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Optical isomerism pdf

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Optical isomerism. Task: ➢ Build a molecule using: ➢ Black moly mod in the centre. ➢ Attach a green, blue, red and white moly mod to the central black moly. STEREOISOMERISM - OPTICAL ISOMERISM. Optical isomerism is a form of stereoisomerism. This page explains what stereoisomers are and how you. is a single compound without isomer, while pentane has 3 isomers, a linear .. molecules rotate the light polarised in plane. this property is called as optical.


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Isomerism of Organic Molecules: Optical Isomerism. Structural isomers: Different compounds; all properties like boiling points, melting points, and solubilities are. Optical Isomerism occurs around a chiral center. If an atom is bonded to four different groups, its mirror image can not be rotated and superimposed onto. Explains what optical isomerism is and how you recognise the possibility of it in a molecule.

In that case, there must be 2 hydrogens attached, and so there can't possibly be 4 different groups attached. Unless your visual imagination is reasonably good, this is much easier to understand if you have actually got some models to play with. 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. Unless your visual imagination is reasonably good, this is much easier to understand if you have actually got some models to play with. A skeletal formula is the most stripped-down formula possible. An introductory organic set is more than adequate.

A skeletal formula is the most stripped-down formula possible. Look at the structural formula and skeletal formula for butanol. 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. We have already discussed the butanol 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.

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.

The diagrams show an uncluttered skeletal formula, and a repeat of it with two of the carbons labelled. 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. 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. 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. 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.

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. 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.

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. 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.

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! 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.

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-. 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.

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. For a detailed discussion of this, you could have a look at the page on the addition of HCN to aldehydes.

A skeletal formula is the most stripped-down formula possible. Look at the structural formula and skeletal formula for butanol. 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. We have already discussed the butanol 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.

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. The diagrams show an uncluttered skeletal formula, and a repeat of it with two of the carbons labelled. 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. This has a methyl group below it, an ethyl group to the right, and a more complicated hydrocarbon group to the left.

What is Optical Isomerism Notes pdf ppt

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. 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. 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. 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. 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. 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.

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.

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! 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. 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. 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. 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.

Isomerism pdf optical

If there isn't a plane of symmetry, it will be a chiral centre. The skeletal diagram shows the structure of cholesterol. Some of the carbon atoms have been numbered for discussion purposes below.

Pdf optical isomerism

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.

I am being deliberately unkind here!

optical isomerism

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! 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. If this is the first set of questions you have done, please read the introductory page before you start. What is stereoisomerism? What are isomers? What are stereoisomers? Optical isomerism Why optical isomers?

Optical isomers are named like this because of their effect on plane polarised light. 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: Chiral and achiral molecules The essential difference between the two examples we've looked at lies in the symmetry of the molecules.

Where there are four groups attached, there is no symmetry anywhere in the molecule. 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. Some real examples of optical isomers Butanol The asymmetric carbon atom in a compound the one with four different groups attached is often shown by a star. The two enantiomers are: Structurally, it is just like the last example, except that the -OH group is replaced by -NH 2 The two enantiomers are: Identifying chiral centres in skeletal formulae A skeletal formula is the most stripped-down formula possible.

In a skeletal diagram of this sort: