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DICTIONARY OF CHEMISTRY-ENGLISH

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ORGANIC BASES

Why are primary amines basic? Ammonia as a weak base All of the compounds we are concerned with are derived from ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water. Ammonia in solution sets up this equilibrium: An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left. The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water. Comparing the strengths of weak bases The strengths of weak bases are measured on the pKb scale. The smaller the number on this scale, the stronger the base is. Three of the compounds we shall be looking at, together with their pKb values are: Remember - the smaller the number the stronger the base. Comparing the other two to ammonia, you will see that methylamine is a stronger base, whereas phenylamine is very much weaker. Methylamine is typical of aliphatic primary amines - where the -NH2 group is attached to a carbon chain. All aliphatic primary amines are stronger bases than ammonia. Phenylamine is typical of aromatic primary amines - where the -NH2 group is attached directly to a benzene ring. These are very much weaker bases than ammonia. Explaining the differences in base strengths The factors to consider Two of the factors which influence the strength of a base are: the ease with which the lone pair picks up a hydrogen ion, the stability of the ions being formed. Why are aliphatic primary amines stronger bases than ammonia? Methylamine Methylamine has the structure: The only difference between this and ammonia is the presence of the CH3 group in the methylamine. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions. Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion: In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there isn't any way of spreading the charge. To summarise: The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily. The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again. Taken together, these mean that methylamine is a stronger base than ammonia.
	Note:  This is a bit of a simplification for A' level purposes. As bases get more complex, another factor concerning the stability of the ions formed becomes important. That concerns the way they interact with water molecules in the solution. You don't need to worry about that at this level.
The other aliphatic primary amines The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of the other aliphatic primary amines are very similar to methylamine.
	Note:  If you want more information about the inductive effect of alkyl groups, you could read about carbocations (carbonium ions) in the mechanism section of this site.
For example: pKb CH3NH2 3.36 CH3CH2NH2 3.27 CH3CH2CH2NH2 3.16 CH3CH2CH2CH2NH2 3.39 Why are aromatic primary amines much weaker bases than ammonia? An aromatic primary amine is one in which the -NH2 group is attached directly to a benzene ring. The only one you are likely to come across is phenylamine. Phenylamine has the structure: The lone pair on the nitrogen touches the delocalised ring electrons . . . . . . and becomes delocalised with them: That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalised electrons will be attracted towards it, but the intensity of charge around the nitrogen is nothing like what it is in, say, an ammonia molecule. The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalisation. That means that the delocalisation would have to be disrupted if the phenylamine acts as a base. Delocalisation makes molecules more stable, and so disrupting the delocalisation costs energy and won't happen easily. Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalisation - this means that phenylamine is a very weak base indeed.

THE ACIDITY OF ORGANIC ACIDS

Why are organic acids acidic? Organic acids as weak acids For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water. An acid in solution sets up this equilibrium:
	Note:  We are writing the acid as AH rather than HA, because, in all the cases we shall be looking at, the hydrogen we are interested in is at the right-hand end of a molecule.
A hydroxonium ion is formed together with the anion (negative ion) from the acid. This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid. If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions. The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left. Comparing the strengths of weak acids The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is. Three of the compounds we shall be looking at, together with their pKa values are: Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all! Why are these acids acidic? In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X":
	Note:  If you aren't sure about coordinate covalent (dative covalent) bonding, you might like to follow this link. It isn't, however, particularly important to the rest of the current page. Use the BACK button on your browser to return to this page later.
So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths? Differences in acid strengths between carboxylic acids, phenols and alcohols The factors to consider Two of the factors which influence the ionisation of an acid are: the strength of the bond being broken, the stability of the ions being formed. In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar.
	Note:  You've got to be a bit careful about this. The bonds won't be identically strong, because what's around them in the molecule isn't the same in each case.
The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case. Ethanoic acid Ethanoic acid has the structure: The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH3COO-. You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond. To understand why this is, you have to look in some detail at the bonding in the ethanoate ion.
	Warning!  If you don't already understand about the bonding in the carbon-oxygen double bond, you would be well advised to skip this next bit - all the way down to the simplified structure of the ethanoate ion towards the end of it. It goes beyond anything that you are likely to want for UK A level purposes. If you do choose to follow this link, it will probably take you to several other pages before you are ready to come back here again. Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page.
Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen. In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them. This leads to a delocalised pi system over the whole of the -COO- group, rather like that in benzene. All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms. So where is the negative charge in all this? It has been spread around over the whole of the -COO- group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as: The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid. Phenol Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group. When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C6H5O-.
	Warning!  You need to understand about the bonding in benzene in order to make sense of this next bit. If your syllabus says that you need to know about the acidity of phenol, then you will have to understand the next few paragraphs - which in turn means that you will have to understand about benzene. If it doesn't mention phenol, skip it! If you follow this link, you may have to explore several other pages before you are ready to come back here again. Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page.
Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring. This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion. Why then is phenol a much weaker acid than ethanoic acid? Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this: But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram). If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge. That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic. In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge. What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent. However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic. Ethanol Ethanol, CH3CH2OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed: This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form. Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all. Variations in acid strengths between different carboxylic acids You might think that all carboxylic acids would have the same strength because each depends on the delocalisation of the negative charge around the -COO- group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion. In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids: pKa HCOOH 3.75 CH3COOH 4.76 CH3CH2COOH 4.87 CH3CH2CH2COOH 4.82 Remember that the higher the value for pKa, the weaker the acid is. Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid. The methanoate ion (from methanoic acid) is: The only difference between this and the ethanoate ion is the presence of the CH3 group in the ethanoate. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO- group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions. Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions. The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid.
	Note:  If you want more information about the inductive effect of alkyl groups, you could read about carbocations (carbonium ions) in the mechanism section of this site. Use the BACK button on your browser to return to this page if you choose to follow this link.
The acids can be strengthened by pulling charge away from the -COO- end. You can do this by attaching electronegative atoms like chlorine to the chain. As the next table shows, the more chlorines you can attach the better: pKa CH3COOH 4.76 CH2ClCOOH 2.86 CHCl2COOH 1.29 CCl3COOH 0.65 Trichloroethanoic acid is quite a strong acid. Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO- end and so strengthening the acid. pKa CH2FCOOH 2.66 CH2ClCOOH 2.86 CH2BrCOOH 2.90 CH2ICOOH 3.17 The effect is there, but isn't as great as you might expect. Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid. pKa CH3CH2CH2COOH 4.82 CH3CH2CHClCOOH 2.84 CH3CHClCH2COOH 4.06 CH2ClCH2CH2COOH 4.52 The chlorine is effective at withdrawing charge when it is next-door to the -COO- group, and much less so as it gets even one carbon further away.

STEREOISOMERISM - OPTICAL ISOMERISM

Optical isomerism is a form of stereoisomerism. This page explains what stereoisomers are and how you recognise the possibility of optical isomers in a molecule. What is stereoisomerism? What are isomers? Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, and is dealt with on a separate page.
	Note:  If you aren't sure about structural isomerism, it might be worth reading about it before you go on with this page.
What are stereoisomers? In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement. Optical isomerism is one form of stereoisomerism. Optical isomerism Why optical isomers? Optical isomers are named like this because of their effect on plane polarised light.
	Help!  If you don't understand about plane polarised light, follow this link before you go on with this page.
Simple substances which show optical isomerism exist as two isomers known as enantiomers. A solution of one enantiomer rotates the plane of polarisation in a clockwise direction. This enantiomer is known as the (+) form. For example, one of the optical isomers (enantiomers) of the amino acid alanine is known as (+)alanine. A solution of the other enantiomer rotates the plane of polarisation in an anti-clockwise direction. This enantiomer is known as the (-) form. So the other enantiomer of alanine is known as or (-)alanine. If the solutions are equally concentrated the amount of rotation caused by the two isomers is exactly the same - but in opposite directions. When optically active substances are made in the lab, they often occur as a 50/50 mixture of the two enantiomers. This is known as a racemic mixture or racemate. It has no effect on plane polarised light.
	Note:  One of the worrying things about optical isomerism is the number of obscure words that suddenly get thrown at you. Bear with it - things are soon going to get more visual! There is an alternative way of describing the (+) and (-) forms which is potentially very confusing. This involves the use of the lowercase letters d- and l-, standing for dextrorotatory and laevorotatory respectively. Unfortunately, there is another different use of the capital letters D- and L- in this topic. This is totally confusing! Stick with (+) and (-).
How optical isomers arise The examples of organic optical isomers required at A' level all contain a carbon atom joined to four different groups. These two models each have the same groups joined to the central carbon atom, but still manage to be different: Obviously as they are drawn, the orange and blue groups aren't aligned the same way. Could you get them to align by rotating one of the molecules? The next diagram shows what happens if you rotate molecule B. They still aren't the same - and there is no way that you can rotate them so that they look exactly the same. These are isomers of each other. They are described as being non-superimposable in the sense that (if you imagine molecule B being turned into a ghostly version of itself) you couldn't slide one molecule exactly over the other one. Something would always be pointing in the wrong direction.
	Note:  Unless your visual imagination is reasonably good, this is much easier to understand if you have actually got some models to play with. If your school or college hasn't given you the opportunity to play around with molecular models in the early stages of your organic chemistry course, you might consider getting hold of a cheap set. The models made by Molymod are both cheap and easy to use. An introductory organic set is more than adequate. Google molymod to find a supplier and more about them, or have a look at this set or something similar from Amazon. Share the cost with some friends, keep it in good condition and don't lose any bits, and resell it via eBay or Amazon at the end of your course. Alternatively, get hold of some coloured Plasticene (or other children's modelling clay) and some used matches and make your own. It's cheaper, but more difficult to get the bond angles right.
What happens if two of the groups attached to the central carbon atom are the same? The next diagram shows this possibility. The two models are aligned exactly as before, but the orange group has been replaced by another pink one. Rotating molecule B this time shows that it is exactly the same as molecule A. You only get optical isomers if all four groups attached to the central carbon are different. Chiral and achiral molecules The essential difference between the two examples we've looked at lies in the symmetry of the molecules. If there are two groups the same attached to the central carbon atom, the molecule has a plane of symmetry. If you imagine slicing through the molecule, the left-hand side is an exact reflection of the right-hand side. Where there are four groups attached, there is no symmetry anywhere in the molecule. A molecule which has no plane of symmetry is described as chiral. The carbon atom with the four different groups attached which causes this lack of symmetry is described as a chiral centre or as an asymmetric carbon atom. The molecule on the left above (with a plane of symmetry) is described as achiral. Only chiral molecules have optical isomers. The relationship between the enantiomers One of the enantiomers is simply a non-superimposable mirror image of the other one. In other words, if one isomer looked in a mirror, what it would see is the other one. The two isomers (the original one and its mirror image) have a different spatial arrangement, and so can't be superimposed on each other. If an achiral molecule (one with a plane of symmetry) looked in a mirror, you would always find that by rotating the image in space, you could make the two look identical. It would be possible to superimpose the original molecule and its mirror image. Some real examples of optical isomers Butan-2-ol The asymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star. It's extremely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimensional arrangement around the asymmetric carbon atom. Then draw the mirror to show the examiner that you know what you are doing, and then the mirror image.
	Help!  If you don't understand this bond notation, follow this link to drawing organic molecules before you go on with this page.
Notice that you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse large groups - look, for example, at the ethyl group at the top of the diagram. It doesn't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image is drawn accurately, you will automatically have drawn the two isomers. So which of these two isomers is (+)butan-2-ol and which is (-)butan-2-ol? There is no simple way of telling that. For A'level purposes, you can just ignore that problem - all you need to be able to do is to draw the two isomers correctly. 2-hydroxypropanoic acid (lactic acid) Once again the chiral centre is shown by a star. The two enantiomers are: It is important this time to draw the COOH group backwards in the mirror image. If you don't there is a good chance of you joining it on to the central carbon wrongly. If you draw it like this in an exam, you won't get the mark for that isomer even if you have drawn everything else perfectly. 2-aminopropanoic acid (alanine) This is typical of naturally-occurring amino acids. Structurally, it is just like the last example, except that the -OH group is replaced by -NH2 The two enantiomers are: Only one of these isomers occurs naturally: the (+) form. You can't tell just by looking at the structures which this is. It has, however, been possible to work out which of these structures is which. Naturally occurring alanine is the right-hand structure, and the way the groups are arranged around the central carbon atom is known as an L- configuration. Notice the use of the capital L. The other configuration is known as D-. So you may well find alanine described as L-(+)alanine. That means that it has this particular structure and rotates the plane of polarisation clockwise. Even if you know that a different compound has an arrangement of groups similar to alanine, you still can't say which way it will rotate the plane of polarisation. The other amino acids, for example, have the same arrangement of groups as alanine does (all that changes is the CH3 group), but some are (+) forms and others are (-) forms. It's quite common for natural systems to only work with one of the enantiomers of an optically active substance. It isn't too difficult to see why that might be. Because the molecules have different spatial arrangements of their various groups, only one of them is likely to fit properly into the active sites on the enzymes they work with. In the lab, it is quite common to produce equal amounts of both forms of a compound when it is synthesised. This happens just by chance, and you tend to get racemic mixtures.
	Note:  For a detailed discussion of this, you could have a look at the page on the addition of HCN to aldehydes
Identifying chiral centres in skeletal formulae A skeletal formula is the most stripped-down formula possible. Look at the structural formula and skeletal formula for butan-2-ol. Notice that in the skeletal formula all of the carbon atoms have been left out, as well as all of the hydrogen atoms attached to carbons. In a skeletal diagram of this sort: there is a carbon atom at each junction between bonds in a chain and at the end of each bond (unless there is something else there already - like the -OH group in the example); there are enough hydrogen atoms attached to each carbon to make the total number of bonds on that carbon up to 4. We have already discussed the butan-2-ol case further up the page, and you know that it has optical isomers. The second carbon atom (the one with the -OH attached) has four different groups around it, and so is a chiral centre. Is this obvious from the skeletal formula? Well, it is, provided you remember that each carbon atom has to have 4 bonds going away from it. Since the second carbon here only seems to have 3, there must also be a hydrogen attached to that carbon. So it has a hydrogen, an -OH group, and two different hydrocarbon groups (methyl and ethyl). Four different groups around a carbon atom means that it is a chiral centre. A slightly more complicated case: 2,3-dimethylpentane The diagrams show an uncluttered skeletal formula, and a repeat of it with two of the carbons labelled. Look first at the carbon atom labelled 2. Is this a chiral centre? No, it isn't. Two bonds (one vertical and one to the left) are both attached to methyl groups. In addition, of course, there is a hydrogen atom and the more complicated hydrocarbon group to the right. It doesn't have 4 different groups attached, and so isn't a chiral centre. What about the number 3 carbon atom? This has a methyl group below it, an ethyl group to the right, and a more complicated hydrocarbon group to the left. Plus, of course, a hydrogen atom to make up the 4 bonds that have to be formed by the carbon. That means that it is attached to 4 different things, and so is a chiral centre. Introducing rings - further complications At the time of writing, one of the UK-based exam boards (Cambridge International - CIE) commonly asked about the number of chiral centres in some very complicated molecules involving rings of carbon atoms. The rest of this page is to teach you how to cope with these. We will start with a fairly simple ring compound: When you are looking at rings like this, as far as optical isomerism is concerned, you don't need to look at any carbon in a double bond. You also don't need to look at any junction which only has two bonds going away from it. In that case, there must be 2 hydrogens attached, and so there can't possibly be 4 different groups attached. In