Identify the compound Y in the following reaction. ![Aniline reacting via NaNO2/HCl (273-278 K) to benzenediazonium chloride, then Cu2Cl2 to give Y + N2, drawn as real benzene rings matching the NCERT Exemplar page](/figures/haloalkanes-and-haloarenes/diagram-6-exemplar-q3-aniline-diazonium.svg) ![Options (i)-(iv): chlorobenzene, benzene, 1,3-dichlorobenzene, and 1,4-dichlorobenzene, drawn as real benzene rings matching the NCERT Exemplar page](/figures/haloalkanes-and-haloarenes/diagram-6-exemplar-q3-options.svg)

> [!TLDR] > The reaction is the Sandmeyer reaction: the diazonium group is replaced by chlorine using $\mathrm{Cu_2Cl_2}$, giving chlorobenzene ($\mathrm{C_6H_5Cl}$) as product Y. ![Diazotization and Sandmeyer route](/figures/haloalkanes-and-haloarenes/diagram-6-5-2.svg) The key to this question is recognising the **Sandmeyer reaction** — a classic method for replacing the diazonium group ($-\mathrm{N_2^+}$) with a halogen using a copper(I) halide. Let’s walk through the chemistry step by step. 1. **First step: Diazotisation** Aniline ($\mathrm{C_6H_5NH_2}$) reacts with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at low temperature (273–278 K). This converts the amino group into a diazonium group: $$\mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\,K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O}$$ The product is **benzenediazonium chloride**, a key intermediate in aromatic substitution. The low temperature is critical — diazonium salts decompose above about 5°C. 2. **Second step: The Sandmeyer reaction** The diazonium salt is then treated with $\mathrm{Cu_2Cl_2}$ (copper(I) chloride). This is the classic **Sandmeyer reaction**, where the diazonium group is replaced by a chlorine atom. The mechanism involves a single-electron transfer from Cu(I) to the diazonium ion, generating an aryl radical, which then abstracts chlorine from Cu(II) to form the aryl chloride. $$\mathrm{C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2} C_6H_5Cl + N_2}$$ The nitrogen gas ($\mathrm{N_2}$) bubbles off, driving the reaction forward. 3. **What about the options?** - **(i) Chlorobenzene** — This is the direct product of the Sandmeyer reaction with $\mathrm{Cu_2Cl_2}$. - **(ii) Benzene** — This would require reduction of the diazonium group (e.g., with $\mathrm{H_3PO_2}$), not with $\mathrm{Cu_2Cl_2}$. - **(iii) 1,3-Dichlorobenzene** and **(iv) 1,4-Dichlorobenzene** — These would require two chlorine substitutions, but the reaction conditions only introduce one c

Identify the compound Y in the following reaction.

NCERT Exemplar · Q3

Q. Identify the compound Y in the following reaction.

Aniline reacting via NaNO2/HCl (273-278 K) to benzenediazonium chloride, then Cu2Cl2 to give Y + N2, drawn as real benzene rings matching the NCERT Exemplar page — Figure

Options (i)-(iv): chlorobenzene, benzene, 1,3-dichlorobenzene, and 1,4-dichlorobenzene, drawn as real benzene rings matching the NCERT Exemplar page — Figure

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✓ Free question

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.

Watch out

Do not confuse inductive effect with resonance effect. Inductive effect operates through sigma bonds and is distance-dependent. Resonance effect operates through pi bonds and can act over long distances. For example, $-\text{NO}_2$ is both strongly electron-withdrawing inductively (through sigma bonds) and by resonance (through pi bonds). But $-\text{Cl}$ is electron-withdrawing inductively but electron-donating by resonance — the net effect on acidity depends on which dominates.

The Key Takeaway

Inductive effect on acidity: Electron-withdrawing groups (EWGs) increase acidity by stabilising the conjugate base through sigma-bond polarisation. Electron-donating groups (EDGs) decrease acidity. The effect is strongest when the group is closest to the acidic site and diminishes with distance.

