ChemistryISC Class 12 · Chapter 7

Alcohols, Phenols and Ethers ISC Class 12 Chemistry

Q: Do I need properties of secondary and tertiary alcohols?

The ISC 2027 syllabus asks for physical and chemical properties of primary alcohols only, plus the differences between primary, secondary and tertiary alcohols and the Lucas test to distinguish them.

Q: How much detail on the Bosch process and fermentation?

Only an outline of the method, but the chemical equations are required.

Q: Which mechanisms are required in this chapter?

Esterification and dehydration of alcohols are listed with mechanism.

Q: What tests for phenol are in the syllabus?

The FeCl₃ test (violet colour) and the azo dye test (orange dye with benzenediazonium chloride in alkali).

Q: Which aryl ether reactions should I know?

Preparation of anisole by Williamson's synthesis and its electrophilic substitution: halogenation, nitration and Friedel-Crafts reactions.

Q: How many marks does this chapter carry?

It is part of the Organic Chemistry group, 31 of the 70 theory marks, shared with four other organic chapters.

Updated 15 October 2026 · Fact-checked

Alcohols, Phenols and Ethers is Chapter 7 of ISC Class 12 Chemistry. It covers classification and naming, preparation of alcohols (hydration, Grignard, reduction), Bosch and fermentation processes, reactions with mechanisms of esterification and dehydration, Lucas test, phenol preparation and reactions (Kolbe, Reimer-Tiemann), and ethers by Williamson's synthesis.

What Alcohols, Phenols and Ethers is about

Alcohols contain -OH on an sp³ carbon, phenols -OH directly on a benzene ring, and ethers an oxygen joining two alkyl or aryl groups (R-O-R'). The chapter is part of the Organic Chemistry group (31 of the 70 theory marks) and is a rich source of conversions, distinguishing tests and 'identify A, B, C' problems.

Alcohols are classified as monohydric, dihydric (ethane-1,2-diol) and polyhydric (glycerol), and monohydric ones as primary, secondary or tertiary. They are prepared by hydration of alkenes (direct, indirect via H₂SO₄, and hydroboration-oxidation), from Grignard reagents and carbonyl compounds, by hydrolysis of alkyl halides, by reduction of aldehydes and ketones, and from primary amines with nitrous acid. Industrially, methanol is made by the Bosch process (CO + 2H₂) and ethanol by fermentation. The syllabus asks for properties of primary alcohols: weak acidity (reaction with Na), esterification (with mechanism), reaction with HX, PCl₃, PCl₅ and SOCl₂, acylation, oxidation, and dehydration (with mechanism), which gives ethene at 443 K but diethyl ether at 413 K. The Lucas test distinguishes 1°, 2° and 3° alcohols.

Phenol is made from diazonium salts, chlorobenzene (Dow's process), benzene sulphonic acid and cumene. It is more acidic than alcohols and water because the phenoxide ion is resonance-stabilised. The -OH group strongly activates the ring, so phenol undergoes easy bromination (2,4,6-tribromophenol with bromine water), nitration (picric acid with concentrated acid) and sulphonation, plus Kolbe's reaction (salicylic acid) and the Reimer-Tiemann reaction (salicylaldehyde). Tests: violet colour with neutral FeCl₃ and an orange azo dye.

Ethers are made by Williamson's synthesis (sodium alkoxide + alkyl halide). They react with chlorine, form explosive peroxides in air, and are cleaved by HI. Anisole undergoes ortho-para electrophilic substitution (halogenation, nitration, Friedel-Crafts).

