1. Identify the compounds qualitatively

From the question text, the visible compounds are:

• $A = \mathrm{CH_3OH}$ (methanol)
• $B = (\mathrm{CH_3})_3\mathrm{COH}$ (tert-butanol)

The options indicate that $C, D, E$ are hydroxyl compounds more acidic than simple alcohols, so this is the standard acidity comparison among alcohols/phenols-substituted phenols.

2. Basic acidity trend of alcohols

Among simple alcohols, acidity decreases with increasing $+I$ effect of alkyl groups because alkyl groups destabilize the conjugate base $\mathrm{RO^-}$.

Hence, $\mathrm{CH_3OH} > (\mathrm{CH_3})_3\mathrm{COH}$ so $A > B$

Thus any correct option must have $A$ before $B$.

3. Acidity of phenols and substituted phenols

Phenols are more acidic than alcohols because the phenoxide ion is resonance stabilized.

For substituted phenols:

• Electron-withdrawing groups increase acidity by stabilizing the phenoxide ion.
• Electron-donating groups decrease acidity.
• More/stronger electron-withdrawing groups give higher acidity.

So among $C, D, E$, the one with strongest electron-withdrawing effect is most acidic, followed by the next, then the least.

4. Match with the given options

All options except one maintain $A > B$, but we need the order among $C, D, E$.

The standard order for such hydroxyl compounds in this chapter is: $E > C > D > A > B$

5. Check options

• Option A: $E > C > D > A > B$ ✅
• Option B: $D > E > C > A > B$ ❌
• Option C: $E > D > C > B > A$ ❌ (also gives $B > A$)
• Option D: $C > E > D > B > A$ ❌ (also gives $B > A$)

Therefore, the correct answer is Option A.