NEET/JEE Coaching Scholarship
ApplyGet up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
Some Basic Principles and Techniques is one of the most important chapters in Class 11 Chemistry as it forms the foundation of organic Chemistry which is going to help students throughout their career in Chemistry. This chapter helps students to understand topics like the shapes of compounds, their representation, classification of compounds, nomenclature of compounds and purification of organic compounds. Some Basic Principles And Techniques is an important chapter for final exams as well as competitive exams like JEE, NEET, and BITSAT. Chemistry chapter 8 is a blend of theoretical and practical concepts. The concepts provided in this chapter help us to understand some phenomena from our day-to-day lives, like the purification of organic compounds by distillation.
New: Get up to 90% Scholarship on NEET/JEE Coaching from top Coaching Institutes
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | ALLEN
Chapter 8 solutions of Class 11 Chemistry are prepared by subject experts in a very comprehensive and systematic way, which helps students develop a clear understanding of the topics used to solve particular problems. Solutions provided help students in their CBSE class 11 as well as competitive exams like JEE, NEET, and BITSAT. By referring to the NCERT solutions for class 11, students can understand all the important concepts well enough before their examination.
To perform well in upcoming exams like CBSE exams, state board exams, and competitive exams, students need to solve every question in NCERT class 11 chapter 8. The solutions provided below cover key concepts in a very concise and systematic way which helps students to gain confidence in their preparation.
Also Read:
Organic Chemistry Some Basic Principles and Techniques Class 11 Chemistry NCERT Exemplar Solutions |
Organic Chemistry Some Basic Principles and Techniques Class 11 Chemistry Chapter Notes |
Q8.1(i) What are the hybridization states of each carbon atom in the following compounds?
Answer:
c1 c2
Here
Q 8.1(ii) What are the hybridisation states of each carbon atom in the following compounds?
Answer:
c1 c2 c3
Here
Q 8.1(iii) What are the hybridisation states of each carbon atom in the following compounds?
Answer:
Here is the structure
Here
Q 8.1(iv) What are the hybridisation states of each carbon atom in the following compounds ?
Answer:
In benzene, all the six carbon atoms are
Q 8.1(v) What are the hybridisation states of each carbon atom in the following compounds ?
Answer:
1 2 3
Here, in the above compound the
Q 8.2(i) Indicate the
Answer:
There are six C-C sigma bonds, six C-H sigma bonds and three
Q 8.2(ii) Indicate the
Answer
In
Q 8.2(iii) Indicate the
Answer:
In
There are two sigma C-H bonds and two C-Cl sigma bonds.
Q 8.2 (iv) Indicate the
Answer:
In this compound, there are two C-C sigma bonds, four C-H sigma bonds and two
Q 8.2(v) Indicate the
There are three C-H sigma bonds, one C-N sigma bond, one N-O sigma bond and one N-O
Q 8.2(vi) Indicate the
Answer:
Two C-N sigma bonds, four C-H sigma bonds, one N-H sigma bond and one
Q 8.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4- one.
Answer:
Bond line formula of the following compounds-
Isopropyl alcohol | ![]() |
2, 3-Dimethyl butanal | ![]() |
Heptan-4-one | ![]() |
Q 8.4(a) Give the IUPAC names of the following compounds:
Answer:
The IUPAC name of the above compound is 1-phenylpropane
Q 8.4(c) Give the IUPAC names of the below compound :
Answer:
The IUPAC name of the above structure is 2, 5-dimethyl heptane.
Q 8.4(f) Give the IUPAC names of the following compounds :
Answer:
The IUPAC name of the compound is 1, 1-dichloro-2-ethanol
Q 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 .
Answer:
(a) The prefix
(b) In IUPAC nomenclature the sum of locant number should be minimum. Here the sum of 2,4,7 (=13)is less than the sum of 2, 5, 7 (=14). Thus the correct IUPAC name is 2,4,7- Trimethyloctane
(c) In IUPAC nomenclature, if the substituent group acquires the equivalent position of the parent chain then the lower number is given to one that comes first in alphabetical order. Hence 2-chloro-4-methyl pentane is the correct IUPAC name of the compound.
(d) If the two functional groups are present in the parent chain then the suffix of the IUPAC name depends on the principal functional group. Here alcohol is the principal function group so the suffix should be
Answer:
the first five members of each homologous series beginning with the following compound are shown as-
Answer: the first five members of each homologous series beginning with the following compound are shown as-
(propanone)
Answer: The first five members of each homologous series beginning with the following compound are shown as-
Answer: 2,2,4-Trimethylpentane
Condensed formula-
Bond line formula-
Answer:
The condensed formula of 2-Hydroxy-1,2,3-propanetricarboxylic acid
The bond line structure -
Answer:
the condensed formula of hexanediol
The bond line structure is-
Q 8.8 Identify the functional groups in the following compounds
Answer:
The functional groups in the above structure are-
(b)
The following functional groups are present-
Here,
Q 8.9 Which of the two:
Answer: Since the Nitro group is an electron-withdrawing group. So, it shows
Hence
Q 8.10 Explain why alkyl groups act as electron donors when attached to a
Answer: When an alkyl group is attached to the
for example in propene,
In the figure, you can see that the sigma electrons of C-H bonds of the alkyl group are delocalised because of the partially overlapping of the
Answer: The resonating structure of the phenol--
The lone pair of electrons start shifting to an oxygen-carbon bond and form a double bond character and
Answer: The resonating structure of the nitrobenzene-
Here the electrons of N-O bonds shift to the Oxygen atom (more electronegative) and then the
Answer: The resonating structure of the But-2-ene-1-al-
Here the
Answer: The resonating structure of the benzaldehyde--
Here the
Answer: The resonating structure of Benzyl carbocation-
The
Answer: The resonating structure of But-2-ene-1-yl carbocation-
The
Q 8.12 What are electrophiles and nucleophiles? Explain with examples.
Answer:
Electrophile- It is an electron-deficient species, which seeks an electron pair. This reagent takes away an electron pair. It is denoted as
A nucleophile is a reagent that brings an electron pair. In other words, it is nucleus- seeking reagent called nucleophile.
For examples-
Q 8.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
Answer:
Electrophiles are electron deficient species and can be seeking for an electron pair. On the other hand, the nucleophile is electron rich reagents and they can electron donor(nucleus seeking).
Therefore,
(a)
(b)
(c)
Q 8.14. (a) Classify the following reactions in one of the reaction types studied in this unit.
Answer:
It is a substitution reaction as in this above equation the bromine group is replaced by the
Q 8.14.(b) Classify the following reactions in one of the reaction types studied in this unit.
Answer:
The given reaction is an example of an addition reaction because in this reaction the two reactant molecules combined to form a single product. Also, we can say that the Hydrogen and chlorine of
Q 8.14(c) Classify the following reactions in one of the reaction types studied in this unit.
Answer:
In this reaction, the hydrogen and bromine are removed from the original compound and formed ethene. So, this is an elimination reaction.
Q 8.14(d) Classify the following reactions in one of the reaction type studied in this unit.
Answer:
In the above reaction, the substitution takes place between the two reactants, followed by the rearrangement of atoms and the groups of atoms.
Answer:
These pairs are structural isomers.
Compounds having the same molecular formula but different structures are called structural isomers. The above compounds have the same molecular formula but the structures are different due to the difference in the position of the carbonyl group.
In structure one -CO group is present at
(a)
Answer:
Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as
It is an example of hemolysis because the two electrons are equally divided into the products. (see in fig)
Answer:
Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as;
It is an example of heterolytic cleavage as the bond breaks in such a manner that the electron pair will remain with the carbon of propanone. The reaction intermediate is carbanion.
Answer: Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as;
It is heterolysis as shared pair of an electron is distributed to only bromine ion. Here the reaction intermediate is carbocation.
Answer: Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as
The reaction intermediate is carbocation. It is a heterolytic cleavage as the bonds break in such a manner that the shared; electron pair will remain with the one species.
