THE REACTION BETWEEN UNSYMMETRICAL ALKENES AND THE HYDROGEN HALIDES
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between the hydrogen halides and alkenes like propene. Hydrogen halides include hydrogen chloride and hydrogen bromide. If you want the mechanisms explained to you in detail, there is a link at the bottom of the page.
An unsymmetrical alkene is one like propene in which the groups or atoms attached to either end of the carbon-carbon double bond are different.
For example, in propene there are a hydrogen and a methyl group at one end, but two hydrogen atoms at the other end of the double bond. But-1-ene is another unsymmetrical alkene.
Electrophilic addition reactions involving hydrogen bromide
As with all alkenes, unsymmetrical alkenes like propene react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other.
In the case of propene, 2-bromopropane is formed.
This would normally be written in a more condensed form as
The product is 2-bromopropane.
Note: There is another possible reaction between unsymmetrical alkenes and hydrogen bromide (but not the other hydrogen halides) unless the hydrogen bromide and alkene are absolutely pure.
A different mechanism happens (a free radical chain reaction - not on UK A' level syllabuses) which leads to the hydrogen and bromine adding the opposite way round. For A' level purposes, you don't need to worry about that. However, if you are interested, you will find the free radical addition mechanism by following this link.
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This is in line with Markovnikov's Rule which says:
In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group.
Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant.
Warning! Markovnikov's Rule is a useful guide for you to work out which way round to add something across a double bond, but it isn't the reason why things add that way. As a general principle, don't quote Markovnikov's Rule in an exam unless you are specifically asked for it.
This is an example of electrophilic addition.
The addition is this way around because the intermediate carbocation (previously called a carbonium ion) formed is secondary. This is more stable (and so is easier to form) than the primary carbocation which would be produced if the hydrogen became attached to the centre carbon atom and the bromine to the end one.
Electrophilic addition reactions involving the other hydrogen halides
Hydrogen fluoride, hydrogen chloride and hydrogen iodide all add on in exactly the same way as hydrogen bromide.
The only differences lie in the rates of reaction:
This is because the hydrogen-halogen bond gets weaker as the halogen atom gets bigger. If the bond is weaker, it breaks more easily and so the reaction is faster.
If the halogen is given the symbol X, the equation for the reaction with propene is:
Notice that the product is still in line with Markovnikov's Rule.
These are still examples of electrophilic addition.
Again using X to stand for any halogen:
Again, the intermediate carbocation formed is secondary. This is more stable than the primary carbocation ion which would be formed if the hydrogen attached to the centre carbon atom and the X to the end one. If it is more stable it will be easier to make.