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Current time:0:00Total duration:5:35

- [Voiceover] We've been
talking about bonds as springs and so far we've ignored the
idea of the dipole moment but the dipole moment is going to effect the intensity of the signal and so we need to review
it really quickly. And so if you have a dipole. If you have two charges separated so two opposite charges separated you could express the dipole moment as equal to the magnitude
of charge in your dipole. So I'm going to write
a lower case delta here our partial charge times the distance
between your charge is d. So if you have a partial
positive charge right separated by some distance
by a partial negative charge and if we're thinking
of our bonds as springs I could draw a spring in here, right. So we have a magnitude of charge one positive and one negative. So there's our magnitude of charge separated by some distance, d and so obviously we're going
to have a dipole moment. Right, so we have a magnitude of charge separated by distance so
we get a dipole moment. If we get a stretching vibration. So if this bond stretches,
if it stretches this way we're obviously going
to change the distance and if we're changing the distance we're changing the dipole moment and that's important because only stretching vibrations that produce a change in the dipole moment are observed as signals on your IR spectrum. And so let's look at an example of this. Let's look at cyclohexanone. So here's cyclohexanone. We know that the carbonyl
has a dipole moment. The oxygen is more
electronegative than the carbon. So the oxygen gets a partial negative. This carbon here gets a partial positive and so we have a pretty
large dipole moment associated with our carbonyl. And so we would expect to
see a pretty strong signal for that carbonyl bond stretch. If we look at our spectrum here and we go ahead and divide our diagnostic from our fingerprint region here's a very very strong signal. So just past 1700 wave numbers. This is approximately 1715 and this is in our double-bond region that we talked about earlier and so this must be the
carbonyl bond stretch. So this represents let me go ahead and use a different color. So this signal on our spectrum represents the carbonyl bond stretch and this is the partial
negative, partial positive. So large dipole moment
means a strong signal and if a large dipole
moment means a strong signal that means that a smaller dipole moment would be a weaker signal. So let's look at an example of that next. So we're going to compare. We're going to compare
these two IR spectra. And so now we're looking
at a different molecule. We're looking at the IR
spectrum for 1-Hexene here and once again let's go
ahead and divide our regions alright and if we look
in the double-bond region we see this signal right here. So if we drop down, the
signal is about half way between 1600 and 1700, so
we'll say approximately 1650. So that's the signal in
the double-bond region and of course that is the carbon
carbon double bond stretch. So that's this double-bond
here on 1-hexene. And notice that it's not
as intense as the one that we talked about before, right. So this signal, let me go ahead and use a different color here. This signal is much
stronger than this signal. Alright so this is a weaker signal. So this must not have as
strong of a dipole moment and indeed that's the case. So if we think about this
double-bond right here put some hydrogen's on so it's a little bit easier to think about it's going to have a
very small dipole moment. It has a dipole moment because
this alkyl group right here. Remember alkyl groups
are electron donating and so because it's not
symmetric you're going to get a weak dipole moment and so because you have
a weak dipole moment you're not going to get an intense signal. You get this weaker signal here. So this, going back up
here to this carbonyl again this is really important when
you're looking at IR spectrum. This super-intense signal for a carbonyl often helps you figure out what functional groups
that you're dealing with. Alright, let's do one more. Let's compare this alkyn to another alkyn. So let's look at this one now. So over here we have a
2,3-Dimethyl-2-butene. Right so we go one, two , three, four. That's 2,3-Dimethyl-2-Butene and if we once again draw our line at around 1500 and we look
in the double bond region so somewhere in here
we don't see a signal. And the reason we don't
see a signal is because this is a symmetrical alkyn. This is symmetric about
the double bond here so it's the same on both sides. And since it's a symmetrical alkyn there's no dipole moment right. The electron donating effect of the alkyl groups would cancelled. There's no dipole moment and so therefore there's
no change in dipole moment when the carbon
carbon-double bond stretches and so therefore we don't see a signal. And so this signal is
absent on our IR spectrum. This is important to think about if you have something
that's symmetrical right you could talk about an alkyn too. You're not going to see a
signal on your IR spectrum. So that's something to think about.