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.

COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{NaHCO}_3$
›Reveal solutionSolution
The question asks for the incorrect statement about acetic acid. Option (A) is wrong because the reaction of acetic acid with HI and red P at 473 K yields ethane, not iodoethane. The correct answer is (A).

The key here is to recall the specific reactions and properties of acetic acid. Each option tests a different fact: a reduction reaction, acid strength comparison, a test for acidity, and a test for reducing sugars. We need to spot the one that doesn't match reality.

Option (A): Acetic acid on reaction with HI and red P at 473 K gives iodoethane

This describes the reduction of a carboxylic acid to an alkane. Red phosphorus and hydroiodic acid (HI) at high temperature are a strong reducing agent. They first convert the –COOH group to –CH₃, but the mechanism involves replacing the –OH with iodine, then reducing the iodine to hydrogen. For acetic acid (CH₃COOH), the product is ethane (CH₃CH₃), not iodoethane (CH₃CH₂I). Iodoethane would require stopping at the alkyl iodide stage, but under these conditions, the reduction goes all the way to the alkane. So this statement is false.

Option (B): Acetic acid is a weaker acid than formic acid

This is true. Formic acid (HCOOH) has a pKa of about 3.75, while acetic acid (CH₃COOH) has a pKa of about 4.76. The methyl group in acetic acid is electron-donating (inductive effect), which destabilizes the conjugate base (acetate ion) by increasing electron density, making it a weaker acid. Formic acid’s hydrogen has no such donating effect, so it is stronger.

Option (C): Acetic acid gives effervescence with aqueous NaHCO₃ solution

This is true. Acetic acid is a carboxylic acid, and it reacts with sodium bicarbonate to produce carbon dioxide gas, which causes effervescence:
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COMEDK 2025Set 2025-M1 markMCQ
$[A]$
›Reveal solutionSolution
[!TLDR]

The stated acidity order (Cl>F>Br>CH $_3$ ) is wrong because para-fluorobenzoic acid is actually a weaker acid than para-bromobenzoic acid, while the Reason (F has the largest $-I$ effect) is a correct statement — so option (C).

Concept

In CBSE/NCERT aromatic chemistry, a para substituent affects benzoic-acid acidity through both its inductive ( $-I$ ) and resonance ( $\pm M$ ) effects. Electron-withdrawing groups strengthen the acid (stabilise the carboxylate); electron-donating groups (like $-CH_3$ ) weaken it. Halogens are $-I$ (acid-strengthening) but also weak $+M$ donors, and at the para position the $+M$ donation is felt directly.

Solution

The four acids are para-substituted benzoic acids with substituents Br [A], Cl [B], CH $_3$ [C], F [D]. Their measured strengths (pK $_a$ in water) are approximately:
$\text{p-Cl }3.98,\quad \text{p-Br }3.97,\quad \text{p-F }4.14,\quad \text{p-CH}_3\ 4.37.$
Lower pK $_a$ = stronger acid, giving the real order
$\text{Cl}\approx\text{Br} > \text{F} > \text{CH}_3.$
The Assertion claims Cl $>$ F $>$ Br $>$ CH $_3$ , i.e. it places F above Br. This is incorrect: although fluorine has the strongest $-I$ pull, at the para position its lone pairs donate electron density into the ring by resonance ( $+M$ ), destabilising the carboxylate and making p-F-benzoic acid a weaker acid than p-Br-benzoic acid. So the Assertion order is wrong. …
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COMEDK 2024Set 2024-A1 markMCQ
$Q. Among the following compounds, the most acidic is : (A) 3, 4-dinitrobenzoic acid (B) Benzoic acid (C) 4-methoxybenzoic acid (D) 4-nitrobenzoic acid$
›Reveal solutionSolution
The acidity of benzoic acid derivatives is governed by the electron-withdrawing or electron-donating nature of substituents. The most acidic compound here is 3,4-dinitrobenzoic acid because two nitro groups strongly stabilize the conjugate base by resonance and induction.

Concept & Intuition

Acidity in carboxylic acids depends on the stability of the conjugate base (the carboxylate anion). Electron-withdrawing groups (EWGs) like –NO₂ pull electron density away from the carboxylate, dispersing its negative charge and making the acid stronger. Electron-donating groups (EDGs) like –OCH₃ do the opposite, destabilizing the anion and weakening the acid. The more EWGs and the closer they are to the –COOH group, the greater the effect.

Step-by-step reasoning

Identify the substituent effects

Benzoic acid (B) has no substituent — it’s the reference.

4-Methoxybenzoic acid (C) has –OCH₃ at the para position. Methoxy is an electron-donating group (resonance donor), so it decreases acidity relative to benzoic acid.

4-Nitrobenzoic acid (D) has –NO₂ at the para position. Nitro is a strong electron-withdrawing group (both inductive and resonance), so it increases acidity.

3,4-Dinitrobenzoic acid (A) has two –NO₂ groups at the meta and para positions. Both withdraw electrons, and their effects are additive.

Compare acid strengths qualitatively

(C) is the weakest because of electron donation.

(B) is stronger than (C) but weaker than any nitro-substituted acid.

(D) is stronger than (B) because one –NO₂ stabilizes the carboxylate.

(A) has two –NO₂ groups, so it should be stronger than (D). The meta nitro also withdraws electrons inductively, and the para nitro does so via both induction and resonance. Together, they create a highly stabilized conjugate base.

