The Hook: Breaking Bad Chemistry (But Legal!)
Remember the hand sanitizer shortage during 2020? Everyone was scrambling for bottles containing 70% ethanol. But have you ever wondered why exactly 70% and not 100%? Or how that same compound that sanitizes your hands can also power cars, make perfumes, and be the starting material for hundreds of organic reactions?
The JEE Question: In the 2023 JEE Advanced paper, students who understood alcohol reactions scored 8 marks in just 3 minutes. Those who didn’t? They stared at the paper wondering why their ROH wasn’t behaving like they expected!
The Core Concept: What Makes Alcohols Special?
General Formula & Functional Group
$$\boxed{\text{R-OH (Hydroxyl group attached to saturated carbon)}}$$In simple terms: Alcohols are like hydrocarbons that got upgraded with an -OH group. Think of it as adding a “smart feature” to a basic phone - the carbon chain is the phone, and -OH is the feature that makes everything interesting!
Critical Distinction:
- Alcohol: -OH attached to sp³ carbon (C-OH)
- Phenol: -OH attached to sp² carbon of benzene ring (not covered here - see phenols)
- Enol: -OH attached to carbon with C=C (unstable, tautomerizes)
Classification: The 1°, 2°, 3° Family
Based on Carbon Bearing -OH
Primary (1°) Secondary (2°) Tertiary (3°)
| | |
R-CH₂-OH R₂CH-OH R₃C-OH
| | |
1 R group 2 R groups 3 R groups
Memory Trick: “Primary has Pathetic friends (only 1), Secondary is Social (2 friends), Tertiary is Too popular (3 friends)”
Interactive Demo: Visualize Reaction Mechanisms
Watch SN1, SN2, and E1, E2 mechanisms in alcohol reactions.
Reactivity Pattern
$$\boxed{\text{Reactivity Order: } 3° > 2° > 1° \text{ (for substitution/elimination)}}$$Why? Carbocation stability: 3° carbocation is most stable (hyperconjugation + inductive effect)
For SN1/E1 reactions:
- 3° alcohols form 3° carbocations (most stable) → fastest reaction
- 1° alcohols struggle to form 1° carbocations → slowest/different mechanism
For oxidation:
- 1° → can be oxidized to aldehyde, then carboxylic acid
- 2° → can be oxidized to ketone (stops here)
- 3° → cannot be oxidized (no H on C-OH carbon!)
JEE Gold: This reactivity flip is tested heavily!
Preparation Methods: Your Reaction Toolkit
Method 1: Hydration of Alkenes (Markovnikov’s Rule)
Reaction:
H₂SO₄/H₂O
CH₃-CH=CH₂ ────────→ CH₃-CH(OH)-CH₃ (2° alcohol)
Propene Isopropyl alcohol
Mechanism (Markovnikov Addition):
Step 1: Protonation
CH₃-CH=CH₂ + H⁺ → CH₃-CH⁺-CH₃ (2° carbocation - more stable!)
NOT CH₃-CH₂-CH₂⁺ (1° - less stable)
Step 2: Nucleophilic attack
CH₃-CH⁺-CH₃ + H₂O → CH₃-CH(OH)-CH₃
Memory Trick: “Markovnikov is Mighty Mean - gives -OH to the More substituted carbon!”
Want the -OH on the less substituted carbon? Use hydroboration-oxidation:
(1) BH₃/THF
CH₃-CH=CH₂ ────────────→ CH₃-CH₂-CH₂-OH (1° alcohol)
(2) H₂O₂/OH⁻
JEE Pattern: They’ll give you an alkene and ask which reagent gives which alcohol isomer!
Method 2: Reduction of Carbonyl Compounds
The Big Picture:
Reduction
Carboxylic acid ──────→ 1° alcohol
↑ ↓
↓
Aldehyde ──────────────→ 1° alcohol
↑ ↓
↓
Ketone ────────────────→ 2° alcohol
Reagents:
| Starting Material | Reagent | Product | Mechanism Type |
|---|---|---|---|
| Aldehyde | LiAlH₄ or NaBH₄ | 1° alcohol | Nucleophilic addition |
| Ketone | LiAlH₄ or NaBH₄ | 2° alcohol | Nucleophilic addition |
| Ester | LiAlH₄ (not NaBH₄!) | 1° alcohol | Nucleophilic acyl substitution |
| Carboxylic acid | LiAlH₄ (not NaBH₄!) | 1° alcohol | Reduction |
Memory Trick: “LiAlH₄ is Like a Lion - powerful, reduces everything! NaBH₄ is a Nice Baby - only reduces aldehydes/ketones”
Mechanism (LiAlH₄ reduction of aldehyde):
H H H⁻
| | |
R-C=O + H⁻ (from LiAlH₄) → R-C-O⁻ ──H₃O⁺──→ R-CH₂-OH
| |
(Aldehyde) (Alkoxide) (1° alcohol)
Method 3: Grignard Reaction (THE JEE FAVORITE!)
Grignard Reagent: R-MgX (behaves like R⁻)
Pattern Recognition:
| Carbonyl + Grignard | Product Alcohol Class | Example |
|---|---|---|
| Formaldehyde (HCHO) + RMgX | 1° alcohol | CH₃MgBr + HCHO → CH₃CH₂OH |
| Other aldehyde (R’CHO) + RMgX | 2° alcohol | CH₃MgBr + CH₃CHO → (CH₃)₂CHOH |
| Ketone (R’COR") + RMgX | 3° alcohol | CH₃MgBr + CH₃COCH₃ → (CH₃)₃COH |
| Ester (RCOOR’) + 2 RMgX | 3° alcohol (2 same R groups!) | 2 CH₃MgBr + HCOOCH₃ → (CH₃)₃COH |
Memory Trick: “Formaldehyde makes First-degree (1°), Aldehyde makes Almost-middle (2°), Ketone makes King (3°)”
Mechanism:
δ+ δ- O⁻MgBr OH
| | | |
R'-C=O + R-MgBr ──→ R'-C-R ──H₃O⁺──→ R'-C-R
| | |
R" R" R"
Question Type: “Which Grignard reagent + carbonyl gives this specific alcohol?”
Strategy: Work backward!
- Identify alcohol class (1°, 2°, 3°)
- Check symmetry - if two groups are identical in 3° alcohol → ester route
- Draw the carbonyl and Grignard components
Example: To make (CH₃)₂CH-CH(OH)-CH₃
- 2° alcohol → use aldehyde + Grignard
- Options: CH₃CHO + (CH₃)₂CHMgBr OR (CH₃)₂CHCHO + CH₃MgBr
Method 4: Hydrolysis of Alkyl Halides
Reaction:
NaOH/H₂O
R-X ────────────→ R-OH + NaX
(or KOH)
Problem: This competes with elimination (E2 mechanism)!
Decision Table:
| Alkyl Halide | Aqueous NaOH (dilute) | Alcoholic KOH (conc.) |
|---|---|---|
| 1° (like CH₃CH₂Br) | Mainly substitution → alcohol | Mainly elimination → alkene |
| 2° | Competition | Mainly elimination |
| 3° | Mainly elimination (SN1 + E1) | Mainly elimination |
Memory Trick: “Bulky bases Boost elimination, Watery Wants substitution”
Reactions of Alcohols: The Main Event
Reaction 1: Reaction with Hydrogen Halides (ROH → R-X)
Order of Reactivity:
$$\boxed{\text{HI} > \text{HBr} > \text{HCl (very slow)}}$$ $$\boxed{3° \text{ alcohol} > 2° > 1°}$$Mechanism for 3° alcohol (SN1):
Step 1: Protonation
R₃C-OH + HBr → R₃C-OH₂⁺
Step 2: Loss of H₂O (good leaving group!)
R₃C-OH₂⁺ → R₃C⁺ + H₂O
Step 3: Nucleophilic attack
R₃C⁺ + Br⁻ → R₃C-Br
Mechanism for 1° alcohol (SN2):
R-CH₂-OH₂⁺ + Br⁻ → R-CH₂-Br + H₂O
(Backside attack in one step)
Reagent Tricks:
| Reagent Combination | Purpose | Alcohol Type |
|---|---|---|
| Lucas Reagent (conc. HCl + ZnCl₂) | Distinguishes 1°, 2°, 3° | See below |
Lucas Test:
- 3° alcohol: Turbidity immediately (ZnCl₂ helps stabilize carbocation)
- 2° alcohol: Turbidity in 5 minutes
- 1° alcohol: No turbidity at room temp (needs heating)
Memory Trick: “Lucas Loves Large alcohols - 3° reacts Lickety-split!”
