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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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# Bonds as springs

Learn how frequency of bond vibration can be compared to the oscillation of a spring by using Hooke's law. Explore how the strength of the bond and the mass of the atoms affect the frequency of vibration. The concept of reduced mass is introduced for situations where both atoms in a bond are moving. Created by Jay.

## Want to join the conversation?

- At11:30, it is said that stronger bonds vibrate faster. So does a C double bond vibrate faster than a C-H bond?(8 votes)
- No. A C-H bond vibrates almost twice as fast as a C=C bond — 3000 cm⁻¹ vs. 1700 cm⁻¹.

If you have atoms of the same type, the stronger bond vibrates faster.

C≡C > C=C > C-C (2200 cm⁻¹ > 1700 cm⁻¹ > 1200 cm⁻¹)

The vibrational frequency of a vibrating two-body system depends on both the force constant and the reduced mass of the system: ν ∝ √(k/µ).

The force constant depends on the strength of the bond. A C=C bond is about 1.5 times as strong as a C=C bond. That would make it vibrate faster than a C-H bond.

But µ(C-C) = 6, while µ(C-H) = 0.92. The reduced mass of a C-C pair is about 6.5 times that of a C-H pair. The greater reduced mass makes the C=C bond vibrate much more slowly than a C-H bond.

A C-H bond vibrates faster by a factor of about √(6.5/1.5) ≈ 2.(15 votes)

- I don't understand why at8:07we are we using a reduced mass instead of say, m1 + m2(5 votes)
- In a vibration, it's not just one atom that moves back and forth. The centre of mass must stay in the same place.

Thus, as one atom moves in one direction, the other atom must move in the opposite direction.

This is what the physicists call a "two-body vibration" problem, in which the two atoms vibrate about a common centre of mass.

The position of the centre of mass is determined by the reduced mass of the system.(15 votes)

- At10:05, if increasing the reduced mass results in a lower frequency of vibration, why does an O-H bond stretch at a higher wavenumber than C-H? Shouldn't the increased reduced mass of O-H make the wavenumber go lower?(3 votes)
- The frequency also depends on the force constant k, which corresponds roughly to the bond strength.

An OH bond is stronger than a CH bond. Its effect overpowers the effect of the reduced mass.(8 votes)

- How true is the assumption that a double bond is twice the strength of a single bond?(4 votes)
- From looking at a tables of bond dissociation energies --

Alkane C-C 347–356 kJ/mol

Alkene C=C 611–632 kJ/mol ~1.8x stronger on*average*

Bond energies for carbon oxygen bonds are similar:

Typical C-O ~360 kJ/mol

Typical C=O ~750 kJ/mol ~2.1x stronger

However for bonds between nitrogen atoms:

N-N ~160 kJ/mol

N=N ~420 kJ/mol ~2.6x stronger

So, twice is a very rough approximation.

Note, however, that the context in which a bond occurs can have a large effect on bond strength (e.g. a carbonyl in an acyl chloride is much stronger than the carbonyl in an amide).

Some useful sources of data:

https://chem.libretexts.org/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_Chemistry%3A_The_Central_Science_(Brown_et_al.)/08._Basic_Concepts_of_Chemical_Bonding/8.8%3A_Strength_of_Covalent_Bonds

https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf(4 votes)

- Would it be correct to assume that sound wave resonance works on the same principal? As a musician, main instrument being the guitar, I see it as being analogous to guitar string physics and for me personally, it would seem to be an easy way to make sense of it all. A heavier(thicker) string has a lower frequency and the tighter the string, the higher the pitch. If so, does harmonic resonance play a part in the resulting noise in the spectrograph?(5 votes)
- Can someone please direct me to the full calculations which resulted to the equation presented at3:10(a name, or a link)? Thanks.(3 votes)
- I can understand your interest here but it probably wouldn't be very helpful to know how that equation was derived. This is because in this series of videos Jay is explaining everything in terms of classical physics but if you were studying this at a much deeper level then you would need to learn about the quantum physics of the vibrations and this would involve different, and more complex, equations which are derived from the Schrodinger equation. The classical physics approach serves as a good model for the quantum physics of vibrations but it has its limitations. You would not gain anything to help you interpret IR spectra by understanding how the classical equation at3:10is derived.(5 votes)

- I don't quite understand what the K-value is representing in regards to the bonds between atoms. When talking about springs, k is the spring constant, right, meaning the stiffness of the spring and what not. What is it really for bonds though?(4 votes)
- I'm guessing that it is related to both the density and focus of the electron "cloud(s)" that form the bond(s).

