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

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

29% · 43/147 Questions

✓ 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.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$pK_a$
›Reveal solutionSolution
Ranking pKa by substituent electronics: $-NO_2$ (EWG) lowers pKa below benzoic acid's, $-OCH_3$ (net EDG at para) raises it above — giving A-II, B-III, C-I.

Concept and Intuition

For substituted benzoic acids, acid strength (and hence pKa) tracks how well the ring substituent stabilises (or destabilises) the negative charge on the conjugate-base carboxylate. Electron-withdrawing groups (like $-NO_2$ ) stabilise the anion, increasing acidity (lower pKa); electron-donating groups (like para $-OCH_3$ , whose resonance-donation dominates its inductive withdrawal at that position) destabilise the anion slightly, decreasing acidity (higher pKa) relative to unsubstituted benzoic acid.

Step-by-Step Solution

A. $C_6H_5COOH$ (no substituent) is the reference acid: pKa $\approx 4.19$ → matches II.

C. $p\text{-}O_2N\text{-}C_6H_4COOH$ : $-NO_2$ is a strong electron-withdrawing group (both inductively and by resonance at para), stabilising the carboxylate anion strongly, so this is the strongest acid of the three, i.e. the lowest pKa: $3.41$ → matches I.

B. $p\text{-}CH_3O\text{-}C_6H_4COOH$ : $-OCH_3$ at the para position donates electron density into the ring by resonance, which (net) destabilises the carboxylate anion slightly, making this the weakest acid of the three, i.e. the highest pKa: $4.46$ → matches III.

So the correct matches are A–II, B–III, C–I.

Common Mistakes

Assuming $-OCH_3$ always behaves as an electron-withdrawing group because oxygen is electronegative — at the para position, its resonance electron-donation into the ring dominates over its inductive withdrawal, so it is a net deactivator of acidity (raises pKa) here.

Mixing up which direction pKa moves — lower pKa means a stronger acid, so the electron-withdrawing nitro compound must get the smallest number (3.41).

✓Final answer
$The correct option is (B) — A-II, B-III, C-I.$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$C_6H_5COOH \xrightarrow[Conc.\ H_2SO_4]{Conc.\ HNO_3} Y$
›Reveal solutionSolution
Both transformations install an electron-withdrawing group near the carboxylic acid, which stabilises the conjugate base and increases acidity (lowers $pK_a$ ): $Y < X$ and $B < A$ .

Concept and Intuition

$pK_a$ decreases (acidity increases) whenever an electron-withdrawing group is introduced close to a $-COOH$ group, because it stabilises the resulting carboxylate anion through the inductive effect. Distance and number of such groups matter — closer substituents have a bigger effect — but even a single EWG anywhere on the ring or chain will make the acid stronger than the parent.

Step-by-Step Solution

Reaction I: X = $C_6H_5COOH$ (benzoic acid). Nitration with conc. $HNO_3$ /conc. $H_2SO_4$ substitutes a ring hydrogen with $NO_2$ ; since $-COOH$ is a meta-director, Y = m-nitrobenzoic acid. The electron-withdrawing $NO_2$ group (even at the meta position) pulls electron density away, stabilising the carboxylate and making Y a stronger acid than X. So $pK_a(Y) < pK_a(X)$ , i.e. $Y < X$ .

Reaction II: A = $C_6H_5CH_2COOH$ (phenylacetic acid). The Hell-Volhard-Zelinsky (HVZ) reaction — $Br_2$ with a catalytic amount of red phosphorus, followed by hydrolysis with water — replaces the $\alpha$ -hydrogen (the $CH_2$ adjacent to $-COOH$ ) with bromine, giving B = 2-bromo-2-phenylacetic acid ( $C_6H_5CHBrCOOH$ ).

The bromine sits directly on the $\alpha$ -carbon, right next to the carboxyl group, so its inductive electron-withdrawal has a strong stabilising effect on the carboxylate — B is more acidic (lower $pK_a$ ) than A. So $B < A$ .

