Important Questions Class 12 Chemistry Chapter 7: Alcohols, Phenols and Ethers

Alcohols, phenols and ethers are organic compounds containing oxygen-linked functional groups such as –OH, alkoxy and aryloxy groups.
Important Questions Class 12 Chemistry Chapter 7 help students revise nomenclature, preparation, reactions, mechanisms and conversions.

Class 12 Chemistry Chapter 7 connects functional group chemistry with practical organic reactions used in medicines, detergents, solvents, antiseptics and fragrances. Alcohols contain –OH attached to an aliphatic carbon, phenols contain –OH attached to an aromatic ring, and ethers contain an alkoxy or aryloxy group. Students should revise IUPAC naming, classification, preparation methods, hydrogen bonding, acidity of phenols, Lucas test, oxidation, dehydration, Kolbe reaction, Reimer-Tiemann reaction, Williamson synthesis and ether cleavage carefully. This chapter is important for 2026-27 because CBSE often asks conversions, reasoning questions, mechanisms and product-prediction questions.

Key Takeaways

• Alcohols: Alcohols contain one or more –OH groups attached to aliphatic carbon atoms.
• Phenols: Phenols are more acidic than alcohols because phenoxide ion is resonance stabilised.
• Ethers: Ethers can be prepared by acid dehydration of alcohols or Williamson synthesis.
• Reactions: Phenols undergo electrophilic substitution mainly at ortho and para positions.

Important Questions Class 12 Chemistry Chapter 7 Structure 2026-27

Section A: MCQs from Important Questions Class 12 Chemistry Chapter 7

Section A tests nomenclature, classification, acidity, boiling points, reaction conditions and product prediction. Alcohols phenols and ethers important questions in this section usually need exact NCERT facts.

Q1. Alcohols contain the –OH group attached to:

1. Carbon atom of an aliphatic system
   b. Only nitrogen atom
   c. Only oxygen atom
   d. Only aromatic oxygen

Answer: a. Carbon atom of an aliphatic system

Alcohols contain hydroxyl group attached to aliphatic carbon.

Q2. Phenols contain –OH group attached directly to:

1. sp³ carbon of alkyl group
   b. sp² carbon of aromatic ring
   c. Carbonyl carbon
   d. Ether oxygen

Answer: b. sp² carbon of aromatic ring

Phenol has –OH directly attached to benzene ring carbon.

Q3. Ethers contain the functional group:

1. R–OH
   b. Ar–OH
   c. R–O–R or R–O–Ar
   d. R–CHO

Answer: c. R–O–R or R–O–Ar

Ethers have alkoxy or aryloxy linkage.

Q4. The IUPAC name of CH3OH is:

1. Methyl alcohol
   b. Methanol
   c. Ethanol
   d. Methoxy methane

Answer: b. Methanol

The suffix “ol” is used for alcohols.

Q5. The IUPAC name of C6H5OCH3 is:

1. Ethoxybenzene
   b. Methoxybenzene
   c. Phenoxyethane
   d. Benzyl alcohol

Answer: b. Methoxybenzene

C6H5OCH3 is also called anisole.

Q6. Primary alcohols on oxidation with strong oxidising agents give:

1. Ethers
   b. Carboxylic acids
   c. Alkenes only
   d. Phenols

Answer: b. Carboxylic acids

Acidified KMnO4 can oxidise primary alcohols directly to acids.

Q7. Secondary alcohols on oxidation generally form:

1. Aldehydes
   b. Ketones
   c. Carboxylic acids directly
   d. Ethers only

Answer: b. Ketones

Secondary alcohols are oxidised to ketones.

Q8. Tertiary alcohols resist oxidation because they:

1. Have no oxygen atom
   b. Lack α-hydrogen on the carbon bearing –OH
   c. Are always gases
   d. Cannot form carbocations

Answer: b. Lack α-hydrogen on the carbon bearing –OH

Strong oxidising conditions may cause C–C bond cleavage.

