How to Use This Page
This page covers all exercise question types from Chapter 11 — Hydroxy Compounds and Ethers. Work through the sections in order:
1. Key Concepts — quick-revision table of essential terms and definitions. 2. MCQs — test your conceptual clarity; the correct answer is bolded with a brief reason. 3. Fill in the Blanks / True or False / Match the Following — drill factual recall. 4. Brief Answers (2-mark) — practice short, pointed responses. 5. Detailed Answers (5-mark) — structure your long answers with clear steps and equations. 6. Common Mistakes & Exam Checklist — avoid frequent errors and verify readiness.
Write chemical equations in ``text`` blocks so structures and conditions are visible at a glance.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Alcohol | Organic compound containing –OH group bonded to a saturated carbon atom |
| Phenol | –OH group directly attached to a benzene ring |
| Ether | Compound with an oxygen atom bridging two alkyl/aryl groups (R–O–R') |
| Primary (1°) alcohol | –OH group on a carbon bonded to only one other carbon |
| Secondary (2°) alcohol | –OH group on a carbon bonded to two other carbons |
| Tertiary (3°) alcohol | –OH group on a carbon bonded to three other carbons |
| Dehydration | Elimination of water from an alcohol to form an alkene |
| Lucas test | Distinction test using conc. HCl / ZnCl₂; 3° alcohols react fastest |
| Victor Meyer test | Distinction of 1°, 2°, 3° alcohols using KI / NaNO₂ / NaOH |
| Esterification | Reaction of alcohol with carboxylic acid in acid catalyst to form an ester |
| Williamson synthesis | Preparation of ethers from alkoxide and alkyl halide (SN₂ mechanism) |
| Reimer-Tiemann reaction | Phenol + CHCl₃ / NaOH → salicylaldehyde (ortho-hydroxybenzaldehyde) |
| Kolbe's reaction | Phenol + CO₂ / NaOH → salicylic acid (ortho-hydroxybenzoic acid) |
| Resonance stabilisation | Delocalisation of negative charge in phenoxide ion over the benzene ring |
| Iodoform test | Positive for CH₃CH(OH)– group; gives yellow CHI₃ precipitate |
I. Choose the Best Answer (MCQs)
1. Which type of alcohol gives turbidity fastest in the Lucas test? (a) 1° alcohol (b) 2° alcohol (c) 3° alcohol (d) Allyl alcohol
Answer(c) — Tertiary alcohols form stable tertiary carbocations immediately, so turbidity appears within seconds.
2. Phenol is more acidic than ethanol because: (a) Phenol has a higher molecular weight (b) The phenoxide ion is resonance-stabilised (c) Ethanol has a higher boiling point (d) Phenol can form hydrogen bonds
Answer(b) — The phenoxide ion delocalises the negative charge over the benzene ring, making proton release more favourable.
3. Williamson synthesis is preferred for preparing: (a) Symmetrical ethers only (b) Both symmetrical and unsymmetrical ethers (c) Esters (d) Aldehydes
Answer(b) — Williamson synthesis uses an SN₂ mechanism and can prepare both types of ethers, provided the alkyl halide is primary.
4. The product of dehydration of ethanol at 443 K with conc. H₂SO₄ is: (a) Diethyl ether (b) Ethane (c) Ethene (d) Acetaldehyde
Answer(c) — At 443 K, intramolecular dehydration dominates, producing ethene. At 413 K, intermolecular dehydration gives diethyl ether.
5. Which test distinguishes a primary alcohol from a secondary alcohol? (a) Tollens' test (b) Victor Meyer test (c) Fehling's test (d) Iodoform test
Answer(b) — Victor Meyer test gives different colour solutions: blood red for 1°, blue for 2°, and colourless for 3° alcohols.
