ChemistryISC Class 12 · Alcohols, Phenols and Ethers

Reactions of Alcohols: ISC Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Alcohols react by breaking the O–H bond (with sodium, acids and acylating agents, as in esterification) or the C–O bond (with HX, PCl₃, PCl₅, SOCl₂ and in dehydration). Primary alcohols oxidise to aldehydes and acids, secondary to ketones; tertiary resist. The Lucas test tells the three classes apart.

Ask which bond breaks: O–H gives alkoxides and esters; C–O gives alkyl halides and alkenes.

Illustration drawn for QuestionBuddy.

Understand Reactions of Alcohols

Alcohols (R–OH) react in two ways, and sorting reactions this way makes them easy to remember.

A. Reactions breaking the O–H bond (alcohol as a nucleophile or acid).
• Acidic nature: alcohols react with active metals to give alkoxides and hydrogen, $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ . They are weaker acids than water, because the electron-releasing alkyl group (+I) increases electron density on oxygen and destabilises the alkoxide ion; they do not react with NaOH. Acid strength: 1° > 2° > 3°.
• Esterification: with a carboxylic acid and a little concentrated H₂SO₄, a reversible reaction gives an ester: $C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$ . Removing water as it forms drives the equilibrium forward. Mechanism: (1) H⁺ protonates the carbonyl oxygen of the acid, making the carbonyl carbon more electrophilic; (2) the lone pair of the alcohol oxygen attacks this carbon (nucleophilic addition) to give a tetrahedral intermediate; (3) a proton moves to one –OH, turning it into a good leaving group, –OH₂⁺; (4) water leaves and (5) H⁺ is lost to give the ester and regenerate the catalyst. Labelling studies show that the ester's alkoxy oxygen comes from the alcohol. With acid chlorides or anhydrides the reaction is fast and not reversible: $C_{2} H_{5} O H + C H_{3} C O C l \to C H_{3} C O O C_{2} H_{5} + H C l$ (pyridine removes HCl).

B. Reactions breaking the C–O bond.
• With HX: $C_{2} H_{5} O H + H C l anhyd. ZnCl_{2} C_{2} H_{5} C l + H_{2} O$ . Reactivity: 3° > 2° > 1° (via carbocations). This is the basis of the Lucas test: Lucas reagent is concentrated HCl with anhydrous ZnCl₂. Tertiary alcohols give turbidity (an insoluble alkyl chloride) at once; secondary in about five minutes; primary show no turbidity at room temperature.
• With phosphorus halides and thionyl chloride: $3 C_{2} H_{5} O H + P C l_{3} \to 3 C_{2} H_{5} C l + H_{3} P O_{3}$ ; $C_{2} H_{5} O H + P C l_{5} \to C_{2} H_{5} C l + P O C l_{3} + H C l$ ; $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$ (both by-products are gases, so the product is easily obtained pure).
• Dehydration with concentrated H₂SO₄: at 443 K ethanol gives ethene (intramolecular), $C_{2} H_{5} O H \to C_{2} H_{4} + H_{2} O$ ; at 413 K with excess alcohol it gives diethyl ether (intermolecular), $2 C_{2} H_{5} O H \to C_{2} H_{5} O C_{2} H_{5} + H_{2} O$ . Alumina at about 633 K also gives alkenes. Mechanism to the alkene: (1) protonation of the –OH oxygen, $C_{2} H_{5} O H + H^{+} \to C_{2} H_{5} O H_{2}^{+}$ (fast); (2) loss of water to form a carbocation (slow, rate-determining); (3) loss of H⁺ from the neighbouring carbon to give the C=C bond. Ease of dehydration: 3° > 2° > 1°, following carbocation stability. If two alkenes can form, the more substituted one is major (Saytzeff rule): butan-2-ol gives mainly but-2-ene.

C. Oxidation and dehydrogenation. Oxidation removes H from O–H and from the carbon carrying –OH, so the result depends on how many H atoms that carbon has. Primary: aldehyde, then carboxylic acid with strong oxidants (acidified K₂Cr₂O₇ or KMnO₄); pyridinium chlorochromate (PCC) or CrO₃ in anhydrous medium stops at the aldehyde. Secondary: ketone. Tertiary: not oxidised under ordinary conditions (no H on the carbinol carbon). Over heated copper at 573 K, 1° alcohols give aldehydes and 2° alcohols ketones by dehydrogenation ( $C H_{3} C H_{2} O H \to C H_{3} C H O + H_{2}$ ), while 3° alcohols are dehydrated to alkenes. Acidified dichromate turns from orange to green when it oxidises a 1° or 2° alcohol.

