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

CBSEMCQ· 1mImportance★★★★★

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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 25 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$Q. Assertion (A) : Phenol is less acidic than 4-methylphenol. Reason (R) : The presence of an electron releasing group in phenol makes it less acidic. Options : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
Assertion (A) is false because electron-releasing groups like methyl decrease acidity, making 4-methylphenol less acidic than phenol. Reason (R) is true as electron-releasing groups indeed decrease acidity. The correct option is (D).

Understanding the acidity of organic compounds, especially phenols, hinges on the stability of their conjugate bases. When an acid donates a proton ( $H^+$ ), it forms a conjugate base. The more stable this conjugate base is, the more readily the acid will donate its proton, and thus, the stronger the acid.

In the case of phenols, the acidity arises from the resonance stabilization of the phenoxide ion (the conjugate base). Any factor that stabilizes this negative charge on the oxygen atom will increase acidity, while any factor that destabilizes it will decrease acidity.

Electron-releasing groups (ERGs) or electron-donating groups (EDGs) push electron density towards the benzene ring. This increased electron density is then delocalized onto the oxygen atom of the phenoxide ion. By intensifying the negative charge on the oxygen, ERGs destabilize the conjugate base, making the parent phenol less acidic. Conversely, electron-withdrawing groups (EWGs) pull electron density away from the benzene ring, thereby stabilizing the negative charge on the oxygen of the phenoxide ion, making the parent phenol more acidic.

Analyze Assertion (A): "Phenol is less acidic than 4-methylphenol."

Let's consider phenol ( $C_6H_5OH$ ) and 4-methylphenol ( $CH_3-C_6H_4-OH$ ).

The acidity of these compounds depends on the stability of their respective conjugate bases: the phenoxide ion and the 4-methylphenoxide ion.

In 4-methylphenol, a methyl group ( $CH_3$ ) is present at the para position. The methyl group is an electron-releasing group (ERG) due to both its positive inductive effect ( $+I$ effect) and hyperconjugation.

This electron-releasing effect of the methyl group pushes electron density into the benzene ring. This increased electron density is then delocalized onto the oxygen atom of the 4-methylphenoxide ion.

By increasing the electron density on the already negatively charged oxygen, the methyl group destabilizes the 4-methylphenoxide ion compared to the phenoxide ion (which lacks this additional electron-donating group).

Since the conjugate base of 4-methylphenol is less stable, 4-methylphenol is less acidic than phenol.

Therefore, phenol is more acidic than 4-methylphenol.

The assertion states "Phenol is less acidic than 4-methylphenol," which is the opposite of our finding.

Thus, Assertion (A) is false.

Analyze Reason (R): "The presence of an electron releasing group in phenol makes it less acidic."

As discussed in the introductory concept, electron-releasing groups (ERGs) destabilize the conjugate base (phenoxide ion) by intensifying the negative charge on the oxygen atom.

A less stable conjugate base means the parent acid is less likely to donate a proton, hence it is less acidic.

This statement is a fundamental principle of how substituents affect the acidity of phenols.

Thus, Reason (R) is true.

Evaluate the relationship between Assertion (A) and Reason (R).

We found that Assertion (A) is false, while Reason (R) is true.

Therefore, Reason (R) cannot be the correct explanation for Assertion (A), as Assertion (A) itself is incorrect.

Watch out

Remember that electron-releasing groups decrease acidity, while electron-withdrawing groups increase acidity. A common mistake is to confuse these effects or their direction.

The correct option is the one that states Assertion (A) is false and Reason (R) is true.

✓Final answer
$Assertion (A) is false, but Reason (R) is true, so the correct option is (D) .$
CBSE 2026Set 56/1/11 markMCQ
$-OH$
›Reveal solutionSolution
The $-OH$ group in phenols is an ortho-para director (not meta), but it does activate the ring toward electrophilic substitution through resonance donation of its lone pair. Assertion is false, Reason is true → (D).

Understanding Directing Effects in Electrophilic Aromatic Substitution

When a substituent is already present on a benzene ring, it controls two things: where the next electrophile attacks (orientation) and how fast the reaction proceeds (reactivity). The hydroxyl group in phenols is one of the most instructive examples because it showcases the interplay between resonance and inductive effects.

Why the $-OH$ Group Activates the Ring

The oxygen in $-OH$ carries two lone pairs. Through resonance, one of these lone pairs delocalizes into the aromatic $\pi$ -system:
$\ce{Ph-OH <-> Ph^{+}=O^{-}}$
This resonance pushes electron density into the ring, making it more nucleophilic and thus more reactive toward electrophiles (which are electron-seeking species). The ring becomes "electron-rich" compared to benzene itself, so electrophilic substitution happens faster. This is what we mean by activation.

