Organic Chemistry — Some Basic Principles and Techniques

Welcome to HSLC Guru. This page provides clear, exam-ready notes and complete question–answer practice for ASSEB Class 11 Chemistry Chapter 12, “Organic Chemistry — Some Basic Principles and Techniques.” Organic chemistry is the chemistry of carbon compounds, and this chapter introduces the basic vocabulary, structural ideas, electronic effects, reaction types, and laboratory techniques that you will use throughout the rest of the course. Use the summary, Q&A, MCQ, fill-in-the-blanks, true/false, and glossary sections together for thorough revision.

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Chapter Summary

Carbon is unique among elements because of its tetravalency and catenation — the ability to form four covalent bonds and to link with itself in long chains, branched chains, and rings. The four bonds around carbon point to the corners of a regular tetrahedron when carbon is sp3 hybridised, giving rise to the three-dimensional shapes of organic molecules. Carbon also forms multiple bonds (double, triple) by sp2 and sp hybridisation. Organic structures may be shown as Lewis (electron-dot) structures, complete structural formulae, condensed formulae such as CH3CH2OH, or bond-line (zig-zag) formulae, where each vertex and end represents a carbon and hydrogens are implicit. Three-dimensional shapes are shown using wedge–dash notation.

Organic compounds are classified first as acyclic (open-chain) or cyclic; cyclic compounds are further divided into alicyclic, aromatic (benzenoid and non-benzenoid), and heterocyclic (containing N, O, S in the ring). They are also classified by the functional group present — alcohol (–OH), aldehyde (–CHO), ketone (>C=O), carboxylic acid (–COOH), amine (–NH2), halide (–X), nitro (–NO2), nitrile (–CN), ether (–O–), ester (–COO–) and so on. Compounds having the same functional group and differing by –CH2– form a homologous series. IUPAC nomenclature assigns each compound a systematic name built from a root (longest chain), suffix (principal functional group), prefix (substituents), and locants chosen to give the lowest set of numbers; substituents are listed alphabetically and numbered to give the principal group the lowest locant.

Organic reactions are classified as substitution, addition, elimination, and rearrangement. They proceed via reactive intermediates — carbocations (positively charged, electron-deficient C, sp2, stabilised by +I and hyperconjugation; order 3° > 2° > 1° > methyl), carbanions (negatively charged C, sp3 pyramidal, stabilised by –I groups; order methyl > 1° > 2° > 3°), free radicals (one unpaired electron, sp2, stability 3° > 2° > 1° > methyl) and carbenes (neutral six-electron carbon). Bond cleavage may be homolytic (one electron to each fragment, giving radicals) or heterolytic (both electrons to one fragment, giving ions). Reagents are classified as electrophiles (electron-loving, e.g. H+, NO2+, AlCl3) or nucleophiles (electron-pair donors, e.g. OH–, CN–, NH3).

Permanent and temporary electron displacement effects control reactivity: inductive effect (+I, –I — permanent, transmitted through sigma bonds, decreases with distance), electromeric effect (+E, –E — temporary, occurs at the moment of attack of a reagent on a multiple bond), resonance/mesomeric effect (+M, –M — delocalisation of pi electrons in conjugated systems), and hyperconjugation (sigma-pi or “no-bond resonance,” involving C–H sigma bonds adjacent to a multiple bond or positive carbon — explains the stability order of carbocations and free radicals and the heat of hydrogenation trend in alkenes). For purification we use crystallisation (for solids of differing solubility), sublimation (for solids that pass directly to vapour, e.g. naphthalene, camphor), simple distillation (large boiling-point gap), fractional distillation (close boiling points), distillation under reduced pressure (heat-sensitive liquids), steam distillation (water-immiscible volatile substances such as aniline, nitrobenzene), and chromatography (adsorption — TLC and column — or partition — paper). For qualitative analysis, the Lassaigne sodium fusion test converts N, S and halogens into ionic forms (NaCN, Na2S, NaX); these are detected as Prussian blue (N), violet/black PbS or sodium nitroprusside violet colour (S), and silver halide precipitates (Cl white, Br pale yellow, I yellow). Quantitative analysis includes Liebig’s combustion (C and H), Dumas and Kjeldahl methods (N), Carius (halogens, S), and oxygen by difference.

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1-Mark Questions

Q1. What is the valency of carbon in organic compounds?

Answer: Four (tetravalent).

Q2. Define catenation.

Answer: The self-linking of atoms of the same element through covalent bonds to form long chains or rings; carbon shows it strongly.

Q3. Write the bond-line formula of butane.

Answer: A simple zig-zag with four carbons: /\/.

Q4. Give the IUPAC name of (CH3)2CHCH2OH.

