JEE MainChemistry

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Ordering substituted phenols by $p K_{a}$

Q: Why does higher acidity mean lower pKa?

Because pKa is defined as -log10(Ka). A higher Ka means a stronger acid, which yields a more negative/smaller pKa value.

The increasing order of pK_()a for the following phenols is (A) 2,4 -Dinitrophenol (B) 4 -Nitrophenol (C) 2,4,5 -Trimethylphenol (D) Phenol (E) 3 -Chlorophenol Choose the correct answer from the option given below:

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A(A), (B), (E), (D), (C)B(C), (D), (E), (B), (A)C(A), (E), (B), (D), (C)D(C), (E), (D), (B), (A)

Step-by-step solutionView

Correct answer

The correct increasing order of $p K_{a}$ is (A) < (B) < (E) < (D) < (C), because stronger acids have lower $p K_{a}$ values.

Option analysis

Why each option works or fails

A · (A), (B), (E), (D), (C)

This is the correct order. Electron-withdrawing groups increase acidity (lower $p K_{a}$ ), while electron-donating groups decrease acidity (raise $p K_{a}$ ). Ranking from strongest acid to weakest acid yields (A) < (B) < (E) < (D) < (C).

B · (C), (D), (E), (B), (A)

Confusing acid strength ( $K_{a}$ ) with $p K_{a}$ , ranking from weakest acid to strongest acid instead of lowest $p K_{a}$ to highest $p K_{a}$ . Remember that $p K_{a} = - \log K_{a}$ ; a stronger acid has a smaller (more negative) $p K_{a}$ value.

C · (A), (E), (B), (D), (C)

Believing inductive electron withdrawal from meta-chloro is stronger than the combined resonance and inductive electron withdrawal of para-nitro. A para- $- N O_{2}$ group stabilizes the phenoxide ion through strong $- M$ and $- I$ effects, making 4-nitrophenol significantly more acidic than 3-chlorophenol (which only has a $- I$ effect).

D · (C), (E), (D), (B), (A)

Reversing the relation between acidity and $p K_{a}$ while simultaneously misplacing meta-chloro relative to unsubstituted phenol. Electron-withdrawing groups like chlorine increase acidity and lower $p K_{a}$ relative to unsubstituted phenol, and increasing $p K_{a}$ means ordering from strongest acid to weakest acid.

Reviewed route

Solution

StepWorking

01Concept

Acidic strength of substituted phenols is enhanced by electron-withdrawing groups ( $- M, - I$ ) which stabilize the phenoxide anion, and decreased by electron-donating groups ( $+ I, + H$ ) which destabilize it. Since $p K_{a} = - \log_{10} K_{a}$ , a stronger acid has a lower $p K_{a}$ . Therefore, the increasing order of $p K_{a}$ corresponds directly to the decreasing order of acidic strength (from strongest acid to weakest acid).

02Option verdict

(A) 2,4-Dinitrophenol has two strong $- M$ and $- I$ nitro groups $⟹$ strongest acid $⟹$ lowest $p K_{a}$ . (B) 4-Nitrophenol has one $- M, - I$ group $⟹$ second strongest acid. (E) 3-Chlorophenol has meta-Cl exerting only $- I$ effect $⟹$ moderately acidic. (D) Phenol is unsubstituted. (C) 2,4,5-Trimethylphenol has three $+ I, + H$ methyl groups destabilizing the phenoxide $⟹$ weakest acid $⟹$ highest $p K_{a}$ . Thus, increasing order of $p K_{a}$ is (A) < (B) < (E) < (D) < (C).

03Option verdict

This ranks the compounds in increasing order of acidic strength (or decreasing order of $p K_{a}$ ), confusing high acidity with high $p K_{a}$ .

04Option verdict

Places (E) 3-chlorophenol as more acidic than (B) 4-nitrophenol, mistakenly overestimating the $- I$ effect of chlorine relative to the resonance electron-withdrawing ( $- M$ ) effect of the para-nitro group.

05Option verdict

Inverts the $p K_{a}$ relationship and also misplaces 3-chlorophenol relative to phenol.

✓Discriminator

The strongest acid (2,4-dinitrophenol, A) must have the lowest $p K_{a}$ , eliminating Options 1 and 3 immediately. Since 4-nitrophenol (B) has strong $- M$ stabilization while 3-chlorophenol (E) only has weak $- I$ , B is more acidic than E, meaning $p K_{a} (B) < p K_{a} (E)$ , picking Option 0.

