MCQ Chapter 2 1 Which of the Following Cycloalkanes Has the Lowest Heat of Combustion Per CH2 Group? a Cyclopropane B Cyclobutan

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MCQ Chapter 2

1 Which of the following cycloalkanes has the lowest heat of combustion per CH2 group? A Cyclopropane B Cyclobutane C Cyclopentane D Cyclohexane Solution D

2 The general formula of a cycloalkane is

A CnH2n+2

B CnH2n

C CnH2n-2

D CnHn 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 C4H9Cl? A 3 B 4 C 5 D 6 Solution B

6 Which of the following C6H12 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, C10H22 and hexadecane, C16H34. 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 CH4; II CO; III CO2; IV H2O; V NH3 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?

CHCl3 CH2Cl2 CH3Cl CCl4

A B C D Solution D 11 Which of the following molecules HAS a permanent dipole moment?

A CH4

B CO2 C Cyclohexane

D NH3 Solution D

12 What is the IUPAC name for

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 H2? A Hydrogen bonding B Dispersion forces C Dipole-dipole interactions D Electrostatic interactions Solution B 15 What is the IUPAC name for

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 sp3 hybridization at the central atom

A CH4

B CH3F

C C2H4

D C2H6 Solution C

18 The angles between sp3 orbitals are approximately A 90 o B 109 o C 120 o D 180 o Solution B

19 The stick drawing of CH3CH2CH(CH3)CH(CH3)CH2CH3 is A B

Solution A

20 The stick diagram for 2,5-dichloro-2,5-dimethylhexane is A Cl Cl

B Cl

C Cl Cl

D Cl

Cl Solution D

21 Which of the following cycloalkanes would have the LOWEST heat of combustion per CH2 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 CH4

C CF4

D CH3F Solution D Chapter 2 – Structure, bonding and nomenclature of alkanes

Test questions - solutions • 1 What are the most important intermolecular forces in

• (a) He (b) N2 (c) O2 (d) Br2 (e) HBr (f) H2O (g) Chloroethane (h) hexane?

• (a) He dispersion forces

• (b) N2 dispersion forces

• (c) O2 dispersion forces

• (d) Br2 dispersion forces • (e) HBr dipole-dipole interactions

• (f) H2O hydrogen bonding

• (g) Chloroethane dipole-dipole interactions

• (h) Hexane dispersion forces • 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

• CO2 • (c) A molecule that forms strong hydrogen bonds

•H2O

• (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) CHCl3 • Approx. tetrahedral, sp3 hybridization

• (b) CF4 • Perfect tetrahedron, sp3 hybridization • (c) CH3CH2OH

• CH3, CH2 - approx. tetrahedral. • Bent at O, with two lone pairs. • All atoms are sp3 hybridized

• (d) CH3NHCH2CH3

• CH3, CH2 approx. tetrahedral. • Pyramidal at N, with one lone pair. • All atoms are sp3 hybridized • 4 Give a systematic name for each of the following molecules • (a) Cl

Cl • 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 • (b)

• 3-ethyl-3-methylpentane • (c)

4 8 2 3 6 5 7

• This time we must be careful to find the longest chain in a less conventional drawing – this will be named as an octane • The substituents are at C-3 and C-6 • So 3,6-diethyloctane 2 • (d) 1

4 3

Br 5

• We need to check the longest chain carefully as this a less conventional drawing

• 3-(bromomethyl)pentane • (e)

• cis-1,3-diethylcyclohexane • (f) F

• 1-fluoro-1,2,2-trimethylcyclobutane • (g) F

• cis-1-fluoro-3-ethylcyclohexane • 1R,2S-1-fluoro-3-ethylcyclohexane • (h) 3 1 2 4 Br

• We number the molecule to put the bromine on the lowest numbered carbon atom

• 1-bromo-2,2-dimethylbutane • 5 Draw a stick structure for each of the following molecules

• (a) 5-bromodecane

Br • (b) 2,2-dimethylbutane • (c) 2-iodoethylcyclohexane

I • (d) 4,4-difluoroheptane

F F • (e) 3-ethyl-4-methylnonane • (f) 1-chloro-1-methylcyclopentane

Cl • (g) cis-1,2-dichlorocyclooctane

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

Cl • 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 oC, cyclohexane 81 oC. Cyclohexane stacks better in the liquid phase • (b) hexane and 2,2-dimethylbutane • Hexane 69 oC, 2,2-dimethylbutane 49 oC. • Substituents make packing more difficult in the liquid phase, and reduce intermolecular interactions

