The Hook: Antiseptics and Explosives - Same Molecule?
Remember those old “Dettol” antiseptic bottles in hospitals? The active ingredient was phenol (carbolic acid). But here’s the twist - the same compound that kills bacteria is also used to make explosives like picric acid and plastics like Bakelite!
Even more surprising: In Oppenheimer (2023), when they showed the Manhattan Project labs, those brown bottles? Many contained phenol-based compounds used in early nuclear chemistry!
The JEE Question: Why is phenol acidic (reacts with NaOH) while alcohols are not? Why does phenol give colored complexes with FeCl₃ but benzyl alcohol doesn’t? Get this right, and you’ll crack 6+ marks in JEE Advanced organic section!
The Core Concept: Phenol vs Alcohol - The Great Divide
General Structure
$$\boxed{\text{C}_6\text{H}_5\text{-OH (hydroxyl on benzene ring)}}$$In simple terms: Phenol is what you get when you attach -OH directly to a benzene ring. Think of it as giving an aromatic ring a “smart power-up” that completely changes its personality!
Critical Distinction:
| Compound | Structure | Carbon Type | Acidic? |
|---|---|---|---|
| Phenol | C₆H₅-OH | sp² (aromatic) | YES (pKₐ ≈ 10) |
| Alcohol | R-OH | sp³ (aliphatic) | NO (pKₐ ≈ 16) |
| Benzyl alcohol | C₆H₅-CH₂-OH | sp³ (methylene) | NO (it’s an alcohol!) |
Wrong: C₆H₅-CH₂-OH is a phenol Right: This is benzyl alcohol - the -OH is on sp³ carbon (CH₂), NOT directly on the benzene ring!
Test:
- Phenol + FeCl₃ → violet color ✓
- Benzyl alcohol + FeCl₃ → no color ✗
JEE loves this trap!
Why is Phenol Acidic? The Resonance Story
Acidity Comparison
$$\boxed{\text{Carboxylic acid} >> \text{Phenol} > \text{Water} > \text{Alcohol}}$$ $$\boxed{\text{pK}_a: \text{R-COOH (4-5)} << \text{PhOH (10)} < \text{H}_2\text{O (15.7)} < \text{R-OH (16)}}$$The Resonance Explanation
When phenol loses H⁺:
C₆H₅-OH ⇌ C₆H₅-O⁻ + H⁺
(Phenoxide ion - STABLE due to resonance!)
Resonance structures of phenoxide ion:
O⁻ O O O
| ‖ | ‖
/═\\ /═\\ /═\\ /═\\
| | ↔ | | ↔ | | ↔ | |
\═/ \═/ \═/ \═/
(Negative charge delocalized into benzene ring via resonance)
In alcohols (R-O⁻):
- No resonance stabilization
- Negative charge localized on oxygen
- Less stable → less willing to release H⁺ → less acidic
Memory Trick: “Phenoxide has Perfect Places for negative charge to Park (resonance!) - makes it stable, so phenol is acidic!”
Interactive Demo: Visualize Phenol Structure
Explore the resonance structures of phenoxide ion and understand acidity.
Effect of Substituents on Acidity
Electron-Withdrawing Groups (EWG): Increase acidity
- NO₂ (nitro) - strongest
- CN (cyano)
- CHO (formyl)
- X (halogens: F > Cl > Br > I)
Electron-Donating Groups (EDG): Decrease acidity
- OH (hydroxyl)
- OCH₃ (methoxy)
- NH₂ (amino)
- CH₃, C₂H₅ (alkyl groups)
Acidity Order:
$$\boxed{p\text{-NO}_2\text{-C}_6\text{H}_4\text{OH} > m\text{-NO}_2\text{-C}_6\text{H}_4\text{OH} > \text{C}_6\text{H}_5\text{OH} > p\text{-CH}_3\text{-C}_6\text{H}_4\text{OH}}$$Why?
- NO₂ at para/ortho: Stabilizes phenoxide ion by resonance (pulls electron density)
- CH₃ at para/ortho: Destabilizes phenoxide ion by +I effect (pushes electron density)
Positional Effect:
- For EWG: ortho/para > meta (direct resonance stabilization)
- For EDG: All positions decrease acidity, but ortho/para more so
Memory Trick: “Nitro Normalizes Negative charge - makes phenoxide stable - increases acidity!”
Common Question: Arrange in increasing order of acidity: I. Phenol II. p-Nitrophenol III. p-Cresol (p-methylphenol) IV. m-Nitrophenol
Strategy:
- Identify EWG (NO₂) vs EDG (CH₃)
- NO₂ increases acidity, CH₃ decreases
- For NO₂: para/ortho > meta (resonance)
Answer: III < I < IV < II (increasing acidity)
Why?
