Aromaticity Video Lecture | Chemistry for JEE Main & Advanced

hi guys, it's professor Dave. let's talk
about aromaticity.
in organic chemistry we frequently refer to certain compounds
as being
aromatic. this terminology came about in
a time when we noticed
pleasant odors emanating from certain
compounds isolated from natural oils
produced by plants
it wasn't until much later that we
understood that this property is
actually due to the presence
of fully conjugated unsaturated ring systems
so when we use that terminology now we are referring to just that
we're referring to the molecular structure
and so the most common of these structures is benzene
and so let's take a look at what we mean by that
if we remember from what we were
discussing about resonance
we have this molecule here, we have three pi bonds and we notice that they
are conjugated, meaning we have this
alternating double bond single bond
system and so these are now
fully delocalized they're not just
delocalized about a portion of the
molecule
they are actually delocalized around the entire molecule
so we have these two different
resonance structures for benzene and we can see
that given that in each of the two
resonance structures there's a double
bond in every single possible position
between
and so that means that recalling the
fact that these
actually do not exist but it is actually
the composite resonance structure
that is real. also benzene really looks
like this, we have partial pi electron
density
all about the ring so that's what we
mean when we talk about a ring system
that is fully conjugated unsaturated
so that is a prerequisite for aromaticity. it's not only benzene that is
aromatic but it's the most common example
and so what we want to do now is take a
look at some rules to help us determine
whether a molecule is or is not aromatic
so let's take a look at the rules for aromaticity
number one the ring system in question
must be fully planar
so looking at something like this
there's a portion here that is sp3
hybridized
and therefore has tetrahedral geometry so
there are groups that are projecting
into and out of the board so that is not fully
planar
we must remember that only the carbons
that are sp2 hybridized
will have trigonal planar geometry therefore can be
the molecule could be considered planar
so the entire molecule must be planar
number two the system must be fully
conjugated
so while in addition the fact that these
are sp3 hybridized carbons
also these pi bonds are not situated in
a way that makes the system
conjugated, remember that at least here
there is some
delocalization of these pi electrons
because they are arranged in the
appropriate fashion alternating double
bond single bond here
these are too far apart for these two be conjugated
so this could not qualify as aromatic
and the last thing
is that the number of pi electrons
present that are delocalized must obey
huckel's rule and huckel's rule is as
follows: we have this term 4n + 2
where n could be any integer and
for whatever integer you put into
huckel's rule it spits out a value for a
number of pi electrons
that are delocalized that will generate
an aromatic system
so that sounds more complicated than it
is, let's put in any integer
let's say zero. if n equals zero, four times zero plus two equals two
that means that a system with two pi electrons
could be aromatic. let's put in the
number one
then we'll have four plus two or six
that means that six
is a number of pi electrons in the system that could
promote aromaticity so we end up with
this 2, 6, 10
14 and and so forth and so
those are the values that allow for aromaticity when we're looking at the number
of pi electrons
that are delocalized within the system so for
8, 12 et cetera those even if the molecule
is fully planar and fully
conjugated will not be aromatic so
let's take a look at a couple examples
and apply these rules and see if we can
predict which are going to be aromatic
and which are going to be antiaromatic
so let's take a look at some examples so
first of all we have this
three-membered ring here and this is
fully planar and it is fully conjugated
because we have a
carbocation up here so that makes
that sp2 hybridized and planar
and it is fully conjugated because we
can move this single pi bond
around the molecule, we can push that
here which would neutralize this
charge and leave the cation there
and then push the pi bond over here
neutralizing leaving that there so
we could
have three resonance structures of this
molecule but with the cation moving
amongst each carbon therefore this pi
bond is truly delocalized
around the entire molecule so a single pi bond here
is actually fully conjugated around the molecule so
the first two criteria are satisfied and then in addition
and is only two pi electrons
that are involved in resonance, two is a
number that satisfies huckel's rule
so this is an aromatic compound so that works
now over here we look at a couple
examples involving four
pi electrons that are delocalized so
over here this is a fully planar
molecule these are all
sp2 hybridized carbons so it is a
completely flat molecule
and it is fully conjugated. there is an
additional resonance structure
where we push these pi bonds to these two positions
so it is planar and fully conjugated however
there are four pi electrons involved in resonance
and that does not satisfy hackle's rule so this is not
going to be aromatic, it will be antiaromatic
and then the same goes over here for
this five-membered ring
there's a cation here so we can draw
up five actually five different
resonance structures
where the cation is on all of the different carbons
and the pi bonds move accordingly however once again
there are
only four pi electrons involved, that does not satisfy huckel's rule
so this is not going to be aromatic.
then we look over here
we already know about benzene but just to reiterate
we do have a situation that is fully
planar these are all
sp2 hybridized carbons so the implied
hydrogens are actually in the plane of
the board, fully planar and is fully
conjugated. there's one more
resonance structure here where the
pi bonds are in the other positions
and looking at the number of pi electrons
we have 2, 4, 6 pi electrons
remembering that every single bond
every single covalent bond has two
electrons so when we're looking at one
pi bond we know that there are two pi electrons involved so
2 per bond, we have three bonds, that's six pi electrons
that does satisfy huckel's rule so that
is going to be aromatic as we already
knew
and then over here we have the same
situation as before but now we have an
anion instead of a cation and so
remember that is not just
bonds that can participate in resonance
it is also lone pairs so this negative charge implies
the presence of an additional pair of electrons
that are also delocalizable because
they are pi electrons
and so this negative charge could go
here and generate a pi bond there
pushing this here, pushing this around the
center so we again could have five
different resonance structure for this molecule
and so it is fully
delocalized, it is fully conjugated and
now we have 2, 4
6 pi electrons instead of 4, this is a cation
an absence of a lone pair this is an
anion, the presence of a lone pair
so there are six pil electrons here and
that does satisfy huckel's rule
and now let's take a look at some of these
heterocyclic compound, so heterocyclic
meaning that there are elements
other than carbon within the actual ring structure
so some of them have nitrogen, furan has oxygen
thiophene has sulfur and now these can also be aromatic
because these atoms have lone pairs that may
or may not be involved in resonance if
they are needed to be
in order to generate aromaticity. aromaticity is a highly stabilizing
feature
for a molecule so if it can do it will but
the tricky part here is that available
lone pairs
only contribute to resonance if they
need to do so for example here with
pyridine
this lone pair doesn't need to be
involved in resonance in order to
generate
aromaticity so we just ignore it, these three pi bonds
are going to be able to delocalize and generate aromaticity on their own
however in this situation and actually
all the situations
one lone pair, the only one available
here for this nitrogen atom
will be needed in order to generate aromaticity
over here one of the two and in this
case as well, one of the two
lone pairs will be incorporated into resonance
in order to generate aromaticity and
in each of these cases
it will then therefore be six pi electrons
two pi bonds and one lone pair that will be involved
and therefore that does satisfy huckel's
rule and that's why all of these
compounds
are also aromatic.
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