Weightage: Alcohols, Phenols and Ethers is part of the Organic Chemistry group, which carries 31 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Classification, nomenclature and physical properties: Monohydric (one -OH, $C_{n} H_{2 n + 1} O H$ ), dihydric (glycols, e.g. ethane-1,2-diol), polyhydric (glycerol, propane-1,2,3-triol). Monohydric alcohols are primary (-CH₂OH, e.g. ethanol), secondary (>CHOH, propan-2-ol) or tertiary (≡C-OH, 2-methylpropan-2-ol). IUPAC: replace -e of alkane by -ol, number the chain to give -OH the lowest locant: CH₃CH(OH)CH₂CH₃ butan-2-ol. Structure: O is sp³, C-O-H angle about 108.9°. Boiling points are much higher than those of alkanes, haloalkanes and ethers of similar mass because of intermolecular hydrogen bonding; they rise with chain length and fall with branching (1° > 2° > 3° for isomers, less surface area). More -OH groups mean more hydrogen bonds: ethane-1,2-diol boils far above ethanol. Lower alcohols are miscible with water (hydrogen bonding with water); solubility falls as the hydrophobic alkyl part grows. Differences between 1°, 2°, 3°: structure (H atoms on the carbinol carbon: 2, 1, 0); oxidation products (aldehyde then acid; ketone; no easy oxidation); rate with Lucas reagent; ease of dehydration (3° > 2° > 1°).; Worked example
Arrange in increasing order of boiling point and explain: propane-1,2,3-triol, propan-1-ol, 1-chloropropane, propane.
Try it first, then show the solutionSolution
Propane has only weak dispersion forces: lowest.
1-Chloropropane adds dipole-dipole attraction but has no hydrogen bonding.
Propan-1-ol forms hydrogen bonds through one -OH; propane-1,2,3-triol has three -OH groups per molecule and forms far more extensive hydrogen bonding.
Order: propane < 1-chloropropane < propan-1-ol < propane-1,2,3-triol.
Answer: propane < 1-chloropropane < propan-1-ol < propane-1,2,3-triol
Preparation of alcohols: Direct hydration of alkenes: $C_{2} H_{4} + H_{2} O \to C_{2} H_{5} O H$ (H₃PO₄, about 573 K, 60-70 atm); follows Markovnikov's rule for unsymmetrical alkenes. Indirect hydration: absorb the alkene in cold conc. H₂SO₄, then boil with water: $C_{2} H_{4} + H_{2} S O_{4} \to C_{2} H_{5} H S O_{4}$ ; $C_{2} H_{5} H S O_{4} + H_{2} O \to C_{2} H_{5} O H + H_{2} S O_{4}$ . Hydroboration-oxidation: diborane adds, then alkaline H₂O₂ oxidises, giving anti-Markovnikov alcohols: $3 C_{3} H_{6} + B H_{3} \to (C_{3} H_{7})_{3} B$ ; $(C_{3} H_{7})_{3} B + 3 H_{2} O_{2} \to 3 C_{3} H_{7} O H + H_{3} B O_{3}$ (propene gives propan-1-ol). From Grignard reagents: methanal gives 1° alcohols, other aldehydes 2°, ketones 3°; the adduct is hydrolysed by dilute acid: $C H_{3} C H O + C H_{3} M g B r \to (C H_{3})_{2} C H O M g B r$ ; $(C H_{3})_{2} C H O M g B r + H_{2} O \to (C H_{3})_{2} C H O H + M g (O H) B r$ . Hydrolysis of alkyl halides with aqueous KOH. Reduction of carbonyl compounds with H₂/Ni, LiAlH₄ or NaBH₄: $C H_{3} C H O + 2 [H] \to C_{2} H_{5} O H$ ; ketones give 2° alcohols. From primary amines with nitrous acid (NaNO₂ + HCl): $C_{2} H_{5} N H_{2} + H N O_{2} \to C_{2} H_{5} O H + N_{2} + H_{2} O$ . Methanol (Bosch process, outline): water gas from steam over hot coke ( $C + H_{2} O \to C O + H_{2}$ ), enriched with H₂, passed over ZnO-Cr₂O₃ at 573-673 K and 200-300 atm: $C O + 2 H_{2} \to C H_{3} O H$ . Ethanol by fermentation (outline): molasses (sucrose) with yeast; invertase: $C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$ (glucose + fructose); zymase: $C_{6} H_{12} O_{6} \to 2 C_{2} H_{5} O H + 2 C O_{2}$ . Anaerobic conditions are kept so ethanol is not oxidised to ethanoic acid.; Worked example
How would you prepare 2-methylpropan-2-ol from propanone?
Try it first, then show the solutionSolution
Add methylmagnesium bromide in dry ether: $C H_{3} C O C H_{3} + C H_{3} M g B r \to (C H_{3})_{3} C O M g B r$
Hydrolyse the adduct with dilute acid: $(C H_{3})_{3} C O M g B r + H_{2} O \to (C H_{3})_{3} C O H + M g (O H) B r$
Answer: Grignard addition of CH₃MgBr to the ketone, then hydrolysis (a ketone always gives a 3° alcohol)
Acidity, reaction with sodium and esterification (mechanism): Acidic nature: the O-H bond is polar, so alcohols release H⁺ to very strong bases and react with active metals: $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ . Alcohols are weaker acids than water: the +I effect of the alkyl group increases electron density on O, destabilising the alkoxide; so 1° > 2° > 3° in acidity. Electron-withdrawing groups increase acidity: CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH. Ethers have no O-H and do not react with sodium. Esterification: an alcohol and a carboxylic acid with a little conc. H₂SO₄ form an ester and water reversibly: $C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$ . Mechanism: (1) H⁺ protonates the C=O oxygen of the acid, making carbon more positive; (2) the alcohol oxygen (nucleophile) attacks this carbon, giving a tetrahedral intermediate; (3) a proton moves to one of the -OH groups; (4) water leaves; (5) loss of H⁺ gives the ester and regenerates the catalyst. The oxygen of the ester's C-O-C comes from the alcohol (shown with ¹⁸O labelling). Excess alcohol or removal of water pushes the equilibrium forward. Acylation: with acid chlorides (in pyridine to remove HCl) or anhydrides: $C_{2} H_{5} O H + C H_{3} C O C l \to C H_{3} C O O C_{2} H_{5} + H C l$ ; $C_{2} H_{5} O H + (C H_{3} C O)_{2} O \to C H_{3} C O O C_{2} H_{5} + C H_{3} C O O H$ .; Worked example
Give a reason: sodium metal can be used to remove the last traces of water from diethyl ether, but not from ethanol.
Try it first, then show the solutionSolution
Ethanol has a hydrogen attached to oxygen (an active, acidic H), so sodium reacts with the ethanol itself: $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ .
Diethyl ether has no O-H bond, so sodium reacts only with the water present and leaves the ether unchanged.
Answer: Ethanol has an active O-H hydrogen that reacts with Na; the ether has none.
Reactions with HX and phosphorus halides, and the Lucas test: With hydrogen halides: $C_{2} H_{5} O H + H B r \to C_{2} H_{5} B r + H_{2} O$ . Reactivity of HX: HI > HBr > HCl; of alcohols: 3° > 2° > 1°. Primary alcohols need HCl with anhydrous ZnCl₂ (Groves' process). With PCl₃, PCl₅, SOCl₂: $3 C_{2} H_{5} O H + P C l_{3} \to 3 C_{2} H_{5} C l + H_{3} P O_{3}$ ; $C_{2} H_{5} O H + P C l_{5} \to C_{2} H_{5} C l + P O C l_{3} + H C l$ (white fumes of HCl show an -OH group); $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$ . Lucas test: Lucas reagent is concentrated HCl with anhydrous ZnCl₂. Alcohols dissolve in it, but the alkyl chloride formed is insoluble and makes the solution cloudy (turbid). Tertiary: turbidity immediately at room temperature. Secondary: turbidity within about 5 minutes. Primary: no turbidity at room temperature (only on heating). The order follows carbocation stability (SN1 route). The test works for alcohols up to about six carbons (soluble in the reagent).; Worked example
Using one reagent, how would you distinguish butan-1-ol, butan-2-ol and 2-methylpropan-2-ol?
Try it first, then show the solutionSolution
Add Lucas reagent (conc. HCl + anhydrous ZnCl₂) to each at room temperature.
2-Methylpropan-2-ol (3°) turns cloudy immediately; butan-2-ol (2°) turns cloudy in about 5 minutes; butan-1-ol (1°) stays clear at room temperature.
Answer: Lucas test: immediate (3°), about 5 min (2°), no turbidity (1°)
Oxidation, dehydration (mechanism) and interconversion: Oxidation: with acidified K₂Cr₂O₇ or KMnO₄, a primary alcohol gives an aldehyde, then a carboxylic acid: $C_{2} H_{5} O H + [O] \to C H_{3} C H O + H_{2} O$ ; $C H_{3} C H O + [O] \to C H_{3} C O O H$ . Secondary gives a ketone; tertiary resists oxidation (no H on the carbinol carbon). PCC stops primary alcohols at the aldehyde. Over hot copper at 573 K, 1° alcohols are dehydrogenated to aldehydes ( $C_{2} H_{5} O H \to C H_{3} C H O + H_{2}$ ), 2° to ketones, 3° dehydrated to alkenes. Dehydration with conc. H₂SO₄: at 443 K (170 °C) with excess acid, ethene: $C_{2} H_{5} O H \to C_{2} H_{4} + H_{2} O$ ; at 413 K (140 °C) with excess alcohol, diethyl ether: $2 C_{2} H_{5} O H \to C_{2} H_{5} O C_{2} H_{5} + H_{2} O$ . Mechanism (alkene): (1) the -OH is protonated: $C H_{3} C H_{2} O H + H^{+} \to C H_{3} C H_{2} O H_{2}^{+}$ ; (2) water leaves giving a carbocation (slow); (3) a β-hydrogen is lost as H⁺, forming the C=C. For the ether, a second alcohol molecule attacks the protonated alcohol (SN2) and loses H⁺. Ease of dehydration: 3° > 2° > 1° (carbocation stability). Conversions: *methanol to ethanol (ascent)*: CH₃OH → CH₃I (red P/I₂) → CH₃CN (KCN) → CH₃CH₂NH₂ (Na/ethanol or LiAlH₄) → C₂H₅OH (HNO₂). *Ethanol to methanol (descent)*: C₂H₅OH → CH₃COOH ([O]) → CH₃CONH₂ (via NH₃, heat) → CH₃NH₂ (Br₂/KOH, Hofmann) → CH₃OH (HNO₂). *Propan-1-ol to propan-2-ol*: dehydrate to propene, add H₂SO₄ (Markovnikov), hydrolyse.; Worked example
