22 Hydrocarbons Topic 1 Saturated Hydrocarbons Objective Questions I (Only one correct option) 1. In the following skew conformation of ethane, H¢ ¾ C ¾ C ¾ H¢¢ dihedral angle is H H H¢ 29° H H¢¢ H (a) 58° (b) 149° (d) 120° conformation in acyclic compounds? (b) (c) (d) into three distinct classes as shown in the figure. and II. chlorination of 2-methyl butane ? (a) 2 (c) 6 (b) 4 (d) 8 · 3. Isomers of hexane, based on their branching, can be divided and (a) 7. Consider the following reaction Electrostatic forces of interaction Torsional strain Angle strain Steric interactions I. equivalents of Na, in the presence of ether which of the following will be formed? Cl Br 6. How many chiral compounds are possible on mono (c) 151° 2. Which of these factors does not govern the stability of a (a) (b) (c) (d) 5. 1-bromo-3-chlorocyclobutane when treated with two H3 C ¾ CH ¾ CH ¾ CH3 + Br ¾® X + HBr ½ ½ CH3 D Identify the structure of the major product X · (a) H3C¾ CH¾ CH ¾ CH2 ½ ½ CH3 D · (c) H3C¾ C ¾ CH ¾ CH3 ½ ½ D CH3 III. The correct order of their boiling point is (a) (b) (c) (d) 4. 8. Benzyl chloride (C6H5CH2Cl) can be prepared from toluene (a) SO2Cl2 (c) Cl2 (b) SOCl2 (d) NaOCl 9. (CH3)3CMgCl on reaction with D2O produces Cl 2 , hv ¾¾¾® N (isomeric products) C5 H11 Cl (a) (CH3)3CD (c) (CD3)3CD (b) (CH3)3OD (d) (CD3)3OD 10. When cyclohexane is poured on water, it floats because H 3C CH3 Fractional distillation ¾¾¾¾¾¾¾® M (isomeric products) What are N and M ? (a) 6, 6 (c) 4, 4 · (d) H3C¾ CH ¾ CH ¾ CH3 ½ CH3 by chlorination with I > II > III III > II > I II > III > I III > I > II CH3 · (b) H3C¾ CH¾ C ¾ CH3 ½ ½ CH3 D (b) 6, 4 (d) 3, 3 (a) cyclohexane is in ‘boat’ form (b) cyclohexane is in ‘chair’ form (c) cyclohexane is in ‘crown’ form (d) cyclohexane is less dense than water 11. The C ¾ H bond distance is the longest in (a) C2H2 (c) C2H6 (b) C2H4 (d) C2H2Br2 Hydrocarbons 12. The compound which has one isopropyl group, is (a) 2,2,3,3-tetramethyl pentane (b) 2,2-dimethyl pentane (c) 2,2,3-trimethyl pentane (d) 2-methyl pentane (a) 2-methyl butane (c) 2, 2-dimethyl propane (b) n -pentane (d) n-hexane 16. Marsh gas mainly contains 13. The highest boiling point is expected for (a) C2H2 (c) H2S (a) iso-butane (b) n-octane (c) 2, 2, 3, 3-tetramethyl butane (d) n-butane 14. Which of the following compounds does not dissolve in conc. H2 SO4 even on warming ? (a) Ethylene (c) Hexane 15. The compound with highest boiling point is (b) CH4 (d) CO Integer Answer Type Question 17. The maximum number of isomers (including stereoisomers) (b) Benzene (d) Aniline that are possible on mono-chlorination of the following compound, is CH3 CH2 CH( CH3 ) CH2 CH2 Topic 2 Unsaturated Hydrocarbons Objective Questions I (Only one correct option) yields majorly an anti Markownikov product? 1. Consider the following reactions, Ag2O ppt D A Hg2+/H+ NaBH4 B C 5. Which one of the following alkenes when treated with HCl ZnCl2 Conc. HCl Turbidity within 5 minutes (a) Cl ¾ CH == CH2 (c) CH3O ¾ CH == CH2 (b) H2N ¾ CH == CH2 (d) F3C¾ CH == CH2 6. The correct order for acid strength of compounds CH ºº CH, CH3 ¾ C ºº CH and CH 2 == CH 2 is as follows : A is (a) CH ºº CH (c) CH3 ¾ C ºº CH (b) CH3 ¾ C ºº C ¾ CH3 (d) CH2 == CH2 2. But-2-ene on reaction with alkaline KMnO4 at elevated temperature followed by acidification will give (a) CH3 ¾ CH¾ CH ¾ CH3 ½ ½ OH OH (b) one molecule of CH3CHO and one molecule of CH3COOH (c) 2 molecules of CH3COOH (d) 2 molecules of CH3CHO 3. The major product of the following addition reaction is Cl 2 / H 2O H3C ¾ CH == CH 2 ¾¾® (a) CH3 ¾ CH ¾ CH2 ½ ½ Cl OH (a) CH3 ¾ Cºº CH > CH 2 == CH 2 > HCºº CH (b) CH3 -- C ºº CH > CH ºº CH > CH 2 == CH 2 (c) HC ºº CH > CH3 -- C ºº CH > CH 2 == CH 2 (d) CHºº C H > CH2 == CH2 > CH3 -- Cºº CH 7. The trans-alkenes are formed by the reduction of alkynes with (a) H2-Pd/C, BaSO4 (c) Na/liq. NH3 (b) NaBH4 (d) Sn-HCl 8. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate + (a) CH3 ¾ CH ¾ CH2 ¾ Cl + (c) CH3 ¾ CHCl ¾ CH2 + (b) CH3 ¾ CH(OH) ¾ CH2 + (d) CH3 ¾ CH ¾ CH2 ¾ OH 9. Which of the following compounds will exhibit geometrical isomerism? (a) 1-phenyl-2-butene (c) 2-phenyl-1-butene (b) CH3 ¾ CH ¾ CH2 ½ ½ OH Cl (b) 3-phenyl-1-butene (d) 1, 1-diphenyl-1-propane 10. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis? CH3 O (c) H3C CH3 CH3 O (a) (d) H 3C (b) CH3 CH3 CH3 4. The major product of the following reaction is l equiv.) ¾ (¾ ¾¾® CH 3C ºº CH ¾(i)¾DCl (ii) DI (a) CH3CD(Cl)CHD(I) (c) CH3CD(I)CHD(Cl) (b) CH3CD2CH(Cl)(I) (d) CH3C(I)(Cl)CHD2 (c) CH3 (d) CH3 Hydrocarbons 11. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is (a) acetylene (c) 2-butyne Hg 2+ 20. Ph ¾ C ºº C ¾ CH3 ¾¾® A; A is H+ (b) ethene (d) 2-butene 12. The number of optically active products obtained from the CH3 CH CH C (b) (b) 1 Ph CH C CH CH (c) 2 distinguish between 1-butyne and 2-butyne. (d) 4 (a) BrCH2CH2CH2CH2CH3 and CH3CH2C ºº CH (b) BrCH2CH2CH3 and CH3CH2CH2C ºº CH 14. The number of stereoisomers obtained by bromination of trans-2-butene is (b) H2 , Lindlar catalyst (d) ammoniacal CuCl 2 solution 22. In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markownikoff’s addition to alkenes because 23. The reaction of propene with HOCl proceeds via the addition of (a) H+ in the first step (c) OH– in the first step (b) 2 (d) 4 H (b) Cl+ in the first step (d) Cl+ and OH– single step 24. Hydrogenation of the adjoining compound in the presence of 15. The reagent(s) for the following conversion, Br is/are (a) bromine, CCl 4 (c) dilute H2SO4 , HgSO4 (a) both are highly ionic (b) one is oxidising and the other is reducing (c) one of the steps is endothermic in both the cases (d) all the steps are exothermic in both the cases (c) BrCH2CH2CH2CH2CH3 and CH3C ºº CH (d) BrCH2CH2CH2CH3 and CH3CH2C ºº CH ? —OH H3C H3C 21. Identify a reagent from the following list which can easily CH3 bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are Br Ph (d) CH2 13. The synthesis of 3-octyne is achieved by adding a (a) 1 (c) 3 H3C OH (c) CH ==O H 3C H H (a) 0 Ph (a) complete ozonolysis of the given compound, is CH3 O Ph H (a) alcoholic KOH (b) alcoholic KOH followed by NaNH2 (c) aqueous KOH followed by NaNH2 (d) Zn/CH3OH 16. CH3 — CH == CH2 + NOCl ¾® P; Identify the adduct. poisoned palladium catalyst gives H H 3C H3C CH3 H H (a) an optically active compound (b) an optically inactive compound (c) a racemic mixture (d) a diastereomeric mixture 25. Propyne and propene can be distinguished by (a) CH3 —CH—CH2 ½ ½ Cl NO Cl ½ (c) CH3—CH2 ¾ CH ½ NO (b) CH3— CH—CH2 ½ ½ NO Cl (a) conc. H2SO4 (c) dil. KMnO4 26. Which one of the following alkenes will react fastest with H2 (d) CH2 —CH2—CH2 ½ ½ Cl NO 17. Cyclohexene is best prepared from cyclohexanol by which of under catalytic hydrogenation condition ? R (b) conc. HCl / ZnCl2 (d) conc. HBr 18. 2-hexyne gives trans-2-hexene on treatment with (a) Li/NH3 (c) LiAlH4 (b) Pd/BaSO4 (d) Pt/H2 19. 2-phenyl propene on acidic hydration, gives (a) 2-phenyl-2-propanol (b) 2-phenyl-1-propanol (c) 3-phenyl-1-propanol (d) 1-phenyl-2-propanol R (a) the following? (a) conc. H3PO4 (c) conc. HCl (b) Br2 in CCl4 (d) AgNO 3 in ammonia R H R R H R R R (b) H R H R R H (c) (d) 27. The product(s) obtained via oxymercuration (HgSO4 + H2 SO4 ) of 1-butyne would be O ½½ (a) CH3 ¾ CH2 ¾ C ¾ CH3 (b) CH3 ¾ CH2 ¾ CH2 ¾ CHO (c) CH3 ¾ CH2 ¾ CHO + HCHO (d) CH3 ¾ CH2 ¾ COOH + HCOOH Hydrocarbons 28. In the compound, H , the C2-C3 bond is of the type (a) sp - sp2 (b) sp3 - sp3 (c) sp - sp3 (d) sp2 - sp3 29. The reaction of CH3 CH==CH OH with HBr (a) (M and O) and (N and P) are two pairs of enantiomers (b) Bromination proceeds through trans-addition in both the reactions (c) O and P are identical molecules (d) (M and O) and (N and P) two pairs of diastereomers gives (a) CH3CHBrCH2 OH (b) CH3CHBrCH2 Br (c) CH3CH2CHBr OH (d) CH3CH2CHBr Br Assertion and Reason Read the following questions and answer as per the direction given below : (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. 30. Which one of the following has the smallest heat of (a) 1-butene (c) cis-2-butene 38. Statement I Addition of bromine to trans-2-butene yields meso-2, 3-dibromo butane. Statement II Bromine addition to an alkene is an electrophilic addition. hydrogenation per mole ? (b) trans-2-butene (d) 1, 3-butadiene 31. The number of structural and configurational isomers of a 39. Statement I Dimethyl sulphide is commonly used for the bromo compound, C5 H9 Br, formed by the addition of HBr to 2-pentyne respectively, are reduction of an ozonide of an alkene to get the carbonyl compound. Statement II It reduces the ozonide giving water soluble dimethyl sulphoxide and excess of it evaporates. (a) 1 and 2 (b) 2 and 4 (c) 4 and 2 (d) 2 and 1 (c) benzene (d) ethane 32. Acidic hydrogen is present in (a) ethyne (b) ethene 33. Baeyer’s reagent is of a peroxide produces 1-bromobutane. Statement II It involves the formation of a primary radical. (a) alkaline permanganate solution (b) acidified permanganate solution (c) neutral permanganate solution (d) aqueous bromine solution 41. Statement I Addition of Br2 to 1-butene gives two optical 34. When propyne is treated with aqueous H2 SO4 in the presence of HgSO4 , the major product is (a) propanal (c) acetone 40. Statement I 1-butene on reaction with HBr in the presence isomers. Statement II The product contains one asymmetric carbon. (b) propyl hydrogen sulphate (d) propanol 35. The compound 1, 2-butadiene has Passage Based Questions (a) only sp-hybridised carbon atoms (b) only sp2-hybridised carbon atoms (c) both sp and sp2-hybridised carbon atoms (d) sp, sp2 and sp3-hybridised carbon atoms Passage 1 Pd-BaSO4 (i) B2H6 C8H8 X H2 (ii) H2O2,NaOH, H2O H2O HgSO4,H2SO4 (i) EtMgBr, H2O C8H8 Y (ii) H+, Heat C8H6 36. Which of the following will decolourise alkaline KMnO4 solution? (a) C3H8 (b) CH4 (c) CCl 4 (d) C2H4 42. Compound X is Objective Questions II (One or more than one correct option) OH (a) CH3 (b) CH3 (d) CHO 37. The correct statement(s) for the following addition reactions is (are) H3C H H3C Br2/CHCl3 (i) H CH3 (c) H Br2/CHCl3 M and N (i) H CH3 O and P OH Hydrocarbons 43. The major compound Y is (a) CH3 (i) NaNH2 (excess) (b) (ii) CH3CH2I (1 equivalent) H CH3 HO M (iv) H2, Lindlar's catalyst CH3 CH3 CH3 (c) (d) (i) NaNH2 (2 equivalent) OH (ii) Br CH3 H Passage 2 (i) dil. H2SO4/HgSO4 P (C6H10) Q (ii) NaBH4/ethanol (iii) dil. acid Y (Scheme 