Q.Assertion: Acylation of amines gives a monosubstituted product whereas alkylation of amines gives polysubstituted product.
Reason: Acyl group sterically hinders the approach of further acyl groups.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Inductive Effect on Acidity
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.
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:
R-COOHwhere R = substituent
The acid dissociation constant Ka changes as:
- If R is electron-withdrawing (e.g., −Cl, −NO2, −CF3): Ka increases → stronger acid.
- If R is electron-donating (e.g., −CH3, −C2H5): Ka 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 | pKa | Explanation |
|---|---|---|
| CH3COOH | 4.76 | Reference (no EWG) |
| ClCH2COOH | 2.86 | Cl withdraws through one bond |
| Cl2CHCOOH | 1.29 | Two Cl atoms, stronger withdrawal |
| Cl3CCOOH | 0.65 | Three Cl atoms, strongest withdrawal |
| CH3CH2COOH | 4.87 | Ethyl group is electron-donating (slightly weaker acid) |
Notice: ClCH2COOH is about 100 times stronger than acetic acid (ΔpKa≈1.9). But if the Cl is moved further away:
| Compound | pKa |
|---|---|
| ClCH2CH2COOH | 4.08 |
| ClCH2CH2CH2COOH | 4.52 |
The effect fades as the chlorine moves farther from the carboxyl group. …
Why this formula?
Inductive Effect on Acidity: Why It Works
The inductive effect is a through-bond electron displacement caused by differences in electronegativity. When we ask why it affects acidity, we must first understand what acidity means at the molecular level.
The Core Idea: Stabilising the Conjugate Base
Acidity is governed by the equilibrium:
HA⇌H++A−
The stronger the acid, the more it favours the right side. This happens when the conjugate base A− is more stable. The inductive effect directly influences this stability.
Why Electron-Withdrawing Groups (EWG) Increase Acidity
Consider a carboxylic acid with an electronegative atom (like Cl) attached to the carbon chain:
Cl−CH2−COOH
- The inductive pull: The Cl atom is more electronegative than carbon. It pulls electron density toward itself through the sigma bonds.
- Effect on the O–H bond: This electron withdrawal travels along the carbon chain, reducing electron density around the O–H bond. The bond becomes more polarised, making the H⁺ easier to remove.
- Stabilising the conjugate base: After losing H⁺, the negative charge on the carboxylate ion (RCOO−) is delocalised by resonance. But the inductive effect further stabilises this negative charge by pulling electron density away from the oxygen atoms. This makes the conjugate base less reactive (more stable), shifting equilibrium toward dissociation.
Key insight: The inductive effect doesn't just weaken the O–H bond — it stabilises the anion that forms after deprotonation.
The Quantitative Relationship: Hammett Equation
For substituted benzoic acids, the effect is quantified by the Hammett equation:
log(Ka0Ka)=σρ
Where:
- Ka = acid dissociation constant of substituted acid
- Ka0 = acid dissociation constant of unsubstituted benzoic acid
- σ = substituent constant (measures inductive + resonance effect)
- ρ = reaction constant (sensitivity of the reaction to substituent effects)
Why This Formula Holds
The derivation comes from linear free-energy relationships:
- Free energy change: For any acid dissociation:
ΔG∘=−RTlnKa
- Effect of substituent: A substituent changes ΔG∘ by an amount proportional to its electronic effect:
Δ(ΔG∘)=−RTln(Ka0Ka)
-
Separability assumption: The total effect of a substituent on any reaction can be factored into:
- A substituent-specific term (σ) — how strongly it pulls/pushes electrons
- A reaction-specific term (ρ) — how sensitive the reaction is to electronic effects
-
Empirical validation: Hammett found that for meta and para substituted benzoic acids, plotting log(Ka/Ka0) against σ gives a straight line. This confirms the additive nature of inductive effects.
The Inductive Effect Constant (σI) …
Concept: Steric hindrance and resonance effects in acylation vs. alkylation of amines.
Reasoning:
- Alkylation of amines produces a more nucleophilic product (e.g., RNH2→R2NH), so further alkylation occurs readily, giving polysubstitution.
- Acylation introduces an acyl group (−COR). The lone pair on nitrogen is delocalised into the carbonyl π-system, making the amide nitrogen much less nucleophilic than the starting amine. …
The key idea is that acylation of amines stops cleanly at the monoacylated stage because the acyl group is strongly electron-withdrawing, which drastically reduces the nucleophilicity of the nitrogen — not because of steric hindrance. The reason given (steric hindrance) is therefore incorrect. The correct option is (C).
-
Understanding the assertion.
When an amine reacts with an alkyl halide (alkylation), the product is itself a more nucleophilic amine, so it reacts again — leading to a mixture of mono-, di-, and tri-alkylated products. This is a classic problem in organic synthesis.
In contrast, when an amine reacts with an acyl chloride or anhydride (acylation), the reaction stops cleanly at the monoacylated stage. The product is an amide, and further acylation does not occur under normal conditions. So the assertion is correct.
-
Why does acylation stop at one step?
