addition to unsymmetrical alkenes

Electrophilic addition to alkenes

Addition reactions involving alkenes always proceed via the same mechanism:

An example of a typical addition reaction to an alkene is shown below. Here the electrophile is the partial positively charged hydrogen atom (δ+) in a molecule of hydrogen bromide. This is attacked by the pi electrons to form a carbocation (which is itself an electrophile) and a negatively charged bromide ion. This bromide ion then uses one of its lone pairs of electrons to form a covalent bond to the positively charged carbon atom, that is it is acting as a nucleophile. This is outlined below:

Depending on the alkene molecule you start with the intermediate carbocation formed could be a primary, secondary or tertiary carbocation, as shown below:

In the example above a molecule of hydrogen bromide adds to the alkene ethene. Ethene is a symmetrical molecule and it makes no difference to the final product which carbon atom in the molecule ends up forming the intermediate carbocation. However if we had started with an unsymmetrical alkene, such as 2-methylpropene-2-ene, shown below then it does make a difference to the final product of the reaction which of the carbon atoms in the carbon carbon double bond (C=C) ends up forming the intermediate carbocation. In the example shown below a molecule of hydrogen chloride gas adds to our unsymmetrical alkene. Addition of the electrophilic hydrogen from the hydrogen chloride molecule can in theory result in the formation of two intermediate carbocations. One of the carbocations is a primary carbocation and the other is a tertiary carbcation.

However the 1-chloro-2-methylpropane, one of the possible products of the addition reaction is NOT produced. Why is this? The simplest explanation is that the intermediate carbocations are not all equally stable and so some will require more energy to form than others, that is they will have a higher activation energy and more energy will be required to form them. Primary carbocations are much less stable than secondary carbocations which in turn are less stable than tertiary carbocations. One possible explanation for this is due to the inductive effect of alkyl groups. Alkyl groups appear to "push" or release electrons towards carbocations and so help stabilise the positive charge on the carbon atom. Obviously the more alkyl groups the stronger will be this inductive effect more electron density will be "pushed" towards the carbocation which should help stabilise it.

So in our example above, only one product, the 2-chloro-2-methylpropane is produced. The reason is that this compound is formed by a route that has as an intermediate a relatively stable tertiary carbonium ion, whereas the other product, the 1-chloro-2-methylpropane is formed via the unstable primary carbonium ion.

Marovnikov's rule

By studying many addition reactions the Russian chemist Vladimir Markovnikov suggested that in the addition of hydrogen halides to alkenes the electrophilic hydrogen becomes attached to the carbon atom with the fewer alkyl substituents. We could also express Markovnikov rule by saying that in the addition of HX to an alkene the more substituted carbocations are more likely to form as intermediates, since the more substituted the carbocation the more stable it will be.

In the example above using HCl and 2-methylprop-2-ene the reaction can proceed via either a tertiary or a primary carbocation. The tertiary carbocation being much more stable than the primary carbocation resulted in the formation of only one product. However if we start with propene rather than 2-methylprop-2-ene then the addition of hydrogen chloride gas will produce as an intermediate a primary and a secondary carbocation. This reaction will produce a mixture of two products, 1-chloroprpane and 2-chloropropane, as shown below. The major product will be the 2-chloropropane since it is produced by the more stable secondary carbocation will the minor product will be the 1-chloropropane since it is produced from the less stable primary carbocation.

Key Points

Practice questions

Check your understanding - Questions on addition reactions