The Hook: From Vinegar to Aspirin - The Acid Story
Ever used vinegar (acetic acid) to clean? Or taken aspirin for a headache? Both are carboxylic acids! Vinegar is 5% acetic acid (CH₃COOH), and aspirin is acetylsalicylic acid - a modified carboxylic acid.
Here’s the fascinating part: Carboxylic acids are weak acids in chemistry class (don’t fully dissociate in water), but they’re strong enough to donate H⁺ to NaHCO₃, causing fizzing (CO₂ gas). Try that with phenol or alcohol - nothing happens!
In 3 Idiots, Rancho says “Follow excellence, success will chase you” - think of carboxylic acids as the “excellence” of oxidation. Start with an alcohol, oxidize it excellently, and you’ll reach the carboxylic acid - the final stable oxidation product!
The JEE Question: Why is acetic acid (pKₐ ~ 4.76) more acidic than phenol (pKₐ ~ 10) despite both having -OH? Why do electron-withdrawing groups increase acidity while electron-donating groups decrease it? Crack this, and you’ll master 6-8 marks in JEE!
The Core Concept: The Carboxyl Group (-COOH)
General Structure
$$\boxed{\text{R-COOH or R-C(=O)-OH}}$$Structure Breakdown:
O
‖
R-C-O-H
In simple terms: Carboxylic acids combine carbonyl (C=O) and hydroxyl (-OH) groups into one powerful functional group called the carboxyl group!
Why “Carboxyl”? Carb (carbonyl C=O) + oxyl (hydroxyl -OH) = Carboxyl!
Nomenclature
IUPAC System: Replace “-e” of alkane with "-oic acid"
| Common Name | IUPAC Name | Formula |
|---|---|---|
| Formic acid | Methanoic acid | HCOOH |
| Acetic acid | Ethanoic acid | CH₃COOH |
| Propionic acid | Propanoic acid | CH₃CH₂COOH |
| Butyric acid | Butanoic acid | CH₃(CH₂)₂COOH |
| Benzoic acid | Benzoic acid | C₆H₅COOH |
| Oxalic acid | Ethanedioic acid | HOOC-COOH |
Position Numbering:
- Carboxyl carbon is always position 1
- Greek letters (α, β, γ, δ…) used in common names
Example:
4 3 2 1
CH₃-CH₂-CH₂-COOH → Butanoic acid
γ β α
CH₃-CH₂-CH₂-COOH → γ-carbon, β-carbon, α-carbon
Acidity: Why Carboxylic Acids are Acidic
Acidity Comparison
$$\boxed{\text{Strong acids (HCl)} >> \text{Carboxylic acids} >> \text{Phenol} > \text{Water} > \text{Alcohols}}$$ $$\boxed{\text{pK}_a: \text{RCOOH (~4-5)} << \text{PhOH (~10)} < \text{H}_2\text{O (~15.7)} < \text{ROH (~16)}}$$Question: Why is acetic acid (CH₃COOH) so much more acidic than ethanol (CH₃CH₂OH)?
The Resonance Stabilization Story
When carboxylic acid loses H⁺:
O O⁻
‖ ‖
R-C-O-H ⇌ R-C-O⁻ + H⁺
| |
(Acid) (Carboxylate ion - VERY STABLE!)
Resonance in carboxylate ion:
O⁻ O
‖ ↔ ‖
R-C-O R-C-O⁻
Two equivalent resonance structures!
Negative charge equally shared between both oxygens.
Key Point: The two resonance structures are equivalent (equal contributors), making the carboxylate ion extremely stable!
In alcohols (R-O⁻):
- No resonance stabilization
- Negative charge localized on single oxygen
- Less stable → less willing to release H⁺ → less acidic
In phenol (C₆H₅-O⁻):
- Resonance exists, but charge delocalizes into benzene ring
- Less effective than carboxylate (non-equivalent structures)
- Moderately stable → moderately acidic
Memory Trick: “Carboxylate has Complete Charge distribution (two equivalent O atoms) - most stable - most acidic!”
Interactive Demo: Visualize Reaction Mechanism
Watch the mechanism of esterification and other carboxylic acid reactions.
Effect of Substituents on Acidity
Electron-Withdrawing Groups (EWG): Increase acidity
- Stabilize carboxylate ion (pull electron density away from negative charge)
- Examples: -NO₂, -CN, -F, -Cl, -Br, -I, -CHO
Electron-Donating Groups (EDG): Decrease acidity
- Destabilize carboxylate ion (push electron density toward negative charge)
- Examples: -CH₃, -C₂H₅, -OCH₃, -NH₂
Acidity Order Examples:
$$\boxed{\text{CCl}_3\text{COOH} > \text{CHCl}_2\text{COOH} > \text{CH}_2\text{Cl-COOH} > \text{CH}_3\text{COOH}}$$Why? More Cl atoms (EWG) → more electron withdrawal → more stable carboxylate → more acidic!
Inductive Effect Distance:
- Closer to -COOH → stronger effect
- α-position > β-position > γ-position
Example:
Acidity order:
Cl-CH₂-COOH > Cl-CH₂-CH₂-COOH > Cl-CH₂-CH₂-CH₂-COOH
(α-Cl) (β-Cl) (γ-Cl)
Closer Cl → stronger -I effect → more acidic!
Memory Trick: “EWG Enhances acidity, EDG Erodes it!”
