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