Answer: 2-Methylpropan-1-ol.

Q5. Name a reactive intermediate having an unpaired electron.

Answer: Free radical.

Q6. Which method is used to separate two miscible liquids with very close boiling points?

Answer: Fractional distillation.

Q7. What colour confirms the presence of nitrogen in Lassaigne’s test?

Answer: Prussian blue.

Q8. Which reagent in Lassaigne’s test detects sulphur?

Answer: Sodium nitroprusside (gives violet colour) or lead acetate (gives black PbS).

Q9. Name an electrophile and a nucleophile.

Answer: Electrophile: NO2+; Nucleophile: OH– (or CN–).

Q10. Which method is used to purify aniline?

Answer: Steam distillation.

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2 to 3-Mark Questions

Q1. Differentiate between homolytic and heterolytic bond cleavage with one example each.

Answer: In homolytic cleavage the shared pair is split equally and each atom takes one electron, giving free radicals (e.g. Cl–Cl → Cl• + Cl• under UV light). In heterolytic cleavage both electrons go to one atom, producing a cation and an anion (e.g. CH3–Br → CH3+ + Br– in polar solvents). Homolysis is favoured by non-polar solvents and high-energy radiation; heterolysis is favoured by polar solvents and polar bonds.

Q2. Explain the inductive effect with examples of +I and –I groups.

Answer: The inductive effect is the permanent partial polarisation of a sigma bond caused by an attached atom or group. +I groups push electrons toward the chain (alkyl groups, –CH3, –C2H5). –I groups withdraw electrons (–NO2, –CN, –COOH, –X, –OH, –NH3+). The effect decreases rapidly with distance; it influences acid and base strength (e.g. CCl3COOH is more acidic than CH3COOH because three –I chlorines stabilise the carboxylate anion).

Q3. Give the IUPAC names of (i) CH3CH(CH3)CH2CH2OH and (ii) CH3COCH2CH3.

Answer: (i) 3-Methylbutan-1-ol; (ii) butan-2-one.

Q4. What is hyperconjugation? How does it stabilise carbocations?

Answer: Hyperconjugation is the delocalisation of sigma electrons of a C–H bond on the carbon adjacent to an sp2 centre (carbocation, alkene, radical) into the empty p orbital or the pi system. The greater the number of alpha C–H bonds available, the greater the stabilisation; this explains the order 3° > 2° > 1° > methyl for carbocations.

Q5. Briefly describe TLC (thin-layer chromatography).

Answer: A small spot of the mixture is applied near the bottom of a glass plate coated with silica gel or alumina (stationary phase). The plate is dipped into a solvent (mobile phase), which rises by capillary action carrying the components at different rates. After drying, spots are visualised by UV light or iodine vapour. The Rf value = (distance moved by component)/(distance moved by solvent) is characteristic of a substance under given conditions.

Q6. State Markovnikov’s rule and give one example.

Answer: When an unsymmetrical reagent HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon already bearing more hydrogens, and X to the carbon with fewer hydrogens. Example: propene + HBr → 2-bromopropane (major), explained by the greater stability of the secondary carbocation intermediate.

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5 to 7-Mark Questions

Q1. Describe the IUPAC rules for naming branched-chain alkanes and substituted compounds with examples.

Answer: (i) Identify the longest continuous carbon chain that contains the principal functional group — this gives the parent name. (ii) Number the chain so the principal functional group gets the lowest locant; if there is no functional group, number from the end nearer to the first substituent (lowest set of locants). (iii) Identify substituents and prefix them with their locants, listing alphabetically (di-, tri- prefixes are ignored for alphabetisation). (iv) Use the suffix corresponding to the highest-priority functional group (–oic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine), replacing the “e” of the parent alkane (e.g. propan-1-ol). (v) Multiple bonds get suffixes -ene/-yne with locants. Examples: CH3CH(CH3)CH2CH3 = 2-methylbutane; CH3CH=CHCH2OH = but-2-en-1-ol; (CH3)2C(Cl)CH2COOH = 3-chloro-3-methylbutanoic acid.

Q2. Explain resonance with reference to benzene and the carboxylate ion. Give the rules for writing canonical structures.

Answer: When a single Lewis structure cannot represent a molecule, two or more contributing structures (canonical or resonance forms) are drawn that differ only in the position of pi or lone-pair electrons. The actual molecule is a hybrid of these forms and is more stable than any single one. Rules: atoms must remain in the same positions; only pi or lone-pair electrons are moved; each canonical form must have the same number of paired electrons; structures with more covalent bonds, complete octets, and minimum charge separation contribute more. Benzene: two Kekulé structures with alternating double bonds give a planar hexagonal hybrid with all C–C bonds equal (139 pm) — exceptional stability (resonance energy ≈ 152 kJ/mol). Carboxylate ion (RCOO–): the negative charge is shared equally between the two oxygen atoms, both C–O bonds are equivalent (≈127 pm), and the anion is greatly stabilised — explaining the acidity of carboxylic acids.

