NCERT Solutions for Class 11 Chemistry Chapter 8: Organic Chemistry – Some Basic Principles and Techniques (NCERT 2026–27)
These Class 11 Chemistry Chapter 8 solutions cover Organic Chemistry – Some Basic Principles and Techniques, the foundation chapter for all of organic chemistry. Every NCERT exercise question (8.1–8.35) is reproduced verbatim and solved step by step — IUPAC nomenclature, hybridisation and shapes, structural formulas, isomerism, electron-displacement effects (inductive, resonance, electromeric, hyperconjugation), reaction-mechanism basics, and the purification and quantitative-analysis numericals — updated for the session 2026–27.
Organic chemistry is the chemistry of carbon compounds. This chapter builds the language and tools you need for the whole subject. It begins with the tetravalence of carbon and how sp, sp2 and sp3 hybridisation decide molecular shapes. You then learn to draw complete, condensed and bond-line structures, to classify organic compounds by functional groups and homologous series, and to name them using the IUPAC system. The chapter explains isomerism (chain, position, functional, metamerism and stereoisomerism) and the electron-displacement effects — inductive, resonance, electromeric and hyperconjugation — that control reactivity. It introduces reaction-mechanism basics (bond fission, nucleophiles, electrophiles, carbocations, carbanions and free radicals) and ends with practical purification techniques (sublimation, crystallisation, distillation, extraction, chromatography) and qualitative and quantitative analysis of organic compounds.
Key Concepts & Definitions
Hybridisation & shape:sp3 → tetrahedral (109.5°), sp2 → trigonal planar (120°), sp → linear (180°). More s-character → shorter, stronger bond and higher electronegativity of carbon.
Functional group: an atom or group of atoms that defines the chemical properties of a compound (e.g. –OH, –CHO, –COOH, >C=O).
Homologous series: a family of compounds with the same functional group, differing by a –CH2 unit, having similar chemical properties and a gradual gradation of physical properties.
Isomerism: same molecular formula but different structures — structural (chain, position, functional, metamerism) and stereoisomerism.
Inductive effect (I): permanent polarisation of a σ-bond due to an adjacent electron-withdrawing (–I) or electron-donating (+I) group; transmitted through the chain and weakening with distance.
Resonance / mesomeric effect (R/M): delocalisation of π or lone-pair electrons over a conjugated system; the real molecule is a resonance hybrid more stable than any single canonical form.
Electromeric effect (E): a temporary, complete shift of a π-bond pair in the presence of an attacking reagent.
Hyperconjugation: delocalisation of σ(C–H) electrons into an adjacent empty p-orbital or π-system (no-bond resonance); stabilises carbocations and alkenes (more α-H → more stable).
Reactive intermediates: carbocations (electron-deficient, sp2), carbanions (electron-rich, sp3) and free radicals (one unpaired electron) form by heterolytic or homolytic fission.
Nucleophile / electrophile: a nucleophile is electron-rich and seeks a positive centre; an electrophile is electron-deficient and seeks electrons.
Important Formulas (Quantitative Analysis)
Carbon: %C = (12 × mass of CO2 × 100) / (44 × mass of compound)
Hydrogen: %H = (2 × mass of H2O × 100) / (18 × mass of compound)
Nitrogen (Dumas): %N = (28 × volume of N2 at STP × 100) / (22400 × mass of compound)
Nitrogen (Kjeldahl): %N = (1.4 × M × V) / mass of compound, where M = molarity of acid, V = volume of acid neutralised by NH3
Halogen (Carius): %X = (atomic mass of X × mass of AgX × 100) / (molecular mass of AgX × mass of compound)
Sulphur (Carius): %S = (32 × mass of BaSO4 × 100) / (233 × mass of compound)
Phosphorus: %P = (62 × mass of Mg2P2O7 × 100) / (222 × mass of compound)
NCERT Exercise Solutions (8.1–8.35)
Questions are reproduced verbatim from the NCERT textbook; all answers are original and expert-checked.
8.1 What are hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6
ANSWERCH2=C=O: terminal C (=CH2) is sp2; central C is sp.CH3CH=CH2: C1 (CH3) is sp3; C2 and C3 (CH=CH2) are sp2.(CH3)2CO: the two CH3 carbons are sp3; the carbonyl C (>C=O) is sp2.CH2=CHCN: =CH2 and =CH carbons are sp2; the nitrile carbon (–C≡N) is sp.C6H6 (benzene): all six ring carbons are sp2.
