00:00
So, let’s have a look at the formation of
alkyl halides from alcohols.
00:07
If we take our ideal ROH, which is our alcohol
and we react it in the presence of a hydro-iodic
or a hydro-halide acid, we can form a haloalkane.
However, contrary to what you might imagine,
it tends to work best when you have an iodide
as the attacking nucleophile. This is nucleophilic
substitution.
00:34
The mechanisms for this are both SN2 and SN1
depending on the substrate. And let’s talk
a little bit more about that.
00:43
Substitution mechanism 1.
00:45
In a SN1, it is necessary to have a good leaving
group and the conjugate base of a strong acid.
00:51
And OH- is the base of a very weak acid, which
means it’s not a particularly good leaving
group.
00:57
If we look, for example, at the following,
the reaction of a nucleophile with our alcohol
doesn’t actually yield a good outcome. The
reason for this is that OH- is a particularly
poor leaving group and that’s something
that you’ll find throughout when you’re
looking at synthetic chemistry that whenever
a reaction step involves loss of OH-, it’s
usually something that is not going to be
thermodynamically favourable.
01:23
So, in this particular scenario, the substitution
directly of the carbon which bears the OH
group can’t take place in the presence of
a nucleophile. And that reaction does not
happen.
01:36
It’s necessary, therefore, in the case of
our OH group, to modify it slightly in order
to make it a better leaving group. And in
this case, when we use acid, represented here
as H+, which forms there an oxonium ion, as
shown, with the positive charge on the oxygen,
what we have done there is essentially create
a good leaving group in the form of water.
02:03
And so, under certain conditions, for example,
in the presence of hydro-iodic acid, it is
possible to carry out a direct SN2 reaction
by leaving or allowing to leave the water
group and having the iodide ion attack the
carbon. This results in the formation of our
haloalkane, shown there as RX, and the production
of water.
02:26
H2O is the conjugate base of the strong acid
H3O+ and is, therefore, a good leaving group.
02:35
Let’s have a look at some other conditions
here because in an ideal world, you don’t
always want to make the halo-iodide. And this
is where a number of other reactions can take
place to give you the desired haloalkane.
02:49
In the case of an alcohol, you can react it
with thionyl chloride in the presence of pyridine
which gives rise to the appropriate chloroalkane.
This reaction is particularly easy to carry
out and is also relatively clean producing,
as it does, sulphur dioxide and hydrochloric
acid.
03:10
Another way of generating a haloalkane from
an alcohol is to react it with phosphorus
tribromide shown there as PBr3 or phosphorus
triiodide.
03:20
Phosphorus tribromide is usually prepared
in situ by the addition of bromine to red
phosphorus and alcohol with warming and, in
both cases, results in the production of the
bromo- or iodo-alkane and the production,
in this case, of phosphoric acid.
03:38
Now, let’s have a quick look at elimination
reactions.