Common Question: Arrange in increasing order of acidity: I. CH₃COOH II. FCH₂COOH III. ClCH₂COOH IV. BrCH₂COOH
Strategy:
- All have same structure (X-CH₂-COOH)
- Compare EWG strength: F > Cl > Br > H (electronegativity order)
- Stronger EWG → more acidic
Increasing acidity: I < IV < III < II (CH₃COOH < BrCH₂COOH < ClCH₂COOH < FCH₂COOH)
Answer: F is most electronegative → strongest EWG → most acidic!
Preparation of Carboxylic Acids
Method 1: Oxidation of Primary Alcohols and Aldehydes
Pattern:
1° Alcohol ──[O]──→ Aldehyde ──[O]──→ Carboxylic Acid
Reagents: K₂Cr₂O₇/H⁺, KMnO₄/H⁺, or even air (for aldehydes)
Examples:
(a) From 1° Alcohol:
K₂Cr₂O₇/H⁺ O
CH₃CH₂-OH ──────────────→ CH₃-C-OH
(Acetic acid)
(b) From Aldehyde:
K₂Cr₂O₇/H⁺ O
or air! ‖
CH₃CHO ──────────→ CH₃-C-OH
(Acetic acid)
Note: Aldehydes are easily oxidized - even air slowly oxidizes them to carboxylic acids!
Memory Trick: “Primary alcohols Proceed to acids Perfectly with oxidants!”
Method 2: Oxidation of Alkylbenzenes
Reaction:
KMnO₄/H⁺
C₆H₅-CH₃ ──────────→ C₆H₅-COOH
(Toluene) (Heat) (Benzoic acid)
General Pattern:
KMnO₄/H⁺
C₆H₅-R ──────────→ C₆H₅-COOH (R oxidized to -COOH, regardless of length!)
(Heat)
Examples:
C₆H₅-CH₂-CH₃ ──KMnO₄/H⁺──→ C₆H₅-COOH (ethyl group → COOH)
C₆H₅-CH(CH₃)₂ ──KMnO₄/H⁺──→ C₆H₅-COOH (isopropyl → COOH)
C₆H₅-C(CH₃)₃ ──KMnO₄/H⁺──→ C₆H₅-COOH (tert-butyl → COOH)
Key Point: Any alkyl group attached to benzene ring gets oxidized to -COOH, no matter how big!
Exception: If carbon attached to benzene has no H atoms (like C₆H₅-C(CH₃)₃), oxidation is very slow/difficult.
Method 3: Hydrolysis of Nitriles
Reaction:
H₃O⁺ or OH⁻/H₂O O
R-C≡N ──────────────────→ R-C-OH + NH₃ (or NH₄⁺)
(Heat) (Carboxylic acid)
Example:
H₃O⁺/Heat
CH₃-C≡N ──────────────→ CH₃-COOH + NH₄⁺
(Acetonitrile) (Acetic acid)
Mechanism (Acidic Hydrolysis):
Step 1: Protonation
R-C≡N + H⁺ → R-C≡N⁺-H
Step 2: Water attack (nucleophilic addition)
R-C≡N⁺-H + H₂O → R-C(=NH)-OH (Amide intermediate)
Step 3: Further hydrolysis
R-C(=NH)-OH + H₂O → R-COOH + NH₄⁺
Importance: This is a chain extension method!
- R-X → R-CN (SN2 with CN⁻) → R-COOH (hydrolysis)
- Net result: R-X → R-COOH (one more carbon!)
Method 4: Grignard Reagent + CO₂
Reaction:
(1) CO₂ O
R-MgX + ───────────→ R-C-O⁻ Mg⁺X ──H₃O⁺──→ R-C-OH
(2) H₃O⁺ |
OH
Example:
(1) CO₂
CH₃-MgBr ──────────→ CH₃-COOH
(2) H₃O⁺ (Acetic acid)
Importance: Another chain extension method!
- Adds one carbon to Grignard reagent
Method 5: Hydrolysis of Esters (Saponification)
Reaction:
NaOH/H₂O
R-COO-R' ──────────→ R-COO⁻Na⁺ + R'-OH
(Heat) (Sodium carboxylate)
Then: R-COO⁻Na⁺ + HCl → R-COOH + NaCl
Example:
NaOH/H₂O
CH₃-COO-C₂H₅ ──────────→ CH₃COO⁻Na⁺ + C₂H₅-OH
(Heat)
Then: CH₃COO⁻Na⁺ + HCl → CH₃COOH
Note: This is called saponification (soap-making process uses this!)
Physical Properties
Hydrogen Bonding & Dimerization
Unique Property: Carboxylic acids exist as dimers in vapor phase and non-polar solvents!
Dimer Structure:
O---H-O
‖ ‖
R-C C-R
O-H---O
Two molecules held together by TWO H-bonds!
Consequence:
- Higher boiling points than expected
- Molecular mass appears doubled in vapor density measurements
Boiling Point Comparison:
| Compound | Molecular Mass | Boiling Point |
|---|---|---|
| Ethanol (CH₃CH₂OH) | 46 | 78°C |
| Acetic acid (CH₃COOH) | 60 | 118°C |
Why? Strong H-bonding in dimers requires more energy to break!
Memory Trick: “Carboxylic acids are Clingy - form dimers via double H-bonds!”
Reactions of Carboxylic Acids
Reaction 1: Acidity - Salt Formation
With NaOH:
R-COOH + NaOH → R-COO⁻Na⁺ + H₂O
(Sodium carboxylate, ionic salt)
With NaHCO₃ (distinguishes from phenol!):
R-COOH + NaHCO₃ → R-COO⁻Na⁺ + H₂O + CO₂↑ (Effervescence!)
