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NCERT Solutions · Class 11 Chemistry · NCERT Chemistry Part I and II Part II · Chapter 8

Chapter 8: Organic Chemistry: Some Basic Principles and Techniques (Organic Chemistry)

Step-by-step solutions to all 40 exercises of Unit 8, Organic Chemistry: Some Basic Principles and Techniques (NCERT Class 11 Chemistry Part II, 2026-27): hybridisation, sigma and pi bonds, bond-line formulas and IUPAC names, inductive, resonance and hyperconjugation effects, electrophiles and nucleophiles, types of reactions and bond cleavage, purification methods, Lassaigne's test and quantitative estimation of elements. All 40 questions are answered, with the key answer highlighted.

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Hybridisation of carbon: four single bonds sp³, one double bond sp², a triple bond or two double bonds sp. −I groups (NO₂, halogens, COOH) stabilise anions; +I groups (alkyl) stabilise cations. Carbocation stability: 3° > 2° > 1° > CH₃⁺. Estimation: ; Kjeldahl ; Carius , .

Exercises

8.1
What are the hybridisation states of each carbon atom in the following compounds? CH₂=C=O, CH₃CH=CH₂, (CH₃)₂CO, CH₂=CHCN, C₆H₆
Solution

Rule: four single bonds → sp³; one double bond → sp²; two double bonds or a triple bond → sp.

CH₂=C=O: CH₂ sp², central C sp. CH₃CH=CH₂: sp³, sp², sp². (CH₃)₂CO: CH₃ sp³, C=O carbon sp². CH₂=CHCN: sp², sp², sp (C≡N). C₆H₆: all six carbons sp².

As listed

8.2
Indicate the σ and π bonds in the following molecules: C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃
Solution

C₆H₆: 12σ (6 C–C + 6 C–H), 3π. C₆H₁₂ (cyclohexane): 18σ (6 C–C + 12 C–H), no π. CH₂Cl₂: 4σ. CH₂=C=CH₂: 6σ (2 C–C + 4 C–H), 2π. CH₃NO₂: 6σ (3 C–H, 1 C–N, 2 N–O), 1π (N=O). HCONHCH₃: 8σ, 1π (C=O).

C₆H₆ 12σ 3π; C₆H₁₂ 18σ; CH₂Cl₂ 4σ; allene 6σ 2π; CH₃NO₂ 6σ 1π; HCONHCH₃ 8σ 1π

8.3
Write bond-line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Solution

Isopropyl alcohol (propan-2-ol, CH₃CH(OH)CH₃): a "V" of two lines with –OH on the middle vertex.

2,3-Dimethylbutanal (OHC–CH(CH₃)–CH(CH₃)–CH₃): a four-carbon zig-zag with the –CHO at C1 and methyl branches on C2 and C3.

Heptan-4-one (CH₃CH₂CH₂COCH₂CH₂CH₃): a seven-carbon zig-zag with =O on the fourth (middle) carbon.

As described

8.4
Give the IUPAC names of the following compounds (bond-line formulas): (a) a benzene ring with a three-carbon chain (b) CH₃CH₂CH(CH₃)CH₂CN (c) a seven-carbon chain with methyls at C2 and C5 (d) a seven-carbon chain with Br and Cl on C3 (e) ClCH₂CH₂CHO (f) Cl₂CHCH₂OH
Solution

(b) The nitrile carbon is C1. (d) Prefixes in alphabetical order: bromo before chloro.

(a) Propylbenzene (b) 3-methylpentanenitrile (c) 2,5-dimethylheptane (d) 3-bromo-3-chloroheptane (e) 3-chloropropanal (f) 2,2-dichloroethanol

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
Solution

(a) Each methyl needs its own locant. (b) Lowest locant set: 2,4,7 is lower than 2,5,7 at the first point of difference. (c) Locant sets {2, 4} tie, so the group first in alphabetical order (chloro) gets the lower number. (d) –OH is the principal group (suffix) and gets the lowest number.

(a) 2,2-Dimethylpentane (b) 2,4,7-Trimethyloctane (c) 2-Chloro-4-methylpentane (d) But-3-yn-1-ol

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
Solution

Each next member adds –CH₂–.

(a) HCOOH, CH₃COOH, CH₃CH₂COOH, CH₃(CH₂)₂COOH, CH₃(CH₂)₃COOH (b) CH₃COCH₃, CH₃COCH₂CH₃, CH₃CO(CH₂)₂CH₃, CH₃CO(CH₂)₃CH₃, CH₃CO(CH₂)₄CH₃ (c) CH₂=CH₂, CH₃CH=CH₂, CH₃CH₂CH=CH₂, CH₃(CH₂)₂CH=CH₂, CH₃(CH₂)₃CH=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
Solution

(a) (CH₃)₃CCH₂CH(CH₃)₂: a five-carbon zig-zag with two methyls on C2 and one on C4; no functional group (alkane).

(b) HOOCCH₂C(OH)(COOH)CH₂COOH (citric acid): three carboxyl groups and one hydroxyl group.

(c) OHC(CH₂)₄CHO: a six-carbon chain with –CHO at both ends: two aldehyde groups.

(a) none (b) 3 –COOH + 1 –OH (c) 2 –CHO

8.8
Identify the functional groups in the following compounds: (a) vanillin (benzene ring with –CHO, –OCH₃ and –OH) (b) procaine (H₂N–C₆H₄–COOCH₂CH₂N(C₂H₅)₂) (c) C₆H₅CH=CHNO₂
Solution

(a) Aldehyde, ether (methoxy), phenolic –OH (b) primary aromatic amine, ester, tertiary amine (c) nitro group and C=C (alkene), with a phenyl ring

8.9
Which of the two: O₂NCH₂CH₂O⁻ or CH₃CH₂O⁻ is expected to be more stable and why?
Solution

O₂NCH₂CH₂O⁻. The –NO₂ group has a strong −I effect: it withdraws electron density and disperses the negative charge on oxygen. In CH₃CH₂O⁻ the ethyl group (+I) pushes electrons towards O⁻, intensifying the charge.

O₂NCH₂CH₂O⁻ (−I effect of NO₂ disperses the charge)

8.10
Explain why alkyl groups act as electron donors when attached to a π system.
Solution

By hyperconjugation: the σ electrons of the C–H bonds on the carbon next to the π system delocalise into the π orbital. For propene, H–CH₂–CH=CH₂ ↔ H⁺ CH₂=CH–CH₂⁻ (a "no-bond" resonance form). This raises the electron density of the π system, so the alkyl group acts as an electron donor.

Hyperconjugation (σ C–H electrons delocalise into the π system).

8.11
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. (a) C₆H₅OH (b) C₆H₅NO₂ (c) CH₃CH=CHCHO (d) C₆H₅–CHO (e) C₆H₅–CH₂⁺ (f) CH₃CH=CHCH₂⁺
Solution

(a) Phenol: a lone pair on O moves into the ring (+R): three structures with O⁺ and a negative charge at the two ortho carbons and the para carbon, plus the two Kekulé forms.

(b) Nitrobenzene: ring π electrons move towards NO₂ (−R): structures with a positive charge at the two ortho and the para carbons (O⁻ on the nitro group).

(c) CH₃CH=CH–CH=O ↔ CH₃CH⁺–CH=CH–O⁻ (the C=C π pair moves to C–C, the C=O π pair onto O).

(d) Benzaldehyde: the C=O π pair moves onto O, then the ring donates: positive charge at the ortho and para carbons.

(e) Benzyl cation: ring π electrons move towards CH₂⁺: positive charge at the two ortho and the para carbons (four structures).

(f) CH₃CH=CH–CH₂⁺ ↔ CH₃CH⁺–CH=CH₂ (allylic resonance).

As described (curved arrows move π pairs or lone pairs one step at a time)

8.12
What are electrophiles and nucleophiles? Explain with examples.
Solution

Electrophile ("electron-loving"): an electron-deficient reagent that accepts an electron pair; positive ions (H⁺, NO₂⁺, CH₃⁺) or neutral molecules with an electron-poor centre (BF₃, AlCl₃, SO₃, the carbon of C=O). It attacks electron-rich sites.

Nucleophile ("nucleus-loving"): a reagent that donates an electron pair; negative ions (OH⁻, CN⁻, Cl⁻, RO⁻) or neutral molecules with lone pairs (H₂O, NH₃, R₃N). It attacks electron-poor centres, e.g. OH⁻ + CH₃Cl → CH₃OH + Cl⁻.

Electrophiles accept, nucleophiles donate an electron pair.

8.13
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O (b) CH₃COCH₃ + ⁻CN → (CH₃)₂C(CN)(OH) (c) C₆H₆ + CH₃C⁺O → C₆H₅COCH₃
Solution

(a) HO⁻ donates a pair to the acidic H. (b) ⁻CN attacks the carbonyl carbon. (c) The acylium ion CH₃CO⁺ attacks the π electrons of benzene.

(a) Nucleophile (b) nucleophile (c) electrophile

8.14
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)₂ClC–CH₃ (c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻ (d) (CH₃)₃C–CH₂OH + HBr → (CH₃)₂CBrCH₂CH₃ + H₂O
Solution

(a) HS⁻ replaces Br⁻. (b) H⁺ adds first (Markovnikov). (c) H and Br are removed from adjacent carbons. (d) The 1° carbocation rearranges (methyl shift) to a 3° carbocation, then Br⁻ attacks.

(a) Nucleophilic substitution (b) electrophilic addition (c) β-elimination (d) nucleophilic substitution with rearrangement

8.15
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors? (a) pentan-3-one and pentan-2-one (b) the two forms of DHC=CHD (D atoms on the same side and on opposite sides) (c) H–C(=O⁺H)–OH and H–C⁺(OH)–OH
Solution

(a) Same formula C₅H₁₀O, C=O in different positions. (b) Same connectivity, different arrangement about C=C. (c) Same positions of nuclei; only the π electrons differ.

(a) Structural (position) isomers (b) geometrical isomers (cis–trans) (c) resonance contributors

8.16
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) a ketone + ⁻OH → carbanion (enolate) + H₂O (c) a tertiary bromide → carbocation + Br⁻ (d) benzene + E⁺ → arenium ion
Solution

(a) Two half-arrows (fish-hooks) from the O–O bond, one to each O. (b) OH⁻ takes the α-H; the C–H pair stays on carbon. (c) The C–Br pair goes with Br. (d) A ring π pair forms a bond to E⁺.

(a) Homolysis: free radicals (b) heterolysis: carbanion (c) heterolysis: carbocation (d) heterolysis: carbocation (arenium ion)

8.17
Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH (b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH
Solution

Inductive effect: permanent displacement of σ electrons along a chain due to a polar bond (−I: halogens, NO₂; +I: alkyl groups).

Electromeric effect: temporary, complete transfer of a π pair to one atom of a multiple bond when an attacking reagent is present.

Both orders are explained by the inductive effect: (a) each Cl (−I) withdraws electrons and stabilises the carboxylate ion, so more Cl makes a stronger acid. (b) Alkyl groups (+I) push electrons and destabilise the carboxylate, so more alkyl groups make a weaker acid.

Inductive effect in both

8.18
Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography
Solution

(a) Crystallisation: the compound and the impurities have different solubilities in a solvent; the compound is sparingly soluble cold but quite soluble hot. A hot saturated solution is filtered and cooled, and pure crystals form (benzoic acid from hot water).

(b) Distillation: liquids with different boiling points vaporise at different temperatures; the vapours are condensed separately (chloroform, b.p. 334 K, from aniline, b.p. 457 K).

(c) Chromatography: the components distribute differently between a stationary phase and a mobile phase, so they move at different rates (separating the dyes of ink on paper).

Difference in solubility; in boiling point; in distribution between two phases.

8.19
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Solution

Fractional crystallisation: dissolve the mixture in hot S to make a saturated solution and cool it. The less soluble compound crystallises first and is filtered off; the mother liquor is concentrated and cooled again to obtain the more soluble one. Repeating gives each pure.

Fractional crystallisation

8.20
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Solution

Simple distillation: for liquids that boil without decomposition and differ widely in boiling point, at atmospheric pressure.

Under reduced pressure: the pressure is lowered so the liquid boils below its normal boiling point; used for liquids that decompose at their boiling points (e.g. glycerol).

Steam distillation: for steam-volatile, water-immiscible substances (aniline, essential oils). Steam is passed through the mixture, which boils when the sum of the vapour pressures equals atmospheric pressure, below 373 K.

Normal pressure; lowered pressure (decomposing liquids); with steam (steam-volatile, immiscible liquids).

8.21
Discuss the chemistry of Lassaigne's test.
Solution

The compound is fused with sodium, converting covalent N, S and X into ionic NaCN, Na₂S and NaX. The fused mass is boiled with water and filtered (sodium fusion extract).

N: boil with FeSO₄, acidify with H₂SO₄: ; (Prussian blue).

S: acetic acid + lead acetate → black PbS; or sodium nitroprusside → violet colour.

N and S together: NaSCN forms; with Fe³⁺ a blood-red colour ([Fe(SCN)]²⁺).

Halogens: boil with HNO₃, add AgNO₃: white AgCl (soluble in NH₃), pale yellow AgBr (sparingly soluble), yellow AgI (insoluble).

Sodium fusion converts N, S, X into ions that give characteristic colour or precipitate tests.

