NCERT Solutions Class 11 Chemistry Chapter 12
NCERT Solution for Class 11 Chemistry Chapter 12
Class 11 Chemistry Chapter 12 NCERT Solutions are prepared by subject experts who have many years of experience in their respective fields. The solutions by Extramarks are the best online learning aids for students who want to score high marks in board exams, pursue higher studies or crack competitive exams.
Based on the updated syllabus, the solutions would help you score high marks in the chemistry exam and gain more knowledge. All the solutions are written in a very simple language so that you can understand the basics of chemistry with ease.
NCERT Solution for Class 11 Chemistry Chapter 12 – Organic Chemistry – Some Principles and Techniques
Chapter 12 is about all the concepts related to the composition of the variety of non-living things that exist in our environment. Students will understand the concepts deeply and chemical compositions and reactions that take place from the combination of different elements. Class 11 Chemistry NCERT Solutions cover chapters with all important questions and answers explained in a detailed way.
NCERT Solution for Class 11 Chemistry Chapter 12 Free Download
NCERT Solution for Class 11 Chemistry Chapter 12 Free Download
Chapter 12 – Organic Chemistry Some Basic Principles and Techniques
Here’s an overview of the concepts covered in Chapter 12 – Organic Chemistry Some Basic Principles and Techniques
12.1 General Introduction
In this section students will read about – what is meant by the catenation of carbon elements and that this property is the reason why carbon forms covalent bonds with other elements. Students will get to know why organic chemistry is a separate branch and the concepts it focuses on. They will read about the history of organic chemistry and how the electronic theory of covalent bonds led to the concept of organic chemistry.
12.2 Tetravalency of Carbon
This section talks about the shapes of organic compounds. Hybridisation involves the s and p orbital that result in three types of hybridisation:
- sp3 found in alkanes
- sp2, found in alkenes. It is tetrahedral in shape
- sp, found in alkynes. It is the planar structure
12.4 Classification of Organic Compounds
Since organic compounds are always increasing, it has become important to classify them based on their structures. Here’s the broad classification of organic compounds:
- Open chain compounds (Acyclic) – Straight or branched chain compounds such as Ethane (CH3CH3).
- Closed Chain Compounds (Cyclic or Ring) – Carbon atoms in these compounds are joined in the shape of a ring. Atoms other than carbon atoms can also be present in these compounds. Cyclohexane is an example of such a compound.
- Homocyclic (Carbocyclic) Compounds – These consist only of carbon atoms.
- Alicyclic compounds
- Aromatic compounds
- Benzenoid Compounds – Benzene, Aniline, etc.
- Non-Benzenoid Compounds – Tropolone
- Heterocyclic Compounds – They have atoms other than carbon atoms in them
12.5 Nomenclature of Organic Compounds
The International Union of Pure and Applied Chemistry (IUPAC) system of nomenclature is used for naming organic compounds. In IUPAC, names are related to their structures, so anyone can deduce the structures by reading the names. The IUPAC system follows the following method:
- Organic compounds are named after their parent hydrocarbon, which is the word root.
- The parent name is modified using prefixes and suffixes.
- The word root is the denominator of the number of carbon atoms present in the principal chain. For example – C1 is meth, C2 is eth, C3 is prop and C4 is but.
- The primary suffix is denoted by the type of bond in the carbon atoms. For instance, for a C-C bond, the primary suffix is ane and for a C=C bond, it is ene.
- The secondary suffix is the functional group that is considered as a substituent. For example, F is Fluoro, Cl is Chloro and NO is nitroso.
Key Features of NCERT Solution for Class 11 Chemistry Chapter 12
Chapter 12 covers techniques and principles of organic chemistry. Chemistry Class 11 NCERT Solutions Chapter 12 have answers to all the questions that students might need help in finding solutions to. Students get the updated solutions that are in line with the latest syllabus.
Q.1 What are hybridisation states of each carbon atom in the following compounds?
CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6
Q.2 Indicate the σ and p bonds in the following molecules:
C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3
Q.3 Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Q.4 Give the IUPAC names of the following compounds:
d) 3-Bromo 3-chloroheptane
Q.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 methyl pentane (d) But-3-yn-1-ol or But-4-ol-1-yne.