Hints that build this answer step by step

How does acid strength correlate with $p K_{a}$ values?
Stronger acids have lower $p K_{a}$ values, so increasing $p K_{a}$ means ordering from strongest acid to weakest acid.
How do the substituents on these phenols affect acid strength via electronic effects?
Two $- N O_{2}$ groups > one $- N O_{2}$ group > meta- $- C l$ > $- H$ > three $- C H_{3}$ groups in stabilizing the conjugate base.
Combining the acidity order (A > B > E > D > C) with the inverse relation to $p K_{a}$ , what is the increasing order of $p K_{a}$ ?
(A) < (B) < (E) < (D) < (C)

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why does higher acidity mean lower pKa?

Because pKa is defined as -log10(Ka). A higher Ka means a stronger acid, which yields a more negative/smaller pKa value.

Answer

The correct increasing order of $p K_{a}$ is (A) < (B) < (E) < (D) < (C), because stronger acids have lower $p K_{a}$ values.

Why each option works or fails

A · correct: (A), (B), (E), (D), (C) - This is the correct order. Electron-withdrawing groups increase acidity (lower $p K_{a}$ ), while electron-donating groups decrease acidity (raise $p K_{a}$ ). Ranking from strongest acid to weakest acid yields (A) < (B) < (E) < (D) < (C).
B: (C), (D), (E), (B), (A) - Confusing acid strength ( $K_{a}$ ) with $p K_{a}$ , ranking from weakest acid to strongest acid instead of lowest $p K_{a}$ to highest $p K_{a}$ . Remember that $p K_{a} = - \log K_{a}$ ; a stronger acid has a smaller (more negative) $p K_{a}$ value.
C: (A), (E), (B), (D), (C) - Believing inductive electron withdrawal from meta-chloro is stronger than the combined resonance and inductive electron withdrawal of para-nitro. A para- $- N O_{2}$ group stabilizes the phenoxide ion through strong $- M$ and $- I$ effects, making 4-nitrophenol significantly more acidic than 3-chlorophenol (which only has a $- I$ effect).
D: (C), (E), (D), (B), (A) - Reversing the relation between acidity and $p K_{a}$ while simultaneously misplacing meta-chloro relative to unsubstituted phenol. Electron-withdrawing groups like chlorine increase acidity and lower $p K_{a}$ relative to unsubstituted phenol, and increasing $p K_{a}$ means ordering from strongest acid to weakest acid.

Step-by-step solution

Concept: Acidic strength of substituted phenols is enhanced by electron-withdrawing groups ( $- M, - I$ ) which stabilize the phenoxide anion, and decreased by electron-donating groups ( $+ I, + H$ ) which destabilize it. Since $p K_{a} = - \log_{10} K_{a}$ , a stronger acid has a lower $p K_{a}$ . Therefore, the increasing order of $p K_{a}$ corresponds directly to the decreasing order of acidic strength (from strongest acid to weakest acid).
Option verdict: (A) 2,4-Dinitrophenol has two strong $- M$ and $- I$ nitro groups $⟹$ strongest acid $⟹$ lowest $p K_{a}$ . (B) 4-Nitrophenol has one $- I$ group $⟹$ second strongest acid. (E) 3-Chlorophenol has meta-Cl exerting only $- I$ effect $⟹$ moderately acidic. (D) Phenol is unsubstituted. (C) 2,4,5-Trimethylphenol has three $+ I, + H$ methyl groups destabilizing the phenoxide $⟹$ weakest acid $⟹$ highest $p K_{a}$ . Thus, increasing order of $p K_{a}$ is (A) < (B) < (E) < (D) < (C).
Option verdict: This ranks the compounds in increasing order of acidic strength (or decreasing order of $p K_{a}$ ), confusing high acidity with high $p K_{a}$ .
Option verdict: Places (E) 3-chlorophenol as more acidic than (B) 4-nitrophenol, mistakenly overestimating the $- I$ effect of chlorine relative to the resonance electron-withdrawing ( $- M$ ) effect of the para-nitro group.
Option verdict: Inverts the $p K_{a}$ relationship and also misplaces 3-chlorophenol relative to phenol.
Discriminator: The strongest acid (2,4-dinitrophenol, A) must have the lowest $p K_{a}$ , eliminating Options 1 and 3 immediately. Since 4-nitrophenol (B) has strong $- M$ stabilization while 3-chlorophenol (E) only has weak $- I$ , B is more acidic than E, meaning $p K_{a} (B) < p K_{a} (E)$ , picking Option 0.

Ordering substituted phenols by $p K_{a}$

What feels right?

The correct increasing order of $p K_{a}$ is (A) < (B) < (E) < (D) < (C), because stronger acids have lower $p K_{a}$ values.

Why each option works or fails

Solution

01Concept

02Option verdict

03Option verdict

04Option verdict

05Option verdict

✓Discriminator

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