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Organic chemistry
Course: Organic chemistry > Unit 14
Lesson 1: Infrared spectroscopy- Introduction to infrared spectroscopy
- Bonds as springs
- Signal characteristics - wavenumber
- IR spectra for hydrocarbons
- Signal characteristics - intensity
- Signal characteristics - shape
- Symmetric and asymmetric stretching
- IR signals for carbonyl compounds
- IR spectra practice
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IR signals for carbonyl compounds
The IR signals of various carbonyl-containing compounds. How differences in wavenumber relate to differences in compound structure. Created by Jay.
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- How do you know whether inductive effect is stronger than resonance, like at? 6:40(22 votes)
- when we look at a carboxylic acid derivative like the one at the end of this vedio, how do we compare the effects of induction and resonance? at what case will induction beats resonance or just induction always wins?(3 votes)
- at 7.17 why the value of k is more in the anhydride(4 votes)
- I think if you watch the previous video this will make sense. But the major reason is due to the fact that the carbonyl bonds in the anhydride are high energy bonds due to induction from the central oxygen. In other words the central oxygen is pulling electron density away from the carbonyl carbon requiring the bond between each carbonyl carbon and their respective oxygens much stronger, higher energy and leading to a larger k value.(4 votes)
- Around, why would Y have a positive charge? 4:00(4 votes)
- Y had a free pair of electrons and in the new configuration this free pair goes between the C and Y to form a double bond. Therefore there is less negative charge on the Y (because it is shared with the C now). Since the positive charge of the Y hasn't decreased, it becomes positively charged. I hope that explanation makes sense.(3 votes)
- At, how is the inductive effect more important for the acid anhydride than the ester? Both of them have the same electronegative atom, O. 7:25(3 votes)
- In an acid anhydride you have an ester group (O-C=OR) pulling electrons away from each carbonyl. In an ether, you have an alkoxy group -OR group pulling electrons away from the carbonyl. An ester group is "stronger" than an alkoxy group so it has more effect. (The carbonyl carbon in the ester has a partial positive charge that will pull on electrons -- in contrast the alkyl in an alkoxy group donates electron density.)
There is also a contribution from resonance to the acid anhydride having stronger carbonyl bonds.
In the acid anhydride there are two resonance structures with each carbonyl oxygen getting a partial negative charge. These structures are competing with each other for the electron density (i.e. lone pairs) on the oxygen between them. This means that each of these structures makes less of a contribution to the overall 'hybrid' structure and therefore that there is a smaller disruption of each carbonyls double bond when compared to an ester, where the lone carbonyl has no competition.(2 votes)
- Why does it matter if it's an acidic anhydride?(1 vote)
- Is it possible to tell if the compound is a carboxylic acid or simply an alcohol with a ketone somewhere within the alkane chain?(1 vote)
- 1) The carbonyl in a ketone would have a stronger dipole and therefore would have a higher wavenumber than in an acid.
2) The absorption for -OH in an alcohol has a higher wavenumber and narrower shape compared to that in an acid.(1 vote)
- Atyou discussed how to identify an aldehyde. My professor mentioned that aldehydes should have two peaks due to something he called fermi resonance but I'm not sure how you can have that. 9:18(1 vote)
- can you explain the resonance structure of aldehyde and keton ?(1 vote)
- For a generic ketone (or aldehyde) you can draw a resonance form where the electrons in the double bond move out onto the oxygen. This gives you a structure where there is a positive charge on the carbonyl carbon and a negative charge on the carbonyl oxygen. This form is a minor contributor to the overall (hybrid) structure.