Combining: $Y < X$ and $B < A$ , matching option (A).

Common Mistakes

Thinking a meta-substituent has no acid-strengthening effect — inductive effects act through space/bonds regardless of the substitution pattern, just with reduced magnitude compared to ortho.

Forgetting that HVZ specifically brominates the $\alpha$ -carbon (not the ring), which is exactly why it has such a strong acidifying effect on B.

✓Final answer
$Y < X$

ANSWER: A
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\mathrm{C_6H_5-CO_2H}$
›Reveal solutionSolution
Ranking acidity requires comparing functional group (carboxylic acid vs phenol) first, then substituent effects (EWG strengthens, EDG weakens); the order is IV > I > II > III > V.

Concept and Intuition

Carboxylic acids are inherently far more acidic than phenols because the carboxylate anion is stabilised by resonance across two equivalent C–O bonds, whereas the phenoxide ion delocalises charge onto a less electronegativity-matched ring system. Within each class, an electron-withdrawing substituent (like $-NO_2$ ) further stabilises the conjugate base and increases acidity, while an electron-donating group (like $-CH_3$ ) destabilises the conjugate base and decreases acidity.

Step-by-Step Solution

Separate into carboxylic acids (I, IV) and phenols (II, III, V). All carboxylic acids are more acidic than all these phenols.

Among carboxylic acids: p-nitrobenzoic acid (IV, EWG $NO_2$ ) is more acidic than plain benzoic acid (I). So IV > I.

Among phenols: p-nitrophenol (II, EWG) is most acidic, followed by plain phenol (III), followed by p-cresol (V, EDG $CH_3$ , least acidic). So II > III > V.

Combine: IV > I > II > III > V.

Common Mistakes

Ranking a highly substituted phenol (like p-nitrophenol) as more acidic than an unsubstituted carboxylic acid — carboxylic acids are always more acidic than phenols regardless of ring substitution, because of the fundamentally different (and stronger) resonance stabilisation of the carboxylate anion.

✓Final answer
$The correct option is (A) — IV > I > II > III > V.$

ANSWER: A
AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$pK_a$
›Reveal solutionSolution
Among the four acids, 4-nitrobenzoic acid has the lowest pKₐ (is the strongest acid) because the powerful electron-withdrawing $-NO_2$ group at the para position strongly stabilises the carboxylate conjugate base via both induction and resonance.

Concept and Intuition

Acid strength of a carboxylic acid is governed by how well its conjugate base (carboxylate anion) is stabilised. Electron-withdrawing groups (EWGs) on the ring pull electron density away from the $-COO^-$ , spreading out (delocalising) the negative charge and stabilising the anion — this lowers pKₐ (increases acidity). Electron-donating groups (EDGs) do the opposite, destabilising the anion and raising pKₐ (decreasing acidity). At the para position specifically, resonance donation/withdrawal is transmitted efficiently through the ring to the carboxylate.

Step-by-Step Solution

Benzoic acid ( $C_6H_5COOH$ ): the baseline reference, pKₐ ≈ 4.2.

Phenylacetic acid ( $C_6H_5CH_2COOH$ ): the extra $CH_2$ spacer insulates the ring from the carboxyl group, so the phenyl ring's (mild) inductive withdrawal is felt less — this acid is slightly WEAKER (higher pKₐ) than benzoic acid.

4-Nitrobenzoic acid ( $4-O_2N-C_6H_4-COOH$ ): the nitro group is a strong EWG both inductively and by resonance (it can pull electron density directly through the conjugated ring from the para position, delocalising the negative charge of the carboxylate into the nitro group). This makes it a much STRONGER acid — the LOWEST pKₐ among the four.

4-Methoxybenzoic acid ( $4-CH_3O-C_6H_4-COOH$ ): the methoxy group is an EDG by resonance (lone pair on O conjugates into the ring, pushing electron density toward the carboxylate, destabilising the anion) — making this acid WEAKER (highest pKₐ among the four).

