In epoxidation reactions, and in electrophilic additions of bromine, each end of the alkene is joined to the same sort of atom (Br or O). But in the addition reactions of other electrophiles, H–Br for example, there is a choice: which carbon gets the H and which gets the Br? You will need to be able to predict, and to explain, reactions of unsymmetrical alkenes with HBr, but we should start by looking at the reaction with a symmetrical alkene—cyclohexene. This is what happens. When H–Br reacts as an electrophile, it is attacked at H, losing Br −. Unlike a bromine atom, a hydrogen atom can’t form a three-membered ring cation—it has no lone pairs to use. So electrophilic addition of a proton (which is what this is) to an alkene gives a product best represented as a carbocation. This carbocation rapidly reacts with the bromide ion just formed. Overall, H–Br adds across the alkene. This is a useful way of making simple alkyl bromides.

Here are two more syntheses of alkyl bromides, but this time we need to ask our question about which end of the alkene is attacked because the alkenes are unsymmetrical (they have different substituents at each end). First, the results.

In each case, the bromine atom ends up on the more substituted carbon, and the mechanism explains why. There are two possible outcomes for protonation of styrene by HBr, but you should immediately be able to spot which is preferred, even if you don’t know the outcome of the reaction. Protonation at one end gives a stabilized benzylic cation, with its positive charge delocalized into the benzene ring.

Protonation at the other end would give a highly unstable primary cation, and therefore does not take place.

You get the same result with isobutene (2-methylpropene): the more stable tertiary cation leads to the product; the alternative primary cation is not formed.

Markovnikov’s rule There is a traditional guideline called Markovnikov’s rule for electrophilic additions of H–X to alkenes, which can be stated as: ‘The hydrogen ends up attached to the carbon of the double bond that had more hydrogens to start with.’ We don’t suggest you learn this rule, although you may hear it referred to. As with all ‘rules’ it is much more important to understand the reason behind it. For example, you can now predict the product of the reaction below. With all due respect, Markovnikov couldn’t.

The protonation of alkenes to give carbocations is quite general. The carbocations may trap a nucleophile, as you have just seen, or they may simply lose a proton to give back an alkene. This is just the same as saying the protonation is reversible, but it needn’t be the same proton that is lost. A more stable alkene may be formed by losing a different proton, which means that acid can catalyse the isomerization of alkenes—both between Z and E geometrical iso mers and between regioisomers.

E1 and isomerization

The isomerization of alkenes in acid is probably a good part of the reason why E1 eliminations in acid generally give E alkenes. In Chapter 17 we explained how kinetic control could lead to E alkenes: interconversion of E and Z alkenes under the conditions of the reaction allows the thermodynamic product to prevail. This was also discussed in Chapter 12.

Other nucleophiles may also intercept the cation, for example alkenes can be treated with HCl to form alkyl chlorides, with HI to form alkyl iodides, and with H₂S to form thiols.

مواضيع ذات صلة

The intermediate in the base-catalysed reaction is an enolate ion

Enolization is catalysed by acids and bases

Evidence for the equilibration of carbonyl compounds with enols

Tautomerism: formation of enols by proton transfer

Hydration of alkynes

Breaking a double bond completely: periodate cleavage and ozonolysis

Adding two hydroxyl groups: dihydroxylation

Electrophilic addition to alkenes can produce stereoisomers

Hydration of alkynes

Breaking a double bond completely: periodate cleavage and ozonolysis

Adding two hydroxyl groups: dihydroxylation

Electrophilic addition to alkenes can produce stereoisomers