Class 11 Chemistry

Chapter 8 Solutions — Organic Chemistry — Some Basic Principles and Techniques

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Overview

Step-by-step NCERT solutions for Organic Chemistry — Some Basic Principles and Techniques (Chapter 8, NCERT Class 11 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the Organic Chemistry — Some Basic Principles and Techniques textbook chapter.

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All 40 questions in Organic Chemistry — Some Basic Principles and Techniques are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

What are the hybridisation states of each carbon atom in the following compounds? CH₂=C=O, CH₃CH=CH₂, (CH₃)₂CO, CH₂=CHCN, C₆H₆
Indicate the sigma and pi bonds in the following molecules: C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃
Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
4.
Give the IUPAC names of the following compounds shown as bond-line formulas:
- (a) phenyl attached to a 3-carbon chain
- (b) a nitrile with a methyl branch
- (c) a branched alkane with two methyl branches
- (d) a 6-carbon chain bearing Cl and Br on the same (middle) carbon
- (e) a chloro-aldehyde chain
- (f) Cl₂CHCH₂OH.
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
6.
Draw the formulas for the first five members of each homologous series beginning with the following compounds:
- (a) H-COOH
- (b) CH₃COCH₃
- (c) H-CH=CH₂.
7.
Give condensed and bond-line structural formulas and identify the functional groups present, if any, for:
- (a) 2,2,4-Trimethylpentane
- (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
- (c) Hexanedial
8.
Identify the functional groups in the following compounds:
- (a) a benzene ring bearing -CHO, -OCH₃ (OMe) and -OH
- (b) a benzene ring bearing -NH₂ and -COOCH₂CH₂N(C₂H₅)₂
- (c) a benzene ring with a -CH=CH-NO₂ substituent.
Which of the two: O₂N-CH₂-CH₂-O⁻ or CH₃-CH₂-O⁻ is expected to be more stable and why?
Explain why alkyl groups act as electron donors when attached to a pi system.
11.
Draw the resonance structures for the following compounds and show the electron shift using curved-arrow notation:
- (a) C₆H₅OH
- (b) C₆H₅NO₂
- (c) CH₃CH=CHCHO
- (d) C₆H₅-CHO
- (e) C₆H₅-CH₂⁺ (benzyl cation)
- (f) CH₃CH=CH-CH₂⁺ (allyl cation).
What are electrophiles and nucleophiles? Explain with examples.
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) C₆H₅ + CH₃CO⁺ → C₆H₅COCH₃
14.
Classify the following reactions in one of the reaction types 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)₂CBrCH₂CH₂CH₃ + H₂O
15.
What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
- (a) two ketones — one with the C=O at the middle of a 5-carbon chain and one with the C=O next to the end
- (b) two C=C structures differing only in the D/H arrangement about the double bond
- (c) H-C(=O+H)-OH and H-C+(OH)-OH.
16.
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediate produced as free radical, carbocation, or carbanion:
- (a) CH₃-O-CH₃ → CH₃-O^ + ^CH₃
- (b) (CH₃)₂C=O + HO⁻ →
- (c) (CH₃)₃C-Br → (CH₃)₃C⁺ + Br⁻
- (d) benzene ring + E⁺ →
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
18.
Give a brief description of the principles of the following techniques taking an example in each case:
- (a) Crystallisation
- (b) Distillation
- (c) Chromatography
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
What is the difference between distillation, distillation under reduced pressure, and steam distillation?
Discuss the chemistry of Lassaigne's test.
22.
Differentiate between the principle of estimation of nitrogen in an organic compound by
- (i) Dumas method and
- (ii) Kjeldahl's method.
Discuss the principle of estimation of halogens, sulphur, and phosphorus present in an organic compound.
Explain the principle of paper chromatography.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur, and halogens.
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Explain why an organic liquid vaporises at a temperature below its boiling point in its steam distillation.
Will CCl₄ give a white precipitate of AgCl on heating with AgNO₃? Hint: CCl₄ does not give silver chloride precipitate with AgNO₃ whereas CH₃Cl does. Give reason for your answer.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
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.
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 NaOH solution for neutralisation. Find the percentage composition of nitrogen in the compound.
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.
In the 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.
36.
In the organic compound CH₂=CH-CH₂-CN, the pair of hybridised orbitals involved in the formation of the C₂-C₃ bond is:
- (a) sp-sp²
- (b) sp²-sp³
- (c) sp³-sp²
- (d) sp³-sp
37.
In the Lassaigne's test for 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) Fe₃[Fe(CN)₆]
- (d) Fe₄[Fe(CN)₆]
38.
Which of the following carbocations is most stable?
- (a) (CH₃)₃C⁺
- (b) CH₃CH₂⁺
- (c) CH₃CH₂CH₂⁺
- (d) (CH₃)₂CH⁺
39.
The best and latest technique for isolation, purification, and separation of organic compounds is:
- (a) Crystallisation
- (b) Distillation
- (c) Sublimation
- (d) Chromatography
40.
The reaction: CH₃CH₂I + OH⁻ → CH₃CH₂OH + I⁻ is classified as:
- (a) electrophilic substitution
- (b) nucleophilic substitution
- (c) elimination
- (d) addition

Sample

Worked examples

3 questions from Organic Chemistry — Some Basic Principles and Techniques, solved step by step.