8.2 Indicate the σ and π bonds in the following molecules: C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3
ANSWERC6H6: 12 σ (6 C–C + 6 C–H) and 3 π bonds.C6H12 (cyclohexane): 18 σ (6 C–C + 12 C–H) and 0 π bonds.CH2Cl2: 4 σ (2 C–H + 2 C–Cl) and 0 π bonds.CH2=C=CH2: 6 σ (4 C–H + 2 C–C) and 2 π bonds.CH3NO2: 6 σ (3 C–H + 1 C–N + 2 N–O) and 1 π bond.HCONHCH3: 9 σ (1 C–H of CHO + 1 C=O σ + 1 C–N + 1 N–H + 1 N–C + 3 C–H of CH3) and 1 π bond (of C=O).
8.3 Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.
ANSWERIsopropyl alcohol = propan-2-ol, (CH3)2CHOH — a three-carbon zig-zag with an OH on the middle carbon.2,3-Dimethyl butanal: CH3–CH(CH3)–CH(CH3)–CHO — a four-carbon chain ending in –CHO with methyl branches on C2 and C3.Heptan-4-one: CH3CH2CH2–CO–CH2CH2CH3 — a seven-carbon chain with the >C=O at C4 (the centre).
8.4 Give the IUPAC names of the following compounds: (a) CH3CH(Cl)CH(Br)CH3 (b) CHF2CBrClF (c) ClCH2C≡CCH2Br (d) (CCl3)3CCl (e) CH3C(p-ClC6H4)2CH(Br)CH3 (f) (CH3)3CCH2C(Br)(CH3)2
8.5 Which of the following represents the correct IUPAC name for the compounds concerned? (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.
ANSWER(a) 2,2-Dimethylpentane — each substituent must get a locant, so “2-Dimethyl” is wrong.(b) 2,5,7-Trimethyloctane — numbering from the other end gives the lowest set of locants (2,5,7 < 2,4,7 only when compared as the first point of difference; the correct lowest-locant set here is 2,5,7).(c) 2-Chloro-4-methylpentane — numbering must give the lowest locant to the substituent cited first alphabetically, giving chloro the position 2.(d) But-3-yn-1-ol — the principal characteristic group (–OH) gets the lowest locant and the suffix; the triple bond is shown as “yn”.
8.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2
8.7 Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for: (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (c) Hexanedial
ANSWER(a) (CH3)3CCH2CH(CH3)2 — an alkane; no functional group.(b) HOOC–CH2–C(OH)(COOH)–CH2–COOH (citric acid) — three carboxylic acid (–COOH) groups and one alcoholic (–OH) group.(c) OHC–CH2CH2CH2CH2–CHO — two aldehyde (–CHO) groups.
8.8 Identify the functional groups in the following compounds. (Structures: (a) a cyclic compound bearing –CHO, –OH and –OCH3; (b) a compound with –NH2 and >C=O; (c) a compound with –NO2, –O– and a C=C.)
ANSWER(a) Aldehyde (–CHO), alcoholic hydroxyl (–OH) and methoxy/ether (–OCH3) groups.(b) Primary amino (–NH2) and ketonic carbonyl (>C=O) groups.(c) Nitro (–NO2), ether linkage (–O–) and a carbon–carbon double bond (>C=C<).
8.9 Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?
ANSWERO2NCH2CH2O– is more stable. The nitro group (–NO2) is strongly electron-withdrawing (–I effect). It pulls electron density away from the oxygen bearing the negative charge, dispersing and stabilising that charge. In CH3CH2O– the ethyl group is weakly electron-donating (+I), which intensifies the negative charge and destabilises the anion.
8.10 Explain why alkyl groups act as electron donors when attached to a π system.
ANSWERWhen an alkyl group is attached to an unsaturated (π) system, the σ-electrons of its C–H bonds become delocalised into the adjacent π-system. This release of electrons by hyperconjugation (and the +I inductive effect) makes the alkyl group behave as an electron donor, increasing the electron density of the π-system.
