)

Now we examine each option.

A. Aldehyde = $$\text{HCHO}$$ (least hindered, most electrophilic); Alcohol = $$\text{MeOH}$$ (least hindered, most nucleophilic).
B. Aldehyde = $$CH_3CHO$$ (more hindered than HCHO); Alcohol = $$\text{t-BuOH}$$ (very hindered).
C. Aldehyde = $$CH_3CHO$$; Alcohol = $$\text{MeOH}$$ (good alcohol but aldehyde slightly less reactive).
D. Aldehyde = $$\text{HCHO}$$; Alcohol = $$\text{t-BuOH}$$ (excellent aldehyde but very poor, hindered alcohol).

Clearly, Option A combines the most electrophilic carbonyl with the most nucleophilic alcohol, minimising steric hindrance in both partners. This pair will give the fastest and highest-yielding acetal formation under $$HCl$$ catalysis.

Hence, the correct answer is Option A.