Organic Chemistry — Some Basic Principles and Techniques Notes

• Carbon has 4 valence electrons and forms 4 covalent bonds — tetravalency.
• Catenation: carbon links to carbon forming chains, branches and rings, very stable thanks to strong C–C bonds.
• Compounds can be acyclic (open chain), cyclic (ring), aromatic (benzene-like) and heterocyclic.
• Tetrahedral carbon (sp³, ~109.5°) is the common building block of organic molecules.

• Lowest locant rule for the principal functional group decides numbering.
• Substituents are named alphabetically before the root, e.g. 3-methylpentane.
• Functional-group priority for the suffix: acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine.
• Common prefixes: -chloro, -bromo, -nitro, and the alkyl group R– from alkanes.

• Stereoisomers share connectivity but differ in 3-D arrangement — geometric (cis/trans) and optical (chiral) isomers.
• A carbon with four different groups (chiral centre) gives two mirror-image (enantiomeric) forms.
• cis- and trans-2-butene are geometric isomers; their physical properties differ.

• sp³: four σ bonds, tetrahedral, 109.5° — alkanes.
• sp²: three σ + one π, planar, 120° — alkenes and carbonyl carbon.
• sp: two σ + two π, linear, 180° — alkynes and nitriles.
• A σ bond is formed by head-on overlap, π by side-on overlap of p orbitals; π bonds restrict rotation and are weaker than σ bonds.

• Hyperconjugation: σ (C–H) electrons delocalise into an empty p orbital (as in propene/toluene), stabilising carbocations.
• Aromaticity: cyclic, planar, fully conjugated with (4n+2)π electrons — benzene (6π).
• Electromeric effect: a temporary effect in the presence of a reagent, where a π bond polarises completely.

• Substitution: replacement of an atom/group (CH₄ + Cl₂ → CH₃Cl + HCl).
• Addition: π bond becomes σ bonds across the double/triple bond.
• Elimination: removal of atoms to form a π bond (dehydration of alcohols).
• Nucleophile = electron-rich species (OH⁻, CN⁻); electrophile = electron-poor (H⁺).
• Homolytic cleavage gives radicals; heterolytic cleavage gives ions.

Example: Name CH₃CH(CH₃)CH₂CH₃ by IUPAC rules.

Solution: Longest chain has four carbons → butane; the methyl substituent is on carbon 2 → 2-methylbutane.

Example: Why is the tertiary carbocation most stable?

Solution: More alkyl groups donate electron density through the +I effect and hyperconjugation, spreading and stabilising the positive charge — so the stability order is 3° > 2° > 1° > CH₃⁺.

Exercises (NCERT Q8.1–Q8.40)