8.22
Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl's method.
Solution

(i) Dumas: the compound is heated with CuO in an atmosphere of CO₂; its nitrogen is set free as N₂ gas, collected over KOH and its volume measured. Works for all nitrogen compounds.

(ii) Kjeldahl: the compound is digested with conc. H₂SO₄, converting N into ammonium sulphate; NH₃ is liberated with NaOH, absorbed in a known excess of standard acid and the excess is back-titrated. Not suitable for nitro, azo or ring nitrogen.

Dumas: N₂ volume; Kjeldahl: NH₃ absorbed in standard acid.

8.23
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Solution

Halogens (Carius method): heat a known mass with fuming HNO₃ and AgNO₃ in a sealed tube; the halogen forms AgX, which is filtered, dried and weighed.

Sulphur: heat with fuming HNO₃: S → H₂SO₄; precipitate as BaSO₄ with BaCl₂ and weigh.

Phosphorus: heat with fuming HNO₃: P → H₃PO₄; precipitate as ammonium phosphomolybdate, or as MgNH₄PO₄ ignited to Mg₂P₂O₇, and weigh.

X as AgX, S as BaSO₄, P as phosphomolybdate or Mg₂P₂O₇ (gravimetric).

8.24
Explain the principle of paper chromatography.
Solution

It is partition chromatography. Water held in the paper is the stationary phase; a solvent rising by capillary action is the mobile phase. The components partition differently between the two phases, so they travel different distances. Spots are located directly, by a spraying reagent or under UV light, and identified by .

Partition between stationary water and a moving solvent.

8.25
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Solution

If N or S is present, the extract contains NaCN and Na₂S, which would also give precipitates with AgNO₃ (AgCN white, Ag₂S black). Boiling with HNO₃ decomposes them: ↑; ↑.

To remove CN⁻ and S²⁻, which would interfere.

8.26
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Solution

In organic compounds these elements are covalently bonded and give no ions in water, so ionic tests fail. Fusion with Na converts them into ionic, water-soluble NaCN, Na₂S and NaX.

To convert covalent N, S, X into ionic sodium salts.

8.27
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Solution

Camphor sublimes on heating and collects on a cool surface; CaSO₄ does not sublime.

Sublimation

8.28
Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Solution

The liquid and water are immiscible, so each exerts its own vapour pressure. The mixture boils when = atmospheric pressure, so the organic liquid's vapour pressure need only be well below 1 atm: the mixture boils below its boiling point.

The two vapour pressures add to reach atmospheric pressure.

8.29
Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Solution

No. Chlorine in CCl₄ is covalently bonded; CCl₄ does not give Cl⁻ ions, so no AgCl forms.

No (no free Cl⁻ ions)

8.30
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Solution

CO₂ is acidic and is absorbed completely by the strong base: . The increase in mass of the KOH tube gives the mass of CO₂.

KOH absorbs CO₂ completely; its gain in mass is weighed.

8.31
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Solution

With H₂SO₄, lead acetate would give a white precipitate of PbSO₄, masking the black PbS. Acetic acid acidifies without adding sulphate ions.

H₂SO₄ would precipitate white PbSO₄.

8.32
An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Solution

C g; CO₂ g.

H g; H₂O g.

CO₂ 0.506 g; H₂O 0.0864 g

8.33
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 residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Solution

H⁺ taken mmol; NaOH used mmol, so 30 mmol H⁺ were left. H⁺ neutralised by NH₃ mmol mmol NH₃.

N g; .

56%

8.34
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.
Solution

37.57%

8.35
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.
Solution

19.6% (NCERT key 19.66%)

8.36
In the organic compound CH₂=CH–CH₂–CH₂–C≡CH, the pair of hydridised orbitals involved in the formation of: C2 – C3 bond is: (a) sp – sp² (b) sp – sp³ (c) sp² – sp³ (d) sp³ – sp³
Solution

C2 is a double-bond carbon (sp²); C3 is saturated (sp³).

(c)

8.37
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
Solution

Prussian blue is iron(III) hexacyanoferrate(II).

(b)

8.38
Which of the following carbocation is most stable? (a) (CH₃)₃C.C⁺H₂ (b) (CH₃)₃C⁺ (c) CH₃CH₂C⁺H₂ (d) CH₃C⁺HCH₂CH₃
Solution

The tertiary cation is stabilised by three +I methyl groups and nine hyperconjugating C–H bonds.

(b)

8.39
The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography
Solution

Chromatography works even for very small amounts and for compounds with very similar properties.

(d)

8.40
The reaction: CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI is classified as: (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition
Solution

OH⁻ (a nucleophile) replaces I⁻.

(b)

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