Is it correct to say: Expecting hydroboration–oxidation to follow Markovnikov's rule.?

It gives the anti-Markovnikov alcohol: propene → propan-1-ol. Acid-catalysed hydration gives propan-2-ol.

Is it correct to say: Using a tertiary alkyl halide in Williamson synthesis.?

A 3° halide + alkoxide gives an alkene by elimination. Put the bulky group on the alkoxide and use a 1° or methyl halide.

Is it correct to say: Writing iodobenzene + methanol as the products of anisole with HI.?

The O–aryl bond does not break. Products are phenol + CH₃I.

Is it correct to say: Writing the acid strength of alcohols as 3° > 2° > 1°.?

It is 1° > 2° > 3°. More alkyl groups (+I effect) destabilise the alkoxide ion.

Is it correct to say: Mixing up the Kolbe and Reimer–Tiemann products.?

Kolbe (phenoxide + CO₂) → salicylic acid (–COOH). Reimer–Tiemann (CHCl₃ + NaOH) → salicylaldehyde (–CHO). Both substitute at the ortho position.

Is it correct to say: Using bromine water to make p-bromophenol.?

Bromine water gives 2,4,6-tribromophenol. For monobromination use Br₂ in CS₂ or CHCl₃ at low temperature.

📄 Formula sheet (PDF)

Alcohols, Phenols and Ethers

Preparation and reactions of alcohols, phenols and ethers — acidity, Lucas test, dehydration mechanism, Kolbe, Reimer–Tiemann, Williamson synthesis and ether cleavage — NCERT Class 12 Chemistry Ch 7

📐 20 formulas✏️ 4 examples🎯 6 practice⚖️ 6 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→The syllabus names the mechanism of dehydration of alcohols. Practise writing all three steps: protonation, carbocation formation (slow step), loss of H⁺.
→Learn each named reaction as reagent → product: Kolbe (phenoxide + CO₂ → –COOH), Reimer–Tiemann (CHCl₃ + NaOH → –CHO), Williamson (alkoxide + 1° alkyl halide), cumene → phenol + acetone.
→Acidity: phenol > water > alcohol. –NO₂ at ortho/para raises the acidity of phenol, –CH₃ lowers it. Always justify with the stability of the phenoxide ion.
→To tell 1°, 2° and 3° alcohols apart, use the Lucas test, or heated Cu at 573 K (aldehyde / ketone / alkene).
→Temperature changes the product: ethanol with conc. H₂SO₄ gives ethoxyethane at 413 K but ethene at 443 K.

📐 Formulas(20)

Acid-Catalysed Hydration of Alkenes

CH_{3} CH=CH_{2} + H_{2} O H^{+} CH_{3} CH(OH)CH_{3}

Hydroboration–Oxidation

CH_{3} CH=CH_{2} (i) B_{2} H_{6} (ii) H_{2} O_{2}, OH^{-} CH_{3} CH_{2} CH_{2} OH

Reduction of Aldehydes and Ketones

RCHO H_{2} / Pd RCH_{2} OH; RCOR^{'} NaBH_{4} RCH(OH)R^{'}

Alcohols from Grignard Reagents

R_{2} C=O + R^{'} MgX \to R_{2} C(OMgX)R^{'} H_{2} O R_{2} C(OH)R^{'} + Mg(OH)X

Phenol from Cumene★ Board fav

C_{6} H_{5} CH(CH_{3})_{2} O_{2} (air) C_{6} H_{5} C(CH_{3})_{2} OOH H^{+} / H_{2} O C_{6} H_{5} OH + CH_{3} COCH_{3}

Phenol from a Diazonium Salt

C_{6} H_{5} NH_{2} NaNO_{2} + HCl 273 - 278 K C_{6} H_{5} N_{2}^{+} Cl^{-} H_{2} O, warm C_{6} H_{5} OH + N_{2} + HCl

Alcohols and Phenols with Active Metals

2 R-OH + 2 Na \to 2 R-O^{-} Na^{+} + H_{2} ↑

Acid Strength — Phenols vs Ethanol

p-nitrophenol (7.1) > phenol (10.0) > p-cresol (10.2) > ethanol (15.9) [p K_{a}]

Lucas Test (1°, 2°, 3° alcohols)★ Board fav

R-OH + HCl anhyd. ZnCl_{2} R-Cl + H_{2} O

Dehydration of Ethanol (443 K)★ Board fav

C_{2} H_{5} OH conc. H_{2} SO_{4} 443 K CH_{2} =CH_{2} + H_{2} O

Oxidation of Alcohols (PCC / CrO₃)

RCH_{2} OH PCC RCHO; R_{2} CHOH CrO_{3} R_{2} CO

Dehydrogenation over Heated Copper (573 K)

RCH_{2} OH Cu 573 K RCHO + H_{2}; R_{2} CHOH Cu 573 K R_{2} CO + H_{2}

Bromination of Phenol

C_{6} H_{5} OH + 3 Br_{2} H_{2} O 2,4,6-Br_{3} C_{6} H_{2} OH ↓ + 3 HBr

Nitration of Phenol

C_{6} H_{5} OH dil. HNO_{3} 298 K o-nitrophenol + p-nitrophenol

Kolbe's Reaction★ Board fav

C_{6} H_{5} O^{-} Na^{+} (i) CO_{2} (ii) H^{+} o-HOC_{6} H_{4} COOH (salicylic acid)

Reimer–Tiemann Reaction★ Board fav

C_{6} H_{5} OH (i) CHCl_{3} + aq. NaOH (ii) H^{+} o-HOC_{6} H_{4} CHO (salicylaldehyde)

Esterification (acetylation)

R-OH + R^{'} COOH ⇌ H^{+} R^{'} COOR + H_{2} O

Williamson Ether Synthesis★ Board fav

R-X + R^{'} -O^{-} Na^{+} \to R-O-R^{'} + NaX

Ethers by Dehydration of Alcohols (413 K)

2 C_{2} H_{5} OH conc. H_{2} SO_{4} 413 K C_{2} H_{5} -O-C_{2} H_{5} + H_{2} O

Cleavage of Ethers by HI

R-O-R^{'} + HI \to R-I + R^{'} -OH; C_{6} H_{5} OCH_{3} + HI \to C_{6} H_{5} OH + CH_{3} I

✏️ Solved Examples

1Solved Exampleeasy2 steps

How will you convert propene into (a) propan-2-ol and (b) propan-1-ol?

