Explain why is p-nitrophenol more acidic than phenol.

NCERT Exemplar · Q48

Q. Explain why is p-nitrophenol more acidic than phenol.

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Concept understanding — Acidity Of Phenol

Inductive Effect on Acidity – From Intuition to Precision

Imagine you are holding a rope tied to a heavy box. If you pull the rope, the box moves toward you. Now imagine the rope is made of rubber bands — the pull still reaches the box, but it gets weaker the farther away you are. That is exactly how the inductive effect works inside a molecule.

The Core Intuition

An acid donates a proton ( $H^+$ ). After it does, the remaining part (the conjugate base) carries a negative charge. The stability of that negative charge determines how willing the molecule is to give up the proton. More stable conjugate base → stronger acid.

Now, some atoms or groups are electron-withdrawing — they pull electron density toward themselves through the sigma bonds. If such a group is attached near the acidic proton, it pulls some electron density away from the negative charge on the conjugate base. That spreads out (delocalises) the negative charge, making the conjugate base more stable. The acid becomes stronger.

Conversely, electron-donating groups push electron density toward the negative charge, concentrating it and making the conjugate base less stable. The acid becomes weaker.

Note

The inductive effect operates through sigma bonds only. It does not involve pi bonds or resonance. It is a permanent, through-bond polarisation.

The Precise Statement

Inductive effect on acidity: The acidity of a compound increases with the presence of electron-withdrawing groups (EWGs) near the acidic site, and decreases with electron-donating groups (EDGs). The effect is strongest when the group is closest to the acidic proton, and diminishes rapidly with distance.

Mathematically, for a series of substituted carboxylic acids:

\text{R-COOH} \quad \text{where R = substituent}

The acid dissociation constant $K_a$ changes as:

If R is electron-withdrawing (e.g., $-\text{Cl}$ , $-\text{NO}_2$ , $-\text{CF}_3$ ): $K_a$ increases → stronger acid.
If R is electron-donating (e.g., $-\text{CH}_3$ , $-\text{C}_2\text{H}_5$ ): $K_a$ decreases → weaker acid.

Why Distance Matters

The inductive effect falls off with distance because sigma bonds are localised. Each bond attenuates the effect by roughly a factor of 2–3. For example, compare:

Compound	$pK_a$	Explanation
$\text{CH}_3\text{COOH}$	4.76	Reference (no EWG)
$\text{ClCH}_2\text{COOH}$	2.86	Cl withdraws through one bond
$\text{Cl}_2\text{CHCOOH}$	1.29	Two Cl atoms, stronger withdrawal
$\text{Cl}_3\text{CCOOH}$	0.65	Three Cl atoms, strongest withdrawal
$\text{CH}_3\text{CH}_2\text{COOH}$	4.87	Ethyl group is electron-donating (slightly weaker acid)

Notice: $\text{ClCH}_2\text{COOH}$ is about 100 times stronger than acetic acid ( $\Delta pK_a \approx 1.9$ ). But if the Cl is moved further away:

Compound	$pK_a$
$\text{ClCH}_2\text{CH}_2\text{COOH}$	4.08
$\text{ClCH}_2\text{CH}_2\text{CH}_2\text{COOH}$	4.52

The effect fades as the chlorine moves farther from the carboxyl group. …

Why this formula?

Acidity of Phenol: Why It's More Acidic Than Alcohols

Let's build this from first principles — understanding why phenol is acidic is the key to mastering organic chemistry.

1. The Core Observation

Phenol ( $\ce{C6H5OH}$ ) has a pKa ≈ 10, while ethanol ( $\ce{CH3CH2OH}$ ) has a pKa ≈ 16.

This means phenol is about 1 million times more acidic than a typical alcohol.

The question: Why does the O–H bond in phenol break so much more easily?

2. The Key: Stability of the Conjugate Base

Acidity is determined by the stability of the conjugate base after losing $\ce{H+}$ .

Alcohol conjugate base: $\ce{CH3CH2O-}$ (alkoxide ion) — negative charge is localized on oxygen.
Phenol conjugate base: $\ce{C6H5O-}$ (phenoxide ion) — negative charge is delocalized into the benzene ring.

