class 12 chemistry case study questions pdf

CBSE Expert

CBSE Class 12 Chemistry Case Study Questions PDF

Case studies play a pivotal role in CBSE Class 12 Chemistry, as they enable students to apply theoretical knowledge to real-life scenarios. CBSE Class 12 Chemistry Case Study Questions PDF section introduces the significance of case studies in enhancing analytical skills and understanding complex chemical reactions.

Case studies challenge students to think critically, analyze experimental data, and devise problem-solving strategies. They provide a deeper understanding of chemical principles and their practical applications, fostering a holistic learning experience. Familiarize yourself with the structure of case study questions to streamline your preparation. Each case study presents a unique chemical problem, encouraging students to identify relevant concepts and devise accurate solutions.

Table of Contents

Class 12 Chemistry Case Study Questions

CBSE Class 12 Chemistry question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 12 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 12 Chemistry

Class 12 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 12 Chemistry examinations. Our expert faculty for standard 12 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 12 students understand the concepts and also easy-to-learn solutions.

Tips to Excel in CBSE Class 12 Chemistry Examinations

Excel in your Chemistry exams with these practical tips.

A. Regular Practice with Case Studies

Consistent practice with case study questions enhances your ability to tackle complex problems. Dedicate time to solving various case studies to build confidence.

B. Understanding Analytical Skills

Develop strong analytical skills to approach case studies logically. Break down complex problems into simpler components and analyze them step-by-step.

C. Time Management Strategies

Allocate sufficient time for each case study during the exam. Practice time management in mock tests to complete the paper within the stipulated time.

Best Books for Class 12 Chemistry

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2024 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2024 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2024. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th Chemistry.

Mastering CBSE Class 12 Chemistry case study questions is crucial for excelling in the exams. Embrace case studies as a valuable learning tool, and with practice, you’ll ace your Chemistry exams with confidence.

Benefits of Utilizing the CBSE Class 12 Chemistry Case Study PDF

• Enhanced Learning Experience : The case study PDF offers practical examples and scenarios, making the learning process engaging and relatable for students.
• Application of Theoretical Concepts : It enables students to apply theoretical knowledge to practical situations, honing their problem-solving and analytical skills.
• Real-World Relevance : By connecting classroom learning to real-life applications, students can grasp the practical significance of chemistry in various industries.
• Critical Thinking Development : Analyzing case studies encourages students to think critically and make informed decisions based on chemical principles.
• Exam Preparation : Exposure to case studies aids in better preparation for chemistry examinations by providing a comprehensive understanding of the subject.

The CBSE Class 12 Chemistry case study PDF brings a refreshing perspective to the world of education. By intertwining theoretical knowledge with practical applications, it equips students to face real-world challenges with confidence. The diverse case studies provide invaluable insights, encouraging students to explore chemistry beyond the classroom and make a positive impact on society.

What is the CBSE Class 12 Chemistry case study PDF?

The CBSE Class 12 Chemistry case study PDF is a curated document by CBSE, presenting real-life applications of chemistry concepts for students to understand the subject’s practical relevance.

How does the case study PDF benefit students?

The case study PDF enhances the learning experience, fosters critical thinking, promotes application-based learning, and prepares students for examinations.

Are the case studies diverse in content?

Yes, the case studies cover various branches of chemistry, including organic, inorganic, physical, environmental, and analytical chemistry.

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Download India's best Exam Preparation App Now.

Key Features

• Revision Notes
• Important Questions
• Previous Years Questions
• Case-Based Questions
• Assertion and Reason Questions

No thanks, I’m not interested!

• CBSE Class 12

CBSE Class 12 Chemistry Case Study Based Important Questions with Solutions, Download PDF Here

Cbse class 12 chemistry case study based questions: for the cbse class 12 chemistry exam that is scheduled for february 27, 2024, important case study-based questions in chemistry are provided in this article. also, get information related to the chemistry exam pattern and marking scheme in the article..

CBSE Class 12 Chemistry Paper Pattern

Important Case Study Based Questions for Class 12th Chemistry with Solutions

The following questions are case-based questions. Each question has an internal choice and carries 4 marks each. Read the passage carefully and answer the questions that follow.

Q.1. Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, DNA was first identified in the late 1860s by Swiss chemist Friedrich Miescher. Then, in the decades following Miescher's discovery, other scientists--notably, Phoebus Levene and Erwin Chargaff--carried out a series of research efforts that revealed additional details about the DNA molecule, including its primary chemical components and the ways in which they joined with one another. Without the scientific foundation provided by these pioneers, Watson and Crick may never have reached their groundbreaking conclusion of 1953: that the DNA molecule exists in the form of a three-dimensional double helix. Chargaff, an Austrian biochemist, as his first step in this DNA research, set out to see whether there were any differences in DNA among different species. After developing a new paper chromatography method for separating and identifying small amounts of organic material, Chargaff reached two major conclusions: (i) the nucleotide composition of DNA varies among species. (ii) Almost all DNA, no matter what organism or tissue type it comes from maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is similar to the amount of thymine (T), and the amount of guanine (G) approximates the amount of cytosine (C). In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. This conclusion is now known as "Chargaff's rule." Chargaff’s rule is not obeyed in some viruses. These either have single- stranded DNA or RNA as their genetic material. Answer the following questions: a. A segment of DNA has 100 adenine and 150 cytosine bases. 

a. What is the total number of nucleotides present in this segment of DNA? 

Ans. A = 100 so T = 100

C=150 so  G = 150

Total nucleotides = 100+100+150+150 =500 

b. A sample of hair and blood was found at two sites. Scientists claim that the samples belong to the same species. How did the scientists arrive at this conclusion? 

Ans. They studied the nucleotide composition of DNA. It was the same so they concluded that the samples belong to the same species. 

c. The sample of a virus was tested and it was found to contain 20% adenine, 20% thymine, 20 % guanine and the rest cytosine. Is the genetic material of this virus (a) DNA- double helix (b) DNA-single helix (c) RNA? What do you infer from this data? 

Ans.  A = T = 20%

But G is not equal to C so double helix is ruled out. 

 The bases pairs are ATGC and not AUGC so it is not RNA.

 The virus is a single helix DNA virus.

Importance of Solving Case Study Questions for Class 12 Chemistry

• Case study-based questions in Class 12 are pivotal for good scores. Practicing these questions equips students to tackle these types of questions effectively.
• Solving case study-based questions helps students develop an efficient time management strategy.
• Excelling in Case study-based questions cultivates a holistic subject comprehension essential for future pursuits.

Important Study Resources for CBSE Class 12 Chemistry Exam 2024

• CBSE Class 12 Chemistry Syllabus 2023-24
• CBSE Chemistry Sample Paper Class 12 2023-24 
• Chemistry Additional Questions for CBSE Class 12
• CBSE Chemistry Previous Year Question Paper Class 12
• Top 30+ CBSE Class 12 Chemistry MCQs for Board Exam 2024 with Answers to Secure Good Marks
• CBSE Board Exam Mistakes To Avoid During Exam Preparations For 100% Success
• CBSE Class 12 Preparation Tips: TOP 10 Ways to Score High in CBSE Class 12th Board Exam

Get here latest School , CBSE and Govt Jobs notification and articles in English and Hindi for Sarkari Naukari , Sarkari Result and Exam Preparation . Download the Jagran Josh Sarkari Naukri App .

• India Post GDS Merit List 2024
• India Post GDS Result 2024
• UP Police Constable Admit Card 2024
• UGC NET Exam Analysis 2024
• UP Police Constable Mock Test
• India GDS Merit List 2024 PDF
• UP Police Exam Analysis 2024 Live Updates
• National Space Day Speech
• National Space Day Essay
• National Space Day Quiz
• CBSE Class 12 Study Material

Latest Education News

National Space Day Poster Making and Drawing Ideas for Students with Images

YVU Result 2024 OUT at yvu.edu.in; Download Yogi Vemana University UG and PG Marksheet

UP Police Question Paper 2024: Download Constable Set Wise PDF (All Shifts)

Maharashtra HSC Supplementary Result 2024 Today @mahresult.nic.in, Download Class 12 Marksheet PDF Online

Maharashtra SSC Supplementary Result 2024 Releasing Today: Check MSBSHSE Class 10th Supply Results at mahresult.nic.in, Check by Roll Number

Short and Long Speech on National Space Day Speech 2024

UP Police Constable Exam Review 2024 Live Update: यहाँ देखें कैसा रहा यूपी पुलिस परीक्षा का पहला दिन, क्या रहा पेपर का कठिनाई स्तर

National Space Day 2024: Everything You Need to Know About Chandrayaan 1, 2, and 3

Bihar Sahayak Urdu Anuvadak Admit Card 2024 घोषित, bssc.bih.nic.in से करें डाउनलोड मेन्स हॉल टिकट

NEET UG Counselling 2024 Round 1 Seat Allotment Today at mcc.nic.in, Check Details Here

Maharashtra Bandh 24 August 2024: What’s Open and Closed on August 24 in Maharashtra? Check Status of Schools, Colleges, and Banks

NEET Counselling 2024 Result LIVE: MCC NEET Round 1 Seat Allotment/Seat Matrix Results Today at mcc.nic.in, Check Online by Roll Number

Maharashtra Board Supplementary Result 2024: MSBSHSE HSC, SSC Supply Results Today at mahresult.nic.in, Check Online by Roll Number

ICMAI CMA June 2024 Intermediate, Final Result Declared at icmai.in

National Space Day 2024: Theme and Know All About ISRO Chandrayaan-3 Mission

[PDF लिंक जारी] India Post GDS Result 2024: यूपी, एमपी, बिहार और राजस्थान सहित जीडीएस भर्ती के नतीजे indiapostgdsonline.gov.in पर घोषित

National Space Day Quiz Questions and Answers For School Students

OSSSC Forest Guard Result 2024 OUT: Download Stock Inspector, Forester Marks And What's Next

चंद्रयान 3 पर हिंदी निबंध और भाषण: Chandrayaan 3 Essay in Hindi for School Students

UP Police Constable Admit Card 2024 Out: यूपी पुलिस कांस्टेबल भर्ती परीक्षा का एडमिट कार्ड जारी, ये रहा Direct Link

Case Study Questions Haloalkanes and Haloarenes Class 12 Chemistry

        Case Study Questions Haloalkanes and Haloarenes Class 12 Chemistry

1. Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. The reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state, and polarity of the solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of the solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in the case of S N 1 reactions. (i) Which of the following is most reactive towards nucleophilic substitution reaction? (a) C 6 H 5 Cl (b) CH 2 = CHCl (c) ClCH 2 CH = CH 2 (d) CH 3 CH = CHCl

(ii) Isopropyl chloride undergoes hydrolysis by (a) S N 1 mechanism (b) S N 2 mechanism (c) S N 1 and S N 2 mechanism (d) Neither S N 1 nor S N 2 mechanism

(iii) Tertiary alkyl halides are practically inert to substitution by S N 2 mechanism because of (a) Insolubility (b) Instability (c) Inductive effect (d) Steric Hindrance

(iv) Which of the following is the correct order of decreasing S N 2 reactivity? (a) RCH 2 X > R 2 CHX > R 3 CX (b) R 3 CX > R 2 CHX >RCH 2 X (c) R 2 CHX > R 3 CX > RCH 2 X (d) RCH 2 X > R 3 CX > R 2 CHX

(v) An organic molecule necessarily shows optical activity if it- a) Contains asymmetric carbon atoms b) Is non-polar c) Is non-superimposable on its mirror image d) Is superimposable on its mirror image.

2. Read the passage given below and answer the following questions: The replacement of hydrogen atom in a hydrocarbon, aliphatic or aromatic results in the formation of haloalkanes and haloarenes respectively. Haloalkanes contain a halogen atom attached to sp 3 hybridized carbon atom of an alkyl group whereas haloarenes contain a halogen atom attached to sp 2 hybridized carbon atom of an aryl group. Haloalkanes and haloarenes may be classified on the basis of the number of halogen atoms in their structures as mono, di, or poly halogen compounds and also on the basis of the state of hybridization of the carbon atom to which the halogen atom is bonded. (i) Which of the following halide is 2°? (a) Isopropyl chloride (b) Isobutyl chloride (c) n-propyl chloride (d) n-butyl chloride

(ii) Which of the following is a Gem-dibromide is: (a) CH 3 CH(Br)CH 2 (Br) (b) CH 3 CBr 2 CH 3 (c) CH 2 (Br)CH 2 CH 2 (d) CH 2 BrCH 2 Br

(iii) IUPAC name of (CH 3 ) 3 CCl is: (a) 3-Chlorobutane (b) 2-Chloro-2-methylpropane (c) t-butyl chloride (d) n-butyl chloride

(iv) Which of the following is a primary halide? (a) Isopropyl iodide (b) Secondary butyl iodide (c) Tertiarybutyl bromide (d) Neohexyl chloride

(v) Which one of the following is not an allylic halide? (a) 4-Bromopent-2-ene (b) 3-Bromo-2-methylbut-1-ene (c) 1-Bromobut-2-ene (d) 4-Bromobut-1-ene

3. Read the passage given below and answer the following questions: Alkyl halides are prepared by the free radical halogenation of alkanes, addition of halogen acids to alkenes, replacement of -OH group of alcohols with halogens using phosphorus halides, thionyl chloride, or halogen acids. Aryl halides are prepared by electrophilic substitution to arene. Fluorine and iodides are best prepared by the halogen exchange method. These compounds find wide applications in industry as well as in day-to-day life. These compounds are generally used as solvents and as starting materials for the synthesis of a large number of organic compounds. (i) The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with (a) PCl 5 (b) Dry HCl in the presence of anhydrous ZnCl 2 (c) SOCl 2 in presence of pyridine (d) None of these

(ii) The catalyst used in the preparation of an alkyl chloride by the action of dry HCl on alcohol is (a) anhydrous AlCl 3 (b) FeCl 3 (c) anhydrous ZnCl 2 (d) Cu

(iii) An alkyl halide reacts with metallic sodium in dry ether. The reaction is known as: (a) Frankland’s reaction (b) Sandmeyer’s reaction (c) Wurtz reaction (d) Kolbe’s reaction

(iv) Fluorobenzene (C 6 H 5 F) can be synthesized in the laboratory (a) By direct fluorination of benzene with F 2 gas (b) By reacting bromobenzene with NaF solution (c) By heating phenol with HF and KF (d) From aniline by diazotization followed by heating the diazonium salt with HBF 4

Related Posts

CBSE Sample Paper Session 2022-23 (Chemistry) PDF

Haloalkanes and Haloarenes Class 12 – Questions & Answers

CBSE Important Questions Electrochemistry

Save my name, email, and website in this browser for the next time I comment.

Type above and press Enter to search. Press Esc to cancel.

myCBSEguide

• Case Study Questions Class...

Case Study Questions Class 12 Chemistry

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

CBSE will ask Case Study Questions class 12 Chemistry in session 2020-21. These will be the first two questions in the board exam question paper. The first question will have 5 MCQs out of which students will attempt any 4 questions. The second question will carry 5 Assertion & Reason type questions with the choice to attempt any four.

Case Study Questions

As you know, CBSE will hold exams in May-June this year. There is already a reduction of 30% in the syllabus. Now, the case study questions have been added. So, this year the question paper is going to be a bit easier. Although it is easy yet these case study questions need special attention and regular practice.

We have added around 10 sample questions based on the latest pattern in myCBSEguide App. These all questions include two case study questions.

Class 12 Chemistry Question Bank

If you go through the previous year question papers, you will analyze that many questions are repeated word by word and many others are almost similar. So, it is always recommended to check all questions asked in previous years. This will not only help you to get an idea about the question pattern but also help you to understand the difficulty level of the questions.

myCBSEguide App has the previous year’s question bank. These questions are arranged chapter-wise. If you are preparing a particular chapter, you will get all questions asked from that chapter in the last 10 years.

Case Study Questions Examples

Here are two examples of case study questions. To get more such questions download the myCBSEguide App and browse Sample Papers there.

Read the passage given below and answer any four out of the following questions: Ammonia is present in small quantities in air and soil where it is formed by the decay of nitrogenous organic matter e.g., urea. On a large scale, ammonia is manufactured by Haber’s process. In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. Ammonia is a colourless gas with a pungent odour. Its freezing and boiling points are 198.4 and 239.7 K respectively. In the solid and liquid states, it is associated through hydrogen bonds as in the case of water and that accounts for its higher melting and boiling points than expected on the basis of its molecular mass. Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH– ions. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

• caustic soda
• calcium chloride
• sodium hydroxide
• sodium chloride
• ​200 10 5 Pa
• 400 10 5 Pa
• 100 10 5 Pa
• 300 10 5 Pa
• Mg 2 O 3  + K 2 O
• Al 2 O 3  + K 2 O
• NaO 3  + K 2 O
• None of these
• five bond pair and two lone pair
• four lone pair and one bond pair
• three bond pair and one lone pair
• three bond pair and two lone pair

Read the passage and answer any four out of the following questions: Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be either positive or negative. The charge on the sol particles is due to one or more reasons, viz., due to electron capture by sol particles during electrodispersion of metals. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place. When silver nitrate solution is added to the potassium iodide solution, the precipitated silver iodide adsorbs iodide ions from the dispersion medium, and negatively charged colloidal solution results. acquired a positive or a negative charge by selective adsorption on the surface of a colloidal particle The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The presence of equal and similar charges on colloidal particles is largely responsible for providing stability to the colloidal solution.

In these questions, 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 correct explanation for assertion
• Assertion and reason both are correct statements but reason is not correct explanation for assertion
• Assertion is correct statement and reason is wrong statement
• Assertion is wrong statement but reason is correct statement
• Assertion:  The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution. Reason:  The repulsive forces between charged particles having the same charge prevent them from aggregating and provide stability.
• Assertion: The first layer is mobile in Helmholtz electrical double layer. Reason:  The potential difference between the fixed layer and the diffused layer of opposite charges is called zeta potential.
• Assertion: The sol particle in colloid has a charge. Reason:  The charge in sol is due to electron capture by sol particles during the electrodispersion of metals.
• Assertion:  Methylene blue sol is a negatively charged sol. Reason: When KI solution is added to AgNO 3 solution, positively charged sol formed.
• Assertion:  If FeCl3 is added to an excess of hot water, a positively charged sol of hydrated ferric oxide is formed. Reason: When ferric chloride is added to NaOH a negatively charged sol is obtained with adsorption of OH- ions.

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

• Competency Based Learning in CBSE Schools
• Class 11 Physical Education Case Study Questions
• Class 11 Sociology Case Study Questions
• Class 12 Applied Mathematics Case Study Questions
• Class 11 Applied Mathematics Case Study Questions
• Class 11 Mathematics Case Study Questions
• Class 11 Biology Case Study Questions
• Class 12 Physical Education Case Study Questions

1 thought on “Case Study Questions Class 12 Chemistry”

Answer for questions

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

• New QB365-SLMS
• 12th Standard Materials
• 11th Standard Materials
• 10th Standard Materials
• 9th Standard Materials
• 8th Standard Materials
• 7th Standard Materials
• 6th Standard Materials
• 12th Standard CBSE Materials
• 11th Standard CBSE Materials
• 10th Standard CBSE Materials
• 9th Standard CBSE Materials
• 8th Standard CBSE Materials
• 7th Standard CBSE Materials
• 6th Standard CBSE Materials
• Tamilnadu Stateboard
• Scholarship Exams
• Scholarships

CBSE 12th Standard CBSE Chemistry question papers, important notes , study materials , Previuous Year questions, Syllabus and exam patterns. Free 12th Standard CBSE Chemistry books and syllabus online. Practice Online test for free in QB365 Study Material. Important keywords, Case Study Questions and Solutions. Updates about latest education news and Scholorships in one place.

Latest 12th Standard CBSE Case study Questions Update

12th chemistry the d and f block elements chapter case study question with answers cbse, 12th chemistry haloalkanes and haloarenes chapter case study question with answers cbse, 12th chemistry coordination compounds chapter case study question with answers cbse, 12th chemistry chemical kinetics chapter case study question with answers cbse, 12th chemistry biomolecules chapter case study question with answers cbse, 12th chemistry amines chapter case study question with answers cbse, 12th aldehydes ketones and carboxylic acids chapter case study question with answers cbse, 12th chemistry alcohols phenols and ethers chapter case study question with answers cbse, cbse 12th standard chemistry subject biomolecules chapter case study questions 2021, cbse 12th standard chemistry subject amines chapter case study questions 2021, cbse 12th standard chemistry subject aldehydes , ketones and carboxylic acids chapter case study questions 2021, cbse 12th standard chemistry subject alcohols , phenols and ethers chapter case study questions 2021, cbse 12th standard chemistry subject haloalkanes and haloarenes chapter case study questions 2021, cbse 12th standard chemistry subject coordination compounds chapter case study questions 2021, cbse 12th standard chemistry subject the p-block elements chapter case study questions 2021, cbse 12th standard chemistry subject chemical kinetics chapter case study questions 2021, cbse 12th standard chemistry subject electrochemistry chapter case study questions 2021, cbse 12th standard chemistry subject solution chapter case study questions 2021, cbse 12th standard chemistry subject the solid state chapter case study questions 2021, cbse 12th standard chemistry subject biomolecules chapter case study questions with solution 2021, 12th standard cbse study materials.

12th Standard CBSE Subjects

Study Materials for Other CBSE Board Standards

Class VI to XII

Tn state board / cbse, 3000+ q&a's per subject, score high marks.

Latest CBSE 12th Standard CBSE Study Material Updates

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 12 Chemistry Chapter 14 Biomolecules

• Last modified on: 3 years ago
• Reading Time: 5 Minutes

There is Case Study Questions in class 12 Chemistry in session 2020-21. For the first time, the board has introduced the case study questions in the board exam. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. The first question will have 5 MCQs out of which students will have to attempt any 4 questions. The second question will carry 5 Assertion & Reason type questions with the choice to attempt any four. Here are the questions based on case study.

Case Study Question 1:

Read the passage given below and answer the following questions:

When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.

The denaturation causes change in secondary and tertiary structures but primary structures remains intact.

Examples of denaturation of protein are coagulation of egg white on boiling, curding of milk, formation of cheese when an acid is added to milk.

The following questions are multiple choice question. Choose the most appropriate answer:

(i) Mark the wrong statement about denaturation of proteins. (a) The primary structure of the protein does not change. (b) Globular proteins are converted into fibrous proteins. (c) Fibrous proteins are converted into globular proteins. (d) The biological activity of the protein is destroyed.

(ii) Which statement(s) of protein remain(s) intact during denaturation process? (a) Both secondary and tertiary structures (b) primary structure only (c) secondary structure only (d) tertiary structure

(iii) α-helix and β-pleated structures of proteins are classified as (a) primary structure (b) secondary structures (c) tertiary structure (d) quaternary structure

Cheese is a (a) globular protein (b) conjugated protein (c) denatured protein (d) derived protein

(iv) Secondary structure of protein refers to v(a) mainly denatured of proteins and structures of prosthetic groups (b) three-dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain (c) linear sequence of amino acid residues in the polypeptide chain (d) regular folding patterns of continuous portions of the polypeptide chain.

Download CBSE Books

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Leave a Reply Cancel reply

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

Passage Based Questions: Determination of Income and Employment | Economics Class 12 - Commerce PDF Download

Passage - 1

Direction: read the hypothetical case study given below and answer the questions that follow:.

Passage: An economy is in short-run equilibrium when the aggregate amount of output demanded is equal to the aggregate amount of output supplied. In the ADAS model, the short-run equilibrium is found at the point where the Aggregate Demand (AD) curve intersects the Short-Run Aggregate Supply (SRAS) curve. This equilibrium identifies the equilibrium price level and equilibrium output. A useful way to think about the short-run equilibrium is by considering how much real GDP is being produced and what the Consumer Price Index (CPI) is at that point in time. When an economy is not in equilibrium, prices adjust until the market reaches an equilibrium. In a short-run macroeconomic context, if the aggregate output demanded exceeds the output produced, the scarcity of goods drives up the price level. As prices rise, producers increase output until aggregate production matches aggregate demand.

Q1: What is the main focus of the article above? Ans: The above article is talking about the short-run equilibrium.

Q2: According to the article, under what conditions is the economy in short-run equilibrium? Ans: According to the article, the economy is in short-run equilibrium when the aggregate amount of output demanded equals the aggregate amount of output supplied.

Q3: Based on the information in the article, what would happen if aggregate demand is less than aggregate supply in the economy? Ans: When aggregate demand in the economy is less than aggregate supply, it leads to excess supply, causing prices to fall and potentially creating a deflationary situation. The government may need to increase expenditure to inject cash into the economy.

Passage - 2

Direction:  read the news article given below and answer the questions that follow:.

Passage: The Concept of Investment Multiplier: The theory of multiplier occupies an important place in the modern theory of income and employment. The concept of multiplier was first of all developed by F.A. Kahn in the early 1930s. But Keynes later further refined it. F.A. Kahn developed the concept of multiplier with reference to the increase in employment, direct as well as indirect, as a result of initial increase in investment and employment. Keynes, however, propounded the concept of multiplier with reference to the increase in total income, direct as well as indirect, as a result of original increase in investment and income. Therefore, whereas Kahn’s multiplier is known as ’employment multiplier’, Keynes’ multiplier is known as investment or income multiplier. The essence of multiplier is that total increase in income, output or employment is manifold the original increase in investment. For example, if investment equal to Rs. 100 crores is made, then the income will not rise by Rs. 100 crores only but a multiple of it. If as a result of the investment of Rs. 100 crores, the national income increases by Rs. 300 crores, multiplier is equal to 3. If as a result of investment of ` 100 crores, total national income increases by ` 400 crores, multiplier is 4. The multiplier is, therefore, the ratio of increment in income to the increment in investment. If ΔI stands for increment in investment and ΔY stands for the resultant increase in income, then multiplier is equal to the ratio of increment in income (ΔK) to the increment in investment (ΔI). Therefore k = ΔY/ΔI where k stands for multiplier. Now, the question is why the increase in income is many times more than the initial increase in investment. It is easy to explain this. Suppose Government undertakes investment expenditure equal to Rs. 100 crores on some public works, say, the construction of rural roads. For this Government will pay wages to the labourers engaged, prices for the materials to the suppliers and remunerations to other factors who make contribution to the work of road-building.

Q1: What topic does the article above discuss? Ans: The above article is about the Investment Multiplier.

Q2: How was the concept described in the article originally developed? Ans: The concept of the multiplier was first developed by Kahn, focusing on the increase in employment relative to investment. Keynes expanded this concept by considering the change in income relative to investment.

Passage - 3

Direction:  read the hypothetical case study given below and answer the questions that follow:.

Passage: Central banks use tools such as interest rates to adjust the supply of money to keep the economy stable. Monetary policy has taken on various forms over time, but its core objective remains to regulate the money supply within the economy to achieve a balance between inflation and output stabilization.

Most economists agree that, in the long run, the output of an economy—typically measured by Gross Domestic Product (GDP)—is constant, meaning that any variations in the money supply will only influence price levels. However, in the short term, because prices and wages do not immediately adjust, changes in the money supply can impact the actual production of goods and services. This makes monetary policy, generally managed by central banks like the U.S. Federal Reserve (Fed) or the European Central Bank (ECB), a crucial tool for managing both inflation and economic growth.

During a recession, for instance, consumer spending decreases, business production declines, leading to layoffs and reduced investments, and foreign demand for exports may also diminish. In other words, there is a reduction in overall, or aggregate, demand, which the government can counter with policies that counteract the economic downturn. Monetary policy often serves as this countercyclical tool of choice.

Implementing such a countercyclical policy would stimulate the desired increase in output (and employment). However, since it involves increasing the money supply, it would also lead to rising prices. As the economy nears full capacity, growing demand will drive up input costs, including wages. Workers, with higher incomes, will then spend more on goods and services, further driving up prices and wages, which could lead to higher overall inflation—something policymakers typically aim to prevent. Monetary Policy: Stabilizing Prices and Output

Q1: What is the main focus of the article? Provide a detailed explanation. Ans: The above article is discussing Monetary Policy. Monetary policy is utilized by the central bank to manage inflation and deflation within the economy through various credit control measures.

Q2: According to the article, how does monetary policy manage inflationary pressures? Ans: Countercyclical measures are employed to achieve the desired expansion or contraction of employment and money supply. In the case of inflation, the central bank implements monetary policy to decrease the money supply, which in turn reduces aggregate demand, thereby stabilizing prices in the economy.

Top Courses for Commerce

FAQs on Passage Based Questions: Determination of Income and Employment - Economics Class 12 - Commerce

Objective type Questions

Shortcuts and tricks, previous year questions with solutions, passage based questions: determination of income and employment | economics class 12 - commerce, extra questions, important questions, viva questions, semester notes, mock tests for examination, study material, sample paper, video lectures, past year papers, practice quizzes.

Passage Based Questions: Determination of Income and Employment Free PDF Download

Importance of passage based questions: determination of income and employment, passage based questions: determination of income and employment notes, passage based questions: determination of income and employment commerce, study passage based questions: determination of income and employment on the app.

Welcome Back

Create your account for free.

Forgot Password

Unattempted tests, change country, practice & revise.

Talk to our experts

1800-120-456-456

CBSE Biomolecules Class 12 Notes PDF

Biomolecules Class 12 Chemistry Notes: Free PDF Download

CBSE Biomolecules Class 12 Notes PDF can be downloaded for free from Vedantu. The revision notes for Chapter 14 are written easy to understand language and all the important topics such as carbohydrates, proteins, enzymes, nucleic acids, etc., are covered in these notes.

These revision notes make the best study material for students’ last-minute exam preparations. Download the Biomolecules Chemistry Class 12 Notes PDF for free from Vedantu to develop a better understanding of the chapter.

Key Concepts Covered in Biomolecules Class 12 Chemistry Notes

Before preparing for Class 12 Chemistry Chapter 14 - Biomolecules, it is important to be well aware of the concepts discussed in this chapter. So, here is the list of concepts you will learn from Biomolecules Class 12 Notes PDF.   

14.1 Carbohydrates 

14.1.1 Classification of Carbohydrates

14.1.2 Monosaccharides

14.1.2.1 Glucose

14.1.2.2 Fructose

14.1.3 Disaccharides

14.1.4 Polysaccharides

14.1.5 Importance of Carbohydrates

14.2 Proteins

14.2.1 Amino Acids

14.2.2 Classification of Amino Acids 

14.2.3 Structure of Proteins

14.2.4 Denaturation of Proteins 

14.3 Enzymes

14.3.1 Mechanism of Enzyme Action

14.4 Vitamins

14.4.1 Classification of Vitamins

14.5 Nucleic Acids

14.5.1 Chemical Composition of Nucleic Acids

14.5.2 Structure of Nucleic Acids

14.5.3 Biological Functions of Nucleic Acids 

14.6 Hormones

Download CBSE Class 12 Chemistry Notes 2024-25 PDF

Also, check CBSE Class 12 Chemistry revision notes for other chapters:

Biomolecules Class 12 Chemistry Notes - Basic Subjective Questions

Section-a (1 mark questions).

1. Which carbohydrate is the main constituent of plant cell wall?

Ans. Cellulose.

2. What are the common types of secondary structure of proteins?

Ans.   (i) α-helix

(ii) β-pleated sheet

3. Name two α-amino acids which form a dipeptide which is 100 times more sweet than cane sugar.

Ans. Aspartic acid and phenylalanine.

4. What is the basic structural difference between starch and cellulose?

Ans. Starch is polymer of α-D glucose units whereas cellulose is linear polymer of β-D glucose units.

5. Why are carbohydrates generally optically active?

Ans. Carbohydrates have chiral or asymmetric carbon atom.

6. During curdling of milk, what happens to sugar present in it?

Ans. When milk is curdled, its sugar get oxidize to form lactic acid.

7. If one strand of a DNA has the sequence −ATGCTTCA−, what is the sequence of the bases is the complementary strand. 

Ans. As we know that in DNA molecules, adenine (A) always pairs with thymine (T) and cytosine (C) always pairs with guanine (G). Thus,

Sequence of bases in one strand : ATGCTTCA

Sequence of bases in the complementary strand : TACGAAGT.

8. Which one of the following is a disaccharide: Starch, Maltose, Fructose, Glucose?

Ans. Maltose is a disaccharide as it consists of two α-D-glucose units.

9. Define the following term: Anomers

Ans. The pair of stereoisomers which differ only in the configuration of the hydroxyl group at C1 are called anomers.

10. Define the following term: Polysaccharides

Ans. Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides.

Section-B (2 Mark Questions)

11. What are nucleic acids? Mention their two important functions.

Ans.  Nucleic acids are biomolecules which are found in the nuclei of all living cells. They are polymers composed of repeating units called nucleotides. Nucleic acids play a very vital role in the transmission of the hereditary characteristics and the biosynthesis of proteins.

12. What is the difference between a nucleoside and a nucleotide?

Ans. Nucleoside is formed by condensation of a base with pentose sugar at position 1. When nucleoside is linked to phosphoric acid at 5 position of sugar moiety, we get a nucleotide. Hence, a nucleotide has their units–phosphate group, pentose sugar and a base, whereas nucleoside has two units–pentose sugar and a base.

13. How do you explain the presence of an aldehydic group in a glucose molecule ?

Ans. Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin so it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a six carbon carboxylic acid. Thus indicates that carbonyl group present in glucose is an aldehydic group.

14. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.

Ans. Amino acids behave like salts rather than simple amines or carboxylic acids. This behavior is due to the presence of both acidic and basic groups in the same molecule. In aqueous solution, the carbonyl group loses a proton and amino group accepts a proton to form a zwitter ion. 

15. Name two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet.

Ans. Examples of fat-soluble vitamins are vitamin A and D.

16. List two characteristic features of enzymes.

Ans. (i) Enzymes are highly specific for a particular reaction and for a particular substrate.

(ii) Very small amount of enzyme is required for the process of a reaction.

17. Describe what you understand by primary structure and secondary structure of proteins?

Ans. Primary structure: The specific sequence in which the various amino acids present in a protein are linked to one another is called its primary structure. Any change in the primary structure creates a different protein.

Secondary structure: The conformation of the polypeptide chain is known as secondary structure. The two types of secondary structure are α-helix and β-pleated sheet structure.

In α-helix structure, the polypeptide chain forms all the possible hydrogen bonds by twisting into a right-handed screw (helix) with the –NH groups of each amino acid residue hydrogen bonded to the >C=0 group of an adjacent turn of the helix. In β-pleated sheet structure, all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds.

18. Explain what is meant by a peptide linkage.

Ans. Proteins are the polymers of α-amino acids linked by amide formation between carboxyl and amino group. This is called peptide linkage or peptide bond e.g.,

19. Give the preparation of glucose from starch.

Ans. Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute $H_{2}SO_{4}$ at 393 K under pressure.

$\underset{Starch\;cellulose}{(C_{6}H_{10}O_{5})_{n}}\;+nH_{2}O\xrightarrow[393K;2-3atm]{H^{+}}\;\underset{Glucose}{nC_{6}H_{12}O_{6}}$

20. Explain what is meant by the following: pyranose structure of glucose?

Ans. The six membered cyclic structure of glucose is called pyranose structure (α- or β-), in analogy with heterocyclic compound pyran.

PDF Summary - Class 12 Chemistry Biomolecules Notes (Chapter 14)

1. introduction.

Biomolecules are complex organic compounds that govern the common activities of living organisms. Carbohydrates, proteins, nucleic acids, lipids, and other complex biomolecules make up living systems. Furthermore, simple molecules such as vitamins and mineral salts play an important role in the functions of organisms. 

2. Carbohydrates 

Plants are the primary producers of carbohydrates, which comprise a large group of naturally occurring organic compounds. Cane sugar, glucose, starch, and other common examples The majority of them have the general formula \[{{\text{C}}_x}{{\text{H}}_{2y}}{{\text{O}}_y}\] and were thought to be carbon hydrates, hence the name carbohydrate. The molecular formula of glucose \[\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}} \right){\text{,}}\] for example, fits into this general formula, \[{{\text{C}}_{\text{6}}}\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right){\text{6}}{\text{.}}\] However, not all of the compounds that fit into this formula are carbohydrates. Rhamnose, \[{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{5}}}{\text{,}}\] is a carbohydrate, but it does not fall under this definition. Carbohydrates can be defined chemically as optically active polyhydroxy aldehydes or ketones, or compounds that produce such units upon hydrolysis. Some carbohydrates with a sweet taste are also known as sugars. Sucrose is the most common sugar found in our homes, whereas lactose is the sugar found in milk. 

2.1 Classification of Carbohydrates 

Carbohydrates are classified based on how they react to hydrolysis. They have been broadly classified into the three groups listed below: 

2.1.1 Monosaccharides 

A monosaccharide is a carbohydrate that cannot be hydrolyzed further to produce simpler units of polyhydroxy aldehyde or ketone. Examples include glucose, fructose, ribose, and others.

Monosaccharides are further classified based on the number of carbon atoms and the functional group they contain. When a monosaccharide contains an aldehyde group, it is referred to as an aldose, and when it contains a keto group, it is referred to as a ketose. As shown by the examples, the number of carbon atoms constituting the monosaccharide is also introduced in the name.

Different Types of Monosaccharides

2.1.2 Oligosaccharides 

Oligosaccharides are carbohydrates that, when hydrolyzed, yield two to ten monosaccharide units. They are further classified as disaccharides, trisaccharides, tetrasaccharides, and so on, based on the number of monosaccharides produced during hydrolysis. Disaccharides are the most common of these. The two monosaccharide units formed by hydrolysis of a disaccharide may be identical or dissimilar. Sucrose, for example, yields one molecule of glucose and one molecule of fructose when hydrolyzed, whereas maltose yields only two molecules of glucose.

2.1.3 Polysaccharides 

Polysaccharides are carbohydrates that when hydrolyzed yield a large number of monosaccharide units. Starch, cellulose, glycogen, gums, and other common examples of Polysaccharides do not have a sweet taste, so they are also known as non-sugars. 

2.1.4 Reducing and Non-Reducing Sugars 

Carbohydrates are also classified as reducing or non-reducing sugars. Reducing sugars are all carbohydrates that reduce Fehling's solution and Tollens' reagent. Sugars are reduced by all monosaccharides, whether aldose or ketose.

Nonreducing sugars, such as sucrose, are formed when the reducing groups of monosaccharides, such as aldehyde or ketonic groups, are bonded in disaccharides. Reducing sugars, on the other hand, are sugars that have these functional groups free, such as maltose and lactose.

3. Glucose (Aldohexose) 

3.1 preparation of glucose .

From Sucrose (Cane Sugar): 

When sucrose is boiled in alcoholic solution with dilute \[{\text{HCl or }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{,}}\] glucose and fructose are obtained in equal amounts.

\[\mathop {{{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}}\limits_{{\text{Sucrose}}}  + {{\text{H}}_{\text{2}}}{\text{O}}\xrightarrow{{{{\text{H}}^ \oplus }}}\mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}\limits_{{\text{Glucose}}}  + \mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}\limits_{{\text{Fructose}}} \]

From Starch 

Glucose is commercially obtained by hydrolyzing starch with dilute \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] at 393 K under pressure.

\[\mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{10}}{{\text{O}}_{\text{5}}}}\limits_{{\text{Starch or cellulose}}}  + n{{\text{H}}_{\text{2}}}{\text{O}}\xrightarrow[{393{\text{ K,2 - 3 atm}}}]{{{{\text{H}}^ \oplus }}}\mathop {{\text{n}}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}\limits_{{\text{Glucose}}} \]

3.2 Reactions 

Oxidation Reactions of Glucose

Reduction Reactions of Glucose

Note: Reduction with HI yields n-hexane, demonstrating that all six carbons of glucose are arranged in a straight chain.

Oxime Formation 

Oxime Formation by Glucose

Cyanohydrin Formation

Formation by Glucose Cyanohydrin

Note: Since carbonyl C has become chiral, two diastereomers are formed.

Acetylation

Acetylation of Glucose

Reaction with phenylhydrazine (formation of osazone)

Reaction of Glucose with Phenylhydrazine

Note: To produce osazone, one mole consumes three moles of PhNHNH2. Two moles produce hydrazone, and one is used to convert a CHOH group to a carbonyl group.

3.3 Configuration in Monosaccharides 

D(+)-glucose is the correct name for glucose. The letter ‘D' before the name of glucose represents the configuration, whereas the letter ‘(+)' represents the molecule's dextrorotatory nature. It should be noted that the letters 'D' and 'L' have no bearing on the compound's optical activity. The following is a definition of D– and L– notations. The letters 'D' or 'L' before the name of any compound indicate the stereoisomer's relative configuration. This refers to their affinity for a specific isomer of glyceraldehyde. Glyceraldehyde has one asymmetric carbon atom and comes in two enantiomeric forms, as shown. 

Configuration in Glyceraldehyde

All compounds that can be chemically correlated to the (+) isomer of glyceraldehyde have D-configuration, whereas those that can be correlated to the (–) isomer of glyceraldehyde have L-configuration. The lowest asymmetric carbon atom (as shown below) is compared when assigning the configuration of monosaccharides. As with (+) glucose, –OH on the lowest asymmetric carbon is on the right side, similar to (+) glyceraldehyde, so it is assigned D-configuration. The structure is written in such a way that the most oxidised carbon is at the top for this comparison.

(+) Isomer of D-configuration in Glyceraldehyde and Glucose

3.4 Cyclic Structure of Glucose 

Glucose has been discovered to exist in two different crystalline forms, which are known as \[\alpha \]and \[\beta .\] The \[\alpha \]-form of glucose (m.p. 419 K) is obtained by crystallisation from a concentrated glucose solution at 303 K, whereas the \[\beta .\]-form (m.p. 423) is obtained by crystallisation from a hot and saturated aqueous solution at 371 K. In aqueous solution, both \[\alpha \]-D-glucose and \[\beta \]-D-glucose undergo mutarotation. Although the crystalline forms of \[\alpha \]- and \[\beta \]-D (+)-glucose are quite stable in aqueous solution, each form gradually transforms into an equilibrium mixture of the two. 

This is demonstrated by the fact that the specific rotation of a freshly prepared aqueous solution of \[\alpha \]-D(+)-glucose decreases with time from \[{{ + 111^\circ  to  + 52}}{{.5^\circ ,}}\] whereas that of \[\beta \]-D(+)-glucose increases from \[{\text{ + 19}}{{.2^\circ  to 52}}{{.5^\circ }}{\text{.}}\] Thus, mutarotation refers to the spontaneous change in specific rotation of an optically active compound with time to an equilibrium value. It was discovered that glucose forms a six-membered ring, with –OH at C-5 playing a role in ring formation. This explains the lack of a –CHO group as well as the presence of glucose in two forms, as shown below. These two cyclic forms coexist with an open chain structure.

Cyclic Structure of Glucose

The only difference between the two cyclic hemiacetal forms of glucose is the configuration of the hydroxyl group at C1, which is referred to as anomeric carbon (the aldehyde carbon before cyclisation). Such isomers, namely -form and -form, are referred to as anomers. In analogy with pyran, the six-membered cyclic structure of glucose is known as the pyranose structure (– or –). Pyran is a cyclic organic compound with a ring containing one oxygen atom and five carbon atoms. The Haworth structure more accurately represents the cyclic structure of glucose.

3.4.1 How to Draw a Haworth Projection 

A ring of 6 atoms (5 ‘C' and 1 ‘O') is drawn, with the ‘O' atom in the upper right hand corner, as shown below.

Numbering in the Haworth Projection

Number one is assigned to the carbon atom on the right side of oxygen. Then, in a clockwise fashion, other carbon atoms are assigned numbers 2, 3,........... . In Fischer projection, groups attached to a carbon on the right side are placed below the ring, while groups attached to the carbon on the left side are placed above the ring. However, by convention, the \[{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\] group of carbon 5 is placed above the plane of the ring.

(image will be uploaded soon)

4. Fructose (Ketohexose)  

Fructose has the molecular formula \[{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{,}}\]and based on its reactions, it was discovered to have a ketonic functional group at carbon number 2 and six carbons in a straight chain, just like glucose. It is a levorotatory compound from the D-series. D-(–)-fructose is the correct spelling. Its open chain structure is depicted below.

It also exists in two cyclic forms, which are obtained by adding –OH to the carbonyl group at C5. The resulting ring is a five-membered ring known as furanose, after the compound furan. Furan is a five-membered cyclic compound that contains one oxygen atom and four carbon atoms.

Anomers of Fructose

Haworth structures are used to represent the cyclic structures of two anomers of fructose.

Haworth Structures of Anomers of Fructose

5. Comparison of Glucose And Fructose

6. Disaccharides 

An oxide linkage formed by the loss of a water molecule connects the two monosaccharides. The linkage of two monosaccharide units via an oxygen atom is known as glycosidic linkage. 

6.1 Sucrose 

Sucrose is a common disaccharide that, when hydrolyzed, yields an equimolar mixture of D-(+)-glucose and D-(–)-fructose.

${\begin{matrix} C_{12}H_{22}O_{11} & + & H_{2}O \longrightarrow & C_{6}H_{12}O_{6} & + & C_{6}H_{12}O_{6} \\ Sucrose & & & D-(+)-Glucose & & D-(-)-Fructose \\ [\alpha]_D=+66.5^\circ & & & [\alpha]_D=+52.5^\circ & & [\alpha]_D=-92.4^\circ\end{matrix}} $

A glycosidic linkage between C1 of \[\alpha \]-glucose and C2 of \[\beta \]-fructose holds these two monosaccharides together. Sucrose is a non-reducing sugar because the reducing groups of glucose and fructose are involved in the formation of glycosidic bonds.

Sucrose is a dextrorotatory compound that when hydrolyzed yields an equimolar solution of glucose and fructose.

Because the laevo rotation of fructose is greater than the dextro rotation of glucose, this solution is laevorotatory.

Thus, hydrolysis of sucrose causes a change in the sign of rotation, from dextro (+) to laevo (–), and the product is known as invert sugar, and the phenomenon is known as sugar inversion.

6.2 Maltose

Maltose, another disaccharide, is made up of two -Dglucose units, with C1 of one glucose (I) linked to C4 of another glucose unit (II). Because the free aldehyde group can be produced at C1 of second glucose in solution and has reducing properties, it is classified as a reducing sugar.

6.3 Lactose

It is more commonly known as milk sugar since this disaccharide is found in milk. It is composed of \[\alpha \]-D-galactose and \[\beta \]-D-glucose. Fischer projections of -D-Glucose and -D-Galactose are drawn below:

Except for C-4, the configurations of all the carbon atoms in \[\beta \]-D-Glucose and \[\beta \]-D-Galactose are the same. Such stereoisomers that differ in configuration at only one carbon other than anomeric carbon are referred to as epimers, and the C atom in question is referred to as an epimeric carbon atom. As a result, \[\beta \]-DGlucose and \[\beta \]-D-Galactose are epimers, and C-4 is an epimeric carbon atom. The linkage in lactose is between C1 of galactose and C4 of glucose. As a result, it also reduces sugar.

Glycosidic Linkage in Lactose is Between C1 of Galactose and C4 of Glucose

7. Polysaccharides 

Polysaccharides are composed of many monosaccharide units linked together by glycosidic linkages. They primarily serve as food storage or structural materials.

7.1 Starch 

Plants' primary storage polysaccharide is starch. It is the most important source of nutrition for humans. Cereals, roots, tubers, and some vegetables have a high starch content. It is a -glucose polymer composed of two components: Amylose and Amylopectin. Amylose is a water-soluble component that accounts for about 15-20% of starch. Amylose is a long unbranched chain containing 200-1000 \[\alpha \]-D-(+)-glucose units held together by a C1-C4 glycosidic linkage. Amylopectin is insoluble in water and accounts for approximately 80-85 percent of starch. It is a branched chain polymer of \[\alpha \]-D-glucose units with a chain formed by C1-C4 glycosidic linkage and branching caused by C1-C6 glycosidic linkage.

Amylopectin

7.2 Cellulose 

Cellulose is found only in plants and is the most abundant organic substance in the plant kingdom. It is a major component of the cell wall of plant cells. Cellulose is a straight-chain polysaccharide made up entirely of \[\beta \]-D glucose units linked together by a glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

7.3 Glycogen 

Carbohydrates are stored as glycogen in the animal body. It is also known as animal starch due to its structure, which is similar to amylopectin but more highly branched. It can be found in the liver, muscles, and brain. When the body requires glucose, enzymes convert glycogen to glucose. Glycogen is found in yeast and fungi as well.

7.4 Summary

Note: 

Reducing sugars are all carbohydrates that contain the CHO group, the \[\alpha \]-Hydroxy ketonic group, or the hemiacetal group.

Mutarotation is a phenomenon that occurs in all reducing sugars.

8. Proteins

Protein is derived from the Greek word "proteios," which means "primary" or "of primary importance." Proteins are all polymers of \[\alpha \]-amino acids.

8.1 Amino Acids

Amino acids have both amino (–$NH_2$) and carboxyl (–COOH) functional groups. The amino acids are classified as \[\alpha ,\beta ,\gamma ,\delta \] and so on based on the relative position of the amino group with respect to the carboxyl group. When proteins are hydrolyzed, only \[\alpha \]-amino acids are produced. They may also contain other functional groups.

Amino Acids

All \[\alpha \]-amino acids have innocuous names that usually reflect the compound's or its source's property. Glycine is named after its sweet taste (glykos means sweet in Greek), and tyrosine was first obtained from cheese (tyros means cheese in Greek).

8.2 Classification of Amino Acids 

The relative number of amino and carboxyl groups in an amino acid molecule determines whether it is acidic, basic, or neutral.

Non-essential amino acids are amino acids that can be synthesized in the body. Essential amino acids, on the other hand, are those that cannot be synthesized in the body and must be obtained through diet.

8.3 Properties of Amino Acids 

Amino acids are crystalline, colorless solids. These are water-soluble, high-melting-point solids that act more like salts than simple amines or carboxylic acids. The presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule causes this behavior.

Amino Acids Exhibit Amphoteric Behavior in Zwitter Ionic Form

In aqueous solution, the carboxyl group can lose a proton and the amino group can accept a proton, resulting in zwitter ion, a dipolar ion. This is neutral, but it does have both positive and negative charges. Amino acids exhibit amphoteric behavior in zwitter ionic form, reacting with both acids and bases.

Except for glycine, all other naturally occurring \[\alpha \]-amino acids are optically active due to the asymmetry of the -carbon atom.

These are available in both 'D' and 'L' forms. The majority of naturally occurring amino acids have an L-configuration. The –NH 2 group on the left side is used to represent L-amino acids.

8.4.1 Isoelectric Point 

An amino acid's isoelectric point (pI) is the pH at which it has no net charge. In other words, it is the pH at which the negative charge on an amino acid exactly balances the positive charge. pI (isoelectric point) = pH at which no net charge exists. The pI of an amino acid with no ionizable side chain, such as alanine, is halfway between its two pK a values. This is due to the fact that at pH = 2.34, half of the molecules have a negatively charged carboxyl group and half have an uncharged carboxyl group, and at pH = 9.69, half of the molecules have a positively charged amino group and half have an uncharged amino group. As the pH rises from 2.34, more molecules' carboxyl groups become negatively charged; as the pH falls from 9.69, more molecules' amino groups become positively charged.

As a result, the number of negatively charged groups equals the number of positively charged groups when the two pKa values are averaged.

Isoelectric Point of Alanine

If an amino acid has an ionizable side chain, its pI is calculated by averaging the pKa values of similarly ionizing groups (positive ionizing to uncharged or uncharged ionizing to positive). For example, the pI of lysine is the average of the pKa values of two groups that are positively charged in their acidic form but uncharged in their basic form. The pI of glutamate, on the other hand, is the average of the pKa values of the two groups, which are uncharged in their acidic form and negatively charged in their basic form.

Isoelectric Point of Lysine

Isoelectric Point of Glutamic Acid

8.5 Structure of Proteins – 

The Peptide Bond Proteins are polymers of \[\alpha \]-amino acids that are linked together by peptide bonds or peptide linkage. Peptide linkage is chemically defined as an amide formed between the – COOH group and the –NH 2 group. 

Peptide Linkage in Protein

The reaction between two molecules of similar or different amino acids is initiated by the combination of one molecule's amino group with the carboxyl group of the other. This causes the removal of a water molecule and the formation of a peptide bond –CO–NH–.

8.6 Classification of Proteins 

Proteins can be classified into two types on the basis of their molecular shape. 

Fibrous Proteins  

A fiber like structure is formed when polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. In general, such proteins are insoluble in water.

Keratin (found in hair, wool, and silk) and myosin (found in muscles) are two common examples.

Globular Proteins 

This structure is formed when polypeptide chains coil around to form a spherical shape. These are usually water soluble. Insulin and albumin are two common globular proteins. Protein structure and shape can be studied at four different levels: primary, secondary, tertiary, and quaternary.

Primary Structure 

One or more polypeptide chains can be found in proteins. Each polypeptide in a protein contains amino acids that are linked together in a specific sequence, and this sequence of amino acids is referred to as the protein's primary structure. Any change in this primary structure, i.e., the amino acid sequence, results in a different protein.

Secondary Structure 

A protein's secondary structure is the shape in which a long polypeptide chain can exist. They have been discovered to exist in two different structures: \[\alpha \]-helix and \[\beta \]-pleated sheet. These structures form as a result of the regular folding of the polypeptide chain's backbone caused by hydrogen bonding between the peptide bond's C = O and –NH– groups.

The –NH group of each amino acid residue is hydrogen bonded to the C = O of an adjacent turn of the \[\alpha \]helix in one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix). 

All peptide chains in \[\beta \]-structure are stretched to nearly maximum extension and then laid side by side, held together by intermolecular hydrogen bonds. 

Tertiary Structure 

Protein tertiary structure represents overall folding of polypeptide chains, i.e., further folding of the secondary structure. It produces two major molecular shapes: fibrous and globular. Hydrogen bonds, disulphide linkages, van der Waals forces, and electrostatic forces of attraction are the main forces that keep proteins' 2° and 3° structures stable. 

Quaternary Structure 

Some proteins are made up of two or more polypeptide chains known as sub-units. The relationship of subunits to one another is referred to as quaternary structure.

8.7 Denaturation of Proteins 

A native protein is a protein found in a biological system that has a distinct three-dimensional structure and biological activity. When a protein in its native form is subjected to physical change, such as temperature change, or chemical change, such as pH change, the hydrogen bonds are disrupted. As a result, globules unfold, helixes uncoil, and protein loses biological activity. This is referred to as protein denaturation. A common example of denaturation is the coagulation of egg white when heated. Another example is milk curdling, which is caused by the bacteria present in milk producing lactic acid.

9. Nucleic Acids 

Every generation of every species resembles its ancestors in a variety of ways. How are these traits passed down from one generation to the next? It has been discovered that the nucleus of a living cell is in charge of the transmission of inherent characteristics, also known as heredity. The particles in the nucleus of the cell that are responsible for heredity are known as chromosomes, and they are made up of proteins and another type of biomolecule known as nucleic acids. These are primarily of two kinds: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) (RNA). Because nucleic acids are long chain polymers of nucleotides, they are also referred to as polynucleotides. 

9.1 Chemical Composition of Nucleic Acids 

When DNA (or RNA) is completely hydrolyzed, it produces a pentose sugar, phosphoric acid, and nitrogen-containing heterocyclic compounds (called bases). The sugar moiety in DNA molecules is \[\beta \]-D-2-deoxyribose, whereas the sugar moiety in RNA molecules is \[\beta \]-D-ribose.

Structure of Beta -D-ribose and Beta -D-2-deoxyribose

Adenine (A), guanine (G), cytosine (C), and thymine are the four bases found in DNA (T). The first three bases of RNA are the same as those of DNA, but the fourth is uracil (U).

Structure of Adenine (A) and Guanine (G)

Structure of Cytosine(C), Thymine(T) and Uracil(C)

9.2 Structure of Nucleic Acids 

A nucleoside is a unit formed by the attachment of a base to the 1' position of a sugar. The sugar carbons in nucleosides are labeled as 1', 2', 3', and so on. To differentiate these from the bases. A nucleotide is formed when a nucleoside is linked to phosphoric acid at the 5'-position of the sugar moiety.

Structure of (a) a Nucleoide and (b) a Nucleotide

The phosphodiester linkage between the pentose sugar's 5' and 3' carbon atoms connects nucleotides. The following is a simplified version of the nucleic acid chain.

Simplified Version of the Nucleic Acid Chain

The primary structure of a nucleic acid is information about the sequence of nucleotides in its chain. Secondary structure is also present in nucleic acids. The double strand helix structure of DNA was proposed by James Watson and Francis Crick. Two nucleic acid chains are wrapped around each other and held together by hydrogen bonds formed by base pairs. Because hydrogen bonds form between specific pairs of bases, the two strands are complementary to one another. Adenine and thymine form hydrogen bonds, whereas cytosine and guanine form hydrogen bonds. There are helices in the secondary structure of RNA that are only single stranded. They can fold in on themselves to form a double helix structure. There are three types of RNA molecules, each of which serves a different purpose. They are known as messenger RNA (m-RNA), ribosomal RNA (r-RNA), and transfer RNA, respectively (t-RNA).

Double Helix Strand of DNA

9.3 DNA Vs RNA

9.4 biological functions of nucleic acids .

DNA is the chemical basis of heredity and can be thought of as a repository of genetic information. Over millions of years, DNA has been solely responsible for maintaining the identity of various species of organisms. During cell division, a DNA molecule can self-replicate, and identical DNA strands are transferred to daughter cells. Protein synthesis in the cell is another important function of nucleic acids. Actually, proteins are synthesised by various RNA molecules in the cell, but the message for a specific protein's synthesis is present in DNA. 

10. Enzymes 

Enzymes function as biological catalysts. Enzymes are all globular proteins chemically. The following are some important enzymes and their functions.

11. Vitamins

12. Hormones 

Hormones are biomolecules that are produced in the ductless (endocrine) glands and transported by the bloodstream to various parts of the body where they control various metabolic processes. These are required in minute amounts and, unlike fats and carbohydrates, are not stored in the body but are constantly produced. 

12.1 Steroidal Hormones

12.2 Peptide Hormones

12.3 Amine Hormones

13. Test For Biomolecules 

13.1 test of carbohydrates .

13.1.1 Molish Test 

The Molisch test detects all types of carbohydrates, including monosaccharides, disaccharides, and polysaccharides. Molisch reagent (1 percent alcoholic solution of \[\alpha \]-naphthol) is added to an aqueous carbohydrate solution, followed by conc. \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] along the test tube's sides. At the intersection of the two layers, a violet ring forms.

13.2 Test for Proteins 

13.2.1 biuret test .

An alkaline solution of a protein when treated with a few drops of 1% \[{\text{CuS}}{{\text{O}}_{\text{4}}}\] solution, produces a violet colouration. The colour is due to the formation of a coordination complex of \[{\text{C}}{{\text{u}}^{{\text{ + 2}}}}\] with carbonyl group and –NH– groups of the peptide linkages.

13.2.2 Xanthoproteic Test 

When a protein is treated with concentrated \[{\text{HN}}{{\text{O}}_{\text{3}}}{\text{,}}\] it turns yellow. This test is performed by a protein composed of \[\alpha \]-amino acids containing a benzene ring, such as tyrosine, phenylalanine, and others, and the yellow color is caused by benzene ring nitration. An important example of this test is when concentrated \[{\text{HN}}{{\text{O}}_{\text{3}}}{\text{,}}\] is spilled on your hands, the skin turns yellow due to nitration of the benzene ring of the amino acids in your skin's proteins. 

13.2.3 Millon’s Test 

Millon's reagent is a nitric acid solution of mercurous nitrate and mercuric nitrate with little nitrous acid. When Millon's reagent is added to an aqueous protein solution, a white ppt. is formed. All proteins containing phenolic \[\alpha \]-amino acids, such as tyrosine, pass this test. As a result, gelatin that does not contain phenolic \[\alpha \]-amino acids fails this test.

13.2.4 Ninhydrin Test 

A blue-violet color is produced when proteins are boiled in a dilute aqueous solution of ninhydrin (2, 2-dihydroxyindane-1,3-dione). This test is provided by all \[\alpha \]-amino acids. Proteins and peptides both pass this test because they give \[\alpha \]-amino acids when hydrolyzed.

Sub-Topics Covered Under Biomolecules

Let us find out the sub-topics that fall under the topic, Biomolecules.

Carbohydrates - This refers to sugars, fibres, and starches which exist in different kinds of grains, fruits, vegetables, and milk products

Monosaccharides - It refers to the simple sugar, and it happens undoubtedly to be the simplest form of sugar

Disaccharides - The term disaccharide points to a sugar whose formation takes place when the joining of two monosaccharides takes place by the glycosidic linkage

Polysaccharides - This happens to be a particular type of carbohydrate, whose molecules comprised to be the number of sugar molecules

Structure of Proteins - The students can be able to learn about the structure of proteins in detail here

Amino Acids - These refer to the organic compounds, whereas their combination indeed results in the formation of proteins

Enzymes - These molecules happen to be the biological molecules that enhance up the rate of all chemical reactions which occurs inside the cells

Vitamins - Vitamins means the organic compounds which are vital in small quantities for the sustenance of life

Nucleic Acids - These refer to either biopolymers or small biomolecules whose presence is crucial to all forms of life

Structure of Nucleic Acids - Here, the students will study the structure of nucleic acids in detail under this sub-unit

Importance of Biomolecules in Competitive Exams

Biomolecules and Polymers are very important chapters in the JEE Main Chemistry. In general, 1 question is asked from each chapter, Biomolecules and Polymers, which carries about 8 marks in total. Thus, their overall weightage in JEE Main is about 2-3 per cent, and also these chapters cannot be skipped. So, most of the questions will be asked from Biomolecules and Polymers, and it is relatively easy to score. 

Some of the topics tested under the Polymers can be listed as - About Polymerization, Different Types of Polymers, Lipids, Proteins, and Carbohydrates, are a few topics tested under Biomolecules. Reading to the study or revision notes for biomolecules and polymers will help you cover JEE Main Chemistry Syllabus.

Tricks to Solve the Questions from Biomolecules and Polymers

As these questions are mostly theory-based, it is highly recommended you draw all the diagrams and list the equations on paper of every concept on multiple times

Attempt the questions from within a specific text and those after going through the chapters

Primarily focus on preparing from NCERT books. This is because most of the questions will be asked in JEE Main Chemistry belongs to these

While attempting the MCQs (Multiple Choice Questions), take a moment to mentally visualize the concepts, equations, and diagrams that you have learnt pertaining to a specific question. Then, go for the correct answer

Benefits of Class 12 Chemistry Revision Notes on Chapter 14 - Biomolecules

The following are some of the prime benefits of Class 12 Chemistry Revision Notes for Chapter 14 - Biomolecules.

The revision notes include detailed explanations of all the concepts of Chapter 14 Biomolecules of CBSE Class 12 Chemistry.

Going through the revision notes on CBSE Class 12 Chemistry Chapter 14 Biomolecules will help improve students’ understanding of the topics. 

These revision notes are prepared by our Chemistry experts having years of experience in this field to provide quality study material.

Each and every concept is explained in a step-by-step manner, which will help the students to secure good marks in the Class 12 board exams.

Tricks to solve the questions from Biomolecules and Polymers are also covered in these revision notes.

Why Choose Vedantu?

Because, Vedantu is a more significant online platform that provides all types of information including the revision notes, question papers of previous years, and many more. You can find the direct documents in the PDF format such as class 12 chemistry chapter 14 revision notes, revision notes class 12 chapter 14, revision notes class 12 chemistry chapter 14, class 12 notes biomolecules and more. These are some examples of the keywords where you can find many like these and can download them on the official website of Vedantu.

Related Topic Links for Biomolecules Class 12 Notes PDF

Explore a comprehensive collection of resources on Biomolecules Class 12 Notes PDF through our curated list of related topics. Dive deeper into the fascinating world of biomolecules, including carbohydrates, proteins, lipids, nucleic acids, and enzymes. Gain insights into their structures, functions, and biological significance, as well as their roles in living organisms. Whether you're a student seeking to enhance your understanding or an educator looking for supplementary materials, these links provide valuable resources to support your learning journey.

Carbohydrates

Nucleic Acids

DNA and RNA structure

Michaelis-Menten Kinetics

Factors affecting enzyme activity

For More Information on Biomolecules Class 12, Watch Youtube Video on 

You Can Also Watch 

Important Biomolecules Related Links

Explore a compilation of valuable links related to Biomolecules topic, offering comprehensive study materials, solved examples, and practice questions for Class 12 students studying chemistry.

Important Class 12 Study Materials Links

Find a curated selection of study resources for Class 12 subjects, helping students prepare effectively and excel in their academic pursuits.

If you are preparing for Class 12 Chemistry and got stuck in Chapter 14, you can always refer to Vedantu’s Revision Notes on Class 12 Chemistry for help. Biomolecules Class 12 Chemistry Notes has been the best study guide for the students for several years. Therefore, by going through the concepts of Vedantu’s revisions notes of Chapter 14 of Class 12 Chemistry, students will be well-prepared for the exam and can ace their board examinations with remarkable marks. You can also find the other study materials like Class 12 NCERT Solutions, important questions, syllabus, and previous years’ question papers from our website. 

FAQs on CBSE Biomolecules Class 12 Notes PDF

1. List Out Some Advantages of Preparing on Biomolecules Class 12 Chemistry Notes?

Since Biomolecules Class 12 Chemistry Notes is one of the most better options for the students in the final preparation stage of the exam, they have the advantages as given below.

Easy and quick understanding of the concepts

Can find the only topics, diagrams, or equations which are seemed to be challenged

Can reach out to necessary topic soon compared to the guide or some other books

Physically, it is handy

2. Mention Some Important Polymers?

Some Important Polymers Include:

Natural rubber

Synthetic rubbers (Neoprene or Polychloroprene)

Buna rubbers

General Purpose Styrene Rubber (GRS) or Buna - s

Nylon 6 or Perelon

Phenol-formaldehyde polymer

3. Where Can I Download the NCERT Class 12 Revision Notes Chemistry Chapter 14 Solution?

There are many different ways to download the revision notes for the chapters like Biomolecules. You can download these from the website of Vedantu by reaching out to www.vedantu.com , from the official website of NCERT and many other third party sites.

4. Mention the topics discussed in the Biomolecules Chemistry Class 12 Notes PDF?

The Biomolecules Chemistry Class 12 Notes PDF includes the following crucial topics:

1. Carbohydrates

Classification of Carbohydrates

Monosaccharides, Disaccharides, Polysaccharides

Glucose- Preparation and Structure

Structure of fructose

Importance of carbohydrates 

2. Proteins

Amino Acids- Classification

Structure of Protein 

Mechanism of an enzyme

4. Vitamins

Classification of vitamins 

5. Nucleic Acids

Chemical composition of Nucleic acid

Structure of Nucleic acids

Biological actions of Nucleic acids

5. How can I make my own revision notes for Class 12 Chemistry Chapter 14?

When you start to prepare your revision notes, there are few things that you should definitely consider. First off, go through all the main topics that you want to include in your notes. Divide all the topics into the main topics and sub-topics. Do not include too much information as it can mess up your entire notes. instead write the points that are significant in bullets. Keep your notes well-organised and underline all the important terms and definitions you come across. Refer to the Revision notes of this Chapter on the Vedantu website for free.

6. How can I prepare Class 12 Chemistry Chapter 14 for the exams?

Chapter 12 ‘Biomolecules’ of Class 12 Chemistry is a very essential chapter included in the syllabus. While preparing this chapter, make sure you are well-versed with all the important terms such as monosaccharides, polysaccharides, enzymes, etc. Since this chapter contains many chemical illustrations, you should focus on practicing all these chemical structures regularly. Practise the questions given at the end of the NCERT textbook. You can take help of revision notes of Chapter 14 for revising the chapter thoroughly from the official website of Vedantu or from the Vedantu app at free of cost.

7. Can Biomolecules Notes Class 12 PDF be used by educators?

Yes, Biomolecules Notes Class 12 PDF can be used by educators as a supplementary teaching resource. It offers clear explanations and illustrative diagrams that can aid in classroom instruction and reinforce concepts covered in the curriculum.

Previous Year Question Papers CBSE Class 12

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Important Question
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• Question Bank
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• JEE Previous Year Paper
• JEE Mock Test
• JEE Crash Course
• JEE Sample Papers
• Important Info
• SRM-JEEE Previous Year Paper
• SRM-JEEE Mock Test
• VITEEE Previous Year Paper
• VITEEE Mock Test
• BITSAT Previous Year Paper
• BITSAT Mock Test
• Manipal Previous Year Paper
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• AP EAMCET Mock Test
• COMEDK Previous Year Paper
• COMEDK Mock Test
• GUJCET Previous Year Paper
• GUJCET Mock Test
• KCET Previous Year Paper
• KCET Mock Test
• KEAM Previous Year Paper
• KEAM Mock Test
• MHT CET Previous Year Paper
• MHT CET Mock Test
• TS EAMCET Previous Year Paper
• TS EAMCET Mock Test
• WBJEE Previous Year Paper
• WBJEE Mock Test
• AMU Previous Year Paper
• AMU Mock Test
• CUSAT Previous Year Paper
• CUSAT Mock Test
• AEEE Previous Year Paper
• AEEE Mock Test
• UPSEE Previous Year Paper
• UPSEE Mock Test
• CGPET Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• Verbal Ability & Reading Comprehension
• Logical Reasoning & Data Interpretation
• CAT Mock Test
• CAT Important Question
• CAT Vocabulary
• CAT English Grammar
• MBA General Knowledge
• CAT Mind Map
• CAT Study Planner
• CMAT Mock Test
• SRCC GBO Mock Test
• SRCC GBO PYQs
• XAT Mock Test
• SNAP Mock Test
• IIFT Mock Test
• MAT Mock Test
• CUET PG Mock Test
• CUET PG PYQs
• MAH CET Mock Test
• MAH CET PYQs
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

Case Study on Coordination Compounds Class 12 Chemistry PDF

Better preparation of Case Study on Coordination Compounds Class 12 Chemistry can help students score good marks in the CBSE Class 12 Board examination. Additionally, it helps build confidence and enables students to deepen their knowledge of Coordination Compounds. Because case-based questions are equally important for learning and board exam preparation, our team has prepared Case Study on Coordination Compounds Class 12 Chemistry in a PDF file for free distribution among students.

Links to download the PDF file of the Coordination Compounds Case Study for Class 12 Chemistry free of cost are mentioned on this page.

Coordination Compounds Case Study for Class 12 Chemistry with Solutions in PDF

The PDF file of the Coordination Compounds Case Study for Class 12 Chemistry with Solutions is a very important study resource that can help students better prepare for the exam and boost conceptual learning.

The solutions are in the hint manner as well as contain full examples too, refer to the link to access the Case Study on Coordination Compounds Class 12 Chemistry with Solutions in PDF - it’s free.

Features of Class 12 Coordination Compounds Case Study Questions  

Certain features of Class 12 Coordination Compounds Case Study Questions are -

• Available for free 24×7
• Based on CBSE Class 12 Chemistry Syllabus
• Case Study Questions on Coordination Compounds with Answers are given

Step-wise Method to Download Case Study on Coordination Compounds Class 12 Chemistry

The below-given steps are helpful for students wanting to download Coordination Compounds Case Study for Class 12 Chemistry with Solutions in a PDF file.

• Search Selfstudys.com using the internet browser
• After loading the website, click on the navigation button

• Click on CBSE from the list of categories

• Find and Click on Case Study

• Select Class 12 and Click on Chemistry to download the Coordination Compounds case study questions in PDF. 

4 Tips to Answer Class 12 Chemistry Coordination Compounds Case Study Questions?

There are 4 solid tips to answer Class 12 Chemistry Coordination Compounds Case Study questions that we are discussing in this section.

• Read the Case Carefully: To start gathering insights from the given case-based questions, it is vital to read the Coordination Compounds case carefully and identify the key facts, figures, and units of measurement. Pay attention to any diagrams or graphs related to Coordination Compounds provided, as they may contain important information.
• Identify the Problem: While reading the Coordination Compounds case it is essential to consider the possible causes and effects. This will help you determine the appropriate approach to solving the problem in Class 12 Chemistry Coordination Compounds.
• Use Elimination Methods Too: Since case study questions of Class 12 Chemistry Coordination Compounds, are often framed in Multiple choice questions, students should have the knowledge of elimination methods in MCQs to better answer the questions.
• Before All, Master the Concept of Coordination Compounds: If the above two methods are not working for you to answer Case Study on Coordination Compounds Class 12 Chemistry then you need to revisit the lesson and master the concepts explained in the Coordination Compounds Class 12 Chemistry.

What is the Benefit of Practising Class 12 Chemistry Coordination Compounds Case Study Questions?

When a student decides to practise Class 12 Chemistry Coordination Compounds case study questions before the board examination they generally get these 3 benefits:

• Quick Conceptual Revision: Nothing is better for revision than solving relevant questions and therefore, those who involve in solving the Coordination Compounds Case study questions before the Class 12 Chemistry exam are able to better revise their conceptual learnings from the lesson.
• Better Board Exam Preparation: No doubt, the more you practise Coordination Compounds case study questions the better your exam preparation will be so, solving Case-based questions prior to the board examination helps a lot in the preparation. It also enables the students to know what are the Coordination Compounds questions which have the possibility to be asked in the board examination.
• Confident in Using Analytical or Critical Thinking Skills: The Case-based questions on Coordination Compounds are all about using analytical or critical thinking skills where students are required to solve problems based on the situations or data given. Thus, practising Case Study on Coordination Compounds Class 12 Chemistry benefits students to feel confident and comfortable in using analytical skills.

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences

One Last Step...

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.