\text{Acidity} \propto \text{Number and strength of EWGs near acidic site}

The inductive effect on acidity is a recurring theme across the NCERT Class 11 and 12 Organic Chemistry chapters, and ‘inductive effect and acidity of carboxylic acids’ is one of the most common important-question types in CBSE boards, JEE Main and NEET organic chemistry. Comparing acid strengths using electron-withdrawing and electron-donating substituents is a skill tested in nearly every organic reasoning-based MCQ.

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 ( $\sigma_I$ )

For purely inductive effects (no resonance), we use Taft's separation:

\sigma = \sigma_I + \sigma_R

Where $\sigma_I$ is the inductive component. The formula for $\sigma_I$ itself comes from comparing rates of hydrolysis of esters — reactions where resonance effects are minimal.

Why $\sigma_I$ Values Are Additive

For a substituent X at distance $n$ bonds from the reaction centre:

\sigma_I(\text{X at position } n) = \frac{\sigma_I(\text{X at position } 1)}{2.7^n}

This fall-off factor (2.7 ≈ $e$ ) arises because:

Inductive effect propagates through sigma bonds
Each bond attenuates the effect by a factor related to bond polarisability
The exponential decay is a consequence of successive polarisation of each bond

Practical Exam Tip

When comparing acidity of two compounds:

Draw the conjugate base of each
Identify which has more electron-withdrawing groups near the negative charge
More EWG → more stabilised conjugate base → stronger acid

The formulas above are quantitative tools, but the qualitative reasoning — stabilising the anion — is what you need for most exam questions.

Remember: The inductive effect is distance-dependent and additive. Two Cl atoms at the same position have roughly twice the effect of one. But a Cl at the $\beta$ -carbon has much less effect than one at the $\alpha$ -carbon.

The reaction is the Sandmeyer reaction: the diazonium group is replaced by chlorine using $\mathrm{Cu_2Cl_2}$ , giving chlorobenzene ( $\mathrm{C_6H_5Cl}$ ) as product Y.

The key to this question is recognising the Sandmeyer reaction — a classic method for replacing the diazonium group ( $-\mathrm{N_2^+}$ ) with a halogen using a copper(I) halide. Let’s walk through the chemistry step by step.

First step: Diazotisation Aniline ( $\mathrm{C_6H_5NH_2}$ ) reacts with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at low temperature (273–278 K). This converts the amino group into a diazonium group:

\mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\,K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O}

The product is benzenediazonium chloride, a key intermediate in aromatic substitution. The low temperature is critical — diazonium salts decompose above about 5°C.

Second step: The Sandmeyer reaction The diazonium salt is then treated with $\mathrm{Cu_2Cl_2}$ (copper(I) chloride). This is the classic Sandmeyer reaction, where the diazonium group is replaced by a chlorine atom. The mechanism involves a single-electron transfer from Cu(I) to the diazonium ion, generating an aryl radical, which then abstracts chlorine from Cu(II) to form the aryl chloride.

\mathrm{C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2} C_6H_5Cl + N_2}

The nitrogen gas ( $\mathrm{N_2}$ ) bubbles off, driving the reaction forward.

What about the options?
- (i) Chlorobenzene — This is the direct product of the Sandmeyer reaction with $\mathrm{Cu_2Cl_2}$ .
- (ii) Benzene — This would require reduction of the diazonium group (e.g., with $\mathrm{H_3PO_2}$ ), not with $\mathrm{Cu_2Cl_2}$ .
- (iii) 1,3-Dichlorobenzene and (iv) 1,4-Dichlorobenzene — These would require two chlorine substitutions, but the reaction conditions only introduce one chlorine. No further chlorination occurs here.

Watch out

A common mistake is to think that $\mathrm{Cu_2Cl_2}$ causes a second substitution or that the reaction is a simple displacement. It is not — it’s a radical mechanism specific to the Sandmeyer reaction, and only one chlorine is introduced.

