Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine? (i

NCERT Exemplar · Q6

Q. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?

(i)

H_2

(excess)/Pt

(ii)

LiAlH_4

in ether

(iii) Fe and HCl

(iv) Sn and HCl

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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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The key idea is that aryl nitro compounds are unusually resistant to reduction because the nitro group is conjugated with the aromatic ring. Among the given reagents, LiAlH₄ is too mild to overcome this resonance stabilization, so it fails to reduce the nitro group to an amine. The correct answer is (B).

The reduction of an aryl nitro compound (like nitrobenzene) to an amine (like aniline) is a classic transformation in organic chemistry. But not every reducing agent is equally effective here. The reason lies in the electronic structure of the nitro group when it's attached to an aromatic ring.

The nitro group ( $-NO_2$ ) is strongly electron-withdrawing by both induction and resonance. When attached to a benzene ring, the lone pairs on the oxygen atoms participate in resonance with the ring, making the N–O bonds partial double bonds. This resonance stabilizes the nitro group significantly. To break it down, you need a reducing agent that can deliver electrons forcefully enough to overcome this stability.

Let’s examine each option.

Option (A): $H_2$ (excess)/Pt

Catalytic hydrogenation with platinum is a powerful method. The metal surface adsorbs hydrogen and activates it, allowing it to attack the nitro group. The reaction proceeds through a series of intermediates (nitroso, hydroxylamine) and finally gives the amine. This works well for aryl nitro compounds because the catalyst provides enough energy to disrupt the resonance. So this is a good choice.
Option (B): $LiAlH_4$ in ether

Lithium aluminium hydride is a strong reducing agent for many functional groups — carbonyls, esters, even some nitroalkanes. But here’s the catch: with aryl nitro compounds, the resonance stabilization makes the nitro group much less electrophilic. LiAlH₄ works by hydride attack on an electron-deficient centre. The nitro group’s electron density is delocalized into the ring, so it’s not sufficiently electrophilic for hydride to attack effectively. In practice, LiAlH₄ reduces aryl nitro compounds very slowly or not at all, often giving complex mixtures or no reaction. So this is not a good choice.
Option (C): Fe and HCl …

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Here is a breakdown of the common mistakes students make when tackling this specific reduction question, along with the conceptual fixes.

The Core Concept: Why Aryl Nitro Groups are Special

The key to this question is understanding that the nitro group ( $-NO_2$ ) is attached directly to an aromatic ring. This changes the reduction chemistry compared to an aliphatic (non-aromatic) nitro compound.

The Goal: Reduce $-NO_2$ to $-NH_2$ (aniline).
The Trap: Some strong reducing agents will also reduce the benzene ring itself, destroying the aromatic system.

Mistake #1: Assuming $LiAlH_4$ is a "Universal" Reducing Agent

The Error: Students see $LiAlH_4$ and think, "It's the strongest reducing agent, so it must reduce everything, including the nitro group." They then assume it works perfectly.

Why it's Wrong:

$LiAlH_4$ is indeed a powerful reducing agent for polar bonds (like $C=O$ , $C\equiv N$ ). However, it is poor at reducing isolated, non-polar multiple bonds like the $C=C$ bonds in a benzene ring. More importantly, it is inefficient at reducing an aryl nitro group. The reaction is slow, messy, and often gives poor yields of the desired aniline. It is not the "go-to" reagent for this specific job.

How to Avoid:

Memorize the "Job Description": $LiAlH_4$ is for reducing carbonyls, carboxylic acids, esters, and nitriles. It is not the reagent of choice for reducing aromatic nitro groups.
Think "Specificity": For aryl nitro reduction, think of reagents that are selective for the nitro group without touching the ring. $LiAlH_4$ lacks this selectivity in practice.

Mistake #2: Forgetting that $H_2/Pt$ Can Reduce the Ring

The Error: Students see catalytic hydrogenation ( $H_2$ /metal) and assume it will only reduce the nitro group. They forget that under forcing conditions (excess $H_2$ , high pressure, active catalyst like Pt), the benzene ring itself can be hydrogenated to a cyclohexane ring.

Why it's Wrong:

The question specifies $H_2$ (excess)/Pt. "Excess" is the critical clue. While controlled hydrogenation can stop at the aniline stage, using excess hydrogen with a powerful catalyst like platinum will eventually reduce the aromatic ring to cyclohexylamine.

\text{Ar-NO}_2 \xrightarrow{H_2 (\text{excess})/Pt} \text{Cyclohexyl-NH}_2

This is not the desired product (aniline).

How to Avoid:

Read the Fine Print: Always note if the reagent is "excess" or "catalytic." "Excess $H_2$ " is a red flag for over-reduction.
Know Your Catalysts: Pt and Pd are very active and can reduce rings. Ni (Raney Nickel) is often milder and more selective for the nitro group. If the question asked for $H_2$ /Ni, it would be a good choice.

Mistake #3: Confusing "Reducing Agent" with "Good Choice"

The Error: Students know that Fe/HCl and Sn/HCl do reduce aryl nitro groups to anilines. They see both options and think, "Both work, so the question must be about something else."

Why it's Wrong:

Both Fe/HCl and Sn/HCl are excellent and classic choices for this specific reduction. They are selective, high-yielding, and do not touch the benzene ring. The question asks for the reagent that is not a good choice.

How to Avoid:

Identify the "Odd One Out": In this list, (A) and (B) are problematic (over-reduction or inefficiency), while (C) and (D) are the textbook correct answers. The question is testing your ability to spot the bad reagents. …

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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.Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?

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