2. Chapter 3 Hydrocarbons: Nomenclature and Reactions

3. Chapter 3 Problems Review Section 3.9 Read Essays on petroleum (p 204-207) and gasoline (p 221-223) for interest. I recommend that you do all problems except Prob. 8, 14, 33, 34 and 41 Skip ethenyl, 2-propenyl etc on p. 200 Read Section 3.15 and 3-16, but don’t take them too seriously.

4. I nternational U nion of P ure and A pplied C hemistry Sect. 3.1: IUPAC nomenclature systematic nomenclature “eye-you-pac” colloquially:

5. Sect. 3.2: the alkanes Hydrocarbons “Paraffins” Alkanes: formula C n H 2n+2

6. The alkanes: table 3-2

7. Sect. 3.4: IUPAC nomenclature of alkanes 1. Find the longest continuous chain of carbon atoms and name it (use linear names). 3. Give the substituent a name based on the number of carbon atoms it has. replace the -ane ending with -yl 2. Number the chain starting from the end nearest a branch. 4. Give the substituent a number determined on its location on the chain. 5. Assemble the name. Single substituent group

8. Finding the longest continuous chain of carbon atoms is not always simple C-C-C-C-C-C-C-C-C C C-C C-C-C C-C C -C-C-C-C-CC- C-C-C-C-C-C C-C C-C-C- C all possibilites must be examined it won’t always be the horizontal one as shown here try these also …….. 9 68

9. Sect. 3.3: Common alkyl groups (C 1 through C 4 ): table 3-3

10. Name this alkane 4 3 2 1 2-methylbutane

11. Find the longest continuous carbon chain 1 2 3 4 5 3-methylpentane

12. You must choose the longest continuous carbon chain 4 3 2 1 5 6 7 4-ethylheptane

13. Two different substituents number chain from end closest to a group, regardless of alphabetical order locate where groups are on chain with numbers place groups in alphabetical order, with the appropriate number assemble the complete name, using hyphens to separate numbers from “text”

14. Number from the end nearest the first substituent 7 6 5 4 3 2 1 4-ethyl-3-methylheptane

15. Number from the end nearest the first substituent 8 7 6 5 4 3 2 1 3-ethyl-5-methyloctane

16. Two or more identical substituent groups for two identical groups, use prefix di with the name of the group: dimethyl, diethyl, etc. dimethyl alphabetized as methyl, not dimethyl use numbers to locate groups on chain use commas to separate numbers prefixes: di = 2tri = 3 tetra = 4 penta = 5

17. Use “di-” with two substituents 1 2 3 4 2,3-dimethylbutane

18. Every substituent must get a number 1 2 3 4 5 6 3,3-dimethylhexane

19. You need numbers, even though it appears on the same carbon! 5 4 3 2 1 2,2,4-trimethylpentane

20. Number from the end nearest first substituent 10 9 8 7 6 5 4 3 2 1 2,7,8-trimethyldecane

21. Number from the end which has the “first difference” 1 2 3 4 5 6 7 8 9 10 3,4,8-trimethyldecane

22. Number from the end nearest the “first difference” Dimethyl alphabetized as methyl, not dimethyl 1 2 3 4 5 6 7 8 6-ethyl-3,4-dimethyloctane

23. If you can name this, you can name almost anything! 1 2 3 4 5 6 7 8 9 4-isopropyl-2,6,6-trimethylnonane

24. The isopropyl group can be named as a “complex” substituent 1-methylethyl

25. Now, rename the isopropyl group. Notice the alphabetical order! 1 2 3 4 5 6 7 8 9 2,6,6-trimethyl-4-(1-methylethyl)nonane

26. Deciding on alphabetical order for complex groups Complex groups are alphabetized under the first letter of the name (1,3-dimethylbutyl) = d (1,1,2-trimethylpropyl) = t (1-ethyl-1,2-dimethylbutyl) = e

27. Naming complex substituents -- this one is aphabetized under d 1,3-dimethylbutyl

28. Naming complex substituents 2-ethyl-1,1-dimethylbutyl

29. Name this compound! 1 2 3 1 2 3 4 5 6 7 8 9 5-(1-ethyl-1-methylpropyl)-5-propylnonane

30. Name this two ways -- (the complex group) 7-(1,1-dimethylethyl)-3-ethyl-7-methyldecane 7-tert-butyl-3-ethyl-7-methyldecane

31. Sect. 3.5: Common names of alkanes butane isobutane pentane isopentane neopentane

32. Sect. 3.6: the cycloalkanes The names of the cycloalkanes always contain the prefix cyclo Cycloalkanes have the general formula C n H 2n

33. Cyclic molecules

34. Nomenclature of the substituted cycloalkanes If there is only one substituent, do not use the “1”. If there is more than one substituent, you must use all numbers, including “1”! Number around the ring in a direction to get from the first substituent to the second substituent by the shorter path. For equivalent degrees of substitution, number in a direction that follows the alphabetical sequence. A carbon with greater substitution has precedence in numbering.