Important

Activating groups increase the electron density of the aromatic ring, making it more susceptible to attack by electrophiles.

Where Does the Electrophile Attack? Ortho and Para Positions

Now let's see where this extra electron density concentrates. When we draw the resonance structures of phenol, the negative charge (representing excess electron density) appears at the ortho and para positions:

Resonance Structure Negative Charge Location
Structure I ortho (C-2)
Structure II para (C-4)
Structure III ortho (C-6)

The meta positions (C-3 and C-5) never carry the negative charge in any resonance form. This means the ortho and para carbons are electron-rich and preferentially attacked by electrophiles.

Watch out

A common mistake is to confuse the inductive effect (electron-withdrawing through $\sigma$ -bonds due to oxygen's electronegativity) with the dominant resonance effect (electron-donating through $\pi$ -overlap). For $-OH$ , resonance wins, making it an activating, ortho-para director.

Step-by-Step Analysis

Evaluate the Reason (R): The statement " $-OH$ group in phenols activates the aromatic ring towards electrophilic substitution reaction" is true. The lone pair on oxygen donates electron density via resonance, increasing the nucleophilicity of the ring.

Evaluate the Assertion (A): The statement "The presence of $-OH$ group in phenols directs the incoming group to meta position in the ring" is false. As shown by resonance, the $-OH$ group directs incoming electrophiles to the ortho and para positions, not meta.

Check the relationship: Since the Assertion is false, we don't need to evaluate whether R explains A. The only option where A is false but R is true is (D).

Tip

A quick mnemonic: electron-donating groups (by resonance) are ortho-para directors and activating; electron-withdrawing groups (by resonance, like $-NO_2$ , $-CN$ ) are meta directors and deactivating. The exception is halogens (weakly deactivating but ortho-para directing due to lone-pair resonance overcoming inductive withdrawal).

Experimental Confirmation

When phenol undergoes nitration, bromination, or sulfonation, the major products are always ortho- and para-substituted phenols, never the meta isomer as the primary product. For instance, phenol reacts with bromine water at room temperature to give 2,4,6-tribromophenol (all ortho/para positions substituted).

✓Final answer
$-OH$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Weakest acid among the following is$
(a) HCOOH
(b) CH3COOH
(c) FCH2COOH
(d) ClCH2COOH
›Reveal solutionSolution
Electron-withdrawing groups (like halogens) near the -COOH group stabilise the conjugate base and increase acid strength; electron-donating groups (like -CH3) do the opposite and decrease acid strength.

HCOOH (formic acid, no alkyl group): pKa ≈ 3.75

CH3COOH (acetic acid, +I methyl group): pKa ≈ 4.76 - weakest here

FCH2COOH (F is strongly electron-withdrawing, -I effect): pKa ≈ 2.66 - strongest

ClCH2COOH (Cl also -I, slightly weaker than F): pKa ≈ 2.86

So the order of acid strength is CH3COOH < HCOOH < ClCH2COOH < FCH2COOH, making acetic acid the weakest.

✓Final answer
$(b) CH3COOH.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which is most acidic?$
(a) CF3COOH
(b) CCl3COOH
(c) CBr3COOH
(d) CH3COOH
›Reveal solutionSolution

$CF_3COOH$ is the most acidic because fluorine is the most electronegative halogen and exerts the strongest electron-withdrawing (-I) effect, best stabilising the carboxylate anion.

The acidity of a substituted acetic acid, $CX_3COOH$ , increases with the electron-withdrawing (-I) power of the halogen $X$ , because a stronger -I effect stabilises the negative charge on the conjugate base $CX_3COO^-$ more effectively (inductive electron withdrawal, since the halogens are attached to the alkyl carbon, not the carboxyl carbon directly). Electronegativity order: $F > Cl > Br > I$ ; the observed acidity order for these trihalomethyl acids is $CF_3COOH > CCl_3COOH > CBr_3COOH \gg CH_3COOH$ . $CH_3COOH$ (plain acetic acid, no halogens) is the weakest since $-CH_3$ is electron-donating, not withdrawing.

✓Final answer
$CF_3COOH$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Ethanoic acid (pKa = 4.76) on treatment with Cl₂ / red phosphorus gives a derivative of ethanoic acid. The ethanoic acid derivative will have -$
(a) pKa > 4.76
(b) pKa < 4.76
(c) pKa = 4.76
(d) pKa = 0.00
›Reveal solutionSolution

α-Chlorination (HVZ) gives chloroacetic acid; the –I effect of Cl stabilises the conjugate base, increasing acid strength, so pKa < 4.76 — option (B).

