Q.Arrange the following compounds in increasing order of their acidity – O2N–C6H4–COOH, HO–C6H4–COOH, H3C–O–C6H4–COOH, C6H5–COOH

Showing the 12 most recent of 25 on this concept.

1. 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.

   1. 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. …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set 56/1/11 markMCQ
   Q.Assertion (A) : The presence of $-OH$ group in phenols directs the incoming group to meta position in the ring. Reason (R) : $-OH$ group in phenols activates the aromatic ring towards electrophilic substitution reaction. (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

   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

   1. 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. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 …

   🔒 Start your 14-day free trial to see the full solution →
4. 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.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. 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, $ClCH_2COOH$.

   …

   🔒 Start your 14-day free trial to see the full solution →
6. 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.

   1. 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.

   2. 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 ($\ce{-COCH3}$).

   3. 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.

   4. Comparing electron density …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set 56/5/11 markMCQ
   Q.$CH_3CH_2CHO$ and $CH_3CH_2COOH$ can be distinguished by : (A) Sodium bicarbonate test (B) Hinsberg test (C) Iodoform test (D) Lucas test

   ›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) …

   🔒 Start your 14-day free trial to see the full solution →
8. 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).

   1. 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.

   2. 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.

   3. 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.

      Resonance structures of phenoxide ion:

      $C_6H_5O^- \leftrightarrow \text{structures with } - \text{charge on ortho/para carbons}$

      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.

   4. Connecting stability to acidity …

   🔒 Start your 14-day free trial to see the full solution →
9. 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)

   …

   🔒 Start your 14-day free trial to see the full solution →
10. 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. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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. …

    🔒 Start your 14-day free trial to see the full solution →