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.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$pK_a$
›Reveal solutionSolution
Ranking pKa by substituent electronics: $-NO_2$ (EWG) lowers pKa below benzoic acid's, $-OCH_3$ (net EDG at para) raises it above — giving A-II, B-III, C-I.

Concept and Intuition

For substituted benzoic acids, acid strength (and hence pKa) tracks how well the ring substituent stabilises (or destabilises) the negative charge on the conjugate-base carboxylate. Electron-withdrawing groups (like $-NO_2$ ) stabilise the anion, increasing acidity (lower pKa); electron-donating groups (like para $-OCH_3$ , whose resonance-donation dominates its inductive withdrawal at that position) destabilise the anion slightly, decreasing acidity (higher pKa) relative to unsubstituted benzoic acid.

Step-by-Step Solution

A. $C_6H_5COOH$ (no substituent) is the reference acid: pKa $\approx 4.19$ → matches II.

C. $p\text{-}O_2N\text{-}C_6H_4COOH$ : $-NO_2$ is a strong electron-withdrawing group (both inductively and by resonance at para), stabilising the carboxylate anion strongly, so this is the strongest acid of the three, i.e. the lowest pKa: $3.41$ → matches I. …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$C_6H_5COOH \xrightarrow[Conc.\ H_2SO_4]{Conc.\ HNO_3} Y$
›Reveal solutionSolution
Both transformations install an electron-withdrawing group near the carboxylic acid, which stabilises the conjugate base and increases acidity (lowers $pK_a$ ): $Y < X$ and $B < A$ .

Concept and Intuition

$pK_a$ decreases (acidity increases) whenever an electron-withdrawing group is introduced close to a $-COOH$ group, because it stabilises the resulting carboxylate anion through the inductive effect. Distance and number of such groups matter — closer substituents have a bigger effect — but even a single EWG anywhere on the ring or chain will make the acid stronger than the parent.

Step-by-Step Solution

Reaction I: X = $C_6H_5COOH$ (benzoic acid). Nitration with conc. $HNO_3$ /conc. $H_2SO_4$ substitutes a ring hydrogen with $NO_2$ ; since $-COOH$ is a meta-director, Y = m-nitrobenzoic acid. The electron-withdrawing $NO_2$ group (even at the meta position) pulls electron density away, stabilising the carboxylate and making Y a stronger acid than X. So $pK_a(Y) < pK_a(X)$ , i.e. $Y < X$ .

Reaction II: A = $C_6H_5CH_2COOH$ (phenylacetic acid). The Hell-Volhard-Zelinsky (HVZ) reaction — $Br_2$ with a catalytic amount of red phosphorus, followed by hydrolysis with water — replaces the $\alpha$ -hydrogen (the $CH_2$ adjacent to $-COOH$ ) with bromine, giving B = 2-bromo-2-phenylacetic acid ( $C_6H_5CHBrCOOH$ ). …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\mathrm{C_6H_5-CO_2H}$
›Reveal solutionSolution
Ranking acidity requires comparing functional group (carboxylic acid vs phenol) first, then substituent effects (EWG strengthens, EDG weakens); the order is IV > I > II > III > V.

Concept and Intuition

Carboxylic acids are inherently far more acidic than phenols because the carboxylate anion is stabilised by resonance across two equivalent C–O bonds, whereas the phenoxide ion delocalises charge onto a less electronegativity-matched ring system. Within each class, an electron-withdrawing substituent (like $-NO_2$ ) further stabilises the conjugate base and increases acidity, while an electron-donating group (like $-CH_3$ ) destabilises the conjugate base and decreases acidity.

Step-by-Step Solution

Separate into carboxylic acids (I, IV) and phenols (II, III, V). All carboxylic acids are more acidic than all these phenols.

Among carboxylic acids: p-nitrobenzoic acid (IV, EWG $NO_2$ ) is more acidic than plain benzoic acid (I). So IV > I. …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$pK_a$
›Reveal solutionSolution
Among the four acids, 4-nitrobenzoic acid has the lowest pKₐ (is the strongest acid) because the powerful electron-withdrawing $-NO_2$ group at the para position strongly stabilises the carboxylate conjugate base via both induction and resonance.

Concept and Intuition

Acid strength of a carboxylic acid is governed by how well its conjugate base (carboxylate anion) is stabilised. Electron-withdrawing groups (EWGs) on the ring pull electron density away from the $-COO^-$ , spreading out (delocalising) the negative charge and stabilising the anion — this lowers pKₐ (increases acidity). Electron-donating groups (EDGs) do the opposite, destabilising the anion and raising pKₐ (decreasing acidity). At the para position specifically, resonance donation/withdrawal is transmitted efficiently through the ring to the carboxylate.

