Geometrical Isomerism: cis–trans and E–Z

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two samples of but two ene have the same atoms, joined in the same order. Yet one boils at two hundred and seventy seven kelvin, and the other at two hundred and seventy four kelvin. Why can't one simply twist and become the other? The answer is the double bond, and a surprise waits for you at the end. This comes up often in JEE, so let's begin.

0:27Why the double bond locks the shape

So, look at the first drawing, but two ene. The equals sign is the double bond. The two methyl groups on top are written H three C and C H three, where the small three written low is a subscript, for three hydrogens. Now, why can't the right carbon just spin around the double bond? Here is the reason. One of the two bonds in a double bond is a pi bond, the Greek letter pi on screen. Rotating one carbon would break this pi bond. Clear? So at room temperature, it does not rotate. With two different groups on each doubly bonded carbon, we get two fixed arrangements that cannot be interconverted. These are geometrical isomers. With me? And not only for the carbon carbon double bond. In rings, the ring itself restricts rotation, and it also happens about C double bond N, and N double bond N, where N is nitrogen. Got it?

1:25Naming the two forms: cis and trans

So, how do we name them? In the first drawing, both methyl groups are on top. When the identical, or similar, groups sit on the same side, we call it cis, so this is cis but two ene, okay? In the second drawing, the methyls are on opposite sides, so it is trans but two ene. Under each of the first two drawings is mu, the Greek letter for dipole moment. The slashed equals sign means not equal to, so mu is not zero for cis, and it is zero for trans. And b p, the boiling point in kelvin, differs too, our puzzle from the start! More on these in the next lesson. Right?

2:04When is geometrical isomerism possible?

So, when exactly is it possible? Write the alkene as small a, small b, C, double bond, C, small d, small e. Isomerism appears only if small a is not equal to small b, and small d is not equal to small e. Remember, the slashed equals sign means not equal to. Okay? So a C H two end, with two identical hydrogens, rules it out, and so does any carbon with two identical groups. Swap them, and you get the same molecule back. Clear? Now, here is a trap. Do all four groups, small a, b, d and e, have to be different from one another? No, no, wait, not at all! We compare only the two groups on the same carbon, so small a may equal small d. In but two ene, each carbon has hydrogen and C H three, and still we got cis and trans. Got it?

3:01The E–Z system: priorities decide

Now the third drawing, two bromo one chloro propene. The left carbon has C l, chlorine, and hydrogen. The right carbon has B r, bromine, and a methyl group. So, is this cis, or trans? Hard to say, right? All four groups differ, so there is no identical pair. So we use the E Z system, which works for every alkene. First, on each carbon, find the higher priority group. Going by atomic number, as the rules coming up will show, chlorine beats hydrogen, and bromine beats the methyl carbon. Okay? If the two higher priority groups are on the same side, the isomer is Z, from the German zusammen, meaning together. Chlorine and bromine are both on top, so the third drawing is the Z isomer. With me? And if they are on opposite sides, it is called E, from entgegen, meaning opposite. In the fourth drawing, bromine has moved to the bottom, so that one is the E isomer.

4:02Sequence (Cahn–Ingold–Prelog) rules

Right? To decide priority, we use the sequence rules, the Cahn Ingold Prelog rules. The atom with the higher atomic number has the higher priority. Read the greater than signs: iodine, bromine, chlorine, sulphur, fluorine, oxygen, nitrogen, carbon, and hydrogen last. Okay? For isotopes, the heavier one wins. D is deuterium, heavy hydrogen, so it beats ordinary hydrogen. Right? Now, if the first atoms are the same, compare the sets of atoms attached to them, highest first, and decide at the first point of difference. Clear? Now, the dash in front of each group is its bond to the molecule. Compare C H two C l with a carbon carrying three methyl groups, shown by the brackets and the subscript three. Which group has the higher priority? Many students pick the three carbons, but the chlorine group wins. The brackets list the attached atoms: chlorine, hydrogen, hydrogen, against carbon, carbon, carbon. At the first point, chlorine beats carbon, and it is settled. In the same way, carbon, carbon, carbon beats ethyl's carbon, hydrogen, hydrogen, and ethyl beats methyl's three hydrogens. Got it? Last, a multiple bond counts as the same number of single bonds to that atom. So CHO is oxygen, oxygen, hydrogen. And C N with three lines, a triple bond, is nitrogen, nitrogen, nitrogen. With me?

5:37Example 1: which of four alkenes show geometrical isomerism

Let us try the first example. Which of these show geometrical isomerism: pent two ene, two methyl but two ene, three methyl pent two ene, and hex three ene? Pause the video and try it first. Done? Okay, let's check. In pent two ene, the dashes are single bonds. One doubly bonded carbon carries hydrogen and methyl, the other carries hydrogen and C two H five, the ethyl group. Different groups on each, so yes. Right? Two methyl but two ene has two methyl groups, the brackets with subscript two, both on C two, that is, carbon two. Two identical groups on one carbon, so no. Clear? Now, three methyl pent two ene. Carbon two has a methyl, and carbon three has a methyl in brackets. Both doubly bonded carbons carry a methyl group, so no isomers, right? Oh no, careful, remember our trap! Compare within each carbon: carbon two has hydrogen and methyl, and carbon three has methyl and ethyl. Different on each, so yes, it shows geometrical isomerism. Okay? And hex three ene has hydrogen and ethyl on each carbon, so yes. So the answer is pent two ene, three methyl pent two ene and hex three ene. See how quick that was?

7:00Example 2: why cis 1-bromo-1,2-dichloroethene is E, not Z

Now a harder one, and here is the surprise I promised. In one bromo one two dichloro ethene, carbon one carries bromine and chlorine, and carbon two carries chlorine and hydrogen. The two chlorines are on the same side, so it is called cis. Is it the Z isomer? Pause the video and try it first. Done? Here is the tempting answer. Cis means the same side, and Z means together. So it must be the Z isomer, right? No, no, wait, never guess, apply the rules! On carbon one, bromine has the higher atomic number, so bromine beats chlorine. On carbon two, chlorine beats hydrogen. Okay? Now let us picture it. The two chlorines are on the same side. So bromine, the other group on carbon one, sits on the opposite side to the chlorine on carbon two. With me? So the two higher priority groups, bromine and the chlorine on carbon two, are on opposite sides. This cis compound is actually the E isomer, not the Z isomer! That is our surprise, and a common trap: cis does not always mean Z, okay? Cis and trans compare identical groups, while E and Z compare priorities. Got it?

8:20Quick revision: the condition, cis-trans, E-Z by priority

So, quick revision. Geometrical isomers need two different groups on each doubly bonded carbon. Cis and trans: the identical groups on the same side, or on opposite sides. And in E Z, higher priority groups together give Z, and on opposite sides give E, okay? Now go and try the practice problems on this page. All the best!