StudyOSStudyOS CBSE Study Material, Class 5–12

Alcohols, Phenols and Ethers

NCERT Class 12 · Chemistry Based on NCERT Class 12 Chemistry textbook · Free CBSE study kit

Everything you need to ace this chapter, in one place.
Notes · Mind Map · Quiz · Assertion-Reason · Case Studies · Question Bank · Past Papers

Get Full Study Kit →
Studying with friends? Send them this chapter.

Chapter Notes

Overview
This chapter covers the chemistry of three important classes of organic compounds: alcohols, phenols, and ethers. It explains their classification, IUPAC nomenclature, methods of preparation, physical properties, and chemical reactions. The big idea is that the hydroxyl group (-OH) in alcohols and phenols and the alkoxy/aryloxy group (-OR/-OAr) in ethers determine their unique reactivity, including acidity, nucleophilic substitution, and electrophilic aromatic substitution. A student must understand how the structure of the functional group influences properties and reactions to predict products and mechanisms in board exam questions.
Key Concepts (23)
Alcohol
An organic compound containing one or more hydroxyl (-OH) group(s) directly attached to carbon atom(s) of an aliphatic system.
e.g. Methanol (CH3OH) is the simplest alcohol, used as a solvent in paints and varnishes.
Phenol
An organic compound containing -OH group(s) directly attached to carbon atom(s) of an aromatic system.
e.g. Phenol (C6H5OH) is used as an antiseptic and in the manufacture of plastics.
Ether
An organic compound formed by substituting the hydrogen atom of the hydroxyl group of an alcohol or phenol by an alkyl or aryl group, having the general formula R-O-R' or Ar-O-R.
e.g. Diethyl ether (C2H5OC2H5) was used as an inhalation anaesthetic.
Primary alcohol
An alcohol in which the -OH group is attached to a primary carbon atom (a carbon atom bonded to only one other carbon atom).
e.g. Propan-1-ol (CH3CH2CH2OH) is a primary alcohol.
Secondary alcohol
An alcohol in which the -OH group is attached to a secondary carbon atom (a carbon atom bonded to two other carbon atoms).
e.g. Propan-2-ol (CH3CH(OH)CH3) is a secondary alcohol.
Tertiary alcohol
An alcohol in which the -OH group is attached to a tertiary carbon atom (a carbon atom bonded to three other carbon atoms).
e.g. 2-Methylpropan-2-ol ((CH3)3COH) is a tertiary alcohol.
Allylic alcohol
An alcohol in which the -OH group is attached to a sp3 hybridised carbon adjacent to a carbon-carbon double bond (allylic carbon).
e.g. Prop-2-en-1-ol (CH2=CH-CH2OH) is an allylic alcohol.
Benzylic alcohol
An alcohol in which the -OH group is attached to a sp3 hybridised carbon atom next to an aromatic ring.
e.g. Phenylmethanol (C6H5CH2OH) is a benzylic alcohol.
Symmetrical ether
An ether in which the alkyl or aryl groups attached to the oxygen atom are the same.
e.g. Diethyl ether (C2H5OC2H5) is a symmetrical ether.
Unsymmetrical ether
An ether in which the two alkyl or aryl groups attached to the oxygen atom are different.
e.g. Ethylmethyl ether (C2H5OCH3) is an unsymmetrical ether.
Lucas test
A test to distinguish between primary, secondary, and tertiary alcohols based on their reactivity with Lucas reagent (conc. HCl and ZnCl2). Tertiary alcohols produce turbidity immediately, secondary alcohols within 5 minutes, and primary alcohols do not produce turbidity at room temperature.
e.g. When 2-methylpropan-2-ol (tertiary) is treated with Lucas reagent, turbidity appears immediately.
Markovnikov's rule
In the addition of a protic acid (HX) to an unsymmetrical alkene, the hydrogen atom adds to the carbon with the greater number of hydrogen atoms, and the halide (or other group) adds to the carbon with fewer hydrogen atoms.
e.g. In the hydration of propene with dilute H2SO4, the product is propan-2-ol (not propan-1-ol).
Hydroboration-oxidation
A reaction in which diborane (BH3)2 adds to an alkene to form trialkyl borane, which is then oxidised by hydrogen peroxide in the presence of aqueous NaOH to give an alcohol. The addition of boron occurs to the less substituted carbon, giving an anti-Markovnikov product.
e.g. Propene on hydroboration-oxidation gives propan-1-ol.
Grignard reagent
An organomagnesium compound (R-MgX) that acts as a nucleophile and adds to the carbonyl group of aldehydes and ketones to form alcohols after hydrolysis.
e.g. Methylmagnesium bromide (CH3MgBr) reacts with methanal to give ethanol after hydrolysis.
Williamson synthesis
A laboratory method for preparing ethers by reacting an alkyl halide with a sodium alkoxide or aryloxide. The reaction follows an SN2 mechanism and is best suited for primary alkyl halides.
e.g. Sodium ethoxide reacts with ethyl bromide to give diethyl ether.
Kolbe's reaction
An electrophilic substitution reaction in which sodium phenoxide reacts with carbon dioxide (a weak electrophile) under pressure, followed by acidification, to give ortho-hydroxybenzoic acid (salicylic acid).