Butan-2-ol is heated with conc. H₂SO₄ at about 443 K. Name the major product and show how it forms.
Try it first, then show the solutionSolution
Protonation: $C H_{3} C H (O H) C H_{2} C H_{3} + H^{+} \to C H_{3} C H (O H_{2}^{+}) C H_{2} C H_{3}$
Loss of water (slow) gives the secondary carbocation $C H_{3} C H^{+} C H_{2} C H_{3}$ .
Loss of H⁺ from C3 gives the more substituted alkene but-2-ene (major, Saytzeff); loss from C1 gives but-1-ene (minor).
Answer: But-2-ene (major) via a secondary carbocation; but-1-ene minor
Phenol: preparation and acidity: From diazonium salts: warm aqueous benzenediazonium chloride: $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ . From chlorobenzene (Dow's process): fuse with NaOH at 623 K and 300 atm, then acidify: $C_{6} H_{5} C l + 2 N a O H \to C_{6} H_{5} O N a + N a C l + H_{2} O$ ; $C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ . From benzene sulphonic acid: neutralise to sodium benzenesulphonate and fuse with NaOH: $C_{6} H_{5} S O_{3} N a + 2 N a O H \to C_{6} H_{5} O N a + N a_{2} S O_{3} + H_{2} O$ , then acidify. From cumene (industrial): benzene + propene give cumene ( $C_{6} H_{6} + C_{3} H_{6} \to C_{6} H_{5} C H (C H_{3})_{2}$ ); air oxidation gives cumene hydroperoxide ( $C_{6} H_{5} C H (C H_{3})_{2} + O_{2} \to C_{6} H_{5} C (C H_{3})_{2} O O H$ ), which dilute acid splits: $C_{6} H_{5} C (C H_{3})_{2} O O H \to C_{6} H_{5} O H + C H_{3} C O C H_{3}$ ; acetone is a valuable by-product. Physical: colourless crystalline solid (turns pink in air), moderately soluble in water (hydrogen bonding, limited by the large phenyl group, so less soluble than lower aliphatic alcohols). Acidity: phenol (pKa ≈ 10) > water > ethanol. The phenoxide ion is stabilised by resonance (negative charge spread to ortho and para carbons); the sp² carbon also withdraws electrons. Phenol turns blue litmus red faintly, reacts with NaOH ( $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O N a + H_{2} O$ ) and Na ( $2 C_{6} H_{5} O H + 2 N a \to 2 C_{6} H_{5} O N a + H_{2}$ ), but not with NaHCO₃ (weaker acid than carbonic acid). Electron-withdrawing groups (-NO₂) at o/p increase acidity; electron-donating groups (-CH₃) decrease it.; Worked example
Phenol is acidic but ethanol is practically neutral. Explain.
Try it first, then show the solutionSolution
Both lose H⁺ from O-H. The phenoxide ion is stabilised by resonance, the negative charge being delocalised over the ring (ortho and para positions).
The ethoxide ion has no resonance, and the +I effect of the ethyl group concentrates the negative charge on O, destabilising it.
So phenol ionises more readily.
Answer: Resonance stabilisation of phenoxide
Reactions of phenol: With zinc dust (heated): $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ . Acylation: $C_{6} H_{5} O H + C H_{3} C O C l \to C H_{3} C O O C_{6} H_{5} + H C l$ (phenyl ethanoate; anhydride also works). With PCl₅ (poor yield): $C_{6} H_{5} O H + P C l_{5} \to C_{6} H_{5} C l + P O C l_{3} + H C l$ . Electrophilic substitution: -OH is strongly activating and ortho-para directing (+R). Bromination: with bromine water a white precipitate of 2,4,6-tribromophenol forms at once: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ ; in CS₂ or CHCl₃ at low temperature, mainly 4-bromophenol (with some 2-bromophenol). Nitration: dilute HNO₃ at 298 K gives o- and p-nitrophenol ( $C_{6} H_{5} O H + H N O_{3} \to C_{6} H_{4} (O H) N O_{2} + H_{2} O$ ), separated by steam distillation (ortho is volatile due to intramolecular H-bonding); conc. HNO₃ with conc. H₂SO₄ gives picric acid (2,4,6-trinitrophenol): $C_{6} H_{5} O H + 3 H N O_{3} \to C_{6} H_{2} (N O_{2})_{3} O H + 3 H_{2} O$ . Sulphonation: conc. H₂SO₄ gives o-hydroxybenzenesulphonic acid at 288-293 K and the para isomer at 373 K. Kolbe's reaction: sodium phenoxide + CO₂ at about 400 K and 4-7 atm, then acid: $C_{6} H_{5} O N a + C O_{2} \to C_{6} H_{4} (O H) C O O N a$ ; $C_{6} H_{4} (O H) C O O N a + H C l \to C_{6} H_{4} (O H) C O O H + N a C l$ (salicylic acid). Reimer-Tiemann reaction: phenol + chloroform + aqueous NaOH at 340 K, then acid, gives salicylaldehyde (dichlorocarbene :CCl₂ is the electrophile): overall $C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$ . Tests: neutral FeCl₃ gives a violet colour; benzenediazonium chloride in alkali gives an orange azo dye (p-hydroxyazobenzene): $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N_{2} C_{6} H_{4} O H + H C l$ . Uses: antiseptic (dilute) and disinfectant, Bakelite, dyes, drugs (aspirin via salicylic acid), picric acid.; Worked example
Give a chemical test to distinguish phenol from benzyl alcohol (C₆H₅CH₂OH).
Try it first, then show the solutionSolution
Add neutral FeCl₃ solution: phenol, with -OH directly on the ring, gives a violet colour; benzyl alcohol gives no violet colour.
Or add bromine water: phenol gives a white precipitate of 2,4,6-tribromophenol at once; benzyl alcohol does not.
Answer: Neutral FeCl₃: violet with phenol only
Ethers and anisole: General formula $C_{n} H_{2 n + 2} O$ (isomeric with alcohols). Structure: O is sp³, C-O-C angle about 110°, slightly polar. Naming: alkoxyalkane (ethoxyethane, methoxybenzene = anisole). Williamson's synthesis: sodium alkoxide + primary alkyl halide (SN2): $C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$ ; anisole: $C_{6} H_{5} O N a + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$ . With a 3° halide elimination dominates, so the alkoxide should carry the bulky group. Also by dehydration of alcohols at 413 K. Physical: low boiling points (no hydrogen bonding between ether molecules), miscible with water only slightly (diethyl ether about as soluble as butanol, H-bonds with water). Reactions: *chlorine*: in the dark α-hydrogens are substituted ( $C_{2} H_{5} O C_{2} H_{5} + C l_{2} \to C H_{3} C H C l O C_{2} H_{5} + H C l$ ); in light all H are replaced ( $C_{2} H_{5} O C_{2} H_{5} + 10 C l_{2} \to C_{2} C l_{5} O C_{2} C l_{5} + 10 H C l$ ). *Oxidation*: on standing in air and light ethers form explosive peroxides ( $C_{2} H_{5} O C_{2} H_{5} + O_{2} \to C H_{3} C H (O O H) O C_{2} H_{5}$ ), so old ether is not distilled to dryness. *HI*: cold: $C_{2} H_{5} O C_{2} H_{5} + H I \to C_{2} H_{5} O H + C_{2} H_{5} I$ ; hot excess: $C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$ . In mixed ethers I⁻ attacks the smaller alkyl group (SN2): $C H_{3} O C_{2} H_{5} + H I \to C H_{3} I + C_{2} H_{5} O H$ ; with a tertiary group, SN1 gives the tertiary iodide. Anisole gives phenol and CH₃I (the strong aryl C-O bond is not broken). *PCl₅*: $C_{2} H_{5} O C_{2} H_{5} + P C l_{5} \to 2 C_{2} H_{5} C l + P O C l_{3}$ . Anisole: -OCH₃ activates and directs ortho-para. Bromination (Br₂ in ethanoic acid) gives mainly p-bromoanisole; nitration gives 2- and 4-nitroanisole; Friedel-Crafts alkylation ( $C_{6} H_{5} O C H_{3} + C H_{3} C l \to C_{6} H_{4} (C H_{3}) O C H_{3} + H C l$ , AlCl₃) and acylation ( $C_{6} H_{5} O C H_{3} + C H_{3} C O C l \to C_{6} H_{4} (C O C H_{3}) O C H_{3} + H C l$ ). Uses: diethyl ether as a solvent and formerly an anaesthetic; anisole in perfumes.; Worked example
Explain why methoxyethane with HI gives iodomethane and ethanol, not iodoethane and methanol.
Try it first, then show the solutionSolution
HI protonates the ether oxygen; I⁻ then attacks a carbon by SN2.
SN2 is fastest at the least hindered carbon, the methyl group, so CH₃-O breaks: CH₃I + C₂H₅OH.
Answer: Iodide attacks the smaller (methyl) group by SN2.

