EG_Chemistry-12_(22-07-2022) a

• Tertiary Structure of Proteins: It represents spatial arrangement of these polypeptide chains the overall folding of the polypeptide chain, i.e. or subunits with respect to each other is known further folding of the secondary structure. as quaternary structure of proteins. Types of bonding which stabilises the secondary • Denaturation of Proteins: The phenomena and tertiary structure of proteins are: of loss of biological activity of proteins when a protein in its native form is subjected to a (i) Disulphide bridge (–S—S–) physical change like change in temperature or (ii) H–bonding (–C = O ..... H – N) a chemical change like change in pH is called (iii) Ionic bond (COO– ..... +NH3) denaturation of proteins. During denaturation 2° (iv) Hydrophobic interactions and 3° structures are destroyed but 1° structure remains intact. Examples: Coagulation of egg (v) van der Waals forces white on boiling, Curdling of milk. Due to tertiary structure of proteins two shapes Enzymes are possible for them: • Enzymes are the biological catalysts, which are Fibrous Proteins Globular Proteins required to catalyse various biological reactions, e.g. lactase, amylase, etc. Those enzymes which When the polypeptide This structure is resulted catalyses the oxidation of one substrate with chains runs parallel and when the polypeptide simultaneous reduction of the another substrate are held together by chains of the protein coils are called oxidoreductase enzymes. Improper hydrogen and disulphide around to give a spherical functioning of enzymes may disturb several bonds, then a fibre–like shape. important biological processes and hence leads structure is formed. to body-disorders. S u c h p r o t e i n s a r e These are usually soluble Hormones generally insoluble in in water. water. Hormones are chemical messengers produced by endocrine glands in our body. They are of three types: E x a m p l e s : K e r a t i n Examples: Insulin and (present in hair, wool, albumin. • Steroids: Eg. estrogens, progesterones. silk) and myosin (present • Polypeptides: Eg. insulin, glucogen. in muscles), etc. • Amino acid derivatives: Eg. thyroxin. • Quaternary Structure of Proteins: Some of the proteins are composed of two or more polypeptide chains, referred to as sub-units. The ExErCIsE 10.2 Multiple Choice Questions (MCQs) (1 Mark) 5. What type of bonding provides stability to a-helix 1. Curdling of milk is an example of: structure of protein? (CBSE S.P. 2020-21) (i) Intermolecular H-Bonding (i) breaking of peptide linkage (ii) Disulphide bridge (ii) hydrolysis of lactose (iii) Intramolecular H-Bonding (iii) breaking of protein into amino acids (iv) Ionic bonding (iv) denaturation of protein 6. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence 2. What are the products obtained on digestion of of amino acids is said to be .................... . proteins? (i) Carboxylic acid (ii) Nitrous acid (NCERT Exemplar) (iii) Amino Acid (iv) Steroids (i) primary structure of proteins. 3. Which of the following is the correct zwitter ionic (ii) secondary structure of proteins. structure for glycine? HHHH3233NNNN++–– CCCCHHHH2222CC––HHCC22OO––OOCC+–OOOO+– (iii) tertiary structure of proteins. (i) – (ii) – (iv) quaternary structure of proteins. (iii) – (iv) – 7. Lysine, is ................ . 4. What type of bonding occurs in globular proteins? (i) a-Amino acid (ii) Acidic amino acid (iii) Amino acid synthesised in body (i) Intramolecular H-bonding (iv) b-Amino acid 8. Proteins can be classified into two types on the (ii) Disulphide bridging basis of their molecular shape i.e., fibrous proteins (iii) van der Waals interactions (iv) All of these 350 Chemistry-12

and globular proteins. Examples of globular 16. Assertion : Glycine must be taken through diet. proteins are : Reason : It is a non-essential amino acid. (i) Collagen (ii) Keratin 17. Assertion : The pH at which an amino acid shows no tendency to move towards any of the electrode (iii) Albumin (iv) Myosin when placed in an electric field is called isoelectric point. 9. Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Reason : At isoelectric point, amino acids are in the equlibrium state. Which of the following are acidic? (i) Very Short Answer Type Questions (1 Mark) 18. What are the products obtained on digestion of (ii) proteins? (iii) H2N CH2 CH2 CH2 COOH 19. Differentiate between keratin and insulin. (iv) NH2 CH2 CH COOH [A.I. 2008 (C)] NH2 20. Write the zwitter ionic structure for glycine. 10. Which of the following is fibrous protein? OR (i) Albumin (ii) Keratin Give one example of zwitter ion. (iii) Insulin (iv) Globin [CBSE 2022] 21. Name the type of enzymes present in the saliva of human beings. 11. How many peptide linkages are present in a tripeptide? 22. Define the term: Isoelectric point. 23. What type of bonding occurs in globular proteins? (i) 2 (ii) 3 24. What happens on denaturation of proteins? 25. What is meant by the term biocatalysts? (iii) 4 (iv) 5 [Delhi 2012, Foreign 2014] Assertion-Reason Type Questions (1 Mark) 26. What are hormones? Note: In the following questions, a statement of 27. What kind of bonding helps in stabilising the assertion followed by a statement of reason is given. Choose the correct answer out of the following choices a-helix structure of proteins? on the basis of the above passage. [Delhi 2013] [NCERT] (i) Assertion and reason both are correct statements Short Answer Type Questions-I (2 Marks) and the reason is correct explanation of assertion. 28. Name the essential amino acids which are supplied through diet for the proper health and (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. growth of the body. (iii) Assertion is correct statement but reason is wrong 29. Explain how the amino acids in water acts as statement. amphoteric compound? (iv) Assertion is wrong statement but reason is correct OR statement. Amino acids may be acidic, alkaline or neutral. 12. Assertion: The two strands of DNA are How does this happen? complementary to each other. OR Reason : The hydrogen bonds are formed between What are zwitter ions? [A.I. 2010, Delhi 2010] specific pairs of bases. 30. Explain the term enzymes and give one example. 13. Assertion: Keratin is a fibrous protein. [Delhi 2010, 12] Reason: Insulin is a globular protein. 31. What is the difference between acidic amino acids and basic amino acids? [Delhi 2015] 14. Assertion: The protein loses its biological activity due to change in its secondary and tertiary 32. What is meant by tertiary structure of proteins? structures during denaturation process. 33. Where does the water present in the egg go after boiling the egg? [NCERT] Reason: Primary structure of the protein remains intact during denaturation process. 34. What is secondary structure of proteins? What 15. Assertion : All naturally occurring a-amino acids are the common type of secondary structure of except glycine are optically active. proteins? Reason : Most naturally occurring amino acids [Delhi, Foreign 2011; AI 2008] [NCERT] have L-configuration. (NCERT Exemplar) 35. What is the effect of denaturation on the structure of proteins? [Delhi 2014 (C)] (NCERT) biomoleCules 351

Short Answer Type Questions-II (3 Marks) 37. Define the following as related to proteins: 36. What are essential and non-essential amino acids? (i) Peptide linkage [Delhi 2008; 11, 12] Give one example of each type. (ii) Primary structure [AI 2008; Delhi 2008, 12] [A.I. 2008 (C); Delhi 2010] (iii) Denaturation [A.I. 2015, Delhi 2014; AI 2008; Delhi 2008] [NCERT] Answers 1. (iv) 2. (iii) 3. (i) aqueous solution, the carboxylic group can lose a 4. (iv) 5. (iii) 6. (i) proton and the amino group can accept a proton, 7. (i) 8. (iii) 9. (ii) thus giving rise to a dipolar ion known as zwitter 10. (ii) 11. (i) 12. (i) ion. It is neutral but contains both positive and 13. (ii) 14. (ii) 15. (ii) negative charges. 16. (iv) 17. (ii) 18. a-Amino acids H3 N+ CHRCOOH + OH– Acting Cation as an acid 19. Keratin is a fibrous protein while insulin is a globular protein. H3 + – + H2 O Acting as a base NCHRCOO Amino acid 20. H3N+ —CH2—COO– 21. Amylase. (zwitter ion neutral) (Glycine) H3 O+ + NH2 CHRCOO– 22. The pH at which an amino acid shows no tendency Anion to migrate towards any of the electrode, cathode or anode and exists in the equilibrium state when 30. Most of the reactions in living system proceeds placed in an electric field is known as isoelectric under mild conditions which are catalysed by a group of biomolecules called enzymes. Thus, point. enzymes are called biocatalysts and are highly 23. Intramolecular H-bonding, disulphide bridging, specific in their reactions. For example, maltase dipolar interactions (ionic), hydrophobic enzyme catalyses hydrolysis of maltose into interactions and van der Waals forces. glucose. 24. The protein loses its biological activity due to 31. When the side chain of an amino acid have change in its secondary and tertiary structures. —COOH group then it has more —COOH group However, during denaturation process, primary gathrcioadun.pI—ftthNheHens2iigtdreioscuhspaaaiidnndtoofiatbniesabsmaasiidincotoaambceiidnahocaiadvciecid—a.mNiHno2 structure of the protein remains intact. 25. The catalysts which are produced by the living 32. Further folding, twisting and bending of secondary cells and catalyses the biochemical reactions in structure of proteins in the three dimensional living organisms are known as biocatalysts or space can give rise to a different specific compact enzymes, e.g. maltase, lactase, etc. The action shape of proteins, which is referred to as the of enzymes can be illustrated in the following tertiary structure of proteins. examples: Lactose Lactase Glucose + Galactose 33. When the egg is boiled, denaturation of globular protein present in the egg takes place into an Sucrose Invertase Glucose + Fructose insoluble fibrous protein and then coagulation 26. Hormones are the molecules that act as chemical takes place probably through H-bonding. Water messengers produced by endocrine glands in our present in the egg gets adsorbed or absorbed body. during the denaturation process and so it 27. The a-helix structure of proteins is stabilised by disappears away. the intramolecular H-bonding between carboxyl oxygen (C=O) of one amino acid residue and the 34. The conformation of polypeptide chains attained N—H group of the fourth amino acid unit within as a result of hydrogen bonding is called secondary the same chain. As such the polypeptide chain structure of proteins. There are two common coils up into a spiral structure called the right types of secondary structures, namely a-helix and handed a-helix structure. b-pleated sheet structure. 28. Valine (Val), Leucine (Leu), Isoleucine (Ile), 35. The protein loses its biological activity due to Phenylalanine (Phe), Tryptophan (Trp), Threonine changes in its secondary and tertiary structures. (Thr), Methionine (Met), Lysine (Lys), Arginine The coagulation of egg-white on boiling is a (Arg) and Histidine (His) are the essential amino common example of denaturation of proteins. As acids. They are required in regular diet. a result of denaturation, the globular proteins in egg (soluble in water) gets converted into fibrous 29. In an amino acid, the —NH2 (amino) group is proteins (insoluble in water) and their biological basic and —COOH (carboxyl) group is acidic. In activity is lost. 352 Chemistry-12

