Carboxylic Acidic Strength

8 MCQs3 revision cards9-step worked example
Source: NCERT BiomoleculesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Carboxylic acids (RCOOH) are among the most acidic organic compounds you encounter in NEET — more acidic than phenols and far more acidic than alcohols. The high-frequency confusion: confusing the factors that increase or decrease acidity, particularly when electron-withdrawing groups (EWGs) and electron-donating groups (EDGs) are involved.

Why carboxylic acids are acidic. On ionisation, RCOOH → RCOO⁻ + H⁺. The carboxylate anion is stabilised by resonance delocalisation of the negative charge over two equivalent oxygen atoms. This equivalent resonance (both C–O bonds identical, bond order 1.5) is the key distinction from phenol, where the negative charge delocalises over the ring — less effectively.

pKa benchmark (NCERT Class 12 Chemistry Chapter 12, page 22): Acetic acid pKa ≈ 4.76. For comparison, phenol pKa ≈ 10, aliphatic alcohols pKa ≈ 16–18.

Substituent effects — the factor NEET tests most:

  • EWG at α-position (–Cl, –NO₂, –F, –CF₃): stabilise the carboxylate anion by withdrawing electron density → lower pKa → stronger acid. Multiple halogens amplify: CH₃COOH (4.76) → ClCH₂COOH (2.86) → Cl₂CHCOOH (1.29) → Cl₃CCOOH (0.65).
  • EDG at α-position (–CH₃, –C₂H₅, alkyl groups via +I effect): destabilise the anion → higher pKa → weaker acid. Formic acid (pKa 3.75) is stronger than acetic acid (4.76) because the H atom has no +I effect.
  • Distance effect: EWG influence falls off rapidly with distance from –COOH. α > β > γ.

Watch-out for NEET: When ranking substituted acids, check (1) nature of substituent (EWG vs EDG), (2) number of substituents, (3) position relative to –COOH. The distractor that ignores resonance-vs-induction balance is a common trap in phenol acidity comparison questions that also appear in this topic context.


Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The pKa of acetic acid in aqueous solution is approximately:

MCQ 2Easy RecallPractice

Carboxylic acids are more acidic than phenols primarily because:

MCQ 3Easy RecallPractice

Among formic acid, acetic acid, and propionic acid, the correct order of acid strength is:

MCQ 4Direct ApplicationPractice

Arrange the following in decreasing order of acid strength: (I) CH₃COOH, (II) ClCH₂COOH, (III) Cl₂CHCOOH, (IV) Cl₃CCOOH

MCQ 5Direct ApplicationPractice

Which of the following is the strongest acid?

MCQ 6Direct ApplicationPractice

p-Nitrobenzoic acid is a stronger acid than benzoic acid because the –NO₂ group:

MCQ 7Concept TrapPractice

Consider 3-chlorobutanoic acid and 4-chlorobutanoic acid. Which statement correctly compares their acid strengths?

MCQ 8CalculationPractice

Arrange the following in increasing order of acid strength: (I) Benzoic acid, (II) p-Methoxybenzoic acid, (III) p-Nitrobenzoic acid, (IV) p-Methylbenzoic acid

Quick recall before you leave

Worked Example

  1. 1

    Given

    Four compounds to rank by acidity in aqueous solution: - p-Cresol: phenol with para –CH₃ (+I, EDG) - Phenol: pKa ≈ 10 - p-Nitrophenol: phenol with para –NO₂ (EWG, –I and –R) - Acetic acid: pKa ≈ 4.76

  2. 2

    Required

    Decreasing order of acid strength (lowest pKa first).

  3. 3

    Concept

    Acid strength depends on conjugate base stability. For substituted phenols, EWG stabilises phenoxide (increases acidity), EDG destabilises it (decreases acidity). Carboxylic acids are inherently more acidic than phenols due to equivalent resonance.

  4. 4

    Formula

    pKa comparison: RCOOH (≈4–5) < substituted phenols with strong EWG (≈7–8) < phenol (≈10) < substituted phenols with EDG (>10).

