NCERT Class 11 Chemistry Chapter 8 Organic Chemistry-Some Basic Principles and Techniques Solutions to each chapter is provided in the list so that you can easily browse through different chapters NCERT Class 11 Chemistry Chapter 8 Organic Chemistry-Some Basic Principles and Techniques and select need one. NCERT Class 11 Chemistry Chapter 8 Organic Chemistry-Some Basic Principles and Techniques Question Answers Download PDF. NCERT Chemistry Class 11 Solutions.
NCERT Class 11 Chemistry Chapter 8 Organic Chemistry-Some Basic Principles and Techniques
Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 11 Chemistry Chapter 8 Organic Chemistry-Some Basic Principles and Techniques Solutions are part of All Subject Solutions. Here we have given NCERT Class 11 Chemistry Part: I, Part: II Notes. NCERT Class 11 Chemistry Chapter 8 Organic Chemistry-Some Basic Principles and Techniques Notes, NCERT Class 11 Chemistry Textbook Solutions for All Chapters, You can practice these here.
Organic Chemistry-Some Basic Principles and Techniques
Chapter: 8
| Part – II |
1. What are hybridisation states of each carbon atom in the following compounds?
CH2 = C = O, CH3 CH = CH2, (CH3) 2CO, CH2 = CHCN, C6H6
Ans: The hybridization states of each carbon atom in the following compounds are as follows:
2. Indicate the σ and π bonds in the following molecules :
C6H6, C6H12, CH2Cl2, CH2 = C = CH2, CH3NO2, HCONHCH3
Ans: (i) C6H6,
(ii) C6H12
(iii) CH2Cl2
(iv) CH2=C=CH2
(v) CH3NO2
(vi) HCONHCH3
3. Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Ans:
Isopropyl alcohol
2,3- Dimethy Ibutanal
Heptan-4-one
4. Give the IUPAC names of the following compounds:
(a)
Ans: Propylbenzene.
(b)
Ans: 3- Methylpentanenitrile.
(c)
Ans: 2,5- Dimethylheptane.
(d)
Ans: 3-Bromo- 3- Chloroheptane.
(e)
Ans: 3- Chloropropanal.
(f) Cl2CHCH2OH
Ans: 2, 2- Dichloroethanol.
5. Which of the following represents the correct IUPAC name for the compounds concerned?
(a) 2, 2-Dimethylpentane or 2-Dimethylpentane.
Ans: 2,2-Dimethylpentane: When two alkyl groups are attached to the same carbon atom, the locant for that carbon is repeated twice.
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane.
Ans: 2,4,7-Trimethyloctane: The locant set 2,4,7 is lower than 2,5,7, making it the correct IUPAC name.
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
Ans: 2-Chloro-4-methylpentane. Alphabetical order of substituents.
(d) But-3-yn-1-ol or But-4-ol-1-yne.
Ans: But-3-yn-1-ol. Lower locant for the principal functional group, i.e., alcohol.
6. Draw formulas for the first five members of each homologous series beginning with the following compounds.
(a) H–COOH.
Ans: The first five members of each homologous series starting with the given compounds are as follows:
(i) CH₃COOH
(ii) CH₃CH₂COOH
(iii) CH₃CH₂CH₂COOH
(iv) CH₃CH₂CH₂CH₂COOH
(v) CH₃(CH₂)₄COOH
(b) CH3COCH3
Ans: The first five members of each homologous series starting with the given compounds are as follows:
(i) CH₃COCH₂CH₃
(ii) CH₃COCH₂CH₂CH₃
(iii) CH₃COCH₂CH₂CH₂CH₃
(iv) CH₃COCH₂CH₂CH₂CH₂CH₃
(v) CH₃CO(CH₂)₄CH₃
(c) H–CH = CH2
Ans: The first five members of each homologous series starting with the given compounds are as follows:
CH₃CH = CH₂
CH₃CH₂CH = CH₂
CH₃CH₂CH₂CH = CH₂
CH₃CH₂CH₂CH₂CH = CH₂
CH₃CH₂CH₂CH₂CH₂CH = CH₂
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.
Ans:
8. Identify the functional groups in the following compounds:
(a)
Ans:
(b)
Ans:
(c)
Ans:
9. Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?
Ans: The stability of ONCH₃CHO is greater than CH₃CH₂O because the presence of the NO₂ group stabilizes the molecule through resonance and inductive effects, increasing overall stability.
has-1-effect and hence it tends to disperse the- charge on the O-atom. In contrast, CH3CH2 has effect. It, therefore, tends to intensify change and hence destabilizes it.
10. Explain why alkyl groups act as electron donors when attached to a π system.
Ans: Alkyl groups donate electrons to a π system via hyperconjugation, stabilizing the system by delocalizing electron density.