Q3. Write a detailed note on Lassaigne’s test for the detection of nitrogen, sulphur and halogens.

Answer: A small amount of organic compound is fused with metallic sodium in a fusion tube. Covalent N, S and X atoms are converted to ionic NaCN, Na2S and NaX (a single S+N compound forms NaSCN). The red-hot tube is plunged into distilled water and crushed; the filtrate (sodium fusion extract) is divided into portions. Nitrogen: the extract is boiled with FeSO4, acidified with dilute H2SO4 — a Prussian blue colour (Fe4[Fe(CN)6]3) confirms N. Sulphur: (a) freshly prepared sodium nitroprusside gives a violet colour with S2–; (b) lead acetate after acidification with acetic acid gives a black PbS precipitate. Halogens: the extract is acidified with dilute HNO3 (boiled to remove HCN/H2S if N or S also present) and treated with AgNO3 — white curdy precipitate soluble in NH4OH = Cl; pale yellow partially soluble = Br; yellow insoluble = I. Where N and X are both present, sodium thiocyanate forms blood-red colour with FeCl3 (or sodium fusion extract is boiled with concentrated HNO3 to destroy CN– before AgNO3 test).

Q4. Describe the Carius method for the estimation of halogen and the Dumas method for the estimation of nitrogen, with the working formulae.

Answer: Carius method (halogen): a known mass of organic compound is heated with fuming HNO3 and a little AgNO3 in a sealed Carius tube at ~300 °C. The halogen is converted to AgX, which is filtered, washed, dried and weighed.

$\%X = \frac{\text{atomic mass of X} \times \text{mass of AgX}}{\text{molar mass of AgX} \times \text{mass of organic substance}} \times 100$

Dumas method (nitrogen): a weighed sample is heated with excess CuO in an atmosphere of CO2 in a combustion tube. C → CO2, H → H2O, N → N2 (any oxides of nitrogen formed are reduced to N2 by the hot copper gauze placed at the end of the tube). The N2 is collected over concentrated KOH (which absorbs CO2) and its volume measured at known T and P.

$\%N = \frac{28 \times \text{volume of N}_2 \text{ at STP}}{22400 \times \text{mass of organic substance}} \times 100$

The Dumas method is applicable to all nitrogen-containing compounds. The Kjeldahl method (heating with concentrated H2SO4, distilling NH3 into standard acid) is simpler but fails for compounds containing N in rings or –NO2 groups.

Q5. Compare crystallisation, simple distillation, fractional distillation, steam distillation and column chromatography, giving one application of each.

Answer: Crystallisation separates a solid impurity-free product from a saturated hot solution that deposits crystals on cooling — used for purifying benzoic acid. Simple distillation separates a volatile liquid from a non-volatile solid (or two liquids whose boiling points differ by >25 °C) — purifying chloroform from impurities. Fractional distillation uses a fractionating column with multiple plates/packing to separate liquids of close boiling points — separation of crude oil into petroleum fractions; separation of acetone (56 °C) and methanol (65 °C). Steam distillation applies to substances volatile in steam, immiscible with water and having low solubility in water — purification of aniline, nitrobenzene and essential oils. Column chromatography uses a glass column packed with adsorbent (silica gel/alumina); the mixture in solvent is poured at the top and eluted with a suitable solvent — components emerge at different times and are collected separately. Used in purifying organic and biological compounds in the laboratory and industry.

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Q6. Discuss the four common types of organic reactions with one example each.

Answer: (i) Substitution — one atom or group of an organic molecule is replaced by another. Example: methane + Cl2 (light) → chloromethane + HCl, a free-radical substitution. (ii) Addition — atoms or groups add across a multiple bond, increasing the saturation of the molecule. Example: ethene + Br2 → 1,2-dibromoethane, an electrophilic addition. (iii) Elimination — two atoms or groups leave the same or adjacent carbons, forming a multiple bond. Example: ethanol → ethene + H2O on heating with concentrated H2SO4 at 170 °C. (iv) Rearrangement — atoms within the molecule are reorganised, often through carbocation migration. Example: 1-butene rearranges in acid to give 2-butene (more stable). Each type has a typical mechanism, intermediate, and product distribution that you can predict from electronic effects and steric factors.

Q7. Distinguish between inductive effect and electromeric effect.

