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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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.

KEAM 2026Set eng-2026-04174 marksMCQ
$O_2N$
›Reveal solutionSolution
Among the choices, HCOOH lacks any strong electron-withdrawing $\alpha$ -substituent, so its conjugate base is least stabilised — it is the weakest acid and thus has the highest $pK_a$ .

Acidity of a carboxylic acid rises (i.e. $pK_a$ falls) when an electron-withdrawing group near the $-COOH$ stabilises the carboxylate anion.

$O_2N{-}CH_2COOH$ : strong $-I$ nitro, very low $pK_a$ .

$F{-}CH_2COOH$ , $Cl{-}CH_2COOH$ , $CN{-}CH_2COOH$ : all bear strong $-I$ groups, low $pK_a$ .

HCOOH (formic acid): no additional electron-withdrawing substituent; $pK_a \approx 3.75$ , the highest of the set.

Therefore formic acid has the highest $pK_a$ .

✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04224 marksMCQ
$FCH_2COOH > NCCH_2COOH > NO_2CH_2COOH > ClCH_2COOH$
›Reveal solutionSolution

The stronger the electron-withdrawing substituent, the more acidic the acetic acid: $NO_2 > CN > F > Cl$ , giving $O_2NCH_2COOH > NCCH_2COOH > FCH_2COOH > ClCH_2COOH$ .

Electron-withdrawing groups stabilise the carboxylate and increase acidity. The $-I$ /effect order is $NO_2 > CN > F > Cl$ , matched by pKa values ( $\approx1.68,\ 2.47,\ 2.59,\ 2.87$ ). Hence the decreasing acid-strength order is $NO_2CH_2COOH > NCCH_2COOH > FCH_2COOH > ClCH_2COOH$ .

✓Final answer
$The correct option is (C) .$
KEAM 2025Set eng-2025-04254 marksMCQ
$Q. Which of the following acid is highly acidic? (A) Fluoroacetic acid (B) Formic acid (C) Dichloroacetic acid (D) Benzoic acid (E) Acetic acid.$
›Reveal solutionSolution
Acid strength increases with electron-withdrawing substituents that stabilise the conjugate base. Two chlorines in dichloroacetic acid ( $pK_a \approx 1.3$ ) make it the most acidic here.

Comparing approximate $pK_a$ values (lower = more acidic):

Dichloroacetic acid: ~1.3 (two $-I$ Cl atoms) — strongest.

Fluoroacetic acid: ~2.6.

Formic acid: ~3.75.

Benzoic acid: ~4.2.

Acetic acid: ~4.76.

The two strongly electron-withdrawing chlorine atoms in dichloroacetic acid best stabilise the carboxylate anion, so it is the most acidic.

✓Final answer
$The correct option is (C) .$
KEAM 2025Set eng-2025-04294 marksMCQ
$FCH_2COOH$
›Reveal solutionSolution
Acid strength rises with the electron-withdrawing power near the $\text{-}COOH$ . Three fluorines in $CF_3COOH$ give the strongest $\text{-}I$ effect and the most stabilised conjugate base, so it is the strongest acid.

Substituted acetic acids become stronger as the electron-withdrawing (–I) effect of the substituent increases, because a more stabilised carboxylate anion means a more readily released proton.

$CH_3COOH$ (E): electron-donating methyl, weakest acid.

$BrCH_2COOH$ (D), $FCH_2COOH$ (A), $NCCH_2COOH$ (C): a single electron-withdrawing group each — stronger than acetic acid, F and CN being quite effective.

$CF_3COOH$ (B): three fluorine atoms exert a very strong cumulative $\text{-}I$ effect, giving the greatest stabilisation of the anion ( $pK_a \approx 0.23$ ).

Therefore $CF_3COOH$ is the strongest acid.

✓Final answer
$The correct option is (B) .$
KEAM 2025Set pha-2025-0424F4 marksMCQ
$FCH_2COOH$
›Reveal solutionSolution
Acidity rises with electron-withdrawing power near the –COOH. Comparing the groups, mono-fluoro on one $\alpha$ -C ( $FCH_2COOH$ , $pK_a\approx2.6$ ) is the least acid-strengthening, so it is the weakest acid.

Comparison (approximate $pK_a$ ):

$Cl_3C{-}COOH$ (trichloroacetic): $\approx 0.7$ — three Cl directly on the carbonyl carbon, strongest.

$Cl_2CH{-}COOH$ (dichloroacetic): $\approx 1.3$ .

$O_2N{-}CH_2COOH$ (nitroacetic): $\approx 1.7$ .

$NC{-}CH_2COOH$ (cyanoacetic): $\approx 2.5$ .

$FCH_2COOH$ (fluoroacetic): $\approx 2.6$ — a single F on one $\alpha$ -carbon gives the weakest inductive stabilisation of the conjugate base here.

Highest $pK_a$ = weakest acid $\Rightarrow FCH_2COOH$ .

✓Final answer
$The correct option is (A) .$
KEAM 2024Set eng-2024-06054 marksMCQ
$Q. Which of the following carboxylic acid has the highest pKa? (A) ethanoic acid (B) chloroethanoic acid (C) fluoroethanoic acid (D) dichloroethanoic acid (E) triflouroethanoic acid$
›Reveal solutionSolution

Electron-withdrawing halogens stabilise the carboxylate and lower pKa; unsubstituted ethanoic acid, having none, has the highest pKa.

Acid strength increases (pKa decreases) with the number and electronegativity of $\alpha$ -halogen substituents, which stabilise the conjugate base by the $-I$ effect. The order of increasing acidity is roughly:

ethanoic $<$ chloroethanoic $<$ fluoroethanoic $<$ dichloroethanoic $<$ trifluoroethanoic.

Thus ethanoic acid (no halogen) is the weakest and has the highest pKa (~4.76).

✓Final answer
$The correct option is (A) .$
KEAM 2024Set eng-2024-06064 marksMCQ
$FCH_{2}COOH > ClCH_{2}COOH > NO_{2}CH_{2}COOH > CNCH_{2}COOH$
›Reveal solutionSolution

Acidity follows the electron-withdrawing ( $-I$ ) strength: $NO_2>CN>F>Cl$ , matching the pKa order.

Electron-withdrawing substituents on the $\alpha$ -carbon stabilise the carboxylate and raise acidity. The experimental pKa values are: nitroacetic $\approx1.68$ , cyanoacetic $\approx2.47$ , fluoroacetic $\approx2.59$ , chloroacetic $\approx2.87$ . Lower pKa means stronger acid, so the decreasing acid strength is
$NO_2CH_2COOH>CNCH_2COOH>FCH_2COOH>ClCH_2COOH.$
✓Final answer
$The correct option is (E) .$
KEAM 2024Set eng-2024-06094 marksMCQ
$Q. The increasing order of acid strength of the following carboxylic acids is$
(i) $(CH_3)_3C\text{-}COOH$
(ii) $(CH_3)_2CH\text{-}COOH$
(iii) $CH_3CH_2COOH$ (A)
(ii) <
(i) <
(iii) (B)
(i) <
(iii) <
(ii) (C)
(ii) <
(iii) <
(i) (D)
(iii) <
(ii) <
(i) (E)
(i) <
(ii) < (iii)
›Reveal solutionSolution

Electron-donating alkyl groups destabilise the carboxylate; more branching = weaker acid.

The acids are (i) $(CH_3)_3C\text{-}COOH$ (tert-butyl, three methyls), (ii) $(CH_3)_2CH\text{-}COOH$ (isopropyl, two methyls), (iii) $CH_3CH_2\text{-}COOH$ (ethyl, one methyl on the $\alpha$ -carbon side). Electron-donating (+I) alkyl groups intensify the negative charge on the carboxylate anion, destabilising it and weakening the acid. The more/bulkier the electron-donating alkyl substitution, the weaker the acid:
$\text{acidity: }(i)<(ii)<(iii).$
✓Final answer
$The correct option is (E) .$
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$ClCH_2\text{-}CH_2\text{-}COOH < ClCH_2COOH < NC\text{-}CH_2COOH < CHCl_2COOH$
›Reveal solutionSolution
Acidity increases: $ClCH_2CH_2COOH < ClCH_2COOH < NCCH_2COOH < CHCl_2COOH$ .

Concept and Intuition

Electron-withdrawing groups stabilise the carboxylate anion (–I effect), increasing acidity; the effect is stronger when the group is closer to –COOH and when there are more/stronger withdrawing groups.

Step-by-Step Solution

$ClCH_2CH_2COOH$ : Cl is one carbon farther (β to COOH), weak effect ⇒ pKa ≈ 4.0 (weakest acid).

$ClCH_2COOH$ : one α-Cl ⇒ pKa ≈ 2.87.

$NCCH_2COOH$ : –CN is a stronger –I group than Cl ⇒ pKa ≈ 2.47.

$CHCl_2COOH$ : two α-Cl ⇒ pKa ≈ 1.29 (strongest).

Common Mistakes

Ranking chloroacetic above cyanoacetic; the cyano group withdraws electrons more strongly, so cyanoacetic acid is stronger.

✓Final answer
$ClCH_2\text{-}CH_2\text{-}COOH < ClCH_2COOH < NC\text{-}CH_2COOH < CHCl_2COOH$

ANSWER: A

←Previous questionQ2 — Which of the following alcohols will yield the corresponding alkyl chl…→Next questionQ4 — Toluene reacts with a halogen in the presence of iron (III) chloride g…

Q.Identify the compound Y in the following reaction.

Concept understanding — Inductive Effect on Acidity

Inductive Effect on Acidity – From Intuition to Precision

The Core Intuition

The Precise Statement

Why Distance Matters

The Key Takeaway

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 ( $\sigma_I$ )

Why $\sigma_I$ Values Are Additive

Practical Exam Tip

Concept: Sandmeyer Reaction

Method: Sandmeyer Reaction for Chlorination

Final Answer

Common Mistakes in This Diazonium Reaction Problem

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

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

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

✓ Quick Summary Table

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Q.Identify the compound Y in the following reaction.

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