this case, that means that you only need to look at the carbon with the -OH group attached. It has an -OH group, a hydrogen (to make up the total number of bonds to four), and links to two carbon atoms. How does the fact that these carbon atoms are part of a ring affect things? You just need to trace back around the ring from both sides of the carbon you are looking at. Is the arrangement in both directions exactly the same? In this case, it isn't. Going in one direction, you come immediately to a carbon with a double bond. In the other direction, you meet two singly bonded carbon atoms, and then one with a double bond. That means that you haven't got two identical hydrocarbon groups attached to the carbon you are interested in, and so it has 4 different groups in total around it. It is asymmetric - a chiral centre. What about this near-relative of the last molecule? In this case, everything is as before, except that if you trace around the ring clockwise and anticlockwise from the carbon at the bottom of the ring, there is an identical pattern in both directions. You can think of the bottom carbon being attached to a hydrogen, an -OH group, and two identical hydrocarbon groups. It therefore isn't a chiral centre. The other thing which is very noticeable about this molecule is that there is a plane of symmetry through the carbon atom we are interested in. If you chopped it in half through this carbon, one side of the molecule would be an exact reflection of the other. In the first ring molecule above, that isn't the case. If you can see a plane of symmetry through the carbon atom it won't be a chiral centre. If there isn't a plane of symmetry, it will be a chiral centre. A seriously complicated example - cholesterol The skeletal diagram shows the structure of cholesterol. Some of the carbon atoms have been numbered for discussion purposes below. These are not part of the normal system for numbering the carbon atoms in cholesterol. Before you read on, look carefully at each of the numbered carbon atoms, and decide which of them are chiral centres. The other carbon atoms in the structure can't be chiral centres, because they are either parts of double bonds, or are joined to either two or three hydrogen atoms.
	Note:  I am being deliberately unkind here! Normally when a molecule like cholesterol is discussed in this context, extra detail is often added to the skeletal structure. For example, important hydrogen atoms or methyl groups are drawn in. It is good for you to have to do it the hard way!
So . . . how many chiral centres did you find? In fact, there are 8 chiral centres out of the total of 9 carbons marked. If you didn't find all eight, go back and have another look before you read any further. It might help to sketch the structure on a piece of paper and draw in any missing hydrogens attached to the numbered carbons, and write in the methyl groups at the end of the branches as well. This is done for you below, but it would be a lot better if you did it yourself and then checked your sketch afterwards. Starting with the easy one - it is obvious that carbon 9 has two methyl groups attached. It doesn't have 4 different groups, and so can't be chiral. If you take a general look at the rest, it is fairly clear that none of them has a plane of symmetry through the numbered carbons. Therefore they are all likely to be chiral centres. But it's worth checking to see what is attached to each of them. Carbon 1 has a hydrogen, an -OH and two different hydrocarbon chains (actually bits of rings) attached. Check clockwise and anticlockwise, and you will see that the arrangement isn't identical in each direction. Four different groups means a chiral centre. Carbon 2 has a methyl and three other different hydrocarbon groups. If you check along all three bits of rings , they are all different - another chiral centre. This is also true of carbon 6. Carbons 3, 4, 5 and 7 are all basically the same. Each is attached to a hydrogen and three different bits of rings. All of these are chiral centres. Finally, carbon 8 has a hydrogen, a methyl group, and two different hydrocarbon groups attached. Again, this is a chiral centre. This all looks difficult at first glance, but it isn't. You do, however, have to take a great deal of care in working through it - it is amazingly easy to miss one out.

STEREOISOMERISM - GEOMETRIC ISOMERISM

Geometric isomerism (also known as cis-trans isomerism or E-Z isomerism) is a form of stereoisomerism. This page explains what stereoisomers are and how you recognise the possibility of geometric isomers in a molecule. Further down the page, you will find a link to a second page which describes the E-Z notation for naming geometric isomers. You shouldn't move on to that page (even if the E-Z notation is what your syllabus is asking for) until you are really confident about how geometric isomers arise and how they are named on the cis-trans system. The E-Z system is better for naming more complicated structures but is more difficult to understand than cis-trans. The cis-trans system of naming is still widely used - especially for the sort of simple molecules you will meet at this level. That means that irrespective of what your syllabus might say, you will have to be familiar with both systems. Get the easier one sorted out before you go on to the more sophisticated one! What is stereoisomerism? What are isomers? Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, and is dealt with on a separate page.
	Note:  If you aren't sure about structural isomerism, it might be worth reading about it before you go on with this page.
What are stereoisomers? In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement. Geometric isomerism is one form of stereoisomerism. Geometric (cis / trans) isomerism How geometric isomers arise These isomers occur where you have restricted rotation somewhere in a molecule. At an introductory level in organic chemistry, examples usually just involve the carbon-carbon double bond - and that's what this page will concentrate on. Think about what happens in molecules where there is unrestricted rotation about carbon bonds - in other words where the carbon-carbon bonds are all single. The next diagram shows two possible configurations of 1,2-dichloroethane. These two models represent exactly the same molecule. You can get from one to the other just by twisting around the carbon-carbon single bond. These molecules are not isomers. If you draw a structural formula instead of using models, you have to bear in mind the possibility of this free rotation about single bonds. You must accept that these two structures represent the same molecule: But what happens if you have a carbon-carbon double bond - as in 1,2-dichloroethene? These two molecules aren't the same. The carbon-carbon double bond won't rotate and so you would have to take the models to pieces in order to convert one structure into the other one. That is a simple test for isomers. If you have to take a model to pieces to convert it into another one, then you've got isomers. If you merely have to twist it a bit, then you haven't!
	Note:  In the model, the reason that you can't rotate a carbon-carbon double bond is that there are two links joining the carbons together. In reality, the reason is that you would have to break the pi bond. Pi bonds are formed by the sideways overlap between p orbitals. If you tried to rotate the carbon-carbon bond, the p orbitals won't line up any more and so the pi bond is disrupted. This costs energy and only happens if the compound is heated strongly. If you are interested in the bonding in carbon-carbon double bonds, follow this link. Be warned, though, that you might have to read several pages of background material and it could all take a long time. It isn't necessary for understanding the rest of this page.
Drawing structural formulae for the last pair of models gives two possible isomers. In one, the two chlorine atoms are locked on opposite sides of the double bond. This is known as the trans isomer. (trans : from latin meaning "across" - as in transatlantic). In the other, the two chlorine atoms are locked on the same side of the double bond. This is know as the cis isomer. (cis : from latin meaning "on this side") The most likely example of geometric isomerism you will meet at an introductory level is but-2-ene. In one case, the CH3 groups are on opposite sides of the double bond, and in the other case they are on the same side. The importance of drawing geometric isomers properly It's very easy to miss geometric isomers in exams if you take short-cuts in drawing the structural formulae. For example, it is very tempting to draw but-2-ene as CH3CH=CHCH3 If you write it like this, you will almost certainly miss the fact that there are geometric isomers. If there is even the slightest hint in a question that isomers might be involved, always draw compounds containing carbon-carbon double bonds showing the correct bond angles (120°) around the carbon atoms at the ends of the bond. In other words, use the format shown in the last diagrams above. How to recognise the possibility of geometric isomerism You obviously need to have restricted rotation somewhere in the molecule. Compounds containing a carbon-carbon double bond have this restricted rotation. (Other sorts of compounds may have restricted rotation as well, but we are concentrating on the case you are most likely to meet when you first come across geometric isomers.) If you have a carbon-carbon double bond, you need to think carefully about the possibility of geometric isomers. What needs to be attached to the carbon-carbon double bond?
	Note:  This is much easier to understand if you have actually got some models to play with. If your school or college hasn't given you the opportunity to play around with molecular models in the early stages of your organic chemistry course, you might consider getting hold of a cheap set. The models made by Molymod are both cheap and easy to use. An introductory organic set is more than adequate. Google molymod to find a supplier and more about them, or have a look at this set or something similar from Amazon. Share the cost with some friends, keep it in good condition and don't lose any bits, and resell it via eBay or Amazon at the end of your course. Alternatively, get hold of some coloured Plasticene (or other children's modelling clay) and some used matches and make your own. It's cheaper, but more difficult to get the bond angles right.