2) (iii) H3O+, (mild) (iv) H2, Pd/C (v) CrO3 N An acyclic hydrocarbon P, having molecular formula C 6H10, gave acetone as the only organic product through the following sequence of reactions, in which Q is an intermediate organic compound. X (Scheme 1) (iii) CH3I (1 equivalent) 46. The correct statement with respect to product Y is (a) (b) (c) (d) it gives a positive Tollen’s test and is a functional isomer of X it gives a positive Tollen’s test and is a geometrical isomer of X it gives a positive iodoform test and is a functional isomer of X it gives a positive iodoform test and is a geometrical isomer of X 47. The product X is H3CO O (i) dil. H2SO4(catalytic amount) (–H2O) C 2 CH3 H3C (ii) O3 /ethanol (iii) Zn/H2O H (a) (b) H H H3CO H H CH3CH2O 44. The structure of the compound Q is OH H3C (a) H3C ½ OH ½ H ¾ C¾ C ¾ CH2CH3 (b) H3C¾ C¾ C ¾ CH3 ½ ½ H3C H3C H H OH H3C ½ (c) H ¾ C — CH2 CH CH3 (c) (d) H3C (d) H H CH3CH2O H Fill in the Blanks 48. 1,3-butadiene with bromine in molar ratio of 1 : 1 generate predominantly …… . 49. Addition of water to acetylene compounds is catalysed by …… OH and …… ½ CH3CH2CH2 CH CH2CH3 50. Kolbe’s electrolysis of potassium succinate gives CO2 and …… . 45. The structure of compound P is (a) CH3CH2CH2CH2 ¾ C ºº C¾ H 51. The terminal carbon atom in 2-butene is …… hybridised. (b) H3CH2C¾ C ºº C¾ CH2CH3 52. Acetylene is treated with excess sodium in liquid ammonia. The H3C (c) H ¾ C¾ C ºº C¾ CH3 H3C product is reacted with excess methyl iodide. The final product is …… . 53. …… is most acidic. (Ethane, Ethene, Ethyne) H3C (d) H3C¾ C¾ C ºº C¾ H H3C True/False 54. Moist ethylene can be dried by passing it through concentrated Passage 3 Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both schemes. sulphuric acid. Integer Answer Type Question 55. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4 H6 is Hydrocarbons Subjective Questions 56. Monomer A of a polymer on ozonolysis yields two moles of HCHO and one mole of CH3 COCHO. (a) Deduce the structure of A. (b) Write the structure of all cis form of polymer of compound A. this information two isomeric structures can be proposed for alkene (A). Write their structures and identify the isomer which on catalytic hydrogenation (H2/Pd - C) gives a racemic mixture. 61. What would be the major product in the following reaction? H2 57. A biologically active compound, Bombykol (C16H30O) is obtained from a natural source. The structure of the compound is determine by the following reactions. (a) On hydrogenation, Bombykol gives a compound A, C16H34O, which reacts with acetic anhydride to give an ester. (b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidative ozonolysis (O3/H2O2) gives a mixture of butanoic acid, oxalic acid and 10-acetoxy decanoic acid. Determine the number of double bonds in Bombykol. Write the structures of compound A and Bombykol. How many geometrical isomers are possible for Bombykol? 58. Identify X, Y and Z in the following synthetic scheme and Lindlar’s catalyst CH3 62. Complete the following reactions with appropriate reagents : (a) (b) == OH D D CH3 C==C H3C D—C—C—H H3C D CH3 CH3 H3C CH3 63. Complete the following reactions with appropriate structures of products/reagents : Br2 (i) NaNH (3 equivalent) 2 C6H5CH == CH2 ¾® [ A ] ¾¾¾¾¾¾¾¾¾® [B] (ii) CH3I write their structures. CH3 CH2 ¾ C ºº C ¾ H (i) NaNH 64. Write the intermediate steps for each of the following reactions : 2 ¾¾¾¾® (ii) CH3CH2Br H 3O + Alkaline KMnO4 H2 /Pd BaSO4 X ¾¾¾¾¾® Y¾¾¾¾¾¾® Z Is the compound Z optically active? Justify your answer. (i) C6 H5 CH(OH)C ºº CH ¾® C6 H5 ¾ CH == CH ¾ CHO H+ (ii) 59. (a) Identify (A), (B), (C), (D) and (E) in the following schemes Br2/Cl4 H2 NHNCONH2 65. The hydrocarbon A, adds one mole of hydrogen in the presence of a platinum catalyst to form n-hexane. When A is oxidised vigorously with KMnO4 , a single carboxylic acid, containing three carbon atoms, is isolated. Give the structure of A and explain. and write their structures : NaNH2 CH3 O OH A CH3 B HgSO4/H2SO4 C 66. Complete the following, giving the structures of the principal organic products : NaOD/D2 O (excess) C ¾¾¾¾¾® D ¾¾¾¾¾¾¾® E (b) Identify (X), (Y) and (Z) in the following synthetic scheme and write their structures. Explain the formation of labelled formaldehyde (H2C*O) as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed. Mark the C* carbon in the entire scheme. BaC*O3 + H2SO4 ® (X) gas [C* denotes C14] (i) Mg/ether Ph Ph R product (B) C8H8O. Compound (B) on reaction with NaOH/I2 yields sodium benzoate. Compound (B) reacts with KOH/NH2NH2 yielding a hydrocarbon (C) C8H10. Write the structures of compounds (B) and (C). Based on Br R + HClO4 (ii) B [R = nPr] H2C == CH ¾ Br ¾¾¾¾® (Y ) ¾¾¾® (Z ) 60. An alkene (A) C16H16 on ozonolysis gives only one A + KNH2 (i) LiAlH4 (ii) X (iii) H3 O+ H O (iii) + CHBr3 + t-BuOK C 67. An alkyl halide, X, of formula C6 H13 Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C6 H12 ) . Both alkenes on hydrogenation gives 2, 3-dimethyl butane. Predict the structures of X, Y and Z. Hydrocarbons 68. Give the structure of the major organic products obtained (iii) CH3 COCH3 from 3-ethyl-2-pentene under each of the following reaction conditions : (iv) CH3 ¾ CH ¾ CHO | CH3 What are the structures of A, B and C ? (a) HBr in the presence of peroxide (b) Br2 / H2O 76. How would you convert acetylene to acetone? (c) Hg(OAc)2 / H2O, NaBH4 77. Give the chemical test to distinguish between 2-butyne and 1-butyne. 69. Write down the structure of the stereoisomers formed when cis-2-butene is reacted with bromine. 78. Following statements are true, only under some specific 70. An organic compound E (C5 H8 ) on hydrogenation gives conditions. Write the condition for each subquestion in not more than two sentences : compound F (C5 H12 ). Compound E on ozonolysis gives formaldehyde and 2-keto propanal. Deduce the structure of compound E. (i) 2-methyl propene can be converted into isobutyl bromide by hydrogen bromide. (ii) Ethyne and its derivatives will give white precipitate with ammoniacal silver nitrate solution. 71. When gas A is passed through dry KOH at low temperature, a deep red coloured compound B and a gas C are obtained. The gas A, on reaction with but-2-ene, followed by treatment with Zn / H2 O yields acetaldehyde. Identify A, B and C. 79. Give reasons for the following in one or two sentences : (i) Methane does not react with chlorine in the dark. (ii) Propene reacts with HBr to give isopropyl bromide but does not give n-propyl bromide. 72. Give the structures of A, B and C (explanation are not required) A (C4H8 ) which adds on HBr in the presence and in the absence 80. State with balanced equation, what happens when “propene of peroxide to give same product. (ii) B (C4H8 ) which when treated with H2O / H2SO4 gives C4H10O which cannot be resolved into optical isomers. 81. One mole of a hydrocarbon A reacts with one mole of (i) is bubbled through a hot aqueous solution of potassium permanganate.”? bromine giving a dibromo compound, C5 H10 Br2 . Compound A on treatment with cold dilute alkaline potassium permanganate solution forms a compound, C5 H12 O2 . On ozonolysis A gives equimolar quantities of propanone and ethanal. Deduce the structural formula of A. (iii) C (C6H12 ), an optically active hydrocarbon which on catalytic hydrogenation gives an optically inactive compound C6H14. 73. Write the balanced chemical equation for the following “Ethylene glycol is obtained by the reaction of ethylene with potassium permanganate.” 82. Write the structural formula of the major product in each of the following cases 74. Give a chemical test and the reagents used to distinguish (i) Ethene mixed with air is passed under pressure over a silver catalyst. (ii) The compound obtained by hydration of ethyne is treated with dilute alkali. between cyclohexane and cyclohexene. 75. A white precipitate was formed slowly when silver nitrate was added to compound A with molecular formula C6 H13 Cl. Compound A on treatment with hot alcoholic potassium hydroxide gave a mixture of two isomeric alkenes B and C, having formula C6 H12 . The mixture of B and C, on ozonolysis, furnished four compounds (i) CH3 CHO (ii) C2 H5 CHO 83. Outline the reaction sequence for the conversion of ethene to ethyne (the number of steps should not be more than two). 84. Give one characteristic test which would distinguish CH4 from C2 H2 . Answers Topic 1 1. 5. 9. 13. 17. (b) (d) (a) (b) (8) 2. 6. 10. 14. (c) (c) (d) (c) 3. 7. 11. 15. (b) (b) (c) (d) 4. 8. 12. 16. (b) (c) (d) (b) 2. 6. 10. 14. (c) (c) (b) (a) 3. 7. 11. 15. (b) (c) (c) (b) 4. 8. 12. 16. (d) (a) (a) (a) Topic 2 1. 5. 9. 13. (c) (d) (a) (d) 17. 21. 25. 29. 33. 37. 41. 45. 48. (a) 18. (a) (d) 22. (c) (d) 26. (a) (c) 30. (b) (a) 34. (c) (a,b,d) 38. (b) (a) 42. (c) (d) 46. (c) 3,4-dibromo-1-butene 50. ethene 51. sp 3 53. Terminal alkyne (ethyne) 19. 23. 27. 31. 35. 39. 43. 47. 49. (a) 20. (b) 24. (a) 28. (b) 32. (d) 36. (a) 40. (d) 44. (a) H 2SO 4 , HgSO 4 (a) (b) (d) (a) (d) (c) (b) 52. 2-butyne 54. False 55. (5) Hints & Solutions Topic 1 Saturated Hydrocarbons 1. A dihedral angle is the angle between two C ¾ H bonds projected on a plane orthogonal to the C ¾ C bond. In the given skew conformation, having Newman’s projection the dihedral angle is and H¢¾ C ¾ C ¾ H¢¢ III II = (H¢¾ C ¾ C ¾ Ha ) + (Ha ¾ C ¾ C ¾ H¢¢) = 29°+120° = 149° and H H I H¢ 29º (ethane, C2H6) Ha On moving left to right (III to I) · branching increases · surface area decreases · boiling point decreases Hence, correct choice is (b). 120º H¢¢ H b 2. The four types of strains viz (a) electrostatic force of attraction, (b) torsional strain, (c) angle strain, (d) steric stain, are responsible for the stability or energy barriers of conformers. In cyclic compounds, all types of strains may be present. Compound CH3 4. CH3 Cl2 CH3 Type of strains/forces CH2Cl hn H3C CH3 I a+ c CH3 CH3 H3C a+ b+ c+ d F CH3 + H3C Cl Types of strains/forces a H CH3 Cl (V + VI) chiral Since, fractional distillation cannot separate enantiomers (II + III and V + VI), M = 4 and N = 6. In a cyclic or open-chain compounds, angle strain (c) is absent. e.g. H CH3 IV (II + III) chiral Compound CH2Cl CH3 + a+ b+ c OH Cl Na / D 5. Wurtz’s reaction Br H H 1-bromo-3-chlorocyclobutane H H a+ b CH3 CH3 H CH3 6. CH3 H 3C + Cl2 hn I 2-methyl butane H H a+ b+ d OH H OH H H Cl Cl H H 3. + PLAN This problem is based on boiling