- p-Cresol (CH₃ is EDG) < Phenol (no substituent) < m-Nitrophenol (NO₂ but meta) < p-Nitrophenol (NO₂ at para - maximum resonance)
Preparation of Phenols
Method 1: From Chlorobenzene (Dow Process)
Reaction:
NaOH (aq), 623 K
C₆H₅-Cl ────────────────→ C₆H₅-ONa ──H⁺──→ C₆H₅-OH
350 atm (Sodium phenoxide) (Phenol)
Conditions: High temperature (350°C) and high pressure (350 atm) required!
Mechanism: Nucleophilic aromatic substitution (NAS)
Step 1: SNAr mechanism (addition-elimination)
Cl Cl OH
| | |
/═\\ /═\\ /═\\
| | + OH⁻ → | | → -Cl⁻ → | |
\═/ \═/ \═/
OH
(Meisenheimer complex intermediate)
Why harsh conditions? Benzene ring is electron-rich, resistant to nucleophilic attack! Need high energy to overcome activation barrier.
Memory Trick: “Dow process Demands Drastic conditions - high temp, high pressure!”
Method 2: From Benzenesulfonic Acid (Alkali Fusion)
Reaction:
NaOH (fused)
C₆H₅-SO₃H ─────────────→ C₆H₅-ONa ──H⁺──→ C₆H₅-OH
623 K
Advantage: Easier than Dow process (SO₃H is better leaving group than Cl)
Method 3: From Diazonium Salts (Most Important for JEE!)
Reaction:
H₂O, warm
C₆H₅-N₂⁺Cl⁻ ─────────→ C₆H₅-OH + N₂↑ + HCl
or H⁺/H₂O
Full Sequence (from aniline):
Step 1: Diazotization
C₆H₅-NH₂ + NaNO₂ + 2HCl ──0-5°C──→ C₆H₅-N₂⁺Cl⁻ + NaCl + 2H₂O
(Aniline) (Benzenediazonium chloride)
Step 2: Hydrolysis
C₆H₅-N₂⁺Cl⁻ + H₂O ──warm──→ C₆H₅-OH + N₂↑ + HCl
Key Points:
- Diazonium salt is stable only at 0-5°C (cold!)
- Warming releases N₂ gas (good leaving group!)
- This is the best lab method for phenol
Memory Trick: “Diazonium is Delicate - keep it Damn cold (0-5°C)! Warm it up to get phenol!”
Question Pattern: “Convert aniline to phenol”
Answer (2 steps):
- NaNO₂ + HCl at 0-5°C (diazotization)
- Warm with H₂O (hydrolysis)
Why this is tested: Tests knowledge of diazonium chemistry + functional group conversion
Method 4: From Cumene (Industrial Process)
Reaction (Cumene-Phenol Process):
Step 1: Oxidation
CH(CH₃)₂ C(CH₃)₂-OOH
| |
/═\\ O₂ /═\\
| | ─────────→ | |
\═/ \═/
(Cumene) (Cumene hydroperoxide)
Step 2: Acid-catalyzed rearrangement
C(CH₃)₂-OOH OH
| H⁺ |
/═\\ ─────────→ /═\\ + CH₃-CO-CH₃
| | | | (Acetone)
\═/ \═/
(Phenol)
Industrial Importance: Produces both phenol and acetone (two valuable chemicals!)
Memory Trick: “Cumene Cracks to give phenol + acetone - Commercially Clever!”
Reactions of Phenols
Reaction 1: Acidity - Reaction with Bases
With NaOH (distinguishes phenol from alcohol):
C₆H₅-OH + NaOH → C₆H₅-O⁻Na⁺ + H₂O
(Sodium phenoxide)
With NaHCO₃ (distinguishes phenol from carboxylic acid):
C₆H₅-OH + NaHCO₃ → NO REACTION (phenol not acidic enough!)
R-COOH + NaHCO₃ → R-COO⁻Na⁺ + H₂O + CO₂↑ (effervescence!)
Acidity Test Decision Tree:
| Compound | NaOH | NaHCO₃ | Inference |
|---|---|---|---|
| Carboxylic acid | Dissolves | CO₂ evolves | Most acidic |
| Phenol | Dissolves | No reaction | Weakly acidic |
| Alcohol | No reaction | No reaction | Not acidic |
Memory Trick: “Phenol is Picky - only reacts with strong bases (NaOH), not weak (NaHCO₃)!”