Think about this case: Although we've swapped the right-hand groups around, these are still the same molecule. To get from one to the other, all you would have to do is to turn the whole model over. You won't have geometric isomers if there are two groups the same on one end of the bond - in this case, the two pink groups on the left-hand end. So . . . there must be two different groups on the left-hand carbon and two different groups on the right-hand one. The cases we've been exploring earlier are like this: But you could make things even more different and still have geometric isomers: Here, the blue and green groups are either on the same side of the bond or the opposite side. Or you could go the whole hog and make everything different. You still get geometric isomers, but by now the words cis and trans are meaningless. This is where the more sophisticated E-Z notation comes in. Summary To get geometric isomers you must have: restricted rotation (often involving a carbon-carbon double bond for introductory purposes); two different groups on the left-hand end of the bond and two different groups on the right-hand end. It doesn't matter whether the left-hand groups are the same as the right-hand ones or not.
	Note:  The rest of this page looks at how geometric isomerism affects the melting and boiling points of compounds. If you are meeting geometric isomerism for the first time, you may not need this at the moment. If you need to know about E-Z notation, you could follow this link at once to the next page. (But be sure that you understand what you have already read on this page first!) Alternatively, read to the bottom of this page where you will find this link repeated.
The effect of geometric isomerism on physical properties The table shows the melting point and boiling point of the cis and trans isomers of 1,2-dichloroethene. isomer melting point (°C) boiling point (°C) cis -80 60 trans -50 48 In each case, the higher melting or boiling point is shown in red. You will notice that: the trans isomer has the higher melting point; the cis isomer has the higher boiling point. This is common. You can see the same effect with the cis and trans isomers of but-2-ene: isomer melting point (°C) boiling point (°C) cis-but-2-ene -139 4 trans-but-2-ene -106 1 Why is the boiling point of the cis isomers higher? There must be stronger intermolecular forces between the molecules of the cis isomers than between trans isomers. Taking 1,2-dichloroethene as an example: Both of the isomers have exactly the same atoms joined up in exactly the same order. That means that the van der Waals dispersion forces between the molecules will be identical in both cases. The difference between the two is that the cis isomer is a polar molecule whereas the trans isomer is non-polar.
	Note:  If you aren't sure about intermolecular forces (and also about bond polarity), it is essential that you follow this link before you go on. You need to know about van der Waals dispersion forces and dipole-dipole interactions, and to follow the link on that page to another about bond polarity if you need to. Use the BACK button on your browser to return to this page.
Both molecules contain polar chlorine-carbon bonds, but in the cis isomer they are both on the same side of the molecule. That means that one side of the molecule will have a slight negative charge while the other is slightly positive. The molecule is therefore polar. Because of this, there will be dipole-dipole interactions as well as dispersion forces - needing extra energy to break. That will raise the boiling point. A similar thing happens where there are CH3 groups attached to the carbon-carbon double bond, as in cis-but-2-ene. Alkyl groups like methyl groups tend to "push" electrons away from themselves. You again get a polar molecule, although with a reversed polarity from the first example.
	Note:  The term "electron pushing" is only to help remember what happens. The alkyl group doesn't literally "push" the electrons away - the other end of the bond attracts them more strongly. The arrows with the cross on (representing the more positive end of the bond) are a conventional way of showing this electron pushing effect.
By contrast, although there will still be polar bonds in the trans isomers, overall the molecules are non-polar. The slight charge on the top of the molecule (as drawn) is exactly balanced by an equivalent charge on the bottom. The slight charge on the left of the molecule is exactly balanced by the same charge on the right. This lack of overall polarity means that the only intermolecular attractions these molecules experience are van der Waals dispersion forces. Less energy is needed to separate them, and so their boiling points are lower. Why is the melting point of the cis isomers lower? You might have thought that the same argument would lead to a higher melting point for cis isomers as well, but there is another important factor operating. In order for the intermolecular forces to work well, the molecules must be able to pack together efficiently in the solid. Trans isomers pack better than cis isomers. The "U" shape of the cis isomer doesn't pack as well as the straighter shape of the trans isomer. The poorer packing in the cis isomers means that the intermolecular forces aren't as effective as they should be and so less energy is needed to melt the molecule - a lower melting point.