6. Kolbe's reaction produces: (a) p-hydroxybenzoic acid (b) o-hydroxybenzoic acid (salicylic acid) (c) p-hydroxybenzaldehyde (d) o-hydroxybenzaldehyde
Answer(b) — Phenol reacts with CO₂ and NaOH at 125 °C and 100 atm to give salicylic acid (ortho product).
7. Glycerol is a: (a) Monohydric primary alcohol (b) Dihydric secondary alcohol (c) Trihydric alcohol (d) Trihydric phenol
Answer(c) — Glycerol (propane-1,2,3-triol) has three –OH groups on adjacent carbons.
8. The Reimer-Tiemann reaction produces which product from phenol? (a) Salicylic acid (b) o-hydroxybenzaldehyde (c) Phenetole (d) Cresol
Answer(b) — Phenol + CHCl₃ + NaOH gives o-hydroxybenzaldehyde (salicylaldehyde) as the major product.
9. Which alcohol gives a positive iodoform test? (a) Methanol (b) Ethanol (c) Propan-1-ol (d) Butan-2-ol only
Answer(b) — Ethanol has the CH₃CH(OH)– group required for the iodoform test. Butan-2-ol also gives a positive test, but among the options ethanol is the most direct answer.
10. Ethers are cleaved by HI to give: (a) Two molecules of alcohol (b) An alcohol and an alkyl iodide (c) Two molecules of alkene (d) An alcohol and an alkyl chloride
Answer(b) — Ethers undergo cleavage with HI to produce an alcohol and an alkyl iodide; excess HI converts both to alkyl iodides.
11. The boiling point of ethanol is higher than that of diethyl ether because: (a) Ethanol has higher molecular weight (b) Ethanol molecules form intermolecular hydrogen bonds (c) Diethyl ether is more polar (d) Ethanol is less soluble in water
Answer(b) — Hydrogen bonding in ethanol requires more energy to break, giving it a higher boiling point than the non-H-bonding ether.
12. Hydroboration-oxidation of an alkene gives: (a) An aldehyde (b) A ketone (c) An anti-Markovnikov alcohol (d) A Markovnikov alcohol
Answer(c) — BH₃ adds to the less substituted carbon (anti-Markovnikov), and oxidation with H₂O₂/NaOH gives the corresponding alcohol.
II. Fill in the Blanks
1. The general formula of aliphatic alcohols is ______. CₙH₂ₙ₊₁OH (or CₙH₂ₙ₊₂O)
2. In the Lucas test, tertiary alcohols produce turbidity within ______. seconds (immediately)
3. Phenol reacts with bromine water to give a ______ precipitate of 2,4,6-tribromophenol. white
4. Williamson synthesis follows an ______ mechanism. SN₂
5. The IUPAC name of CH₃OCH₂CH₃ is ______. methoxyethane
6. Salicylic acid is produced from phenol by ______ reaction. Kolbe's
7. Dehydration of ethanol at 413 K with conc. H₂SO₄ gives ______ as the major product. diethyl ether
8. The –OH group in phenol is ______ hybridised. sp²
9. Glycerol has the IUPAC name ______. propane-1,2,3-triol
10. Phenol is also called ______ acid. carbolic
III. True or False
1. Statement: Phenol is less acidic than carbonic acid. True. Phenol (pKa ≈ 10) is weaker than carbonic acid (pKa ≈ 6.5), so phenol does not decompose NaHCO₃.
2. Statement: Williamson synthesis uses a tertiary alkyl halide for best yield. False. Tertiary alkyl halides favour elimination (E2) over substitution, giving alkenes. Primary alkyl halides give best yield in Williamson synthesis.
3. Statement: Ethers can form hydrogen bonds with water molecules. True. The oxygen atom in ethers acts as a hydrogen bond acceptor with water, though ethers cannot form H-bonds with each other.
4. Statement: Dehydration of a primary alcohol always gives an alkene. False. At lower temperature (413 K), intermolecular dehydration gives an ether. At higher temperature (443 K), intramolecular dehydration gives an alkene.
