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.

Showing the 12 most recent of 14 on this concept.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Which acid has lowest pKa?$
(a) C6H5COOH
(b) HCOOH
(c) C6H5CH2COOH
(d) CH3CH2COOH
›Reveal solutionSolution
Among simple carboxylic acids, acidity decreases as alkyl/aryl substituents (electron donors) are added onto the carbon bearing -COOH; formic acid (no such substituent) is the strongest.

Acid strength of a carboxylic acid depends on how well the conjugate base (carboxylate, RCOO⁻) is stabilised — electron-donating alkyl groups destabilise the negative charge (reduce acidity), while the acid gets stronger as such donation is minimised or offset by electron-withdrawing character:

HCOOH (formic acid) — H directly attached to the carbonyl carbon, no alkyl group to donate electron density; the strongest of the four, pKa ≈ 3.75 (lowest pKa).

C6H5COOH (benzoic acid) — the ring can donate some electron density by resonance and is only mildly electron-withdrawing overall; pKa ≈ 4.2. …
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GUJCET 2025Set 031 markMCQ
$CH_{3}CH_{2}COOH$
›Reveal solutionSolution
[!TLDR]

Propanoic acid is the weakest acid here, so it has the highest $pK_a$ .

Concept

A higher $pK_a$ means a weaker acid. Electron-withdrawing groups (like $-Cl$ , phenyl) stabilise the carboxylate and increase acidity (lower $pK_a$ ); electron-donating alkyl groups reduce acidity (raise $pK_a$ ).

Solution

Compare approximate $pK_a$ values:

(A) $HCOOH$ (formic acid): $\approx 3.75$ (strongest, no destabilising alkyl chain).

(B) $CH_3CH_2COOH$ (propanoic acid): $\approx 4.87$ (electron-donating ethyl group, no withdrawing group) — weakest acid, highest $pK_a$ . …
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GUJCET 2024Set 131 markMCQ
$NO_2\cdot CH_2\cdot COOH$
›Reveal solutionSolution

Strongest electron-withdrawing group → strongest acid → lowest pKa. $NO_2CH_2COOH$ wins.

Concept. An electron-withdrawing substituent stabilises the carboxylate anion, raising acid strength (lowering pKa). …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Which of the following compound has highest Ka Value?$
(a) NO2CH2COOH
(b) BrCH2COOH
(c) CCl3COOH
(d) CH3COOH
›Reveal solutionSolution
Trichloroacetic acid (CCl3COOH, pKa about 0.7) has the highest Ka because three chlorine atoms together exert the strongest cumulative electron-withdrawing effect. Answer: (c).

Acid strength is governed by how well the conjugate-base carboxylate is stabilised by the electron-withdrawing (-I) substituents. Comparing the four:

CCl3COOH: pKa about 0.66 (three Cl atoms, very strong cumulative -I)

NO2CH2COOH: pKa about 1.68

BrCH2COOH: pKa about 2.9

CH3COOH: pKa about 4.76 (no EWG) …
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GUJCET 2022Set 171 markMCQ
$CH_2FCH_2CH_2COOH < CH_3CHFCH_2COOH$
›Reveal solutionSolution
$HCOOH$ (pK $_a$ 3.75) is a stronger acid than $C_6H_5COOH$ (pK $_a$ 4.20), so (D)'s order is wrong.

Concept. "Increasing acidic strength" means the item on the right must be the stronger acid.

(A) F on β-C (closer to COOH) is more acidic than F on γ-C → order correct.

(B) F is more electronegative than Cl → $CH_2FCOOH$ stronger than $CH_2ClCOOH$ → correct.

(C) $CH_2ClCOOH$ stronger than $CH_3COOH$ → correct. …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. Which acid has the lowest pKa?$
(a) CH3COOH
(b) C6H5CH2COOH
(c) C6H5COOH
(d) CH3CH2COOH
›Reveal solutionSolution

Lower pKa = stronger acid; acidity here is governed by how well the conjugate base (carboxylate anion) is stabilised.

Approximate pKa values: benzoic acid (C6H5COOH) ≈ 4.2 (the phenyl ring is directly conjugated to -COOH, and the -I effect of the sp2 ring stabilises the carboxylate); phenylacetic acid (C6H5CH2COOH) ≈ 4.3 (the CH2 spacer partially insulates the ring's effect); acetic acid (CH3COOH) ≈ 4.76; propanoic acid (CH3CH2COOH) ≈ 4.87 (the extra electron-donat …

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GUJCET 2021Set 151 markMCQ
$o-O_2N-C_6H_4-OH$
›Reveal solutionSolution

Fewer/no electron-withdrawing groups → weaker acid → highest pKa = plain phenol.