The acyl group (−COR) is strongly electron-withdrawing by resonance. Once it attaches to the nitrogen, the lone pair on nitrogen is delocalised into the carbonyl π-system:
R−NHX2+RX′COClR−NH−CO−RX′+HCl
The resulting amide nitrogen’s lone pair is partially tied up in resonance with the carbonyl, making it much less nucleophilic than the original amine. In fact, amides are essentially non-basic and do not undergo further acylation under mild conditions.
- Now examine the reason given. …
Concept: Steric Hindrance in Acylation vs. Alkylation of Amines
Method: Steric and Electronic Effect Analysis
Step 1 — Understand the Assertion
- Acylation of amines (e.g., with acid chlorides or anhydrides) gives monosubstituted amides.
- Alkylation of amines (e.g., with alkyl halides) gives polysubstituted products (mixtures of mono-, di-, and trialkylated amines).
Step 2 — Understand the Reason
The reason claims that the acyl group is sterically hindering, preventing further acylation.
Step 3 — Evaluate the Reason
- The acyl group (−COR) does introduce some steric bulk, but the primary reason for monosubstitution in acylation is electronic:
- The amide formed (RCONHRX′) has a resonance-stabilized N atom — the lone pair is delocalized into the carbonyl, making it much less nucleophilic than the original amine. …
Common Mistakes & How to Avoid Them
Mistake 1: Thinking the Reason is Correct
Many students assume that steric hindrance is the main reason why acylation stops at the monosubstituted stage.
Why this is wrong:
The actual reason is electronic deactivation — after the first acylation, the amide formed (RCONHR) has a carbonyl group that withdraws electrons via resonance, making the nitrogen less nucleophilic. This prevents further acylation.
How to avoid:
Always check the electronic effect of the substituent. Acyl groups are electron-withdrawing (via resonance), not just bulky. Steric hindrance plays a minor role here.
Mistake 2: Confusing Alkylation with Acylation
Students often think both reactions behave similarly because both involve adding a group to the nitrogen.
Key difference to remember:
- Alkylation: Alkyl groups are electron-donating → makes nitrogen more nucleophilic → polysubstitution occurs.
- Acylation: Acyl groups are electron-withdrawing → makes nitrogen less nucleophilic → stops at monosubstitution.
How to avoid:
Memorise this contrast:
Alkyl = donor → polysubstitution
Acyl = acceptor → monosubstitution
Mistake 3: Assuming the Reason Explains the Assertion
Even if you think the reason is correct, you must check if it logically explains the assertion.
Why it fails here:
The assertion is correct (acylation gives monosubstituted product), but the reason given (steric hindrance) is not the correct explanation. The correct explanation is electronic deactivation.
How to avoid:
For assertion-reason questions, always ask:
"Does this reason directly cause the observed effect?"
If the mechanism is different, mark it as "reason not correct explanation."
Mistake 4: Overlooking the "Polysubstitution" in Alkylation …
- KEAM 2026Set eng-2026-04174 marksMCQQ.Which of the following carboxylic acid has the highest pKa value? (A) O2N-CH2-COOH (B) F-CH2-COOH (C) HCOOH (D) CN-CH2COOH (E) Cl-CH2COOH
›Reveal solutionSolution
Among the choices, HCOOH lacks any strong electron-withdrawing α-substituent, so its conjugate base is least stabilised — it is the weakest acid and thus has the highest pKa.
Acidity of a carboxylic acid rises (i.e. pKa falls) when an electron-withdrawing group near the −COOH stabilises the carboxylate anion.
- O2N−CH2COOH: strong −I nitro, very low pKa. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.The decreasing order of acid strength of the following is (A) FCH2COOH>NCCH2COOH>NO2CH2COOH>ClCH2COOH (B) CNCH2COOH>O2NCH2COOH>FCH2COOH>ClCH2COOH (C) NO2CH2COOH>NCCH2COOH>FCH2COOH>ClCH2COOH (D) NO2CH2COOH>FCH2COOH>NCCH2COOH>ClCH2COOH (E) ClCH2COOH>FCH2COOH>NCCH2COOH>NO2CH2COOH
›Reveal solutionSolution
The stronger the electron-withdrawing substituent, the more acidic the acetic acid: NO2>CN>F>Cl, giving O2NCH2COOH>NCCH2COOH>FCH2COOH>ClCH2COOH. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.Which of the following acid is highly acidic? (A) Fluoroacetic acid (B) Formic acid (C) Dichloroacetic acid (D) Benzoic acid (E) Acetic acid.
›Reveal solutionSolution
Acid strength increases with electron-withdrawing substituents that stabilise the conjugate base. Two chlorines in dichloroacetic acid (pKa≈1.3) make it the most acidic here.
Comparing approximate pKa values (lower = more acidic):
- Dichloroacetic acid: ~1.3 (two −I Cl atoms) — strongest.
- Fluoroacetic acid: ~2.6.
- Formic acid: ~3.75.
- Benzoic acid: ~4.2.