MAJOR FORM:
R1
\
C=O
/
R2
MINOR FORM:
R1
\
+ C—O –
/
R2(1 vote)
- At, I think that there is a third contributing resonance structure involved, with the positive charge localized (in LE theory terms, not in reality) on the carbon bonded to the oxygen atom. Right? 1:49(1 vote)
- Since the structure you describe is used for simple ketones, I guess it must be valid here. I believe that it is a relatively minor contributor to the overall structure.(1 vote)
Video transcript
- If you look at the dot
structure for a ketone, we already know we're gonna see a signal for the ketone carbonyl
on our IR spectrum, because the oxygen is partially negative, and this carbon here
is partially positive. So, a relatively large
separation of charge means a relatively large
change in the dipole moment of the carbonyl when it stretches. And so therefore we get
a very strong signal on our IR spectrum. And the signal for the ketone carbonyl shows up at a wave number of
approximately 1715 or 1720. It's a little different
for a conjugated ketone, so down here we have the dot structure for a conjugated ketone. Conjugation lowers the
signal for the wave number, and let's see why. So we can think about resonance here, so if I move these electrons into here, push those electrons off onto oxygen, I can draw a resonance
structure, so over here, we would now have a double bond in here, we would now have a single
bond of carbon to oxygen, and three lone pairs of
electrons around the oxygen. Formal charge is negative one
formal charge for the oxygen, and this carbon right here
has a plus one formal charge. So, if the electron's in
magenta move over to here, like that, then the
electrons in blue right here could move off onto your oxygen. So remember, the actual
structure of the molecule's a hybrid of your resonance structures. And so, the resonance
structure on the left, this looks like a
carbon-oxygen double bond. The one on the right looks like
a carbon-oxygen single bond. So in reality, it's more
of a hybrid of those two. And so, over here I'm
going to draw a carbon bonded to an oxygen, I'm gonna give it a
partial bond right here. So I'm saying it's stronger
than a single bond, and not quite as strong as a double bond. The idea is resonance
lowers the double bond character of your carbonyl. So that's weakening, that's
weakening the carbonyl. And so, if you're
weakening the bonds, right, remember the value for K, the spring or the force
constant goes down, and from earlier videos
we've seen if you decrease K you're going to decrease the frequency, you're going to decrease the wave number where you find the signal. And so the signal for
the carbonyl, alright, let me go ahead and use green here, so the signal for our
carbonyl moves down to approximately 1680 or so, so
definitely less than 1700. And once again that's
because resonance, right, that's weakening, you can think about that decreasing the bond strength. Alright, what happens when you add in an electronegative atom? So now we're talking about a generic carboxylic acid derivative here. So why is some electronegative atom, like oxygen, for example, right? We could also have resonance here, right? So we could draw a resonance structure, we could move the lone
pair of electrons in here, and push those electrons onto our oxygen, and we could go ahead and
draw a resonance structure. So now, once again, the
top oxygen would have a negative one formal charge, and now the Y would have a positive one formal charge, like that. So, for keeping with our colors here, so the lone pair of electrons
moves in here, right? And then the pi electrons
in here in our carbonyl move off onto our oxygen. So once again, for our
carboxylic acid derivative we have resonance. And when you're thinking
about the carbonyl, what is that doing to the carbonyl, right? In your hybrids, it's
decreasing the double bond character of the carbonyl. So you're decreasing the
strength of the carbonyl, and, we've already seen, right? Decrease K, right, decreasing
the strength of the carbonyl, you decrease K, you decrease the signal where you find the wave number. You decrease the frequency
of bond vibration, alright? So that's the idea of resonance. So resonance is present. But when you're talking
about a carboxylic acid, you also have an inductive effect. So we talked about this in the video on carboxylic acid derivatives. There's a competition between
resonance and induction. So let's think about induction here. And if we have our R group,
our carbon, our oxygen, and our eletronegative atom Y, like this. If we are thinking about
the double bond as being, as being decreased by resonance, right? So decreased double bond
character because of resonance, what is induction? Induction, of course, refers
to the electronegativity of this atom here. So if we have a very electronegative atom relative to carbon, like oxygen, right? This is going to withdraw electron density in this direction. So we can think about a
lone pair of electrons in this oxygen, right, some
electron density I should say, moving in here to strengthen the carbonyl. So if you're withdrawing electron density, you can think about some electron density from the oxygen moving in to increase the strength of the carbonyl. And so this has the effect of giving us more of a double bond again, right? So that's just how I like
to think about it here. And so, induction, I'm gonna
go ahead and write this here, induction, by withdrawing
electron density, we're increasing the
strength of the carbonyl, so we're increasing the
spring constant or K. So we're increasing the
frequency of bond vibration, we expect to find the signal
at a higher wave number. And so, once again, these
two competing effects are something we have to think about for carboxylic acid derivatives, right? Resonance is going to lower the signal, and induction is going
to increase the signal. So let's look at a whole bunch
of carbonyl compounds here. Let's look at the different IR signals and see if we can explain the signals. Here we have a bunch of compounds
that contain the carbonyl, and we would expect to find
the signal for the carbonyl at a wave number in a
range of 1850 to 1650, so somewhere in that range
we'd expect to find the signal. And the way I like to think about this is to kind of divide that in half. So 1850 and 1650, so
somewhere in the middle would be about 1750, and if you go a little