• (c) hexane and 1-fluorohexane • Hexane 69 oC, 1-fluorohexane 93 oC. • In 1-fluorohexane there are dipole-dipole forces to overcome as well as the dispersion forces that exist between all molecules. Chapter 2 Alkanes

Test questions

1 What are the most important intermolecular forces in

(a) He (b) N2 (c) O2 (d) Br2 (e) HBr

(f) H2O (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) CHCl3

(b) CF4

(d) CH3NHCH2CH3

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

(b) (c)

(d)

(e)

(f)

(g) F

(h)

5 Draw a stick structure for each of the following molecules

(a) 5-bromodecane (b) 2,2-dimethylbutane (c) 2-iodoethylcyclohexane (d) 4,4-difluoroheptane (e) 3-ethyl-4-methylnonane (f) 1-chloro-1-methylcyclopentane (g) cis-1,2-dichlorocyclooctane (h) 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) hexane and cyclohexane (b) hexane and 2,2-dimethylbutane (c) hexane and 1-fluorohexane Chapter 2

Alkanes and friends – structure, bonding, properties, nomenclature • 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. • 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. • 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. • 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. • What orbitals are available to make the four identical C-H bonds in methane?

• Hydrogen provides four 1s orbitals

2 2 1 1 0 • C 1s 2s 2px 2py 2pz

• How can this give us the four identical C-H bonds in methane? • What orbitals do we need to make four identical C-H bonds in methane?

• Promote one electron from 2s to 2pz to give

2 1 1 1 1 • C 1s 2s 2px 2py 2pz

• 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 • So we mix, or hybridize, 2s, 2px, 2py and 2pz to give four identical hybrid orbitals, called sp3 orbitals.

• These point towards the corners of a tetrahedron

• Why do we get a tetrahedron? • 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 109.5 o from first principles. • Although we don’t always draw sp3 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. • sp3 orbitals • Each of these sp3 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 sp3 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. 3 sp hybridization and bonding in CH4 • Ammonia also has a structure based on sp3 hybridization at nitrogen

• We know that it is a base, and that it has a non-bonding or lone pair of electrons.

2 2 1 1 1 • N 1s 2s 2px 2py 2pz

• You might initially think that we have three unpaired electrons, and need to make three bonds, so all is well. • But

• The orbitals are pointing in the wrong directions

• Electron diffraction studies tell us that ammonia is non-planar

• Ammonia has a dipole moment • 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 sp3 hybridization for ammonia. • If we consider filling the sp3 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 sp3 orbital.

lone pair form bonds to hydrogen • VSEPR tells us that lone pairs have a greater repulsive effect than bonding pairs, so the HNH angles are “squeezed down” from 109.5 o to 107 o. :

N H H H 107 o 2.1 • Moving on to water, oxygen has the electronic configuration:

2 2 2 1 1 • 1s 2s 2px 2py 2pz

• 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. • So again we need to make four sp3 hybrid orbitals.

• Two will be filled and two half-filled

lone pairs form bonds to hydrogen • Again VSEPR tells us that the lone pairs repel more than the bonding pairs

• So in the structure of water

• the HOH angle is 105o

: : O

H H

o 2.2 105 • 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. • 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. • The structures of species with lower symmetry can usually be deduced from these three basic structures.

• So CH3Cl, chloromethane, is like methane, and approximately tetrahedral

C H H H • 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. • In methanol, CH3OH, the carbon of the CH3 is like methane, and the oxygen is like that in water. OH

: : C H O H H H CH3

• So we can see methanol either as a substituted methane, or a substituted water molecule. • Similarly, we can view methylamine,

CH3NH2, either as a substituted methane, or a substituted ammonia.