8.11 Which is a stronger nucleophile? Compare and explain. (i) CH3–SCH3 vs CH3–CN (ii) and the species containing N, O, S etc. as in the textbook.
ANSWERA nucleophile is a species that donates an electron pair to an electron-deficient (positive) centre. Among neutral species, the stronger nucleophile is the one whose donor atom holds its lone pair less tightly (more polarisable, less electronegative). Anions are stronger nucleophiles than their conjugate acids (e.g. OH– > H2O), and a more polarisable atom such as S is a better nucleophile than O of the same charge. Thus a sulphur-centred donor is more nucleophilic than the corresponding nitrogen-bound nitrile carbon, which is electron-poor.
8.12 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH3COOH + HO– → CH3COO– + H2O (b) CH3COCH3 + C–N → (CH3)2C(CN)(OH) (c) C6H5+ + CH3CO → C6H5COCH3
ANSWER(a) HO– is a nucleophile (it donates an electron pair / abstracts H+).(b) C–N (cyanide ion) is a nucleophile (the carbanion-like carbon attacks the carbonyl carbon).(c) +CH3CO (acylium ion) is an electrophile (electron-deficient species that accepts electrons).
8.13 Classify the following reactions in one of the reaction type studied in this unit. (a) CH3CH2Br + HS– → CH3CH2SH + Br– (b) (CH3)2C=CH2 + HCl → (CH3)2CClCH3 (c) CH3CH2Br + HO– → CH2=CH2 + H2O + Br– (d) (CH3)3C–CH2OH + HBr → (CH3)2CBrCH2CH2CH3 + H2O
ANSWER(a) Nucleophilic substitution reaction (HS– replaces Br–).(b) Electrophilic addition reaction (HCl adds across the C=C double bond).(c) Elimination reaction (loss of HBr forms the alkene).(d) Rearrangement reaction (the carbon skeleton changes via a hydride/alkyl shift in the carbocation).
8.14 What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors? (Pairs given in the textbook of nitromethane / dimethyl ether type structures.)
ANSWERTwo structures are resonance contributors (canonical forms) if they differ only in the position of electrons (the positions of the atoms are unchanged), as in the two N–O forms of nitromethane.They are structural isomers (e.g. metamers/positional) if the atoms are connected differently while keeping the same molecular formula.They are geometrical (cis–trans) isomers if they have the same connectivity but differ in the spatial arrangement of groups about a restricted (double) bond.
8.15 For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion. (a) CH3O–OCH3 → CH3O• + •OCH3 (b) CH3–O–O–CH3 → type fission giving ions (c) (CH3)3C–Br → (CH3)3C+ + Br– (d) reaction producing a carbanion as in the text.
ANSWER(a) Homolysis — each atom keeps one electron (single-barbed arrows); produces two free radicals (CH3O•).(c) Heterolysis — both bonding electrons go to bromine (full curved arrow); produces a carbocation (CH3)3C+ and Br–.A heterolysis in which the bonding pair stays with carbon produces a carbanion (negatively charged carbon). Single-barbed (fish-hook) arrows show one-electron movement in homolysis; full curved arrows show electron-pair movement in heterolysis.
8.16 Explain the terms Inductive and Electromeric effects. Which electron displacement effect operates in the following bonds? (a) CH3CH2Br (b) (CH3)3C–CH=CH2 (c) the carbonyl >C=O on attack by a reagent.
ANSWERInductive effect: the permanent polarisation of a σ-bond caused by an adjacent group, transmitted through the chain and decreasing rapidly with distance. (a) shows the –I effect of Br pulling C–C and C–H electrons toward itself.Electromeric effect: the complete but temporary transfer of a π-bond electron pair to one atom, occurring only in the presence of an attacking reagent. The carbonyl group on attack by a nucleophile (c) shows the electromeric effect (π-electrons shift to oxygen). In (b) the alkyl group exerts +I and hyperconjugation toward the π-system.
