Alcohols, Phenols & Ethers — Preparation, Acidity & Reactions

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Alcohols, phenols and ethers all contain a C–O bond, but where that oxygen sits decides everything. On an sp³ carbon (alcohol) the O–H behaves like water; on a benzene ring (phenol) the ring pulls on the oxygen’s lone pair and makes the O–H acidic; between two carbons (ether) there is no O–H at all.

What you will be able to do

• Prepare alcohols from alkenes, carbonyl compounds and Grignard reagents, and phenol by four routes
• Rank the acidity of alcohols and substituted phenols with reasons
• Distinguish 1°, 2° and 3° alcohols by the Lucas test, oxidation and dehydration
• Predict products of electrophilic substitution of phenol, Kolbe and Reimer–Tiemann reactions
• Choose the right Williamson combination and predict products of ether cleavage with HI

The big picture

Where the oxygen sits decides the chemistry.

In one line: Alcohols react at the O–H and the C–O bond; phenols are acidic and their ring is strongly activated toward ortho/para substitution; ethers are unreactive except toward hot HI, which cleaves the C–O bond.

	Alcohol R–OH	Phenol Ar–OH	Ether R–O–R′
Carbon bearing O	sp³	sp² (ring)	sp³ (or sp² in anisole)
Acidity	Very weak (pKa ≈ 16)	Weak acid (pKa ≈ 10)	Not acidic
C–O bond	Breaks easily (with acid)	Partial double bond — hard to break	Breaks only with HI/HBr, heat
Typical reactions	Substitution, dehydration, oxidation	Ring substitution (o/p)	Cleavage, ring substitution (anisole)

Think of it like this: Think of the oxygen’s lone pair as cash. In an alcohol it stays in the oxygen’s pocket, so the O–H holds its hydrogen tightly. In phenol the benzene ring borrows that cash (resonance), the oxygen is left short, and it lets the H⁺ go more easily. The ring, now richer, attracts electrophiles.

Structure and physical properties

Hydrogen bonding explains almost every physical property.

Feature	Methanol	Phenol	Methoxymethane
Bond angle	C–O–H 108.9°	C–O–H about 109°	C–O–C 111.7°
C–O bond length	142 pm	136 pm (partial double bond)	141 pm
Why	Lone pairs squeeze the angle below 109.5°	Lone pair conjugated with the ring; sp² carbon	Two bulky alkyl groups repel, opening the angle

Compound	Molar mass	b.p. / K	Reason
Propane	44	231	Only weak van der Waals forces
Methoxymethane	46	248	Dipole–dipole, no H bonding
Ethanol	46	351	Intermolecular H bonding
Butan-1-ol	74	391	Unbranched: largest surface
Butan-2-ol	74	373	Some branching
2-Methylpropan-2-ol	74	355	Most branched: smallest surface

• Boiling point rises with molar mass and falls with branching; alcohols boil far above ethers and hydrocarbons of similar mass because of H bonding.
• Solubility in water comes from H bonding with water; it falls as the hydrocarbon part grows. Lower alcohols mix in all proportions; ethers are about as soluble as alcohols of similar mass (they accept H bonds from water).
• Phenol is fairly soluble in water and has a much higher b.p. (455 K) than toluene (384 K) of similar mass.

Check yourself

• C1. Why does ethanol boil 103 K higher than methoxymethane, though both have molar mass 46?
• C2. Why is the C–O bond in phenol shorter than in methanol?

Answers at the end.

Preparing alcohols

Four reliable routes.

Route	Reagent	Product	Note
Acid-catalysed hydration of alkenes	dil. H₂SO₄	Markovnikov alcohol	Via carbocation — can rearrange
Hydroboration–oxidation	(i) B₂H₆ (ii) H₂O₂/NaOH	Anti-Markovnikov alcohol	No rearrangement; syn addition
Reduction of carbonyls	NaBH₄ or LiAlH₄; H₂/Pd	Aldehyde → 1°; ketone → 2°	LiAlH₄ also reduces acids and esters to 1° alcohols
Grignard reagent + carbonyl	RMgX, then H₃O⁺	HCHO → 1°; RCHO → 2°; R₂CO → 3°	Adds one new C–C bond

Solved example 01 — Choosing the Grignard partner

Which carbonyl compound and Grignard reagent give 2-methylbutan-2-ol?

