ISOMERISM

STRUCTURAL ISOMERISM

This page explains what structural isomerism is, and looks at some of the various ways that structural isomers can arise. What is structural isomerism? 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. For example, both of the following are the same molecule. They are not isomers. Both are butane. There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds.
If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule.
What are structural isomers? In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise. The names of the various forms of structural isomerism probably don't matter all that much, but you must be aware of the different possibilities when you come to draw isomers. Types of structural isomerism Chain isomerism These isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, C4H10. In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.
	Note:  Although the chain is drawn as straight, in reality it's anything but straight. If you aren't happy about the ways of drawing organic molecules, follow this link.
Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond. You could easily see this with a model. This is the example we've already used at the top of this page. Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models. Position isomerism In position isomerism, the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton. For example, there are two structural isomers with the molecular formula C3H7Br. In one of them the bromine atom is on the end of the chain, whereas in the other it's attached in the middle. If you made a model, there is no way that you could twist one molecule to turn it into the other one. You would have to break the bromine off the end and re-attach it in the middle. At the same time, you would have to move a hydrogen from the middle to the end. Another similar example occurs in alcohols such as C4H9OH These are the only two possibilities provided you keep to a four carbon chain, but there is no reason why you should do that. You can easily have a mixture of chain isomerism and position isomerism - you aren't restricted to one or the other. So two other isomers of butanol are:
You can also get position isomers on benzene rings. Consider the molecular formula C7H7Cl. There are four different isomers you could make depending on the position of the chlorine atom. In one case it is attached to the side-group carbon atom, and then there are three other possible positions it could have around the ring - next to the CH3 group, next-but-one to the CH3 group, or opposite the CH3 group. Functional group isomerism In this variety of structural isomerism, the isomers contain different functional groups - that is, they belong to different families of compounds (different homologous series). For example, a molecular formula C3H6O could be either propanal (an aldehyde) or propanone (a ketone). There are other possibilities as well for this same molecular formula - for example, you could have a carbon-carbon double bond (an alkene) and an -OH group (an alcohol) in the same molecule. Another common example is illustrated by the molecular formula C3H6O2. Amongst the several structural isomers of this are propanoic acid (a carboxylic acid) and methyl ethanoate (an ester).

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.
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 isunrestricted 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.
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?
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.
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.
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.

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.
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.
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.
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 aschiral. 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.
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 thecapital 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.
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 directionsexactly 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.
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.