Ethanoic acid ( $CH_3COOH$ ) treated with $Cl_2$ /red phosphorus undergoes the Hell–Volhard–Zelinsky (HVZ) reaction, substituting a hydrogen on the α-carbon by chlorine to give chloroacetic acid, $ClCH_2COOH$ .

The chlorine atom is strongly electron-withdrawing (−I effect). It disperses/stabilises the negative charge of the carboxylate ion ( $ClCH_2COO^-$ ) formed after the acid loses $H^+$ . A more stable conjugate base means the acid ionises more readily — chloroacetic acid is a stronger acid than acetic acid.

Since a stronger acid has a lower $pK_a$ ( $pK_a = -\log K_a$ ; larger $K_a$ ⇒ smaller $pK_a$ ), the derivative has $pK_a < 4.76$ (chloroacetic acid $pK_a \approx 2.86$ ).

✓Final answer
$(B) pKa < 4.76 — the chloro derivative is a stronger acid than ethanoic acid.$
CBSE 2025Set 56/4/11 markMCQ
$Q. For the following question, two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) given below. Assertion (A) : Acetanilide is less basic than aniline. Reason (R) : Acetylation of aniline results in decrease of electron density on nitrogen. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
Acetanilide is less basic than aniline because the acetyl group withdraws electron density from the nitrogen via resonance, reducing its ability to donate a lone pair. The reason given is correct and directly explains the assertion.

Understanding basicity in amines

Basicity of an amine depends on how readily the nitrogen atom can donate its lone pair of electrons to a proton (or a Lewis acid). Anything that increases electron density on nitrogen makes it more basic; anything that decreases electron density makes it less basic.

Structure of aniline vs. acetanilide

Aniline is $\ce{C6H5NH2}$ — the nitrogen lone pair is partially delocalised into the benzene ring, which already reduces its basicity compared to aliphatic amines.

Acetanilide is $\ce{C6H5NHCOCH3}$ — here the nitrogen is attached to an acetyl group ( $\ce{-COCH3}$ ).

The key effect: resonance in acetanilide

The acetyl group contains a carbonyl ( $\ce{C=O}$ ). The nitrogen lone pair can participate in resonance with the carbonyl $\pi$ system, forming a structure like:

$\ce{C6H5-NH-C(=O)CH3 \longleftrightarrow C6H5-NH+=C(-O-)CH3}$
This resonance delocalises the nitrogen lone pair onto the oxygen, drastically reducing electron density on nitrogen.

Comparing electron density

In aniline, the lone pair is delocalised only into the benzene ring (moderate reduction). In acetanilide, the lone pair is delocalised into both the ring and the carbonyl group (much stronger reduction).

Hence, the nitrogen in acetanilide is far less able to donate its lone pair — it is less basic.

Evaluating the Reason

The Reason states: “Acetylation of aniline results in decrease of electron density on nitrogen.” This is exactly the resonance-based argument above — it is true and it is the direct cause of the lower basicity.

Conclusion

Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

✓Final answer
$The correct option is (A) — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).$
CBSE 2025Set 56/5/11 markMCQ
$CH_3CH_2CHO$
›Reveal solutionSolution
Aldehydes and carboxylic acids differ fundamentally in acidity: only the carboxylic acid will react with sodium bicarbonate (a weak base) to liberate $CO_2$ gas, giving a clear visual distinction.

The question asks us to distinguish between an aldehyde ( $CH_3CH_2CHO$ , propanal) and a carboxylic acid ( $CH_3CH_2COOH$ , propanoic acid). The key lies in recognizing their different functional groups and the chemical properties that flow from them.

Carboxylic acids are acidic enough to react with weak bases like sodium bicarbonate, while aldehydes are not acidic at all in the Brønsted sense. This difference in acidity is the most straightforward way to tell them apart.

Let me walk through each option:

1. Sodium bicarbonate test

Carboxylic acids contain the $-COOH$ group, which readily donates a proton. The $pK_a$ of propanoic acid is around 4.9, making it acidic enough to react with sodium bicarbonate ( $NaHCO_3$ ):
$CH_3CH_2COOH + NaHCO_3 \longrightarrow CH_3CH_2COONa + H_2O + CO_2 \uparrow$
The evolution of $CO_2$ gas produces effervescence (brisk bubbling), a clear positive test.

Propanal, on the other hand, has no acidic proton. The aldehyde hydrogen is not ionizable, and the $\alpha$ -hydrogens are far too weakly acidic ( $pK_a \sim 17$ ) to react with bicarbonate. No reaction occurs, no gas is evolved.