1. 2-Methylbutan-2-ol is (CH₃)₂C(OH)CH₂CH₃ — a 3° alcohol, so start from a ketone.
2. Option 1: acetone + C₂H₅MgBr. Option 2: butan-2-one + CH₃MgBr.

Answer: Either acetone + ethylmagnesium bromide, or butan-2-one + methylmagnesium bromide (then H₃O⁺).

Solved example 02 — Markovnikov or not?

Give the alcohol from propene with (a) dil. H₂SO₄, (b) B₂H₆ then H₂O₂/NaOH.

1. (a) H⁺ adds to C1 to form the 2° carbocation; water attacks C2.
2. (b) Boron adds to the less substituted C1; oxidation replaces B by OH.

Answer: (a) Propan-2-ol. (b) Propan-1-ol.

Mechanism of acid-catalysed hydration

1. Protonation: the alkene’s π electrons take H⁺ from H₃O⁺, giving the more stable carbocation (Markovnikov).
2. Nucleophilic attack: water attacks the carbocation, giving a protonated alcohol.
3. Deprotonation: water removes H⁺, giving the alcohol and regenerating H₃O⁺.

Commercial alcohols

• Methanol (wood spirit): CO + 2H₂ → CH₃OH over ZnO–Cr₂O₃ at 573–673 K and 200–300 atm. Highly poisonous — small amounts cause blindness, larger amounts death.
• Ethanol: fermentation of molasses or starch. Sucrose → glucose + fructose (enzyme invertase); glucose → 2C₂H₅OH + 2CO₂ (zymase, from yeast). Air must be kept out, or ethanol is oxidised to ethanoic acid.
• Denatured alcohol: ethanol made unfit to drink by adding copper sulphate (colour) and pyridine (smell).

Preparing phenol

Four routes, one industrial favourite.

• Cumene process (industrial): cumene (isopropylbenzene) + O₂ → cumene hydroperoxide; with dilute acid it rearranges to phenol + acetone (a phenyl group migrates to oxygen). Acetone is a valuable by-product.
• From benzenesulphonic acid: fuse the sodium salt with NaOH, then acidify.
• From chlorobenzene (Dow process): NaOH at 623 K and high pressure, then acidify.
• From diazonium salts: warm aqueous C₆H₅N₂⁺Cl⁻ → C₆H₅OH + N₂ + HCl.

Acidity

Why phenol beats alcohol, and what substituents do.

Key idea: Phenol is acidic because the phenoxide ion is resonance-stabilised: its negative charge spreads over the ring at the ortho and para positions. An alkoxide ion has no such stabilisation. Phenol (pKa ≈ 10) turns blue litmus red only faintly and dissolves in NaOH but not in NaHCO₃; carboxylic acids (pKa ≈ 4–5) dissolve in both.

• Alcohols: acidity 1° > 2° > 3° — alkyl groups push electron density onto O⁻ and destabilise it (and hinder solvation). Alcohols are weaker acids than water.
• Electron-withdrawing groups (–NO₂, –CN, –X) at ortho/para stabilise phenoxide strongly by resonance and increase acidity; at meta only by induction.
• Electron-donating groups (–CH₃, –OCH₃) decrease acidity.

Compound	pKa (approx.)
Picric acid (2,4,6-trinitrophenol)	0.4
2,4-Dinitrophenol	4.0
p-Nitrophenol	7.1
o-Nitrophenol	7.2
m-Nitrophenol	8.3
Phenol	10.0
p-Cresol	10.2
Ethanol	15.9

Solved example 03 — Ranking acidity

Arrange phenol, p-nitrophenol, m-nitrophenol and p-cresol in decreasing acidity.

1. p-NO₂ stabilises phenoxide by resonance and induction — strongest.
2. m-NO₂ stabilises by induction only — next.
3. –CH₃ is electron-donating — p-cresol is weaker than phenol.