Q9. Lucas reagent is:

1. Conc. HCl and anhydrous ZnCl2
   b. Dilute HCl and NaOH
   c. Br2 water only
   d. NaBH4 and LiAlH4

Answer: a. Conc. HCl and anhydrous ZnCl2

Lucas test class 12 distinguishes primary, secondary and tertiary alcohols.

Q10. In Lucas test, tertiary alcohols show turbidity:

1. Immediately
   b. After heating only
   c. Never
   d. After several days

Answer: a. Immediately

Tertiary alcohols form alkyl halides easily.

Q11. Phenol is more acidic than ethanol because:

1. Phenol is less soluble
   b. Phenoxide ion is resonance stabilised
   c. Ethanol has no oxygen
   d. Ethoxide ion is resonance stabilised

Answer: b. Phenoxide ion is resonance stabilised

Delocalisation of negative charge stabilises phenoxide ion.

Q12. Phenol reacts with bromine water to form:

1. Bromobenzene
   b. 2,4,6-tribromophenol
   c. Nitrobenzene
   d. Benzaldehyde

Answer: b. 2,4,6-tribromophenol

A white precipitate of 2,4,6-tribromophenol forms.

Q13. Kolbe reaction of sodium phenoxide gives mainly:

1. Benzoic acid
   b. Salicylic acid
   c. Benzyl alcohol
   d. Anisole

Answer: b. Salicylic acid

Carbon dioxide attacks mainly at the ortho position.

Q14. Reimer-Tiemann reaction introduces which group in phenol?

1. –COOH
   b. –CHO
   c. –NO2
   d. –Br

Answer: b. –CHO

Phenol gives salicylaldehyde with CHCl3 and NaOH.

Q15. Assertion: Alcohols have higher boiling points than ethers of comparable molecular masses.

Reason: Alcohols form intermolecular hydrogen bonds.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

Intermolecular hydrogen bonding raises boiling point.

Q16. Assertion: Williamson synthesis gives better results with primary alkyl halides.

Reason: Secondary and tertiary alkyl halides favour elimination with alkoxides.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

Alkoxides are strong bases as well as nucleophiles.

Section B: Very Short Answer Questions from Class 12 Chemistry Chapter 7 Important Questions

Section B questions usually test definitions, naming, reasons and direct product prediction. Keep answers brief and use correct reaction terms.

Q17. Define alcohol, phenol and ether.

Alcohol is an organic compound in which –OH group is attached to an aliphatic carbon atom.

Phenol is an organic compound in which –OH group is attached directly to an aromatic ring carbon.

Ether is an organic compound in which oxygen is bonded to two alkyl or aryl groups.

Q18. What are simple and mixed ethers?

Simple ethers have the same alkyl or aryl groups on both sides of oxygen.

Example: C2H5OC2H5 is ethoxyethane.

Mixed ethers have different alkyl or aryl groups on oxygen.

Example: CH3OC2H5 is methoxyethane.

Q19. Why is phenol more acidic than ethanol?

Phenol is more acidic than ethanol because phenoxide ion is resonance stabilised.

In ethanol, ethoxide ion has the negative charge localised on oxygen. This makes ethoxide ion less stable than phenoxide ion.

Q20. What is Williamson synthesis class 12?

Williamson synthesis class 12 is a method of preparing ethers by reacting an alkyl halide with sodium alkoxide or sodium phenoxide.

General reaction:

R–X + NaOR′ → R–OR′ + NaX

It gives best results with primary alkyl halides.

Q21. What is denatured alcohol?

Denatured alcohol is ethanol made unfit for drinking by adding substances such as copper sulphate and pyridine.

Copper sulphate gives colour, and pyridine gives a foul smell.

Section C: Short Answer Questions from Alcohols, Phenols and Ethers Important Questions

Section C includes product prediction, reasoning, comparison and named reactions. Use equations wherever possible.

Q22. Explain hydroboration-oxidation of alkenes with an example.

Hydroboration-oxidation converts alkenes into alcohols.

In the first step, diborane adds to the alkene. In the second step, oxidation with hydrogen peroxide in alkaline medium gives alcohol.

Example:

CH3–CH=CH2 + BH3, followed by H2O2/OH⁻ → CH3–CH2–CH2OH

The alcohol forms opposite to Markovnikov addition.

Q23. How are alcohols prepared from aldehydes and ketones?

Aldehydes and ketones are reduced to alcohols.

Aldehydes give primary alcohols.

R–CHO + 2[H] → R–CH2OH

Ketones give secondary alcohols.

R–CO–R′ + 2[H] → R–CHOH–R′

Common reducing agents are NaBH4, LiAlH4 or H2 with Ni, Pt or Pd.

Q24. Explain acidity of phenols class 12.

Acidity of phenols class 12 depends on stability of phenoxide ion.

Phenol ionises to give phenoxide ion. The negative charge in phenoxide ion is delocalised over the benzene ring by resonance. This makes phenoxide ion stable.

Electron-withdrawing groups such as –NO2 increase phenol acidity, especially at ortho and para positions. Electron-releasing groups such as –CH3 decrease acidity.

Q25. Give the order of dehydration of alcohols and explain the reason.

The ease of dehydration follows:

Tertiary alcohol > Secondary alcohol > Primary alcohol

Dehydration of alcohols class 12 usually occurs through carbocation formation in acid medium.

Tertiary carbocations are more stable than secondary and primary carbocations. Hence, tertiary alcohols dehydrate under milder conditions.

Q26. Distinguish between ethanol and phenol using chemical tests.

Phenol is more acidic and more reactive towards electrophilic substitution.

Q27. Explain Kolbe reaction class 12.

Kolbe reaction class 12 is the reaction of sodium phenoxide with carbon dioxide under pressure, followed by acidification.

Sodium phenoxide reacts with CO2 to form sodium salicylate. Acidification gives salicylic acid.

Reaction:

C6H5ONa + CO2 → o-HOC6H4COONa

o-HOC6H4COONa + H+ → o-HOC6H4COOH

The product is ortho-hydroxybenzoic acid or salicylic acid.

Q28. Explain Reimer-Tiemann reaction class 12.

Reimer-Tiemann reaction class 12 introduces a formyl group at the ortho position of phenol.

Phenol reacts with chloroform and aqueous sodium hydroxide. After hydrolysis and acidification, salicylaldehyde is obtained.

Reaction:

C6H5OH + CHCl3 + NaOH → o-HOC6H4CHO

The main product is salicylaldehyde.

Section D: Case-Based Questions from Class 12 Chemistry Alcohols Phenols and Ethers

Case-based questions from this chapter usually combine reagent choice, product prediction and reaction reasoning. Identify the functional group first.

Q29. Case Study: Identifying Three Alcohols

A student is given three alcohols: butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. The teacher asks the student to use Lucas reagent.

Q29(a). Which reagent should be used?

Lucas reagent should be used.

It contains concentrated HCl and anhydrous ZnCl2.

Q29(b). Which alcohol gives immediate turbidity?

2-methylpropan-2-ol gives immediate turbidity.

It is a tertiary alcohol.

Q29(c). Which alcohol gives turbidity slowly?

Butan-2-ol gives turbidity slowly.

It is a secondary alcohol.

Q29(d). Which alcohol does not give turbidity at room temperature?

Butan-1-ol does not give turbidity at room temperature.

It is a primary alcohol.

Q30. Case Study: Choosing a Method for Ether Preparation

A student wants to prepare ethoxybenzene. The teacher suggests using phenol and ethyl bromide in the presence of a base.

Q30(a). Which ether is being prepared?

Ethoxybenzene is being prepared.

It is also called phenetole.

Q30(b). Which named reaction is used?

Williamson synthesis is used.

Phenoxide ion reacts with ethyl bromide.

Q30(c). Why is ethyl bromide suitable?

Ethyl bromide is a primary alkyl halide.

Primary halides favour SN2 substitution with phenoxide ion.

Q30(d). Why should tertiary alkyl halide be avoided?

Tertiary alkyl halide favours elimination.

Alkoxides and phenoxides are strong bases, so alkenes may form instead of ethers.

Section E: Long Answer Questions from Important Questions Class 12 Chemistry Chapter 7

Long answers in this chapter usually ask for preparation methods, mechanisms and reaction comparisons. Write equations and reasons clearly.

Q31. Explain the preparation of phenols class 12 by four methods.

Phenols can be prepared from haloarenes, benzene sulphonic acid, diazonium salts and cumene.

1. From haloarenes:

Chlorobenzene is fused with NaOH at 623 K and 320 atm.

C6H5Cl + NaOH → C6H5ONa

C6H5ONa + H+ → C6H5OH

1. From benzene sulphonic acid:

Benzene is first sulphonated with oleum to form benzene sulphonic acid. It is then fused with molten NaOH to form sodium phenoxide. Acidification gives phenol.

1. From diazonium salts:

Benzene diazonium chloride is warmed with water to give phenol.

C6H5N2+Cl⁻ + H2O → C6H5OH + N2 + HCl

1. From cumene:

Cumene is oxidised in air to cumene hydroperoxide. Acid treatment gives phenol and acetone.

Cumene → Cumene hydroperoxide → Phenol + Acetone

Most worldwide phenol production uses the cumene method.

Q32. Explain the chemical reactions of alcohols.

Alcohols show reactions involving O–H bond cleavage and C–O bond cleavage.

1. Reaction with sodium:

Alcohols react with sodium to form sodium alkoxide and hydrogen gas.

2ROH + 2Na → 2RONa + H2

1. Esterification:

Alcohols react with carboxylic acids, acid chlorides or acid anhydrides to form esters.

R–OH + R′COOH → R′COOR + H2O

1. Reaction with hydrogen halides:

Alcohols react with HX to form alkyl halides.

ROH + HX → RX + H2O

1. Dehydration:

Alcohols lose water in the presence of concentrated H2SO4 or H3PO4 to form alkenes.

CH3CH2OH → CH2=CH2 + H2O

1. Oxidation:

Primary alcohols give aldehydes with mild oxidising agents and acids with strong oxidising agents.

Secondary alcohols give ketones.

Tertiary alcohols resist oxidation under mild conditions.

These reactions make alcohols useful intermediates in organic synthesis.

Q33. Explain the preparation and reactions of ethers.

Ethers are prepared mainly by acid dehydration of alcohols and Williamson synthesis.

1. Dehydration of alcohols:

Primary alcohols form ethers with concentrated H2SO4 at lower temperature.

2C2H5OH → C2H5OC2H5 + H2O

This method is suitable mainly for primary alcohols.

1. Williamson synthesis:

Alkyl halide reacts with sodium alkoxide.

R–X + NaOR′ → R–OR′ + NaX

It can prepare symmetrical and unsymmetrical ethers. It gives best results with primary alkyl halides.

Chemical reactions of ethers:

1. Cleavage with HI or HBr:

Ethers react with concentrated HI or HBr at high temperature.

R–O–R′ + HI → R–I + R′OH

With excess HI, alcohol may further convert to alkyl iodide.

1. Cleavage of anisole:

C6H5OCH3 + HI → C6H5OH + CH3I

Phenol forms because aryl-oxygen bond has partial double bond character.

1. Electrophilic substitution:

Alkoxy group activates the benzene ring and directs incoming electrophiles to ortho and para positions.

Anisole gives mainly para-bromoanisole on bromination.

Formula and Reaction-Based Revision for Important Questions Class 12 Chemistry Chapter 7

Important questions class 12 chemistry chapter 7 should be revised through classification, preparations, named reactions and product prediction.

Alcohols, Phenols and Ethers Class 12

Alcohol:

R–OH

Phenol:

Ar–OH

Ether:

R–O–R′ or Ar–O–R

Preparation of Alcohols Class 12

From alkenes by acid hydration:

Alkene + H2O / H+ → Alcohol

From alkenes by hydroboration-oxidation:

Alkene + BH3, then H2O2/OH⁻ → Alcohol

From aldehydes:

R–CHO + 2[H] → R–CH2OH

From ketones:

R–CO–R′ + 2[H] → R–CHOH–R′

From Grignard reagent:

Methanal gives primary alcohol.

Other aldehydes give secondary alcohol.

Ketones give tertiary alcohol.

Preparation of Phenols Class 12

From chlorobenzene:

C6H5Cl + NaOH → C6H5ONa

C6H5ONa + H+ → C6H5OH

From diazonium salt:

C6H5N2+Cl⁻ + H2O → C6H5OH + N2 + HCl

From cumene:

Cumene → Cumene hydroperoxide → Phenol + Acetone

Acidity of Phenols Class 12

Phenol is more acidic than alcohol because phenoxide ion is resonance stabilised.

Electron-withdrawing groups increase acidity.

Electron-releasing groups decrease acidity.

Acid strength order example:

2,4,6-trinitrophenol > 3,5-dinitrophenol > nitrophenol > phenol > cresol > alcohol

Dehydration of Alcohols Class 12

Ease of dehydration:

Tertiary > Secondary > Primary

Ethanol on heating with conc. H2SO4 at 443 K gives ethene.

CH3CH2OH → CH2=CH2 + H2O

Oxidation of Alcohols Class 12

Primary alcohol + mild oxidising agent → Aldehyde

Primary alcohol + strong oxidising agent → Carboxylic acid

Secondary alcohol → Ketone

Tertiary alcohol → Resistant to oxidation under mild conditions

Phenol Reactions Class 12

Bromination with bromine water:

Phenol → 2,4,6-tribromophenol

Nitration with dilute HNO3:

Phenol → o-nitrophenol + p-nitrophenol

Nitration with concentrated HNO3:

Phenol → 2,4,6-trinitrophenol

Kolbe reaction:

Phenoxide ion + CO2 → Salicylic acid

Reimer-Tiemann reaction:

Phenol + CHCl3 + NaOH → Salicylaldehyde

Williamson Synthesis Class 12

R–X + NaOR′ → R–OR′ + NaX

Best reactant choice:

Primary alkyl halide + sodium alkoxide

Avoid tertiary alkyl halides because elimination dominates.

Cleavage of Ethers Class 12

R–O–R′ + HI → R–I + R′OH

For anisole:

C6H5OCH3 + HI → C6H5OH + CH3I

Order of hydrogen halide reactivity:

HI > HBr > HCl

Chapter-Wise Revision for Alcohols Phenols and Ethers Important Questions

Important questions class 12 chemistry chapter 7 should be revised in six parts: nomenclature, preparation, physical properties, acidity, named reactions and conversions.

Start with IUPAC names of alcohols, phenols and ethers. Naming questions are common in Section A and Section B.

Next, revise preparation of alcohols class 12. Focus on hydration of alkenes, hydroboration-oxidation, reduction of aldehydes and ketones and Grignard reagent reactions.

Then revise preparation of phenols class 12. Learn haloarene, benzene sulphonic acid, diazonium salt and cumene methods.

After that, revise acidity of phenols class 12. Phenoxide ion stability and substituent effects are important reasoning questions.

Next, practise named reactions. Kolbe reaction class 12, Reimer-Tiemann reaction class 12, Lucas test class 12 and Williamson synthesis class 12 are high-yield topics.

Finally, revise ether reactions. Cleavage of ethers class 12 and electrophilic substitution in anisole are commonly asked as product-prediction questions.

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