This test cleanly distinguishes the two compounds.

2. Hinsberg test

This test is specific for distinguishing primary, secondary, and tertiary amines using benzenesulfonyl chloride. Neither an aldehyde nor a carboxylic acid will give a meaningful Hinsberg reaction. This test is irrelevant here.

3. Iodoform test

Both compounds contain the $CH_3CH_2-$ group. The iodoform test is positive for:

Methyl ketones ( $RCOCH_3$ )

Compounds with $CH_3CH(OH)-$ structure

Ethanol and acetaldehyde (special cases)

Neither propanal ( $CH_3CH_2CHO$ ) nor propanoic acid ( $CH_3CH_2COOH$ ) has a methyl ketone or secondary alcohol structure. Both would give a negative iodoform test, so this cannot distinguish them.

Watch out

A common mistake is thinking all aldehydes give the iodoform test. Only acetaldehyde ( $CH_3CHO$ ) does, because it has the required $CH_3CO-$ structure after oxidation. Propanal lacks this.

4. Lucas test

The Lucas test (conc. $HCl$ + anhydrous $ZnCl_2$ ) is used to distinguish primary, secondary, and tertiary alcohols based on the rate of formation of alkyl chlorides. Neither an aldehyde nor a carboxylic acid is an alcohol, so this test is not applicable.

✓Final answer
$The correct option is (A) — sodium bicarbonate test distinguishes the carboxylic acid (effervescence) from the aldehyde (no reaction).$
CBSE 2025Set 56/6/11 markMCQ
$Q. For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Phenol is strongly acidic as compared to ethanol. Reason (R) : Phenoxide ion is more stable than ethoxide ion.$
›Reveal solutionSolution
The key idea is that the phenoxide ion is resonance-stabilised, making phenol a stronger acid than ethanol. The Assertion is true, the Reason is true, and the Reason correctly explains the Assertion — so the answer is (A).

Understanding the Assertion: Acidity of Phenol vs Ethanol

Acidity is the ability to donate a proton ( $H^+$ ). Phenol ( $C_6H_5OH$ ) is indeed a stronger acid than ethanol ( $CH_3CH_2OH$ ). The $pK_a$ of phenol is about 10, while that of ethanol is about 16 — a difference of six orders of magnitude. So the Assertion is true.

Understanding the Reason: Stability of the Conjugate Base

The strength of an acid depends on the stability of its conjugate base after losing $H^+$ . For phenol, the conjugate base is the phenoxide ion ( $C_6H_5O^-$ ). For ethanol, it is the ethoxide ion ( $CH_3CH_2O^-$ ). The Reason claims that phenoxide is more stable than ethoxide — this is true.

Why is phenoxide more stable? The role of resonance

In the phenoxide ion, the negative charge on oxygen can be delocalised into the aromatic ring through resonance. The lone pair on oxygen interacts with the $\pi$ -electron system of the benzene ring, spreading the negative charge over the ortho and para positions. This delocalisation lowers the energy of the ion, making it more stable.
$C_6H_5O^- \leftrightarrow \text{structures with } - \text{charge on ortho/para carbons}$
In contrast, the ethoxide ion has no such resonance — the negative charge is localised entirely on the oxygen atom. This makes ethoxide a high-energy, less stable species.

Connecting stability to acidity

A more stable conjugate base means the acid dissociates more readily. Since phenoxide is stabilised by resonance, phenol loses its proton more easily than ethanol does. Therefore, the Reason directly explains the Assertion.

Watch out

A common mistake is to think that the inductive effect of the benzene ring (electron-withdrawing) is the main reason. While the inductive effect does play a small role, the dominant factor here is resonance stabilisation of the phenoxide ion. The inductive effect alone cannot account for the huge $pK_a$ difference.

Evaluating the options

Both Assertion and Reason are true.

The Reason correctly explains why phenol is more acidic than ethanol.

Hence, the correct choice is (A).

✓Final answer
$The correct option is (A) — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Strongest acid among the following is -$
(a) FCH2COOH
(b) ClCH2COOH
(c) BrCH2COOH
(d) CH3COOH
›Reveal solutionSolution

Acid strength of halo-acetic acids XCH2COOH increases with the electronegativity of the halogen X, because a more electronegative halogen withdraws electron density more strongly (-I effect), stabilising the conjugate base (carboxylate anion) better.

