Now we’re ready to
start considering
heteroaromatic systems.
Systems that contain
nitrogen, oxygen and sulfur,
and the first thing that we want
to do is to be able to count
the number of π electrons
that that atom contributes
to the π system.
The number of electrons
that an atom contributes
is going to depend not
only on the atom type,
but also on the mode
of connectivity.
What we’re going
to end up doing is
taking the number of
valence electrons
and deciding how many of
those valence electrons
belong to the sp2
atomic orbitals.
And by the way,
we’re always going to
assume in these systems
that every atom is
sp2 hybridized
if it’s part of the π system.
So how many electrons contribute
to the sp2 atomic orbitals,
and then the number
of valence electrons
that are left over will
be part of the π system.
Let’s start with carbon since
we’re familiar with that.
It’s going to look like this
in its sp2 hybridized
bonding scheme.
Carbon will contribute of
its four valence electrons,
three electrons to the sp2
hybridized atomic orbitals.
Of the four valence
electrons then,
that leaves one left
over for the π system.
Every carbon atom that’s going
to be part of the π system
will contribute one
electron to the π system.
Nitrogen can be two connected
as you can see it here
in the case of pyridine.
It’s got two bonding
connections to other atoms.
Nitrogen has five
valence electrons,
and so it’s going
to end up having-
in that sp2 hybridized system
since there are two bonds
two electrons that will part of
that sp2 hybridized σ bond,
and then a pair of electrons
that will be making up the third
of the sp2 hybridized orbitals.
Altogether, four of
the five electrons
are going to end up being in the
sp2 hybridized atomic orbitals.
That leaves one left
over for the π system.
An N2 nitrogen will end up
contributing one electron
to the π system.
What about an N3
connected nitrogen?
An N3 connected
nitrogen makes
all three of its sp2 hybridized
orbitals involved in bonding,
and so the three electrons of
the five valence electrons
in nitrogen will be
part of the sp2 system.
That leaves two electrons
left over for the π system.
So an N3 nitrogen
contributes two π electrons.
Oxygen can be singly
connected as in a carbonyl,
and it will have two lone pairs
as part of the sp2
hybridized orbitals
and one sing, one σ bond.
So one electron will
go in the σ bond,
and then four electrons will go
in those nonbonding orbitals
that are part of
the sp2 system.
Five of the six electrons in O1
will be part of the sp2 system,
and that leaves
one electron over,
left over for the π system.
O1 contributes one
electron to the π system.
O2 is an oxygen that’s
connected to two other atoms,
and that’s going to have just
one lone pair in the sp2 system.
All together, the sp2 system
has 2, 3, 4 electrons.
That leaves two electrons
left over for the π system.
So an O2 is going to contribute
two electrons to the π system.
By these rules, knowing
the atom and the num -
and the mode in which is
connected, we can decide
how many electrons that
atom is going to contribute
to the π system in a
heteroaromatic system.
I can tell you that these atoms
don’t have to be exactly
a part of the π system.
For example, I will just draw …
We could have a system in
which there is a carbonyl
that is not part
of the π system,
and that oxygen is still going
to be an O1 type oxygen.
So something like
this, for example,
we have oxygen as contributing
one electron in each of the -
for each of the
O1 bound oxygens.
And so even though it’s not
part of the cyclical system,
it is part of the π system
because every sp2 atom
is adjacent to
another sp2 atom,
and so we need to account for
all of the adjacent sp2 atoms
when we count the total number
of electrons in π systems.
The last point I’ll make
is that in these more
complicated systems
which involve atoms that
are not part of the, ah,
formal cyclic connectivity,
we don’t need to worry about
the Huckel rule or
anything like that.
All we’re really interested
in is how many electrons
00:04:57.023,00:00:00.000
are part of this
contiguous π system.
€ systems.
The last point I’ll make
is that in these more
complicated systems
which involve atoms that
are not part of the, ah,
formal cyclic connectivity,
we don’t need to worry about
the Huckel rule or
anything like that.
All we’re really interested
in is how many electrons
00:04:57.023,00:00:00.000
are part of this
contiguou