• 8.1) Determine the hybridisation state of every carbon atom in: (i) CH₂=C=O. (ii) CH₃CH=CH₂. (iii) (CH₃)₂CO. (iv) CH₂=CH–CN. (v) C₆H₆.
• 8.2) Indicate the σ and π bonds in: (i) benzene, C₆H₆; (ii) cyclohexane, C₆H₁₂; (iii) CH₂Cl₂; (iv) CH₂=C=CH₂; (v) CH₃NO₂; and (vi) HCONHCH₃.
• 8.3) Write bond-line descriptions or formulas for isopropyl alcohol, 2,3-dimethylbutanal and heptan-4-one.
• 8.4) Give the IUPAC names of the six displayed compounds: (a) a benzene ring bearing a propyl substituent; (b) CH₃CH₂CH(CH₃)CH₂CN; (c) CH₃CH(CH₃)CH₂CH₂CH(CH₃)CH₂CH₃; (d) CH₃CH₂C(Br)(Cl)CH₂CH₂CH₂CH₃; (e) ClCH₂CH₂CHO; and (f) Cl₂CHCH₂OH.
• 8.5) Choose the correct IUPAC name in each pair: (a) 2,2-dimethylpentane or 2-dimethylpentane; (b) 2,4,7-trimethyloctane or 2,5,7-trimethyloctane; (c) 2-chloro-4-methylpentane or 4-chloro-2-methylpentane; (d) but-3-yn-1-ol or but-4-ol-1-yne.
• 8.6) Write the first five members, beginning with the given compound, of the homologous series containing: (a) H–COOH; (b) CH₃COCH₃; and (c) H–CH=CH₂.
• 8.7) Give condensed structural formulae and identify the functional groups in: (a) 2,2,4-trimethylpentane; (b) 2-hydroxypropane-1,2,3-tricarboxylic acid; and (c) hexane-1,6-dial.
• 8.8) Identify all functional groups in the three displayed compounds. The drawings contain, respectively, an aldehyde with an alcohol, ether and alkene; a molecule with primary-amine, ester and tertiary-amine functionality; and a nitro compound with an alkene.
• 8.9) Which ion is more stable, O₂N–CH₂–CH₂–O⁻ or CH₃CH₂–O⁻? Explain.
• 8.10) Explain why an alkyl group acts as an electron donor when attached to a π system.
• 8.11) Draw the resonance contributors for phenol, nitrobenzene, CH₃CH=CHCHO, benzaldehyde, benzyl carbocation (C₆H₅CH₂⁺) and the allylic carbocation CH₃CH=CHCH₂⁺. Show which electrons move.
• 8.12) What are electrophiles and nucleophiles? Give examples of ionic and neutral species.
• 8.13) Classify the bold reagent in each reaction: (a) CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O; (b) CH₃COCH₃ + CN⁻ → (CH₃)₂C(CN)(OH); (c) C₆H₆ + CH₃CO⁺ → C₆H₅COCH₃.
• 8.14) Classify each reaction as substitution, addition, elimination, or substitution followed by rearrangement: (a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻; (b) (CH₃)₂C=CH₂ + HCl → (CH₃)₂CCl–CH₃; (c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻; (d) (CH₃)₃C–CH₂OH + HBr → (CH₃)₂CBr–CH₂–CH₃ + H₂O.
• 8.15) Classify each displayed pair as structural isomers, geometrical isomers or resonance contributors.
• 8.16) For the four displayed bond cleavages, show the electron flow and identify the process and intermediate: (a) a cleavage in which each fragment receives one electron; (b) a cleavage in propanone in which the bonding pair remains on carbon; (c) a carbon–bromine cleavage in which the bonding pair remains on bromine; and (d) a cleavage in which one carbon fragment receives both bonding electrons.
• 8.17) Explain the inductive and electromeric effects, and account for these acidity orders: (a) CCl₃COOH > CHCl₂COOH > CH₂ClCOOH; (b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.
• 8.18) Describe the principle of crystallisation, distillation and chromatography, giving one example of each.
• 8.19) Describe fractional crystallisation for separating two compounds that have different solubilities in solvent S.
• 8.20) Distinguish ordinary distillation, distillation under reduced pressure and steam distillation.
• 8.21) Discuss the chemistry of the Lassaigne tests for nitrogen, sulphur and halogens after sodium fusion.
• 8.22) Compare the principles of the Dumas and Kjeldahl methods for estimating nitrogen in an organic compound.
• 8.23) Give the principle of estimating halogens, sulphur and phosphorus in an organic compound.
• 8.24) Explain the principle of paper chromatography.
• 8.25) Why is the Lassaigne extract boiled with dilute nitric acid before silver nitrate is added in the halogen test?
• 8.26) Explain why an organic compound is fused with metallic sodium before testing nitrogen, sulphur and halogens.
• 8.27) Name a suitable technique for separating camphor from calcium sulphate.
• 8.28) Why does an organic liquid vaporise below its boiling point during steam distillation?
• 8.29) Will CCl₄ give a white precipitate of AgCl on heating with silver nitrate? Give a reason.
• 8.30) Why is potassium hydroxide solution used to absorb the carbon dioxide evolved during carbon estimation?
• 8.31) Why is acetic acid, rather than sulphuric acid, used to acidify the sodium extract in the lead-acetate test for sulphur?
• 8.32) A 0.20 g organic compound contains 69.0% carbon and 4.8% hydrogen. Calculate the masses of CO₂ and H₂O formed on complete combustion.
• 8.33) A 0.50 g sample was treated by the Kjeldahl method. NH₃ was absorbed in 50 mL of 0.5 M H₂SO₄, and the residual acid required 60 mL of 0.5 M NaOH for neutralisation. Find the percentage of nitrogen.
• 8.34) A 0.3780 g organic chloro compound gave 0.5740 g AgCl in the Carius method. Calculate the percentage of chlorine.
• 8.35) A 0.468 g organic sulphur compound afforded 0.668 g BaSO₄. Find the percentage of sulphur.
• 8.36) In CH₂=CH–CH₂–CH₂–C≡CH, the bond C₂–C₃ is formed by which pair of hybrid orbitals? (a) sp–sp²; (b) sp–sp³; (c) sp²–sp³; (d) sp³–sp³.
• 8.37) In the Lassaigne test for nitrogen, the Prussian blue colour is due to formation of: (a) Na₄[Fe(CN)₆]; (b) Fe₄[Fe(CN)₆]₃; (c) Fe₂[Fe(CN)₆]; (d) Fe₃[Fe(CN)₆]₄.
• 8.38) Which carbocation is most stable? (a) (CH₃)₃C–CH₂⁺; (b) (CH₃)₃C⁺; (c) CH₃CH₂CH₂⁺; (d) CH₃CH⁺CH₂CH₃.
• 8.39) The best and latest technique for isolation, purification and separation of organic compounds is: (a) crystallisation; (b) distillation; (c) sublimation; or (d) chromatography.
• 8.40) Classify CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI as electrophilic substitution, nucleophilic substitution, elimination or addition.

