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?

Andhra Pradesh BieapShort· 2mImportance★★★★★

71% · 96/135 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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)

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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 |

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

Start your 14-day free trial to unlock the full solution →

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

AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$pK_a$
›Reveal solutionSolution
Matching known $pK_a$ values: p-Nitrophenol (7.1, most acidic) < Phenol (10.0) < p-Cresol (10.2) < Ethanol (15.9, least acidic) — giving A-II, B-III, C-I, D-IV.

Concept and Intuition

Acidity here is governed by how well the conjugate base (the corresponding phenoxide/alkoxide anion) is stabilised:

p-Nitrophenol: the para $-$ NO $_2$ group is strongly electron-withdrawing by resonance, delocalising and stabilising the negative charge of the phenoxide ion extensively $\Rightarrow$ most acidic of the four (lowest $pK_a$ ).

Phenol: the phenoxide anion is stabilised by resonance delocalisation into the ring, but there is no additional electron-withdrawing substituent $\Rightarrow$ moderately acidic, $pK_a \approx 10.0$ .

p-Cresol: the para $-$ CH $_3$ group is electron-donating (+I/hyperconjugation), which destabilises the phenoxide anion slightly relative to phenol $\Rightarrow$ marginally less acidic than phenol, $pK_a$ slightly higher ( $\approx 10.2$ ).

Ethanol: the ethoxide anion has no aromatic ring to delocalise charge into at all, so it is far less stabilised $\Rightarrow$ ethanol is a much weaker acid, with a much higher $pK_a$ ( $\approx 15.9$ ).

Step-by-Step Solution

Rank acidity qualitatively: p-Nitrophenol (most acidic, EWG) > Phenol > p-Cresol (EDG lowers acidity slightly vs phenol) > Ethanol (no ring stabilisation, weakest acid, highest $pK_a$ ). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$Q. Assertion (A): Carboxylic acids are more acidic than Phenols Reason (R): Resonance structures of carboxylate ion are equivalent, while resonance structures of phenoxide ion are not equivalent (A) Both (A) and (R) are correct and (R) is the correct explanation of (A) (B) Both (A) and (R) are correct But (R) is not the correct explanation of (A) (C) (A) is correct but (R) is incorrect (D) (A) is incorrect but (R) is correct$
›Reveal solutionSolution
Both the assertion (carboxylic acids more acidic than phenols) and the reason (equivalent vs non-equivalent resonance structures of the conjugate bases) are true, and the reason correctly explains the assertion — option (A).

Concept and Intuition

Acid strength of an -OH-bearing compound correlates with how well its conjugate base (the anion formed after losing $\mathrm{H^+}$ ) is stabilized. Stabilization by resonance is most effective when the contributing resonance structures are equivalent (same energy) — equivalent structures each contribute equally and substantially to the true (hybrid) structure, spreading the negative charge symmetrically and lowering the energy a great deal.

In a carboxylate ion ( $\mathrm{RCOO^-}$ ), the negative charge is delocalized over the two oxygen atoms via two resonance structures that are mirror images of each other — completely equivalent, both localizing charge on the (highly electronegative) oxygen. This gives maximal stabilization.

In a phenoxide ion, resonance delocalizes the negative charge onto oxygen (one structure) and onto ring carbons (at ortho and para positions, three additional structures). These carbon-centred structures are not equivalent to the oxygen-centred one — they put negative charge on a less electronegative atom (carbon) and are higher in energy, contributing less to overall stabilization.

Step-by-Step Solution

Confirm Assertion: carboxylic acids (pKₐ ≈ 3–5) are indeed markedly more acidic than phenols (pKₐ ≈ 10) — true. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
$Q. Arrange the following in the correct order of their acidic strength (I = phenol; II = p-cresol, i.e. 4-methylphenol; III = m-nitrophenol, i.e. 3-nitrophenol; IV = p-nitrophenol, i.e. 4-nitrophenol) (A) III > IV > I > II (B) IV > III > I > II (C) II > I > III > IV (D) I > IV > III > II$
›Reveal solutionSolution
Nitro groups (especially at para, via resonance) raise phenol's acidity; a methyl group

lowers it — giving the order p-nitrophenol > m-nitrophenol > phenol > p-cresol.

Concept and Intuition

The acidity of a substituted phenol is governed by how well the ring substituent stabilises

the phenoxide (conjugate base) anion. Electron-withdrawing groups (like $\mathrm{-NO_2}$ )

stabilise the negative charge and increase acidity; this effect is strongest when the group is

at the ortho/para position, where it can directly delocalise the negative charge through

resonance. At the meta position, $\mathrm{-NO_2}$ can only act inductively (no direct

resonance path to the phenoxide oxygen), so its acid-strengthening effect is smaller than at

para. Electron-donating groups (like $\mathrm{-CH_3}$ ) destabilise the phenoxide anion (push

electron density in, the opposite of what's needed) and so decrease acidity relative to plain

phenol.

Step-by-Step Solution

p-Nitrophenol (IV): $\mathrm{-NO_2}$ at para position — full resonance stabilisation of the phenoxide ion — strongest acid of the four.

m-Nitrophenol (III): $\mathrm{-NO_2}$ at meta — only inductive stabilisation, weaker than para but still more acidic than unsubstituted phenol. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$Q. Two statements are given below Statement - I : C - O bond length in methanol is 136 pm and in phenol 142 pm. Statement - II : The C -O- H bond angle in methanol and phenol is almost same correct answer is (A) Both are correct statements (B) Both are incorrect statements (C) Statement - I is incorrect but Statement - II is correct (D) Statement - I is correct but Statement- II is incorrect$
›Reveal solutionSolution
The key idea is that resonance in phenol gives the C–O bond partial double-bond character, shortening it compared to a pure single bond, while the C–O–H bond angle is affected by the same resonance and hybridization changes. The correct answer is that Statement I is incorrect (phenol’s C–O bond is actually shorter, not longer) and Statement II is incorrect (the bond angles differ), so both statements are false.