e.g. Phenol is converted to salicylic acid via Kolbe's reaction.
Reimer-Tiemann reaction
A reaction in which phenol is treated with chloroform in the presence of aqueous NaOH to introduce a -CHO group at the ortho position of the benzene ring, yielding salicylaldehyde.
e.g. Phenol + CHCl3 + NaOH → Salicylaldehyde.
Acidity of phenols
Phenols are stronger acids than alcohols and water because the phenoxide ion is stabilised by resonance delocalisation of the negative charge into the aromatic ring. Electron-withdrawing groups (e.g., -NO2) increase acidity, while electron-releasing groups (e.g., -CH3) decrease it.
e.g. 2,4,6-Trinitrophenol (picric acid) is a strong acid due to three -NO2 groups.
Esterification
The reaction of an alcohol or phenol with a carboxylic acid, acid chloride, or acid anhydride to form an ester. The reaction with carboxylic acid requires a small amount of conc. H2SO4 and is reversible; water is removed to shift equilibrium.
e.g. Ethanol reacts with acetic acid in the presence of conc. H2SO4 to give ethyl acetate.
Dehydration of alcohols
The removal of a molecule of water from an alcohol in the presence of a protic acid (conc. H2SO4 or H3PO4) or catalyst (anhydrous ZnCl2 or Al2O3) to form an alkene. The ease of dehydration follows the order: tertiary > secondary > primary.
e.g. Ethanol is dehydrated to ethene at 443 K with conc. H2SO4.
Oxidation of alcohols
Primary alcohols are oxidised to aldehydes (with mild oxidising agents like PCC) and then to carboxylic acids (with strong oxidising agents like acidified KMnO4). Secondary alcohols are oxidised to ketones. Tertiary alcohols resist oxidation under normal conditions.
e.g. Propan-1-ol is oxidised by acidified KMnO4 to propanoic acid.
Electrophilic aromatic substitution in phenols
The -OH group activates the benzene ring and directs incoming electrophiles to ortho and para positions due to resonance. Common reactions include nitration, halogenation, and Friedel-Crafts reactions.
e.g. Phenol with bromine water gives 2,4,6-tribromophenol as a white precipitate.
Cleavage of C-O bond in ethers
Ethers are cleaved by concentrated HI or HBr at high temperature. The reaction involves protonation of the ether, followed by nucleophilic attack by the halide ion. In mixed ethers, the smaller alkyl group forms the alkyl halide (SN2) unless a tertiary group is present (SN1).
e.g. Anisole (C6H5OCH3) with HI gives phenol and methyl iodide.
Formulas & Equations (15)
RCH=CH2 + H2O → (H+) RCH(OH)CH3 (Markovnikov) or RCH2CH2OH (anti-Markovnikov via hydroboration)
Reads as: Alkene plus water in the presence of acid gives alcohol following Markovnikov's rule; or via hydroboration-oxidation gives anti-Markovnikov alcohol.
Variables: R = alkyl group; H2O = water; H+ = acid catalyst; (BH3)2 = diborane; H2O2/NaOH = hydrogen peroxide in sodium hydroxide.
Use when: When asked to prepare an alcohol from an alkene, or to predict the product of hydration or hydroboration-oxidation.
⚠ Students often forget that hydroboration-oxidation gives the anti-Markovnikov product, not the Markovnikov product.
RCHO + H2 → (catalyst) RCH2OH (primary alcohol); RCOR' + H2 → (catalyst) RCH(OH)R' (secondary alcohol)
Reads as: Aldehyde plus hydrogen in the presence of a catalyst gives primary alcohol; ketone plus hydrogen gives secondary alcohol.
Variables: R, R' = alkyl or aryl groups; H2 = hydrogen gas; catalyst = Pt, Pd, or Ni.
Use when: When asked to reduce a carbonyl compound to an alcohol using catalytic hydrogenation or NaBH4/LiAlH4.
⚠ Students sometimes write the wrong product for reduction of a ketone (e.g., writing a primary alcohol instead of secondary).
RCOOH → (i) LiAlH4, (ii) H2O RCH2OH
Reads as: Carboxylic acid is reduced by lithium aluminium hydride followed by hydrolysis to give primary alcohol.
Variables: R = alkyl group; LiAlH4 = lithium aluminium hydride; H2O = water.
Use when: When asked to prepare a primary alcohol from a carboxylic acid.
⚠ Students forget that LiAlH4 reduces carboxylic acids to alcohols, not to aldehydes.
RMgX + HCHO → (i) dry ether, (ii) H2O RCH2OH (primary alcohol); RMgX + R'CHO → (i) dry ether, (ii) H2O RCH(OH)R' (secondary alcohol); RMgX + R'COR'' → (i) dry ether, (ii) H2O R-C(OH)(R')(R'') (tertiary alcohol)
Reads as: Grignard reagent reacts with methanal to give primary alcohol, with other aldehydes to give secondary alcohol, and with ketones to give tertiary alcohol after hydrolysis.
Variables: R, R', R'' = alkyl or aryl groups; MgX = magnesium halide; HCHO = methanal; R'CHO = aldehyde; R'COR'' = ketone.
Use when: When asked to prepare an alcohol using a Grignard reagent, or to predict the product of a Grignard reaction with a carbonyl compound.
⚠ Students often confuse the product: with methanal it's primary, with other aldehydes secondary, with ketones tertiary.
C6H5Cl + NaOH → (623 K, 320 atm) C6H5ONa → (H+) C6H5OH
Reads as: Chlorobenzene is fused with sodium hydroxide at high temperature and pressure to give sodium phenoxide, which on acidification yields phenol.
Variables: C6H5Cl = chlorobenzene; NaOH = sodium hydroxide; C6H5ONa = sodium phenoxide; H+ = acid.
Use when: When asked to prepare phenol from a haloarene.
⚠ Students forget the harsh conditions (623 K, 320 atm) and write the reaction at room temperature.
C6H5SO3H + NaOH → (fusion) C6H5ONa → (H+) C6H5OH
Reads as: Benzene sulphonic acid is fused with sodium hydroxide to give sodium phenoxide, which on acidification yields phenol.
Variables: C6H5SO3H = benzene sulphonic acid; NaOH = sodium hydroxide; C6H5ONa = sodium phenoxide; H+ = acid.
Use when: When asked to prepare phenol from benzene sulphonic acid.
⚠ Students often write the intermediate as sodium sulphonate instead of sodium phenoxide.
C6H5N2+Cl- + H2O → (warm) C6H5OH + N2 + HCl
Reads as: Benzene diazonium chloride is hydrolysed by warming with water to give phenol, nitrogen gas, and hydrochloric acid.
Variables: C6H5N2+Cl- = benzene diazonium chloride; H2O = water; C6H5OH = phenol; N2 = nitrogen; HCl = hydrochloric acid.
Use when: When asked to prepare phenol from a diazonium salt.
⚠ Students forget that the reaction requires warming and produces N2 gas.
C6H5CH(CH3)2 + O2 → (air) C6H5C(CH3)2OOH → (dil. H+) C6H5OH + CH3COCH3
Reads as: Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, which on treatment with dilute acid gives phenol and acetone.
Variables: C6H5CH(CH3)2 = cumene; O2 = oxygen; C6H5C(CH3)2OOH = cumene hydroperoxide; dil. H+ = dilute acid; C6H5OH = phenol; CH3COCH3 = acetone.
Use when: When asked about the industrial preparation of phenol from cumene.
⚠ Students often forget that acetone is a by-product and write only phenol.
ROH + HX → R-X + H2O
Reads as: Alcohol reacts with hydrogen halide to give alkyl halide and water.
Variables: ROH = alcohol; HX = hydrogen halide (HCl, HBr, HI); R-X = alkyl halide; H2O = water.
Use when: When asked to convert an alcohol to an alkyl halide.
⚠ Students forget that the reactivity order of HX is HI > HBr > HCl, and that tertiary alcohols react fastest.
ROH → (conc. H2SO4, 443 K) alkene + H2O
Reads as: Alcohol undergoes dehydration in the presence of concentrated sulphuric acid at 443 K to give an alkene and water.
Variables: ROH = alcohol; conc. H2SO4 = concentrated sulphuric acid; alkene = product; H2O = water.
Use when: When asked to prepare an alkene from an alcohol by dehydration.
⚠ Students often forget the temperature (443 K) and write the reaction at lower temperature, which gives ether instead.
RCH2OH → (CrO3/H+) RCHO → (CrO3/H+) RCOOH
Reads as: Primary alcohol is oxidised by chromic acid to aldehyde, which is further oxidised to carboxylic acid.
Variables: RCH2OH = primary alcohol; CrO3/H+ = chromic acid; RCHO = aldehyde; RCOOH = carboxylic acid.
Use when: When asked to oxidise a primary alcohol to an aldehyde or carboxylic acid.
⚠ Students often write the aldehyde as the final product when using strong oxidising agents like KMnO4.
RCH(OH)R' → (CrO3) RCOR'
Reads as: Secondary alcohol is oxidised by chromic anhydride to a ketone.
Variables: RCH(OH)R' = secondary alcohol; CrO3 = chromic anhydride; RCOR' = ketone.
Use when: When asked to oxidise a secondary alcohol to a ketone.
⚠ Students sometimes write a carboxylic acid as the product for secondary alcohol oxidation.
2 ROH → (conc. H2SO4, 413 K) R-O-R + H2O
Reads as: Two molecules of alcohol undergo dehydration in the presence of concentrated sulphuric acid at 413 K to give an ether and water.
Variables: ROH = alcohol; conc. H2SO4 = concentrated sulphuric acid; R-O-R = ether; H2O = water.
Use when: When asked to prepare a symmetrical ether from an alcohol by acid dehydration.
⚠ Students often confuse the temperature: 413 K gives ether, 443 K gives alkene.
R-X + NaOR' → R-O-R' + NaX
Reads as: Alkyl halide reacts with sodium alkoxide to give an ether and sodium halide.
Variables: R-X = alkyl halide; NaOR' = sodium alkoxide; R-O-R' = ether; NaX = sodium halide.
Use when: When asked to prepare an ether by Williamson synthesis.
⚠ Students often forget that the reaction is SN2 and works best with primary alkyl halides; secondary/tertiary halides give elimination products.
R-O-R' + HI → (excess, heat) R-I + R'-OH
Reads as: Ether reacts with excess hydrogen iodide on heating to give alkyl iodide and alcohol.
Variables: R-O-R' = ether; HI = hydrogen iodide; R-I = alkyl iodide; R'-OH = alcohol.
Use when: When asked to cleave an ether with HI.
⚠ Students often forget that the smaller alkyl group forms the alkyl iodide (SN2), unless a tertiary group is present (SN1).
How It Works
Acid-catalysed hydration of alkenes to alcohols
  1. Step 1: Protonation of the alkene by H3O+ to form a carbocation (electrophilic attack).
