Dipole moment of phenol is smaller than that of methanol. Why?

NCERT Exemplar · Q44

Q. Dipole moment of phenol is smaller than that of methanol. Why?

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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 dipole moment of phenol is smaller than that of methanol because the lone pair on oxygen in phenol is delocalised into the aromatic ring via resonance, reducing the net charge separation, whereas in methanol the dipole arises from a localised O–H bond with no such delocalisation.

The core idea: what dipole moment measures

A dipole moment ( $\mu$ ) is a measure of charge separation in a molecule. It depends on two things: the magnitude of the partial charges and the distance between them. For an O–H bond, the oxygen is more electronegative, so it pulls electron density away from hydrogen, creating a dipole pointing from H to O.

But here’s the twist: the size of that dipole is not fixed — it changes if the oxygen’s lone pairs get involved in resonance. That’s exactly what happens in phenol but not in methanol.

Step-by-step reasoning

1. Methanol: a simple, localised dipole

In methanol ( $\mathrm{CH_3OH}$ ), the oxygen has two lone pairs and is bonded to a methyl group and a hydrogen. The O–H bond is polar, and the lone pairs are localised on oxygen. There is no other functional group to pull electron density away. So the dipole moment is essentially that of the O–H bond plus a small contribution from the C–O bond. The measured value is about $1.70\ \mathrm{D}$ .

2. Phenol: the oxygen’s lone pairs are shared with the ring

In phenol ( $\mathrm{C_6H_5OH}$ ), the oxygen is attached directly to an aromatic ring. The lone pairs on oxygen can participate in resonance with the $\pi$ -system of the benzene ring. This is a key point: the oxygen donates electron density into the ring.

\mathrm{C_6H_5OH \leftrightarrow \overset{+}{O}H-C_6H_4^-}

This delocalisation has two effects on the dipole:

The partial negative charge on oxygen is reduced because some of its electron density is now spread over the ring.
The positive charge on hydrogen is also somewhat reduced because the O–H bond becomes slightly less polar.

3. The net result: smaller charge separation

Because the negative charge is no longer concentrated on oxygen alone, the charge separation between O and H is smaller in phenol than in methanol. The dipole moment of phenol is about $1.55\ \mathrm{D}$ — noticeably less than methanol’s $1.70\ \mathrm{D}$ .

Watch out

A common mistake is to think that because phenol has a larger $\pi$ -system, it should have a larger dipole. But resonance delocalises charge, which reduces the dipole, not increases it. The dipole is about separation of charge, not total charge.

4. A helpful analogy …

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Common Mistakes: Dipole Moment of Phenol vs. Methanol

Students often struggle with this comparison because they focus on the number of polar bonds rather than the net vector sum. Here are the most frequent errors and how to avoid them.

✗ Mistake 1: Assuming More Polar Bonds = Higher Dipole Moment

The error: Phenol has an O–H bond and a C–O bond, while methanol has only one O–H bond. Students conclude phenol must have a larger dipole moment.

Why it's wrong: Dipole moment is a vector sum, not a count of bonds. In phenol, the dipole of the C–O bond and the dipole of the O–H bond are not aligned — they point in different directions and partially cancel.

How to avoid: Always draw the vector diagram:

In methanol: C–O and O–H dipoles are roughly aligned (both pointing toward oxygen), so they add.
In phenol: The C–O dipole points from C to O, but the O–H dipole points from O to H (away from the ring). The ring’s $\pi$ -electron cloud also contributes a small opposing dipole.

Key takeaway: Vector addition, not bond count, determines the net dipole.

✗ Mistake 2: Ignoring the Role of the Aromatic Ring

The error: Treating phenol as just "methanol with a benzene ring attached" — forgetting that the ring’s $\pi$ -electrons create their own dipole.

Why it's wrong: The benzene ring has a delocalised $\pi$ -electron cloud that is polarisable. The oxygen atom withdraws electron density from the ring via resonance, creating a small dipole from ring to oxygen. This opposes the O–H dipole.

How to avoid: Remember that in phenol, the oxygen’s lone pairs participate in resonance with the ring. This:

Reduces the partial negative charge on oxygen (compared to methanol)
Creates an opposing dipole from the ring toward oxygen

Key takeaway: Resonance in phenol reduces the effective polarity of the O–H bond.

✗ Mistake 3: Confusing Acidity with Dipole Moment

The error: "Phenol is more acidic than methanol, so its O–H bond must be more polar — hence a larger dipole moment."

Why it's wrong: Acidity depends on the stability of the conjugate base (phenoxide ion), not on the bond polarity of the neutral molecule. Phenol’s higher acidity comes from resonance stabilisation of phenoxide, not from a more polar O–H bond.

How to avoid: Separate the concepts:

Dipole moment → property of the neutral molecule (vector sum of bond dipoles)
Acidity → property of the conjugate base (stability of the anion)

Key takeaway: A more acidic O–H does not imply a larger dipole moment.

✓ Correct Explanation (Summary)

| Molecule | Dipole Moment (D) | Reason |

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

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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.Dipole moment of phenol is smaller than that of methanol. Why?

Concept understanding — Acidity Of Phenol

Inductive Effect on Acidity – From Intuition to Precision

The Core Intuition

The Precise Statement

Why Distance Matters

Why this formula?

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

1. The Core Observation

2. The Key: Stability of the Conjugate Base

The Resonance Explanation

3. Why Alcohols Can't Do This

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

5. The Quantitative Picture (pKa Values)

The core idea: what dipole moment measures

Step-by-step reasoning

Concept: Acidity of Phenol — Dipole Moment Comparison

Method: Resonance and Inductive Effect Analysis

Step 1: Recall the definition of dipole moment

Step 2: Analyse the structure of methanol ( $\text{CH}_3\text{OH}$ )

Step 3: Analyse the structure of phenol ( $\text{C}_6\text{H}_5\text{OH}$ )

Step 4: Compare and conclude …

Common Mistakes: Dipole Moment of Phenol vs. Methanol

✗ Mistake 1: Assuming More Polar Bonds = Higher Dipole Moment

✗ Mistake 2: Ignoring the Role of the Aromatic Ring

✗ Mistake 3: Confusing Acidity with Dipole Moment

✓ Correct Explanation (Summary)

Step 1 — Conjugate base of a carboxylic acid

Step 2 — Conjugate base of phenol

Step 3 — Why the other options are wrong …

Concept

Solution

Q.Dipole moment of phenol is smaller than that of methanol. Why?

The core idea: what dipole moment measures

Step-by-step reasoning

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