The Resonance Explanation

The phenoxide ion has multiple resonance structures:

\ce{C6H5O-} \leftrightarrow \text{(several resonance forms where negative charge moves to ortho/para carbons)}

Draw the structures mentally:

One structure has the negative charge on oxygen.
Other structures show the negative charge on carbon atoms at the ortho and para positions of the ring.

This delocalization spreads the negative charge over more atoms, making the ion more stable.

Key principle: The more stable the conjugate base, the stronger the acid.

3. Why Alcohols Can't Do This

In an alkoxide ion ( $\ce{RO-}$ ), the negative charge is stuck on oxygen.

There are no empty p-orbitals or conjugated $\pi$ systems nearby to accept the charge.

Result: The alkoxide is less stable, so the alcohol is less acidic.

4. The Inductive Effect Also Helps (But Resonance Dominates)

The benzene ring is slightly electron-withdrawing (due to its $sp^2$ carbons being more electronegative than $sp^3$ ).

This inductive effect pulls electron density away from the O–H bond, making the proton slightly more positive and easier to remove.

However, resonance stabilization of the conjugate base is the dominant factor — inductive effects alone cannot explain the million-fold difference.

5. The Quantitative Picture (pKa Values)

Compound	pKa	Conjugate base stability
Ethanol	~16	Localized charge on O
Phenol	~10	Delocalized charge via resonance
Acetic acid	~4.76	Even more resonance (two O atoms)

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The nitro group at the para position stabilises the phenoxide ion through both an electron-withdrawing inductive effect and, more importantly, resonance delocalisation of the negative charge, making p-nitrophenol a stronger acid than phenol.

Why This Question Matters

Acidity of phenols is a classic topic in organic chemistry because it beautifully illustrates how substituents influence reactivity through electronic effects. The key is always the stability of the conjugate base — the phenoxide ion. A stronger acid has a more stable conjugate base. Here, we compare phenol (no substituent) with p-nitrophenol (a nitro group at the para position). The nitro group is famously electron-withdrawing, but how it withdraws electrons makes all the difference.

Step-by-Step Reasoning

Recall the acid-base equilibrium for phenols Phenol donates a proton to form the phenoxide ion. The equilibrium constant $K_a$ measures acidity.

\text{ArOH} \rightleftharpoons \text{ArO}^- + H^+

The more stable the phenoxide ion, the more the equilibrium shifts right, and the stronger the acid.

The phenoxide ion in phenol: negative charge is localised on oxygen

In phenol itself, the negative charge on the phenoxide ion is delocalised into the benzene ring by resonance. This gives some stability, but the charge density remains highest on the oxygen atom. The $pK_a$ of phenol is about 10.
Introduce the nitro group: a powerful electron-withdrawing group

The nitro group ( $-NO_2$ ) has a nitrogen with a formal positive charge and two oxygen atoms. It pulls electron density through two mechanisms:
- Inductive effect: The electronegative atoms pull $\sigma$ -electron density through the sigma bonds. This is a through-bond effect that weakens with distance.
- Resonance (mesomeric) effect: The nitro group can accept electron density from the ring through its $\pi$ -system. This is a through-space effect that is strongest when the group is ortho or para to the site of negative charge.
Why the para position is special

When the nitro group is at the para position relative to the -OH group, the negative charge on the phenoxide oxygen can be delocalised directly onto the nitro group via resonance. Draw the resonance structures: the lone pair from oxygen moves into the ring, and the $\pi$ -electrons shift so that the negative charge ends up on the electronegative oxygen atoms of the nitro group. This creates additional resonance contributors that are not possible in phenol.

Tip

A quick way to see this: draw the phenoxide ion with the negative charge on oxygen. Push electrons toward the nitro group. You'll get a structure where the negative charge is on one of the nitro oxygens — a very stable arrangement because oxygen is highly electronegative.
Compare the stability of the conjugate bases
- In phenoxide (from phenol), the negative charge is delocalised over the ring and oxygen — about 5 resonance structures.
- In p-nitrophenoxide (from p-nitrophenol), the negative charge can be delocalised onto the nitro group as well, giving additional resonance structures where the charge sits on oxygen atoms. This extra delocalisation dramatically stabilises the anion.
The $pK_a$ values tell the story:
- Phenol: $pK_a \approx 10.0$
- p-Nitrophenol: $pK_a \approx 7.2$ …

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Method: Resonance and Inductive Effect Analysis

This method explains relative acidity by comparing the stability of the conjugate base (phenoxide ion) formed after losing the proton.