Answer: Inductive effect- The permanent displacement of sigma (
Electromeric effect- The complete transfer of the shared pair of
(a) The order of acidity can be explained by the negative inductive effect (
(b) This can be explained by the
Crystallisation
Answer: Crystallisation- It is one of the most commonly used techniques for the purification of solid organic compounds. Its principle is based on the difference in the solubilities of the compounds and the solvent's impurities. The impure compounds are dissolved in solvent but they are sparingly soluble at room temperature but soluble at the higher temperature. On cooling the compound, the pure compounds crystallise and are removed by filtration.
For example - pure aspirin is obtained by recrystallising crude aspirin. Around 2 - 4 g of crude aspirin is dissolved in 20 mL of ethyl alcohol and the solution is heated for complete dissolution. Then after crystal formation, they can filter out and dried.
Distillation
Answer: Distillation- This method is used for the purification of liquids from non-volatile impurities. It is based on the fact that fluids of having different boiling points vaporise at different temperatures. The impure liquid is boiled in a flask, and initially, the vapours of lower boiling points component are formed. The vapours are condensed by using a condenser, and the liquid is collected in a receiver.
For example- Organic liquids such as benzene, toluene, xylene etc can be purified by this method.
Q 8.18 (c) Give a brief description of the principles of the following techniques taking an example in each case.
Chromatography
Answer: Chromatography- It is one of the extensive methods for the separation and purification of organic compounds. It is based on the difference in the movements of components of mixtures through the stationary phase under the influence of the mobile phase. The stationary phase can be solid or liquid. While the mobile phase is only liquid or gas.
For example- This technique can separate a mixture of blue and red ink.
Answer: Fractional distillation can be used to separate the two compounds with different solubilities in a solvent S. The following steps are carried out in this process-
Answer: The differences between distillation, distillation under reduced pressure and steam distillation are-
Distillation | Distillation under reduced pressure | Steam Distillation |
1. Used for the purification of compounds that are non-volatile impurities or liquids, which don't decompose on boiling. | 1. Used to purify liquids, which tend to decompose on boiling. Under the condition of reduced pressure. | 1. Used to purify the organic compounds, which are steam volatile and immiscible with water. |
Simply, to separate the volatile liquids from non-volatile impurities. Or a mixture of liquids having sufficient Boiling point difference. | The liquid will boil at a lower temperature than its boiling points, therefore, it does not decompose. | The mixture of water and aniline can be separated by this method. |
The mixture of petrol and kerosene is separated by this method. | Glycerol is purified by this method. |
Q 8.21. Discuss the chemistry of Lassaigne’s test.
Answer: Lassaigne’s test.-
This test is used to detect the presence of nitrogen, sulphur, halogens and sulphur in organic compounds. These elements are present in the covalent form in an organic compound. So, they are converted into the ionic form by fusing the compound with the sodium metal.
Cyanide, sulphide or halide of sodium are extracted from the fused mass by boiling it with distilled water. This Is known as sodium extract method or Lassaigne's extract.
Q 8.22(i) Differentiate between the principle of estimation of nitrogen in an organic compound by
Dumas method
Answer: In the Dumas method, Nitrogen-containing organic compound is heated with copper oxide in a carbon dioxide atmosphere, yielding free nitrogen in addition to
Traces of nitrogen oxides can also be formed in the reaction, which can be reduced to nitrogen by passing the gaseous mixture over the heated copper gauge. The produced mixture of gases is collected over by an aqueous solution of
Q 8.22(ii) Differentiate between the principle of estimation of nitrogen in an organic compound by
Kjeldahl’s method.
Answer:
In Kjeldahl’s method, the nitrogen-containing organic compound is heated with concentrated sulphuric acid. Nitrogen is converted into ammonium sulphate. It is then passed into the known volume of sulphuric acid. The amount of ammonia produced can be estimated by the amount of
It is done by estimating the amount of unreacted
Answer:
Estimation of halogen is done by the Carius method. In this method, a known quantity of organic compound is heated with fuming
Let the mass of the organic compound be
Mass of
1 mol of
Therefore, Mass of halogen inm1 g of AgX = (Atomic mass of
Thus, % of halogen will be = (Atomic mass of
Estimation of sulphur- In this method, Organic compound is heated either fuming nitric acid or sodium peroxide in a hard glass tube called carius tube. Sulphur present in the compound is oxidised to form sulphuric acid. It is precipitated by as barium sulphate by adding barium chloride solution in water. Then ppt is filtered, washed and weighed.