35. 1,1-dimethylcyclohexane

36. 4-ethyl-1,1-dimethylcyclohexane

37. CH 3 CH 3 CH 3 CH 3 CH 3 CH 2 CH 3 CH 3 CH 3 CH 3 CH 2 1,3-dimethylcyclopentane 1-ethyl-4-methylcyclohexane 3-ethyl-1,1-dimethylcyclobutane Some cycloalkanes The more substituted carbon takes precedence even though E comes before M. Drawn differently but same name. = E before M 1 2 3 1 2 3 1 2 3 1 2 3 4

38. Two ways of naming this 1-isopropyl-2-methylcyclohexane 1-methyl-2-(1-methylethyl)cyclohexane

39. Numbering starts at the most highly-substituted carbon 2 1 3 7 4 6 5 2-chloro-1,1,6-trimethylcycloheptane

40. Sect. 3.7: cycloalkyl groups

41. 3-cyclobutyl-3-methylpentane

42. (1-methylpropyl)cyclohexane or 2-cyclohexylbutane No locant is needed. With one substituent on a ring, it is automatically on carbon 1. 1-(1-methylpropyl)cyclohexane is overkill, but OK! Rings with one substitutent

43. Another name of a group

44. 3-methyl-2-phenylpentane

45. Sect. 3.8: Degree of Substitution methyl methylene methine

46. QUATERNARY PRIMARY TERTIARY SECONDARY A hydrocarbon containing carbon atoms with differing degrees of substitution All of the methyl groups (CH 3 ) are primary.Example

47. Sect. 3.9 -- review We already did this in Chapter 1

48. Sect. 3.10 and 3.11: nomenclature of halides and nitro compounds

49. bromoethane (IUPAC) ethyl bromide (common)

50. bromocyclopropane

51. 2-chloro-2-methylpropane (IUPAC) tert-butyl chloride (common)

52. 2-bromo-3-methylpentane

53. iodocyclohexane (IUPAC) cyclohexyl iodide (common)

54. 1-bromo-2-chlorocyclohexane

55. 2-nitropropane

56. Sect. 3.12: Block diagram for nomenclature

57. Sect. 3.13: alkene nomenclature ending is ene identify the longest chain with the C=C number from the end closest to the C=C and assign a number - - i.e. 2-pentene C=C is more important than groups! now number the attached groups and place them in alphabetical order

58. ethene (IUPAC)propene (IUPAC) 1-butene2-butene ethylene (common)propylene (common)

59. 2-methyl-2-butene

60. 6-methyl-2-heptene

61. trans-6-methyl-3-propyl-2-octene (Don’t worry about “trans” until Chapter 4)

62. 4,4-dimethylcyclohexene

63. 2-methyl-1,3-cyclohexadiene

64. 2,5-diethyl-1,3-cyclooctadiene

65. Very important! benzene It is never cyclohexatriene!!!

66. Sect. 3.14: nomenclature of alkynes similar system used as with alkenes ending is yne identify the longest chain with the triple bond everything else is the same as alkenes

67. ethynepropyne 1-butyne2-butyne “acetylene”

68. ALKYNES ( -YNE ) 2-hexyne 4-methyl-2-pentyne The suffix has precedence over any substituents The functional group has precedence in numbering. functional group

69. 4-chloro-4-methyl-2-pentyne 5-bromo-2-methyl-3-heptyne

70. Number from the end closest to either the double bond or the triple bond, whichever is closest to the end. ene vs. yne: which one wins? CH 3 -CH 2 -C C - CH 2 -CH=CH-CH 3 2-octen-5-yne 8 7 6 5 4 3 2 1 Compounds are named: en-yne.