36. Essential amino acids: They cannot be O synthesised in the body and must be obtained H2 N—CH2 —C—NH—CH—COOH through diet, e.g. Valine. Peptide bond Non-essential amino acids: They can be synthesised in the body and need not to be taken CH3 through regular diet, e.g. Glycine. Glycylalanine (Gly-Ala) 37. (i) The (—CO—NH—) bond which link together (ii) It refers to the number and sequence of various amino acid units in proteins is called a-amino acids in the polypeptide chains of a protein. Any change in the sequence of peptide linkage. Chemically, a peptide bond a-amino acids creates a different protein. is an amide linkage formed between —COOH (iii) The process that brings about the changes in pH or temperature etc. which leads to group of one a-amino acid and —losNsHo2f group of unfolding of globules or uncoiling of helix i.e., another a-amino acid (by the a water change in 2° or 3° structure of protein due to which it loses its biological activity is called molecule). For example, denaturation of protein. O H2 N—CH2 —C—OH + H2 NCH—COOH –H2 O Glycine CH3 Alanine Topic 3. Vitamins Introduction • Vitamins are the organic compounds required in our diet in very small amounts to perform specific biological functions for normal maintenance, optimum growth and health of an organism. Classification of Vitamins • Vitamins have been classified into two groups depending upon their solubility in water or fat. Fat soluble vitamins Water soluble vitamins These vitamins are soluble in fats and oils, but they are These vitamins are water soluble, but insoluble in fats and insoluble in water. oils. They are stored in liver and adipose (fat storing) tissues. Water soluble vitamins must be supplied regularly in diet Examples: Vitamin A, D, E and K. because they are readily excreted out through urine and cannot be stored (except vitamin B12) in our body. Examples: Vitamin C, B group vitamins. Important vitamins, their sources and their deficiency diseases Name of vitamin Sources Deficiency diseases Vitamin A Fish liver oil, carrot, butter and milk Xerophthalmia (hardening of cornea of eye), Night blindness Vitamin B1 (Thiamine) Yeast, milk, green vegetables and cereals Beri beri (loss of appetite, retarded growth) Vitamin B2 (Riboflavin) Milk, eggwhite, liver, kidney Cheilosis (fissuring at the corners of mouth and lips), digestive disorders and burning sensation of the skin. Vitamin B6 Yeast, milk, egg-yolk, Convulsions (Pyridoxine) cereals and grams Vitamin B12 Meat, fish, egg and curd Pernicious anaemia (RBC deficiency in haemoglobin) Vitamin C (Ascorbic acid) Citrus fruits, amla and green leafy vegetables Scurvy (bleeding gums) Vitamin D Exposure to sunlight, fish and egg-yolk Rickets (bone deformities in children) and osteomalacia (soft bones and joint pain in adults) Vitamin E Vegetable oils like wheat –germ oil, sunflower Increased fragility of RBCs and muscular oil, etc. weakness Vitamin K Green leafy vegetables Increase in time for blood clotting. biomoleCules 353

ExErCIsE 10.3 Multiple Choice Questions (MCQs) (1 Mark) Reason : Sources of thiamine are yeast, milk and green vegetables. 1. Which of the following vitamins is a water soluble vitamin? 5. Assertion: Fat soluble vitamins are stored in liver. (i) Vitamin A (ii) Vitamin K (iii) Vitamin E (iv) Vitamin C Reason : Water soluble vitamins are stored in kidneys. 2. Which of the following vitamins is named as riboflavin? Very Short Answer Type Questions (1 Mark) (i) Vitamin B1 (ii) Vitamin B2 6. Explain why vitamin C cannot be stored in our (iii) Vitamin B6 (iv) Vitamin B12 3. Deficiency of which vitamin increase in time for body? [Delhi 2016] [NCERT] blood clothing? Short Answer Type Questions-I (2 Marks) (i) Vitamin B (ii) Vitamin C 7. What is the significance of vitamin C in our daily life? (iii) Vitamin E (iv) Vitamin K 8. Fresh tomatoes are better source of vitamin C Assertion-Reason Type Questions (1 Mark) than those which have been stored for sometime. Explain why? Note: In the following questions, a statement of assertion followed by a statement of reason is given. 9. Name the central metal ion present in (i) Choose the correct answer out of the following choices Haemoglobin, (ii) vitamin B12. Short Answer Type Questions-II on the basis of the above passage. (3 Marks) (i) Assertion and reason both are correct statements 10. Name two water soluble vitamins, their sources and the reason is correct explanation of assertion. and the diseases caused due to their deficiency in (ii) Assertion and reason both are correct statements diet. [Delhi 2009] but reason is not correct explanation of assertion. 11. How are vitamins classified? Name the vitamin (iii) Assertion is correct statement but reason is wrong responsible for coagulation of blood. statement. [Delhi 2015(C)][NCERT] (iv) Assertion is wrong statement but reason is correct 12. Why vitamin A and vitamin C are essential for statement. us? Give their important sources. 4. Assertion: Deficiency of thiamine causes Beri beri. [Delhi 2014(C)] [NCERT] Answers 1. (iv) 2. (ii) 3. (iv) Name Sources Deficiency of vitamins diseases 4. (i) 5. (iii) Yeast, milk, Vitamin B1 green leafy Beri-beri 6. Vitamin C is mainly constituted of ascorbic acid, vegetables, (paralysis which is water soluble. It can be readily excreted fruits and of legs and cereals general through urine and hence cannot be stored in our weakness) and loss of appetite body. Scurvy 7. Vitamin C increases the resistance of our body Vitamin C Citrus fruits (bleeding towards diseases. It maintains healthy skin like lemon, of gums), amla and Pyorrhoea and also helps in proper healing after cuts and green leafy (loosening and vegetables. bleeding of abrasions. teeth) 8. On prolonged exposure to air vitamin C present in 11. On the basis of solubility in water or fats, Vitamins tomatoes is destroyed due to oxidation by oxygen, are generally classified into two categories: present in air. (i) Water soluble vitamins: These include vitamin B-complex (B1, B2, B6, B12 and 9. (i) Haemoglobin : Fe (ii) Vitamin B12 : Co nicotinic acid, etc.) and vitamin C. 10. Examples of water soluble vitamins includes vitamin B-complex — B1, B2, B6, B12, etc. and vitamin C. 354 Chemistry-12

(ii) Fat-soluble vitamins: These include xerophthalmia (hardening of cornea of eye). vitamins A, D, E and K. Liver cells are rich in fat soluble vitamins. Sources: Carrot, Milk, Butter, Fish liver oil. Vitamin K is responsible for the coagulation Vitamin C: It is essential for us because its of blood. deficiency causes scurvy (bleeding gums), tooth decay, pyorrhoea, etc. 12. Vitamin A: It is essential for us because its deficiency causes night blindness and Sources: Lemon, orange (citrus fruits), amla, tomato and green leafy vegetables. Topic 4. Nucleic Acids Sugar present in DNA is Sugar present in RNA is Introduction • Nucleoside: Base + Sugar b-D-2-deoxyribose. b-D-ribose. • Nucleotide: Base + sugar + phosphate group: Bases: Bases: • Nucleic Acid (or Polynucleotides): It is Purine bases: Adenine (A) Purine bases: Adenine (A) long chain polymer of nucleotides. Nucleotides and and are joined together by phosphodiester linkage between 5¢C and 3¢C– atoms of a pentose sugar. Guanine (G) Guanine (G) Types of Nucleic Acids Pyrimidine bases: Thymine Pyrimidine bases: Uracil (T) and (U) and Deoxyribonucleic Acid Ribonucleic Acid (RNA) (DNA) Cytosine (C) Cytosine (C) It has a double stranded It has a single stranded It mainly occurs in the It mainly occurs in the a-helix structure, in which a-helix structure. two strands are coiled in a nucleus of the cell. cytoplasm of the cell. spiral manner in opposite directions. It is responsible for the It helps in synthesis of transmission of heredity proteins. characteristics. In DNA, two nucleic acid chains are wound around each other which are complementary to each other due to hydrogen bonds between specific pairs of bases. Adenine forms two hydrogen bonds with thymine whereas cytosine forms three hydrogen bonds with guanine. RNA are single stranded though sometimes they fold back on themselves to form a double helix structure. RNA molecules are of 3 types viz. messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA) and they perform different functions. ExErCIsE 10.4 Multiple Choice Questions (MCQs) (1 Mark) 3. Which of the following is the polymer of nucleic acids? 1. Select the nucleic acid which has double helix structure. (i) Nucleosides (ii) Nucleotides (i) Nucleoside (ii) DNA (iii) Bases (iv) Sugars (iii) RNA (iv) None of these 4. Dinucleotides is obtained by joining two nucleotides together by phosphodiester linkage. 2. What type of linkage hold together the monomer Between which carbon atoms of pentose sugars units of DNA? of nucleotides are these linkages present? (i) Disulphide bridging (ii) Ionic bonds [NCERT Exemplar] (iii) van der Waals interaction (i) 5’ and 3’ (ii) 1’ and 5’ (iv) Hydrogen bonds (iii) 5’ and 5’ (iv) 3’ and 3’ biomoleCules 355

5. DNA and RNA contain four bases each. Which of 14. Assertion: Aldopentoses named as ribose and the following bases is not present in RNA? 2-deoxyribose are found in nucleic acids. (i) Adenine (ii) Uracil Reason: Their relative configuration is D-configuration. (iii) Thymine (iv) Cytosine 6. Which of the following are purine bases? 15. A s s e r t i o n : D e o x y r i b o s e , C 5H 10O 4 i s a carbohydrate. (i) Cytosine (ii) Adenine Reason : Carbohydrates are hydrates of carbon (iii) Thymine (iv) Uracil so compounds which follow Cx(H2O)y formula are carbohydrates. 7. Which of the following product would be formed when a nucleotide from DNA containing thymine 16. Assertion : Transcription is the process of copying base undergo hydrolysis? of DNA base sequences into a RNA molecules (m–RNA) (i) Thymine (ii) 2-Deoxy-D-ribose (iii) Phosphoric acid (iv) All of these Reason : The bases of m-RNA are complementary to those of the DNA strand. 8. Which of the following is the possible pairing of bases in DNA double helical structure? 17. Assertion : DNA has unique property of self- replication. (i) Guanine and thymine (ii) Cytosine and thymine Reason : RNA does not replicate on its own. (iii) Adenine and thymine Very Short Answer Type Questions (1 Mark) 18. Name the nucleic acid which has double helix (iv) All of these structure? 9. The sugar present in RNA is (i) D-2-deoxyribose (ii) D-ribose 19. What type of sugars are present in RNA and DNA? [Delhi 2011] (iii) Glucose (iv) Fructose 10. Which part of the cell mainly contains RNA? 20. What products would be formed when a nucleotide (i) Nucleus (ii) Cytoplasm from DNA containing thymine base undergo (iii) Mitochondria (iv) None of these hydrolysis? [NCERT] 11. Ribose in RNA and deoxyribose in DNA differ in 21. When RNA is hydrolysed, there is no relationship the structure around which carbon atom? among the quantities of different bases obtained, (i) C-1 (ii) C-2 what does this fact suggest about the structure of (iii) C-3 (iv) C-4 RNA? [NCERT] Assertion-Reason Type Questions (1 Mark) 22. What are the different types of RNA found in the Note: In the following questions, a statement of cell? [Delhi 2013 (C)][NCERT] assertion followed by a statement of reason is given. 23. What type of linkage hold together the monomer Choose the correct answer out of the following choices units of DNA? on the basis of the above passage. 24. What are the possible pairing of bases in DNA double helical structure? (i) Assertion and reason both are correct statements and the reason is correct explanation of assertion. Short Answer Type Questions-I (2 Marks) (ii) Assertion and reason both are correct statements 25. What are the differences between a nucleoside but reason is not correct explanation of assertion. and nucleotide? [Delhi 2016, 2014 (C)] [NCERT] (iii) Assertion is correct statement but reason is wrong 26. The two strands in DNA are not identical but they statement. are complementary. Explain. (iv) Assertion is wrong statement but reason is correct [Delhi 2010 (C), 2009] statement. Short Answer Type Questions-II (3 Marks) 12. Assertion: The possible pairing in DNA double helical structure is in between guanine and 27. Name the chemical components which constitutes cytosine through 2 hydrogen bonds. a nucleotide. Write any three functions of Reason : The possible pairing in DNA double nucleotides in a cell? helical structure is in between adenine and thymine through 2 hydrogen bonds. 28. What are nucleic acids? Mention their two important functions. [NCERT] 13. Assertion: DNA and RNA contain four bases 29. Write the structural and functional differences each. between DNA and RNA. Reason: Thymine base is not present in RNA [Delhi 2013(C)] [NCERT] 356 Chemistry-12

Answers 1. (ii) 2. (iv) 3. (ii) bonds are formed between purine base of one strand and pyrimidine base of the other strand 4. (i) 5. (iii) 6. (ii) and vice-versa. Because of the size and geometry of the bases, the only possible pairing in DNA are 7. (iv) 8. (iii) 9. (ii) between G (Guanine) and C (Cytosine) through three H-bonds and between A (Adenosine) and 10. (ii) 11. (ii) 12. (iv) T (Thymine) through two H-bonds. Therefore, GC base pair has more stability as compared 13. (ii) 14. (ii) 15. (iii) to AT base pair. Thus, the two strands are complementary but not identical. 16. (ii) 17. (ii) 18. DNA 27. Bases (Purine or Pyrimidines) + Sugar (Ribose 19. A five carbon atom sugar (pentose), i.e. the sugar of deoxyribose) + Phosphate group constitutes a is D-ribose in RNA and 2-deoxy-D-ribose in DNA. nucleotide. 20. Upon hydrolysis, the nucleotide from DNA will The three functions are: form thymine along with 2-deoxy-D-ribose sugar (i)As ATP, they acts as the energy carrier. and phosphoric acid. (ii)Acts as a co-enzyme. 21. When RNA is hydrolysed, there is no relationship between the quantities of four bases, i.e. Cytosine (iii)As monomer of RNA and DNA. (C), Guanine (G), Adenine (A) and Uracil (U) bases 28. Nucleic acids are the biologically important obtained, because the base-pairing principle, i.e. polymers which are present in the nuclei of (A) pairs with (T) and (C) pairs with (G) is not all living cells in the form of nucleoproteins or chromosomes (responsible for heredity). They followed in case of RNA. This suggests that unlike play a vital role in the biosynthesis of proteins, including enzymes. Thus they control the overall DNA (a double stranded structure), RNA has a metabolic activity of the living organisms. single stranded molecular structure. Nucleic acids are of two types: 22. There are three types of RNA: (i) Deoxyribonucleic acid (DNA) and (i) Messenger RNA (m-RNA) (ii) Ribonucleic acid (RNA). (ii) Ribosomal RNA (r-RNA) (i) DNA has a unique property to replicate itself, i.e. it can bring about the synthesis of (iii) Transfer RNA (t-RNA) another DNA molecule on itself. As a result of replication of DNA two double stranded DNA 23. The monomeric units of DNA are held together molecules can be formed called two daughter by H-bonds. There are 3-hydrogen bonds between DNA molecules. guanine and cytosine (G∫C) base pairs and two (ii) RNA and DNA are responsible for synthesis H-bonds between adenine and thymine (A=T) base of all the proteins needed for the growth and maintenance of our body. Transcription pairs. is the process, which involves copying of DNA base sequences into a RNA molecule 24. The possible pairing is in between guanine and called the messenger RNA (m-RNA). The cytosine through 3 hydrogen bonds and adenine bases of m-RNA are complementary to those and thymine through 2 hydrogen bonds. of the DNA strand. Actually the proteins are synthesised by various RNA molecules 25. Nucleoside is a combination of a base and sugar. (m-RNA, t-RNA and r-RNA) in the cell but Nucleosides are obtained when a purine (adenine the message for the synthesis of a particular protein is present only in DNA. or guanine) base or a pyrimidine base (uracil, cytosine or thiamine) is attached to 1′ position of a sugar (ribose or deoxyribose) moeity through b-linkage. Nucleotides are nucleoside plus phosphate group (a combination of base, sugar and phosphoric acid moeity). Nucleotides are obtained when 5′ position of sugar of a nucleoside is esterified using phosphoric acid. 26. The two strands in DNA molecules are complementary to each other because hydrogen 29. Structural Differences (i) The pentose sugar present in RNA is D-ribose. (i) The pentose sugar present in DNA is D-2-deoxyribose. (ii) RNA contains uracil and cytosine as pyrimidine (ii) DNA contains cytosine and thymine as pyrimidine bases and adenine and guanine are present as bases and adenine and guanine are present as purine bases. purine bases. (iii) RNA is a single chain of polynucleotides and (iii) DNA is a double chain of polynucleotides (i.e. it has therefore has a single stranded structure. double stranded a-helix structure). biomoleCules 357

(iv) RNA molecules are relatively short with low (iv) DNA molecules are very large with molecular mass molecular mass ranging from 20,000 to 40,000 u. ranging from 6 × 106 to 16 × 106 u. Functional Differences (i) RNA controls protein synthesis. (i) DNA controls the transmission of heredity as well as protein synthesis. (ii) RNA does not replicate on its own. (ii) DNA has unique property of self-replication. Case based questions 1. Case Study (a) (b) CHO H—C—OH H—C—OH Monosaccharides, the simplest carbohydrates H—C—OH HO—C—H existing, are polyhydroxycarbonyls. The carbonyl O H—C—OH H—C—OH group may be an aldehyde group or a ketone group. HO—C—H CH OH Monosaccharides have between three to nine carbon H—C—OH 2 atoms, although the most abundant ones are those H—C (d) COOH H—C—OH with five carbon atoms called pentoses (D-ribose; CH2OH (c) HO—C—H D-2-deoxyribose; L-arabinose; D-xylose) and six carbon atoms called hexoses (glucose, fructose, galactose and mannose). The cosmetic relevance of monosaccharides is based on their thickening, emollient (having the quality of softening or soothing the skin) and filmogenic properties, which are due to their structure, arrangement and high content of hydrophilic groups. H—C—OH H—C—OH (Source: A. Benaigas, P. Guillen, in Analysis of Cosmetic Products, 2007, 8.5 Botanical Extracts, 345-363) CH OH 2 The following questions are multiple choice 2. Case Study questions. Choose the most appropriate answer: Carbohydrates are the main source of energy (i) Which of the following have free carbonyl that is ingested by the human body (Caffall et al., group to give brick red ppt with Fehling’s 2009). Brain mainly utilizes the glucose. Red blood solution? cells also use glucose only. Fiber in the diet is not digested by human body due to lack of cellulase (a) glucose (b) fructose enzyme. Glucose is the major energy source in the body. Glycogen is the storage form of glucose (c) Both (a) and (b) (d) None of the above and glycogen is stored in skeletal muscles and liver. If glucose intake exceeds than it is utilized (ii) Which of the following is not the reason for in the body then it is converted into fat. Riboses the cosmetic properties in monosaccharides? are utilized in formation of deoxyribonucleic acid (Hou et al., 2009). Carbohydrates are (a) Structure (b) Arrangement polyhydroxy alcohol with potentially active carbonyl group which may be aldehyde or keto (c) High content of hydrophilic groups group. Carbohydrates can be classified on the basis of carbon atom present in the carbohydrates. (d) Carbonyl group Carbohydrates are classified into four types monosaccharides, disaccharides, oligosaccharides, OR polysaccharides. Monosaccharides cannot be hydrolyzed further into simpler form. Disaccharides Which property of monosaccharides gives give two monosaccharides on hydrolysis. Polysaccharides may be homopolysaccharides them cosmetic relevance? and heteropolysaccharides. (a) Thinning (b) Whitening (c) Emollient (d) Non-sticky (iii) Which of the following is a monosaccharide with five-carbon atoms? (a) L-arabinose (b) Fructose (c) Mannose (d) Galactose (iv) Which of the following is not the structure of glucose? 358 Chemistry-12

(Source: H.M. Asif, Muhammad Akram, Tariq Reason: ‘D’ represents its dextrorotatory Saeed, M. Ibrahim Khan, Naveed Akhtar, Riaz nature. ur Rehman, S.M. Ali Shah, Khalil Ahmed, Ghazala Shaheen, 2011, Carbohydrates, (ii) Assertion: In human system, starch can act International Research Journal of Biochemistry as nutrient but cellulose does not. and Bioinformatics Vol. 1(1) pp. 001-005) In these questions, a statement of assertion Reason: Human body does not contain followed by a statement of reason is given. enzyme to hydrolyse cellulose. Choose the correct answer out of the following choices. (iii) Assertion: Glucose has an open structure. (a) Assertion and reason both are correct statements Reason: Glucose on reacting with HI gives n-hexane. and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements (iv) Assertion: Sucrose is a non-reducing sugar. Reason: A free aldehyde group is present is but reason is not correct explanation for sucrose. assertion. (c) Assertion is correct statement but reason is OR wrong statement. Assertion: Glycogen is called animal starch. (d) Assertion is wrong statement but reason is Reason: Glycogen is stored in animal body. correct statement. (i) Assertion : D (+)-Glucose is dextrorotatory Answers in nature. 1. (i) (c) (ii) (d) or (c) (iii) (a) or (a) (d) (a) 2. (i) (c) (ii) (a) (iii) (b) (iv) (c) or (a) Analogy based questions 2. Complete the following analogy: Glucose : A : : 1. Which of the following analogies is correct? Fructose : B (i) a-Helix : 2° structure of protein : : b-pleated sheet : 3° structure of protein (i) A : Pyranose, B : Furanose (ii) A : Reducing sugar, B : Non-reducing sugar (ii) Mannose : Epimer of glucose : : Fructose : (iii) A : Ketose, B : Aldose Anomer of glucose (iv) A : Soluble in water, B : Insoluble in water (iii) Proteins : Polymer of amino acids : : DNA : 1. (iii) Answers Polymer of nucleotides 2. (i) (iv) Essential amino acids : Required by body : : Non-essential amino acids : Not required by body. Matching type questions 1. Column I Column II A. Fructose (1) Non-essential amino acid B. Valine (2) Polyhydroxy ketone C. Glycine (3) Polymer of nucleotides D. RNA (4) Essential amino acid Code: C (1) D (2) (ii) A (3) B (1) C (4) D (2) (i) A (4) B (3) (iv) A (4) B (1) C (3) D (2) (iii) A (2) B (1) C (3) D (4) Column II 2. Column I D (3) (1) Reducing sugar (2) Thymine A. DNA (3) Insulin B. Glucose (4) Uracil C. RNA D. Globular protein (ii) A (4) B (2) C (1) D (3) Code: (iv) A (2) B (1) C (4) D (3) (i) A (2) B (4) C (1) Answers (iii) A (3) B (1) C (2) D (4) 1. (i) 2. (iv) biomoleCules 359

Quick revision notes • Structure of proteins can be studied at different • The molecules which are naturally occurring in levels, namely primary structure; secondary the biological systems are called biomolecules. structure, tertiary structure and quarternary structure. • Carbohydrates, proteins, vitamins, enzymes, nucleic acids, hormones, etc. are a few examples • Any change in the temperature, pH or conditions of biomolecules. can alter the functions of proteins by disturbing the secondary and tertiary structure of proteins. • Carbohydrates are the optically active polyhydroxy This process is called denaturation of proteins. aldehydes and ketones, which on hydrolysis yield simpler molecules like glucose, fructose, etc. • Enzymes referred to as biocatalysts, are actually globular proteins associated with a moeity called • Depending upon the number and type of smaller cofactor which increase the rate of bio-chemical units produced by the hydrolysis of carbohydrates, reactions. they are classified as monosaccharides (Simple sugars) which cannot be hydrolysed, • Enzymes are highly specific and selective in their oligosachharides that breaks into two to ten mode of action and are necessary for carrying smaller units and polysaccharides (that breaks out the chemical reactions within the biological down into large number of smaller units). systems. • Glucose and fructose are a few examples of • Vitamins are the organic compounds, required; in monosaccharides, sucrose and raffinose are very small amounts for the proper functioning of oligosaccharides and starch and cellulose are our body and keeps us healthy. polysaccharides. • Depending upon the solubility in different • Carbohydrates act as biofuel and provide energy solvents, vitamins have been classified into two for the proper growth and functioning of living groups, namely – Fat soluble Vitamins (A, D, systems. E and K group of vitamins) and water-soluble vitamins (B – complex – vitamins and vitamin C). • Commercially, glucose is obtained by hydrolysis of starch. • Deficiency or sometimes even excess of vitamins may lead to several diseases. • The monosaccharides are held together by the glycosidic linkage to form oligosaccharides and • Nucleic acids are the polymers of monomeric polysaccharides. units-nucleotides, which in turn consist of a base, a pentose sugar and phosphate ion (from • Glucose is a six carbon straight chain aldose, phosphinic acid). having a –CHO group, one primary –OH group and four secondary –OH groups. • Nucleic acids are of two types – DNA and RNA. • Glucose has the molecular formula of C6H12O6. It • DNA consists of 2 – deoxyribose sugar, whereas is considered to have a cyclic structure which can RNA consists of ribose sugar. be explained as it does not give 2, 4 –DNP test, although having a –CHO group. • DNA forms the chemical basis of heredity and it has a double helix-structure. • Proteins are the polymers of different types of a-amino acids. • RNA is responsible for the synthesis of proteins in the cell. • Ten amino acids are considered to be essential, as they cannot synthesise in our body and thus • RNA has a single strand structure. needs to be provided through regular diet, the rest of the amino acids that can be synthesised • Both DNA and RNA consists of purine and and stored in our body are considered as non- pyrimidine bases, with difference in only essential amino acids. pyrimidine bases. Thymine is present in DNA, while Uracil is present in RNA. • Amino acids are the monomeric units of proteins O • All the other three bases, cytosine, guanine and adenine are common to both DNA and RNA. and are held together by peptide (—C—NH—) bonds. • There are three types of RNA, m-RNA (messenger RNA), t-RNA (transfer RNA) and r-RNA • Proteins perform a variety of structural and (ribosomal RNA) dynamic functions to maintain a living system. • DNA fingerprinting helps in identifying the • Proteins on the basis of molecular shape can biological parents of children. be classified as (a) fibrous proteins (b) globular proteins. 360 Chemistry-12

Important reactIons 1. C12 H22 O11 + H2 O H+ C6 H12 O6 + C6 H12 O6 2. (C6 H10 O5) n + nH2 O H+ nC6 H12 O6 Sucrose Glu cos e Fructose Starch 393 K, 2–3 atm Glu cos e Invert Sugar HI CH3—CH2—CH2—CH2—CH2—CH3 ∆ n-hexane CH=NOH NH2OH (CHOH)4 CHO Br2(aq) CH2OH CN CH 3. (CHOH)4 (Glucose monoxime) HCN OH CH2OH (CHOH)4 COOH (Glucose) CH2OH (CHOH)4 (Glucose cyanohydrin) CH2OH COOH (CHOH)4 (Gluconic acid) conc. HNO3 (CH3CO)2O CHO COOH (CHOCOCH3)4 Acetic (Sachharic acid) anhydride CH2OCOCH3 (Glucose pentaacetate) OO 6. C12 H22 O11 + H2 O Maltase 2C6 H12 O6 Maltose 4. R—CH—C—O—H R—CH—C—O6 D - D - Glucose NH2 N5 H3 7. C12 H22 O11 + H2 O Lactase C6 H12 O6 + C6 H12 O6 Lactose Zwitter ion b - D - Glucose b - D - Galactose 5. H2 N—CH2 —COOH + H2 N—CH—COOH –H2 O 8. C12 H22 O11 + H2 O Invertase C6 H12 O6 + C6 H12 O6 Sucrose CH3 D(+) glucose D(–) fructose O 9. (C6 H10 O5) n + nH2 O H+ nC6 H12 O6 Starch or 393 K, 2 - 3 atm Glu cos e Cellulose H2 N—CH2 —C—NH —CH—COOH Peptide linkage CH3 Glycylalanine (Gly-Ala) common errors errors correctIons (i) Students commit mistake in representation of (i) Groups projected to the right in the Fischer Haworth structure of carbohydrates. Projection are written below the plane while those on left are written above the plane in Haworth (ii) Students commit mistake and write thiamine as structure. base in case of RNA and uracil as base in case of DNA. (ii) Mnemonic to learn bases in RNA is Uhi Aya Chala Gaya (U, A, C, G) and for DNA is TiAChali Gayi (T, A, C, G). biomoleCules 361

revIsIon chart Classification Polysaccharides • On the basis of having or not having free- • Starch is an important polysaccharide and constitutes of amylase and aldehydic/ketonic groups, they are classified as: amylopectin. (i) Reducing sugars (having free aldehydic/ ketonic group), and • Cellulose a straight chain polymer of b-D-Glucose joined by b-Links (ii) Non-reducing sugars (do not have free • Glycogen is similar to amylopectin structurally though more branched. aldehydic/ketonic group) CarBoHydrates Monosaccharides • On the basis of hydrolysis reaction; they are classified as: They are polyhydroxyaldoses • Glucose is the most important mono-saccharide; (i) M onosaccharides; e.g. Glucose, Fructose. or ketones with general it is a 6-C-atom molecule, and can be represented (ii) O ligosaccharides; e.g. Sucrose, Lactose, formula Cx(H2O)y by open chain structure (having free –CHO Maltose. group) and cyclic structure. (iii) Polysaccharides; e.g. Starch, Cellulose • Similarly fructose is also represented by open chain (free keto group) and ring structure. • In the ring structure the ethereal O-atom linkage is called the glycosidic linkage. Oligosaccharides • Disaccharides (e.g. Sucrose) and polysaccharides (e.g. Starch) on hydrolysis yeild monosaccharides. • Among disaccharides; Sucrose is non-reducing, while Maltose and Lactose are reducing sugars. • On hydrolysis they yeild monosaccharides units with different values of specific rotation as in the equilibrium mixture and this process is called mutarotation. Amino Acids Denaturation Classification of Amino acids If temperature and pH of proteins is distrubed They have been classified as: • Amino acids are the organic compounds it may lead to denaturation of proteins, i.e. loss • Essential Amino Acids: Cannot be containing both –COOH and –NH2 groups. in its biological activity. • Most naturally occurring amino acids have synthesised by our body on its own and L-configuration. Proteins needs to be taken through daily diet. E.g. Valine. • Amino acids can exists as a dipolar ion These are the biomolecules • Non-essential Amino acids: Can be ahmavininoga cNidHa3+nadntdheCpOHOa–t ions withi n the in a living system which synthesised by our body on its own. E.g. which they exists is made of nitrogenous Glycine. as dipolar (or Zwitter) ion is called the organic compounds isolectronic point. resulted by joining together Structure of Proteins of a-amino acids. The linear sequence of amino acids in proteins • Amino acids join together by CO-NH bond forms primary structure, folding of peptide ofof rtmweod a bmeintwoe aecnid Cs O(sOuHch a an bdo NnHd 2isg croaullepds Enzymes and Hormones chain forms their secondary structure (either peptide bond) to form a peptide chain • Enzymes are also proteins, which catalyses a-helix or b-pleated sheets), overall folding which polymerises to form polypeptides. If of polypeptide chains form tertiary structure mass of polypeptide chain is greater than several biological processes. and spatial arrangement of polypeptides in 10000 u, then it form proteins. • Hormones are the molecules that can act 3D-space forms the quaternary structure. as messengers between the cells and are released directly into the blood stream by ductless (endocrine) glands. 362 Chemistry-12

Fat soluble vitamins revIsIon chart Water soluble vitamins They are soluble in fats and oils, e.g. vitaMins They are readily soluble in water Vitamins A, D, E and K. and get excreted out through urine, They have been classified into two therefore needs to be taken through groups: regular diet (Except vitamin B12), e.g. Group-B vitamins. Excess or deficiency of vitamins may lead to several diseases. RNA • Three types of RNA: Messenger RNA (m-RNA) Ribosomal RNA (r-RNA) and transfer RNA (t-RNA) • These RNA molecules are responsible for the synthesis of proteins in a cell and is controlled by DNA molecule for the protein synthesis. Classification nuCleiC aCids DNA • RNA: It is ribonucleic acid and consists They are the polymers of nucleotides, • DNA has a double helical structure of a 5-C-sugar D-ribose. which in turn itself is constituted of a while RNA has a single strand base, a sugar and a phosphate moeity. structure. • DNA: It is deoxyribonucleic acid and it also consists of a 5-C-sugar • DNA has a unique property of self- D-deoxyribose. replication while RNA cannot replicate itself on its own. Linkage In RNA & DNA • Both DNA and RNA consists of purine and pyrimidine bases which are specifically paired as follows: In RNA: Purine Bases Link: Pyrimidine bases Adenine (A) 2-H bonds Uracil (U) Guanine (G) 3-H bonds Cytosine (C) In DNA: Adenine (A) 2-H bonds Thymine (T) Guanine (G) 3-H bonds Cytosine (C) biomoleCules 363

chapter trend—Based on past Years’ cBse exams ➣ It has been observed from this chapter that the weightage of topics ‘Proteins’ and ‘Carbohydrates’ are maximum. Hence, these are most important topics. ➣ From this chapter , generally 1 and 2 marks questions were asked from the topic ‘Carbohydrates’. ➣ Most of the 3 marks questions asked from this chapter belong to the topic ‘Proteins’. Questions For practice Very Short Answer Type Questions (1 Mark) 1. Which one of the following is a disachharide: Starch, maltose, fructose, glucose? [Delhi 2014] 2. Which one of the following is oligosachharide: Starch, maltose, fructose, glucose? 3. Which component of starch is a branched polymer of a-glucose insoluble in water? Short Answer Type Questions-I (2 Marks) 4. Differentiate between the following: (i) Essential and non-essential amino acids. (ii) Fibrous and globular proteins. [Delhi 2017 (C)] 5. (i) Write the product obtained when D-glucose reacts with H2NOH? (ii) Amino acids show amphoteric behaviour, why? Short Answer Type Question-II (3 Marks) 6. (i) What is denaturation of proteins? Is formation of cheese is an example of this process? (ii) Why are enzymes are called biocatalysts? 364 Chemistry-12

assIgnment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. Which of the following is the name of the linkage joining two nucleotides? (i) Peptide linkage (ii) Phosphodiester linkage (iii) Amide linkage (iv) Glycosidic linkage 2. The helical structure of protein is stabilized by: (i) Peptide bond (ii) Dipeptide bond (iii) Hydrogen bonds (iv) van der Waals forces 3. Glucose reacts with acetic anhydride to form (i) Monoacetate (ii) Tetraacetate (iii) Penta-acetate (iv) Hexa-acetate Assertion-Reason Type Questions (1 Mark) Note: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (i) Assertion and reason both are correct statements and the reason is correct explanation of assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 4. Assertion: Pentose sugar present in DNA is D-ribose. Reason: DNA has thymine as pyrimidine base. 5. Assertion: The spontaneous change in specific rotation of an optically active compound is called mutarotation. Reason: When alpha or beta form of D-glucose dissolved in water and allow to stand then the specific rotation of solution becomes +52.5°. Very Short Answer Type Questions (1 Mark) 6. Write the structure of zwitter ion of alanine. 7. Write the name of vitamin whose deficiency causes bleeding of gums. 8. What is isoelectric point? 9. Name the pentose sugar present in DNA. 10. What is mutarotation. Short Answer Type Questions-I (2 Marks) 11. Explain the differences between a-helix and b-pleated sheet structure of proteins? 12. Which sugar is called invert sugar? Why is it called so? [NCERT Exemplar Problems] 13. Coagulation of egg white on boiling is an example of denaturation of protein. Explain in terms of structural changes under what conditions, denaturation of protein will take place. Short Answer Type Questions-II (3 Marks) 14. What do you understand by primary, secondary and tertiary structure of proteins? 15. (i) Which vitamin deficiency causes rickets? (ii) Name the base that is found in nucleotide of RNA only. (iii) Glucose on reaction with acetic acid gives glucose pentaacetate. What does it suggest about the structure of glucose? [CBSE 2017-18 Sample Paper] 16. Name the sugar present in milk. How many monosachharide units are present in it? What are such oligosaccharide called? [NCERT Exemplar Problems] Answers 1. (ii) 2. (iii) 3. (iii) 4. (ii) 5. (ii) biomoleCules 365

Sample Paper-1 (Solved) Time: 3 Hours Maximum Marks: 70 General Instructions: (i) All questions are compulsory. (ii) Section-A : Question number 1 to 9 are objective type questions and carry 1 mark each. (iii) Section-B : Question number 10 to 16 are short answer questions and carry 2 marks each. (iv) Section-C : Question number 17 to 27 are also short answer questions and carry 3 marks each. (v) Section-D : Question number 28 to 29 are long answer questions and carry 5 marks each. (vi) Section-E : Question number 30 is case based question and carry 4 marks. (vi) There is no overall choice. However, an internal choice has been provided. You have to attempt only one of the choices in such questions. (vii) Use log table, if necessary. Use of calculators is not allowed. SECTION-A 1. Which of the following conditions is correct for an ideal solution? 1 (i) DHmix = 0 and DVmix = 0 (ii) DHmix > 0 and DVmix > 0 (iii) DHmix < 0 and DVmix < 0 (iv) DHmix > 0 and DVmix < 0 Ans. (i) 2. iOsnhaigdhdliytieoxnpolfossimvealilnanmaotuunret.oIfdKeMntnifOy 4thtoeccoonmcepnoturnadtefdroHm2StOhe4,faolglorweeinngo:ily compound is obtained which 1 (i) Mn2O7 (ii) MnO2 (iii) MnSO4 (iv) Mn2O3 Ans. (i) 3. Which of the following analogies is correct? 1 (i) Chloroform-acetone : Positive deviation :: Ethanol-H2O : Negative deviation. (ii) pA = pA° . xA : Henry’s law : : p = KH . x : Raoult’s law. (iii) PTotal = pA + pB : Non-ideal solution : : PTotal > pA + pB : Ideal solution. (iv) p = CRT : Osmotic pressure :: P > p : Reverse osmosis. Ans. (iv) 4. Complete the following analogy : o-nitrophenol : A :: o-cresol : B 1 (i) A : more acidic than phenol, B : less acidic than phenol (ii) A : less acidic than phenol, B : more acidic than phenol (iii) A : more acidic than phenol, B : more acidic than phenol (iv) A : less acidic than phenol, B : less acidic than phenol Ans. (i) 5. Match the items of Column I and Column II. 1 Column I Column II A. Mathematical expression for rate of reaction (1) rate constant B. Rate of reaction for zero order reaction is equal to (2) rate law C. Units of rate constant for zero order reaction is same as that of (3) order of slowest step D. Order of a complex reaction is determined by (4) rate of a reaction Codes: (i) A (1) B (2) C (3) D (4) (ii) A (4) B (3) C (2) D (1) (iii) A (2) B (1) C (4) D (3) (iv) A (2) B (4) C (1) D (3) Ans. (iii) Note: In the question no. 6 and 7, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are correct and the reason is correct explanation of assertion. (ii) Both assertion and reason are correct but reason does not explain assertion. 366

(iii) Assertion is correct but reason is incorrect. (iv) Assertion is incorrect but reason is correct. 6. Assertion: The highest oxidation state of osmium is +8. Reason: Osmium is a 5d-block element. 1 Ans. (ii) 7. Assertion: The enthalpy of reaction remains constant in the presence of a catalyst. Reason: A catalyst participating in the reaction, forms difference activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same. 1 Ans. (i) 8. Arrange the following in decreasing order of solubility in water : 1 (CH3)3N, (CH3)2NH, CH3NH2 1 Ans. (CH3)3N < (CH3)2NH < CH3NH2 9. What is the basic structural difference between glucose and fructose? OR Write the products obtained after hydrolysis of lactose. Ans. Glucose has aldehyde group whereas fructose has ketone group. OR Lactose on hydrolysis gives glucose and galactose SECTION-B 10. cWomhepnouMnndO(A2 )i.sCfoumsepdouwnidth(AK) OdiHsprinoptohretiopnreasteesncine aocfidKiNc sOo3luatsioanntoogxiivdeizpiunrgpalegceonmt,piotugnivde(sB)a. dark green An alkaline solution of compound (B) oxidises KI to compound (C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C) and (D). 2 Ans. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O (A) → 2MnO4– + MnO2 +2H2O 3MnO42– + 4H+ (B) [62OHH2O– ++I–3e– + MnO4– → IMOn3–O+ 2 6+e –4O+ H3H– 2.O... ...........(..i.)](i×i)2 → 2MnO4–+ I–+ 2OH– → 2MnO2+ IO–3 (C) 2MnO4–+10I–+16H+ → 2Mn2+ + 5I2 + 8H2O (D) ‘A’ is K2MnO4, ‘B’ is KMnO4, ‘C’ is HIO3, ‘D’ is I2 2 11. State Henry’s law and write its two applications. Ans. Henry’s Law : It states that ‘the partial pressure of gas dissolved in liquid in vapour phase is directly proportional to mole fraction of gas in solution. Applications: p ∝ xgas p = KH × xgas (i) Cold drink bottles are filled at high pressure to increase solubility of CO2 in water. (ii) uMnidxeturrpereosf sOu2reanofdwHaeteirs taken by deep sea divers because helium is less soluble in blood than N2 and does not cause pain or bends. 12. Write IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex. 2 OR Sample paper-1 (Solved) 367

Using IUPAC norms write the formulae for the following: (i) Hexaamminecobalt(III) sulphate (ii) Potassium trioxalatochromate(III) Ans. Dichloridobis(ethane-1,2-diamine)platinum(II) OR (i) [Co(NH3)6]2(SO4)3 (ii) K3[Cr(C2 O4)3] 2 13. Carry out the following conversions: (i) Methanamine to N-Methylbenzamide (ii) Aniline to sulphanilic acid OR Complete the following given chemical equations: (i) C6H5NH2 + CHCl3 + alc. KOH –→ (ii) C6H5NH2 + Br2 (aq) –→ Ans. (i) CH3 NH2 + C6 H5 COCl CH3 NHCOC6 H5 + HCl Methanamine Benzoylchloride N - Methylbenzamide NH2 NH+3 [HSO4]– NHSO3H NH2 (ii) (conc.) H2SO4 Heat (D) Rearrangement 455 K (–H2O) 455 K Aniline OR SO3H Sulphanilic acid (i) C6 H5 NH2 + CHCl3 + 3KOH (alc.) T C6 H5 N == C + 3KCl + 3H2 O Phenyl isocyanide NH2 NH2 Br Br 3Br2 (aq) + 3HBr (ii) Br Aniline 2, 4, 6-Tribromo- aniline 14. For a reaction 2 2H2 O2 alkaline medium 2H2 O + O2 the proposed mechanism is as given below: 1. H2O2 + I– → H2O + IO– (slow) 2. H2O2 + IO– → H2O+I– + O2 (fast) (i) Write rate law for the reaction. (ii) Write the overall order of reaction. (iii) Out of steps (1) and (2). Which one is rate determining step? Ans. (i) Rate = k [H2O2] [I–] (ii) 1 + 1 = 2 (iii) Step I is slow step, therefore, rate determining step 15. Write the hybridization and magnetic character of following complexes: 2 (Atomic no. of Fe = 26) (i) [Fe(H2O)6]2+ (ii) [Fe(CO)5] Ans. (i) sp3d2, paramagnetic [Fe(H2O)6]2+ .. .. .. .. .. .. (ii) dsp3, diamagnetic [Fe(CO)5] sp3d2 .. .. .. .. .. dsp3 2 16. Write structures of main compounds A and B in each of the following reactions : (i) 368 ChemiStry-12

(ii) Ans. (i) (ii) 17. Complete the following reactions: SECTION-C 3 (i) (ii) (iii) OR Write chemical equations for the following reactions: (i) Propanone is treated with dilute Ba(OH)2. (ii) Acetophenone is treated with Zn(Hg)/Conc. HCl (iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO4. Ans. (i) OR (ii) (iii) (i) (ii) (iii) Sample paper-1 (Solved) 369