bit lower than that, 1740, I like to think about that
as being the average signal. So therefore, an average value for K, or an average carbonyl strength, alright? So if we go a little bit higher than that, so we're looking at an estronal, in the video on the reactivity or carboxylic acid derivatives, I told you how to think
about the competing effects of resonance and induction. And for an ester, the inductive effect is a little bit stronger than the resonance. So if the inductive effect
is a little bit stronger than resonance we've just
seen that has the effect of increasing the
strength of the carbonyl. So we're going to increase
the value for K a little bit, and if you increase the value for K, you increase the wave number
where you find the signal. So if you increase K, you
increase the frequency of bond vibration, so you
increase the wave number, and so the wave number
will go up a little bit from 1740 to approximately 1745. So again, this is approximately
where you'd find the signal for the carbonyl for an ester. Next, let's do an acid anhydride. So for an acid anhydride,
the inductive effect is even more important. And if the inductive effect
is even more important, so if we think about this oxygen here as being very electronegative, that means that we're gonna
get an even higher value for K. So an even higher value for K, an even stronger carbonyl. And so if we have a higher value for K, we have a higher frequency of vibration, we expect to find the signal
at a higher wave number. And so the wave number goes up to approximately 1760 to 1790. And this is the first signal
for an acid anhydride. So this is actually the
symmetrical stretch. So we have two carbonyls, right? And if they're stretching
in phase with each other, that's the signal that
we would expect to find. So we also have an asymmetric stretch, so we're actually going
to see a second signal for an acid anhydride, we talked about this
in the previous video, it takes a little bit more energy for the asymmetric stretch, and so you find that at
a higher wave number. So approximately 1810 for
the asymmetric stretch. So two signals for the acid anhydride. For an acyl or acid chloride, the inductive effect completely dominates the resonance effect,
so even more strongly than the previous two examples. So the chlorine is withdrawing
some electron density, we're strengthening
the carbonyl even more, so the force constantly goes up even more. We have a higher frequency of vibration, we get a higher wave number. So the signal for this carbonyl appears at a higher wave number,
so approximately 1800, or even a little bit higher
than that, so 1815 or so. So once again, that's
the approximate value for the signal of the carbonyl
for an acyl or acid chloride. Alright, let's go in the other direction. So if this is our average spring constant, and pretty close to our average
signal for our wave number, let's look at an aldehyde next. So here we have an aldehyde, right, we have a hydrogen here,
so we don't have an electronegative atom to think about. In the previous examples we had oxygen, we had oxygen, we had
chlorine to think about, but here we have hydrogen, so we're not really worried
about the inductive effect here. We're more concerned with
the electron donating of the alkyl group, right? So we have an alkyl group right here, and alkyl groups are electron donating. So if we're donating some electron density in this direction, we're
gonna lost a little bit of electron density from our carbonyl. So, like charges repel
here, and therefore, we're decreasing the
strength of the carbonyl. So we're decreasing the force constant K, and if we decrease the
force constant of K, we would expect the signal to be at a lower wave number here. So the average value would go
down to approximately 1725. And again, all these are
just approximate values, I'm not saying it has to be exactly that, just somewhere around that. We would expect to find the signal for the carbonyl for an aldehyde. Comparing an aldehyde to a ketone, a ketone has two alkyl groups. So we have even more electron donating, and so we're going to weaken
our carbonyl even more. And so if we're weakening our carbonyl, we can think about K decreasing
even more than before, and so we should expect
the signal for the ketone to be at a lower wave number. And it does go down to
approximately 1715 or 1720. And so, once again, this is pretty close to where you'd find the
signal for a ketone here. Alright, let's talk about
carboxylic acids next. And so, if we have a carboxylic acid, let's think about it, let's
draw another one here. If we have a carboxylic
acid in the dimeric form, so we have another carobxylic acid here, we can get some pretty
strong hydrogen bonding. So we get some pretty strong
hydrogen bonding here. And that's gonna have the effect
of weakening the carbonyl. So you can think about
some electron density moving into here because of
your strong hydrogen bonding. And so that turns out
to decrease the value for the force constant even more compared to our average value. And so, for the carboxylic
acid in the dimeric form, we would expect to see
it somewhere around 1710, although the signal can definitely vary for a carboxylic acid, so you
might not see this exact value somewhere else because
everything depends on, on things like what form is it in. Here we're talking about the dimeric form. Finally, we have our amide here. And, a carboxylic acid derivative, so we think about
resonance versus induction, and this is the example
where resonance dominates. So resonance dominates induction, and we have to think
about resonance structure. Move these electrons in here,
we push those electrons off, so we have the effect of
weakening the carbonyl. So we're decreasing the
strength of the carbonyl, and resonance dominates, so
we're gonna decrease the value for K a lot here, so we're gonna decrease the value for K a lot compared to our made-up average value. So we're gonna decrease the
signal for the wave number. So we're gonna get a much
lower wave number here. So approximately 1650's to 1690. So again, these are all
just approximate values for where you would find your signal. But, thinking about
resonance and induction, and electron, and donating alkyl groups, and hydrogen bonding, allows
you to approximate the signal, and therefore approximate
the strength of the carbonyl.