Ranking (strongest to weakest acid / lowest to highest pKₐ): 4-nitrobenzoic acid < benzoic acid < phenylacetic acid < 4-methoxybenzoic acid.

Lowest pKₐ = 4-nitrobenzoic acid.

Common Mistakes

Forgetting that the para-nitro group can donate directly into resonance with the carboxylate (unlike meta-nitro, which acts purely inductively) — this makes para-nitrobenzoic acid even more acidic than meta-nitrobenzoic acid.

Assuming the extra $CH_2$ in phenylacetic acid makes it MORE acidic than benzoic acid — it actually makes it slightly less acidic, since the ring's influence is now one bond further away.

✓Final answer
$4-O_2N-C_6H_4-COOH$

ANSWER: C
AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{C_6H_5CO_2H}$
›Reveal solutionSolution
Among benzoic acid, phenylacetic acid, formic acid and acetic acid, formic acid is the strongest (most acidic) since it has no electron-donating alkyl/aryl group to destabilise its conjugate base.

Concept and Intuition

Carboxylic acid strength tracks the stability of the carboxylate anion formed on deprotonation. Any electron-donating group (+I effect, e.g. an alkyl group) attached to the $-\mathrm{COOH}$ carbon pushes electron density onto the already-negative carboxylate, destabilising it and weakening the acid. Formic acid is the unique case where the group attached is just a hydrogen atom — no +I donor at all — so its carboxylate is the least destabilised, making $\mathrm{HCOOH}$ noticeably more acidic than acetic acid. Aromatic acids (benzoic, phenylacetic) sit in between: the phenyl ring is mildly electron-withdrawing by induction but resonance/conjugation effects are modest, and phenylacetic acid's $-\mathrm{CH_2}-$ spacer partially insulates the ring's effect from the carboxyl, making it slightly weaker than benzoic acid.

Step-by-Step Solution

List approximate $\mathrm{p}K_a$ values (lower $\mathrm{p}K_a$ = stronger acid): $\mathrm{HCOOH} \approx 3.75$ ; benzoic acid $\approx 4.20$ ; phenylacetic acid $\approx 4.31$ ; $\mathrm{CH_3COOH} \approx 4.76$ .

Compare acetic vs formic: the $\mathrm{CH_3}$ group in acetic acid is +I (electron donating), destabilising the carboxylate and making it weaker than formic acid, which has only an H there.

Compare the two acids to the aromatic ones: benzoic/phenylacetic acids are stronger than acetic acid (phenyl's inductive pull beats an alkyl's donation) but still weaker than formic acid, since formic acid has literally no electron-donating substituent at all.

Therefore formic acid is the most acidic of the four.

Common Mistakes

Assuming aromatic acids are always the strongest because "conjugation" sounds stabilising — but the resonance stabilisation of the carboxylate is similar in both benzoic and simple aliphatic acids; the dominant effect distinguishing these four is the +I/no-I difference, favouring formic acid.

Confusing formic acid's lack of a substituent with it being "weaker" (mistakenly reasoning fewer groups = less acidic) — here it is precisely the ABSENCE of a destabilising +I group that makes it stronger.

✓Final answer
$\mathrm{HCOOH}$

ANSWER: C
AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$pK_a$
›Reveal solutionSolution
This tests how substituents affect the acidity of benzoic acid via inductive/resonance electron withdrawal or donation. The strongest electron-withdrawing group gives the lowest $pK_a$ (strongest acid); the electron-donating group gives the highest $pK_a$ (weakest acid). Answer: III, IV.

Concept and Intuition

The acidity of a substituted benzoic acid depends on how well the ring substituent stabilises (or destabilises) the resulting carboxylate anion, $\mathrm{ArCOO^-}$ .

An electron-withdrawing group (EWG) pulls electron density away from the carboxylate, spreading out (stabilising) the negative charge. This makes the conjugate base more stable, so the acid ionises more readily $\Rightarrow$ stronger acid $\Rightarrow$ lower $pK_a$ .

