Assertion: In monohaloarenes, further electrophilic substitution occurs at ortho and para positions. Reason: H

NCERT Exemplar · Q90

Q. Assertion: In monohaloarenes, further electrophilic substitution occurs at ortho and para positions. Reason: Halogen atom is a ring deactivator.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.

(ii) Assertion and reason both are wrong statements.

(iii) Assertion is correct but reason is wrong statement.

(iv) Assertion is wrong but reason is correct statement.

(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.

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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.

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?

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:

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

The stronger the acid, the more it favours the right side. This happens when the conjugate base $\text{A}^-$ 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:

\text{Cl} - \text{CH}_2 - \text{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 ( $\text{RCOO}^-$ ) 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 \left( \frac{K_a}{K_a^0} \right) = \sigma \rho

Where:

$K_a$ = acid dissociation constant of substituted acid
$K_a^0$ = acid dissociation constant of unsubstituted benzoic acid
$\sigma$ = substituent constant (measures inductive + resonance effect)
$\rho$ = 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:

\Delta G^\circ = -RT \ln K_a

Effect of substituent: A substituent changes $\Delta G^\circ$ by an amount proportional to its electronic effect:

\Delta (\Delta G^\circ) = -RT \ln \left( \frac{K_a}{K_a^0} \right)

Separability assumption: The total effect of a substituent on any reaction can be factored into:
- A substituent-specific term ( $\sigma$ ) — how strongly it pulls/pushes electrons
- A reaction-specific term ( $\rho$ ) — how sensitive the reaction is to electronic effects
Empirical validation: Hammett found that for meta and para substituted benzoic acids, plotting $\log(K_a/K_a^0)$ against $\sigma$ gives a straight line. This confirms the additive nature of inductive effects.

The Inductive Effect Constant ( $\sigma_I$ ) …

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Both statements are individually true — further electrophilic substitution on a monohaloarene really does go to ortho/para, and a halogen really is a ring deactivator — but the reason does not explain the assertion: deactivation on its own would predict meta-direction (as it does for other deactivating groups like $-NO_2$ ), not ortho/para. What actually decides the ortho/para direction is the halogen's separate resonance (lone-pair donation) effect, which the reason never mentions. The correct option is (v): both correct, reason is not the correct explanation.

Why the assertion is true

In a monohaloarene undergoing a second electrophilic aromatic substitution (nitration, sulfonation, Friedel-Crafts, etc.), the new group goes overwhelmingly to the ortho and para positions, with very little meta product.

Why the reason is also true, on its own

Compared to benzene, a haloarene reacts more slowly with electrophiles. The halogen is more electronegative than carbon and withdraws electron density inductively, making the ring less electron-rich and less reactive overall — so yes, a halogen genuinely is a ring deactivator.

Why the reason does not explain the assertion

Deactivation on its own says nothing about WHERE substitution happens — it only affects HOW FAST the reaction goes. In fact, most other deactivating groups ( $-NO_2$ , $-CN$ , $-CHO$ ) are meta-directing, not ortho/para-directing. If deactivation alone decided direction, halogens would be meta-directors too — but they are not. …

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Method: Inductive Effect + Resonance Effect Analysis

This method resolves the apparent contradiction between the deactivating nature of halogens and their ortho/para directing behaviour.

Step 1: Identify the Inductive Effect of Halogen

Halogen atoms are highly electronegative.
They withdraw electron density through the sigma (σ) bond via −I effect.
This deactivates the benzene ring toward electrophilic substitution — Reason is correct.

Step 2: Identify the Resonance Effect of Halogen

Halogens have lone pairs of electrons.
These lone pairs participate in +R effect (resonance donation) with the benzene ring.
Resonance structures show negative charge density at ortho and para positions.

Step 3: Compare the Two Effects

Effect	Direction	Impact on Ring
−I (Inductive)	Withdraws electrons	Deactivates (slows reaction)
+R (Resonance)	Donates electrons	Directs to ortho/para

The −I effect dominates → overall ring is deactivated.
But the +R effect controls orientation → substitution occurs at ortho and para positions.