Exercises

Indicate the sigma and pi bonds in the following molecules: C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃

Sigma (σ) bonds: all single bonds + the sigma component of every double/triple bond.

Pi (π) bonds: the extra bond(s) in each double bond (1π) or triple bond (2π).

C₆H₆ (benzene): 6 C-C σ (ring) + 6 C-H σ + 3 π (from the 3 formal double bonds in the Kekulé structure) → σ = 12, π = 3

C₆H₁₂ (cyclohexane): 6 C-C σ (ring) + 12 C-H σ; no double bonds → σ = 18, π = 0

CH₂Cl₂ (dichloromethane): 2 C-H σ + 2 C-Cl σ → σ = 4, π = 0

CH₂=C=CH₂ (propadiene / allene): 4 C-H σ + 2 C-C σ (each C=C contributes 1 σ) + 2 π (one per cumulated double bond) → σ = 6, π = 2

CH₃NO₂ (nitromethane): 3 C-H σ + 1 C-N σ + 2 N-O σ (one N-O single, one N=O σ) + 1 π (from N=O) → σ = 6, π = 1

HCONHCH₃ (N-methylformamide): H-C σ, C=O (σ+π), C-N σ, N-H σ, N-CH₃ σ, and 3 C-H σ on the methyl → σ = 8, π = 1

Answer: C₆H₆: 12σ, 3π; C₆H₁₂: 18σ, 0π; CH₂Cl₂: 4σ, 0π; CH₂=C=CH₂: 6σ, 2π; CH₃NO₂: 6σ, 1π; HCONHCH₃: 8σ, 1π

Exercises

Which of the two: O₂N-CH₂-CH₂-O⁻ or CH₃-CH₂-O⁻ is expected to be more stable and why?

Both species are alkoxide anions (O⁻ at the end of a chain).

CH₃-CH₂-O⁻ (ethoxide ion):

The ethyl group (-CH₂CH₃) is electron-donating by inductive effect.

This INCREASES the electron density on the already negatively charged oxygen, making it less stable (destabilisation).

O₂N-CH₂-CH₂-O⁻ (2-nitroethoxide ion):

The nitro group (-NO₂) at the other end is strongly electron-withdrawing by inductive effect (-I).

This DECREASES the electron density on the O⁻ end, dispersing/delocalising the negative charge and STABILISING the anion.

The electron-withdrawing -NO₂ group stabilises the anion by pulling electron density away from the O⁻ through the inductive effect.

Therefore: O₂N-CH₂-CH₂-O⁻ is more stable than CH₃-CH₂-O⁻.

Answer: O₂N-CH₂-CH₂-O⁻ is more stable. The -NO₂ (nitro) group is strongly electron-withdrawing (-I inductive effect), which disperses the negative charge and stabilises the anion. CH₃CH₂O⁻ has an electron-donating ethyl group (+I) that intensifies the charge and destabilises the anion.

Exercises

Explain why alkyl groups act as electron donors when attached to a pi system.

Alkyl groups donate electrons to adjacent pi systems through two effects:

1. Hyperconjugation (Baker-Nathan effect):

The C-H sigma bonding electrons in the alkyl group can overlap with the adjacent p-orbital (part of the pi system).

This is called hyperconjugation — the sigma electrons delocalise into the pi system.

For example, in propene (CH₃-CH=CH₂), the 3 C-H bonds of CH₃ can each donate electron density into the C=C pi bond.

Greater the number of alpha-C-H bonds → greater the hyperconjugation → greater electron donation.

2. Inductive effect:

Carbon is slightly less electronegative than the central sp²/sp carbon of the pi system.

Alkyl groups push electrons toward the pi system through the sigma framework (weak inductive donation).

Result: Alkyl-substituted alkenes and carbocations are more stable than unsubstituted ones because the alkyl group donates electron density, stabilising the electron-deficient pi system or cation.

Answer: Alkyl groups donate electrons to adjacent pi systems via hyperconjugation (C-H sigma bond electrons overlap with the adjacent p-orbital) and inductive effect. The more C-H bonds adjacent to the pi system, the greater the electron donation and stabilisation.

All 40 questions in Organic Chemistry — Some Basic Principles and Techniques are solved step by step in the Solutions PDF above.

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Chapter 8 Solutions — Organic Chemistry — Some Basic Principles and Techniques

Overview

What these solutions cover

Exercises

Worked examples

Indicate the sigma and pi bonds in the following molecules: C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃

Which of the two: O₂N-CH₂-CH₂-O⁻ or CH₃-CH₂-O⁻ is expected to be more stable and why?

Explain why alkyl groups act as electron donors when attached to a pi system.

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