point of isomeric alkanes. As we know more the branching in an alkane, lesser will be its surface area and lesser will be the boiling point + + II Cl + Cl III IV Out of the four products formed above, II and IV are chiral, produced in pairs, giving total of six mono-chlorination products. Hydrocarbons Due to this method of formation, methane is also known as marsh gas. 7. Bromination is highly selective, occur at the carbon, where the most stable free radical is formed : CH3 ¾ CHD ¾ CH — CH3 + Br · ¾® ½ CH3 17. (a tertiary free radical) 8. Toluene on treatment with Cl 2 in the presence of heat or light undergo free-radical chlorination at benzylic position, giving benzyl chloride CH3 CH2Cl + Cl2 * Cl + * + HCl or heat Benzyl chloride CH3 ½ 9. (CH3)3 CMgCl + D2O ¾® CH3 ¾ C — D + Mg(OD)Cl ½ CH3 10. Alkanes are all less dense than water, floats over water. II III IV Cl I has one chiral carbon = two isomers II has two chiral carbons and no symmetry = four isomers. III and IV have no chiral carbon, no stereoisomers. Topic 2 Unsaturated Hydrocarbons alkyne with triple bond present at the terminal. The chemical reactions involved are as follows: Step 1 CH3 C 2, 2, 3, 3-tetramethyl pentane (no isopropyl group) CH3 CH3 ½ ½ CH3 ¾ C — CH— CH2CH3 ½ CH3 2, 2, 3-trimethyl pentane (no isopropyl group) CH3 ½ CH3 ¾ C — CH2CH2CH3 ½ CH3 CH Ag2O CH3 C C Ag (Precipitates) Prop-1-yne ( A) Step 2 CH3 C 11. C—H bond with sp3-C will be longest in C2H6. CH3 CH3 ½ ½ 12. CH3 ¾ C — C — CH2 ¾ CH3 ½ ½ CH3 CH3 + 1. According to the given conditions, the compound should be hv Toluene I Cl · CH3 ¾ CHD ¾ C— CH3 + HBr ½ CH3 Cl + * Hg2+ CH dil. H SO 2 4 OH CH3 C CH2 O Tautomerisation CH3 C CH3 (Propan-2-one) (B) NaBH4 2, 2-dimethyl pentane (no isopropyl group) CH3 CH CH3 OH CH3 ½ CH3 ¾ CH — CH2CH2CH3 (C) (2° alcohol) Conc. HCl + ZnCl2 Isopropyl group 2-methyl pentane CH3 CH CH3 Cl Turbidity within 5 minutes (Insoluble in Lucas reagent) 13. Boiling point of alkane increases with molar mass. Among isomeric alkanes, branching decreases boiling point. Therefore, n-octane has highest boiling point, higher than 2, 2, 3, 3-tetramethyl-butane (an isomer of n-octane). 14. Ethylene absorb H2SO4 forming CH3 ¾ CH2OSO3H and dissolve. Benzene, with warm H2SO4, undergo sulphonation and dissolve. Aniline, with H2SO4, forms anilinium sulphate salt and dissolve. Hexane, a hydrophobic molecule, does not react with H2SO4, remains insoluble. 15. Among alkanes, boiling point increases with molar mass. Among isomeric alkanes, branching decreases boiling point. Therefore, n-hexane has highest boiling point among these. 16. Methane is produced due to the decay of vegetables or animal organisms present in swamps and marsh, by the action of bacteria. In step-1, prop-1-yne reacts with Ag2O to form CH3 ¾ C ºº C ¾ Ag, that forms white precipitates. In step 2, prop-1-yne in presence of mercuric sulphate and dil × H2SO4 produces carbonyl compound (CH3 )2 C == O which produces (CH3 )2CH ¾ OH in presence of NaBH4. 2ºalcohol on reaction with Lucas reagent produces turbidity in about 5 min. 2. But-2-ene on reaction with alkaline KMnO4 at elevated temperature followed by acidification will give acetic acid (CH3COOH). Hot alkaline solution of potassium permanganate followed by acidification oxidatively cleaved alkenes. The reaction proceed as follows : CH3 ¾ CH == CH ¾ CH3 But-2 -ene Alk. KMnO4 , heat ¾¾¾¾¾¾® 2CH3COOH H3O+ Acetic acid Hydrocarbons 3. The major product of the given addition reaction is +R H3C ¾ CH ¾ CH2. ½ ½ OH Cl d+ d– Cl – fast r Cl — CH — — CH2 H—Cl Cl — — CH — CH3 Slow Cl In this reaction, H2O is used as a solvent and the major product of the reaction will be a vicinal halohydrin. A halohydrin is an organic molecule that contains both OH group and a halogen. In a vicinal halohydrin, the OH and halogen are bonded to adjacent carbons. Cl — CH — CH3 Similarly, d+ d– (+R) Cl 2/H2O H3 C ¾ CH == CH2 ¾¾¾® H3C ¾ CH ¾ CH2 + ½ ½ OH Cl Slow CH3—CH—CH2 H2 O – CH3—CHCH2—Cl | + OH | H Fast CH3CH—CH2—Cl + H3O | OH Chlorohydrin 4. The major product obtained in the given reaction is CH3C (I) (Cl) CHD2. 1. DCl (1 equiv.) CH3C ºº CH ¾¾¾¾® CH3C (I)(Cl)CHD2 2. DI Addition in unsymmetrical alkynes takes place according to Markovnikov’s rule. Reaction proceed as follows : Cl ½ DCl (1 equiv.) DI CH3 C ºº CH ¾¾¾¾¾® CH3 C == CHD ¾¾ Prop-1-yne Cl ½ CH3 ¾ C ¾ CHD2 ½ I Product 5. Attachment of electron donating group (+ R or + I) with sp2-carbon of an unsymmetrical alkene supports Markownikov’s addition rule through electrophilic-addition-pathway. But, attachment of electron-withdrawing group (- R or - I) for the same will follow anti-Markownikov’s pathway (even in absence of organic peroxide which favours free radical addition) through electrophilic addition pathway. The product formed by given alkenes when treated with HCl. CH3O — CH — CH3 Cl d– d+ Cl — H (– I) – Cl Fast + + Cl Fast r F3C — CH2 — CH2 s s Cl Slow – Chloronium + Cl ion H2 O NH2 — CH — CH3 CH3O — CH — CH3 (+R) F3C — < CH — — CH2 Slow Cl CH3—CH= =CH2 + Cl—Cl H—Cl CH3O — CH — — CH2 HCl + Cl Fast d+ d– CH3 ¾ CH ¾ CH2 ½ ½ Cl Cl The reaction proceeds through following mechanism : + NH2 — CH — CH3 Cl – A chlorohydrin (major product) (Minor