- Acetic acid: ~4.76. …
- KEAM 2025Set eng-2025-04294 marksMCQQ.Which of the following is the strongest acid? (A) FCH2COOH (B) CF3COOH (C) NC-CH2COOH (D) Br-CH2COOH (E) CH3COOH
›Reveal solutionSolution
Acid strength rises with the electron-withdrawing power near the -COOH. Three fluorines in CF3COOH give the strongest -I effect and the most stabilised conjugate base, so it is the strongest acid.
Substituted acetic acids become stronger as the electron-withdrawing (–I) effect of the substituent increases, because a more stabilised carboxylate anion means a more readily released proton.
- CH3COOH (E): electron-donating methyl, weakest acid. …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.Which of the following is the weakest acid? (A) FCH2COOH (B) NC−CH2COOH (C) Cl3C−COOH (D) O2N−CH2COOH (E) Cl2CHCOOH
›Reveal solutionSolution
Acidity rises with electron-withdrawing power near the –COOH. Comparing the groups, mono-fluoro on one α-C (FCH2COOH, pKa≈2.6) is the least acid-strengthening, so it is the weakest acid.
Comparison (approximate pKa):
- Cl3C−COOH (trichloroacetic): ≈0.7 — three Cl directly on the carbonyl carbon, strongest.
- Cl2CH−COOH (dichloroacetic): ≈1.3.
- O2N−CH2COOH (nitroacetic): ≈1.7.
- NC−CH2COOH (cyanoacetic): ≈2.5. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.Which of the following carboxylic acid has the highest pKa? (A) ethanoic acid (B) chloroethanoic acid (C) fluoroethanoic acid (D) dichloroethanoic acid (E) triflouroethanoic acid
›Reveal solutionSolution
Electron-withdrawing halogens stabilise the carboxylate and lower pKa; unsubstituted ethanoic acid, having none, has the highest pKa.
Acid strength increases (pKa decreases) with the number and electronegativity of α-halogen substituents, which stabilise the conjugate base by the −I effect. The order of increasing acidity is roughly: …
- KEAM 2024Set eng-2024-06064 marksMCQQ.The order of decreasing acid strength of carboxylic acids is (A) FCH2COOH>ClCH2COOH>NO2CH2COOH>CNCH2COOH (B) CNCH2COOH>FCH2COOH>NO2CH2COOH>ClCH2COOH (C) NO2CH2COOH>FCH2COOH>ClCH2COOH>CNCH2COOH (D) FCH2COOH>NO2CH2COOH>ClCH2COOH>CNCH2COOH (E) NO2CH2COOH>CNCH2COOH>FCH2COOH>ClCH2COOH
›Reveal solutionSolution
Acidity follows the electron-withdrawing (−I) strength: NO2>CN>F>Cl, matching the pKa order.
Electron-withdrawing substituents on the α-carbon stabilise the carboxylate and raise acidity. The experimental pKa values are: nitroacetic ≈1.68, cyanoacetic ≈2.47, fluoroacetic ≈2.59, chloroacetic ≈2.87. Lower …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The increasing order of acid strength of the following carboxylic acids is(i) (CH3)3C-COOH(ii) (CH3)2CH-COOH(iii) CH3CH2COOH (A)(ii) <(i) <(iii) (B)(i) <(iii) <(ii) (C)(ii) <(iii) <(i) (D)(iii) <(ii) <(i) (E)(i) <(ii) < (iii)
›Reveal solutionSolution
Electron-donating alkyl groups destabilise the carboxylate; more branching = weaker acid.
The acids are (i) (CH3)3C-COOH (tert-butyl, three methyls), (ii) (CH3)2CH-COOH (isopropyl, two methyls), (iii) CH3CH2-COOH (ethyl, one methyl on the α-carbon side). Electron-donating (+I) alkyl groups intensify the negative charge on the carboxylate anion, destabilising it and …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.The increasing order of acid strength of the following carboxylic acids is (A) ClCH2-CH2-COOH<ClCH2COOH<NC-CH2COOH<CHCl2COOH (B) ClCH2-COOH<NC-CH2COOH<ClCH2CH2COOH<CHCl2COOH (C) ClCH2-CH2-COOH<CHCl2-COOH<ClCH2-COOH<NC-CH2-COOH (D) NC-CH2-COOH<Cl-CH2COOH<CH-Cl2COOH<Cl-CH2CH2COOH (E) ClCH2CH2-COOH<CHCl2COOH<ClCH2COOH<NC-CH2COOH
›Reveal solutionSolution
Acidity increases: ClCH2CH2COOH<ClCH2COOH<NCCH2COOH<CHCl2COOH.
Concept and Intuition
Electron-withdrawing groups stabilise the carboxylate anion (–I effect), increasing acidity; the effect is stronger when the group is closer to –COOH and when there are more/stronger withdrawing groups.
Step-by-Step Solution
- ClCH2CH2COOH: Cl is one carbon farther (β to COOH), weak effect ⇒ pKa ≈ 4.0 (weakest acid).
- ClCH2COOH: one α-Cl ⇒ pKa ≈ 2.87.
- NCCH2COOH: –CN is a stronger –I group than Cl ⇒ pKa ≈ 2.47.
- CHCl2COOH: two α-Cl ⇒ pKa ≈ 1.29 (strongest). …
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