NH2 :

C N H CH H H 3 H H 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. • If we consider ethane, C2H6, 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. • 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. H H

H H H H H H H

H H H

eclipsed staggered • Free rotation is possible about the carbon- carbon 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. • 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. • 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. H H H H H

H H H H H H H

eclipsed staggered • 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. • 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. • 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. • So

• is the shorthand way H 3 C H H H of writing C H C C H

C C H H C H H H H • 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. • Alkanes CnH2n+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. • Alkanes are the simplest of hydrocarbons, with a

general formula CnH2n+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. Low molecular weight alkanes

Name Formula bpt Comments (oC)

Methane CH4 -164 Main constituent of natural gas, marsh gas, firedamp. An important greenhouse gas

Ethane C2H6 -89 1-6 % of natural gas, from which it is isolated. Thought to exist as lakes and precipitation on Titan (one of the moons of Saturn)

Propane C3H8 -42 Main constituent of LPG, widely used in fuel for heating, portable stoves, barbecues

Butane C4H10 -0.5 Fuel for outdoor cooking, cigarette lighters. Propellant in aerosols

Pentane C5H12 36 Blowing agent for polystyrene foam. • 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. • Early names came from origins • Lactic acid is isolated from milk (Latin, lactis, of milk). OH

COOH H CH3

lactic acid • 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 • 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 Buckminsterfullerene • There are also a number of jokes….

housane fenestrane broken window snoutane basketane

[6]-Ivyane (CH2)7CO2Me ladderane • 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. • 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. • 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. • 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. • 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 C74H150 is tetraheptacontane; tetra gives us the 4, hepta the 7 and contane is used for the decades between 30 and 100. Nomenclature of alkanes Formula of linear alkane Name

CH4 methane

C2H6 ethane

C3H8 propane

C4H10 butane

C5H12 pentane

C6H14 hexane

C7H16 heptane

C8H18 octane

C9H20 nonane

C10H22 decane

C11H24 undecane

C12H26 dodecane • 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. • 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. • 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! • With pentane, three structures are possible - n-pentane (normal pentane), iso-pentane and neo-pentane, the new pentane.

n-pentane iso-pentane neo-pentane • 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. • 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 C8. So we will name this as an octane.

• Number from one end so that the substituents are on the lowest number carbon atom. 7 1 8 5 7 5 4 3 2 2 3 4 6 8 1 6

2.21a  2.21b x • Name substituents and say where they are • The suffix for a substituent is “yl”

• CH3 = methyl, Me •C2H5 = ethyl, Et • -CH2CH2CH3 = n-propyl, n-Pr • -CH(CH3)2 = iso-propyl, i-Pr • -CH2CH2CH2CH3 = n-butyl, n-Bu • Order substituents alphabetically (not a very rigid rule, because of language differences)

• So this is 2,3-dimethyl-5-ethyloctane • How do we deal with a substituent that is itself branched?

• provides an example. • 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).

1 6 11 10 9 8 7 5 4 3 2 12 • 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. 1 11 10 9 8 7 6 4 3 2 12 5 1

• 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. • 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. • 3-bromoheptane

2 3 5 4 7 6 1

Br • 2-chloro-7-methyloctane

8 5 7 2 3 4 6 1 • Cycloalkanes

• Cycloalkanes contain a carbocyclic ring, and because there are no methyl group “ends” to the molecule, they have the general

formula CnH2n.

• The names still derive from the number of carbon atoms, but now the whole name has the prefix “cyclo” to indicate the ring structure. cyclopropane cyclohexane • Substituted cycloalkanes are named as derivatives of the ring.

• is methylcyclohexane • 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. • 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. • These terms are taken from Latin, and are normally italicized. So we have

cis-1,2-dimethylcyclohexane trans-1,3-dimethylcyclopentane • 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 sp3 hybridized, but clearly can’t attain a CCC angle of 109.5 o. • 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. Strain energy in cycloalkanes

Cycloalkane Strain energy kJ mol-1 cyclopropane 115.5 cyclobutane 110.5 cyclopentane 27.2 cyclohexane 0 cycloheptane 26.4 cyclooctane 40.2 cyclononane 52.7 cyclodecane 50.2 cycloundecane 46 cyclododecane 10 • 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. • 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. • 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

•CnH2n+2 + XS O2  nCO2 + (n + 1)H2O + heat • 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. • 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 Heats of combustion of the lower alkanes

-1 Alkane ΔHcombustion kJ mol

CH4 891

C2H6 1560

C3H8 2219 2877 2868

3535 3529

3514 • 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. • This is quite general for alkanes, and indeed quite a number of other compound classes. • 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. Heats of combustion of cycloalkanes per carbon atom