But: C₆H₅-OH + NaHCO₃ → No reaction (phenol not acidic enough!)
Acidity Test:
| Compound | NaOH | NaHCO₃ |
|---|---|---|
| RCOOH (Carboxylic acid) | Dissolves ✓ | CO₂ evolves ✓ |
| PhOH (Phenol) | Dissolves ✓ | No reaction ✗ |
| ROH (Alcohol) | No reaction ✗ | No reaction ✗ |
Memory Trick: “Carboxylic acids are Champions - react even with weak base (NaHCO₃)!”
Question: How to distinguish between benzoic acid and phenol?
Answer: NaHCO₃ test
- Benzoic acid: Effervescence (CO₂ gas) ✓
- Phenol: No reaction ✗
Reason: Carboxylic acids are stronger acids than carbonic acid (H₂CO₃), so they displace CO₂ from bicarbonate!
Acid strength: RCOOH > H₂CO₃ > PhOH
Reaction 2: Formation of Acid Derivatives
(a) Acid Chloride (Acyl Chloride) Formation
Reagents: SOCl₂ (thionyl chloride), PCl₃, PCl₅
Reaction:
SOCl₂ O
R-COOH ──────→ R-C-Cl + SO₂↑ + HCl↑
(Acid chloride)
Example:
SOCl₂
CH₃COOH ──────→ CH₃COCl + SO₂ + HCl
(Acetyl chloride)
Why SOCl₂ is preferred? Byproducts (SO₂ and HCl) are gases - easy to remove!
With PCl₅:
R-COOH + PCl₅ → R-COCl + POCl₃ + HCl
Memory Trick: “SOCl₂ is Smart - gaseous byproducts Separate easily!”
(b) Ester Formation (Esterification)
Fischer Esterification:
R'-OH, conc. H₂SO₄
R-COOH + ──────────────────────→ R-COO-R' + H₂O
(Heat) (Ester)
Example:
CH₃OH, H₂SO₄
CH₃COOH ────────────────→ CH₃-COO-CH₃ + H₂O
(Heat) (Methyl acetate)
Mechanism:
Step 1: Protonation of carbonyl oxygen
O OH⁺
‖ H⁺ |
R-C-OH ──────→ R-C-OH
Step 2: Nucleophilic attack by alcohol
OH⁺ OH
| ROH |
R-C-OH ──────────→ R-C-OH
|
OR'
Step 3: Proton transfer + loss of H₂O
OH O
| -H₂O ‖
R-C-OH ──────────→ R-C-OR' + H₂O
|
OR'
Key Points:
- Equilibrium reaction (reversible!)
- Acid (H₂SO₄) acts as catalyst and dehydrating agent
- To shift equilibrium toward ester: use excess alcohol or remove water
Reverse Reaction (Ester Hydrolysis):
H₂O, H⁺ (or OH⁻)
R-COO-R' ──────────────────→ R-COOH + R'-OH
(Heat)
(c) Amide Formation
Reaction (via Acid Chloride):
(1) R-COCl
NH₃ + ──────────→ R-CO-NH₂ + HCl
(2) or heat with RCOOH (Amide)
Example:
CH₃COCl
NH₃ + ─────────→ CH₃-CO-NH₂
(Acetamide)
Note: Direct heating of RCOOH + NH₃ gives ammonium salt first, which must be heated strongly to get amide!
Heat
RCOOH + NH₃ ────→ RCOO⁻NH₄⁺ ──Strong heat──→ RCONH₂ + H₂O
(Ammonium salt) (Amide)
Memory Trick: “Acid chloride + Ammonia = Amide (easily!)”
(d) Acid Anhydride Formation
Reaction:
P₂O₅ (dehydrating agent)
2 R-COOH ────────────────────────→ R-CO-O-CO-R + H₂O
(Heat) (Acid anhydride)
Example:
P₂O₅
2 CH₃COOH ──────→ (CH₃CO)₂O + H₂O
(Acetic anhydride)
Reaction 3: Reduction
Reagent: LiAlH₄ (powerful reducer)
Reaction:
LiAlH₄/Ether
R-COOH ──────────────→ R-CH₂-OH
then H₃O⁺ (1° alcohol)
Example:
LiAlH₄
CH₃COOH ──────────→ CH₃-CH₂-OH
(Ethanol)
Note: NaBH₄ does NOT reduce carboxylic acids (too weak)!
Memory Trick: “LiAlH₄ is Like a Lion - reduces even stubborn acids!”
Reaction 4: Decarboxylation (Loss of CO₂)
Sodalime Decarboxylation:
NaOH + CaO
R-COOH ──────────────→ R-H + Na₂CO₃
(Heat, 600 K) (Alkane, one less carbon!)
Example:
Sodalime
CH₃COOH ─────────→ CH₄ + Na₂CO₃
(Heat) (Methane)
Use: Convert carboxylic acid to alkane with one less carbon
Kolbe Electrolysis (for carboxylate salts):
Electrolysis
2 R-COO⁻Na⁺ ───────────→ R-R + 2 CO₂ + 2 Na⁺
(Alkane with 2R groups)
Example:
Electrolysis
2 CH₃COO⁻Na⁺ ──────────────→ CH₃-CH₃ + 2 CO₂
(Ethane)
Memory Trick: “Decarboxylation Decrements carbon count!”
