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### 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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# Signal characteristics - wavenumber

Learn how to use equations to predict the wavenumber for different types of chemical bonds. See how the frequency of bond vibration in molecules, likened to a spring oscillation, depends on the force constant and reduced mass. Learn how to calculate wave numbers for different bonds using these factors, as you develop your understanding of where signals appear in IR spectroscopy. Created by Jay.

## Want to join the conversation?

- At6:55, it is mentioned that the k value for the single bond between C - O will be 5x10^5, which was the same k value for the bond C - H. Could anyone tell me where this number was taken from? I understand that a double bond would be double the constant but where does the original approximation come from?(21 votes)
- The approximation comes from modelling the bonds as a spring: https://www.khanacademy.org/science/organic-chemistry/spectroscopy-jay/infrared-spectroscopy-theory/v/bonds-as-springs

The amount of energy that a spring can hold varies depending on the type of spring. To account for this, springs are multiplied by a constant value.

The potential energy stored in a spring is given by the equation:

U=½kx^2

where U is equal to potential energy, x is displacement, and**k is the spring constant**.

k = "stiffness" of the spring.

Stiffness describes the rigidity of an object. A double bond is stiffer than a single bond, thus k increases with increasing bond order.

**References**

single bond between carbon and oxygen, k = 5 x 10^5 dyne/cm

double bond between carbon and oxygen, k =10 x 10^5 dyne/cm

triple bond between carbon and oxygen, k = 15 x 10^5 dyne/cm.(16 votes)

- at1:59, why do we divide by avagadro's number to turn grams to amu? If we have grams written as k/grams (spring constant/reduced mass) then shouldn't we multiply by 1gram/6.022 X 10^23 amu to cancel out the grams and be left with only amu?(10 votes)
- I had the same question before, so I watched this and the last video several times. Then I noticed he did make a mistake, but it is not about dividing or multipling the Avogadro number. On the last video https://www.khanacademy.org/test-prep/mcat/physical-processes/infrared-and-ultraviolet-visible-spectroscopy/v/bonds-as-springs at8:25, he said that the unit of the reduced mass is amu, not gram. However, he said it is gram in this video by mistake. Actually what he intended to do is to convert amu to gram, which requires dividing the Avogadro number. That makes sense because the unit of K is N/cm, N is g*cm/s^2, so the actual unit of K is g/s^2, and if we convert the unit of M from amu to gram, the only thing left inside the square root is 1/s^2. finally, the unit of 1/c is s/cm, so we could get the correct unit for the wave-number which is 1/cm.(8 votes)

- why do we use the reduced mass? Perhaps I missed it, but how is this derived? Thank you!(5 votes)
- This comes from the previous video, the link is below

https://www.khanacademy.org/science/organic-chemistry/spectroscopy-jay/infrared-spectroscopy-theory/v/bonds-as-springs

It's essentially because a Force is being applied to both masses attached to each end of the spring and not just one mass attached to one end of a spring.(7 votes)

- How can we know the value of K the force constant?(7 votes)
- Where can we get more ks for various different bonds?(6 votes)
- The speed of light in this video was recorded as 3x10^10 instead of 3x10^8.(3 votes)
- I believe he was using units of cm/s. Therefore, since the speed of light is 3x10^8 m/s it would follow that the speed of light is 3x10^10 cm/s.(7 votes)

- where is the region for a double bond to an atom that's not carbon?(3 votes)
- The point of IR spectroscopy is to identify functional groups. So an alkene or the carbonyl of a carboxylic acid would be the two double bond groups frequently seen. Carbonyl's typically show up at 1700 1/cm.(2 votes)

- why use wavenumber instead of wavelength ?(3 votes)
- As we look for the electromagnetic radiation spectrum and compared different regions of it with respect to energy, frequency, and wavelength.

Then, the energy, frequency both are directly proportional to each other whereas energy, frequency both are inversely proportional to wavelength.

e.g. In the case of UV-Visible spectroscopy, at 200 nm (shorter wavelength) the energy and frequency of UV-Visible radiation are high as compared to 800 nm (Longer wavelength).

Now, wavenumber expressed as reciprocal to wavelength is expressed in centimeters. When wavelength converted to wavenumber, then, Wavenumber is directly proportional to energy and frequency (In case of wavelength, it is inversely proportional to energy and frequency).

So, to use direction correlation, between wavenumber, energy, and frequency, wavenumber to be used in IR

eg. In the case of the IR region, the range is 2.5 to 25 micron. At 2.5 micron (wavenumber 4000 cm-1) the energy and frequency of IR radiation are high as compared to 25 microns (wavenumber 400 cm-1).(1 vote)

- so what is the value of k?(2 votes)
- The units of K is in dynes/cm because the reduced mass is in grams, correct?(2 votes)

## Video transcript

- [Voiceover] In the last
video we saw that the frequency of bond vibration
can be thought about like a spring oscillation and so
it's dependent on two things. It's dependent upon K, which
is the spring constant, or the force constant. And the reduced mass, where the reduced mass
is equal to the mass m1, at times m2, over m1 plus m2 and so here's your bond as a spring, with m1 and m2 on either end. And so, if you increase
the force constant, increase the spring constant, so that's like increasing
the strength of the bond, obviously you would increase the frequency just looking at the map here. So you increase the
frequency of bond vibration. So a stronger bond vibrates
faster than a weaker bond. And also, if you decrease
the reduced mass, right? So if you decrease the reduced mass, mathematically that's
also going to increase the frequency of
vibration here, like that. In the first video, we talked about relating the
frequency to wave numbers. So a wave number, you take the frequency
divided by the speed of light, in centimeters per second. So if you take this over here and just divide by the speed of light, which is C, you're going
to get wave numbers. So this is good to use to
approximate the wave number, where you would find your
signals in IR spectroscopy. All right, there's one more
thing we have to do though, to get the proper units for wave number and that's because here, the reduced mass is in grams and so we need to make that
atomic mass units, or AMU. So we can divide by
Avogadro's number to get that. So I'm not too concerned with units here. I'm just going to show you, real quickly, where this equation comes from. Wave number is equal to one over two pi C, times the square root of
K, over the reduced mass and to get our proper units, we have to divide by Avogodro's number which is 6.02 times 10 to the 23. All right, so let's continue here, that'll be one over two pi
C times the square root. We'll make a little bit of
a long square root sign here because we still have the force constant over the reduced mass. And now we can just move
Avogadro's number up to the numerator here, so that's
6.02 times 10 to the 23. Let's go ahead and find these square roots of Avogadro's number. So the square root of Avogadro's number is equal to... So we have square root of
6.02 times 10 to the 23 and we get, that number is
7.76 times 10 to the 11. So let's go ahead and
write that down here. So now we wave numbers equal
to 7.76 times 10 to the 11, divided by 2 pi, C is the speed of light and centimeters per second. That's 3 times 10 to the tenth. And then we still have the square root of the force constant
over the reduced mass. So let's go ahead and
solve it even further here. We need to divide that number by two. We need to divide that number by pi. And we need to divide that
number by the speed of light in centimeters per second. So divide by 3 times 10 to the tenth and we get 4.12 and so we arrive at this wave number as equal to 4.12 times the square root of K over the reduced mass. So here we have a nice little equation where we can approximate the wave number for different bonds. So where we would expect to see the signal for different bonds stretching. And let me use a different color to point these things out here so, K, call this the spring
constant or the force constant. The units are dynes per
order, dynes per centimeter and then we have the
reduced mass here in AMUs. So let's do some calculations and let's see if we can
approximate where we would find the signals for some bonds. And so let's get some more room down here. And let's start with a
carbon hydrogen bond. So we have a carbon hydrogen bond here. First thing we could do is
calculate the reduced mass, so reduced mass is equal to... Atomic mass of carbon is 12, mass of hydrogen is one, so 12 times 1, over 12 plus 1. And we did this calculation
in the previous video and we got .923. So we're going to plug this into our equation for wave number. So wave number is equal to
4.12 times the square root of the force constant, for a single bond. We're talking about a
carbon hydrogen bond here. It's a single bond. You can use a force constant
of 5 times 10 to the 5, dynes per centimeter. And we're going to divide that by 0.923, the reduced mass and so let's get out the calculator and do that calculation. All right, so we have 5 times 10 to the 5. We're going to divide
that number by 5.923. We're going to take the
square root of our answer here and then we're going to
multiply that by 4.12. And this gives us 3032,
so 3032 wave numbers. 3032, units would be 1 over centimeters. So this is what we calculated, so this is the approximate wave number. So this is approximately
where you would find the signal, for a carbon
hydrogen bond stretching. So if you're looking on an IR spectrum and the actual signals is
pretty close to this number, so this is a good approximation. And that's why it's kind of useful to use this equation here. Let's do it again for carbon oxygen. So it's still a single bond but this is carbon oxygen this time. So the reduced mass is going to change. But the force constant
is going to stay the same because in both cases we're talking about a single bond here. We're going to pretend like they're exactly the same strength. Which, just to help us
for our calculations to approximate things. All right, so the reduced
mass would be equal to, carbon is 12, oxygen is 16. So we have 12 times 16 over 12 plus 16 and let's do that math really quickly. So we have 12 times 16, which is 192, divided by 28, gives us about 6.9. So we're going to say the reduced mass is approximately 6.9. And let's calculate the
approximate wave number. So where would we expect
to find this signal? So 4.12 times the square root-- We're going to use the same number. We're going to use 5 times 10 to the 5. So we're saying we have
a single bond here, so again, these are just approximations. So this is 5 times 10 to the 5. We're going to divide
that by the reduced mass which is 6.9 and let's see what we get for the wave number. So let's do that math. So we have 5 times 10 to the 5, going to divide that by 6.9. We're going to take the
square root of our answer and then we're going to multiply by 4.12 to get the approximate wave number, which is 1109. So we have 1109, 1 over centimeters, for our wave number here. And let's compare these two wave numbers. So what did we do in our calculations? Let me use blue here. So what did we do? We switched out a hydrogen for an oxygen. So we increased the