Concept: An $-NO_2$ group withdraws electron density and stabilises the phenoxide anion, increasing acidity (lowering pKa). Removing it makes the phenol the weakest acid, i.e. the largest pKa. …
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GUJCET 2021Set 151 markMCQ
$Q. Which compound having maximum acidic strength of the following? (A) 4-methoxy benzoic acid (B) 2-methoxy benzoic acid (C) Benzoic acid (D) 4-nitrobenzoic acid$
›Reveal solutionSolution
Electron-withdrawing $-NO_2$ (para) most stabilises the anion → strongest acid.

Concept: Groups that withdraw electron density stabilise the carboxylate and raise acidity; electron-donating groups (like $-OCH_3$ ) lower it.

(A) 4-methoxy and (B) 2-methoxybenzoic acid — $-OCH_3$ donates by resonance, weaker acids. …
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GUJCET 2020Set 071 markMCQ
$FCH_2COOH$
›Reveal solutionSolution

Highest pKa = weakest acid; $\text{C}_6\text{H}_5\text{CH}_2\text{COOH}$ has the least electron-withdrawing substituent.

Concept — inductive stabilisation of the carboxylate. Stronger electron-withdrawing groups ( $-\text{NO}_2 > -\text{CN} > -\text{F}$ ) stabilise the conjugate base and lower pKa. The phenyl …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Conjugate base of which of the following acid is weak?$
(a) CH3CH2CH(I)COOH
(b) CH3CH2CH(F)COOH
(c) CH3CH2CH(Br)COOH
(d) CH3CH2CH(Cl)COOH
›Reveal solutionSolution

The stronger the acid, the weaker (more stable, less basic) its conjugate base; among these halo-substituted acids, the most electronegative halogen (F) gives the strongest acid and hence the weakest conjugate base.

A strong acid ionises readily because its conjugate base is comparatively stable and has little tendency to re-accept a proton (i.e. it is a WEAK base). Acid strength here is controlled by the -I (electron-withdrawing inductive) effect of the halogen substituent close to -COOH: this effect is strongest for the most electronegative halogen and weakens down the group, F …

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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. For which acid the value of pKa is highest? (para-substituted benzoic acids)$
(a) p-Nitrobenzoic acid (4-NO2-C6H4-COOH)
(b) p-Toluic acid / p-methylbenzoic acid (4-CH3-C6H4-COOH)
(c) p-Anisic acid / p-methoxybenzoic acid (4-OCH3-C6H4-COOH)
(d) p-Chlorobenzoic acid (4-Cl-C6H4-COOH)
›Reveal solutionSolution
pKa is highest for the weakest acid; electron-donating para substituents raise pKa (weaken acidity) while electron-withdrawing substituents lower pKa (strengthen acidity).

Acid strength of a substituted benzoic acid depends on how the para substituent affects stability of the carboxylate anion (its conjugate base) via induction and resonance:

p-NO2 (-NO2 is strongly electron-withdrawing by both induction and resonance) stabilises the anion most -> strongest acid -> LOWEST pKa.

p-Cl (weak electron-withdrawing by induction, small resonance donation) -> mildly increases acidity -> pKa close to/slightly below benzoic acid.

p-CH3 (weak electron-donating by hyperconjugation) -> mildly decreases acidity -> pKa slightly above benzoic acid. …
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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Which of the following compound has highest acidic strength?$
(a) p-methylbenzoic acid (COOH with para CH3)
(b) o-nitrobenzoic acid (COOH with ortho NO2)
(c) benzoic acid
(d) p-nitrobenzoic acid (COOH with para NO2)
›Reveal solutionSolution
o-nitrobenzoic acid is the most acidic because the ortho -NO2 group withdraws electrons most strongly (ortho effect).

Acidity of substituted benzoic acids depends on the substituent:

Electron-withdrawing groups (like -NO2) stabilise the carboxylate anion -> increase acidity.

Electron-donating groups (like -CH3) decrease acidity.

Ranking:

p-CH3 (p-toluic acid): weakest (EDG).

benzoic acid: reference.

p-NO2: strong EWG, more acidic. …
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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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