D. Converting one alcohol into another. To climb one carbon, change the alcohol to a bromide, make a Grignard reagent and add methanal: $C_{2} H_{5} B r + M g \to C_{2} H_{5} M g B r$ ; with HCHO then water, propan-1-ol. To make a secondary alcohol, add a Grignard reagent to an aldehyde (other than methanal); for a tertiary alcohol, to a ketone. To move –OH from C-1 to C-2, dehydrate to the alkene and add water (acid-catalysed hydration, Markovnikov).

Reactions of Alcohols: formulas and key facts

Esterification: $R C O O H + R^{'} O H ⇌ H^{+} R C O O R^{'} + H_{2} O$; Reversible; conc. H₂SO₄ catalyst; acid chlorides react faster and irreversibly.
Dehydration of ethanol: $C_{2} H_{5} O H conc. H_{2} S O_{4}, 443 K C_{2} H_{4} + H_{2} O; 2 C_{2} H_{5} O H 413 K C_{2} H_{5} O C_{2} H_{5} + H_{2} O$; Higher temperature favours the alkene; excess alcohol at 413 K gives the ether.
Chlorination reagents: $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$; Also PCl₃ (gives H₃PO₃) and PCl₅ (gives POCl₃ + HCl).
Oxidation pattern: $1^{\circ} \to R C H O \to R C O O H; 2^{\circ} \to R_{2} C O; 3^{\circ} \to no reaction (mild)$; Cu at 573 K: 1° → aldehyde, 2° → ketone, 3° → alkene.
Lucas test timing: $3^{\circ} : immediate; 2^{\circ} : \sim 5 min; 1^{\circ} : none at room temperature$; Conc. HCl + anhydrous ZnCl₂; turbidity = alkyl chloride.

How to solve Reactions of Alcohols questions

ISC questions on alcohols are conversions, distinguishing tests, mechanisms (esterification, dehydration) and identification of unknowns. Start every answer by classifying the alcohol as 1°, 2° or 3°.

1Classify the alcohol by the number of carbons attached to the carbinol carbon.
2Decide which bond must break: O–H (Na, acids, acylation) or C–O (HX, PX₃, SOCl₂, dehydration).
3For oxidation, predict the product from the class (1° → aldehyde/acid, 2° → ketone, 3° → none) and give the reagent.
4For tests, use Lucas (class), iodoform (CH₃CH(OH)– group), or acidified dichromate (orange to green for 1° and 2°); state results for both compounds.
5For mechanisms, show protonation, the key intermediate (tetrahedral intermediate for esterification, carbocation for dehydration) and the final deprotonation.
6For conversions that change the carbon chain, plan through halide → Grignard → carbonyl compound.

Reactions of Alcohols in real life

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.

Ethanol is oxidised to ethanoic acid

Suppose a home winemaker leaves a 1.0 L bottle of wine open to the air for weeks and it turns sour. Imagine the wine contained 92 g of ethanol (M = 46) and that air oxidation, helped by bacteria, converts it to ethanoic acid (M = 60): $C H_{3} C H_{2} O H + O_{2} \to C H_{3} C O O H + H_{2} O$ .

What is the greatest mass of ethanoic acid that could form? Which class of alcohol does ethanol belong to, and why can it be oxidised all the way to an acid?

Try it first, then show the solutionSolution

Moles of ethanol $= \frac{92}{46} = 2.0$ mol.
1 mol ethanol gives 1 mol ethanoic acid, so at most $2.0 \times 60 = 120$ g.
Ethanol is a primary alcohol: the carbinol carbon carries two H atoms.
Oxidation removes H to give ethanal, which has an aldehyde H that is oxidised further to the –COOH group.

Answer: At most 120 g of ethanoic acid. Ethanol is primary, so it is oxidised via ethanal to ethanoic acid.

Memory hook: Primary alcohols go all the way: alcohol → aldehyde → acid.