Reaction 2: Electrophilic Aromatic Substitution (The Big One!)
Why phenol is highly reactive:
- -OH group is strongly activating (+M effect)
- Donates electron density into benzene ring via resonance
- Makes ring electron-rich → loves electrophiles!
Activating Effect of -OH:
OH O⁺H O⁺H O⁺H
| | | |
/═\\ ↔ /═\\ ↔ /═\\ ↔ /═\\
| | | | | | | |
\═/ \═/ \═/ \═/
⊕ ⊕ ⊕
(ortho) (para) (meta)
Electron density increases at ortho and para positions!
Result: Electrophilic substitution occurs at ortho and para positions, and is much faster than benzene!
(a) Nitration
With dilute HNO₃:
OH OH OH
| | |
/═\\ HNO₃ /═\\ /═\\
| | ─────────→ | | NO₂ + NO₂ | |
\═/ (dilute) \═/ \═/
(o-Nitrophenol) (p-Nitrophenol)
(major, steam (minor,
volatile) sublimes)
With concentrated HNO₃ + H₂SO₄:
OH
|
/═\\ conc. HNO₃/H₂SO₄
| | ───────────────────→
\═/
NO₂
|
OH /═\\ NO₂
| | | |
NO₂
(2,4,6-Trinitrophenol, Picric acid)
Memory Trick: “Phenol is so Powerful, dilute HNO₃ is enough! Benzene needs concentrated acid + catalyst!”
JEE Pattern:
- Phenol + dil. HNO₃ → mononitration (ortho + para)
- Phenol + conc. HNO₃/H₂SO₄ → trinitration (picric acid)
(b) Halogenation
With Br₂ in CS₂ (non-polar solvent):
OH OH
| |
/═\\ Br₂/CS₂ /═\\
| | ─────────────→ | | Br (monosubstitution)
\═/ \═/
(mixture of o- and p-bromophenol)
With Br₂ water (no catalyst needed!):
OH OH
| |
/═\\ Br₂/H₂O /═\\
| | ─────────────→ Br | | Br (white ppt!)
\═/ \═/
Br
(2,4,6-Tribromophenol)
Key Point: Phenol is SO activated that it doesn’t need FeBr₃ catalyst (benzene does!)
Observation: White precipitate of tribromophenol → qualitative test for phenol!
Memory Trick: “Bromine water + phenol = Bright white precipitate - Best test for phenol!”
Benzene + Br₂:
- Needs FeBr₃ catalyst (Lewis acid)
- Gives monosubstitution (C₆H₅Br)
- Slow reaction
Phenol + Br₂ water:
- No catalyst needed!
- Gives trisubstitution (2,4,6-tribromophenol)
- Very fast reaction (immediate white ppt)
Why? -OH group activates ring via +M effect, making it electron-rich and reactive!
(c) Sulfonation
Reaction:
OH OH
| |
/═\\ conc. H₂SO₄ /═\\
| | ─────────────→ | | SO₃H
\═/ 100°C \═/
(o-Phenolsulfonic acid, major)
Temperature effect:
- 100°C → ortho product (kinetic control)
- Higher temp → para product (thermodynamic control)
Reaction 3: Kolbe’s Reaction (CO₂ insertion)
Reaction:
Step 1: Form sodium phenoxide
C₆H₅-OH + NaOH → C₆H₅-O⁻Na⁺
Step 2: Kolbe's reaction
CO₂, 400 K
C₆H₅-O⁻Na⁺ ─────────────→
4-7 atm
COO⁻Na⁺
|
/═\\ + NaOH
| |
\═/
OH
(Sodium salicylate)
Step 3: Acidification
COOH
|
/═\\ (Salicylic acid)
| |
\═/
OH
Mechanism (Simplified):
Step 1: Electrophilic attack by CO₂
O⁻ O⁻
| |
/═\\ + CO₂ → /═\\
| | | |
\═/ \═/
C=O
|
O⁻
(ortho position - sterically favorable)
Product: Salicylic acid (precursor to aspirin!)
Memory Trick: “Kolbe’s reaction Kicks in CO₂ at ortho position - Kritical for aspirin synthesis!”
JEE Application:
- Salicylic acid + CH₃COCl → Aspirin (acetylsalicylic acid)
- Salicylic acid + CH₃OH → Methyl salicylate (oil of wintergreen)
Reaction 4: Reimer-Tiemann Reaction (CHCl₃ insertion)
Reaction:
CHCl₃ + NaOH
C₆H₅-OH ─────────────→
CHO
|
/═\\ (Salicylaldehyde, o-hydroxybenzaldehyde)
| |
\═/
OH
Mechanism:
Step 1: Formation of dichlorocarbene (:CCl₂)
CHCl₃ + OH⁻ → :CCl₂ + H₂O + Cl⁻
Step 2: Electrophilic attack on phenoxide
O⁻ O⁻
| |
/═\\ + :CCl₂ → /═\\
| | | | CHCl₂
\═/ \═/
Step 3: Hydrolysis
O⁻ OH
| |
/═\\ /═\\
| | CHCl₂ → | | CHO + 2Cl⁻
\═/ \═/
Product: Salicylaldehyde (ortho-substituted)
Memory Trick: “Reimer-Tiemann Reacts to give Tiny aldehyde group at ortho!”
JEE Pattern: Kolbe vs Reimer-Tiemann
- Kolbe: CO₂ → COOH at ortho
- Reimer-Tiemann: CHCl₃ + NaOH → CHO at ortho
Reaction 5: Coupling with Diazonium Salts (Azo Dye Formation)
Reaction:
+ C₆H₅-N₂⁺Cl⁻
C₆H₅-OH ─────────────────→
NaOH, 273 K
OH
|
/═\\
| |
\═/
|
N=N
|
/═\\
| |
\═/
(p-Hydroxyazobenzene, orange dye)
Conditions:
- Basic medium (pH 9-10) - converts phenol to phenoxide ion (more reactive!)
- Cold temperature (0-5°C)
- Diazonium salt must be freshly prepared
Mechanism:
Step 1: Formation of phenoxide (more nucleophilic)
C₆H₅-OH + NaOH → C₆H₅-O⁻Na⁺
Step 2: Electrophilic attack by diazonium cation
O⁻ O⁻
| |
/═\\ + ⁺N₂-C₆H₅ → /═\\ N=N-C₆H₅
| | | |
\═/ \═/
(para position - major)
Product: Azo dye (colored compound!)
Color: Orange to red (depends on substituents)
Use: Textile dyes, food coloring, indicators
Memory Trick: “Coupling makes Colorful azo dyes - always at para position!”
All these reactions occur at ortho/para positions (activating -OH group):
| Reagent | Product | Key Point |
|---|---|---|
| Dil. HNO₃ | Mono-nitrophenol | No catalyst needed! |
| Conc. HNO₃/H₂SO₄ | Picric acid (trinitro) | Explosive |
| Br₂/H₂O | 2,4,6-Tribromophenol | White ppt, test for phenol |
| Br₂/CS₂ | Mono-bromophenol | Non-polar solvent |
| CO₂/NaOH (Kolbe) | Salicylic acid | ortho-COOH |
| CHCl₃/NaOH (Reimer-Tiemann) | Salicylaldehyde | ortho-CHO |
| C₆H₅-N₂⁺Cl⁻ | Azo dye | para-N=N-C₆H₅ |
Pattern: Phenol is SO reactive, it often doesn’t need catalysts that benzene requires!
Reaction 6: Oxidation
With oxidizing agents (K₂Cr₂O₇, FeCl₃):
[O]
C₆H₅-OH ──→ Quinone (colored products, complex mixtures)
No clear product! Unlike alcohols, phenol oxidation is complex and not useful synthetically.
Reaction 7: Esterification (Different from alcohols!)
With acid chloride (Schotten-Baumann reaction):
C₆H₅COCl
C₆H₅-OH ──────────→ C₆H₅-O-CO-C₆H₅ (Phenyl benzoate)
NaOH
OR with acetyl chloride:
CH₃COCl
C₆H₅-OH ──────────→ C₆H₅-O-CO-CH₃ (Phenyl acetate)
NaOH/Pyridine
Note: Phenol does NOT easily undergo Fischer esterification with carboxylic acids (unlike alcohols!)
- Reason: C-O bond in phenol has partial double-bond character (resonance), less reactive
Must use acid chloride or anhydride!
Special Test: Ferric Chloride (FeCl₃) Test
Reaction:
3 C₆H₅-OH + FeCl₃ → [Fe(OC₆H₅)₃] + 3 HCl
(Violet/Blue complex)
Observation: Violet or blue-green coloration
Specificity:
- Phenols: Give color ✓
- Alcohols: No color ✗
- Enols (tautomeric): Give color ✓
Memory Trick: “FeCl₃ Finds Fenols (phenols) - Fantastic violet color!”