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–C + H2
(f) CH3CH=CH C + H2
Ans:
(b)
12. What are electrophiles and nucleophiles ? Explain with examples.
Ans: Nucleophiles are reagents with lone pairs of electrons or negative charges, making them electron-rich. They donate electrons to electron-deficient sites (nucleus-loving). According to Lewis theory, nucleophiles act as Lewis bases. Examples include OH− and NH3.
Electrophiles: A positively charged on neutral species which are deficient of electrons and can accept a pair of electrons are called electrophiles. These are also called electron loving (philic) species.
For example:
H+, H3O+, CI+, CH3+, NO2+ (Positively charged)
AICI3,BF3 SO3 (Neutral spevied)
Aluminium and boron, with six electrons, seek to complete their octet, making them electrophiles. According to Lewis theory, they are classified as Lewis acids due to this electron deficiency.
13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
(a) CH3COOH + HO → CH3COO– + H2O
Ans: HO− is a nucleophile, CH3COOH is an electrophile.
(b) CH3COCH3+ C – N → (CH3) 2C(CN)(OH)
Ans: C–N is a nucleophile CH3COCH3 is an electrophile.
(c) C6H6 + CH3C + O → C6H5COCH3
Ans: CH3CO is an electrophile, C6H6 is a nucleophile.
14. Classify the following reactions in one of the reaction types studied in this unit.
(a) CH3CH2Br + HS– → CH3CH2SH + Br–.
Ans: Nucleophilic substitution.
(b) (CH3) 2C = CH2 + HCI → (CH3) 2CIC – CH3
Ans: Electrophilic addition.
(c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br–
Ans: Elimination.
(d) (CH3) 3C– CH2OH + HBr → (CH3) 2CBrCH2CH2CH3 + H2O
Ans: Substitution.
15. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
(a)
(b)
(c)
Ans: (a) These are position isomers.
(b) These are geometrical isomers.
(c) These are resonance structures because they differ in the distribution of electron pairs, while the positions of the atoms remain unchanged.
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)
Ans: Use curved arrows to show electron flow. Homolysis produces free radicals, heterolysis produces carbocations or carbanions.
(b)
Ans: Bond cleavage with curved arrows shows electron flow: homolysis yields free radicals, while heterolysis yields carbocations or carbanions.
(c)
Ans: The bond cleavage using curved arrows to show electron flow can be represented as follows:
(d)
Ans: The bond cleavage using curved arrows shows electron flow: homolysis forms free radicals, and heterolysis forms carbocations and carbanions.
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
Ans: Inductive effect is the permanent shift of electron density in a molecule due to the difference in electronegativity between atoms, like in a C-X bond. This effect diminishes as it moves along the carbon chain, weakening with distance from the substituent atom.
Electromeric effect is a temporary shift of electrons within a molecule with a multiple bond, triggered by an approaching electrophile. This results in the complete transfer of electrons from one atom to another, creating temporary polarization in the molecule.
Inductive effect explains the correct orders of acidity of the carboxylic acid.
In the first case, the inductive effect in di and tri halogen substituted acids is more marked with the result these acids are progressively more stronger than the corresponding mono halogeno substituted acid. (- I effect)
In the second case, the decreasing acidic strength is due to increase in + I effect due to alky1 groups. (+ I effect).
18. Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation.
Ans: Crystallisation: This is the most common method for the purification of solid organic compounds. It is based on the fact that certain organic compounds are partly soluble in a solvent at room temperature and solubility increases with increase in temperature.
Example: Separation of sugar from a mixture of sugar and common salt by using C2H5OH.
(b) Distillation.
Ans: Distillation: This method is based on the principle that at constant pressure every pure liquid boils at a definite temperature called its boiling point. The method is used for the purification of those liquids which boil without decomposition provided the impurities are non-volatile. The method is applied for the purification when the two liquids differ in their boiling points by 30-50 Κ.
Example: Distillation of crude oil to separate it into fractions like gasoline, diesel, and kerosene.
(c) Chromatography.
Ans: Chromatography: Chromatography is a technique used to separate and analyse components of a mixture based on their movement through a stationary phase under the influence of a mobile phase. Different components travel at different speeds, leading to separation.
Example: Separation of plant pigments using paper chromatography, where different pigments move at different rates on filter paper based on their solubility in the solvent.
19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Ans: To separate two compounds with different solubilities in a solvent S, extraction can be used. Add the solvent S to the mixture, which selectively dissolves the more soluble compound. Then, separate the two compounds by filtration or decanting, leaving the less soluble compound behind.