Answer: p-Nitrophenol > m-nitrophenol > phenol > p-cresol.

Check yourself

• C3. Why does phenol not dissolve in NaHCO₃ solution?
• C4. Which is more acidic: ethanol or water?

Answers at the end.

Reactions of alcohols

Break O–H, break C–O, or oxidise.

Breaking O–H

• With Na: 2ROH + 2Na → 2RONa + H₂ (shows alcohols are weak acids).
• Esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O (conc. H₂SO₄); with acid chlorides it goes to completion (pyridine removes HCl).

Breaking C–O

• With HX: reactivity 3° > 2° > 1° (via carbocation for 3°); HI > HBr > HCl. ZnCl₂ catalyses HCl.
• With PCl₅, PCl₃, SOCl₂: give alkyl chlorides; SOCl₂ is best (by-products SO₂ and HCl are gases).
• Dehydration with conc. H₂SO₄: ethanol at 443 K → ethene; at 413 K with excess alcohol → diethyl ether. Ease of dehydration: 3° > 2° > 1° (E1 via carbocation, Saytzeff product, rearrangement possible).

Mechanism of dehydration (ethanol → ethene)

1. Protonation of –OH: CH₃CH₂–OH + H⁺ → CH₃CH₂–OH₂⁺ (fast).
2. Loss of water gives a carbocation: CH₃CH₂⁺ + H₂O (slow, rate-determining).
3. Loss of H⁺ from the β-carbon gives CH₂=CH₂.

At 413 K with excess ethanol, a second ethanol molecule attacks the protonated alcohol instead (SN2), giving diethyl ether: this route works only for primary alcohols — 2° and 3° alcohols dehydrate to alkenes.

Alcohol	Reaction with HX	Mechanism
3°	Fast, even with conc. HCl alone	SN1 via a stable carbocation
2°	Moderate; needs ZnCl₂ with HCl	SN1 / SN2
1°	Slow; needs ZnCl₂ and heat with HCl	SN2 on R–OH₂⁺

Oxidation

Alcohol	KMnO₄ / K₂Cr₂O₇ (acid)	PCC (mild)	Cu at 573 K
1°	Carboxylic acid (via aldehyde)	Aldehyde	Aldehyde (dehydrogenation)
2°	Ketone	Ketone	Ketone
3°	Resists oxidation	No reaction	Alkene (dehydration)

Solved example 04 — Rearrangement on dehydration

3,3-Dimethylbutan-2-ol is heated with conc. H₂SO₄. Give the major alkene.

1. Loss of water gives a 2° carbocation next to a tert-butyl carbon.
2. A methyl group shifts to give a more stable 3° carbocation.
3. Loss of H⁺ gives the most substituted alkene (Saytzeff).

Answer: 2,3-Dimethylbut-2-ene (not 3,3-dimethylbut-1-ene).

Solved example 05 — Oxidation as a test of class

Three isomeric alcohols C₄H₁₀O are treated with acidified K₂Cr₂O₇. A gives butanoic acid, B gives butan-2-one, C is not oxidised. Identify A, B and C.

1. A → acid: a 1° alcohol with an unbranched chain: butan-1-ol.
2. B → ketone: a 2° alcohol: butan-2-ol.
3. C not oxidised: a 3° alcohol: 2-methylpropan-2-ol.

Answer: A = butan-1-ol; B = butan-2-ol; C = 2-methylpropan-2-ol.

Check yourself

• C5. Which step is rate-determining in the acid dehydration of ethanol?
• C6. Why does 2-methylpropan-2-ol react with conc. HCl at room temperature without ZnCl₂?

Answers at the end.

Reactions of phenol

A strongly activated ring.

The –OH group donates electrons into the ring (+R), so phenol undergoes electrophilic substitution much faster than benzene, at ortho and para positions.

