2 The general formula of a cycloalkane is A C nh 2n+2. D C nh n. C C nh 2n-2. Solution B

Transcription

1 MCQ Chapter 2 1 Which of the following cycloalkanes has the lowest heat of combustion per CH 2 group? A Cyclopropane B Cyclobutane C Cyclopentane D Cyclohexane Solution D 2 The general formula of a cycloalkane is A C nh 2n+2 B C nh 2n C C nh 2n-2 D C nh n Solution B 3 In the presence of light, alkanes react with molecular chlorine to give substitution products. So methane can be converted to chloromethane etc. How many dichlorinated structural isomers can be formed by halogenation of butane with chlorine in the presence of light? A 2 B 3 C 5 D 6 Solution D 4 What is the IUPAC name for the following compound? A Trimethylpentane B 2,4,4-trimethylpentane C 2,2,4-trimethylpentane D 2,2,4,4-tetramethylbutane Solution C 5 How many constitutional isomers are there with formula C 4H 9Cl? A 3 B 4

2 C 5 D 6 Solution B 6 Which of the following C 6H 12 isomers has the highest heat of combustion? A Cyclohexane B Methylcyclopentane C cis-1,3-dimethylcyclobutane D Trimethylcyclopropane Solution D 7 Two hydrocarbons found in crude oil are decane, C 10H 22 and hexadecane, C 16H 34. Which of the following statements about the properties of hexadecane compared with the equivalent property of decane is correct? A Hexadecane has a lower boiling point B Hexadecane is less viscous C Hexadecane is more volatile D Hexadecane catches fire less easily Solution D 8 Which two of the following products are produced by the complete combustion of fuels? I CH 4; II CO; III CO 2; IV H 2O; V NH 3 A I and III B I and II C III and IV D IV and V Solution C 9 Water has a higher boiling point than compounds of similar molecular weight. What is the best explanation for this? A Extensive hydrogen bonding B Water is a polar covalent compound C Water is largely dissociated leading to large electrostatic forces D London dispersion forces exist between the molecules Solution A 10 Which of the following molecules does NOT have a permanent dipole moment? CHCl 3 CH 2 Cl 2 CH 3 Cl CCl 4 A B C D Solution D

3 11 Which of the following molecules HAS a permanent dipole moment? A CH 4 B CO 2 C Cyclohexane D NH 3 Solution D 12 What is the IUPAC name for Cl A chlorotrimethylbutane B 2-chloro-2,4-dimethylpentane C 4-chloro-2,4-dimethylpentane D 1-chloro-1,1-3trimethylbutane Solution B 13 What is the IUPAC name for A cis-1,3-dimethylcyclopentane B trans-1,3-dimethylcyclopentane C 1,3-dimethylcyclopentane D 1,3-dimethylcyclopentene Solution A 14 What are the most important intermolecular forces in H 2? A Hydrogen bonding B Dispersion forces C Dipole-dipole interactions D Electrostatic interactions Solution B

4 15 What is the IUPAC name for Cl A chloroethylcycloexane B 2-chloro-1-ethylcyclohexane C 1-chloro-1-ethylcyclohexane D 1-ethyl-1-chlorocyclohexane Solution C 16 Cyclohexane boils at a higher temperature than n-hexane. What is the best explanation for this observation? A Cyclohexane stacks better in the liquid phase B Cyclohexane has a lower molecular weight than hexane C Cyclohexane can hydrogen bond D Dipolar forces are stronger in cyclohexane Solution A 17 Which of the following molecules does NOT show sp 3 hybridization at the central atom A CH 4 B CH 3F C C 2H 4 D C 2H 6 Solution C 18 The angles between sp 3 orbitals are approximately A 90 o B 109 o C 120 o D 180 o Solution B 19 The stick drawing of CH 3CH 2CH(CH 3)CH(CH 3)CH 2CH 3 is A

5 B C D Solution A 20 The stick diagram for 2,5-dichloro-2,5-dimethylhexane is A Cl Cl B Cl Cl C

6 Cl Cl D Cl Solution D Cl 21 Which of the following cycloalkanes would have the LOWEST heat of combustion per CH 2 group? A Cyclopropane B Cyclopentane C Cyclohexane D Cyclodecane Solution C 22 Which of the following hydrocarbons would be expected to have the LOWEST boiling point? A n-hexane B 2-methylpentane C 2,2-dimethylbutane D They would all have the same boiling point Solution C 23 For which of the following molecules are dispersion forces NOT the main source of intermolecular attraction? A He B CH 4 C CF 4 D CH 3F Solution D

