Introduction
To be honest, ‘destruction’ sounds cool enough to be used in the title of this article, but really, it’s not a destruction or disintegration of the molecule; it’s more like a separation, or breaking a molecule into smaller, individual parts. I would point out that it’s something like a chemical digestion of molecules, or a hydrolysis reaction (which is, indeed, an example of a cleavage reaction). Before we start taking a look at all the examples of ‘molecular destruction’, let us talk about what we mean by ‘separation’, and the classes of cleavage reactions that exist. By separation, we mean that two parts of a molecule, which are joined by a bond (usually a single bond), are separated from each other by the breakage of that bond (bond cleavage). A bond is formed by two electrons, and as such, the question arises of where the electrons will go after the bond is broken, and this is where the classes of cleavage reactions come into play. There are only two classes of cleavage reactions, which are heterolytic cleavage and homolytic cleavage. This is shown in Fig. 1.
Fig. 1: Homolytic and heterolytic cleavage.
Heterolytic cleavage (right in Fig. 1) occurs when we use something to break the bond between two atoms or molecules such that both electrons go to one atom (giving it a negative charge) while the other atom effectively loses two electrons (that it had once possessed while a covalent bond was still formed between the two atoms). Heterolytic cleavage can occur widely in many chemical reactions. When we want to destroy the bond holding the two molecules together, we would of course use a weapon, like a sword. It is, of course, very likely that we will not cut the bond cleanly into two, but perhaps one atom gets ‘more of the bond’ than the other, leading to one atom grabbing both electrons and leaving the other with none. Electrons really love being in pairs, and thus in the case of heterolytic cleavage, both molecules, or atoms, will be stable, and do not desire to react further at all. However, the atom that gets both the electrons will likely be ‘happier’, and thus more stable, than the atom that does not obtain the electrons. It is also possible for a win-win scenario, where one atom likes hogging the electrons while the other atom does not. This allows both atoms to be stable and results in a ‘win-win’ situation.
Now, let’s imagine that we want to break the bond, but this time, we want to break it right down the middle, so that each of the atoms or molecules gets one electron. It is simply not possible or viable for us to use a sword to do this, because it is impossible for us to precisely split the bond equally, such that both atoms get one electron. To break this bond equally, we must turn to other methods. One possible solution would be to increase the tension in the bond so that it snaps of its own accord, and to do this, we need to increase the energies of the two atoms that form the bond. How do we do this? There are two ways. The first way is the photochemical method, where the molecule is exposed to UV light. The second way is simply to heat the molecule until the bond breaks equally. When the bond splits equally, it is known as homolytic cleavage (left in Fig. 1). Remember that electrons like being in pairs the most. When the bond is split equally such that each atom gets one electron only, it is much worse for them; both of them will be highly reactive and will stop at nothing to gain the other electron back and form a molecule. Such atoms are known as free radicals. Of course, the only way for an atom with an unpaired electron to become stable is by reacting with another atom with an unpaired electron, i.e. another free radical, to form a stable molecule.
In this article, we will only be discussing free radical substitution; and in later articles, we will start explaining more of heterolytic cleavage and prominent examples of both homolytic and heterolytic cleavage reactions.
Free Radical Substitution
Let’s say we wanted to ‘cut’ a molecule in half by breaking the bond between two portions of the molecule. The only way for us to destroy the molecule would be to use a method such that the products of the cleavage are stable enough, to ensure that they will not recombine. As such, homolytic cleavage is typically not a good idea, because the molecule will likely end up breaking into radicals, which subsequently react and form larger molecules, including the starting material. Because of this, we will be discussing heterolytic and homolytic cleavage differently. Heterolytic cleavage usually is a reaction in itself, because the products generated are generally stable enough so as to prevent further reaction. However, homolytic cleavage is rarely a reaction by itself because the cleaved products are both radicals and are highly reactive, thus they will continue to react to form the final, stable products. Here, we will explain what will happen to the radicals after homolytic cleavage occurs, and the mechanism by which the radicals react to form stable products.
So, what is the mechanism whereby the free radicals react with each other so that the stable products can be formed? Radicals, in fact, are so reactive that there may be dozens of side products formed in the course of the reaction, as we will see later. Free radicals may react with basically any molecule it encounters, so for any molecule in a reaction mixture, there is a likely chance that the radical will encounter it and react with it. We will worry about such products later. Before that, let us first take a look at the mechanism of free radical substitution. The first step is known as the initiation step; in this step, homolytic cleavage occurs to molecules to break them apart and form radicals. Of course, it is not easy to break strong bonds by heat or photochemical methods, and as such, it is usually only weak bonds that are targeted and broken to form radicals. Examples of such weak bonds include halogen-halogen bonds, as well as peroxides, which are most commonly used for the generation of radicals. To facilitate this example, we will use the example of a reaction mixture with methane (CH4) and chlorine (Cl2), which is a very simple example. In this initiation step, then, the Cl2 breaks apart to form two chlorine free radicals.
The step after the initiation step is known as the propagation step (Fig. 2). As suggested by the name of this step, it is simply to propagate the radicals by reacting the radical with a stable molecule (that has paired electrons), after the reaction, there will still continue to be an odd number of electrons, and thus, one of the electrons will still not be paired, and there will be a radical. The reaction between the radical and the stable molecule will lead to the radical abstracting an atom from the stable molecule, which homolytically cleaves the stable molecule. In the case of the example we raised earlier, between methane (CH4) and the Cl radical, the chlorine radical abstracts a hydrogen from the methane, forming the stable molecule HCl along with the CH3 radical. Why is Cl able to combine with a hydrogen to form a stable molecule? This is because the hydrogen atom is also a radical, since it also possesses only one, unpaired electron. However, it is not as reactive as a free radical. In summary, the propagation step involves the ‘propagation’ of free radicals, because in this step more radicals are formed via reaction of the original, homolytically cleaved radicals.
Fig. 2: Propagation step of free radical substitution.
Finally, there is one last step in the free radical substitution mechanism, the termination step (Fig. 3). Again, the name is very obvious; in this step, the radicals are terminated. Recall that we stated earlier that there is only one way for a free radical to be terminated: it must combine with another radical. Let us again take a look at the example we raised previously, of the methane radical. In the reaction mixture, there is likely to be a mixture of free radicals, the methane radical and the chlorine radical. As such, it is possible for many different termination reactions to occur and for a mixture of products to form. For the methane radical, it may combine with another methane radical to form ethane or with another chlorine radical to form chloromethane, CH3Cl. The latter example is showcased below in Fig. 3. Chloromethane seems stable enough, so now we can ask: can it participate in a propagation reaction? The answer is yes. Chloromethane can continue to participate in propagation and more of the hydrogens will be substituted with chlorine atoms.
Fig. 3: Termination step of free radical substitution.
So that’s it for the free radical substitution mechanism, and also for this article (sadly). In the next article, we will take a look at some cleavage reactions of both homolytic and heterolytic cleavages.