-1 Cycloalkane ΔHcombustion/n kJ mol cyclopropane 697 cyclobutane 680.1 cyclopentane 658.2 cyclohexane 655 cycloheptane 655.2 cyclooctane 658.2 cyclodecane 660

n is the number of carbon atoms in the ring • 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 • 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. • Physical properties of alkanes

Alkane Mpt (oC) Bpt (oC) methane -183 -162 ethane -184 -88 propane -188 -42 butane -137 0 iso-butane(2-methylpropane) -133 -11 pentane -130 36 iso-pentane (2-methylbutane) -60 28 neo-pentane (2,2-dimethylpropane) -18 9 hexane -95 69 iso-hexane (2-methylpentane) -153 61 neo-hexane (2,2-dimethylbutane) -100 49 octane -53 126 decane -30 174 hexadecane 18 281 triacontane (C30) 66 550 Chapter 2

Solutions to review problems • 1 0.543 g of a molecule containing C, H

and O is burned in air to give 1.039 g CO2 and 0.6369 g H2O. Calculate its empirical formula

• 1.039 g CO2 contains 12/44 x 1.039 = 0.2834 g C

• 0.6368 g H2O contains 2/18 x 0.638 = 0.07089 g H • So the molecule contains • 52.3 % C • 13.05 % H • 34.65 % O

• Relative # atoms C = 52.3/12 = 4.36 • Relative # atoms H = 13.05 • Relative # atoms O = 34.65/16 = 2.17 • Dividing through by the lowest number gives

• C 2 • H 6 • O 1

• So formula is C2H6O • 2 Vitamin C has the following percentage composition. • Derive its empirical formula. • 40.9 % C • 4.5 % H • 54.5 % O • 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 • We must multiply by 3 to make whole numbers of H atoms

• So empirical formula is C3H4O3

• Vitamin C is actually C6H8O6 – this is the molecular formula • 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 • 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 5 1 4 2 3 6

• 2,3,4-trimethylhexane • 3(c)

• nonane • 3(d) Cl Cl

1 2

• In numbering the ring the chlorine atoms take precedence over the methyl substituent.

• So this is 1,1-dichloro-3-methylcyclohexane • 3(e) 1 2

7 6 5 4 3 8

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

• Here we will need to check carefully which is the longest chain – it is not the most obvious one • So we have 11 1 10 2

9 8 7 6 5 3 4

• We will name this as an undecane, and the methyl substituents at positions 3 and 9 are obvious • We now need to name the substituents at positions 5 and 6

• We number the side chain outwards from the

main chain 2 1

1 2

3 • 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 • 3(g)

CH3 CH3

CH3 CH2 CH CH2 CH2 CH CH3 7 6 5 4 3 2 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 • 3(h)

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

2 4 6 8 10 12 13 1 3 5 7 9 11

• First number the chain from the left so that the substituent is on C-6 • So we will name this as 6-(…)tridecane • Next we number the substituent out from the main chain 5

4 3 Cl

2 1 • So we will name this as

• 6-(3-chloro-3,4,4-trimethylpentyl)tridecane • 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 10

Br 9

2 3 4 5 6 7 8 1

• 4-bromo-5,8-dimethyl-6-ethyldecane • 4 Draw a line structure for each of the following molecules: • (a) 2-methylheptane • First draw heptane

2 4 6 1 3 5 7

Number Add the methyl group at C-2 • 4(b) ethylcyclopropane • 4(c) 2,3,4-trimethylheptane

• First draw heptane, and number it • Then put in the substituents

6 3 5 1 2 4 7 • 4(d) 2,8-dimethyl-4-ethyl-3-(1-methylethyl)decane

• First draw and number decane

10 3 2 4 5 6 7 8 9 1

• Then add substituents at the appropriate points 10 3 2 4 5 6 7 8 9 1 • 4(e) • 4-methyl-2,6-diphenylheptane • First draw and number heptane

Ph Ph

2 4 6 1 3 5 7

• Then add the substituents • 4(f) cis-1-ethyl-2-propylcyclohexane • 4(g) 2,4,6-trimethylnonane

• Draw and number nonane

5 2 3 4 6 7 8 1 9

• Add in the substituents • 4(h) trans-1,4-dichlorocyclohexane

• Draw cyclohexane Cl • Add the substituents • Add the stereochemistry

Cl • 4(i) 6-methyl-4(1-methylpropyl)nonane • Draw and number nonane

3 5 7 2 4 6 8 9 1

• Add the 6-methyl substituent • We now need to add the 1-methylpropyl substituent at position 4.