(a) Acid-catalysed hydration follows Markovnikov's rule — OH goes to C-2

CH_{3} CH=CH_{2} + H_{2} O H^{+} CH_{3} CH(OH)CH_{3}

2Solved Exampleboard2 steps

Convert propanone into 2-methylpropan-2-ol.

The target (CH₃)₃C–OH is a 3° alcohol with one more carbon, so add CH₃ from a Grignard reagent to the ketone

CH_{3} COCH_{3} + CH_{3} MgBr dry ether (CH_{3})_{3} C-OMgBr

3Solved Exampleboard3 steps

Convert phenol into aspirin (acetylsalicylic acid).

Make sodium phenoxide with NaOH

C_{6} H_{5} OH + NaOH \to C_{6} H_{5} O^{-} Na^{+} + H_{2} O

4Solved ExampleHOTS3 steps

Two students try to make 2-methoxy-2-methylpropane, (CH₃)₃C–O–CH₃, by Williamson synthesis. A uses (CH₃)₃C–Br + CH₃ONa; B uses (CH₃)₃C–ONa + CH₃Br. Which route works? What forms when the ether is heated with HI?

Route A: a 3° halide with a strong base undergoes elimination, not substitution

(CH_{3})_{3} C-Br + CH_{3} ONa \to (CH_{3})_{2} C=CH_{2} + CH_{3} OH + NaBr

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Expecting hydroboration–oxidation to follow Markovnikov's rule.
✓It gives the anti-Markovnikov alcohol: propene → propan-1-ol. Acid-catalysed hydration gives propan-2-ol.
2
Using a tertiary alkyl halide in Williamson synthesis.
✓A 3° halide + alkoxide gives an alkene by elimination. Put the bulky group on the alkoxide and use a 1° or methyl halide.
3
Writing iodobenzene + methanol as the products of anisole with HI.
✓The O–aryl bond does not break. Products are phenol + CH₃I.
4
Writing the acid strength of alcohols as 3° > 2° > 1°.
✓It is 1° > 2° > 3°. More alkyl groups (+I effect) destabilise the alkoxide ion.
5
Mixing up the Kolbe and Reimer–Tiemann products.
✓Kolbe (phenoxide + CO₂) → salicylic acid (–COOH). Reimer–Tiemann (CHCl₃ + NaOH) → salicylaldehyde (–CHO). Both substitute at the ortho position.
6
Using bromine water to make p-bromophenol.
✓Bromine water gives 2,4,6-tribromophenol. For monobromination use Br₂ in CS₂ or CHCl₃ at low temperature.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Write the product when phenol is treated with bromine water.
Q2
Arrange in increasing order of acid strength: ethanol, phenol, p-nitrophenol, p-cresol.
Q3
Give a chemical test to distinguish butan-1-ol from 2-methylpropan-2-ol.
Q4
What are the products when anisole is heated with HI?
Q5
Ethanol is heated with conc. H₂SO₄ at (a) 413 K and (b) 443 K. Name the main product in each case.
Q6
Name the reaction and the product when phenol is treated with CHCl₃ and aqueous NaOH, followed by acid.

📝 Notes

Alcohols, Phenols and Ethers

All three classes contain a C–O single bond. Alcohols have –OH on an sp³ carbon, phenols have –OH on a benzene ring, and ethers have O between two carbon groups. Most questions test which bond breaks and why.

Which bond breaks?

O–H cleavage: reaction with Na, acidity, esterification. Phenols do this more readily than alcohols.
C–O cleavage (alcohols only): reaction with HCl/ZnCl₂ (Lucas test) and dehydration. Both are easiest for alcohols that form stable carbocations, so the order is 3° > 2° > 1°.
Ring substitution (phenols): –OH activates the ortho and para positions, so bromination and nitration happen under mild conditions, and phenoxide is reactive enough to substitute with the weak electrophile CO₂ (Kolbe). Reimer–Tiemann also substitutes at the ortho position.

Phenol is acidic, alcohol hardly

Phenoxide spreads its negative charge over the ring, so phenol (pKa 10.0) is far more acidic than ethanol (pKa 15.9). Groups that pull electrons away (–NO₂) stabilise the phenoxide further; groups that push electrons in (–CH₃) do the opposite. Phenol reacts with NaOH but ethanol does not.

Ethers: make with Williamson, break with HI

In Williamson synthesis the alkyl halide should be primary: with a 2° halide elimination competes, and a 3° halide gives only the alkene. When an ether is cleaved by HI, decide the mechanism first: with 1°/2° groups (SN2) iodide goes to the smaller group; with a 3° group (SN1) iodide goes to the 3° carbon; an aryl group always stays as phenol.

Conversion questions

Count the carbons in the start and target, find the functional-group change, then pick the reagent: a Grignard reagent adds carbon, PCC stops at the aldehyde, and B₂H₆ or H₂O/H⁺ decides which alcohol forms from an alkene.