Strength of Acids and Bases Explained: pKa, Trends & Factors Affecting Acidity For NEET, IIT-JEE, IIT-JAM, BITSAT, TGT, PGT Exam Preparation

Chapter 3 · Peter Sykes Series

The Strengths of Acids and Bases
A Complete Guide for Competitive Exams

Concepts, Trends, Structures and Exam Strategies — Explained from First Principles

JEE Main & Advanced NEET IIT-JAM CSIR-NET

Strength of Acids and Bases Explained: pKa, Trends & Factors Affecting Acidity

Acid–Base Definitions: From Arrhenius to Lewis

Three major frameworks describe acid–base behaviour, and competitive exams test all three. Understanding why each exists — and its limitations — is far more powerful than memorising definitions.

1.1 Arrhenius Definition (Limited but Foundational)

  • Acid: A substance that releases H⁺ (hydrogen ions) in aqueous solution. Example: HCl → H⁺ + Cl⁻
  • Base: A substance that releases OH⁻ ions in aqueous solution. Example: NaOH → Na⁺ + OH⁻
  • Limitation: Only works in water; cannot explain NH₃ as a base (no OH⁻ released).

1.2 Brønsted–Lowry Definition (Most Used in Organic Chemistry)

Brønsted defined an acid as a proton donor and a base as a proton acceptor. This elegantly explains the concept of conjugate pairs.

H₂SO₄ Acid + H₂O Base H₃O⁺ Conj. Acid + HSO₄⁻ Conj. Base Stronger acid + Stronger base → Weaker conj. acid + Weaker conj. base

Fig 1.1 — First ionisation of H₂SO₄ showing conjugate acid–base pairs (Brønsted–Lowry)

1.3 Lewis Definition (Most General)

  • Lewis Acid: Electron-pair acceptor. Examples: BF₃, AlCl₃, SnCl₄, ZnCl₂, FeCl₃
  • Lewis Base: Electron-pair donor. Examples: NH₃, amines, ethers, water
Me₃N: Lewis Base + BF₃ Lewis Acid Me₃N⁺–BF₃⁻ Adduct (m.p. 128°C)

Fig 1.2 — Lewis acid–base adduct formation: BF₃ (acid) + N(CH₃)₃ (base)

⚡ JEE/NEET Exam Point

Lewis acids must have empty orbitals to accept electrons. BF₃ (empty 2p on B), AlCl₃ (empty 3p on Al), carbocations (empty orbital), transition metal ions — all are Lewis acids. NH₃, H₂O, ROH, R₂S, CN⁻ are classic Lewis bases. This comes up in Friedel–Crafts reactions and coordination chemistry questions.

Understanding pKₐ — The Master Acidity Scale

For an acid HA dissociating in water, the equilibrium is:

Dissociation Equilibrium HA + H₂O ⇌ H₃O⁺ + A⁻

Kₐ = [H₃O⁺][A⁻] / [HA]

pKₐ = –log₁₀(Kₐ)

Relation to free energy: –ΔG° = 2.303 RT log Kₐ
Also: ΔG° = ΔH° – TΔS°

The Golden Rule of pKₐ

  • Lower pKₐ = Stronger Acid (more dissociated, larger Kₐ)
  • Higher pKₐ = Weaker Acid
  • Acids with pKₐ > ~16 cannot be detected as acids in water (they produce less H₃O⁺ than water's own autolysis)
  • Very strong acids (HCl, HNO₃, HClO₄) all appear equally strong in water — this is the levelling effect of water
–1 3 5 10 16 43 HClO₄ HCOOH CH₃COOH PhOH H₂O / MeOH CH₄ ◄ STRONGER ACID WEAKER ACID ►

Fig 2.1 — Relative pKₐ scale for common organic and inorganic acids

💡 Key Concept — Entropy Matters Too!

Many students assume Ka depends only on bond energy (ΔH°). But the data reveals something remarkable: for ethanoic acid, ΔH° ≈ –0.5 kJ/mol (almost zero!) while TΔS° = –27.6 kJ/mol dominates ΔG°. The real driver of acid strength differences in carboxylic acids and haloacids is entropy of solvation, not bond enthalpy. This is a favourite IIT-JAM concept.

Why Organic Compounds Are Acidic — Four Governing Factors

The acidity of any organic compound HA is controlled by four factors working together. Understanding each separately — then in combination — is the key to predicting relative acidities without memorisation.

  1. Bond strength of H–A: Rarely the deciding factor among organic compounds of the same family
  2. Electronegativity of A: Higher electronegativity → stronger acid (better at retaining the negative charge after proton loss)
  3. Stabilisation of conjugate base A⁻ vs HA: This is the MOST IMPORTANT factor in organic chemistry
  4. Nature of solvent: Solvation can override structural factors entirely

Resonance Stabilisation of Carboxylate Anion — The Key Principle

Formic acid (methanoic acid, pKₐ = 3.77) is a relatively strong organic acid because its conjugate base, the formate ion, is stabilised by complete charge delocalisation over two equivalent oxygen atoms.

H–C O O⁻ Structure A H–C O⁻ O Structure B Both structures have EQUAL energy → Max. Stabilisation

Fig 3.1 — Resonance stabilisation of formate (HCOO⁻) anion. Both canonical structures are equivalent, giving maximum delocalisation and hence maximum stability.

Why Alcohols Are Far Less Acidic Than Carboxylic Acids

In alkoxide ions (RO⁻), the negative charge sits entirely on one oxygen atom — no delocalisation is possible. So the alkoxide is not significantly more stable than the parent alcohol, and the equilibrium strongly disfavours ionisation.

CH₄pKₐ ≈ 43
CH₃OHpKₐ ≈ 16
C₆H₅OHpKₐ = 9.95
HCOOHpKₐ = 3.77
← Increasing Acidity
⚠️ Common Misconception

Students often think "electronegativity of O explains carboxylic acid strength." Wrong order of importance! The resonance stabilisation of the carboxylate anion is the dominant effect. Both carboxylic acids and phenols have an O–H bond, but carboxylic acids are ~10,000× stronger because their anion has two equivalent O atoms sharing the charge, while phenoxide distributes charge onto less electronegative carbon atoms of the ring.

The Role of the Solvent — Water as the Ionising Medium

Despite all the structural reasoning above, the solvent is often the dominant factor determining actual acid strength. Water is uniquely effective because of two properties:

  • High dielectric constant (ε = 80): Reduces the electrostatic energy holding ion pairs together — ions stabilise more easily in high-ε solvents
  • Ion-solvating ability: Water molecules wrap around ions, delocalising charge via hydrogen bonding (especially powerful for anions)

Hydrogen-Bonded Solvation of H⁺

H₃O⁺ H₂O H₂O H₂O H₂O --- = H-bond

Fig 4.1 — Solvation shell around H₃O⁺ via hydrogen bonding with surrounding water molecules

💡 Levelling Effect — Important for JEE Advanced

When a very strong acid dissolves in water, it ionises completely regardless of its intrinsic strength. HCl, HBr, HI, HNO₃, HClO₄ all appear equally strong in water. To distinguish their true strengths, one must use a weaker base solvent (like glacial acetic acid) where they don't fully ionise.

Simple Aliphatic Acids — Alkyl Groups and Acidity

Replacing the H in formic acid by an alkyl group introduces an electron-donating inductive effect (+I effect), which pushes electrons toward the carboxylate, destabilising the anion and reducing acidity.

AcidStructurepKₐKey Feature
Methanoic (Formic)HCOOH3.77Reference — strongest simple acid
Ethanoic (Acetic)CH₃COOH4.76–CH₃ donates electrons (+I)
PropanoicCH₃CH₂COOH4.88Slightly weaker — longer chain
2-Methylpropanoic(CH₃)₂CHCOOH4.86Branching, minor effect
2,2-Dimethylpropanoic(CH₃)₃CCOOH5.05Bulky, but mainly steric/solvation
Propenoic (Acrylic)CH₂=CHCOOH4.25sp² carbon — weaker +I than sp³
Propynoic (Propiolic)HC≡CCOOH1.84sp carbon — strong –I, very acidic

Why Does sp Hybridisation Increase Acidity So Dramatically?

The hybridisation state of the carbon adjacent to –COOH determines how strongly it pulls electrons away from (or donates electrons to) the carboxyl group.

CH₃CH₂COOH sp³ — pKₐ 4.88 25% s-character More acidic CH₂=CHCOOH sp² — pKₐ 4.25 33% s-character More acidic HC≡CCOOH sp — pKₐ 1.84 50% s-character

Fig 5.1 — Increasing s-character in hybrid orbitals causes stronger electron withdrawal (–I effect), increasing acidity. sp has 50% s-character vs 25% for sp³.