Let the mass of organic compound taken =
and the mass of barium sulphate formed =
1 mol of
Percentage (%)of sulphur =
Estimation of phosphorus- In this process, a known mass of an organic compound is heated with fuming nitric acid and the phosphorus gets oxidised to phosphoric acid. By adding ammonia and ammonium molybdate, phosphorus can be precipitated as ammonium phosphomolybdate,
Let the mass of organic compound taken =
Molar mass of
Percentage(%) of phosphorus =
If phosphorus is estimated as
Percentage(%) of phosphorus =
Q 8.24 Explain the principle of paper chromatography.
Answer: In paper chromatography, chromatography paper is used. It contains water trapped in it, which acts as the stationary phase. The solution of the mixture is spotted on this base of chromatography paper. The strip of paper is suspended in a suitable solvent, which is the mobile phase. Due to capillary action, the solvent rises up in the paper and it flows over the spot. The spots of the different component travel with the mobile phase to a different level of heights. The obtained paper is called chromatogram.
Q 8.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer: Nitric acid was added to sodium extract before adding silver nitrate for testing halogens to decompose the
Q 8.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
In Organic compounds, nitrogen, sulphur and halogens are covalently bonded. To detect them, they have to convert into ionic form. This is done by the fusing of an organic compound by the sodium metal. This is known as Lassaigne's test.
The following chemical reactions are-
Here X = halogen atoms
Answer:
To separate a mixture of calcium sulphate and camphor, we use the sublimation method. In this method, the sublimable compound is converted to vapour state from the solid state without achieving its liquid states. Here camphor is sublimable compound and calcium sulphate is not.
Answer:
In steam distillation, Organic liquid starts boiling when the total sum of the vapour pressure of an organic liquid (P') and the of water (P'') becomes equal to the atmospheric pressure (P),
it means P = P' + P''. since P'< P'', the organic liquid will vaporise at a lower temperature than its boiling point.
Q 8.29. Will
Answer:
No,
Answer:
Carbon dioxide is acidic in nature and
Thus, due to the increase in KOH the mass of U-tube is also increases. This increased mass of the tube gives the mass of carbon dioxide produced. From its mass, the % of carbon in the Organic compound can be estimated.
Answer:
It is necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test because in case of sulphuric acid the complete ppt formation of lead sulphate does not take place. While the addition of acetic acid will ensure a complete ppt. formation of sulphur in the form of lead sulphate due to the common ion effect.
Answer:
Given,
Percentage of carbon = 69%
% of Hydrogen = 4.8%, and
% of Oxygen = 26.2 %
So, 0.2 g of OC have
we know that,
molecular weight of
12 g of C is present in 44 g of
Therefore, 0.138 g of C present in
Hence, from 0.2 g of OC 0.506 g of
Similarly,
100 g of OC contains 4.8 g of H
So, 0.2 g of OC contains
We know the mol. wt. of water = 18 g
Therefore 0.0096 g of H will present in
Hence 0.2 gram of OC will produce 0.0864 g of water on complete combustion of
Answer:
Given that,the
total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of
60 mL of 0.5 M
Therefore,
Acid consumed in the absorption of evolved ammonia is (50-30) mL = 20 mL
Again, 20 mL of 0.5 M
Also, since 1000 mL of 1 M
So, 40 mL of 0.5 M
Therefore, percentage(%) of nitrogen in 0.50 g of organic compound =
Answer:
Given that,
Mass of organic compound = 0.3780 g.
Mass of AgCl formed = 0.5740 g
It is known that,
1 mol of
Thus, the mass of chlorine in 0.5740 g of
= 0.1421 g
∴ Percentage(%) of chlorine =
Hence, the percentage of chlorine present in the given organic compound is 37.59%.
