1. Reaction involved: formation of acetal

An aldehyde reacts with an alcohol in the presence of acid to form first a hemiacetal and then an acetal:

$RCHO + 2ROH \xrightarrow{H^+} RCH(OR)_2 + H_2O$

This reaction is favored when:

• the aldehyde is more reactive,
• the alcohol is less hindered,
• acetal formation is not sterically difficult.

2. Compare the aldehydes

The two aldehydes are:

• $\mathrm{HCHO}$ (formaldehyde)
• $\mathrm{CH_3CHO}$ (acetaldehyde)

Formaldehyde is more reactive than acetaldehyde toward nucleophilic addition because it has:

• less steric hindrance,
• no electron-donating alkyl group.

So, for acetal formation:

$\mathrm{HCHO > CH_3CHO}$

3. Compare the alcohols

The two alcohols are:

• $\mathrm{MeOH}$ (methanol)
• $\mathrm{tBuOH}$ (tert-butanol)

Methanol is much better for acetal formation because:

• it is less hindered,
• it attacks the carbonyl carbon more easily.

Tert-butanol is highly hindered and under acidic conditions may also undergo side reactions/dehydration, so it is a poor choice for forming acetals.

Thus:

$\mathrm{MeOH \gg tBuOH}$

4. Evaluate the options

Option A: $\mathrm{CH_3CHO}$ and $\mathrm{MeOH}$

Possible, but acetaldehyde is less reactive than formaldehyde.

Option B: $\mathrm{HCHO}$ and $\mathrm{MeOH}$

This is the best combination because:

• $\mathrm{HCHO}$ is the most reactive aldehyde,
• $\mathrm{MeOH}$ is the least hindered alcohol.

Option C: $\mathrm{CH_3CHO}$ and $\mathrm{tBuOH}$

Both components are less favorable; especially $\mathrm{tBuOH}$ is very poor for acetal formation.

Option D: $\mathrm{HCHO}$ and $\mathrm{tBuOH}$

Formaldehyde is good, but $\mathrm{tBuOH}$ is highly hindered, so this is inferior to option B.

5. Conclusion

The best aldehyde–alcohol pair for acetal formation is:

$\boxed{\mathrm{HCHO \ and \ MeOH}}$

So the correct option is B.