Dehydrohalogenation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Heat two bromobutane with alcoholic potassium hydroxide, and you get an alkene. But two different alkenes can form, and not fifty fifty. About eighty percent is just one of them! So why does the reaction prefer that one? Interesting, right? Questions on the major alkene come up often in JEE papers. So, let's find out.

0:27Dehydrohalogenation: losing H and X

So, first the name. Dehydrohalogenation means taking away a hydrogen and a halogen. Beta, the Greek letter on screen, marks the beta hydrogen, a hydrogen on the carbon next to C, dash, X, the carbon halogen bond. In the figure, that halogen carbon is alpha, and its neighbour is beta. Clear? We heat R X, the alkyl halide, with alcoholic KOH, potassium hydroxide in alcohol. Note the brackets: not aqueous. That is a classic trap, because aqueous KOH substitutes and gives an alcohol instead. Got it? Here is the whole change in one line. B minus, with the minus written high, is a base. It pulls the hydrogen off C beta, while X leaves C alpha as X minus; both Greek letters are written small and low, and the short dashes are single bonds. The arrow means gives: B H, then C double bond C, where the equals sign is the new double bond, and X minus. With me? Follow the orange arrows in the figure. As the base takes the hydrogen, the electrons of that C H bond move in between the two carbons, and the halide leaves. So an alkene forms, and since the hydrogen came from the beta carbon, this is called beta elimination. Right?

1:48Saytzeff's rule: the more substituted alkene

Now, back to our puzzle, at the top of the figure: two bromobutane. The small numbers written low count the hydrogens. Both neighbours of the bromine carbon, C one and C three, carry hydrogens, so we get two alkenes: but two ene, about eighty percent, and but one ene, about twenty percent. Okay? So why does but two ene win? Let us count. How many alkyl groups sit on the two doubly bonded carbons in but two ene, and how many in but one ene? Yes, it is two versus one, exactly! In but two ene there is one methyl group on each doubly bonded carbon, so two alkyl groups. In but one ene there is only one, the ethyl group. Right? That is Saytzeff's rule. When more than one alkene can form, the more substituted alkene, with more alkyl groups on the double bond, is the major product. So our puzzle is solved. Clear? One more detail. But two ene has two geometrical isomers, cis and trans. Trans is formed more, because its methyl groups are farther apart, which makes it more stable. Got it?

2:58E2: one step, with H and X anti

Now, the mechanism. The usual path is E two, where E stands for elimination. It is one step: the base removes the beta hydrogen while the halide leaves, as the arrows show. With me? So let me ask you this. In that single step, both the halide and the base take part. Then what should the rate depend on? It depends on both, absolutely! Rate equals k times the concentration of R X, times the concentration of B minus. The square brackets mean concentration, and k is the rate constant. Okay? In the figure, the hydrogen is at the top, and X leaves from the bottom. In E two, H and X must be anti, on opposite sides. And E two needs a strong base.

3:48E1: through a carbocation

Right? Now the second path, E one, in two steps. First the halide leaves, giving the same carbocation as in S N one, with the N written small and low. Then the carbocation loses a beta hydrogen as a proton. So the rate is k times the concentration of R X only. Clear? E one goes along with S N one when a tertiary halide is warmed in a protic solvent, without a strong base. The small circle written high is the degree sign, so three degree means tertiary. Okay? And one warning. Just as in S N one, the carbocation may rearrange before it loses the proton, so E one may give rearranged alkenes. Got it?

4:30Which halides eliminate faster? And a bulky base

Now, the ease of elimination. Tertiary halides eliminate most easily, then secondary, then primary, as the greater than signs show. The more substituted halide gives the more stable alkene, through a lower energy transition state. Right? For the halogen, the order is R I, then R B r, then R C l, then R F, so iodides eliminate most easily, and fluorides least. Clear? Now a trap, because Saytzeff's rule is not the whole story. Potassium tert butoxide is a bulky base. It cannot easily reach a buried hydrogen, so it takes the more exposed one. So which alkene do we get now, the more substituted or the less substituted? The less substituted one, yes, exactly! With a bulky base, the main product is the less substituted alkene, the Hofmann product. So always check the base before applying Saytzeff's rule.

5:26Example 1: alkenes from 2-bromopentane with sodium ethoxide

Okay? Now our first example. Two bromopentane is heated with sodium ethoxide in ethanol. Name the alkenes formed, and the major one. Here is a hint: find the beta carbons first. Pause the video and try it first. Done? Okay, let's check. Here is two bromopentane, with the bromine on carbon two; the dashes are bonds. So the beta carbons are carbon one and carbon three, both carrying hydrogens. Right? Take a hydrogen from carbon one, and the double bond forms at the end of the chain: C H two, double bond, C H, and the rest. That is pent one ene. Okay? Take a hydrogen from carbon three instead, and the double bond sits between carbons two and three. That is pent two ene. So which one is the major product? Pent two ene, yes, exactly! It has two alkyl groups on the double bond, a methyl and an ethyl, while pent one ene has only one. So by Saytzeff's rule it is major, mostly as the more stable trans isomer. Got it?

6:34Example 2: 2-bromo-2,3-dimethylbutane, small vs bulky base

Now a harder one, with a twist. Two bromo two, three dimethylbutane is heated, in part a with sodium ethoxide in ethanol, and in part b with potassium tert butoxide in tert butyl alcohol. Name the major alkene in each case. Pause the video and try it first. Done? Okay, let's check. C H three in brackets with a small two written low means two methyl groups. So carbon two carries the bromine and two methyls, and carbon three, one hydrogen and two methyls. That gives six beta hydrogens on the two methyls of carbon two, and one inner one on carbon three. The methyls on carbon three are not beta, so they do not count. With me? In part a, sodium ethoxide is a small base, so Saytzeff's rule wins. It takes the inner hydrogen on carbon three, putting four methyl groups on the double bond: two, three dimethylbut two ene. Okay? Now part b, with potassium tert butoxide. Same halide, so we get the same alkene again, right? No, wait, this base is bulky, and that changes everything. It cannot easily reach the inner hydrogen, so it takes one of the six beta hydrogens on the methyls of carbon two instead. Got it? So the double bond forms at the end: C H two, double bond, C, and the rest. That is two, three dimethylbut one ene, the less substituted alkene, the Hofmann product. Same halide, different base, different alkene.

8:10Quick revision: Saytzeff vs Hofmann, E2 vs E1 rate laws

Clear? So, quick revision. Alcoholic KOH removes a beta hydrogen and the halogen, giving an alkene. Saytzeff's rule makes the more substituted alkene major, but a bulky base gives the Hofmann product. E two is one step and anti, its rate depending on the halide and the base; E one goes through a carbocation, its rate depending on the halide alone. Now try the practice problems on this page. All the best!