Which of the following is most acidic? (i) benzyl alcohol (ii) cyclohexanol (iii) phenol (iv) m-chlorophenol

NCERT Exemplar · Q13

Q. Which of the following is most acidic?

(i) benzyl alcohol

(ii) cyclohexanol

(iii) phenol

(iv) m-chlorophenol

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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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Acidity is governed by the stability of the conjugate base after losing $H^+$ . Phenol is more acidic than alcohols because its conjugate base (phenoxide) is resonance-stabilized. Adding an electron-withdrawing group like $-Cl$ at the meta position further increases acidity. So m-chlorophenol is the most acidic.

Why this question is about conjugate base stability

Acidity is not about how willing a molecule is to give away a proton — it’s about how stable the resulting anion is. The stronger the conjugate base, the weaker the acid. So to compare acidity, we compare the stability of the anions formed after deprotonation.

For alcohols and phenols, the acidic proton is the one on the $-OH$ group. The key difference: in alcohols, the negative charge on the oxygen is localised. In phenols, the negative charge can be delocalised into the aromatic ring via resonance. That makes phenoxide far more stable than an alkoxide ion.

Now let’s look at each option.

Step-by-step reasoning

1. Benzyl alcohol (i)

Structure: $C_6H_5CH_2OH$ . The $-OH$ is on a carbon that is one bond away from the benzene ring. After losing $H^+$ , the negative charge sits on oxygen — there is no resonance with the ring because the oxygen is not directly attached to the ring. The benzyl group is electron-donating by hyperconjugation, which destabilises the anion. So benzyl alcohol is a weak acid, comparable to a typical primary alcohol.

2. Cyclohexanol (ii)

A saturated alcohol. The conjugate base is an alkoxide ion with the negative charge fully localised on oxygen. No resonance, no special stabilisation. This is the least acidic of the four.

3. Phenol (iii)

Here the $-OH$ is directly attached to the benzene ring. After deprotonation, the negative charge on oxygen can be delocalised into the ring via resonance — the phenoxide ion has several contributing structures where the negative charge appears on ortho and para carbons. This dramatically stabilises the anion, making phenol about $10^6$ times more acidic than cyclohexanol.

pK_a \text{ values: cyclohexanol } \approx 16,\; \text{phenol } \approx 10

4. m-Chlorophenol (iv) …

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Method: Acidity-Structure Analysis (Inductive & Resonance Effects)

This method compares acid strength by analyzing how well the conjugate base (after losing H⁺) is stabilized.

Step 1: Recall the acidity trend for alcohols vs. phenols

Alcohols (benzyl alcohol, cyclohexanol) are weakly acidic — their conjugate base (alkoxide ion) has the negative charge localized on oxygen.
Phenols are more acidic because the phenoxide ion can delocalize the negative charge into the aromatic ring via resonance.

Step 2: Compare the given compounds

Compound	Type	Conjugate base stabilization
Benzyl alcohol (A)	Primary alcohol	No resonance delocalization of charge
Cyclohexanol (B)	Secondary alcohol	No resonance delocalization of charge
Phenol (C)	Simple phenol	Resonance delocalization into ring
m-Chlorophenol (D)	Substituted phenol	Resonance + inductive withdrawal by Cl

Step 3: Identify the strongest acid …

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Here are the common mistakes students make when tackling this question on acidic strength trends (often confused with boiling point trends), along with how to avoid each.

Mistake 1: Confusing Acidity with Boiling Point

The Error: Students see "alcohol" and "phenol" and immediately rank by boiling point (H-bonding strength). They pick benzyl alcohol or cyclohexanol as most acidic because they think "more H-bonding = more acidic."
Why It’s Wrong: Acidity depends on the stability of the conjugate base (the anion after losing $H^+$ ), not on the strength of the O–H bond or boiling point. Phenol is far more acidic than alcohols because its conjugate base (phenoxide) is resonance-stabilized.
How to Avoid: Always ask: "How stable is the negative charge after losing $H^+$ ?" Draw the conjugate base. If the negative charge can be delocalized (resonance), acidity increases.

Mistake 2: Ignoring Resonance in Phenol

The Error: Students treat phenol like a simple alcohol, forgetting that the lone pair on oxygen can delocalize into the benzene ring.
Why It’s Wrong: In phenol, the phenoxide ion has resonance structures that spread the negative charge over the ortho and para carbons. This makes it much more stable than an alkoxide ion (from cyclohexanol or benzyl alcohol), which has no such delocalization.
How to Avoid: For any compound with an –OH group attached to an aromatic ring, always draw the resonance structures of the conjugate base. If you see charge moving into the ring, acidity is higher.

Mistake 3: Forgetting the Effect of Electron-Withdrawing Groups (EWG)

The Error: Students pick phenol as the most acidic, forgetting that m-chlorophenol has an electron-withdrawing group (Cl) that further stabilizes the conjugate base.
Why It’s Wrong: Chlorine is an EWG (inductive effect, –I). Even at the meta position, it pulls electron density away from the ring, making the phenoxide ion even more stable than unsubstituted phenoxide. Hence, m-chlorophenol is more acidic than phenol.
How to Avoid: When comparing substituted phenols, check for EWGs (like –Cl, –NO₂, –CN). They increase acidity regardless of position (ortho/meta/para). Electron-donating groups (like –CH₃, –OCH₃) decrease acidity.

Mistake 4: Misjudging the Effect of Benzyl Alcohol

The Error: Students think the benzene ring in benzyl alcohol gives resonance stabilization similar to phenol. …

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Q.Which of the following is most acidic?

Why this question is about conjugate base stability

Step-by-step reasoning

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