Difficulty: Intermediate to Advanced | Prerequisites: IUPAC nomenclature basics, functional groups, hybridisation, resonance.
This is the largest study-notes set because the CBSE 2025 paper draws from five organic chapters: haloalkanes and haloarenes (Ch. 6), alcohols, phenols and ethers (Ch. 7), aldehydes, ketones and carboxylic acids (Ch. 8), amines (Ch. 10), and stereochemistry concepts from Ch. 6. Together these carry roughly 20 marks. The paper tests named reactions by name and equation, mechanism reasoning (SN1 vs SN2), IUPAC naming, distinguishing tests, acid/base strength ordering, and reaction-sequence identification.
TL;DR
SN1 favours tertiary substrates (stable carbocation); SN2 favours primary (least steric hindrance). Named reactions (Wolff-Kishner, Etard, Cannizzaro, Reimer-Tiemann) each have a specific substrate, reagent and product to memorise. Amine basicity depends on electron density on nitrogen: electron-donating groups increase it, electron-withdrawing groups and aromatic rings decrease it.
SN1 reaction
A two-step nucleophilic substitution: the leaving group departs first (forming a carbocation), then the nucleophile attacks. Rate depends only on substrate concentration. Favoured by tertiary substrates, polar protic solvents and weak nucleophiles.
SN2 reaction
A one-step nucleophilic substitution: the nucleophile attacks while the leaving group departs simultaneously (backside attack). Rate depends on both substrate and nucleophile concentration. Favoured by primary substrates, polar aprotic solvents and strong nucleophiles.
Carbocation
A positively charged carbon. Stability order: 3 degrees > 2 degrees > 1 degrees > methyl. Stabilised by hyperconjugation and inductive effects.
Enantiomers
Non-superimposable mirror-image molecules. They have identical physical properties except for the direction they rotate plane-polarised light.
Racemic mixture
An equal mixture of two enantiomers. Net optical rotation is zero.
Reimer-Tiemann reaction
Phenol + CHCl3 + NaOH --> salicylaldehyde (ortho-hydroxybenzaldehyde). Introduces a -CHO group ortho to -OH on the ring.
Wolff-Kishner reduction
Reduces a carbonyl (C=O) to a methylene (CH2) using hydrazine (H2NNH2) and a strong base (KOH) with heat.
Etard reaction
Oxidation of a methyl group on an aromatic ring to an aldehyde using CrO3 in acetic anhydride. Toluene --> benzaldehyde.
Cannizzaro reaction
A disproportionation of aldehydes without alpha-hydrogens in the presence of strong base. One molecule is oxidised to a carboxylate, the other reduced to an alcohol.
Tollens' test
Ammoniacal silver nitrate (Tollens' reagent) produces a silver mirror with aldehydes but not ketones. Used to distinguish the two.
Iodoform test
Yellow precipitate of CHI3 forms with compounds containing the CH3CO- or CH3CH(OH)- group when treated with I2/NaOH.
SN1 rate depends on carbocation stability. Order of reactivity: tertiary > secondary > primary. The compound with the more substituted halogenated carbon reacts faster via SN1.
SN2 rate depends on steric accessibility. Order of reactivity: primary > secondary > tertiary (opposite of SN1). Less steric hindrance means faster backside attack.
CBSE example: arrange 2-bromo-2-methylbutane, 1-bromopentane, 2-bromopentane in increasing SN2 reactivity. Answer: 2-bromo-2-methylbutane (tertiary, slowest) < 2-bromopentane (secondary) < 1-bromopentane (primary, fastest).
Substituted benzenes: list substituents alphabetically, assign lowest possible locants. A benzene ring with -CH3 at position 1 and -Cl at position 4 is 1-chloro-4-methylbenzene (not 1-methyl-4-chlorobenzene, because C comes before M alphabetically).
Unsaturated aldehydes: CH3=CH-CH=CH-CHO is pent-2,4-dienal. The aldehyde carbon is C1; double bonds at positions 2 and 4; the "-al" suffix indicates aldehyde.
Wolff-Kishner reduction: R-CO-R' + H2NNH2, KOH, heat --> R-CH2-R'. Converts carbonyl to methylene.
Etard reaction: C6H5CH3 + CrO3 (in acetic anhydride) --> C6H5CHO. Toluene to benzaldehyde.
Cannizzaro reaction: 2R-CHO + strong base --> R-CH2OH + R-COO-. Works only with aldehydes lacking alpha-hydrogens (e.g. formaldehyde, benzaldehyde).
Reimer-Tiemann reaction: Phenol + CHCl3 + NaOH --> salicylaldehyde. Introduces -CHO ortho to -OH.
Primary alcohols + CrO3 (or K2Cr2O7) --> aldehydes (mild oxidation) or carboxylic acids (strong oxidation).
Secondary alcohols + CrO3 --> ketones. They stop at the ketone stage; ketones are not oxidised further by these reagents.
CrO3 (chromic anhydride) selectively oxidises secondary alcohols to ketones.
Propanal (aldehyde) vs propanone (ketone): use Tollens' test. Propanal gives a silver mirror; propanone does not.
Butan-1-ol vs butan-2-ol: oxidation test. Butan-1-ol oxidises to butanoic acid (carboxylic acid); butan-2-ol oxidises to butanone (ketone). Alternatively, both give iodoform (since butan-2-ol has a CH3CH(OH)- group), but the oxidation products differ.
Ethanol < water < phenol. Phenol is the strongest acid because its conjugate base (phenoxide) is resonance-stabilised across the aromatic ring.
Chlorobenzene resists nucleophilic substitution for two reasons:
The C-Cl bond acquires partial double-bond character through resonance between the lone pair on Cl and the aromatic ring, making it stronger.