7 Chapter 2 Structure, bonding and nomenclature of alkanes Test questions - solutions

8 1 What are the most important intermolecular forces in (a) He (b) N 2 (c) O 2 (d) Br 2 (e) HBr (f) H 2 O (g) Chloroethane (h) hexane? (a) He (b) N 2 (c) O 2 (d) Br 2 dispersion forces dispersion forces dispersion forces dispersion forces

9 (e) HBr dipole-dipole interactions (f) H 2 O hydrogen bonding (g) Chloroethane dipole-dipole interactions (h) Hexane dispersion forces

10 2 Give an example of each of the following (a) A substituted methane with a permanent dipole moment Chloromethane (b) A linear triatomic molecule without a permanent dipole moment CO 2

11 (c) A molecule that forms strong hydrogen bonds H 2 O (d) A cis-disubstituted cycloalkane

12 3 What shape would you predict for each of the following molecules? Consider geometry at nitrogen and oxygen as well as carbon. (a) CHCl 3 Approx. tetrahedral, sp 3 hybridization (b) CF 4 Perfect tetrahedron, sp 3 hybridization

13 (c) CH 3 CH 2 OH CH 3, CH 2 - approx. tetrahedral. Bent at O, with two lone pairs. All atoms are sp 3 hybridized (d) CH 3 NHCH 2 CH 3 CH 3, CH 2 approx. tetrahedral. Pyramidal at N, with one lone pair. All atoms are sp 3 hybridized

14 4 Give a systematic name for each of the following molecules (a) Cl Cl

15 This is a 1,4-disubstituted cyclohexane - no relative stereochemistry is indicated. If the two carbons attached to the ring had been methyl groups, it would be 1,4-dimethylcyclohexane But each of the methyl groups is substituted by chlorine, so 1,4-bis(chloromethyl)cyclohexane

16 (b) 3-ethyl-3-methylpentane

17 (c) This time we must be careful to find the longest chain in a less conventional drawing this will be named as an octane

18 The substituents are at C-3 and C-6 So 3,6-diethyloctane

19 (d) Br 5 We need to check the longest chain carefully as this a less conventional drawing 3-(bromomethyl)pentane

20 (e) cis-1,3-diethylcyclohexane

21 (f) F 1-fluoro-1,2,2-trimethylcyclobutane

22 (g) F cis-1-fluoro-3-ethylcyclohexane 1R,2S-1-fluoro-3-ethylcyclohexane

23 (h) Br We number the molecule to put the bromine on the lowest numbered carbon atom 1-bromo-2,2-dimethylbutane

24 5 Draw a stick structure for each of the following molecules (a) 5-bromodecane Br

25 (b) 2,2-dimethylbutane

26 (c) 2-iodoethylcyclohexane I

27 (d) 4,4-difluoroheptane F F

28 (e) 3-ethyl-4-methylnonane

29 (f) 1-chloro-1-methylcyclopentane Cl

30 (g) cis-1,2-dichlorocyclooctane Cl Cl

31 (h) 1-dichloromethyl-1-methylcyclohexane Cl Cl

32 6 For each of the following pairs of molecules, predict which would have the higher boiling point. Give a brief reason (a) hexane and cyclohexane hexane 69 o C, cyclohexane 81 o C. Cyclohexane stacks better in the liquid phase

33 (b) hexane and 2,2-dimethylbutane Hexane 69 o C, 2,2-dimethylbutane 49 o C. Substituents make packing more difficult in the liquid phase, and reduce intermolecular interactions (c) hexane and 1-fluorohexane Hexane 69 o C, 1-fluorohexane 93 o C. In 1-fluorohexane there are dipole-dipole forces to overcome as well as the dispersion forces that exist between all molecules.

34 Chapter 2 Alkanes Test questions 1 What are the most important intermolecular forces in (a) He (b) N 2 (c) O 2 (d) Br 2 (e) HBr (f) H 2O (g) Chloroethane (h) hexane 2 Give an example of each of the following (a) A substituted methane with a permanent dipole moment (b) A linear triatomic molecule without a permanent dipole moment (c) A molecule that forms strong hydrogen bonds (d) A cis-disubstituted cycloalkane 3 What shape would you predict for each of the following molecules? Consider geometry at nitrogen and oxygen as well as carbon. (a) CHCl 3 (b) CF 4 (c) CH 3CH 2OH (d) CH 3NHCH 2CH 3 4 Give a systematic name for each of the following molecules (a) Cl Cl (b)

35 (c) (d) Br (e) (f) F (g)

36 F (h) Br 5 Draw a stick structure for each of the following molecules (a) (b) (c) (d) (e) (f) (g) (h) 5-bromodecane 2,2-dimethylbutane 2-iodoethylcyclohexane 4,4-difluoroheptane 3-ethyl-4-methylnonane 1-chloro-1-methylcyclopentane cis-1,2-dichlorocyclooctane 1-dichloromethyl-1-methylcyclohexane 6 For each of the following pairs of molecules, predict which would have the higher boiling point. Give a brief reason (a) (b) (c) hexane and cyclohexane hexane and 2,2-dimethylbutane hexane and 1-fluorohexane

37 Chapter 2 Alkanes and friends structure, bonding, properties, nomenclature

38 At the end of the last chapter, we had learned how to combine atomic orbitals into covalent bonds for simple systems two s-orbitals in molecular hydrogen and an s and a p-orbital to make a C-H bond, and p-orbitals to make either σ- or π-bonds. Unfortunately, these unsophisticated pictures do not allow us to understand even the simplest organic molecule, methane.

39 Carbon is four-valent, and the methane molecule has been long known, and convincingly shown, to be a perfect tetrahedron, but none of the orbitals we have seen so far are directed towards the corners of a tetrahedron. In order to make the right orbitals, we will need to mix, or hybridize s- and p-orbitals.

40 Organic chemists need a model that reliably produces and predicts what we need to understand organic molecules and their reactions. Physical chemistry involves more serious mathematics and needs other parameters; it may be more accurately descriptive but it is less usable for the complex molecules involved in organic chemistry.

41 It s probably best to think of hybridization not so much as something that happens (as though the molecule has a little think, and then decides which orbitals to mix together ) but as something that has happened in a ground state molecule.

42 What orbitals are available to make the four identical C-H bonds in methane? Hydrogen provides four 1s orbitals C 1s 2 2s 2 2p x1 2p y1 2p z 0 How can this give us the four identical C-H bonds in methane?

43 What orbitals do we need to make four identical C-H bonds in methane? Promote one electron from 2s to 2p z to give C 1s 2 2s 1 2p x1 2p y1 2p z 1 This gives us four unpaired electrons with which to make bonds, but they are not all identical, nor do they point in the right directions

44 So we mix, or hybridize, 2s, 2p x, 2p y and 2p z to give four identical hybrid orbitals, called sp 3 orbitals. These point towards the corners of a tetrahedron Why do we get a tetrahedron?

45 To say that this comes from complex mathematics that we don t need to understand is true, if a cop-out. It is consistent with VSEPR a tetrahedral arrangement of electron pairs minimizes repulsion between them Those of you who are mathematically inclined can demonstrate using vector algebra that a tetrahedron is defined by four corners of a cube (the proof is not trivial), and can thus calculate the tetrahedral angle as o from first principles.

46 Although we don t always draw sp 3 orbitals in the form that matches physical reality, organic chemistry is very visual and it s useful to be aware of what they actually look like.

47 sp 3 orbitals

48 Each of these sp 3 orbitals combines with a 1s orbital to give a σ-bond. This is the conventional drawing, but we should not forget that each of the sp 3 orbitals has a back lobe that we are not showing, and that every time we combine atomic orbitals we generate not only a bonding but also an antibonding orbital.

49 sp3 hybridization and bonding in CH4

50 Ammonia also has a structure based on sp 3 hybridization at nitrogen We know that it is a base, and that it has a non-bonding or lone pair of electrons. N 1s 2 2s 2 2p x1 2p y1 2p z 1 You might initially think that we have three unpaired electrons, and need to make three bonds, so all is well.

51 But The orbitals are pointing in the wrong directions Electron diffraction studies tell us that ammonia is non-planar Ammonia has a dipole moment

52 Electron counting and VSEPR tell us that there are 4 pairs of electrons around nitrogen in ammonia, so the shape should be approximately tetrahedral. Thus we also need sp 3 hybridization for ammonia.

53 If we consider filling the sp 3 orbitals, then there are three unpaired electrons, which will pair up with the 1s electrons of the three hydrogens to make bonds, and a lone pair of electrons in an sp 3 orbital. lone pair form bonds to hydrogen

54 VSEPR tells us that lone pairs have a greater repulsive effect than bonding pairs, so the HNH angles are squeezed down from o to 107 o. : H N 2.1 H H 107 o

55 Moving on to water, oxygen has the electronic configuration: 1s 2 2s 2 2p x 2 2p y 1 2p z 1 As with ammonia, there are two unpaired electrons to make the two bonds to hydrogen, but the orbitals that they occupy point in the wrong directions for the known structure of water.

56 So again we need to make four sp 3 hybrid orbitals. Two will be filled and two half-filled lone pairs form bonds to hydrogen

57 Again VSEPR tells us that the lone pairs repel more than the bonding pairs So in the structure of water the HOH angle is 105 o : O : H H o

58 You may be wondering about the way in which the bonds in ammonia and water have been drawn. The use of what are called wedge and hashed bonds indicate stereochemistry, the 3-dimensional shape of the molecule. The solid wedge indicates that the bond is coming out of the plane of the paper, towards the viewer, and the hashed wedge that the bond is behind the plane of the diagram, away from the viewer.