  2. Step 2: Nucleophilic attack of water on the carbocation to form a protonated alcohol.
  3. Step 3: Deprotonation of the protonated alcohol to give the alcohol and regenerate H+.
💡 The reaction follows Markovnikov's rule: the hydrogen adds to the carbon with more hydrogens, so the -OH ends up on the more substituted carbon.
Hydroboration-oxidation of alkenes to alcohols
  1. Step 1: Addition of diborane (BH3)2 to the alkene to form trialkyl borane. Boron attaches to the less substituted carbon (anti-Markovnikov).
  2. Step 2: Oxidation of trialkyl borane with hydrogen peroxide in the presence of aqueous NaOH to give the alcohol.
💡 This gives the anti-Markovnikov product, i.e., the -OH ends up on the less substituted carbon.
Mechanism of dehydration of ethanol to ethene
  1. Step 1: Protonation of ethanol by H2SO4 to form protonated alcohol (ethyl oxonium ion).
  2. Step 2: Formation of carbocation by loss of water (slow, rate-determining step).
  3. Step 3: Elimination of a proton from the carbocation to form ethene.
💡 The acid used in step 1 is regenerated in step 3; ethene is removed to drive the equilibrium to the right.
Mechanism of ether cleavage by HI
  1. Step 1: Protonation of the ether oxygen by HI to form an oxonium ion.
  2. Step 2: Nucleophilic attack by iodide ion on the less substituted carbon of the oxonium ion via SN2 (or SN1 if a tertiary carbocation can form), displacing an alcohol molecule.
  3. Step 3: If HI is in excess, the alcohol formed reacts further with HI to give a second alkyl iodide.
💡 In mixed ethers, the smaller alkyl group forms the alkyl iodide (SN2), unless a tertiary group is present (SN1).
Williamson synthesis of ethers
  1. Step 1: Formation of alkoxide ion by reaction of alcohol with sodium metal.
  2. Step 2: SN2 attack of the alkoxide ion on a primary alkyl halide, displacing the halide ion to form the ether.
💡 The method is best for primary alkyl halides; secondary or tertiary halides give elimination products (alkenes) instead of ethers.
Kolbe's reaction
  1. Step 1: Phenol is treated with NaOH to form the more reactive phenoxide ion.
  2. Step 2: Phenoxide ion undergoes electrophilic substitution with CO2 (a weak electrophile) at the ortho position.
  3. Step 3: Acidification gives ortho-hydroxybenzoic acid (salicylic acid).
💡 The reaction requires high pressure and temperature; the product is mainly the ortho isomer.
Reimer-Tiemann reaction
  1. Step 1: Phenol is treated with chloroform in the presence of aqueous NaOH to form an intermediate (dichlorocarbene is generated in situ).
  2. Step 2: The intermediate attacks the ortho position of the phenoxide ion.
  3. Step 3: Hydrolysis of the intermediate (substituted benzal chloride) gives salicylaldehyde.
💡 The -CHO group is introduced only at the ortho position; no para product is formed.
Electrophilic aromatic substitution in phenol (nitration example)
  1. Step 1: The -OH group activates the benzene ring via resonance, increasing electron density at ortho and para positions.
  2. Step 2: The electrophile (NO2+) attacks the ortho or para position.
  3. Step 3: Loss of H+ restores aromaticity, giving ortho- and para-nitrophenol.
💡 With dilute HNO3 at low temperature, a mixture of ortho and para isomers is obtained; with conc. HNO3, picric acid (2,4,6-trinitrophenol) is formed.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Markovnikov and anti-Markovnikov products
✗ Students write propan-1-ol as the product of acid-catalysed hydration of propene.
→ Acid-catalysed hydration follows Markovnikov's rule, so the -OH adds to the more substituted carbon, giving propan-2-ol, not propan-1-ol.
✓ Propan-2-ol is the product of acid-catalysed hydration of propene; propan-1-ol is obtained via hydroboration-oxidation.
⚠ Forgetting the temperature in alcohol dehydration
✗ Students write that ethanol on heating with conc. H2SO4 gives ethene, without specifying the temperature.
→ At 413 K, ethanol gives ethoxyethane (ether); at 443 K, it gives ethene. The temperature determines the product.
✓ Ethanol with conc. H2SO4 at 443 K gives ethene; at 413 K, it gives ethoxyethane.
⚠ Wrong product in Grignard reaction with methanal
✗ Students write that methylmagnesium bromide reacts with methanal to give a secondary alcohol.
→ Methanal (HCHO) has no alkyl group attached to the carbonyl carbon, so the product is a primary alcohol (ethanol), not secondary.
✓ Methylmagnesium bromide with methanal gives ethanol (primary alcohol) after hydrolysis.
⚠ Confusing acidity order of phenols
✗ Students write that phenol is more acidic than 2,4,6-trinitrophenol.
→ Electron-withdrawing groups like -NO2 increase acidity by stabilising the phenoxide ion. 2,4,6-Trinitrophenol has three -NO2 groups, making it a much stronger acid than phenol.
✓ 2,4,6-Trinitrophenol (picric acid) is a strong acid, much more acidic than phenol.
⚠ Forgetting that Williamson synthesis fails with tertiary alkyl halides
✗ Students propose using a tertiary alkyl halide in Williamson synthesis to prepare a tertiary ether.
→ Alkoxides are strong bases; with tertiary alkyl halides, elimination (E2) predominates over substitution, giving an alkene instead of an ether.
✓ Williamson synthesis is suitable only for primary alkyl halides; for tertiary ethers, use a different method (e.g., alkoxide with a primary halide and a tertiary alcohol).
⚠ Incorrect cleavage product of anisole with HI
✗ Students write that anisole (C6H5OCH3) with HI gives iodobenzene and methanol.
→ The C-O bond in anisole is cleaved at the alkyl-oxygen bond because the aryl-oxygen bond has partial double bond character (sp2 carbon). Iodide attacks the methyl carbon, giving methyl iodide and phenol.
✓ Anisole with HI gives phenol and methyl iodide.
Compare & Contrast
Acidity
Alcohol
Phenol
Phenols are stronger acids than alcohols because the phenoxide ion is stabilised by resonance delocalisation of the negative charge into the aromatic ring, whereas the alkoxide ion has the negative charge localised on oxygen.
Oxidation product
Primary alcohol
Secondary alcohol
Primary alcohols are oxidised to aldehydes (with mild oxidising agents) and then to carboxylic acids (with strong oxidising agents), while secondary alcohols are oxidised to ketones and resist further oxidation.
Boiling point
Ether
Alcohol
Ethers have much lower boiling points than alcohols of comparable molecular mass because alcohols form intermolecular hydrogen bonds, while ethers do not.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks), 1 LA (5 marks) — total approximately 10 marks in the board exam.
Most asked: IUPAC nomenclature of alcohols, phenols, and ethers · Preparation of alcohols from alkenes (hydration and hydroboration-oxidation) · Preparation of phenols (from cumene, diazonium salts, haloarenes) · Acidity of alcohols and phenols (including effect of substituents) · Dehydration of alcohols and mechanism · Oxidation of alcohols · Williamson synthesis of ethers · Cleavage of ethers by HI · Electrophilic substitution in phenols (nitration, halogenation, Kolbe's, Reimer-Tiemann) · Lucas test
Don't skip: Mechanism of dehydration of ethanol to ethene (5 marks); Comparison of acidity of alcohols and phenols with resonance structures (3-5 marks); Preparation of phenol from cumene (3 marks); Williamson synthesis and its limitations (3 marks); Cleavage of ethers by HI with mechanism (3 marks)
One-Line Revision (read aloud before the exam)
  1. Alcohols have -OH attached to sp3 carbon of an aliphatic system; phenols have -OH attached to sp2 carbon of an aromatic system.
  2. Ethers have the general formula R-O-R' or Ar-O-R, where the oxygen is bonded to two alkyl/aryl groups.
  3. Primary alcohols have the -OH on a carbon bonded to one other carbon; secondary on a carbon bonded to two; tertiary on a carbon bonded to three.
  4. Acid-catalysed hydration of alkenes follows Markovnikov's rule; hydroboration-oxidation gives anti-Markovnikov alcohols.
  5. Phenol is prepared industrially from cumene via cumene hydroperoxide, yielding phenol and acetone.
  6. Phenols are stronger acids than alcohols because the phenoxide ion is resonance-stabilised.
  7. Electron-withdrawing groups (e.g., -NO2) increase the acidity of phenols; electron-releasing groups (e.g., -CH3) decrease it.
  8. Lucas test distinguishes primary, secondary, and tertiary alcohols based on turbidity formation with conc. HCl/ZnCl2.
  9. Dehydration of alcohols to alkenes requires conc. H2SO4 at 443 K; at 413 K, ethers are formed.
  10. Primary alcohols are oxidised to aldehydes (with PCC) and then to carboxylic acids (with KMnO4); secondary alcohols give ketones.
  11. Williamson synthesis prepares ethers by SN2 reaction of an alkoxide with a primary alkyl halide; tertiary halides give elimination.
  12. Ethers are cleaved by excess HI at high temperature; the smaller alkyl group forms the alkyl iodide (SN2).
  13. In anisole, the C-O bond to the methyl group is cleaved by HI, giving phenol and methyl iodide.
  14. Kolbe's reaction introduces a -COOH group at the ortho position of phenol via CO2 and NaOH.
  15. Reimer-Tiemann reaction introduces a -CHO group at the ortho position of phenol using CHCl3 and NaOH.
  16. Phenol with bromine water gives a white precipitate of 2,4,6-tribromophenol.
  17. The -OH group in phenol activates the benzene ring and directs electrophiles to ortho and para positions.
  18. Methanol is produced by catalytic hydrogenation of CO; ethanol is produced by fermentation of sugars.
  19. Denatured alcohol is made unfit for drinking by adding copper sulphate and pyridine.
  20. The boiling points of alcohols are higher than ethers and hydrocarbons of comparable mass due to hydrogen bonding.

MCQs — 10 Questions with Answers

Q1. Which of the following compounds is correctly named according to IUPAC nomenclature?

  • A. CH₃CH(OH)CH₂CH₃ is named butane-2-ol ✓
  • B. CH₃CH(OH)CH₂CH₃ is named butane-3-ol
  • C. CH₃CH(OH)CH₂CH₃ is named 2-hydroxybutane
  • D. CH₃CH(OH)CH₂CH₃ is named sec-butyl alcohol (IUPAC)

Answer: A — The longest chain is 4 carbons; numbering from the end nearest to −OH gives −OH at position 2, hence butan-2-ol (IUPAC). Option B is wrong because position 3 is farther from the −OH; option C uses 'hydroxy' as prefix which is incorrect for primary functional groups; option D is a common name, not IUPAC.

Q2. An alcohol has the structure: CH₂=CH−CH₂−CH₂−OH. Which statement correctly describes this compound?

  • A. It is a primary alcohol with a terminal C=C double bond
  • B. It is a vinylic alcohol because −OH is attached to the sp² carbon of the C=C
  • C. It is an allylic alcohol because −OH is on a carbon adjacent to the C=C double bond ✓
  • D. It is a secondary benzylic alcohol

Answer: C — The −OH is attached to sp³ C adjacent to the C=C (allylic position), making this an allylic alcohol. Option A ignores the unsaturation location; option B confuses vinylic alcohols (−OH on C=C) with allylic alcohols; option D is irrelevant as there is no aromatic ring.

Q3. The IUPAC name of HO−CH₂−CH₂−OH is:

  • A. 1,2-ethanediol
  • B. ethane-1,2-diol ✓
  • C. ethylene diol
  • D. 2-hydroxyethanol

Answer: B — For polyhydric alcohols, the 'e' of the alkane is retained and 'diol' is added with locants; the correct IUPAC name is ethane-1,2-diol. Option A omits the hyphen; option C is a common name; option D is incorrect nomenclature.

Q4. Classify the following alcohol: (CH₃)₂CH−OH

  • A. Primary alcohol
  • B. Secondary alcohol ✓
  • C. Tertiary alcohol
  • D. Cannot be classified without more information

Answer: B — In (CH₃)₂CH−OH, the carbon bearing −OH is bonded to two other carbons (two methyl groups), making it a secondary carbon and secondary alcohol. Option A (primary) would be R−CH₂−OH; option C (tertiary) would be R₃C−OH.

Q5. Which of the following is an unsymmetrical ether?

  • A. CH₃−O−CH₃
  • B. C₂H₅−O−C₂H₅
  • C. CH₃−O−C₂H₅ ✓
  • D. C₆H₅−O−C₆H₅

Answer: C — An unsymmetrical ether has different alkyl or aryl groups attached to oxygen. Options A and B are symmetrical (dimethyl and diethyl ethers); option D is also symmetrical (diphenyl ether); only CH₃−O−C₂H₅ (ethyl methyl ether) has different groups.

Q6. The common name for C₂H₅−O−C₂H₅ is:

  • A. ethylene ether
  • B. diethyl ether ✓
  • C. ethyl ether
  • D. diethyl oxide

Answer: B — The compound has two identical ethyl groups attached to oxygen, so the common name is diethyl ether. Option A refers to a different compound; option C is incomplete; option D uses non-standard terminology.

Q7. Which pair of names refers to the same compound?

  • A. Propan-1-ol and isopropyl alcohol
  • B. Propan-2-ol and isopropyl alcohol ✓
  • C. Butan-1-ol and sec-butyl alcohol
  • D. Butan-2-ol and tert-butyl alcohol

Answer: B — Propan-2-ol (IUPAC) is CH₃CH(OH)CH₃, which is the same as isopropyl alcohol (common name). Option A: propan-1-ol is n-propyl alcohol, not isopropyl; option C: butan-1-ol is n-butyl; option D: tert-butyl is 2-methylpropan-2-ol.

Q8. A compound has two −OH groups at positions 1 and 3 on a benzene ring. Which name is correct?

  • A. 1,3-benzenediol
  • B. meta-benzenediol
  • C. 1,3-dihydroxybenzene
  • D. Resorcinol (common name) — all above are acceptable ✓

Answer: D — The IUPAC name is 1,3-benzenediol (positions 1 and 3 on benzene); meta-benzenediol refers to 1,3-disubstitution; the common name is resorcinol. All three naming methods are correct and accepted.

Q9. Assertion: Vinylic alcohols are stable compounds that can be isolated and studied. Reason: The C−OH bond in vinylic alcohols is strong due to sp² hybridization of the carbon.

  • A. Both assertion and reason are true, and reason is the correct explanation
  • B. Both assertion and reason are true, but reason is not the correct explanation
  • C. Assertion is false; vinylic alcohols are unstable and tautomerize to carbonyl compounds ✓
  • D. Reason is false; sp² hybridization leads to weaker C−O bonds

Answer: C — Vinylic alcohols (enols) are highly unstable and immediately tautomerize to more stable aldehydes or ketones through keto-enol tautomerism. The assertion is false because these compounds cannot be easily isolated under normal conditions.

Q10. HOTS: Consider three compounds: (1) CH₃CH₂OH, (2) C₆H₅OH, and (3) HC≡C−OH. Rank them in order of increasing acidity and explain the basis for your ranking.

  • A. (1) < (2) < (3) because the more substituted the carbon, the weaker the acid
  • B. (1) < (2) < (3) because sp > sp² > sp³ hybridization increases acidity ✓
  • C. (3) < (1) < (2) because vinylic alcohols are unstable and phenols are stabilized by resonance
  • D. (1) < (3) < (2) because only phenols show resonance stabilization of the conjugate base

Answer: B — Acidity increases as the carbon bearing −OH becomes more sp-hybridized (greater s-character stabilizes the negative charge). Alkynes (sp C) are most acidic (~10⁻²⁵ pKa region for terminal alkynes), phenols (sp²) are weakly acidic (~10 pKa), and alcohols (sp³) are essentially neutral. Option D is incomplete; vinylic alcohols are unstable and do not persist to demonstrate acidity.

Flashcards

What is the key difference between an alcohol and a phenol in terms of carbon hybridization?

Alcohols have −OH attached to sp³ hybridized carbon (aliphatic), while phenols have −OH attached to sp² hybridized carbon (aromatic ring).

IUPAC nomenclature rule for alcohols: how do you name CH₃CH₂CH(OH)CH₃?

Identify the longest chain containing −OH, number from the end nearest to −OH (giving −OH position 2), name as butan-2-ol.

Define a primary alcohol with an example.

A primary alcohol has the −OH group attached to a carbon bonded to only one other carbon (R−CH₂−OH); example: CH₃CH₂OH (ethanol).

What is an allylic alcohol? Give an example.

An allylic alcohol has −OH attached to sp³ carbon adjacent to a C=C double bond; example: CH₂=CH−CH₂−OH (allyl alcohol).

Distinguish between symmetrical and unsymmetrical ethers.

Symmetrical ethers have identical alkyl or aryl groups on both sides of oxygen (R−O−R), while unsymmetrical ethers have different groups (R−O−R').

Name the compound C₂H₅−O−C₂H₅ by both common and IUPAC methods.

Common name: diethyl ether; IUPAC name: ethoxyethane (treating one ethyl as main chain and other as ethoxy substituent).

Why are vinylic alcohols (C=C−OH) unstable?

Vinylic alcohols tautomerize immediately to aldehydes or ketones because the enol form is thermodynamically unfavourable due to poor orbital overlap.

Classify CH₃CH(OH)CH₂CH₃ and state whether it is primary, secondary or tertiary.

This is butan-2-ol, a secondary alcohol, because −OH is attached to a secondary carbon bonded to two other carbons.

What functional group difference makes phenols acidic while alcohols are essentially neutral?

In phenols, the aromatic ring delocalizes the negative charge of the phenoxide anion through resonance, stabilizing it; alcohols lack this resonance stabilization.

Name the common name for HO−CH₂−CH₂−OH and its IUPAC name.

Common name: ethylene glycol; IUPAC name: ethane-1,2-diol (two −OH groups at positions 1 and 2 of a two-carbon chain).

Important Board Questions

Define a primary alcohol and write the IUPAC name of CH₃CH₂CH₂OH. [2 marks]

Primary alcohol = −OH attached to carbon bonded to only one other carbon (R−CH₂−OH). For IUPAC: identify longest chain (3 carbons), replace final 'e' with 'ol', number from −OH end (position 1).

Explain why phenols are more acidic than alcohols. (Show the role of resonance in your explanation.) [5 marks]

Phenols ionize to phenoxide anion: C₆H₅−OH ⇌ C₆H₅−O⁻ + H⁺. The negative charge on oxygen is delocalized into the aromatic π-system through resonance, stabilizing the conjugate base. Alcohols lack this aromatic ring, so the alkoxide anion (R−O⁻) cannot be resonance-stabilized, making loss of H⁺ much less favorable. Write resonance structures to show charge distribution in phenoxide.

A compound has the molecular formula C₅H₁₀O₂ and contains two −OH groups. One −OH is on a primary carbon and the other is on a secondary carbon. Draw the structure and give both the common and IUPAC names. Also classify this compound as monohydric, dihydric, or polyhydric. [6 marks]

Dihydric alcohol with 5-carbon main chain: primary −OH at end (position 1), secondary −OH at position 2 or higher. Structure: HO−CH₂−CH(OH)−CH₂−CH₂−CH₃ or similar. IUPAC: pentane-1,2-diol (number from primary −OH end). Common: 1,2-pentanediol or 2-hydroxypentanol (older style). Classification: dihydric because it contains exactly 2 −OH groups.

Assertion-Reason Questions for Alcohols, Phenols and Ethers

Q1. Assertion (A):

Phenol is a stronger acid than ethanol.

Reason (R):

The OH group in phenol is directly attached to an aromatic carbon, while in ethanol it is attached to an aliphatic carbon, and aromatic carbons are more electronegative.

Answer: C — A is true but R is false — The assertion is true, but the reason is incorrect. Phenol is indeed more acidic than ethanol, but NOT because aromatic carbons are more electronegative. The actual reason is that the phenoxide ion (C₆H₅O⁻) is stabilized by resonance with the benzene ring, making it a better leaving group. The OH attachment to aromatic vs aliphatic carbons is noted correctly in terms of location, but the explanation of WHY this makes phenol more acidic is missing.

Q2. Assertion (A):

Diethyl ether (C₂H₅OC₂H₅) is a symmetrical ether.

Reason (R):

Both alkyl groups attached to the oxygen atom are identical ethyl (C₂H₅) groups.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why diethyl ether is symmetrical. A symmetrical (or simple) ether has identical groups on both sides of the oxygen atom, which is exactly the case here.

Q3. Assertion (A):

The IUPAC name of CH₃CH(OH)CH(CH₃)CH₃ is 3-methylbutan-2-ol.

Reason (R):

The longest carbon chain contains 4 carbons, the OH group is at position 2, and the methyl substituent is at position 3 when numbered from the end nearest to the hydroxyl group.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. Following IUPAC nomenclature rules, we identify the longest chain (4 carbons = butane), number from the end closest to OH (placing it at C-2), and locate substituents (methyl at C-3). The name correctly reflects all these features.

Unlock more Assertion-Reason questions in Premium — unlock →

Case-Based Questions (CBQ) for Alcohols, Phenols and Ethers

Case Study:

A pharmaceutical company in India manufactures hand sanitizers containing two key active components. The first is ethanol (CH₃CH₂OH), commonly used as a disinfectant, which is miscible with water due to strong hydrogen bonding between the hydroxyl groups and water molecules. The second component is glycerol (propane-1,2,3-triol, HOCH₂CH(OH)CH₂OH), used as a humectant to prevent skin drying. Both compounds are polyhydric alcohols derived from natural sources. The company receives regular shipments of these compounds and must verify their purity using physical properties such as boiling points and miscibility tests. Ethanol has a boiling point of 78°C, while glycerol has a boiling point of 290°C due to extensive intermolecular hydrogen bonding.

Q1. Classify ethanol and glycerol according to the number of hydroxyl groups they contain.

  • A. Both are monohydric alcohols
  • B. Ethanol is monohydric; glycerol is trihydric ✓
  • C. Ethanol is dihydric; glycerol is polyhydric
  • D. Both are polyhydric alcohols with different carbon chains

Answer: B

Q2. The significantly higher boiling point of glycerol (290°C) compared to ethanol (78°C) can be best explained by:

  • A. Greater molecular weight of glycerol
  • B. More extensive hydrogen bonding due to three OH groups in glycerol vs one in ethanol ✓
  • C. Aromatic nature of glycerol making it less volatile
  • D. Glycerol being a secondary alcohol while ethanol is primary

Answer: B

Q3. Write the IUPAC name of ethanol according to the nomenclature rules for alcohols.

  • A. Ethan-1-ol ✓
  • B. Ethyl alcohol
  • C. Methyl alcohol
  • D. Ethane-1,1-diol

Answer: A

Q4. If a sample of hand sanitizer contains both ethanol and glycerol, which test would best distinguish between them based on their chemical structure?

  • A. Boiling point determination only
  • B. Miscibility with water only
  • C. Boiling point determination combined with the number of carbon atoms in the molecule ✓
  • D. Density measurement and molecular weight calculation

Answer: C

Unlock more case studies in Premium — unlock →

Previous Year Questions (PYQ) for Alcohols, Phenols and Ethers

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

Propan-2-ol on oxidation with CrO3 gives :

  • Propanal
  • Propanoic acid
  • Propene
  • Propanone

CBSE 2026 · Set 56(B) · 1 mark

Assertion (A) : p-nitrophenol is less acidic than phenol. Reason (R) : Nitro group is an electron-withdrawing group, which stabilises the phenoxide ion.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56(B) · 4 marks

The boiling points of alcohols and phenols increase with increase in the number of carbon atoms due to increase in van der Waals forces. The − OH group in alcohols and phenols is involved in intermolecular hydrogen bonding, due to which their boiling points are higher than those of ethers, haloalkanes and haloarenes of comparable molecular masses. Alcohols and phenols are acidic in nature. Electron-withdrawing groups in phenol increase its acidic strength, while electron-releasing groups decrease it. The presence of − OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to the ortho and para positions due to resonance effect. The Reimer-Tiemann reaction of phenol yields salicylaldehyde, whereas Kolbe's reaction of phenol yields salicylic acid. Aspirin is formed by the acetylation of salicylic acid.

(i) Write the name of the major product(s) formed in : (I) Reimer-Tiemann reaction (II) Kolbe's reaction (ii) Name the method by which aspirin is formed from salicylic acid.

CBSE 2026 · Set 56/1/1 · 1 mark

Aspirin is obtained by acetylation of

  • Phenol
  • Salicylaldehyde
  • 2-Hydroxybenzoic acid
  • Benzoic acid

CBSE 2026 · Set 56/1/1 · 1 mark

At low temperature, phenol reacts with dil. HNO3 to yield

  • 2, 4, 6-Trinitrophenol
  • o-Nitrophenol only
  • p-Nitrophenol only
  • ortho-and para-nitrophenol

CBSE 2026 · Set 56/1/1 · 1 mark

Assertion (A): The presence of –OH group in phenols directs the incoming group to meta position in the ring. Reason (R): –OH group in phenols activates the aromatic ring towards electrophilic substitution reaction.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/1 · 3 marks

Write the reaction involved in the following: (a) Reimer-Tiemann reaction (b) Kolbe's reaction (c) Friedal-Crafts acylation of anisole

CBSE 2026 · Set 56/1/2 · 1 mark

Assertion (A): The presence of –OH group in phenols directs the incoming group to ortho and para positions. Reason (R): –OH group in phenols activates the aromatic ring towards electrophilic substitution reaction.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/3 · 1 mark

Which of the following is 'not' true about enantiomers ?

  • They have the same chemical reactivity.
  • They have the same specific rotation.
  • They have the same melting or boiling point.
  • They have the same refractive index.

CBSE 2026 · Set 56/1/3 · 1 mark

At low temperature, phenol on reaction with Br2 in CS2 gives

  • 2, 4, 6-Tribromophenol
  • a mixture of ortho-and para-bromophenol
  • ortho-bromophenol only
  • para-bromophenol only

CBSE 2026 · Set 56/1/3 · 1 mark

Assertion (A) : The presence of –OH group in phenols directs the incoming group at ortho- and para- positions. Reason (R) : –OH group in phenols deactivates the aromatic ring.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/3 · 3 marks

Write the reaction involved in the following : (a) Reimer-Tiemann reaction (b) Kolbe's reaction (c) Friedal-Crafts acylation of anisole

See all 152 board PYQs for this chapter — open PYQ tab →

Class 12 Chemistry Formula Sheet

Every formula and named reaction from all 10 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

View the formula sheet →

Class 12 Chemistry Previous Year Questions, Chapter-wise

All 1241 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

See all past board questions →

Next chapterAldehydes, Ketones and Carboxylic Acids →
Found this useful? Share it with your class group.

You've seen the free preview. Open the full interactive study kit — mind map, instant-graded quiz, progress tracking — free, no sign-in required.

Open Full Study Kit — Free →

Premium unlocks the full Question Bank, Daily Practice Papers, Chapter Quiz and Full Mock Papers for every chapter.