Question: How to distinguish between: (a) Phenol and benzyl alcohol (b) Phenol and benzoic acid
Answer: (a) FeCl₃ test:
- Phenol → Violet color ✓
- Benzyl alcohol → No color ✗
(b) NaHCO₃ test:
- Phenol → No CO₂ ✗
- Benzoic acid → CO₂ evolution (effervescence) ✓
OR NaOH + CO₂ test:
- Phenol + NaOH → sodium phenoxide; add CO₂ → phenol regenerated
- Benzoic acid + NaOH → sodium benzoate; add CO₂ → benzoic acid regenerated (less acidic than carbonic acid from CO₂!)
Memory Tricks & Patterns
The Master Mnemonic: “PHENOLS REACT BOLDLY EVERYDAY”
Phenoxide is stable (resonance) - hence phenol is acidic
Halogenation without catalyst (Br₂/H₂O → white ppt)
Electrophilic substitution at ortho/para
Nitration with dilute HNO₃ (no catalyst!)
Ortho-para directing (-OH group)
Less acidic than carboxylic acids (doesn’t react with NaHCO₃)
Sulfonation at ortho/para
Reimer-Tiemann gives CHO (aldehyde)
Esterification needs acid chloride (not acid!)
Azo coupling gives colored dyes
CO₂ insertion (Kolbe) gives COOH
Trisubstitution with Br₂/H₂O (2,4,6-tribromophenol)
Bases (NaOH) dissolve phenol
Oxidation gives complex mixtures
Lab test: FeCl₃ gives violet color
Diazonium salt hydrolysis → phenol
EWG increases acidity, EDG decreases
Very reactive in electrophilic substitution
Electron-donating via +M effect
Resonance stabilizes phenoxide
Yields salicylic acid (Kolbe’s reaction)
Day and night difference from alcohols!
Activating group for aromatic ring
Yellow-orange color with bromine water → white ppt
Acidity Ranking Shortcut
Step 1: Identify substituents Step 2: EWG (NO₂, CN, CHO, X) → increases acidity; EDG (OH, OCH₃, NH₂, CH₃) → decreases acidity Step 3: For EWG: ortho/para > meta > unsubstituted phenol > phenol with EDG
Example:
Rank: p-NO₂-phenol, m-NO₂-phenol, phenol, p-CH₃-phenol
Answer (most to least acidic):
p-NO₂-phenol > m-NO₂-phenol > phenol > p-CH₃-phenol
Why?
- p-NO₂: EWG at para (strong resonance) ✓✓
- m-NO₂: EWG at meta (inductive only) ✓
- Phenol: baseline
- p-CH₃: EDG at para (destabilizes phenoxide) ✗
Common Mistakes to Avoid
Wrong: Treating C₆H₅-CH₂-OH as phenol Right: This is benzyl alcohol (aliphatic -OH)!
Test to remember:
- Phenol: -OH on sp² carbon (benzene ring directly)
- Benzyl alcohol: -OH on sp³ carbon (CH₂ group)
Consequence: Benzyl alcohol doesn’t give FeCl₃ test, doesn’t react with dil. HNO₃, etc.
Wrong: Expecting phenol to react with NaHCO₃ (like carboxylic acids) Right: Phenol is NOT acidic enough for NaHCO₃!
Remember:
- Carboxylic acid + NaHCO₃ → CO₂ (effervescence) ✓
- Phenol + NaHCO₃ → No reaction ✗
Acidity: R-COOH > H₂CO₃ > phenol
Wrong: Mixing up products of these two reactions Right:
- Kolbe: CO₂ + NaOH → ortho-COOH (salicylic acid)
- Reimer-Tiemann: CHCl₃ + NaOH → ortho-CHO (salicylaldehyde)
Memory: Kolbe has “C” → COOH; Reimer has “R” → aRoma (aldehyde)
Wrong: Thinking -OH is meta-directing (like NO₂) Right: -OH is ortho/para directing (activating group!)
Mechanism: +M (mesomeric) effect dominates, donates electrons to ortho/para positions
Groups:
- Ortho/para: -OH, -OR, -NH₂, -R, -X (halogens are special!)
- Meta: -NO₂, -CN, -CHO, -COOH, -SO₃H
Wrong: Using carboxylic acid + H₂SO₄ to esterify phenol (like alcohols) Right: Phenol needs acid chloride or acid anhydride (not carboxylic acid!)
Reason: Phenol C-O bond has partial double-bond character (less reactive than alcohol)
Correct:
- C₆H₅-OH + CH₃COCl → C₆H₅-O-CO-CH₃ ✓
- C₆H₅-OH + (CH₃CO)₂O → C₆H₅-O-CO-CH₃ ✓
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Why is phenol more acidic than ethanol?
Solution: Phenol is more acidic because the phenoxide ion (C₆H₅-O⁻) is stabilized by resonance, whereas the ethoxide ion (C₂H₅-O⁻) has no such stabilization.
Resonance in phenoxide: The negative charge on oxygen delocalizes into the benzene ring via resonance (4 resonance structures), making phenoxide more stable.
No resonance in ethoxide: The negative charge remains localized on oxygen in C₂H₅-O⁻.
Result: Since phenoxide is more stable, phenol more readily donates H⁺ → more acidic!
pKₐ values: Phenol ≈ 10, Ethanol ≈ 16 (lower pKₐ = more acidic)
Question: How will you distinguish between phenol and benzyl alcohol?
Solution: FeCl₃ test:
- Add neutral FeCl₃ solution to both samples
- Phenol: Gives violet or blue-green color ✓
- Benzyl alcohol: No color change ✗
Explanation: FeCl₃ forms a colored complex with phenols (enols), not with aliphatic alcohols.
Alternative test:
- Br₂/H₂O: Phenol gives white ppt (tribromophenol), benzyl alcohol doesn’t
Question: Complete the reaction:
OH
|
/═\\ + Br₂/H₂O → ?
| |
\═/
Solution:
OH OH
| |
/═\\ Br₂/H₂O /═\\
| | ─────────────→ Br | | Br (white precipitate)
\═/ \═/
Br
(2,4,6-Tribromophenol)
Observation: White precipitate forms immediately (qualitative test for phenol!)
Why trisubstitution? -OH group strongly activates all ortho/para positions, and Br₂ is a strong electrophile.
Level 2: JEE Main Style
Question: Arrange the following in increasing order of acidity: I. p-Nitrophenol II. Phenol III. p-Methylphenol IV. p-Methoxyphenol
(A) IV < III < II < I (B) I < II < III < IV (C) II < I < III < IV (D) III < IV < II < I
Solution: (A) IV < III < II < I
Analysis:
- p-Nitrophenol (I): NO₂ is strong EWG → increases acidity (most acidic)
- Phenol (II): Baseline acidity
- p-Methylphenol (III): CH₃ is weak EDG → decreases acidity
- p-Methoxyphenol (IV): OCH₃ is strong EDG → decreases acidity more (least acidic)
Increasing order: IV < III < II < I ✓
Memory: EWG makes stronger acid, EDG makes weaker acid!
Question: Which of the following will NOT give a positive test with neutral FeCl₃?
(A) Phenol (B) p-Cresol (C) Benzyl alcohol (D) Salicylic acid
Solution: (C) Benzyl alcohol
Explanation: FeCl₃ test is positive for phenols (and enols).
- (A) Phenol: Has -OH on benzene ring → gives violet color ✓
- (B) p-Cresol: Has -OH on benzene ring (methylphenol) → gives violet color ✓
- (C) Benzyl alcohol (C₆H₅-CH₂-OH): Has -OH on aliphatic carbon (NOT phenol!) → no color ✗
- (D) Salicylic acid: Has both -OH (phenolic) and -COOH → gives violet color ✓
Answer: (C)
Question: Phenol is treated with CO₂ at 400 K under pressure, followed by acidification. The major product is:
(A) Benzoic acid (B) Salicylic acid (C) Phthalic acid (D) Terephthalic acid
Solution: (B) Salicylic acid
Reaction (Kolbe’s reaction):
Step 1: NaOH + phenol → sodium phenoxide
Step 2: CO₂ insertion at ortho position (400 K, pressure)
Step 3: H⁺ → salicylic acid (ortho-hydroxybenzoic acid)
COOH
|
/═\\
| |
\═/
OH
(Salicylic acid)
Why ortho? Electrophilic attack by CO₂ favored at ortho due to chelation with -O⁻
Answer: (B)
Level 3: JEE Advanced Style
Question: An organic compound (A) with molecular formula C₇H₈O gives violet color with FeCl₃. (A) reacts with Br₂/H₂O to give white precipitate (B). On treating (A) with CHCl₃ + NaOH, compound (C) is formed which has a pleasant smell. Identify (A), (B), and (C) with reactions.
Solution:
Analysis:
- Violet color with FeCl₃ → (A) is a phenol
- Molecular formula C₇H₈O → (A) = C₆H₅-OH + CH₂ → could be methylphenol (cresol) OR benzyl alcohol
- Br₂/H₂O gives white ppt → confirms (A) is phenol (tribromophenol ppt)
- CHCl₃ + NaOH → pleasant smell → Reimer-Tiemann reaction → aldehyde
But wait! Molecular formula C₇H₈O for methylphenol (cresol) is correct!
Let’s check:
- p-Cresol (p-methylphenol): C₇H₈O ✓
- o-Cresol: C₇H₈O ✓
- m-Cresol: C₇H₈O ✓
Most likely: p-Cresol (most common isomer)
Reactions:
(A) = p-Cresol (4-methylphenol)
CH₃
|
/═\\
| |
\═/
OH
(B) = 2,4,6-Tribromo-4-methylphenol (white ppt)
CH₃
|
Br₂/H₂O
(A) ────────→
CH₃
|
Br /═\\ Br
| |
\═/
OH
Br
(C) = 2-Hydroxy-5-methylbenzaldehyde (pleasant smell)
CHCl₃ + NaOH
(A) ─────────────────→
CHO
|
/═\\
| | CH₃
\═/
OH
(salicylaldehyde derivative)
Note: Reimer-Tiemann reaction gives ortho-aldehyde derivative!
Question: Compare and explain the acidity order:
p-Nitrophenol > m-Nitrophenol > Phenol > p-Methylphenol
Solution:
Acidity order: p-NO₂-phenol > m-NO₂-phenol > phenol > p-CH₃-phenol
Explanation:
1. p-Nitrophenol (most acidic):
NO₂ at para position stabilizes phenoxide ion via RESONANCE:
O⁻ O⁻
| |
/═\\ /═\\
| | NO₂ ↔ | | N⁺=O
\═/ \═/ |
O⁻
Direct resonance stabilization of negative charge!
2. m-Nitrophenol:
NO₂ at meta position stabilizes via INDUCTIVE effect only (no direct resonance)
Less effective than para position
3. Phenol (baseline):
Resonance stabilization of phenoxide within benzene ring only
4. p-Methylphenol (least acidic):
CH₃ at para position DESTABILIZES phenoxide via +I effect:
O⁻
|
/═\\
| | CH₃
\═/
Electron-donating CH₃ pushes more electron density onto already negative O⁻
Makes phenoxide LESS stable → phenol LESS acidic
Summary:
- EWG (NO₂): Withdraws electron density → stabilizes O⁻ → more acidic
- ortho/para position: Direct resonance > meta (inductive only)
- EDG (CH₃): Donates electron density → destabilizes O⁻ → less acidic
Question: Starting from benzene, how would you synthesize: (a) p-Nitrophenol (b) Salicylic acid
Solution:
(a) p-Nitrophenol from benzene:
Route 1 (via nitration first):
Step 1: Nitration
conc. HNO₃/H₂SO₄
C₆H₆ ──────────────────→ C₆H₅-NO₂
(Nitrobenzene)
Step 2: Sulfonation
conc. H₂SO₄
C₆H₅-NO₂ ─────────→ NO₂-C₆H₄-SO₃H (p-isomer, major)
Step 3: Alkali fusion
NaOH, fused, 623 K
NO₂-C₆H₄-SO₃H ──────────────→ p-NO₂-C₆H₄-O⁻Na⁺
Step 4: Acidification
p-NO₂-C₆H₄-O⁻Na⁺ ──H⁺──→ p-NO₂-C₆H₄-OH
(p-Nitrophenol)
Route 2 (via phenol first - BETTER!):
Step 1: Chlorination
Cl₂/FeCl₃
C₆H₆ ─────────→ C₆H₅-Cl
(Chlorobenzene)
Step 2: Dow process
NaOH, 623 K, 350 atm
C₆H₅-Cl ──────────────────→ C₆H₅-OH
(Phenol)
Step 3: Nitration
dil. HNO₃
C₆H₅-OH ──────────→ p-NO₂-C₆H₄-OH + o-NO₂-C₆H₄-OH
(p-Nitrophenol) (o-isomer)
Separate by steam distillation (o-isomer more volatile due to intramolecular H-bonding)
Route 2 is better because phenol is highly reactive toward electrophiles!