Reaction	Conditions	Main product
Bromination	Bromine water	2,4,6-Tribromophenol (white precipitate)
Bromination	Br₂ in CS₂ or CHCl₃, low temperature	Mixture of o- and p-bromophenol (para major)
Nitration	Dilute HNO₃, 298 K	o- and p-Nitrophenol (separated by steam distillation)
Nitration	Conc. HNO₃	2,4,6-Trinitrophenol (picric acid)
Kolbe reaction	Sodium phenoxide + CO₂, about 400 K, 4–7 atm; then H⁺	Salicylic acid (o-hydroxybenzoic acid)
Reimer–Tiemann	CHCl₃ + aq. NaOH, then H⁺	Salicylaldehyde (o-hydroxybenzaldehyde); electrophile is dichlorocarbene, :CCl₂
Zinc dust distillation	Heat with Zn	Benzene
Oxidation	Air or chromic acid	Benzoquinone

More reactions of phenol

• With NaOH: C₆H₅OH + NaOH → C₆H₅ONa + H₂O (phenol is acidic enough). With Na: 2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂.
• Esterification (acylation): phenol + acetyl chloride (or acetic anhydride) in pyridine → phenyl acetate.
• Aspirin: salicylic acid + acetic anhydride → acetylsalicylic acid (aspirin) + CH₃COOH.
• Coupling with diazonium salts: phenol + C₆H₅N₂⁺Cl⁻ in mild alkali → p-hydroxyazobenzene, an orange dye (azo coupling at the para position).
• FeCl₃ test: neutral FeCl₃ gives a violet colour (iron–phenoxide complex).

Solved example 06 — Phenol to aspirin

Show how aspirin is made from phenol.

1. Kolbe reaction: sodium phenoxide + CO₂ (about 400 K, 4–7 atm), then acid → salicylic acid.
2. Acetylation: salicylic acid + (CH₃CO)₂O → acetylsalicylic acid + CH₃COOH.

Answer: Phenol → (Kolbe) salicylic acid → (acetylation) aspirin.

Key idea: o-Nitrophenol is steam-volatile; p-nitrophenol is not. The ortho isomer forms an intramolecular hydrogen bond, so its molecules do not hold each other; the para isomer forms intermolecular H bonds with other molecules and with water.

Check yourself

• C7. What is the electrophile in the Reimer–Tiemann reaction?
• C8. Why does bromine water give a tribromo product with phenol but not with benzene?

Answers at the end.

Ethers

Easy to make, hard to break.

• Ethers have a bent C–O–C unit (about 111.7° in methoxymethane) and a small dipole moment.
• No O–H, so no H bonding between ether molecules: b.p. close to alkanes of similar mass, far below alcohols.
• Ethers are good solvents (diethyl ether, THF) because they are unreactive toward most reagents.
• From alcohols: 2CH₃CH₂OH → CH₃CH₂OCH₂CH₃ + H₂O (conc. H₂SO₄, 413 K). Only for symmetrical ethers of 1° alcohols: 2° and 3° alcohols give alkenes, and two different alcohols give a mixture of three ethers.

Williamson synthesis

R–O⁻Na⁺ + R′–X → R–O–R′ + NaX

An SN2 reaction: the alkyl halide must be methyl or primary

Key idea: Put the bulky group on the alkoxide, never on the halide. For tert-butyl methyl ether use (CH₃)₃CO⁻Na⁺ + CH₃Br. The other pair, CH₃O⁻Na⁺ + (CH₃)₃CBr, gives mainly 2-methylpropene by E2. For aryl ethers, use phenoxide + alkyl halide (an aryl halide will not undergo SN2).

Cleavage with HI

• Ethers are cleaved by hot conc. HI (HI > HBr ≫ HCl).
• Methyl + 1° or 2° alkyl: I⁻ attacks the less hindered carbon by SN2 → CH₃I + ROH.
• One group 3°: the C–O bond to the 3° group breaks by SN1 → (CH₃)₃C–I + CH₃OH.
• Alkyl aryl ether (anisole): gives phenol + CH₃I — never iodobenzene, because the aryl C–O bond has partial double-bond character and an sp² carbon resists attack.
• With excess HI, the alcohol formed is converted to a second alkyl iodide.

Solved example 07 — Cleaving an ether

Predict the products of (a) CH₃OC₂H₅, (b) CH₃OC(CH₃)₃, (c) C₆H₅OCH₃ with one equivalent of hot HI.