Step 1: Write the dissociation equilibrium

For any phenol, acidity is determined by the equilibrium:

\text{ArOH} \rightleftharpoons \text{ArO}^- + H^+

The more stable the phenoxide ion ( $\text{ArO}^-$ ), the stronger the acid.

Step 2: Draw resonance structures of phenoxide ion (phenol case)

Phenol’s conjugate base (phenoxide) has negative charge delocalised into the benzene ring:

\chemfig{*6(-=-(-[::+90]O^{-})-=)} \quad \longleftrightarrow \quad \chemfig{*6(-(-[::+90]O^{-})?(-[::+90]?)-=)}

The negative charge is spread over the ortho and para positions of the ring. This delocalisation stabilises the phenoxide ion, making phenol more acidic than alcohols.

Step 3: Add the nitro group effect (p-nitrophenol case)

In p-nitrophenol, the nitro group ( $-\text{NO}_2$ ) is at the para position. The nitro group is a strong electron-withdrawing group by both:

Inductive effect (pulls electron density through sigma bonds)
Resonance effect (delocalises negative charge further)

Draw the extra resonance structures possible only when the nitro group is present:

\chemfig{*6(-(-[::+90]O^{-})?(-[::+90]NO_2)?-=)} \quad \longleftrightarrow \quad \chemfig{*6(-(-[::+90]O^{-})?(-[::+90]?(-[::+90]O^{-})?)-=)}

The negative charge can now be delocalised onto the oxygen atoms of the nitro group itself. This provides additional stabilisation not possible in phenol.

Step 4: Compare stability of conjugate bases

| Factor | Phenoxide (phenol) | p-Nitrophenoxide (p-nitrophenol) |

|--------|-------------------|----------------------------------| …

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Common Mistakes: Acidity of p-Nitrophenol vs Phenol

Students often lose marks on this question due to conceptual errors. Here are the most frequent mistakes and how to avoid each.

✗ Mistake 1: Saying "nitro group withdraws electrons by induction"

Why it's wrong:

The nitro group ( $-NO_2$ ) is a strong electron-withdrawing group, but its inductive effect alone is weak at the meta and para positions. The real reason is resonance (mesomeric effect).

How to avoid:

Always check: for para and ortho substituents, resonance dominates. Draw the resonance structures:

In phenol, the phenoxide ion ( $C_6H_5O^-$ ) is stabilised by resonance into the ring.
In p-nitrophenol, the nitro group participates in resonance — it can accept the negative charge from the phenoxide oxygen, delocalising it onto the electronegative oxygen atoms of $-NO_2$ .

Key point: The nitro group stabilises the conjugate base by extending the resonance — not just by induction.

✗ Mistake 2: Forgetting to compare the conjugate bases

Why it's wrong:

Acidity is about the stability of the conjugate base (phenoxide ion), not the acid itself. Students compare the neutral molecules and get confused.

How to avoid:

Always shift focus to the anion after deprotonation:

Acid	Conjugate base	Stability
Phenol	Phenoxide ion	Stabilised by resonance into ring
p-Nitrophenol	p-Nitrophenoxide ion	Extra stabilisation — negative charge delocalised onto $-NO_2$ group

The more stable the conjugate base, the stronger the acid.

✗ Mistake 3: Ignoring the ortho/para directing nature

Why it's wrong:

Some students think the nitro group withdraws electrons equally from all positions. In reality, $-NO_2$ is a meta-directing group for electrophilic substitution, but for resonance withdrawal of electron density, it works best at ortho and para positions.

How to avoid:

Remember: resonance structures require the substituent to be conjugated with the ring. For p-nitrophenol, the $-NO_2$ is directly conjugated with the $-O^-$ via the ring — this is only possible at ortho or para positions.

✗ Mistake 4: Writing incomplete resonance structures

Why it's wrong:

Examiners expect you to draw the resonance structures showing delocalisation onto the nitro group. Simply stating "resonance stabilisation" without diagrams loses marks.

How to avoid:

In your answer, draw:

Phenoxide ion resonance (3 structures with negative charge on ortho/para carbons).
For p-nitrophenoxide: additional structures where the negative charge moves onto the oxygen atoms of $-NO_2$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{C}_4 \mathrm{H}_7 \mathrm{OH}$
›Reveal solutionSolution
The question asks for the incorrect statement among four organic chemistry claims. The key is to recall that dehydration ease follows Tertiary > Secondary > Primary, opposite to option (C), making (C) the false statement.

Concept & Intuition

This problem tests your grasp of alcohol reactivity, isomerism, and acidity trends. The trick is to spot the reversal of a well-known order: dehydration of alcohols with acid favors more substituted carbocations (tertiary > secondary > primary) because the reaction proceeds via a carbocation intermediate. Option (C) states the opposite, so it’s the clear outlier. The other options require checking structural isomer counts, the electron-withdrawing effect of nitro vs. methyl groups, and the role of acid in esterification.

Step-by-Step Reasoning

Evaluate option (A): Cyclic $\mathrm{C}_4 \mathrm{H}_7 \mathrm{OH}$ — a cyclobutanol derivative. The formula suggests a four-carbon ring with one OH group. Structural isomers include:

Cyclobutanol (OH on a 4-membered ring)

1-Methylcyclopropanol (OH on a 3-membered ring with a methyl)

2-Methylcyclopropanol (OH on a different carbon of the cyclopropane)

Cyclopropylmethanol (OH on a side chain) These are four distinct structural isomers, so (A) is correct.

Evaluate option (B): p-Nitrophenol vs. p-cresol. The nitro group ( $-\mathrm{NO}_2$ ) is strongly electron-withdrawing, stabilizing the phenoxide ion by resonance, making p-nitrophenol more acidic. The methyl group ( $-\mathrm{CH}_3$ ) in p-cresol is electron-donating, destabilizing the phenoxide, so p-cresol is less acidic. Thus (B) is correct. …
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KCET 2026Set D31 markMCQ
$Q. Carboxylic acids are more acidic than phenols because (A) Formation of dimers (B) Intermolecular hydrogen bonding (C) More covalent nature (D) More resonance stabilisation of their conjugate base$
›Reveal solutionSolution

Acid strength tracks how well the conjugate base's negative charge is stabilised; the carboxylate ion is stabilised more effectively by resonance than the phenoxide ion is.

Step 1 — Conjugate base of a carboxylic acid

Deprotonating $R-COOH$ gives the carboxylate ion $R-COO^-$ . Its two C–O bonds are equivalent by resonance: the negative charge and the double-bond character are shared equally between both oxygens, so each oxygen carries only half a unit of negative charge. This symmetric, low-energy delocalisation is very effective at stabilising the anion.

Step 2 — Conjugate base of phenol

Deprotonating phenol gives the phenoxide ion, $C_6H_5-O^-$ . Its negative charge can delocalise into the benzene ring through resonance, but this places some negative charge on ring carbons (less electronegative than oxygen) and, more importantly, disrupts the ring's aromatic stability in some resonance contributors — a less favourable and less effective stabilisation than the symmetric carboxylate case.

Step 3 — Why the other options are wrong …
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KCET 2025Set D-41 markMCQ
$Q. Phenol can be distinguished from propanol by using the reagent (A) Bromine water (B) Iron metal (C) Iodine in alcohol (D) Sodium metal$
›Reveal solutionSolution
Phenol's ring is so activated by the –OH that bromine water alone brominates it three times, precipitating white 2,4,6-tribromophenol; propanol does nothing.

Step 1 — What a distinguishing test must do.

It must give an observable change with one compound and no change (or a clearly different change) with the other. Both phenol and propanol contain an –OH group, so any test that keys on the –OH alone will fail. The reagent must exploit what makes them different: phenol's –OH is attached to a benzene ring; propanol's is on an alkyl chain.

Step 2 — Why phenol reacts with bromine water.