The carbon bearing the chlorine is less electrophilic because electron density from Cl resonance is delocalised into the ring.
Basicity depends on how available the nitrogen lone pair is for protonation.
Electron-donating groups (-CH3, -NH2) on an aromatic ring increase basicity by pushing electron density toward nitrogen.
Electron-withdrawing groups (-NO2, -CN) decrease basicity by pulling electron density away.
Para substituents have a stronger effect than meta substituents due to direct resonance communication.
In aqueous solution, the basicity order of aliphatic amines is: (CH3)2NH > CH3NH2 > (CH3)3N > NH3. The anomaly (secondary > tertiary) arises because tertiary amines have greater steric hindrance to solvation, reducing their effective basicity in water.
Aniline (C6H5NH2) is a weaker base than methylamine (CH3NH2). In aniline, the nitrogen lone pair is delocalised into the aromatic ring by resonance, making it less available for protonation. In methylamine, the methyl group is electron-donating by the inductive effect, increasing the lone pair's availability. Consequently, pKb of aniline > pKb of methylamine.
Ammonolysis of alkyl halides produces a mixture of primary, secondary and tertiary amines (plus quaternary ammonium salts) because the product amine is itself nucleophilic and reacts further with excess alkyl halide. This makes it unsuitable for preparing pure primary amines. Gabriel synthesis is preferred for that purpose.
Students often confuse the Reimer-Tiemann reaction (phenol --> salicylaldehyde) with the Kolbe reaction (phenol --> salicylic acid). Both involve phenol and an alkali, but the reagents and products differ.
Students assume SN1 and SN2 reactivity orders are the same. They are opposite: SN1 favours tertiary, SN2 favours primary.
The Cannizzaro reaction only works with aldehydes lacking alpha-hydrogens. Students sometimes apply it to acetaldehyde (which has alpha-hydrogens and undergoes aldol condensation instead).
Students assume amine basicity always increases with alkyl substitution. In water, secondary amines are more basic than tertiary due to solvation effects.
⚠️ Named reactions (Wolff-Kishner, Etard, Cannizzaro, Reimer-Tiemann) are guaranteed 3-mark questions. Memorise the substrate, reagent, product and equation for each.
⚠️ SN1 vs SN2 reactivity ordering is a standard 2-mark question. State the mechanism, then rank substrates.
⚠️ Amine basicity ordering appears in case-based questions (Section D, 4 marks). Know how -NO2, -NH2, -CH3 substituents affect basicity at para and meta positions.
⚠️ Distinguishing tests (Tollens', iodoform, 2,4-DNP) appear as 2-mark or 3-mark questions. State the reagent, what you observe, and why.
⚠️ IUPAC naming of aromatic compounds: alphabetical order of substituents, lowest set of locants.
True or false: SN2 reactions are faster with tertiary substrates. ______
The Cannizzaro reaction works only with aldehydes that lack ______ hydrogens.
Phenol is a stronger acid than ethanol because the ______ ion is resonance-stabilised.
Aniline is a weaker base than methylamine because the nitrogen lone pair is delocalised into the ______.
Chromic anhydride (CrO3) oxidises secondary alcohols to ______.
Q: Arrange 2-bromo-2-methylbutane, 1-bromopentane and 2-bromopentane in increasing order of SN2 reactivity.
A: 2-bromo-2-methylbutane (tertiary, slowest) < 2-bromopentane (secondary) < 1-bromopentane (primary, fastest). SN2 favours the least sterically hindered substrate.
Q: Write the chemical equation for the Wolff-Kishner reduction.
A: R-CO-R' + H2NNH2, KOH, heat --> R-CH2-R' + N2 + H2O. The carbonyl is reduced to a methylene group.
Q: Why is chlorobenzene resistant to nucleophilic substitution?
A: The lone pair on chlorine is in resonance with the aromatic ring, giving the C-Cl bond partial double-bond character (stronger, harder to break). Additionally, this resonance reduces the electrophilicity of the carbon attached to chlorine, making nucleophilic attack less favourable.
Q: Why is the pKb of aniline greater than that of methylamine?
A: In aniline, the nitrogen lone pair is delocalised into the benzene ring by resonance, reducing its availability for protonation (weaker base, higher pKb). In methylamine, the methyl group donates electron density to nitrogen by the inductive effect, making the lone pair more available (stronger base, lower pKb).
Q: Give a simple chemical test to distinguish propanal from propanone.
A: Use Tollens' test. Propanal (aldehyde) reduces ammoniacal silver nitrate to metallic silver, forming a silver mirror. Propanone (ketone) does not react.
SN1 and SN2 mechanisms connect to Coordination Compounds (Ch. 9): the Gabriel phthalimide synthesis, which avoids the over-alkylation problem of ammonolysis, relies on SN2 at an alkyl halide. The acid-strength order (ethanol < water < phenol) connects to the concept of resonance stabilisation taught in Chemical Bonding (Ch. 2, Class 11). Amine basicity connects to the Lewis acid-base theory covered in General Principles (Ch. 7, Class 11).
Tags: organic chemistry, SN1, SN2, nucleophilic substitution, carbocation, enantiomers, racemic mixture, Reimer-Tiemann, Kolbe, Wolff-Kishner, Etard, Cannizzaro, Tollens test, iodoform test, IUPAC naming, haloalkanes, haloarenes, alcohols, phenols, aldehydes, ketones, carboxylic acids, amines, basicity, aniline, methylamine, chlorobenzene, acid strength, CBSE Class 12 Chemistry, Chapter 6, Chapter 7, Chapter 8, Chapter 10