Tricks to remember Reactions of Alcohols

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

Count the H on the carbinol carbon
Two H: oxidise twice (aldehyde, then acid). One H: once (ketone). No H: not at all.
Use it when: Predicting oxidation products.
Hot = alkene, warm = ether
Ethanol + conc. H₂SO₄: 443 K (hotter) gives ethene; 413 K (cooler, excess alcohol) gives ether.
Use it when: The dehydration fill-in.
Lucas: Tertiary Turns at once
Tertiary: Turbid instantly; Secondary: in Several minutes; Primary: Patience won't help (no turbidity at room temperature).
Use it when: Lucas test observations.

Reactions of Alcohols: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Fastest Lucas test · 1 mark · MCQBoard level

Which alcohol gives turbidity immediately with Lucas reagent at room temperature?
(a) butan-1-ol
(b) butan-2-ol
(c) 2-methylpropan-2-ol
(d) 2-methylpropan-1-ol

Try it first, then show the solutionSolution

Lucas reagent reacts via carbocations; tertiary alcohols react fastest.
2-Methylpropan-2-ol, (CH₃)₃COH, is the only tertiary alcohol listed.

Answer: (c) 2-methylpropan-2-ol.

Example 2: Ethanol with sodium and with PCl₃ · 2 marks · Short answerBoard level

Write equations for the reaction of ethanol with (a) sodium metal and (b) phosphorus trichloride. What does reaction (a) show about ethanol?

Try it first, then show the solutionSolution

(a) $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ (½ mark).
(b) $3 C_{2} H_{5} O H + P C l_{3} \to 3 C_{2} H_{5} C l + H_{3} P O_{3}$ (1 mark).
(a) shows ethanol is weakly acidic: it has a replaceable H on oxygen (½ mark).

Answer: (a) Sodium ethoxide + H₂ (acidic O–H). (b) Chloroethane + phosphorous acid.

Example 3: Mechanism of esterification · 3 marks · Short answerHarder board level

Write the mechanism of the acid-catalysed esterification of ethanoic acid with ethanol.

Try it first, then show the solutionSolution

Step 1: H⁺ (from conc. H₂SO₄) protonates the C=O oxygen of CH₃COOH, giving CH₃C(OH)₂⁺; the carbonyl carbon becomes more electrophilic (½ mark).
Step 2: the oxygen lone pair of C₂H₅OH attacks this carbon to give a tetrahedral intermediate, CH₃C(OH)₂(O⁺HC₂H₅) (1 mark).
Step 3: a proton moves from the ethanol oxygen to one –OH, forming –OH₂⁺, a good leaving group; Step 4: water leaves, giving protonated ester (1 mark).
Step 5: loss of H⁺ gives CH₃COOC₂H₅ and regenerates the catalyst (½ mark).

Answer: Protonation of C=O, nucleophilic attack by ethanol, proton transfer, loss of water, deprotonation to ethyl ethanoate.

Example 4: Oxidation of the three classes · 3 marks · Short answerHarder board level

Give the product formed when each of the following is (i) oxidised with acidified K₂Cr₂O₇ and (ii) passed over heated copper at 573 K: propan-1-ol, propan-2-ol and 2-methylpropan-2-ol.

Try it first, then show the solutionSolution

Propan-1-ol (1°): (i) propanal, then propanoic acid with excess oxidant; (ii) propanal (1 mark).
Propan-2-ol (2°): (i) propanone; (ii) propanone (1 mark).
2-Methylpropan-2-ol (3°): (i) no oxidation under ordinary conditions; (ii) dehydration to 2-methylpropene (1 mark).

Answer: 1°: propanal/propanoic acid; propanal. 2°: propanone; propanone. 3°: no reaction; 2-methylpropene.

Example 5: Identify a pentanol · 5 marks · Long answerHOTS / competency

An unbranched alcohol X (C₅H₁₂O) gives turbidity with Lucas reagent after about five minutes. On oxidation with acidified K₂Cr₂O₇, X gives Y (C₅H₁₀O), which gives a yellow precipitate with I₂ and NaOH. Heating X with concentrated H₂SO₄ gives Z (C₅H₁₀) as the major product.
(a) Identify X, Y and Z, with reasons. [3]
(b) Write the equation for the oxidation of X. [1]
(c) State the rule that decides the major product Z. [1]