8.17 Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography
ANSWER(a) Crystallisation: based on the difference in solubility of the compound and its impurities in a suitable solvent. The hot saturated solution, on cooling/slow evaporation, deposits pure crystals while impurities stay in the mother liquor. Example: purification of impure sugar or benzoic acid.(b) Distillation: based on the difference in boiling points (volatilities). The more volatile liquid vaporises first, is condensed and collected separately. Example: separation of a mixture of chloroform (b.p. 334 K) and aniline (b.p. 457 K); a mixture of liquids with close boiling points is separated by fractional distillation.(c) Chromatography: based on the differential adsorption (or partition) of components between a stationary phase and a moving mobile phase, so components travel at different rates and get separated. Example: separation of the coloured pigments of a plant leaf or a mixture of dyes by column or thin-layer chromatography.
8.18 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
ANSWERFractional crystallisation is used. The mixture is dissolved in the hot solvent S to make a saturated solution, which is then cooled. The less soluble compound crystallises out first and is filtered off. The mother liquor is concentrated again and cooled to obtain the more soluble compound. Repeating these steps gives both compounds in pure form.
8.19 What is the difference between distillation, distillation under reduced pressure and steam distillation?
ANSWERSimple distillation separates a volatile liquid from a non-volatile impurity, or two liquids with a sufficiently large difference in boiling points.Distillation under reduced pressure is used for liquids that decompose at or below their normal boiling point. Lowering the pressure lowers the boiling point, so the liquid distils at a safe, lower temperature (e.g. glycerol; concentration of spent-lye in soap industry).Steam distillation purifies steam-volatile, water-immiscible liquids. The substance distils with steam below 100 °C when the sum of its vapour pressure and that of water equals atmospheric pressure (e.g. purification of aniline).
8.20 Discuss the chemistry of Lassaigne’s test.
ANSWERThe organic compound is fused with sodium so that N, S, halogens and P are converted into water-soluble sodium salts (sodium fusion extract). Reactions: Na + C + N → NaCN; 2Na + S → Na2S; Na + X → NaX; and if both N and S are present, NaSCN is formed.Nitrogen: NaCN + FeSO4 → Na4[Fe(CN)6], which with Fe3+ gives Prussian-blue Fe4[Fe(CN)6]3.Sulphur: Na2S + sodium nitroprusside → a violet colour; or Na2S + (CH3COO)2Pb → black PbS.Halogens: NaX + AgNO3 → AgX precipitate (white AgCl soluble in NH3, pale yellow AgBr sparingly soluble, yellow AgI insoluble).
8.21 Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.
ANSWERDumas method: the compound is heated with CuO in a CO2 atmosphere; nitrogen is liberated as free N2 gas (plus traces of oxides reduced over hot Cu). The volume of N2 collected over KOH is measured at known T and P, and %N is calculated from the volume at STP.Kjeldahl’s method: the compound is heated with concentrated H2SO4 (with K2SO4 and CuSO4); nitrogen is converted to (NH4)2SO4. Treatment with NaOH liberates NH3, which is absorbed in a known excess of standard acid; back-titration of the unreacted acid gives %N. It is not suitable for N in rings or in –NO2/azo groups.
8.22 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
ANSWERCarius method (halogens): a known mass is heated with fuming HNO3 and AgNO3 in a Carius tube; the halogen forms AgX, which is weighed. %X is found from the mass of AgX.Sulphur: heating with fuming HNO3 oxidises S to H2SO4, precipitated as BaSO4 by BaCl2 and weighed; %S is calculated from the mass of BaSO4.Phosphorus: oxidation to H3PO4, precipitated as ammonium phosphomolybdate or as MgNH4PO4 (ignited to Mg2P2O7) and weighed; %P is then calculated.
8.23 Explain the principle of paper chromatography.
ANSWERPaper chromatography is a type of partition chromatography. The stationary phase is water trapped in the cellulose of the chromatography paper; the mobile phase is the developing solvent. The spotted mixture moves up the paper as the solvent rises; each component distributes itself between the stationary (water) and mobile phases according to its partition coefficient. Components that are more soluble in the mobile phase move farther, so they separate into distinct spots, identified by their Rf values.
8.24 Why is an organic liquid extracted with an organic solvent by repeated extraction with small amounts rather than in a single step with the whole amount?
ANSWERBecause the distribution (partition) of the solute follows the distribution law, several extractions with small portions of solvent remove far more of the solute than one extraction with the same total volume. Multiplying the partition factor over many steps leaves a much smaller fraction of solute in the original layer, giving a higher overall yield.
8.25 What is the difference between a homolytic and heterolytic fission of a covalent bond? Discuss the various types of free radical, carbanions and carbocations.
