Important Questions for CBSE Class 11 Chemistry Chapter 12 – Organic Chemistry – Some Basic Principles and Techniques

CBSE Class 11 Chemistry Chapter 12 Important Questions with Answers

Students may find Chemistry an exciting subject if they receive proper study material, guidance and step-by-step solutions to the various problems covered in the NCERT textbook. The Extramarks team has created the most precise and reliable NCERT based solutions for important questions in Class 11 Chemistry Chapter 12. Since these questions and their answers are based on the CBSE syllabus, they can help you achieve good marks in Chemistry exams. So our Chemistry Class 11 Chapter 12 important questions and their solutions have proven to be very helpful from an exam perspective.

In this chapter, students will learn how organic compounds are crucial for maintaining life on earth and contain complex molecules like DNA, which carries genetic information, and proteins, which are constituents of our blood, muscles, and skin. Materials like textiles, fuels, polymers, dyes, and medications all contain organic molecules. These are a few crucial fields in which these chemicals are used. The special ability of the element carbon to make covalent bonds with other carbon atoms is known as catenation, as you learned in the last chapter. This chapter includes some fundamental analytical ideas and methods needed to comprehend the synthesis and characteristics of organic molecules.

.At Extramarks, we highlight key ideas and exam-oriented questions from every chapter to assist students with their studies before exams. Students will be familiar with the questions likely to come in final exams by completing  Class 11 Chemistry Chapter 12 important questions. Since Chemistry is a subject that demands in-depth conceptual understanding, only memorising the structures and formulas will not be enough.

Students must therefore understand the core concepts and theories to answer the questions. These questions are a combination of the NCERT textbook, NCERT exemplar, and other reference books. To create a solid basis for the chapter, students should try to solve all questions from our important questions Class  11 Chemistry Chapter 12. To study the fundamentals of Organic Chemistry and some frequent terms used in this area of Chemistry, students are recommended to regularly refer to these Class 11 Chemistry Chapter 12 important questions.

 One of the top e-learning platforms in India, Extramarks has earned the trust of millions of primary and secondary school students. The credibility of Extramarks lies in providing the best study material to students through its repository of resources. To speed up their learning and improve their academic performance, students must sign up at the Extramarks website now to begin their preparation without any further delay.

Access CBSE Class 11 Chemistry Important Questions and Answers

Sign up and get complete access to CBSE Class 11 Chemistry Important Questions for other chapters too:

CBSE Class 11 Chemistry Important Questions
Sr No Chapters Chapter Name
1 Chapter 1 Some Basic Concepts of Chemistry
2 Chapter 2 Structure of Atom
3 Chapter 3 Classification of Elements and Periodicity in Properties
4 Chapter 4 Chemical Bonding and Molecular Structure
5 Chapter 5 States of Matter
6 Chapter 6 Thermodynamics
7 Chapter 7 Equilibrium
8 Chapter 8 Redox Reactions
9 Chapter 9 Hyrdogen
10 Chapter 10 The S-Block Elements
11 Chapter 11 The P-Block Elements
12 Chapter 12 Organic Chemistry Some Basic Principles and Techniques
13 Chapter 13 Hydrocarbons
14 Chapter 14 Enviornmental Chemistry

Organic Chemistry Class 11 Important Questions and Answers

The Important Questions Class 11 Chemistry Chapter 12 have been compiled by the Extramarks team from various sources. The problems span many topics, including classifying organic compounds, homologous series, and reactivity series. Students can better grasp IUPAC naming using these real-world questions and their solutions. Students may enjoy the clarity of concepts which in turn is useful in answering the most difficult questions. 

The NCERT books are well designed such that it covers all the concepts along with their clarity up to the student irrespective of their intelligence level and it is also meant to clear their doubts and polish their basics as well.

A few Important Questions Class 11 Chemistry Chapter 12 are provided here, along with their answers:

Question 1. The reaction:

CH3CH2I + KOH(aq) → CH3CH2OH + KI

is classified as :

(a) electrophilic substitution 

(b) nucleophilic substitution

(c) elimination 

(d) addition

Answer 1. (b) Nucleophilic substitution


It is a nucleophilic substitution reaction. As a nucleophile, the hydroxyl group of KOH with a lone pair of itself replaces the iodide ion in CH3CH2I to produce ethanol.

Question 2. Which of the following is the correct IUPAC name?

(i) 3-Ethyl-4, 4-dimethylheptane

(ii) 4,4-Dimethyl-3-ethylheptane

(iii) 5-Ethyl-4, 4-dimethylheptane

(iv) 4,4-Bis(methyl)-3-ethylheptane

Answer 2. (i) 3-Ethyl-4, 4-dimethylheptane


The various alkyl groups present in a compound are listed in the IUPAC name in alphabetical order. Di, tri, and tetra are left out of the alphabetical list. Therefore, the ethyl group (e) comes before the methyl group (m). As a result, the ethyl group is assigned the number 3 and is written first.

Question 3. Benzene hexachloride is prepared from benzene and chlorine in sunlight by

(a) Substitution reaction

(b) Elimination reaction

(c) Addition reaction

(d) Rearrangement reaction

Answer 3. (c) Addition reaction

Explanation: In addition to the reaction, benzene hexachloride is prepared from benzene and chlorine in sunlight. It occurs in the absence of oxygen, and a catalyst may be involved to increase the reaction rate.

C6H6    +    3Cl2     →   C6H6Cl6

Question 4. 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

Answer 4. (b) Fe4[Fe(CN)6]3


In Lassaigne’s test for nitrogen in an organic molecule, the sodium fusion extract is heated with iron (II) sulphate. It is then acidified with sulphuric acid. Iron (II) sulphate and sodium cyanide first combine to create sodium hexacyanoferrate (II). Sulphuric acid is subsequently used to oxidise the iron (II), resulting in the production of iron (III)hexacyanoferrate (II) or Prussian blue.

The following are the chemical reactions:

6CN  +  Fe2+ ⟶ [Fe (CN)6]4-

3[Fe(CN)6]4-      +    4Fe3+    xH2O⟶      Fe4[Fe(CN)6]3.xH2O

                                                             (Prussian blue)

Question 5. The fragrance of flowers is due to the presence of some steam volatile organic

compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in the vapour phase. A suitable method for the extraction of these oils from the flowers is:

(i) Distillation

(ii) Crystallisation

(iii) Distillation under reduced pressure

(iv) Steam distillation

Answer 5. (iv) Steam distillation


Steam distillation is used to distinguish between compounds that are immiscible with water and are steam volatile. In steam distillation, heated flasks carrying the liquid to be distilled are passed through the steam produced by a generator. The steam and the volatile organic compound mixture are then condensed and collected. The separating funnel is afterwards used to separate the chemical from the water.

Question 6. Which of the following compounds will not exist as a resonance hybrid? Give reasons for your answer.

(i) CH3OH

(ii) R−CONH2

(iii) CH3CH=CHCH2NH2

Answer 6. (i) CH3OH   and  (iii) CH3CH=CHCH2NH2


(i) CH3OH does not contain pi-electrons and thus cannot form a resonance hybrid. 

(iii) In CH3CH=CHCH2NH2, the lone pair of electrons on the nitrogen atom is not coupled to the pi -electrons. Thus, the formation of a resonance hybrid is not possible.

Question 7. Which of the two:

 O2NCH2CH2O–    or CH3CH2O–     is expected to be more stable, and why?

Answer 7.  O2NCH2CH2O is more stable.


Nitrogen being an electron-withdrawing group shows -I effect. It stabilises the molecule by lowering its negative electron charge. The methyl group, on the other hand,  is an electron donor group and exhibits a +I effect. This makes the molecule more negatively charged and makes it less stable.

Question 8. In which of the following compounds, the Carbon marked with asterisk is

expected to have the greatest positive charge?

(i) *CH3—CH2—Cl

(ii) *CH3—CH2—Mg+Cl

(iii) *CH3—CH2—Br

(iv) *CH3—CH2—CH3

Answer 8. (i) *CH3—CH2—Cl

Explanation: According to the order of electronegativity: 

Cl > Br > C > Mg.

The more electronegative group attached to the Carbon will give a more positive charge. Therefore, the C-atom having -Cl group attached to it will have the greatest positive charge.

Question 9. Which of the following compounds contain all the carbon atoms in the same

hybridisation state?

(i) H—C ≡ C—C ≡ C—H

(ii) CH3—C ≡ C—CH3

(iii) CH2 = C = CH2

(iv) CH2 = CH—CH = CH2

Answer 9. (i) H—C ≡ C—C ≡ C—H   and (iv) CH2 = CH—CH = CH2


In these two compounds, all the carbon atoms have the same hybridisation state. In (i), all the carbon atoms are sp-hybridised, and in (iv), all the carbon atoms are sp2-hybridized. 

Question 10. In the organic compound CH2= CH – CH2– CH2– C ≡ CH, the pair of hybridised

orbitals involved in the formation of the C2– C3 bond is:

(a) sp – sp2

(b) sp – sp3

(c) sp2– sp3

(d) sp3– sp3

Answer 10. (b) sp– sp3


In the given compound, the carbon atoms can be numbered as:

6        5        4       3      2     1

CH2= CH – CH2– CH2– C ≡ CH

The hybridisation of carbon atoms 1, 2, 3, 4, 5, and 6 results in sp, sp, sp3, sp3, sp2, and sp2 atoms, respectively. Therefore, the pair of hybridised orbitals involved in forming the C2-C3 bond is sp-sp3.

Question 11. Explain why alkyl groups act as electron donors when attached to a π system.

Answer 11. When connected to a -system, an alkyl group acts as an electron-donor group due to hyperconjugation. Take propane as an illustration.


Due to hyperconjugation, the sigma electrons of the C-H bond become delocalised. An unsaturated system is immediately joined to the alkyl. Delocalisation occurs when an sp3-s sigma bond orbital partially overlaps with an empty p orbital of a nearby carbon atom’s n-bond.

orbital diagram showing hyperconjugation in propene

This kind of overlap causes the electrons to be delocalised, also known as no-bond resonance, which increases the stability of the molecule.

no-bond resonance

Question 12. Explain, how is the electronegativity of carbon atoms related to their state of

hybridization in an organic compound?

Answer 12. As the s-character grows, electronegativity rises. This is due to the nucleus’s stronger attraction to s-electrons than p-electrons.

sp3 – 25% s-character, 75% p-character

sp2 – 33% s-character, 67% p-character 

sp – 50% s-character, 50% p-character

Thus, the order of electronegativity is sp3 < sp2 < sp

Question 13. What are electrophiles and nucleophiles? Explain with examples.

Answer 13. A reagent known as an electrophile, or an electron-loving pair, requires an electron pair—for instance, carbonyl groups, CH3CH2+, C=0 (due to the lone pair).

A reagent that contains an electron pair and is willing to contribute is known as a nucleophile. It is also referred to as a nucleus-loving reagent. These include NC, OH, and R3C (carbanions).

Question 14. In DNA and RNA, nitrogen atoms are present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.

Answer 14. No, the heterocyclic rings in DNA and RNA include nitrogen that cannot be eliminated as ammonia.

As nitrogen contained in these structures cannot be transformed into ammonium sulfate, the Kjeldahl method cannot be used to estimate nitrogen present in rings, azo, or nitro groups.

Question 15. Explain the principle of paper chromatography.

Answer 15. Partition chromatography includes paper chromatography. Chromatography paper is a unique kind of paper that is used in paper chromatography. Water that has been trapped inside a chromatography paper serves as the stationary phase.

The solution of the mixture is spotted at the base of a strip of chromatography paper primarily suspended in a suitable solvent or mixture of solvents. The mobile phase is this solvent. Capillary action causes the solvent to ascend up the paper and run over the spot. According to how they were divided differently into the two phases, the paper preserves particular components only in some cases. The produced paper strip is referred to as a chromatogram.

paper chromatography paper in two different shapes.

On the chromatogram, the spots representing the separated coloured compounds are visible at various heights from the location of the initial spot. Using a suitable spray reagent, as described under thin layer chromatography, or by using UV light, one can see the spots of the separated colourless chemicals.

Question 16. Give a brief description of the principles of the following techniques taking an

example in each case.

(a) Crystallisation (b) Distillation (c) Chromatography

Answer 16.


