case study questions on organic chemistry class 11

Case Study Questions Class 11 Chemistry Organic Chemistry

Case study questions class 11 chemistry chapter 12 organic chemistry.

CBSE Class 11 Case Study Questions Chemistry Organic Chemistry. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Organic Chemistry.

CBSE Case Study Questions Class 11 Chemistry Organic Chemistry

Case study – 1.

F Wohler synthesised an organic compound, urea from an inorganic compound, ammonium cyanate.

Characteristic Features of π Bonds In a π (pi) bond formation, parallel orientation of the two p orbitals on adjacent atoms is necessary for a proper sideways overlap. Thus, in H 2 C=CH 2 molecule all the atoms must be in the same plane. The p orbitals are mutually parallel and both the p orbitals are perpendicular to the plane of the molecule. Rotation of one CH 2 fragment with respect to other interferes with maximum overlap of p orbitals and, therefore, such rotation about carbon-carbon double bond (C=C) is restricted. The electron charge cloud of the π bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents. In general, π bonds provide the most reactive centres in the molecules containing multiple bonds.

a) Zig zag structure

b) Lewis structure

c) Triple bond

d) Equivalent bond

2) What is condensed structural formula?

Case Study – 2

b) Aromatic compounds Aromatic compounds are special types of compounds. These include benzene and other related ring compounds (benzenoid). Like alicyclic compounds, aromatic comounds may also have hetero atom in the ring. Such compounds are called hetrocyclic aromatic compounds. Some of the examples of various types of aromatic compounds are:

The IUPAC name for a homologous series of such compounds is alkane. Paraffin (Latin: little affinity) was the earlier name given to these compounds. Unsaturated hydrocarbons are those, which contain at least one carbon- carbon double or triple bond. IUPAC Nomenclature of Alkanes Straight chain hydrocarbons: The names of such compounds are based on their chain structure, and end with suffix ‘-ane’ and carry a prefix indicating the number of carbon atoms present in the chain (except from CH4 to C4H10, where the prefixes are derived from trivial names). The IUPAC names of some straight chain saturated hydrocarbons are given in Table. The alkanes in table differ from each other by merely the number of – CH 2 groups in the chain. They are homologues of alkane series.

a) C 4 H 10

 Case Study – 3

Case study – 4.

b) Negative Electromeric Effect (–E effect) -In this effect the π – electrons of the multiple bond are transferred to that atom to which the attacking reagent does not get attached. For example: When inductive and electromeric effects operate in opposite directions, the electomeric  effect predominates.

Ans- a) nucleophile

a) hindrance

Case Study – 5

iii) Distillation

In general, greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and stabilisation of the cation. Thus, we have the following relative stability of carbocations :

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Class 11 Chemistry Case Study Questions

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Are you perplexed? Are you having trouble finding different Class 11 Chemistry Case Study Questions? If the answer is affirmative! You’ve arrived at the perfect place. The myCBSEguide App offers case study questions for CBSE class 11 Chemistry.

CBSE class 11 Chemistry students have always been under a lot of pressure to perform well in their board exams. This pressure has increased in recent years, with the inclusion of case study questions and other Higher Order Thinking Skills Questions. myCBSEguide provides with all the resources that class 11 Chemistry students need to prepare for their class 11 Chemistry exams. This includes sample papers, question banks, case study questions and revision notes prepared by experienced teachers.

The logical reason for Chemistry

Higher Secondary is the most important stage of schooling because it is at this point that specialized discipline-based, content-oriented courses are introduced. Students reach this stage after completing ten years of general education and choose Chemistry to pursue a career in basic sciences or professional courses such as medicine, engineering, or technology, as well as tertiary study courses in applied science and technology. As a result, learners must be provided with a sufficient conceptual background in Chemistry in order to be prepared to meet the challenges of academic and professional courses after high school.

Chemistry in Class 11

Chemistry studies the composition, structure, properties, and reactions of matter. Chemistry is included as one of the major disciplines in CBSE class 11 for a variety of reasons.

• First and foremost, chemistry is essential for understanding the basic structure and composition of matter.
• In addition, chemistry plays a critical role in many other disciplines, such as biology, environmental science, and engineering.
• Furthermore, chemistry is a fascinating subject in its own right, and studying it can provide a deeper understanding of the world around us.
• Finally, chemistry is a practical subject that can be used in everyday life, and it is therefore important for students to have a good understanding of it.

Integrating Case Study Questions in Chemistry for Class 11

Case study questions can be a great way to integrate real-world applications of chemistry into your class 11 curriculum. By posing questions that require students to analyze and apply their knowledge of chemical concepts, you can help them to see how those concepts are relevant to their lives outside of the classroom. Additionally, case study questions can be used to spark lively classroom discussion and debate.

Sample Class 11 Case Study Questions

The recent addition of Case Study Questions in Chemistry for Class 11 by CBSE has been welcomed by students and teachers alike. In Class 11 Chemistry, the case study approach is seen as an engaging and effective way to learn, and the questions included in Class 11 Chemistry is designed to challenge and stimulate students’ thinking. Class 11 Chemistry Case Study Questions cover a wide range of topics, from the basics of atomic structure to more complex concepts such as chemical reactions and equilibrium. The addition of this resource is sure to enhance the learning experience for Class 11 Chemistry students of all abilities while solving Class 11 Chemistry Case Study Questions.

myCBSEguide is the only app that provides a wide range of Class 11 Chemistry Case Study Questions curated by expert teachers. By providing access to this wealth of resources, myCBSEguide makes it easy for Class 11 Chemistry students to learn about and understand complex chemical concepts. In addition, the app’s comprehensive collection of Class 11 Chemistry Case Study Questions covers all major topics in the subject, making it an invaluable resource for students preparing for Class 11 Chemistry exams.

Have a look at few samples of Class 11 Chemistry Case Study Questions for better understanding.

Class 11 Chemistry Case Study Question 1

Read the passage and answer the following question: In 1830, Michael Faraday showed that if electricity is passed through a solution of an electrolyte, chemical reactions occurred at the electrodes, which resulted in the liberation and deposition of matter at the electrodes. In mid-1850s Faraday began to study electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes.  When sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles moving in the tube from the negative electrode to the positive electrode. These were called cathode rays or cathode ray particles. J.J. Thomson measured the ratio of electrical charge (e) to the mass of the electron (m e ) by using a cathode ray tube and applying electrical and magnetic fields perpendicular to each other as well as to the path of electrons. A positively charged particle was characterised in 1919. Later, a need was felt for the presence of an electrically neutral particles as one of the constituents of the atom.

In this question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices

• Assertion and reason both are correct statements and reason is the correct explanation for assertion.
• Assertion and reason both are correct statements and reason is not the correct explanation for assertion.
• Assertion is the correct statement but reason is wrong statement.
• Assertion is the wrong statement but reason is correct statement.
• Assertion: The cathode rays start from cathode and move towards the anode. Reason:  In the absence of electrical or magnetic field, cathode rays travel in straight lines.
• Assertion: Thomas argued that the lighter the particle, greater the deflection. Reason: Deflection depends upon the mass of the particle.
• Assertion: Television picture tubes are anode ray tubes. Reason: Electrons are the basic constituent of all the atoms.
• Assertion: The charge to mass ratio of the particles depends on the gas from which these originate. Reason: The smallest and lightest positive ion was obtained from hydrogen and was called proton.

Assertion: A cathode ray tube is made of glass containing two thin pieces of metal electrodes. Reason: The value of e/m e  is 2.758820 × 10 11 C kg -11 .

Answer Key:

• (b) Assertion and reason both are correct statements and reason is not the correct explanation for assertion.
• (a) Assertion and reason both are correct statements and reason is the correct explanation for assertion.
• (d) Assertion is the wrong statement but reason is correct statement.
• (c) Assertion is the correct statement but reason is wrong statement.

Class 11 Chemistry Case Study Question 2

Read the passage and answer the following question: The existing large number of organic compounds and their ever-increasing numbers has made it necessary to classify them on the basis of their structures. Organic compounds are broadly classified as open-chain compounds which are also called aliphatic compounds. Aliphatic compounds further classified as homocyclic and heterocyclic compounds. Aromatic compounds are special types of compounds. Alicyclic compounds, aromatic compounds may also have heteroatom in the ring. Such compounds are called heterocyclic aromatic compounds. Organic compounds can also be classified on the basis of functional groups, into families or homologous series. The members of a homologous series can be represented by general molecular formula and the successive members differ from each other in molecular formula by a –CH 2  unit.

In these questions, a statement of assertion followed by the statement of reason is given. Choose the correct answer out of the following choices

• Assertion: Tetrahydrofuran is aliphatic compounds Reason: Sometimes atoms other than carbon are also present in the ring known as heterocyclic.
• Assertion: Hydroxyl group (–OH) is a functional group. Reason: The functional group is defined as an atom or group of atoms joined in a specific manner with characteristic chemical properties of the organic compounds.
• Assertion: Non-benzenoid compound is a classification as the alicyclic compound. Reason: Aniline is a benzenoid compound.
• Assertion: H 2 C=CH 2  is a condensed structural formula. Reason: Condensed structural formula is represented by omitting some or all of the dashes representing covalent bonds.

Assertion: Cyclic compound is classified as a carbocyclic and heterocyclic compound. Reason: Thiophene is a homocyclic compound.

Class 11 Chemistry Case Study Question 3

Read the passage and answer the following questions: A large number of orbitals are possible in an atom. Qualitatively these orbitals can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along with certain directions than along others. The principal quantum number determines the size and to large extent the energy of the orbital. Azimuthal quantum number, ‘l’ is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. Each shell consists of one or more subshells or sub-levels. The number of sub-shells in a principal shell is equal to the value of n. Magnetic orbital quantum number. ‘ml ’ gives information about the spatial orientation of the orbital with respect to a standard set of co-ordinate axis. The fourth quantum number is known as the electron spin quantum number (m s ). An electron spins around its own axis, much in a similar way as the earth spins around its own axis while revolving around the sun. In these questions, a statement of assertion followed by the statement of reason is given. Choose the correct answer out of the following choices:

• Assertion:  Each orbital is designated by three quantum numbers labelled as n, l and m l . Reason:   ‘n’ is a positive integer with value of n = 1,2,3.
• Assertion:  The principal quantum number identifies the shell. Reason:  Size of an orbital decrease with the increase of principal quantum number ‘n’.
• Assertion:  For n = 2, the possible value of l can be 0 and 1. Reason:  For a given value of n, l can have n values ranging from 0 to n – 1.
• Assertion:  Each orbital in an atom, is defined by a set of values for n, l and m l . Reason:  m l  designates the orientation of the orbital. OR Assertion:  Spin quantum numbers m ​​​​​​s  can take the values of +½ or –½. Reason:  Two spin states of the electron and are normally represented by two arrows, ↑ (spin down) and ↓ (spin up).
• (b) Assertion and reason both are correct statements and reason is not the correct explanation for assertion. OR (c) Assertion is the correct statement but reason is wrong statement.

Significance of Class 11 Chemistry case study questions

Case study questions in Class 11 Chemistry are important because:

• They help Class 11 Chemistry students understand basic chemistry facts and concepts while keeping the excitement alive.
• It prepares Class 11 Chemistry students to study chemistry in tertiary courses such as medicine, engineering, and technology. • introduce students to a variety of emerging new areas of chemistry and inform them of their importance in future studies as well as their applications in various fields of chemical science and technology.
• It prepares Class 11 Chemistry students to face a variety of health, nutrition, environmental, population, weather, industry, and agriculture challenges.
• It also improves Class 11 Chemistry students’ problem-solving abilities.
• It helps in familiarizing Class 11 Chemistry students with various industrial processes as well as their technological applications.

Glancing the Class 11 Chemistry syllabus

The revised and updated Class 11 Chemistry syllabus is built on a disciplinary approach with rigour and depth, while keeping the Class 11 Chemistry syllabus light and comparable to the worldwide level. Greater emphasis has been placed in the Class 11 Chemistry syllabus on the use of new nomenclature, symbols, and formulations, the teaching of fundamental concepts, the application of concepts in chemistry to industry/technology, logical sequencing of units, the removal of obsolete content and repetition, and so on.

CBSE Class 11 Chemistry (Code No. 043) Syllabus (2022-23)

myCBSEguide: Balanced equation for Class 11 Chemistry students

For Class 11 Chemistry students, myCBSEguide is always a balanced equation. It includes a plethora of class 11 chemistry case study questions to help students better understand the ideas in Class 11 Chemistry. Class 11 chemistry case study questions are intended to assess students’ comprehension while also providing a platform for them to strengthen their problem-solving abilities. myCBSEguide also provides a number of other tools that students can utilise to deepen their grasp of the subject. Additionally, the answer keys and explanations provided by myCBSEguide are always clear and concise, making it an excellent tool for students who want to improve their grades. So, what are you waiting for? Get myCBSEguide today and start studying for your future.

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Case Study Questions for Class 11 Chemistry PDF Download

We have provided here Case Study questions for the Class 11 Chemistry for final board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your exams.

CBSE 11th Standard CBSE Chemistry question papers, important notes, study materials, Previous Year questions, Syllabus, and exam patterns. Free 11th Standard CBSE Chemistry books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Chemistry Case Study Questions

CBSE Class 11 Chemistry question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 Chemistry students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 11 Chemistry

• Case Study Based Questions on Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry
• Case Study Based Questions on Class 11 Chemistry Chapter 2 Structure of Atom
• Case Study Based Questions on Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
• Case Study Based Questions on Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure 
• Case Study Based Questions on Class 11 Chemistry Chapter 6 Thermodynamics
• Case Study Based Questions on Class 11 Chemistry Chapter 7 Equilibrium
• Case Study Based Questions on Class 11 Chemistry Chapter 8 Redox Reactions
• Case Study Based Questions on Class 11 Chemistry Chapter 12 Organic Chemistry -Some Basic Principles and Techniques
• Case Study Based Questions on Class 11 Chemistry Chapter 13 Hydrocarbons

Class 11 students should go through important Case Study problems for Chemistry before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 11 Chemistry examinations. Our expert faculty for standard 11 Chemistry have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 11 students understand the concepts and also easy to learn solutions.

Class 11 Chemistry Books

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

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

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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.

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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.

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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:

CH 3 CH 2 I + KOH (aq) → CH 3 CH 2 OH + KI

is classified as :

(a) electrophilic substitution 

(b) nucleophilic substitution

(c) elimination 

(d) addition

Answer 1. (b) Nucleophilic substitution

Explanation:

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 CH 3 CH 2 I 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.

C 6 H 6     +    3Cl 2     →   C 6 H 6 Cl 6

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) Na 4 [Fe(CN) 6 ] 

(b) Fe 4 [Fe(CN) 6 ] 3

(c) Fe 2 [Fe(CN) 6 ] 

(d) Fe 3 [Fe(CN) 6 ] 4

Answer 4. (b) Fe 4 [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 –   +  Fe 2+ ⟶ [Fe (CN) 6 ] 4-

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

(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:

 O 2 NCH 2 CH 2 O –    or CH 3 CH 2 O –     is expected to be more stable, and why?

Answer 7.   O 2 NCH 2 CH 2 O – 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) *CH 3 —CH 2 —Cl

(ii) *CH 3 —CH 2 —Mg + Cl –

(iii) *CH 3 —CH 2 —Br

(iv) *CH 3 —CH 2 —CH 3

Answer 8. (i) *CH 3 —CH 2 —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) CH 3 —C ≡ C—CH 3

(iii) CH 2 = C = CH 2

(iv) CH 2 = CH—CH = CH 2

Answer 9. (i) H—C ≡ C—C ≡ C—H   and (iv) CH 2 = CH—CH = CH 2

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 sp 2 -hybridized. 

Question 10. In the organic compound CH 2 = CH – CH 2 – CH 2 – C ≡ CH, the pair of hybridised

orbitals involved in the formation of the C 2 – C 3 bond is:

(a) sp – sp 2

(b) sp – sp 3

(c) sp 2 – sp 3

(d) sp 3 – sp 3

Answer 10. (b) sp– sp 3

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

6        5        4       3      2     1

CH 2 = CH – CH 2 – CH 2 – C ≡ CH

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

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.

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

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.

sp 3 – 25% s-character, 75% p-character

sp 2 – 33% s-character, 67% p-character 

sp – 50% s-character, 50% p-character

Thus, the order of electronegativity is sp 3 < sp 2 < 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, CH 3 CH 2 + , 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 R 3 C – (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.

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

Crystallisation: 

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.

Distillation:

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.

Chromatography:

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.

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?

Molecular formula = n × empirical formula,

 Therefore,  Molecular formula = 42 14 *  CH 2 = C 3 H 6 

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 NaHCO 3 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.

S 2+     +Pb 2+     →    PbS

                                 (Black)

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.

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

                                                     (Violet)

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

Na + C + N + S    →   NaSCN

Fe 3+     + 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  +  Na 2 S

(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:

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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 + ( 2 x + y 2 ) C u O x C O 2 + y 2 H 2 O + z 2 N 2 + ( 2 x + y 2 ) C u

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 CO 2 , water vapours etc.

G i v e n , M a s s 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 m m , P 2 = 760 m m , V 1 = 100 m L , 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 P e r c e n t a g e o f n i t r o g e n = 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 C H = C H C H 3 + B r 2 C H 3 C H B r C H B r C H 3 ( B ) C H 3 C H 2 C l + H S C H 3 C H 2 S H + 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 sp 2 hybridisation.

Q.4 Given below are two statements labelled as Assertion (A) and Reason (R) Assertion (A): (CH 3 ) 3 C + is more easily formed than (CH 3 ) 2 HC + . 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.

Explanation : Tertiary carbocations are more stable than the secondary carbocations due to +I effect of three CH 3 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.

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Cbse class 11 chemistry important questions, chapter 1 - some basic concepts of chemistry.

Chapter 2 - Structure of Atom

Chapter 3 - classification of elements and periodicity in properties, chapter 4 - chemical bonding and molecular structure, chapter 5 - states of matter, chapter 6 - thermodynamics, chapter 7 - equilibrium, chapter 8 - redox reactions, chapter 9 - hyrdogen, chapter 10 - the s-block elements, chapter 11 - the p-block elements, chapter 13 - hydrocarbons, chapter 14 - enviornmental chemistry, 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.

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Class 11th Chemistry - Organic Chemistry: Some Basic Principles and Techniques Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Chemistry Subject - Organic Chemistry: Some Basic Principles and Techniques, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Organic chemistry: some basic principles and techniques case study questions with answer key.

11th Standard CBSE

Final Semester - June 2015

It is essential to purify an organic compound. The method used for purification depends upon nature of compound and impurity present in it. The common methods to purify a solid is sublimation and crystallisation. Crystallisation is most common method applicable to most of solid organic compounds. Liquids are purified by simple distillation, fractional distillation, distillation under reduced pressure, steam distillation. Differential extraction is used to extract organic compound from aqueous solution. Chromatography is used to separate coloured substances from plants. Column chromatography, thin layer chromatography and partition chromatography are types of chromatography used for isolation and purification of organic compounds. (a) Which method is used to purify camphor? (b) How is unwanted colour from organic compounds removed? (c) How is chloroform (Boiling point 334 K) and aniline (b.pt. 457 K) are separated? Why? (d) Which is condensed first in fractionating column, vapours of higher boiling point liquid or lower boiling point liquid? (e) How is glycerol purified? (f) How is Aniline purified? Why? (g) How are o-nitrophenol and p-nitrophenol separated? Why?

Organic compounds are formed by covalent bonding. The nature of covalent bonding can described with the help of hybridisation, sp, Sp 2 and sp 3 . The structure and reactivity depends upon type of bonds present in organic compounds. Organic compound can be represented by various structural formulae, Wedge and Dash formula is 3-D representation. Organic compounds can be classified on the basis of functional groups. Organic reactions mechanism are based on structure of substrate and the attacking reagent. The intermediate formed can be free radical, carbocation, carbanion or carbene. The attacking reagent can be electrophile or nucleophile. The inductive, electromeric, resonance and hyper conjugative effect may help in polarisation of covalent bond. Organic reactions may be regarded as substitution, addition, elimination and rearrangement, oxidation and reduction reaction. After the compound is obtained in pure state, qualitative analysis helps to detect elements present in organic compounds whereas quantitative analysis helps to find percentage of various elements. Dumas and Kjeldahl method help to determine percentage of nitrogen, Carius method for halogens and sulphur. Carbon and hydrogen are estimated by the amount of CO 2 and H 2 Oformed. Phosphorus estimation is done by oxidising it to H 3 PO 4 , sulphur to H 2 SO 4 , The percentage of oxygen is determined by taking the difference of 100 and percentage of all elements. Empirical formula gives simple ratios of elements whereas molecular formula gives exact number of atoms of each element present in a compound. (a) What are free radicals? (b) Write the order of stability of carbocation. (c) An organic compounds has 8% sulphur. What is minimum molar mass of compound? (d) If C is 75%, H = 25%, what is molecular formula of compound? (e) In estimation of sulphur, which compound of sulphur is formed? (f) f Lassaigne extract of organic compound give blood-red colour with FeCI 3 , what does it show? (g) Why should we add HNO 3 to Lassaigne extract before testing for halogens.

*****************************************

Organic chemistry: some basic principles and techniques case study questions with answer key answer keys.

(a) Sublimation. (b) It is removed by adsorbing over activated charcoal. (c) Simple distillation because there is large difference in boiling points. (d) Vapours of higher boiling point liquid are condensed first. (e) It is purified by distillation under reduced pressure. (f) It is purified by steam distillation because it is steam volatile and insoluble in water. (g) They are separated by steam distillation because o-nitrophenol is steam volatile where as p-nitrophenol is not.

(a) Free radicals are atoms or group of atoms having one electron in excited state (b) 3° > 2° > 1°. (c) 400 g mol -1 , because it will contain at least one atom 'S', i.e., 32 g of sulphur. (d) CH 4 . (e) It is estimated by taking mass of BaSO 4 formed. (f) It is done to convert NaCN to HCN and N 2 S to H 2 S if nitrogen and sulphur are present because they will interfere with the AgNO 3 test for halogens, H 2 S and HCN are removed as gases.

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Important Questions for CBSE Class 11 Chemistry 2024-25

• Class 11 Important Question

CBSE Class 11 Chemistry Chapter wise Important Questions - Free PDF Download

Class 11 Chemistry has an excellent syllabus that covers various important topics. The chapters are all important to prepare to develop a conceptual foundation and to score well in the exams. You will need the Important Questions for Class 11 Chemistry Board Exam to prepare well. These questions have been formulated by the subject experts of Vedantu along with the solutions. Develop your concepts and knowledge about Class 11 Chemistry by practising these questions and score well in the exams.

Most students belonging to the science stream after the CBSE 10 board exam possess an interest in Chemistry . This subject provides you with a chance to work out different chemical hypotheses. You can learn about various in-depth problems of both inorganic and organic Chemistry. However, you should be aware that the syllabus of Class 11 Chemistry is vast and you need to solve the important questions in order to get good marks. You can refer to the Vedantu site and download the important questions for Class 11 Chemistry . This can also help you grip the basics of different chapters in the 12th standard and enhance the score in the final board examination of the CBSE board . 

Download CBSE Class 11 Chemistry Important Questions 2024-25 PDF

Also, check CBSE Class 11 Chemistry Important Questions for All chapters:

Chapter wise Important Questions for CBSE Class 11 Chemistry

Important Questions for Class 11 Chemistry CBSE Board

Importance of class 11 chemistry syllabus.

As mentioned earlier, all the chapters present in the Class 11 Chemistry syllabus act as the base of the branches that are in the next higher standard. If you have thoroughly studied Chemistry till the 10th standard , you can observe a notable difference in the chapters. You will read the same thing but can expect to find new concepts and scientific facts. The 11th Chemistry essential questions can help you to seek the answers in the right way. 

Important Chapters in Class 11 Chemistry Syllabus

Some important chapters in Class 11 Chemistry are mentioned here in detail. You need to study these chapters thoroughly to grip the concepts to have a clear understanding of the same.

Atomic Structure

The Atomic Structure is one of the entry-level chapters of Class 11 Chemistry . You need to study it thoroughly after grasping the basic concepts. Here you can learn in-depth about the atoms and practice to draw the diagrams of the particles. 

Chemical Bonding and Structure of Molecule

The chemical bonding and molecular structure chapter can help you understand how the compounds make bonds and retain their stability. The Chemistry important question Class 11 is based majorly upon the concepts present in this chapter and will help you to score good marks.

Hydrocarbons: The chapter on Hydrocarbons consists of all primary concepts of organic Chemistry. You can thoroughly learn about organic structures and IUPAC names.

Concepts of Organic Chemistry: The chapter named Concepts of Organic Chemistry mainly covers the organic structures’ reactions. You can also learn about how these compounds change bondings while reacting with other compounds. Moreover, you can gain knowledge about these compounds’ properties and whether it can help you in any way or not.

Significance of CBSE Class 11 Chemistry Term 2 Important Questions with Solutions

There are 14 chapters that cover the important topics in this syllabus. The prime topics are the modern periodic table, chemical reactions, states of matter, thermodynamics, chemical equilibrium, etc. Hence, all the chapters are important to study and complete the syllabus .

Students will need the complete material to prepare all these chapters well. Apart from the exercise questions, they will need a set of important questions to solve and test their knowledge. The questions compiled by the subject experts of Vedantu will fit into this requirement quite well. These questions come with respective solutions. Solving these questions will add a new perspective on the exam format and probable question patterns followed by the CBSE board.

The solutions will enable the students to find out the right answers to these questions. They will find the exact format of the answers to follow and learn how to accurately answer such questions and score well in the exams.

Advantages of Class 11 Chemistry Important Questions with Answers PDF

Solving these questions will help you realise how the concepts and scientific principles explained in a chapter of Class 11 Chemistry have been used to answer. You will also become capable of following the same format and score more in the exams.

The answering formats are simple. They are designed in such a way that you can easily comprehend and practice them. This step will sharpen your Chemistry answering skills.

Resolving doubts will not be a problem anymore when you have the solutions added to your study material. Focus on how the concepts and formulas are used to answer such questions and resolve doubts instantly.

Assess your preparation level by solving these questions at home. Compare your answers to the solutions and find out which part of the specific chapters you need to study more.

Solved Examples

Q1. what do you mean by isobars.

Answer. Elements with the same mass number but different atomic numbers are known as isobars. In the case of isobars, the number of nucleons of the components remains the same, but the number of electrons differs. In accordance with the isobars and their positions, the properties of elements are determined. 

Q2. What is the Definition of Isotone?

Answer: The elements in which there are similar amounts of protons, but the number of neutrons differs as isotones. These elements are considered for studying nuclear structure. 

Download CBSE Class 11 Chemistry Important Questions with Answers PDF 2024

Get the free PDF versions of these important questions and answers for all the chapters of Class 11 Chemistry. Take your practice sessions to the next level by following the question patterns and answering formats in the solutions. Sharpen your answering skills and increase your knowledge to score well in the Class 11 Chemistry exams.

Other CBSE Class 11 Important Questions Links

Important Related Links for CBSE Class 11 Chemistry

FAQs on Important Questions for CBSE Class 11 Chemistry 2024-25

1. Why do you need to choose Vedantu for the questions of 11th standard Chemistry?

2. What is the right way to prepare the questions of Chemistry?

Ans: The right way to prepare the questions of Chemistry starts with going through the chapter thoroughly. Then you should go through the numerical and reactions properly. Make sure you are following conceptual study and not blindly memorising the reactions and facts. You can always draw the diagrams wherever needed and you should learn all the labels properly. While studying in the 11th standard, you should remember that you are preparing for the boards and the joint entrance examination. So, you must make proper notes and read them at regular intervals. This can increase the chances for you to get higher ranks in the entrance examinations.

3. How many chapters are there in Class 11 Chemistry? 

Ans: In Class 11, there are 14 chapters. These chapters are listed as- 

Some Basic Concepts of Chemistry

Structure of Atom

Classification of Elements and Periodicity in Properties

Chemical Bonding and Molecular Structure

States of Matter

Thermodynamics

Equilibrium

Redox Reactions

The s-block elements

The p-block elements

Organic Chemistry- Some Basic Principles and Techniques

Hydrocarbons

Environmental Chemistry

These solutions are available on Vedantu's official website( vedantu.com ) and mobile app free of cost.

4. What are the benefits of solving Class 11 Chemistry sample papers?

Ans: Solving Class 11 Chemistry sample papers help you in several ways to write impressive answers in your answers and getting a good result. Here are a few benefits of solving Class 11 Chemistry. 

Effective Revision- Class 11 Chemistry sample papers will help you to revise the entire syllabus before the examination. These sample papers cover each and every important chapter of the book along with their concepts. Thus, are good study materials to consider during exams. 

Time Management- Sample papers will enable you to get familiar with the latest exam paper pattern of chemistry. If you solve them then you’ll be able to know your accuracy and speed of solving the paper. 

5. What is the chapter-wise weightage of Class 11 Chemistry? 

Ans: The chapter-wise weightage of Class 11 Chemistry is as follows- 

Some Basic Concepts of Chemistry- three marks

Structure of Atom- six marks

Classification of Elements and Periodicity in Properties- four marks

Chemical Bonding and Molecular Structure- five marks

States of Matter- four marks 

Thermodynamics- six marks

Equilibrium- six marks 

Redox Reactions- three marks

Hydrogen- three marks 

The s-block elements- five marks 

The p-block elements- seven marks 

Organic Chemistry- Some Basic Principles and Techniques- seven marks 

Hydrocarbons- eight marks 

Environmental Chemistry- three marks 

6. How to score 90% above in Class 11 Chemistry? 

Ans: If you are a Class 11 Student and want to achieve more than 90% marks in Class 11 then follow these suggestions. 

Start preparing for the Chemistry examination from the NCERT books. 

Read each chapter clearly. Do not hassle to complete chapters at a faster pace. You must understand that this subject deals with practical concepts, reactions and theory so reading them just for the sake of completing the syllabus won’t help you to get good marks. 

Try to understand concepts and ask questions from your mentors. This way, you can comprehend topics easily.

Q7.  How can you get Important Questions of Class 11 Chemistry?

Ans: You can get the Class 11 Chemistry notes, summaries, important questions, textbook questions and answers on the Vedantu app. You can download the Class 11 Chemistry study material by following the mentioned steps. 

Begin with the link of Important Questions of Class 11 Chemistry. 

This link will take you to the important question webpage of Vedantu. 

There you will find all the chapters that are mentioned in the Chemistry NCERT book. 

Choose those chapters for which you are finding the study material and click on that chapter’s tab. 

Link on the download icon to get the content. 

Important Questions for CBSE Class 11

Cbse class 11 study materials, home tuitions in india.

• Important Questions
• Class 11 - Chemistry
• Chapter 12: Organic Chemistry Basic Principles Technique

Important Questions for Class 11 Chapter 12 - Organic Chemistry Some Basic Principles and Technique.

Class 11 chemistry important questions with answers are provided here for Chapter 12 – Organic Chemistry Some Basic Principles and Technique. These important questions are based on the CBSE board curriculum and correspond to the most recent Class 11 chemistry syllabus. By practising these Class 11 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 11 Annual examinations as well as other entrance exams such as NEET and JEE.

Download Class 11 Chemistry Chapter 12 – Organic Chemistry Some Basic Principles and Technique Important Questions with Answers PDF by clicking on the button below.

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Recommended Video

General organic chemistry class 11 one-shot.

Class 11 Organic Chemistry Some Basic Principles and Technique Important Questions with Answers

Short answer type questions.

Based on the structure (I to VII), Answer the questions given below.

Q1. Which of the above compounds form pairs of metamers?

V and VI, VI and VII and V and VII form pairs of metamers because they have different alkyl chains on either side of ethereal oxygen (-O-).

Q2. Identify the pairs of compounds which are functional group isomers.

I and V, I and VI, I and VII; II and V, II and VI, II and VII; III and V, III and VI; III and VII; IV and V, IV and VI, IV and VI are functional group isomers.

Isomerism is a phenomenon in which two compounds have the same molecular formula, but different structural formulas and such compounds are called isomers. Functional isomers have the same molecular formula but a different functional group.

Q3. Identify the pairs of compounds that represents position isomerism.

I and II, III and IV, VI and VII are position isomers as they have different positions of -OH group and -O- group.

Q4. Identify the pairs of compounds that represent chain isomerism.

Compounds I and III, I and IV, II and III and IV represent chain isomerism.

Q5. For testing halogens in an organic compound with AgNO 3 solution, sodium extract (Lassaigne’s test) is acidified with dilute HNO 3 . What will happen if a student acidifies the extract with dilute H 2 SO 4 in place of dilute HNO 3 ?

On adding dilute H 2 SO 4 for testing halogens in an organic compound with AgNO 3 , a white precipitate of Ag 2 SO 4 is formed. This will interfere with the chlorine test, and this Ag 2 SO 4 may be mistaken for the white precipitate of chlorine as AgCl. Hence, dilute HNO 3 is used instead of dilute H 2 SO 4 .

Q6. What is the hybridisation of each carbon in H 2 C = C = CH 2 ?

In H 2 C = C = CH 2, the terminal carbons are sp 2 hybridised as they form three (two with H and one with C) sigma bonds and one pi bond(between carbons), while the centre carbon is sp hybridised (as it forms two sigma bonds, one with each carbon and two pi bonds, one with each carbon).

Q7. Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?

Since s-electrons are more firmly attached by the nucleus than p-electrons, therefore, electronegativity increases as the s-character of the hybridised orbital increases, i.e., in the order

sp 3 < sp 2 < sp.

Q8. Show the polarisation of carbon-magnesium bonds in the following structure.

CH 3 – CH 2 -CH 2 -CH 2 -Mg-X

CH 3 – CH 2 -CH 2 – δ− CH 2 −Mg δ+ −X

Carbon is more electronegative than magnesium; therefore, Mg has a partial positive charge, and C has a partial negative charge because bonded pairs of electrons are attracted more towards carbon.

Q9. Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by

The two isomers that differ in the functional group’s position on the carbon skeleton are called position isomers, and this phenomenon is position isomerism.

Thus, (A) and (B) may be position isomers. Further, they cannot be regarded as metamers since metamers are those isomers with different numbers of carbon atoms on either side of the functional group. Here, the number of carbon atoms on either side of the sulphur atom (functional group) is the same, i.e., 1 and 3.

Q10, Which of the following selected chains is correct to name the given compound according to the IUPAC system.

The 4-carbon chain is the correct chain for the IUPAC name since it contains the maximum number of functional groups (two) while the rest have only one functional group.

Q11. 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.

In DNA and RNA, nitrogen is present in heterocyclic rings. We cannot use the Kjeldahl method to estimate nitrogen present in rings because nitrogen present in these systems/groups cannot be converted entirely into (NH 4 ) 2 SO 4 during digestion. Therefore, we can not use the Kjeldahl method to estimate nitrogen present in DNA and RNA.

Q12. If a liquid compound decomposes at its boiling point, which method (s) can you choose for its purification. It is known that the compound is stable at low pressure, steam volatile and insoluble in water.

If the compound decomposes at its boiling point, steam is volatile and insoluble in water and stable at low pressure. Steam distillation can be used for its purification.

Q13. Draw the possible resonance structures for CH-O-CH 2 + , and predict which of the structures is more stable. Give a reason for your answer.

The resonance structures are :

Structure (II) is more stable because both the carbon & oxygen atoms have an octet of electrons (shared pairs/lone pair).

Q14. Which of the following ions is more stable? Use resonance to explain your answer.

Structure A is more stable because carbocation A is more Planar, and pi electrons from the ring shift to the side group and are stabilised by resonance. Thus, increasing the stability of carbocation A. Moreover, the double bond is more stable within the ring than the side chain.

Q15. The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can be stored for months. Explain the cause of the high stability of this cation.

Three benzene rings are attached to triphenyl methyl cation’s positively charged carbon atom. Several conjugated structures are possible.

Therefore, triphenyl methyl cation is relatively stable and is isolated from its solution in the solid state. And we can even store some of its for months.

Q16. Write structures of various carbocations that can be obtained from 2-methylbutane.

Arrange these carbocations in order of increasing stability.

We can obtain four structures from 2-methylbutane.

The stability order according to increasing stability is (III) > (II) > (IV) > (I).

Because (I) and (IV) are primary, (II) is secondary carbocation, and (III) is tertiary carbocation.

Q17. Three students, Manish, Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne’s extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid FeSO 4 and dilute sulphuric acid to a part of Lassaigne’s extract. Manish and Rajni obtained

prussian blue colour but Ramesh got red colour. Ramesh repeated the test with the same

Lassaigne extract but again got red colour only. They were surprised and went to their

teacher and told him about their observation. The teacher asked them to think over the

reason for this. Can you help them by giving the reason for this observation? Also, write

the chemical equations to explain the formation of compounds of different colours.

In Lassaigne’s test, SCN – ions are formed due to the presence of sulphur and nitrogen. These give red colour with Fe 3+ ions. With an excess of sodium, the thiocyanate ion, if formed, is decomposed.

NaSCN + 2Na ⎯→ NaCN + Na 2 S

Q18. Name the compounds whose line formulae are given below :

(i) 3-ethyl-4-methylhept-5-en-2-one.

(ii) 3-nitrocyclohex-1-en

Q19. Write structural formulae for compounds named as-

(a) 1-Bromoheptane

(b) 5-Bromoheptanoic acid

(i) CH 3 −CH 2 −CH 2 −CH 2 −CH 2 −CH 2 −CH 2 Br

(ii) CH 3 −CH 2 −CH(Br)−CH 2 −CH 2 −CH 2 −COOH

Q20. Draw the resonance structures of the following compounds.

The resonance structures of the following compounds are.

(i) CH 3 + is the most stable species because replacing H atom by Br (-Inductive effect) increases +ve charge on carbon atom and destabilised the species.

(ii) – CCI 3 is most stable because the -ve charge on the carbon atom is dispersed due to the -I effect of Cl. More Cl atoms correspond to more dispersal of-ve charge and thus more stabilisation.

Q22. Give three points of differences between inductive effect and resonance effect.

Q23. Which of the following compounds will not exist as a resonance hybrid. Give reason for your answer

(i) CH 3 OH

(ii) R−CONH 2

(iii) CH 3 CH=CHCH 2 NH 2

(i) CH 3 OH does not contain pi-electrons; hence, it cannot exist as a resonance hybrid.

(ii) Due to the presence of pi-electrons in C = O bond and lone pair of electrons on N amide can be represented by the following resonating structures.

(iii) In CH 3 CH=CHCH 2 NH 2 , the lone pair of electrons on the nitrogen atom is not conjugated with the pi -electrons. Therefore, resonance is not possible.

Q24. Why does SO 3 act as an electrophile?

Three highly electronegative oxygen atoms are attached to the sulphur atom. It makes sulphur atom electron deficient. Due to resonance, sulphur also acquires a positive charge. Both these factors make SO 3 an electrophile.

Q25. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give a reason for your answer.

Structure I, propenal, is more stable than structure II because it has more covalent bonds in its resonating structures. In contrast, Structure II has only six electrons on its terminal carbon, making it less stable.

Q26. Alcohol (boiling point 97°C) was mixed with a hydrocarbon (boiling point 68°C) by mistake. Suggest a suitable method to separate the two compounds. Explain the reason for your choice.

A mixture of alcohol and hydrocarbon can be separated by simple distillation because both components have a significant difference in their boiling points(more than 25°C). When the temperature is raised to the boiling point of the low boiling liquid, i.e., hydrocarbon, the vapours will consist only of the hydrocarbon without any contamination of alcohol as their boiling points are far apart. We can condense the vapours back to get the hydrocarbon.

Q27. Which of the two structures (A) and (B) given below is more stabilised by resonance?

Both CH 3 COOH and CH 3 COO – ion exhibit resonance as follows:

Structure (B) is more resonance stabilised since both the contributing structures for the carboxylate ion are equivalent.

Structure [A] is comparatively less stabilised since the contributing structures for the carboxylic acid are not equivalent.

Long Answer Type Questions

Q1. What is meant by hybridisation? Compound CH 2 = C = CH 2 contains sp or sp 2 hybridised carbon atoms. Will it be a planar molecule?

The atomic orbitals combine to form a new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation, which can be defined as the process of intermixing the orbitals of slightly different energies to redistribute their energies, resulting in forming a new set of orbitals of equivalent energies and shape.

Carbon atom three are sp 2 hybridized as each one has 3σ bonds while carbon atom 2 has 2σ bonds, and it is sp hybridized. Thus, CH 2 = C = CH 2 (allene) as a whole is non-planar.

Q2. Benzoic acid is an organic compound. Its crude sample can be purified by crystallisation from hot water. What characteristic differences in the properties of benzoic acid and the impurity make this process of purification suitable?

Benzoic acid can be purified by hot water because of the following characteristics.

(i) Benzoic acid is more soluble in hot water and less soluble in cold water.

(ii) Impurities in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation.

Q3. Two liquids (A) and (B) can be separated by the method of fractional distillation. The boiling point of a liquid (A) is less than the boiling point of a liquid (B). Which of the liquids do you expect to come out first in the distillate? Explain.

Fractional distillation is used to separate the components of a mixture if their boiling points differ by 20 ∘ or less. In this method, a fractionating column is used in between the flask and the condenser. The fractionating column aims to provide hurdles for the ascending vapours and provide ample surface area for condensing the high boiling liquid. Consequently, the low boiling liquid (A) vapours will move up while those of the high boiling liquid will condense and fall back into the flask. As a result of this fractionation, liquid A with a lower boiling point will distil first, and liquid B with a higher boiling point afterwards.

Q4. You have a mixture of three liquids A, B and C. There is a large difference in the boiling points of A and the rest of the two liquids i.e., B and C. Boiling points of liquids B and C are quite close. Liquid A boils at a higher temperature than B and C and the boiling point of B is lower than C. How will you separate the components of the mixture. Draw a diagram showing the setup of the apparatus for the process.

Since the boiling point of liquid A is much higher than those of liquids B and C, whose boiling points are pretty close. Therefore, we can separate liquid A by simple distillation. The boiling points of liquids B and C are pretty close but much lower than liquid A. Thus, the mixture of liquids B and C will distil together, leaving behind liquid A. On further heating liquid A will distil over.

Now place the mixture of liquids B and C in a flask fitted with a fractionating column as shown in Fig. Since the b.p. of liquid B is lower than that of liquid C, therefore, on fractional distillation, first liquid B will distil over and then liquid C.

Q5. Draw a diagram of a bubble plate type fractionating column. When do we require such type of a column for separating two liquids? Explain the principle involved in the separation of components of a mixture of liquids by using a fractionating column. What industrial applications does this process have?

Fractional distillation is used for the separation and purification of organic liquids from non-volatile impurities or for separating two or more volatile liquids from a liquid mixture that have boiling points close to each other. Since, in this process, the distillate is collected in fractions under different temperatures, it is known as fractional distillation.

When the liquids present in the mixture have their boiling points close to each other, separating pure liquids is difficult. The fractions collected are always contaminated. Repeated distillations further purify the fractions. To decrease the number of distillations, the separation is done by fitting the distillation flask with a fractionating column which, in turn, is connected to the condenser.

Various types of fractionating columns are given in Fig. Liquids forming a constant boiling mixture (azeotropic mixture) cannot be separated by this method. Fractional distillation is used these days in the industry, primarily in the distillation of petroleum, coal tar and crude alcohol. A mixture of methanol (b.pt. 338K) and propanone (b.pt. 330K) or a mixture of benzene and toluene may be separated by fractional distillation.

Steam distillation is a particular type of separation process for temperature-sensitive materials like natural organic compounds. Some organic compounds tend to decompose at higher temperatures, and ordinary distillation does not suit this purpose. So, steam/water is added to the apparatus, and the temperature of the compounds are depressed, evaporating them at a lower temperature. Once the distillation is accomplished, the vapours are condensed, and constituents are separated at ease.

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Organic Chemistry Class 11 Notes PDF (Handwritten & Short Notes)

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• Eye Catching Format: These notes in the PDF are organised in a way so that students can easily be attracted to the Organic Chemistry notes. These eye-catching formats can attract more and more students to cover the topics of Organic Chemistry. 

How the Organic Chemistry Class 11 Chemistry Notes Are Prepared?

The process of creating a well structured Organic Chemistry Class 11 Chemistry Notes starts with understanding the Class 11 Chemistry Syllabus. After understanding the Syllabus experts refer to the prescribed NCERT Class 11 Chemistry textbooks to begin creating the revision notes of Organic Chemistry.

What is the Importance of Organic Chemistry Class 11 Notes for Class 11 Students?

It is very important for Class 11 students to refer to the Organic Chemistry Class 11 notes. Because with the help of notes, students can understand all topics and concepts of Organic Chemistry in a creative manner. Covering the topic and concept of Organic Chemistry in a creative manner can aid students to improvise their creative skills. The creative skills will help students to answer the questions in the most effective manner.

Tips to Cover the Chapter Organic Chemistry With the Help of Class 11 Notes

Those who are referring to the Class 11 Organic Chemistry Notes to revise their CBSE syllabus can use the below-given tips to utilise the notes better.

• Look Through the Topics and Concepts: Before covering the chapter Organic Chemistry, students need to look through the topics and concepts. It can help students to get a brief idea about facts, figures, topics, concepts, etc. 
• Finish the Chapter: After getting a brief idea, students need to finish off the chapter with the help of Organic Chemistry Class 11 notes. Through this, students can identify all types of topics and can study accordingly. 
• Solving of Doubts: With the help of Class 11 Organic Chemistry notes, students can solve all their doubts. This can help Class 11 students to remove any confusion regarding the chapter.  
• Diagrams are Given: Some of the topics and concepts included in the chapter Organic Chemistry are explained through diagrams. With the help of diagrams, students can also understand difficult topics in a better way. 
• Practising Questions: After covering the Class 11 notes of the chapter Organic Chemistry, students need to practise all kinds of questions. By practising questions, students can improve their marks for the chapter Organic Chemistry. 
• Acts as a Revision Tool: These Class 11 Organic Chemistry notes can act as a revision tool. Revising all the topics while preparing for the chapter Organic Chemistry can help students to improvise their marks. 

Ways Organic Chemistry Class 11 Notes Can Improve One’s Preparation

With the help of Organic Chemistry Class 11 notes, students can change their preparation level. Students can easily track their skills and flaws with the help of Class 11 Chemistry notes. According to strengths and weaknesses, students can improve and change their preparation for the chapter Organic Chemistry. This can help students to solve questions related to the chapter Organic Chemistry. 

What Are Organic Chemistry Class 11 Notes and Why Should You Care?

The Organic Chemistry Class 11 notes are considered to be the short summary of the chapter. With the help of a short summary, Class 11 students can smoothly understand all the topics included in the chapter Organic Chemistry. Better understanding in the chapter can help students to improve their answering methodology and in maintaining the higher accuracy. Because of these reasons the students should care about Class 11 Chemistry Organic Chemistry Notes.

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Class 11 Chemistry Case Study Questions – Chapter 1

  Class 11 Chemistry Case Study Questions

In these questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices: (a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion. (b) Assertion and Reason both are correct statements but Reason is not a correct explanation for Assertion. (c) Assertion is a correct statement but Reason is an incorrect statement. (d) Assertion is an incorrect statement but Reason is a correct statement.

Read the passage given below and answer the following questions: Passage 1. Stoichiometry is a section of chemistry that involves a calculation based on chemical equations. Chemical equations are governed by laws of chemical combination. The mass of reactants is equal to the mass of products. The compound obtained from different methods contains the same elements in the fixed ratio by mass. A mole is a counting unit, equal to 6.022 × 10 23 particles. One mole is also equal to molar mass expressed in grams. One mole of every gas at STP has a volume equal to 22.4 L. The reacting species which are consumed in the reaction completely is called limiting reagent which decides the amount of products formed. (i) Assertion: 22.4 L of N 2 at NTP and 5.6 L of O 2 at NTP contain an equal number of molecules. Reason: Under similar conditions of temperature and pressure, all gases contain an equal number of molecules. Ans (i). Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. Molar volume (at NTP) = 22.4 L Now 22.4 L of N 2 = volume occupied by 1 mole of N 2 = 28 g = 6.023 × 10 23 molecules. Similarly, 1 mole of O 2 = 2 × 16 = 32 g = 6.023 × 10 23 molecules = 22.4 L ∴22.4 L = 6.023 × 10 23 molecules

(ii) Assertion: A reactant that is entirely consumed when a reaction goes to completion is known as a limiting reagent. Reason: The amount of limiting reactant limits the amount of product formed. Ans (ii). Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(iii) Assertion: Both 44 g CO 2 and 16 g CH 4 have the same number of carbon atoms. Reason: Both contain 1 g atom of carbon which contains 6.023 ×10 23 carbon atoms. Ans (iii). Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. 44 g of CO 2 =1 mole ≡1 g atom of C 16 g of CH 4 =1 mole ≡1 g atom of C 1 g atom of C =12 g of C 12 g of C contains 6.023 × 10 23 carbon atoms.

(iv) Assertion: As mole is the basic chemical unit, the concentration of the dissolved solute is usually specified in terms of the number of moles of solute. Reason: The total number of molecules of reactants involved in a balanced chemical equation is known as the molecularity of the reaction. Ans (iv). Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. The number of moles of a solute present in a litre of solution is known as molarity (M). The total number of molecules of reactants present in a balanced chemical equation is known as molecularity. For example, PCl5 → PCl 3 + Cl 2 (Unimolecular) 2HI → H 2 + I 2 (Bimolecular) Therefore, molarity and molecularity are used in a different sense.

Read the passage given below and answer the following questions: Passage 2. Most of the reactions occurring in the laboratories are carried out in solutions. In solutions, generally, two components are present. The one which is a user in amount is called the solute and the other one which is in higher amount is called the solvent. The amount of solute present in a given quantity of solvent or solution is expressed in terms of concentration. The concentration of the solution is expressed in many ways, i.e. in mass percent, volume percent, parts per million, mole fraction, molarity, morality, and normality. (i) Assertion: The mass percentage of an element is used to determine the percentage composition of each element in a compound. Reason: Mass percentage depends on the molar mass of the compound. Ans (i). Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

(ii) Assertion: The sum of the mole fraction of all the components of a solution is unity. Reason : Mole fraction is a temperature-dependent mode of concentration. Ans (ii). The assertion  is correct but the Reason is incorrect.

(iii) Assertion Molarity of a solution represents its concentration. Reason: Molarity is the number of moles of solute per litre of solution. Ans (iii). Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. Concentration means how much amount of substance solute is present in a given volume of a solution now, as the amount can be measured in terms of moles so molarity means the concentration of the solution.

(iv) Assertion: The molality of the solution does not change with a change in temperature. Reason: Molality depends on the mass of the solvent. Ans (iv). The molality of the solution does not change with temperature as it depends on the mass and mass remains unaffected by temperature.

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Important Questions for Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques - PDF Download

class 11 chemistry chapter 8, organic chemistry some basic principles and techniques are listed in this article. These important questions have been developed by the experts in the subject. They cover all the topics given in the chapter. You can download and reference the PDF of the Organic Chemistry class 11 important questions for free.

The main purpose of developing these important questions was to help students in organizing their study materials before the tests. The latest CBSE rules have been used in the development of these important questions. The organic chemistry class 11 important questions in this chapter will help students to understand the kind of questions asked during the competitive exams. For better exam preparation, students can consult additional study materials for Class 11 chemistry chapter 12 found on eSaral.

Class 11 Chemistry Important Questions with Answers for Organic Chemistry Chapter 12 – Some Basic Principles & Technique. These Class 11 Chemistry Important Questions are based on CBSE Board Curriculum and according to the latest Class 11 Chemistry Syllabus. Practicing these Class 11 Important Questions will help you to review all the ideas in this chapter in a short time and will help you in preparing for the Annual examinations of Class 11 and other entrance exams like NEET & JEE.

Important Topics & Sub-topics of Organic Chemistry - Some Basic Principles and Techniques Class 11 Chemistry

Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques Weightage

In class 11 CBSE exam, every subject exam paper consist of 70 marks. Class 11 chemistry CBSE exam paper also consist of 70 marks. Every single chapter of chemistry have different weightage in the CBSE chemistry exam. Class 11 chemistry chapter 8 Organic chemistry - some basic principles and techniques have 11 marks weightage in the class 11 chemistry CBSE exam paper. The weightage percentage of chemistry chapter 8 Organic chemistry - some basic principles and techniques in the CBSE chemistry exam is 16%. Hence, chemistry chapter 8 Organic chemistry - some basic principles and techniques is consider as important and scoring chapter in the class 11 chemistry CBSE exam.

Tips to Solve Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques 

Time management strategies to solve class 11 chemistry chapter 8 important questions.

Effectively managing your time to solve important questions in Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques is crucial for efficient and thorough preparation. Here are some time management strategies to help you tackle these questions:

Understand the Chapter: Before you start solving questions, ensure you have a solid understanding of the chapter's concepts. Read the chapter thoroughly and make concise notes to serve as quick references.

Prioritize Important Questions: Identify the most important questions. Your teacher or study guide may provide a list of such questions. Focus on these first.

Break It Down: Break your study session into manageable chunks. Don't try to solve all the questions in one go. Allocate specific time slots for different types of questions.

Special Focus on Challenging Topics

Special focus on challenging topics in Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques can yield several benefits for students. 

Comprehensive Understanding: Focusing on challenging topics allows you to delve deeper into the subject matter, ensuring a more comprehensive understanding of organic chemistry.

Problem-Solving Skills: Complex topics often involve intricate problem-solving. Tackling challenging areas helps hone your analytical and problem-solving skills, which are valuable not only in chemistry but also in various other fields.

Self Test Checklist

A self test checklist for class 11 chemistry chapter 8 Organic Chemistry - Some Basic Principles and Techniques can help you assess your understanding and preparedness for class 11 annual exam. Here's a checklist of important topics and concepts you should cover in this chapter:

Understand the distinction between organic and inorganic compounds.

Know the importance of carbon and its bonding properties.

Comprehend the concept of carbon's tetravalency and its ability to form diverse organic compounds.

Benefits of Solving Class 11 Chemistry Chapter 8 Important Questions with Answers

Solving important questions with answers from Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques offers several advantages for students. Here are the benefits of using these questions as part of your study and exam preparation:

Conceptual Clarity: Solving important questions reinforces your understanding of the fundamental concepts in organic chemistry. 

Application of Knowledge: By answering these questions, you can apply what you've learned to real-world scenarios and problem-solving. 

Problem-Solving Skills: Practicing important questions enhances your problem-solving skills, enabling you to tackle complex problems with confidence.

Frequently Asked Questions

Question 1: What are the topics of chapter 8 Organic Chemistry - Some Basic Principles and Techniques introduced in Class 11 Chemistry?

Answer 1: Following topics have been introduced in class 11 chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques:

General Introduction

Tetravalence of carbon: shapes of organic compounds

Structural representations of organic compounds

Classification of organic compounds

Nomenclature of organic compounds

Fundamental concepts in organic reaction mechanism

Methods of purification of organic compounds

Qualitative analysis of organic compounds

Quantitative analysis

Question 2: How to score good marks in Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles and Techniques? 

Answer 2: Scoring good marks in Class 11 Chemistry Chapter 8 (Organic Chemistry - Some Basic Principles and Techniques) requires a systematic approach and consistent effort. Start by building a strong foundation in organic chemistry. Understand the basics of bonding, isomerism, and the structure of organic compounds.

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Class 11 Chemistry Case Study Questions Chapter 1 Some Basic Concepts of Chemistry

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In Class 11 Final Exams there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 11 Chemistry Chapter 1 Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 11 Chemistry Case Study Questions Some Basic Concepts of Chemistry to know their preparation level.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 11 Chemistry Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Some Basic Concepts of Chemistry Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 11 Chemistry  Chapter 1 Some Basic Concepts of Chemistry

Case Study/Passage-Based Questions

Case Study 1: The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures. Significant figures are meaningful digits that are known with certainty plus one which is estimated or uncertain. The uncertainty is indicated by writing the certain digits and the last uncertain digit. there are certain rules for determining the number of significant figures. These are stated below:

• All non-zero digits are significant. For Example in 285 cm, there are three Significant figures and in 0.25 mL, there are two significant figures.
• Zeros preceding to first non-zero digit are not significant. such zero indicates the position of the decimal point. thus, 0.03 has one significant figure and 0.0052 has two significant figures.
• Zeros between two non-zero digits are significant. thus, 2.005 has four Significant figures.
• Zeros at the end or right of a number are significant, provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures. But, if otherwise, the terminal zeros are not significant if there is no decimal point.

Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result.

LAWS OF CHEMICALCOMBINATIONS- The combination of elements to form compounds is governed by the following five basic laws.

1) Law of Conservation of Mass-This law was put forth by Antoine Lavoisier in 1789. He performed careful experimental studies for combustion reactions and reached the conclusion that in all physical and chemical changes, there is no net change in mass during the process. Hence, he reached the conclusion that matter can neither be created nor destroyed. This is called the ‘Law of Conservation of Mass’.

2) Law of Definite Proportions-This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight.

3) Law of Multiple Proportions-This law was proposed by John Dalton. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. For example, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.

Hydrogen + Oxygen→ Water

    2g                  16g       18g

Hydrogen + Oxygen → Hydrogen Peroxide

    2g                  32g        34g

Here, the masses of oxygen (i.e., 16 g and 32 g), which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e., 16:32 or 1:2.

4) Gay Lussac’s Law of Gaseous Volumes-This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure.

5) Avogadro’s Law – In 1811, Avogadro proposed that equal volumes of all gases at the same temperature and pressure should contain an equal number of molecules.

In 1808, Dalton published ‘A New System of Chemical Philosophy, in which he proposed the following :

1.) Matter consists of indivisible atoms.

2.) All atoms of a given element have identical properties, including identical mass. Atoms of different elements differ in mass.

3.) Compounds are formed when atoms of different elements combine in a fixed ratio.

4.) Chemical reactions involve the reorganization of atoms. These are neither created nor destroyed in a chemical reaction.

[A[ Multiple Choice Question

1) … refers to the closeness of various measurements for the same quantity.

• a) Accuracy
• b) Reliability
• c) Precision
• d) Uncertainty

Ans – c) Precision

2) Law of Conservation of mass was put forth by ….in 1789.

• a) Joseph Proust
• b) Antoine Lavoisier
• c) Joseph Louis
• d) Gay Lussac

Ans – b) Antoine Lavoisier

3) Which of the following number has two significant figures.

Ans – d) 0.0052

4) … is the agreement of a particular value to the true value of the result.

Ans – a) Accuracy

5) Law of Multiple Proportions proposed by.

• d) John Dalton

Ans – d) John Dalton

Case Study 2: The identity of a substance is defined not only by the types of atoms or ions it contains but by the quantity of each type of atom or ion. The experimental approach required the introduction of a new unit for the number of substances, the mole, which remains indispensable in modern chemical science. A mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) as the number of atoms in a sample of pure 12C weighing exactly 12g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, the number of atoms, molecules, and so forth. The number of entities composing a mole has been experimentally determined to be 6.02214179 × 10 23 .

6.02214179 × 10 23 , a fundamental constant named Avogadro’s number (NA ) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022×10 23 /mol. Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol).

(i) A sample of copper sulfate pentahydrate contains 8.64 g of oxygen. How many grams of Cu is present in the sample? (a) 0.952g (b) 3.816g (c) 3.782g (d) 8.64g

Answer: (b) 3.816g

(ii) A gas mixture contains 50% helium and 50% methane by volume. What is the percent by \ weight of methane in the mixture? (a) 19.97% (b) 20.05% (c) 50% (d) 80.03%

Answer: (d) 80.03%

(iii) The mass of oxygen gas which occupies 5.6 liters at STP could be (a) gram atomic mass of oxygen (b) one-fourth of the gram atomic mass of oxygen (c) double the gram atomic mass of oxygen (d) half of the gram atomic mass of oxygen

Answer: (b) one fourth of the gram atomic mass of oxygen

(iv) What is the mass of one molecule of yellow phosphorus? (Atomic mass of phosphorus = 30) (a)1.993 x 10 -22  mg (b)1.993 x 10 -19  mg (c) 4.983 x 10 -20  mg (d) 4.983 x 10 -23  mg

Answer: (d) 4.983 x 10-23 mg

(v) The number of moles of oxygen in 1L of air containing 21% oxygen by volume, in standard conditions is (a) 0.186 mol (b) 0.21 mol (c) 2.10 mol (d) 0.0093 mol

Answer: (d) 0.0093 mol

Case Study 3: Chemistry plays an important role in human needs for food, health care products, and improving life. Cis platin and taxol are used in chemotherapy, and AZT (Azidothymidine) is used for AIDS. SI units are international units of measurement. The matter is classified into elements, compounds, and mixtures, which can be homogeneous as well as heterogeneous. A mixture can be separated by physical methods, compounds can be separated by chemical methods only. Atomic mass is the average of masses of isotopes depending upon their natural abundance. The empirical formula is calculated with the help of the percentage composition of elements in a compound and molecular mass helps to calculate the molecular formula. A chemical equation must be balanced so as to follow the laws of chemical combination.

Which of the following are used in chemotherapy? A) Taxol B) AZT C) Cis platin D) A and C E) A, B, and C

What are SI units? A) Chemical formulas B) Units of time C) International units of measurement D) Isotopes

How can a compound be separated? A) Physical methods B) Chemical methods C) Both physical and chemical methods D) None of the above

What does the atomic mass of an element represent? A) Mass of a single atom B) Mass of all isotopes combined C) Average mass of isotopes based on natural abundance D) Mass of the most common isotope

Which of the following statements is true regarding a chemical equation? A) It does not need to be balanced B) It must be balanced according to the laws of chemical combination C) It represents only the physical states of the reactants D) It includes only empirical formulas

Hope the information shed above regarding Case Study and Passage Based Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 11 Chemistry Some Basic Concepts of Chemistry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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