Reaction 5: Hell-Volhard-Zelinsky Reaction (α-Halogenation)
Reaction:
Cl₂ or Br₂ O
P (catalyst) ‖
R-CH₂-COOH ──────────────→ R-CH(X)-C-OH
(α-Halo acid)
Example:
Br₂/P
CH₃-CH₂-COOH ──────────→ CH₃-CH(Br)-COOH
(2-Bromopropanoic acid)
Mechanism (Simplified):
Step 1: Acid → Acid chloride (by PCl₃/P + Cl₂)
R-CH₂-COOH → R-CH₂-COCl
Step 2: Enolization + halogenation
R-CH₂-COCl ⇌ R-CH=C(OH)-Cl ──X₂──→ R-CH(X)-COCl
Step 3: Hydrolysis back to acid
R-CH(X)-COCl + H₂O → R-CH(X)-COOH + HCl
Key Point: Halogenation occurs specifically at α-position (carbon next to -COOH)!
Use: Introduces halogen at α-position for further substitution (SN2 to introduce other groups)
Carboxylic Acid Derivatives: The Big Four
Order of Reactivity (Nucleophilic Acyl Substitution):
$$\boxed{\text{Acid chloride} > \text{Acid anhydride} > \text{Ester} > \text{Amide}}$$Why this order?
- Leaving group ability: Cl⁻ > RCOO⁻ > RO⁻ > NH₂⁻
- More stable leaving group → more reactive derivative
Interconversion of Derivatives
General Pattern: Higher reactivity derivative can make lower reactivity derivatives!
Acid Chloride → Acid Anhydride → Ester → Amide
(Most reactive) (Least reactive)
Examples:
(1) Acid Chloride → Ester:
R-COCl + R'-OH → R-COO-R' + HCl
(2) Acid Chloride → Amide:
R-COCl + NH₃ → R-CO-NH₂ + HCl
(3) Acid Anhydride → Ester:
(RCO)₂O + R'-OH → R-COO-R' + R-COOH
(4) Ester → Amide:
R-COO-R' + NH₃ → R-CO-NH₂ + R'-OH (slow, needs heat!)
Memory Trick: “Acid chloride is Almighty - makes all other derivatives!”
Question Type: “Identify the most reactive derivative toward nucleophilic attack.”
Answer: Acid chloride (best leaving group Cl⁻)
Reactivity Order:
R-COCl > (RCO)₂O > R-COOR' > R-CONH₂
↑ ↑ ↑ ↑
(Fastest) (Fast) (Slow) (Slowest)
Why? Better leaving group → faster nucleophilic acyl substitution!
Special Carboxylic Acids
Formic Acid (HCOOH) - The Unique One
Structure:
O
‖
H-C-O-H
Special Property: Has both aldehyde and acid functional groups!
Why?
- Aldehyde: Has H on carbonyl carbon (like R-CHO)
- Acid: Has -COOH group
Unique Reactions:
(1) Reduces Tollens’ Reagent (like aldehydes):
HCOOH + [Ag(NH₃)₂]⁺ → CO₂ + Ag↓ + H₂O
(Silver mirror!)
No other carboxylic acid does this!
(2) Reduces Fehling’s Solution:
HCOOH + Fehling's → CO₂ + Cu₂O↓
(Red precipitate)
(3) Decomposes on heating:
conc. H₂SO₄
HCOOH ─────────────→ CO + H₂O (at lower temp)
Heat
or
HCOOH ──────→ CO₂ + H₂ (at higher temp)
Memory Trick: “Formic acid is Fancy - acts like aldehyde too!”
Oxalic Acid (HOOC-COOH) - The Dicarboxylic Acid
Structure:
O O
‖ ‖
HO-C-C-O-H
Special Properties:
(1) Reduces KMnO₄ (gets oxidized itself):
5 (COOH)₂ + 2 KMnO₄ + 3 H₂SO₄ → 10 CO₂ + 2 MnSO₄ + K₂SO₄ + 8 H₂O
Use: Oxalic acid used in titrations to standardize KMnO₄!
(2) Decarboxylation on heating:
Heat
(COOH)₂ ──────→ HCOOH + CO₂
(>150°C)
Heat (>200°C)
HCOOH ──────────────→ CO + H₂O
Overall:
(COOH)₂ ──Heat──→ CO₂ + CO + H₂O
Memory Tricks & Patterns
The Master Mnemonic: “ACIDS CREATE DERIVATIVES EFFICIENTLY”
Acidity (strongest organic acid, reacts with NaHCO₃)
Carboxylate (resonance-stabilized, two equivalent O atoms)
Inductive effect (EWG increases, EDG decreases acidity)
Dimer formation (H-bonding in vapor phase)
SODCl₂ (makes acid chloride cleanly)
Chloride (most reactive derivative)
Reduction (LiAlH₄ → 1° alcohol)
Esterification (Fischer: acid + alcohol + H₂SO₄)
Amide (from acid chloride + NH₃)
Terminal oxidation product (from 1° alcohol or aldehyde)
EWG (increases acidity via -I effect)
Decarboxylation (sodalime → alkane, -1 carbon)
Electrophilic acyl carbon (in derivatives)
Reactivity order (acid chloride > anhydride > ester > amide)
Interconversion (higher reactivity → lower reactivity)
Vapor density (doubled due to dimers!)
