Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with two questions from Set two. Hepta two five diene has two double bonds, each cis or trans. One three dimethyl cyclopentane has two chiral carbons, each one R or S, two choices again. So two times two, four isomers each, right? But the answer to both is three! So where did the fourth one go? The answer is one word: symmetry. Counting questions like these come up often in JEE Main, so let's begin.

0:33Which double bonds count?

So, first, geometrical isomers. On the left of the figure, cis but two ene has both C H three groups on the same side of the double bond; the small three written low is a subscript. Trans but two ene, next to it, has them on opposite sides. The Greek letter mu is the dipole moment, the slashed equals sign means not equal, and b p is the boiling point, with capital K for kelvin. The drawings on the right show E and Z names. Okay? Now let me ask you this. Why can the cis form not simply twist about the double bond and become trans? Because rotation about the C double bond C on screen would break the pi bond, the Greek letter pi. That does not happen at room temperature, so the two forms stay separate. Right? Now the condition. Say one carbon carries the groups small a and small b, and the other carries small d and small e. Isomers exist only if a is not equal to b, and d is not equal to e. Clear? So an end like C H two, double bond, with two hydrogens on the same carbon, rules it out. Count only the double bonds that pass this test, okay?

1:46Counting geometrical isomers in a chain

Now, counting. Take a chain with n double bonds that all pass our test. Each one can be cis or trans. So how many arrangements do we get in all? Yes, exactly, two times two, n times, which is two to the power n; the small n written high is the power. That is the answer when the two ends of the chain are different. Got it? But when both ends are the same group, careful! Read the cis, trans isomer from the other end, and it becomes trans, cis: the same molecule, counted twice. With me? So for identical ends, we use two to the power n minus one, plus two to the power p minus one. Okay? And what is p? For even n, p is n over two, and for odd n, p is n plus one, over two. Right?

2:38Chiral carbons, meso forms and the count

Now the second idea. Each chiral carbon is bonded to four different groups. In the figure, skip D glyceraldehyde on the far left; the other three Fischer projections are tartaric acids, with COOH at both ends. Each crossing point is a chiral carbon, and labels like two R, three R give their configurations. Clear? The first two sit on either side of the grey dashed line, a mirror. They are non-superimposable mirror images, a pair of enantiomers, both optically active. Right? Now look at the third form, marked meso, with the label two R, three S in brackets. What do you notice about its top half and its bottom half? Yes, exactly, the top half mirrors the bottom half! The coloured dashed line marked plane is a plane of symmetry. Any molecule with a plane or a centre of symmetry is achiral and optically inactive, even with chiral carbons, hence the label inactive. Got it? Such a compound is called meso: chiral carbons, plus a plane of symmetry. Its mirror image is the same compound, so tartaric acid has three forms, not four. With me? Now the count. Let n be the number of stereogenic units: chiral carbons, and double bonds showing geometrical isomerism. With no two identical halves, there are two to the power n stereoisomers. Okay? If it is symmetrical and n is even, there are two to the power n minus one active forms, and two to the power, n over two, minus one, meso forms. For tartaric acid, n is two: two active, one meso, three in all. Right?

4:23Example 1: geometrical isomers of hepta-2,5-diene

Now, Example one, our first question. How many geometrical isomers does hepta two five diene have? Here is a hint: write the structure and check the ends. Pause the video and try it first. Done? Okay, let's check. Hepta means seven carbons, with double bonds starting at carbons two and five. So, C H three, C H double bond C H, C H two, C H double bond C H, C H three; the dashes are single bonds and the equals signs are double bonds. With me? Now the test. Every doubly bonded carbon carries a hydrogen and a different group, so both double bonds count, and n is two. Okay? So the answer is two squared, which is four isomers, right? No, no, careful, look at the ends! Both ends are C H three, so the ends are the same, and two to the power n would count some molecules twice. Clear? So we use the second formula. Here n is two, which is even, so p is two over two, which gives one. Two to the power one, plus two to the power zero, is two plus one, which gives three. Got it? And here are the three: cis cis, trans trans, and cis trans. Trans cis is the same molecule read from the other end, as the brackets say. So our fourth form was only cis trans, counted twice!

5:54Example 2: stereoisomers of 1,3-dimethylcyclopentane

And now, Example two, our second question. How many stereoisomers does one three dimethyl cyclopentane have? Pause the video and try it first. Done? Okay, let's check. Cyclopentane is a ring of five carbons, with methyls on carbons one and three; C dash one means carbon one. Carbon one carries a hydrogen, a methyl, and two ring paths, a short one through carbon two and a long one through carbons five and four. They differ, so carbons one and three are both chiral, and n is two. Okay? So, two to the power two, which is four stereoisomers, right? Wait, not so fast, think back to the meso form. Put both methyl groups on the same face of the ring: that is the cis form. There is a plane of symmetry through carbon two and the middle of the bond between carbons four and five, and one half of the ring mirrors the other. So the cis form is meso, and optically inactive. Got it? Now the trans form, with the methyls on opposite faces. It has no plane and no centre of symmetry, so it is chiral, and its mirror image is a different molecule: a pair of enantiomers. With me? So, in total, one meso form plus two enantiomers gives three stereoisomers, not four. That answers our second question. Clear? And a quick check with our rule. The molecule has two identical halves, and 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, so three again, perfect.

7:34Quick revision: cis-trans test, chain formula, meso forms

So, quick revision. First, a double bond counts only when each of its carbons carries two different groups. Second, for a chain, two to the power n when the ends differ, and two to the power n minus one, plus two to the power p minus one, when they are the same. Third, a plane of symmetry gives an inactive meso form, so a symmetrical molecule can have fewer than two to the power n. Now open Set two, give yourself twenty five minutes, and try all fifteen problems before you look at the solutions. All the best!