Organic Chemistry Class 12 Notes

The parent chain (longest continuous carbon chain) is numbered to give the lowest locant to the principal functional group. The suffix reflects the functional group; substituents are listed alphabetically with their locants.

• Priority order for suffixes:
• $-COOH>-S{O}_{3}H>-COOR>-COCl>-CON{H}_2>-CN>-CHO>-CO->-OH>-N{H}_2$
• .
• For multiple bonds use -ene, -yne; for functional groups use the appropriate suffix.
• Numbering: principal functional group gets the lowest possible number.

• $\text{Carbocation:}$
• sp², planar, electron-deficient — stability order
• $3^{\circ }>2^{\circ }>1^{\circ }>{\text{CH}}_3^{+}$
• .
• $\text{Carbanion:}$
• electron-rich, pyramidal — stability order
• ${\text{CH}}_3^{-}>1^{\circ }>2^{\circ }>3^{\circ }$
• .
• $\text{Free radical:}$
• stability
• $3^{\circ }>2^{\circ }>1^{\circ }$
• .
• SN1 favours tertiary substrates (carbocation intermediate); SN2 favours primary (concerted backside attack).

Haloalkanes are formed by the replacement of H by halogen. Reactivity for SN2: $\text{R-I}>\text{R-Br}>\text{R-Cl}>\text{R-F}$. Haloarenes are much less reactive toward nucleophilic substitution than haloalkanes.

• Finkelstein (NaI in acetone) converts R-Cl to R-I.
• Swarts reaction (AgF/Hg₂F₂) converts R-Cl/R-Br to R-F.
• Sandmeyer reaction: aryl diazonium + CuCl/CuBr gives halobenzenes.
• Grignard reagent (RMgX) — a strong nucleophile and base.

Alcohols are distinguished by their boiling points (H-bonding) and acidity. Phenols are acidic because the phenoxide ion is resonance-stabilised; electron-withdrawing groups increase phenol acidity.

• Carbonyl carbon is attacked by nucleophiles; aldehydes are more reactive than ketones (less steric hindrance, more polar C=O).
• Tollens' reagent (ammoniacal AgNO₃) — aldehydes give a silver mirror, ketones do not.
• Fehling's test — aldehydes give a red Cu₂O precipitate; aromatic aldehydes do not.
• Iodoform test — positive for CH₃-CO- group and secondary alcohols that oxidise to it.
• Clemmensen (Zn-Hg/HCl) and Wolff-Kishner (NH₂NH₂/KOH) reduce a carbonyl to a methylene group.

Carboxylic acids are stronger acids than phenols and alcohols because the carboxylate anion is resonance stabilised. Electron-withdrawing groups (like -NO₂, -Cl) increase acidity; electron-donating groups decrease it.

• Decarboxylation: heating a sodium salt with soda-lime gives an alkane.
• Reduction with LiAlH₄ gives a primary alcohol.
• Acid chloride formation via PCl₅, PCl₃ or SOCl₂.
• Hell-Volhard-Zelinsky (HVZ) reaction: α-halogenation of acids with halogen + red phosphorus.

Amines are classified as primary, secondary or tertiary. Aliphatic amines are more basic than ammonia because alkyl groups donate electron density to nitrogen; aromatic amines (aniline) are much less basic.

• Basicity in gas phase:
• $3^{\circ }>2^{\circ }>1^{\circ }>{\text{NH}}_3$
• ; in aqueous phase solvation effects change the order.
• Hinsberg test distinguishes 1°, 2° and 3° amines using benzenesulphonyl chloride.
• Carbylamine reaction: a primary amine + CHCl₃ + KOH gives a foul-smelling isocyanide.
• Diazotisation of aromatic primary amines with NaNO₂ + HCl gives diazonium salts.

Example: Arrange in order of basicity: aniline, methylamine, dimethylamine, trimethylamine (aqueous).

Solution: In aqueous medium the order is $2^{\circ }>1^{\circ }>3^{\circ }>{\text{NH}}_3>\text{aniline}$. Dimethylamine (2°) is most basic; aniline is least basic due to resonance delocalisation of the N lone pair into the ring.

Exercises (NCERT Q1–Q4)

• 1) Write the IUPAC names of the following compounds: (i) (CH3)3CCH2CH2OH (ii) CH3CH2CH(Br)CH2CH2CH3 (iii) CH3CH=CHC(O)CH3
• 2) Arrange the following compounds in increasing order of their boiling points: CH3CH2CH2CH3, CH3CH2OCH2CH3, CH3CH2CH2OH
• 3) Explain why o-nitrophenol is more acidic than o-methoxyphenol.
• 4) Give a chemical test to distinguish between ethanol and ethanoic acid.

Answers at a glance

1. 1
   Q1 → (i) 3,3-Dimethylbutan-1-ol (ii) 3-Bromohexane (iii) Pent-3-en-2-one
2. 2
   Q2 → Butane < Ether < Propanol
3. 3
   Q3 → NO2 group stabilizes phenoxide ion via -I effect; OCH3 group destabilizes it via +I effect
4. 4
   Q4 → Ethanoic acid gives brisk effervescence with NaHCO3 due to CO2 evolution; ethanol does not react

MCQ Questions

1 Mark Questions

1. 1
   Write the structure of 4-methylpent-3-en-2-one.
2. 2
   Why is electrophilic substitution in benzene difficult despite presence of three double bonds?
3. 3
   Arrange in increasing order of basicity: NH3, CH3NH2, (CH3)2NH, (CH3)3N.
4. 4
   Give a chemical test to distinguish between phenol and benzoic acid.

2 Mark Questions

1. 1
   Explain why chloroacetic acid is stronger acid than acetic acid.
2. 2
   Describe the mechanism of esterification of acetic acid with ethanol in presence of concentrated H2SO4.
3. 3
   How will you convert benzoic acid to benzaldehyde?

3 Mark Questions

1. 1
   An organic compound A (C4H10O) on oxidation with acidified potassium dichromate gives acid B (C4H8O2). Compound B on reaction with phosphorus pentachloride gives compound C (C4H7OCl) which on treatment with ammonia gives compound D (C4H9NO). Identify A, B, C and D.
2. 2
   Explain the following: (i) Kolbe's reaction (ii) Reimer-Tiemann reaction (iii) Williamson ether synthesis
3. 3
   How will you bring about the following conversions: (i) Propene to propan-2-ol (ii) Ethanol to but-1-yne

5 Mark Questions

1. 1
   Explain the mechanism of acid-catalyzed dehydration of ethanol to yield ethene.
2. 2
   An organic compound (A) with molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen's reagent and undergoes Cannizzaro's reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify compound A and explain the reactions involved.

Assignments walk the chapter's topics in order; the tests follow the current CBSE blueprint (MCQ/VSA, 2M, 3M and 5M questions) with answer keys. They will appear here as direct download links.