CBSE · NCERT · Class 11 Chemistry · Chapter 8

NCERT Solutions: Class 11 Chemistry Chapter 8 - Organic Chemistry - Some Basic Principles and Techniques

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Chapter-wise NCERT intext questions and exercise answers for Organic Chemistry - Some Basic Principles and Techniques, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 30
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1Exercises30 questions
Q.8.1What are hybridisation states of each carbon atom in the following compounds ? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6v
Solution

A carbon with four sigma bonds is sp3, one double bond is sp2, and either one triple bond or two cumulated double bonds gives sp hybridisation. Applying this rule gives the listed hybridisations in the order of carbon atoms written.

Answer:

CH2=C=O: sp2, sp; CH3CH=CH2: sp3, sp2, sp2; (CH3)2CO: methyl carbons sp3 and carbonyl carbon sp2; CH2=CHCN: sp2, sp2, sp; C6H6: all carbons sp2.

Q.8.5Which 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.v
Solution

Identical substituents require repeated locants, so 2,2-dimethylpentane is correct. The lowest set of locants gives 2,4,7-trimethyloctane. When locant sets tie, the substituent cited first alphabetically gets the lower number, so chloro gets 2. The alcohol suffix has priority for numbering, so the -OH group is at C-1 in but-3-yn-1-ol.

Answer:

(a) 2,2-Dimethylpentane; (b) 2,4,7-trimethyloctane; (c) 2-chloro-4-methylpentane; (d) but-3-yn-1-ol.

Q.8.6Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H-COOH (b) CH3COCH3 (c) H-CH=CH2v
Solution

Successive members of a homologous series differ by -CH2-. The series are carboxylic acids, ketones and alkenes respectively.

Answer:

(a) HCOOH, CH3COOH, CH3CH2COOH, CH3CH2CH2COOH, CH3(CH2)3COOH. (b) CH3COCH3, CH3COCH2CH3, CH3COCH2CH2CH3, CH3COCH2CH2CH2CH3, CH3CO(CH2)4CH3. (c) CH2=CH2, CH3CH=CH2, CH3CH2CH=CH2, CH3CH2CH2CH=CH2, CH3CH2CH2CH2CH=CH2.

Q.8.9Which of the two: O2NCH2CH2O- or CH3CH2O- is expected to be more stable and why?v
Solution

The nitro group has a strong -I effect and withdraws electron density through sigma bonds. This disperses and stabilises the negative charge on oxygen in O2NCH2CH2O-. The ethyl group in CH3CH2O- has a +I effect and does not stabilise the negative charge.

Answer:

O2NCH2CH2O- is more stable.

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

Alkyl groups are less electronegative than sp2 carbon and push electron density through sigma bonds. In addition, C-H sigma bonds on the alpha carbon can overlap with the adjacent π system or empty p orbital; this hyperconjugation delocalises electron density toward the π system.

Answer:

They donate electron density through +I effect and hyperconjugation.

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

Electrophiles are electron-deficient species that attack electron-rich centres, for example H+, NO2+, BF3, carbocations and acyl cations. Nucleophiles are electron-rich species that donate an electron pair, for example OH-, CN-, Cl-, NH3, H2O and alkenes.

Answer:

Electrophiles are electron-pair acceptors; nucleophiles are electron-pair donors.

Q.8.13Identify 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 + CH3CO+ → C6H5COCH3v
Solution

HO- and CN- donate electron pairs, so they are nucleophiles. CH3CO+ is electron deficient at carbon and accepts electron density from benzene, so it is an electrophile.

Answer:

(a) HO- is a nucleophile; (b) CN- is a nucleophile; (c) CH3CO+ is an electrophile.

Q.8.17Explain 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.COOHv
Solution

The inductive effect is permanent displacement of sigma electrons through a chain due to electronegativity differences. The electromeric effect is temporary complete transfer of π electrons under the influence of an attacking reagent. In (a), more chlorine atoms exert a stronger -I effect, stabilising the carboxylate ion and increasing acidity. In (b), alkyl groups exert +I effect, destabilising the carboxylate ion; more alkyl substitution lowers acidity.

Answer:

Both orders are explained mainly by the inductive effect.

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

Crystallisation purifies a solid by dissolving it in a suitable hot solvent and crystallising it on cooling; impurities remain in solution or are filtered off. Distillation separates liquids or purifies a liquid based on different boiling points, for example separating acetone and water. Chromatography separates components because they adsorb or partition differently between stationary and mobile phases, for example separating plant pigments on paper or a column.