Crystallisation is a method used to purify solid organic molecules. The difference in the solubility of the substance and impurities in a particular solvent is the basis on which it operates. Since the impure product is only sparingly soluble at lower temperatures, it is made to dissolve in the solvent at a higher temperature. We keep doing this until the solution is practically saturated. We obtain its crystals by cooling and filtering the mixture. For instance, we can obtain pure aspirin by crystallising 2-4g of crude aspirin in 20mL of ethyl alcohol. If necessary, it is heated and then left alone until it crystallises. After being separated, the crystals are dried.


Non-volatile liquids are separated from volatile components using this technique. Additionally, it is employed when the boiling temperatures of the constituents differ significantly.

It operates under the premise that liquids with various boiling points vaporise at various temperatures. The generated liquids are then separated after they have been cooled.

Ex: In a flask with a circular bottom and a condenser, aniline (b.p. = 457 K) and chloroform (b.p = 334 K) are combined. Due to its high volatility, chloroform vaporises first when heated and is then forced to pass through a condenser where it cools. The flask with a circular bottom still holds the aniline.


It is popularly used for the separation and purification of organic compounds.

It operates on the principle that each component of a mixture moves through the stationary phase under the influence of the mobile phase at a distinct rate.

Ex: A mixture of blue and red ink can be separated using chromatography. The component of the mixture less absorbed by the chromatogram is placed on the chromatogram. It causes the almost immobile component to travel more quickly up the paper than the other component.

Question 17. Give three points of differences between inductive effect and resonance effect.

Answer 17.

Inductive Effect Resonance Effect
It involves the displacement of σ-electrons. It involves the displacement of π-electrons or any lone pair of electrons.
This effect moves up to three carbon atoms before becoming insignificant after the fourth carbon atom. The resonance effect moves all along the length of the conjugated system.
This operates in saturated compounds. This operates only in unsaturated conjugated systems.


Question 18. Why does S03 act as an electrophile?

Answer 18. The sulphur atom is attached to three highly electronegative oxygen atoms, resulting in the Sulphur atom becoming electron-deficient. Additionally, Sulphur also takes up a positive charge due to resonance. Both these factors make S03 an electrophile.


Question 19. The empirical formula of a compound is CH2. Its one mole has a mass of 42 g. What is its molecular formula?

Answer 19.

Molecular formula = n × empirical formula,

 Therefore,  Molecular formula =4214 *  CH2 = C3H

Question 20. A mixture containing benzoic acid and nitrobenzene is given to you. Using an appropriate chemical reagent, how will you proceed to separate them?

Answer 20. When nitrobenzene reaches the organic layer, the mixture is agitated with a diluted NaHCO3 solution and extracted with ether or chloroform. The distillation of this will result in nitrobenzene. Dilute HCl is used to acidify the aqueous layer, and the solution is then cooled.

The final product is benzoic acid.

C6H5COOH + NaHCO3 → C6H5COONa + CO2 + H2O.

C6H5COONa + HCl (dilute) → C6H5COOH + NaCl.

Question 21. Discuss the chemistry of Lassaigne’s test.

Answer 21. Lassaigne’s test is used to identify the presence of phosphorus, halogens, nitrogen, and sulphur in an organic molecule. In an organic compound, these components are found in the covalent state. The chemical is fused with sodium metal to create the ionic form of these.


Boiling the fused mixture in distilled water releases the produced cyanide, sulfide, and sodium halide. The obtained extract is known as Lassaigne’s extract. Then, the extract from Lassaigne is examined for the presence of phosphorus, sulphur, nitrogen, and halogens.

(a)Test for nitrogen

A Lassaigne test is performed on sodium fusion extract by heating it with iron (II) sulphate and then acidifying it with sulfuric acid. The first reaction involves sodium cyanide interacting with iron sulfate (II) to form sodium hexacyanoferrate (II). Then, some iron (II) is heated with sulphuric acid to create iron (III) hexacyanoferrate (II), which is Prussian blue. Here are the chemical equations involved in the reaction:


(b) Test for sulphur

Acetic acid is used to make the sodium fusion extract acidic to perform Lassaigne’s test for sulphur in an organic compound. Next, lead acetate is added. Sulfur is present in the molecule because lead sulphide, which is black, precipitates out.

S2+     +Pb2+     →    PbS


Sodium nitroprusside is used to treat the sodium fusion extract. The compound’s appearance also indicates the presence of sulphur in the compound as violet.

S2+    +   [Fe(CN)5NO]2-   →  [Fe(CN)5NOS]4-


When both nitrogen and sulphur are present in an organic molecule, NaSCN is formed instead of NaCN.

Na + C + N + S    →   NaSCN

Fe3+    + SCN   →    [Fe(SCN)]2+

                                     (Blood red)

The colour of this NaSCN (sodium thiocyanate) is blood red. The lack of free cyanide ions prevents the formation of Prussian colour.

NaSCN   +   2Na  →  NaCN  +  Na2S

(c)Test for halogens

To determine whether an organic molecule contains halogens, sodium fusion extracts are treated with silver nitrate after being acidified with nitric acid.

X  +    Ag+   → AgX

Here, X represents a halogen Cl, Br or I.

As a result of heating Lassaigne’s extract, the nitrogen and sulphur are removed, which may interfere with the determination of halogens in organic compounds containing both elements.

Benefits of Solving Organic Chemistry Class 11 Important Questions

Organic Chemistry is essential because the majority of chemicals used in daily life are powered by organic chemistry. It also plays an essential role in producing common household chemicals, foods, polymers, and medications. Chemistry may be a tough subject and it becomes challenging for many students to comprehend a few advanced topics. To overcome this, students need to practice a lot of questions to self-assess their knowledge of the chapter. Extramarks organic chemistry class 11 extra questions along with our solutions is one of the best resources to practice exam-oriented questions. All key areas from the chapter are covered in our important questions and solutions. Our subject matter experts team have curated these questions with the CBSE exams in mind and many questions are most likely to be asked in exams. Students can gain a competitive advantage by regularly practising our Chapter 12 Class 11 Chemistry Important Questions.

 Extramarks believes in incorporating joyful learning experiences through its repository of resources.

 Its credibility lies in providing reliable and trusted comprehensive learning solutions for students from Class 1 to Class 12. We recommend students also register on reliable online learning platforms such as Extramarks which strictly follows books and provides solved exercises and practice questions NCERT to step up their learning experience Following are some of the benefits for students to practise from our set of Important Questions Class 11 Chemistry Chapter 12:

  • It is a one-stop solution, meaning students will get questions from different sources. Consequently, they will save time during preparation by having the questions and responses in advance. Students can entirely rely upon these important questions as these are made following all the latest guidelines and curricula laid by CBSE.
  •  Chemistry faculty experts review and verify the questions and answers before providing them to the students. It ensures superior knowledge and understanding of the subject because it provides all the information and is complete in every way. Students need not look for answers elsewhere. That itself reduces stress and anxiety to a great extent S.
  • After studying the updated CBSE curriculum, experienced Chemistry teachers have created these solutions. So they adhere to the latest NCERT guidelines.
  • Since every question and answer is created with the exam in mind, students can confidently take exams and come out with flying colours.

Extramarks provides students with the best study materials and constantly strives to upgrade its products year after year to meet the changing demands of the curriculum and present its millennial generation with very simple and easy solutions for each and every student irrespective of their level.

Its credibility lies in providing reliable and trusted comprehensive learning solutions for students from Class 1 to Class 12. Extramarks offers high-quality resources like CBSE revision notes, CBSE sample papers, CBSE past years’ question papers, CBSE extra questions, and CBSE mock examinations. To access some of these resources, students must click on the URLs provided below:

  • CBSE revision notes
  • CBSE syllabus
  • CBSE sample papers
  • CBSE past years’ question papers
  • CBSE extra questions and solutions

Q.1 (i) How can you estimate nitrogen in Dumas method

(ii) In Dumas method for the estimation of nitrogen, 0.60 g of an organic compound gave 100 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)



(i) The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.

C x H y N z +(2x+ y 2 )CuO xC O 2 + y 2 H 2 O+ z 2 N 2 +(2x+ y 2 )Cu

However, traces of nitrogen oxides are formed that are reduced to nitrogen by passing the gaseous mixture over heated copper gauze. The nitrogen thus formed is collected over an aqueous solution of KOH which absorbs other gases like CO2, water vapours etc.

Given, Mass of the substance taken = 0.60 g Volume of nitrogen collected = 100 mL Atmospheric pressure = 715 mm Hg Room temperature = 300 K Aqueous tension at 300 K = 15 mm So, the actual pressure of the gas = 715 – 15 = 700 mm Hg Volume of nitrogen at STP, P 1 =700 mm, P 2 =760 mm, V 1 =100 mL, V 2 =, T 1 =300 K, T 2 =273 K Substituting the values in the gas equation, P 1 V 1 T 1 = P 2 V 2 T 2 , 700×100 300 = 760× V 2 273 V 2 = 700×273×100 300×760 =83.8 mL 22400 mL of nitrogen at STP weigh = 28 g 83.8 mL of nitrogen at STP will weigh = 28×83.8 22400 g Percentage of nitrogen= Mass of N 2 at STP Mass of the substance taken ×100 = 28×83.8×100 22400×0.60 ×100=17.46

Q.2 (i) What are electrophiles and nucleophiles

(ii) Classify the following transformations according to the reaction type:

(A) C H 3 CH=CHC H 3 +B r 2 C H 3 CHBrCHBrC H 3 (B) C H 3 C H 2 Cl+H S C H 3 C H 2 SH+C l



(i) Electrophiles are the electron seeking chemical species and have an electron deficient atom in them. They may be positively charged or neutral chemical species.

Nucelophiles are the electron rich chemical species containing at least one lone pair of electrons and thus they are nucleus loving species. They may be negatively charged or neutral chemical species.


(A) Electrophilic addition reaction

(B) Nucleophilic substitution reaction

Q.3 The homolytic fission of CC bond in ethane gives an intermediate. Name this reaction intermediate. Explain the hybridisation of carbon in this intermediate.



Homolytic fission is the bond breaking whereby the bond breaks to give two similar species each keeping an electron. The intermediates so formed are known as free radicals.
In case of methyl free radical the carbon atom has 3 bond pairs and one odd electron which is present in one of the p-orbital and this p-orbital does not involve in hybridisation. Thus, the carbon in methyl radical involves in sp2 hybridisation.

Q.4 Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): (CH3)3C+ is more easily formed than (CH3)2HC+.
Reason (R): Tertiary butyl carbocation is more stable than isopropyl carbocation.
Select the most appropriate answer from the options given below:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true but R is not correct explanation of A.
C. A is true but R is false.
D. A is false bur R is true.



A. Both A and R are true, and R is the correct explanation of A.

Tertiary carbocations are more stable than the secondary carbocations due to +I effect of three CH3 groups.

Q.5 Identify the incorrect statement about electromeric effect.
A. It is a temporary effect.
B. This effect is shown by saturated carbon compounds.
C. This effect is shown in the presence of an attacking species only.
D. Complete transfer of electron takes place in this effect.



B. This effect is shown by saturated carbon compounds.

Please register to view this section

FAQs (Frequently Asked Questions)

1. How to start preparing for Organic Chemistry in Class 11?

Organic Chemistry is one best-scoring chapter in Class 11 Chemistry. However students need to have good study resources. To get excellent grades in this chapter, students must adhere to these tips:

  • Start their studies with NCERT textbook and refer exemplars for solving questions.
  • Simple chemical processes and theories form the foundation of organic Chemistry. Instead of only memorising facts, students should understand the concepts.
  • Ask the teacher questions and try to solve important questions Class 11 Chemistry Chapter 12.
  • Make an effort to comprehend the ideas behind the chemical reactions. It will guarantee a thorough understanding of the chapter.
  • Students can get comprehensive guidance from the subject experts  and have doubt clearing sessions  by signing up on the official website.

2. Can I learn Organic Chemistry Chapter 12 just by reading NCERT books?

 One can learn Organic Chemistry Chapter 12  just by reading NCERT books, but you will need some good class notes for better conceptual clarity. Students can further refer to additional reliable study resources such as Extramarks platform to gain a comprehensive understanding of the subject. 

Students can take advantage by practising from our Extramarks important questions Class 11 Chemistry Chapter 12, which includes questions from NCERT textbook, NCERT exemplars, and many other reference sources.

3. Which elements make up the organic compounds?

 Carbon and Hydrogen are always present in the formation of organic compounds. Different organic molecules may also contain other elements such as oxygen, nitrogen, and phosphorus.