Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle from Set two. An optically pure halide is hydrolysed by the S N one path, and the alcohol comes out with a ten percent excess of the inverted form. So how many molecules reacted with inversion? Only ten percent? No, and by the end of this video you will see why. This is a typical JEE Main pattern question, so let's begin.

0:26SN2: one step, attack from the back

So, first the S N two path: a capital S, a small N written low as a subscript, and a two, for substitution, nucleophilic, bimolecular. In the picture, H O minus, the hydroxide ion, is the nucleophile; the small minus written high is its charge. The orange arrow shows it attacking the carbon from the side opposite to the bromine, the leaving group. The black wedge is a bond towards you, the dashed line a bond going away, and C H three, with a subscript three, is a methyl group. Okay? Now, the bromine leaves at the same moment the new bond forms. That is the middle drawing in square brackets, with dashed bonds half made and half broken. Delta minus, with the Greek letter delta, means a partial negative charge, and the double dagger marks a transition state. So, one step, no intermediate, and back-side attack. With me? Both the halide and the nucleophile are in that single step. So tell me, the rate depends on whose concentration? Both of them, yes, exactly! Rate equals k, times the concentration of R X, the alkyl halide, times the concentration of N u minus, the nucleophile. The square brackets mean concentration, and k is the rate constant, so it is second order overall. Right? In that transition state, five groups crowd round one carbon, so bulky groups slow the reaction. That gives this order: C H three X, with X any halogen, then one degree, two degree and three degree halides. The small circle means degree, and the greater than sign means faster than. Got it? Now, a point many students miss. The beta carbon, beta the Greek letter, is the carbon next to the one carrying the halogen. Branching there also slows S N two, and neopentyl halides, with three methyl groups on the beta carbon, are very slow. Okay? And on the right of the picture, we get ethanol and B r minus, the bromide ion. The three other bonds have turned inside out, like an umbrella in the wind, so the configuration is inverted. This is called Walden inversion. Clear? And when does S N two win? It is favoured by a strong nucleophile in high concentration, and by a polar aprotic solvent, such as acetone, DMSO or DMF, got it?

2:45SN1: two steps, through a flat carbocation

Now the S N one path, where the one means unimolecular. On the left of the picture, the carbon carries a bromine and three groups, R one, R two and R three; the small numbers written high just label different groups. Okay? Step one is slow: R X gives R plus and X minus. The dash is the bond, the arrow means gives, and the small plus and minus are charges. In the picture, minus B r minus under the arrow means a bromide ion leaves, and we get C plus, a planar s p two carbocation, with the two written high. Right? Then step two is fast: the nucleophile, N u minus, attacks the carbocation, along the orange arrows. Only R X takes part in the slow first step. So which concentration decides the rate here? Only R X, that is correct! Rate equals k, times the concentration of R X, so it is first order, and the nucleophile does not appear at all. Got it? And reactivity follows carbocation stability, so the order flips: three degree, then two degree, then one degree, then methyl. It is exactly the opposite of S N two. And S N one is favoured by a polar protic solvent, such as water or an alcohol, which solvates both ions. With me? Now the stereochemistry. The carbocation is completely flat. So from which side can the nucleophile attack it? From both faces, yes, exactly! One face gives the inverted product, and the other gives the retained one, with the arrangement unchanged; see the two labels on the right. So an optically active halide gives a largely racemised product. Clear? So S N one always gives exactly fifty to fifty, inverted and retained, doesn't it? Not quite, because in practice there is slightly more of the inverted product. The departing halide ion stays close to the face it just left, as an ion pair, and partly shields that side. Remember this for our puzzle.

4:48Counting the products

Okay? Now, two words for counting. Racemic means a fifty to fifty mixture of the two enantiomers, the mirror image forms; the colon is read as to. Their rotations are equal and opposite and cancel, so it is optically inactive. Right? And the enantiomeric excess is the percent of the major form, minus the percent of the minor form; the percent sign is written before each word.

5:11Example 1: 1-bromopropane vs 1-bromo-2-methylpropane by SN2

Clear? Now, Example one, from Set two. Which reacts faster by the S N two mechanism, one bromopropane or one bromo two methylpropane? Pause the video and try it first. Done? Okay, let's check. One bromopropane is C H three, C H two, C H two, B r, and the dashes are single bonds. The bromine is on an end carbon with one carbon attached, so it is primary. Right? One bromo two methylpropane has C H three in brackets with a subscript two, meaning two methyl groups, on a C H, then C H two, B r. The bromine is again on an end C H two, so it is also primary. Okay? So both are primary halides. Then they should react at the same speed by S N two, shouldn't they? No, no, wait, and look one carbon further along the chain! In the second halide, the beta carbon carries two methyl groups. That is branching. Clear? And branching at the beta carbon hinders the approach of the nucleophile: those two methyl groups get in the way of the attack from the back. Got it? So, one bromopropane reacts faster by S N two. Remember to check the carbon with the halogen, and then check its neighbour too!

6:28Example 2: 10% excess inverted, find inversion and retention

And now, Example two, our puzzle from the start. An optically pure halide is hydrolysed by the S N one mechanism, and the alcohol formed has an enantiomeric excess of ten percent of the inverted form. What percentages of the molecules reacted with inversion, and with retention? Pause the video and try it first. Done? Okay, let's check. The excess is ten percent inverted, so ten percent inversion and ninety percent retention, right? Oh no, careful, that is the trap! Ten percent is only the excess of the inverted form. The rest, one hundred minus ten, which is ninety percent, is racemic. With me? And racemic means fifty to fifty. So half of ninety, forty five percent, is inverted, and the other forty five percent is retained. Okay? Now add the excess to the inverted side: forty five plus ten gives fifty five percent inversion. And retention is just the forty five percent from the racemic part. Right? Let us check with the definition: fifty five minus forty five gives ten percent, the given excess. So, fifty five percent inversion and forty five percent retention: slightly more inversion, as we saw for S N one, and our puzzle is solved!

7:48Quick revision: inversion, racemisation, excess counting

So, quick revision. First, S N two is one step with back-side attack, so the configuration is inverted, and branching, even at the beta carbon, slows it. Second, S N one goes through a flat carbocation, attacked from both faces, so the product is largely racemised. Third, the racemic part is one hundred minus the excess, split equally, and the excess joins the major form. Now try all fifteen problems of Set two in twenty five minutes, then check the solutions. All the best!