32 Biomolecules and Chemistry in Everyday Life

32 Biomolecules and Chemistry in Everyday Life Topic 1 Biomolecules Objective Questions I (Only one correct option) 1. Which of the given statements is incorrect about glycogen? (a) It is straight chain polymer similar to amylose (b) Only a-linkages are present in the molecule (c) It is present in animal cells (d) It is present in some yeast and fungi 2. Which of the following statement is not true about RNA? (a) It controls the synthesis of protein (b) It has always double stranded a-helix structure (c) It usually does not replicate (d) It is present in the nucleus of the cell 8. Maltose on treatment with dilute HCl gives (a) D-glucose and D-fructose (b) D-fructose (c) D-galactose (d) D-glucose 9. The correct structure of histidine in a strongly acidic solution ( pH = 2 ) is r (a) H3N s COO r NH2 CH r (c) H3N COOH r NH2 N s COO CH r (d) H3N CH COOH + NH NH NH Nr H 10. The correct match between Item I and Item II is 4. Amylopectin is composed of (a) (b) (c) (d) CH N 3. Number of stereo-centers present in linear and cyclic structures of glucose are respectively (a) 4 and 5 (b) 4 and 4 (c) 5 and 4 (d) 5 and 5 r (b) H3N b-D-glucose, C1-C4 and C2-C6 linkages a-D-glucose, C1-C4 and C2-C6 linkages b-D-glucose, C1-C4 and C1-C6 linkages a-D-glucose, C1-C4 and C1 -C6linkages Item I Item II A. Ester test P. Tyr B. Carbylamine test Q. Asp C. Phthalein dye test R. Ser S. Lys 5. The peptide that gives positive ceric ammonium nitrate and carbylamine tests is (a) Lys-Asp (c) Gln-Asp (b) Ser-Lys (d) Asp-Gln 6. Which of the following statement is not true about sucrose? (a) It is also named as invert sugar. (b) The glycosidic linkage is present between C1 of a-glucose and C1 of b-fructose (c) It is a non-reducing sugar (d) On hydrolysis, it produces glucose and fructose (a) A ® Q; B ® S; C ® R (b) A ® R, B ® Q; C ® P (c) A ® R; B ® S; C ® Q (d) A ® Q; B ® S; C ® P 11 Among the following compounds, which one is found in RNA? O NH (a) N H (b) Barfoed’s test (d) Seliwanoff’s test (c) NH (b) N O N N Me (d) N H O O H O NH2 7. Fructose and glucose can be distinguished by (a) Fehling’s test (c) Benedict’s test O CH3 N Me O 454 Biomolecules and Chemistry in Everyday Life 12. Which of the following tests cannot be used for identifying amino acids ? (a) Barfoed test (c) Xanthoproteic test (b) Ninhydrin test (d) Biuret test 13. The correct sequence of amino acids present in the Me H N H2 N O Me O OH (a) Thr - Ser - Leu (c) Val - Ser - Thr OH C N H (a) Quinoline (c) Cytosine OH 21. Synthesis of each molecule of glucose in photosynthesis involves 14 The increasing order of pK a of the following amino acids in] aqueous solution is Gly, Asp, Lys, Arg H OH H O HO OH HO H HO H H (a) a ketohexose (c) an a-furanose CH2OH O H2COH O H H H H H HO OH H OCH2CH2O CH2OH HO OH H H OH (a) (b) 15. Glucose on prolonged heating with HI gives (b) 1-hexene (d) 6-iodohexanal 16. The predominant form of histamine present in human blood is (pK a , Histidine = 6.0 ) NH2 (a) H N (b) H N (c) r NH3 r N N H r NH3 (d) r N H N 17. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution? HOH2C O (a) CH2OH HOH2C O (b) HO (a) (b) (c) (d) Ring (a) is pyranose with a-glycosidic link Ring (a) is furanose with a-glycosidic link Ring (b) is furanose with a-glycosidic link Ring (b) is pyranose with b-glycosidic link 24. Two forms of D-glucopyranose, are called (a) enantiomers (c) epimers H N NH2 (b) an aldohexose (d) an a-pyranose 23. The correct statement about the following disaccharide is (a) Asp < Gly < Arg < Lys (b) Arg < Lys < Gly < Asp (c) Gly < Asp < Arg < Lys (d) Asp < Gly < Lys < Arg H N CH2OH (b) anomers (d) diastereomers 25. Which of the following pairs give positive Tollen’s test ? (a) Glucose, sucrose (b) Glucose, fructose (c) Hexanal, acetophenone (d) Fructose, sucrose Objective Question II (One or more than one correct option) 26. The Fischer presentation of D-glucose is given below. CHO HO OCOCH3 OH HOH2C (c) O CH OCH 2 3 CH2OH (d) HO OCH3 CH2OH OH D-glucose OH The correct structure(s) of b-L-glucopyranose is (are) OH OH H 18. Thiol group is present in (a) cystine (c) methionine (b) cysteine (d) cytosine 19. Which of the vitamins given below is water soluble? (a) Vitamin C (c) Vitamin E OH H OH OH H HO H H OH HOH2C O (b) 10 molecules of ATP (d) 6 molecules of ATP 22. The following carbohydrate is O (b) Leu - Ser - Thr (d) Thr - Ser - Val (a) n-hexane (c) Hexanoic acid (b) Adenine (d) Thymine (a) 18 molecules of ATP (c) 8 molecules of ATP tripeptide given below is Me 20. Which one of the following bases is not present in DNA? (b) Vitamin D (d) Vitamin K HO (a) H H O CH2OH H OH HO H OH HO (b) H H O CH2OH OH H H OH OH H Biomolecules and Chemistry in Everyday Life CH2OH H O H H (c) HO HO CH2OH (d) OH HO H OH O H H H H HO OH Integer Answer Type Questions H 30. The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is OH O H O O 27. For ‘invert sugar’, the correct statement(s) is (are) (Given: specific rotations of ( + ) - sucrose, (+) - maltose, L-( - ) -glucose and L-( + ) -fructose in aqueous solution are + 66 °, +140 °, - 52 °and 92° , respectively) (a) Invert sugar is prepared by acid catalysed hydrolysis of maltose (b) Invert sugar is an equimolar mixture of D-(+) -glucose and D-(-)- fructose (c) Specific rotation of invert sugar is - 20 ° (d) On reaction with Br2 water, invert sugar forms saccharic acid as one of the products 28. The correct statement(s) about the following sugars X and Y is/are: CH2OH H O H OH H HOH2C H HO OH OH CH2 O N H N N H OH H O H OH 31. A tetrapeptide has ¾ COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with ¾ NH2 group attached to a chiral centre is 32. The substituents R1 and R2 for nine peptides are listed in the CH2OH H H H OH H OH H (a) (b) (c) (d) H OH H OH R2 R1 ¾ CO ¾ NH ¾ CH ¾ COO O H O H CH2OH O CH2OH CH2 O X H H H N N O H N O H3N ¾ CH ¾ CO ¾ NH ¾ CH ¾ CO ¾ NH ¾ CH H O H N H O table given below. How many of these peptides are positively charged at pH = 7.0 ? O H OH N H Y X is a reducing sugar and Y is a non-reducing sugar X is a non-reducing sugar and Y is a reducing sugar The glucosidic linkages in X and Y are a and b, respectively The glucosidic linkages in X and Y are b and a, respectively Peptide R1 I H H II H CH 3 III CH 2COOH H IV CH 2CONH 2 (CH 2 ) 4 NH 2 CH 2CONH 2 V CH 2CONH 2 VI (CH 2 ) 4 NH 2 (CH 2 ) 4 NH 2 VII CH 2COOH CH 2CONH 2 VIII CH 2OH (CH 2 ) 4 NH 2 IX (CH 2 ) 4 NH 2 CH 3 33. When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form, is Assertion and Reason CHO Read the following questions and answer as per the direction given below : CH2 (a) Statement I is correct; Statement II is correct; Statement II is a correct explanation of Statement I. (b) Statement I is correct; Statement II is correct; Statement II is not the correct explanation of Statement I. (c) Statement I is correct; Statement II is incorrect. (d) Statement I is incorrect; Statement II is correct. 29. Statement I Glucose gives a reddish-brown precipitate with Fehling’s solution. Statement II Reaction of glucose with Fehling’s solution gives CuO and gluconic acid. R2 CHOH CHOH CHOH CH2OH 34. A decapeptide (Molecular weight 796) on complete hydrolysis gives glycine (Molecular weight 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is Biomolecules and Chemistry in Everyday Life 35. The total number of basic groups in the following form of (i) Draw the structure of L-glucose. (ii) Give the reaction of L-glucose with Tollen’s reagent. lysine is Å O H3N—CH2—CH2—CH2—H2C 38. Following two amino acids lysine and glutamine form CH—C dipeptide linkage. What are two possible dipeptides? s O H 2N H2 N ¾ CH ¾ COOH ½ CH2 CH2 CH2 CH2 NH2 Subjective Questions 36. Which of the following disaccharide will not reduce Tollen’s reagent? CH2OH CH2OH O HO (a) H HO H O H H H O O OH (b) H HO OH following structure CH2 C6H5 ½ H2 N ¾ CH ¾ CONH ¾ CH ¾ COOCH3 ½ CH2 ¾ COOH OH O O 39. Aspartame, an artificial sweetener, is a peptide and has the H CH2OH OH H H H HO H OH H CH2OH H2 N ¾ CH ¾ COOH ½ CH2 CH2 COOH OH OH H H OH H H (i) Identify the four functional groups. (ii) Write the Zwitter ionic structure. (iii) Write the structures of the amino acids obtained from the hydrolysis of aspartame. (iv) Which of the two amino acids is more hydrophobic? 37. The structure of D-glucose is as follows : CHO H OH HO H 40. Give the structures of the products in the following reaction H H OH H+ Sucrose ¾® A + B OH OH 41. Write the structure of alanine at pH = 2 and pH = 10. Topic 2 Chemistry in Everyday Life Objective Questions I (Only one correct option) 1. Which of the following is a thermosetting polymer? (2019 Main, 12 April I) (a) Bakelite (c) Nylon-6 (b) Buna-N (d) PVC High density polythene I. Peroxide catalyst B. Polyacrylonitrile II. Condensation at high temperature and pressure C. Novolac III. Ziegler-Natta catalyst D. Nylon-6 IV. Acid or base catalyst Codes B I IV (b) Neoprene (d) Buna - S (a) antidepressant (c) neurotransmitter (b) antihistamine (d) antacid Item-II A. A (a) III (c) II (a) Nylon-6, 6 (c) Teflon 4. Noradrenaline is a/an 2. The correct match between Item-I and Item-II is Item-I 3. Which of the following is a condensation polymer? C IV I D II III A (b) IV (d) III B II I C I II D III IV 5. Which of the following compounds is a constituent of the O ½½ polymer — [ HN — C — NH — CH 2— ]n ? (a) (b) (c) (d) N-methyl urea Methylamine Ammonia Formaldehyde Biomolecules and Chemistry in Everyday Life 6. The structure of nylon-6 is 12. The correct match between item (I) and item (II) is O H O H || | || | (a) ¾ [ (CH 2 )6 ¾ C ¾ N ¾ ]n (b) ¾ [ C ¾ (CH 2 )5 ¾ N ¾ ]n O H O H || | || | (c) ¾ [ (CH 2 )4 ¾ C ¾ N ¾ ]n (d) ¾ [ C(CH 2 )6 ¾ N ¾ ]n Norethindrone (P) (B) Ofloxacin (Q) Antifertility (C) Equanil (R) Hypertension (S) Analgesics (a) (b) (c) (d) 7. The two monomers for the synthesis of nylon 6, 6 are (a) HOOC(CH 2 )4 COOH, H 2N(CH 2 )4 NH 2 (b) HOOC(CH 2 )6 COOH, H 2N(CH 2 )4 NH 2 (c) HOOC(CH 2 )4 COOH, H 2N(CH 2 )6 NH 2 (d) HOOC(CH 2 )6 COOH, H 2N(CH 2 )6 NH 2 Item - II Item - I (A) Antibiotic (A) ® (Q); (B) ® (R); (C) ® (S) (A) ® (Q); (B) ® (P); (C) ® (R) (A) ® (R); (B) ® (P); (C) ® (S) (A) ® (R); (B) ® (P); (C) ® (R) 13. The correct match between Item - I and Item - II is Item I (Drug) Item II (Test) A. Chloroxylenol P. Carbylamine test B. Norethindrone Q. Sodium hydrogen carbonate test copolymer of …… C. Sulphapyridine R. Ferric chloride test (a) 3-hydroxybutanoic acid and 2-hydroxypentanoic acid (b) 2-hydroxybutanoic acid and 3-hydroxypentanoic acid (c) 3-hydroxybutanoic acid and 4-hydroxypentanoic acid (d) 3-hydroxybutanoic acid and 3-hydroxypentanoic acid D. Penicillin S. Bayer’s test 8. Poly-b-hydroxybutyrate-Co-b-hydroxyvalerate (PHBV) is a 9. The homopolymer formed from 4-hydroxybutanoic acid is O O (a) —OC(CH2)3—O— n O hydrolysis reaction? n O (a) Nylon-6 (b) Bakelite (c) Nylon-6, 6 (d) Terylene (d) —C(CH2)2C—O—n (c) —C(CH2)3—O— n 10. The correct match between Item I and Item II is Item I Allosteric effect P. Molecule binding to the active site of enzyme. B. Competitive inhibitor Q. Molecule crucial for communication in the body. C. Receptor R. Molecule binding to a site other than the active site of enzyme. D. Poison S. Molecule binding to the enzyme covalently. A ® P; B ® R; C ® S; D ® Q A ® P, B ® R; C ® Q; D ® S A ® R; B ® P; C ® S; D ® Q A ® R; B ® P; C ® Q; D ® S + HOOC (a) It is a poor conductor of electricity (b) Its synthesis required dioxygen or a peroxide initiator as a catalyst (c) It is used in the manufacture of buckets, dustbins etc. (d) Its synthesis requires high pressure 16. Which of the following is an anionic detergent?(2016 Main) (a) Sodium lauryl sulphate (b) Cetyltrimethyl ammonium bromide (c) Glyceryl oleate (d) Sodium stearate 17. On complete hydrogenation, natural rubber produces 11. The polymer obtained from the following reaction is: NH2 15. Which of the following statements about low density polythene is false? Item II A. (a) (b) (c) (d) A®R;B®P;C®S;D®Q A®R;B®S;C®P;D®Q A®Q;B®P;C®S;D®R A®Q;B®S;C®P;D®R 14. The formation of which of the following polymers involves (b) —C(CH2)2C— O O (a) (b) (c) (d) (i) NaNO2 /H3O (ii) Polymerisation (a) ethylene-propylene copolymer (b) vulcanised rubber (c) polypropylene (d) polybutylene 18. Which polymer is used in the manufacture of paints and O O lacquers? H (a) —C—(CH2)4—N— n O (b) —O—(CH2)4—C—n O (c) —OC(CH2)4O— n O H (d) —HNC(CH2)4—C—N— n (a) Bakelite (b) Glyptal (c) Polypropene (d) Polyvinyl chloride Biomolecules and Chemistry in Everyday Life AcO 19. Match the polymers in Column I with their main uses in Column II and choose the correct answer: (c) Column I (A) (B) (C) (D) Column II Polystyrene Glyptal Polyvinyl chloride Bakelite 1. 2. 3. 4. 20. Paints and lacquers Raincoats Manufacture of toys Computer discs HH OAc H O H OAc A B C D (b) 3 1 2 4 (d) 3 4 2 1 HH H O OH H AcO H H AcO H OH H OAc H O AcO (d) Codes A B C D (a) 2 1 3 4 (c) 2 4 3 1 AcO H OH OAc OAc OAc OH H H O O OO H OAc H H H OAc AcO H H OO H H O H H OAc OAc OAc OAc Objective Question II (One or more than one correct option) OCOCH3 COOH is used as 25. Under hydrolysis conditions, the compounds used for preparation of linear polymer and for chain termination, respectively are (a) Insecticide (c) Analgesic (b) Antihistamine (d) Antacid 21. Which of the following is not an antacid? (a) Aluminium hydroxide (c) Phenelzine (a) CH3 SiCl3 and Si(CH3 )4 (b) (CH3 )2 SiCl2 and (CH3 )3 SiCl (c) (CH3 )SiCl2 and CH3 SiCl3 (d) SiCl4 and (CH3 )3 SiCl 26. The correct functional group X and the reagent/reaction (b) Cimetidine (d) Ranitidine conditions Y in the following schemes are 22. Which one is classified as a condensation polymer? X—(CH2)4—X (i) Y O O (a) Dacron (c) Teflon (b) Neoprene (d) Acrylonitrile 23. Among cellulose, poly (vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is (a) nylon (b) poly (vinyl chloride) (c) cellulose (d) natural rubber AcO O O HH OAcH O H O H OAcH H H H HH O OH H O H OH O H OAc OH X = COOCH3, Y = H2/Ni/heat X = CONH2, Y = H2/Ni/heat X = CONH2, Y = Br2/NaOH X = CN, Y = H2/Ni/heat H OH H H H OH Column II Column I A. B. C. D. Cellulose Nylon-66 Protein Sucrose p. q. r. s. Natural polymer Synthetic polymer Amide linkage Glycoside linkage Subjective Questions 28. Monomer A of a polymer on ozonolysis yields two moles of (a) Deduce the structure of A. (b) Write the structure of ‘all cis’ – form of polymer of compound A. 29. Name the heterogeneous catalyst used in the polymerisation O OO H OH H H H H OH 30. Give the structures of the products in the following reaction. H O polymers/type of bond in Column II. HCHO and one mole of CH3COCHO. AcO AcO AcO H OAc OAc (b) O OO H OAcH H H AcO H (a) (b) (c) (d) heat 27. Match the chemical substances in Column I with type of H2 SO4 (catalytic) gives cellulose triacetate whose structure is AcO C—(CH2)4—C HO Match the Columns 24. Cellulose upon acetylation with excess acetic anhydride/ (a) (ii) Condensation polymer of ethylene. NOH H+ C Polymerisation *—D—n* Answers Topic 1 1. (a) Topic 2 2. (b) 3. (a) 4. (d) 1. (a) 2. (a) 3. (a) 4. (c) 5. (b) 6. (b) 7. (d) 8. (d) 5. (d) 6. (b) 7. (c) 8. (d) 9. (a) 10. (d) 11. (a) 12. (a) 9. (c) 10. (d) 11. (b) 12. (b) 13. (c) 14. (d) 15. (a) 16. (d) 13. (b) 14. (a) 15. (c) 16. (a) 17. (a) 18. (b) 19. (a) 20. (a) 17. (a) 18. (b) 19. (b) 20. (c) 21. (a) 22. (b) 23. (a) 24. (b) 21. (c) 22. (a) 23. (d) 24. (a) 26. (a, b, c, d) 25. (b) 26. (d) 27. (b,c) 28. (b,c) 25. (b) 29. (c) 30. (1) 31. (4) 32. (4) 27. A ® p, s; B ® q, r; C ® p, r; D ® s 33. (8) 34. (6) 35. (2) Hints & Solutions Topic 1 Biomolecules 1. Statement (a) is incorrect. Glycogen is not a straight chain polymer similar to amylose. It is highly branched structure similar to amylopectin. It is known to be the storage material of animals. It is found in liver, muscles and brain. It breaks down to glucose by the action of enzymes when body needs a glucose. It is also found in yeast and fungi. 2. RNA does not have double stranded a-helix structure. Helixes a - D- glucose units and the branches are connected to the main chain by a - 1,6¢- glycosidic linkages. Its structure can be represented as: 6 4 CH2OH O HO 3 Linear structure of glucose is as follows : CHO HO H H OH H OH CH2OH CH2OH O O H H H H H OH HO OH OH a-D-glucose HO 6 4 O 2 5 3 H OH CH2OH H O 2 5 HO H 1 3 OH O O a-1, 4¢-glycosidic linkage (Chain) Structure of Amylopectin 5. The peptide that gives positive cerric ammonium nitrate and Serine Fischer formula Number of stereo-centre (C* ) = 4. Cyclic structure of glucose are as follows : HO OH a-1, 6¢-glycosidic linkage (Branch) O 6 4 CH2 Lysine 6. Statement-(b) is not true for sucrose. It is linked through a OH CH2OH H H 1 carbylamine tests is ser - lys. The structures of serine and lysine are, HO ¾ CH2 ¾ C H ¾ COOH H2N ¾ (CH2)4 ¾ C H ¾ COOH | | NH2 NH2 stereogenic center. H H OH present in RNA are single-stranded but sometimes they fold back on themselves to form a double helix structure. RNA usually does not replicate. It is present in the nucleus of the cell. It controls the synthesis of protein. RNA molecules are of three types, i.e. messenger’s RNA (m-RNA), ribosomal RNA (rRNA), transfer RNA (t-RNA). 3. Key Idea Chiral centre is also called stereo-centre or O 2 5 glycosidic linkage between C-1 of a-glucose and C-2 of b-fructose. Since, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar. CH2OH H OH H H OH HO O H OH H H OH 4. Amylopectin is the water-soluble component of starch. It is a branched-chain polymer of a-D-glucose. The main chain consists of an a - 1, 4¢- glycosidic linkages between a - D- HOH2C b O Glycosidic linkage H HO CH2OH H a-D-glucose OH H b-D-fructose b-D-glucose Haworth formula Number of stereo-centre (C* ) in each anomer = 5. H a On hydrolysis with acids or enzyme, sucrose gives equimolar mixture of D-(+)-glucose and D-(–)-fructose. HCl C12H22O11 + H2O ¾¾® C6H12O 6 + D-(+) -glucose C6H12O 6 D-(–)-fructose Biomolecules and Chemistry in Everyday Life 7. Both fructose and glucose give following test positive. R (i) Fehling’s test (red ppt. of Cu 2O is obtained). (ii) Barfoed’s test (red ppt. of Cu 2O is obtained) (iii) Benedict’s test (red ppt. of Cu 2O is obtained) (1%) (92%) Benedict’s solution : CuSO4 + Na-citrate + Na 2CO3 8. Maltose on treatment with dil. HCl gives D-glucose. Hydrolysis of maltose yields two moles of a- D-glucose. Thus, it is composed of two a-D-glucose units in which C-1 of one glucose unit (I) is linked to C-4 of another glucose unit (II). The free aldehyde group can be produced at C-1 of second glucose in solution and it shows reducing properties. So, it is a reducing sugar. H OH They have the tendency to loose H+ of their ¾COOH group at alkaline (higher) pH while the ¾ NH2 group present in them have the tendency to gain H+ at acidic (lower) pH. 10. The correct match is : Seliwanoff’s test is used to differentiate between ketose and aldose. The reagent is a solution of resorcinol in concentrated HCl. The reagent when heated along with a sugar will produce furfural or hydroxy-methylfurfural, which further reacts to give red color. Ketose (fructose) reacts more quickly than aldose (glucose). CH2OH A ® (Q); (B) ® (S) (C) ® (P) R Þ CH2—COOH C2H5OH/H2SO4/D Asp [Aspartic acid] [Ester test] (A) (Q) Sweet smell of ester R Þ (CH2)4—NH2 CHCl /alc.KOH 3 Lys [Lysine] [Carbylamine test] (B) (S) OOC Foul smell CH— R of isocyanide O H3N General formula of O, Conc. H2SO4/D amino acid OH R Þ ¾CH2 O Tyr [Tyrosine] (P) CH2OH O H H H H O OH H H OH O H OH H H OH OH CH2OH Dil. HCl H O H 2 OH OH H OH H H OH Lone pairs of both of these nitrogen are freely available for donation C CH2 O N N (U). In the given options, O NH OH + N H is uracil (present in RNA only) O O These lone pairs are not free for donation as H these are delocalised with-p bonds. At highly acidic pH, i.e. 2 both the nitrogens with lone pairs will accept one H+ each and ¾ C ¾ OH will not loose its H+ . Thus, ½½ O the final structure of histidine at pH = 2 will be H 3N Resultant colourless solution turns pink 11. RNA contains, adenine (A), guanine (G), cytosine (C) and uracil 9. Histidine has following structure in CH [Phthalein dye test] (C) (A) Ester test confirms the presence of ¾ COOH group. (B) Carbylamine test confirms the presence of ¾ NH2 group (1°). (C) Phthalein dye test confirms the presence of phenolic ¾ OH group. a-D-glucose H 2N COOH NH2 Fehling’s solution :CuSO4 + Na, K-tartrate (Rochelle salt) Barfoed’s reagent (CH3COO )2 Cu + CH3COOH + H 2O (7%) C CH C CH2 O OH H3 C NH N H is thymine (present in DNA) O NH2 N is cytosine (present in both DNA and RNA) N H O O H N¾Me is not a pyrimidine base but a derivative of uracil. + N NH Thus, option (d) is the correct answer. Note Amino acids have following generalise structure: N Me O 12. (i) Barfoed test is used for detecting the presence of monosaccharides like glucose, fructose etc. Barfoed reagents is Cu (II) acetate solution. Biomolecules and Chemistry in Everyday Life CHO OH H + Cu2++ 2H2O OH (Cu (II)-acetale) OH H HO H H 15. HI is a strong reducing agent. It reduces both primary and COOH OH H + Cu2O +4H+ OH (Red ppt.) OH H HO H H CH2OH secondary alcoholic groups of glucose along with the carbonyl group to produce n-hexane as (D-gluconic acid) Glucose (Ninhydrin Test) Violet colouration Ninhydrin/ D Amino acid (Xanthoprotic Test) (Protein) Conc. HNO3 Me CH CH O OH H HN H 2N CH O CH2 Valine (Val) Me –2H2O O Me CH CH NH H 2N CH O CH2 Val OH O Threonine (Thr) Me CH It is produced by decarboxylation of histidine having following structure. It is clearly visible from the above structure that histamine has two basic centres namely aliphatic amino group and basic nitrogen of imidazole ring. The aliphatic amino group has pK a around 9.4. In blood with pH around 7.4 the aliphatic amino group of histamine become protonated to give a single charged cation as shown below H N OH CH OH 17. Sugars that have an aldehyde, a ketone, a hemiacetal or a O OH Ser hemiketal group is able to reduce an oxidising agent. These sugars are classified as reducing sugars. Thr O HOH2C 14 Amino acid molecules can be represented as, CH2OH HO COOH R + NH3 N C NH NH2 Basic nitrogen of imidazole ring OH C OH C C CH OH H HN Serine (Ser) Aliphatic amino group N CH C C n-hexane H Imidazole N ring Violet colouration Me CH3 7.35-7.4. The structure of histamine is given below : 13. Formation of the tripeptide (Val-Ser-Thr) can be shown as: Me NH2 OH + ¾ (CH2)3¾ NH ¾ C NH2 Nature of R Basic NH2 O CH2 C Acidic 3. ¾H (Glycine) : Gly Hemiacetal can be easily reduced by oxidising agent such as Tollen’s reagent. O HOH2C 4. ¾ ( CH2 ) 4 ¾ NH3 + (Lysine : Lys) CH2OH O Neutral Neutral Basic Basic CH2OH O HO OH (a-hydroxyketone) O– (Aspartic acid: Asp) O HOH2C More basic (due to the presence of acetamidine group) Acidic O– OH Hemiacetal Nature of the amino acids (Arginine : Arg) 2. CH2OH OH –+ –CH3COOK Nature of the ‘R’ group will determine the basicity (hence, pK a ) of an amino acid. 1. O HOH2C aq·KOH OCOCH3 CH ‘R’ in the amino group (CH2)4 16. Our blood is slightly basic in nature with pH range from Yellow colouration (Biuret Test) CuSO4 (aq)+NaOH HI D (Prolonged) Primary alcoholic group CH2OH (ii) CH3 Secondary alcoholic group (CHOH)4 CH2OH (D-glucose) Carbonyl group CHO OH OH (Reducing sugar) Tollen’s reagent positive silver mirror test. Biomolecules and Chemistry in Everyday Life NH2 18. Cystine HO H S C Cysteine COOH HS Methionine Cytosine N O S C OH , H 3C O NH2 C2 OH HO C3 H H C4 H C5 NH2 20. Quinoline is an alkaloid, it is not present in DNA. DNA has four nitrogen bases in adenine, guanine, cytosine and thymine. 21. 18 ATPs are involved in the formation of 1 glucose molecule as shown in the reaction below : 6CO2 + 12NADPH + 18 ATP ¾® C6 H12 O6 + 12NADP + 18 ADP H 5 H OH HO C4 H OH HO C5 H 6CH2OH L-glucose 24. “a” and “b” cyclic hemiacetals of D-glucose having difference in configuration at C-1 only are called anomers. 25. Both glucose and fructose are reducing sugars, reduces Tollen’s reagent to metallic silver. 3 OH and —C—OH looks like OH O C Hemiacetal or Hemiketal 1C O H 4 H O H CH2OH 6 H 1 OH OH 2 3 H OH H b-L-glucopyranose The a-L-glucopyranose has configurational change at C1 only and looks like H O HO CH2OH HO Configuration at 1 H OH H OH H this carbon atom is opposite in a and b-forms a-L-glucopyranose 27. If there is inversion of specific rotation from (+ ) to (-), then invert sugar is formed. (a ) C12H22O11 + H2O ¾® Glucose (+) Maltose 140° D(+) 52 ° (b) C12H22O11 + H2O ¾® Glucose + Fructose (+) Sucrose + 66 ° 26. (d) A pyranose ring is a 6 membered ring having 5 carbon atoms and one oxygen atom. In glucose, it is formed by the reaction between >C=O group at position 1 and —OH group at 5th carbon atom. In general reaction between >C=O group 5 H 2 H the five membered cyclic ether is known as furanose. Hence, ring (a) is a pyranose and it has ether linkage at a-position that is known as a-glycosidic linkage in carbohydrate chemistry. Alcoholic group OH OH H 23. The six-membered cyclic ether is known as pyranose while Carbonyl group C3 HO 4 is a b-pyranose of an aldohexose. C H H O HO 6 CH OH 2 22. Here, the — OH of hemiacetal group is equatorial therefore, it OH H So, b-L glucopyranose is formed as K are fat soluble or water insoluble. C C2 D-glucose 19. Vitamin B and C are water soluble while vitamin A,D,E and O+ HO 6CH2OH N H Thiol group (SH) is present in cysteine. C C1 H NH2 NH2 O C1 S O H O COOH D(+) L(-) 52 ° -92 ° -40 ° for 2 moles mixture -20 ° for 1 mole mixture There is formation of invert sugar. Thus, correct. (c) Specific rotation of invert sugar is -20 ° per mole. Thus, correct. (d) Br2 water is a weak oxidising agent. It oxidises ¾ CHO to ¾ COOH. ¾ CH2OH group is not affected. COOH CHO COOH ½ ½ ½ Br water 2 HNO (CHOH)4 ¬¾¾3¾ (CHOH)4 ¾¾¾® (CHOH)4 ½ ½ ½ COOH CH2OH CH2OH The product formed in called hemiacetal Saccharic acid (if>C=O group belongs to an aldehyde) or hemiketal (if >C=O group belongs to a ketone). L- glucose has the mirror image configuration of D-glucose i.e., HNO3 (a strong oxidising agent) oxidises invert sugar to saccharic acid. Thus, incorrect. Gluconic acid (one of the products) Biomolecules and Chemistry in Everyday Life 28. X is acetal, has no free hemiacetal, hence a non-reducing sugar 32. The amino acid remain completely in Zwitter ionic form at its while Y has a free hemiacetal group, it is reducing sugar. Also, glucosidic linkage of X is ‘a’ while that of Y is b-linkage. isoelectric point. Amino acids with additional acidic group have their isoelectric pH less than 7.0 and increasing pH above isoelectric point makes them anionic. 29. Statement I is correct Presence of — CHO group in glucose is tested by Fehling’s solution test where a reddish-brown precipitate of Cu 2O is formed. Hence, Statement II is incorrect. 30. PLAN This problem can be solved by performing hydrolysis of peptide and deciding the nature of product. Chemical reaction and product formed after hydrolysis of given peptide can be represented as O O NH—C N H CH2 N O H O H O N NH Peptide IV VI H VIII IX N N O O N On the other hand, amino acids with additional basic group have their isoelectric pH greater than 7.0 and decreasing pH below isoelectric point (by adding acid solution) makes them cationic. The given peptide with followings R1 and R2 are basic, will remain protonated (cationic) at pH = 7.0. CH2 H O R1 R2 CH 2CONH 2 (CH 2 ) 4 NH 4 (CH 2 ) 4 NH 2 (CH 2 ) 4 NH 4 CH 2OH (CH 2 ) 4 NH 4 (CH 2 ) 4 NH 2 CH 3 Thus, 4 is the correct integer. 33. The D-form of given sugar is O CHO H3O+ CMe3 (A) + H 3N CO2 CH2 + (B) CHOH HO—C—CHNH—CH2—C—OH O CH2 O CH2OH CH2 + + – H3N—CH2—CO2+ + O2C—CH—NH3 O H Cyclisation H H * CHOH * OH H * OH H OH H CH2OH D-Pyranose D- Sugar (C ) (D ) (A) is glycine which is only naturally occurring amino acid. While (B), (C) and (D) are not the naturally occurring amino acids. Hence, correct integer is (1). 31. PLAN A peptide linkage is hydrolysed to two free amino acids. O || H2N ¾ CH ¾ C ¾ NH ¾ CH ¾ COOH | | R R¢ ­ Peptide O || * * H2N ¾ CH ¾ C ¾ OH + H2N ¾ CH ¾ COOH | | R R¢ C* is chiral carbon tetrapeptide has four amino acids joined by three peptide linkage. ¾ COOH group is on alanine part, thus it is at fixed C-terminal position in each combination. Glycine is optically inactive thus it cannot be on the N¾ terminal side. Thus, possible combinations are Phe-Gly-Val-Ala, Phe-Val-Gly-Ala, Val-Gly-Phe-Ala, Val-Phe-Gly-Ala Thus, in all four combinations are possible. Configurations at the three chiral carbons (starred) can be changed maintaining D-configuration. Hence, the total number of steroisomers of D-pyranose = 23 = 8 Thus, the correct integer is 8. 34. A decapeptide has nine peptide (amide) linkage as O O O O O O O O O O —CONH— bond Therefore, on hydrolysis, it will absorb nine water molecules. Hence, total mass of hydrolysis product = 796 +18 × 9 = 958 Þ mass of glycine in hydrolysis product = 958 ´ 47 = 450 100 Þ number of glycine molecule in one 450 molecule of decapeptide = =6 75 35. — OO- and —NH2 are two basic groups in lysine. 36. In structure (a), one ring has a free hemiacetal group, will hydrolyse into open chain in aqueous solution and therefore will reduce Tollen’s reagent. Structure (b) has only acetal groups, will not hydrolyse in aqueous solution into open chain, will not reduce Tollen’s reagent Biomolecules and Chemistry in Everyday Life 37. (i) D-glucose and L-glucose are enantiomers, hence Mirror CHO H OH HO OH H CHO HO H H CHO HO H H C H H OH H OH OH H HO H HO pH = 10 ½ CH3 ½ CH3 Alanine COOH H OH H (ii) H H OH H OH H OH H OH 1. Bakelite is a thermosetting polymer. These polymers are CH2OH gluconic acid cross-linked or heavily branched molecules which on heating undergo extensive cross linking in moulds and become infusible. Once they get set, they cannot be reshaped and reused. CH2OH 38. The dipeptides are HOOC ¾ (CH2 )2 ¾ CH ¾ CO ¾ NH ¾ CH ¾ (CH2 )4 ¾ NH2 ½ ½ NH3 COO + and HOOC ¾ (CH2 )2 ¾ CH ¾ NH ¾ CO ¾ CH ¾ (CH2 )4 ¾ NH2 ½ ½ NH3 COO+ O CH2C6H5 ½ ½½ 39. H2N ¾ CH ¾ C ¾ NH ¾ CH ¾ COOCH3 ½ CH2 ¾ COOH Topic 2 Chemistry in Everyday Life 2. (A) [TiCl + (C H ) Al] 4 2 5 3 n CH2 == CH2 ¾¾¾¾¾¾® Ziegler-Natta catalyst (III) Ethylene ¾ [ CH2 ¾] n High density polythene (HDPE) CN CN | ½ O22 - /O2 (B) nCH2 == CH ¾¾¾¾® ¾ [ CH2 ¾ C H ]¾ n (Catalyst) (I) Acrylonitrile Polyacrylonitrile (PAN) (C) OH OH + HCHO Formaldehyde Phenol CH2OH sOH Polymerisation/ H Or Acid or base catalyst (IV) Aspartame 3 OH OH (i) Aspartame has amine, acid, amide and ester groups. O CH2C6H5 ½ + ½½ (ii) H3 N ¾ CH ¾ C ¾ NH ¾ CH ¾ COOCH3 ½ CH2 ¾ COO H+ (iii) Aspartame ¾¾® H2N ¾ CH ¾ COOH + H2 O ½ CH2 COOH CH2 n Novolac (D) CH2C6H5 ½ H2N ¾ CH ¾ COOH + CH3OH II (iv) II is more hydrophobic due to the presence of phenyl group. H+ Sucrose ¾¾® D-glucose + D-fructose H2 O O NH O Caprolactum I 40. + pH = 2 ¾¾® H3N ¾ CH ¾ COOH ½ CH3 OH + Ag(NH3)2 HO OH CH2OH 41. H2N ¾ CH ¾ COO- ¾¾® H2N ¾ CH ¾ COOH CH2OH L-glucose CHO OH H H CH2OH D-glucose H H CH2OH H O HO OH HO CO2OH OH 550 K 20 atm –H2O —C—NH—(CH2)5— (II) n Nylon-6 Thus, the correct match is as follows : (A) ® (III), (B) ® (I), (C) ® (IV), (D) ® (II) 3. Nylon-6, 6 (an amide) is a condensation copolymer because it is obtained by condensation hexamethylenediamine. between adipic acid and Biomolecules and Chemistry in Everyday Life O O of adipic acid and hexamethylene diamine is the correct answer. Both of these units react as follows to form nylon-6, 6. H O nHO—C—(CH2)4—C—OH + nN—(CH2)6—NHH Adipic acid O –nH2O Hexamethylene diamine H O ——C—(CH2)4—C—NH—(CH2)6—NH—— nH (CH2)6 N H Hexamethylene diamine n 1 mol nylon-6,6 plays a major role in mood changes. If the level of noradrenaline is low for some reason, then signal-sending activity becomes low and the person suffers from depression. O || 5. Monomer of ¾[NH ¾ C ¾ NH ¾ CH2¾ ]n is formaldehyde. The polymer is also known as urea-formaldehyde resin. It is made from urea (NH2CONH2) and formaldehyde (HCHO). Polymerisation NH 2CONH 2 + HCHO ¾¾¾¾¾® O || ¾(NH ¾ C ¾ NH ¾ CH 2 )¾ n Urea-formaldehyde resin O NH C OH Adipic acid O (CH2)6 H N C H O (CH2)4 C Nylon-6, 6 n 8. Poly-b-hydroxy butyrate Co-b-hydroxyvalerate (PHBV) is a copolymer of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid. It is used in speciality packaging, orthopaedic devices and in controlled release of drugs. PHBV undergoes bacterial degradation in the environment. The reaction involved is as follows : OH OH ½ ½ CH3 ¾ CH ¾ CH2COOH + CH3CH2 CHCH2COOH ¾® 3-hydroxypentanoic acid CH2CH 3 æ ö O ç ÷ ½ ½½ ç ÷ ¾ç O ¾ CH ¾ C H2 C ¾ O ¾ CH ¾ CH2 ¾ C ÷¾ ½ ½½÷ ç O÷ n ç CH3 è ø It is used for making unbreakable cups and laminated sheets. 6. (CH2)4 C –nH2O H 3-hydroxybutanoic acid Formaldehyde Urea H N Neoprene, teflon and buna-S are addition polymers. 4. Noradrenaline is one of the example of neurotransmitters. It H + OH N (PHBV) 9. On polymerisation, 4-hydroxy butanoic acid will produce a condensation homopolymer by loss of H2O molecules. Polymerisation O O HO—C—CH2CH2CH2—O H + HO —C—(CH2)3—O H Caprolactum O 4-hydroxy butanoic acid –H2O O O —C—NH—(CH2)5 —C—NH—(CH3)5— -----—C—(CH2)2CH2—O—C—(CH2)2CH2—O---- or Ester linkage O —NH—(CH2)5—C— n Nylon-6 Nylon-6 is prepared by ring opening polymerisation of caprolactum. It is heated about 533 K in an inert atmospheric nitrogen about 4-5 hrs. Nylon-6 fibres are tough, possessing high tensile strength, as well as elasticity and lustre. They are wrinkle proof and highly resistant to abrasion and chemicals such as acids and alkalis. 7. Nylon-6,6 has following structure: O O ù é ê ½½ ½½ ú ê N ¾ (CH2)6 ¾ N ¾ C ¾ (CH2)4 ¾ C ¾ ú ¾ ê ½ ú ½ ê H ú H ë ûn 6 C-atoms O 6 C-atoms As it is a condensation polymer hence, each of its monomeric unit must contain 6 carbon atoms in them. Hence, a combination The homopolymer obtained can also be represented as O ½½ — [ C — (CH2)3 O —]n 10. (A) Molecule binding to a site other than the active site of enzyme is called allosteric effect. (B) Molecule binding to the active site of enzyme is called competitive inhibitor. (C) Molecule crucial for communication in the body is called receptor. (D) Molecule binding to the enzyme covalently is called poison. Thus, the correct match is : A ® R, B ® P, C ® Q, D ® S 11. Given amino acid on reaction with NaNO 2 / H3O+ gives diazotisation reaction which further evolves ¾ N 2 gas along with formation of carbocation. On further reaction with water, it form HOOC ¾ (CH2 ) ¾ OH that undergoes polymerisation to give polymer. Biomolecules and Chemistry in Everyday Life HOOC—(CH2)4—NH2 O NaNO2/H3O+ (Diazotisation) HOOC—(CH2)4—N2 (D) R Aliphatic diazonium ion (Unstable) –N2 C H HOOC—(CH2)3—CH2 Polymerisation C—(CH2)4—O H H NaHCO3(aq) (Sodium hydrogen carbonate) (Q) HOOC—(CH2)4—OH OH OH –H2O H O—(CH2)4—C == O O O -----O—(CH2)4—C—O—(CH2)4—C---- or —O—(CH2)4—C— n 14. Nylon-6 or perlon is prepared by polymerisation of amino caproic acid at high temperature. Caprolactam is first hydrolysed with water to form amino acid which on heating undergoes polymerisation to give nylon-6. (Polymer) 12. The correct match is: A ® (Q) B ® (P) C ® (R) (A) Norethindrone It is an antifertility drug(Q ) containing synthetic progesterone derivative. [Other similar drug, is ethinylestradiol (novestrol)]. (B) Ofloxacin It is an antibiotic (P ), i.e produced wholly or partly by chemical synthesis with low concentration of microorganism.[Some other similar drugs : Penicillin, chloramphenicol, salvarsan etc.] (C) Equanil (meprobamate) It is a mild tranquilizer for relieving hypertension. It relieve anxiety, stress, excitement by inducing a sense of well being. (Other similar drug is chlordiazepoxide.) O [ HN— (CH2)5—C — ]n Nylon-6 15. High density polythene is used in the manufacture of buckets, dustbins etc. 16. Sodium lauryl sulphate [(CH3 (CH2 )10 CH2OSO-3 Na + )] = Anionic detergent Cetyltrimethyl ammonium + é ù CH3 ê ú ½ ê CH (CH ) ¾ N ¾ CH ú Br - = Cationic detergent 3 2 15 3 ê ú ½ ê ú CH 3 ë û Glyceryl oleate [(C17 H32COO)3 C3H5 ] = Non-ionic detergent Sodium stearate [C17 H35COO- Na + ] = Anionic soap (R) Violet colouration CH3 H3 C D Polymerisation O Caprolactam Phenolic OH –OH group FeCl3 (Ferric chloride) O + – H3N—(CH2)5—C—O NH Hydrolysis 13 (A) Becomes soluble with effervescence of CO2 gas Thus, the correct match is: A ® R; B ® S; C ® P; D ® Q O (An ester group) CH3 COOH Penicillin –H –H2O O CH3 S N O + H NH Cl Chloroxylenol (Dettol) 17. Natural rubber is formed by polymerisation of isoprene. H OH CººCH Dil. KMnO4/5°C/OH – H (Baeyer’s test) (B) H O CH2 CH2—C CH—CH2 CH3 Polymerise Ethylene Norethindrone C (Double bond) O (C) C—CH CH3 Propylene C (S) H CH2 Pink colour of KMnO4 gets discharged — H 3C H 2N S H2/Pt N H Ethylene-propylene copolymer CH—CH2— Natural rubber n N Sulphapyridine CHCl3 + KOH (alc.) (Carbylamine test) (P) —CH2—C CH3 CH3 n O Primary amine (–NH2) —CH2—CH— CH2—CH2— This co-polymer is formed from propylene and ethylene. Foul smell of isocyanide n CH2 == CH + nCH2 == CH2 ¾® —CH2—CH ½ CH3 CH3 CH2—CH2— n Biomolecules and Chemistry in Everyday Life 18. (a) Bakelite is used for making gears, protective coating and electrical fittings. (b) Glyptal is used in the manufacture of paints and lacquers. (c) PP is used in the manufacture of textile, packaging materials etc. (d) Polyvinyl chloride (PVC) is used in the manufacture of rain coats, hand bags, leather clothes etc. 19. (a) Polystyreme- manufacturing toys (b) Glyptal- Paints and lacquers (c) Polyvinyl chloride (PVC)- Raincoats (d) Bakelite- computer discs Thus, the correct match is A ® (1), B ® (1), C ® (2), D ® (4) 20. The given structure is of aspirin which is used as analgesic. 21. Aluminium hydroxide Al(OH)3, cimetidine and ranitidine are antacids while phenelzine is not. H N CH(NO2) N S N H N S Me2N C C N H O NHCN NHMe Ranitidine Cimetidine Phenelzine is a tranquilizer, not an antacid. H N ¾ NH2 Phenelzine is used as antidepressant drug. 22. Dacron is a condensation polymer of ethylene glycol and methyl terepthalate. Formation of dacron can be shown as C—OMe + nHO — CH2—CH2—OH MeO — C O Polymerisation —O—C C—OCH2—CH2— (n-1) MeOH O O O n Dacron Here, elimination of MeOH occurs as a by product. So, this reaction is known as condensation polymerisation. 23. Cellulose and nylons have H-bonding type of intermolecular attraction while poly (vinyl chloride) is polar. Natural rubber is hydrocarbon and has the weakest intermolecular force of attraction, i.e. van der Waals’ force of attraction. 24. C H OH C C OH H CH2OH H C H C C O H C CHOH OH C O O C H H O H C H OH Tri-acetylated cellulose 25. HO—Si—OH CH2OAc O O O H H H CH2OAc H O OAc H H OAc H H OAc H OAc H O H O OAc H H OAc Tri-acetylated cellulose CH3 CH3 (CH3)2SiCl2 + H2O H (CH3CO)2O CH2OAc Polymerisations CH3 CH3 CH3 HO—Si—O —Si—O —Si—OH CH3 CH3 n CH3 Linear chain polymer CH3 (CH3)3 XSiCl CH3 CH3 CH3 CH3 CH3 —Si—O —Si— O—Si—O —Si—O —Si— CH3 CH3 CH3 CH3 n CH3 Chain terminated linear chain silicone CH3 26. (a) When X = COOCH3 H 2 / Ni HOOC—(CH 2 ) 4 —COOH Heat Heat CH3OOC — (CH2 )4 — COOCH3 ¾ ¾¾® HOCH2 — (CH2 )4 — CH2OH + 2CH3OH ¾¾¾¾¾¾¾¾® O O —O—(CH2)6—O—C—(CH2)4—C— Ester, condensation polymer n Biomolecules and Chemistry in Everyday Life (b) When X = CONH2 O O H / Ni HOOC—(CH 2 ) 4 —COOH Heat Heat 2 H2NOC — (CH2 )4 — CONH2 ¾ ¾¾ ® H2N ¾ (CH2 )6 — NH2 ¾¾¾¾¾¾¾¾® —HN—(CH2)6—NH—C—(CH2)4—C— O Br n Nylon, condensation polymer O HOOC—(CH ) —COOH 2 2 4 (c) H2NOC — (CH2 )4 — CONH2 ¾¾® H2N— (CH2 )4 — NH2 ¾¾¾¾¾¾¾¾® —HN—(CH2)4—NH—C—(CH2)4—C— NaOH Heat Hofmann's bromamide reaction n Nylon, condensation polymer O O HOOC—(CH 2 ) 4 —COOH H 2 / Ni (d) When X = CN NC — (CH2 )4 — CN ¾ ¾¾ ® H2N(CH2 )6 NH2 ¾¾¾¾¾¾¾® —HN—(CH2)6—NH—C—(CH2)4—C— Heat Heat Nylon, condensation polymer In author’s opinion (a) and (b) should also be the answer. O O O (a) HOCH2—(CH2)4—CH2OH + HO—C— (CH2)4—C—OH Heat n O —O—(CH2)6 —O—C—(CH2)4—C— n Polyester, a condensation polymer O O H2/Ni Heat (b) H2N—C—(CH2)4—C—NH2 H2NCH2—(CH2)4—CH2NH2 O O O H2NCH2—(CH2)4—CH2NH2 + HO—C—(CH2)4—C—OH Heat O —NH—(CH2)6 —NH—C—(CH2)4—C— n Nylon, a condensation polymer 27. (A) Cellulose—a natural polymer of a-D-glucose, linked by glycoside linkage. (B) Nylon-6, 6—a synthetic polymer of adipic acid and 1,6-diaminohexane. The diacid is linked with diamine through amide linkage. (C) Protein—a natural polymer of a-amino acids where individual amino acid units are linked by amide linkage. (D) Sucrose—has glycoside linkage, a disaccharide. 28. CH3 O ½ ½½ O3 (a) H2C == C ¾ CH == CH2 ¾¾¾® 2HCHO + CH3 ¾ C ¾ CHO Zn - H2 O ‘ A ’ Isoprene CH2 ¾ CH2 H2C (b) Isoprene ¾® C == C H3C CH2 C == C H H3C H natural rubber 29. Zeigler-Natta catalyst, which is a mixture of triethylaluminium ‘(C2H5 )3 Al’ and TiCl 4, is used as heterogeneous catalyst in polymerisation of ethylene. 30. N—OH O H+ NH Caprolactum (C) O Polymerisation [–C—(CH2)5—NH–] n Nylon-6 (D)