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Organic Compounds Containing Oxygen: Chemistry | JEE Main

The correct order of

\text{pK}_\text{a}

values for the following compounds is:

Four substituted phenol compounds labelled a, b, c, d: a is phenol, b is 4-(dimethylamino)phenol, c is 4-nitrophenol, d is 4-isopropylphenol.

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\text{c} > \text{a} > \text{d} > \text{b}

\text{b} > \text{a} > \text{d} > \text{c}

\text{b} > \text{d} > \text{a} > \text{c}

\text{a} > \text{b} > \text{c} > \text{d}

Step-by-step solutionView

Correct answer

The correct $\text{pK}_\text{a}$ order is $\text{b} > \text{d} > \text{a} > \text{c}$ , which is the exact inverse of the acidity order ( $\text{c} > \text{a} > \text{d} > \text{b}$ ).

Option analysis

Why each option works or fails

A ·

\text{c} > \text{a} > \text{d} > \text{b}

Confusing $\text{pK}_\text{a}$ with acid strength (thinking higher $\text{pK}_\text{a}$ means stronger acid). Remember that $\text{pK}_\text{a} = -\log K_\text{a}$ ; a stronger acid has a larger $K_\text{a}$ but a smaller $\text{pK}_\text{a}$ .

B ·

\text{b} > \text{a} > \text{d} > \text{c}

Assuming that the strongly electron-withdrawing $-\text{NO}_2$ group increases acidity less at the meta position (d) than unsubstituted phenol (a). Even at the meta position, $-\text{NO}_2$ exerts a strong $-I$ electron-withdrawing effect, making $m$ -nitrophenol (d) more acidic than phenol (a), meaning phenol has the higher $\text{pK}_\text{a}$ .

C ·

\text{b} > \text{d} > \text{a} > \text{c}

None. This correctly ranks the least acidic compound with the highest $\text{pK}_\text{a}$ and the most acidic with the lowest $\text{pK}_\text{a}$ . Correctly determined the acidity order as $p$ -nitrophenol (c) > $m$ -nitrophenol (d) > phenol (a) > $p$ -cresol/methoxyphenol (b), leading to the $\text{pK}_\text{a}$ order $\text{b} > \text{a} > \text{d} > \text{c}$ or $\text{b} > \text{d} > \text{a} > \text{c}$ depending on whether (b) is electron-donating.

D ·

\text{a} > \text{b} > \text{c} > \text{d}

Believing unsubstituted phenol (a) has a higher $\text{pK}_\text{a}$ than a phenol with an electron-donating group (b). Electron-donating groups destabilize the phenoxide conjugate base, decreasing acidity and increasing $\text{pK}_\text{a}$ above that of unsubstituted phenol.

Reviewed route

Solution

StepWorking

01concept

Acidic strength of phenols depends on conjugate base (phenoxide) stability: electron-withdrawing groups ( $-M, -I$ ) increase acidity, while electron-donating groups ( $+M, +I, +H$ ) decrease acidity. $\text{pK}_\text{a} = -\log K_\text{a}$ , so $\text{pK}_\text{a} \propto \frac{1}{\text{Acidic strength}}$ . The substituents are: (a) unsubstituted phenol (reference), (b) 4-methoxyphenol ( $-\text{OCH}_3$ at para gives strong $+M > -I$ ), (c) 4-nitrophenol ( $-\text{NO}_2$ at para gives strong $-M, -I$ ), (d) 4-methylphenol ( $-\text{CH}_3$ at para gives $+H, +I$ ).

02option_verdict

This is the order of acidic strength ( $c > a > d > b$ ), not $\text{pK}_\text{a}$ . Higher acidity corresponds to lower $\text{pK}_\text{a}$ .

03option_verdict

Places unsubstituted phenol (a) as less acidic (higher $\text{pK}_\text{a}$ ) than 4-methylphenol (d), ignoring that $+H/+I$ of $-\text{CH}_3$ destabilizes the phenoxide and decreases acidity below phenol.

04option_verdict

Acidity order is: c ( $-M$ of $-\text{NO}_2$ ) > a (phenol reference) > d ( $+H$ of $-\text{CH}_3$ ) > b ( $+M$ of $-\text{OCH}_3$ ). Therefore, $\text{pK}_\text{a}$ order is the exact inverse: $b > d > a > c$ .

05option_verdict

Places phenol as the weakest acid (highest $\text{pK}_\text{a}$ ) without accounting for the $+M$ destabilization of methoxy in b.

✓discriminator

Identify the extreme ends: $-\text{NO}_2$ makes (c) the strongest acid (lowest $\text{pK}_\text{a}$ ), and $-\text{OCH}_3$ makes (b) the weakest acid (highest $\text{pK}_\text{a}$ ). Only Option (2) has $b > \dots > c$ with $d > a$ .

Hints that build this answer step by step

What is the relationship between the acidity of a substituted phenol and its $\text{pK}_\text{a}$ value?
Higher acidity corresponds to a lower $\text{pK}_\text{a}$ value.
How do electron-donating groups (EDGs) and electron-withdrawing groups (EWGs) on the aromatic ring affect the acidity of phenols?
EWGs stabilize the phenoxide anion (increasing acidity), while EDGs destabilize it (decreasing acidity).
Based on the electronic effects of the substituents, what is the order of acidity from most acidic to least acidic, and the corresponding $\text{pK}_\text{a}$ order?
Acidity: $\text{c} > \text{a} > \text{d} > \text{b}$ ; therefore $\text{pK}_\text{a}$ : $\text{b} > \text{d} > \text{a} > \text{c}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is 4-methoxyphenol (b) less acidic than 4-methylphenol (d) even though oxygen is more electronegative than carbon?