• Start by putting in a propyl side chain and numbering it 3

2 1

• Finally add a methyl group at position-1 of the side chain • 4(j) bromocyclobutane Br • 4(k) 2,3,4-trimethyloctane

• Draw and number octane

8 2 3 4 5 6 7 1

• Add the methyl substituents • 4(l) cis-1-bromo-3-methylcyclopentane

• Draw cyclopentane • Add the substituents • Add the stereochemistry

Me • 5 What hybridization and what shape would you predict for each of the following molecules?

• CH2Cl2 (CH3)3N CH3CH2OCH2CH3

• As far as the carbon atoms are concerned,

CH2Cl2, (CH3)3N and CH3CH2OCH2CH3 can all be considered as substituted methanes.

• This means we have sp3 carbon, and an approximately tetrahedral shape • (CH3)3N can also be considered to be a substituted ammonia – the hydrogen

atoms of NH3 being replaced by methyl groups.

• So at nitrogen we expect the structure to resemble that of ammonia. • sp3 hybridisation, pyramidal geometry, and a lone pair of electrons

N CH3 H3C CH3 • In the same way we can consider diethyl

ether, (CH3CH2)2O, 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 – sp3 hybridisation, and a bent, or angular structure O

H3CCH2 CH2CH3 • 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 • 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 • 7(c) hydrogen fluoride • Hydrogen bonding

• 7(d) butane • London dispersion forces • 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

•O2 • (c) An sp3 hybridized molecule with a permanent dipole moment

• CH3F

• (d) A triatomic molecule with a permanent dipole moment

•H2O Chapter 2

Worked examples • Problem 2.1

• Each of the following molecules contains one or more sp3 hybridized atom.

• Sketch the structures you predict, focusing in turn on each of the C, N, and O atoms

• CF3Cl CH3OCH3 (CH3)2NH • Solution

• CF3Cl 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 108.6 o C l

C F F F • CH3OCH3 can be viewed either as a substituted methane or as a disubstituted water.

• As with water the COC angle will be about 105 o. OCH3 : : C O H H H3C CH3 H • In Me2NH, 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 o NHCH be about 107 3 :

C N H CH3 H H H CH3 • Problem 2.2 • For each of the following compounds, give a name according to the IUPAC nomenclature rules.

• 2.2(a)

CH3 CH2CH3

CH3 CH2 CH CH CH2 CH2 CH2 CH3 • Solution • We first need to number the chain so that the substituents have the lowest numbers – in this case from the left.

CH3 CH2CH3 1 2 5 6 7 8 CH3 CH2 CH CH CH2 CH2 CH2 CH3 3 4

• So this is 4-ethyl-3-methyloctane • 2.2(b)

8 7 6 5 4 3 2 9 1 • This time we will number from the right

• So this is 2-methylnonane • 2.2(c)

Br • Solution

• This problem is trickier.

• The longest continuous chain is 3 carbons long, but there are two chains of three carbons that we could select 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 3

1 2

• So this is 1-bromo-2-methylpropane • Problem 2.3

• For each of the following compounds draw a structural formula, using stick-type diagrams.

• 2.3(a) 2-bromo-2-methylpropane • 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)

2 1 3 • We now add the substituents Br at the 2-position • 2.3(b) 2,2,4-trimethylpentane • Solution • First draw and number pentane

3 2 4 1 5

• Then add the substituents • 2.3(c) 2,3-dimethyl-4-ethyloctane • Solution • Draw and number octane 8 3 2 4 5 6 7 1

• Add the substituents • Problem 2.4 Give a structure for each of the following molecules • 2.4(a) trans-1,3-dimethylcyclobutane

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

Cl • Problem 2.5

• Give a name for the following molecules, according to IUPAC rules • 2.5(a)

• trans-1,3-dimethylcyclohexane • 2.5(b)

• 1-bromo-1-methylcyclopentane