5. Statement: Phenol reacts with dilute NaOH to form sodium phenoxide. True. Phenol is acidic enough to react with NaOH: C₆H₅OH + NaOH → C₆H₅ONa + H₂O.
6. Statement: Glycerol is a dihydric alcohol. False. Glycerol is a trihydric alcohol (three –OH groups). Ethylene glycol is a dihydric alcohol.
7. Statement: The iodoform test is positive for ethanol. True. Ethanol contains the CH₃CH(OH)– group and gives a yellow CHI₃ precipitate with NaOH/I₂.
8. Statement: Ethers are highly reactive compounds. False. Ethers are relatively inert. They undergo mainly cleavage reactions with strong acids like HI. Their low reactivity is why they are used as solvents.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Lucas test | (c) Distinguishes 1°, 2°, 3° alcohols |
| 2. Williamson synthesis | (e) Prepares ethers from alkoxide + alkyl halide |
| 3. Reimer-Tiemann reaction | (a) Converts phenol to salicylaldehyde |
| 4. Kolbe's reaction | (f) Converts phenol to salicylic acid |
| 5. Victor Meyer test | (b) Distinguishes 1°, 2°, 3° alcohols by colour |
| 6. Iodoform test | (g) Yellow precipitate with CH₃CH(OH)– or CH₃CO– groups |
| 7. Dehydration at 413 K | (d) Gives diethyl ether from ethanol |
| 8. Dehydration at 443 K | (h) Gives ethene from ethanol |
Answer1–c, 2–e, 3–a, 4–f, 5–b, 6–g, 7–d, 8–h
V. Answer Briefly (2-mark)
Q1. Why is phenol more acidic than ethanol?
Answer
Phenol releases H⁺ to form the phenoxide ion (C₆H₅O⁻), which is stabilised by resonance — the negative charge is delocalised over the entire benzene ring across multiple positions. Ethanol forms the ethoxide ion (C₂H₅O⁻) which has no such stabilisation. Greater stability of the conjugate base makes phenol more acidic.
Q2. Write the equation for Kolbe's reaction.
C₆H₅ONa + CO₂ →(125 °C, 100 atm) C₆H₄(OH)COONa →(H⁺) C₆H₄(OH)COOH
Sodium phenoxide + CO₂ → sodium salicylate → salicylic acid (o-hydroxybenzoic acid)
Q3. Distinguish between primary and tertiary alcohols using the Lucas test.
Answer
A primary alcohol shows no turbidity at room temperature (reacts very slowly). A tertiary alcohol produces immediate turbidity because it forms a stable tertiary carbocation rapidly with Lucas reagent (conc. HCl / ZnCl₂). Secondary alcohols produce turbidity within 5–10 minutes.
Q4. What is the product when phenol reacts with bromine water?
Answer
Phenol reacts with bromine water to give 2,4,6-tribromophenol as a white precipitate and HBr. The activating –OH group directs substitution to ortho and para positions, and both positions are substituted by excess bromine.
C₆H₅OH + 3Br₂ → C₆H₂Br₃OH↓ + 3HBr
Q5. Why are ethers used as solvents?
Answer
Ethers are relatively inert chemically, have a low tendency to react with most reagents, and have moderate polarity that dissolves a wide range of organic compounds. Their oxygen atom can also coordinate with cations, making them good solvents for Grignard reactions.
VI. Answer in Detail (5-mark)
Q1. Explain the preparation of alcohols from alkenes (hydration, hydroboration-oxidation, and oxymercuration-demercuration).
Answer
Alcohols are prepared from alkenes by three important methods:
(a) Acid-catalysed hydration:
CH₃CH=CH₂ + H₂O →(H⁺) CH₃CH(OH)CH₃
Propene → propan-2-ol (Markovnikov product)
The reaction follows Markovnikov's rule — the –OH attaches to the more substituted carbon. A carbocation intermediate is involved, so rearrangement is possible.
