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

Gujarat GsebMCQ· 1mImportance★★★★★

31% · 34/108 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 ( $\sigma_I$ ) …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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. …
Start your 14-day free trial to see the full solution →
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 $pK_a$ . …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →
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) …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →

←Previous questionQ5 — Benzylamine may be alkylated as shown in the following equation: C₆H₅C…→Next questionQ7 — In order to prepare a 1° amine from an alkyl halide with simultaneous …

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

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: Reduction of Aryl Nitro Compounds to Amines

Method: Reagent Suitability Analysis

Steps:

The Core Concept: Why Aryl Nitro Groups are Special

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

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

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

Concept

Solution

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

🎓Unlock everything free for 14 days

Unlock everything free for 14 days