Assertion: Hoffmann's bromamide reaction is given by primary amines. Reason: Primary amines are more basic tha

NCERT Exemplar · Q72

Q. Assertion: Hoffmann's bromamide reaction is given by primary amines. Reason: Primary amines are more basic than secondary amines.

(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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Hoffmann bromamide reaction is given by amides (not amines), so the assertion is false. The reason is also false because secondary amines are more basic than primary amines in the gas phase, and the usual teaching order in aqueous solution is secondary > primary > tertiary > ammonia. Neither statement is correct.

Let’s unpack this carefully — the question tests two separate ideas: the Hoffmann bromamide reaction and the basicity order of amines. Each needs to be examined on its own merit.

1. What is the Hoffmann bromamide reaction?

This is a classic name reaction where an amide ( $\ce{RCONH2}$ ) is treated with bromine and a strong base (like $\ce{NaOH}$ ) to give a primary amine with one fewer carbon atom. The reaction proceeds through a rearrangement (the Hofmann rearrangement) and is famously not given by amines themselves. The starting material must be an amide, not an amine.

So the assertion says: "Hoffmann's bromamide reaction is given by primary amines." That is factually incorrect. Primary amines are the product of the reaction, not the reactant. The assertion is wrong.

Watch out

A common mistake is to confuse "amines" with "amides" — they sound similar but are entirely different functional groups. Amides have a carbonyl group ( $\ce{-CONH2}$ ), amines do not. The Hoffmann reaction starts with an amide.

2. What about the basicity of primary vs secondary amines?

Basicity depends on the availability of the lone pair on nitrogen for protonation. In the gas phase, alkyl groups are electron-donating (through the inductive effect), so more alkyl groups on nitrogen increase electron density and thus basicity. The order in the gas phase is:

\text{tertiary} > \text{secondary} > \text{primary} > \text{ammonia}

However, in aqueous solution, solvation effects complicate things. The ammonium cation formed after protonation is stabilized by hydrogen bonding with water. More hydrogen atoms on the nitrogen (as in primary amines) allow better solvation, but this does not make primary amines the most basic — for simple alkyl amines the commonly taught aqueous order is:

$\text{secondary} > \text{primary} > \text{tertiary} > \text{ammonia}$ …

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Here are the common mistakes students make with this Assertion-Reason question, along with how to avoid each.

Mistake 1: Confusing the Reactant in Hoffmann Bromamide Reaction

The Mistake: Students often think the reaction starts with a primary amine (R-NH₂). They see "Hoffmann's bromamide reaction is given by primary amines" and assume it's correct because the product is a primary amine.

The Correction: The reactant is an amide (R-CONH₂), not an amine. The reaction converts an amide into a primary amine with one less carbon atom.

Reactant: Amide ( $\ce{R-CONH2}$ )
Reagent: Bromine ( $\ce{Br2}$ ) in aqueous/alkaline medium ( $\ce{NaOH}$ )
Product: Primary amine ( $\ce{R-NH2}$ )

How to Avoid: Memorize the exact starting material. Write the reaction equation every time you revise:

\ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O}

Mistake 2: Misjudging the Assertion's Truth Value

The Mistake: Because the reactant is an amide, students incorrectly mark the Assertion as "wrong."

The Correction: The Assertion says: "Hoffmann's bromamide reaction is given by primary amines." This is false. The reaction is given by amides, not amines. The product is a primary amine, but the reactant is not.

How to Avoid: Read Assertions literally. The statement says the reaction is "given by" (i.e., performed on) primary amines. That is incorrect. Do not confuse the product with the reactant.

Mistake 3: Assuming All Amines are More Basic Than Others

The Mistake: Students accept the Reason ("Primary amines are more basic than secondary amines") as correct without checking the actual trend.

The Correction: In aqueous solution, the basicity order is:

\text{Secondary} > \text{Primary} > \text{Tertiary} > \text{Ammonia}

So, secondary amines are more basic than primary amines. The Reason is wrong.

How to Avoid: Memorize the correct order. Use the logic of inductive effect and solvation:

Alkyl groups are electron-donating (+I effect), which increases electron density on nitrogen.
More alkyl groups = more electron density = stronger base (in gas phase).
In water, solvation of the conjugate acid matters. Tertiary amines have poor solvation, so secondary amines win in aqueous medium.

Mistake 4: Choosing Option (B) — "Both correct, Reason not the explanation"

The Mistake: Students think both statements are true but unrelated, so they pick (B).

The Correction: Both statements are actually false, so (B) is invalid. Evaluate each part independently:

Assertion: "Hoffmann's bromamide reaction is given by primary amines." → False (it is given by amides).
Reason: "Primary amines are more basic than secondary amines." → False (secondary amines are more basic in aqueous medium).

Both statements are wrong, so the correct answer is (A) Both assertion and reason are wrong. …

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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: Hoffmann's bromamide reaction is given by primary amines.
Reason: Primary amines are more basic than secondary amines.

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 ( $\sigma_I$ ) …

Concept: Hoffmann Bromamide Reaction & Basicity of Amines

Method: Factual Verification + Logical Linkage Check

Step 1: Verify the Assertion

Step 2: Verify the Reason

Mistake 1: Confusing the Reactant in Hoffmann Bromamide Reaction

Mistake 2: Misjudging the Assertion's Truth Value

Mistake 3: Assuming All Amines are More Basic Than Others

Mistake 4: Choosing Option (B) — "Both correct, Reason not the explanation"

Concept

Solution

Q.Assertion: Hoffmann's bromamide reaction is given by primary amines. Reason: Primary amines are more basic than secondary amines.

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Q.Assertion: Hoffmann's bromamide reaction is given by primary amines.
Reason: Primary amines are more basic than secondary amines.

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