Electronegativity order: F > Cl > Br (I not in options)

So the -I effect (and hence acid strength) order is:

FCH2COOH > ClCH2COOH > BrCH2COOH > CH3COOH (no halogen, weakest, since CH3 is even electron-donating by comparison)

FCH2COOH has fluorine, the most electronegative substituent among the choices, pulling electron density away from the carboxylate oxygen most strongly and best stabilising the resulting anion (F-CH2-COO-) after loss of H+. This makes it the strongest acid of the four.

✓Final answer
$(a) FCH2COOH.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the relation between acidic strength of A and B? A and B are shown as two substituted benzoic acids.$
(a) A = B
(b) A > B
(c) A < B
(d) A >> B
›Reveal solutionSolution
Comparing acid strength of substituted benzoic acids depends on whether the substituent stabilises (EWG, increases acidity) or destabilises (EDG, decreases acidity) the carboxylate anion; here OCH3 is a net electron donor and NO2 is a strong electron acceptor.

A carries a -OCH3 group at the para position and B carries a -NO2 group at the para position, both relative to -COOH.

-OCH3 (compound A): although inductively electron-withdrawing at short range, at the para position its dominant effect is resonance electron-donation into the ring (+M), which pushes electron density towards the carboxylate, destabilising the conjugate base. This makes p-methoxybenzoic acid a weaker acid than benzoic acid.

-NO2 (compound B): a powerful electron-withdrawing group both inductively and by resonance (-M); it stabilises the carboxylate anion strongly, making p-nitrobenzoic acid a much stronger acid than benzoic acid.

So acid strength order is A (weaker, electron-donating substituent) < B (stronger, electron-withdrawing substituent).

✓Final answer
$(c) A < B.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. In the following, strongest Acid is:$
(a) CH₃CH₂COOH
(b) CH₃COOH
(c) C₆H₅COOH
(d) C₆H₅CH₂COOH
›Reveal solutionSolution
Among the four acids given, benzoic acid (C₆H₅COOH) is the strongest because the –COOH group is attached directly to the electron-withdrawing benzene ring.

Acid strength of a carboxylic acid depends on how well the conjugate base (carboxylate ion) is stabilised.

$CH_3CH_2COOH$ (propanoic acid) and $CH_3COOH$ (acetic acid) — alkyl groups are electron-donating (+I effect), which destabilises the carboxylate anion, making these comparatively weaker acids (propanoic acid, with an extra +I-donating CH₂, is even weaker than acetic acid).

$C_6H_5CH_2COOH$ (phenylacetic acid) — the phenyl ring is one CH₂ away from –COOH, so its electron-withdrawing effect on the carboxyl group is weak.

$C_6H_5COOH$ (benzoic acid) — here the ring is directly attached to –COOH; its electron-withdrawing inductive effect (and resonance stabilisation of the carboxylate through the ring) stabilises the conjugate base most effectively, making benzoic acid the strongest acid of the four (lowest pKa, ≈ 4.2).

✓Final answer
$(c) C₆H₅COOH — benzoic acid is the strongest acid.$
CBSE 2025Set ANNUAL1 mark
$Q. Arrange the following acids in decreasing order of their acidic strength: CHCl2COOH, CHI2COOH, CHF2COOH, CHBr2COOH$
›Reveal solutionSolution
Acidity of a haloacetic acid rises with the electronegativity of the halogen, because a more electronegative halogen exerts a stronger -I (electron-withdrawing inductive) effect, which stabilizes the conjugate-base carboxylate anion more.

All four acids have the same skeleton (CHX2COOH) differing only in the halogen X, so the comparison is purely about each halogen's inductive (-I) effect on the -COOH group.

A stronger -I effect pulls electron density away from the O-H bond and helps disperse the negative charge on the carboxylate ion (CHX2COO⁻) formed after ionisation, making that ion more stable and the acid stronger.

The -I effect of halogens follows their electronegativity order: F > Cl > Br > I.

Therefore acid strength follows the same order: CHF2COOH is the strongest and CHI2COOH is the weakest among these four.

So, in decreasing order of acidic strength:

CHF2COOH > CHCl2COOH > CHBr2COOH > CHI2COOH

✓Final answer
$CHF2COOH > CHCl2COOH > CHBr2COOH > CHI2COOH (acidity decreases as the halogen's electronegativity, and hence its -I effect, decreases from F to I).$

Resonance Structure	Negative Charge Location
Structure I	ortho (C-2)
Structure II	para (C-4)
Structure III	ortho (C-6)

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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

Understanding Directing Effects in Electrophilic Aromatic Substitution

Why the $-OH$ Group Activates the Ring

Where Does the Electrophile Attack? Ortho and Para Positions

Step-by-Step Analysis

Experimental Confirmation

Q.Identify the compound Y in the following reaction.

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