Step-by-Step Solution

Benzoic acid ( $C_6H_5COOH$ ): the baseline reference, pKₐ ≈ 4.2.

Phenylacetic acid ( $C_6H_5CH_2COOH$ ): the extra $CH_2$ spacer insulates the ring from the carboxyl group, so the phenyl ring's (mild) inductive withdrawal is felt less — this acid is slightly WEAKER (higher pKₐ) than benzoic acid.

4-Nitrobenzoic acid ( $4-O_2N-C_6H_4-COOH$ ): the nitro group is a strong EWG both inductively and by resonance (it can pull electron density directly through the conjugated ring from the para position, delocalising the negative charge of the carboxylate into the nitro group). This makes it a much STRONGER acid — the LOWEST pKₐ among the four. …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{C_6H_5CO_2H}$
›Reveal solutionSolution
Among benzoic acid, phenylacetic acid, formic acid and acetic acid, formic acid is the strongest (most acidic) since it has no electron-donating alkyl/aryl group to destabilise its conjugate base.

Concept and Intuition

Carboxylic acid strength tracks the stability of the carboxylate anion formed on deprotonation. Any electron-donating group (+I effect, e.g. an alkyl group) attached to the $-\mathrm{COOH}$ carbon pushes electron density onto the already-negative carboxylate, destabilising it and weakening the acid. Formic acid is the unique case where the group attached is just a hydrogen atom — no +I donor at all — so its carboxylate is the least destabilised, making $\mathrm{HCOOH}$ noticeably more acidic than acetic acid. Aromatic acids (benzoic, phenylacetic) sit in between: the phenyl ring is mildly electron-withdrawing by induction but resonance/conjugation effects are modest, and phenylacetic acid's $-\mathrm{CH_2}-$ spacer partially insulates the ring's effect from the carboxyl, making it slightly weaker than benzoic acid.

Step-by-Step Solution

List approximate $\mathrm{p}K_a$ values (lower $\mathrm{p}K_a$ = stronger acid): $\mathrm{HCOOH} \approx 3.75$ ; benzoic acid $\approx 4.20$ ; phenylacetic acid $\approx 4.31$ ; $\mathrm{CH_3COOH} \approx 4.76$ .

Compare acetic vs formic: the $\mathrm{CH_3}$ group in acetic acid is +I (electron donating), destabilising the carboxylate and making it weaker than formic acid, which has only an H there. …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$pK_a$
›Reveal solutionSolution
This tests how substituents affect the acidity of benzoic acid via inductive/resonance electron withdrawal or donation. The strongest electron-withdrawing group gives the lowest $pK_a$ (strongest acid); the electron-donating group gives the highest $pK_a$ (weakest acid). Answer: III, IV.

Concept and Intuition

The acidity of a substituted benzoic acid depends on how well the ring substituent stabilises (or destabilises) the resulting carboxylate anion, $\mathrm{ArCOO^-}$ .

An electron-withdrawing group (EWG) pulls electron density away from the carboxylate, spreading out (stabilising) the negative charge. This makes the conjugate base more stable, so the acid ionises more readily $\Rightarrow$ stronger acid $\Rightarrow$ lower $pK_a$ .

An electron-donating group (EDG) pushes electron density toward the carboxylate, concentrating (destabilising) the negative charge. This makes the acid ionise less readily $\Rightarrow$ weaker acid $\Rightarrow$ higher $pK_a$ .

Step-by-Step Solution

Classify each substituent:

I) $-I$ (para-iodo): halogens are net electron-withdrawing by induction (despite weak $+M$ donation from lone pairs) — mildly acid-strengthening.

II) $-CN$ (para-cyano): strong $-I$ and $-M$ withdrawing (conjugated nitrile) — strongly acid-strengthening.

III) $-CH_3$ (para-methyl): alkyl groups are weakly electron-donating ( $+I$ , hyperconjugation) — acid-weakening.

IV) $-NO_2$ (para-nitro): the strongest $-I$ and $-M$ withdrawing group of the four (fully conjugated, highly electronegative) — most acid-strengthening.

Rank electron-withdrawing power (acid strength, lowest $pK_a$ to highest): $NO_2 > CN > I > CH_3$ . …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$\mathrm{C_6H_5COOH}$
›Reveal solutionSolution
Comparing an aromatic acid with three trihalo/dihalo-substituted acetic acids, the strongest acid is trifluoroacetic acid, because fluorine's high electronegativity gives it the strongest -I (inductive electron-withdrawing) effect.