Reaction 2: Dehydration to Alkenes (Elimination)
General Reaction:
conc. H₂SO₄
R-CH₂-CH₂-OH ──────────→ R-CH=CH₂ + H₂O
Heat (170°C)
Mechanism (E1 for 2°/3°):
Step 1: Protonation
CH₃-CH(OH)-CH₃ + H₂SO₄ → CH₃-CH(OH₂⁺)-CH₃
Step 2: Loss of H₂O
CH₃-CH(OH₂⁺)-CH₃ → CH₃-CH⁺-CH₃ + H₂O
Step 3: Loss of H⁺ (E1)
CH₃-CH⁺-CH₃ → CH₂=CH-CH₃ (propene)
Saytzeff’s Rule (Zaitsev): “In elimination, the more substituted alkene is the major product”
Example:
conc. H₂SO₄
CH₃-CH(OH)-CH₂-CH₃ ──────────→ CH₃-CH=CH-CH₃ (major, more substituted)
+
CH₂=CH-CH₂-CH₃ (minor, less substituted)
Memory Trick: “Saytzeff Says Stability Succeeds - more substituted = more stable = major”
140°C with H₂SO₄: Forms ether (2 R-OH → R-O-R + H₂O) 170°C with H₂SO₄: Forms alkene (R-OH → R-CH=CH₂ + H₂O)
JEE Trap: They’ll change temperature and ask for product!
Reaction 3: Oxidation - The Degree Matters!
Reagents: KMnO₄ (alkaline), K₂Cr₂O₇/H₂SO₄ (acidified), PCC, Jones reagent
Pattern:
| Alcohol | Mild Oxidation (PCC) | Strong Oxidation (K₂Cr₂O₇/H⁺) |
|---|---|---|
| 1° (R-CH₂-OH) | R-CHO (aldehyde) | R-COOH (carboxylic acid) |
| 2° (R₂CH-OH) | R₂C=O (ketone) | R₂C=O (ketone - stops here!) |
| 3° (R₃C-OH) | No reaction | No reaction |
Mechanism (simplified):
1° alcohol oxidation:
H O O
| ‖ ‖
R-C-OH ──[O]──→ R-C-H ──[O]──→ R-C-OH
|
H Aldehyde Carboxylic acid
(1° alcohol)
PCC (Pyridinium Chlorochromate) - The Stopper:
- Stops oxidation at aldehyde stage
- Used in anhydrous conditions (CH₂Cl₂ solvent)
Memory Trick: “PCC is Perfect for Partial oxidation - stops at aldehyde!”
JEE Pattern:
Question: CH₃CH₂CH₂OH ──Reagent X──→ CH₃CH₂CHO (propanal, NOT propanoic acid)
Answer: PCC (not K₂Cr₂O₇!)
Reaction 4: Esterification (Fischer Esterification)
Reaction:
conc. H₂SO₄
R-OH + R'-COOH ⇌ R'-COO-R + H₂O
(alcohol) (acid) (ester)
Mechanism:
Step 1: Protonation of acid
O OH⁺
‖ |
R'-C-OH + H⁺ → R'-C-OH
Step 2: Nucleophilic attack by alcohol
OH⁺ OH
| |
R'-C-OH + R-OH → R'-C-OH
|
O-R
Step 3: Loss of H₂O
OH O
| ‖
R'-C-OH → R'-C-O-R + H₂O
|
O-R
Important: This is an equilibrium reaction!
To shift toward ester:
- Use excess alcohol or acid
- Remove water (use Dean-Stark apparatus)
- Use conc. H₂SO₄ (dehydrating agent)
Quick Recognition:
- Alcohol + Carboxylic acid + acid catalyst → Ester + Water
- Ester + Water + acid catalyst → Alcohol + Carboxylic acid (hydrolysis)
Both use H₂SO₄, so check if water is added or removed!
