ChemistryCBSE Class 12 · Alcohols, Phenols and Ethers

Reactions of Alcohols: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Alcohols react in two ways: by breaking the O–H bond (acidity with sodium, esterification) or the C–O bond (with HX, $PCl_5$, $SOCl_2$, and dehydration to alkenes). Tertiary alcohols react fastest with HX and on dehydration, and resist oxidation. Primary alcohols oxidise to aldehydes, then acids; secondary to ketones.

Ask which bond breaks: O–H gives alkoxides and esters, C–O gives alkyl halides and alkenes; tertiary reacts fastest by C–O, primary is most acidic.

Illustration drawn for QuestionBuddy.

Understand Reactions of Alcohols

An alcohol molecule $R - O - H$ has two polar bonds. Its reactions either break O–H (the alcohol acts as a nucleophile or a weak acid) or break C–O (the alcohol behaves as an electrophile once the OH is protonated or converted into a better leaving group).

1. Reactions with cleavage of O–H.

*Acidity.* Alcohols react with active metals such as sodium to give alkoxides and hydrogen: $2 R - O H + 2 N a \to 2 R - O^{-} N a^{+} + H_{2}$ . Their acidic strength is in the order $1° > 2° > 3°$ : alkyl groups push electron density onto oxygen (+I effect), which makes the O–H bond less polar and the alkoxide ion less stable. Alcohols are weaker acids than water, so an alkoxide ion is a stronger base than hydroxide: sodium ethoxide reacts with water to regenerate ethanol and NaOH.

*Esterification.* Alcohols react with carboxylic acids, acid chlorides and anhydrides to form esters. With a carboxylic acid the reaction is reversible and needs a little concentrated $H_{2} S O_{4}$ : $R - O H + R^{'} C O O H ⇌ R^{'} C O O R + H_{2} O$ ; water is removed as it forms to push the equilibrium forward. With an acid chloride, a base such as pyridine is added to neutralise the HCl. Acetylation of salicylic acid gives aspirin.

2. Reactions with cleavage of C–O.

*With hydrogen halides:* $R O H + H X \to R X + H_{2} O$ . The Lucas test uses concentrated HCl with anhydrous $Z n C l_{2}$ . Alkyl chlorides are insoluble, so the solution turns cloudy. Tertiary alcohols give turbidity immediately; primary alcohols give none at room temperature. Reactivity follows carbocation stability: $3° > 2° > 1°$ .

*With phosphorus halides and thionyl chloride:* $3 R O H + P B r_{3} \to 3 R B r + H_{3} P O_{3}$ ; $R O H + P C l_{5} \to R C l + P O C l_{3} + H C l$ ; $R O H + S O C l_{2} \to R C l + S O_{2} + H C l$ (the by-products are gases, so the alkyl chloride is easy to purify).

*Dehydration:* heating with a protic acid removes water and gives an alkene. Ethanol needs concentrated $H_{2} S O_{4}$ at 443 K; a secondary alcohol, 85% phosphoric acid at 440 K; a tertiary alcohol only 20% phosphoric acid at 358 K. So ease of dehydration is $3° > 2° > 1°$ . The mechanism has three steps: (i) protonation of the alcohol, $C_{2} H_{5} O H + H^{+} ⇌ C_{2} H_{5} O H_{2}^{+}$ (fast); (ii) loss of water to form a carbocation, the slow, rate-determining step; (iii) loss of a proton from the neighbouring carbon to give the alkene. Where more than one alkene can form, the more substituted alkene is the major product.

3. Oxidation and dehydrogenation. Oxidation removes hydrogen from the O–H and from the carbon carrying the OH (α-hydrogen). A primary alcohol gives an aldehyde, which is easily oxidised further to a carboxylic acid by strong oxidants such as acidified $K M n O_{4}$ or $K_{2} C r_{2} O_{7}$ . To stop at the aldehyde, use $C r O_{3}$ in an anhydrous medium or PCC (pyridinium chlorochromate). A secondary alcohol gives a ketone (for example with chromic anhydride, $C r O_{3}$ ). A tertiary alcohol has no α-hydrogen and resists oxidation; under strong conditions its C–C bonds break and a mixture of acids with fewer carbons forms. With hot copper at 573 K, primary and secondary alcohols are dehydrogenated to aldehydes and ketones, while tertiary alcohols are dehydrated to alkenes: 2-methylpropan-2-ol gives 2-methylpropene.

Reactions of Alcohols: formulas and key facts

Acidic nature: $2 R - O H + 2 N a \to 2 R - O N a + H_{2}$; Acidity: $1° > 2° > 3°$ ; alcohols are weaker acids than water.
Esterification: $R - O H + R^{'} C O O H ⇌ H^{+} R^{'} C O O R + H_{2} O$; Reversible; remove water to increase yield.
Halogenation of alcohols: $3 R O H + P B r_{3} \to 3 R B r + H_{3} P O_{3}; R O H + S O C l_{2} \to R C l + S O_{2} + H C l$; Lucas test: conc. HCl/ $Z n C l_{2}$ ; 3° turbid at once, 1° not at room temperature.
Dehydration conditions: $1° : conc. H_{2} S O_{4}, 443 K; 2° : 85% H_{3} P O_{4}, 440 K; 3° : 20% H_{3} P O_{4}, 358 K$; Ease: $3° > 2° > 1°$ (carbocation stability).
Oxidation: $1° R C H_{2} O H P C C R C H O K M n O_{4} R C O O H; 2° R_{2} C H O H C r O_{3} R_{2} C O$; 3° alcohols resist oxidation (no α-H).
Dehydrogenation over Cu, 573 K: $1° \to aldehyde; 2° \to ketone; 3° \to alkene$; Tertiary alcohols lose water instead of hydrogen.

How to solve Reactions of Alcohols questions

Most questions give an alcohol and a reagent, ask for a mechanism, or ask you to identify an alcohol from test results. Start by classifying the alcohol.

1Classify the alcohol: count the carbon atoms attached to the carbon bearing OH (1°, 2°, 3°).
2Identify the reagent: Na (O–H cleavage, $H_{2}$ ), acid/acid chloride (ester), HX/ $Z n C l_{2}$ , $P X_{3}$ , $P C l_{5}$ , $S O C l_{2}$ (alkyl halide), conc. acid + heat (alkene), oxidant (aldehyde, ketone, acid), hot Cu (dehydrogenation).
3For oxidation, check α-hydrogens: two (1°) → aldehyde or acid; one (2°) → ketone; none (3°) → no reaction under normal conditions.
4For dehydration, find the β-hydrogens and choose the more substituted alkene as the major product.
5For mechanisms, write protonation, carbocation formation (slow) and deprotonation, with curved-arrow descriptions in words.
6For identification puzzles, match each observation: Lucas speed gives 1°/2°/3°; oxidation product gives the carbon skeleton; chirality narrows the choice.
7Balance equations and write conditions (temperature, catalyst) above the arrow.

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.

Alcohols are weaker acids than water

Suppose a laboratory assistant must safely destroy a few small scraps of sodium metal left after an experiment. Instead of dropping them into water, a senior chemist advises adding them slowly to a beaker of ethanol.

Write the reaction of sodium with ethanol, and explain why it is gentler than the reaction with water.

Try it first, then show the solutionSolution

$2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ : sodium ethoxide and hydrogen form.
Water: $2 H_{2} O + 2 N a \to 2 N a O H + H_{2}$ , which is very vigorous.
The ethyl group releases electrons (+I) onto oxygen, making the O–H bond of ethanol less polar; ethanol is a weaker acid than water.
So hydrogen is released more slowly and the heat is produced gradually, reducing the risk of fire.

Answer: $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ ; ethanol is a weaker proton donor than water, so the reaction is slower and safer.

Memory hook: Alkyl groups calm the O–H: ethanol gives up its proton less eagerly than water.

Tricks to remember Reactions of Alcohols

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

Primary Prefers to go all the way
1° → aldehyde → acid with strong oxidants; only PCC (or $C r O_{3}$ anhydrous) stops at the aldehyde. 2° stops at ketone. 3° says no.
Use it when: Oxidation questions.
3° first for HX and heat, last for acidity and oxidation
Tertiary: fastest Lucas, easiest dehydration; weakest acid and no oxidation.
Use it when: Ordering questions.
P-C-D for dehydration mechanism
Protonation, Carbocation (slow), Deprotonation.
Use it when: Writing the 3-step mechanism.