59 We will be using this notation throughout it is a way for us to show the three-dimensional shape of a molecule projected onto a two-dimensional page.

60 The structures of species with lower symmetry can usually be deduced from these three basic structures. So CH 3 Cl, chloromethane, is like methane, and approximately tetrahedral Cl H C H H

61 Note that this will not be a perfect tetrahedron like methane chlorine is much larger than hydrogen, and much more electronegative, so a C-H bond is not identical to a C-Cl bond. Look carefully at how this is drawn everyone develops their own style, but for a tetrahedron two bonds should be in the plane of the page, one in front, and one behind.

62 In methanol, CH 3 OH, the carbon of the CH 3 is like methane, and the oxygen is like that in water. OH H C H H : O : H CH 3 So we can see methanol either as a substituted methane, or a substituted water molecule.

63 Similarly, we can view methylamine, CH 3 NH 2, either as a substituted methane, or a substituted ammonia. NH 2 : H C H H H N H CH 3

64 Drawing molecules Organic molecules are three-dimensional, so organic chemists need to have methods of drawing in two dimensions that reflect the original structures. For example, it is possible to focus on one specific bond.

65 If we consider ethane, C 2 H 6, and look at it from the side, we see this view H H C C H H H H The carbon-carbon bond and one carbon-hydrogen bond at each atom are in the plane of the paper, with one hydrogen in front and one behind the plane at each carbon. This is called a wedge form.

66 If we now move around the molecule about 45 o, then an oblique view, called a sawhorse form, is seen. There are two extreme forms of this type of diagram, shown in the next slide In the first all the C-H bonds are eclipsed or parallel hence we call this the eclipsed form. In the other, the C-H bonds on the adjacent carbons are as far apart as we can make them this is called the staggered form.

67 H H H H H H H H H H H H eclipsed staggered

68 Free rotation is possible about the carboncarbon single bond every intermediate form is also accessible. Intuitively you might expect the staggered form to be the more stable one and indeed it is, something we will explore in detail in Chapter 7. Notice that in these drawings we show the two carbon atoms simply as vertices.

69 As we progress to larger molecules it s important to keep diagrams uncluttered, and many of the types of representations we use do not show every atom.

70 Finally we move further around the molecule so that we are looking along the carbon-carbon bond the next slide shows the eclipsed and staggered forms of ethane. Here, again we don t show the carbon atom specifically the two aligned carbon atoms are represented by the circle. These are called Newman projections.

71 H H H H H H H H H H H H eclipsed staggered

72 Sometimes a line drawing of a molecule, without showing any three-dimensional geometry, is used, as in H H H C C H H H It s not ideal as it implies that the bonds around carbon point towards the corner of a square rather than a tetrahedron, but it is a layout you will encounter in your reading.

73 The stereochemistry of the molecule may not be relevant to the reaction being studied, or we simply wish a shorthand representation for a part of the molecule that remains unchanged.

74 The representations generally used by organic chemists are even simpler, however they can be described as stick drawings. These show only carbon-carbon bonds, with no hydrogen or carbon atoms (other atoms, oxygen, nitrogen, etc., are shown). Although most students initially find this idea alarming, it is the only practicable way to represent large molecules.

75 So is the shorthand way of writing H H H 3 C C C H C H H H H C H C H C H H

76 Although this is difficult to begin with, a little practice, and you will soon find it s both quicker and easier than writing out every atom. Don t forget that carbon will always be 4-valent and any bonds not shown at a particular vertex are bonds to hydrogen.

77 Alkanes C n H 2n+2 Organic chemists tend to have a limited interest in alkanes, despite their being some of the most abundant organic molecules on the planet. Abundant yes, reactive no their most important reaction is combustion. Much of the energy used in the developed world derives from the burning of hydrocarbons, in one form or another.

78 Alkanes are the simplest of hydrocarbons, with a general formula C n H 2n+2, and no multiple bonds. They are non-polar and largely insoluble in water. The lower molecular weight alkanes are gaseous, the next group are non-polar liquids, and the highest members of the family are oils or waxes. Boiling point and melting point increase with molecular weight.

79 Low molecular weight alkanes Name Formula bpt ( o C) Comments Methane CH Main constituent of natural gas, marsh gas, firedamp. An important greenhouse gas Ethane C 2 H % of natural gas, from which it is isolated. Thought to exist as lakes and precipitation on Titan (one of the moons of Saturn) Propane C 3 H 8-42 Main constituent of LPG, widely used in fuel for heating, portable stoves, barbecues Butane C 4 H Fuel for outdoor cooking, cigarette lighters. Propellant in aerosols Pentane C 5 H Blowing agent for polystyrene foam.

80 How are these and other compounds to be named? With some 60 million compounds recorded into the Chemical Abstracts Services database by June 2011 there has to be a system. The CAS registry has been in existence for 40 years, and it took 33 years to register the first 10 million compounds. The last 10 million were registered between September 2009 and June 2011.

81 Early names came from origins Lactic acid is isolated from milk (Latin, lactis, of milk). OH H CH 3 COOH lactic acid

82 Others were named by their discoverers thus barbituric acid (an anti-epilepsy drug also used in some sleeping pills) was supposedly discovered on the feast of St Barbara. O HN NH O O barbituric acid

83 Yet others are names for their shapes, such as cubane and buckminsterfullerene, named for its resemblance to the geodesic domes designed by the architect Richard Buckminster Fuller cubane

84 Buckminsterfullerene

85 There are also a number of jokes. housane fenestrane broken window

86 basketane snoutane [6]-Ivyane

87 ladderane (CH 2 ) 7 CO 2 Me

88 Formal nomenclature Formal nomenclature is not an exciting subject no-one could pretend otherwise, even when it s enlivened by stories of the origins of the names. However, it is important, and something that you will need to keep working on throughout the course.

89 It s analogous to learning the regular and irregular verbs of a language; you may learn much other vocabulary, but if you can t conjugate to be, to have and to go, which are irregular in most languages, you will always make mistakes. The rules for formal nomenclature are defined (in all languages, not just English) by IUPAC, the International Union of Pure and Applied Chemistry.

90 However, many common names are so well entrenched that they are still widely used, so you will need to recognize these, and they will be discussed as we meet the various organic functional groups. A complete list of common names that you should know (at the very least because this is how the bottles in your lab course will be labeled) is in Appendix 2 of the text.

91 Returning to the simple alkanes, the names from pentane onwards derive from the number of carbon atoms. For now we ll just look at the linear alkanes, and the names are shown below in the table below. Each name is composed of a root, which tells us how many carbon atoms we have, plus a suffix which tells us what type of molecule it is.

92 For alkanes our suffix is ane. Most of the prefixes are Greek in origin, and many will be familiar from other words that you know. You would not easily deduce the names of very long alkanes, but most will be understandable, once seen. For example C 74 H 150 is tetraheptacontane; tetra gives us the 4, hepta the 7 and contane is used for the decades between 30 and 100.

93 Nomenclature of alkanes Formula of linear alkane Name CH 4 methane C 2 H 6 ethane C 3 H 8 propane C 4 H 10 butane C 5 H 12 pentane C 6 H 14 hexane C 7 H 16 heptane C 8 H 18 octane C 9 H 20 nonane C 10 H 22 decane C 11 H 24 undecane C 12 H 26 dodecane

94 Structural isomers and formal nomenclature With methane, ethane and propane, there is only one way the molecule can be put together that conforms to the requirements of valency. However, when we reach butane there are two possibilities.

95 The carbon atoms can be arranged in a linear 4-carbon chain, or a branched chain. n-butane iso-butane These were initially called n-butane, normal butane, and iso-butane.

96 We describe these as structural isomers the carbon atoms are joined together in a different pattern. Notice that I used stick drawings; if you draw out the same structures with all the atoms there, you will quickly see why these are better!

97 With pentane, three structures are possible - n-pentane (normal pentane), iso-pentane and neo-pentane, the new pentane. n-pentane iso-pentane neo-pentane

98 However, moving on to higher alkanes, clearly a different system is needed new prefixes cannot be endlessly invented. If we want to name this molecule, we need to use the IUPAC system.

99 The rules are defined by IUPAC, the International Union of Pure and Applied Chemistry. First find the longest chain (don t be fooled by the drawing) in this case C 8. So we will name this as an octane. Number from one end so that the substituents are on the lowest number carbon atom.

100 a 2.21b x

101 Name substituents and say where they are The suffix for a substituent is yl CH = methyl, Me 3 C H = ethyl, Et 2 5 -CH CH CH = n-propyl, n-pr CH(CH ) = iso-propyl, i-pr 3 2 -CH CH CH CH = n-butyl, n-bu

102 Order substituents alphabetically (not a very rigid rule, because of language differences) So this is 2,3-dimethyl-5-ethyloctane

103 How do we deal with a substituent that is itself branched? provides an example.

104 We first note that the chain has 12 carbons, so it will be named as a dodecane, and we number from the right, so that the substituent is on C-6 (it would be on C-7 if we numbered from the left)

105 So this will be 6-(???)dodecane. In naming the substituent we find the longest chain working outwards from the main chain in this case 3 carbons long, so this is a propyl group. Finally we add in the substituents on that propyl chain to get 6-(1,1-dimethylpropyl)dodecane.

106 Students often worry about the location of the parentheses you will get it right if you bear in mind that their purpose is simply to make things clear.

107 Haloalkanes When an alkane is substituted with one of more halogen atoms, then the halogens are specified as substituents In numbering, they take precedence over alkyl substituents Described as chloro, bromo, etc.

108 3-bromoheptane Br

109 2-chloro-7-methyloctane Cl

110 Cycloalkanes Cycloalkanes contain a carbocyclic ring, and because there are no methyl group ends to the molecule, they have the general formula C n H 2n. The names still derive from the number of carbon atoms, but now the whole name has the prefix cyclo to indicate the ring structure.

111 cyclopropane cyclohexane

112 Substituted cycloalkanes are named as derivatives of the ring. is methylcyclohexane

113 If there is more than one substituent on the ring, then we need to state their relative positions. One of the substituents positions will be designated as carbon-1, and the ring will then be numbered so that the other(s) are on the lowest numbered carbon atoms. So is 1,2-dimethylcyclohexane.

114 One more specification needs to be added with cycloalkanes if there is more than one substituent are the two on the same or on opposite sides of the ring? We will defer considering what the real three-dimensional shapes of various sized rings are for the present We describe two groups on the same side of a ring as being cis and those on opposite sides as being trans.

115 These terms are taken from Latin, and are normally italicized. So we have cis-1,2-dimethylcyclohexane trans-1,3-dimethylcyclopentane

116 Stability, strain and properties of alkanes and cycloalkanes Although we have not yet studied the shapes of cycloalkanes in detail, there are some simple observations that we can make. Cyclopropane and cyclobutane must be strained molecules the carbon atoms are sp 3 hybridized, but clearly can t attain a CCC angle of o.

117 So these cannot be very happy molecules. The table shows the strain in various cycloalkanes. In cyclopropane and cyclobutane most of the strain is angle strain resulting from their inability to attain true tetrahedral angles.

118 Strain energy in cycloalkanes Cycloalkane Strain energy kj mol -1 cyclopropane cyclobutane cyclopentane 27.2 cyclohexane 0 cycloheptane 26.4 cyclooctane 40.2 cyclononane 52.7 cyclodecane 50.2 cycloundecane 46 cyclododecane 10

119 The other cycloalkanes are not planar (in fact cyclobutane is not quite planar, but the distortion is small), and hence the tetrahedral angles can be accommodated. Notice that the six-membered ring is unstrained, and this gives rise to a theme that you will see repeated throughout your studies.

120 Six-membered rings are the easiest to make, and the most stable of ring structures, in organic chemistry. What may be more surprising is the increase in strain in the 7-12 carbon range, usually referred to as medium rings. With rings with more than 13 carbon atoms, there is no strain.

121 Where do the figures for strain energy come from? We get data about alkane/cycloalkane stability from their heats of combustion and formation Heats of combustion are easier to measure C n H 2n+2 + XS O 2 nco 2 + (n + 1)H 2 O + heat

122 The greater the number of hydrogen and carbon atoms, the greater the heat of combustion. You might think that all alkanes with the same molecular formula burn with the same heat of combustion, but this is not so.

123 The strengths of O=O, C=O and O-H bonds determine the energy content of the products, and these are fixed and invariant. So differences in heats of combustion must reflect differences in stability of the starting alkanes

124 Heats of combustion of the lower alkanes Alkane ΔH combustion kj mol -1 CH C 2 H C 3 H

125 We can see the obvious conclusion that more carbon means more heat, but the differences between the isomeric butanes, and the isomeric pentanes are instructive. In each case the more branched the isomer, the less energy we get on burning it. Since the products are identical each time this means that the branched isomers are lower in energy than the linear ones.

126 This is quite general for alkanes, and indeed quite a number of other compound classes.

127 With cycloalkanes the heat of combustion/carbon atom gives a good measure of stability The more energy is generated by combustion, the more strained the original ring, since the strain energy is completely released on combustion.

128 Heats of combustion of cycloalkanes per carbon atom Cycloalkane ΔH combustion /n kj mol -1 cyclopropane 697 cyclobutane cyclopentane cyclohexane 655 cycloheptane cyclooctane cyclodecane 660 n is the number of carbon atoms in the ring

129 Physical properties of alkanes depend on molecular weight, and molecular shape. Boiling points increase with molecular weight. Melting points also generally increase, but the increase is less regular, as it depends on the crystalline form of the material

130 Branching lowers the boiling point of liquids. Boiling depends on overcoming the energy of interaction between molecules for hydrocarbons these are dispersion forces. Dispersion forces are maximized when the molecules can stack in the liquid phase. Substituents disrupt stacking, and lower the interactions.

