00:01
So, nucleophilic substitution.
00:04
Now, this is a different take on what we originally
know to happen with carbonyl compounds. Carbonyl
compounds, as we saw in the previous lecture,
undergo nucleophilic addition, that is to
say the carbon-oxygen double bond is broken
and you have a tetrahedral structure of an
addition product formed. However, what makes
these carboxylic acids derivates so useful
and so reactive is that they undergo a nucleophilic
addition followed by elimination reaction
which is otherwise known as a nucleophilic
substitution.
00:40
So, let’s have a look at this very sort
of basic example where we have got a carboxylic
acid derivative where Z could be any of those
chaps shown underneath that particular equation;
let’s just say, for the sake of argument,
it is a chloride and we have a nucleophile
which is shown above the double headed arrows
showing dynamic equilibrium as Nu-.
01:02
First thing that happens, of course, is it
being negatively charged, it attacks the carbonyl
carbon, it opens up that carbon-oxygen double
bond and formally puts a negative charge on
that oxygen. However, what then happens is
that you have got the reformation of that
carbon-oxygen double bond and the loss of
whatever group Z is. In this case, if we say
it was chlorine, it would mean that we would
lose chloride.
01:34
So, if we react an acetyl chloride or an acid
chloride with a nucleophile, we would usually
displace the chloride and introduce the nucleophile
on that carbonyl carbon. And this is why I
say, you can call it substitution or as some
people prefer it, an addition elimination
reaction.
01:57
So, to start off with, the bond is opened,
then it is reformed again. And in this regard,
the oxygen is regarded to serve as an electron
sink where we need to put the electrons back
in to kick off Z which is… well, has to
be a better leaving group than the nucleophile
you started with. This reaction happens, to
stress only if the nucleophile is a strong…
Nu- is a stronger nucleophile than the Z-
leaving group. So, Z- must be a leaving group.
02:27
So, let’s look at an example reaction, not
with something simple like an acid chloride,
but with, for example, an ester shown here.
Note the same format: we have here a carbonyl
carbon to which the carbon of that… where
the oxygen is, is attached to an OR group
and now, what we are doing is we are reacting
it with a Grignard reagent.
02:50
Now, I think we touched briefly on Grignard
reagents earlier on. To all intents and purposes
and as far as you are concerned, all you should
need to know is in a case of a Grignard reagent,
the alkyl group, in this case a methyl, is
said to carry a full negative charge. This
effectively means that Me- is the nucleophile,
Me- is the nucleophile can attack the carbonyl
carbon, open up that carbon-oxygen double
bond.
03:20
And remember what I said before, once it’s
opened up then what it can do is it can reform
and kick out the leaving group Z, which, in
this case, happens to be an alkoxide group;
this is an alcohol which has been deprotonated.
Now, of course, what this gives us here is
now as you can hopefully see, a ketone. What
we have done is we have actually substituted
that OR prime group with a methyl group and
so, we have generated a ketone. At the same
time, we have also generated magnesium alkoxide.
03:59
What I would like you to do is, well, then,
is consider what the effects would be if we
actually had an excess of the Grignard reagent
because as we have already said, carbonyl
compounds, even ketones, also react with nucleophiles
via nucleophilic addition. And so, in an excess
of methyl magnesium bromide, you would find
that this product would not remain much around
for longer and will be converted into corresponding
dimethyl alcohol.