18. Compound (A), having an empirical formula, C7H8 is chlorinated in the presence of sunlight to form product (B), which on hydrolysis gives a compound (C). Compound (C) gives a positive test with Tollens’ reagent. Identify A, B and C. 3 CH3 CHCl2 CHO Ans. +2Cl2 H3O+ Sunlight Toluene Benzyl dichloride Benzaldehyde (A) (B) (C) 19. A 4% solution (w/w) of sucrose (M = 342 g mol–1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water. (Given : Freezing point of pure water = 273.15 K) 3 WB = 4g; MB = 342g; WA = 100–4 = 96g, Ans. DTf = 273.15K – 271.15K = 2K WB 1000 DTf = Kf × MB × WA 2= K × 4 × 1000 f 342 96 Kf = 2 × 24 × 342 = 16416 = 16.416 K kg mol-1 1000 1000 WB = 5g, MB = 180g, WA = 100 – 5 = 95g WB 1000 DTf = Kf × MB × WA = 16.416 × 5 × 1000 = 16416 = 4.8K 180 95 3420 Freezing point of glucose solution = 273.15 – 4.80 = 268.35 K 20. Arrange the following compounds in the order of their increasing acidic strength: 3 (i) Propanoic acid, chloroethanoic acid, 3-Bromopropanoic acid and Trichloroacetic acid. (ii) 2-Fluorobutanoic acid, 2-Iodobutanoic acid, 2-Bromobutanoic acid and Butanoic acid. (iii) Acetic acid, 2-Methylpropanoic acid, 2,2-Dimethylpropanoic acid. Ans. (i) CH3CH2COOH < BrCH2CH2COOH < ClCH2COOH < Cl3CCOOH (ii) CH3CH2CH2COOH < CH3CH2CHICOOH < CH3CH2CHBrCOOH < CH3CH2CHFCOOH (iii) (CH3)3CCOOH < (CH3)2CHCOOH < CH3COOH 21. Provide a suitable example and explain each the following: (i) Crystal field splitting (ii) Linkage isomerism (iii) Ambidentate ligand 3 Ans. (i) The splitting of d orbitals under the influence of approaching ligands is called crystal field splitting. For example, d4 configuration in the presence of weak field ligand is t23g eg1 while the d4 configuration in the presence of a strong field ligand is t24g eg0. (ii) Linkage isomers are the ones which have ambidentate ligand. They have the same molecular formula but they differ in the linkage of ligand atom to the central metal atom, e.g. [Co(NH3)5 NO2] Cl2, [Co(NH3)5 ONO]Cl2 (iii) A unidentate ligand which coordinates to the central atom through more than one coordinating site but only one site at a time. e.g. NO2– or ONO–, SCN– or NCS–. 3 22. Give reasons for the following: (i) The transition metals generally form coloured compounds. (ii) E° value for (Mn3+Mn2+) is highly positive than that for (Cr3+Cr2+) couple. (iii) The chemistry of actinoids elements is not so smooth as that of the lanthanoids. Ans. (i) It is due to presence of unpaired electrons, they undergo d–d transition by absorbing light from visible region and radiate complementary colour. 370 ChemiStry-12

(ii) Mn2+ (3d5) is more stable than Mn3+(3d4) because it has half filled d-orbital ∴ E°Mn3+/Mn2+ is more positive. Cr2+ (3d4) is less stable than Cr3+(t2g3) having half filled t2g orbitals that is why E°Cr3+/Cr2+ is less positive. (iii) The chemistry of actinoids is not smooth because they show greater number of oxidation states due to comparable energies of 5f, 6d and 7s orbitals. 23. What are nucleic acids? Mention their two important functions. 3 OR Write the structural and functional differences between DNA and RNA. Ans. Nucleic acids are the biologically important polymers which are present in the nuclei of all living cells in the form of nucleoproteins or chromosomes (responsible for heredity). They play a vital role in the biosynthesis of proteins, including enzymes. Thus they control the overall metabolic activity of the living organisms. Nucleic acids are of two types: (i) Deoxyribonucleic acid (DNA), and (ii) Ribonucleic acid (RNA). (i) DNA has a unique property to replicate itself, i.e. it can bring about the synthesis of another DNA molecule on itself. As a result of replication of DNA two double stranded DNA molecules can be formed called two daughter DNA molecules. (ii) RNA and DNA are responsible for synthesis of all the proteins needed for the growth and maintenance of our body. Transcription is the process, which involves copying of DNA base sequences into a RNA molecule called the messenger RNA (m-RNA). The bases of m-RNA are complementary to those of the DNA strand. Actually the proteins are synthesised by various RNA molecules (m-RNA, t-RNA and r-RNA) in the cell but the message for the synthesis of a particular protein is present only in DNA. OR Structural Differences (i) The pentose sugar present in RNA is D-ribose. (i) The pentose sugar present in DNA is D-2- deoxyribose. (ii) RNA contains uracil and cytosine as pyrimidine (ii) DNA contains cytosine and thymine as pyrimidine bases and adenine and guanine are present as bases and adenine and guanine are present as purine bases. purine bases. (iii) RNA is a single chain of polynucleotides and (iii) DNA is a double chain of polynucleotides (i.e. it therefore has a single stranded structure. has double stranded a-helix structure). (iv) RNA molecules are relatively short with low (iv) DNA molecules are very large with molecular mass molecular mass ranging from 20,000 to 40,000 u. ranging from 6 × 106 to 16 × 106 u. Functional Differences (i) RNA controls protein synthesis. (i) DNA controls the transmission of heredity as well as protein synthesis. (ii) RNA does not replicate on its own. (ii) DNA has unique property of self-replication. 24. (i) Out of (CH3)3C–Br and (CH3)3C–I, which one is more reactive towards SN1 and why? 3 (ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification. (iii) Why dextro and laevo – rotatory isomers of butan – 2 – ol are difficult to separate by fractional distillation? Ans. (i) (CH3)3C–I is more reactive because C – I bond is weaker, has lower bond dissociation enthalpy than C–Br bond due to longer bond length. (ii) Sample paper-1 (Solved) 371

(iii) It is because they have less difference in their boiling points, therefore, cannot be separated by fractional distillation. 25. Write equations of the following reactions: (ii) Coupling reaction 3 (i) Acetylation of aniline (iii) Carbylamine reaction Ans. (i) Acetylation (ii) Coupling reaction (iii) Carbylamine reaction 26. The decomposition of tNokHr=e3do4un×cpe1lta0ht–ie3nMiunmsit–i1sau;ltrcf=oanc?ceeinstzreartoioonrdoefrNrHea3cftrioonm. If rate constant (k) is 4 × 10–3 Ms–1, Ans. how long will it take 0.1 M to 0.064 M. 3 [R]° = 0.1 M; [R] = 0.064 M k = [R]0 – [R] t t = [R]O – [R] = 0.1–0.064 = 0.036 = 36 = 9s k 4×10–3 4×10–3 4 27. Define the following with a suitable example in each: 3 (i) Oligosaccharides (ii) Denaturation of protein (iii) Vitamins OR Write the reactions involved when D-glucose is treated with the following reagents: (i) Br2 water (ii) H2N-OH (iii) O(ClHig3oCsOa)c2Ocharides: Ans. (i) Those carbohydrates which give 2 to 10 molecules of monosachharides on hydrolysis e.g. sucrose, raffinose. (ii) Denaturation of protein: When proteins are subjected to heat or change in pH, their secondary and tertiary structure is ruptured but primary structure remains same. It leads to loss of biological activity. This process is called denaturation of proteins. (iii) Vitamins: These are complex organic molecules required in the diet in small amounts to perform specific biological functions for maintaining optimum health of organism. OR (i) (ii) (iii) 372 ChemiStry-12

SECTION-D 28. (i) How do you convert the following: 5 (a) Phenol to Anisole (b) Ethanol to Propan-2-ol (ii) Write mechanism of the following reaction: (iii) Why phenol undergoes electrophilic substitution more easily than benzene? OR (i) Account for the following: (a) o-nitrophenol is more steam volatile than p-nitrophenol. (b) t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butylmethylether. (ii) Write the reaction involved in the folliowing: (a) Reimer-Tiemann reaction (b) Friedel-Crafts alkylation of phenol (iii) Give simple chemical test to distinguish between ethanol and phenol. Ans. (i) (a) (b) (ii) (a) Protonation of alcohol (b) Formation of carbocation (c) Deprotonation (iii) In phenol, —OH group is electron releasing, activating due to +R effect, therefore, it undergoes electrophilic substitution reactions more readily than benzene. OR (i) (a) o—Nitrophenol has intramolecular H-bonding while intermolecular H–bonding exists in p– nitrophenol, therefore, ortho isomer has lower melting point and is steam volatile. (b) It is because tertiary butyl chloride undergoes nucleophilic elimination reaction with stronger nucleophile CH3O and do not undergo substitution reaction. (ii) (a) Riemer Tiemann Reaction Sample paper-1 (Solved) 373

(b) Friedel-Crafts Alkylation of phenol (iii) Add neutral FeCl3 to each, ethanol will not react while phenol gives violet colour. 5 Add I2 and NaOH to each, phenol does not react while ethanol gives yellow ppt of iodoform. 29. E°cell for the given redox reaction is 2.71 V Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is (i) less than 2.71 V and (ii) greater than 2.71 V OR (i) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. (Molar mass : Fe = 56 g mol–1, Zn = 65.3 g mol–1, 1F = 96500 C mol–1) (ii) In the plot of molar conductivity (∧m) vs square root of concentration (c1/2), following curves are obtained for two electrolytes A and B: Answer the following: (i) Predict the nature of electrolytes A and B. (ii) What happens on extrapolation of ∧m to concentration approaching zero for electrolytes A and B? Ans. Mg (s) \" Mg2+ (aq) + 2e— Cu2+ (aq) + 2e– \" Cu (s) Ecell = E0 – 0.0591 log 6Mg2+@ cell 2 5Cu ?2+ = 2.71V – 0.0591 log 10 –3 2 10 –2 = 2.71 V – 0.0295 log 10–1 = 2.71 V + 0.0295 V = 2.7395 V (i) When external opposite e.m.f is less than 2.71 V, the flow of current is from cathode to anode i.e. from Cu to Mg. (ii) When external emf is greater than 2.71 V, current will flow from Mg to Cu. OR (i) I = 2 A mFe = 2.8 g m= Z×I×t Equivalent weight = Atomic weight = 56 = 28 Valency 2 Z= Eq. Wt = 28 96500 96500 t= 96500 × 2.8 28 × 2 t = 4825 seconds 374 ChemiStry-12

mZn = mFe EZn EFe mZn = 2.8 65.3 56 22 ⇒ mZn = 1 × 65.3 = 3.265 g 10 2 (ii) (a) ‘A’ is strong electrolyte, ‘B’ is weak electrolyte. (b) When concentration approaches zero, intercept on Y-axis gives ∧° (limiting molar conductivity) of ‘A’ but not of ‘B’ because a curve cannot be extropolated. Therefore ∧° for weak electrolyte cannot be obtained by extrapolation. SECTION-E 30. The galvanic replacement method was applied for the synthesis of gold nanotubes from sacrificial silver nanorods. The progress of the replacement reaction was monitored by obtaining UV-Vis spectra and by measuring the silver and gold ion concentrations. The replacement reaction was promoted by increased gold precursor concentration as well as by increased reaction temperature. The dissolved silver concentration increased linearly with the gold ion concentration. The replacement reaction was promoted with increased temperature, which enhances the reaction rate constant. The reaction rate depends on the gold precursor concentration nonlinearly. This study determined kinetic information about gold replacement with sacrificial silver nanorod templates through the galvanic replacement reaction. 4 (Reference: : Sunil Kwon, Hyunbae Dong, and Sang-Yup Lee . Study of the Reaction Rate of Gold Nanotube Synthesis from Sacrificial Silver Nanorods through the Galvanic Replacement Method . Journal of Nanomaterials Volume 2010, Article ID 819279, 7 pages) In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion: Rate of reaction increases with increase in temperature. Reason: Kinetic energy of molecules and effective collisions between molecules increases on increasing temperature. (ii) Assertion: The reaction rate depends on the gold precursor concentration linearly. Reason: The dissolved silver ion concentration increased linearly with gold ion concentration. (iii) Assertion: Arrhenius equation, k = Ae–Ea/RT shows the temperature dependence of rate constant. −Ea Reason: The graph between log k and 1 has slope of value 2.303 R T (iv) Assertion: According to collision theory, rate is equal to PZABe–Ea/RT Reason: The factor e–Ea/RT represents the fraction of molecules with energy less than Ea. OR Assertion: Rate of an irrevesible reaction depends on concentration of products. Reason: Rate of a reaction depends on presence of catalyst. Ans. (i) (a) (ii) (d) (iii) (b) (iv) (c) OR (d) Sample paper-1 (Solved) 375

Time: 3 Hours Sample Paper-2 (Unsolved) Maximum Marks: 70 General Instructions: Refer to Sample Paper-1. SECTION-A 1. Increasing the temperature of an aqueous solution will cause 1 (i) Increase in Molarity (ii) Increase in Molality (iii) Decrease in Molarity (iv) Decrease in Molality 2. Phenol is more acidic than ethanol because 1 (i) Ethoxide ion is more stable than Phenoxide ion. (ii) Phenoxide ion is more stable than Ethoxide ion. (iii) Phenol undergoes electrophilic substitution reaction. (iv) Phenol undergoes protonation easily. 3. Complete the following analogy: 1 time–1 : x : : mol lit –1 time–1 : y (i) x : unit of zero order, y : unit of first order (ii) x : unit of zero order, y : unit of pseudo first order (iii) x : unit of first order, y : unit of zero order (iv) x : unit of second order, y : unit of first order 4. Complete the following analogy 1 Nickel-Cadmium Cell : A :: Lead storage Battery : B (i) A : Longer life, B : Shorter life (ii) A : Cheaper, B : Expensive (iii) A : Rechargeable, B : Non-Rechargeable (iv) A : Zinc Anode, B : Lead Anode 5. Match the items of Column I and Column II. 1 Column I Column II A. Diamond (1) short interval of time B. Instantaneous rate (2) ordinarily rate of conversion is imperceptible C. Average rate (3) long duration of time Codes: (i) A (1) B (2) C (3) (ii) A (2) B (3) C (1) (iii) A (3) B (1) C (2) (iv) A (2) B (1) C (3) Note: In the question no. 6 and 7, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are correct and the reason is correct explanation of assertion. (ii) Both assertion and reason are correct but reason does not explain assertion. (iii) Assertion is correct but reason is incorrect. (iv) Assertion is incorrect but reason is correct. 6. Assertion: Relative lowering in vapour pressure is a colligative property. 1 Reason: Relative lowering in vapour pressure depends upon mole fraction of pure solvent. 7. Assertion: ECell should have a positive value for the cell to function. 1 Reason: Ecathode < Eanode 8. Out of chlorobenzene and cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why? 1 OR Write one stereochemical difference between SN1 and SN2 reactions. 9. Among lanthanides, Ln (III) compounds are more predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also seen. Explain. 1 376

SECTION-B 2 10. Define order of reaction. Predict the order of reaction in the given graphs: (i) (ii) Where [R]O is the initial concentration of reactant and t1/2 is half-life. 2 11 Give reasons: (i) An increase in temperature is observed on mixing chloroform and acetone. (ii) Aquatic animals are more comfortable in cold water than in warm water. 12. CWohmenpoFuenCdr(2AO)4 is afucisdeidficwaittihonNgai2vCesOc3oimn ptohuenpdr(eBse).nCceomofpaouirnidt gives a yellow solution of compound (A). on (B) on reaction with KCl forms an orange coloured compound (C). An acidified solution of compound (C) oxidises Na2SO3 to (D). Identify (A), (B), (C) and (D). 2 13. Answer the following: 2 (i) Carbon tetrachloride is insoluble in water. Why? (ii) Write about two industrial applications of chloroform and freons. OR Explain why vinyl halide is less reactive while allyl halide is more reactive than alkyl halides? 14. Define the following terms with a suitable example of each: 2 (i) Polydentate ligand (ii) Homoleptic complex OR Using IUPAC norms, write the formulae for the following complexes: (i) Potassium tri(oxalato) chromate(III) (ii) Hexaaquamanganese(II) sulphate 15. (i) Although both [NiCl4]2– and [Ni(CO)4] have sp3 hybridisation yet [NiCl4]2– is paramagnetic and [Ni(CO)42] is diamagnetic. Give reason. (Atomic no. of Ni = 28) (ii) Write the electronic configuration of d5 on the basis of crystal field theory when (a) ∆o < P and (b) ∆o > P 2 16. Write structures of main compounds A and B in each of the following reactions: (i) (ii) SECTION-C 3 17. The following data were obtained for the reaction: A + 2B → C Experiment [A]/M [B]/M Initial rate of formation of C /M min–1 (i) Find the order of reaction with 1 0.2 0.3 4.2 × 10–2 respect to A and B. 2 0.1 0.1 6.0 × 10–3 3 0.4 0.3 1.68 × 10–1 (ii) Write the rate law of overall 4 0.1 0.4 2.40 × 10–2 order of reaction. (iii) Calculate the rate constant (k). 18. From analysis and molecular weight determination, the molecular formula of a compound (A) is found to be C3H7NO. The following information is given for this compound: 3 (i) On hydrolysis it gives an amine (B) and a carboxylic acid (C). (ii) Amine (B) reacts with benzene sulphonyl chloride and gives a product which is insoluble in aqueous sodium hydroxide solution. (iii) Acid (C) on reaction with Tollens’ reagent gives silver mirror test. What are the name and molecular structures of (A), (B) and (C)? Explain the reactions with the help of equations. 19. A solution containing 1.9 g per 100 mL of KCl (M = 74.5 g mol–1) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol–1). Calculate the degree of dissociation of KCl solution. Assume that both the solutions have same temperature. 3 Sample paper-2 (UnSolved) 377

20. Two half cell reactions of an electrochemical cell are given as: 3 MnO4– (aq) + 8H+ (aq) + 5e– → Mn2+ + 4H2O(l) E° = +0.151V Sn2+(aq) → Sn4+(aq) + 2e– E° = + 0.15 V Construct the redox equation from the two half-cell reactions and predict if this reaction favours the formation of reactants or products as shown in the equation. 21. Give reasons for the following: 3 (i) Transition metals have high enthalpies of atomization. (ii) Manganese has lower melting point even though it has a higher number of unpaired electrons for bonding. (iii) Ce4+ is a strong oxidising agent. 22. Explain the following: 3 (i) Low spin octahedral complexes of nickel are unknown. (ii) p complexes are known for transition metals only. (iii) CO is strong ligand than NH3 for most of the metals. 3 23. Complete the following reactions: (i) (ii) (iii) OR How do you convert the following: (i) N-phenylethanamide to p-bromoaniline (ii) Benzene diazonium chloride to nitrobenzene (iii) Benzoic acid to aniline 24. Give the reagent used to bring about the following transformations: 3 (i) Pent -3-en-2-ol to pent -3-en-2-one (ii) But-1-yne to Butan-2-one 3 3 (iii) p-nitrotoluene to p-nitrobenzaldehyde. 3 OR Carry out the following conversions? (i) Acetone from ethanol (ii) Methanal to ethanal (not more than 3 steps) (iii) Propanal to propane 25. Among all the isomers of molecular formula C4H9Br, identify (i) the one isomer which is optically active. (ii) the one isomer which is highly reactive towards SN2. (iii) the two isomers which give same product on dehydrohalogenation with alcoholic KOH. 26. (i) Write the product when D-glucose reacts with Br2 (aq). (ii) What type of bonding provides stability to a-helix structure of protein? (iii) Name the vitamin whose deficiency causes pernicious anaemia. OR Define the following terms: (i) Invert sugar (ii) Native protein (iii) Nucleotide 27. (i) Give reasons: (a) Benzoic acid is a stronger acid than acetic acid. (b) Methanal is more reactive towards nucleophilic addition reaction than ethanal. (ii) Give a simple chemical test to distinguish between propanal and propanone. SECTION-D 28. (i) The conductivity of 0.001 mol L–1 acetic acid is 4.95 ×10–5 S cm–1. Calculate the dissociation constant if for acetic acid is 390.5 S cm2 mol–1. 5 / o m (ii) Write Nernst equation for the reaction at 25°C: 2 Al (s) + 3 Cu2+ (aq) → 2 Al3+ (aq) + 3 Cu (s) 378 ChemiStry-12

(iii) What are secondary batteries? Give an example. OR (i) Represent the cell in which the following reaction takes place: 2 Al (s) + 3 Ni2+ (0.1 M) → 2Al3+ (0.01 M) + 3 Ni (s) (ii) Calculate its emf cifonEd0cuelclt=iv1it.y41vVar.y with increase in concentration for strong electrolyte and weak How does molar obtain conductivity ( electrolyte? How can you limiting molar ) for weak electrolyte? / o m 29. (i) Give equations of the following reactions: 5 (a) Phenol is treated with conc. HNO3. (b) Propene is treated with B2H6 followed by H2O2/OH–. (c) Sodium t-butoxide is treated with CH3Cl. (ii) How will you distinguish between butan-1-ol and butan-2-ol? (iii) Arrange the following in increasing order of acidity: Phenol, ethanol, water OR (i) How can you obtain phenol from (a) Cumene, (b) Benzene sulphonic acid, (c) Benzene diazonium chloride? (ii) Write the structure of the major product obtained from dinitration of 3-methylphenol. (iii) Write the reaction involved in Kolbe’s reaction. SECTION-E 30. Carbohydrates are the main source of energy that is ingested by the human body (Caffall et al., 2009). Brain mainly utilizes the glucose. Red blood cells also use glucose only. Fiber in the diet is not digested by human body due to lack of cellulase enzyme. Glucose is the major energy source in the body. Glycogen is the storage form of glucose and glycogen is stored in skeletal muscles and liver. If glucose intake exceeds than it is utilized in the body then it is converted into fat. Riboses are utilized in formation of deoxyribonucleic acid (Hou et al., 2009). Carbohydrates are polyhydroxy alcohol with potentially active carbonyl group which may be aldehyde or keto group. Carbohydrates can be classified on the basis of carbon atom present in the carbohydrates. Carbohydrates are classified into four types monosaccharides, disaccharides, oligosaccharides, polysaccharides. Monosaccharides cannot be hydrolyzed further into simpler form. Disaccharides give two monosaccharides on hydrolysis. Polysaccharides may be homopolysaccharides and heteropolysaccharides. (Source: H.M. Asif, Muhammad Akram, Tariq Saeed, M. Ibrahim Khan, Naveed Akhtar, Riaz ur Rehman, S.M. Ali Shah, Khalil Ahmed, Ghazala Shaheen, 2011, Carbohydrates, International Research Journal of Biochemistry and Bioinformatics Vol. 1(1) pp. 001-005) 4 In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : D (+)-Glucose is dextrorotatory in nature. Reason: ‘D’ represents its dextrorotatory nature. (ii) Assertion: In human system, starch can act as nutrient but cellulose does not. Reason: Human body does not contain enzyme to hydrolyse cellulose. (iii) Assertion: Glucose has an open structure. Reason: Glucose on reacting with HI gives n-hexane. (iv) Assertion: Sucrose is a non-reducing sugar. Reason: A free aldehyde group is present is sucrose. OR Assertion: Glycogen is called animal starch. Reason: Glycogen is stored in animal body. Answers 1. (iii) 2. (ii) 3. (iii) 4. (i) 5. (iv) 6. (iii) 7. (iii) 17. (i) Order = 2 w.r.t. A and 1 w.r.t. B, (iii) 3.5 L2 mol–2 s–1 19. a = 96% OR (a) 28. (i) K = 1.84 × 10–5 OR (i) Ecell = 1.4196 V 30. (i) (c) (ii) (a) (iii) (b) (iv) (c) Sample paper-2 (UnSolved) 379

Time: 3 Hours Sample Paper-3 (Unsolved) Maximum Marks: 70 General Instructions: Refer to Sample Paper-1. SECTION-A 1. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature? 1 (i) 2-Bromobutane (ii) 1-Bromobutane (iii) 2-Bromopropane (iv) 2-Bromopropan-2-ol 2. We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the order .................... . 1 (i) iA < iB < iC (ii) iA > iB > iC (iii) iA = iB = iC (iv) iA < iB > iC 1 3. Which of the following analogies is correct? (i) Haloalkanes with KCN : Alkyl cyanide : : Haloalkanes with AgCN : Alkyl isocyanide (ii) Wurtz reaction : For alkyl halides only : : Wurtz-Fittig reaction : For aryl halides only (iii) Cl– ion : meta-directing : : NO2− ion : ortho-, para-directing (iv) Friedel-Crafts reaction : Nucleophilic substitution : : Replacement of halide ion by hydroxyl group from aryl halide : Electrophilic substitution 4. Complete the following analogy: Glucose : A : : Fructose : B 1 (i) A : Pyranose, B : Furanose (ii) A : Reducing sugar, B : Non-reducing sugar (iii) A : Ketose, B : Aldose (iv) A : Soluble in water, B : Insoluble in water 5. Match the compounds given in Column I with the items given in Column II. 1 Column I Column II A. Benzenesulphonyl chloride (1) Zwitter ion B. Sulphanilic acid (2) Hinsberg reagent C. Alkyl diazonium salts (3) Dyes D. Aryl diazonium salts (4) Conversion to alcohols Codes: (i) A (2) B (1) C (4) D (3) (ii) A (1) B (2) C (4) D (3) (iii) A (2) B (4) C (3) D (1) (iv) A (3) B (1) C (2) D (4) Note: In the question no. 6 and 7, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are correct and the reason is correct explanation of assertion. (ii) Both assertion and reason are correct but reason does not explain assertion. (iii) Assertion is correct but reason is incorrect. (iv) Assertion is incorrect but reason is correct. 6. Assertion: Boiling points of alcohols and ethers are high. 1 Reason: Alcohols and ethers both can form hydrogen-bonding with water molecules. 7. Assertion: Hydrolysis of (–)-2-bromooctane proceeds with inversion of configuration. 1 Reason: This reaction proceeds through the formation of a carbocation. 8. What does stability constant indicate? 1 OR Name an organometallic compound that is used as a homogeneous catalyst. 9. HCl does not give an acidic solution in benzene why? 1 380

SECTION-B 10. Write the reagent required in the following reactions: 2 CH = CH — CH2OH CH = CH — CHO (i) ? (ii) CH3COOH ? → CH3COONH+4 2 11. Complete the following reactions: (ii) FeCr2O4 + Na2CO3 + O2 Heat → ? (i) KMnO4 + NO2– H+ ? OR Transition metals form complex compounds, give suitable reason. 12. What will be your comment for the following solutions on variability from ideal behaviour. 2 (i) Solution of A (10 mL) mixed with B (10 mL) and the resultant solution is 18 mL. (ii) H2SO4 (conc.) and distilled water are at same temperature but on mixing temperature start rising. 13. What will be the mass of Al deposited on the cathode when a current of 20 amperes is passed for 30 minutes in Al2(SO4)3 molten liquid. 2 14. Write one similarity and one difference between the chemistry of lanthanoids and that of actinoids. 2 15. The possible mechanism for the reaction: 2H2 (g) + 2NO (g) → N2(g) + 2H2O is 2 (i) 2NO  N2O2 (ii) N2O2 + H2 → N2O + H2O (Slow) (iii) N2O + H2 → N2 + H2O (Fast) What is (a) rate law of the reaction? (b) the order of the reaction? 16. Write IUPAC names of the following compounds. 2 CH3 OH CH3 (i) CH3—CH—CH—C—CH3 (ii) CH3 (iv) (iii) OR Explain why alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. SECTION-C 17. Carry out the following conversions: 3 (i) Aniline to p-bromoaniline (ii) Chlorobenzene to 2-chloroacetophenone (iii) 2-Iodopropane to 2, 3-dimethylbutane 18. Answer the following questions: 3 (i) Semicarbazide posses 2 NH2 group but only one reacts with carbonyl group which is attached with NH group. Why? (ii) Carboxylic acids are stronger acids than phenol. Why? (iii) Comment: HCHO can get oxidised as well as reduced in basic medium. 19. Define the following as related to proteins: 3 (i) Peptide linkage (ii) Primary structure (iii) Denaturation 20. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements? 3 OR Why only dilute sulphuric acid and not dilute HCl or dilute HNO3 is used to acidify the permaganate solution during volumetric analysis? Sample paper-3 (UnSolved) 381

21. Both C = O and C = C bonds can undergo addition reactions. Write about the basic differences in their 3 nature. 22. (i) Methoxy group present on benzene ring activates it for SE reaction at ortho and para position with respect to –OCH3 group. Explain. 3 (ii) Complete the following reaction O—CH3 HI (iii) Explain Kolbe’s reaction with a suitable example. OR Name the reagents used in the following reactions: (i) Benzyl alcohol to Benzoic acid (ii) Dehydration of propan-2-ol to propene (iii) Butan-2-one to butan-2-ol 23. (i) Discuss hybridisation and magnetic behaviour of [Ni(CO)4] according to VBT. 3 (ii) On the basis of CFT write the electronic configuration for d6 ion if ∆0 < P. 24. Benzoic acid is completely dimerised in benzene. What will the pressure of the solution containing 61 g of benzoic acid per 500 g benzene when the vapour pressure of pure benzene at that temperature of experiment is 66.6 torr? What will be the vapour pressure in normal state? 3 OR At 27°C the osmotic pressure of 7 g protein per 100 mL solution is 25 mm Hg. Calculate the molecular mass of protein. 25. When a conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 ohm at 25°C. When the same cell was filled with aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohm–1. Calculate the molar conductivity of the unknown electrolyte at this concentration (specific conductance of 0.1 M KCl = 1.29 × 10–2 ohm–1 cm–1). 3 26. Draw crystal field energy level diagram and predict the number of unpaired electrons for the following complexes. [Atomic number of Fe (26), Ni (28), Cr (24)] 3 (i) [FeCl4]– (Tetrahedral) (ii) [NiCl4]2– (Tetrahedral) (iii) [Cr(CN)6]3+ (Octahedral) 27. Name the chemical components which constitutes a nucleotide. Write any three functions of nucleotides in a cell? 3 SECTION-D 28. An aromatic compound ‘A’ of molecular formula C7H6O2 undergoes a series of reaction as shown below. Write the structure of A, B, C, D and E in the following reaction. 5 (C7H6O2) A NH3 C6H5CONH2 Br2 + NaOH B heat Br2(aq) (CH3CO)2O LiAlH4 Ether D EC OR (i) Write the structures of main products formed when benzene diazonium chloride reacts with the following: (a) Phenol/OH– (b) C2H5OH (c) H3PO2 + H2O and heat (ii) Account for the following: (a) Ethylamine is soluble in water, whereas aniline is not. (b) Aniline does not undergo Friedel-Crafts reaction. 382 ChemiStry-12

29. Answer the following questions on the basis of the given plot of potential energy is reaction co-ordinate: 5 (i) What is threshold energy for the reaction? (ii) What is the activation energy for forward reaction? (iii) What is the activation energy for backward reaction? (iv) What is the enthalpy change for the reaction? (v) What is effect of catalyst on ∆G and Ea (activation energy) of reaction? OR (i) The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table. Experiment [A]/mol L–1 [B]/mol L–1 Initial rate (mol L–1 min–1) I 0.1 0.1 2.0 × 10–2 II — 0.1 4.0 × 10–2 III 0.4 0.4 — IV — 0.2 2.0 × 10–2 (ii) For a reaction, A + H2O → B, rate ∝ [A] what is its (i) molecularity (ii) order of reaction? SECTION-E 30. A semiempirical model with five parameters suitable for both pure and mixed solvents systems was proposed based on the influence of the electrolyte concentration and temperature on the number of free ions and ion mobility: K = (P1T + P2 ) mn exp  − P3m  . The  T − P4  fcHroo2nOmdbu2icn8tai3vr.iy1t5iseostlovoef3nN3t3sa.wC1e5lraeKnm.dTeCahaseuCrflee2adisnaitbptirleoimtpypylaeernnadetuacracecrsubvroaanrcayytineo−gf the proposed model were verified by the experimental data and the (k, m, T) data of 28 electrolyte solutions from the literature. The results showed that in a wide temperature and concentration range, the proposed model can fit the experimental data well for both pure and mixed solvents systems. The parameters in the proposed model are related to the type of solvent, and the value of n is closely relevant to the dielectric constant of the solvent. In a solvent with a high dielectric constant, the value of n is close to 1. In addition, compared with the conductivity model reported in the literature, the proposed model can obtain good accuracy with fewer parameters, which is of great value to practical engineering applications. (Reference: Weitao Zhang, Xia Chen, Yan Wang, Lianying Wu, and Yangdong Hu. Experimental and Modeling of Conductivity for Electrolyte Solution Systems. ACS Omega 2020,5,22465-22474) 4 The following questions are multiple choice questions. Choose the most appropriate answer: (i) What is the effect of change in temperature on conductivity of electrolyte? (a) Conductivity increases with increase in temperature. (b) Conductivity decreases with increase in temperature. (c) Conductivity increases initially and then becomes constant on increasing temperature. (d) No change in conductivity on change in temperature. (ii) What are the factors that influence conductivity of electrolytes? (a) Number of free ions (b) Ion mobility (c) Both (a) and (b) (d) Number of free electrons Sample paper-3 (UnSolved) 383

(iii) In the proposed model, the value of ’n’ is closely related to which property of the solvent? (a) Molar mass (b) Charge (c) Size (d) Dielectric constant OR How conductivity and molar conductivity of a weak electrolyte vary with increase in concentration? (a) Conductivity decreases, molar conductivity decreases (b) Conductivity increases, molar conductivity increases (c) Conductivity increases, molar conductivity decreases (d) Conductivity decreases, molar conductivity increases (iv) What is the correct unit of molar conductivity of an electrolyte solution? (a) S cm–1 (b) S cm (c) S cm–1mol–1 (d) S cm2 mol–1` Answers 1. (i) 2. (ii) 3. (i) 4. (i) 5. (i) 6. (iv) 7. (iii) 13. 1.68 g 15. (b) Order = 3 24. P = 69.01 torr OR mB = 156.06 g mol–1 25. Ÿm = 219.61 ohm–1 cm2 mol–1 30. (i) (a) (ii) (c) (iii) (d) OR (c) (iv) (d) 384 ChemiStry-12