In the following reaction:
carbonyl compound + MeOH $$\overset{\mathrm{HCl}}{\rightleftharpoons}$$ acetal
Rate of the reaction is the highest for:

We are dealing with the acid-catalysed conversion of a carbonyl compound to an acetal in methanol. The simplified mechanism proceeds through these key elementary steps:

$$\text{(1) Protonation of the carbonyl: }R_{2}C=O + H^+ \longrightarrow R_{2}C=OH^+$$

$$\text{(2) Nucleophilic attack by methanol: }R_{2}C=OH^+ + MeOH \longrightarrow R_{2}C(OMe)OH + H^+$$

The overall rate, to a very good approximation in the initial stages, is proportional to the concentrations of the species actually colliding in the slow step. After the fast proton-transfer of step (1), the slow or rate-determining step is generally step (2). Therefore we can write

$$\text{Rate} \propto [\text{protonated carbonyl}]\,[\text{MeOH}].$$

Now we compare the options on the basis of two independent factors that control this rate expression:

Factor 1: Nature of the carbonyl electrophile.

An aldehyde is more electrophilic than a ketone because:

(i) it has only one electron-donating alkyl group instead of two, so the carbonyl carbon bears a larger partial positive charge, and

(ii) it suffers less steric hindrance to nucleophilic attack.

Hence $$\text{Rate}_{\text{aldehyde}} > \text{Rate}_{\text{ketone}}.$$

Between the substrates listed, propanal (an aldehyde) is thus inherently faster than acetone (a ketone).

Factor 2: Concentration of the nucleophile MeOH.

The rate expression contains $$[\text{MeOH}],$$ so increasing the concentration of methanol directly raises the rate. Using methanol in excess means a higher $$[\text{MeOH}]$$ than in a stoichiometric amount.

Therefore $$\text{Rate}_{\text{excess MeOH}} > \text{Rate}_{\text{stoichiometric MeOH}}.$$

Let us rank the four options by systematically combining these two factors.

Step 1 - Substrate effect:

Propanal options (B and C) > Acetone options (A and D).

Step 2 - Methanol concentration effect (within each substrate type):

Excess MeOH (B and D) > Stoichiometric MeOH (A and C).

Combining both steps:

The single option that has the more reactive aldehyde and methanol in excess is Option B.

Mathematically, if we denote the proportionality constant as $$k$$ and use subscripts to distinguish situations, we can write

$$\text{Rate}_{\text{B}} = k[\text{propanal}\!-\!\text{H}^+][\text{MeOH}]_{\text{excess}},$$

$$\text{Rate}_{\text{C}} = k[\text{propanal}\!-\!\text{H}^+][\text{MeOH}]_{\text{stoich}},$$

and clearly $$[\text{MeOH}]_{\text{excess}} > [\text{MeOH}]_{\text{stoich}},$$ so $$\text{Rate}_{\text{B}} > \text{Rate}_{\text{C}}.$$

Any rate involving acetone will already be slower due to the less electrophilic carbonyl, cementing the conclusion that

$$\text{Rate}_{\text{B}} > \text{Rate}_{\text{D}} \gt \text{Rate}_{\text{C}} \gt \text{Rate}_{\text{A}}.$$

Hence, the correct answer is Option B.