Why does acetylation of -NH₂ group of aniline reduce its activating effect?

NCERT Exemplar · Q35

-NH_2

Andhra Pradesh BieapShort· 2mImportance★★★★★

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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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Acetylation converts the strongly activating $-NH_2$ group into a moderately activating $-NHCOCH_3$ group by delocalising the nitrogen lone pair into the carbonyl $\pi$ -system, reducing its availability for resonance donation to the benzene ring.

The key to understanding this lies in the resonance effect — specifically, how the lone pair on nitrogen interacts with the rest of the molecule.

Aniline’s $-NH_2$ group is a powerful activating and ortho/para-directing group because the nitrogen lone pair is directly conjugated with the benzene ring. This lone pair can delocalise into the ring, increasing electron density at the ortho and para positions. The more freely this lone pair is available, the stronger the activation.

When you acetylate aniline, you replace one hydrogen on the $-NH_2$ with an acetyl group ( $-COCH_3$ ), forming acetanilide. The product now has an amide linkage: $-NHCOCH_3$ .

The critical structural change: The acetyl group contains a carbonyl ( $C=O$ ) bond. The carbon of this carbonyl is $sp^2$ hybridised and is directly attached to the nitrogen. This creates a new, competing resonance interaction.
The competition for the lone pair: The nitrogen lone pair can now delocalise in two directions:
- Into the benzene ring (as in aniline), activating the ring.
- Into the carbonyl group of the acetyl group, forming a resonance structure where the nitrogen has a partial positive charge and the carbonyl oxygen has a partial negative charge.
This second resonance is very significant. The carbonyl group acts as an electron sink, pulling electron density away from the nitrogen.
The net effect on activation: Because the lone pair is now partially tied up in resonance with the carbonyl, it is less available to donate into the benzene ring. The electron-donating capacity of the $-NHCOCH_3$ group is therefore much weaker than that of the free $-NH_2$ group.
$\text{Ph} - \overset{\ominus}{\text{N}} - \overset{\oplus}{\text{C}} = \overset{\ominus}{\text{O}} \quad \longleftrightarrow \quad \text{Ph} - \text{N} = \text{C} - \overset{\ominus}{\text{O}}$

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Concept: Acidity of Phenol

Method: Resonance Stabilisation Analysis of Conjugate Base

Why this method?

The acidity of any compound is determined by the stability of its conjugate base after losing a proton ( $H^+$ ). For phenol, we compare the phenoxide ion (conjugate base) with the alkoxide ion (from alcohols) to understand why phenol is more acidic.

Steps:

Write the dissociation reaction Phenol loses $H^+$ from its $-OH$ group to form the phenoxide ion:

C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+

Draw resonance structures of the phenoxide ion

The negative charge on oxygen can be delocalised into the benzene ring. Draw the contributing structures:
- One structure with negative charge on oxygen.
- Three structures where the negative charge moves to ortho and para positions of the ring (via π-bond shifts).
Key result: The negative charge is spread over multiple atoms, making the ion more stable.
Compare with alcohol (e.g., ethanol)

In alkoxide ion ( $CH_3CH_2O^-$ ), the negative charge is localised only on oxygen — no resonance delocalisation possible. This makes it less stable.
Conclusion

Greater stability of phenoxide ion → phenol loses $H^+$ more easily → phenol is more acidic than alcohols.

Why does acetylation of $-NH_2$ group of aniline reduce its activating effect?

Method: Resonance & Electron-Withdrawing Effect of Amide Group

Steps:

Recall the activating effect of $-NH_2$

In aniline, the lone pair on nitrogen is delocalised into the benzene ring (resonance), making the ring electron-rich and highly reactive toward electrophilic substitution.
What happens during acetylation?

The $-NH_2$ group reacts with acetyl chloride/acetic anhydride to form an amide:

C_6H_5NH_2 + CH_3COCl \rightarrow C_6H_5NHCOCH_3 + HCl

Analyse the new group ( $-NHCOCH_3$ )
- The lone pair on nitrogen is now shared with the carbonyl group ( $C=O$ ) via resonance: …

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Common Mistakes & How to Avoid Them

1. Confusing “Activating Effect” with “Basicity”

Mistake: Students think acetylation increases activation because the $-\text{NHCOCH}_3$ group still donates electrons via resonance.
Why it’s wrong: Activating effect in electrophilic aromatic substitution (EAS) depends on electron density on the ring. Acetylation converts a strong activator ( $-\text{NH}_2$ ) into a moderate activator ( $-\text{NHCOCH}_3$ ). The resonance donation is weaker because the lone pair on N is partially delocalised into the carbonyl group ( $\text{C=O}$ ), reducing its availability to the ring.
How to avoid: Always compare the resonance structures:
- In aniline: lone pair on N directly donates into the ring → strong activation.
- In acetanilide: lone pair is shared with the $\text{C=O}$ group → less electron density reaches the ring.

2. Forgetting the Role of the Carbonyl Group

Mistake: Treating $-\text{NHCOCH}_3$ as if it were just $-\text{NH}_2$ with an extra carbon.
Why it’s wrong: The carbonyl group is electron-withdrawing by induction ( $-I$ effect) and also participates in resonance with the N lone pair. This reduces the net electron-donating ability of the nitrogen.
How to avoid: Draw the resonance hybrid of acetanilide. Notice that the lone pair on N is delocalised into the $\text{C=O}$ bond, forming a partial double bond between N and C. This decreases the lone pair’s availability for donation to the benzene ring.

3. Misinterpreting “Reduced Activating Effect” as “Deactivating”

Mistake: Saying acetylation makes the group deactivating.
Why it’s wrong: $-\text{NHCOCH}_3$ is still activating (though weaker than $-\text{NH}_2$ ). It is an ortho/para director but less powerful.
How to avoid: Remember the order: $-\text{NH}_2$ (strong activator) > $-\text{OH}$ > $-\text{NHCOCH}_3$ (moderate activator) > $-\text{OCH}_3$ (moderate). Acetylation reduces but does not reverse the activating nature.

4. Ignoring the Inductive Effect of the Acetyl Group

Mistake: Only considering resonance, forgetting the $-I$ effect of the carbonyl.
Why it’s wrong: The $\text{C=O}$ group pulls electron density through sigma bonds, further decreasing the electron density on the nitrogen and hence on the ring.
How to avoid: Always evaluate both resonance and inductive effects. For $-\text{NHCOCH}_3$ $- NHCOCH_{3}$ :
- Resonance: $+M$ (donation to ring) but weaker than $-\text{NH}_2$ .
- Inductive: $-I$ (withdrawal) from the carbonyl → net activation is reduced.

5. Not Relating to Exam-Specific Examples …

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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 $pK_a$ ). …
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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. …
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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. …
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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: …
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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. …
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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. …
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Q.Why does acetylation of −NH2-NH_2−NH2​ group of aniline reduce its activating effect?

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Q.Why does acetylation of $-NH_2$ group of aniline reduce its activating effect?

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