Concept and Intuition

The acidity of a carboxylic acid depends on how well the conjugate base (carboxylate anion) is stabilised. Electron-withdrawing groups near the -COOH (via the inductive effect) pull electron density away, stabilising the negative charge on the carboxylate and increasing acid strength. Benzoic acid has only a mild inductive/resonance effect from the phenyl ring, making it much weaker than the halogenated acetic acids. Among the halogens, electronegativity decreases down the group ( $F > Cl > Br$ ), so per-atom, fluorine withdraws electron density most strongly through the sigma-bond framework.

Step-by-Step Solution

Benzoic acid: only the phenyl ring's weak inductive/resonance effect on -COOH; the weakest acid of the four (pKa around 4.2).

Dichloroacetic acid ( $Cl_2CHCOOH$ ): two chlorine atoms provide a substantial -I effect, but fewer/less electronegative than trihalomethyl analogues.

Tribromoacetic acid ( $Br_3CCOOH$ ): three halogens give a strong -I effect, but bromine is less electronegative than fluorine, so the effect per atom is weaker than fluorine's. …
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AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$\mathrm{H_2SO_4}$
›Reveal solutionSolution
Both statements are individually true, but the reason given doesn't actually explain the assertion — hence (D).

Concept and Intuition

Acid-base behaviour is relative: a substance can act as an acid towards a weaker acid/base but as a base towards a stronger acid. $HClO_4$ is one of the strongest known Brønsted acids (a "superacid" relative to $H_2SO_4$ ), so when the two are mixed, $HClO_4$ protonates $H_2SO_4$ (forming $H_3SO_4^+$ and $ClO_4^-$ ) — here $H_2SO_4$ is accepting a proton, i.e. acting as a base.

Step-by-Step Solution

Check Assertion (A): Since $HClO_4$ is a stronger acid than $H_2SO_4$ , it can protonate $H_2SO_4$ , making $H_2SO_4$ act as a base in that mixture. This is true.

Check Reason (R): $H_3PO_4$ (orthophosphoric acid) is indeed a weaker acid than $H_2SO_4$ — this is also a true standalone fact. …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$pK_a$
(a) 4-nitrobenzoic acid: benzene ring with COOH substituent and a para $NO_2$ substituent
(b) 4-methoxybenzoic acid: benzene ring with COOH substituent and a para $OCH_3$ substituent
(c) 4-nitrophenol: benzene ring with OH substituent and a para $NO_2$ substituent
(d) benzoic acid: benzene ring with only a COOH substituent (A) c < b < a < d (B) b < d < c < a (C) a < d < b < c (D) a < b < c < d
›Reveal solutionSolution
Comparing electron-withdrawing/donating substituent effects and the carboxylic-acid-vs-phenol acidity gap gives the increasing $pK_a$ order a < d < b < c.

Concept and Intuition

$pK_a$ tracks inversely with acid strength: a lower $pK_a$ means a stronger (more dissociated) acid. Two effects combine here:

Substituent electronic effect on the benzoic-acid series: an electron-withdrawing group (like $-NO_2$ ) stabilizes the carboxylate conjugate base, increasing acidity (lower $pK_a$ ) relative to plain benzoic acid; an electron-donating group (like $-OCH_3$ ) destabilizes the carboxylate (relative to H), decreasing acidity (higher $pK_a$ ).

Functional group identity: carboxylic acids are intrinsically far more acidic than phenols, because the carboxylate anion is resonance-stabilized symmetrically over two oxygens, while phenoxide delocalizes charge into the (less stabilizing) aromatic ring. So even a nitro-activated phenol remains a much weaker acid (higher $pK_a$ ) than any of the benzoic acid derivatives here.

Putting these together (approximate literature $pK_a$ values):

(a) 4-nitrobenzoic acid: $\approx 3.4$ (EWG $-NO_2$ boosts acidity of the –COOH)

(d) benzoic acid: $\approx 4.2$ (reference/parent)

(b) 4-methoxybenzoic acid: $\approx 4.5$ (EDG $-OCH_3$ reduces acidity of the –COOH)

(c) 4-nitrophenol: $\approx 7.2$ (phenol is intrinsically much weaker acid, even with the activating nitro group)

Increasing $pK_a$ (weakest-acid-last order): a < d < b < c.

Step-by-Step Solution …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\mathrm{C_6H_5-C\equiv CH}$
›Reveal solutionSolution
A para $-NO_2$ group withdraws electron density (by resonance and induction), stabilising the acetylide anion and increasing acidity; a para $-NH_2$ group donates electron density, decreasing acidity. So II (nitro) > I (plain) > III (amino).