At the para position, resonance (+M) dominates over induction (-I) for -OCH3. This strongly donates electron density to the ring. It destabilizes the phenoxide ion more than hyperconjugation (+H) from -CH3.

Answer

The correct $\text{pK}_\text{a}$ order is $\text{b} > \text{d} > \text{a} > \text{c}$ , which is the exact inverse of the acidity order ( $\text{c} > \text{a} > \text{d} > \text{b}$ ).

Why each option works or fails

A: $\text{c} > \text{a} > \text{d} > \text{b}$ - Confusing $\text{pK}_\text{a}$ with acid strength (thinking higher $\text{pK}_\text{a}$ means stronger acid). Remember that $\text{pK}_\text{a} = -\log K_\text{a}$ ; a stronger acid has a larger $K_\text{a}$ but a smaller $\text{pK}_\text{a}$ .
B: $\text{b} > \text{a} > \text{d} > \text{c}$ - Assuming that the strongly electron-withdrawing $-\text{NO}_2$ group increases acidity less at the meta position (d) than unsubstituted phenol (a). Even at the meta position, $-\text{NO}_2$ exerts a strong $-I$ electron-withdrawing effect, making $m$ -nitrophenol (d) more acidic than phenol (a), meaning phenol has the higher $\text{pK}_\text{a}$ .
C · correct: $\text{b} > \text{d} > \text{a} > \text{c}$ - None. This correctly ranks the least acidic compound with the highest $\text{pK}_\text{a}$ and the most acidic with the lowest $\text{pK}_\text{a}$ . Correctly determined the acidity order as $\text{c} > \text{a} > \text{d} > \text{b}$ -nitrophenol (c) > $m$ -nitrophenol (d) > phenol (a) > $p$ -cresol/methoxyphenol (b), leading to the $\text{pK}_\text{a}$ order $\text{b} > \text{a} > \text{d} > \text{c}$ or $\text{b} > \text{d} > \text{a} > \text{c}$ depending on whether (b) is electron-donating.
D: $\text{a} > \text{b} > \text{c} > \text{d}$ - Believing unsubstituted phenol (a) has a higher $\text{pK}_\text{a}$ than a phenol with an electron-donating group (b). Electron-donating groups destabilize the phenoxide conjugate base, decreasing acidity and increasing $\text{pK}_\text{a}$ above that of unsubstituted phenol.

Step-by-step solution

concept: Acidic strength of phenols depends on conjugate base (phenoxide) stability: electron-withdrawing groups ( $-M, -I$ ) increase acidity, while electron-donating groups ( $+M, +I, +H$ ) decrease acidity. $\text{pK}_\text{a} = -\log K_\text{a}$ , so $\text{pK}_\text{a} \propto \frac{1}{\text{Acidic strength}}$ . The substituents are: (a) unsubstituted phenol (reference), (b) 4-methoxyphenol ( $-\text{OCH}_3$ at para gives strong $+M > -I$ ), (c) 4-nitrophenol ( $-\text{NO}_2$ at para gives strong $-M, -I$ ), (d) 4-methylphenol ( $-\text{CH}_3$ at para gives $+H, +I$ ).
option_verdict: This is the order of acidic strength ( $c > a > d > b$ ), not $\text{pK}_\text{a}$ . Higher acidity corresponds to lower $\text{pK}_\text{a}$ .
option_verdict: Places unsubstituted phenol (a) as less acidic (higher $\text{pK}_\text{a}$ ) than 4-methylphenol (d), ignoring that $+H/+I$ of $-\text{CH}_3$ destabilizes the phenoxide and decreases acidity below phenol.
option_verdict: Acidity order is: c ( $-M$ of $-\text{NO}_2$ ) > a (phenol reference) > d ( $+H$ of $-\text{CH}_3$ ) > b ( $+M$ of $-\text{OCH}_3$ ). Therefore, $-\text{OCH}_3$ order is the exact inverse: $b > d > a > c$ .
option_verdict: Places phenol as the weakest acid (highest $\text{pK}_\text{a}$ ) without accounting for the $+M$ destabilization of methoxy in b.
discriminator: Identify the extreme ends: $-\text{NO}_2$ makes (c) the strongest acid (lowest $\text{pK}_\text{a}$ ), and $-\text{OCH}_3$ makes (b) the weakest acid (highest $\text{pK}_\text{a}$ ). Only Option (2) has $b > \dots > c$ with $d > a$ .

Organic Compounds Containing Oxygen: Chemistry | JEE Main

What feels right?

The correct $\text{pK}_\text{a}$ order is $\text{b} > \text{d} > \text{a} > \text{c}$ , which is the exact inverse of the acidity order ( $\text{c} > \text{a} > \text{d} > \text{b}$ ).

Why each option works or fails

Solution

01concept

02option_verdict

03option_verdict

04option_verdict

05option_verdict

✓discriminator

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