Anhydride (two acids lose H₂O)
Titration (oxalic acid standardizes KMnO₄)
Isomerism (positional for substituted acids)
Very high BP (strong H-bonding dimers)
Esterification equilibrium (reversible, Le Chatelier)
Saponification (ester hydrolysis, soap-making)
Electron-withdrawing (F, Cl, NO₂) increases acidity
Formic acid (unique, reduces Tollens’/Fehling’s)
Fischer esterification (acid + alcohol + H₂SO₄ catalyst)
Increase chain (nitrile hydrolysis, Grignard + CO₂)
Chain decrease (decarboxylation)
Ionic salt (with NaOH/NaHCO₃)
Electrophile (acid chloride in acylation)
Nucleophilic acyl substitution (derivative reactions)
Two O atoms (carboxylate resonance)
LiAlH₄ (powerful, reduces to alcohol)
Yield (esterification: excess ROH or remove H₂O)
Acidity Ranking Strategy
Step 1: Identify substituents on carboxylic acid Step 2: Classify as EWG or EDG Step 3: Consider distance from -COOH (α > β > γ) Step 4: For halogens, electronegativity order: F > Cl > Br > I
Example:
Rank: CH₃COOH, FCH₂COOH, F₃CCOOH, CH₃CH₂COOH
Analysis:
- CH₃COOH: weak EDG (baseline)
- FCH₂COOH: 1 F (strong EWG)
- F₃CCOOH: 3 F atoms (very strong EWG!)
- CH₃CH₂COOH: EDG (weaker acid than CH₃COOH)
Increasing acidity: CH₃CH₂COOH < CH₃COOH < FCH₂COOH < F₃CCOOH
Common Mistakes to Avoid
Wrong: Using NaBH₄ to reduce RCOOH to R-CH₂-OH Right: Use LiAlH₄ (much more powerful!)
Remember:
- NaBH₄: Reduces aldehydes/ketones only
- LiAlH₄: Reduces aldehydes, ketones, esters, AND carboxylic acids
JEE Impact: This is tested frequently!
Wrong: Expecting 100% conversion to ester Right: It’s an equilibrium reaction!
To shift toward ester (Le Chatelier’s Principle):
- Use excess alcohol
- Remove water (use dehydrating agent or Dean-Stark trap)
- Use concentrated H₂SO₄ (acid catalyst + dehydrating agent)
Remember: RCOOH + R’OH ⇌ RCOOR’ + H₂O
Wrong: Thinking all carboxylic acids reduce Tollens’/Fehling’s Right: Only HCOOH (formic acid) does this!
Why? Formic acid has H on carbonyl carbon (aldehyde-like behavior)
Structure:
O
‖
H-C-OH (H on C=O carbon, like aldehydes!)
JEE Pattern: “Which carboxylic acid reduces Tollens’ reagent?” → HCOOH
Wrong: Thinking amides are most reactive (they have N!) Right: Amides are LEAST reactive in nucleophilic acyl substitution!
Reactivity Order:
Acid chloride > Anhydride > Ester > Amide
(Most reactive) (Least)
Why? NH₂⁻ is poorest leaving group (very basic, unstable)
Remember: Better leaving group → more reactive derivative!
Wrong: Thinking -OCH₃ increases acidity (it has oxygen!) Right: -OCH₃ is an EDG (electron-donating) → decreases acidity!
EWG (increase acidity): -NO₂, -CN, -F, -Cl, -Br, -I, -CHO, -COR EDG (decrease acidity): -CH₃, -C₂H₅, -OCH₃, -NH₂, -OH
Why? +M effect of -OCH₃ dominates (donates electrons via lone pair resonance)
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Why is acetic acid more acidic than ethanol?
Solution:
Acidity: CH₃COOH (pKₐ ~ 4.76) » CH₃CH₂OH (pKₐ ~ 16)
Reason: Resonance stabilization of conjugate base
Acetic acid:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
Acetate ion (CH₃COO⁻) has resonance:
O⁻ O
‖ ↔ ‖
CH₃-C-O CH₃-C-O⁻
Two equivalent resonance structures!
Negative charge equally distributed between both O atoms.
VERY STABLE conjugate base → readily releases H⁺ → MORE ACIDIC
Ethanol:
CH₃CH₂OH ⇌ CH₃CH₂O⁻ + H⁺
Ethoxide ion (CH₃CH₂O⁻) has NO resonance.
Negative charge localized on single O atom.
LESS STABLE conjugate base → reluctant to release H⁺ → LESS ACIDIC
In short: Resonance stabilization of carboxylate makes carboxylic acids much more acidic!
Question: How will you distinguish between: (a) Benzoic acid and phenol (b) Acetic acid and acetone
Solution:
(a) Benzoic acid (C₆H₅COOH) vs Phenol (C₆H₅OH)
Test: NaHCO₃ (sodium bicarbonate)
- Benzoic acid: Effervescence (CO₂ gas evolves) ✓
C₆H₅COOH + NaHCO₃ → C₆H₅COO⁻Na⁺ + H₂O + CO₂↑ - Phenol: No reaction ✗ (not acidic enough to displace CO₂ from bicarbonate)
(b) Acetic acid (CH₃COOH) vs Acetone (CH₃COCH₃)
Test 1: NaHCO₃
- Acetic acid: CO₂ evolves ✓
- Acetone: No reaction ✗ (ketone, not acidic)
Test 2: Tollens’ Reagent
- Acetic acid: No reaction ✗ (acids don’t reduce Tollens')
- Acetone: No reaction ✗ (ketones don’t reduce Tollens')
Test 3: 2,4-DNP
- Acetic acid: No reaction ✗ (no C=O with just carbonyl, has -COOH)
- Acetone: Orange precipitate ✓ (ketone, has C=O)
Best for (b): NaHCO₃ or 2,4-DNP test
Question: Complete the reactions:
(a) CH₃COOH + NaOH → ?
(b) CH₃COOH + SOCl₂ → ?
(c) CH₃COOH + LiAlH₄ → ?
Solution:
(a) Neutralization (Salt formation):
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
(Sodium acetate)
(b) Acid chloride formation:
CH₃COOH + SOCl₂ → CH₃COCl + SO₂↑ + HCl↑
(Acetyl chloride)
(c) Reduction to alcohol:
LiAlH₄
CH₃COOH ──────────→ CH₃CH₂OH
then H₃O⁺ (Ethanol, 1° alcohol)
Level 2: JEE Main Style
Question: Arrange the following in increasing order of acidity: I. CH₃COOH II. ClCH₂COOH III. Cl₂CHCOOH IV. Cl₃CCOOH
(A) I < II < III < IV (B) IV < III < II < I (C) I < IV < III < II (D) II < I < III < IV
Solution: (A) I < II < III < IV
Analysis:
All compounds have same basic structure: X-COOH where X changes.
Effect of Cl (electron-withdrawing group):
- Cl pulls electron density away from -COO⁻ (via -I effect)
- Stabilizes carboxylate ion
- Makes acid more acidic
More Cl atoms → stronger -I effect → more stable carboxylate → more acidic
I. CH₃COOH:
- No Cl, only CH₃ (weak EDG)
- Least acidic
II. ClCH₂COOH:
- 1 Cl atom → moderate -I effect
III. Cl₂CHCOOH:
- 2 Cl atoms → stronger -I effect
IV. Cl₃CCOOH:
- 3 Cl atoms → strongest -I effect
- Most acidic
Increasing acidity: I < II < III < IV ✓
Answer: (A)
Question: Which of the following carboxylic acid derivatives is MOST reactive toward nucleophilic attack?
(A) Amide (RCONH₂) (B) Ester (RCOOR’) (C) Acid anhydride ((RCO)₂O) (D) Acid chloride (RCOCl)
Solution: (D) Acid chloride
Reactivity Order:
$$\text{Acid chloride} > \text{Anhydride} > \text{Ester} > \text{Amide}$$Reason: Leaving group ability
(D) Acid chloride (RCOCl):
- Leaving group: Cl⁻ (weak base, stable)
- Best leaving group → Most reactive ✓
(C) Anhydride ((RCO)₂O):
- Leaving group: RCOO⁻ (carboxylate, stable)
- Good leaving group → reactive
(B) Ester (RCOOR’):
- Leaving group: RO⁻ (alkoxide, less stable)
- Moderate leaving group → less reactive
(A) Amide (RCONH₂):
- Leaving group: NH₂⁻ (very basic, very unstable!)
- Poorest leaving group → Least reactive ✗
Answer: (D)
Memory: Better leaving group → more reactive derivative!
Question: Benzoic acid can be prepared from toluene by:
(A) Oxidation with O₂ (B) Oxidation with KMnO₄/H⁺ (C) Reduction with LiAlH₄ (D) Both (A) and (B)
Solution: (B) Oxidation with KMnO₄/H⁺
Reaction:
KMnO₄/H⁺
C₆H₅-CH₃ ──────────→ C₆H₅-COOH
(Toluene) (Heat) (Benzoic acid)
Analysis:
(A) Oxidation with O₂:
- Toluene does NOT easily oxidize with just O₂ at normal conditions
- Needs catalyst and high temperature (industrial process)
- Not the standard lab method ✗
(B) Oxidation with KMnO₄/H⁺:
- Standard lab method ✓
- Complete oxidation of methyl group to -COOH
- This works! ✓
(C) Reduction with LiAlH₄:
- LiAlH₄ is a reducing agent, not oxidizing!
- Cannot convert CH₃ to COOH (opposite direction!)
- Wrong direction ✗
(D) Both (A) and (B):
- (A) doesn’t work under normal lab conditions
- Only (B) works
Answer: (B)
Level 3: JEE Advanced Style
Question: A carboxylic acid (A) with molecular formula C₃H₆O₂ reacts with SOCl₂ to give (B). (B) reacts with benzene in the presence of AlCl₃ to give (C), which has molecular formula C₉H₁₀O. Identify (A), (B), and (C) with reactions.
Solution:
Analysis:
Compound A (C₃H₆O₂):
- Carboxylic acid with 3 carbons → CH₃CH₂COOH (propanoic acid)
Reaction 1: A + SOCl₂ → B
SOCl₂
CH₃CH₂COOH ──────→ CH₃CH₂COCl + SO₂ + HCl
(A) (B, Propanoyl chloride)
Compound B = Propanoyl chloride (CH₃CH₂COCl)
Reaction 2: B + Benzene + AlCl₃ → C (Friedel-Crafts Acylation)
AlCl₃
CH₃CH₂COCl + C₆H₆ ──────→ C₆H₅-CO-CH₂-CH₃ + HCl
(B) (C, Propiophenone)
Compound C = Propiophenone (C₆H₅-CO-CH₂-CH₃, C₉H₁₀O)
Verification:
- A = C₃H₆O₂ = CH₃CH₂COOH ✓
- B = acid chloride (from SOCl₂) = CH₃CH₂COCl ✓
- C = C₉H₁₀O = C₆H₅-CO-C₂H₅ ✓ (6C from benzene + 3C from acid chloride = 9C total)
Answer:
- A = Propanoic acid (CH₃CH₂COOH)
- B = Propanoyl chloride (CH₃CH₂COCl)
- C = Propiophenone (C₆H₅-CO-CH₂-CH₃)
Question: How would you carry out the following conversions? (a) Ethanol → Propanoic acid (chain extension by 1 carbon) (b) Benzene → Benzoic acid
Solution:
(a) Ethanol → Propanoic acid (C₂ → C₃)
Strategy: Chain extension via nitrile
Step 1: Convert alcohol to alkyl halide
PBr₃ or HBr
CH₃CH₂OH ──────────→ CH₃CH₂Br
(Bromoethane)
Step 2: Nucleophilic substitution with CN⁻ (chain extension!)
KCN (or NaCN)
CH₃CH₂Br ────────────→ CH₃CH₂CN
(DMF/DMSO) (Propanenitrile)
Step 3: Hydrolysis of nitrile
H₃O⁺ or OH⁻/H₂O
CH₃CH₂CN ────────────────→ CH₃CH₂COOH
(Heat) (Propanoic acid)
Overall: CH₃CH₂OH → CH₃CH₂Br → CH₃CH₂CN → CH₃CH₂COOH
Alternative Route (via Grignard):
Step 1: CH₃CH₂OH → CH₃CH₂Br (as above)
Step 2: Grignard reagent formation
Mg/dry ether
CH₃CH₂Br ──────────────→ CH₃CH₂MgBr
Step 3: CO₂ insertion
(1) CO₂
CH₃CH₂MgBr ─────────→ CH₃CH₂COOH
(2) H₃O⁺
(b) Benzene → Benzoic acid
Route 1: Via Friedel-Crafts Alkylation + Oxidation
Step 1: Friedel-Crafts alkylation
CH₃Cl/AlCl₃
C₆H₆ ────────────→ C₆H₅-CH₃
(Toluene)
Step 2: Oxidation
KMnO₄/H⁺
C₆H₅-CH₃ ──────────→ C₆H₅-COOH
(Heat) (Benzoic acid)
Route 2: Via Grignard + CO₂
Step 1: Halogenation
Br₂/FeBr₃
C₆H₆ ──────────→ C₆H₅-Br
(Bromobenzene)
Step 2: Grignard formation
Mg/dry ether
C₆H₅-Br ──────────→ C₆H₅-MgBr
Step 3: CO₂ insertion
(1) CO₂
C₆H₅-MgBr ─────────→ C₆H₅-COOH
(2) H₃O⁺
Route 1 is simpler and more commonly used!
Question: A compound (X) with molecular formula C₄H₆O₄ reacts with NaHCO₃ to evolve CO₂ gas. (X) also reduces KMnO₄ solution. On heating, (X) loses CO₂ to form (Y) with formula C₃H₄O₃. Identify (X) and (Y).
Solution:
Analysis:
- Reacts with NaHCO₃ → CO₂ evolves → (X) is a carboxylic acid
- Reduces KMnO₄ → (X) has reducing property (unusual for carboxylic acids!)
- C₄H₆O₄ with reducing property + acid → likely oxalic acid derivative OR formic acid derivative
- Loses CO₂ on heating (decarboxylation) → C₄H₆O₄ → C₃H₄O₃
Key Insight: Only formic acid (HCOOH) among carboxylic acids reduces KMnO₄!
Working out molecular formula:
- If (X) contains HCOOH units…
- C₄H₆O₄ could be malonic acid (HOOC-CH₂-COOH)? No, that’s C₃H₄O₄
- Wait! Malic acid or Succinic acid? Check formulas…
Better approach: (X) reduces KMnO₄ + has two -COOH groups (reacts with NaHCO₃, loses CO₂)
Compound X = Malonic acid (HOOC-CH₂-COOH)? No, that’s C₃H₄O₄…
Actually: Let’s recalculate:
- Oxalic acid: (COOH)₂ = C₂H₂O₄ (reduces KMnO₄!)
- Malonic acid: HOOC-CH₂-COOH = C₃H₄O₄
- Succinic acid: HOOC-CH₂-CH₂-COOH = C₄H₆O₄ ✓
But succinic acid doesn’t reduce KMnO₄!
Rethinking: What if (X) is malic acid (has -OH group too)?
HOOC-CH(OH)-CH₂-COOH = C₄H₆O₅ (No, wrong formula!)
Correct Answer: Tartaric acid? No…
Actually, the answer is MALEIC ACID or FUMARIC ACID (have C=C!):
Compound X = Maleic acid (cis-butenedioic acid)
COOH
/
C=C
\
COOH
Molecular formula: C₄H₄O₄ (Hmm, not C₄H₆O₄...)
Let me reconsider: C₄H₆O₄ and reduces KMnO₄…
Final Answer (most likely):
X = Malic acid (2-Hydroxybutanedioic acid, HOOC-CH(OH)-CH₂-COOH)
- Formula: C₄H₆O₅ (Wait, extra O!)
Actually, given constraints, X is probably:
X = Succinic acid (HOOC-CH₂-CH₂-COOH, C₄H₆O₄)
But it doesn’t reduce KMnO₄… There’s a contradiction in the question!
Most consistent answer:
- X = Oxalic acid (HOOC-COOH, C₂H₂O₄) - but wrong formula!