Reaction 5: Reaction with Metals (Na, K)
Reaction:
2 R-OH + 2 Na → 2 R-O⁻Na⁺ + H₂↑
(Sodium alkoxide)
Test for -OH group: Effervescence (H₂ gas) with Na metal
Uses:
- Confirms presence of -OH
- Prepares alkoxides (strong bases for organic synthesis)
Comparison with Water:
2 H₂O + 2 Na → 2 NaOH + H₂↑ (vigorous)
2 R-OH + 2 Na → 2 R-ONa + H₂↑ (less vigorous, since R-OH is weaker acid than H₂O)
Acidity Order:
$$\boxed{\text{Water} > 1° \text{ alcohol} > 2° > 3°}$$Why? +I effect of R groups destabilizes R-O⁻ (makes it less willing to donate H⁺)
Memory Tricks & Patterns
The Master Mnemonic: “GRILLED DUCKS OFTEN ESCAPE HEAT”
- Grignard → Alcohols of all degrees
- Reduction → Carbonyl to alcohol
- Identification → Lucas test (1°, 2°, 3°)
- Lucas → Distinguishing test
- LiAlH₄ → Powerful reducer
- Elimination → Alkenes (Saytzeff rule)
- Dehydration → 170°C for alkene, 140°C for ether
- Oxidation → 1° → aldehyde → acid, 2° → ketone, 3° → no reaction
- Fischer → Esterification
- Tertiary → Most reactive for substitution/elimination
Reactivity Pattern
For SN1/E1 (with HX, dehydration):
3° > 2° > 1° (carbocation stability)
For SN2 (with NaOH/H₂O):
1° > 2° > 3° (steric hindrance)
For Oxidation:
1° (goes furthest) > 2° (stops at ketone) > 3° (no reaction)
Memory Trick: “SN1 loves Stability (3° wins), SN2 loves Space (1° wins)”
Common JEE Patterns
| Pattern Type | Example | Key Point |
|---|---|---|
| Reagent swap | PCC vs K₂Cr₂O₇ | PCC stops at aldehyde |
| Temperature change | 140°C vs 170°C with H₂SO₄ | Ether vs alkene |
| Mechanism switch | 1° vs 3° with HBr | SN2 vs SN1 |
| Grignard backward | Given alcohol, find reagents | Check symmetry! |
| Rearrangement | 2° carbocation to 3° | Hydride/methyl shift |
Special Case: Carbocation Rearrangements
In SN1/E1 reactions (3° and 2° alcohols), carbocation rearrangements can occur!
Types:
- Hydride shift (H⁻ with 2 e⁻)
- Methyl shift (CH₃⁻ with 2 e⁻)
Rule: Rearrangement happens if it creates a more stable carbocation
Example:
CH₃-CH(OH)-CH(CH₃)₂ ──HBr──→ ?
Step 1: CH₃-CH⁺-CH(CH₃)₂ (2° carbocation)
Step 2: Hydride shift from adjacent carbon
CH₃-CH₂-C⁺(CH₃)₂ (3° carbocation - MORE STABLE!)
Step 3: Br⁻ attacks
CH₃-CH₂-C(Br)(CH₃)₂ (major product)
Memory Trick: “Carbocations Crave Comfort - they’ll rearrange to 3° if possible!”
JEE Pattern: They give you a 2° alcohol with HBr and expect you to predict rearrangement!
Common Mistakes to Avoid
Wrong: Treating benzyl alcohol (C₆H₅-CH₂-OH) like phenol Right: Benzyl alcohol is an alcohol (aliphatic -OH), NOT phenol (aromatic -OH)
Test: Phenol gives violet color with FeCl₃; benzyl alcohol doesn’t!
Wrong: Using NaBH₄ to reduce esters or carboxylic acids Right: NaBH₄ only reduces aldehydes and ketones. Use LiAlH₄ for esters/acids!
Mnemonic: “NaBH₄ is Not strong enough for esters!”
Wrong: Writing direct product without checking for possible carbocation rearrangement Right: Always check if 2° carbocation can become 3° through H⁻ or CH₃⁻ shift!
JEE Impact: This single mistake costs 4 marks in JEE Advanced!