Try it first, then show the solutionSolution

Turbidity in about 5 min: X is a secondary alcohol. Y gives the iodoform test, so it has a CH₃CO– group; X must have CH₃CH(OH)–. Unbranched: X = pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃ (1 mark).
Y = pentan-2-one, CH₃COCH₂CH₂CH₃ (1 mark).
Dehydration of pentan-2-ol can give pent-1-ene or pent-2-ene; the more substituted alkene, pent-2-ene (CH₃CH=CHCH₂CH₃), is major: Z (1 mark).
(b) $C H_{3} C H (O H) C H_{2} C H_{2} C H_{3} + [O] \to C H_{3} C O C H_{2} C H_{2} C H_{3} + H_{2} O$ (K₂Cr₂O₇/H₂SO₄) (1 mark).
(c) Saytzeff (Zaitsev) rule: the more highly substituted alkene is the major product (1 mark).

Answer: X pentan-2-ol, Y pentan-2-one, Z pent-2-ene (Saytzeff rule).

Where students lose marks in Reactions of Alcohols

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

Mistake: Swapping the products of dehydration
ISC examiners reported ethene written for the 140 °C (413 K) product, and diethyl ether or ethyl carbonate for 170 °C (443 K) (2024).
Do this: 413 K: diethyl ether (intermolecular). 443 K: ethene (intramolecular).
Usually costs: 1 mark.
Mistake: Confusing Lucas test observations
Examiners noted confused observations for secondary and tertiary alcohols, and tests described for only one compound (2024).
Do this: Give the reagent (conc. HCl + anhydrous ZnCl₂) and both results: tertiary immediate, secondary after about 5 minutes.
Usually costs: 1 mark per pair.
Mistake: Wrong alcohol in identification problems
Examiners reported butan-1-ol written instead of butan-2-ol in a C₄H₁₀O problem where the oxidation product gave the iodoform test (2024, 2025).
Do this: An iodoform-positive ketone means the alcohol has CH₃CH(OH)–: it is a 2-ol, not a 1-ol.
Usually costs: Up to all the identification marks.
Mistake: Omitting hydrolysis in Grignard conversions
Examiners noted the hydrolysis step of the Grignard adduct and the by-product left out (2024).
Wrong working
$C H_{3} C H O + C H_{3} M g B r \to (C H_{3})_{2} C H O H$
Right working
$C H_{3} C H O + C H_{3} M g B r \to (C H_{3})_{2} C H O M g B r$ , then $(C H_{3})_{2} C H O M g B r + H_{2} O \to (C H_{3})_{2} C H O H + M g (O H) B r$
Do this: Always write the adduct, then the hydrolysis with water or dilute acid.
Usually costs: 1 mark.

Exam tips

Write temperatures in kelvin as the paper does (413 K, 443 K, 573 K) and give reagents in full.
In mechanism answers, show curved arrows and mark the slow step.
For distinguishing tests, use the format: reagent → observation with A → observation with B.
In identification problems, use every clue: Lucas (class), iodoform (CH₃CH(OH)–), Tollens' (aldehyde) and molecular formulae.
Mention that esterification is reversible and catalysed by acid; examiners reward the equilibrium arrow.