Chemistry in real life: Alcohols, Phenols and Ethers

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Fermentation yield

Suppose a small distillery ferments 1.0 kg of glucose completely with yeast.

What mass of ethanol and what volume of CO₂ at STP (22.4 L mol⁻¹) are formed?

Try it first, then show the solutionSolution

$C_{6} H_{12} O_{6} \to 2 C_{2} H_{5} O H + 2 C O_{2}$ : 180 g gives 92 g ethanol and 2 mol CO₂.
Ethanol = 1000 × 92/180 = 511 g; CO₂ = (1000/180) × 2 × 22.4 = 249 L.

Answer: About 511 g of ethanol and 249 L of CO₂.

Memory hook: One glucose, two ethanols, two CO₂.

Peroxides in ether

Imagine a laboratory technician finds a half-empty bottle of diethyl ether that has stood in sunlight for two years.

Why should it not be distilled to dryness?

Try it first, then show the solutionSolution

Ethers slowly react with air in light to form peroxides: $C_{2} H_{5} O C_{2} H_{5} + O_{2} \to C H_{3} C H (O O H) O C_{2} H_{5}$ .
Peroxides are less volatile and concentrate in the residue on distillation, where they may explode.

Answer: Accumulated peroxides could explode; test and remove them first (e.g. with FeSO₄).

Memory hook: Old ether, hidden peroxide.

Phenol as antiseptic

Suppose a hospital uses dilute (about 0.2%) phenol as an antiseptic and a 1% solution as a floor disinfectant.

Why does concentration matter, and what property of phenol makes it act on microbes?

Try it first, then show the solutionSolution

Phenol denatures proteins and disrupts microbial cell membranes.
Dilute solutions kill microbes without harming living tissue (antiseptic); stronger solutions are used on non-living surfaces (disinfectant) because they burn skin.

Answer: Dilute phenol is antiseptic; about 1% is disinfectant.

Memory hook: Weak on skin, strong on floors.

Topics in this chapter

Alcohols, Phenols and Ethers formula sheet

Hydration of ethene: $C_{2} H_{4} + H_{2} O \to C_{2} H_{5} O H$; H₃PO₄, 573 K, 60-70 atm.
Hydroboration-oxidation: $(C_{3} H_{7})_{3} B + 3 H_{2} O_{2} \to 3 C_{3} H_{7} O H + H_{3} B O_{3}$; Anti-Markovnikov: propan-1-ol from propene.
Bosch process (methanol): $C O + 2 H_{2} \to C H_{3} O H$; ZnO-Cr₂O₃, 573-673 K, 200-300 atm.
Fermentation: $C_{6} H_{12} O_{6} \to 2 C_{2} H_{5} O H + 2 C O_{2}$; Zymase; sucrose first inverted by invertase.
Ethanol + sodium: $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$; Active H; ethers do not react.
Esterification: $C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$; Conc. H₂SO₄; mechanism required.
Oxidation of primary alcohol: $C_{2} H_{5} O H + [O] \to C H_{3} C H O + H_{2} O$; Then $C H_{3} C H O + [O] \to C H_{3} C O O H$ .
Dehydration to alkene: $C_{2} H_{5} O H \to C_{2} H_{4} + H_{2} O$; Conc. H₂SO₄, 443 K.
Dehydration to ether: $2 C_{2} H_{5} O H \to C_{2} H_{5} O C_{2} H_{5} + H_{2} O$; Conc. H₂SO₄, 413 K, excess alcohol.
Amine to alcohol: $C_{2} H_{5} N H_{2} + H N O_{2} \to C_{2} H_{5} O H + N_{2} + H_{2} O$; NaNO₂ + HCl.
Cumene process: $C_{6} H_{5} C (C H_{3})_{2} O O H \to C_{6} H_{5} O H + C H_{3} C O C H_{3}$; Dilute acid; acetone by-product.
Bromination of phenol: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$; Bromine water: white precipitate.
Picric acid: $C_{6} H_{5} O H + 3 H N O_{3} \to C_{6} H_{2} (N O_{2})_{3} O H + 3 H_{2} O$; Conc. HNO₃ + conc. H₂SO₄.
Kolbe's reaction: $C_{6} H_{5} O N a + C O_{2} \to C_{6} H_{4} (O H) C O O N a$; 400 K, 4-7 atm; acidify to salicylic acid.
Reimer-Tiemann reaction: $C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$; 340 K then acid; salicylaldehyde.
Phenol with zinc dust: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$; Heat.
Williamson's synthesis: $C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$; Use sodium alkoxide, not alcohol.
Ether + HI (cold): $C_{2} H_{5} O C_{2} H_{5} + H I \to C_{2} H_{5} O H + C_{2} H_{5} I$; Hot excess HI: $2 C_{2} H_{5} I + H_{2} O$ .
Ether + PCl₅: $C_{2} H_{5} O C_{2} H_{5} + P C l_{5} \to 2 C_{2} H_{5} C l + P O C l_{3}$; Heat.

Alcohols, Phenols and Ethers: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Mechanism of acid-catalysed dehydration of ethanol

'Write the mechanism of dehydration of ethanol to ethene'.

Draw

Step 1: curved arrow from O lone pair of CH₃CH₂-OH to H⁺, giving CH₃CH₂-OH₂⁺ (fast).
Step 2: curved arrow from the C-O bond to O, losing H₂O and giving CH₃CH₂⁺ (slow).
Step 3: curved arrow from a β C-H bond to the C⁺, H⁺ leaving, giving CH₂=CH₂.

Label

protonation
loss of water (slow)
carbocation
loss of H⁺
ethene

Watch out: State the temperature (443 K) and that H₂SO₄ is regenerated.

Mechanism of esterification

'Explain the mechanism of esterification of ethanoic acid with ethanol'.

Draw

Protonation of the C=O of CH₃COOH.
Attack of the ethanol O on the carbonyl carbon, giving a tetrahedral intermediate.
Proton transfer to an -OH, loss of H₂O, and loss of H⁺ giving CH₃COOC₂H₅.

Label

H⁺ (catalyst)
nucleophilic attack
tetrahedral intermediate
−H₂O
−H⁺
ester

Watch out: Mark that the ester's alkoxy O comes from the alcohol.

Resonance in phenoxide ion

Explaining the acidity of phenol or its ortho-para directing effect.

Draw

Phenoxide with negative charge on O.
Three structures with C=O and the negative charge on the two ortho carbons and the para carbon.
The second Kekulé structure.

Label

O⁻
negative charge at ortho
at para
resonance arrows

Watch out: Compare with ethoxide, which has no resonance.

How the board asks Alcohols, Phenols and Ethers

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Conversions	Acetaldehyde → propan-2-ol, ethyl bromide → ether, phenol → salicylaldehyde/salicylic acid/picric acid, methanol ↔ ethanol.	Write each step with reagent, conditions and by-products, balanced.
Distinguishing tests	Ethanol/phenol, ethanol/methanol, ethanol/ethanal, 2°/3° alcohols.	Reagent + observation for one compound + 'no change' for the other.
Mechanisms	Esterification; dehydration of alcohols to alkenes and ethers.	Show protonation, attack/loss and deprotonation with curved arrows.
Named reactions	Kolbe, Reimer-Tiemann, Williamson, Dow, cumene, Bosch, fermentation.	Complete balanced equations with conditions.
Identify A, B, C	C₄H₁₀O/C₃H₈O oxidised to iodoform-positive ketones, then Grignard; phenol sequences.	Use the iodoform test to locate CH₃CH(OH)- or CH₃CO-; count carbons after Grignard.
Reasons and orders	Acidity (phenol vs alcohols, CF₃CH₂OH etc.), boiling points, solubility, HI cleavage.	Use resonance, inductive effects and hydrogen bonding; write orders with '<'.

Tricks to remember Alcohols, Phenols and Ethers

Short memory aids. Say each one aloud once, then use it on a question the same day.

140 for Ether, 170 for Ethene
Fourteen-oh, the Ether is lOw; seventy, the Ethene is lighter (gas) and needs more heat.
Use it when: Dehydration temperatures.
Lucas: 3 is quick, 2 is five, 1 is never
Tertiary: immediate; secondary: about 5 min; primary: no turbidity at room temperature.
Use it when: Lucas test.
Methanal 1°, aldehyde 2°, ketone 3°
Grignard + HCHO → primary; + RCHO → secondary; + R₂CO → tertiary.
Use it when: Grignard syntheses.
Kolbe gives aCid, Reimer gives aldehyde
Kolbe: CO₂ → salicylic acid. Reimer-Tiemann: CHCl₃ → salicylaldehyde.
Use it when: Phenol named reactions.
HI hits the small one
In mixed ethers, iodide goes to the smaller alkyl (SN2); tertiary groups go via SN1.
Use it when: Ether cleavage.

Alcohols, Phenols and Ethers: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Nitration of phenol · 1 mark · Very short answerBoard level

Fill in the blanks: Dilute nitric acid at 298 K converts phenol into a mixture of o- and p-nitrophenol, which are separated by ______; the ______ isomer distils over.

Try it first, then show the solutionSolution

o-Nitrophenol has intramolecular hydrogen bonding, so it is steam volatile; the para isomer is associated by intermolecular hydrogen bonds and stays behind.

Answer: steam distillation; ortho

Example 2: Strongest acid · 1 mark · MCQBoard level

Which is the most acidic?
(a) ethanol
(b) 2-methylpropan-2-ol
(c) phenol
(d) methanol

Try it first, then show the solutionSolution

Phenoxide is resonance-stabilised; alkoxides are not.

Answer: (c) phenol

Example 3: Two conversions · 2 marks · Short answerHarder board level

Write equations for: (a) phenol to anisole; (b) propene to propan-1-ol. [2]

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O N a + H_{2} O$ ; then Williamson: $C_{6} H_{5} O N a + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$ .
(b) Hydroboration-oxidation: $3 C H_{3} C H = C H_{2} + B H_{3} \to (C H_{3} C H_{2} C H_{2})_{3} B$ ; $(C H_{3} C H_{2} C H_{2})_{3} B + 3 H_{2} O_{2} \to 3 C H_{3} C H_{2} C H_{2} O H + H_{3} B O_{3}$ (alkaline H₂O₂).

