2. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 1 Alkanesand Free Radical Substitution

3. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 2 Preparation of Alkanes

4. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 3 Petroleum Refining The first step is fractional distillation  separate petroleum into fractions based on the volatility of its components  performed in a petroleum refinery  a fractionating tower is used Crude Oil as a Source of Alkanes

5. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 4 Petroleum Refining A petroleum refinery

6. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 5 A simplified diagram of a fractionating tower Contains H 2 S Removed by treatment with a weak base

7. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 6 Typical fractions obtained by fractional distillation of petroleum FractionBoiling point range ( o C) Number of carbon atoms that the molecules contain Major uses Fuel gas (e.g. LPG) < 401 – 4Gaseous fuel, raw materials for manufacture of chemicals Petrol40 – 1505 – 10Fuel for motor vehicles Naphtha10 – 14For making town gas Gas oil (Diesel oil) 220 – 35014 – 25Fuel for trucks, lorries and locomotives

8. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 7 Typical fractions obtained by fractional distillation of petroleum FractionBoiling point range ( o C) Number of carbon atoms that the molecules contain Major uses Fuel oil> 350> 25Fuel for cargo ships and generating electricity in power stations Lubricating oil > 350> 25Lubricating oil for moving parts of machinery Bitumen> 350> 25For surfacing roads and roofs

9. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 8 Cracking of Petroleum

10. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 9 Cracking of Petroleum Convert hydrocarbons from heavier fractions into lighter fractions Performed in the absence of air Providing lighter fractions of petroleum, alkenes and sometimes H 2 and C C 11 H 24  C 9 H 20 + CH 2 =CH 2 C 14 H 30  C 8 H 18 + 2CH 2 =CH 2 + 2C + 2H 2

11. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 10 Thermal Cracking The process is done at high temperature in the absence of catalysts Hydrocracking The process is done at very high pressure of H 2 in the absence of catalysts Alkanes obtained are mainly unbranched

12. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 11 Catalytic Cracking When a mixture of alkanes from the heavier fractions is heated  at relatively low T & P  in the presence of catalysts The molecules break down and rearrange into smaller, highly branched hydrocarbons

13. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 12 Unbranched petrol is not a good motor fuel It burns before it is ignited by the spark plug of the petrol engine (pre-ignition). The shock wave creates the characteristic metallic pinging sound (engine knocking) A waste of energy Branched (anti-knocking) petrol can be produced by reforming.

14. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 13 Reforming Straight-chain alkanes are heated under pressure in the presence of a platinum catalyst reforming

15. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 14 Laboratory Preparation of Alkanes

16. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 15 1.Hydrogenation of alkenes Ni, 200°C high P Pt, 25°C

17. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 16 Methods 2 and 3 below are specific to the preparation of methane and benzene.

18. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 17 2.Decarboxylation of sodium ethanoate or sodium benzoate by heating with soda lime (NaOH + CaO) NaOH(s) from soda lime fusion CH 3 COONa(s) CH 4 + Na 2 CO 3 NaOH(s) from soda lime fusion + Na 2 CO 3

19. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 18 More difficult Not easy to separate mixture into individual alkanes. Not suitable for preparing alkanes other than methane.

20. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 19 3.Heating aluminium carbide with HCl (Optional) Al 4 C 3 + 12HCl 3CH 4 + 4AlCl 3 heat

21. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 20 4.Reduction of halogenoalkanes (optional)

22. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 21 Reactions of Alkanes

23. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 22 Reactivity of Alkanes Inertness to chemical reactions  strong C  C and C  H bonds C and H have nearly the same electronegativity  C  H bonds are only slightly polarized  not easily attacked by charged particles Alkanes  paraffin  little affinity

24. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 23 Two Ways of Breaking Covalent Bonds : 1.Homolytic fission 2.Heterolytic fission

25. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 24 Homolytic fission (homolysis) A  + B  Symmetrical breaking Free radicals each containing an unpaired electron

26. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 25 Homolytic fission (homolysis) A  + B  The movement of single electron is represented by a half-curly arrow Free radicals are very reactive(unstable). They tend to look for an additional electron to complete the octet.

27. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 26 Homolytic fission is favoured 1.when the bonding atoms have equal or similar electronegativities; 2.at high temperature; 3.when the reaction mixture is exposed to UV light; 4.in the presence of peroxides ( - O – O - ) (All favour the formation of free radicals)

28. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 27 Heterolytic fission (heterolysis) Unsymmetrical breaking A + + B  when B is more electronegative A  + B + when A is more electronegative

29. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 28 Heterolytic fission (heterolysis) The movement of an electron pair is represented by a curly arrow. A + + B  A  + B +

30. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 29 Hetrolytic fission is favoured 1.when the bonding atoms have significantly different electronegativities; 2.in the presence of a polar solvent which can stabilize the ions formed by ion-dipole interaction.

31. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 30 Q.23 Homolytic fission

32. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 31 Q.23 carbanion

33. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 32 Q.23 carbocation or carbonium ion

34. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 33 1.Pyrolysis (Cracking) Homolytic fission of C – C and/or C – H bonds heat catalyst e.g. C 6 H 14 C 3 H 8 + C 3 H 6

35. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 34 Q.24 C 6 H 14 C 2 H 6 + C 4 H 8 C 6 H 14 C 4 H 10 + C 2 H 4 C 6 H 14 CH 4 + C 5 H 10 C 6 H 14 C 2 H 4 + C 4 H 8 + H 2 C 6 H 14 C 2 H 6 + C 3 H 8 + C

36. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 35 Q.25

37. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 36 2.Combustion When alkanes react with sufficient oxygen  carbon dioxide and water are formed  release a large amount of heat  known as complete combustion

38. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 37 Combustion General equation for the complete combustion of an alkane: C n H 2n+2 + ( )O 2  nCO 2 + (n + 1) H 2 O

39. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 38 Combustion Alkanes are a common fuel Methane is the main component of natural gas Butane is a component of bottle gas Butane is used as the fuel for a portable gas burner

40. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 39 Q.26 In limited supply of oxygen, Large alkanes with high carbon contents

41. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 40 3. Halogenation (free radical substitution) X 2 = F 2, Cl 2, Br 2 or I 2 Alkanes react with halogens to give haloalkanes and hydrogen halide RH + X 2 RX + HX UV light or heat or peroxide

42. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 41 Halogenation One or more hydrogen atoms in methane are substituted by chlorine atoms  depend on the relative amounts of methane and chlorine

43. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 42 Halogenation CH 4 (g) + Cl 2 (g)  CH 3 Cl(g) + HCl(g) UV light CH 3 Cl(g) + Cl 2 (g)  CH 2 Cl 2 (g) + HCl(g) UV light CH 2 Cl 2 (g) + Cl 2 (g)  CHCl 3 (g) + HCl(g) UV light CHCl 3 (g) + Cl 2 (g)  CCl 4 (g) (g) + HCl(g) UV light

44. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 43 Halogenation When methane is in excess  chloromethane predominates in the products When chlorine is in excess  tetrachloromethane predominates in the products

45. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 44 Halogenation The reactivity of halogens decreases in the order: F 2 > Cl 2 > Br 2 > I 2 Not a good method to prepare haloalkanes since it is difficult to separate the resulting mixture into individual haloalkanes.Not a good method to prepare haloalkanes since it is difficult to separate the resulting mixture into individual haloalkanes.

46. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 45 Q.27(a) C n H 2n+2 = 72 12n + 2n + 2 = 72 n = 5 All 12 H atoms are identical and only ONE kind of monochlorinated product can be obtained.

47. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 46 Q.27(b) * 1C 2C 3C * *

48. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 47 ** Q.27(b)

49. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 48 1.Which piece of evidence suggests that the reaction is a chain reaction ? C For each photon of light absorbed, many thousands of molecules of chloromethane are formed.

50. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 49 2.Which of the following is the first step of the mechanism? CH 4  CH 3  + H  Cl 2  2Cl  CH 4  CH 3 + + H  CH 4  CH 3  + H + Cl 2  Cl + + Cl  homolytic heterolytic

51. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 50 For elementary steps, the structure of the transition state resembles that of the final products.  the stability of the transition state is determined by the stability of the final products.  the activation energy can be estimated by the enthalpy change of the elementary step The more endothermic reaction proceeds more slowly.

52. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 51 CH 4  CH 3  + H  Cl 2  2Cl   H = +435 kJ mol  1  H = +242 kJ mol  1

53. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 52 CH 4 CH 3 + + H  CH 3  + H  +435 +949  78  H = +435 + 949 -78 = +1306 kJ mol  1 HH

54. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 53 CH 4 CH 3  + H + CH 3  + H  +435 -7.7 +1310  H = +435 -7.7 +1310 = +1737.3 kJ mol  1 HH

55. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 54 Cl 2 Cl + + Cl  Cl  + Cl  +242 +1260  355  H = +242 + 1260 -355 = +1147 kJ mol  1 HH

56. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 55 Cl 2  2Cl   H = +242 kJ mol  1 First step of the mechanism

57. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 56 3.Second step of the mechanism Cl  + CH 4  CH 3 Cl + H  Cl  + CH 4  CH 3  + HCl

58. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 57 Cl  + CH 4  CH 3 Cl + H  Cl  + CH 3  + H  +435  350 HH  H = +435 -350 = +85 kJ mol  1

59. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 58 Cl  + CH 4  CH 3  + HCl HH Cl  + H  + CH 3  +435  431  H = +435 -431 = +4 kJ mol  1 Less endothermic and does not lead to formation of H 2

60. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 59 4.The 3rd step of the mechanism It gives the other product  CH 3 Cl It gives another Cl  to complete the chain cycle. CH 3  + Cl 2  CH 3 Cl + Cl 

61. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 60 5.Termination steps CH 3  +  CH 3  CH 3  CH 3 Cl  +  Cl  Cl  Cl CH 3  +  Cl  CH 3  Cl

62. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 61 Reaction Mechanism: Free Radical Substitution Reaction 1.Chain initiation Homolytic fission of a chlorine molecule Two chlorine radicals are formed

63. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 62 Reaction Mechanism: Free Radical Substitution Reaction 2.Chain propagation

64. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 63 Reaction Mechanism: Free Radical Substitution Reaction 3.Chain termination

65. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 64 7.Which step is rate-determining ? The propagation stage is the most time- consuming. Cl  + CH 4  CH 3  + HCl CH 3  + Cl 2  CH 3 Cl + Cl   H = +4 kJ mol  1 CH 3  +  Cl +  Cl +242  350  H = +242 -350 =  108 kJ mol  1

66. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 65 7.Which step is rate-determining ? The propagation stage is the most time- consuming. 1. Cl  + CH 4  CH 3  + HCl 2. CH 3  + Cl 2  CH 3 Cl + Cl  Or, E(C – H) > E(Cl – Cl) 435 kJ mol  1 242 kJ mol  1 Step 1 has higher E a  rate-determining

67. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 66 Example 27-5A Example 27-5A

68. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 67 Stability of Alkyl Radicals According to the number of alkyl groups directly attached to the carbon that bears the unpaired electron  alkyl radicals are classified as primary, secondary or tertiary

69. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 68 Stability of Alkyl Radicals Methyl radical (no alkyl group attached to the carbon that bears the unpaired electron) Primary radical (one alkyl group attached to the carbon that bears the unpaired electron) Secondary radical (two alkyl groups attached to the carbon that bears the unpaired electron) Tertiary radical (three alkyl groups attached to the carbon that bears the unpaired electron)

70. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 69 Stability of Alkyl Radicals The stability of the alkyl radicals decreases in the order: R 3 C Tertiary alkyl radical  R 2 CH Secondary alkyl radical  RCH 2 Primary alkyl radical  CH 3 Methyl radical  > > >

71. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 70 Stability of Alkyl Radicals Alkyl radical has an unpaired electron Electron-deficient Stabilized by substituents (e.g. alkyl groups)  release electrons to the carbon atom bearing the unpaired electron

72. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 71 Electron-donating property of alkyl groups arises from the polarization of C – H  bonds ++ ++ ++ 

73. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 72 Electron-donating groups(e.g. alkyl groups) cause positive inductive effect on adjacent groups. Electron-withdrawing groups(e.g. -F) cause negative inductive effect on adjacent groups. Inductive effect arises from polarization of  bonds.

74. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 73 Q.28(a)

75. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 74 No. of reaction sites 2 reaction sites 6 reaction sites Stability of free radical intermediates is more important than the no. of reaction sites available

76. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 75 Q.28(a) Stability : -   > 2°2° 1°1°

77. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 76 Q.28(b)

78. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 77 No. of reaction sites 9 reaction sites 1 reaction site Stability of free radical intermediates is more important than the no. of reaction sites available

79. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 78 Q.28(b) Stability : - < 1°1°3°3°  

80. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 79 Check Point 27-5B Check Point 27-5B Example 27-5B Example 27-5B

81. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 80 The END

82. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 81 27.2 Nomenclature of Alkanes (SB p.148) Draw the structures of the following compounds. Are the names provided correct or incorrect? If they are incorrect, give the correct name. (a)2,4-Dimethylpentane Answer (a) The name is correct.

83. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 82 27.2 Nomenclature of Alkanes (SB p.148) Draw the structures of the following compounds. Are the names provided correct or incorrect? If they are incorrect, give the correct name. (b)1,3-Dimethylpentane Answer (b) The name is incorrect. The correct name is 3-methylhexane.

84. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 83 27.2 Nomenclature of Alkanes (SB p.148) Draw the structures of the following compounds. Are the names provided correct or incorrect? If they are incorrect, give the correct name. (c) 4-Ethyl-1-methylheptane Answer (c) The name is incorrect. The correct name is 4-ethyloctane. Back

85. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 84 27.2 Nomenclature of Alkanes (SB p.149) Draw the structures of all the isomers of C 6 H 14, and give the IUPAC names for all the structures drawn. Answer Back

86. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 85 27.4 Preparation of Alkanes (SB p.155) Define each of the following terms and give an example of each: (a)Catalytic cracking Answer (a)Catalytic cracking is the process in which a mixture of alkanes from the heavier fractions is heated at very high temperatures, in the presence of catalysts and the absence of air. Smaller and highly branched hydrocarbons are formed. For example, C 8 H 18  C 5 H 12 + C 3 H 6

87. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 86 27.4 Preparation of Alkanes (SB p.155) Define each of the following terms and give an example of each: (b)Thermal cracking Answer (b)Thermal cracking is the breakdown of large alkane molecules in the heavier fractions into lighter fractions of smaller molecules in the absence of catalysts and air. Straight-chain alkanes are usually formed in this process. For example, C 14 H 30  C 8 H 18 + 2CH 2 = CH 2 + 2C + 2H 2

88. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 87 27.4 Preparation of Alkanes (SB p.155) Define each of the following terms and give an example of each: (c)Reforming Answer (c)Reforming is a process in which straight-chain alkanes are heated under pressure in the presence of a platinum catalyst. The chains break up and reform to give branched-chain molecules. For example, Back

89. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 88 27.5 Reactions of Alkanes (SB p.161) (a)Explain why cracking has to be carried out in the absence of air. Answer (a)Cracking must be carried out in the absence of air because combustion occurs instead of cracking in the presence of air.

90. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 89 (b)An alkane has a relative molecular mass of 72 and produces one product only on monochlorination. Deduce its structure. Answer 27.5 Reactions of Alkanes (SB p.161)

91. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 90 27.5 Reactions of Alkanes (SB p.161) (b)Let the molecular formula of the alkane be C n H 2n+2. Relative molecular mass of the alkane = 72 12.0 × n + 1.0 × (2n + 2) = 72 n = 5 ∴ The alkane has the molecular formula of C 5 H 12. As the alkane produces one product only on monochlorination, all hydrogen atoms of the alkane molecule must be equivalent. ∴ It must be Back

92. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 91 27.5 Reactions of Alkanes (SB p.161) Two isomeric compounds A and B have the molecular formula C 3 H 7 Cl. Chlorination of A gave a mixture of two dichlorides of formula C 3 H 6 Cl 2. Chlorination of B gave three different compounds of formula C 3 H 6 Cl 2 (they may not all be different from the dichlorides from A). What are the structural formulae of A and B and the dichlorides obtained from each? Answer

93. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 92 27.5 Reactions of Alkanes (SB p.161) A: B: Dichlorides formed from A:

94. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 93 27.5 Reactions of Alkanes (SB p.161) Back Dichlorides formed from B:

95. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 94 27.5 Reactions of Alkanes (SB p.162) The chlorination of butane gives a mixture of products as shown below: Explain their formation and also their relative abundances. Answer

96. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 95 27.5 Reactions of Alkanes (SB p.162) The two products arise because in one of the propagation steps of the reaction, a chlorine atom may abstract a hydrogen atom from either C1 or C2 of butane to give two different radicals.

97. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 96 The resulting free radicals react with chlorine in a succeeding propagation step to give the corresponding haloalkane. Butyl radical gives only 1-chlorobutane while secondary butyl radical gives only 2- chlorobutane. Since the more stable secondary radical is formed at a higher rate, 2-chlorobutane is formed as the major product of the reaction. 27.5 Reactions of Alkanes (SB p.162) Back

98. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 97 27.5 Reactions of Alkanes (SB p.162) Give the structure of the major product formed by free radical bromination of each of the following: (a)Methylcyclopentane Answer (a) Bromomethylcyclopentane

99. New Way Chemistry for Hong Kong A-Level Book 3A New Way Chemistry for Hong Kong A-Level 3A 98 27.5 Reactions of Alkanes (SB p.162) Give the structure of the major product formed by free radical bromination of each of the following: (b)2,2,4-Trimethylpentane Answer (b) 4-Bromo-2,2,4-trimethylpentane Back