Haloalkanes : Introduction

00:01 Okay, now, I want to bring you on to the lecture that deals with Haloalkanes. Over the course of the Module III, you’ll see that we’re going to be covering a number of different so called “functional groups”. These are groups which are important in terms of not only synthetic pathways, but also in terms of biological application, medical application that results from that. The general formula for haloalkanes is R-X, where X is one of the halogens either fluorine, chlorine, bromine or iodine. But, before we move on to how they react and with what, let’s just have a quick look at their nomenclature. As you can see, at the bottom of the board, we’ve got two different types of haloalkanes. We have 2-bromobutane and we have also 3-bromo-4-ethylheptane. And I want to briefly talk to you a little bit about priority, when it comes to assigning them a nomenclature. This is important not just for the compounds that you produce, but also when you’re using starting material that is labelled accordingly. Typically speaking, when you’re looking at something which has a large atomic mass in comparison to carbon, it takes priority. So, if you look at Bromine, for example, with a mass of over 80 grams per mole, it takes priority. And the numbering goes from the closest terminal carbon to that substituent; the closest being the 1-carbon away, which we’ve denoted 1 in the case of 2-bromobutane.

01:28 We then count along 1, 2, 3 and 4. And so, we can therefore say, bromine, where it’s positioned is 2. And therefore, as a halogen, the halogen component appears as a prefix to the alkane chain itself: Halo-ane. If we look at 3-bromo-4-ethylheptane, we’ll see the same thing. The nearest terminal carbon to where the bromine is substituted is the one that’s indicated as 1. And that, if you were unsure as to where the number went, is how you would try to determine what the numbering sequence through your longest chain of alkane would be. Therefore, we have here 3-bromo-4-ethylheptane where the bromine, because it is the largest atom, takes priority over the ethyl substituent we see in the 4-position.

02:19 Now, let us talk about reactivity. This is predicated on three different things - the electronegativity; there’s that phrase again “of a halogen - X”, the bond strength between the carbon and a halogen X and also, the stability of the leaving group, which we will call the for the sake of argument here, X- correlating to fluoride, chloride, bromide or iodine. So, let’s first have a look at electronegativity.

02:51 If we have a look at the electronegativity of carbon on the Pauling scale, it says 2.550.

02:58 If we look at the electronegativity of our halogens, you can see that it varies - 3.98; one of the most electronegative elements we have is fluorine, then chlorine, bromine and iodine. If you recall when I talked about inductive effect, you recall I had said that where you have an element which is more electronegative than another element in a single sigma bond, it will pull electron density towards it.

03:25 This is what we see in the little green square that shows R-X, where we have a delta- or partial negative charge on our halogen X and a partial positive charge on our alkyl unit, which we’ll designate R. The C-X bond is polarised and this gives rise to a partial delta+ charge on the alkane part and a partial negative charge onto the halogen part. Let’s have a quick look at bond strength as well because whilst the size of the dipole can influence how attractive that delta+ is to attacking nucleophiles, for example, bond strength is also an important factor. As you can see, the strongest bond in this case is the Carbon-Fluorine bond. It’s by virtue of the fact that the fluorine, being a very small atom, is able to get closer to the carbon as it shares its electrons in a covalent bond. Iodine, on the other hand, C-I, has the weakest bond strength. This is because it exists in a lower set of pe... a lower period within the periodic table. As a consequence, it has more shells of electrons outside; meaning that the covalent bond radius that it has with carbon is substantially larger.

04:42 Let’s also look at the stability of the leaving group. It’s based on the relative acidity, going back to what we said in acids and bases in Module II of the hydroiodic, hydrochloric, hydrobromic or hydrofluoric acid in question. The more accepting that negatively charged halide is, of existing on its own, the better it is as a leaving group.

05:07 And in this scenario, whilst the carbon-iodine bond provides you with a relatively small dipole, it is actually one of the most stable leaving groups. It is assumed that the stronger the acid, the more stable the conjugate base. And apart from hydrofluoric acid, all of the other acids are strong acids with small differences in terms of their dissociation.

05:33 Now, let’s look at what that means in terms of synthesis. There are two principle synthetic steps or synthetic reactions that you’ll be considering in this lecture. One is Nucleophilic Substitution and the other is an Elimination Reaction, otherwise known as SN - Substitution Nucleophilic or E - Elimination. And here we have our example of our halogen and alkane or haloalkane; often sometimes referred to as an alkyl halide, which is the old common name for the same species. And what we’re going to focus on mostly is the nucleophilic attack at the alpha position shown on this model haloalkane.