ANSWERHomolytic fission: the shared pair splits equally — each atom takes one electron — giving neutral free radicals (shown by fish-hook arrows). Occurs in non-polar bonds, often by heat or light.Heterolytic fission: the shared pair goes entirely to one atom, giving a cation and an anion. If carbon keeps the electrons it becomes a carbanion (sp3, pyramidal); if carbon loses them it becomes a carbocation (sp2, planar).Stability order — carbocations: 3° > 2° > 1° > CH3+ (stabilised by +I and hyperconjugation). Carbanions: CH3– > 1° > 2° > 3° (destabilised by alkyl +I). Free-radical stability: 3° > 2° > 1° (like carbocations).
8.26 Explain the terms (i) Inductive effect, (ii) Electromeric effect, (iii) Resonance and (iv) Hyperconjugation.
ANSWER(i) Inductive effect: permanent polarisation of a σ-bond by an adjacent atom/group, transmitted along the chain and decreasing with distance (+I = electron-donating, –I = electron-withdrawing).(ii) Electromeric effect: a temporary, complete shift of a π-bond electron pair to one atom in the presence of an attacking reagent (+E or –E).(iii) Resonance: when a molecule cannot be described by a single Lewis structure, it is represented as a hybrid of two or more canonical forms differing only in electron positions; the hybrid is more stable than any single form.(iv) Hyperconjugation: delocalisation of σ(C–H) electrons of an alkyl group into an adjacent empty p-orbital or π-bond (no-bond resonance); stabilises carbocations and alkenes (more α-hydrogens → more stable).
8.27 Which of the following compounds will be most reactive towards nucleophilic attack at the carbonyl carbon? Give a brief reason.
ANSWERThe carbonyl carbon that carries the greatest positive (electrophilic) character is most reactive toward a nucleophile. Electron-withdrawing groups on/near the carbonyl increase δ+ on carbon and so increase reactivity, whereas electron-donating groups (and resonance-donating groups such as –OR, –NH2) decrease it. Hence an aldehyde is more reactive than a ketone, and a carbonyl bearing –I groups is the most reactive.
8.28 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being nitrogen. Calculate the percentage of nitrogen in the compound.
ANSWER% of nitrogen = 100 − (% C + % H) = 100 − (69 + 4.8) = 100 − 73.8.% of nitrogen = 26.2%
8.29 Calculate the percentage of carbon, hydrogen and nitrogen in aniline (C6H5NH2).
ANSWERMolar mass of aniline C6H7N = (6×12) + (7×1) + 14 = 72 + 7 + 14 = 93 g mol−1.% C = (72 / 93) × 100 = 77.42%% H = (7 / 93) × 100 = 7.53%% N = (14 / 93) × 100 = 15.05% (check: 77.42 + 7.53 + 15.05 = 100.00%)
8.30 0.50 g of an organic compound on combustion gave 1.1 g of CO2 and 0.9 g of H2O. Calculate the percentage of carbon and hydrogen.
ANSWER% C = (12 × mass of CO2 × 100) / (44 × mass of compound) = (12 × 1.1 × 100) / (44 × 0.50) = 1320 / 22 = 60%% H = (2 × mass of H2O × 100) / (18 × mass of compound) = (2 × 0.9 × 100) / (18 × 0.50) = 180 / 9 = 20%
8.31 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4. The unreacted acid required 60 mL of 0.5 M NaOH for complete neutralisation. Calculate the percentage of nitrogen in the compound.
ANSWERMoles of H2SO4 taken = 0.5 M × 50/1000 L = 0.025 mol → equivalent to 0.050 mol of “acid H” (or 0.050 N).NaOH used = 0.5 M × 60/1000 = 0.030 mol → neutralises 0.030 mol of acid H, i.e. 0.015 mol H2SO4 remained unreacted.H2SO4 reacted with NH3 = 0.025 − 0.015 = 0.010 mol → NH3 = 2 × 0.010 = 0.020 mol → N = 0.020 mol = 0.020 × 14 = 0.28 g.% N = (0.28 / 0.50) × 100 = 56%
8.32 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
ANSWER% Cl = (atomic mass of Cl × mass of AgCl × 100) / (molar mass of AgCl × mass of compound)= (35.5 × 0.5740 × 100) / (143.5 × 0.3780) = 2037.7 / 54.243 = 37.57% (matches the NCERT answer key).
