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NCERT Class 11 Chemistry Organic Chemistry: Some Basic Principles and Techniques exercises cover definitions, derivations, numerical problems, and application-based questions. The MyAiSchool solution set provides full step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should master core formulas, key concepts, and problem-solving methodology to score full marks.

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For numerical problems in NCERT Class 11 Chemistry Organic Chemistry: Some Basic Principles and Techniques: (1) list all given data with proper units, (2) write the relevant formula(e), (3) substitute values carefully checking units, (4) calculate with correct significant figures, (5) state the final answer with the appropriate unit. The MyAiSchool solutions follow this CBSE-aligned step-by-step format.

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NCERT Exercises and Solutions: Organic Chemistry: Some Basic Principles and Techniques

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In NCERT Class 11 Chemistry Chapter 8 Organic Chemistry: Some Basic Principles and Techniques, NCERT Exercises and Solutions: Organic Chemistry: Some Basic Principles and Techniques is taught with definitions, examples, and interactive simulations on MyAiSchool aligned with CBSE 2025-26.

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NCERT Exercises and Solutions: Organic Chemistry Basic Principles

Chapter Summary

Chapter 8 laid the foundations of organic chemistry. Below is a compact recap before you tackle the exercises.

Key Concepts at a Glance

Carbon is tetravalent; its bonding geometry is decided by hybridisation — sp³ (tetrahedral, 109.5°), sp² (trigonal planar, 120°), sp (linear, 180°).
σ bonds form by head-on overlap, π bonds by lateral overlap. Ethene has 5 σ + 1 π, ethyne 3 σ + 2 π.
Three principal ways to draw an organic molecule: Lewis (complete), condensed, bond-line (skeletal). Wedge–dash indicates 3-D geometry.
Compounds are classified as acyclic / cyclic (alicyclic, aromatic, heterocyclic) and by functional group. Members of a homologous series share a formula and differ by a –CH₂– unit.
IUPAC names follow a fixed five-step recipe. Priority order for principal suffix: –COOH > ester > amide > nitrile > –CHO > ketone > –OH > –NH₂ > C=C / C≡C > halide / nitro.
Isomerism: structural (chain, position, functional, metamerism, tautomerism) and stereo (geometrical cis–trans, optical).
Bond fission is homolytic (→ radicals) or heterolytic (→ ions). Key intermediates: carbocation, carbanion, free radical.
Electronic effects: inductive (−I / +I through σ), resonance (π delocalisation), hyperconjugation (σ-C-H into π / empty p), electromeric (temporary, induced).
Purification: sublimation, crystallisation, distillation (simple, fractional, reduced pressure, steam), differential extraction, chromatography. R_f = distance of solute / distance of solvent front.
Qualitative: CuO test for C and H; Lassaigne's test for N (Prussian blue), S (violet with nitroprusside), halogens (AgX precipitates).
Quantitative: % C and % H from CO₂ and H₂O masses after combustion.

IUPAC Rules — Quick Reference

Step	Rule
1	Identify the longest chain containing the principal functional group.
2	Number it so that the principal group gets the lowest locant.
3	Break ties using the lowest set of locants for all substituents.
4	Add substituents as prefixes with locants, in alphabetical order (ignore di/tri).
5	Combine: [locants-substituents]-[parent][locant of suffix]-[suffix].

Functional Groups — Master Table

Group	Class	Suffix	Example (IUPAC)
–COOH	Carboxylic acid	-oic acid	CH₃COOH ethanoic acid
–COOR	Ester	alkyl -oate	CH₃COOC₂H₅ ethyl ethanoate
–CN	Nitrile	-nitrile	CH₃CN ethanenitrile
–CHO	Aldehyde	-al	CH₃CHO ethanal
>C=O	Ketone	-one	CH₃COCH₃ propan-2-one
–OH	Alcohol / phenol	-ol	CH₃OH methanol
–NH₂	Primary amine	-amine	CH₃NH₂ methanamine
–X	Haloalkane	halo- (prefix)	CH₃Cl chloromethane
–NO₂	Nitro	nitro- (prefix)	CH₃NO₂ nitromethane
–O–	Ether	alkoxy- (prefix)	CH₃OCH₃ methoxymethane

Keywords Grid

Catenation: self-linking of atoms.

