Reaction Mechanisms

This is yet another revision blogpost. I might as well prepare for exams this way.

There are alkanes and alkenes - both hydrocarbons, differing mainly by the bonds that are present. Alkanes contain only single bonds, which consist of only sigma bonds that form from the overlapping of the carbon and carbon/hydrogen orbitals. Alkenes, on the other hand, contain one C=C double bond, which also has a single pi bond. This comes about due to the overlapping of orbitals around this double bond. 

Should you be able to break up these bonds, you can go on to form more complex hydrocarbons - for example, get a halogen atom (like chlorine) in there and you can form a haloalkane or haloalkene. Get a hydroxyl group involved and you've got an alcohol, and any further reactions could result in an ester. 

But to do that, you need to break the bonds first, and once you've managed that, you can represent this through a series of reaction mechanisms. These are equations which break down each step of the equation to show how each bond is broken, and how the resultant products - whether free radicals, ions or atoms - then later react with each other to form a desired product.

The first type of reaction mechanism that I will focus on is radical substitution - or in other words, the breaking of covalent bonds to form free radicals which later randomly collide to form new products. This in itself is a subset of substitution reactions, which I might blog about in the future. Anyways, say you want to react ethane and chlorine. The main reaction will look like this:

C₂H₆ + Cl₂ → C₂H₅Cl + HCl

Except that's not entirely true. Perhaps it will look like this:

C₂H₆ + Cl₂ → C₂H₄Cl₂ + H₂

It all comes down to what radicals react and in which way. And it all begins with UV light, which provides the energy needed to break the covalent bonds present.

Initiation is the first stage, in which the bonds present in the chlorine molecule will be broken through homolytic fission. In general, this will occur with the substituent of the reaction (the molecule providing the electrons to be used in the reaction). In homolytic fission, the electrons that would have been shared between the chlorine atoms will be distributed equally between the chlorine atoms. As a result, two radicals will form - they are now more likely to react, as they need one electron to form a full outer shell:

 

The crudely drawn curly arrows in this case represent the equal distribution of electrons - the half arrows represent this homolytic fission.

Then there's the propagation stage, which can be represented simply as two reactions. In this case, ethane may react with a chlorine radical to form an ethyl radical and hydrochloric acid, as in the first reaction. Or, the ethyl radical could then react with chlorine to form chloroethane and another chlorine radical, as such:

From here, however, the process could continue - should more chlorine radicals react with chloroethane, you could end up with a hydrocarbon with fewer and fewer hydrogen atoms - to its extreme, you could form hexachloroethane like this (C₂Cl₆). Note to oneself - when writing out these equations, please make sure there's only one radical on either side.

Eventually, though, the reaction will stop once two radicals collide with each other. Obviously, though, there are many radicals that will be present, so the overall reaction could still carry on. This is the termination stage, and with the example of chlorine and ethane, could look like this:

As you can see, there will also be many different products forming, due to the radicals randomly colliding with each other. Therefore, the product you want to form the most (that being chloroethane in this case) will be produced in a much lower yield than intended. That's less leaded petrol for all of us, then.

And this is obviously a very simplified model, one that can be easily written out during exams.

This process mainly occurs with alkanes. The main reaction mechanism that I learned involving alkenes was the addition reaction. In this case, the example I'll use will be the reaction of ethene and chlorine.

As mentioned earlier, alkenes contain a double bond which can easily open up. This is as it is a region of high electron density, so is more likely to attract electrophiles (substances which form bonds by taking up a pair of electrons), so alkenes are more keen to react (as a friend told me).

As always, the first stage will include the breaking up of the chlorine molecule. This time, however, the pair of electrons that the two atoms shared will be transferred completely to one of the chlorine atoms. This is what is known as heterolytic fission, and there is an uneven distribution present. This also induces a dipole, as seen here:

 

This time, the arrowhead is whole to show the full distribution of electrons. In diagrams, this would be shown connected by an arrow to the ethene molecule, but I've decided to include just this to simplify the process.

Once the double bond has opened up, however, the chlorine which lost out on the electrons will bond with the ethene, as such: 

 

In this process, it's important to show that a carbocation (a positively charged ion forming after the double bond broke up) forms - from here, the chlorine with all the electrons can then come in and react, as per crude arrow, forming a C₂H₄Cl₂ molecule. Those are the basics of it, but I'd rather not delve into them right now. Hopefully I don't forget these notes come the exam, however.