Assertion: N,N-Diethylbenzene sulphonamide is insoluble in alkali. Reason: Sulphonyl group attached to nitroge

NCERT Exemplar · Q74

Q. Assertion: N,N-Diethylbenzene sulphonamide is insoluble in alkali. Reason: Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.

(i) Both assertion and reason are wrong.

(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.

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

(iv) Both assertion and reason are correct statements and reason is 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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N,N-Diethylbenzenesulphonamide has no N–H bond at all (both hydrogens on nitrogen are replaced by ethyl groups), so it cannot form a soluble salt with alkali — it is genuinely insoluble. The reason given (the sulphonyl group is a strong electron-withdrawing group) is a true statement on its own, but it is not why the compound is insoluble — the real cause is the missing N–H proton. Both statements are true, but the reason does not explain the assertion. The correct option is (ii).

Understand the structure.

N,N-Diethylbenzenesulphonamide has the formula $\ce{C6H5SO2N(C2H5)2}$ . The nitrogen is bonded to two ethyl groups and the sulphonyl group — there is no hydrogen attached to nitrogen.
Recall the acid–base behaviour of sulphonamides.

A sulphonamide like $\ce{RSO2NH2}$ or $\ce{RSO2NHR'}$ has an N–H bond. The sulphonyl group ( $\ce{-SO2-}$ ) is strongly electron withdrawing, which makes that N–H bond polarised enough that the hydrogen can be removed by a strong base like $\ce{NaOH}$ . That is why primary and secondary sulphonamides (one N–H remaining) are soluble in alkali — they form salts.
$\ce{RSO2NH2 + NaOH -> RSO2N^-Na+ + H2O}$
Apply to the given compound.

In N,N-diethylbenzenesulphonamide, the nitrogen has no hydrogen to donate — both N–H's of the parent sulphonamide have already been replaced by ethyl groups. Even though the sulphonyl group withdraws electrons strongly, there is no acidic proton left to remove. So the compound cannot react with alkali and remains insoluble. The assertion ("insoluble in alkali") is therefore correct.
Evaluate the reason. …

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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: N,N-Diethylbenzene sulphonamide is insoluble in alkali.
Reason: Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.

Concept understanding — Inductive Effect on Acidity

Inductive Effect on Acidity – From Intuition to Precision

The Core Intuition

The Precise Statement

Why Distance Matters

Why this formula?

Inductive Effect on Acidity: Why It Works

The Core Idea: Stabilising the Conjugate Base

Why Electron-Withdrawing Groups (EWG) Increase Acidity

The Quantitative Relationship: Hammett Equation

Why This Formula Holds

The Inductive Effect Constant ( $\sigma_I$ ) …

Method: Inductive Effect Analysis for Acidity / Basicity of Sulphonamides

🧪 The Core Concept: Inductive Effect on Acidity of Sulphonamides

✓ Correct Answer

✗ Common Mistakes Students Make

1. Thinking “electron-withdrawing” always means acidic

2. Ignoring the structure — missing that both H’s on N are replaced

3. Confusing “reason is correct” with “reason explains assertion”

Concept and Intuition

Step-by-Step Solution

Q.Assertion: N,N-Diethylbenzene sulphonamide is insoluble in alkali. Reason: Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.

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Q.Assertion: N,N-Diethylbenzene sulphonamide is insoluble in alkali.
Reason: Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.

Unlock everything free for 14 days