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) Propan-1-ol (b) Butan-2-ol (c) 2-Methylpropan-2-ol (d) Methanol

Try it first, then show the solutionSolution

Lucas test speed follows carbocation stability; 2-methylpropan-2-ol is tertiary.

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

Example 2: Two oxidising agents · 2 marks · Very short answerBoard level

What is formed when propan-1-ol is oxidised with (a) PCC (b) acidified potassium permanganate? Write the structures.

Try it first, then show the solutionSolution

(a) PCC is a mild oxidant that stops at the aldehyde: $C H_{3} C H_{2} C H_{2} O H \to C H_{3} C H_{2} C H O$ (propanal) (1 mark).
(b) Acidified $K M n O_{4}$ oxidises fully to the carboxylic acid: $C H_{3} C H_{2} C O O H$ (propanoic acid) (1 mark).

Answer: (a) Propanal, $C H_{3} C H_{2} C H O$ . (b) Propanoic acid, $C H_{3} C H_{2} C O O H$ .

Example 3: Mechanism of dehydration of a secondary alcohol · 3 marks · Short answerHarder board level

Write the mechanism of the acid-catalysed dehydration of propan-2-ol to propene. Why does this need milder conditions (85% $H_{3} P O_{4}$ , 440 K) than ethanol (conc. $H_{2} S O_{4}$ , 443 K)?

Try it first, then show the solutionSolution

Step 1 (fast): protonation of the OH: $(C H_{3})_{2} C H - O H + H^{+} ⇌ (C H_{3})_{2} C H - O H_{2}^{+}$ (½ mark).
Step 2 (slow, rate-determining): loss of water gives a secondary carbocation: $(C H_{3})_{2} C H - O H_{2}^{+} \to (C H_{3})_{2} C H^{+} + H_{2} O$ (1 mark).
Step 3 (fast): loss of a proton from a neighbouring carbon: $C H_{3} - C H^{+} - C H_{3} \to C H_{3} C H = C H_{2} + H^{+}$ ; the acid is regenerated (½ mark).
The slow step forms a carbocation; a secondary carbocation is more stable than a primary one (inductive effect and hyperconjugation of two methyl groups), so it forms more easily and milder conditions suffice (1 mark).

Answer: Protonation → secondary carbocation (slow) → loss of $H^{+}$ to $C H_{3} C H = C H_{2}$ ; the more stable 2° carbocation makes dehydration easier than for ethanol.

Example 4: Three reasons about alcohol class · 3 marks · Short answerHarder board level

Give reasons: (a) The acidic strength of alcohols is in the order $1° > 2° > 3°$ . (b) Tertiary alcohols are dehydrated most easily. (c) Tertiary alcohols are not oxidised by acidified $K_{2} C r_{2} O_{7}$ under normal conditions.

Try it first, then show the solutionSolution

(a) Alkyl groups release electrons (+I), raising electron density on oxygen; the O–H bond becomes less polar and the alkoxide less stable. More alkyl groups (3°) means weaker acid (1 mark).
(b) Dehydration goes through a carbocation; the tertiary carbocation is the most stable (inductive effect and hyperconjugation), so it forms fastest (1 mark).
(c) Oxidation needs a hydrogen on the carbon bearing OH; tertiary alcohols have none, so the C–OH carbon cannot form C=O without breaking C–C bonds (1 mark).

Answer: (a) +I effect of alkyl groups. (b) Most stable 3° carbocation. (c) No α-hydrogen.

Example 5: Identifying four C₄H₁₀O alcohols · 5 marks · Long answerHOTS / competency

Four alcohols A, B, C and D all have the formula $C_{4} H_{10} O$ . A gives turbidity immediately with Lucas reagent and is not oxidised by acidified dichromate. B is chiral and is oxidised to a ketone. C and D are oxidised to carboxylic acids; C gives 2-methylpropanoic acid and D gives butanoic acid. Identify A to D, and write what A and B give with hot copper at 573 K.

Try it first, then show the solutionSolution

A: tertiary (immediate Lucas, no oxidation): 2-methylpropan-2-ol, $(C H_{3})_{3} C O H$ (1 mark).
B: secondary (ketone) and chiral: butan-2-ol, $C H_{3} C H (O H) C H_{2} C H_{3}$ (C2 has H, OH, $C H_{3}$ , $C_{2} H_{5}$ ) (1 mark).
C: primary with a branched chain: 2-methylpropan-1-ol, $(C H_{3})_{2} C H C H_{2} O H$ → $(C H_{3})_{2} C H C O O H$ (1 mark).
D: primary, straight chain: butan-1-ol, $C H_{3} C H_{2} C H_{2} C H_{2} O H$ → $C H_{3} C H_{2} C H_{2} C O O H$ (1 mark).
Hot Cu, 573 K: A (3°) is dehydrated to 2-methylpropene, $(C H_{3})_{2} C = C H_{2}$ ; B (2°) is dehydrogenated to butanone, $C H_{3} C O C H_{2} C H_{3}$ (1 mark).

Answer: A 2-methylpropan-2-ol; B butan-2-ol; C 2-methylpropan-1-ol; D butan-1-ol. With Cu/573 K: A → 2-methylpropene, B → butanone.

Where students lose marks in Reactions of Alcohols

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

Mistake: Oxidising a tertiary alcohol to a ketone
Do this: Tertiary alcohols have no α-H; write 'no reaction under ordinary conditions'.
Usually costs: 1 mark.
Mistake: Reversing the acidity order
Wrong working
Acidity: 3° > 2° > 1°
Right working
Acidity: 1° > 2° > 3°
Do this: More alkyl groups mean more +I effect and a weaker acid.
Usually costs: 1 mark.
Mistake: Missing the slow step in the dehydration mechanism
Do this: Label carbocation formation as the slow, rate-determining step.
Usually costs: ½ to 1 mark.
Mistake: Writing an aldehyde as the final product with acidified KMnO₄
Do this: Strong oxidants take a primary alcohol to the carboxylic acid; only PCC or anhydrous $C r O_{3}$ stops at the aldehyde.
Usually costs: 1 mark.

Exam tips

State the class of the alcohol before every reaction; most answers depend on it.
Give reagents with conditions: 'conc. $H_{2} S O_{4}$ , 443 K', 'Cu, 573 K', 'conc. HCl + anhyd. $Z n C l_{2}$ '.
In the dehydration mechanism, show the regeneration of $H^{+}$ to show catalysis.
For identification puzzles, write one line linking each observation to the conclusion.
Use PCC for 'convert a primary alcohol into an aldehyde' questions.

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
Which reagent converts a primary alcohol to an aldehyde without further oxidation?
(a) Acidified $K M n O_{4}$ (b) Acidified $K_{2} C r_{2} O_{7}$ (c) PCC (d) Hot conc. $H_{2} S O_{4}$
Hint
A mild, anhydrous oxidant.
Answer
(c) PCC.
Q2 · 2 marks · Very short answerBoard level
Write equations for the reaction of ethanol with (a) sodium metal and (b) thionyl chloride. Why is thionyl chloride preferred for making alkyl chlorides?
Hint
Look at the by-products.
Answer
(a) $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ . (b) $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$ . The by-products $S O_{2}$ and HCl are gases that escape, leaving pure alkyl chloride.
Q3 · 2 marks · Very short answerHarder board level
What happens when (a) propan-2-ol and (b) 2-methylpropan-2-ol are passed over heated copper at 573 K? Write equations.
Hint
Secondary dehydrogenates; tertiary dehydrates.
Answer
(a) $C H_{3} C H (O H) C H_{3} \to C H_{3} C O C H_{3} + H_{2}$ (propanone). (b) $(C H_{3})_{3} C O H \to (C H_{3})_{2} C = C H_{2} + H_{2} O$ (2-methylpropene).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Alcohols are weaker acids than water.
Reason (R): The alkyl group releases electrons and increases the electron density on oxygen, decreasing the polarity of the O–H bond.
(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
+I effect.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
How would you distinguish propan-1-ol from 2-methylpropan-2-ol using the Lucas test? Explain the chemistry behind the observation.
Hint
Turbidity comes from an insoluble alkyl chloride.
Answer
Add conc. HCl/anhydrous $Z n C l_{2}$ . 2-Methylpropan-2-ol gives turbidity at once (stable 3° carbocation forms quickly, giving insoluble $(C H_{3})_{3} C C l$ ); propan-1-ol gives no turbidity at room temperature (1° carbocation does not form easily).
Q6 · 3 marks · Short answerHarder board level
Write the equation for the esterification of ethanol with ethanoic acid. Why is a little concentrated $H_{2} S O_{4}$ added, and how can the yield of ester be increased?
Hint
Reversible reaction.
Answer
$C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$ . $H_{2} S O_{4}$ acts as a catalyst (and absorbs water). Because the reaction is reversible, removing water as it forms (or using excess of one reactant) shifts the equilibrium towards the ester.
Q7 · 3 marks · Short answerHOTS / competency
Butan-2-ol is heated with concentrated $H_{2} S O_{4}$ . (a) Name the possible alkenes and the major product. (b) Show the carbocation formed. (c) Why is this alkene the major one?
Hint
β-hydrogens on both sides.
Answer
(a) But-1-ene and but-2-ene; but-2-ene is major. (b) $C H_{3} - C H^{+} - C H_{2} C H_{3}$ (secondary carbocation) after protonation and loss of water. (c) Loss of H⁺ from C3 gives the more substituted, more stable alkene (but-2-ene), which is preferred (Saytzeff rule).
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student warms three alcohols separately with acidified potassium dichromate solution: ethanol, propan-2-ol and 2-methylpropan-2-ol. With the first two, the orange solution turns green; with the third, it stays orange.
(a) What causes the colour change from orange to green? (1 mark)
(b) Name the organic products from ethanol if the product is distilled off as it forms, and if the mixture is refluxed for a long time. (1 mark)
(c) Name the product from propan-2-ol. (1 mark)
(d) Why does 2-methylpropan-2-ol not react? (1 mark)
Hint
$C r_{2} O_{7}^{2 -}$ is reduced to $C r^{3 +}$ .
Answer
(a) Orange $C r_{2} O_{7}^{2 -}$ (Cr +6) is reduced to green $C r^{3 +}$ as the alcohol is oxidised. (b) Distilled at once: ethanal ( $C H_{3} C H O$ ); refluxed: ethanoic acid ( $C H_{3} C O O H$ ). (c) Propanone. (d) It is tertiary: the carbon bearing OH has no hydrogen, so it cannot be oxidised under these conditions.

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.Lucas reagent?Show answer
Conc. HCl with anhydrous $Z n C l_{2}$ .
2.Conditions to dehydrate ethanol to ethene?Show answer
Conc. $H_{2} S O_{4}$ at 443 K.
3.Product of a 2° alcohol over hot Cu?Show answer
A ketone.
4.Reagent that stops oxidation at the aldehyde?Show answer
PCC (or $C r O_{3}$ in anhydrous medium).
5.Rate-determining step in alcohol dehydration?Show answer
Formation of the carbocation.
6.Acid strength order of alcohols?Show answer
$1° > 2° > 3°$ .

Reactions of Alcohols: frequently asked questions

What are the main reactions of alcohols in Class 12?

Reactions breaking O–H (with sodium, esterification) and reactions breaking C–O (with HX, PX_3, PCl_5, SOCl_2, and dehydration), plus oxidation and dehydrogenation.

Why do tertiary alcohols react fastest with Lucas reagent?

They form the most stable (tertiary) carbocation, so the alkyl chloride forms almost immediately and makes the solution cloudy.

Why are tertiary alcohols not oxidised easily?

The carbon carrying the OH group has no hydrogen atom, so it cannot form a C=O bond without breaking a C–C bond.

What is the mechanism of dehydration of alcohols?

Protonation of the OH group, slow loss of water to form a carbocation, and loss of a proton from the neighbouring carbon to form the alkene.

Why are alcohols weaker acids than water?

The alkyl group releases electrons towards oxygen, making the O–H bond less polar and the alkoxide ion less stable than hydroxide.