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.

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.

C6H5CH2COOH (phenylacetic acid) — the CH2 spacer insulates the ring's effect, closer to a simple alkyl acid; pKa ≈ 4.3.

CH3CH2COOH (propanoic acid) — an ordinary alkyl acid, with the alkyl group's +I effect destabilising the carboxylate the most; pKa ≈ 4.87 (weakest, highest pKa).

Lowest pKa = strongest acid = HCOOH.

✓Final answer
$(b) HCOOH.$
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 $pK_a$ .

(C) $C_6H_5CH_2COOH$ (phenylacetic acid): $\approx 4.3$ (phenyl mildly withdraws).

(D) $ClCH_2CH_2COOH$ : $\approx 4.0$ ( $-Cl$ is electron-withdrawing, increasing acidity).

The highest $pK_a$ is propanoic acid.

[!ANSWER]

(B) $CH_3CH_2COOH$
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).

Approximate pKa values: nitroacetic acid $\approx1.7$ , formic $\approx3.75$ , benzoic $\approx4.2$ , acetic $\approx4.76$ . The powerful $-NO_2$ inductive effect makes nitroacetic acid the strongest acid, so it has the least pKa.

✓Final answer
$NO_2\cdot CH_2\cdot COOH$

ANSWER: (A)
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)

Although a single -NO2 is a stronger individual withdrawing group than a single -Cl, the cumulative effect of THREE chlorine atoms in CCl3COOH outweighs the single nitro group, giving it the lowest pKa / highest Ka.

✓Final answer
$(c) CCl3COOH.$
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.

(D) Claims benzoic acid stronger than formic acid, but formic acid is actually the stronger of the two → incorrect order.

✓Final answer
$HCOOH < C_6H_5COOH$

ANSWER: (D)
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-donating alkyl group is destabilising for the anion, making it the weakest acid here). Benzoic acid, having its ring directly attached to -COOH, has the lowest pKa.

✓Final answer
$(c) C6H5COOH has the lowest pKa.$
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.

(A) o-, (B) p-, (C) m-nitrophenol all bear $-NO_2$ → more acidic, smaller pKa.

(D) $C_6H_5OH$ has no $-NO_2$ → least acidic → maximum pKa.

✓Final answer
$C_6H_5OH$

ANSWER: (D)
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.

(C) Benzoic acid — reference.

(D) 4-nitrobenzoic acid — $-NO_2$ withdraws strongly → maximum acidic strength.

✓Final answer
$(D) 4-nitrobenzoic acid$

ANSWER: (D)
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 group withdraws least, so phenylacetic acid is the weakest acid and has the highest pKa.

✓Final answer
$C_6H_5CH_2COOH$

ANSWER: (D)
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 > Cl > Br > I. So CH3CH2CH(F)COOH (the alpha-fluoro acid) is the STRONGEST acid of the four, and correspondingly its conjugate base (the fluoro-substituted butanoate anion) is the WEAKEST base/conjugate base of the set.

✓Final answer
$(b) CH3CH2CH(F)COOH.$
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.

p-OCH3 (strong electron-DONATING by resonance, lone pair on O conjugates into the ring and pushes electron density toward -COO-) destabilises the anion the MOST -> weakest acid -> HIGHEST pKa.

Ranking (approx. real pKa values): p-NO2 (3.42) < p-Cl (3.98) < unsubstituted (4.20) < p-CH3 (4.34) < p-OCH3 (4.47).

✓Final answer
$(c) p-Anisic acid / p-methoxybenzoic acid (4-OCH3-C6H4-COOH) has the highest pKa.$
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.

o-NO2: the ortho position gives an additional 'ortho effect' (steric + inductive proximity), making o-nitrobenzoic acid the most acidic (pKa ~ 2.2, lower than p-nitrobenzoic acid ~ 3.4).

✓Final answer
$(b) o-nitrobenzoic acid.$

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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

Solution

Q.Identify the compound Y in the following reaction.

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