Hybridisation: mixing of atomic orbitals.

σ bond: head-on overlap.

π bond: lateral overlap.

Bond-line formula: skeletal zig-zag.

Wedge–dash: 3-D projection notation.

Alicyclic: non-aromatic ring.

Aromatic: (4n+2) π stable ring.

Heterocyclic: ring with N/O/S.

Homologous series: differ by –CH₂–.

IUPAC: systematic naming body.

Locant: position number.

Chain isomerism: different skeletons.

Position isomerism: group location differs.

Metamerism: alkyl split around heteroatom.

Tautomerism: keto–enol interconversion.

cis / trans: geometrical isomers.

Homolysis: → radicals.

Heterolysis: → ions.

Electrophile: e⁻ acceptor.

Nucleophile: e⁻ donor.

Inductive effect: through σ.

Resonance: π delocalisation.

Hyperconjugation: σ(C-H)→π.

Sublimation: solid→vapour→solid.

Fractional distillation: close BPs.

Steam distillation: co-distil with steam.

Chromatography: mobile / stationary.

R_f: solute / solvent distance.

Lassaigne's test: Na-fusion for N,S,X.

NCERT Exercises (with Solutions)

Click "Show solution" under each question to reveal a detailed working.

A · Hybridisation and Shapes (Q 8.1–8.5)

Q 8.1

State the hybridisation of each carbon in the following: (i) propane, CH₃CH₂CH₃; (ii) propene, CH₂=CHCH₃; (iii) propyne, HC≡C–CH₃; (iv) benzaldehyde, C₆H₅CHO.

(i) All three carbons are sp³ (only single bonds).
(ii) C1 & C2 sp² (C=C), C3 sp³ (CH₃).
(iii) C1 & C2 sp (C≡C), C3 sp³.
(iv) Six ring carbons sp² (aromatic); the –CHO carbon sp² (C=O).

Q 8.2

How many σ and π bonds are present in each of the following: (a) CH₃–CH=CH–C≡CH, (b) CH₂=C=CHCH₃?

(a) CH₃–CH=CH–C≡CH: σ bonds = 3(C3–H in CH₃) + 1(C1–C2) + 1(C2–H) + 1(C2–C3) + 1(C3–H) + 1(C3–C4) + 1(C4–C5) + 1(C5–H) = 10 σ. π bonds: 1 (from C=C) + 2 (from C≡C) = 3 π.
(b) Buta-1,2-diene has: C1(H₂)=C2=C3(H)–C4(H₃). σ count = 2(C1–H) + 1(C1=C2 σ) + 1(C2=C3 σ) + 1(C3–H) + 1(C3–C4) + 3(C4–H) = 9 σ. π = 2 (one per double bond) = 2 π.

Q 8.3

Predict the shape and bond angle at every carbon in CH₃–CH=CH–C≡CH.

CH₃ carbon: sp³, tetrahedral, 109.5°. Two CH=CH carbons: sp², trigonal planar, 120°. Two C≡CH carbons: sp, linear, 180°.

Q 8.4

Explain why the C–C bond length in ethane (154 pm) is longer than the C–C bond length in ethene (134 pm) and that in ethyne (120 pm).

As s-character in hybrid orbitals increases (sp³ 25% → sp² 33% → sp 50%), the orbitals are held closer to the nucleus. The hybrid carbon becomes more electronegative, and the internuclear distance shortens. Additionally, π bonds (absent in ethane, one in ethene, two in ethyne) pull the two carbons even closer.

Q 8.5

Describe the orbital picture of ethyne, specifying the number and type of σ and π bonds and the geometry.

Each C uses sp hybrid orbitals — one forms a C–H σ bond, the other forms the C–C σ bond. Two unhybridised p orbitals on each carbon (p_y, p_z) overlap laterally with the matching orbitals of the other carbon, producing two mutually perpendicular π bonds. Total: 2 C–H σ + 1 C–C σ + 2 C–C π. Molecule is linear, 180°.

B · Structural Representation (Q 8.6–8.10)

Q 8.6

Write bond-line formulas for: (a) isopropyl alcohol, (b) 2,3-dimethylbutanal, (c) heptan-4-one.

(a) A three-carbon zig-zag with an –OH on the central carbon.
(b) A four-carbon chain ending in –CHO; methyl branches on C2 and C3.
(c) A seven-carbon zig-zag with =O on C4 (middle).

Q 8.7

Give the condensed and bond-line structural formulas of 2,2,4-trimethylpentane.

Condensed: (CH₃)₃C–CH₂–CH(CH₃)–CH₃.
Bond-line: five-carbon zig-zag with two methyl branches on C2 and one methyl on C4. (Also known as iso-octane — the reference fuel with octane number 100.)

Q 8.8

Expand the following condensed formulas into full (Lewis) structures: (a) CH₃(CH₂)₄CH₃, (b) CH₃CH=CHCH₂OH.

(a) Six carbons linked in a row, each internal carbon having 2 H's, each terminal having 3 H's → H₃C–CH₂–CH₂–CH₂–CH₂–CH₃ (n-hexane).
(b) H₃C–CH=CH–CH₂–OH (but-2-en-1-ol); double bond between C2 and C3.

Q 8.9

What do wedge and dash bonds represent? Illustrate using a tetrahedral carbon with four different substituents W, X, Y, Z.

A solid wedge (▲) indicates a bond coming toward the viewer; a dashed wedge indicates a bond going away. Plain lines lie in the paper. A stereocentre C with W, X, Y, Z is drawn with two substituents in the plane, one on a wedge (out) and one on dashes (behind). Mirror-image arrangements are the two enantiomers.

Q 8.10

Classify each: acetic acid, cyclopentane, pyridine, naphthalene, ethanal.

Acetic acid — acyclic aliphatic carboxylic acid.
Cyclopentane — alicyclic saturated.
Pyridine — aromatic heterocyclic (N in 6-ring).
Naphthalene — aromatic (two fused benzene rings, benzenoid).
Ethanal — acyclic aliphatic aldehyde.

C · IUPAC Nomenclature (Q 8.11–8.15)

Q 8.11

Write IUPAC names for: (a) (CH₃)₂CHCH₂CH₂Br, (b) CH₃CH(OH)CH₂CH₃, (c) HOCH₂CH₂CH₂COOH, (d) C₆H₅–COCH₃.

(a) 1-bromo-3-methylbutane (parent = butane; locants: 1-bromo, 3-methyl).
(b) butan-2-ol.
(c) 4-hydroxybutanoic acid (–COOH is C1; –OH subordinate, hence prefix).
(d) 1-phenylethan-1-one (acetophenone).

Q 8.12

Draw structures for: (a) 3-ethyl-2,2-dimethylpentane, (b) 4-ethyl-2-methylhex-2-ene, (c) 2-methylcyclohexan-1-one.

(a) Pentane backbone with two methyls on C2 (geminal) and one ethyl on C3.
(b) Hex-2-ene (C=C between C2–C3) with a methyl on C2 and an ethyl on C4.
(c) Cyclohexanone with a methyl on the carbon adjacent to the carbonyl.

Q 8.13

Point out the mistake in each "name" and give the correct IUPAC name. (a) 2-ethyl-3-methylpentane for CH₃CH₂–CH(C₂H₅)–CH(CH₃)–CH₃; (b) 3-methylbut-3-enol for (CH₃)₂C=CHCH₂OH.

(a) The parent chain was not the longest. Including the ethyl gives a 6-C chain — name should be 3,4-dimethylhexane.
(b) The –OH must get the lowest locant; numbering should start from the CH₂OH end. Correct name: 3-methylbut-2-en-1-ol.

Q 8.14

Name the following using IUPAC rules: CH₃–CH(Cl)–CH(Br)–CH₃; CH₃COCH₂CHO; HOOC–CH₂–CH₂–COOH.

2-bromo-3-chlorobutane (alphabetical order; lowest-locant set {2,3}).
3-oxobutanal (aldehyde is principal, ketone is "oxo" prefix).
Butanedioic acid (succinic acid; both –COOH groups are the principal termini).