Tip

Remember the mnemonic: Sandmeyer for Cl, Br, CN using $\mathrm{CuX}$ or $\mathrm{CuCN}$ ; Schiemann for F using $\mathrm{HBF_4}$ ; and Gattermann for Cl, Br using $\mathrm{Cu}$ + $\mathrm{HX}$ .

Confirming the product The reaction is clean: one diazonium group, one chlorine atom replaces it, and nitrogen is lost. The product is chlorobenzene, $\mathrm{C_6H_5Cl}$ .

✓Final answer

The compound Y is chlorobenzene, option (i) .

Common Mistakes in This Diazonium Reaction Problem

This question tests your understanding of the Sandmeyer reaction — specifically the replacement of the diazonium group ( $-\mathrm{N}_2^+$ ) with chlorine using $\mathrm{Cu_2Cl_2}$ .

✗ Mistake 1: Thinking $\mathrm{Cu_2Cl_2}$ gives substitution on the ring

Why students make it:

They see $\mathrm{Cu_2Cl_2}$ and assume it chlorinates the benzene ring directly (like electrophilic substitution), producing dichlorobenzenes.

How to avoid:

Remember: $\mathrm{Cu_2Cl_2}$ in the Sandmeyer reaction replaces the diazonium group ( $-\mathrm{N}_2^+$ ) with a chlorine atom at the same position. It does not add extra chlorines to the ring.

Correct result: Only one chlorine replaces the $-\mathrm{N}_2^+$ group → chlorobenzene ( $\mathrm{C_6H_5Cl}$ ).

✗ Mistake 2: Choosing benzene ( $\mathrm{C_6H_6}$ )

Why students make it:

They recall that diazonium salts can be reduced to benzene using $\mathrm{H_3PO_2}$ (hypophosphorous acid) or ethanol, and confuse the reagent.

How to avoid:

Memorise the reagent–product mapping:

Reagent	Product
$\mathrm{Cu_2Cl_2}$	Chlorobenzene
$\mathrm{Cu_2Br_2}$	Bromobenzene
$\mathrm{CuCN}$	Benzonitrile
$\mathrm{H_3PO_2}$ / $\mathrm{C_2H_5OH}$	Benzene

Here, $\mathrm{Cu_2Cl_2}$ cannot give benzene — it gives chlorobenzene.

✗ Mistake 3: Forgetting that $\mathrm{N_2}$ gas is released

Why students make it:

They focus only on the product structure and ignore the stoichiometric clue.

How to avoid:

The equation shows $\mathrm{N_2}$ is evolved. This means the diazonium group ( $-\mathrm{N}_2^+$ ) leaves completely. The only thing that can replace it is a single atom or group from the reagent — here, $\mathrm{Cl}$ from $\mathrm{Cu_2Cl_2}$ .

✓ Quick Summary Table

Mistake	Why it happens	How to avoid
Choosing dichlorobenzenes	Confusing Sandmeyer with electrophilic chlorination	Sandmeyer replaces — does not add
Choosing benzene	Confusing $\mathrm{Cu_2Cl_2}$ with $\mathrm{H_3PO_2}$	Memorise reagent–product pairs
Ignoring $\mathrm{N_2}$ evolution	Overlooking reaction stoichiometry	$\mathrm{N_2}$ means the group is replaced, not modified

Final correct answer: $\boxed{\text{(i) Chlorobenzene}}$

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:

$\text{CH}_3\text{COOH} + \text{NaHCO}_3 \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$
This is a standard test for carboxylic acids.

Option (D): Acetic acid does not reduce Fehling's solution This is true. Fehling's solution is used to test for aldehydes (and some reducing sugars). Acetic acid is a carboxylic acid, not an aldehyde, and it lacks the aldehyde group needed to reduce Cu²⁺ to Cu⁺. So it gives no reaction.

Watch out

A common mistake is to think that HI and red P produce an alkyl iodide. In fact, red P + HI is a powerful reducing system that converts carboxylic acids all the way to alkanes, not stopping at the alkyl halide.