20. What is the difference between distillation, distillation under reduced pressure and steam distillation?
Ans: Distillation: This technique separates liquids with significantly different boiling points by heating and condensing the vapour.
Distillation under reduced pressure: It is used to purify those liquids which decompose at or below their normal boiling point.
Steam distillation: This technique separates steam-volatile substances from non-volatile impurities by using steam to vaporise the desired substances, which are then condensed and collected separately.
21. Discuss the chemistry of Lassaigne’s test.
Ans: Preparation of Lassaigne’s filtrate or sodium fusion extract: In Lassaigne’s test, heat a small piece of sodium in a fusion tube until it melts. Add the organic compound, reheat until red hot, and immerse the tube in cold water. Break the tube, boil the mixture, and filter. The filtrate, called Lassaigne’s extract, is used to detect N, S, and halogens.
The chemical reactions are as follows:
(a) If N is Present
(b) If S is present
(c) If Both N and S Are present
(d) If X is Present
22. Differentiate between the principle of estimation of nitrogen in an organic compound by:
(i) Dumas method. and
Ans: Principle of Dumas’ method: A known mass of the organic compound is heated strongly with excess cupric oxide in a CO₂ atmosphere. Carbon is oxidized to CO₂ and hydrogen to H₂O. Nitrogen in the compound is released as nitrogen gas. Any nitrogen oxides formed are reduced back to nitrogen by passing over hot reduced copper gauze.
The gaseous mixture is passed through concentrated KOH solution, which absorbs CO₂ and H₂O vapors. Nitrogen (N₂) is not absorbed and is collected separately. The volume of nitrogen evolved is measured, allowing the calculation of nitrogen content or percentage in the original compound.
(ii) Kjeldahl’s method.
Ans: Principle of Kjeldahl’s Method: A known mass of the organic compound is digested (heated strongly) with concentrated sulfuric acid (H₂SO₄), a small amount of potassium sulfate (to raise the boiling point of H₂SO₄), and a catalyst such as mercury or copper sulfate in a long-necked flask called a Kjeldahl flask. This digestion converts the nitrogen in the organic compound into ammonium sulfate. The resulting solution is then treated with excess sodium hydroxide (NaOH). The ammonia gas liberated is absorbed in a known volume of a standard acid solution, such as sulfuric acid (H₂SO₄), with a known molarity.
The volume of the acid left unused is determined by titrating it against a standard alkali solution such as NaOH.
23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Ans: Principle of Estimation of Halogens: The Carius method involves converting the halogen in an organic compound into the corresponding silver halide. By measuring the mass of the organic compound and the mass of the silver halide produced, the percentage of halogen in the compound can be determined.
Principle of estimation of sulphur: Sulphur present in any organic compound is oxidized to sulphuric acid by digesting with fuming nitric acid. Sulphuric acid so formed is then quantitatively precipitated as barium sulphate by the addition of excess barium chloride. The precipitate of barium sulphate is filtered, washed, dried and weighed.
Principle of Estimation of Phosphorus: Phosphorus in an organic compound is oxidized to phosphoric acid by heating with fuming nitric acid. The resulting phosphoric acid is then precipitated as MgNH4PO4, which, upon ignition, is converted into Mg2P2O7. From the mass of Mg2P2O7 obtained, the amount of phosphorus in the organic compound can be determined.
24. Explain the principle of paper chromatography.
Ans: Paper chromatography is a special type of partition chromatography.
It is based on the differential (continuous) partitioning of components of a mixture between stationary phase and mobile phase.
In paper chromatography, a strip of paper is spotted with a solution of the mixture and then placed in a solvent or a mixture of solvents. This solvent moves up the paper by capillary action, carrying the components of the mixture with it. Different components move at different rates and get separated as the solvent rises. The paper with the separated components is called a chromatogram. Colored spots appear at various heights on the paper, while colorless spots can be seen using ultraviolet light or a special spray reagent.
25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Ans: When testing Lassaigne’s extract for halogens, it is first boiled with dilute nitric acid. This step decomposes NaCN into HCN and Na2S into H2S, allowing these gases to be expelled. Essentially, if nitrogen and sulfur are present as NaCN and Na2S, they are removed. The chemical reactions involved are represented as follows:
NaCN + HNO3 → NaNO3 + HCI ↑
Na2 + 2HNO3 → NaNO3 + H2S ↑
Therefore, dil nitric acid is added before testing halogens to expel all the gases if evolved.
26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Ans: In organic compounds, nitrogen, sulphur, and halogens are covalently bonded to carbon atoms. To simplify their detection, these elements must first be converted into ionic forms. This conversion facilitates their analysis by transforming them into more easily detectable ionic species.