1. (a) SN2 at the less hindered methyl: CH₃I + C₂H₅OH.
2. (b) SN1 at the 3° carbon: (CH₃)₃CI + CH₃OH.
3. (c) Aryl C–O does not break: C₆H₅OH + CH₃I.

Answer: (a) CH₃I + ethanol. (b) tert-butyl iodide + methanol. (c) Phenol + CH₃I.

Reaction of anisole	Reagent	Main product
Bromination	Br₂ in ethanoic acid (no catalyst)	p-Bromoanisole (major) + o-bromoanisole
Friedel–Crafts alkylation	CH₃Cl, anhydrous AlCl₃	2-Methoxytoluene + 4-methoxytoluene
Friedel–Crafts acylation	CH₃COCl, anhydrous AlCl₃	4-Methoxyacetophenone (major) + 2-methoxyacetophenone
Nitration	conc. H₂SO₄ + conc. HNO₃	4-Nitroanisole (major) + 2-nitroanisole

• Anisole undergoes electrophilic substitution at ortho/para (–OCH₃ activates the ring): bromination in acetic acid gives mainly p-bromoanisole; Friedel–Crafts gives o- and p-products.

Tests and distinctions

How to tell them apart in the lab or on paper.

Test	Result	Distinguishes
Lucas reagent (conc. HCl + anhydrous ZnCl₂)	3°: turbidity at once; 2°: in about 5 min; 1°: no turbidity at room temperature	1°, 2°, 3° alcohols
Neutral FeCl₃	Violet colour	Phenol (and enols) from alcohols
Bromine water	White precipitate	Phenol from alcohols
NaHCO₃	No effervescence	Phenol from carboxylic acids
Iodoform (I₂ + NaOH)	Yellow CHI₃	Ethanol and alcohols with CH₃CH(OH)– (e.g. propan-2-ol) from methanol, propan-1-ol

Solved example 08 — Distinguishing two isomers

How would you distinguish propan-1-ol from propan-2-ol?

1. Iodoform test: propan-2-ol has CH₃CH(OH)– and gives yellow CHI₃; propan-1-ol does not.
2. Lucas test: propan-2-ol (2°) turns turbid in about 5 min; propan-1-ol (1°) stays clear.

Answer: Iodoform test (or Lucas test).

Conversions

The step-by-step problems examiners love.

Solved example 09 — Methanol to ethanol

Convert methanol into ethanol.

1. CH₃OH + PBr₃ → CH₃Br.
2. CH₃Br + Mg (dry ether) → CH₃MgBr.
3. CH₃MgBr + HCHO, then H₃O⁺ → CH₃CH₂OH.

Answer: Methanol → bromomethane → Grignard → (with HCHO) ethanol.

Solved example 10 — Ethanol to propan-2-ol

Convert ethanol into propan-2-ol.

1. Oxidise ethanol to ethanal with PCC.
2. Add CH₃MgBr to ethanal; hydrolyse with H₃O⁺.

Answer: CH₃CH₂OH → CH₃CHO → (CH₃MgBr, then H₃O⁺) CH₃CH(OH)CH₃.

Solved example 11 — Benzene to phenol

Convert benzene into phenol.

1. Sulphonate: benzene + conc. H₂SO₄ (oleum) → benzenesulphonic acid.
2. Fuse the sodium salt with NaOH → sodium phenoxide.
3. Acidify → phenol.

Answer: Benzene → C₆H₅SO₃H → C₆H₅ONa → C₆H₅OH (or via chlorobenzene and the Dow process).

Solved example 12 — Propan-2-ol to 2-methylpropan-2-ol

Convert propan-2-ol into 2-methylpropan-2-ol.