Q 8.15

Write IUPAC names for the four dibromobenzenes.

1,2-dibromobenzene (o-), 1,3-dibromobenzene (m-), 1,4-dibromobenzene (p-). There are only three isomers — 1,5- would be identical to 1,3- and 1,6- to 1,2- due to the ring's symmetry.

D · Isomerism (Q 8.16–8.20)

Q 8.16

Draw and name all isomers of C₅H₁₀ (alkenes only).

Acyclic C₅H₁₀ alkenes: pent-1-ene, pent-2-ene (cis & trans = two geometrical isomers), 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene. Total 6 structural/geometric isomers.

Q 8.17

Identify the type of isomerism in each pair: (a) CH₃CH₂OH and CH₃OCH₃; (b) propan-1-ol and propan-2-ol; (c) n-butane and 2-methylpropane; (d) CH₃COCH₃ and CH₃C(OH)=CH₂.

(a) Functional group isomerism (alcohol vs ether).
(b) Position isomerism.
(c) Chain isomerism.
(d) Tautomerism (keto ↔ enol).

Q 8.18

Which of these can exhibit cis–trans isomerism? (a) CH₂=CH₂, (b) CH₃CH=CHCH₃, (c) (CH₃)₂C=CHCH₃, (d) ClCH=CHBr.

(a) No — each C has two identical H's.
(b) Yes — but-2-ene has cis & trans.
(c) No — left C has two identical methyls.
(d) Yes — each C has two different groups.

Q 8.19

Draw the metamers of C₅H₁₂O that are ethers.

Ether metamers: CH₃–O–C₄H₉ (three C₄H₉ possibilities: n-butyl, iso-butyl, sec-butyl, tert-butyl give four isomers), C₂H₅–O–C₃H₇ (two C₃H₇: n- and iso-, two isomers). In total six ethers of formula C₅H₁₂O.

Q 8.20

Explain keto–enol tautomerism with an example. Which form usually predominates?

A carbonyl compound with an α-hydrogen can rearrange into its enol: the α-H migrates to the oxygen, and the C=O becomes C=C with a new –OH. Example: propan-2-one ⇌ prop-1-en-2-ol (CH₃COCH₃ ⇌ CH₃C(OH)=CH₂). The keto form dominates strongly at equilibrium because C=O is significantly stronger than C=C + O–H.

E · Electronic Effects (Q 8.21–8.25)

Q 8.21

Arrange in increasing order of acid strength: formic acid, acetic acid, benzoic acid, chloroacetic acid.

Acetic < benzoic < formic < chloroacetic. The electron-donating methyl (+I) destabilises the acetate anion most; formate has only an H (no +I); benzoic's phenyl shows mild −I/resonance and is slightly stronger than acetic; chloroacetic's −I is the largest — Cl strongly stabilises the anion, making it the most acidic.

Q 8.22

Rank the following in order of decreasing stability: (a) CH₃⁺, (b) C₂H₅⁺, (c) (CH₃)₂CH⁺, (d) (CH₃)₃C⁺.

(d) > (c) > (b) > (a). Greater alkyl substitution = stronger +I and more hyperconjugating α-H's (9, 6, 3, 0 respectively). 3° carbocations are the most stable.

Q 8.23

Draw the significant resonance structures of the carboxylate anion RCOO⁻ and explain its stability.

Two equivalent structures exist — the negative charge sits on either oxygen while the C=O / C–O bonds exchange roles. The true anion is the hybrid: the two C–O bonds are equivalent (bond length ~127 pm), and the charge is spread equally on the two oxygens. This delocalisation lowers the anion's energy markedly, which is why carboxylic acids are much more acidic than alcohols (in alcohols, the RO⁻ cannot delocalise).

Q 8.24

Explain hyperconjugation. How does it stabilise the tertiary butyl cation?