Step 4: Conclusion for Assertion & Reason

Assertion: True — substitution occurs at ortho/para.
Reason: True — halogen is a ring deactivator. …

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Here are the common mistakes students make on this exact question, along with how to avoid each.

Mistake 1: Confusing “Deactivator” with “Meta-Director”

The error:

Students assume that if a group deactivates the ring, it must be a meta-director.

Here, the reason says “halogen is a ring deactivator” — which is true — but they incorrectly conclude that this means the assertion (ortho/para substitution) must be false.

Why it’s wrong:

Halogens are unique: they are deactivating (due to strong -I effect) yet ortho/para directing (due to +R effect). Deactivation does not automatically imply meta direction.

How to avoid:

Memorise the two exceptions to the “deactivator = meta-director” rule:

Halogens (F, Cl, Br, I)
(No other deactivator is ortho/para directing)

Key fact: For halogens, the resonance effect (+R) overrides the inductive effect (-I) in deciding orientation.

Mistake 2: Thinking “Deactivator” Means “No Substitution Possible”

The error:

Some students think a deactivator makes the ring so unreactive that substitution cannot occur at ortho/para positions.

Why it’s wrong:

Deactivation only means the reaction is slower than benzene, not impossible. Ortho/para positions are still more reactive than meta positions in halobenzene.

How to avoid:

Remember:

Deactivator → slower reaction
Ortho/para director → substitution occurs at ortho/para positions These two facts can coexist.

Mistake 3: Misreading the Assertion-Reason Options

The error:

Students pick option (iii) — “Assertion is correct but reason is wrong” — because they think the reason should say “halogen is a ring activator” for ortho/para substitution.

Why it’s wrong:

The reason is correct — halogen is a deactivator. The assertion is also correct. The only catch: the reason does not explain the assertion (deactivation does not cause ortho/para direction — resonance does). So the correct answer is not (iii).

How to avoid:

Check two things separately:

Is the assertion factually correct? → Yes
Is the reason factually correct? → Yes
Does the reason explain the assertion? → No (deactivation ≠ ortho/para direction)

Since both statements are individually correct but the reason does not correctly explain the assertion, the correct choice is:

(v) Both assertion and reason are correct, but the reason is NOT the correct explanation of the assertion. (Careful: option (ii) as actually printed in this question means "both are WRONG statements" -- a different claim entirely. The situation described here -- both true, reason doesn't explain -- matches option (v).)

Mistake 4: Forgetting the Resonance Stabilisation Argument

The error: …

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

KEAM 2026Set eng-2026-04174 marksMCQ
$O_2N$
›Reveal solutionSolution
Among the choices, HCOOH lacks any strong electron-withdrawing $\alpha$ -substituent, so its conjugate base is least stabilised — it is the weakest acid and thus has the highest $pK_a$ .

Acidity of a carboxylic acid rises (i.e. $pK_a$ falls) when an electron-withdrawing group near the $-COOH$ stabilises the carboxylate anion.

$O_2N{-}CH_2COOH$ : strong $-I$ nitro, very low $pK_a$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$FCH_2COOH > NCCH_2COOH > NO_2CH_2COOH > ClCH_2COOH$
›Reveal solutionSolution

The stronger the electron-withdrawing substituent, the more acidic the acetic acid: $NO_2 > CN > F > Cl$ , giving $O_2NCH_2COOH > NCCH_2COOH > FCH_2COOH > ClCH_2COOH$ . …

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KEAM 2025Set eng-2025-04254 marksMCQ
$Q. Which of the following acid is highly acidic? (A) Fluoroacetic acid (B) Formic acid (C) Dichloroacetic acid (D) Benzoic acid (E) Acetic acid.$
›Reveal solutionSolution
Acid strength increases with electron-withdrawing substituents that stabilise the conjugate base. Two chlorines in dichloroacetic acid ( $pK_a \approx 1.3$ ) make it the most acidic here.

Comparing approximate $pK_a$ 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. …
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KEAM 2025Set eng-2025-04294 marksMCQ
$FCH_2COOH$
›Reveal solutionSolution
Acid strength rises with the electron-withdrawing power near the $\text{-}COOH$ . Three fluorines in $CF_3COOH$ give the strongest $\text{-}I$ 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.

$CH_3COOH$ (E): electron-donating methyl, weakest acid. …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$FCH_2COOH$
›Reveal solutionSolution
Acidity rises with electron-withdrawing power near the –COOH. Comparing the groups, mono-fluoro on one $\alpha$ -C ( $FCH_2COOH$ , $pK_a\approx2.6$ ) is the least acid-strengthening, so it is the weakest acid.

Comparison (approximate $pK_a$ ):

$Cl_3C{-}COOH$ (trichloroacetic): $\approx 0.7$ — three Cl directly on the carbonyl carbon, strongest.

$Cl_2CH{-}COOH$ (dichloroacetic): $\approx 1.3$ .

$O_2N{-}CH_2COOH$ (nitroacetic): $\approx 1.7$ .

$NC{-}CH_2COOH$ (cyanoacetic): $\approx 2.5$ . …
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KEAM 2024Set eng-2024-06054 marksMCQ
$Q. Which of the following carboxylic acid has the highest pKa? (A) ethanoic acid (B) chloroethanoic acid (C) fluoroethanoic acid (D) dichloroethanoic acid (E) triflouroethanoic acid$
›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 $\alpha$ -halogen substituents, which stabilise the conjugate base by the $-I$ effect. The order of increasing acidity is roughly: …
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KEAM 2024Set eng-2024-06064 marksMCQ
$FCH_{2}COOH > ClCH_{2}COOH > NO_{2}CH_{2}COOH > CNCH_{2}COOH$
›Reveal solutionSolution

Acidity follows the electron-withdrawing ( $-I$ ) strength: $NO_2>CN>F>Cl$ , matching the pKa order.

Electron-withdrawing substituents on the $\alpha$ -carbon stabilise the carboxylate and raise acidity. The experimental pKa values are: nitroacetic $\approx1.68$ , cyanoacetic $\approx2.47$ , fluoroacetic $\approx2.59$ , chloroacetic $\approx2.87$ . Lower …
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KEAM 2024Set eng-2024-06094 marksMCQ
$Q. The increasing order of acid strength of the following carboxylic acids is$
(i) $(CH_3)_3C\text{-}COOH$
(ii) $(CH_3)_2CH\text{-}COOH$
(iii) $CH_3CH_2COOH$ (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) $(CH_3)_3C\text{-}COOH$ (tert-butyl, three methyls), (ii) $(CH_3)_2CH\text{-}COOH$ (isopropyl, two methyls), (iii) $CH_3CH_2\text{-}COOH$ (ethyl, one methyl on the $\alpha$ -carbon side). Electron-donating (+I) alkyl groups intensify the negative charge on the carboxylate anion, destabilising it and …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$ClCH_2\text{-}CH_2\text{-}COOH < ClCH_2COOH < NC\text{-}CH_2COOH < CHCl_2COOH$
›Reveal solutionSolution
Acidity increases: $ClCH_2CH_2COOH < ClCH_2COOH < NCCH_2COOH < CHCl_2COOH$ .

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

$ClCH_2CH_2COOH$ : Cl is one carbon farther (β to COOH), weak effect ⇒ pKa ≈ 4.0 (weakest acid).

$ClCH_2COOH$ : one α-Cl ⇒ pKa ≈ 2.87.

$NCCH_2COOH$ : –CN is a stronger –I group than Cl ⇒ pKa ≈ 2.47.

$CHCl_2COOH$ : two α-Cl ⇒ pKa ≈ 1.29 (strongest). …
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Q.Assertion: In monohaloarenes, further electrophilic substitution occurs at ortho and para positions. Reason: Halogen atom is a ring deactivator.

Why the assertion is true

Why the reason is also true, on its own

Why the reason does not explain the assertion

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Q.Assertion: In monohaloarenes, further electrophilic substitution occurs at ortho and para positions.
Reason: Halogen atom is a ring deactivator.

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