  5. 5

    Substitution

    - Acetic acid: pKa = 4.76 - p-Nitrophenol: pKa ≈ 7.15 (–NO₂ withdraws electrons, stabilises phenoxide) - Phenol: pKa = 10.0 - p-Cresol: pKa ≈ 10.3 (–CH₃ donates electrons by +I, destabilises phenoxide slightly)

  6. 6

    Calculation

    Ranking by pKa (lower = stronger): 4.76 < 7.15 < 10.0 < 10.3

  7. 7

    Final answer

    Decreasing acid strength: **Acetic acid > p-Nitrophenol > Phenol > p-Cresol**

  8. 8

    Common trap

    Confusing the acidity ordering by assuming "all phenols are more acidic than carboxylic acids" — the opposite is true. Carboxylic acids have pKa 4–5 while even strongly EWG-substituted phenols rarely go below pKa 7. The resonance-vs-induction balance trap (from NEET pattern: phenol electrophilic substitution) applies: students sometimes rank p-nitrophenol above carboxylic acids because –NO₂ "sounds more powerful."

  9. 9

    Similar NEET-style question

    "Arrange in increasing order of acidity: (i) Ethanol, (ii) Phenol, (iii) p-Chlorophenol, (iv) Propanoic acid." (Answer: Ethanol < Phenol < p-Chlorophenol < Propanoic acid) ---

Before solving, remember these

Carboxylic acids (pKa ~4-5) more acidic than alcohols and phenols due to resonance stabilisation of carboxylate ion (delocalised over 2 O atoms). Electron-withdrawing groups increase acidity (Cl₃CCOOH << CCl₃COOH; pKa 0.7 vs CH₃COOH pKa 4.76).

-- NCERT, p. 22

Formulas

pKa of carboxylic acid

Stronger acid than phenol due to more effective resonance over two equivalent oxygens. EWG substituents (Cl, NO2) increase acidity.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • α-substituent effects strongest

pKa of phenol vs aliphatic alcohol

Phenols ~10⁶× more acidic than aliphatic alcohols due to resonance stabilisation of phenoxide ion. Electron-withdrawing substituents lower pKa further.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • Substituent effects shift values

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Category: Organic Reaction Conditions

1° alcohol: PCC/PDC → aldehyde (stops). KMnO4/K2Cr2O7 in acidic → carboxylic acid (continues). 2° alcohol: any oxidiser → ketone. 3° alcohol: not oxidised by ordinary reagents.

When it triggers

Question gives 1° alcohol oxidation with specified reagent.

How to avoid

PCC, PDC, Swern, DMP: mild → stop at aldehyde. KMnO4, K2Cr2O7, CrO3, jones: strong → carboxylic acid. Reagent choice matches desired product.

Past Year Questions

20 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2024Revised key

Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis . In the light of the above statements, choose the correct answer from the options given below:

1Both statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is correct but Statement II is false
4Statement I is incorrect but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2024Revised key

Fehling’s solution ‘A’ is

1aqueous copper sulphate
2alkaline copper sulphate
3alkaline solution of sodium potassium tartrate (Rochelle’s salt)
4aqueous sodium citrate
NTA Answer: Option 1(revised_final)
NEET 2022

Given below are two statements: Statement I: In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnCl , known as Lucas Reagent. 2 Statement II: Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent. In the light of the above statements, choose the most appropriate answer from the options given below:

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2021

Match List-I with List-II List-I List-II CO, HCI (a) (i) Hell-Volhard-Zelinsky Anhyd. AlCl/ Cu3Cl reaction O R — C — CH + (b) (ii) Gattermann-Koch 3 NaOX reaction R — CH — OH (c) (iii) Haloform reaction 2 +R′ COOH Conc. HSO 2 4 (d) R — CH COOH (iv) Esterification 2 ( II) X /Red P 2 (ii) HO 2 Choose the correct answer from the options given below.

1(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
2(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
3(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
4(a) - (i), (b) - (iv), (c) - (iii), (d) - (ii)
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Chemistry Chapter 12, p.22

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