(b) Salicylic acid from benzene:
Step 1: Chlorination
Cl₂/FeCl₃
C₆H₆ ─────────→ C₆H₅-Cl
Step 2: Dow process OR NaOH fusion
NaOH, 623 K, 350 atm
C₆H₅-Cl ──────────────────→ C₆H₅-OH
(Phenol)
Step 3: Kolbe's reaction
(1) NaOH
C₆H₅-OH ──────────────────→ C₆H₅-O⁻Na⁺
(2) CO₂, 400 K, pressure
(3) H⁺
COOH
|
/═\\
| |
\═/
OH
(Salicylic acid)
Key Steps:
- Benzene → Chlorobenzene (Cl₂/FeCl₃)
- Chlorobenzene → Phenol (NaOH, high temp/pressure)
- Phenol → Salicylic acid (Kolbe’s reaction: CO₂, NaOH, then H⁺)
Quick Revision Box
| Situation | Formula/Approach | Key Point |
|---|---|---|
| Distinguish phenol from alcohol | FeCl₃ test | Phenol → violet color; alcohol → no color |
| Distinguish phenol from RCOOH | NaHCO₃ test | RCOOH → CO₂; phenol → no reaction |
| Test for phenol | Br₂/H₂O | White ppt of 2,4,6-tribromophenol |
| Increase acidity | Add EWG (NO₂, CN) at ortho/para | Stabilizes phenoxide |
| Decrease acidity | Add EDG (CH₃, OCH₃) at any position | Destabilizes phenoxide |
| Make phenol from benzene | Cl₂/FeCl₃ → Dow process | Or via diazonium salt |
| Make phenol from aniline | NaNO₂/HCl (0-5°C) → warm with H₂O | Diazotization → hydrolysis |
| Get ortho-COOH (salicylic acid) | Kolbe’s reaction | CO₂ + NaOH, then H⁺ |
| Get ortho-CHO (salicylaldehyde) | Reimer-Tiemann | CHCl₃ + NaOH |
| Get azo dye | Diazonium salt + phenol/NaOH | Cold, basic medium |
| Nitration of phenol | Dil. HNO₃ (no catalyst!) | Much easier than benzene |
| Trisubstitution of phenol | Br₂/H₂O | Immediate white ppt |
| Esterify phenol | Acid chloride (not RCOOH!) | C₆H₅-OH + RCOCl → ester |
Connection to Other Topics
Prerequisites (Review these first):
- Hydrocarbons: Benzene - for electrophilic aromatic substitution
- Alcohols - to understand key differences
- Organic Principles: Resonance - for phenoxide stability
Related Topics:
- Aldehydes & Ketones - products of Reimer-Tiemann reaction
- Carboxylic Acids - product of Kolbe’s reaction, acidity comparison
- Nitrogen Compounds: Diazonium Salts - azo coupling, preparation of phenol
- Nitrogen Compounds: Aniline - precursor to phenol via diazotization
Advanced Applications:
- Synthesis of aspirin (acetylsalicylic acid) from salicylic acid
- Synthesis of dyes (azo dyes, textile industry)
- Preparation of explosives (picric acid)
- Polymer chemistry (Bakelite from phenol + formaldehyde)
Teacher’s Summary
1. Phenol ≠ Alcohol:
- Phenol has -OH on sp² carbon (benzene ring)
- Benzyl alcohol (C₆H₅-CH₂-OH) is an alcohol, NOT phenol!
- FeCl₃ test distinguishes them: phenol → violet; alcohol → no color
2. Acidity is the Game-Changer:
- Phenol is acidic (pKₐ ~ 10) due to resonance-stabilized phenoxide ion
- Reacts with NaOH, but NOT with NaHCO₃
- EWG increases acidity (NO₂ > CN > X), EDG decreases (CH₃, OCH₃, NH₂)
3. Electrophilic Substitution Superstar:
- -OH activates benzene ring strongly (+M effect)
- ortho/para directing in all reactions
- No catalyst needed for many reactions (unlike benzene!)
- Br₂/H₂O → trisubstitution (white ppt) - qualitative test!
4. Name Reactions are High-Yield:
- Kolbe: CO₂ + NaOH → salicylic acid (ortho-COOH)
- Reimer-Tiemann: CHCl₃ + NaOH → salicylaldehyde (ortho-CHO)
- Azo coupling: Diazonium salt → colored azo dye (para position)
- Dow process: C₆H₅Cl + NaOH → C₆H₅OH (industrial phenol prep)
5. JEE Strategy:
- Master acidity ranking (EWG vs EDG, ortho/para vs meta)
- Know qualitative tests (FeCl₃, Br₂/H₂O, NaHCO₃)
- Practice Kolbe vs Reimer-Tiemann distinction
- Remember: phenol esterification needs acid chloride, not acid!
- Watch for benzyl alcohol vs phenol traps!
“Phenol is where resonance meets reactivity - that -OH on the benzene ring isn’t just a spectator, it’s the director of the entire aromatic orchestra. Master phenol’s dual nature (acidic yet reactive), and you’ll ace both JEE Main and Advanced organic chemistry sections!”
Previous Topic: ← Alcohols - The aliphatic -OH cousins Next Topic: Ethers → - When two carbons share an oxygen!