Tip

Remember: The reaction RCOOH + HI + red P → RCH₃ is a classic way to reduce a carboxylic acid to an alkane. For acetic acid (R = CH₃), the product is ethane.

✓Final answer
$The correct option is (A). ANSWER: A$
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.

The Reason states “Fluorine has a larger $-I$ effect than Cl and Br.” Since electronegativity runs F $>$ Cl $>$ Br, fluorine indeed exerts the strongest inductive electron withdrawal — the Reason is a true statement. But it does not correctly rationalise the given (faulty) order, because $-I$ alone would put F first, not third.

Therefore the Assertion is false while the Reason is true.

[!ANSWER]

(C) A is wrong but R is correct.

Note

This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.
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)

3,4-Dinitrobenzoic acid: pKa ≈ 2.82 (most acidic) These values confirm the trend: the more nitro groups, the lower the pKa, hence the stronger the acid.

Why not something else?

A common pitfall is thinking that ortho or meta substitution might be stronger than para for a single nitro group, but here both nitro groups are in positions that allow resonance withdrawal (para) and inductive withdrawal (meta). The combination is unbeatable among the given options.

Watch out

Don’t confuse electron-donating groups like –OCH₃ with electron-withdrawing ones. Methoxy is actually a strong resonance donor, so it decreases acidity — the opposite of what a beginner might guess.

Tip

For substituted benzoic acids, remember the mnemonic: EWG = stronger acid, EDG = weaker acid. And more EWGs = even stronger.

✓Final answer
$The correct option is (A) . ANSWER: A$
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).

Caprolactam is a cyclic amide (lactam) with a 7-membered ring, $\mathrm{\overset{\frown}{CO-(CH_2)_5-NH}}$ . It is not a vinyl monomer at all: heated with water it ring-opens and undergoes condensation polymerisation to Nylon-6:
$-\!\left[\mathrm{\overset{O}{\overset{\|}{C}} - (CH_2)_5 - \underset{H}{N}}\right]_{\!n}\!-$
That is exactly the drawn amide repeat unit (d) — note the five $\mathrm{CH_2}$ groups and the $\mathrm{-CO-NH-}$ amide link, the fingerprint of a polyamide. ⇒ 1-d

Step 6 — Assemble.
$1\text{-d},\quad 2\text{-c},\quad 3\text{-b},\quad 4\text{-a}$
Option (C) is the near-miss (it swaps styrene and propene onto the wrong repeat units); only (D) is fully correct.

✓Final answer
$The correct option is (D) 1-d, 2-c, 3-b, 4-a .$

ANSWER: D
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.

$\mathrm{CF_3-}$ is even more strongly withdrawing because fluorine is more electronegative than chlorine, so it stabilises the alkoxide the most $\Rightarrow$ strongest acid.

Hence increasing acidity: $\mathrm{CH_3CH_2OH}<\mathrm{CCl_3CH_2OH}<\mathrm{CF_3CH_2OH}$ .

✓Final answer
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}<\mathrm{CCl}_3\mathrm{CH}_2\mathrm{OH}<\mathrm{CF}_3\mathrm{CH}_2\mathrm{OH}$
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.

Watch out

A common mistake is to think that because $\mathrm{BF_3}$ is acidic in water (it hydrolyses to give boric acid and $\mathrm{HBF_4}$ ), it must be an Arrhenius acid. That reaction is a chemical change, not a direct proton donation by $\mathrm{BF_3}$ itself. The acidity of $\mathrm{BF_3}$ in water is a consequence of its Lewis acidity triggering hydrolysis — the underlying reason is still the electron deficiency of boron.

Tip

The Lewis concept is the only one that covers non-protonic acids like $\mathrm{BF_3}$ , $\mathrm{AlCl_3}$ , and $\mathrm{SO_3}$ . Whenever you see a molecule with an incomplete octet, a positive charge, or a vacant orbital, think Lewis acid.

✓Final answer
$The correct option is (C) Lewis concept .$