(a) 2, 2-Dimethylpentane is correct name because two methyl groups are located on the same carbon and its locant is repeated on the same carbon.
(b) 2, 4, 7-Trimethyloctane is the correct one; the locant set 2, 4, 7 is lower than 2, 5, 7.
(c) 2-Chloro-4-methylpentane is the correct one as chlorine atom is placed in the lower locant.
(d) But-3-yn-1-ol is the correct one as the principal functional group ‘ol’ is in lower locant in this nomenclature.
Q.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2
Q.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
Bond line formula:
Q.8 Identify the functional groups in the following compounds:
Q.9 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
O2NCH2CH2O– is more stable than CH3CH2O– because NO2 group has –I effect. It tends to disperse the –ve charge on the O– atom.
In CH3CH2O– , CH3CH2 group has +I effect, So, it tends to intensify the negative charge on the oxygen atom and hence it destabilises the structure.
Q.10 Explain why alkyl groups act as electron donors when attached to a p system.
Alkyl groups can act as electron donors when attached to a p- system due to the hyperconjugation.
Hyperconjugation can be easily understood by the following structures:
Q.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) C6H5CH2+ (f) CH3CH=CHCH2+.
a) Resonance structures of C6H5OH:
b) Resonance structures of nitrobenzene:
c) Resonance structures of But-2-en-1-al:
d) Resonance structures of Benzaldehyde:
e) Resonance structures of Benzyl carbocation:
f) Resonance structures of but-2-en-1-yl carbocation:
Q.12 What are electrophiles and nucleophiles? Explain with examples.
Electrophiles: Electrophiles are electron deficient species. They may be positively charged or electrically neutral species. E.g: H+, NO2+, NO+, R+ (carbocation) and alkyl free radicals(R.), carbenes(:CR2)etc.
Nucleophiles: Nucleophiles are electron rich species containing at least one lone pair of electrons. They may be negatively charged or electrically neutral species. E.g: H–(hydride ion), Cl–, Br–, OH–, OR–, CN– ,H2O , NH3,RNH2,etc.
Q.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
a) CH3COOH + OH– → CH3COO– + H2O
b) CH3COOH3 + CN– → (CH3)2C(CN)(OH)
c) C6H6 + CH3CO+ → C6H5COCH3
In reaction (a) hydroxide ion acts as nucleophile and in (b) cyanide ion acts as nucleophile. In reaction (c) the reagent is electrophile.
Q.14 Classify the following reactions in one of the reaction type studied in this unit.
a) CH3CH2Br + HS– → CH3CH2SH + Br–
b) (CH3)2C = CH2 + HCl → (CH3)2CCl — CH3
c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br–
c) (CH3)3C — CH2OH + HBr → (CH3)2CBrCH2CH3 + H2O
a) Nucleophilic substitution reaction
b) Electrophilic addition
c) Bimolecular elimination reaction
d) Nucleophilic substitution reaction with rearrangement
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
a) Structural isomers
b) Geometrical isomers.
c) They are resonance contributors because they differ in position of electrons but not atoms.
Q.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) CH3O — OCH3 → CH3O•OCH3
Q.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
Inductive effect: It is the displacement of electrons along the chain of carbon atoms caused due to the presence of groups or atoms of different electronegativity at one end of the chain. It is a permanent effect and involves sigma bond. Electronegativity difference causes the shifting of electron density towards more electronegative atom.
Electromeric effect: It is temporary effect and occurs only when a reagent is added to a system and involves transfer of shared pair of pi electrons of the multiple bond to one of the linked atoms.
(a) –I effect explains the order of acidity of the given acids.
As the number of halogen (here it is Cl atom) atoms decreases, the –I effect also decreases and the acid strength decreases accordingly. Due to the –I effect of Cl atom, the electron density over the oxygen atom decreases, it helps in the removal of the proton. Thus more the number of chlorine atoms higher will be the acidity.
(b) +I-effect explains the order of acidity of the given acids.
As the number of alkyl group (alkyl group has +I effect) increases, the +I effect also increases, thus the acid strength decreases. Due to the +I effect of methyl group, the charge over the oxygen atom gets intensified, thus the removal of proton becomes difficult. More the number of methyl groups, lesser will be the acidity.
