JEE Main Chemistry · Chapter 16

Alcohols, Phenols, Ethers, Aldehydes, Ketones and Acids

Preparation and reactions, acidity, aldol and Cannizzaro, Reimer–Tiemann, Kolbe, and distinguishing tests.

NCERT Class 12JEE Main: 2–3 Qs per shift 8 min read

1. Chapter at a Glance

This chapter combines two NCERT units: Alcohols, Phenols and Ethers and Aldehydes, Ketones and Carboxylic Acids. All these compounds contain oxygen, and their reactions are driven by the polar C–O and C=O bonds. It is the highest-weightage organic chapter in JEE Main — expect two or three questions on named reactions (aldol, Cannizzaro, Reimer–Tiemann, Kolbe), distinguishing tests and acidity.

What JEE asks most
  • Acidity order of alcohols, phenols and carboxylic acids with substituents.
  • Named reactions and their products; aldol vs Cannizzaro (α-hydrogen rule).
  • Tests: Lucas, iodoform, Tollens', Fehling's, FeCl₃, NaHCO₃.
  • Nucleophilic addition reactivity of aldehydes vs ketones.

2. Alcohols

Preparation

  • Acid-catalysed hydration of alkenes (Markovnikov); hydroboration–oxidation (anti-Markovnikov).
  • Reduction of aldehydes (→ 1°) and ketones (→ 2°) with NaBH₄ or LiAlH₄; acids and esters with LiAlH₄ → 1°.
  • Grignard reagents: HCHO → 1° alcohol; other aldehydes → 2°; ketones → 3°.

Properties and reactions

  • High boiling points due to hydrogen bonding; lower alcohols are water-soluble.
  • Acidity: CH₃OH > 1° > 2° > 3° (alkyl +I destabilises the alkoxide). Alcohols are weaker acids than water.
  • Reaction with HX (breaking C–O): 3° > 2° > 1°. Lucas test (conc. HCl + ZnCl₂): 3° gives turbidity at once, 2° in about 5 minutes, 1° not at room temperature.
  • Dehydration with conc. H₂SO₄: 3° > 2° > 1° (via carbocation).
  • Oxidation: 1° → aldehyde (PCC) → acid (KMnO₄, K₂Cr₂O₇); 2° → ketone; 3° resist mild oxidation. Heated Cu at 573 K dehydrogenates 1° and 2° alcohols and dehydrates 3° alcohols to alkenes.

3. Phenols

  • Preparation: from chlorobenzene (Dow process), benzenesulphonic acid (fusion with NaOH), diazonium salts (warm water) and cumene (air oxidation gives phenol + acetone).
  • Acidity: phenol (pKa ≈ 10) is more acidic than alcohols because the phenoxide ion is resonance-stabilised. It reacts with NaOH but not with NaHCO₃.
  • −I/−R groups at o/p increase acidity (p-nitrophenol > phenol); +I/+R groups decrease it (p-cresol < phenol). Picric acid (2,4,6-trinitrophenol) is strongly acidic.
  • –OH strongly activates the ring (o/p). Bromine water gives a white precipitate of 2,4,6-tribromophenol; in CS₂ at low temperature, monobromophenols form.
  • Kolbe's reaction: sodium phenoxide + CO₂ (400 K, 4–7 atm) then acid → salicylic acid.
  • Reimer–Tiemann: phenol + CHCl₃ + aq. NaOH → salicylaldehyde (electrophile is dichlorocarbene, :CCl₂).
  • With neutral FeCl₃, phenols give a violet colour.
1615.7104.763.75 EtOHH₂OPhOHCH₃COOHHCOOH pKa
Approximate pKa values (bar height ∝ pKa). Shorter bar = stronger acid: HCOOH > CH₃COOH > phenol > water > ethanol.

4. Ethers

  • Williamson synthesis: R–X + R′O⁻Na⁺ → R–O–R′ (SN2). Use a primary halide; a tertiary halide gives an alkene instead. So t-butyl methyl ether is made from CH₃Br + (CH₃)₃CONa.
  • Cleavage by HI: mixed alkyl ethers give the iodide of the smaller group (SN2) unless one group is tertiary (SN1, iodide on the tertiary carbon). Anisole + HI → phenol + CH₃I.
  • –OCH₃ in anisole is o/p directing and activating.