An electron-donating group (EDG) pushes electron density toward the carboxylate, concentrating (destabilising) the negative charge. This makes the acid ionise less readily $\Rightarrow$ weaker acid $\Rightarrow$ higher $pK_a$ .

Step-by-Step Solution

Classify each substituent:

I) $-I$ (para-iodo): halogens are net electron-withdrawing by induction (despite weak $+M$ donation from lone pairs) — mildly acid-strengthening.

II) $-CN$ (para-cyano): strong $-I$ and $-M$ withdrawing (conjugated nitrile) — strongly acid-strengthening.

III) $-CH_3$ (para-methyl): alkyl groups are weakly electron-donating ( $+I$ , hyperconjugation) — acid-weakening.

IV) $-NO_2$ (para-nitro): the strongest $-I$ and $-M$ withdrawing group of the four (fully conjugated, highly electronegative) — most acid-strengthening.

Rank electron-withdrawing power (acid strength, lowest $pK_a$ to highest): $NO_2 > CN > I > CH_3$ .

So the lowest $pK_a$ (strongest acid) is IV (para-nitrobenzoic acid), and the highest $pK_a$ (weakest acid) is III (para-methylbenzoic acid, p-toluic acid), since it's the only electron-donor of the four.

This matches known experimental $pK_a$ values: p-toluic acid $\approx 4.34$ (highest), p-iodobenzoic $\approx 4.00$ , p-cyanobenzoic $\approx 3.55$ , p-nitrobenzoic $\approx 3.42$ (lowest).

Common Mistakes

Assuming halogens are always the strongest withdrawers — nitro and cyano (fully conjugated $-M$ groups) withdraw more strongly than a halogen's inductive-only-dominant effect.

Forgetting that $-CH_3$ is the only donor in the list, making it the automatic "highest $pK_a$ " choice.

✓Final answer
$pK_a$

ANSWER: D
AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$\mathrm{C_6H_5COOH}$
›Reveal solutionSolution
Comparing an aromatic acid with three trihalo/dihalo-substituted acetic acids, the strongest acid is trifluoroacetic acid, because fluorine's high electronegativity gives it the strongest -I (inductive electron-withdrawing) effect.

Concept and Intuition

The acidity of a carboxylic acid depends on how well the conjugate base (carboxylate anion) is stabilised. Electron-withdrawing groups near the -COOH (via the inductive effect) pull electron density away, stabilising the negative charge on the carboxylate and increasing acid strength. Benzoic acid has only a mild inductive/resonance effect from the phenyl ring, making it much weaker than the halogenated acetic acids. Among the halogens, electronegativity decreases down the group ( $F > Cl > Br$ ), so per-atom, fluorine withdraws electron density most strongly through the sigma-bond framework.

Step-by-Step Solution

Benzoic acid: only the phenyl ring's weak inductive/resonance effect on -COOH; the weakest acid of the four (pKa around 4.2).

Dichloroacetic acid ( $Cl_2CHCOOH$ ): two chlorine atoms provide a substantial -I effect, but fewer/less electronegative than trihalomethyl analogues.

Tribromoacetic acid ( $Br_3CCOOH$ ): three halogens give a strong -I effect, but bromine is less electronegative than fluorine, so the effect per atom is weaker than fluorine's.

Trifluoroacetic acid ( $CF_3COOH$ ): three highly electronegative fluorine atoms provide the strongest possible inductive withdrawal among these options, most effectively stabilising the carboxylate anion.

Hence $CF_3COOH$ is the strongest acid.

Common Mistakes

Assuming bromine (larger atomic size) is more electron-withdrawing than fluorine — inductive strength tracks electronegativity, and F is the most electronegative element.

Forgetting that benzoic acid, despite being 'aromatic,' is actually the weakest acid here due to the lack of strong inductive withdrawal.