Answer: Phenoxide + CH₃I (Williamson); hydroboration-oxidation (anti-Markovnikov)

Example 4: Acidity of substituted phenols · 2 marks · Short answerHarder board level

Arrange in increasing order of acidic strength: phenol, 4-nitrophenol, 4-methylphenol, ethanol. Give reasons. [2]

Try it first, then show the solutionSolution

Ethanol is weakest: ethoxide has no resonance and the +I ethyl group destabilises it.
Among the phenols, -CH₃ (+I, hyperconjugation) destabilises the phenoxide and -NO₂ at para (−I, −R) delocalises the negative charge further.
Order: ethanol < 4-methylphenol < phenol < 4-nitrophenol.

Answer: ethanol < 4-methylphenol < phenol < 4-nitrophenol

Example 5: Distinguishing tests · 3 marks · Short answerHarder board level

Give one chemical test each to distinguish: (a) propan-1-ol and propan-2-ol; (b) benzoic acid and benzyl alcohol; (c) ethanol and ethoxyethane. [3]

Try it first, then show the solutionSolution

(a) Iodoform test (I₂ + NaOH, warm): propan-2-ol (has CH₃CH(OH)-) gives a yellow precipitate of CHI₃; propan-1-ol gives none.
(b) Aqueous NaHCO₃: benzoic acid gives brisk effervescence of CO₂; benzyl alcohol, a neutral alcohol, gives none.
(c) Sodium metal: ethanol liberates H₂ gas; ethoxyethane has no O-H and does not react.

Answer: Iodoform; NaHCO₃; sodium

Example 6: Identify A, B and C · 3 marks · Short answerHOTS / competency

An alcohol A (C₃H₈O) is oxidised by acidified K₂Cr₂O₇ to B (C₃H₆O), which gives a yellow precipitate with I₂ and NaOH but does not reduce Tollens' reagent. B reacts with C₂H₅MgBr followed by hydrolysis to give C (C₅H₁₂O). Identify A, B and C and write the equations. [3]

Try it first, then show the solutionSolution

B is a ketone with CH₃CO-: propanone, CH₃COCH₃. So A is propan-2-ol: $C H_{3} C H (O H) C H_{3} + [O] \to C H_{3} C O C H_{3} + H_{2} O$ .
$C H_{3} C O C H_{3} + C_{2} H_{5} M g B r \to (C H_{3})_{2} C (O M g B r) C_{2} H_{5}$
Hydrolysis gives C = 2-methylbutan-2-ol, (CH₃)₂C(OH)C₂H₅ (a tertiary alcohol, C₅H₁₂O).

Answer: A propan-2-ol; B propanone; C 2-methylbutan-2-ol

Example 7: Phenol in industry · 5 marks · Long answerHarder board level

(a) Describe the cumene process for phenol with equations. [2]
(b) Write the equations for Kolbe's reaction and name the product. [2]
(c) Why does phenol not liberate CO₂ from sodium hydrogencarbonate? [1]

Try it first, then show the solutionSolution

(a) $C_{6} H_{6} + C_{3} H_{6} \to C_{6} H_{5} C H (C H_{3})_{2}$ (H₃PO₄); air oxidation: $C_{6} H_{5} C H (C H_{3})_{2} + O_{2} \to C_{6} H_{5} C (C H_{3})_{2} O O H$ .
$C_{6} H_{5} C (C H_{3})_{2} O O H \to C_{6} H_{5} O H + C H_{3} C O C H_{3}$ (dilute acid).
(b) $C_{6} H_{5} O N a + C O_{2} \to C_{6} H_{4} (O H) C O O N a$ (about 400 K, 4-7 atm).
$C_{6} H_{4} (O H) C O O N a + H C l \to C_{6} H_{4} (O H) C O O H + N a C l$ : salicylic acid (2-hydroxybenzoic acid).
(c) Phenol (pKa ≈ 10) is a weaker acid than carbonic acid, so it cannot displace CO₂ from HCO₃⁻.

Answer: Cumene → hydroperoxide → phenol + acetone; salicylic acid; phenol weaker than H₂CO₃

Example 8: Anisole with HI and bromine · 3 marks · Short answerHOTS / competency

Anisole is heated with HI, and separately treated with bromine in ethanoic acid. Give the products and explain each. [3]

Try it first, then show the solutionSolution

$C_{6} H_{5} O C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$ : the aryl C-O bond has partial double-bond character and phenyl cations do not form, so I⁻ attacks the methyl carbon.
-OCH₃ activates the ring and directs ortho-para; bulk favours para.
$C_{6} H_{5} O C H_{3} + B r_{2} \to C_{6} H_{4} B r O C H_{3} + H B r$ : 4-bromoanisole (major) with some 2-bromoanisole.

Answer: Phenol + iodomethane; 4-bromoanisole

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Hand sanitiser chemistry

Suppose a company makes hand sanitiser with 70% ethanol. A chemist compares boiling points: propan-1-ol 370 K, butan-1-ol 391 K, butan-2-ol 373 K, 2-methylpropan-2-ol 356 K, and ethoxyethane 308 K.

(i) · 1 mark
Why does 2-methylpropan-2-ol boil much lower than butan-1-ol though both are C₄H₁₀O?
Answer
Answer: Branching reduces surface area and hydrogen bonding is less effective.
The compact, branched molecule has weaker intermolecular forces.
(ii) · 1 mark
Why does ethoxyethane boil far lower than butan-1-ol?
Answer
Answer: Ether molecules cannot hydrogen-bond with each other.
They lack an O-H bond, so only weak dipole and dispersion forces act.
(iii) · 1 mark
Name the test that would distinguish butan-2-ol from 2-methylpropan-2-ol and give observations.
Answer
Answer: Lucas test.
2-methylpropan-2-ol: immediate turbidity; butan-2-ol: turbidity in about 5 minutes.
(iv) · 2 marks
Write the two equations for industrial preparation of ethanol by fermentation of cane sugar.
Answer
Answer: Invertase then zymase.
$C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$ ; $C_{6} H_{12} O_{6} \to 2 C_{2} H_{5} O H + 2 C O_{2}$ .

From phenol to an explosive

Imagine a student reading that phenol is converted into picric acid, a yellow explosive once used in shells, while phenol itself is used to make plastics such as Bakelite.