This is done by fusing the organic compound with sodium metal. The ionic compounds formed during the fusion are extracted in aqueous solution, and can be detected by simple chemical tests.
If both nitrogen and sulphur are present in an organic compound, sodium thiocyanate (NaSCN) is formed during fusion with excess sodium. This reacts to produce sodium cyanide and sodium sulphide.
27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Ans: Calcium sulphate and camphor present in a mixture can be separated by sublimation. Camphor sublimes in heating and can be recovered in crystalline form as sublimate.
Sublimation separates camphor from calcium sulphate as camphor transitions from solid to vapour, leaving non-sublimable calcium sulphate behind.
28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Ans: In steam distillation, the liquid begins to boil when the sum of the vapour pressures of the organic liquid (p2) and the water vapour (p1) equals the atmospheric pressure (p).
P = p1 + p2
Or
p2 = p – p1
A liquid boils when its vapor pressure equals the atmospheric pressure. Therefore, a mixture of two immiscible liquids will boil at a temperature lower than the normal boiling points of each liquid. The mixture will continue to boil at this temperature until one of the liquids is fully distilled.
29. Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Ans: Carbon tetrachloride does not form a precipitate of AgCl when heated with silver nitrate due to the lack of any electron-donating groups attached to the carbon atom bonded to the Cl atoms.
For covalent compounds, Lassaigne’s extract is prepared to detect ionic forms of elements like chloride ions.
30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Ans: Carbon dioxide reacts with KOH to form soluble potassium carbonate. The amount of carbon dioxide can be estimated by measuring the quantity of potassium carbonate formed in the reaction.
2KOH + CO2 → K2CO3 + H2O.
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?
Ans: Although sulfuric acid precipitates lead sulfate, acetic acid ensures complete precipitation of sulfur as lead sulfate by leveraging the common ion effect, which enhances the formation of the precipitate.Hence, it not H₂SO₄ is used for the acidification of the sodium extract.
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.
Ans: 100 g of the organic compound, there are 69 g of carbon.
C 0.2 g of compound contains 69/100 × 0.2
12 g carbon contains
The molecular mass of carbon dioxide (CO₂) is 44 g (4.8 × 0.2)/100 g, hydrogen is present in (18 × 4.8 × 0.2)/(2 × 100) = 0.0857g
Mass of water formed = 0.0857 g.
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.
Ans: Given that, total mass of organic compound = 0.50 g
Initial volume of acid = 50 ml
Molarity of acid = 0.5 M
60 mL of 0.5 M NaOH was used to neutralize the residual acid after ammonia was absorbed in the Kjeldahl method.
= 30 mL of 0.5 M H2SO4
Volume of H₂SO₄(0.5M) consumed = 20 mL 20 mL of 0.5 M H2SO4 = 40 m of 0.5 M NH3 or 20 mL of 1 M NH3
Also, since 1000 mL of 1 MNH3 contains 17 g of nitrogen,
20 ml of 1MNH3 weights 17 × 20 / 1000 g ammonia
0.5g of organic compound contains 14 × 20 / 1000 gN
100 g of organic compound contains
So, the percentage composition of nitrogen in the organic compound is 56%.
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.
Ans: Mass of AgCl formed = 0.5740 g
Mol. mass of AgCl = 108 + 35.5 = 143.5
143.5 of AgCl contain chlorine = 35.5 g
To find the mass of chlorine in 0.5740 g of AgCl:
Mass of chlorine = (35.5g × 0.5740g) / 143.5g
= 0.142g
Calculation of percentage of chlorine:
(0.142 / 0.3780) ×100
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.
Ans: Molar mass of BaSO₄ = 233 g/mol
Mass of BaSO₄ = 0.668 g
Mass of sulfur (S) in BaSO₄ can be calculated using the ratio of the molar masses of sulfur and BaSO₄.
The molar mass of sulfur (S) = 32 g/mol.
Mass of sulfur in 0.668 g of BaSO₄:
36. 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) sp – sp3
(c) sp2 – sp3
(d) sp3 – sp3
Ans: (b) sp – sp3
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
Ans: (b) Fe4[Fe(CN)6]3
38. Which of the following carbocation is most stable?
(a) (CH3) 3C. C+ H2
(b) (CH3) 3C+
(c) CH3CH2C+ H2
(d) CH3C+ HCH2CH3
Ans: (b) (CH3) 3C+
39. The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation.
(b) Distillation.
(c) Sublimation.
(d) Chromatography.
Ans: (d) Chromatography.
40. The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as:
(a) Electrophilic substitution.
(b) Nucleophilic substitution.
(c) Elimination.
(d) Additio.
Ans: (b) Nucleophilic substitution.
