Exhaustive Guide: Alcohols & Phenols - Reactions | Class 12 Chemistry | ChemCA

Alcohols & Phenols: Properties & Reactions

Module 2 | CBSE Class 12 Chemistry | Organic Chemistry

1. Physical Properties

1.1 Boiling Points & Intermolecular Hydrogen Bonding

The boiling points of alcohols and phenols are significantly higher than those of corresponding hydrocarbons, ethers, and haloalkanes of comparable molecular masses.

Reason: The high boiling points of alcohols and phenols are mainly due to the presence of intermolecular hydrogen bonding. The highly electronegative oxygen atom polarizes the O-H bond, allowing the slightly positive hydrogen of one molecule to form a strong electrostatic attraction with the lone pair on the oxygen of another molecule.

Effect of chain length: Boiling points increase with an increase in the number of carbon atoms (increase in van der Waals forces).
Effect of branching: Boiling points decrease with an increase of branching in the carbon chain (due to a decrease in surface area, lowering van der Waals forces).

Solubility: Lower alcohols are highly soluble in water because they can form hydrogen bonds with water molecules. Solubility decreases as the size of the hydrophobic alkyl/aryl group (which resists H-bonding) increases.

2. Chemical Reactions (Cleavage of O-H Bond)

Alcohols and phenols are versatile; they can react both as nucleophiles (cleavage of O-H bond) and electrophiles (cleavage of C-O bond in protonated alcohols).

2.1 Acidity of Alcohols

Alcohols react with active metals like Sodium, Potassium, and Aluminium to yield corresponding alkoxides and hydrogen gas, showing their acidic nature (Bronsted acids).
2R-O-H + 2Na → 2R-O^-Na⁺ (Sodium alkoxide) + H₂↑

However, alcohols are weaker acids than water. Water is a better proton donor than alcohol. The electron-releasing (+I) alkyl group in alcohols increases electron density on oxygen, destabilizing the alkoxide ion relative to the hydroxide ion.

Order of Acidity of Alcohols: Primary (1°) > Secondary (2°) > Tertiary (3°).
(More +I groups in 3° alcohols destabilize the alkoxide ion the most).

2.2 Acidity of Phenols & Substituent Effects (Board Favorite)

Unlike alcohols, phenols react with aqueous NaOH to form sodium phenoxide, proving that phenols are significantly stronger acids than alcohols and water.

Why is Phenol more acidic than Alcohol?
1. In phenol, the -OH group is attached to an sp² hybridized carbon, which is more electronegative than the sp³ carbon in alcohols, aiding the release of H⁺.
2. Resonance Stabilization: After losing H⁺, phenol forms a phenoxide ion. The negative charge on the phenoxide ion is highly stabilized by delocalization (resonance) over the entire benzene ring. The alkoxide ion of alcohols lacks resonance stabilization.

Effect of Substituents on Acidity of Phenol:

Electron Withdrawing Groups (EWG): Groups like -NO₂, -CN, -X stabilize the phenoxide ion by dispersing the negative charge, thereby increasing acidity. This effect is maximum at the ortho and para positions. (e.g., 2,4,6-Trinitrophenol is a very strong acid, called Picric Acid).
Electron Donating Groups (EDG): Groups like -CH₃, -OCH₃ destabilize the phenoxide ion by intensifying the negative charge, thereby decreasing acidity. (e.g., Cresols are less acidic than phenol).

2.3 Esterification (Synthesis of Aspirin)

Alcohols and phenols react with carboxylic acids, acid chlorides, and acid anhydrides to form esters. The reaction with carboxylic acid and anhydride is carried out in the presence of a small amount of concentrated H₂SO₄. The reaction is reversible, so water must be removed to push the equilibrium forward.

R-OH + R'-COOH ⇔(H⁺) R'-COOR + H₂O

Preparation of Aspirin (Acetylsalicylic Acid):
Salicylic acid (o-Hydroxybenzoic acid) reacts with acetic anhydride in the presence of an acid catalyst to undergo acetylation of the -OH group. This produces Aspirin, a crucial analgesic and antipyretic drug.
Salicylic Acid + (CH₃CO)₂O →(H⁺) Acetylsalicylic acid (Aspirin) + CH₃COOH

3. Reactions Involving Cleavage of C-O Bond in Alcohols

These reactions occur only in alcohols. Phenols do not undergo cleavage of the C-O bond easily because it has a partial double bond character due to resonance.

3.1 Reaction with Hydrogen Halides (The Lucas Test)

Alcohols react with hydrogen halides (HX) to form alkyl halides.

Lucas Test: Used to distinguish between 1°, 2°, and 3° alcohols. The Lucas Reagent is a mixture of concentrated HCl and anhydrous ZnCl₂.
Alcohols are soluble in Lucas reagent, while their halides are immiscible and produce turbidity (cloudiness) in the solution.

Tertiary (3°) alcohols: Produce turbidity immediately (most reactive due to stable 3° carbocation).
Secondary (2°) alcohols: Produce turbidity after 5 minutes.
Primary (1°) alcohols: Do not produce turbidity at room temperature (requires heating).

3.2 Dehydration

Alcohols undergo dehydration (removal of a water molecule) to form alkenes when treated with a protic acid (conc. H₂SO₄ or H₃PO₄). The reaction follows the Saytzeff Rule.

The relative ease of dehydration of alcohols follows the order: 3° > 2° > 1°.
(Because a 3° carbocation intermediate is formed faster due to its high stability).

3.3 Oxidation

Oxidation of alcohols involves the formation of a carbon-oxygen double bond with the cleavage of an O-H and C-H bond (often called dehydrogenation).

Primary Alcohols: Oxidized to aldehydes, which are easily further oxidized to carboxylic acids.
R-CH₂OH →(KMnO₄ or K₂Cr₂O₇/H⁺) R-COOH.
To stop the reaction at the aldehyde stage, a mild oxidizing agent like Pyridinium Chlorochromate (PCC) or CrO₃ in anhydrous medium is used.
Secondary Alcohols: Oxidized to ketones by chromic anhydride (CrO₃).
Tertiary Alcohols: Do not undergo oxidation under normal conditions. Under strong reaction conditions (strong oxidizing agents and high temperatures), cleavage of various C-C bonds takes place to yield a mixture of carboxylic acids containing fewer carbon atoms.

4. Electrophilic Aromatic Substitution in Phenols

The -OH group attached to the benzene ring powerfully activates it towards electrophilic substitution. It also directs the incoming electrophile to the ortho and para positions due to resonance (+R effect).

4.1 Nitration and Halogenation

A. Nitration:

With Dilute HNO₃ at low temperature (298 K): Phenol yields a mixture of ortho and para nitrophenols.
With Concentrated HNO₃: Phenol is completely nitrated to yield 2,4,6-trinitrophenol (Picric Acid).

Separation of o- and p-nitrophenol: The ortho and para isomers formed by dilute HNO₃ can be separated by steam distillation. o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding. p-Nitrophenol is less volatile because of intermolecular hydrogen bonding, which causes the association of molecules.

B. Halogenation (Bromination):

With Br₂ in CS₂ or CHCl₃ (low polarity solvent, low temp): Yields monobromophenols (o- and p-bromophenol).
With Bromine Water (highly polar): The phenol is so strongly activated that it immediately forms a white precipitate of 2,4,6-tribromophenol.

4.2 Name Reactions: Kolbe's & Reimer-Tiemann

A. Kolbe's Reaction:
Phenol is first converted to sodium phenoxide with NaOH (the phenoxide ion is even more reactive than phenol). It undergoes electrophilic substitution with carbon dioxide (CO₂), followed by acidification, to yield ortho-hydroxybenzoic acid (Salicylic acid).

Phenol →(1. NaOH) Sodium Phenoxide →(2. CO₂, H⁺) Salicylic Acid

B. Reimer-Tiemann Reaction:
On treating phenol with chloroform (CHCl₃) in the presence of aqueous sodium hydroxide at 340 K, a -CHO group is introduced at the ortho position of the ring. Acidification yields Salicylaldehyde.

Phenol + CHCl₃ + aq. NaOH → Intermediate →(H⁺) Salicylaldehyde

4.3 Reaction with Zinc Dust & Oxidation

Reduction to Benzene: Phenol is converted to benzene on heating with zinc dust.
C₆H₅OH + Zn(dust) →(heat) C₆H₆ (Benzene) + ZnO

Oxidation: Phenol undergoes oxidation with strong oxidizing agents like sodium dichromate (Na₂Cr₂O₇) and H₂SO₄ to produce a conjugated diketone known as benzoquinone.

5. NCERT Solved Examples (Step-by-Step)

NCERT Example 11.4: Arrange the following compounds in increasing order of their boiling points:
Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.

Solution:
The boiling point of alcohols increases with an increase in the number of carbon atoms (increased van der Waals forces). For isomers, branching decreases the boiling point (decreased surface area).
Increasing order:
Methanol (1C) < Ethanol (2C) < Propan-1-ol (3C) < Butan-2-ol (4C, branched) < Butan-1-ol (4C, straight) < Pentan-1-ol (5C).

NCERT Example 11.5: Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.

Solution:
1. Propan-1-ol is an aliphatic alcohol and is the weakest acid (weaker than water). Phenols are stronger acids.
2. 4-Methylphenol (p-Cresol): The -CH₃ group is an electron-donating group (+I, hyperconjugation), which decreases the acidity of phenol.
3. Phenol: Reference point.
4. Nitrophenols: The -NO₂ group is strongly electron-withdrawing (-I, -R), which stabilizes the phenoxide ion and increases acidity. Acidity increases with the number of -NO₂ groups.
Increasing order:
Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 3,5-dinitrophenol < 2,4,6-trinitrophenol (Picric acid).

6. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2021]

Q1. How do you account for the fact that unlike phenol, 2,4-dinitrophenol is soluble in aqueous sodium carbonate solution?

Answer: Sodium carbonate (Na₂CO₃) is a weak base. Phenol is a relatively weak acid and cannot displace carbonic acid to react with Na₂CO₃. However, the presence of two strong electron-withdrawing -NO₂ groups in 2,4-dinitrophenol drastically increases its acidity (making it a stronger acid than H₂CO₃). Therefore, it readily reacts with aqueous Na₂CO₃ to form a soluble sodium salt and evolves CO₂ gas.

[CBSE 2017, 2019]

Q2. Ortho-nitrophenol is steam volatile whereas para-nitrophenol is not. Explain.

Answer: Ortho-nitrophenol forms intramolecular hydrogen bonds (within the same molecule). This prevents the association of molecules, making it volatile. Para-nitrophenol, on the other hand, forms intermolecular hydrogen bonds (between different molecules), which causes molecular association and heavily restricts volatility.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q3. Assertion (A): The acidic strength of primary alcohols is more than that of secondary alcohols.
Reason (R): The +I effect of alkyl groups increases the electron density on the oxygen atom of the O-H bond.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. A secondary alcohol has more alkyl groups (more +I effect) than a primary alcohol. This increases electron density on the oxygen, destabilizing the resulting alkoxide ion and making the release of the H⁺ ion more difficult.

Part C: Distinguishing Tests & Conversions (3 Marks)

[CBSE 2016, 2022]

Q4. Give a chemical test to distinguish between:
(a) Propan-2-ol and 2-Methylpropan-2-ol
(b) Phenol and Ethanol

Answer:
(a) Lucas Test: Add Lucas Reagent (conc. HCl + anhydrous ZnCl₂). 2-Methylpropan-2-ol is a tertiary (3°) alcohol and will produce immediate turbidity. Propan-2-ol is a secondary (2°) alcohol and will produce turbidity after about 5 minutes.
(b) Ferric Chloride Test: Add neutral FeCl₃ solution. Phenol forms a complex with iron, yielding a characteristic violet/purple coloration. Ethanol does not react with neutral FeCl₃.

[CBSE 2015]

Q5. Write the chemical equations for the following Name Reactions:
(i) Reimer-Tiemann Reaction
(ii) Kolbe's Reaction

Answer:
(i) Reimer-Tiemann: Phenol + CHCl₃ + aq. NaOH → [Intermediate] →(H⁺) 2-Hydroxybenzaldehyde (Salicylaldehyde).
(ii) Kolbe's: Phenol + NaOH → Sodium phenoxide.
Sodium phenoxide + CO₂ (400K, 4-7 atm) → Sodium salicylate →(H⁺) 2-Hydroxybenzoic acid (Salicylic acid).

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Exhaustive Guide: Alcohols & Phenols - Reactions | Class 12 Chemistry