Concept and Intuition

Removing the terminal alkyne proton gives an aryl-acetylide anion, $Ar-C\equiv C^-$ . Any factor that stabilises this negative charge increases the acidity of the C–H bond (lower $pK_a$ , stronger acid). A $-NO_2$ group at the para position is a strong electron-withdrawing group (by both resonance delocalisation into the ring and induction), so it stabilises the anion and raises acidity. Conversely, a $-NH_2$ group at the para position is a strong electron donor (lone pair conjugates into the ring), which destabilises the negative charge (pushes electron density toward an already negative centre) and lowers acidity relative to unsubstituted phenylacetylene.

Step-by-Step Solution

Unsubstituted phenylacetylene (I) is the baseline acidity. …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$Q. Arrange the following in increasing order of their acidic strength$
(a) $\mathrm{CH_3COOH}$
(b) $\mathrm{Ph\text{-}CH_2\text{-}COOH}$
(c) $\mathrm{Br\text{-}CH_2\text{-}COOH}$
(d) $\mathrm{O_2N\text{-}CH_2\text{-}COOH}$ (A) a < c < d < b (B) a < c < b < d (C) d < c < a < b (D) a < b < c < d
›Reveal solutionSolution
This tests how the –I (inductive, electron-withdrawing) effect of a substituent on

the $\alpha$ -carbon changes carboxylic-acid strength. Increasing –I strength gives

increasing acidity: acetic < phenylacetic < bromoacetic < nitroacetic acid.

Concept and Intuition

A carboxylic acid ionises as $RCOOH \rightleftharpoons RCOO^- + H^+$ . Whatever

stabilises the conjugate base $RCOO^-$ makes the acid stronger (lower $pK_a$ ).

An electron-withdrawing group (EWG) attached near the $-COOH$ pulls electron density

away through the sigma-bond framework (the inductive effect), which spreads out

(delocalises) the negative charge on the carboxylate oxygen and stabilises it. The

closer and stronger the EWG, the bigger this stabilisation, and the stronger the acid.

Conversely, an electron-donating or only weakly-withdrawing group leaves the negative

charge more concentrated on oxygen — less stable anion, weaker acid.

Step-by-Step Solution

Identify the substituent replacing one H of the $CH_3$ group in each acid:

(a) $CH_3COOH$ — no substituent (reference acid).

(b) $Ph\text{-}CH_2COOH$ — phenyl group, a mild net electron-withdrawing group by induction (much weaker than a halogen or $NO_2$ ).

(c) $Br\text{-}CH_2COOH$ — bromine, a fairly strong –I halogen substituent.

(d) $O_2N\text{-}CH_2COOH$ — nitro group, one of the strongest –I groups known.

Rank the inductive (–I) strength of the substituents:

$NO_2 > Br > C_6H_5 > (\text{no substituent})$

Stronger –I substituent $\Rightarrow$ better anion stabilisation $\Rightarrow$ stronger acid. So acidic strength increases in the same order as –I strength: $a\ (\text{no EWG}) < b\ (Ph) < c\ (Br) < d\ (NO_2)$ …
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AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$Cl_2 > Br_2 > F_2 > I_2$
›Reveal solutionSolution
This tests periodic trends of halogens/halogen compounds; the hypohalous acid acidity order in (C) is reversed — electronegativity of the halogen (not its size) governs $HOX$ acidity, so $HOCl$ is the strongest, not $HOI$ .

Concept and Intuition

For the hypohalous acids $HOX$ , acid strength depends on how well the halogen atom pulls electron density away from the O–H bond (inductive effect) and stabilizes the resulting $OX^-$ conjugate base. Since electronegativity decreases down the group ( $Cl>Br>I$ ), $HOCl$ withdraws electron density most strongly and is the most acidic; $HOI$ is the least acidic. This is opposite to what one might guess by analogy with the hydrohalic acids $HX$ , where acidity increases down the group (there, bond strength/size dominates, not electronegativity).

Step-by-Step Solution

(A): Bond dissociation enthalpies (kJ/mol) are $Cl_2(242) > Br_2(192) > F_2(159) > I_2(151)$ — matches the given order, so (A) is correctly matched.

(B): For hydrohalic acids, the weaker the H–X bond, the more easily $H^+$ dissociates; bond strength decreases $HF>HCl>HBr>HI$ , so acid strength increases $HI>HBr>HCl>HF$ — matches (B), correctly matched.

(C): For hypohalous acids $HOX$ , acidity depends on halogen electronegativity (inductive withdrawal from the O–H bond), which is $Cl>Br>I$ . So real order is $HOCl>HOBr>HOI$ — but option (C) states $HOI>HOBr>HOCl$ , which is exactly reversed. …
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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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Q.Assertion: Hoffmann's bromamide reaction is given by primary amines.
Reason: Primary amines are more basic than secondary amines.

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