product) H—Cl ·· H2N — CH — — CH2 F3C — CH2CH2Cl (H2C == CH2 ) is sp 2-hybridised and ethyne (HC ºº CH) is sp-hybridised. In ethyne, the sp-hybridised carbon atom possesses maximum s-character and hence, maximum electronegativity. Due to which, it attracts the shared electron pair of C¾H bond to a greater extent and makes the removal of proton easier. Hence, alkyne is much more acidic than alkene. Presence of electron donating group in alkyne (H3C ¾ C ººCH) decreases the acidic strength of compound. Hence, the correct order of acidic strength is: HC ºº CH > H3C ¾ C º CH > CH2 == CH2 6. Ethene 7. Sodium metal in liquid ammonia reduces alkynes with anti stereochemistry to give trans alkenes. The reduction is selectively anti since the vinyl radical formed during reduction is more stable in trans configuration. Mechanism R R Na C C C R¢ C R¢ Na+ H–NH2 –NH2 R H C H C R¢ H–NH2 –NH2 C Vinylic radical Sodium atom donates an electron to alkyne which after H-abstraction from NH3 forms vinylic radical. Transfer of another electron gives a vinylic anion, which is more stable in trans form. This in turn gives trans-alkene after H-abstraction from NH3. R H C Na Na+ R – R¢ H C C Vinylic anion R¢ Hydrocarbons d– 8. d+ 16. NOCl undergo electrophilic addition on alkene as: + HO—Cl — 2 CH3—CH—CH (Electrophilic addition) CH3—CH—CH2—Cl NOCl ¾® (Intermediate) + N==O + Cl - CH3—CH—CH2—Cl CH3 ¾ CH ¾ CH2 ½ ½ Cl NO OH 9. PhCH2 C CH3 C H PhCH2 H C 17. Cyclohexanol on treatment with concentrated H3PO4 undergo CH3 acid catalysed dehydration giving cyclohexene. trans form CH3 10. C H H cis form CH3 O3 OH O CHO Zn/H2O CH3 conc.H3PO4 CH3 cyclohexanol 5-keto-2-methyl hexanal cyclohexene 18. Alkynes on treatment with alkali metals in liquid ammonia gives 11. The reaction is trans hydrogenation product: 6Ag 2CH3 ¾ CCl 3 ¾ ¾ ¾® CH3 ¾ C ºº C ¾ CH3 + 6 AgCl D Li/NH 3 CH3 ¾ C ºº C ¾ CH2 ¾ CH2 ¾ CH3 ¾¾¾® But-2-yne 2-hexyne 12. Ozonolysis of the given triene occur as follows : CH CH C CH CH C H O3 Zn/H2O H3 C 2 CH3 H C==C CH3 CH3 H 3C H CH CH H CH3 19. Reaction proceeds through carbocation intermediate: H+ CH3 CHO + 2 OHC C CH2 CH2 CH3 trans-2-hexene CHO CH3 ¾ C== CH2 ¾® ½ Ph 2-phenyl propene + CH3 ¾ C ¾ CH3 ½ Ph OH ½ CH3 ¾ C — CH3 ½ Ph 13. CH3CH2C ºº CH + Br — CH2CH2CH2CH3 2-phenyl-2-propanol ¾¾® CH3CH2 — C ºº C— CH2CH2CH2CH3 — HBr 20. Reaction proceeds through carbocation intermediate : 3-octyne + 14. Br2 undergo anti-addition on C == C bonds as: H + Br C==C H CH3 H C C Resonance stabilised CH3 H (meso isomer) 15. Br CH2 ==CHBr OH ½ Ph ¾ C == CH ¾ CH3 Unstable enol e (Tautomerisation) O ½½ Ph ¾ C ¾ CH2 ¾ CH3 21. Ammoniacal CuCl 2 forms red precipitate with terminal alkynes, Br Br Alc. KOH H2 O Ph ¾ C ºº C ¾ CH3 + H+ ¾® Ph ¾ C == CH ¾ CH3 ¾® Br CH3 H O 2 ¾® 3 °, resonance stabilised H Since, none of the above dial is chiral, no optically active product is obtained. H3C Cl - + CH3 ¾ CH==CH2 + + NO ¾® CH3 ¾ CH ¾ CH2 ¾® ½ NO – OH NaNH2 H—C ººC—H can be used to distinguish terminal alkynes from internal alkynes: NH ( aq ) 3 CH3 ¾ CH2 ¾ C ºº C ¾ H + CuCl 2 ¾¾® CH3 ¾ CH2 ¾ C ºº C - Cu + ¯ red ppt. Hydrocarbons 22. In addition of HBr to an alkene, in the presence of peroxide, both 28. According to the IUPAC conventions, compound can be the propagation steps are exothermic : HBr + HO· ¾® H2O + Br · numbered as: Propagation Here, C-2 is sp2 and C-3 is sp3-hybridised. 1 · ì · ï CH3 ·¾ CH == CH2 + Br ¾® CH3 ¾ CH ¾ CH2Br; DH < 0 ï íCH3 ¾ CH ¾ CH2Br + HBr ¾® CH3 ¾ CH2 ¾ CH2Br ï + Br ·; DH < 0 ïî In case of addition of HCl and HI, one of the propagation step is endothermic, reaction fail to occur. - 23. HOCl ¾® HO + Cl 2 CH3—CH==CH— —OH H+ + CH3—CH2—CH— —OH CH3—CH2—CH== ==OH Cl + i.e. reaction is initiated by Cl + (chloronium ion electrophile) 24. Hydrogenation with poisoned palladium brings about cis hydrogenation of alkyne and does not affect double bonds : H CH3 H2/Pd CH3 H H H 3C CH3 Br– H meso-compound (optically inactive) 25. Terminal alkynes forms silver salt with Tollen’s reagent while NH3 ( aq ) 26. Ease of catalytic hydrogenation depends upon the size of groups present at the doubly bonded carbon. Larger the size of groups, difficult the hydrogenation. Therefore, in the given situation, disubstituted reacts at faster rate than tri and tetra substituted alkenes. Among disubstituted, the stability order is : cis < geminal < trans. 27. Oxymercuration-demercuration brings about Markownikoff’s addition of water as : — OH 30. Among alkenes-heat of hydrogenation depends on : (a) The number of double bonds-greater, greater the amount of heat evolved in hydrogenation. Hence, 1, 3-butadiene has highest heat of hydrogenation among these. (b) Relative stability of alkenes-greater the stability, smaller the heat evolved in hydrogenation. trans-2-butene is most stable among three given butenes, has least heat of hydrogenation. 31. CH3 ¾ C ºº C ¾ CH2CH3 + HBr ¾® C2H5 C == C H3C + geometrical isomers H3C C == C C2H5 + H3C C == C H Br H Br C2H5 1 44444444244444444 3 geometrical isomers Therefore, two structural and four configurational isomers. 32. Terminal alkynes are slightly acidic, forms salt with very strong base like Na, NaNH2 etc. D H ¾ C ºº C ¾ H + Na ¾® H ¾ C ºº C - Na + + ethyne 1 H2 2 33. Baeyer’s reagent is cold, dilute, alkaline permanganate solution, used to detect presence of olefinic bonds. 