Organic Chemistry – Chapter 18 Alcohols | Interactive Revision

18.1 Introduction & Classification of Alcohols

⚗️ Definition & General Formula

  • Alcohols: Organic compounds containing hydroxyl (-OH) functional group attached to an sp³ hybridized carbon atom
  • General Formula: R-OH where R = alkyl group
  • Monohydric Alcohols: Contain one -OH group (e.g., methanol, ethanol)
  • Polyhydric Alcohols: Contain two or more -OH groups (e.g., ethylene glycol, glycerol)
  • Uses: Solvents, fuels, perfumes, pharmaceuticals, synthesis of organic compounds
📝 Key Point: Alcohols are classified based on the carbon atom to which the -OH group is attached: Primary (1°), Secondary (2°), and Tertiary (3°)

🏷️ Classification of Monohydric Alcohols

Primary (1°)
-OH attached to carbon bonded to only one alkyl group
→
Secondary (2°)
-OH attached to carbon bonded to two alkyl groups
→
Tertiary (3°)
-OH attached to carbon bonded to three alkyl groups
TypeStructureExampleIUPAC Name
Primary (1°) R-CH₂-OH CH₃-CH₂-OH Ethanol
Secondary (2°) R₂CH-OH (CH₃)₂CH-OH 2-Propanol (Isopropyl alcohol)
Tertiary (3°) R₃C-OH (CH₃)₃C-OH 2-Methyl-2-propanol (tert-Butyl alcohol)

🔬 Molecular Structure

R – O – H (sp³ hybridized)
C-O bond length: 143 pm
O-H bond length: 96 pm
Bond angle: C-O-H ≈ 109° (tetrahedral)

18.2 Preparation of Alcohols

⚗️ 1. From Alkenes (Hydration)

Electrophilic Addition of Steam

CH₂=CH₂ + H₂O → CH₃-CH₂-OH

Conditions: H₃PO₄ catalyst, 300°C, 60-70 atm pressure

With KMnO₄ (Oxidation)

R-CH=CH₂ + [O] + H₂O → R-CH(OH)-CH₂(OH)

Conditions: Cold dilute acidified KMnO₄

Product: Diol (Vicinal dihydroxy compound)

⚡ 2. From Haloalkanes (Nucleophilic Substitution)

R-X + OH⁻ → R-OH + X⁻

Conditions: Aqueous NaOH, heat

Mechanism: Sₙ2 (for 1°), Sₙ1 (for 3°)

🔬 3. Reduction of Carbonyl Compounds

a) Aldehydes → Primary Alcohols R-CHO + 2[H] → R-CH₂-OH NaBH₄ or LiAlH₄ b) Ketones → Secondary Alcohols R-CO-R’ + 2[H] → R-CH(OH)-R’ NaBH₄ or LiAlH₄ c) Carboxylic Acids → Primary Alcohols R-COOH + 4[H] → R-CH₂-OH + H₂O LiAlH₄ (strong reducing agent)

🧪 4. Hydrolysis of Esters

R-COOR’ + H₂O ⇌ R-COOH + R’-OH

Acid Hydrolysis: H⁺ catalyst, reversible

Base Hydrolysis (Saponification): OH⁻ catalyst, irreversible, forms carboxylate salt

🎯 Industrial Production of Ethanol:
• Fermentation of sugars: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
• Hydration of ethene: CH₂=CH₂ + H₂O → C₂H₅OH (H₃PO₄, 300°C)

18.3 Acidity & Physical Properties

⚖️ Acidity of Alcohols

  • Alcohols are weaker acids than water
  • Reason: Alkyl group (+I effect) decreases polarity of O-H bond
  • Electron-releasing alkyl groups reduce acidity
  • Acidity Order: H₂O > CH₃OH > 1° > 2° > 3° alcohols

Reaction with Metals

R-OH + Na → R-O⁻Na⁺ + ½H₂↑

H-OH + Na → NaOH + ½H₂↑

🔥 Physical Properties

PropertyExplanationTrend
Boiling Point Higher than alkanes due to hydrogen bonding Decreases with branching
Solubility Soluble in water due to H-bonding with water Decreases with increasing chain length
Density Less than water (float on water) Methanol (0.79), Ethanol (0.79)
Hydrogen Bonding Strong intermolecular forces Decreases with branching
💡 Key Comparison:
• Water: pKa = 15.7 (more acidic)
• Methanol: pKa = 15.5
• Ethanol: pKa = 15.9
• tert-Butanol: pKa = 19.2 (least acidic)

18.4 Reactions of Alcohols

🔥 1. Combustion

CH₃OH + 1.5O₂ → CO₂ + 2H₂O + Heat
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + Heat

Uses: High-octane fuels, racing cars, clean burning

⚗️ 2. Formation of Alkyl Halides

ReagentReactionConditions SOCl₂ (Thionyl chloride) R-OH + SOCl₂ → R-Cl + SO₂ + HCl Pyridine, room temp PCl₃ 3R-OH + PCl₃ → 3R-Cl + H₃PO₃ Room temperature PCl₅ R-OH + PCl₅ → R-Cl + POCl₃ + HCl Room temperature HX (HBr/HCl) R-OH + HX → R-X + H₂O Acid catalyst (ZnCl₂ for Lucas test) KBr + H₂SO₄ R-OH + KBr + H₂SO₄ → R-Br + KHSO₄ + H₂O Heat

🧪 Lucas Test (Distinguishing Alcohols)

Tertiary (3°)
Immediate turbidity
(within seconds)
→
Secondary (2°)
Turbidity in 5-10 min
→
Primary (1°)
Turbidity on heating only

Lucas Reagent: ZnCl₂ + conc. HCl

Reaction: R-OH + HCl → R-Cl + H₂O

➖ 3. Dehydration

Conditions & Products

Intramolecular Dehydration (Alkene formation):

CH₃-CH₂-OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 170°C)

Intermolecular Dehydration (Ether formation):

2C₂H₅OH → C₂H₅-O-C₂H₅ + H₂O (conc. H₂SO₄, 140°C)

🧪 4. Esterification

R-COOH + R’-OH ⇌ R-COO-R’ + H₂O

Conditions: Conc. H₂SO₄ catalyst, heat

Example: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

🔬 5. Oxidation

Alcohol TypeOxidizing AgentProductObservation
Primary (1°) K₂Cr₂O₇/H⁺ Aldehyde → Carboxylic acid Orange → Green
Secondary (2°) K₂Cr₂O₇/H⁺ Ketone Orange → Green
Tertiary (3°) K₂Cr₂O₇/H⁺ No oxidation (dehydration to alkene) No color change

⚡ 6. Iodoform Reaction

Test for:
• Ethanol (CH₃CH₂OH)
• Secondary alcohols with CH₃-CHOH- group
• Acetaldehyde (CH₃CHO)
• Methyl ketones (CH₃COR)
CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃↓ + HCOONa + 5NaI + 5H₂O

Observation: Pale yellow precipitate of triiodomethane (iodoform)

Negative for: Methanol, tertiary alcohols

Exercises – Complete Solutions

Multiple Choice Questions:

1. Ethanol can be converted into ethanoic acid by:

a. Hydrogenation
b. Hydration
c. Oxidation
d. Fermentation

Answer: (c) Oxidation

Explanation: Primary alcohols (like ethanol) are oxidized first to aldehydes and then to carboxylic acids. Ethanol (C₂H₅OH) → Ethanal (CH₃CHO) → Ethanoic acid (CH₃COOH) using oxidizing agents like K₂Cr₂O₇/H⁺ or KMnO₄/H⁺.

2. Which alcohol reacts fastest with HCl/ZnCl₂ on heating?

a. Primary
b. Secondary
c. Tertiary
d. All of these

Answer: (c) Tertiary

Explanation: This is the Lucas test. Tertiary alcohols react immediately with Lucas reagent (ZnCl₂/HCl) at room temperature to form an oily layer of alkyl chloride. Secondary alcohols take 5-10 minutes, while primary alcohols require heating.

3. Which of the following does not undergo iodoform reaction?

a. Ethanol
b. Methanol
c. 2-Propanol
d. Ethanal

Answer: (b) Methanol

Explanation: Iodoform test requires CH₃-CH(OH)- group for alcohols or CH₃-CO- group for carbonyl compounds. Methanol (CH₃OH) lacks the CH₃-CH(OH)- group, so it doesn’t give iodoform test. Ethanol, 2-propanol, and ethanal all give positive iodoform test.

Short Answer Questions:

i. What are alcohols? How are they classified?

Answer:

Alcohols: Organic compounds containing hydroxyl (-OH) functional group attached to an sp³ hybridized carbon atom. General formula: R-OH.

Classification:

1. Based on -OH groups 2. Based on carbon type (Monohydric) • Monohydric: One -OH group (CH₃OH, C₂H₅OH)
• Dihydric: Two -OH groups (Ethylene glycol)
• Trihydric: Three -OH groups (Glycerol)
• Polyhydric: Multiple -OH groups • Primary (1°): -OH on carbon attached to one alkyl group (R-CH₂-OH)
• Secondary (2°): -OH on carbon attached to two alkyl groups (R₂CH-OH)
• Tertiary (3°): -OH on carbon attached to three alkyl groups (R₃C-OH)

ii. Why is an alcohol less acidic than water?

Answer:

Alcohols are weaker acids than water due to:

1. +I Effect
Alkyl group donates electrons
→
2. Reduced Polarity
O-H bond less polar
→
3. Weaker Acid
Harder to lose H⁺

Detailed Explanation:

• In water (H-OH), oxygen is bonded to hydrogen only, making O-H bond highly polar.
• In alcohols (R-OH), alkyl group (R) has electron-donating inductive effect (+I), which reduces the polarity of O-H bond.
• This makes it harder for alcohols to release H⁺ compared to water.
• Acidity order: H₂O > CH₃OH > Primary alcohols > Secondary alcohols > Tertiary alcohols

Reaction with Sodium:
H-OH + Na → NaOH + ½H₂ (vigorous)
R-OH + Na → RONa + ½H₂ (slower)

iii. How will you distinguish between primary, secondary, and tertiary alcohols?

Answer: Using Lucas Test (ZnCl₂ + conc. HCl)

Alcohol TypeObservation with Lucas ReagentTime TakenChemical Reaction Primary (1°) No turbidity at room temperature; turbidity appears only on heating Several minutes with heating R-CH₂-OH + HCl → R-CH₂-Cl + H₂O (slow) Secondary (2°) Turbidity appears within 5-10 minutes at room temperature 5-10 minutes R₂CH-OH + HCl → R₂CH-Cl + H₂O (moderate) Tertiary (3°) Immediate turbidity/oily layer formation at room temperature Instant (within seconds) R₃C-OH + HCl → R₃C-Cl + H₂O (fast)

Additional Tests:

1. Oxidation Test:
• Primary: Orange → Green (forms aldehyde then acid)
• Secondary: Orange → Green (forms ketone)
• Tertiary: No color change (not oxidized)

2. Iodoform Test:
• Ethanol (primary) and 2° alcohols with CH₃-CHOH- give yellow ppt
• Other 1° alcohols (except ethanol) and 3° alcohols don’t give this test

iv. Give at least two methods for the preparation of ethanol.

Answer:

Method 1: Fermentation
Biological method
→
Method 2: Hydration
Industrial method

1. Fermentation of Sugars (Biological Method):

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Conditions: Yeast (enzyme zymase), 25-30°C, anaerobic conditions
Raw Material: Molasses, starch, grapes, barley
Process: Starch → Maltose → Glucose → Ethanol

2. Hydration of Ethene (Industrial Method):

CH₂=CH₂ + H₂O → CH₃-CH₂-OH

Conditions: H₃PO₄ catalyst, 300°C, 60-70 atm pressure
Mechanism: Electrophilic addition
Advantage: Pure ethanol (95%)

Other Methods:

3. From Haloalkanes: C₂H₅Br + NaOH(aq) → C₂H₅OH + NaBr
4. Reduction of Carbonyl: CH₃CHO + 2[H] → C₂H₅OH (NaBH₄)
5. Hydrolysis of Esters: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH

Conceptual Questions:

1. How does ethanol react with:
(i) HCl
(ii) H₂SO₄
(iii) KBr

Answer:

(i) Ethanol with HCl:

C₂H₅OH + HCl → C₂H₅Cl + H₂O

Conditions: Anhydrous ZnCl₂ (Lucas reagent), room temperature
Reaction Type: Nucleophilic substitution
Observation: Slow reaction (primary alcohol) – turbidity appears on heating

(ii) Ethanol with H₂SO₄:

TemperatureProductReaction Type 170°C CH₂=CH₂ (Ethene) Intramolecular dehydration 140°C C₂H₅-O-C₂H₅ (Diethyl ether) Intermolecular dehydration

Reactions:
At 170°C: C₂H₅OH → CH₂=CH₂ + H₂O (conc. H₂SO₄)
At 140°C: 2C₂H₅OH → C₂H₅-O-C₂H₅ + H₂O (conc. H₂SO₄)

(iii) Ethanol with KBr + H₂SO₄:

C₂H₅OH + KBr + H₂SO₄ → C₂H₅Br + KHSO₄ + H₂O

Mechanism:
1. H₂SO₄ reacts with KBr: 2KBr + H₂SO₄ → K₂SO₄ + 2HBr
2. HBr reacts with ethanol: C₂H₅OH + HBr → C₂H₅Br + H₂O
Product: Bromoethane
Conditions: Heat, concentrated H₂SO₄

2. Explain following terms using ethyl alcohol as an example:
(i) Oxidation
(ii) Dehydration
(iii) Esterification
(iv) Iodoform formation

Answer: Using Ethanol (C₂H₅OH) as example:

(i) Oxidation:

Step 1
Ethanol
[O]→
Step 2
Ethanal
[O]→
Step 3
Ethanoic acid

C₂H₅OH + [O] → CH₃CHO + H₂O (Partial oxidation)
CH₃CHO + [O] → CH₃COOH (Complete oxidation)

Oxidizing agent: K₂Cr₂O₇/H⁺ or KMnO₄/H⁺
Observation: Orange (Cr₂O₇²⁻) → Green (Cr³⁺)

(ii) Dehydration:

Intramolecular (170°C): C₂H₅OH → CH₂=CH₂ + H₂O
Intermolecular (140°C): 2C₂H₅OH → C₂H₅-O-C₂H₅ + H₂O

Dehydrating agent: Conc. H₂SO₄ or Al₂O₃
Different products at different temperatures

(iii) Esterification:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Conditions: Conc. H₂SO₄ (catalyst and dehydrating agent), heat
Product: Ethyl ethanoate (ester)
Characteristic: Fruity smell

(iv) Iodoform formation:

C₂H₅OH + 4I₂ + 6NaOH → CHI₃↓ + HCOONa + 5NaI + 5H₂O

Conditions: I₂ + NaOH (warm)
Observation: Pale yellow precipitate of triiodomethane (iodoform)
Specific test for: Ethanol, methyl ketones, acetaldehyde

3. Give reason for the followings:
(i) Ethanol gives different products with conc. H₂SO₄ under different conditions
(ii) Tertiary alcohols are not oxidized by acidified KMnO₄ or K₂Cr₂O₇

Answer:

(i) Different products with conc. H₂SO₄:

Ethanol reacts with concentrated H₂SO₄ differently at different temperatures:

TemperatureReaction TypeProductReason 140°C Intermolecular dehydration Diethyl ether (C₂H₅-O-C₂H₅) Two ethanol molecules react together removing one water molecule 170°C Intramolecular dehydration Ethene (CH₂=CH₂) Single ethanol molecule loses water internally to form alkene

Chemical Reactions:
At 140°C: 2C₂H₅OH → C₂H₅-O-C₂H₅ + H₂O
At 170°C: C₂H₅OH → CH₂=CH₂ + H₂O

Reason: Different temperatures favor different reaction pathways:
• Lower temperature (140°C): Favors nucleophilic substitution (Sₙ2) forming ether
• Higher temperature (170°C): Favors elimination (E2) forming alkene

(ii) Tertiary alcohols not oxidized:

Tertiary alcohols (R₃C-OH) lack α-hydrogen

Reasons:

  1. No α-hydrogen: Tertiary alcohols have no hydrogen on the carbon bearing -OH group (R₃C-OH). Oxidation requires removal of α-hydrogen along with -OH as water.
  2. Strong C-C bonds: Breaking C-C bonds requires harsh conditions not provided by mild oxidizing agents like KMnO₄ or K₂Cr₂O₇.
  3. Steric hindrance: Three alkyl groups create steric hindrance preventing approach of oxidizing agent.
  4. Alternative reaction: Tertiary alcohols undergo dehydration with acidified dichromate to form alkenes rather than oxidation.

Example: (CH₃)₃C-OH + [O] → No reaction (with mild oxidants)
But with heat: (CH₃)₃C-OH → (CH₃)₂C=CH₂ + H₂O (dehydration)

Comparison:
• Primary alcohols: R-CH₂-OH → R-CHO → R-COOH
• Secondary alcohols: R₂CH-OH → R-CO-R’
• Tertiary alcohols: R₃C-OH → No oxidation (requires C-C bond cleavage)

4. How will you distinguish between?
(i) Methanol and ethanol
(ii) 1-Propanol and 2-propanol

Answer:

(i) Methanol (CH₃OH) vs Ethanol (C₂H₅OH):

TestMethanol (CH₃OH)Ethanol (C₂H₅OH) Iodoform Test Negative (no yellow ppt) Positive (pale yellow ppt of CHI₃) Oxidation + Fehling’s Forms HCOOH (formic acid) → No red ppt with Fehling’s Forms CH₃COOH (acetic acid) after full oxidation Lucas Test Slow reaction (primary) Slow reaction (primary) Best Test Iodoform test is definitive

Iodoform Test Reaction (for ethanol only):
C₂H₅OH + 4I₂ + 6NaOH → CHI₃↓ (yellow) + HCOONa + 5NaI + 5H₂O

(ii) 1-Propanol (CH₃CH₂CH₂OH) vs 2-Propanol (CH₃CHOHCH₃):

Test1-Propanol (Primary)2-Propanol (Secondary) Lucas Test No turbidity at room temp; turbidity on heating only Turbidity appears in 5-10 minutes at room temp Oxidation Test Forms propanal then propanoic acid (Orange → Green) Forms propanone (acetone) only (Orange → Green) Iodoform Test Negative Positive (yellow ppt of CHI₃) Best Test Lucas test is fastest and easiest
Summary of Distinguishing Tests:
• Lucas Test: Differentiates 1°, 2°, 3° alcohols by reaction rate
• Iodoform Test: Identifies CH₃-CH(OH)- group in alcohols
• Oxidation Test: Differentiates by oxidation products

5. How can you bring about the following conversions?
(i) A halogenoalkane into an alcohol
(ii) An alcohol into a halogenoalkane
(iii) An alcohol into a carboxylic acid
(iv) An aldehyde into an alcohol
(v) An alcohol into an ester

Answer:

(i) Halogenoalkane → Alcohol
→
(ii) Alcohol → Halogenoalkane
→
(iii) Alcohol → Carboxylic acid

(i) Halogenoalkane → Alcohol:

R-X + NaOH(aq) → R-OH + NaX

Example: C₂H₅Br + NaOH(aq) → C₂H₅OH + NaBr
Conditions: Heat with aqueous NaOH
Mechanism: Nucleophilic substitution (Sₙ2 for 1°, Sₙ1 for 3°)

(ii) Alcohol → Halogenoalkane:

ReagentReactionConditions SOCl₂ R-OH + SOCl₂ → R-Cl + SO₂ + HCl Pyridine, room temp PCl₃ 3R-OH + PCl₃ → 3R-Cl + H₃PO₃ Room temp HX R-OH + HX → R-X + H₂O ZnCl₂ for Lucas test

(iii) Alcohol → Carboxylic acid:

R-CH₂-OH + 2[O] → R-COOH + H₂O

Example: C₂H₅OH + 2[O] → CH₃COOH + H₂O
Oxidizing agent: K₂Cr₂O₇/H⁺ or KMnO₄/H⁺
Note: Only primary alcohols give carboxylic acids

(iv) Aldehyde → Alcohol:

R-CHO + 2[H] → R-CH₂-OH

Example: CH₃CHO + 2[H] → C₂H₅OH
Reducing agent: NaBH₄ or LiAlH₄ or catalytic hydrogenation (H₂/Ni)
Product: Primary alcohol

(v) Alcohol → Ester:

R-COOH + R’-OH ⇌ R-COO-R’ + H₂O

Example: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Conditions: Conc. H₂SO₄ catalyst, heat
Reaction name: Esterification (Fischer esterification)

Project Suggestion:

Create an interactive chart showing the interconversions between alcohols and other organic compound families (alkenes, haloalkanes, carbonyl compounds, carboxylic acids, esters).

Project Outline: Alcohols as Central Compounds in Organic Synthesis

Section 1: Introduction to Alcohols • Definition & classification
• Structure & bonding
• Physical properties
• Industrial importance Section 2: Preparation Methods • From alkenes (hydration)
• From haloalkanes (substitution)
• From carbonyl compounds (reduction)
• Fermentation process
• Industrial production Section 3: Chemical Reactions • Oxidation pathways (flow chart)
• Dehydration mechanisms
• Esterification mechanism
• Halogenation methods
• Acid-base properties Section 4: Interconversions • Alcohol ↔ Alkene (dehydration/hydration)
• Alcohol ↔ Haloalkane (substitution)
• Alcohol ↔ Carbonyl (oxidation/reduction)
• Alcohol ↔ Ester (esterification/hydrolysis)
• Alcohol ↔ Ether (dehydration) Section 5: Applications • Fuels and energy sources
• Solvents in industry
• Pharmaceuticals
• Perfumes and cosmetics
• Environmental considerations
Interactive Elements to Include:
• Color-coded reaction mechanisms with curved arrows
• Flow charts showing synthetic routes
• Molecular models showing structural changes
• Comparison tables of different alcohol types
• Real-world application images with explanations
Central Position of Alcohols:
Alkenes ← Dehydration/Hydration → Alcohols
Haloalkanes ← Substitution → Alcohols
Carbonyls ← Oxidation/Reduction → Alcohols
Esters ← Esterification/Hydrolysis → Alcohols