Substituted Aliphatic Acids — The Power of Electron-Withdrawing Groups

Introducing electron-withdrawing groups (EWG) — especially halogens — near the carboxyl dramatically increases acidity by stabilising the carboxylate anion through spreading negative charge.

Halogen Substituted Acetic Acids

CompoundpKₐRelative Strength vs CH₃COOH
CH₃COOH (Acetic acid)4.76Reference (1×)
FCH₂COOH (Fluoroacetic)2.57~150× stronger
ClCH₂COOH (Chloroacetic)2.86~80× stronger
BrCH₂COOH (Bromoacetic)2.90~70× stronger
ICH₂COOH (Iodoacetic)3.16~40× stronger
Cl₂CHCOOH (Dichloroacetic)1.25~3200× stronger
Cl₃CCOOH (Trichloroacetic)0.65~13,000× stronger
CF₃COOH (Trifluoroacetic)0.23~340,000× stronger

Why Does the Effect Diminish with Distance?

The inductive effect operates through bonds and decreases rapidly (attenuates) as the number of intervening carbon atoms increases.

Distance from COOH : ClCH₂COOH pKₐ = 2.86 α-position ClCH₂CH₂COOH pKₐ = 4.52 β-position CH₃CHClCH₂COOH pKₐ = 4.06 γ-position MeCH₂CH₂COOH pKₐ = 4.82 Unsubstituted ref.

Fig 6.1 — Inductive effect of Cl diminishes rapidly with distance from the –COOH group

⚡ JEE Trick — Ordering Haloacids

For α-halo acids: F > Cl > Br > I in terms of acid-strengthening (matches electronegativity order). For same halogen, more halogens = much stronger acid. For position effect: α > β > γ ≫ unsubstituted. These two rules together can answer most ordering questions in 10 seconds.

Phenols — Where Resonance and Induction Battle

Phenol (pKₐ = 9.95) sits between alcohols (pKₐ ~16) and carboxylic acids (pKₐ ~4–5), because the phenoxide anion is partially stabilised by delocalisation into the ring — but less effectively than carboxylate.

Effect of Substituents on Phenol Acidity

Electron-withdrawing groups (–NO₂) increase acidity; electron-donating groups (–CH₃) decrease acidity slightly. Position matters enormously for mesomeric effects.

PhenolpKₐReason
Phenol (C₆H₅OH)9.95Reference
o-Nitrophenol7.23–I and –M (o-position) both withdraw electrons
m-Nitrophenol8.35Only –I effect; no mesomeric effect at meta
p-Nitrophenol7.14Both –I and –M; strongest withdrawing at para
2,4-(NO₂)₂C₆H₃OH4.01Two nitro groups — very large effect
2,4,6-(NO₂)₃C₆H₂OH (Picric acid)1.02Mineral acid strength!
o-Methylphenol10.28+I effect — slight decrease in acidity
p-Methylphenol10.19+I effect; +M actually destabilises anion
⚡ NEET/JEE Trick — o vs p vs m Nitro Effect

For nitrophenols: p-NO₂ < o-NO₂ < m-NO₂ in terms of pKₐ (both o and p are more acidic than m because they get both inductive AND mesomeric electron withdrawal). Note: o-nitrophenol can form intramolecular H-bonds, making it less soluble in water but stronger as an acid. For o compounds always look for intramolecular H-bonding!

Aromatic Carboxylic Acids — Benzoic Acid and Derivatives

Benzoic acid (pKₐ = 4.20) is stronger than cyclohexane carboxylic acid (pKₐ = 4.87), confirming that the sp² carbon of the ring is less electron-donating than a saturated sp³ carbon. The same logic as acrylic > propionic acid (Section 5).

Effect of Substituents on Benzoic Acid

CompoundpKₐEffect
Benzoic acid (C₆H₅COOH)4.20Reference
o-Nitrobenzoic acid2.17Very strong: –I (short distance) + direct interaction
m-Nitrobenzoic acid3.45–I only at meta
p-Nitrobenzoic acid3.43–I + –M at para
p-Methoxybenzoic acid4.47+M (OMe donates at para) dominates –I
p-Hydroxybenzoic acid4.58+M from OH weakens acid — weaker than benzoic!
o-Hydroxybenzoic acid (Salicylic)2.98Intramolecular H-bond stabilises anion
2,6-Dihydroxybenzoic acid1.30H-bonding from both ortho-OH groups
⚠️ The Salicylic Acid Exception

o-Hydroxybenzoic acid (salicylic acid) is much more acidic than m- or p-isomers because, after ionisation, the carboxylate anion is stabilised by intramolecular hydrogen bonding with the adjacent –OH group. The m- and p-isomers cannot do this. This is a classic JEE question on intramolecular vs intermolecular H-bonding effects.

Dicarboxylic Acids — A Tale of Two pKₐ Values

Every dicarboxylic acid has two ionisation steps: pKₐ₁ (first COOH loss) and pKₐ₂ (second COOH loss). The second step is always harder because you are removing H⁺ from an already-negative species.

AcidpKₐ₁pKₐ₂Reason for large difference
Ethanedioic (Oxalic) HOOC-COOH1.234.19Very close COOH groups: strong EWG mutual effect
Propanedioic (Malonic) HOOCCH₂COOH2.835.691 carbon between — still strong
Butanedioic (Succinic) HOOC(CH₂)₂COOH4.195.482 carbons — inductive effect reduced
Maleic acid (cis)1.926.23Intramolecular H-bond stabilises cis mono-anion
Fumaric acid (trans)3.024.38No intramolecular H-bond possible

Maleic vs Fumaric — Geometric Isomers, Radically Different Acidity

Maleate mono-anion (pKₐ₁=1.92) H H O O⁻ O OH Intramolecular H-bond → Anion stabilised Fumaric acid (pKₐ₁=3.02) H H HOOC COOH Groups on opposite sides — no H-bond → less stable

Fig 9.1 — Maleic acid (cis) mono-anion is stabilised by intramolecular H-bonding, making it far more acidic (pKₐ₁ = 1.92) than fumaric acid (trans, pKₐ₁ = 3.02)


Bases — The pKₐ of the Conjugate Acid

Modern practice measures base strength using the pKₐ of the conjugate acid (BH⁺), rather than the older pKb. This creates one unified scale for both acids and bases.

Base Strength via Conjugate Acid BH⁺ + H₂O ⇌ B: + H₃O⁺

Kₐ(BH⁺) = [B:][H₃O⁺] / [BH⁺]

Rule: Larger pKₐ(BH⁺) = Stronger base (BH⁺ less willing to lose H⁺ → B: more willing to hold it)
💡 Key Relationship

pKₐ(HA) + pKₐ(BH⁺) at 25°C: a stronger conjugate acid (lower pKₐ of BH⁺) means a weaker base. Remember: pKa of NH₄⁺ = 9.25, so NH₃ is the reference aliphatic base.

Aliphatic Amines — The Solvation Paradox

You might predict basicity order: NH₃ < RNH₂ < R₂NH < R₃N (more alkyl groups → more electron density on N → stronger base). Reality in water is more complex due to solvation.

Observed pKₐ Values (Conjugate Acid)

NH₃pKₐ = 9.25
MeNH₂pKₐ = 10.64
Me₂NHpKₐ = 10.77
Me₃NpKₐ = 9.80

The tertiary amine drops back! The two competing effects are:

  • Inductive effect (+I): More alkyl groups → more electrons on N → better proton acceptor → increases basicity
  • Solvation effect: More alkyl groups → fewer N–H bonds in cation BH⁺ → weaker hydrogen bonding with water → less stabilisation of cation → decreases basicity
RNH₃⁺ 3 N–H bonds Strong solvation R₂NH₂⁺ 2 N–H bonds Moderate solvation R₃NH⁺ 1 N–H bond Weak solvation Decreasing stabilisation →

Fig 11.1 — Decreasing H-bond solvation of ammonium ions as N–H bonds are replaced by N–C bonds

⚡ IIT-JAM Favourite — Guanidine

Guanidine HN=C(NH₂)₂ has pKₐ ≈ 13.6 — one of the strongest organic bases. Why? On protonation, the resulting cation H₂N⁺=C(NH₂)₂ has its positive charge symmetrically delocalised over three equivalent nitrogen atoms via three equivalent resonance structures. This exceptional stabilisation of the cation makes protonation highly favourable. Compare: the neutral molecule's resonance involves charge separation, so it's less stabilised.

Amides — Extremely Weak Bases

Amides (RCONH₂) have pKₐ ≈ 0.5 — vastly weaker than amines (pKₐ ~10). The C=O group withdraws the N lone pair via resonance (–M effect), making it unavailable for protonation.

Amide Resonance — N lone pair is delocalised R–C(=O)–NH₂ ↔ R–C(–O⁻)=NH₂⁺

The N lone pair is tied up in π system → NOT available for H⁺

Aromatic Bases — Aniline and the Resonance Trap

Aniline (pKₐ = 4.62) is a dramatically weaker base than cyclohexylamine (pKₐ = 10.68) or ammonia (pKₐ = 9.25). The reason is elegant and fundamental.

In aniline, the nitrogen lone pair is delocalised into the benzene ring via resonance. This stabilises the free amine. When aniline is protonated, this delocalisation is destroyed — the lone pair is consumed by H⁺. So protonation is energetically unfavourable.

Substituent Effects on Aniline Basicity

Aniline derivativepKₐEffect
Aniline (PhNH₂)4.62Reference
p-Nitroaniline0.98–M + –I of NO₂ at para; further stabilises free amine
m-Nitroaniline2.45–I only (no mesomeric effect at meta)
o-Nitroaniline–0.28–M + –I + steric; conjugate acid unstable
p-Methoxyaniline5.29+M from OMe increases electron density on N
p-Methylaniline5.10+I from CH₃ — slightly more basic
Diphenylamine (Ph₂NH)0.8Two rings pulling lone pair away
Triphenylamine (Ph₃N)~–5Not basic by ordinary standards

Heterocyclic Bases — Pyridine and Pyrrole Contrasted

Pyridine — sp² N, Lone Pair in Plane

In pyridine, nitrogen is sp² hybridised. One electron goes into the π system (making it aromatic). The lone pair occupies an sp² orbital in the plane of the ring — it does NOT contribute to aromaticity, so it IS available for protonation. However, sp² orbitals have more s-character than sp³, pulling the lone pair closer to the nucleus → weaker base than aliphatic amines.

Et₃Nsp³ N, pKₐ=10.75
Pyridinesp² N, pKₐ=5.21
MeCNsp N, pKₐ=–4.3

Pyrrole — sp² N, Lone Pair in π System

Pyrrole's nitrogen contributes its lone pair to complete the 6π aromatic system (4n+2 with n=1). This lone pair is fully delocalised and NOT available for protonation. If forced to protonate, pyrrole does so at the α-carbon (not nitrogen), and loses its aromaticity — energetically very costly. Hence pyrrole is actually an extremely weak base (pKₐ = –0.27), but can act as a weak acid (pKₐ ~17) since the pyrrolide anion retains aromaticity!

PYRIDINE N : Lone pair in sp² orbital IN plane → available PYRROLE N H Lone pair in π system → NOT available pKₐ = –0.27 (very weak base)

Fig 13.1 — Pyridine lone pair is in the plane (sp² orbital), available for protonation. Pyrrole lone pair is part of the aromatic π cloud — unavailable for protonation.

Acid/Base Catalysis — Specific vs General

Catalysis in homogeneous solution works by providing an alternative reaction path of lower activation energy. The acid–base distinction determines which step is rate-limiting.

Specific Acid Catalysis

  • Rate depends only on [H₃O⁺], not on other acids present
  • Characteristic when: fast, reversible protonation first → then slow rate-limiting step
  • Example: Acetal hydrolysis — Rate = k[H₃O⁺][acetal]

General Acid Catalysis

  • Rate depends on [H₃O⁺] AND concentration of other proton donors (HA)
  • Characteristic when: protonation itself is the slow, rate-limiting step
  • Example: Orthoester hydrolysis — Rate = k₁[H₃O⁺][substrate] + k₂[HA][substrate]

Specific Base Catalysis

  • Rate depends only on [⁻OH]
  • Characteristic when: fast, reversible proton removal → slow step follows
  • Example: Aldol reversal — Rate = k[⁻OH][substrate]

General Base Catalysis

  • Rate depends on [⁻OH] AND other bases (B:) present
  • Characteristic when: proton removal is slow (rate-limiting)
  • Example: Base-catalysed bromination of acetone in acetate buffer — Rate = k₁[⁻OH][acetone] + k₂[MeCO₂⁻][acetone]
⚡ IIT-JAM / CSIR-NET Exam Tip

The key distinction: if adding extra acid (HA) at constant pH speeds up the reaction → general acid catalysis. If only [H₃O⁺] matters → specific acid catalysis. Same logic applies for bases. The mechanism indicator: specific catalysis → proton transfer is fast; general catalysis → proton transfer is rate-limiting.

🎯 High-Value Exam Tips, Tricks & Key Facts

🏆 Acidity Order (Must Know)

Carboxylic acids ≫ phenols ≫ alcohols ≫ water ≫ C–H acids. Memorise: HCOOH (3.77) < CH₃COOH (4.76) < PhOH (9.95) < MeOH (16) < CH₄ (43).

🏆 Hybridisation & Acidity

sp > sp² > sp³ in acidity (and basicity reversed for N atoms). C–H in acetylene (sp, pKₐ≈25) > ethylene (sp², pKₐ≈44) > ethane (sp³, pKₐ≈50).

🏆 Halogen Effects

α-Halo acid order: F > Cl > Br > I. More halogens = much stronger acid. Inductive effect falls off rapidly with distance: α ≫ β > γ.

🏆 Ortho Effect in Benzoic Acids

Ortho substituents often show anomalous behaviour due to (1) intramolecular H-bonding stabilising anion or (2) steric inhibition of resonance. Always check for H-bonding!

🏆 Amine Basicity in Water

R₂NH > RNH₂ > R₃N > NH₃ is the typical order in water. Tertiary drops because its conjugate acid R₃NH⁺ has only ONE N–H bond for H-bonding with water.

🏆 Pyridine vs Pyrrole

Pyridine = good base (sp² lone pair in plane, pKₐ=5.21). Pyrrole = extremely weak base (lone pair in aromatic π cloud, pKₐ=–0.27) but can act as weak acid (pKₐ≈17, anion is aromatic).

🏆 Guanidine Strength

Guanidine (pKₐ≈13.6) is among the strongest organic bases because its protonated form has charge spread symmetrically over THREE equivalent N atoms — maximum delocalisation.

🏆 Maleic < Fumaric pKₐ₁

Maleic (cis) is more acidic (pKₐ₁=1.92) because intramolecular H-bonding stabilises its mono-anion. For pKₐ₂, fumaric is more acidic (4.38 vs 6.23) because maleate must lose H⁺ from a cyclic intramolecularly H-bonded anion.

🏆 Levelling Effect

HCl, HBr, HI, HNO₃, HClO₄ all appear equally strong in water. Use non-aqueous solvents (glacial acetic acid, DMSO) to distinguish their true intrinsic strengths.

🏆 ΔH° vs ΔS° in Acidity

For carboxylic acids in water, ΔH° ≈ 0 (bond energy ≈ solvation energy). The actual difference in Ka between acids comes from ΔS° — entropy of solvation. This is a standard IIT-JAM theoretical question.

Quick-Reference pKₐ Summary Table

CompoundTypepKₐ
Trifluoroacetic acid (CF₃COOH)Acid0.23
Trichloroacetic acid (CCl₃COOH)Acid0.65
Picric acid (2,4,6-(NO₂)₃PhOH)Acid1.02
Oxalic acid (pKₐ₁)Acid1.23
Propynoic acid (HC≡CCOOH)Acid1.84
Formic acid (HCOOH)Acid3.77
Benzoic acid (PhCOOH)Acid4.20
Acrylic acid (CH₂=CHCOOH)Acid4.25
Acetic acid (CH₃COOH)Acid4.76
Pyridine (conjugate acid)Base5.21
Phenol (PhOH)Acid9.95
Ammonium ion (NH₄⁺)Base ref.9.25
MeNH₂ (methylamine conj. acid)Base10.64
Methanol (MeOH)Acid~16
Guanidine (conjugate acid)Base~13.6
Methane (CH₄)Acid~43

50 Previous Year Questions
Acids & Bases — Organic Chemistry

Scientifically validated answers with detailed explanations

JEE Main & Advanced NEET IIT-JAM CSIR-NET
50Total Questions
20JEE Questions
18NEET Questions
12IIT-JAM / NET
Your Progress 0 / 50 revealed

50 PYQs — Acids & Bases

Q1 JEE Main Medium 2019

The correct order of acid strength among the following is:
(I) CHCl2COOH   (II) CCl3COOH   (III) CH3COOH   (IV) CH2ClCOOH

A III < IV < I < II
B III < IV < I < II
C II < I < IV < III
D IV < III < I < II
✓ Correct Answer Option A: III < IV < I < II

More Cl atoms at the α-carbon = stronger –I (inductive) effect = greater stabilisation of the carboxylate anion = stronger acid.

pKₐ values: CH₃COOH (4.76) > ClCH₂COOH (2.86) > Cl₂CHCOOH (1.25) > Cl₃CCOOH (0.65)

Rule: Each additional Cl spreads the negative charge further over the anion, reducing the tendency of H₃O⁺ to recombine — equilibrium shifts right, stronger acid.

