The Bimolecular Mechanism: S_N2

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two alkyl bromides. In the first, the carbon holding the bromine carries three hydrogen atoms. In the second, that carbon carries three methyl groups. Same bromine, and the same hydroxide ion attacking. Yet on today's path, the first reacts easily, and the second practically does not react at all! So why should three methyl groups matter so much? The answer is the direction of attack. This is the S N two mechanism, and questions on it come up often in JEE, so let's begin.

0:37One step, attack from the back

So, first the name. On screen it is a capital S, then a capital N written small and low, and then a two. S stands for substitution, N for nucleophilic, and the two for bimolecular, because two species, the halide and the nucleophile, meet in its one step. On the right, bromoethane reacts with hydroxide ion. With me? Look at the left of the picture. H O minus, the hydroxide ion, is the nucleophile, written N u minus on screen; the small minus written high is a negative charge. The orange arrow shows it going straight at the carbon, from the side opposite to the bromine, the leaving group, which the screen calls X, for halogen. The solid black wedge is a bond coming towards you, and the dashed line is a bond going away from you. Okay? Now let me ask you this. Does the bromine leave first, before the hydroxide arrives? Or does it leave later? Neither, they happen together! As the new bond to the nucleophile forms, the bond to the halogen breaks. That is the middle of the picture, inside the big square brackets, where the dashed lines to H O and to B r are bonds half made and half broken. Delta minus, the Greek letter delta with a minus sign, is a partial negative charge, and the double dagger at the top right marks a transition state. Clear? So the whole change is one step. There is one transition state, and no intermediate. On the right, after the arrow, we get ethanol, and bromide ion, B r minus. Got it?

2:09Rate law: second order

Alright, now the speed. Both the halide and the nucleophile are in that one step. So tell me, the rate depends on whose concentration? Both of them, yes, exactly! Rate equals k, times the concentration of R X, times the concentration of N u minus. Here R X is the alkyl halide, k is the rate constant, and the square brackets mean concentration, in moles per litre. Okay? So it is first order in each, and second order overall. Example one uses this line. Right?

2:42Crowding decides the speed

Now, the transition state again. The carbon is bonded partly to the nucleophile and partly to the halogen. The other three groups, two hydrogens and C H three, a methyl group, lie in one plane; the small three written low means three hydrogens. So five groups crowd round one carbon, and bulky groups slow the reaction. With me? So, back to our puzzle. Methyl halide, with only hydrogens on the carbon, or tertiary halide, with three alkyl groups on it. Which reacts faster by S N two? The methyl halide, correct! The order is C H three X, then one degree, two degree and three degree halides. The small circle means degree, so these are primary, secondary and tertiary, and the greater than sign means faster than. Each extra alkyl group on the carbon hinders the back side attack. Right? And tertiary halides practically do not react this way. Their three alkyl groups block the back of the carbon. There, our puzzle is solved! Clear? But careful, crowding one carbon further away matters too. The beta carbon, written with the Greek letter beta, is next to the carbon carrying the halogen. Branching there also slows the reaction, so neopentyl halides are very slow. Keep this for Example two. Okay?

4:01The carbon turns inside out

Now compare the two ends of the picture. On the left, the two hydrogens and the methyl group lean to the left, towards the incoming hydroxide. Look at the product on the right. Where do they point now? To the right, yes! The three other bonds have turned inside out, like an umbrella in a strong wind. With me? So the configuration of the carbon is inverted. This is called Walden inversion. Remember it as a pair, S N two means inversion. Got it?

4:31What speeds up SN2: strong nucleophile, aprotic solvent

Now, what helps S N two along? First, a strong nucleophile, in high concentration. Of course, its concentration sits right inside the rate law. Okay? Second, polar aprotic solvents, like acetone, DMSO and DMF, also help. But why should the solvent matter at all? See, a protic solvent like water or methanol hydrogen bonds to the negative ion and holds it back. But polar aprotic solvents like DMSO surround only the positive ion, and leave the anion free to attack. Right? Third, some halides react especially fast. Allylic halides have the halogen on the carbon next to a carbon carbon double bond, and benzylic halides on the carbon next to a benzene ring. In alpha halo ketones, the halogen sits on the carbon next to the carbonyl group; alpha is this Greek letter. Clear? And the reason is that the neighbouring pi system, written with the Greek letter pi, stabilises the transition state. So these react faster. Got it?

5:32Example 1: rate change when both concentrations are doubled

Now, Example one, an easy one. In an S N two reaction, the concentrations of the halide and of the nucleophile are both doubled. By what factor does the rate change? Pause the video and try it first. Done? Okay, let's check. Start from the rate law, rate equals k, times R X, times N u minus. Both are doubled, so the rate doubles too, right? Oh no, careful, that is the trap! Each concentration is its own factor. Doubling the halide doubles the rate, and doubling the nucleophile doubles it again. Two times two gives four, so the rate becomes four times as fast. Clear?

6:18Example 2: SN2 order of four primary bromides

Now, Example two, a harder one. All four of these are primary bromides. They are one bromobutane, one bromo two two dimethylpropane, one bromo two methylbutane, and one bromo three methylbutane. Arrange them in decreasing order of S N two reactivity. Pause the video and try it first. Done? Okay, let's check. All four are primary, so all four react at the same speed, right? No, no, wait, remember the beta carbon. Number the chain from the carbon holding bromine; carbon two is the beta carbon. One bromobutane is a straight chain with no branch, so it is the fastest. In one bromo three methylbutane, the branch is on carbon three, farther away, so it hinders less. Okay? In one bromo two methylbutane, a branch sits right on carbon two, the beta carbon. So the back side gets more crowded, and the reaction is slower. Right? And one bromo two two dimethylpropane has two methyl branches on that beta carbon. This is neopentyl bromide, one of the neopentyl halides we met earlier, and it is very slow. With me? So the order is one bromobutane, then the three methyl, then the two methyl, and the two two dimethyl compound last. The nearer the branch is to the reacting carbon, and the more branches there are, the greater the hindrance. Got it?

7:44Quick revision: rate law, crowding, inversion, solvent

So, quick revision. S N two is one step, with back side attack, and its rate equals k, times R X, times N u minus. Crowding slows it, from methyl down to tertiary, and branching at the beta carbon slows it too. The configuration of the carbon is inverted, and a strong nucleophile and a polar aprotic solvent help. Now go and try the practice problems on this page. You can do them. All the best!