So, what does it mean? You recall when we
started talking about the idea of nucleophiles
and electrophiles we used Nu- just to represent
an ideal nucleophile. In this scenario, what
happens in the case of a good nucleophile
is that it attacks the delta+ charge on the carbon
of our haloalkane in a solvent at a specific
temperature. It, then, kicks off the halogen,
leaving as it does, as a halide. And the possible
parameters that influence this reaction are
manifold - the density of the nucleophile,
the leaving group. And this gives rise to
two possible pathways - the so called SN2
and SN1 pathway. We’re going to discuss those
Let’s start off with the SN2 reaction. All
of that abbreviation means is that it is substitution
nucleophilic bimolecular. And in terms of
the reaction mechanism, you will understand
why it is bimolecular rather than unimolecular.
In the case of, for example, sodium hydroxide,
which produces a good, strong nucleophile
in the form of OH-, can attack at the carbon
which is delta+ on our bromoalkane. In this case,
it is 2-bromobutane, our old friend from before.
And what this can do is invert the stereochemistry;
in fact, SN2 reactions carry out that inversion.
And why I mean inversion is if you look at
the species on your screen, which is S in
steric chemistry, I’d recommend that you go
back to when we talked about stereochemistry
for Cahn-Ingold-Prelog rules. Note what’s
In the case of our S steric chemistry, the
bromine takes priority, but in the case of
sodium hydroxide, when it attacks, it attacks
from behind. In doing so, the priority now
falls on the OH group and the bromide ion
is lost, thus changing the order of priority
on that central carbon, which is stereogenic.
This inversion of the stereochemistry can
be shown here, where, for example, we have
the bromine which, in its current form, where
the current stereochemistry is given, results
in an S enantiomer, we can invert that enantiomer
by a nucleophilic attack by hydroxide from
the rear. This kicks off the bromide ion as
effectively as it is attacking on the opposite
side of the leaving group.
As both the nucleophile and the substrate
(here, the 2-bromobutane) are participating
in the main step in the reaction mechanism,
it is said to be bimolecular. That is to say
the reaction rate, not to do with the thermodynamics
or the equilibrium constant, but the reaction
rate, is dependent upon the concentration
of our haloalkane (bromobutane) and also the
concentration of our hydroxide anion.
The rate is dependent on these and is given
in the equation below where the rate is proportional,
although not necessarily always equal directly
to, R-X (our haloalkane) and OH-. This is
bimolecular. And the reason for this should
be relatively easy to see. The more hydroxide
you have and the more haloalkane you have,
the greater the likelihood of them coming
together. It’s the basic principle of collision
theory. So, the more you have of either, the
more likelihood the reaction is to occur.
And in fact, the reaction is dependent upon
those two coming together.
Let’s have a look at the reaction parameters.