Question8.1(i) What are the hybridization states of each carbon atom in the following compounds?
Answer:
Here
Question 8.1(ii) What are the hybridisation states of each carbon atom in the following compounds?
Answer:
Here
Question 8.1(iii) What are the hybridisation states of each carbon atom in the following compounds?
Answer:
Here is the structure
Here
Question 8.1(iv) What are the hybridisation states of each carbon atom in the following compounds ?
Answer:
Here, in the above compound the
Question 8.1(v) What are the hybridisation states of each carbon atom in the following compounds ?
Answer:
In benzene, all six carbon atoms are
Question 8.2(i) Indicate the
Answer:
There are six C-C sigma bonds, six C-H sigma bonds and three
Question 8.2(ii) Indicate the
Answer
In
Question 8.2(iii) Indicate the
Answer:
In
There are two sigma C-H bonds and two C-Cl sigma bonds.
Question 8.2 (iv) Indicate the
Answer:
In this compound, there are two C-C sigma bonds, four C-H sigma bonds and two
Question 8.2(v) Indicate the
Answer:
There are three C-H sigma bonds, one C-N sigma bond, one N-O sigma bond and one N-O
Question 8.2(vi) Indicate the
Answer:
Two C-N sigma bonds, four C-H sigma bonds, one N-H sigma bond and one
Question 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
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2, 3-Dimethyl butanal
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Heptan-4-one
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Question 8.4(a) Give the IUPAC names of the following compounds:
Answer:
The IUPAC name of the above compound is 1-phenylpropane
Question 8.4(b) Give the IUPAC names of the following compounds:
Answer:
The IUPAC name of the above compound is 3-Methylpentanenitrite.
Question 8.4(c) Give the IUPAC names of the below compound :
Answer:
The IUPAC name of the above structure is 2, 5-dimethyl heptane.
Question 8.4(d) Give the IUPAC names of the following compounds:
Answer:
The IUPAC name of the above compound is 3-Bromo- 3-chloroheptane
Question 8.4(e) Give the IUPAC names of the following compounds:
Answer:
The IUPAC name of the compound is 3-chloropropanal.
Question 8.4(f) Give the IUPAC names of the following compounds :
Answer:
The IUPAC name of the compound is 1, 1-dichloro-2-ethanol
Question 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 numbers should be a 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 the 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 functional group, so the suffix should be
Question 8.6(a) Draw formulas for the first five members of each homologous series beginning with the following compound.
Answer:
The first five members of each homologous series beginning with the following compound are shown as
- ethanoic acid
- propanoic acid
- butanoic acid
- pentanoic acid
Question 8.6 (b) Draw formulas for the first five members of each homologous series beginning with the following compound.
Answer:
The first five members of each homologous series beginning with the following compound are shown as-
(propanone)
- Butanone
- Pentan-2-one
- hexan-2-one
- Hept-2-one
Question 8.6 (c) Draw formulas for the first five members of each homologous series beginning with the following compound.
Answer:
The first five members of each homologous series beginning with the following compound are shown as-
- Propene
- 1-Butene
- 1-Pentene
- 1-Hexene
Question 8.7(a) Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
2,2,4-Trimethylpentane
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 -
Question 8.7(c) Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
Hexanediol
Answer:
The condensed formula of hexanediol
The bond line structure is-
Question 8.8 Identify the functional groups in the following compounds
Answer:
The functional groups in the above structure are-
- Aldehyde (
) - HYdroxyl (
) - Methoxy (
) double bond
(b)
The following functional groups are present-
- Amino, primary amine (
) - Ester (
) - Tertiary amine (
) R = ethyl group
Here,
- Nitro group (
) double bond
Question 8.9 Which of the two:
Answer:
Since the Nitro group is an electron-withdrawing group. So, it shows
Hence
Question 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 partial overlap of the
Question 8.11(a) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
Answer:
The resonating structure of the phenol--
The lone pair of electrons starts shifting to an oxygen-carbon bond and forms a double bond character and
Question 8.11 (b) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
Answer:
The resonating structure of the nitrobenzene-
Here, the electrons of N-O bonds shift to the Oxygen atom (more electronegative), and then the
Question 8.11(c) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
Answer:
The resonating structure of the But-2-ene-1-al-
Here, the
Question 8.11(d) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
Answer:
The resonating structure of the benzaldehyde--
Here the
Question 8.11(e) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
Answer:
The resonating structure of Benzyl carbocation-
The
Question 8.11(f) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
Answer:
The resonating structure of But-2-ene-1-yl carbocation-
The
Question 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 the nucleus, seeking a reagent called a nucleophile.