✓Final answer
$F_3CCOOH$

ANSWER: C
AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$\mathrm{H_2SO_4}$
›Reveal solutionSolution
Both statements are individually true, but the reason given doesn't actually explain the assertion — hence (D).

Concept and Intuition

Acid-base behaviour is relative: a substance can act as an acid towards a weaker acid/base but as a base towards a stronger acid. $HClO_4$ is one of the strongest known Brønsted acids (a "superacid" relative to $H_2SO_4$ ), so when the two are mixed, $HClO_4$ protonates $H_2SO_4$ (forming $H_3SO_4^+$ and $ClO_4^-$ ) — here $H_2SO_4$ is accepting a proton, i.e. acting as a base.

Step-by-Step Solution

Check Assertion (A): Since $HClO_4$ is a stronger acid than $H_2SO_4$ , it can protonate $H_2SO_4$ , making $H_2SO_4$ act as a base in that mixture. This is true.

Check Reason (R): $H_3PO_4$ (orthophosphoric acid) is indeed a weaker acid than $H_2SO_4$ — this is also a true standalone fact.

But does R explain A? R talks about a completely different acid pair ( $H_3PO_4$ vs $H_2SO_4$ ), with no connection to $HClO_4$ at all. It cannot logically account for why $H_2SO_4$ behaves as a base in the presence of $HClO_4$ .

So both statements are true, but R is not the explanation for A.

Common Mistakes

Assuming any two true statements placed together must be causally linked.

Confusing "acid strength relative to $HClO_4$ " with "acid strength relative to $H_3PO_4$ " — these are separate comparisons.

✓Final answer
$The correct option is (D) — Both A and R are correct, but R is not the correct explanation of A.$

ANSWER: D
AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$pK_a$
(a) 4-nitrobenzoic acid: benzene ring with COOH substituent and a para $NO_2$ substituent
(b) 4-methoxybenzoic acid: benzene ring with COOH substituent and a para $OCH_3$ substituent
(c) 4-nitrophenol: benzene ring with OH substituent and a para $NO_2$ substituent
(d) benzoic acid: benzene ring with only a COOH substituent (A) c < b < a < d (B) b < d < c < a (C) a < d < b < c (D) a < b < c < d
›Reveal solutionSolution
Comparing electron-withdrawing/donating substituent effects and the carboxylic-acid-vs-phenol acidity gap gives the increasing $pK_a$ order a < d < b < c.

Concept and Intuition

$pK_a$ tracks inversely with acid strength: a lower $pK_a$ means a stronger (more dissociated) acid. Two effects combine here:

Substituent electronic effect on the benzoic-acid series: an electron-withdrawing group (like $-NO_2$ ) stabilizes the carboxylate conjugate base, increasing acidity (lower $pK_a$ ) relative to plain benzoic acid; an electron-donating group (like $-OCH_3$ ) destabilizes the carboxylate (relative to H), decreasing acidity (higher $pK_a$ ).

Functional group identity: carboxylic acids are intrinsically far more acidic than phenols, because the carboxylate anion is resonance-stabilized symmetrically over two oxygens, while phenoxide delocalizes charge into the (less stabilizing) aromatic ring. So even a nitro-activated phenol remains a much weaker acid (higher $pK_a$ ) than any of the benzoic acid derivatives here.

Putting these together (approximate literature $pK_a$ values):

(a) 4-nitrobenzoic acid: $\approx 3.4$ (EWG $-NO_2$ boosts acidity of the –COOH)

(d) benzoic acid: $\approx 4.2$ (reference/parent)

(b) 4-methoxybenzoic acid: $\approx 4.5$ (EDG $-OCH_3$ reduces acidity of the –COOH)

(c) 4-nitrophenol: $\approx 7.2$ (phenol is intrinsically much weaker acid, even with the activating nitro group)

Increasing $pK_a$ (weakest-acid-last order): a < d < b < c.