Wrong: Thinking “more substituted” means more H atoms on alkene carbon Right: More substituted = more carbon groups attached to C=C carbons
Example: CH₃-CH=CH₂ (less substituted) vs CH₃-CH=CH-CH₃ (more substituted)
Wrong: Thinking 3° alcohols can be oxidized under harsh conditions Right: 3° alcohols have no H on C-OH carbon, so oxidation is impossible without breaking C-C bonds!
What happens: Under very harsh conditions, C-C bond breaks (destructive)
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Classify the following alcohols as 1°, 2°, or 3°: (a) CH₃-CH₂-CH₂-OH (b) (CH₃)₂CH-OH (c) (CH₃)₃C-OH (d) C₆H₅-CH₂-OH
Solution: (a) 1° (primary) - only 1 carbon attached to C-OH (b) 2° (secondary) - 2 carbons attached to C-OH (c) 3° (tertiary) - 3 carbons attached to C-OH (d) 1° (primary) - only 1 carbon (the benzene ring) attached to C-OH; this is benzyl alcohol, not phenol!
Question: Complete the reaction:
CH₃-CH₂-OH + Na → ?
Solution:
2 CH₃-CH₂-OH + 2 Na → 2 CH₃-CH₂-O⁻Na⁺ + H₂↑
(Sodium ethoxide)
Observation: Effervescence of H₂ gas (less vigorous than with water)
Question: What product is formed when propan-2-ol is oxidized with K₂Cr₂O₇/H⁺?
Solution:
K₂Cr₂O₇/H⁺
CH₃-CH(OH)-CH₃ ──────────→ CH₃-CO-CH₃
(Propan-2-ol, 2°) (Propanone/acetone, ketone)
Why? 2° alcohols are oxidized to ketones and stop there (no further oxidation possible).
Level 2: JEE Main Style
Question: Which reagent can be used to convert propene to propan-2-ol?
(A) H₂O/H₂SO₄ (B) BH₃/THF followed by H₂O₂/OH⁻ (C) Both (A) and (B) (D) Neither (A) nor (B)
Solution: (A)
Explanation:
Option A (Markovnikov):
H₂SO₄/H₂O
CH₃-CH=CH₂ ────────→ CH₃-CH(OH)-CH₃ ✓
(propan-2-ol, 2°)
Option B (Anti-Markovnikov):
(1) BH₃/THF
CH₃-CH=CH₂ ────────────→ CH₃-CH₂-CH₂-OH ✗
(2) H₂O₂/OH⁻ (propan-1-ol, 1°)
Key: Markovnikov gives -OH to more substituted carbon!
Question: Arrange the following in increasing order of reactivity towards Lucas reagent: I. Butan-1-ol II. Butan-2-ol III. 2-Methylpropan-2-ol
(A) I < II < III (B) III < II < I (C) I < III < II (D) II < I < III
Solution: (A) I < II < III
Explanation: Lucas reagent (conc. HCl + ZnCl₂) distinguishes 1°, 2°, 3° alcohols via SN1 mechanism.
Reactivity order: 3° > 2° > 1° (carbocation stability)
- I (Butan-1-ol): 1° → slowest (no turbidity at room temp)
- II (Butan-2-ol): 2° → medium (turbidity in ~5 min)
- III (2-Methylpropan-2-ol): 3° → fastest (immediate turbidity)
Therefore: I < II < III ✓
Question: Identify the major product:
conc. H₂SO₄
CH₃-C(OH)(CH₃)-CH₂-CH₃ ──────────→ ?
170°C
(A) CH₃-C(CH₃)=CH-CH₃ (B) CH₂=C(CH₃)-CH₂-CH₃ (C) Both in equal amounts (D) CH₃-O-CH₂-CH(CH₃)₂
Solution: (A) CH₃-C(CH₃)=CH-CH₃
Explanation: This is a dehydration reaction at 170°C → forms alkene (not ether at 140°C!)
Possible alkenes:
Option A: CH₃-C(CH₃)=CH-CH₃ (tetrasubstituted alkene)
Option B: CH₂=C(CH₃)-CH₂-CH₃ (trisubstituted alkene)
Saytzeff’s Rule: More substituted alkene is major product!
Option A has 4 R groups on C=C → major product ✓
Level 3: JEE Advanced Style
Question: A compound (A) with molecular formula C₄H₁₀O gives turbidity immediately with Lucas reagent. On oxidation with K₂Cr₂O₇/H⁺, (A) gives no product. Identify (A) and explain.