1. Oxidise to acetone (K₂Cr₂O₇ / H⁺).
2. Add CH₃MgBr; hydrolyse.

Answer: Propan-2-ol → acetone → (CH₃MgBr, H₃O⁺) 2-methylpropan-2-ol.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	Anisole + HI → iodobenzene + methanol.	Phenol + CH₃I.	The aryl C–O bond does not break.
2	CH₃ONa + (CH₃)₃CBr gives tert-butyl methyl ether.	It gives mainly 2-methylpropene (E2).	Use (CH₃)₃CONa + CH₃Br.
3	Phenol dissolves in NaHCO₃.	It does not; it dissolves in NaOH.	Phenol (pKa 10) is weaker than H₂CO₃.
4	Ethanol is more acidic than water.	Water is slightly more acidic.	The ethyl group donates electrons to O⁻.
5	3° alcohols are oxidised to ketones by acidified K₂Cr₂O₇.	3° alcohols resist oxidation.	No H on the carbon bearing OH.
6	Hydration of propene with dil. H₂SO₄ gives propan-1-ol.	It gives propan-2-ol (Markovnikov); use hydroboration for propan-1-ol.	The 2° carbocation forms.
7	p-Nitrophenol is steam-volatile.	o-Nitrophenol is steam-volatile.	Intramolecular H bond in the ortho isomer.
8	m-Nitrophenol is more acidic than p-nitrophenol.	p-Nitrophenol is more acidic.	Para –NO₂ stabilises phenoxide by resonance; meta only by induction.
9	Branching raises the boiling point of isomeric alcohols.	Branching lowers it: butan-1-ol 391 K > 2-methylpropan-2-ol 355 K.	Smaller surface area, weaker van der Waals forces.
10	Any two alcohols with conc. H₂SO₄ give a single unsymmetrical ether.	They give a mixture of three ethers (and alkenes).	Use the Williamson synthesis instead.
11	Phenol + acetyl chloride gives o-hydroxyacetophenone directly.	It gives phenyl acetate (O-acylation).	Acylation occurs at the oxygen first.
12	Ethers boil higher than isomeric alcohols.	Alcohols boil much higher.	Ethers have no O–H for H bonding.

Memory hook: “Formaldehyde one, aldehyde two, ketone three” — the Grignard alcohol classes. · “Lucas: 3 at once, 2 in five, 1 never.” · “Bulky on the oxygen, small on the halide” — Williamson. · “HI never breaks the aryl bond.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Phenol is more acidic than ethanol because: (Easy)

• (a) phenol is aromatic
• (b) phenoxide is resonance-stabilised
• (c) ethanol has more H atoms
• (d) phenol has a lower molar mass

Q02. HCHO + CH₃MgBr, then H₃O⁺, gives: (Easy)

• (a) methanol
• (b) ethanol
• (c) propan-2-ol
• (d) acetaldehyde

Q03. Lucas test gives immediate turbidity with: (Easy)

• (a) butan-1-ol
• (b) butan-2-ol
• (c) 2-methylpropan-2-ol
• (d) methanol

Q04. Phenol + bromine water gives: (Easy)

• (a) o-bromophenol
• (b) p-bromophenol
• (c) 2,4,6-tribromophenol
• (d) bromobenzene

Q05. Cumene process gives phenol and: (Easy)

• (a) acetaldehyde
• (b) acetone
• (c) benzene
• (d) propene

Q06. Most acidic among these: (Moderate)

• (a) phenol
• (b) p-cresol
• (c) p-nitrophenol
• (d) m-nitrophenol

Q07. Propene + B₂H₆, then H₂O₂/NaOH, gives: (Moderate)

• (a) propan-2-ol
• (b) propan-1-ol
• (c) propanal
• (d) propane

Q08. Ethanol with conc. H₂SO₄ at 413 K (excess alcohol) gives: (Moderate)

• (a) ethene
• (b) diethyl ether
• (c) ethanal
• (d) ethyl hydrogen sulphate only

Q09. Kolbe reaction product is: (Moderate)

• (a) salicylaldehyde
• (b) salicylic acid
• (c) benzoic acid
• (d) picric acid

Q10. The electrophile in Reimer–Tiemann reaction is: (Moderate)

• (a) CHCl₃
• (b) :CCl₂
• (c) CCl₃⁺
• (d) Cl⁺

Q11. Best synthesis of tert-butyl ethyl ether: (Moderate)