34. Alkynes undergo Markownikoff’s addition of water in the presence of H2SO4 / HgSO4 : HgSO4 CH3 ¾ CH2 ¾ C ºº C ¾ H + H2 SO4 ¾¾® é ù OH O ê ú ½ ½½ êCH3 ¾ CH2 ¾ C== CH2 ú ¬® CH3 ¾ CH2 ¾ C ¾ CH3 ê ú butanone ê ú ë û Br C == C H Br H CH 1 44444444244444444235 white ppt. Therefore, Tollen’s reagent can be used to distinguish a terminal alkyne like propyne from alkene as well as from internal alkynes. (resonance stabilised carbocation) CH3CH2—CH— alkene does not react with Tollen’s reagent. CH3 ¾ C ºº C ¾ H + AgNO3 ¾¾® CH3 ¾ C ºº CAg ¯ + Br H3C HH H unstable enol 6 + Cl ½ ¾¾® CH3 ¾ CH — CH2 ½ OH H 5 carbocation: HO- H 3C 4 29. Electrophilic addition on C == C is governed by stability of CH3 ¾ CH==CH2 + Cl + ¾¾® CH3 ¾ CH ¾¾¾ CH2 CH3 3 H2 C== CH ¾ CH2 ¾ CH2 ¾ C ºº C ¾ H HgSO4 CH3 ¾ C ºº C ¾ H + H2SO4 ¾¾® é ê ê ê êë ù OH ú ú ½ CH3 ¾ C == CH2 úúû Unstable enol e (Tautomerisation) O ½½ CH3 ¾ C ¾ CH3 Acetone Hydrocarbons Both Statement I and Statement II are correct and Statement II is correct explanation of Statement I. 35. Structural formula of 1, 2-butadiene is : H2C == C == CH¾ CH3 sp2 sp sp2 sp3 40. CH3 ¾ CH2 ¾ CH== CH2 + Br · ¾® 36. Unsaturated compounds which contain C == C or C ºº C, a secondary radical Therefore, Statement I is correct but Statement II is incorrect. decolourises the purple colour of alkaline KMnO4 solution. CH2 ¾ OH HOCH2 == CH2 + KMnO4 ¾¾® ½ + MnO2 ¯ CH2 ¾ OH purple 41. CH3 ¾ CH2 ¾ CH == CH2 + Br2 ¾® CH3 ¾ CH2 ¾ CH ¾ CH2 Br ½ Br coloured 37. Addition of halogen at double bond occur in antiorientation via has one chiral carbon cyclic halonium ion intermediate. (i) CH3 C Both Statement I and Statement II are correct and Statement II is correct explanation of Statement I. r Br H CH3 H Passage 1 CH3 The reaction condition indicates that starting compound is phenyl acetylene. H Br Meso (M and N) H trans H Br – H H 3C Br CH3 CH3 Br2 C · CH3 ¾ CH2 ¾ CH ¾ CH2 Br 42. (ii) H r Br H C CH3 C cis (ii) H2O2, NaOH, H2O (2-phenyl ethanol) (X) CH3 H Br Hydroboration oxidation brings about anti-Markonikoff’s hydration of alkene. + CH3 Br H P CH3 43. O C ºº CH ó ó ô í ô H Br O CH3 Br H CH3 Br – Enantiomers H2SO4 + H3C H C + Br2 H trans C CH3 CH3 CH3 Br C==C CH3 H2O, HgSO4 Here, (M + O) and (N + P) are pair of diastereomers. 38. OH (i) B2H6 Pd/BaSO4 H2 H H CH3 CH ¾ ¾ CH2 CººC—H CH3 Br2 H CH3 C2H5 MgBr, H2O H OH CH3 CH3 H+/D CH3 Statement I is correct. Statement II is also correct. Meso form of the product is due to anti addition of Br - on cyclic bromonium ion intermediate, hence Statement II is not correct explanation of Statement I. Br CH3 C Br– H C CH3 P H Br (meso) O 39. C==C + O3 C C O O Passage 2 The final ozonolysis product indicates that the alkene before ozonolysis is O H 3C CH3 O3 C 2 C C Zn–H2O CH H C CH3 H 3C 3 3 CH3—S—CH3 ozonide Also P(C6H10) has two degree of unsaturation and oxymercuration demercuration hydration indicates that it is an alkyne. As alkyne, on hydration, gives a carbonyl compound which on reduction with NaBH4 gives a 2º alcohol. O OH —C C— + H2O —C—CH2— (i) NaBH4 + (ii) H — C—CH2— H 2° alcohol C ==O + O == C + (CH3 )2 SO Hydrocarbons The secondary alcohol that can give above shown alkene on acid catalysed dehydration is CH3 OH CH3 — C — CH — CH3 + CH3 — C — CH— CH3 + H –H2O Hence, using the concept of regioselectivity we come on the conclusion that final product is correctly represented by structure (a). 48. 3, 4-dibromo-1-butene : Me-shift + Br2 Br CH3 Q 2° carbocation CH3 –H + CH3 —C—CH—CH 3 Br CH3 + CH3 — C HgSO C — CH3 CH2 ¾ COOK 50. ½ CH3 CH3 4 49. CH ºº CH + H2SO4 ¾¾® CH3CHO CH2 ¾ COOK 44. Explained in the beginning. 45. CH3—C—C HgSO4 CH H2SO4 CH3—C— C—CH3 (i)NaBH4 (ii) H + Q ethene 51. CH3 ¾ CH== CH ¾ CH3 2- butene (P) + - CH I 3 ¾® Na C ºº C - Na + ¾¾® H ¾ C ºº C ¾ H + Na excess Passage 3 46. CH2 == CH2 + 2CO2 52. 2-butyne : CH3 CH3 ¾¾¾® sp 3 CH3 O CH3 Electrolysis excess CH3 ¾ C ºº C ¾ CH3 PLAN This problem can be solved by using the concept of iodoform test and functional isomerism. 53. Terminal alkyne (ethyne) is most acidic among these. Iodoform test The compound containing — COCH3 or ¾CH(OH) group will undergo iodoform test 54. Sulphuric acid undergo addition to alkene. Br H N NaNH2 OH 55. s C NaNH2(1eq) Iequivalent , s H2Pd/C OH H3Os mild O , , CH3 ½ O3 56. (a) CH2 == C — CH == CH2 ¾¾® 2HCHO Zn -H2 O (A ) OH CrO3 Me , O ½½ + CH3 ¾ C ¾ CHO O Y • 47. Thus, X and Y are functional isomers of each other and Y gives iodoform test due to the presence of CH 3CO group as Indicated. Hence, correct choice is (c). PLAN This problem can be solved by using the concept of nucleophilic substitution reaction, oxidation reaction and reduction reaction including strength of nucleophile and regioselectivity. Reaction of Scheme 1 can be completed as H HO C == C H2C II O I Among two nacked nucleophilic group I and II, II is more nucleophilica and then will react selectively as follows CH2 ¾¾ H2C H H 57. From oxidation products, structure of starting compound can be C3 H8 ¾ COOH + HOOC ¾ COOH butanoic acid oxalic acid O ½½ + HOOC ¾ (CH2 )8 ¾ CH2 O ¾ C ¾ CH3 10-acetoxy decanoic