Answer: Inductive effect operates through sigma bonds, is permanent, partial, decreases sharply with distance, and operates in saturated and unsaturated systems. Electromeric effect operates through pi bonds, is temporary (only in the presence of an attacking reagent), involves complete transfer of pi electrons, and operates only in unsaturated systems. +E pushes pi electrons toward the attacking reagent (e.g. addition of HCN to >C=O); –E withdraws them away (e.g. addition of H+ to alkenes).

Multiple Choice Questions

Q1. The most stable carbocation among the following is —
(a) CH3+ (b) CH3CH2+ (c) (CH3)2CH+ (d) (CH3)3C+
Answer: (d) (CH3)3C+

Q2. Hybridisation of carbon in ethyne is —
(a) sp3 (b) sp2 (c) sp (d) sp3d
Answer: (c) sp

Q3. The IUPAC name of (CH3)3CCH2OH is —
(a) 2,2-dimethylpropan-1-ol (b) tert-butyl alcohol (c) 2-methylbutan-1-ol (d) neopentanol
Answer: (a) 2,2-dimethylpropan-1-ol

Q4. Prussian blue colour in Lassaigne’s test confirms —
(a) S (b) N (c) Cl (d) Br
Answer: (b) N

Q5. Which technique purifies camphor?
(a) Distillation (b) Steam distillation (c) Sublimation (d) Crystallisation
Answer: (c) Sublimation

Q6. A reagent that donates an electron pair is called —
(a) electrophile (b) free radical (c) nucleophile (d) carbene
Answer: (c) nucleophile

Q7. Carius method estimates —
(a) C and H (b) N (c) Halogens and S (d) O
Answer: (c) Halogens and S

Q8. Inductive effect is —
(a) temporary and through pi bonds (b) permanent and through sigma bonds (c) temporary and through sigma bonds (d) only in aromatic compounds
Answer: (b) permanent and through sigma bonds

Q9. Number of sigma bonds in propene (CH3CH=CH2) is —
(a) 6 (b) 7 (c) 8 (d) 9
Answer: (c) 8

Q10. Rf value in chromatography lies between —
(a) 0 and 1 (b) 1 and 10 (c) any value (d) negative values
Answer: (a) 0 and 1

Fill in the Blanks

Q1. The geometry of an sp3 carbon is __________.
Answer: tetrahedral

Q2. A homologous series differs in successive members by __________.
Answer: –CH2–

Q3. Hyperconjugation is also called __________ resonance.
Answer: no-bond

Q4. A white precipitate insoluble in HNO3 but soluble in NH4OH formed in halogen test indicates __________.
Answer: chlorine

Q5. The technique used to separate liquids whose boiling points are close is __________.
Answer: fractional distillation

True or False

Q1. The bond-line formula shows every C and H explicitly. Answer: False (vertices and ends are C; H atoms on C are implicit).

Q2. Carbocations are sp2 hybridised. Answer: True.

Q3. Inductive effect involves complete transfer of electrons. Answer: False (only partial polarisation).

Q4. Aniline can be purified by steam distillation. Answer: True.

Q5. The Kjeldahl method works for nitro and azo compounds. Answer: False (it fails for these — Dumas is used).

Glossary

Term	Meaning
Tetravalency	Carbon’s ability to form four covalent bonds.
Catenation	Self-linking of carbon atoms into chains and rings.
Functional group	Atom or group that decides the characteristic chemistry of a compound.
Homologous series	Series of compounds with the same functional group differing by –CH2–.
IUPAC name	Systematic name based on root, suffix and prefix rules.
Carbocation	Positively charged, electron-deficient carbon intermediate (sp2).
Carbanion	Negatively charged, electron-rich carbon intermediate (sp3).
Free radical	Species with an unpaired electron, formed by homolysis.
Electrophile	Electron-deficient species that accepts an electron pair.
Nucleophile	Electron-rich species that donates an electron pair.
Inductive effect	Permanent polarisation of sigma bonds by an attached group.
Resonance	Delocalisation of pi or lone-pair electrons over canonical structures.
Hyperconjugation	Sigma–pi delocalisation of C–H electrons into adjacent p orbitals.
Crystallisation	Purification by selective solubility and recrystallisation from solvent.
Sublimation	Direct conversion of solid to vapour and back to solid.
Fractional distillation	Separation of liquids of close boiling points using a fractionating column.
Steam distillation	Distillation of water-immiscible volatiles with steam.
Chromatography	Separation based on differential adsorption or partition between phases.
Lassaigne’s test	Sodium fusion test for detection of N, S and halogens.
Carius method	Quantitative estimation of halogens or sulphur as AgX or BaSO4.
Dumas method	Quantitative nitrogen estimation by combustion to N2 over hot Cu.
Kjeldahl method	N estimation by digestion with H2SO4 and NH3 distillation.