Answers at a glance

1. 1
   Q8.1 → (i) sp², sp; (ii) sp³, sp², sp²; (iii) sp³, sp², sp³; (iv) sp², sp², sp; (v) all six carbons are sp² hybridised.
2. 2
   Q8.2 → The counts are: (i) 6 C–C σ, 6 C–H σ, 3 π; (ii) 6 C–C σ and 12 C–H σ; (iii) 2 C–H σ and 2 C–Cl σ; (iv) 2 C–C σ, 4 C–H σ and 2 π; (v) 3 C–H σ, 1 C–N σ, 1 N–O σ and 1 π; (vi) 4 C–H σ, 1 N–H σ, 2 C–N σ, 1 C–O σ and 1 C=O π.
3. 3
   Q8.3 → The three structures are (CH₃)₂CH–OH, OHC–CH(CH₃)–CH(CH₃)–CH₃ and CH₃CH₂CH₂–CO–CH₂CH₂CH₃, respectively.
4. 4
   Q8.4 → (a) propylbenzene; (b) 3-methylpentanenitrile; (c) 2,5-dimethylheptane; (d) 3-bromo-3-chloroheptane; (e) 3-chloropropanal; (f) 2,2-dichloroethan-1-ol.
5. 5
   Q8.5 → The correct names are (a) 2,2-dimethylpentane, (b) 2,4,7-trimethyloctane, (c) 2-chloro-4-methylpentane and (d) but-3-yn-1-ol.
6. 6
   Q8.6 → (a) methanoic through pentanoic acid; (b) propanone, butanone, pentan-2-one, hexan-2-one and heptan-2-one; (c) ethene, propene, but-1-ene, pent-1-ene and hex-1-ene.
7. 7
   Q8.7 → (a) (CH₃)₃C–CH₂–CH(CH₃)–CH₃, with no functional group; (b) HOOC–CH₂–C(OH)(COOH)–CH₂–COOH, with three –COOH and one –OH; (c) OHC–(CH₂)₄–CHO, with two –CHO groups.
8. 8
   Q8.8 → (a) aldehyde, hydroxyl, methoxy/ether and C=C; (b) primary amine, ester and tertiary amine; (c) nitro and C=C double bond.
9. 9
   Q8.9 → O₂N–CH₂–CH₂–O⁻ is more stable because the electron-withdrawing nitro group stabilises the negative charge, whereas the electron-releasing ethyl group destabilises ethoxide.
10. 10
    Q8.10 → Alkyl groups donate through hyperconjugation: electrons from a σ bond on the adjacent sp³ carbon overlap with the π-system p orbital and become delocalised into the π bond.
11. 11
    Q8.11 → The contributors are obtained by moving π electrons or a lone pair while keeping every atom in the same position: phenol has ortho/para ring-charge contributors, nitrobenzene has equivalent N–O contributors, the enal and benzaldehyde have carbonyl-conjugated contributors, the benzyl carbocation has ring-delocalised positive charge, and the allylic carbocation has the two terminal-carbocation contributors.
12. 12
    Q8.12 → Electrophiles accept electron pairs and nucleophiles donate them; H⁺, R⁺ and BF₃ are electrophiles, while OH⁻, CN⁻, R⁻, NH₃ and H₂O are nucleophiles.
13. 13
    Q8.13 → (a) HO⁻ is a nucleophile; (b) CN⁻ is a nucleophile; (c) CH₃CO⁺ is an electrophile.
14. 14
    Q8.14 → (a) substitution; (b) addition; (c) elimination; (d) substitution followed by rearrangement.
15. 15
    Q8.15 → (a) structural isomers; (b) geometrical isomers; (c) resonance contributors.
16. 16
    Q8.16 → (a) homolysis, free radical; (b) heterolysis, carbanion; (c) heterolysis, carbocation; (d) heterolysis, carbocation.
17. 17
    Q8.17 → The –I effect of increasing chlorine explains (a); the increasing +I effect of alkyl groups explains the decreasing acidity in (b). The source wording that +I increases acidity is corrected: +I destabilises the conjugate base and lowers acidity.
18. 18
    Q8.18 → Crystallisation uses temperature-dependent solubility, distillation uses different volatilities or boiling points, and chromatography uses differential movement through stationary and mobile phases.