Given the question, most likely: X = Malonic acid (HOOC-CH₂-COOH, C₃H₄O₄) Decarboxylation: HOOC-CH₂-COOH → CH₃COOH + CO₂ (C₃H₄O₄ → C₂H₄O₂, not C₃H₄O₃!)
The question has inconsistencies. For JEE, remember:
- Oxalic acid reduces KMnO₄ (gets oxidized to CO₂)
- Decarboxylation decreases carbon count
Quick Revision Box
| Situation | Formula/Approach | Key Point |
|---|---|---|
| Strongest organic acid | RCOOH (pKₐ ~ 4-5) | Resonance-stabilized carboxylate |
| Distinguish from phenol | NaHCO₃ test | RCOOH → CO₂ evolves; PhOH → no reaction |
| Make acid chloride | SOCl₂ | Byproducts are gases (easy removal) |
| Esterification | ROH + H₂SO₄, heat | Equilibrium: excess ROH or remove H₂O |
| Reduce to alcohol | LiAlH₄ | NaBH₄ too weak! |
| Increase acidity | Add EWG (F, Cl, NO₂) | Stabilizes carboxylate via -I effect |
| Decrease acidity | Add EDG (CH₃, OCH₃) | Destabilizes carboxylate |
| Decarboxylation | Sodalime, heat | RCOOH → RH + CO₂ (-1 carbon) |
| Chain extension | R-X → R-CN → R-COOH | Or RMgX + CO₂ |
| Most reactive derivative | Acid chloride (RCOCl) | Best leaving group (Cl⁻) |
| Least reactive derivative | Amide (RCONH₂) | Poorest leaving group (NH₂⁻) |
| Formic acid special | Reduces Tollens’/Fehling’s | Only RCOOH with H on C=O carbon |
| Dimer formation | H-bonding in vapor | Doubled molecular mass |
Connection to Other Topics
Prerequisites (Review these first):
- Aldehydes & Ketones - oxidation to carboxylic acids
- Alcohols - oxidation pathway (1° alcohol → aldehyde → carboxylic acid)
- Organic Principles: Resonance - for understanding acidity
Related Topics:
- Phenols - acidity comparison, NaHCO₃ test distinction
- Ethers - esterification comparison
- Nitrogen Compounds: Amines - amide formation, basicity vs acidity
Advanced Applications:
- Amino acids (both -COOH and -NH₂ groups)
- Soaps and detergents (long-chain carboxylates)
- Aspirin synthesis (from salicylic acid)
- Polymer chemistry (polyesters from dicarboxylic acids)
Teacher’s Summary
1. Acidity = Resonance Stabilization:
- Carboxylate ion (RCOO⁻) has two equivalent resonance structures
- Negative charge equally shared between both O atoms
- Most stable conjugate base among organic acids → most acidic
2. Substituent Effects (Inductive):
- EWG (F, Cl, Br, NO₂, CN): Increase acidity (stabilize carboxylate)
- EDG (CH₃, C₂H₅, OCH₃, NH₂): Decrease acidity (destabilize carboxylate)
- Distance matters: α > β > γ (closer = stronger effect)
3. Preparation Methods:
- Oxidation: 1° alcohol → aldehyde → carboxylic acid (K₂Cr₂O₇/H⁺)
- Alkylbenzene oxidation: C₆H₅-R → C₆H₅-COOH (KMnO₄/H⁺)
- Nitrile hydrolysis: R-CN → R-COOH (chain extension!)
- Grignard + CO₂: R-MgX + CO₂ → R-COOH (chain extension!)
4. Derivative Formation:
- Acid chloride: SOCl₂ (clean, gaseous byproducts)
- Ester: Fischer esterification (ROH + H₂SO₄, equilibrium!)
- Amide: Via acid chloride + NH₃ (or direct heating with salt)
- Anhydride: P₂O₅ dehydration
5. Reactivity Order of Derivatives:
$$\text{Acid chloride} > \text{Anhydride} > \text{Ester} > \text{Amide}$$Why? Leaving group ability: Cl⁻ > RCOO⁻ > RO⁻ > NH₂⁻
6. Special Reactions:
- Reduction: LiAlH₄ → 1° alcohol (NaBH₄ doesn’t work!)
- Decarboxylation: Sodalime → alkane (-1 carbon)
- α-Halogenation: Hell-Volhard-Zelinsky (Br₂/P)
7. Special Carboxylic Acids:
- Formic acid (HCOOH): Only acid that reduces Tollens’/Fehling’s (has H on C=O)
- Oxalic acid ((COOH)₂): Reduces KMnO₄, used in titrations
8. JEE Strategy:
- Master acidity ranking (EWG/EDG, distance effects)
- Know qualitative test: NaHCO₃ distinguishes RCOOH from PhOH
- Understand derivative reactivity order (acid chloride > anhydride > ester > amide)
- Remember LiAlH₄ (not NaBH₄!) for reduction
- Practice chain extension (nitrile, Grignard routes)
- Know Fischer esterification is reversible (Le Chatelier!)
“Carboxylic acids are the terminal oxidation station - you can’t oxidize them further without breaking bonds! Their acidity comes from perfect resonance in the carboxylate ion, making them the most acidic organic compounds. Master the derivative conversions, and you’ve conquered half of organic functional group transformations!”
Previous Topic: ← Aldehydes & Ketones - The carbonyl powerhouses Next Steps: Explore Nitrogen Compounds to see how -NH₂ behaves with -COOH!