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Current time:0:00Total duration:11:59

Voiceover: Let's look at the
amide, or "amid" functional group, and let's start by
assigning a steric number to this nitrogen. So, the steric number is
equal to the number of sigma bonds; so here's a sigma
bond, here's a sigma bond, and here's a sigma bond;
so three sigma bonds. Plus number of lone pairs
of electrons, so there's one lone pair of electrons on that
nitrogen, so I'll go ahead and highlight them there. So three plus one gives us a
total of four for the steric number, which means four
hybrid orbitals, which implies SP three hybridization for
that nitrogen, and from earlier videos, you know that SP
three hybridization means, a trigonal, pyramidal,
geometry for that nitrogen. And so, that's one way of
looking at this function group, and that lone pair of
electrons being localized to that nitrogen;
however, now that we know resonance structures, we know that, that lone pair of
electrons is not localized to that nitrogen; it's
de-localized in resonance. So we could take that
lone pair of electrons, and move it in here, to form
a pi bond, so that would force us to push some pi
electrons off, onto this oxygen, so let's go ahead and draw
the other resonance structure. So this top oxygen would
now have three lone pairs of electrons around it, a
negative one formal charge, and there'd be a double-bond
between this carbon and this nitrogen, so let's go
ahead and draw in everything. This nitrogen how has a
plus one formal charge, and we can go ahead and complete our resonance bracket here. So let's follow those
electrons along, the electrons in magenta, the lone pair
of electrons moved in here to form our pi bond, and
the pi electrons over here in blue, came off onto the
oxygen, to give the oxygen a negative one formal charge. All right, let's now
calculate a steric number for this nitrogen, in our
second resonance structure. So, steric number is equal
to number of sigma bonds: So here's a sigma bond,
here's a sigma bond, and our double-bond; we know one
of them is a sigma bond, and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen, zero
lone pairs of electrons, so three plus zero gives us
a steric number of three; that implies three hybrid
orbitals, which means SP two hybridization, and a
trigonal planar geometry around that nitrogen,
so a planar geometry. And, here we've shown our
electrons being de-localized, so the lone pair of
electrons are de-localized due to resonance, and
so, experimental studies have shown that the amide
function group is planar, so these atoms actually are
planar here, which means that the electrons in
magenta are not localized to that nitrogen; they
are actually de-localized. And so, that implies this
nitrogen is SP two hybridized, and has a P orbital, and
that allow that lone pair of electrons in magenta
to be de-localized. And so, here's a situation where drawing a resonance structure
helps clue us into what's actually happening: That
lone pair is participating in resonance, which makes
this nitrogen SP two hybridized, so it has a P orbital. All right, let's look at
this example down here, and let's look first at this
left side of the molecule, and so we can see our
amide functional group, and if I look at the
lone pair of electrons on this nitrogen, we've
just talked about the fact that this lone pair of electrons
is actually de-localized, so this lone pair is
participating in resonance. And so, that affects the
geometry, and how you think about the hybridization
of this nitrogen, here. So, the electrons in
magenta are de-localized because they participate in
resonance, and if I think about, let's make this a
different color here; let's make these electrons in here blue,
so the electrons in blue on the amine, these
electrons have nowhere to go; they can't participate in
resonance; that lone pair of electrons in blue is
localized to that nitrogen. And so, this is why you
can think about an amide being different from an amine, in terms of functional group, and in terms of how they
react and how they behave. If we look at another
example, so this molecule right here, and we assume
the lone pair of electrons on that nitrogen is
localized to that nitrogen, let's go ahead and
calculate the steric number. So the steric number'd
be equal to sigma bonds, so that's one, two, and
three; so three sigma bonds. Plus lone pairs of electrons,
there's one lone pair of electrons on that
nitrogen, so three plus one is four; so four hybrid
orbitals, which implies SP three hybridization on that nitrogen. But we know that, that lone
pair of electrons is not localized to that nitrogen;
that lone pair of electrons is de-localized; it
participates in resonance, because we have this pattern here. All right, so this pattern
of a lone pair of electrons, in blue, next to a pi bond,
which I will make magenta, and so we could draw
a resonance structure. So I could take the electrons
in blue, move them into here, too many bonds to this carbon,
so I push the electrons in magenta off, onto this carbon. So we draw the resonance structure, so I have my ring here,
the nitrogen's bonded to a hydrogen, the
electrons in blue moved in to form a pi bond, and
the electrons in magenta, moved off, onto this
carbon right here, to give that carbon a negative one formal charge. All right, let's go ahead
and calculate a steric number for the nitrogen here, which
gets a plus one formal charge. All right, so the steric
number will be equal to number of sigma bonds: So