(i) · 1 mark
Write the equation for preparing picric acid from phenol, with conditions.
Answer
Answer: $C_{6} H_{5} O H + 3 H N O_{3} \to C_{6} H_{2} (N O_{2})_{3} O H + 3 H_{2} O$
Concentrated HNO₃ with concentrated H₂SO₄.
(ii) · 1 mark
Why is picric acid much more acidic than phenol?
Answer
Answer: Three -NO₂ groups at ortho and para positions stabilise the phenoxide ion.
They withdraw electrons by −I and −R effects, delocalising the negative charge further.
(iii) · 1 mark
Picric acid dissolves in aqueous sodium hydrogencarbonate with effervescence, but phenol does not. Explain.
Answer
Answer: Picric acid is a stronger acid than carbonic acid; phenol is weaker.
Only an acid stronger than H₂CO₃ can displace CO₂ from HCO₃⁻; the three -NO₂ groups make picric acid far more acidic than phenol (pKa about 10).
(iv) · 2 marks
Write the equations for making phenol from benzene sulphonic acid.
Answer
Answer: Neutralise, fuse with NaOH, then acidify.
$C_{6} H_{5} S O_{3} H + N a O H \to C_{6} H_{5} S O_{3} N a + H_{2} O$ ; $C_{6} H_{5} S O_{3} N a + 2 N a O H \to C_{6} H_{5} O N a + N a_{2} S O_{3} + H_{2} O$ (fusion); $C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Give a chemical test to distinguish between phenol and cyclohexanol. (2 marks)

Step	Marks
Reagent: neutral FeCl₃ solution (or bromine water)	1
Phenol gives a violet colour (white precipitate with bromine water); cyclohexanol gives neither	1

Write equations for: (a) Kolbe's reaction, (b) the acid cleavage step of the cumene process, (c) Williamson's synthesis of anisole. (3 marks)

Step	Marks
$C_{6} H_{5} O N a + C O_{2} \to C_{6} H_{4} (O H) C O O N a$ (about 400 K, 4-7 atm); acidify to salicylic acid	1
$C_{6} H_{5} C (C H_{3})_{2} O O H H^{+} C_{6} H_{5} O H + C H_{3} C O C H_{3}$	1
$C_{6} H_{5} O N a + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$	1

Explain the mechanism of acid-catalysed esterification of ethanoic acid with ethanol, and state how the yield can be increased. (5 marks)

Step	Marks
Protonation of the carbonyl oxygen of the acid	1
Nucleophilic attack by ethanol oxygen forming a tetrahedral intermediate	1
Proton transfer and elimination of water	1
Loss of H⁺ giving ethyl ethanoate; catalyst regenerated	1
Use excess alcohol or remove water (equilibrium shifts forward)	1

Where students lose marks in Alcohols, Phenols and Ethers

The slips examiners see most often, and what to write instead.

Mistake: Swapping the dehydration products
CISCE examiners' comments, 2024: candidates wrote ethene for the 140 °C product and diethyl ether or ethyl carbonate for the 170 °C product.
Wrong working
140 °C: ethene; 170 °C: diethyl ether.
Right working
140 °C (413 K): diethyl ether; 170 °C (443 K): ethene.
Do this: Lower temperature, bigger molecule (ether).
Usually costs: 1 mark.
Mistake: Half-answers in distinguishing tests
CISCE examiners' comments, 2024: candidates named only the reagent or gave the observation for only one compound, gave orange or pink for phenol with FeCl₃, omitted the yellow precipitate or the negative result in the iodoform test, used the iodoform test for ethanol vs ethanal without explaining both observations, and confused Lucas observations for 2° and 3° alcohols.
Wrong working
Add FeCl₃: phenol turns pink.
Right working
Neutral FeCl₃: phenol gives violet colour; ethanol does not.
Do this: Reagent + positive result + negative result.
Usually costs: 1 mark each.
Mistake: Using alcohol instead of alkoxide, and incomplete named reactions
CISCE examiners' comments, 2024 and 2025: candidates used ethyl alcohol instead of sodium ethoxide in Williamson's synthesis, used CH₃Cl instead of CHCl₃ or gave salicylic acid in Reimer-Tiemann, wrote unbalanced Reimer-Tiemann equations without by-products, omitted the hydrolysis step or by-product of the Grignard adduct, omitted conc. H₂SO₄ and water in nitration, and converted chlorobenzene to phenol without sodium phenoxide and conditions.
Wrong working
$C_{2} H_{5} O H + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5}$
Right working
$C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$
Do this: Write full reagents, conditions and by-products; check balance.
Usually costs: 1 mark each.
Mistake: Wrong isomer in identification problems
CISCE examiners' comments, 2024 and 2025: candidates chose butan-1-ol instead of butan-2-ol (giving wrong B and C), wrote an incorrect oxidation equation, and used molecular hydrogen instead of nascent hydrogen in a reduction step.
Wrong working
A = butan-1-ol, B = butanal (but butanal does not give iodoform).
Right working
Iodoform-positive ketone C₄H₈O is butanone, so A is butan-2-ol.
Do this: Check every clue (iodoform, Tollens, carbon count) against your answer.
Usually costs: 1-2 marks.
Mistake: Wrong direction in orders and reasons for ether cleavage
CISCE examiners' comments, 2024 and 2025: candidates wrote decreasing instead of increasing order for acidity of halogenated ethanols, confused boiling-point order, gave resonance in benzene as the reason for phenol's lower solubility, and could not explain why the halogen goes to the smaller alkyl group in HI cleavage.
Wrong working
CF₃CH₂OH > CCl₃CH₂OH > CH₃CH₂OH (when increasing order was asked).
Right working
Increasing: CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH.
Do this: Read 'increasing' or 'decreasing' and use '<' or '>' accordingly.
Usually costs: 1 mark.

Practice questions on Alcohols, Phenols and Ethers

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Hydroboration-oxidation of propene gives:
(a) propan-2-ol
(b) propan-1-ol
(c) propanal
(d) propanone
Hint
Anti-Markovnikov.
Answer
(b) propan-1-ol
Q2 · 1 mark · MCQBoard level
Phenol with bromine water gives:
(a) 2-bromophenol
(b) 4-bromophenol
(c) 2,4,6-tribromophenol
(d) bromobenzene
Hint
Strong activation in water.
Answer
(c)
Q3 · 1 mark · Very short answerBoard level
Fill in the blanks: Sodium phenoxide heated with CO₂ under pressure and then acidified gives ______; the reaction is called ______ reaction.
Hint
2-hydroxybenzoic acid.
Answer
salicylic acid; Kolbe's
Q4 · 1 mark · Very short answerBoard level
What is Lucas reagent?
Hint
Two components.
Answer
Concentrated HCl with anhydrous ZnCl₂.
Q5 · 3 marks · Short answerHarder board level
How would you convert methanol into ethanol?
Hint
Add a carbon via cyanide.
Answer
CH₃OH + red P/I₂ → CH₃I; CH₃I + KCN → CH₃CN + KI; CH₃CN + 4[H] (Na/C₂H₅OH) → CH₃CH₂NH₂; CH₃CH₂NH₂ + HNO₂ → C₂H₅OH + N₂ + H₂O.
Q6 · 2 marks · Short answerHarder board level
Why is o-nitrophenol steam volatile but p-nitrophenol is not?
Hint
Intra vs inter hydrogen bonding.
Answer
o-Nitrophenol forms intramolecular hydrogen bonds (chelation), so its molecules do not associate and it is volatile. p-Nitrophenol forms intermolecular hydrogen bonds, associating molecules and raising its boiling point.
Q7 · 3 marks · Short answerHarder board level
Write the mechanism of formation of diethyl ether from ethanol at 413 K.
Hint
SN2 on protonated ethanol.
Answer
(1) $C_{2} H_{5} O H + H^{+} \to C_{2} H_{5} O H_{2}^{+}$ ; (2) a second ethanol attacks the α-carbon by SN2, expelling water: C₂H₅-O(H)⁺-C₂H₅; (3) loss of H⁺ gives C₂H₅OC₂H₅.
Q8 · 2 marks · Short answerHOTS / competency
Which ether cannot be made well by Williamson's synthesis from tert-butyl bromide and sodium methoxide? Suggest the correct pair.
Hint
3° halides eliminate.
Answer
tert-Butyl methyl ether: with (CH₃)₃CBr, methoxide acts as a base and gives 2-methylpropene by elimination. Use sodium tert-butoxide with iodomethane (or bromomethane): $(C H_{3})_{3} C O N a + C H_{3} I \to (C H_{3})_{3} C O C H_{3} + N a I$ .
Q9 · 2 marks · Short answerHarder board level
Write equations for reaction of ethoxyethane with (a) excess hot HI, (b) PCl₅.
Hint
Both alkyl groups end up as halides.
Answer
(a) $C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$ (b) $C_{2} H_{5} O C_{2} H_{5} + P C l_{5} \to 2 C_{2} H_{5} C l + P O C l_{3}$
Q10 · 2 marks · Short answerBoard level
Give the Bosch process for methanol in outline with equations.
Hint
Water gas.
Answer
Steam over red-hot coke: $C + H_{2} O \to C O + H_{2}$ (water gas). Enriched with H₂ and passed over ZnO-Cr₂O₃ at 573-673 K and 200-300 atm: $C O + 2 H_{2} \to C H_{3} O H$ .
Q11 · 5 marks · Long answerHOTS / competency
(a) An organic compound A (C₇H₈O) is insoluble in NaHCO₃ but dissolves in NaOH and gives a violet colour with neutral FeCl₃. Suggest a structure. [1]
(b) Another isomer B does not dissolve in NaOH and reacts with hot HI to give phenol and CH₃I. Identify B. [1]
(c) A third isomer C reacts with Na liberating H₂ and is oxidised to benzoic acid. Identify C. [1]
(d) Write one reaction to distinguish A from C. [2]
Hint
Cresol, anisole, benzyl alcohol.
Answer
(a) A: a cresol, e.g. 4-methylphenol (phenolic OH). (b) B: anisole, C₆H₅OCH₃ ( $C_{6} H_{5} O C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$ ). (c) C: benzyl alcohol, C₆H₅CH₂OH. (d) Neutral FeCl₃: A gives violet colour, C does not; or NaOH: A dissolves (forms phenoxide), C does not.
Q12 · 2 marks · Short answerHOTS / competency
Explain why the C-O bond in phenol is shorter than in methanol, and how this affects reaction with HX.
Hint
Resonance and sp².
Answer
In phenol the O lone pair is delocalised into the ring, giving the C-O bond partial double-bond character, and the carbon is sp². So the bond is shorter (about 136 pm vs 142 pm) and stronger, and phenol does not react with HX to replace -OH as alcohols do.
Q13 · 1 mark · Very short answerBoard level
Arrange in increasing order of boiling point: ethoxyethane, butan-1-ol, butane, propan-1-ol.
Hint
Hydrogen bonding wins.
Answer
butane < ethoxyethane < propan-1-ol < butan-1-ol (no hydrogen bonding in the first two; among the alcohols the larger one boils higher)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of propene by hydroboration-oxidation?Show answer
Propan-1-ol.
2.Which carbonyl gives a tertiary alcohol with a Grignard reagent?Show answer
A ketone.
3.Lucas test observation for a 2° alcohol?Show answer
Turbidity in about 5 minutes.
4.Reagent for Reimer-Tiemann reaction?Show answer
CHCl₃ and aqueous NaOH (then acid).
5.Test for phenol with FeCl₃?Show answer
Violet colour with neutral FeCl₃.
6.Name the industrial process making phenol and acetone together.Show answer
Cumene process.
7.Williamson's synthesis reagents?Show answer
Sodium alkoxide (or phenoxide) and a primary alkyl halide.
8.Why must old ether not be distilled to dryness?Show answer
Explosive peroxides may have formed.

Best order to study it

Learn classification, naming and physical properties of alcohols.
Learn all preparations, including Bosch and fermentation outlines.
Learn reactions of primary alcohols with esterification and dehydration mechanisms; Lucas test; interconversions.
Learn phenol preparations, acidity and reactions with Kolbe and Reimer-Tiemann; tests.
Learn ethers: Williamson's synthesis, reactions with Cl₂, O₂, HI, PCl₅; anisole substitution.
Practise identification problems and conversions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you name 1°, 2°, 3° alcohols and explain boiling-point trends?
Can you write all three hydration methods?
Can you make 1°, 2°, 3° alcohols by Grignard reactions?
Can you write esterification and dehydration mechanisms?
Can you explain the Lucas test and iodoform test results?
Can you prepare phenol four ways?
Can you write Kolbe, Reimer-Tiemann, bromination and nitration of phenol?
Can you prepare and cleave ethers, and predict HI products in mixed ethers?

Alcohols, Phenols and Ethers: frequently asked questions

Do I need properties of secondary and tertiary alcohols?

The ISC 2027 syllabus asks for physical and chemical properties of primary alcohols only, plus the differences between primary, secondary and tertiary alcohols and the Lucas test to distinguish them.

How much detail on the Bosch process and fermentation?

Only an outline of the method, but the chemical equations are required.

Which mechanisms are required in this chapter?

Esterification and dehydration of alcohols are listed with mechanism.

What tests for phenol are in the syllabus?

The FeCl₃ test (violet colour) and the azo dye test (orange dye with benzenediazonium chloride in alkali).

Which aryl ether reactions should I know?

Preparation of anisole by Williamson's synthesis and its electrophilic substitution: halogenation, nitration and Friedel-Crafts reactions.

How many marks does this chapter carry?

It is part of the Organic Chemistry group, 31 of the 70 theory marks, shared with four other organic chapters.

Based on: CISCE ISC 2027 Chemistry syllabus; CISCE ISC 2027 Chemistry specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Chemistry, 2025; CISCE Analysis of Pupil Performance, ISC Chemistry, 2024