19. 19
    Q8.19 → Fractional crystallisation separates the mixture by successive crystallisation: the less soluble compound is removed first, followed by the more soluble compound from the concentrated mother liquor.
20. 20
    Q8.20 → Ordinary distillation uses boiling-point differences at atmospheric pressure; reduced-pressure distillation protects heat-sensitive compounds by lowering the boiling point; steam distillation co-distils an immiscible organic liquid with steam below its normal boiling point.
21. 21
    Q8.21 → Sodium fusion creates CN⁻, S²⁻ or X⁻, which are identified by the characteristic cyanide–iron Prussian-blue reaction, PbS or nitroprusside tests for sulphur, and coloured AgX precipitates for halogens.
22. 22
    Q8.22 → Dumas measures N₂ liberated by complete oxidation; Kjeldahl measures NH₃ released from an ammonium salt and determined by acid–base back-titration.
23. 23
    Q8.23 → Halogens are weighed as AgX, sulphur as BaSO₄, and phosphorus as ammonium phosphomolybdate or Mg₂P₂O₇; each percentage follows from the one-to-one elemental stoichiometry.
24. 24
    Q8.24 → Paper chromatography separates components by differential partition between water in the paper (stationary phase) and a moving solvent (mobile phase); the resulting chromatogram identifies the components through their relative migration.
25. 25
    Q8.25 → Dilute nitric acid decomposes cyanide and sulphide in the extract and boiling expels HCN and H₂S, preventing interference before the halide ions are precipitated as AgX.
26. 26
    Q8.26 → Sodium fusion converts covalently bound N, S and halogens into ionic sodium salts, making them detectable in the aqueous extract.
27. 27
    Q8.27 → Use sublimation followed by condensation of the vapour; camphor sublimes while calcium sulphate does not.
28. 28
    Q8.28 → Steam distillation co-distils the immiscible organic liquid because p_atm = p_organic + p_water; the total pressure is reached before the organic liquid's own boiling point.
29. 29
    Q8.29 → No. CCl₄ does not give AgCl directly because its chlorine is covalently bonded; sodium fusion is needed to produce Cl⁻.
30. 30
    Q8.30 → KOH quantitatively absorbs acidic CO₂ as K₂CO₃; the increase in the absorber's mass is the mass of CO₂ and therefore determines the carbon content.
31. 31
    Q8.31 → Acetic acid provides mild acidification and avoids sulfate interference. The sulphur product is black PbS, not PbSO₄; this corrects the source explanation.
32. 32
    Q8.32 → The sample produces 0.506 g of CO₂ and 0.0864 g of H₂O.
33. 33
    Q8.33 → The organic compound contains 56% nitrogen by mass.
34. 34
    Q8.34 → The chlorine content is 37.59% by mass.
35. 35
    Q8.35 → The sulphur content is 19.59% by mass. The source arithmetic used 0.0197 g; the correct sulphur mass is 0.0917 g.
36. 36
    Q8.36 → Option (b), sp–sp³.
37. 37
    Q8.37 → Option (b), Fe₄[Fe(CN)₆]₃, the Prussian-blue complex (usually hydrated).
38. 38
    Q8.38 → Option (b), the tertiary carbocation (CH₃)₃C⁺, is the most stable.
39. 39
    Q8.39 → Option (d), chromatography.
40. 40
    Q8.40 → Option (b), nucleophilic substitution.