• (a) C₂H₅ONa + (CH₃)₃CBr
• (b) (CH₃)₃CONa + C₂H₅Br
• (c) C₂H₅OH + (CH₃)₃COH, heat
• (d) (CH₃)₃CBr + C₂H₅Br + Na

Q12. Which alcohol gives an alkene with Cu at 573 K? (Moderate)

• (a) ethanol
• (b) propan-2-ol
• (c) 2-methylpropan-2-ol
• (d) methanol

Q13. CH₃–O–C(CH₃)₃ + HI (1 equivalent) gives: (Tough)

• (a) CH₃I + (CH₃)₃COH
• (b) CH₃OH + (CH₃)₃CI
• (c) CH₄ + (CH₃)₃COI
• (d) CH₃OH + (CH₃)₂C=CH₂ only

Q14. 3,3-Dimethylbutan-2-ol with conc. H₂SO₄, heat, gives mainly: (Tough)

• (a) 3,3-dimethylbut-1-ene
• (b) 2,3-dimethylbut-2-ene
• (c) 2,3-dimethylbut-1-ene
• (d) 2,2-dimethylbutane

Q15. A compound C₃H₈O gives a positive iodoform test. It is: (Tough)

• (a) propan-1-ol
• (b) propan-2-ol
• (c) methoxyethane
• (d) propanal

Q16. Wood spirit is: (Easy)

• (a) ethanol
• (b) methanol
• (c) propanol
• (d) glycerol

Q17. Phenol gives a violet colour with: (Easy)

• (a) Br₂ water
• (b) neutral FeCl₃
• (c) NaHCO₃
• (d) Tollens’ reagent

Q18. The enzyme converting glucose to ethanol is: (Easy)

• (a) invertase
• (b) zymase
• (c) diastase
• (d) maltase

Q19. Highest boiling point: (Moderate)

• (a) butan-1-ol
• (b) butan-2-ol
• (c) 2-methylpropan-2-ol
• (d) ethoxyethane

Q20. Rate-determining step of acid dehydration of ethanol: (Moderate)

• (a) protonation of OH
• (b) loss of water to form the carbocation
• (c) loss of H⁺
• (d) attack of water

Q21. Phenol + benzenediazonium chloride in mild alkali gives: (Moderate)

• (a) aniline
• (b) p-hydroxyazobenzene
• (c) chlorobenzene
• (d) benzene

Q22. Anisole + CH₃COCl / AlCl₃ gives mainly: (Moderate)

• (a) phenyl acetate
• (b) 4-methoxyacetophenone
• (c) acetophenone
• (d) methyl benzoate

Q23. Salicylic acid + acetic anhydride gives: (Moderate)

• (a) methyl salicylate
• (b) aspirin
• (c) phenyl acetate
• (d) salicylaldehyde

Q24. C–O–C bond angle in methoxymethane is about: (Tough)

• (a) 104.5°
• (b) 109.5°
• (c) 111.7°
• (d) 120°

Q25. CH₃MgBr + ethanal, then H₃O⁺, gives: (Tough)

• (a) ethanol
• (b) propan-1-ol
• (c) propan-2-ol
• (d) 2-methylpropan-2-ol

Answer key with explanations

1. (b) phenoxide is resonance-stabilised — Charge spreads into the ring.
2. (b) ethanol — Formaldehyde → 1° alcohol.
3. (c) 2-methylpropan-2-ol — 3° alcohol.
4. (c) 2,4,6-tribromophenol — Highly activated ring.
5. (b) acetone — By-product.
6. (c) p-nitrophenol — Para –NO₂, resonance.
7. (b) propan-1-ol — Anti-Markovnikov.
8. (b) diethyl ether — Lower temperature → ether.
9. (b) salicylic acid — Phenoxide + CO₂.
10. (b) :CCl₂ — Dichlorocarbene.
11. (b) (CH₃)₃CONa + C₂H₅Br — SN2 on a 1° halide.
12. (c) 2-methylpropan-2-ol — 3° alcohols dehydrate.
13. (b) CH₃OH + (CH₃)₃CI — SN1 at the 3° carbon.
14. (b) 2,3-dimethylbut-2-ene — Methyl shift, then Saytzeff.
15. (b) propan-2-ol — Has CH₃CH(OH)–.
16. (b) methanol — Methanol.
17. (b) neutral FeCl₃ — Iron–phenoxide complex.
18. (b) zymase — From yeast.
19. (a) butan-1-ol — Unbranched alcohol.
20. (b) loss of water to form the carbocation — Slow step.
21. (b) p-hydroxyazobenzene — Azo coupling.
22. (b) 4-methoxyacetophenone — o/p director, para major.
23. (b) aspirin — Acetylation of –OH.
24. (c) 111.7° — Repulsion between methyl groups.
25. (c) propan-2-ol — Aldehyde (not HCHO) → 2° alcohol.

Advanced problems

Each needs two ideas from the chapter.

Solved example 13 — Identify the compound

Compound A (C₄H₁₀O) reacts with Na to give H₂, gives turbidity with Lucas reagent at once, and does not decolourise acidified K₂Cr₂O₇. Identify A.

1. Na → H₂: it is an alcohol (not an ether).
2. Instant Lucas turbidity and resistance to oxidation: 3° alcohol.
3. The only 3° C₄ alcohol is (CH₃)₃COH.

Answer: A is 2-methylpropan-2-ol.

Solved example 14 — Separating nitrophenols

Nitration of phenol gives a mixture of o- and p-nitrophenol. How are they separated, and why does the method work?

1. Steam distil the mixture.
2. o-Nitrophenol has an intramolecular H bond, is less associated, more volatile and passes over with steam.
3. p-Nitrophenol is H-bonded to other molecules and to water, and stays behind.

Answer: Steam distillation: o-nitrophenol distils over; p-nitrophenol remains.

Check-yourself answers

• C1. Ethanol molecules are linked by intermolecular hydrogen bonds; methoxymethane has no O–H.
• C2. The oxygen lone pair is conjugated with the ring, giving the C–O bond partial double-bond character, and the carbon is sp² (more s character).
• C3. Phenol (pKa 10) is a weaker acid than carbonic acid (pKa 6.4), so it cannot displace CO₂ from bicarbonate.
• C4. Water (pKa 15.7) is slightly more acidic than ethanol (pKa ≈ 15.9).
• C5. Loss of water from the protonated alcohol to form the carbocation.
• C6. It forms a stable 3° carbocation easily, so the SN1 reaction is fast without a catalyst.
• C7. Dichlorocarbene, :CCl₂ (formed from CHCl₃ and OH⁻).
• C8. The –OH group strongly activates the ring by donating its lone pair, so even the weak electrophile in bromine water substitutes all three o/p positions; benzene needs a Lewis acid catalyst.

One-page revision

Read this page the night before the exam.

Making alcohols

• dil. H₂SO₄: Markovnikov (may rearrange)
• B₂H₆/H₂O₂: anti-Markovnikov
• NaBH₄: aldehyde → 1°, ketone → 2°
• RMgX: HCHO 1°, RCHO 2°, R₂CO 3°

Making phenol

• Cumene → phenol + acetone
• Benzenesulphonate + NaOH fusion
• Dow: C₆H₅Cl + NaOH, 623 K
• Diazonium salt + warm water

Acidity

• Phenol pKa 10; ethanol ≈ 16
• Alcohols: 1° > 2° > 3°; water > ethanol
• o/p –NO₂ ↑ acidity; –CH₃ ↓
• Phenol: dissolves in NaOH, not NaHCO₃

Alcohol reactions

• Lucas: 3° at once, 2° 5 min, 1° none
• H₂SO₄: 443 K ethene; 413 K ether
• 1° → acid / aldehyde (PCC); 2° → ketone; 3° resists
• Cu, 573 K: 3° → alkene

Phenol reactions

• Br₂ water → 2,4,6-tribromophenol
• dil. HNO₃ → o + p; conc. → picric acid
• Kolbe → salicylic acid
• Reimer–Tiemann (:CCl₂) → salicylaldehyde

Ethers

• Williamson: SN2, 1° halide
• Bulky group on the alkoxide
• HI: SN2 at methyl; SN1 at 3°
• Anisole + HI → phenol + CH₃I