Step-by-Step Solution

Group (a), (b), (d) are all benzoic-acid derivatives — rank them by substituent effect: EWG ( $-NO_2$ , a) lowers $pK_a$ most; EDG ( $-OCH_3$ , b) raises it above the unsubstituted acid (d).

So among the carboxylic acids: a < d < b (increasing $pK_a$ ).

(c) is a phenol, not a carboxylic acid — phenols are always much weaker acids (much higher $pK_a$ ) than any benzoic acid derivative, so c sits at the very top.

Final increasing order: a < d < b < c.

Common Mistakes

Ranking the nitrophenol among the benzoic acids by substituent strength alone, ignoring that its acidic proton (phenolic O–H) is fundamentally weaker than a carboxylic acid's O–H.

Reversing the direction of the EWG/EDG effect on $pK_a$ (remember: EWG → lower $pK_a$ → stronger acid).

✓Final answer
$The correct option is (C) — a < d < b < c.$

ANSWER: C
AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\mathrm{C_6H_5-C\equiv CH}$
›Reveal solutionSolution
A para $-NO_2$ group withdraws electron density (by resonance and induction), stabilising the acetylide anion and increasing acidity; a para $-NH_2$ group donates electron density, decreasing acidity. So II (nitro) > I (plain) > III (amino).

Concept and Intuition

Removing the terminal alkyne proton gives an aryl-acetylide anion, $Ar-C\equiv C^-$ . Any factor that stabilises this negative charge increases the acidity of the C–H bond (lower $pK_a$ , stronger acid). A $-NO_2$ group at the para position is a strong electron-withdrawing group (by both resonance delocalisation into the ring and induction), so it stabilises the anion and raises acidity. Conversely, a $-NH_2$ group at the para position is a strong electron donor (lone pair conjugates into the ring), which destabilises the negative charge (pushes electron density toward an already negative centre) and lowers acidity relative to unsubstituted phenylacetylene.

Step-by-Step Solution

Unsubstituted phenylacetylene (I) is the baseline acidity.

4-nitrophenylacetylene (II): $-NO_2$ (EWG, strong) stabilises the conjugate base → most acidic.

4-aminophenylacetylene (III): $-NH_2$ (EDG, strong) destabilises the conjugate base → least acidic.

Order of decreasing acidity: II > I > III.

Common Mistakes

Forgetting that resonance effects of para-substituents (not just induction) dominate the acid-strengthening/weakening here.

Reversing which substituent increases vs decreases acidity.

✓Final answer
$The correct option is (C) — II > I > III.$

ANSWER: C
AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$Q. Arrange the following in increasing order of their acidic strength$
(a) $\mathrm{CH_3COOH}$
(b) $\mathrm{Ph\text{-}CH_2\text{-}COOH}$
(c) $\mathrm{Br\text{-}CH_2\text{-}COOH}$
(d) $\mathrm{O_2N\text{-}CH_2\text{-}COOH}$ (A) a < c < d < b (B) a < c < b < d (C) d < c < a < b (D) a < b < c < d
›Reveal solutionSolution
This tests how the –I (inductive, electron-withdrawing) effect of a substituent on

the $\alpha$ -carbon changes carboxylic-acid strength. Increasing –I strength gives

increasing acidity: acetic < phenylacetic < bromoacetic < nitroacetic acid.

Concept and Intuition

A carboxylic acid ionises as $RCOOH \rightleftharpoons RCOO^- + H^+$ . Whatever

stabilises the conjugate base $RCOO^-$ makes the acid stronger (lower $pK_a$ ).

An electron-withdrawing group (EWG) attached near the $-COOH$ pulls electron density

away through the sigma-bond framework (the inductive effect), which spreads out

(delocalises) the negative charge on the carboxylate oxygen and stabilises it. The

closer and stronger the EWG, the bigger this stabilisation, and the stronger the acid.

Conversely, an electron-donating or only weakly-withdrawing group leaves the negative

charge more concentrated on oxygen — less stable anion, weaker acid.