(b) Hydroboration-oxidation:
CH₃CH=CH₂ + BH₃ → CH₃CH₂CH₂BH₂ →(H₂O₂/NaOH) CH₃CH₂CH₂OH
Propene → propan-1-ol (anti-Markovnikov product)
BH₃ adds across the double bond with boron attaching to the less substituted carbon. Oxidation replaces B with –OH. This method gives anti-Markovnikov addition with no rearrangement.
(c) Oxymercuration-demercuration:
CH₃CH=CH₂ + Hg(OAc)₂/H₂O → NaBH₄ → CH₃CH(OH)CH₃
Propene → propan-2-ol (Markovnikov product, no rearrangement)
Mercuric acetate adds across the double bond, then sodium borohydride replaces HgOAc with H. This gives Markovnikov addition without carbocation rearrangement.
Q2. Describe the chemical reactions of alcohols — dehydration, oxidation, and esterification with equations.
Answer
(a) Dehydration: Alcohols lose water to form alkenes or ethers depending on conditions:
C₂H₅OH →(conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O
Ethanol → ethene (intramolecular dehydration at high temperature)
2C₂H₅OH →(conc. H₂SO₄, 413 K) C₂H₅–O–C₂H₅ + H₂O
Ethanol → diethyl ether (intermolecular dehydration at lower temperature)
(b) Oxidation:
Primary alcohol →(K₂Cr₂O₇/H⁺) Aldehyde →(further oxidation) Carboxylic acid
CH₃CH₂OH → CH₃CHO → CH₃COOH
Secondary alcohol →(K₂Cr₂O₇/H⁺) Ketone (no further oxidation under these conditions)
CH₃CH(OH)CH₃ → CH₃COCH₃
Tertiary alcohol → does NOT oxidise easily (no α-hydrogen)
(c) Esterification:
CH₃COOH + C₂H₅OH →(conc. H₂SO₄, reflux) CH₃COOC₂H₅ + H₂O
Acetic acid + ethanol → ethyl acetate + water
This is a reversible reaction. Conc. H₂SO₄ acts as a catalyst and dehydrating agent, shifting equilibrium toward the ester.
Q3. Explain the preparation, properties, and reactions of phenol. Write equations for Reimer-Tiemann and Kolbe's reactions.
Answer
Preparation of phenol: From chlorobenzene (Dow's process):
C₆H₅Cl + NaOH →(300 °C, 200 atm) C₆H₅ONa →(H⁺) C₆H₅OH
Chlorobenzene → sodium phenoxide → phenol
From cumene (Hock process):
Cumene →(O₂) Cumene hydroperoxide →(H⁺) Phenol + Acetone
Properties: Phenol is more acidic than alcohol because the phenoxide ion is resonance-stabilised. Phenol reacts with NaOH but NOT with NaHCO₃ (weaker acid than carbonic acid).
Reimer-Tiemann reaction:
C₆H₅OH + CHCl₃ + 3NaOH →(heat) o-HOC₆H₄CHO + 3NaCl + 2H₂O
Phenol + chloroform + NaOH → salicylaldehyde (o-hydroxybenzaldehyde)
This is an electrophilic substitution on the ortho position. Dichlorocarbene (:CCl₂) is the electrophile.
Kolbe's reaction:
C₆H₅ONa + CO₂ →(125 °C, 100 atm) o-HOC₆H₄COONa →(H⁺) o-HOC₆H₄COOH
Sodium phenoxide + CO₂ → sodium salicylate → salicylic acid
CO₂ acts as the electrophile, attacking the ortho position of the phenoxide ring.
Q4. Explain Williamson ether synthesis. Why does it fail with tertiary alkyl halides?