Practice questions on Reactions of Alcohols

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
Ethanol heated with excess concentrated H₂SO₄ at 443 K mainly gives:
(a) diethyl ether
(b) ethene
(c) ethanal
(d) ethyl hydrogensulphate only
Hint
Higher temperature favours elimination.
Answer
(b) ethene.
Q2 · 2 marks · Short answerBoard level
What is Lucas reagent? How does it distinguish primary, secondary and tertiary alcohols?
Hint
Watch the time taken for turbidity.
Answer
Concentrated HCl with anhydrous ZnCl₂. At room temperature, tertiary alcohols give turbidity immediately, secondary alcohols in about five minutes, and primary alcohols do not give turbidity (they react only on heating).
Q3 · 2 marks · Short answerHarder board level
Explain why alcohols are weaker acids than water.
Hint
Compare RO⁻ and HO⁻.
Answer
The alkyl group releases electrons (+I effect) and increases the electron density on oxygen. This makes the O–H bond less polar and destabilises the alkoxide ion compared with the hydroxide ion, so alcohols lose H⁺ less readily than water.
Q4 · 3 marks · Short answerHarder board level
Write the mechanism of dehydration of ethanol to ethene with concentrated H₂SO₄ at 443 K.
Hint
Three steps; the carbocation step is slow.
Answer
(1) Protonation: $C_{2} H_{5} O H + H^{+} \to C_{2} H_{5} O H_{2}^{+}$ (fast). (2) Loss of water: $C_{2} H_{5} O H_{2}^{+} \to C_{2} H_{5}^{+} + H_{2} O$ (slow). (3) Loss of H⁺ from the adjacent carbon: $C H_{3} C H_{2}^{+} \to C H_{2} = C H_{2} + H^{+}$ (fast).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): 2-Methylpropan-2-ol is not oxidised by acidified potassium dichromate under ordinary conditions.
Reason (R): The carbon atom bearing the –OH group in a tertiary alcohol has no hydrogen atom.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Oxidation removes H from the carbinol carbon.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Convert ethanol into propan-1-ol. Write each step with reagents.
Hint
Add one carbon with a Grignard reagent and methanal.
Answer
$3 C_{2} H_{5} O H + P B r_{3} \to 3 C_{2} H_{5} B r + H_{3} P O_{3}$ ; $C_{2} H_{5} B r + M g dry ether C_{2} H_{5} M g B r$ ; $C_{2} H_{5} M g B r + H C H O \to C H_{3} C H_{2} C H_{2} O M g B r$ ; $C H_{3} C H_{2} C H_{2} O M g B r + H_{2} O \to C H_{3} C H_{2} C H_{2} O H + M g (O H) B r$ .
Q7 · 3 marks · Short answerHOTS / competency
Arrange in decreasing order of acidic strength and explain: methanol, 2-methylpropan-2-ol, ethanol, 2-chloroethanol.
Hint
Electron-releasing groups weaken; electron-withdrawing groups strengthen.
Answer
2-Chloroethanol > methanol > ethanol > 2-methylpropan-2-ol. Cl withdraws electrons (−I), stabilising the alkoxide ion. Alkyl groups release electrons (+I): methanol has none, ethanol one, and the tertiary alcohol three, so the alkoxide becomes progressively less stable (and the bulky tertiary alkoxide is also poorly solvated).
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a perfumer makes ethyl ethanoate, which has a fruity smell, by heating 6.0 g of ethanoic acid (M = 60) with excess ethanol and a few drops of concentrated H₂SO₄, and distils off 5.72 g of the ester (M = 88).
(i) Write the equation. [1]
(ii) Calculate the percentage yield. [1]
(iii) Suggest one way to increase the yield, and explain why it works. [1]
Hint
The reaction is an equilibrium.
Answer
(i) $C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$ . (ii) Acid $= 0.10$ mol, theoretical ester $= 0.10 \times 88 = 8.8$ g; yield $= \frac{5.72}{8.8} \times 100 = 65%$ . (iii) Remove water (or the ester) as it forms, or use more ethanol; by Le Chatelier's principle the equilibrium shifts towards the ester.

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.Composition of Lucas reagent?Show answer
Concentrated HCl and anhydrous ZnCl₂.
2.Product of ethanol with conc. H₂SO₄ at 413 K?Show answer
Diethyl ether (ethoxyethane).
3.Why is SOCl₂ convenient for making alkyl chlorides?Show answer
The by-products SO₂ and HCl are gases and escape, leaving a pure product.
4.Product of propan-2-ol over Cu at 573 K?Show answer
Propanone.
5.Order of reactivity of alcohols with HX?Show answer
3° > 2° > 1°.
6.Which alcohol class is dehydrated (not dehydrogenated) by hot copper?Show answer
Tertiary.

Reactions of Alcohols: frequently asked questions

What is the mechanism of esterification?

The acid's carbonyl oxygen is protonated, the alcohol attacks the carbonyl carbon, a proton transfer makes –OH₂⁺, water leaves and the ester is formed after loss of H⁺.

Why do tertiary alcohols react fastest with HCl?

The reaction goes through a carbocation, and tertiary carbocations are the most stable.

How is a primary alcohol oxidised only to an aldehyde?

Use a mild oxidant such as pyridinium chlorochromate (PCC) or CrO₃ in an anhydrous medium, or pass the vapour over hot copper at 573 K.

What happens when ethanol reacts with sodium?

Sodium ethoxide forms and hydrogen gas is evolved, showing that ethanol is weakly acidic.

How can you distinguish ethanol from methanol?

With the iodoform test: ethanol gives a yellow precipitate of iodoform with I₂ and NaOH; methanol does not.