Answer:
Given data,
The total mass of the OC = 0.468 g
Mass of barium sulphate formed = 0.668 g
We know that,
1 mol of BaSO 4 = 233 g of
Thus, 0.668 g of
Therefore, the percentage(%) of sulphur
= 19.59 %
Hence, the percentage of sulphur in the given compound is 19.59 %.
Q 8.36. In the organic compound
&nbnbsp;
Answer:
In the organic compounds-
1 2 3 4 5
In
So, the correct option is (c)
Answer:
The Prussian blue colour is due to the formation of
So, the correct option is (b)
Q 8.38. Which of the following carbocation is most stable?
Answer:
We know the stability of carbocation order is-
So, the correct option is (b)
Q 8.39. The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography
Answer: Chromatography is the latest technique of separation and purification of organic compounds.
So, the correct option is (d)
Q 8.40. The reaction:
(a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition
Answer:
Here,
So, the correct option is (b)
Chapter Number | Chapter name |
1 | |
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 | Organic chemistry- some basic principles and techniques |
9 |
Chapter 8 Some Basic Principles and Techniques is an important chapter for a chemistry student from the perspective of getting hold of basic concepts of organic chemistry which will eventually help in subsequent chapters. Since this is a theoretical chapter and many concepts and facts will be learned from exercise questions directly, hence It is must to go through NCERT solutions for Class 11 Chemistry Chapter 8 . In this chapter, there are a total of 40 questions in the exercise.
This chapter along with NCERT solutions for Class 11 Chemistry Chapter 8 Organic Chemistry Some Basic Principles and Techniques discusses some basic concepts in the reactivity and structure of organic compounds which are formed due to the covalent bonding, IUPAC nomenclature and classification of organic compounds, electromeric effect, inductive effect, hyperconjugation, and resonance. All related information about the heterolytic fission and homolytic fission of a covalent bond- carbocations, carbanions, free radicals, nucleophiles, and electrophiles is discussed in the later part of the NCERT syllabus chapter.
Organic compounds are essential for sustaining life's on the earth and include complex molecules like protein and DNA that constitutes important compounds of our blood, skin, and muscles. Some important areas of application of organic compounds are medicines, dyes, fuels, polymers, and clothing. These compounds can be classified on the basis of their functional group or the structure.
8.1 General Introduction
8.2 Tetravalence of Carbon: Shapes of Organic Compounds
8.3 Structural Representations of Organic Compounds
8.4 Classification of Organic Compounds
8.5 Nomenclature of Organic Compounds
8.6 Isomerism
8.7 Fundamental Concepts in Organic Reaction Mechanism
8.8 Methods of Purification of Organic Compounds
8.9 Qualitative Analysis of Organic Compounds
8.10 Quantitative Analysis
A functional group is defined as an atom or group of atoms bonded together in a specific manner, which gives the chemical and physical properties of the organic compounds and they are the centres of the chemical reactivity.
1. Organic compounds are formed due to covalent bonding.
2. According to orbital hybridization concept carbon can have
3. The 3D representation of organic compounds on paper can be drawn by dash and wedge formula.
4. Compounds having the same molecular formula but differ in their physical and chemical properties are known as isomers and the phenomenon is called isomerism.
5. Addition, substitution, elimination and rearrangement reactions are the types of organic reactions.
- classification and IUPAC nomenclature of organic compounds
- Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyperconjugation
- Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles
- Types of organic reactions.
For complete solutions of NCERT, students can refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry
In Homolytic, the bond between two atoms breaks evenly. While in heterolytic fission bonds between atoms break unevenly.
Nucleophiles are electron-rich species and they tend to donate electrons to electron-deficient molecules.
The inductive effect is important because it helps to explain how electrons move in a molecule. Due to this movement, some molecules have a positive charge while some have a negative charge
Admit Card Date:03 February,2025 - 04 April,2025
Admit Card Date:17 February,2025 - 29 March,2025
Admit Card Date:24 February,2025 - 25 March,2025
Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide