Stereochemistry of Substitution

Transcript

0:00Introduction

Hello everyone! Okay, one strange experiment before we start. Take optically active two iodobutane, which turns plane polarised light, and keep it with sodium iodide in acetone. No new compound is formed. And yet, slowly, its optical activity disappears! How can a compound lose its optical activity without changing into anything new? Strange, right? To answer it, we need to know what a substitution does to a chiral carbon: inversion, retention or racemisation. Questions on these come up often in JEE, so let's begin.

0:39Inversion: the umbrella flips

First, inversion. It means the product has the opposite spatial arrangement at the carbon. The picture on the right is from the course: bromoethane reacting with hydroxide ion. The solid black wedge is a bond coming towards you, and the dashed line is a bond going away from you. Okay? In S N two, written as a capital S, a small capital N written low, and a two, the nucleophile attacks from the side opposite to the leaving group. Here H O minus, the hydroxide ion, comes in along the orange arrow, away from the bromine. In the middle, in square brackets with a double dagger, is the transition state; delta minus, the Greek letter delta with a minus, is a partial negative charge. With me? On the left of the picture, the two hydrogens and the methyl group, C H three, with the small three written low, lean towards the incoming hydroxide. Now look at the product on the right, next to bromide ion, B r minus. Where do those three groups point now? They point the other way, yes, exactly! The three bonds turn inside out, like an umbrella in a strong wind. This carbon has two hydrogens, so it is not chiral; but with four different groups, the same flip gives the opposite arrangement. So S N two at a chiral carbon gives inversion. Got it?

2:00Racemisation: the flat carbocation

Now S N one, with a one in place of the two. In the new picture, the halide carries three different groups, R one, R two and R three; the small numbers written high just label them. In the slow first step, bromide ion leaves, shown as minus B r minus under the arrow, and we get C plus, a carbocation. It is planar and s p two, with the small two written high: all three groups lie flat in one plane. Clear? Now the nucleophile, N u minus, arrives in the fast second step. The carbocation is flat, open from the front and from the back. So tell me, which face will the nucleophile attack? It can attack either face, yes, exactly! See the two orange arrows. Attack from one face gives the product labelled inversion; from the other face, the one labelled retention. These two products are mirror images, a pair of enantiomers. Right? Both faces are equally open, so we get a fifty to fifty mixture of the two enantiomers; the colon on screen means a ratio. Such a mixture is called racemic. Okay? And why is it inactive? The two enantiomers turn the light by equal amounts in opposite directions, so their rotations cancel. This is racemisation, the result of S N one at a chiral carbon. Got it?

3:22Retention: nothing flips

Now the third possibility, retention. Retention means the spatial arrangement at the carbon is unchanged. The course gives two ways to get it. With me? The first way: no bond to the chiral carbon is broken. If the reaction happens at some other carbon, the chiral carbon just goes along for the ride, so its arrangement cannot change. Keep this in mind for Example one. Okay? Now let me ask you this. Suppose a chiral carbon goes through one S N two inversion, and then through a second one. What arrangement do you get at the end? The original arrangement, yes, exactly! Flip the umbrella once and it turns inside out; flip it again and it is back to normal. So two successive inversions give retention.

4:09The sign trap: plus or minus?

Clear? Now one big trap. Before a compound's name you often see a plus or a minus sign in round brackets. Plus means it turns plane polarised light to the right, clockwise, and minus means it turns it to the left. Okay? So suppose a reaction inverts the configuration. Must the sign also change, from plus to minus? No, no, wait, it need not change at all! The sign of rotation is measured in the lab, and it cannot be predicted from the mechanism. Right? The mechanism predicts only the configuration, the spatial arrangement. So never decide inversion or retention by looking at the sign. That is the trap in our first example. Got it?

4:52Example 1: 2-methylbutan-1-ol with HCl, inverted or not

Now, Example one. Two methylbutan one ol, the minus form, is heated with concentrated H C l, hydrochloric acid. It gives one chloro two methylbutane, the plus form. Has the configuration been inverted? Pause the video and try it first. Done? Okay, let's check. The sign changed from minus to plus. So the configuration is inverted, right? Oh no, careful, that is the sign trap we just saw! First find the chiral carbon in the structure on screen; the C H three in round brackets is a branch on the carbon before it. Number the chain from the C H two O H end; C dash two on screen means carbon number two. It holds a hydrogen, a methyl, an ethyl, C two H five, and the C H two O H group. Four different groups, so carbon two is chiral. With me? Now, where does the reaction happen? At carbon one: the O H is replaced by chlorine, so C H two O H becomes C H two C l, as the arrow shows. No bond to carbon two is broken at all. Okay? So the arrangement at carbon two is retained. The answer is no, the configuration has not been inverted; only the sign of rotation changed, from minus to plus. See how the sign told us nothing about the mechanism? Clear?

6:13Example 2: 2-iodobutane racemises in NaI and acetone

Now, remember our puzzle from the start? Let's finish it. Optically active two iodobutane slowly loses its optical activity when kept with sodium iodide in acetone, although no new compound is formed. Explain this. Pause the video and try it first. Done? Okay, let's check. Acetone is a polar aprotic solvent, and that favours S N two. So what can the iodide ion do to two iodobutane? It can attack the carbon holding the iodine, exactly! On screen, I minus is the iodide ion. One iodide comes in from the back, and the other iodide leaves. And the product? Two iodobutane again! That is why no new compound is formed. Right? But this is S N two, so each exchange inverts one molecule. R and S in round brackets are the labels of the two configurations, and the arrow means becomes. Since the incoming and the leaving groups are both iodine, an R molecule becomes an S molecule, and an S molecule becomes an R molecule. Okay? Now suppose we start with only the R form. Will it all turn into the S form in the end? No, it will not, because the flipping works both ways! As soon as some S form appears, iodide flips it back to R as well. So in time the R and S forms become equal: a racemic mixture, optically inactive. Puzzle solved: the rotation dies away, though no new compound is formed. Clear?

7:52Quick revision: inversion, racemisation, retention, sign

So, quick revision. S N two at a chiral carbon gives inversion, and S N one gives racemisation, a fifty to fifty mixture. Retention comes when no bond to the chiral carbon is broken, or after two inversions. And the mechanism predicts the configuration, never the sign, plus or minus. Now go and try the practice problems on this page. You can do them. All the best!