JEE Chemistry 2024 Question with Solution

Given: The question asks for the ascending order of acidity of the –OH group in compounds (A), (B), (C), (D) and (E).

Find: The correct increasing order of acidity.

Phenols are acidic because they can lose a proton from the hydroxyl group to form a phenoxide ion. The acidity depends on the stability of the conjugate base. Electron-withdrawing groups increase acidity by stabilizing the negative charge, while electron-donating groups decrease acidity by destabilizing it.

From the given solution:

• (A) Bu – OH is a simple alcohol, so it is the least acidic.
• (C) $p-\text{OCH}_3\text{-C}_6\text{H}_4\text{-OH}$ contains an electron-donating methoxy group, so it is less acidic than phenol.
• (D) Phenol is the reference compound.
• (B) $p-\text{NO}_2\text{-C}_6\text{H}_4\text{-OH}$ contains a strong electron-withdrawing nitro group, so it is more acidic than phenol.
• (E) $2,4-\text{NO}_2\text{-C}_6\text{H}_3\text{-OH}$ has two nitro groups, so it is the most acidic.

Therefore, the ascending order is:

Hence, the correct option is D.

Given: We compare the acidity of the –OH group in the listed compounds.

Find: The increasing order of acidity.

The key idea is the effect of substituents on the phenoxide ion formed after deprotonation.

• Electron-withdrawing groups such as $-\text{NO}_2$ stabilize the phenoxide ion and increase acidity.
• Electron-donating groups such as $-\text{OCH}_3$ decrease acidity.

So the compounds are ordered as follows:

1. (A) Butanol-type alcohol: no resonance stabilization, least acidic.
2. (C) $p-\text{OCH}_3\text{-C}_6\text{H}_4\text{-OH}$: methoxy donates electron density, reducing acidity.
3. (D) Phenol: more acidic than alcohol and methoxy-substituted phenol.
4. (B) $p-\text{NO}_2\text{-C}_6\text{H}_4\text{-OH}$: nitro group strongly withdraws electron density, increasing acidity.
5. (E) $2,4-\text{NO}_2\text{-C}_6\text{H}_3\text{-OH}$: two nitro groups give maximum stabilization and highest acidity.

Thus,

So the correct option is D.