Solution:
Analysis:
- Immediate turbidity with Lucas reagent → compound is 3° alcohol
- No oxidation product → confirms 3° alcohol (no H on C-OH carbon)
- Molecular formula C₄H₁₀O with 3° alcohol → must be 2-methylpropan-2-ol
Structure of A:
CH₃
|
CH₃-C-OH
|
CH₃
(2-methylpropan-2-ol, also called tert-butyl alcohol)
Explanation:
- 3 methyl groups attached to C-OH → 3° alcohol ✓
- No H on C-OH → cannot be oxidized ✓
- Forms 3° carbocation quickly → immediate turbidity with Lucas ✓
Question: Predict the major product of the following reaction with mechanism:
CH₃-CH(OH)-CH(CH₃)₂ ──HBr──→ ?
Solution:
Mechanism:
Step 1: Protonation
CH₃-CH(OH)-CH(CH₃)₂ + HBr → CH₃-CH(OH₂⁺)-CH(CH₃)₂
Step 2: Loss of H₂O
CH₃-CH(OH₂⁺)-CH(CH₃)₂ → CH₃-CH⁺-CH(CH₃)₂ + H₂O
(2° carbocation)
Step 3: Hydride shift (rearrangement!)
H
|
CH₃-CH⁺-CH(CH₃)₂ → CH₃-CH₂-C⁺(CH₃)₂
(2° carbocation) (3° carbocation - MORE STABLE!)
Step 4: Nucleophilic attack
CH₃-CH₂-C⁺(CH₃)₂ + Br⁻ → CH₃-CH₂-C(Br)(CH₃)₂
Major product: CH₃-CH₂-C(Br)(CH₃)₂ (2-bromo-2-methylbutane)
Key Point: Carbocation rearrangement from 2° to 3° via hydride shift!
Question: A Grignard reagent reacts with formaldehyde to give compound (X). (X) on treatment with PCC gives (Y). (Y) undergoes Cannizzaro reaction. Identify (X) and (Y).
Solution:
Working backward:
- (Y) undergoes Cannizzaro reaction → (Y) must be an aldehyde without α-H (like HCHO, C₆H₅CHO)
- (X) + PCC → (Y, aldehyde) → (X) must be 1° alcohol
- R-MgX + HCHO → 1° alcohol → (X) = R-CH₂-OH
Simplest case:
- Grignard reagent: CH₃MgBr
- Reaction: CH₃MgBr + HCHO → CH₃CH₂OH (X, ethanol)
- Oxidation: CH₃CH₂OH + PCC → CH₃CHO (Y, acetaldehyde)
But wait! CH₃CHO has α-H, so it doesn’t undergo Cannizzaro!
Correct case: Must be aromatic aldehyde without α-H:
- Grignard: C₆H₅MgBr
- (X): C₆H₅-CH₂-OH (benzyl alcohol)
- (Y): C₆H₅-CHO (benzaldehyde - no α-H!)
Answer:
- X = Benzyl alcohol (C₆H₅CH₂OH)
- Y = Benzaldehyde (C₆H₅CHO)
Verification: Benzaldehyde has no α-H (H on carbon next to CHO), so it undergoes Cannizzaro! ✓
Question: Starting from propene, how would you synthesize propan-1-ol? Give reagents for each step.
Solution:
Route 1: Hydroboration-Oxidation (Direct)
Step 1: Hydroboration-oxidation
(1) BH₃/THF
CH₃-CH=CH₂ ────────────→ CH₃-CH₂-CH₂-OH
(2) H₂O₂/OH⁻ (propan-1-ol)
One-step synthesis! ✓
Route 2: Via Markovnikov alcohol (Alternative)
Step 1: Markovnikov hydration
H₂SO₄/H₂O
CH₃-CH=CH₂ ────────→ CH₃-CH(OH)-CH₃
(propan-2-ol, 2°)
Step 2: Oxidation to ketone
K₂Cr₂O₇/H⁺
CH₃-CH(OH)-CH₃ ──────────→ CH₃-CO-CH₃
(propanone)
Step 3: Reduction
LiAlH₄
CH₃-CO-CH₃ ──────→ CH₃-CH(OH)-CH₃
(back to 2° alcohol)
This doesn’t work! Reduction gives back 2° alcohol, not 1°!
Correct answer: Use Route 1 (hydroboration-oxidation) for Anti-Markovnikov addition!
Quick Revision Box
| Situation | Formula/Approach | Key Point |
|---|---|---|
| Distinguish 1°, 2°, 3° alcohols | Lucas reagent (HCl + ZnCl₂) | 3° immediate, 2° in 5 min, 1° no reaction |
| Make 1° alcohol from alkene | BH₃/THF, then H₂O₂/OH⁻ | Anti-Markovnikov |
| Make 2° alcohol from alkene | H₂SO₄/H₂O | Markovnikov |
| Aldehyde → 1° alcohol | LiAlH₄ or NaBH₄ | Both work |
| Ester → 1° alcohol | LiAlH₄ (NOT NaBH₄!) | NaBH₄ too weak |
| Stop oxidation at aldehyde | PCC | Anhydrous conditions |
| Oxidize 1° to carboxylic acid | K₂Cr₂O₇/H₂SO₄ | Aqueous, excess |
| Make alkene from alcohol | Conc. H₂SO₄, 170°C | Saytzeff rule |
| Make ether from alcohol | Conc. H₂SO₄, 140°C | Lower temp than alkene |
| Reactivity with HX | 3° > 2° > 1° | Carbocation stability |
| Check for -OH group | Na metal | Effervescence (H₂ gas) |
| Formaldehyde + RMgX | Gives 1° alcohol | R-CH₂-OH |
| Aldehyde + RMgX | Gives 2° alcohol | R-CH(OH)-R' |
| Ketone + RMgX | Gives 3° alcohol | R-C(OH)(R’)R" |
Connection to Other Topics
Prerequisites (Review these first):
- Hydrocarbons: Alkenes - for addition reactions
- Organic Principles: Reaction Mechanisms - SN1, SN2, E1, E2
- Hydrocarbons: Alkyl Halides - for substitution comparison
Related Topics:
- Phenols - aromatic -OH compounds with different chemistry
- Ethers - alcohol dehydration product
- Aldehydes & Ketones - oxidation products of alcohols
- Carboxylic Acids - final oxidation product of 1° alcohols
Advanced Applications:
- Nitrogen Compounds: Amines - similar H-bonding, basicity comparison
- Grignard reagents (covered in detail in organometallics)
Teacher’s Summary
1. Classification is Key:
- 1°, 2°, 3° determines reactivity in SN1/E1 (3° > 2° > 1°) and oxidation (1° most, 3° none)
- Lucas test distinguishes them in exam: 3° instant, 2° in 5 min, 1° no reaction
2. Preparation Master List:
- Alkenes → alcohols: Markovnikov (H₂SO₄/H₂O) vs Anti-Markovnikov (BH₃, then H₂O₂)
- Carbonyls → alcohols: LiAlH₄ (powerful), NaBH₄ (selective for aldehydes/ketones only)
- Grignard: Formaldehyde → 1°, aldehyde → 2°, ketone/ester → 3°
3. Oxidation Memory:
- 1° alcohol → aldehyde (PCC stops here) → carboxylic acid (K₂Cr₂O₇ goes further)
- 2° alcohol → ketone (stops, can’t go further)
- 3° alcohol → NO REACTION (no H on C-OH)
4. Reaction Mechanisms Matter:
- 3° and 2° alcohols: SN1/E1 with carbocation (watch for rearrangements!)
- 1° alcohols: SN2 (no carbocation)
- Temperature: 140°C → ether, 170°C → alkene
5. JEE Strategy:
- Memorize Lucas test, Grignard patterns, oxidation outcomes
- Always check for carbocation rearrangements in 2°/3° alcohols
- Know when to use PCC vs K₂Cr₂O₇, NaBH₄ vs LiAlH₄
- Practice working backward from products to reagents
“In JEE, alcohols are not just functional groups - they’re the crossroads where substitution meets elimination, where oxidation meets reduction, and where carbocations love to rearrange. Master alcohols, and you’ve mastered organic chemistry’s foundation!”
Next Topic: Phenols → - Discover why that -OH on benzene ring changes everything!