Q.18 Give a brief description of the principles of the following techniques taking an example in each case. (a)Crystallisation (b) Distillation (c) Chromatography.
(a) Crystallisation: Crystallisation is a process of purification by which crystals of pure compound are obtained from impure sample of compound. It is a technique which is used to purify solid compounds. The basic principle used in this technique is the solubility difference of the compound and the impurity present in the suitable solvent. The impure compound is dissolved in a suitable solvent in which it is sparingly soluble at room temperature but soluble in appreciable amount at higher temperature. The solution is then heated and saturated solution of impure compound is prepared. The crystals of the pure compound separate out when the solution is cooled, which are removed/separated by filtration. The remaining solution contains impurity and small amount of the pure compound.
Example: Crystallisation of sugar: A sample of sugar having impurity of common salt. The impure sugar sample is dissolved in ethanol with shaking and heating. The sugar dissolves in the solvent and the salt remains insoluble. The hot solution is filtered out and then concentrated. Then the solution is allowed to cool, the sugar crysytals separate out.
(b) Distillation: This process is used to separate two liquids having difference in boiling point by heating followed by condensation of the vapours. In this process, liquid is converted into vapours by heating, followed by condensation of the vapours by cooling them. The liquids purified by this method, contain non-volatile impurity and are stable at their boiling points.
Example: By this technique, simple organic liquids such as ethanol, benzene, chloroform, acetone, and carbon tetrachloride can be separated from impurities. When the liquid mixture is heated, it gradually starts boiling when the vapour pressure becomes equal to the atmospheric pressure. The more volatile component distills over first, and the less volatile component distills over later. The impurities which are non-volatile are left over the distillation flask.
(c) Chromatography: It is the method of separation in which components of a mixture are separated by differential movement of individual components through a stationary phase under the influence of a mobile phase. In chromatography, the stationary phase can be either solid or tightly bound liquid on solid support, while the mobile phase can be either liquid or gas. Two important types of chromatography which are widely used are column chromatography and thin layer chromatography.
Q.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
The two compounds having different solubility in solvent S can be separated out by fractional crystallization. The saturated solution of the two compounds is heated to prepare saturated solution. Then it is allowed to cool down. The component which is less soluble, crystallizes out first while the more soluble one remains soluble in the solution. The crystals of the first component are separated out from the mother liquor. Then the mother liquor is again concentrated and the hot solution is again cooled down and the crystals of the second component are obtained.
Q.20 What is the difference between distillation, distillation under reduced pressure and steam distillation?
Distillation involves the conversion of the liquid into vapour and condensation of the vapour into liquid. On cooling, we get the pure liquid and the non-volatile impurities remain in the flask. The non-volatile impurities are separated out. This method is applicable for the liquids which are quite stable at their boiling points and contain non-volatile impurities.
Distillation under reduced pressure is used for the liquids which tend to decompose on boiling. In this process, the applied pressure acting on the system is less than the atmospheric pressure; the pressure is reduced by using vacuum pump. This kind of method used to purify the liquids which decompose at or below their boiling points. For example, glycerol is purified by this method.
Steam distillation is quite similar to that of distillation under reduced pressure but there is no reduction in the total pressure acting on the solution. This process is used to purify and separate the liquids which are steam volatile.
The vapour pressure of water is quite high than other liquid, so the mixture will boil at a temperature much lower than its normal boiling point, thus decomposition can be avoided. A mixture of water and aniline is separated using this method.
Q.21 Discuss the chemistry of Lassaigne’s test.
This test is used for the detection of nitrogen, halogens, phosphorus and sulphur in an organic compound. In this test covalent bonds present in organic compounds are converted into ionic bonds. It is done by fusing organic compound with sodium.
The chemistry involved in it is as follows
The C and N present in the organic compound combine to form sodium cyanide.
If S and N both are present in the organic sample, during the fusion they combine to form sodium thiocyanate.
Na + N + S + C → NaSCN
Organic compounds containing halogens react with sodium to give NaX.
Q.22 Differentiate between the principles of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.
i) Dumas method: A known mass of organic compound is heated with excess CuO in CO2 atmosphere, the N of the organic compound is converted into N2 gas. By this reaction, the volume of obtained N2 gas is converted into NTP. Then the volume obtained in N.T.P. further converted into its mass. We will get the percentage of nitrogen by applying the following equation:
ii) Kjeldahl’s method: A known mass of organic sample is heated with conc. H2SO4 in presence of K2SO4 and catalyst Hg or CuSO4 in a long necked flask called Kjeldahl’s flask, N present in the compound is converted into (NH4)2SO4. It is further boiled with excess of NaOH solution to liberate NH3 gas which is absorbed in a known excess volume of a standard acid such as H2SO4 or HCl.
The volume of acid unused (unreacted) is determined by titration against standard alkali solution. The %N can be calculated by the following equation:
Q.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
The principle of estimation of halogens: A known mass of organic sample is heated with fuming nitric acid and few drops of silver nitrate in a sealed hard glass tube (Carius tube). The C and H present in the sample are converted into the CO2 and H2O respectively while the silver salt is converted into silver halide. The precipitates of silver halide formed are filtered out, washed, dried and weighed. Knowing the mass of the substances taken and the mass of the precipitates formed, the % of halogen can be calculated by using the following formula:
X = Halogen present, m=mass of the silver halide,
w = mass of the substance taken.
The principle of estimation of sulphur: A known mass of organic sample is heated with fuming nitric acid or sodium peroxide in Carius glass tube. The C and H present in the sample are converted into the CO2 and H2O respectively while S present is oxidized to sulphuric acid, which then reacts with excess of BaCl2 solution and precipitates as BaSO4.
The ppt. of BaSO4 is collected, washed, dried and weighed. The % of S can be calculated by using the following formula:
The principle of estimation of phosphorus: A known mass of organic sample is heated with fuming nitric acid in a Carius tube; C and H are oxidized to CO2 and H2O respectively, while P present in the compound is oxidized to phosphoric acid, later it reacts with magnesia mixture (MgCl2 + NH4Cl + NH3). A ppt. of magnesium ammonium phosphate is formed which further ignited to obtain magnesium pyrophosphate.
The yellow ppt. is collected, washed and weighed. The % of P can be calculated by knowing the mass of the substance taken and the mass of the ppt., by using the following formula:
Q.24 Explain the principle of paper chromatography.
In paper chromatography, chromatographic paper (water trapped in it), acts as the stationary phase. A spot of the solution of the mixture is made on the paper strip which is then suspended in a suitable solvent, which acts as the mobile phase. By capillary action, this solvent rises up the chromatographic paper. The solvent flows over the spot, the components are selectively retained on the paper (because they have different partition coefficient in these two phases). The different components present in the spot travel with the mobile phase to different heights. The paper so obtained is known as a chromatogram.
Q.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
When we carry out the Lassaigne’s test for the detection of halogens, the sodium extract is first boiled with dilute nitric acid. The acid is added to decompose NaCN to HCN and Na2S to H2S and to expel these gases. So, if any nitrogen and sulphur are present in the form of NaCN and Na2S, they are removed. Otherwise, these will react with AgNO3 and will interfere with the test. The chemical equations involved in the reaction are represented as:
NaCN + HNO3 → NaNO3 + HCN ↑
Na2S + 2HNO3 → 2NaNO3 + H2S ↑
Q.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Nitrogen, sulphur and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This test is called “Lassaigne’s test”. The chemical reactions involved in this test are:
Na + N + S + C → NaSCN
2Na + S → Na2S
Na + X → NaX
X = Halogens
Q.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
The process which is used to separate a mixture of camphor and calcium sulphate is known as sublimation. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is sublimable while calcium sulphate is non-sublimable. Therefore, on heating a mixture containing camphor and calcium sulphate, camphor will form vapours leaving behind calcium sulphate.
Explain, why an organic liquid vapourises at a temperature below its boiling point in its steam distillation?
In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid (p1) and the vapour pressure due to water (p2) becomes equal to atmospheric pressure (p), i.e., p=p1+p2.Since p1< p2 (as water has high specific heat and boiling point than other liquids) organic liquid will vapourise at a lower temperature than its boiling point.
Q.29 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
CCl4 will not give white precipitate of AgCl on heating it with AgNO3 because the chlorine atoms are covalently bonded to carbon atom in CCl4. To get the precipitate Cl atom should be bonded by ionic bond.
Q.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
CO2 is acidic gas and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water. The reaction is as follows:
2KOH + CO2 → K2CO3 + H2O
Q.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?
During the test of sulphur, the sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test. Instead of acetic acid, if H2SO4 is used, lead acetate will react with it to form white precipitate of lead sulphate which will interfere with the test.
(CH3COO)2Pb +H2SO4 → PbSO4 + 2CH3COOH
Q.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.
We know that,
Thus, substituting the values of %C and mass of the substance taken, we will get the mass of carbon dioxide formed.
Substituting the values in the above equation, we have
Q.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.
Total mass of organic compound taken= 0.50 g
60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.
Now 1 mole of H2SO4 neutralises 2 moles of NaOH.
60 mL of 0.5 M NaOH solution =
(60/30) mL of 0.5 M H2SO4 solution º
= 20 mL of 0.5 M H2SO4 solution
Acid consumed in absorption of evolved ammonia is
(50–30) mL = 20 mL
1 mole of H2SO4 neutralizes 2 moles of NH3.
20 mL of 0.5 M H2SO4 neutralises 40 mL of 0.5 M NH3
Since, 1000 mL of 1 M NH3 contains 14 g of nitrogen
Therefore, 40 mL of 0.5 M NH3 will contain=
Therefore, percentage of nitrogen in 0.50 g of organic compound=
Q.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.
The mass of the substance (chloro compound) taken = 0.3780 g
Mass of AgCl formed = 0.5740 g
1 mole of AgCl = 1 g atom of Cl
or (108+35.5) = 143.5 g of AgCl = 35.5 g of Cl
Applying the formula
Thus, the % of Cl present in the organic compound is 37.566%
Q.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.
The mass of the substance taken= 0.468g
Mass of BaSO4 formed= 0.668g
1 mole of BaSO4 contain 1 mole of S
Or, [137 + 32+ 4 x 16) = 233 g] of BaSO4 contain 32 g of S
Applying the formula of % of S, we get
Thus, the % of S present in the organic compound is 19.60%
Q.36 In the organic compound CH2=CH–CH2–CH2–C≡CH, the pair of hybridized orbitals involved in the formation of C2– C3 bond is:
(a) sp – sp2 (b) sp – sp3 (c) sp2– sp3 (d) sp3– sp3
Both the double bond and triple bond are present at equivalent position, and the double bond is given preference over triple bond while numbering the carbon atoms in main carbon chain.
Thus, C2-C3 bond is formed by the overlap of sp2-sp3 orbitals.
Option (c) is the correct answer.
Q.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.
The Prussian blue colour is obtained due to the formation of Fe4[Fe(CN)6]3.
Thus option (b) is the right answer.
Q.38 Which of the following carbocation is most stable?
The stability of the carbocations follows the order: 3°>2°>1°. (CH3)3C+ is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the + I effect of three methyl groups. An increased + I effect by the three methyl groups stabilises the positive charge on the carbocation. Thus, option (b) is the correct answer.
Q.39 The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography
Chromatography is the best and latest technique. Thus, option (d) is the correct answer.
Q.40 The reaction:
CH3CH2I + KOH(aq) → CH3CH2OH + KI
is classified as (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition
It is an example of nucleophilic substitution reaction. The hydroxyl group (OH–) of KOH acts as a nucleophile and substitutes iodide ion (I–) from CH3CH2I and forms ethanol. Thus, option (b) is a correct answer.
FAQs (Frequently Asked Questions)
1. What happens in an organic reaction?
The fundamental concept of organic reaction is related to the fission of covalent bonds. There are two ways in which covalent bonds undergo fission:
Homolytic fission – Where each of the atoms acquires one of the bonding electrons.
Heterolytic fission – When a bond breaks in this fission, one of the atoms acquires both the bonding electrons.
Reaction intermediaries, which are short-lived fragments, are the result of the process of fission. Carbonium ions, carbanions, carbon-free radicals, and carbenes are some examples of intermediaries.
2. Why is Chapter 12 Organic Chemistry of Class 11 important for students?
Chapter 12 is important from the perspective of Class 12 board exams and competitive exams. Many concepts from the chapter are covered in the syllabus of NEET and JEE. Also, understanding the concepts in Class 11 is important because students will have to read more advanced concepts related to this chapter in Class 12.