Q2 NEET Easy 2020

Which of the following is the strongest acid?

A Phenol (pKₐ = 9.95)
B Ethanol (pKₐ ≈ 16)
C Picric acid (pKₐ = 1.02)
D Acetic acid (pKₐ = 4.76)
✓ Correct Answer Option C: Picric acid (2,4,6-trinitrophenol)

Picric acid (2,4,6-trinitrophenol) has pKₐ = 1.02 — far lower than any of the others listed. Three –NO₂ groups exert both –I (inductive) and –M (mesomeric) electron-withdrawing effects, massively stabilising the picrate anion by delocalising negative charge over multiple oxygen atoms of three nitro groups plus the ring.

Remember: lower pKₐ = stronger acid.

Q3 JEE Advanced Hard 2018

Among the following the weakest acid is:

A HC≡C–COOH (pKₐ = 1.84)
B CH₂=CH–COOH (pKₐ = 4.25)
C CH₃CH₂COOH (pKₐ = 4.88)
D (CH₃)₃C–COOH (pKₐ = 5.05)
✓ Correct Answer Option D: (CH₃)₃CCOOH — pKₐ = 5.05

Hybridisation governs the electron-donating ability: sp³ > sp² > sp in +I effect. The tert-butyl group (all sp³ C) has the strongest +I effect, destabilising the carboxylate anion most — weakest acid. sp carbon in propynoic acid has 50% s-character, withdrawing electrons strongly — strongest acid in this set.

Order: HC≡CCOOH < CH₂=CHCOOH < CH₃CH₂COOH < (CH₃)₃CCOOH (acidity decreasing)

Q4 NEET Medium 2021

The correct order of basic strength of the following amines in aqueous solution is:
NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N

A NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
B NH₃ < CH₃NH₂ < (CH₃)₂NH < (CH₃)₃N
C (CH₃)₃N < NH₃ < CH₃NH₂ < (CH₃)₂NH
D (CH₃)₂NH < CH₃NH₂ < NH₃ < (CH₃)₃N
✓ Correct Answer Option A: NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

In water, two effects compete: (1) +I effect of alkyl groups (increases basicity) and (2) solvation of RₙNH(3-n)⁺ cation via H-bonding (decreases as N–H bonds are replaced).

pKₐ values: NH₃ (9.25) < (CH₃)₃N (9.80) < CH₃NH₂ (10.64) < (CH₃)₂NH (10.77)

Tertiary amine drops back because its conjugate acid (CH₃)₃NH⁺ has only one N–H bond — minimal hydrogen bonding with water, poorly solvated, less stable cation.

Q5 JEE Main Medium 2022

Which of the following pairs correctly represents an acid and its conjugate base?

A H₂O / OH⁻ and NH₄⁺ / NH₂⁻
B H₂SO₄ / HSO₄⁻ and H₂O / OH⁻
C HCl / Cl⁻ and NaOH / Na⁺
D NH₃ / NH₄⁺ and H₂O / H₃O⁺
✓ Correct Answer Option B

By Brønsted–Lowry definition: an acid donates H⁺ to give its conjugate base (differs by one H⁺). H₂SO₄ → HSO₄⁻ + H⁺ ✓ and H₂O → OH⁻ + H⁺ ✓. Both pairs are valid conjugate acid–base pairs.

Option D is wrong because NH₃/NH₄⁺ shows base→conjugate acid (reverse direction), and H₂O/H₃O⁺ is also base→conjugate acid. These are conjugate base/acid pairs, not acid/conjugate base.

Q6 NEET Easy 2018

Aniline (C₆H₅NH₂) is a much weaker base than cyclohexylamine (C₆H₁₁NH₂). This is because in aniline:

A The C–N bond is stronger
B The lone pair on N is delocalised into the benzene ring π system
C The N–H bond is weaker
D Steric hindrance by the ring reduces proton access
✓ Correct Answer Option B

In aniline, the N lone pair overlaps with the benzene ring π orbitals (mesomeric donation, +M). Four resonance structures can be drawn, placing positive charge on N and negative charge at ortho/para positions. This stabilises free aniline. On protonation (→ anilinium cation), this delocalisation is completely destroyed — protonation is energetically unprofitable.

pKₐ(aniline) = 4.62 vs pKₐ(cyclohexylamine) = 10.68 — a difference of ~10⁶ in basicity!

Q7 IIT-JAM Hard 2020

For ionisation of acetic acid in water, ΔH° ≈ –0.5 kJ/mol and TΔS° ≈ –27.6 kJ/mol. Which statement is correct?

A Ionisation is enthalpy-driven
B Ionisation is both enthalpy and entropy driven equally
C Ionisation is opposed mainly by the entropy term (TΔS°)
D ΔG° is positive, so ionisation is spontaneous
✓ Correct Answer Option C ΔG° = ΔH° – TΔS° = (–0.5) – (–27.6) = +27.1 kJ/mol

ΔH° is tiny (≈0) because energy to break O–H bond is nearly cancelled by solvation energy of the ions produced. The positive ΔG° (acid only partially ionised) comes almost entirely from the large negative TΔS° term — solvation of RCO₂⁻ and H₃O⁺ severely orders surrounding water molecules, reducing entropy.

This is a defining IIT-JAM concept: acid strength differences in the carboxylic acid series are entropy-controlled, not enthalpy-controlled.

Q8 JEE Main Medium 2017

The pKₐ of maleic acid (cis-butenedioic acid, pKₐ₁ = 1.92) is much lower than that of fumaric acid (trans-butenedioic acid, pKₐ₁ = 3.02). The reason is:

A Fumaric acid has a higher molecular weight
B Intramolecular hydrogen bonding stabilises the maleate mono-anion
C Maleic acid is more symmetric
D The trans isomer has stronger C–H bonds
✓ Correct Answer Option B

In the cis (maleate) mono-anion, the –COO⁻ and –COOH groups are on the same side of the double bond. The O⁻ of the carboxylate can form an intramolecular hydrogen bond with the O–H of the adjacent carboxyl group. This extra stabilisation of the anion lowers ΔG° for ionisation → lower pKₐ₁.

Trans (fumaric) geometry places the two –COOH groups on opposite sides — intramolecular H-bonding is geometrically impossible, so no such stabilisation exists.

Q9 NEET Medium 2022

The order of acidity of the following nitrophenols is:
o-NO₂C₆H₄OH, m-NO₂C₆H₄OH, p-NO₂C₆H₄OH

A m < o < p
B m < p ≈ o (both o and p stronger than m)
C o < m < p
D p < m < o
✓ Correct Answer Option B: m-NO₂ least acidic; o- and p- both stronger

pKₐ values: o-NO₂PhOH = 7.23, m-NO₂PhOH = 8.35, p-NO₂PhOH = 7.14

At ortho and para positions, –NO₂ exerts both –I (inductive) and –M (mesomeric) electron withdrawal — stabilising the phenoxide anion through extended delocalisation. At meta, only –I effect operates (no mesomeric pathway). Hence meta isomer is the weakest acid.

Note: o-isomer has additional intramolecular H-bonding contribution, making it slightly more acidic than p-.

Q10 IIT-JAM Hard 2019

Pyrrole (pKₐ ≈ –0.27 as base) is a much weaker base than pyridine (pKₐ = 5.21) despite both having a ring nitrogen. The correct explanation is:

A Pyrrole N is more electronegative than pyridine N
B Pyrrole has more ring strain
C The N lone pair in pyrrole is part of the aromatic 6π system and unavailable for protonation
D Pyridine has a higher boiling point
✓ Correct Answer Option C

In pyridine: N is sp² hybridised; one electron enters the π system; the lone pair is in an sp² orbital in the plane of the ring — NOT part of aromatic cloud — available for H⁺.

In pyrrole: N contributes both electrons of its lone pair to complete the 6π aromatic system (4n+2, n=1). The lone pair is fully delocalised around the ring — unavailable for protonation. If forced to protonate, it does so at C-2 (α-carbon), destroying aromaticity — extremely unfavourable energetically.

Pyrrole can act as a weak acid (pKₐ ≈ 17): deprotonation at N–H gives pyrrolide anion which retains aromaticity.

Q11 JEE Main Easy 2016

Which of the following is a Lewis acid but NOT a Brønsted acid?

A HCl
B H₂SO₄
C BF₃
D CH₃COOH
✓ Correct Answer Option C: BF₃

BF₃ has an empty 2p orbital on boron — it accepts an electron pair (Lewis acid). However, it has no transferable H⁺ — it cannot donate a proton, so it is NOT a Brønsted acid.

HCl, H₂SO₄, CH₃COOH are all Brønsted acids (proton donors) AND Lewis acids (the H⁺ they release is itself a Lewis acid). BF₃ is the only example that is exclusively a Lewis acid.

Q12 NEET Medium 2019

Salicylic acid (o-hydroxybenzoic acid, pKₐ = 2.98) is much stronger than m-hydroxybenzoic acid (pKₐ = 4.08). The primary reason is:

A –OH at ortho has greater –I effect than at meta
B Intramolecular H-bonding in the salicylate anion stabilises it
C Ortho substituents always increase acid strength
D The molecule has lower molecular weight
✓ Correct Answer Option B

After ionisation of the –COOH group, the resulting –COO⁻ in salicylate forms an intramolecular hydrogen bond with the adjacent –OH group at the ortho position. This extra stabilisation of the anion (not available in the m-isomer) significantly lowers ΔG° of ionisation → lower pKₐ → stronger acid.

The effect is even more dramatic for 2,6-dihydroxybenzoic acid (pKₐ = 1.30) where H-bonding occurs from both ortho positions.

Q13 JEE Advanced Hard 2021

Guanidine [HN=C(NH₂)₂] has pKₐ ≈ 13.6, making it one of the strongest organic bases. The exceptional basicity is best explained by:

A Three N atoms donate electrons by induction
B The guanidinium cation is stabilised by three equivalent resonance structures of identical energy
C Guanidine is a cyclic molecule with extra ring stability
D High molecular weight increases basicity
✓ Correct Answer Option B

On protonation, guanidinium cation [H₂N–C(=NH₂)–NH₂]⁺ has its positive charge symmetrically delocalised over all three nitrogen atoms via three equivalent resonance structures of equal energy. This is far more effective than in the neutral molecule (where charge separation makes two of the three structures higher energy).

The cation is greatly stabilised relative to the neutral base → protonation is strongly energetically favourable → extremely strong base. Compare: amidines RC(=NH)NH₂ are also strong bases (pKₐ ≈ 12) for the same reason but with only 2 N atoms.

Q14 NEET Easy 2017

Which statement about the levelling effect of water is correct?

A Water levels the pH of all solutions to 7
B Very weak acids cannot dissolve in water
C Strong acids (HCl, HNO₃, HClO₄) all appear equally strong in water because they are fully ionised
D All acids have pKₐ = 0 in water
✓ Correct Answer Option C

The levelling effect occurs because water is a sufficiently strong base to completely remove H⁺ from any acid stronger than H₃O⁺ (pKₐ ≈ –1.7). All such acids appear fully ionised and thus identically strong. To distinguish true strengths of HCl, HBr, HI, HNO₃, HClO₄, one must use a weaker base solvent such as glacial acetic acid where they are not fully ionised.

Q15 IIT-JAM Hard 2021

In the series: NH₃ → RNH₂ → R₂NH → R₃N, measurements of basicity in chlorobenzene (a non-H-bonding solvent) give the order BuNH₂ < Bu₂NH < Bu₃N. But in water, Bu₃N (pKₐ = 9.87) is weaker than Bu₂NH (pKₐ = 11.28). This is because:

A In water, solvation of the cation via N–H···O H-bonding stabilises RₙNH(3-n)⁺; tertiary cation has fewest N–H bonds
B Tertiary amines are more sterically hindered towards protonation
C Bu₃N has a lower lone pair energy than Bu₂NH
D Bu₃N is insoluble in water
✓ Correct Answer Option A

In chlorobenzene (no H-bonding), only the +I inductive effect of alkyl groups operates → basicity increases steadily: primary < secondary < tertiary. In water, an additional factor: solvation of the ammonium cation via N–H···OH₂ hydrogen bonds. Tertiary ammonium R₃NH⁺ has only ONE N–H bond → weakest H-bonding → poorest solvation → least stabilised cation → lowest basicity in water. This perfectly explains the reversal and confirms solvation (entropy effect) as the cause.

Q16 JEE Main Medium 2020

The pKₐ of p-hydroxybenzoic acid (4.58) is greater than benzoic acid (4.20). This means p-hydroxybenzoic acid is a weaker acid than benzoic acid. The reason is:

A –OH group at para has –I effect that destabilises the anion
B The +M (mesomeric electron-donating) effect of –OH at para dominates, destabilising the carboxylate anion
C Hydrogen bonding between –OH and –COOH reduces ionisation
D The molecule is more symmetric
✓ Correct Answer Option B

–OH is a +M group (electron donor by resonance) when at ortho or para positions. At para, it donates electron density into the ring, which is then relayed to the –COO⁻ group — increasing negative charge on an already negative carboxylate, destabilising it and making proton removal harder.

The –I effect of –OH would increase acidity, but at para the +M effect dominates the –I effect → net acid-weakening → pKₐ rises above benzoic acid. At meta, only –I operates → m-hydroxybenzoic acid (pKₐ = 4.08) is stronger than benzoic acid (4.20).

Q17 NEET Medium 2023

Acetic acid (CH₃COOH) is a stronger acid than methanol (CH₃OH) primarily because:

A The O–H bond in acetic acid is weaker
B The carboxylate anion is stabilised by resonance involving two equivalent oxygen atoms
C Acetic acid has a higher molecular weight
D The carbonyl group has a dipole moment
✓ Correct Answer Option B

After CH₃COOH loses H⁺, the acetate anion CH₃COO⁻ delocalises its negative charge over both oxygen atoms equally via resonance (two equivalent canonical forms). This massive stabilisation of the anion shifts the ionisation equilibrium to the right.

After CH₃OH loses H⁺, the methoxide CH₃O⁻ has the full negative charge on a single oxygen — no delocalisation possible. Much less stable anion → equilibrium lies far to the left → pKₐ(MeOH) ≈ 16 vs pKₐ(AcOH) = 4.76.

Q18 CSIR-NET Hard 2019

In specific acid catalysis, the rate of reaction depends on:

A [H₃O⁺] only — adding other acids at constant pH has no effect
B Total concentration of all proton donors present
C [H₃O⁺] and [HA] where HA is any weak acid
D [OH⁻] only
✓ Correct Answer Option A

Specific acid catalysis: mechanism involves rapid, reversible protonation of substrate by H₃O⁺ to form a more reactive intermediate, followed by the slow rate-limiting step. Since protonation is fast and equilibrium-controlled, only [H₃O⁺] (which sets the equilibrium position) matters. Adding NH₄⁺ or any other weak acid at the same pH does NOT change the rate.

Rate = k [H₃O⁺][substrate] — e.g., acetal hydrolysis

Contrast with general acid catalysis: protonation IS the slow step → any acid HA can catalyse → Rate = k₁[H₃O⁺][S] + k₂[HA][S]

Q19 JEE Main Medium 2023

Benzoic acid (pKₐ = 4.20) is a stronger acid than cyclohexane carboxylic acid (pKₐ = 4.87). This is due to:

A Phenyl group has +I effect greater than cyclohexyl
B The sp² carbon of the phenyl group attached to –COOH is less electron-donating than sp³ carbon of cyclohexyl
C Benzene ring delocalises the carboxylate anion charge
D Benzoic acid has a higher boiling point
✓ Correct Answer Option B

The C-1 of benzene ring (sp² hybridised, 33% s-character) is less electron-donating (+I) than the sp³ C of cyclohexyl (25% s-character). More s-character = electrons held closer to nucleus = less electron donation to –COOH. Result: benzoic acid's carboxylate is slightly better stabilised → lower pKₐ → stronger acid. Same principle explains why acrylic acid (sp² α-C) is stronger than propionic acid (sp³ α-C).

Q20 NEET Easy 2016

The pKₐ of formic acid (HCOOH) is 3.77 and that of acetic acid (CH₃COOH) is 4.76. Formic acid is stronger because:

A The methyl group in acetic acid has a +I effect that destabilises the acetate anion
B The O–H bond in formic acid is weaker
C Formic acid is more soluble in water
D The formate anion is larger and more stable
✓ Correct Answer Option A

The –CH₃ group in acetic acid exerts a +I (electron-donating inductive) effect. This pushes electrons toward the already-negative carboxylate oxygen atoms, destabilising the acetate anion and promoting recombination with H⁺. Formate (HCOO⁻) lacks this destabilising effect — its anion is better stabilised → lower pKₐ → stronger acid.

The actual difference is dominated by the solvation entropy of the two anions, but the +I destabilisation of the anion is the structural explanation.

Q21JEE AdvancedHard2022

The second dissociation of maleic acid (pKₐ₂ = 6.23) is harder than the second dissociation of fumaric acid (pKₐ₂ = 4.38) because:

A The maleate mono-anion is locked in cyclic intramolecular H-bonded structure, making H⁺ removal from it harder
B Fumaric acid has a higher molecular weight
C Second ionisation always depends on geometry
D maleic acid is more soluble
✓ Correct Answer Option A

The maleate mono-anion forms a cyclic intramolecular H-bond between –COO⁻ and –COOH — this system is more stable (lower energy) and therefore the second H⁺ (from –COOH within this ring) is harder to remove. The fumarate mono-anion has no such cyclic stabilisation → second H⁺ is more accessible → lower pKₐ₂.

Q22NEETEasy2018

Which of the following is the correct relationship? (Kₐ = acidity constant, Kb = basicity constant of conjugate base)

A Kₐ × Kb = Kw (= 10⁻¹⁴ at 25°C)
B Kₐ + Kb = Kw
C pKₐ + pKb = 0
D Kₐ / Kb = Kw
✓ Correct Answer Option A: Kₐ × Kb = Kw Kₐ × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25°C) ∴ pKₐ + pKb = pKw = 14

This is a fundamental thermodynamic identity. A stronger acid (larger Kₐ) has a weaker conjugate base (smaller Kb), and vice versa.

Q23JEE MainMedium2021

Among FCH₂COOH, ClCH₂COOH, BrCH₂COOH, ICH₂COOH, the order of acid strength is:

A F > Cl > Br > I (pKₐ: 2.57 < 2.86 < 2.90 < 3.16)
B I > Br > Cl > F
C Cl > Br > F > I
D All equal strength
✓ Correct Answer Option A: F > Cl > Br > I in acid strength

The inductive (–I) effect order of halogens follows electronegativity: F > Cl > Br > I. More electron withdrawal → better stabilisation of –CH₂COO⁻ anion → lower pKₐ → stronger acid. Fluoroacetic acid (pKₐ = 2.57) is ~100× stronger than acetic acid (4.76).

Q24NEETMedium2022

Triphenylamine (Ph₃N) is essentially non-basic in water. This is because:

A All three phenyl groups withdraw the N lone pair via resonance into their π systems, leaving nothing for protonation
B Steric bulk of three phenyl groups blocks the N
C Ph₃N is insoluble in water
D The three C–N bonds are too strong
✓ Correct Answer Option A

Each phenyl group withdraws the N lone pair via resonance (+M of N into ring = –M on N from ring's perspective). With three phenyl groups, the combined resonance withdrawal of the lone pair is so extensive that essentially nothing remains for proton acceptance. pKₐ ≈ –5. Compare: Ph₂NH (pKₐ = 0.8), PhNH₂ (pKₐ = 4.62).

Q25IIT-JAMHard2018

Water is an effective ionising solvent for organic acids because of its: (Select the MOST complete answer)

A High dielectric constant (ε = 80) only
B Small molecular size only
C High dielectric constant AND powerful ion-solvating ability via H-bonding (especially of anions)
D Its ability to dissolve all organic compounds
✓ Correct Answer Option C

Two synergistic properties: (1) High ε = 80 lowers electrostatic attraction between ions, stabilising separated ion pairs. (2) Small, polar H₂O molecules form hydrogen-bonded solvation shells around both cations (via O lone pairs) and anions (via O–H···A⁻ H-bonds). Anionic solvation is especially powerful. Together these effects stabilise the dissociated ions enormously, driving ionisation equilibrium to the right.

Q26JEE MainMedium2020

The pKₐ of oxalic acid (HOOCCOOH) pKₐ₁ = 1.23 is much lower than malonic acid pKₐ₁ = 2.83. This is primarily because:

A In oxalic acid the two –COOH groups are directly adjacent; the –I inductive effect of one –COOH on the other is very large
B Oxalic acid has higher molecular weight
C Oxalic acid is more symmetric
D Hydrogen bonding between the two COOH groups
✓ Correct Answer Option A

The carboxyl group (–COOH) is itself a powerful electron-withdrawing group (–I and –M). When two –COOH groups are directly bonded (oxalic acid), each strongly withdraws electrons from the other — maximum inductive effect on the ionising O–H. As the chain lengthens (malonic, succinic), the –I effect attenuates rapidly across saturated C–C bonds and the pKₐ₁ rises toward normal carboxylic acid values.

Q27NEETEasy2019

Amides (RCONH₂) are very weak bases (pKₐ ≈ 0.5) compared to aliphatic amines (pKₐ ≈ 10). This is because in amides:

A The C=O group withdraws the N lone pair by resonance (–M), making it unavailable
B The N–H bond strength is increased by the C=O
C Amides are insoluble in water
D The molecular weight of amides is higher
✓ Correct Answer Option A R–C(=O)–NH₂ ↔ R–C(–O⁻)=NH₂⁺

The resonance delocalisation of the N lone pair into the C=O group is so effective that N has essentially partial double bond character with C. The lone pair is tied up in the π system — not available for protonation. If two C=O groups flank N (imides), the compound may even act as an acid (e.g., phthalimide forms alkali metal salts).

Q28JEE AdvancedHard2023

2,4,6-trinitro-N,N-dimethylaniline is about 40,000 times (ΔpKₐ = 4.6) stronger a base than 2,4,6-trinitroaniline. The explanation involves:

A Steric inhibition of resonance: the bulky NMe₂ cannot become coplanar with the ring, blocking –M withdrawal by NO₂ groups
B Methyl groups donate electrons by induction strongly enough to overcome three nitro groups
C NMe₂ forms intramolecular H-bonds
D The compound has lower molecular weight
✓ Correct Answer Option A — Steric Inhibition of Resonance (SIR)

The large NMe₂ group cannot fit coplanar with the ring because its methyl groups clash with the adjacent o-NO₂ groups. The molecule twists — N's p orbital is no longer parallel to ring p orbitals. Mesomeric withdrawal of the N lone pair by NO₂ groups is blocked. The NMe₂ lone pair stays on N, available for protonation. The NH₂ group in trinitroaniline IS coplanar (small enough) → full –M withdrawal → lone pair fully delocalised → non-basic. This is a classic example of SIR (steric inhibition of resonance).

Q29NEETMedium2020

The pKₐ of acetic acid at 25°C is 4.76. If the temperature is raised to 50°C, the pKₐ will:

A Remain exactly the same
B Always decrease
C Change — Kₐ is temperature-dependent (ΔG° = ΔH° – TΔS°)
D Always increase
✓ Correct Answer Option C

Kₐ is an equilibrium constant and must vary with temperature (van't Hoff equation). Since ΔG° = ΔH° – TΔS°, changing T changes ΔG° and hence Kₐ. Even relative acidities reverse: ethanoic acid is weaker than Et₂CHCOOH below 30°C but stronger above 30°C. This is why pKₐ values are always quoted at a specified temperature (usually 25°C).

Q30IIT-JAMMedium2022

General base catalysis (as distinguished from specific base catalysis) is characterised by:

A Rate depending only on [OH⁻]
B Rate depending on [OH⁻] AND concentrations of other bases; proton removal from substrate is the rate-limiting step
C Reaction occurring only in basic solutions
D Rate being independent of base concentration
✓ Correct Answer Option B

In general base catalysis, proton abstraction from the substrate is the slow, rate-limiting step. Any base (B:) present can participate — not just OH⁻. Classic example: base-catalysed bromination of acetone in acetate buffer.

Rate = k₁[⁻OH][CH₃COCH₃] + k₂[CH₃CO₂⁻][CH₃COCH₃]

Specific base catalysis: proton removal is fast and reversible (pre-equilibrium), slow step follows → only [OH⁻] matters.

Q31JEE MainEasy2018

AlCl₃ acts as a Lewis acid because:

A It has an empty 3p orbital on Al and can accept an electron pair
B It releases Al³⁺ ions in water
C It has a high melting point
D It contains chlorine atoms which are electronegative
✓ Correct Answer Option A

Al in AlCl₃ has only 6 electrons around it (three bond pairs) — the 3p orbital is empty. This empty orbital readily accepts an electron pair from a Lewis base (e.g., Cl⁻ from another AlCl₃, or an organic substrate in Friedel–Crafts reactions). AlCl₃ is a classic Lewis acid catalyst; others include FeCl₃, BF₃, SnCl₄, ZnCl₂.

Q32NEETMedium2021

Among the following which has the highest pKₐ (weakest acid)?

A CF₃COOH (pKₐ = 0.23)
B CHCl₂COOH (pKₐ = 1.25)
C CH₂ClCOOH (pKₐ = 2.86)
D CH₃CH₂COOH (pKₐ = 4.88)
✓ Correct Answer Option D: Propanoic acid

Propanoic acid has no electron-withdrawing halogen substituents — only the electron-donating +I ethyl group, which destabilises the anion. CF₃COOH has three F atoms with maximum –I effect → most stable anion → strongest acid. Remember: higher pKₐ = weaker acid.

Q33JEE AdvancedHard2019

The pKₐ of phenol (9.95) is much lower than methanol (~16). Despite both having O–H bonds, phenol is ~10⁶ times more acidic. The BEST explanation is:

A O–H bond in phenol is weaker due to ring strain
B Phenoxide anion is stabilised by delocalisation of negative charge into the benzene ring via four resonance structures
C Phenol is more polar than methanol
D Phenol has a higher molecular weight
✓ Correct Answer Option B

Phenoxide (C₆H₅O⁻) has four resonance contributors: O⁻ on oxygen, and negative charge at ortho (×2) and para positions of the ring. This delocalisation stabilises the anion significantly (though less than carboxylate — ring carbons are less electronegative than O atoms). Methoxide (CH₃O⁻) has no resonance — full charge on one O atom — far less stable. Greater anion stability → lower ΔG° ionisation → lower pKₐ.

Q34NEETEasy2023

The conjugate base of H₂PO₄⁻ is:

A H₃PO₄
B PO₄³⁻
C HPO₄²⁻
D H₂PO₄⁻ itself
✓ Correct Answer Option C: HPO₄²⁻ H₂PO₄⁻ → H⁺ + HPO₄²⁻

Conjugate base is always formed by removing ONE proton (H⁺) from the acid. H₂PO₄⁻ loses H⁺ → HPO₄²⁻. The conjugate acid would be H₃PO₄ (one more H⁺ added). PO₄³⁻ is three deprotonation steps away.

Q35CSIR-NETHard2020

Pyrrolidine (fully saturated pyrrole ring, pKₐ = 11.27) has nearly the same basicity as diethylamine (pKₐ = 11.04). This is because:

A In pyrrolidine, N is sp³ hybridised with lone pair in a fully available sp³ orbital, like a secondary aliphatic amine
B Ring structure always increases basicity
C Pyrrolidine has lower molecular weight
D Pyrrolidine can form more H-bonds
✓ Correct Answer Option A

Pyrrolidine is the fully reduced (saturated) version of pyrrole. All ring carbons are sp³; N is sp³ hybridised with its lone pair in an sp³ orbital — NOT contributing to any aromatic system. It behaves exactly like a normal secondary aliphatic amine (R₂NH). Hence pKₐ ≈ 11.27, very close to diethylamine (11.04). This contrasts sharply with pyrrole (pKₐ = –0.27) where N is sp² and lone pair is in the aromatic π cloud.

Q36JEE MainMedium2022

The inductive effect of a halogen substituent on the acidity of an aliphatic acid diminishes rapidly with chain length. Which set of pKₐ values best illustrates this?

A ClCH₂COOH (2.86) > ClCH₂CH₂COOH (4.52) > MeCH₂CH₂COOH (4.82) in acidity
B pKₐ stays constant regardless of position
C Further Cl always doubles the acidity regardless of position
D Meta position Cl has maximum effect
✓ Correct Answer Option A

The inductive effect operates through bonds and attenuates by roughly a factor of 2.8 per additional C–C bond. α-Cl (adjacent) has dramatic effect (pKₐ drops from 4.82 to 2.86). β-Cl shows much less effect (pKₐ = 4.52 — closer to unsubstituted 4.82). γ-Cl has barely measurable effect. The negative charge in the carboxylate becomes progressively more concentrated (less spread) as Cl moves farther away.

Q37NEETHard2022

o-Nitrobenzoic acid (pKₐ = 2.17) is much stronger than m-nitrobenzoic acid (pKₐ = 3.45) or p-nitrobenzoic acid (pKₐ = 3.43). The extra acidity at ortho is mainly due to:

A The –I inductive effect operates over a very short distance at ortho, and possible direct interaction between –NO₂ and –COOH groups
B Ortho substitution always doubles acidity
C Steric compression of the O–H bond
D Resonance of NO₂ at ortho is stronger than at para
✓ Correct Answer Option A

At the ortho position, the –NO₂ group is very close to –COOH (through-space distance ≈ 2.5 Å). The powerful –I effect of –NO₂ is most effective at short distances. Additionally, direct through-space interaction between the adjacent –NO₂ and –COOH groups cannot be excluded. At para, –M and –I both operate but at longer effective distance. The anomalously high acidity of many o-substituted benzoic acids is a well-established pattern called the ortho effect.

Q38IIT-JAMMedium2017

Which is NOT true about the Lewis definition of acids and bases?

A It is the most general of the three definitions
B Lewis acids need not contain H
C Lewis bases must have a lone pair
D Every Lewis acid is also a Brønsted acid
✓ Correct Answer Option D — This is FALSE

Not every Lewis acid is a Brønsted acid. BF₃, AlCl₃, FeCl₃, carbocations — all are Lewis acids (electron pair acceptors) but have no transferable H⁺, so they are NOT Brønsted acids. The Lewis definition is broader: every Brønsted acid is a Lewis acid (the released H⁺ is a Lewis acid), but not vice versa.

Q39JEE MainMedium2019

Arrange in increasing order of basicity (pKₐ of conjugate acid): pyridine, aniline, ammonia, trimethylamine

A Aniline < Pyridine < NH₃ < (CH₃)₃N
B Pyridine < Aniline < NH₃ < (CH₃)₃N
C NH₃ < Aniline < Pyridine < (CH₃)₃N
D (CH₃)₃N < NH₃ < Aniline < Pyridine
✓ Correct Answer Option A

pKₐ values: PhNH₂ (4.62) < Pyridine (5.21) < NH₃ (9.25) < (CH₃)₃N (9.80)

Aniline: lone pair delocalised into ring → very weak base. Pyridine: sp² N → lone pair less available than sp³. NH₃: sp³ N, no +I groups. (CH₃)₃N: three +I methyl groups, but solvation limits it. In gas phase, (CH₃)₃N would be the strongest by far; in water, solvation effect reduces the advantage.

Q40NEETMedium2018

Which of the following correctly explains why CF₃COOH (pKₐ = 0.23) has ΔG° ≈ 1.3 kJ/mol while CH₃COOH has ΔG° ≈ 27.2 kJ/mol, yet their ΔH° values are nearly identical?

A The CF₃ group delocalises the anion's negative charge, imposing less ordering on surrounding water → more favourable TΔS°
B C–F bonds release heat on ionisation
C CF₃COOH has a shorter C–O bond
D Temperature effects are larger for CF₃COOH
✓ Correct Answer Option A

Both acids have ΔH° ≈ 0 (O–H bond energy ≈ solvation energy). The massive difference in ΔG° is entirely due to ΔS°. The CF₃COO⁻ anion has its charge delocalised over the whole molecule (F atoms spread it) → the anion imposes less rigidity on surrounding water molecules → smaller decrease in entropy on ionisation → TΔS° is less negative → ΔG° is much smaller → Kₐ is much larger → much stronger acid. This is the thermodynamic entropy argument at its most elegant.

Q41JEE MainEasy2017

Which among these is the strongest base in aqueous solution?

A (CH₃)₂NH (dimethylamine, pKₐ = 10.77)
B CH₃NH₂ (methylamine, pKₐ = 10.64)
C (CH₃)₃N (trimethylamine, pKₐ = 9.80)
D NH₃ (ammonia, pKₐ = 9.25)
✓ Correct Answer Option A: Dimethylamine

In water, secondary amines are the strongest aliphatic bases because they benefit from two +I methyl groups (more electron density on N) while their conjugate acid (CH₃)₂NH₂⁺ still has TWO N–H bonds for effective H-bonding solvation. The tertiary amine gains from three methyl groups (+I) but its cation (CH₃)₃NH⁺ has only ONE N–H — poor solvation drops its pKₐ back.

Q42NEETEasy2016

Propiolic acid (HC≡C–COOH, pKₐ = 1.84) is a much stronger acid than propionic acid (CH₃CH₂COOH, pKₐ = 4.88). The reason is:

A The sp-hybridised α-carbon is strongly electron-withdrawing due to 50% s-character
B Triple bond reduces molecular flexibility
C Propiolic acid has a lower molecular weight
D The C≡C bond donates electrons to –COOH
✓ Correct Answer Option A

sp hybridised carbon (50% s-character) is far more electronegative than sp³ (25% s-character). Greater s-character = electrons drawn closer to nucleus = stronger –I effect = better anion stabilisation. This also explains why acetylene (HC≡CH, pKₐ ≈ 25) is much more acidic than ethylene (H₂C=CH₂, pKₐ ≈ 44) or ethane (pKₐ ≈ 50).

Q43IIT-JAMHard2023

Pyrrole can act as a weak acid (pKₐ ≈ 17) when treated with strong bases like NaNH₂. The acidic H is on the N–H, and the pyrrolide anion is stable because:

A The pyrrolide anion retains full aromatic character (6π electrons, 4n+2 with n=1)
B Nitrogen lone pairs are good leaving groups
C The anion is stabilised by H-bonding
D The N–H bond is especially weak in pyrrole
✓ Correct Answer Option A

When pyrrole (18a) loses the N–H proton, the resulting pyrrolide anion (20) has 6 π electrons total (4 from ring carbons + 2 from N lone pair) in a cyclic conjugated system — fully aromatic by Hückel's rule (4n+2, n=1). The anion is thus aromatically stabilised. In contrast, when pyrrole gains H⁺ (acts as a base), it must disrupt this aromatic system → much more unfavourable. This is why pyrrole is a better acid than it is a base.

Q44JEE AdvancedHard2020

Acetal hydrolysis (MeCH(OEt)₂) shows specific acid catalysis. If one adds NH₄⁺ (a weak acid) to the reaction at constant pH, the rate:

A Does not change — only [H₃O⁺] controls the rate in specific acid catalysis
B Increases because NH₄⁺ donates protons to the substrate
C Decreases because NH₄⁺ competes with H₃O⁺
D Doubles because both H₃O⁺ and NH₄⁺ catalyse the reaction
✓ Correct Answer Option A

This is the defining test for specific acid catalysis. The mechanism involves fast, reversible protonation by H₃O⁺ only — the pre-equilibrium concentration of the protonated intermediate depends solely on [H₃O⁺] (pH). NH₄⁺ does not change [H₃O⁺] at constant pH, therefore does not affect the rate. If NH₄⁺ DID accelerate the reaction at constant pH, it would be general acid catalysis instead.

Q45NEETMedium2023

The Brønsted base in the reaction: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ is:

A CH₃COOH
B CH₃COO⁻
C OH⁻
D H₂O
✓ Correct Answer Option B: CH₃COO⁻

CH₃COO⁻ accepts a proton from H₂O → becomes CH₃COOH. Since it accepts H⁺, it is the Brønsted base. H₂O is the acid here (it donates H⁺). OH⁻ is the conjugate base of H₂O, and CH₃COOH is the conjugate acid of CH₃COO⁻. This reaction shows that acetate ion is a base — the basis of buffer chemistry.

Q46CSIR-NETMedium2021

Phthalimide (benzene-1,2-dicarboximide) forms alkali metal salts, indicating it acts as an acid. This is because:

A Two flanking C=O groups withdraw N lone pair so powerfully that N–H becomes acidic (pKₐ ≈ 8.3)
B The benzene ring activates N–H toward ionisation
C The N–H bond is mechanically strained in the ring
D Alkali metals react with any organic N–H compound
✓ Correct Answer Option A

Two C=O groups flanking N each withdraw the N lone pair by –M (resonance) and –I (inductive) effects. The combined withdrawal makes N electron-poor and activates the N–H bond toward ionisation. The resulting phthalimide anion is stabilised by extensive delocalisation over both carbonyl groups and the ring. This makes phthalimide acidic (pKₐ ≈ 8.3) — it reacts with KOH to form potassium phthalimide (key in Gabriel synthesis of primary amines).

Q47JEE MainMedium2021

The correct order of acidity of hydrogen atoms in ethane, ethylene, and acetylene is:

A Ethane < Ethylene < Acetylene (pKₐ ≈ 50 > 44 > 25)
B Acetylene < Ethylene < Ethane
C Ethylene < Ethane < Acetylene
D All equal
✓ Correct Answer Option A: CH₃CH₃ < CH₂=CH₂ < HC≡CH

pKₐ: Ethane (~50) > Ethylene (~44) > Acetylene (~25). As s-character increases (sp³→sp²→sp), C becomes more electronegative → C–H bond becomes more polar → H⁺ more easily released → more acidic. Acetylide anion (sp, 50% s-character) is the most stable carbanion of the three. Terminal alkynes are acidic enough to react with strong bases like NaNH₂ or n-BuLi.

Q48NEETHard2022

The dielectric constant of a solvent affects acid ionisation primarily by:

A Higher ε lowers the electrostatic energy of ion pairs, making them more stable and less likely to recombine
B Higher ε increases the viscosity, slowing ionisation
C Higher ε increases H-bond strength
D ε has no effect on ionisation equilibrium
✓ Correct Answer Option A

Coulomb's law: electrostatic energy between ions ∝ 1/ε. Higher dielectric constant (ε) → lower electrostatic attraction between oppositely charged ions → ion pairs less stable as a pair → they separate more readily → equilibrium favours ionised form. Water (ε = 80) is exceptional; methylbenzene (toluene, ε ≈ 2.4) barely supports ionisation — HCl is almost completely un-ionised in toluene.

Q49IIT-JAMHard2022

Tetraalkylammonium hydroxides (R₄N⁺OH⁻) are as strong as mineral alkalis. This is because:

A There is no tertiary amine available to revert from the ionised form; the hydroxide is fully and permanently ionised
B Four alkyl groups donate maximum electrons to N
C The large size of R₄N⁺ stabilises OH⁻
D Silver oxide acts as a catalyst
✓ Correct Answer Option A

Tertiary amines can revert from their protonated forms: R₃NH⁺ + OH⁻ → R₃N + H₂O (deprotonation equilibrium). But R₄N⁺OH⁻ has NO hydrogen on N — there is literally no pathway for the positive nitrogen to lose its charge and revert to a neutral amine. The hydroxide is completely dissociated — 100% ionised — giving a solution of OH⁻ concentration equivalent to its molar concentration. Hence basic strength = NaOH = KOH.

Q50JEE AdvancedHard2023

FINAL QUESTION — Comprehensive: Arrange the following in decreasing order of acidity:
(I) CF₃COOH   (II) CCl₃COOH   (III) 2,4,6-(NO₂)₃C₆H₂OH   (IV) CH₃COOH   (V) C₆H₅OH

A I > II > III > V > IV (pKₐ: 0.23 < 0.65 < 1.02 < 9.95 < 4.76... wait — see explanation)
B III > I > II > IV > V
C I > III > II > IV > V
D IV > V > II > III > I
✓ Correct Answer & Full Analysis Correct order: I > II > III > IV > V

pKₐ values (lower = stronger acid):

• CF₃COOH (I): pKₐ = 0.23 — three F atoms (highest electronegativity) maximally stabilise anion via –I, plus entropy effect

• CCl₃COOH (II): pKₐ = 0.65 — three Cl atoms, strong but weaker than F (lower electronegativity)

• 2,4,6-(NO₂)₃C₆H₂OH (III) [Picric acid]: pKₐ = 1.02 — three NO₂ groups (–I + –M), phenol acid strengthened to mineral acid level

• CH₃COOH (IV): pKₐ = 4.76 — resonance-stabilised carboxylate, moderate acid

• C₆H₅OH (V): pKₐ = 9.95 — phenoxide partially stabilised by ring, weaker than carboxylic acids

Final order: I > II > III > IV > V

Based on Chapter 3 of "A Guidebook to Mechanism in Organic Chemistry" by Peter Sykes

You May Also Like

Loading...


Reviews



Videos

Sudhir Nama Chemistry Lecture - Video 1
Sudhir Nama Chemistry Lecture - Video 2
Sudhir Nama Chemistry Lecture - Video 3
Sudhir Nama Chemistry Lecture - Video 4

Contact Me

Shoot Your Questions

Academic & Social Profiles