Two or More Chiral Centres

Transcript

0:00Introduction

Hello everyone! Let us begin with a small puzzle. Tartaric acid, which is found in grapes, has two chiral carbons. One chiral carbon gives two mirror image forms, so two chiral carbons should give two times two, four stereoisomers, right? But chemists find only three forms of tartaric acid. So where did the fourth one go? Today we find out, and then learn to count stereoisomers in seconds. In JEE, questions on this come up often. So, let's begin.

0:33Tartaric acid: the missing fourth form

So, here are some Fischer projections. The drawing on the far left is D glyceraldehyde, from the last lesson; today we need only the three tartaric acids. Each crossing point is a chiral carbon, carbon two on top and carbon three below, with COOH at both ends. Horizontal bonds come towards you, and vertical bonds go away. The label under each drawing, like two R, three R, gives the configuration of carbon two and carbon three. With me? Now the first two tartaric acids. The grey dashed line between them is a mirror. The two R, three R form and the two S, three S form are mirror images that cannot be superimposed. So they are a pair of enantiomers, and both are optically active. Okay? Now the third drawing, labelled meso, with two R, three S in brackets. Look at it carefully. What do you notice about its top half and its bottom half? Yes, exactly! The top half is the mirror image of the bottom half: hydrogen on the left and O H on the right, in both rows. The red dashed line, marked plane, runs between carbon two and carbon three. Clear? So this molecule has a plane of symmetry. That makes it achiral and optically inactive, even with two chiral carbons, okay? That is why its label says inactive. We call it a meso compound. Now, the missing form. Draw the mirror image of the meso form, and turn it by one hundred and eighty degrees in the plane of the paper, which never changes a Fischer projection. You get the meso form back! So its mirror image is not a new compound, and that is why tartaric acid has only three forms. That solves our puzzle, got it?

2:22Meso compound or racemic mixture?

So, let us write the definition. Meso compound: it has chiral carbons, and also a plane of symmetry, or a centre of symmetry, as the brackets say. Right? It is one single compound, optically inactive by internal compensation: its two halves cancel each other inside the same molecule. And it cannot be resolved, because there is nothing to separate. Clear? But wait, the racemic mixture from an earlier lesson was also optically inactive. So meso and racemic are the same thing, right? No, no, careful, this is a common trap! The racemic mixture is two compounds, the two enantiomers, mixed together, and it can be resolved. The meso compound is one compound, and it cannot. Both are inactive, but for different reasons. With me?

3:12Diastereomers

Now compare meso tartaric acid with the two R, three R form. Same atoms, same bonds, different arrangement in space, so they are stereoisomers. But are they mirror images of each other? They are not, and that gives us a new name. Stereoisomers that are not mirror images of each other are called diastereomers. Okay? And unlike enantiomers, diastereomers have different physical properties: a different melting point, a different solubility, and a different magnitude of rotation. Got it? Because their properties differ, diastereomers can be separated by ordinary methods. Right? And geometrical isomers, cis and trans, are diastereomers too: they are stereoisomers, but not mirror images. Clear?

4:01Counting stereoisomers

Now the counting shortcut. Let n be the number of stereogenic units: the chiral carbons, and the double bonds that show geometrical isomerism. The two lines between the carbons on screen are a double bond. Okay? If the molecule cannot be divided into two identical halves, there are two to the power n stereoisomers; the small n written high is the power. All of them are optically active, if at least one unit is a chiral carbon. Got it? Now, symmetrical molecules, which divide into two identical halves, like tartaric acid. Will they have more forms than two to the power n, or fewer? Fewer, exactly, because some forms are meso. For n even, when all the units are chiral carbons, there are two to the power, n minus one, optically active forms, and two to the power, n over two, minus one, meso forms. Clear? For n odd, there are two to the power, n minus one, stereoisomers in all, of which two to the power, n minus one, whole over two, are meso. With me? Now check tartaric acid, where n is two, which is even. Two to the power one gives two active forms, and two to the power zero gives one meso form. Three in all, just as the figure showed. Right?

5:24Example 1: stereoisomers of 2,3-dibromopentane

Alright, let us do Example one. How many stereoisomers does two three dibromo pentane have, and how many are optically active? Pause the video and try it first. Done? Okay, let's check. The chain is C H three, CHB r, CHB r, C H two, C H three; each dash is a single bond, and the small numbers written low are subscripts. C dash two means carbon number two. Carbon two and carbon three each carry hydrogen, bromine, and two different carbon groups, so both are chiral, and n is two. Okay? Now, this looks like tartaric acid: two chiral carbons side by side, each carrying a bromine. So there is a meso form, and the answer is three, right? Oh no, careful, look at the two ends! One end is C H three, methyl, and the other is C two H five, ethyl. They differ, so there are no identical halves, and no meso form. Clear? So we use two to the power n, that is two squared, which is four stereoisomers, and all four are optically active, as two pairs of enantiomers. Got it?

6:39Example 2: stereoisomers of pentane-2,3,4-triol

Now a harder one. How many stereoisomers does pentane two three four triol have, and how many are optically active? Here is a hint: check for two identical halves first. Pause the video and try it first. Done? Okay, let's check. The chain is C H three, then three C H units each with O H in brackets, then C H three; each bracket is a branch. Carbons two, three and four all count as stereogenic units here, so n is three. And both ends are C H three, so the molecule is symmetrical about carbon three. With me? So, two to the power three, which is eight stereoisomers, right? No, no, wait, the molecule is symmetrical! And n is three, which is odd. So the total is two to the power, n minus one, that is two squared, which is four stereoisomers. Okay? The meso forms are two to the power, n minus one, whole over two. Three minus one is two, and two over two gives one, so two to the power one, which is two meso forms. Clear? So the optically active ones are four minus two, which leaves two, just one pair of enantiomers. The answer is four stereoisomers, with only two optically active. Got it?

8:05Quick revision: diastereomers, meso forms, counting rules

So, quick revision. First, diastereomers are stereoisomers that are not mirror images, and they have different properties. Second, a meso compound has chiral carbons plus a plane of symmetry, so it is one compound, and inactive. Third, count two to the power n for an unsymmetrical molecule, and fewer for a symmetrical one, because of the meso forms. Now go and try the practice problems on this page. You can do them. All the best!