Assertion: The boiling points of alkyl halides dec – Class 12 Chemistry

NCERT Exemplar · Q86

Q. Assertion: The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. Reason: The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.

(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 — the RI > RBr > RCl > RF boiling-point order is correct, and alkyl halides genuinely do boil higher than comparable hydrocarbons — but the reason does not explain the assertion: comparing halides to hydrocarbons is a different comparison from ranking the halides against each other. The correct option is (v): both correct, reason is not the correct explanation.

Why the assertion is true

Boiling point in alkyl halides is governed mainly by van der Waals (London dispersion) forces, which grow stronger as the halogen atom gets larger and more polarisable. Iodine is the largest, most polarisable halogen and fluorine the smallest, so for a given alkyl group:

\text{R–I} > \text{R–Br} > \text{R–Cl} > \text{R–F}

For example: $CH_3I$ (42°C) > $CH_3Br$ (4°C) > $CH_3Cl$ (−24°C) > $CH_3F$ (−78°C).

Why the reason is also true, on its own

Alkyl halides do boil considerably higher than a hydrocarbon of similar molecular mass — for instance $C_2H_5Cl$ (M = 64.5, b.p. 12°C) boils far above $C_3H_8$ (M = 44, b.p. −42°C). This is a real, correct fact about alkyl halides as a class.

Why the reason does not explain the assertion …

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Common Mistakes Students Make on This Question (Boiling Point Trends)

Mistake 1: Confusing Boiling Point Trends with Inductive Effect

The error: Students think the trend RI > RBr > RCl > RF is due to inductive effect (electronegativity differences).
Why it's wrong: Inductive effect influences acidity/basicity, not boiling points. Boiling points depend on van der Waals (London dispersion) forces, which grow with the halogen's size and polarisability, not its electronegativity.
How to avoid: Remember — boiling point tracks polarisability/size for a series like this. Iodine is heaviest and most polarisable → strongest London forces → highest boiling point.

Mistake 2: Treating the Reason as False Because of an "RF Exception"

The error: Students notice the reason only mentions chlorides/bromides/iodides (not fluorides) and assume this must mean the reason is incomplete or wrong.
Why it's wrong: The reason is scoped correctly — it deliberately excludes RF (whose boiling point is comparable to, not "considerably higher than," the parent hydrocarbon, since fluorine is so small). As stated, covering only RCl/RBr/RI, the reason is fully true.
How to avoid: Read the reason exactly as written — don't test it against a case (RF) it never claims to cover.

Mistake 3: Assuming "Both individually true" must mean the Reason explains the Assertion

The error: Students verify both statements are true and jump straight to option (i), assuming true + true always means "reason explains assertion."
Why it's wrong: The assertion is about the order among halides themselves (why iodide beats bromide beats chloride beats fluoride) — driven by polarisability/London forces increasing with halogen size. The reason is about halides vs. the parent hydrocarbon — a completely different comparison. One true fact doesn't automatically explain a different true fact.
How to avoid: For every AR question, explicitly ask: "does the reason's LOGIC lead to the assertion's specific claim?" Here it doesn't — check for the option that captures "both true, but unrelated" (option v), not (i).

Mistake 4: Missing that option (v) exists

The error: Seeing that neither "(i) reason explains" nor "(iii) reason is wrong" nor "(ii) both wrong" fit cleanly, students force their answer into whichever of those three seems "closest," rather than re-reading the full option list.
Why it's wrong: This question's option list includes exactly the right fit: "(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion" — precisely this situation. …

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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: The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.
Reason: The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.

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 ( $\sigma_I$ ) …

Why the assertion is true

Why the reason is also true, on its own

Why the reason does not explain the assertion …

Method: Analysis of Assertion–Reason Statements Using Factual Verification

Steps:

Common Mistakes Students Make on This Question (Boiling Point Trends)

Mistake 1: Confusing Boiling Point Trends with Inductive Effect

Mistake 2: Treating the Reason as False Because of an "RF Exception"

Mistake 3: Assuming "Both individually true" must mean the Reason explains the Assertion

Mistake 4: Missing that option (v) exists

Concept and Intuition

Step-by-Step Solution

Q.Assertion: The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. Reason: The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.

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: The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.
Reason: The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.

Unlock everything free for 14 days