For examples-
Question 8.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
Answer:
Electrophiles are electron-deficient species and can seek an electron pair. On the other hand, the nucleophile is electron electron-rich reagent and can electron donor(nucleophile seeking).
Therefore,
(a)
(b)
(c)
Question 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 the above equation, the bromine group is replaced by the
Question 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 combine to form a single product. Also, we can say that the Hydrogen and chlorine of
Question 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.
Question 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.
Question 8.15(a) What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

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
Question 8.15(b) What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer:
The given compounds are a pair of geometrical isomers since they have the same molecular formula, sequence of covalent bonds and same constitution but differ in the relative positioning of the atoms in space. These compounds differ in the positioning of the Deuterium and Hydrogen.
Question 8.15(c) What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer:
(c) The given compounds are a pair of contributing structures or canonical structures. They do not represent any real molecule and are purely hypothetical. They are also called resonance isomers.
(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 a 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 the shared pair of an electrons is distributed to only the bromine ion. Here, the reaction intermediate is a 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 a 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
Question 8.18 (a) Give a brief description of the principles of the following techniques taking an example in each case.
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 a solvent, but they are sparingly soluble at room temperature, but soluble at higher temperatures. 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 be filtered out and dried.
Question 8.18 (b) Give a brief description of the principles of the following techniques taking an example in each case.
Distillation
Answer:
Distillation- This method is used for the purification of liquids from non-volatile impurities. It is based on the fact that fluids with different boiling points vaporise at different temperatures. The impure liquid is boiled in a flask, and initially, the vapours of lower lower-boiling-point 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.
Question 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.
Question 8.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
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-
- The powdered mixture is taken in a flask, and the solvent is added to it and stirred simultaneously. We have to prepare a saturated solution and then heat it.
- After heating, this hot saturated solution can be filtered with filter paper in a china dish.
- Now the solution is allowed to cool. The less soluble compounds crystallise first, andthe more soluble compounds remain in the solution. After removing these crystals, the latter is concentrated once again. The hot solution is allowed to cool, and then the crystals of the more soluble compound are obtained.
- The last step is the isolation and drying of crystals from the mother liquor.
Question 8.20 What is the difference between distillation, distillation under reduced pressure and steam distillation ?
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 a sufficient Boiling point difference.
|
The liquid will boil at a lower temperature than its boiling point; 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.
|
|
Question 8.21. Discuss the chemistry of Lassaigne’s test.
Answer: Lassaigne’s test.-
This test is used to detect the presence of nitrogen, 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 the sodium extract method or Lassaigne's extract.
Question 8.22(i) Differentiate between the principle of estimation of nitrogen in an organic compound by
Dumas method
Answer:
In the Dumas method, a 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 an aqueous solution of
Question 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 sulfuric 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
Question 8.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
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, an Organic compound is heated with either fuming nitric acid or sodium peroxide in a hard glass tube called a carius tube. Sulphur present in the compound is oxidised to form sulphuric acid. It is precipitated as barium sulphate by adding barium chloride solution to water. Then the ppt is filtered, washed and weighed.
Let the mass of the 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 =
Question 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 based on 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 components travel with the mobile phase to different levels of heights. The obtained paper is called a chromatogram.
Question 8.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:
Nitric acid was added to the sodium extract before adding silver nitrate for testing halogens to decompose the
Question 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 fusion of an organic compound by sodium metal. This is known as Lassaigne's test.
The following chemical reactions are-
Here X = halogen atoms
Question 8.27. Name a suitable technique for the separation of the components from a mixture of calcium sulphate and camphor.