Answer
Williamson ether synthesis:
R–ONa + R'–X → R–O–R' + NaX
Sodium alkoxide + primary alkyl halide → ether + sodium halide (SN₂ mechanism)
Example:
C₂H₅ONa + CH₃I → CH₃–O–C₂H₅ + NaI
Sodium ethoxide + methyl iodide → ethyl methyl ether
Why it fails with tertiary alkyl halides: Tertiary alkyl halides are highly sterically hindered, making the backside attack required for SN₂ mechanism very difficult. Instead, the strong base (alkoxide) abstracts a β-hydrogen, leading to elimination (E2) and forming an alkene as the major product:
C₂H₅ONa + (CH₃)₃C–Br → CH₂=C(CH₃)₂ + NaBr + C₂H₅OH
Hence, for best yield in Williamson synthesis, the alkyl halide should be primary or methyl, never tertiary.
Conversion Pattern
1. Identify starting compound and target compound. 2. Choose suitable reagent. 3. Write intermediate if needed. 4. Write final product. 5. Mention condition such as heat, acid, or catalyst.
Short Answer Points
- Phenol is more acidic than alcohol due to resonance stabilisation of phenoxide ion.
- Ethers are less reactive but undergo cleavage with HI.
- Williamson synthesis prepares ethers from alkoxide and alkyl halide.
Worked Examples
Example 1 — IUPAC naming
CH₃CH₂OH → ethanol
CH₃CH(OH)CH₃ → propan-2-ol
C₆H₅OH → phenol
Example 2 — Dehydration of ethanol
C₂H₅OH →(conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O
Ethanol gives ethene by intramolecular dehydration.
Example 3 — Williamson ether synthesis
C₂H₅ONa + C₂H₅Br → C₂H₅–O–C₂H₅ + NaBr
Sodium ethoxide + ethyl bromide → diethyl ether.
Example 4 — Why phenol is more acidic than ethanol
The phenoxide ion left after losing H⁺ is resonance-stabilised over the benzene ring, so phenol releases H⁺ more readily than ethanol, whose alkoxide ion has no such stabilisation.
Common Mistakes to Avoid
- ❌ Forgetting reaction conditions (temperature, catalyst, acid) — ✅ Always write the full condition above the arrow.
- ❌ Confusing alcohol acidity with phenol acidity — ✅ Phenol is more acidic due to resonance; alcohols are weaker acids.
- ❌ Using a tertiary alkyl halide in Williamson synthesis — ✅ Use only primary or methyl halides; tertiary halides give elimination.
- ❌ Writing impossible direct conversions — ✅ Plan multi-step routes through logical intermediates.
- ❌ Not naming functional group correctly — ✅ Identify –OH on alkyl chain (alcohol) vs –OH on benzene ring (phenol) vs R–O–R' (ether).
- ❌ Saying all alcohols give iodoform test — ✅ Only alcohols with CH₃CH(OH)– group give the iodoform test (ethanol and secondary alcohols with a methyl group).
- ❌ Confusing intramolecular and intermolecular dehydration — ✅ At 443 K → alkene (intramolecular); at 413 K → ether (intermolecular).
- ❌ Omitting resonance structures for phenoxide ion — ✅ Always draw/show resonance when explaining phenol acidity.
Exam Checklist
- Reagent is correctly named and appropriate for the substrate.
- Product structure is drawn or named clearly.
- Reaction condition (temperature, catalyst, acid/base) is stated.
- Reasoning includes stability, resonance, or inductive effects where required.
- Distinction tests (Lucas, Victor Meyer, iodoform) are matched to correct alcohol class.
- Dehydration product depends on temperature: 413 K → ether; 443 K → alkene.
- Phenol reactions (Kolbe's, Reimer-Tiemann, bromination) give ortho products.
- Williamson synthesis uses SN₂ — only primary alkyl halides work well.
- Esterification is reversible; use excess acid or remove water to drive it forward.
- Hydroboration-oxidation gives anti-Markovnikov alcohol.
- Glycerol's three –OH groups give it distinctive properties (solubility, viscosity).
- Always balance chemical equations and show correct molecular formulae.