Answer:

Crystallisation uses solubility differences; distillation uses volatility/boiling-point differences; chromatography uses differential distribution between stationary and mobile phases.

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

The mixture is shaken with solvent S in a separating funnel. The compound more soluble in S passes preferentially into that layer, while the less soluble compound remains mostly in the other layer or residue. Repeated extraction improves separation, and the solvent is then evaporated to recover the compound.

Answer:

Differential extraction using solvent S can be used.

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

Simple distillation is used for volatile liquids that boil without decomposition and have suitable boiling-point differences. Distillation under reduced pressure is used for liquids that decompose at their normal boiling point; lowering pressure lowers the boiling temperature. Steam distillation is used for substances immiscible with water but volatile in steam; the mixture boils when vapour pressures sum to atmospheric pressure.

Answer:

Simple distillation is at atmospheric pressure; reduced-pressure distillation lowers boiling point; steam distillation co-distils steam-volatile, water-immiscible compounds below their normal boiling points.

Q.8.21Discuss the chemistry of Lassaigne’s test.v
Solution

On sodium fusion, nitrogen forms NaCN, sulphur forms Na2S, and halogens form NaX. Nitrogen is detected by forming sodium ferrocyanide, which with ferric ions gives Prussian blue ferric ferrocyanide. Sulphur is detected by sodium nitroprusside giving a violet colour or by lead acetate giving black PbS. Halogens are detected by acidifying the sodium extract with HNO3 and adding AgNO3 to form AgCl, AgBr or AgI precipitates.

Answer:

Lassaigne's test converts covalently bonded N, S and halogens into ionic sodium salts, which are then detected by characteristic reactions.

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

In Dumas method, the organic compound is heated with CuO; carbon and hydrogen form CO2 and H2O, while nitrogen is liberated as N2 and measured volumetrically. In Kjeldahl's method, the compound is digested with concentrated H2SO4 to convert nitrogen into ammonium sulphate. On treatment with alkali, NH3 is liberated and absorbed in standard acid; the amount of acid neutralised gives nitrogen. Kjeldahl's method is not applicable to nitro, azo and ring nitrogen compounds that do not convert completely to ammonium sulphate.

Answer:

Dumas measures nitrogen gas formed on combustion; Kjeldahl converts nitrogen into ammonium sulphate and estimates liberated ammonia.

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

In Carius method for halogens, the compound is heated with fuming HNO3 and AgNO3 in a sealed tube; halogen is converted into AgX, which is weighed. Sulphur is oxidised to sulphuric acid and precipitated as BaSO4, whose mass gives sulphur. Phosphorus is oxidised to phosphoric acid and precipitated either as ammonium phosphomolybdate or as magnesium ammonium phosphate, which on ignition gives Mg2P2O7 for weighing.

Answer:

Halogens are estimated by Carius method as silver halides, sulphur as BaSO4, and phosphorus as ammonium phosphomolybdate or Mg2P2O7.

Q.8.24Explain the principle of paper chromatography.v
Solution

Paper acts as the stationary phase, usually with water adsorbed on cellulose fibres. The solvent is the mobile phase. Different components distribute differently between these phases and therefore travel different distances, giving separation.

Answer:

Paper chromatography is based on differential partition of solutes between water held on paper and a moving solvent.

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

In Lassaigne's extract, CN- and S2- may also react with AgNO3 to give precipitates. Boiling with nitric acid decomposes these interfering ions and converts them into volatile or soluble products, so any precipitate with AgNO3 is due to halide ions.

Answer:

Nitric acid removes interfering cyanide and sulphide ions and acidifies the extract.

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

Organic compounds usually contain N, S and halogens in covalent form, which cannot be detected directly by ordinary ionic tests. Fusion with sodium converts them into NaCN, Na2S and NaX. These dissolve in water to give ions that can be tested in Lassaigne's extract.

Answer:

Sodium fusion converts covalently bonded elements into water-soluble ionic sodium salts.

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

Camphor sublimes on heating, while calcium sulphate does not. Heating the mixture allows camphor vapours to sublime and condense separately, leaving calcium sulphate behind.

Answer:

Sublimation.

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

For two immiscible liquids, each exerts its own vapour pressure independently. In steam distillation, boiling occurs when pwater + porganic equals atmospheric pressure. Since the organic liquid contributes part of the total pressure, the mixture boils below the normal boiling point of the organic liquid.