Answer:
To separate a mixture of calcium sulphate and camphor, we use the sublimation method. In this method, the sublimable compound is converted to a vapour state from the solid state without achieving its liquid state. Here, camphor is a sublimable compound, and calcium sulphate is not.
Question 8.28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ?
Answer:
In steam distillation, an Organic liquid starts boiling when the total sum of the vapour pressure of the 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.
Question 8.29. Will
Answer:
No,
Answer:
Carbon dioxide is acidic in nature and
Thus, due to the increase in KOH, the mass of the U-tube 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 Organic Compound 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 Organic Compound, 0.506 g of
Similarly,
100 g of Organic Compound contains 4.8 g of H
So, 0.2 g of Organic Compound contains
We know the mol. wt. of water = 18 g
Therefore 0.0096 g of H will present in
Hence, 0.2 grams of Organic Compound will produce 0.0864 g of water on complete combustion of
Answer:
Given that,
The total mass of the organic compound = 0.50 g
60 mL of 0.5 M solution of
60 mL of 0.5 M
= 30 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 =
= 37.59%
Hence, the percentage of chlorine present in the given organic compound is 37.59%.
Answer:
Given data,
The total mass of the Organic Compound = 0.468 g
Mass of barium sulphate formed = 0.668 g
We know that,
1 mol of BaSO4 = 233 g of
= 32 g of sulphur
Thus, 0.668 g of
= 0.0917 g of sulphur
Therefore, the percentage(%) of sulphur
= 19.59 %
Hence, the percentage of sulphur in the given compound is 19.59 %.
Question 8.36. In the organic compound
Answer:
In the organic compounds-
In
Hence, the correct answer is option (3)
Question 8.37. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
Answer:
The Prussian blue colour is due to the formation of
Hence, the correct answer is option (b).
Question 8.38. Which of the following carbocation is most stable?
Answer:
We know the stability of carbocation order is-
Hence, the correct answer is option (b).
Answer:
Chromatography is the latest technique of separation and purification of organic compounds.
Hence, the correct answer is option (d).
Question 8.40. The reaction:
(a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition
Answer:
Here,
Hence, the correct answer is option (b).
NCERT Class 11 Chemistry Chapter 8: Higher Order Thinking Skills (HOTS) Questions
Higher Order Thinking Skills Questions from NCERT Class 11 Chemistry Chapter 8 help students understand concepts in a better way. These questions improve thinking and problem-solving skills. Given below are some important HOTS questions for practice.
Question 1. Identify correct statement/s :
(A)
(B) -CN and -OH are meta directing group
(C) -CN and
(D) Activating groups act as ortho - and para directing groups
(E) Halides are activating groups
Choose the correct answer from the options given below:
Options
(1). (A), (C) and (D) only
(2). (A), (B) and (E) only
(3). (A) only
(4). (A) and (C) only
Answer:
(B) -CN is meta directing, but -OH is ortho/para directing.
(E) Halides are deactivating groups.
Hence, (B) and (E) are incorrect statements
Hence, the correct answer is option (1)
Question 2. Mixture of 1 g each of chlorobenzene, aniline and benzoic acid is dissolved in 50 mL ethyl acetate and placed in a separating funnel, 5 M NaOH ( 30 mL ) was added in the same funnel. The funnel was shaken vigorously and then kept aside. The ethyl acetate layer in the funnel contains :
(1) benzoic acid
(2) benzoic acid and aniline
(3) benzoic acid and chlorobenzene
(4) chlorobenzene and aniline
Answer:
NaOH and benzoic acid will react to form salt. Aniline and chlorobenzene, being organic compounds, will remain in the ethyl acetate layer as they will not react with NaOH.
Hence, the correct answer is option (4).
Question 3. The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula
Answer:
The formula for Degree of Unsaturation (DU) or Index is:
Where,
-
-
-
-
So, Degree of unsaturation
This implies it has one ring with three double bonds. So, the possible isomers are
Hence, the answer is 8.
Approaches to Solve Questions of Class 11 Chapter 8 Chemistry
Sometimes, problems related to organic chemistry seem difficult, but once we understand the basic rules and strategy, it becomes very easy to solve all the questions related to organic chemistry. We can follow the steps given below to the NCERT Class 11 Chemistry Chapter 8 Exercise Solutions.
1) Firstly, understand the basic concepts
Before solving questions, a basic understanding of the topics is a must, like IUPAC Nomenclature, electronic effect and the types of organic reactions, classification of organic compounds, stability and reactivity of intermediates like carbocations, carbanions and free radicals.
2) Read the statement of the question given carefully.
While solving questions, it is very important to read the question carefully and identify exactly what is being asked. Students also pay attention to conditions and reagents.
3) Then identify the function groups, reagent behaviour, and reaction intermediaries.
While solving questions it is important to identify which functional group is present and understand whether the reagent is acting as a nucleophile or an electrophile as these questions related to these topics are frequently asked in exams.
4) Use given data and formulas accurately
While solving numerical problems related to molecular formulas or empirical formulas, make sure that you are using the given data and formulas accurately. Students can follow Chapter 8 Some basic principles and techniques notes.
5) Practice questions
Regularly revise topics and practice questions from NCERT books also practice previous year questions.
Topics and Subtopics Covered in the NCERT Textbook Class 11 Chemistry Chapter 8
All the topics and subtopics covered in the NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry Some Basic Principles and Techniques are listed below:
8.1 General Introduction
8.2 Tetravalence of Carbon: Shapes of Organic Compounds
8.2.1 The Shapes of Carbon Compounds
8.2.2 Some Characteristic Features of
8.3 Structural Representations of Organic Compounds
8.3.1 Complete, Condensed and Bond-line Structural Formulas
8.3.2 Three-Dimensional Representation of Organic Molecules
8.4 Classification of Organic Compounds
8.4.1 Functional Group
8.4.2 Homologous Series
8.5 Nomenclature of Organic Compounds
8.5.1 The IUPAC System of Nomenclature
8.5.2 IUPAC Nomenclature of Alkanes
8.5.3 Nomenclature of Organic Compounds having Functional Group(s)
8.5.4 Nomenclature of Substituted Benzene Compounds
8.6 Isomerism
8.6.1 Structural Isomerism
8.6.2 Stereoisomerism
8.7 Fundamental Concepts in Organic Reaction Mechanism
8.7.1 Fission of a Covalent Bond
8.7.2 Substrate and Reagent
8.7.3 Electron Movement in Organic Reactions
8.7.4 Electron Displacement Effects in Covalent Bonds
8.7.5 Inductive Effect
8.7.6 Resonance Structure
8.7.7 Resonance Effect
8.7.8 Electromeric Effect (E effect)
8.7.9 Hyperconjugation
8.7.10 Types of Organic Reactions and Mechanisms
8.8 Methods of Purification Of Organic Compounds
8.8.1 Sublimation
8.8.2 Crystallisation
8.8.3 Distillation
8.8.4 Differential Extraction
8.8.5 Chromatography
8.9 Qualitative Analysis Of Organic Compounds
8.9.1 Detection of Carbon and Hydrogen
8.9.2 Detection of Other Elements
8.10 Quantitative Analysis
8.10.1 Carbon and Hydrogen
8.10.2 Nitrogen
8.10.3 Halogens
8.10.4 Sulphur
8.10.5 Phosphorus
What Extra Should Students Study Beyond the NCERT for JEE/NEET?
People also search for what extra they have to do beyond the NCERT for JEE or NEET. Beyond the NCERT Class 11 Chemistry Chapter 8 Exercise Solutions, students need to focus on important concepts and theories according to the table given below. The table shows a comparison between topics that are included in the JEE syllabus and topics that are given in the NCERT textbook:
NCERT Solutions for Class 11 Chemistry Chapter-Wise
Besides NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry Some Basic Principles and Techniques, chapter-wise solutions are given below:
NCERT Solutions for Class 11- Subject-Wise
The hyperlinks of the NCERT solution for class 11 are given below:
NCERT Books and NCERT Syllabus:
Students can refer to the links given below for the NCERT books and Syllabus: