00:01
However, which one is the more likely? And
I bring you back to hybridisation from Module
I and the stability of carbocations.
So, the reason why you get one over the other
is by virtue of the stability imparted by
electron donating groups. This is our so called
positive inductive effect, if you go back
to the previous module.
00:24
If we look at what’s happening in the first
instance, when the double bond is split, when
the two electrons are transferred on to H+
and form a sigma bond between the carbon and
the hydrogen, we are left with an empty p
orbital. It was originally hybridised, but
now, it is not. It is a p orbital that has
that dumbbell shape, as we saw before. Also
as a consequence, as you can hopefully appreciate,
it is technically now sp2 hybridised. As
a direct consequence, it adopts a trigonal
planar structure.
00:54
And whilst this may not necessarily seem important
at the moment, from a stereochemical perspective,
it will become important later on. But, more
importantly, it has a formal positive charge
on it. And as you can see, if we look at different
degrees of substitution on that single carbon,
we can see degrees of stability imparted.
Tertiary - what that means is we have three
alkyl groups attached to that positively charged
carbon. Secondary - so we have two alkyl groups
attached to that secondary… that… that
positively charged carbon or primary - where
we have only one alkyl group attached and
then finally, methyl - where we only actually
have hydrogens on the carbon which is positively
charged.
01:39
Alkyl groups, by their nature, are electron
donating. And therefore, the carbocation can
be stabilised by more electron donating alkyl
groups and that’s what we see in the case
of the tertiary, okay, the three, and again,
with that little sort of degree sign above
it, tertiary.
Alkyl groups are electron donating and by
virtue of their inductive effect, can stabilise
better that carbocation. What that means in
the context of our reaction is that the addition
of H+ on to our double bond which results
in a carbocation being present in the central
carbon on our propene molecule is actually
the preferred intermediate resulting in the
two bromopropane being the only product that
results.
So, in this scenario, the electrophilic addition
of any unsymmetrical reagent to an unsymmetrical
double bond proceeds in such a way as to involve
the most stable carbocation. As you can see
the carbocation afforded when the reaction
takes place at C-2, is actually less stable
because it is primary as opposed to secondary.
02:49
The H+ will also add to the carbon that
already has more hydrogens.