To finish this chapter, we consider a reaction that at first sight seems to go against what we have told you so far. It’s an elimination catalysed by a strong base (KOH), so it looks like E2. But the leaving group is hydroxide, which we categorically (and truthfully) stated cannot be a leaving group in E2 eliminations.

The key to what is going on is the carbonyl group. In Chapter 8 you met the idea that negative charges are stabilized by conjugation with carbonyl groups, and the list on p. 176 demon strated how acidic a proton adjacent to a carbonyl group is. The proton that is removed in this elimination reaction is adjacent to the carbonyl group, and is therefore also rather acidic (pKa about 20). This means that the base can remove it without the leaving group departing at the same time—the anion that results is stable enough to exist because it can be delocalized on to the carbonyl group.

Although the anion is stabilized by the carbonyl group, it still prefers to lose a leaving group and become an alkene. This is the next step.

This step is also the rate-determining step of the elimination—the elimination is unimolecular and so is some kind of E1 reaction. The leaving group is not lost from the starting molecule, but from the conjugate base of the starting molecule, so this sort of elimination, which starts with a deprotonation, is called E1cB (cB for conjugate base). Here is the full mechanism, generalized for other carbonyl compounds.

It’s important to note that, while HO− is never a leaving group in E2 reactions, it can be a leaving group in E1cB reactions. The anion it is lost from is already an alkoxide—the oxy anion does not need to be created as the HO− is lost. The establishment of conjugation in the product also assists loss of HO−. As the scheme above implies, other leaving groups are possible too. Here are two examples with methanesulfonate leaving groups.

The fi rst looks E1 (stabilized cation), the second E2—but in fact both are E1cB reactions. The most reliable way to spot a likely E1cB elimination is to see whether the alkene in the product is conjugated with a carbonyl group. If it is, the mechanism is probably E1cB.

β-Halocarbonyl compounds can be rather unstable: the combination of a good leaving group and an acidic proton means that E1cB elimination is extremely easy. This mixture of diastereoisomers is first of all lactonized in acid , and then undergoes E1cB elimination with triethylamine to give a product known as a butenolide. Butenolides are common structures in naturally occurring compounds.

You will have noticed that we have shown the deprotonation step in the last few mecha nisms as an equilibrium. Both equilibria lie rather over to the left-hand side because neither triethylamine (pKa of Et3NH+ about 10) nor hydroxide (pKa of H₂O 15.7) is basic enough to remove completely a proton next to a carbonyl group (pKa > 20). However, because the loss of the leaving group is essentially irreversible, only a small amount of deprotonated car bonyl compound is necessary to keep the reaction going. The important point about substrates that undergo E1cB is that there is some form of anion-stabilizing group next to the proton to be removed—it doesn’t have to stabilize the anion very well but, as long as it makes the proton more acidic, an E1cB mechanism has a chance. Here is an important example with two phenyl rings helping to stabilize the anion, and a carbamate anion (R₂N—CO₂ −) as the leaving group.

The proton to be removed has a pKa of about 25 because its conjugate base is an aromatic cyclopentadienyl anion (we discussed this in Chapter 8). The E1cB elimination takes place with a secondary or tertiary amine as the base. Spontaneous loss of CO2 from the eliminated product gives an amine, and you will meet this class of compounds again in Chapter 23, where we discuss the Fmoc protecting group.

The E1cB rate equation

 The rate-determining elimination step in an E1cB reaction is unimolecular, so you might imagine it would have a first order rate equation. In fact, the rate is also dependent on the concentration of base. This is because the unimolecular elimination involves a species—the anion—whose concentration is itself determined by the concentration of base by the equilibrium we have just been discussing. Using the following general E1cB reaction, the concentration of the anion can be expressed as shown.

مواضيع ذات صلة

Distinguishing acid derivatives by carbon NMR is difficult

E1cB eliminations in context

The regioselectivity of E2 eliminations

E2 elimination from vinyl halides: how to make alkynes

E2 eliminations from cyclohexanes

E2 eliminations can be stereospecific

E2 eliminations have anti-periplanar transition states

E1 reactions can be stereoselective

The role of the leaving group

Substrate structure may allow E1

E1 and E2 mechanisms

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