Step-by-Step Solution

Identify the substituent replacing one H of the $CH_3$ group in each acid:

(a) $CH_3COOH$ — no substituent (reference acid).

(b) $Ph\text{-}CH_2COOH$ — phenyl group, a mild net electron-withdrawing group by induction (much weaker than a halogen or $NO_2$ ).

(c) $Br\text{-}CH_2COOH$ — bromine, a fairly strong –I halogen substituent.

(d) $O_2N\text{-}CH_2COOH$ — nitro group, one of the strongest –I groups known.

Rank the inductive (–I) strength of the substituents:

$NO_2 > Br > C_6H_5 > (\text{no substituent})$

Stronger –I substituent $\Rightarrow$ better anion stabilisation $\Rightarrow$ stronger acid. So acidic strength increases in the same order as –I strength:

$a\ (\text{no EWG}) < b\ (Ph) < c\ (Br) < d\ (NO_2)$

Cross-check with known $pK_a$ values (lower $pK_a$ = stronger acid): acetic $4.76$ > phenylacetic $4.31$ > bromoacetic $2.86$ > nitroacetic $\approx1.68$ . Reading this from weakest to strongest acid gives exactly $a<b<c<d$ .

Common Mistakes

Assuming phenyl is a strong electron-withdrawing group like a halogen — it is only mildly so, much weaker than $Br$ or $NO_2$ , so it must sit just above the unsubstituted acid, not near the top.

Forgetting that "increasing acidic strength" means going from the weakest acid to the strongest, and accidentally reversing the whole order.

Ranking by electronegativity of the atom directly bonded (Br) rather than by the net inductive pull the whole group exerts once bonded through a $CH_2$ spacer.

✓Final answer
$a < b < c < d$

ANSWER: D
AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$Cl_2 > Br_2 > F_2 > I_2$
›Reveal solutionSolution
This tests periodic trends of halogens/halogen compounds; the hypohalous acid acidity order in (C) is reversed — electronegativity of the halogen (not its size) governs $HOX$ acidity, so $HOCl$ is the strongest, not $HOI$ .

Concept and Intuition

For the hypohalous acids $HOX$ , acid strength depends on how well the halogen atom pulls electron density away from the O–H bond (inductive effect) and stabilizes the resulting $OX^-$ conjugate base. Since electronegativity decreases down the group ( $Cl>Br>I$ ), $HOCl$ withdraws electron density most strongly and is the most acidic; $HOI$ is the least acidic. This is opposite to what one might guess by analogy with the hydrohalic acids $HX$ , where acidity increases down the group (there, bond strength/size dominates, not electronegativity).

Step-by-Step Solution

(A): Bond dissociation enthalpies (kJ/mol) are $Cl_2(242) > Br_2(192) > F_2(159) > I_2(151)$ — matches the given order, so (A) is correctly matched.

(B): For hydrohalic acids, the weaker the H–X bond, the more easily $H^+$ dissociates; bond strength decreases $HF>HCl>HBr>HI$ , so acid strength increases $HI>HBr>HCl>HF$ — matches (B), correctly matched.

(C): For hypohalous acids $HOX$ , acidity depends on halogen electronegativity (inductive withdrawal from the O–H bond), which is $Cl>Br>I$ . So real order is $HOCl>HOBr>HOI$ — but option (C) states $HOI>HOBr>HOCl$ , which is exactly reversed.

(D): Oxoacids of chlorine with more oxygens delocalize the negative charge over more O atoms in the conjugate base, increasing acidity: $HClO_4>HClO_3>HClO_2>HClO$ — matches (D), correctly matched.

Only (C) is the incorrect order.

Common Mistakes

Applying the hydrohalic-acid trend (acidity increases down the group) to hypohalous acids, where the opposite trend (electronegativity-driven) applies.

Confusing "oxoacid strength with more oxygens" (D, correct) with "hypohalous acid strength down the group" (C, incorrect) — these are governed by different factors.

✓Final answer
$HOI > HOBr > HOCl$

ANSWER: C

←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.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days