Quantitative check (pKa values)

Benzoic acid: pKa ≈ 4.20

4-Methoxybenzoic acid: pKa ≈ 4.47 (less acidic)

4-Nitrobenzoic acid: pKa ≈ 3.41 (more acidic) …
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KCET 2023Set D-21 markMCQ
Q.Match the List-I with List-II in the following : List-I
1. Caprolactum
2. Vinyl chloride
3. Styrene
4. Propene
List-II
(a)
(b)
(c)
(d) (A) 1-c, 2-d, 3-a, 4-b (B) 1-a, 2-d, 3-c, 4-b (C) 1-d, 2-c, 3-a, 4-b (D) 1-d, 2-c, 3-b, 4-a
›Reveal solutionSolution

In addition polymerisation the C=C opens and the substituent on the monomer's CH stays on the backbone CH — so match each monomer to the repeat unit carrying its substituent; caprolactam is the odd one out (condensation → polyamide).

Step 1 — The concept: what a repeat unit tells you.

For an addition polymer of a vinyl monomer $\mathrm{CH_2{=}CH{-}X}$ , the double bond opens and the chain grows as
$n\,\mathrm{CH_2{=}CH{-}X} \longrightarrow -\!\left(\mathrm{CH_2 - \underset{X}{CH}}\right)_{\!n}\!-$
So the group $X$ hanging off the CH carbon in the drawn repeat unit is the substituent of the monomer. Structures (a), (b) and (c) are all of this $-(\mathrm{CH_2-CHX})_n-$ type, so they must come from the three vinyl monomers; the amide unit (d) must come from the remaining monomer.

Step 2 — Match 4. Propene → (a).

$\mathrm{CH_2{=}CH{-}CH_3}$ has $X = \mathrm{CH_3}$ , so it gives $-(\mathrm{CH_2-CH(CH_3)})_n-$ = polypropene = structure (a) (the $\mathrm{CH_3}$ -bearing unit). ⇒ 4-a

Step 3 — Match 2. Vinyl chloride → (c).

$\mathrm{CH_2{=}CHCl}$ has $X = \mathrm{Cl}$ , giving $-(\mathrm{CH_2-CHCl})_n-$ = PVC = structure (c) (the Cl-bearing unit). ⇒ 2-c

Step 4 — Match 3. Styrene → (b).

$\mathrm{CH_2{=}CH{-}C_6H_5}$ has $X = \mathrm{C_6H_5}$ , giving $-(\mathrm{CH_2-CH(C_6H_5)})_n-$ = polystyrene = structure (b) (the phenyl-bearing unit). ⇒ 3-b

Step 5 — Match 1. Caprolactam → (d). …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}<\mathrm{CCl}_3 \mathrm{CH}_2 \mathrm{OH}<\mathrm{CF}_3 \mathrm{CH}_2 \mathrm{OH}$
›Reveal solutionSolution
Acidity of $\mathrm{RCH_2OH}$ increases with the $-I$ (electron-withdrawing) power of R, which stabilises the conjugate base (alkoxide). $\mathrm{CH_3}$ (electron-donating) gives the weakest acid; $\mathrm{CF_3}$ (most electronegative halogen) gives the strongest, with $\mathrm{CCl_3}$ in between.

The acidic strength of an alcohol is governed by how well the resulting alkoxide $\mathrm{RCH_2O^-}$ is stabilised.

$\mathrm{CH_3-}$ is weakly electron donating ( $+I$ ), destabilising the alkoxide $\Rightarrow$ ethanol is the weakest acid.

$\mathrm{CCl_3-}$ is strongly electron withdrawing ( $-I$ ), stabilising the negative charge. …
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KCET 2018Set A-11 markMCQ
$\mathrm{BF_3}$
›Reveal solutionSolution
$\mathrm{BF_3}$ is an electron-deficient molecule that accepts a lone pair, so its acidity is explained by the Lewis concept — it acts as a Lewis acid. The correct option is (C).

The key is to understand what "acidity" means in each of the three classical theories. Arrhenius and Brønsted-Lowry both define acids in terms of protons ( $\mathrm{H^+}$ ). Lewis, on the other hand, defines an acid as an electron-pair acceptor — a much broader definition that includes molecules like $\mathrm{BF_3}$ .

$\mathrm{BF_3}$ has only six electrons in its valence shell (boron has three, each fluorine contributes one in a single bond). This makes it electron-deficient and highly eager to accept a lone pair from a base (like $\mathrm{NH_3}$ or $\mathrm{F^-}$ ). It has no proton to donate, so it cannot fit the Arrhenius or Brønsted-Lowry definitions.

Let’s check each option systematically.

Arrhenius concept — An acid must produce $\mathrm{H^+}$ in water. $\mathrm{BF_3}$ does not have a hydrogen atom to release, so it is not an Arrhenius acid. This option is out.

Brønsted-Lowry concept — An acid must donate a proton ( $\mathrm{H^+}$ ) to a base. Again, $\mathrm{BF_3}$ has no proton to give. It is not a Brønsted-Lowry acid. This option is also out.

Lewis concept — An acid is any species that can accept an electron pair. $\mathrm{BF_3}$ has an incomplete octet on boron, so it readily accepts a lone pair from a Lewis base (e.g., $\mathrm{BF_3 + NH_3 \rightarrow F_3B-NH_3}$ ). This fits perfectly. $\mathrm{BF_3}$ is a classic example of a Lewis acid.

Brønsted-Lowry as well as Lewis — Since $\mathrm{BF_3}$ fails the Brønsted-Lowry test, this combined option is incorrect. …
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