In phenol, the oxygen's lone pair is delocalised into the ring ( $+M$ effect):
$\mathrm{C_6H_5\ddot{O}H \longleftrightarrow \text{resonance structures with } -\text{ve charge at } o\text{- and } p\text{-positions}}$
This floods the ortho and para carbons with electron density, making the ring enormously more nucleophilic than benzene. It is so activated that it attacks even the weakly polarised Br $_2$ present in bromine water — no Lewis-acid catalyst (FeBr $_3$ /AlBr $_3$ ) is needed, and the reaction does not stop at one substitution:
$\mathrm{C_6H_5OH + 3Br_2 \xrightarrow{\ \text{aq.}\ } 2,4,6\text{-tribromophenol} \downarrow + 3HBr}$
2,4,6-Tribromophenol is insoluble in water, so it appears immediately as a white precipitate — and the brown colour of the bromine water is discharged. Both changes are unmistakable.

Step 3 — Why propanol does not.

$\mathrm{CH_3CH_2CH_2OH}$ is a saturated alcohol: no aromatic ring, no C=C, nothing for an electrophile to attack. Bromine water is not decolourised and no precipitate forms.

The contrast is therefore complete: white ppt ⇒ phenol; no change ⇒ propanol.

Step 4 — Why the other reagents fail.

(D) Sodium metal — the classic trap. Both phenol and propanol have an acidic O–H and both liberate hydrogen gas: …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. Two statements, One Assertion and the other Reason, are given. Which one of the following is the correct option? Assertion: The acid strength of 4 compounds in the descending order is p- Nitrophenol > p-Methoxyphenol > Phenol > p-chlorophenol. Reason: Electron withdrawing groups increase the acid strength while Electron donating groups decrease the acid strength of Phenol and its derivatives. (A) Assertion is correct but Reason is wrong. (B) Both Assertion and Reason are wrong. (C) Assertion is wrong but Reason is correct. (D) Both Assertion and Reason are correct.$
›Reveal solutionSolution
The key idea is that electron-withdrawing groups (EWGs) increase phenol acidity, while electron-donating groups (EDGs) decrease it. The given order is wrong because p-chlorophenol is more acidic than phenol, and p-methoxyphenol is less acidic than phenol. So the Assertion is false, but the Reason is true. The correct option is (C).

The problem asks us to judge both an Assertion (a specific ordering of acid strengths) and a Reason (a general principle about substituent effects). We need to check each separately.

Concept & Intuition

Phenol’s acidity comes from the stability of the phenoxide ion (the conjugate base). The negative charge on oxygen can be delocalized into the benzene ring. Any substituent that pulls electron density away from the oxygen (electron-withdrawing group, EWG) stabilizes the phenoxide ion, making the phenol more acidic. Any substituent that pushes electron density toward the oxygen (electron-donating group, EDG) destabilizes the phenoxide ion, making the phenol less acidic. This is the core principle behind the Reason.

Now, let’s test the Assertion’s order:

p-Nitrophenol > p-Methoxyphenol > Phenol > p-Chlorophenol.

p-Nitrophenol – The nitro group (−NO₂) is a strong EWG (by both inductive and resonance effects). It strongly stabilizes the phenoxide ion. So p-nitrophenol is indeed the most acidic among these four. ✓

p-Methoxyphenol – The methoxy group (−OCH₃) is a strong EDG (by resonance, it donates electrons into the ring). This destabilizes the phenoxide ion, making p-methoxyphenol less acidic than phenol. So placing it second (more acidic than phenol) is already wrong. ✗

Phenol – This is the reference compound. Its acidity is moderate.

p-Chlorophenol – Chlorine (−Cl) is a weak EWG (inductive withdrawal outweighs its weak resonance donation). It slightly stabilizes the phenoxide ion, so p-chlorophenol is more acidic than phenol. Placing it last (least acidic) is also wrong. ✗

So the correct descending order should be:

p-Nitrophenol > p-Chlorophenol > Phenol > p-Methoxyphenol. …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{O}-\mathrm{H}$
›Reveal solutionSolution
The question asks for the incorrect statement. By analyzing each option, we find that (C) contains a factual error in the mechanism of acid-catalyzed hydration of ethene, making it the wrong statement.

Concept & Intuition:

This problem tests your understanding of organic reaction mechanisms, acidity trends, and the dual role of alcohols as nucleophiles and electrophiles. The key is to recall the exact sequence of steps in well-known reactions and to recognize when a statement misrepresents the order of bond-making and bond-breaking. A common pitfall is confusing the role of the oxonium ion (protonated alcohol/water) with the carbocation that forms after it loses water.

Step-by-step analysis:

Option (A): Cresols are methylphenols. The methyl group is an electron-donating group (EDG) via hyperconjugation and induction. EDGs increase electron density on the benzene ring, which destabilizes the phenoxide ion (negative charge is less well dispersed). This makes cresols less acidic than phenol. The reasoning given is correct. So (A) is a true statement.

Option (B): Ethanol has an O–H bond. When the O–H bond breaks heterolytically, the oxygen retains the electron pair, forming an ethoxide ion — this is a nucleophile (donates electrons). When ethanol gets protonated (oxygen accepts a proton), it becomes $\text{CH}_3\text{CH}_2\text{OH}_2^+$ , which is electron-deficient and can act as an electrophile (accepts electrons). The statement is accurate. So (B) is true.

Option (C): In acid-catalyzed hydration of ethene:

Step 1: Ethene reacts with $\text{H}_3\text{O}^+$ (the acid). The proton adds to the double bond, forming a carbocation (not an oxonium ion). …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{pK}_{\mathrm{a}}$
›Reveal solutionSolution
Electron-withdrawing groups lower a phenol's $\mathrm{p}K_a$ and electron-donating groups raise it. Ordering the four para-substituted phenols by decreasing $\mathrm{p}K_a$ gives $A>C>D>B$ , so the correct option is (A).

Concept

The $\mathrm{p}K_a$ of a phenol measures how readily it loses its $-\mathrm{OH}$ proton: the more stable the resulting phenoxide ion, the stronger the acid and the lower the $\mathrm{p}K_a$ . A substituent on the ring shifts this stability:

Electron-withdrawing groups (EWG) such as $-\mathrm{NO_2}$ and $-\mathrm{Cl}$ pull electron density away, spread the negative charge of the phenoxide and stabilise it $\Rightarrow$ stronger acid $\Rightarrow$ lower $\mathrm{p}K_a$ .

Electron-donating groups (EDG) such as $-\mathrm{NH_2}$ and $-\mathrm{OCH_3}$ push electron density in, concentrate the negative charge and destabilise the phenoxide $\Rightarrow$ weaker acid $\Rightarrow$ higher $\mathrm{p}K_a$ .

Because the question asks for decreasing $\mathrm{p}K_a$ , we start with the least acidic (highest $\mathrm{p}K_a$ ) and end with the most acidic (lowest $\mathrm{p}K_a$ ).

Solution

Assign each labelled compound its para substituent and its approximate $\mathrm{p}K_a$ :

A ( $-\mathrm{NH_2}$ ): strong EDG (nitrogen lone pair donates strongly into the ring), most destabilised phenoxide $\Rightarrow$ highest $\mathrm{p}K_a\approx10.3$ .

C ( $-\mathrm{OCH_3}$ ): EDG by resonance, but slightly weaker net donor than $-\mathrm{NH_2}$ here $\Rightarrow$ $\mathrm{p}K_a\approx10.2$ , just below A. …
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COMEDK 2024Set 2024-A1 markMCQ
$Q. Arrange the following compounds in the decreasing order of their acidic strength. (I) m-cresol (II) Phenol (III) m-aminophenol (IV) m-methoxyphenol (A) I > III > IV > II (B) I > II > III > IV (C) IV > III > II > I (D) III > I > II > IV$
›Reveal solutionSolution
At meta only inductive effects operate: –OCH $_3$ and –NH $_2$ are $-I$ (acid-strengthening), –CH $_3$ is $+I$ (acid-weakening), giving IV > III > II > I.

Phenol acidity depends on how well the substituent stabilizes the phenoxide anion. All substituents here are meta to –OH, so their resonance interaction with the oxygen-bearing carbon is not effective; the dominant effect is induction.

(IV) m-methoxyphenol: –OCH $_3$ has a strong $-I$ (electron-withdrawing) inductive effect at meta (oxygen is highly electronegative) → stabilizes the anion most → most acidic (pK $_a\approx 9.65$ ). …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{pK}_{\mathrm{a}}$
›Reveal solutionSolution
The key idea is to evaluate each statement about phenol acidity, substituent effects, steam distillation, and oxidation; the incorrect statement is C, because p‑nitrophenol is less steam volatile than o‑nitrophenol due to stronger intermolecular hydrogen bonding.

Concept & Intuition

This question tests four distinct concepts in organic chemistry:

Acidity of phenols depends on the stability of the phenoxide ion. Electron-withdrawing groups (EWG) increase acidity (lower pKₐ), while electron-donating groups (EDG) decrease acidity (higher pKₐ).

Ortho-substituent effects can involve both electronic and steric factors, plus intramolecular hydrogen bonding.

Steam distillation separates compounds based on volatility; stronger intermolecular forces (like hydrogen bonding) reduce volatility.

Oxidation of phenols with strong oxidants like dichromate can yield quinones (conjugated diketones).

Let’s examine each statement carefully.

Statement A: “Phenol has lower pKₐ than p‑cresol”

p‑Cresol has a methyl group at the para position. Methyl is an electron-donating group (+I effect).

This destabilizes the phenoxide ion (increases negative charge density), making p‑cresol less acidic than phenol.

Hence phenol (pKₐ ≈ 10) is more acidic than p‑cresol (pKₐ ≈ 10.2). So A is correct.

Statement B: “2‑Chlorophenol is more acidic than phenol”

Chlorine is an electron-withdrawing group (−I effect), which stabilizes the phenoxide ion by delocalizing the negative charge.

Additionally, ortho‑chlorophenol can form an intramolecular hydrogen bond between the –OH and –Cl, which further stabilizes the conjugate base.

Therefore, 2‑chlorophenol (pKₐ ≈ 8.5) is indeed more acidic than phenol (pKₐ ≈ 10). So B is correct.

Statement C: “Ortho and para nitrophenols can be separated by steam distillation since p‑Nitrophenol is more steam volatile than o‑Nitrophenol”

o‑Nitrophenol has intramolecular hydrogen bonding (between –OH and –NO₂), reducing its ability to form intermolecular hydrogen bonds with water. This makes it more steam volatile. …
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COMEDK 2024Set 2024-M1 markMCQ
$Q. From the following compounds, identify the one which is most acidic. (A) B (B) D (C) A (D) C$
›Reveal solutionSolution
p-Nitrophenol [C] is most acidic — the para nitro group stabilises the phenoxide anion by resonance/–M, far more than plain phenol or the alcohols.

Compare the four –OH compounds:

[B] dicyclohexyl-carbinol and [D] cyclohexanol are simple alcohols; their alkoxides get no resonance stabilisation, so they are the weakest acids.

[A] phenol: the phenoxide is resonance-stabilised over the aromatic ring, so phenol is markedly more acidic than alcohols. …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{m}$
›Reveal solutionSolution

(A) m-Cresol (3-methylphenol) - the electron-releasing -CH3 group destabilises the phenoxide ion, so m-cresol IS a weaker acid than phenol. CORRECT statement. (B) The acidity of phenol is attributed to the electron-withdrawing (-I) effect of the sp2/attached oxygen plus resonance stabilisation of the phenoxide ion. As stated in NCERT terms this is taken as a correct statement. (C) Acetylation of salicylic acid (with acetic anhydride) gives 2-acetoxybenzoic acid = aspirin. CORRECT statement. (D) Phenol heated with conc. H2SO4 gives PHENOL-4-SULPHONIC ACID (p-hydroxybenzenesulphonic acid), NOT s

Concept: check each statement about phenol.

(A) m-Cresol (3-methylphenol) - the electron-releasing -CH3 group destabilises the phenoxide ion, so m-cresol IS a weaker acid than phenol. CORRECT statement.

(B) The acidity of phenol is attributed to the electron-withdrawing (-I) effect of the sp2/attached oxygen plus resonance stabilisation of the phenoxide ion. As stated in NCERT terms this is taken as a correct statement. …
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Q.Explain why is p-nitrophenol more acidic than phenol.

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