acid CH3I CH3 CH2 CH2 CH==CH ¾ CH==CH ¾ (CH2 )8 ¾ CH2 OCOCH3 Therefore, Bombykol is : CH3O CH3CH2CH2 ¾ CH== CH ¾ CH== CH ¾ (CH2 )8 ¾ CH2OH Bombykol H2 CH3O CH2 O 3 / H 2O 2 O H2 Lindlar’s catalyst (produces cis alkene) C == C natural rubber 1. CH3CH2—I O H H3C deduced as : 1. NaNH2 M H H3C (b) CH3 ¾( CH2 )14 ¾ CH2OH (A ) Hydrocarbons C2H5 Pd/BaSO (i) NaNH H 4 2 58. CH3CH2 ¾ C ºº C ¾ H ¾¾¾¾® CH3CH2 ¾ C ºº C ¾ CH2CH3 ¾¾¾¾® (ii) CH3 CH2 Br NaNH2 —C ººCH B CCl4 Br A C2H5 Y C2H5 H H OH OH C2H5 (meso diol) ‘Z ’ O Br Br2 (a) 59. H2 X alkaline KMnO4 H C== C HgSO4 —C—CH3 H2SO4 C B O CH3 H2NHNCONH2 O —C==N—NH—C—NH2 D CD3 D 2O E D * C NaOD D * (b) Ba CO3 + H2 SO4 ¾® CO2 + BaSO4 X O ½½ LiAlH4 * (i) Mg / ether CH2 == CH ¾ Br ¾¾¾® CH2 ==CH ¾ C ¾ OH ¾¾® CH2 == CH ¾ CH2 ¾ OH (ii) X (iii) H3O+ Y * Z 60. B + NaOH + I2 ¾® C6H5 ¾ COONa (Iodoform reaction) O ½½ \ B is C6H5 ¾ C ¾ CH3 KOH Hydrocarbon (C ) “Wolff-Kishner Reduction” B + N2H4 ¾¾® C8 H10 \C is C6H5 ¾ CH2 ¾ CH3 Hence, A can be one of the following : H5 C6 C6 H5 H3 C H5 C6 CH3 or C==C CH3 I C==C H3 C C6 H5 II I on catalytic hydrogenation, would give meso compound while II on catalytic hydrogenation, would produce racemic mixtures. CH3 61. Lindlar’s H CH3 catalyst H cis hydrogenation O 62. (a) Zn-H2O + O3 conc. H2SO4 LiAlH4 D D D (b) CH3 C==C + BHR2 CH3 H3C OH CH3 R = (CHl 3 )2 CHCH2 H2 O2 HO – D OH CH3 CH3 C—C H3C CH3 H D Hydrocarbons 63. C6H5 ¾ CH == CH2 + Br2 Br ½ ¾® C6H5 ¾ CH ¾ CH2 ½ Br CH2CH3 C2H5 ½ ½ HBr 68. CH3CH == C ¾ CH2 CH3 ¾¾® CH3 ¾ CH ¾ CH ¾ C2H5 peroxide 3-ethyl-2-pentene ½ Br [A] A NaNH2 CH2 CH3 ½ Br 2 /H2O ¾¾¾® CH3 ¾ CH ¾ C — CH2 ¾ CH3 ½ ½ Br OH ¾¾¾® C6H5 ¾ C ºº C ¾ CH3 CH3 I B + OH OH2 ½ ½ H+ 64. (i) C6H5 ¾ CH ¾ C ºº CH ¾® C6H5 ¾ CH ¾ C ºº CH + – H2 O [B] H+ C2H5 ½ ¾¾¾¾¾® CH3 ¾ CH2 ¾ C — CH2 CH3 NaBH4 ½ OH ¾¾® C6H5 ¾ CH ¾ C ºº CH ¾® H2 O C6H5 ¾ CH == CH ¾ CHO e (Tautomerisation) Hg(OAc) 2 /H2 O C6H5 ¾ CH == C == CH ½ OH [C ] unstable enol 69. H+ CH3 H3C + C==C (ii) + OH H CH3 O Br2 H cis H Br H H H C H3C + CH3 O 65. Oxidation product indicates that alkene is symmetrical : + KMnO4 H+ 2CH3CH2COOH 3-hexene ‘A’ H Ph + KNH2 Ph—C C—Ph Br R R Br CH3 H Pair of enantiomers O CH2 ½½ ½½ 70. CH3 ¾ C ¾ CHO + O== C ¾ H ¬¾ CH3 C ¾ CH == CH2 E ½ 2-ketopropanal H Br + CHBr3 + t-BuOK C CH3 Cl CH3 CH2 H 3C CH3 CH3 ‘X’ ‘Y’ CH3 CH3 CH3 CH3 H2/Pt H3C CH3 + B Gas C CH3 B C 72. (i) A must be a symmetrical alkene : Br ½ CH3 CH==CHCH3 + HBr ¾® CH3 CH2 ¾ CH ¾ CH3 CH3 CH3 ½ ½ (ii) CH3 ¾ C == CH2 + H2SO4 + H2O ¾® CH3 ¾ C — CH3 ½ OH Achiral (CH3)3COK H 3C + H 3C \ Gas A is ozone (O3 ) Also; Gas A + KOH (dry) ¾® \ 4O3 + 4KOH ¾® 4KO3 + 2H2O + O2 2R—COOH B O (iii) Zn - H2 O deep red coloured + HClO4 (ii) ‘Z’ Br + Gas A A 67. H C C 71. CH3 ¾ CH == CH ¾ CH3 ¾¾¾® 2CH3 ¾ CHO C==C CH3 CH3 C Formaldehyde Ph 66. (i) CH3 Br CH3 ½ Pt (iii) CH3 ¾ CH2 ¾ C — CH == CH 2 + H2 ¾® D ½ H has one chiral carbon CH3 ½ CH3CH2 ¾ C — CH 2CH 3 ½ H Achiral Hydrocarbons 73. CH2 == CH2 + KMnO4 + H2O ¾® CH2 ¾ CH2 + MnO2 ½ OH ½ OH Ethylene glycol 74. Baeyer’s reagent (cold, dilute, alkaline permanganate) can be used 79. (i) Free radical chlorination of alkane require energy which is supplied either in the form of heat or radiation. (ii) Addition of HBr proceeds through carbocation intermediates. + 2 ° carbocation to distinguish between alkanes and alkenes. Alkenes decolourises purple colour of Baeyer’s reagent while alkanes do not. CH3 ¾ CH ¾ CH3 ½ Br 75. The alkenes are : CH3 ¾ CH == O + O == CH ¾ CH(CH3 )2 CH3 CH3 ½ ½ CH3 ¾ CH == CH ¾ CH — CH3 ; CH3CH2CH ==O + O == C ¾ CH3 - Br CH3 ¾ CH == CH2 + H+ ¾® CH3 ¾ CH ¾ CH3 ¾® Isopropyl bromide D CH ¾ CH ¾ CH + MnO 80. CH3CH == CH2 + KMnO4 (aq) ¾® 2 2 3 ½ OH ½ OH I (B ) CH3 ½ CH3 ¾ CH2 ¾ CH== C — CH3 II (C ) Since, both alkenes I and II are obtained by b-elimination of same halides, the halides must be : CH3 ½ CH3 ¾ CH2 ¾ CH ¾ CH — CH3 ½ Cl (A ) Heat ( D ) CH I 3 76. H ¾ C ºº C ¾ H + Na ¾¾¾® ¾¾® CH3 ¾ C ºº CH O ½½ ¾¾¾® CH3 ¾ C ¾ CH3 H2 SO4 HgSO4 Acetone 77. 1-butyne (terminal) can be distinguished from 2-butyne (internal) by either Tollen’s test or through Fehling’s test. CH3—CH2—C ººC—H AgNO3 CH3—CH2—C ºº CAg¯ NH3(aq) White ppt. 81. Ozonolysis products are the key of identification : CH3 CH3 ½ ½ O3 CH3 ¾ C== O + O == CH ¾ CH3 ¬¾¾ CH3 ¾ C== CH ¾ CH3 Propanone Ethanal Zn -H2 O Other products are: CH3 ½ dil. KMnO4 Br 2 CH3 ¾ C — CH ¾ CH3 ¬¾¾¾¾ A ¾® Cold, OH ½ ½ OH OH CH3 Br ½ ½ CH3 ¾ C — CH — CH3 ½ Br O 1 Ag 82. (i) CH2 == CH2 + O2 ¾¾® H2C ¾¾ CH2 heat 2 Ethylene oxide (oxirane) HgSO dil. OH OH ½ CH3 ¾ CH ¾ CH2 ¾ CHO (Aldol) Peroxide ½ CH3 ½ CH3 2-methyl propene Isobutyl bromide In the absence of peroxide, HBr would be added giving tertiary butyl bromide. (ii) Tertiary alkynes are slightly acidic, forms silver salt with ammoniacal solution of silver nitrate : NH ( aq ) 3 R ¾ C ºº C ¾ H + AgNO3 ¾¾® R ¾ C ºº CAg ¯ White ppt. + 4 (ii) CH ºº CH + H2SO4 ¾¾® CH3 ¾ CHO ¾¾® CuCl2 CH3—CH2—C ºº CCu¯ NH3(aq) Red ppt. 78. (i) CH3 ¾ C == CH2 + HBr ¾¾¾® CH3 ¾ CH ¾ CH2Br A NaNH 2 83. CH2 == CH2 + Br2 ¾® CH2 ¾ CH2 ¾¾¾® ½ Br ½ Br H ¾ C ºº C ¾ H 84. Acetylene can be distinguished from methane using Tollen’s reagent : NH (aq) 3 C2H2 + AgNO3 ¾¾® H ¾ C ºº CAg ¯ White ppt. No such reaction occur with methane.