71. COMPUTER PROGRAM “ORGANIC NOMENCLATURE” Available in Chemistry Computer Lab - CB280 Go to ChemApps Folder : optional, but recommended ….. then choose firstand then Organic Nomenclature Chem Apps Organic

72. Sect. 3.15: physical properties of hydrocarbons the longer the straight chain, the higher the boiling point -- van der Waals forces isomers that are branched have lower boiling points hydrogen bonding increases boiling points Dipole-dipole attractions increase b.p.

73. Sect. 3.16: Combustion of alkanes

74. chlorination examples CH 4 + Cl 2 CH 3 Cl + HCl CH 3 CH 2 CH 3 + Cl 2 CH 3 CHCH 3 + HCl Cl + CH 3 CH 2 CH 2 -Cl takes place at a refinery or a chemical plant - not easy to do in the lab free-radical substitution reaction Sect. 3.17: Halogenation of Alkanes

75. The previous examples given assumed monochlorination (one chlorine added) BUT …the reaction can repeat itself COMMON NAMES CH 4 + Cl 2 CH 3 Cl + HCl methyl chloride CH 3 Cl + Cl 2 CH 2 Cl 2 + HCl methylene chloride CH 2 Cl 2 + Cl 2 CHCl 3 + HCl chloroform CHCl 3 + Cl 2 CCl 4 + HCl carbon tetrachloride fully chlorinated product What are the IUPAC names ?

76. The reaction must be initiated It does not occur in the dark. Exposure to ultraviolet light (sunlight) will start the reaction. Heat will also start the reaction. Once reaction starts, it is exothermic and continues almost explosively. The first step is the dissociation of chlorine : Cl - Cl 2Cl.. :::. h or  chlorine atoms diatomic molecule (radicals)

77. Abstraction of hydrogen atom Cl.. :. BY A CHLORINE “FREE RADICAL” (ATOM) + C. H-Cl.. : + Chlorine takes the hydrogen and one of its electrons unpaired electron = “free” radical HYDROGEN ABSTRACTION.. C H

78. R E P E A T I N G S T E P S Mechanism of chlorination of methane CHAIN REACTION “hydrogen abstraction” “dissociation”

79. 4. Termination Steps These steps stop the chain reaction “recombinations”

80. CH 3 CH 2 CH 3 + Cl 2 CH 3 CHCH 3 Cl + CH 3 CH 2 CH 2 -Cl limited amount QUESTION AB WHAT ARE THE RELATIVE AMOUNTS OF A AND B ? IS IT STATISTICAL ( 2 : 6 ) = (1 : 3 ) ? DOES SOMETHING ELSE CONTROL THE OUTCOME ? Monochlorination of propane: Does one isomer predominate?

81. CH 3 CHCH 3 Cl CH 3 CH 2 CH 2 -Cl AB STATISTICAL PREDICTION25 % 75 % CH 3 -CH 2 -CH 3 = 6 : 2 or 3:1 Monochlorination of propane STATISTICAL VERSUS EXPERIMENTAL RESULTS ACTUALLY FOUND 50 % 50 % Equal amounts Equal amounts

82. Experimental results show: Secondary hydrogens are energenically more easily removed than primary hydrogens CH 3 CHCH 3 H CH 3 CH 2 CH 3 Secondary H Primary H more reactive hydrogen

83. Stability of free radicals explains results!

84. Stability of radicals: TERTIARY > SECONDARY > PRIMARY

85. Another example: isobutane Which product should form in the largest amount? There are 9 primary H’s and only 1 tertiary H Statistically you could predict a 9:1 ratio or a 90% yield of 1-chloro-2-methylpropane! Wrong!!

86. Isobutane gives only 62% of 2-chloro-2-methylpropane! Why? Look at the stability of the intermediate radical.

87. The statistical factor predicts a 9:1 ratio (90%) However, the energy factor predicts that the ratio will be less than 90% and turns out to be 62%.

88. Draw the structure of all of the monochlorinated products. There are 6 total products. The next slide shows the remaining 5 products. Only ONE product is shown here!

89. Here are 5 more isomeric products that are formed!

90. Ethylcyclopentane: monochlorination products MONOCHLORINATION PRODUCTS CH 2 CH 3 CH 2 CH 2 Cl CHCH 3 Cl CH 2 CH 3 Cl CH 2 CH 3 ClCl CH 2 CH 3

91. Hydrochlorofluorocarbons (HCFC’s)

92. Hydrogenation is covered in more detail in Chap 4 - we’ll cover it there. Hydrogenation is included in this chapter (briefly) because it is a method of making ALKANES. Sect. 3.18: hydrogenation of alkenes catalyst = Pt, Pd, Ni