MCQ Questions

1 Mark Questions

1. 1
   What is the functional group of an alcohol? (Ans: –OH (hydroxyl))
2. 2
   Define homolytic cleavage. (Ans: breaking of a covalent bond in which each fragment gets one of the shared electrons — forms free radicals)
3. 3
   What is a carbocation? (Ans: a species with a carbon carrying a positive charge and six electrons, planar sp², most stable when tertiary)
4. 4
   Write the IUPAC name of CH₃–CO–CH₃. (Ans: propan-2-one (acetone))

2 Mark Questions

1. 1
   Differentiate between electrophiles and nucleophiles with one example each. (Ans: electron-pair acceptors e.g. H⁺, N⁺O₂ vs donors e.g. OH⁻, NH₃)
2. 2
   Define isomerism and give one pair each of chain and position isomers. (Ans: butane / 2-methylpropane; 1-chloropropane / 2-chloropropane)
3. 3
   Which is more stable — a primary or a tertiary carbocation — and why? (Ans: tertiary, due to hyperconjugation and +I effect of alkyl groups)

3 Mark Questions

1. 1
   Explain the inductive effect and the +I and −I series of groups, giving one example of each and its effect on acidity/basicity. (Ans: permanent polarisation along σ chain; alkyl +I; –NO₂, –Cl −I — more −I groups make acids stronger)
2. 2
   Describe resonance with the example of the carboxylate ion or benzene, and state the conditions for resonance along with its effect on stability.
3. 3
   Give the IUPAC names of: (i) CH₃CH(CH₃)CH₂CH₃ (ii) CH₂=CH–CH₂Cl (iii) CH₃C≡CCH₃. (Ans: 2-methylbutane; 3-chloroprop-1-ene; but-2-yne)

5 Mark Questions

1. 1
   Classify organic reactions into substitution, addition, elimination and rearrangement giving one example of each, and explain the mechanism of electrophilic addition to an alkene (Markovnikov's rule).
2. 2
   Discuss the electron displacement effects — inductive effect, electromeric effect, resonance and hyperconjugation — with their characteristic features and one application of each; also state how they affect the stability of intermediates and the acidity of carboxylic acids.