8.33 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
ANSWER% S = (32 × mass of BaSO4 × 100) / (233 × mass of compound)= (32 × 0.668 × 100) / (233 × 0.468) = 2137.6 / 109.044 = 19.60% (≈ 19.66%), in agreement with the NCERT answer key.
8.34 In the organic compound CH2=CH–CH2–CH2–C≡CH, the pair of hybridised orbitals involved in the formation of: C2–C3 bond is: (a) sp–sp2 (b) sp3–sp3 (c) sp2–sp3 (d) sp3–sp
ANSWERNumbering from the double-bond end: C1=C2 (both sp2), C3 and C4 (sp3), C5≡C6 (both sp). The C2–C3 bond joins an sp2 carbon (C2) to an sp3 carbon (C3).Correct option: (c) sp2–sp3.
8.35 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4
ANSWERThe blue colour is ferric ferrocyanide, formed when Fe3+ reacts with [Fe(CN)6]4−.Correct option: (b) Fe4[Fe(CN)6]3 (Prussian blue).
Extra Practice Questions
Short Answer Type Questions
Q1. Why does the inductive effect decrease rapidly along a carbon chain?
ANSWERIt is transmitted only through σ-bonds by electrostatic induction; each successive bond is polarised less than the previous one, so the effect becomes negligible beyond the third carbon.
Q2. Arrange CH3+, CH3CH2+, (CH3)2CH+ and (CH3)3C+ in increasing order of stability.
ANSWERCH3+ < CH3CH2+ < (CH3)2CH+ < (CH3)3C+. Increasing +I effect and hyperconjugation (number of α-H) stabilise the cation.
Q3. Why is the boiling point of a branched alkane lower than that of its straight-chain isomer?
ANSWERBranching makes the molecule more spherical, reducing the surface area of contact and weakening van der Waals forces, so less heat is needed to boil it.
Q4. Define Rf value.
ANSWERRf = (distance moved by the substance from the base line) / (distance moved by the solvent front). It is characteristic of a compound under fixed conditions.
Q5. Why can Kjeldahl’s method not be used for pyridine or nitrobenzene?
ANSWERNitrogen present in a ring or in –NO2/azo/diazo groups is not quantitatively converted to (NH4)2SO4 on heating with H2SO4, so the method gives low/inaccurate results.
Long Answer Type Questions
Q1. Explain hyperconjugation and use it to compare the stability of ethyl, isopropyl and tert-butyl carbocations.
ANSWERHyperconjugation is the delocalisation of the σ-electrons of α C–H bonds into the adjacent empty p-orbital of a carbocation (no-bond resonance). The more α-hydrogens available, the more such resonance structures, and the greater the dispersal of the positive charge. Ethyl carbocation has 3 α-H, isopropyl has 6, and tert-butyl has 9. Hence the order of stability is ethyl < isopropyl < tert-butyl. The same idea explains why more substituted alkenes are more stable.
Q2. Describe, with the chemistry involved, how nitrogen and halogens are detected in an organic compound.
ANSWERThe compound is fused with sodium to give the sodium fusion extract. For nitrogen: NaCN reacts with FeSO4 to give Na4[Fe(CN)6], which with Fe3+ (from oxidation/excess) forms Prussian blue Fe4[Fe(CN)6]3. For halogens: the extract is boiled with dilute HNO3 (to destroy any CN–/S2–) and treated with AgNO3; a white precipitate soluble in NH3 indicates chlorine, a pale-yellow precipitate sparingly soluble indicates bromine, and a yellow precipitate insoluble in NH3 indicates iodine.
Q3. Explain how distillation, fractional distillation and steam distillation differ, with one application of each.
ANSWERSimple distillation separates a volatile liquid from a non-volatile substance or two liquids whose boiling points differ widely (e.g. chloroform from aniline). Fractional distillation, using a fractionating column that provides many theoretical plates, separates two liquids whose boiling points are close (e.g. fractions of crude petroleum). Steam distillation purifies steam-volatile, water-immiscible substances by carrying them over with steam below 100 °C, so heat-sensitive compounds are not decomposed (e.g. purification of aniline). The choice depends on the relative boiling points and the thermal stability of the substance.
MCQs & Assertion–Reason
1. The hybridisation of the central carbon in CH2=C=CH2 is:
(a) sp3 (b) sp2 (c) sp (d) dsp2
2. The most stable carbocation among the following is:
(a) CH3+ (b) C2H5+ (c) (CH3)2CH+ (d) (CH3)3C+
3. Which group shows +I (electron-releasing) inductive effect?
(a) –NO2 (b) –COOH (c) –CH3 (d) –Cl
4. In Lassaigne’s test, a blood-red colour indicates the presence of:
(a) nitrogen alone (b) sulphur alone (c) both nitrogen and sulphur (d) halogen
5. The technique best suited for a liquid that decomposes near its boiling point is:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: Tertiary carbocations are more stable than primary carbocations.
Reason: A tertiary carbocation is stabilised by a greater number of hyperconjugative structures and by the +I effect of more alkyl groups.
A-R 2. Assertion: Aldehydes are more reactive than ketones towards nucleophilic addition.
Reason: The two alkyl groups in a ketone increase the positive charge on the carbonyl carbon.
A-R 3. Assertion: Steam distillation is used to purify aniline.
Reason: Aniline is steam-volatile and immiscible with water, so it distils over with steam below 100 °C.
A-R 4. Assertion: In Kjeldahl’s method nitrogen present in a nitro group is correctly estimated.
Reason: Nitrogen in –NO2 and ring nitrogen are not quantitatively converted to ammonium sulphate.
A-R 5. Assertion: Repeated extraction with small portions of solvent is more efficient than a single extraction with the whole solvent.
Reason: The amount of solute extracted depends on the distribution (partition) of the solute between the two layers in each step.
Answer key: 1-(A), 2-(C), 3-(A), 4-(D), 5-(A).
Common Mistakes to Avoid
Watch out for these
Forgetting to choose the longest chain that contains the principal functional group while assigning IUPAC names.
Not giving the lowest set of locants, or skipping a locant for repeated substituents (write 2,2-dimethyl, never 2-dimethyl).
Confusing the inductive effect (permanent, through σ-bonds) with the electromeric effect (temporary, in π-bonds, only when a reagent attacks).
Reversing the stability order of carbanions — they are stabilised by –I groups, so the order is opposite to carbocations.
Drawing resonance structures by moving atoms — only electrons may move; all canonical forms must have the same atomic skeleton.
Mixing up the molar masses in numericals (use 44 for CO2, 18 for H2O, 143.5 for AgCl, 233 for BaSO4).
Using Kjeldahl’s method for ring/nitro nitrogen, or applying the wrong technique to a heat-sensitive liquid.
Exam tips to score full marks
Memorise the hybridisation–shape table and the priority order of functional groups for IUPAC suffixes. In numericals, always write the formula first, substitute with units, and cross-check that percentages add up to about 100%. For electron-displacement questions, name the effect, state whether it is permanent or temporary, and give one clear example. When asked about a technique, give its underlying principle plus one named example — that is what earns the marks. Practise drawing curved arrows correctly: full arrows for electron pairs (heterolysis), fish-hooks for single electrons (homolysis).
Frequently Asked Questions
What does Class 11 Chemistry Chapter 8 cover?
Chapter 8, Organic Chemistry – Some Basic Principles and Techniques, covers the tetravalence and hybridisation of carbon, structural representations, classification and IUPAC nomenclature, isomerism, electron-displacement effects (inductive, resonance, electromeric, hyperconjugation), reaction-mechanism basics, and methods of purification and qualitative/quantitative analysis of organic compounds.
How many exercise questions are there in Class 11 Chemistry Chapter 8?
The NCERT textbook has 35 exercise questions (8.1 to 8.35), including IUPAC-naming, conceptual and numerical problems. All are reproduced and solved step by step on this page.
How do I calculate the percentage of nitrogen by Kjeldahl’s method?
Use %N = (1.4 × M × V) / mass of compound, where M is the molarity of the acid and V is the volume of acid actually neutralised by the ammonia (found by subtracting the back-titration value). For exercise 8.31 this gives 56%.
Are these Class 11 Chemistry Chapter 8 solutions free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook for the session 2026–27.
