Alcohols, Phenols and Ethers

What this chapter is about

This chapter introduces three important families of organic compounds that contain oxygen bonded to carbon: alcohols, phenols and ethers. Alcohols have a hydroxyl group (–OH) attached to a saturated carbon atom; phenols have the –OH group bonded directly to a benzene ring; ethers have an oxygen atom bonded to two carbon groups. All three are widespread in daily life—ethanol in sanitisers, phenol in disinfectants, diethyl ether as an anaesthetic solvent.

A Class 12 student meets this chapter because it builds on the basics of organic chemistry learned earlier: hybridisation, nomenclature, functional-group behaviour and reaction mechanisms. Understanding how the –OH group and the C–O–C linkage behave prepares you for the chemistry of aldehydes, ketones, carboxylic acids and amines that follow. After studying this chapter you should be able to name compounds using IUPAC rules, predict physical properties from hydrogen bonding, write balanced equations for characteristic reactions, and explain mechanisms such as nucleophilic substitution and electrophilic aromatic substitution.

The chapter also covers preparation methods—hydration of alkenes, reduction of carbonyl compounds, Williamson synthesis of ethers—and distinguishing tests that help identify each class in laboratory work.

Key ideas

• Alcohols are classified as primary (1°), secondary (2°) or tertiary (3°) depending on the number of carbon atoms attached to the carbon bearing the –OH group.
• Phenols are more acidic than alcohols because the phenoxide ion (C₆H₅O⁻) is stabilised by resonance with the benzene ring.
• Hydrogen bonding between –OH groups raises the boiling points of alcohols and phenols compared to ethers of similar molar mass.
• Alcohols undergo nucleophilic substitution: the –OH is replaced by halogen when treated with HX, PCl₃, PCl₅ or SOCl₂.
• Dehydration of alcohols with concentrated H₂SO₄ or Al₂O₃ at high temperature gives alkenes; the ease follows the order 3° > 2° > 1°.
• Ethers are relatively inert but undergo cleavage with concentrated HI or HBr to give alcohols and alkyl halides.
• The Williamson synthesis prepares ethers by reacting a sodium alkoxide with an alkyl halide through an Sₙ2 mechanism.
• Phenols show electrophilic aromatic substitution—bromination, nitration, Kolbe and Reimer–Tiemann reactions—due to the activating –OH group.

Formulas and facts to remember

1. General formula of alcohols: CₙH₂ₙ₊₁OH (for saturated, mono-hydroxy alcohols).
2. Acidity order: water (pKₐ ≈ 15.7) < alcohols (pKₐ ≈ 16–18) < phenol (pKₐ ≈ 10).
3. Lucas test: tertiary alcohols react immediately with Lucas reagent (anhydrous ZnCl₂ + conc. HCl) giving a turbid layer; secondary alcohols react in 5–10 minutes; primary alcohols do not react at room temperature.
4. Iodoform test: ethanol and any secondary alcohol with a CH₃–CHOH– unit give a yellow precipitate of CHI₃ when warmed with I₂ and NaOH.
5. Kolbe reaction: phenol + NaOH → sodium phenoxide; then CO₂ at 400 K and 4–7 atm → sodium salicylate → salicylic acid on acidification.
6. Reimer–Tiemann reaction: phenol + CHCl₃ + aqueous NaOH → salicylaldehyde (ortho-hydroxybenzaldehyde).
7. Williamson synthesis: R–O⁻Na⁺ + R′–X → R–O–R′ + NaX (works best when R′–X is a primary halide to favour Sₙ2).
8. Dehydration temperature guide: ethanol requires about 443 K with excess conc. H₂SO₄ to form ethene; at 413 K diethyl ether forms instead.

Worked examples

Example 1 – IUPAC naming

Problem: Write the IUPAC name of CH₃–CH(OH)–CH₂–CH₃.

Solution:

1. Identify the longest chain containing the –OH group: four carbons (butane parent).
2. Number from the end nearest to –OH: the hydroxyl is on carbon 2.
3. Name: butan-2-ol.

Example 2 – Predicting product of dehydration

Problem: What major alkene forms when butan-2-ol is heated with concentrated H₂SO₄ at 443 K?

Solution:

1. Dehydration removes H and OH from adjacent carbons.
2. Two possible products: but-1-ene (less substituted) and but-2-ene (more substituted).
3. Zaitsev's rule: the more substituted alkene is the major product.
4. Major product: but-2-ene (exists as cis and trans isomers; trans is slightly more stable).

Example 3 – Williamson ether synthesis

Problem: How would you prepare methyl propyl ether (CH₃–O–CH₂–CH₂–CH₃) using the Williamson method?

Solution:

1. Choose reagents so that the primary halide undergoes Sₙ2 attack by the alkoxide.
2. Option A: sodium methoxide (CH₃O⁻Na⁺) + 1-bromopropane (CH₃CH₂CH₂Br).
3. Equation: CH₃O⁻Na⁺ + CH₃CH₂CH₂Br → CH₃–O–CH₂–CH₂–CH₃ + NaBr.
4. This choice avoids elimination side-reactions that would occur if a tertiary halide were used.

Common mistakes

• Writing phenol as an alcohol because both have –OH → Remember: in phenol the –OH is directly attached to the benzene ring, giving different acidity and reactions.
• Expecting ethers to react with dilute acids or bases as alcohols do → Ethers lack an acidic hydrogen and are largely unreactive except with strong halogen acids.
• Applying Sₙ2 conditions to tertiary substrates in Williamson synthesis → Tertiary halides favour elimination; always use a primary halide with the bulkier group as the alkoxide.
• Confusing the temperature conditions for alcohol dehydration → Lower temperature (around 413 K) favours ether formation; higher temperature (around 443 K) favours alkene.
• Forgetting resonance stabilisation when comparing acidity of phenol and alcohols → The phenoxide ion delocalises negative charge into the ring, making phenol about 10⁶ times more acidic than ethanol.

Quick revision

• Alcohols: –OH on saturated C; phenols: –OH on benzene; ethers: C–O–C linkage.
• Hydrogen bonding raises boiling points of alcohols and phenols above those of ethers.
• Phenol is acidic enough to dissolve in aqueous NaOH; alcohols are not.
• Lucas test distinguishes 1°, 2° and 3° alcohols by turbidity time.
• Williamson synthesis: alkoxide + primary alkyl halide → ether (Sₙ2 pathway).
• Concentrated HI cleaves ethers; the smaller alkyl group usually leaves as the iodide.