Answer:

Because the sum of vapour pressures of water and the organic liquid reaches atmospheric pressure below the liquid's normal boiling point.

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

CCl4 is a covalent compound and does not furnish Cl- ions in solution. Silver nitrate gives AgCl precipitate only with chloride ions, so CCl4 does not give a white AgCl precipitate under this test.

Answer:

No.

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

During carbon estimation, carbon is oxidised to CO2. Potassium hydroxide reacts with CO2 to form potassium carbonate/bicarbonate, so the increase in mass of the KOH absorber gives the mass of CO2 produced.

Answer:

Because KOH absorbs CO2 quantitatively.

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

Lead acetate detects sulphide as black PbS. If sulphuric acid is used, Pb2+ also forms insoluble white PbSO4, causing interference. Acetic acid acidifies the solution without introducing sulphate ions.

Answer:

Sulphuric acid would give lead sulphate precipitate and interfere with the test.

Q.8.32An 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.v
Solution

Mass of carbon in 0.20 g compound = 0.69 x 0.20 = 0.138 g. CO2 formed = 0.138 x 44/12 = 0.506 g. Mass of hydrogen = 0.048 x 0.20 = 0.0096 g. H2O formed = 0.0096 x 18/2 = 0.0864 g.

Answer:

CO2 produced = 0.506 g; H2O produced = 0.0864 g.

Q.8.340.3780 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.v
Solution

Molar mass of AgCl = 108 + 35.5 = 143.5 g mol^-1. Mass of chlorine in 0.5740 g AgCl = 0.5740 x 35.5/143.5 = 0.1420 g. Percentage chlorine = 0.1420 x 100/0.3780 = 37.6%.

Answer:

37.6% chlorine.

Q.8.35In 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.v
Solution

Molar mass of BaSO4 = 233 g mol^-1, containing 32 g sulphur. Mass of sulphur in 0.668 g BaSO4 = 0.668 x 32/233 = 0.0917 g. Percentage sulphur = 0.0917 x 100/0.468 = 19.6%.

Answer:

19.6% sulphur.

Q.8.36In 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) sp - sp3 (c) sp2 - sp3 (d) sp3 - sp3v
  1. a. sp - sp2
  2. b. sp - sp3
  3. c. sp2 - sp3
  4. d. sp3 - sp3
Solution

In CH2=CH-CH2-CH2-C≡CH, C2 is part of the double bond and is sp2 hybridised. C3 is a saturated -CH2- carbon with four sigma bonds and is sp3 hybridised. Therefore the C2-C3 sigma bond is formed by sp2-sp3 overlap.

Answer:

(c) sp2 - sp3.

Q.8.37In 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]4v
  1. a. Na4[Fe(CN)6]
  2. b. Fe4[Fe(CN)6]3
  3. c. Fe2[Fe(CN)6]
  4. d. Fe3[Fe(CN)6]4
Solution

Nitrogen in the sodium extract forms cyanide, which with iron salts ultimately gives ferric ferrocyanide, Fe4[Fe(CN)6]3, known as Prussian blue.

Answer:

(b) Fe4[Fe(CN)6]3.

Q.8.38Which of the following carbocation is most stable? (a) (CH3)3C-CH2+ (b) (CH3)3C+ (c) CH3CH2CH2+ (d) CH3CH+CH2CH3v
  1. a. (CH3)3C-CH2+
  2. b. (CH3)3C+
  3. c. CH3CH2CH2+
  4. d. CH3CH+CH2CH3
Solution

Carbocation stability increases with alkyl substitution because of +I effect and hyperconjugation: tertiary > secondary > primary. (CH3)3C+ is tertiary and therefore the most stable among the options.

Answer:

(b) (CH3)3C+.

Q.8.39The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatographyv
  1. a. Crystallisation
  2. b. Distillation
  3. c. Sublimation
  4. d. Chromatography
Solution

Chromatography is widely used for separation, purification and identification because it can separate small quantities and closely related components efficiently.

Answer:

(d) Chromatography.

Q.8.40The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) additionv
  1. a. electrophilic substitution
  2. b. nucleophilic substitution
  3. c. elimination
  4. d. addition
Solution

Aqueous KOH supplies OH-, which replaces I- in ethyl iodide to form ethanol. Replacement by a nucleophile is nucleophilic substitution.

Answer:

(b) nucleophilic substitution.