131 Physical properties of alkanes Alkane Mpt ( o C) Bpt ( o C) methane ethane propane butane iso-butane(2-methylpropane) pentane iso-pentane (2-methylbutane) neo-pentane (2,2-dimethylpropane) hexane iso-hexane (2-methylpentane) neo-hexane (2,2-dimethylbutane) octane decane hexadecane triacontane (C 30 )

132 Chapter 2 Solutions to review problems

133 g of a molecule containing C, H and O is burned in air to give g CO 2 and g H 2 O. Calculate its empirical formula g CO 2 contains 12/44 x = g C g H 2 O contains 2/18 x = g H

134 So the molecule contains 52.3 % C % H % O Relative # atoms C = 52.3/12 = 4.36 Relative # atoms H = Relative # atoms O = 34.65/16 = 2.17

135 Dividing through by the lowest number gives C 2 H 6 O 1 So formula is C 2 H 6 O

136 2 Vitamin C has the following percentage composition. Derive its empirical formula % C 4.5 % H 54.5 % O

137 Relative # atoms C = 40.9/12 = 3.41 Relative # atoms H = 4.5 Relative # atoms O = 54.5/16 = 3.4 Dividing through by 3.4 C 1 H 1.32 O 1

138 We must multiply by 3 to make whole numbers of H atoms So empirical formula is C 3 H 4 O 3 Vitamin C is actually C 6 H 8 O 6 this is the molecular formula

139 3 Give a systematic name for each of the following molecules (a) trans-1-ethyl-2-methylcyclopentane If you are using this problem for review you should be able to add the stereochemical designators, R,R

140 3(b) Find the longest chain and number it in this case from the right as drawn, so that the substituents are on the lowest numbered carbon atoms

141 ,3,4-trimethylhexane

142 3(c) nonane

143 3(d) Cl Cl In numbering the ring the chlorine atoms take precedence over the methyl substituent. So this is 1,1-dichloro-3-methylcyclohexane

144 3(e) In this case when we number the longest chain it is not the most obvious one be careful! So 3,4,6-trimethyloctane

145 3(f) Here we will need to check carefully which is the longest chain it is not the most obvious one

146 So we have We will name this as an undecane, and the methyl substituents at positions 3 and 9 are obvious

147 We now need to name the substituents at positions 5 and 6 We number the side chain outwards from the main chain

148 So the complete name is 3,9-dimethyl-5-(1-methyethyl)-6-(2- methylpropyl)undecane The commonest error made here is right at the start, in not finding the longest chain

149 3(g) CH 3 CH 3 CH 2 CH CH 3 CH 2 CH 2 CH CH 3 1 We number the chain from the right hand end as this will give the lowest numbers for the locations of the substituents This is 2,5-dimethylheptane

150 3(h) We need to check carefully which is the longest chain, and from which end we should number

151 This is a decane, bearing 5 methyl groups If we number from the right, the substituents will be on carbons 2,2,3,8 and 9

152 If we numbered from the left the substituents would be on carbons 2,3,8,9 and 9 So we should number from the right, and the full name is 2,2,3,8,9-pentamethyldecane

153 3(i) Cl First number the chain from the left so that the substituent is on C-6

154 So we will name this as 6-( )tridecane Next we number the substituent out from the main chain 5 4 Cl 3 2 1

155 So we will name this as 6-(3-chloro-3,4,4-trimethylpentyl)tridecane

156 3(j) Br Again we need to look carefully at this molecule to find the longest chain And we must number so that the bromine, which takes precedence over the alkyl substituents, is on the lowest numbered carbon

157 Br bromo-5,8-dimethyl-6-ethyldecane

158 4 Draw a line structure for each of the following molecules: (a) 2-methylheptane First draw heptane Number Add the methyl group at C-2

159 4(b) ethylcyclopropane

160 4(c) 2,3,4-trimethylheptane First draw heptane, and number it Then put in the substituents

161 4(d) 2,8-dimethyl-4-ethyl-3-(1-methylethyl)decane First draw and number decane Then add substituents at the appropriate points

162

163 4(e) 4-methyl-2,6-diphenylheptane First draw and number heptane Ph Ph Then add the substituents

164 4(f) cis-1-ethyl-2-propylcyclohexane

165 4(g) 2,4,6-trimethylnonane Draw and number nonane Add in the substituents

166 4(h) trans-1,4-dichlorocyclohexane Draw cyclohexane Add the substituents Add the stereochemistry Cl Cl

167 4(i) 6-methyl-4(1-methylpropyl)nonane Draw and number nonane Add the 6-methyl substituent

168 We now need to add the 1-methylpropyl substituent at position 4. Start by putting in a propyl side chain and numbering it Finally add a methyl group at position-1 of the side chain

169 4(j) bromocyclobutane Br

170 4(k) 2,3,4-trimethyloctane Draw and number octane Add the methyl substituents

171 4(l) cis-1-bromo-3-methylcyclopentane Draw cyclopentane Add the substituents Add the stereochemistry Br Me

172 5 What hybridization and what shape would you predict for each of the following molecules? CH 2 Cl 2 (CH 3 ) 3 N CH 3 CH 2 OCH 2 CH 3 As far as the carbon atoms are concerned, CH 2 Cl 2, (CH 3 ) 3 N and CH 3 CH 2 OCH 2 CH 3 can all be considered as substituted methanes. This means we have sp 3 carbon, and an approximately tetrahedral shape

173 (CH 3 ) 3 N can also be considered to be a substituted ammonia the hydrogen atoms of NH 3 being replaced by methyl groups. So at nitrogen we expect the structure to resemble that of ammonia.

174 sp 3 hybridisation, pyramidal geometry, and a lone pair of electrons.. H 3 C N CH 3 CH 3

175 In the same way we can consider diethyl ether, (CH 3 CH 2 ) 2 O, to be a substituted water; the hydrogen atoms have been replaced by ethyl groups. So at oxygen we expect the structure to resemble that of water sp 3 hybridisation, and a bent, or angular structure

176 O H 3 CCH 2 CH 2 CH 3

177 6 For the following pair of molecules predict which would have the higher boiling point. Give a reason. n-heptane; the linear molecule stacks better

178 7 What are the most important kinds of intermolecular forces in the following atoms/molecules? (a) Krypton London dispersion forces (b) Tetrafluoromethane London dispersion forces

179 7(c) hydrogen fluoride Hydrogen bonding 7(d) butane London dispersion forces

180 8 Give an example of each of the following (a) A diatomic molecule with a permanent dipole moment HCl (b) A diatomic molecule without a permanent dipole moment O 2

181 (c) An sp 3 hybridized molecule with a permanent dipole moment CH 3 F (d) A triatomic molecule with a permanent dipole moment H 2 O

182 Chapter 2 Worked examples

183 Problem 2.1 Each of the following molecules contains one or more sp 3 hybridized atom. Sketch the structures you predict, focusing in turn on each of the C, N, and O atoms CF 3 Cl CH 3 OCH 3 (CH 3 ) 2 NH

184 Solution CF 3 Cl will have three fold symmetry. All the fluorines are identical, but the bond to chlorine will be different chlorine is larger, but less electronegative than fluorine. Structural studies give ClCF = 110 o and FCF as o

185 C l F C F F

186 CH 3 OCH 3 can be viewed either as a substituted methane or as a disubstituted water. As with water the COC angle will be about 105 o. H OCH 3 C H H : O : H 3 C CH 3

187 In Me 2 NH, two of the hydrogens of ammonia have been replaced by methyl groups again we can look at the structure as either a substituted methane, or a substituted ammonia. The bond angles at nitrogen, according to VSEPR, will be about 107 o NHCH 3 : H C H H H N CH 3 CH 3

188 Problem 2.2 For each of the following compounds, give a name according to the IUPAC nomenclature rules. 2.2(a) CH 3 CH 2 CH 3 CH 3 CH 2 CH CH CH 2 CH 2 CH 2 CH 3

189 Solution We first need to number the chain so that the substituents have the lowest numbers in this case from the left. CH 3 CH 2 CH CH 3 CH 2 CH CH CH 2 CH 2 CH 2 CH So this is 4-ethyl-3-methyloctane

190 2.2(b) This time we will number from the right So this is 2-methylnonane

191 2.2(c) Solution Br This problem is trickier. The longest continuous chain is 3 carbons long, but there are two chains of three carbons that we could select

192 Br Br Because of the priority of the bromine atom, we select the red chain, and number it so that the carbon bearing the priority bromine atom is C-1

193 3 2 1 Br So this is 1-bromo-2-methylpropane

194 Problem 2.3 For each of the following compounds draw a structural formula, using stick-type diagrams. 2.3(a) 2-bromo-2-methylpropane

195 Solution First draw and number propane (it s almost always best to start at the end of the name, and get the basic skeleton first) 1 2 We now add the substituents at the 2-position 3 Br

196 2.3(b) 2,2,4-trimethylpentane Solution First draw and number pentane Then add the substituents

197 2.3(c) 2,3-dimethyl-4-ethyloctane Solution Draw and number octane Add the substituents

198 Problem 2.4 Give a structure for each of the following molecules 2.4(a) trans-1,3-dimethylcyclobutane Solution

199 2.4(b) Solution cis-1,4-dichlorocyclohexane Cl Cl

200 Problem 2.5 Give a name for the following molecules, according to IUPAC rules 2.5(a) trans-1,3-dimethylcyclohexane

201 2.5(b) Br 1-bromo-1-methylcyclopentane