Hyperconjugation is the partial donation of σ electrons (from C–H bonds adjacent to an sp² centre) into the vacant p-orbital of a cation or into a π system. In (CH₃)₃C⁺, each of the three methyl groups provides three α-C–H bonds — nine in all. Each can contribute a "no-bond" resonance structure in which the C–H σ pair delocalises into the empty p-orbital. The resulting spread of positive charge is why tertiary cations are markedly more stable than primary.

Q 8.25

Distinguish between inductive and resonance effects with one example each.

Inductive: permanent polarisation transmitted through σ bonds; falls off quickly with distance; example — chlorine in Cl–CH₂–COOH strengthens the acid through −I.
Resonance: delocalisation of π electrons / lone pairs within a conjugated system; distance-independent within the system; example — the –NH₂ group in aniline donates its lone pair into the ring, activating the ring toward electrophilic substitution.

F · Purification & Analysis (Q 8.26–8.30)

Q 8.26

Which method would you use to separate and purify each of the following? Give reasons. (a) Crude naphthalene from common salt; (b) a mixture of benzene (80 °C) and toluene (110 °C); (c) aniline from tarry residues; (d) coloured impurities from sugar.

(a) Sublimation — naphthalene sublimes, NaCl doesn't.
(b) Fractional distillation — BPs close, but separable with a column (both above ~25 °C gap but still benefits from a column to avoid co-distillation).
(c) Steam distillation — aniline is steam-volatile; impurities stay behind.
(d) Crystallisation — recrystallising sugar from hot water leaves coloured impurities in solution; activated charcoal decolourises the hot liquor before cooling.

Q 8.27

Explain the principle of paper chromatography. What does R_f tell us?

Paper chromatography is a partition method. Water adsorbed on the cellulose fibres is the stationary phase; the developing solvent is the mobile phase. A component distributes between the two phases as the solvent rises, and how far it travels depends on its partition coefficient. R_f = distance of spot / distance of solvent front is characteristic of a compound under a given solvent, temperature and paper, and can be used to identify it by comparison with standards.

Q 8.28

In Lassaigne's test for nitrogen, describe what you see and write the reactions.

Fusion with Na converts organic N into NaCN. The aqueous extract is warmed with FeSO₄ (forming Fe(CN)₆⁴⁻), acidified with dilute H₂SO₄, and a drop of FeCl₃ is added. A Prussian-blue precipitate of iron(III) hexacyanoferrate(II) — Fe₄[Fe(CN)₆]₃ — confirms nitrogen.

Na + C + N (in sample) → NaCN FeSO₄ + 6 NaCN → Na₄[Fe(CN)₆] + Na₂SO₄ 3 Na₄[Fe(CN)₆] + 4 FeCl₃ → Fe₄[Fe(CN)₆]₃↓ + 12 NaCl

Q 8.29

0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.

Mass of Cl in AgCl = (35.5 / 143.5) × 0.5740 = 0.1420 g.
% Cl = (0.1420 / 0.3780) × 100 = 37.6 %.

Q 8.30

In a combustion experiment, 0.50 g of a compound containing only C, H and O gave 1.10 g of CO₂ and 0.45 g of H₂O. Compute the empirical formula.

% C = (12/44) × (1.10/0.50) × 100 = 60.0 %.
% H = (2/18) × (0.45/0.50) × 100 = 10.0 %.
% O (by difference) = 100 − 60 − 10 = 30.0 %.
Moles: C = 60/12 = 5.00; H = 10/1 = 10.00; O = 30/16 = 1.875.
Divide by smallest (1.875): C : H : O ≈ 2.67 : 5.33 : 1 ≈ 8 : 16 : 3.
Empirical formula: C₈H₁₆O₃ (the simplest ratio; a small rounding may give C₃H₆O ≈ glyceraldehyde after further data — but from the data above the best integer ratio is 8:16:3).

End-of-Chapter Takeaway

Chapter 8 is the language primer for the rest of your organic chemistry journey. Every topic you will meet in Classes 11 and 12 (hydrocarbons, haloalkanes, alcohols, carbonyl compounds, amines, biomolecules, polymers) is built on the four skills practised here: draw, name, classify, and reason with electrons.

Frequently Asked Questions — NCERT Exercises and Solutions: Organic Chemistry Basic Principles

How do you write IUPAC names in NCERT exercises?