mass of the second atom, or the nucleus, if you want
to think about it that way. So you increase the
mass of the second atom and what happened to the wave number? The wave number decreased. We went from 3032 to 1109 and noticed what we did, right? The reduced mass was .923
and it went up to 6.9. So we increased the reduced mass and we decreased the frequency, or we decreased the wave number. And so again, this is approximately where you would find the signal for a carbon oxygen single bond. Not exact, because
obviously we're just using really simple numbers here but it's pretty close. All right, let's do a carbon
carbon double bond now. So carbon carbon double bond. Let's calculate the reduced mass. The reduced mass would be 12 times 12, over 12 plus 12. And when you do that math,
you're going to get six. So I think we did that in
the previous video too. Let's plug this into our equation. So the approximate wave number would be equal to 4.12
times the square root. All right, so what are
we going to plug in now for the force constant? Well, we have a double bond here. We have a double bond, so a carbon carbon double bond. And if we just approximate and say that a double
bond is twice as strong as a single bond, we can
just take this number and multiply it by two. So that would be 10 times 10 to the 5. So we're just saying that
a double bond is close to being twice as strong and again, not a perfect number, but it's a good approximation and it just gives us
an idea about where do these wave numbers come from. Now, we solve for the wave number, the approximate wave number here. So we have 10 times 10 to the 5. We're going to divide that by six. We're going to take the
square root of our answer and then we're going to multiply by 4.12 and this is going to give
us somewhere around 1682. So our calculation gives us 1682, where we'd find the signal. Again, not perfect, but a decent approximation. Not a perfect number but pretty close to where you would find the signal for a carbon carbon double
bond on an IR spectrum. Let's do one more. One more, let's get some
more room down here. Let's see if we can squeeze in one more calculation. Let's do a carbon triple bond. So a carbon carbon triple bond. The reduced mass obviously
would be the same as above, at six. So we're going to be changing
the force constant here. So the wave number would be
equal to 4.12 at two times. So what would we plug in for
the force constant now, right? This would be... This was double a single bond, so we need to go for triple a single bond. We have a triple bond here. So 3 times 5 times 10 to the 5. So that's 15 times 10 to the 5, so this would be three, we're approximating this,
saying it's three times as strong as a single
bond, 15 times 10 to the 5. We're going to divide that by six. And let's see what happens now. So we have 15 times 10 to the 5. We're going to divide that by six. We're going to take the square root of our answer once again. So the square root of our answer and we get 500, when you
multiply that by 4.12 and we get 2060. So we get 2060 here for the wave number, so however you want to write this. So what happened? What happened between
these last two example? We increased K. That's the only thing that changed between these two calculations. We increased K, we increased
the force constant. We're saying a triple bond is stronger than a double bond. And what happened to the frequency? Or the wave number? The wave number increased. So increase K, you increase the frequency, you increase the wave number here. Finally, let's plot
these on an IR spectrum. So let's first start with the
carbon hydrogen single bond. So somewhere in the 3032, so let's go down here for our spectrum. So 3032, so that would be somewhere around in here. So this would be
approximately where we would find the signal for a
carbon hydrogen bond. And this just allows us to think about regions of our IR spectrum. So somewhere in this region, in this region right in here, is where you would find
a bond to hydrogen. So we just said this was
carbon to hydrogen here but you could generalize
this by saying it's any bond to hydrogen because it's the hydrogen,
it's the smaller mass, that causes the increased frequency. The increased value for the wave number. Let's look at our next one here. Let's look at carbon oxygen. So for carbon oxygen, we have 1109 so that's a big difference. It's a single bond. 1109, that's way down here. So 1109 would be somewhere around in here for this carbon oxygen bond. And so, it turns out that
this region right in here is where youwould find,
it's a single bond region, so this is the single bond region, and when you're not
talking about hydrogen. So this would be the single bond region. This over here would be the
bond to hydrogen region. And then we have two more regions to talk about, right? We have our double bond region, let's use green for that. So carbon carbon double bond, we calculated it's
approximately 1682 wave numbers. So let's find that on our spectrum here. So 1682, so that would
be approximately in here. This would be like six,
somewhere in there, in that range. And so we could say
that it's approximately where we would find the signal for a carbon carbon double bond. And your double bond region... All right, so let me write,
"double bond region." Double bond region is
right around in that area. So double bond region. Expect to see a signal for a double bond in this area here. So somewhere in here. So this is approximately
your double bond region on IR spectrum. And then finally, our last thing that we're going to think
about is our triple bond. So our triple bond, we calculated an approximate value for a
carbon carbon triple bond. We got 2060, so 2060 would
be somewhere in here. So that's approximately where we would find our triple bond region. And that usually does
go about 2100 to 2300, so in here somewhere so pretty small but just to give you an idea of these different regions. So here's the triple
bond region, triple bond. So hopefully that allows
you to think about why you get these different wave numbers. It has to do with two factors. It has to do with the force constant and it has to do with the reduced mass. If you think about those two factors you can think about where the signal should appear for these bonds, for these bonds stretching. So you can figure out the
approximate wave number. And you should know these wave numbers because it's going to help you when you read your IR spectrum.