there's one sigma bond, here's another sigma bond, and then, in our double-bond,
one of them is a sigma, and one of them is a pi; so I'm saying that's our sigma bond. So our steric number is equal
to three, plus the number of lone pairs of electrons, now zero. So that's a steric number of
three, which implies three hybrid orbitals, which
says SP two hybridization. And since we know that, that
lone pair is de-localized, it's going to occupy a P
orbital, and so therefore this nitrogen is SP two
hybridized, because we know SP two hybridization has
three SP two hybrid orbitals, and one P orbital. So that lone pair in blue
is actually de-localized; it's occupying a P orbital,
and so let's go ahead and draw that, down here, so let's
say this is the nitrogen, and you're looking at
it, at a bit of an angle. If that nitrogen is SP two
hybridized, that nitrogen has a P orbital, so we can go
ahead and draw in a P orbital, on that nitrogen. And so, the electrons in
blue, since those electrons are de-localized, those
electrons are going to occupy that P orbital; those electrons can participate in resonance. Alright, so let's go ahead
and draw a hydrogen here, just to finish that off, and
I could have kept on going with resonance structures, so
I actually forgot to put in these electrons, right
here, because I could have kept on going; I could take
these electrons in magenta, move them into here, and
then push these electrons off, onto that carbon,
and I could keep on going. And so there are many
more resonance structures that you could draw; I'm not going to do so for time purposes; here I'm just trying to
show how resonance affects the hybridization, or how
you should think about the hybridization, when
you're drawing these imperfect dot structures. Let's go ahead and finish
our picture of the molecule, because we know that
these four carbons here, all have a double-bond to them. And so those four carbons are
therefore SP two hybridized, and if those carbons are
SP two hybridized, each one of those carbons has a P
orbital, so I can draw in a P orbital on all of my carbons, here. And you could think about
those electrons, let me go ahead an highlight them,
the electrons in red here, so these pi electrons,
so those pi electrons are participating in resonance,
and so we have a total of six pi electrons
participating in resonance: those four in red that I just
highlighted here, and then these electrons here, in blue,
which occupy this P orbital. And so this is going to
be extremely important, when you talk about aromatic
compounds later in the year. So, for right now, just
try to identify the skill of de-localized electrons
versus localized; in this case the electrons on the
nitrogen are de-localized; they participate in resonance. Let's do one more example,
so one final example here. All right, and let's go ahead
and start by calculating the steric number of this nitrogen, so how I've drawn it here. So, steric number is equal
to number of sigma bonds. So here's a sigma bond; I
have a double-bond, one of them is a sigma bond. So I have two sigma bonds,
plus lone pairs of electrons; there's one pair of lone
electrons here, on this nitrogen, so two plus one gives me
a steric number of three; that implies three hybrid
orbitals, or SP two hybridization. So if that nitrogen is SP two hybridized, that nitrogen's going to have a P orbital, so let me go ahead and
sketch in a P orbital here, on this nitrogen, and let's
think about the other carbons too, so let's think about
these carbons, I should say. So I have one, two, three,
four, five carbons; all these carbons have a double bond
to them, so all of those carbons are SP two
hybridized, so I can draw in a P orbital on all of those carbons. So all of those carbons have
three SP two hybrid orbitals, and one un-hybridized P orbital. So, it's the same idea
with this nitrogen here: if this nitrogen is SP two
hybridized, it has three SP two hybrid orbitals, and
one un-hybridized P orbital. I already drew in the
un-hybridized P orbital; let's go ahead and put in
those SP two hybrid orbitals, and I'll use magenta. So, one of those SP two
hybrid orbitals is forming this sigma bond over here;
another one's forming this sigma bond back
here, and then a third SP two hybrid orbital must be here. And that's actually where that
lone pair of electrons is, so this lone pair of electrons
here in blue, is localized to this SP two hybrid
orbital, so this is actually, let me go ahead, and draw an arrow here. So the lone pair of electrons
is actually localized to that SP two hybrid orbital. So that lone pair of
electrons can't participate in resonance, because that
nitrogen already has a pi bond. So, if I go over to here,
these pi electrons are using that P orbital, and since
those pi electrons in green are using this P orbital,
over here on the right, so this P orbital over here; that means that those electrons in blue can't use it, and so, therefore,
those electrons in blue are actually localized to that nitrogen. Which might not be what you first thought because we talked about the pattern, of course, of the electrons
in blue here, lone pair of electrons next to a pi
bond, so you might think we could draw a resonance
structure; however, the reason we can't, is
once again, those electrons in green are actually
participating in resonance, and so they're already
using that P orbital. All right, so this is kind of an exception to what we talked about
before, where we said that there's a lone pair of
electrons next to a pi bond, that lone pair is de-localized, so this is the one thing to think about.