To write IUPAC names in NCERT Class 11 Chemistry Chapter 8 exercises: (1) identify the longest continuous carbon chain that includes the principal functional group; (2) number the chain so the principal functional group gets the lowest locant; (3) list substituents in alphabetical order with their locants as prefixes; (4) write the suffix according to the principal functional group; (5) use Greek prefixes (di, tri, tetra) for multiple substituents. Example: CH₃-CH(Br)-CH(CH₃)-COOH is 3-bromo-2-methylbutanoic acid. Practice with alcohols, aldehydes, ketones, carboxylic acids, amines and halides for board exam fluency.

How do you draw all structural isomers in NCERT problems?

To draw all structural isomers in NCERT Class 11 Chemistry Chapter 8 exercises: (1) count the carbon atoms and identify the molecular formula; (2) draw the straight-chain skeleton first; (3) progressively branch and shorten the parent chain to create all unique chain isomers; (4) for each skeleton, vary the position of the functional group to create position isomers; (5) consider different functional groups for functional isomers. Example: C₄H₁₀O has seven structural isomers (4 alcohols + 3 ethers). Always check for duplicates and write IUPAC names for verification. The MyAiSchool exercise set covers all common patterns up to C₆.

How are electronic effects used to predict acidity and basicity in NCERT exercises?

To predict acidity/basicity from electronic effects in NCERT Class 11 Chemistry Chapter 8: (1) electron-withdrawing groups (-I, -R) increase acidity (stabilise conjugate base) and decrease basicity; (2) electron-donating groups (+I, +R) decrease acidity and increase basicity. Examples: F-CH₂-COOH > Cl-CH₂-COOH > Br-CH₂-COOH > CH₃-COOH (acidity); methylamine > ammonia > aniline (basicity, due to +I from methyl and -R from phenyl). Hyperconjugation also stabilises carbocations and alkenes. Practice with substituted phenols, anilines and carboxylic acids — frequently asked in CBSE board exams.

How do you identify electrophiles and nucleophiles in NCERT problems?

To identify electrophiles and nucleophiles in NCERT Class 11 Chemistry Chapter 8 exercises: (1) electrophiles are electron-deficient — have empty orbitals (BF₃, AlCl₃) or full positive charge (H⁺, NO₂⁺, R⁺); (2) nucleophiles are electron-rich — have lone pairs or negative charge (OH⁻, NH₃, CN⁻, R⁻). Neutral molecules with lone pairs (H₂O, ROH, RNH₂) act as nucleophiles. Carbon species: carbocations are electrophiles, carbanions are nucleophiles, free radicals are neither. In a reaction, electrophiles attack electron-rich sites (double bonds, aromatic rings), nucleophiles attack electron-poor sites (carbonyl C, C bonded to leaving group).

How do you choose a purification method for a given compound?

To choose a purification method in NCERT Class 11 Chemistry Chapter 8 exercises: (1) if the compound is solid and sublimes (camphor, anthracene), use sublimation; (2) if it is a solid soluble in hot solvent and less so in cold, use crystallisation; (3) for two liquids with large ΔT_b, use simple distillation; for small ΔT_b, use fractional distillation; (4) for thermally unstable liquids, use vacuum distillation; (5) for steam-volatile insoluble compounds, use steam distillation (e.g., aniline); (6) for components in a mixture, use chromatography (column or TLC). Match physical/chemical properties to the appropriate method.

What 5-mark questions appear in Chapter 8 CBSE board exams?

Common 5-mark CBSE Class 11 Chemistry Chapter 8 board questions: (1) state IUPAC rules of nomenclature and name 5 given organic compounds; (2) draw and name all structural isomers of a given formula like C₄H₁₀O or C₅H₁₂; (3) explain inductive, resonance, hyperconjugation effects with examples and predict reactivity; (4) distinguish homolytic and heterolytic cleavage; identify electrophiles, nucleophiles, free radicals; (5) describe principle and procedure of any two purification methods (steam distillation, column chromatography). The MyAiSchool exercise set provides model answers, marking schemes and conceptual links.

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