5. Aldehydes and Ketones

Preparation

  • Oxidation of alcohols; ozonolysis of alkenes; hydration of alkynes.
  • Rosenmund reduction: RCOCl + H₂ (Pd/BaSO₄) → RCHO. Stephen reaction: RCN + SnCl₂/HCl then H₂O → RCHO. DIBAL-H reduces nitriles and esters to aldehydes.
  • Etard reaction: toluene + CrO₂Cl₂ → benzaldehyde. Gattermann–Koch: benzene + CO + HCl (AlCl₃) → benzaldehyde.
  • Friedel–Crafts acylation gives aromatic ketones.

Nucleophilic addition

The carbonyl carbon is electrophilic. Aldehydes are more reactive than ketones (less steric hindrance, fewer electron-donating alkyl groups). Order: HCHO > CH₃CHO > CH₃COCH₃. Addition of HCN gives cyanohydrins; NaHSO₃ gives crystalline bisulphite adducts; NH₂–Z derivatives give oximes, hydrazones and 2,4-DNP derivatives (orange precipitate, a test for C=O).

Named reactions

ReactionConditionResult
Aldol condensationCarbonyl with α-H, dil. NaOHβ-hydroxy carbonyl → α,β-unsaturated carbonyl on heating
CannizzaroAldehyde without α-H, conc. NaOHAlcohol + carboxylate salt (disproportionation)
ClemmensenZn–Hg / conc. HClC=O → CH₂
Wolff–KishnerNH₂NH₂, KOH, ethylene glycol, heatC=O → CH₂
HaloformCH₃CO– or CH₃CH(OH)–, X₂/NaOHCHX₃ + carboxylate (iodoform is yellow)
Tollens' test[Ag(NH₃)₂]⁺Silver mirror with aldehydes
Fehling's testCu²⁺ tartrate complexRed Cu₂O with aliphatic aldehydes (not benzaldehyde)

6. Carboxylic Acids

  • Preparation: oxidation of 1° alcohols and aldehydes; alkylbenzenes with KMnO₄ → benzoic acid; hydrolysis of nitriles; Grignard + CO₂.
  • Exist as H-bonded dimers; have higher boiling points than alcohols of similar mass.
  • Acidity: stronger than phenols; they liberate CO₂ from NaHCO₃ (test). HCOOH > CH₃COOH; electron-withdrawing groups increase acidity (F > Cl > Br > I; nearer is stronger).
  • Derivatives: acid chloride (SOCl₂), ester (alcohol + H⁺, Fischer esterification), amide, anhydride. Reactivity: acid chloride > anhydride > ester > amide.
  • Hell–Volhard–Zelinsky (HVZ): Cl₂ or Br₂ with red P gives α-halo acids (needs α-H).
  • Decarboxylation: sodium salt + soda lime → alkane; Kolbe electrolysis → alkane (dimer).

7. Master Table of Tests

TestPositive forObservation
Lucas3° / 2° / 1° alcoholsImmediate / ~5 min / no turbidity
Neutral FeCl₃PhenolsViolet colour
IodoformCH₃CO– and CH₃CH(OH)– (ethanol, acetone, propan-2-ol, acetaldehyde)Yellow CHI₃
2,4-DNPAldehydes and ketonesOrange-yellow precipitate
Tollens'Aldehydes (and HCOOH)Silver mirror
Fehling'sAliphatic aldehydesRed precipitate
NaHCO₃Carboxylic acidsBrisk effervescence (CO₂)

8. Mind Map

Oxygen Compounds
Alcohols
  • Grignard synthesis
  • Lucas test
  • Oxidation 1°/2°/3°
Phenols & ethers
  • Acidity
  • Kolbe, Reimer–Tiemann
  • Williamson, HI cleavage
Aldehydes & ketones
  • Nucleophilic addition
  • Aldol, Cannizzaro
  • Clemmensen, Wolff–Kishner
Acids
  • Acidity trends
  • Derivatives
  • HVZ, decarboxylation

9. Tips and Tricks

Shortcuts
  • Has α-H? → aldol with dilute base. No α-H (HCHO, C₆H₅CHO, (CH₃)₃CCHO)? → Cannizzaro with concentrated base.
  • In a crossed Cannizzaro with HCHO, formaldehyde is always oxidised (to formate) and the other aldehyde reduced.
  • Crossed aldol between two different carbonyls, both with α-H, gives 4 products.
  • Iodoform check: look for CH₃–C=O or CH₃–CH(OH)– attached to H or carbon.
Mistakes students make
  • Saying phenol reacts with NaHCO₃ — it does not (too weak).
  • Using a tertiary halide in Williamson synthesis — it eliminates.
  • Giving methanol or propan-1-ol a positive iodoform test — they lack the CH₃CH(OH)– unit.

10. Important One-liners

  • Phenol is more acidic than ethanol because phenoxide is resonance-stabilised.
  • Cumene process gives phenol and acetone together.
  • Formaldehyde reacts with Grignard reagents to give primary alcohols.
  • Benzaldehyde does not reduce Fehling's solution.
  • Formic acid gives a positive Tollens' test because it has an aldehyde-like H.
  • Acid chlorides are the most reactive acid derivatives.

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

Which alcohol gives immediate turbidity with the Lucas reagent?

  1. Butan-1-ol
  2. Butan-2-ol
  3. 2-Methylpropan-2-ol
  4. Methanol
Show solution

2-Methylpropan-2-ol is tertiary: it forms a stable 3° carbocation at once, giving insoluble t-butyl chloride (turbidity) immediately.

Answer: (C)

Q2Medium

The correct order of acid strength is:

  1. p-cresol > phenol > p-nitrophenol
  2. p-nitrophenol > phenol > p-cresol
  3. phenol > p-nitrophenol > p-cresol
  4. p-nitrophenol > p-cresol > phenol
Show solution

–NO₂ (−R, −I) stabilises the phenoxide ion and increases acidity. –CH₃ (+I, hyperconjugation) destabilises it and decreases acidity.

Answer: (B)

Q3Easy

Phenol reacts with chloroform in aqueous NaOH to give (after acidification):

  1. Salicylic acid
  2. Salicylaldehyde
  3. Benzaldehyde
  4. Chlorobenzene
Show solution

This is the Reimer–Tiemann reaction; dichlorocarbene attacks the ortho position and a –CHO group results. Salicylic acid is from Kolbe's reaction (CO₂).

Answer: (B)

Q4Medium

The best way to make tert-butyl methyl ether by Williamson synthesis is:

  1. (CH₃)₃CBr + CH₃ONa
  2. CH₃Br + (CH₃)₃CONa
  3. (CH₃)₃COH + CH₃OH with H₂SO₄
  4. (CH₃)₃CBr + CH₃OH
Show solution

Williamson synthesis is SN2 and needs an unhindered halide. With t-butyl bromide, the strong base CH₃O⁻ causes elimination to isobutene. So use methyl bromide with sodium t-butoxide.

Answer: (B)

Q5Medium

Formaldehyde heated with concentrated NaOH gives:

  1. Methanol and sodium formate
  2. 3-Hydroxypropanal
  3. Ethanol
  4. Formic acid only
Show solution

HCHO has no α-hydrogen, so it undergoes the Cannizzaro reaction: one molecule is oxidised (HCOONa), the other reduced (CH₃OH).

Answer: (A)

Q6Hard

Which compound does NOT give the iodoform test?

  1. Ethanol
  2. Acetone
  3. Propan-2-ol
  4. Propan-1-ol
Show solution

The test needs CH₃CO– or CH₃CH(OH)–. Ethanol (CH₃CH₂OH → CH₃CHO), acetone and propan-2-ol have it. Propan-1-ol (CH₃CH₂CH₂OH) oxidises to propanal, which lacks a CH₃CO group.

Answer: (D)

Q7Numerical

A mixture of ethanal and propanal is treated with dilute NaOH. How many different aldol (β-hydroxy aldehyde) products can form, ignoring stereoisomers?

Show solution

Both have α-H, so each can act as nucleophile (enolate) and as electrophile: ethanal + ethanal, propanal + propanal, ethanal enolate + propanal, propanal enolate + ethanal. Total 4.

Answer: 4

Q8Hard

Which reagent converts acetophenone (C₆H₅COCH₃) into ethylbenzene?

  1. NaBH₄
  2. Zn–Hg / conc. HCl
  3. LiAlH₄
  4. Tollens' reagent
Show solution

Clemmensen reduction (Zn–Hg/HCl) reduces C=O fully to CH₂. NaBH₄ and LiAlH₄ stop at the alcohol; Tollens' reagent does not reduce ketones at all.

Answer: (B)

12. Quick Revision Summary

Remember before the exam
  • Acidity: RCOOH > PhOH > H₂O > ROH; EWG increase, EDG decrease acidity.
  • Grignard + HCHO → 1°, RCHO → 2°, ketone → 3° alcohol.
  • Kolbe → salicylic acid; Reimer–Tiemann → salicylaldehyde.
  • α-H → aldol; no α-H → Cannizzaro.
  • Clemmensen (acid) and Wolff–Kishner (base): C=O → CH₂.
  • Tests: Lucas, FeCl₃, iodoform, 2,4-DNP, Tollens', Fehling's, NaHCO₃.
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