Understand the claim in Statement I

Statement I says the C–O bond length in methanol is 136 pm and in phenol it is 142 pm — meaning phenol has a longer C–O bond. Intuitively, a longer bond is weaker. But we know from organic chemistry that phenol’s C–O bond has partial double-bond character due to resonance between the oxygen lone pairs and the aromatic ring. A double bond is shorter than a single bond. Therefore, phenol’s C–O bond should be shorter than methanol’s, not longer.

In methanol (CH₃OH), the C–O bond is a pure single bond (sp³ carbon, sp³ oxygen). Typical C–O single bond length is ~143 pm.

In phenol (C₆H₅OH), the oxygen’s lone pairs delocalize into the ring, giving the C–O bond about 30–40% double-bond character. This shortens it to roughly 136–137 pm. So Statement I is incorrect — it has the lengths reversed.

Examine Statement II about bond angles

Statement II claims the C–O–H bond angle in methanol and phenol is “almost the same.” Let’s check:

In methanol, the oxygen is sp³ hybridized (two lone pairs, two sigma bonds), so the ideal bond angle is near the tetrahedral angle (~109.5°). The actual C–O–H angle in methanol is about 108.9°.

In phenol, the oxygen is also sp³ hybridized in the local sense, but one of its lone pairs is partially delocalized into the π system of the ring. This delocalization reduces the electron density on oxygen, which slightly changes the balance of repulsions. More importantly, the oxygen’s hybridization shifts toward sp² character (since the lone pair involved in resonance is in a p orbital). With sp² hybridization, the C–O–H angle would be closer to 120°. Experimentally, the C–O–H angle in phenol is about 109°? Actually, careful measurements show it is around 109° as well — but wait, that seems contradictory. Let’s check data:

Methanol: C–O–H ≈ 108.9°

Phenol: C–O–H ≈ 109.0° (some sources say 109.5°) So they are indeed very close. However, the reason they are close is not because nothing changes — it’s because the oxygen’s hybridization doesn’t fully become sp²; the resonance is partial. But the statement itself is about the fact that they are almost the same, which is true. But — there is a subtlety: The C–O–H angle in phenol is actually slightly larger than in methanol? Some data: methanol 108.9°, phenol 109.2°. That is “almost the same.” So Statement II is correct in its claim.

Re-evaluate with reliable data

Let’s confirm bond lengths: …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\mathrm{C_6H_5OH}$
›Reveal solutionSolution
Electron-donating para-substituents reduce phenol's acidity by destabilizing the phenoxide; $\mathrm{OCH_3}$ donates more strongly by resonance than $\mathrm{CH_3}$ , so methoxyphenol is least acidic: I > II > III.

Concept and Intuition

Phenol is acidic because the phenoxide ion ( $\mathrm{C_6H_5O^-}$ ) is stabilized by resonance delocalization of the negative charge into the ring. Any substituent that makes the ring more electron-rich (an electron-donating group, EDG) works against this delocalization/stabilization of the negative charge, and so decreases acidity. Conversely, an electron-withdrawing group (EWG) stabilizes the anion further and increases acidity.

Step-by-Step Solution

Phenol (I): the reference compound, moderately acidic ( $pK_a \approx 10$ ).

p-Cresol (II): the para $\mathrm{-CH_3}$ group is a weak electron donor (hyperconjugation/+I), pushing electron density into the ring and slightly destabilizing the phenoxide -- makes it less acidic than phenol.

4-Methoxyphenol (III): the para $\mathrm{-OCH_3}$ group is a strong $+M$ (resonance) electron donor -- its lone pair conjugates directly with the ring, pushing even more electron density onto the ring (and especially onto the position para to it, i.e. right where the phenolic oxygen sits), destabilizing the phenoxide anion more than $\mathrm{-CH_3}$ does. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
$\mathrm{C_6H_5OH}$
›Reveal solutionSolution
More electron-withdrawing groups at positions conjugated with the phenoxide oxygen increase phenol acidity; picric acid (three ortho/para $-NO_2$ groups) is the most acidic option. Answer: (C).

Concept and Intuition

A phenol's acidity depends on how stable its conjugate base (the phenoxide ion) is. Electron-withdrawing groups (like $-NO_2$ ) at ortho/para positions delocalize the negative charge of the phenoxide oxygen into the substituent via resonance, spreading out (stabilizing) the charge and making the O–H bond easier to ionize. Electron-donating groups (like $-CH_3$ ) do the opposite — they push electron density onto the ring/oxygen, destabilizing the phenoxide and making the phenol less acidic than plain phenol.

Step-by-Step Solution

Rank the substituent effects: $-NO_2$ (strong EWG, both resonance and inductive) $\gg$ $-H$ (no effect, plain phenol) $>$ $-CH_3$ (weak EDG, destabilizes phenoxide).

Compare the number/position of $-NO_2$ groups: 4-nitrophenol has one $-NO_2$ at the para position (good resonance overlap with the phenoxide oxygen) — more acidic than phenol. 2,4,6-trinitrophenol has THREE $-NO_2$ groups, at both ortho positions AND the para position, all able to resonance-stabilize the phenoxide charge simultaneously. …
Start your 14-day free trial to see the full solution →

←Previous questionQ43 — In Kolbe's reaction, instead of phenol, phenoxide ion is treated with …→Next questionQ45 — Ethers can be prepared by Williamson synthesis in which an alkyl halid…

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)

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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

The core idea: what dipole moment measures

Step-by-step reasoning

🎓Unlock everything free for 14 days

Unlock everything free for 14 days