(I'm thinking of the masses as representing the nuclei with the spring representing the electron "cloud".)(2 votes)

- How should I think of this "bond stretching" in terms of the MOT ? Would it be accurate to think of it as a certain "deformation" of the molecular orbital ( the electron density around the bonded atoms in the molecule) ?

And also how should I think of this in terms of the VBT where bonds are just overlapping between the orbitals ?(3 votes)- I don't know for sure, but I'm thinking of this as the masses corresponding to the nuclei and the spring corresponding to the electron "cloud" (i.e. the bond).(1 vote)

- How are force constants of bonds calculated(2 votes)
- Will someone kindly derive the expression of reduced mass?

P.S.: I know Calculus only for one variable - integration, differentiation, limits, and simple first order differential equations, but no partial derivatives or line integrals. So please keep the derivation simple.(2 votes)

## Video transcript

- [Voiceover] In the last video, we said that certain
frequencies of IR radiation can cause the bond to stretch. And we can think about
the bond as a spring. So this carbon-hydrogen bond
can be modeled as a spring. If it's a spring that obeys Hooke's law, the stretching vibration of the bond is like a spring-mass oscillation. Let's look at some classical
physics here, some mechanics, to see if we can understand
this a little bit better. Here I have a box. So here we have a box of mass m, so this box is mass m here. And there's no friction between
the box and the ground here. But the box is attached to a
wall right here by a spring. So, if you grab this box and
you pull this box to the right, so you're going to apply
a force to the right, and so you're going to stretch the spring. So you move the box over
to here, and the spring would obviously stretch
out tho there, like that. Let's say you moved the
box a distance of delta x. From here to here, we moved
the box a distance delta x. The box is going to feel a
force pulling it to the left. So if you're just holding it there, you're holding it because
there's this force of the spring that wants to
pull the box back to the left. So this is the force of the spring. And according to Hooke's
law, the force of the spring is equal to negative kx. And the negative sign just
means a restoring force. So this negative sign here
means a restoring force, meaning, the force of the spring is back towards this original position right here. K is called the spring constant. So this is the spring constant right here, and it depends on how strong
or weak the spring is. So if you have a very stiff
spring, or a strong spring, you have an increased value for k. If there's a loose
spring, or a weak spring, a decreased value for k. Let me go ahead and write that down. Strong spring or stiff
spring increased value for k. If you have a weak or loose
spring, a decreased value for k. X refers to the displacement from the original position here. We displaced the box delta x here, so that's what we're talking about here. Let's think about what these terms mean for the force of the spring. If you have a strong
spring, the box is going to experience a stronger force, right, according to Hooke's law. If you increase k, you increase
the force of the spring. And also, the more you stretch it ... All right, so if you increase
the value for delta x, also, the stronger the force
of the spring would be. And so, if you get into the physics of ... We're not going to get too far into it, but you could set this equal
to mass times acceleration, because negative kx, F is equal to ma, that's equal to ma, and the acceleration is the second derivative of the position, so you could write the
acceleration as being the second derivative of the position. I won't get into the physics of it, but eventually, you can solve for the frequency of oscillation
for this spring-mass system. And when you do that math ... Let me go ahead and write
down what you would get ... You're going to get the
frequency of oscillation is equal to one over two
pi square root of k over m. Let's think about what the
frequency of oscillation is referring to. If you, once again, you
pull the box to the right, so we're starting with our box
at this position right here, and then, if you release it,
think about what happens. What's the motion of the box? Well, the force of the spring is going to cause the box to go towards
this direction, right? It's going to keep going,
right, because of energy, and it's going to go all
the way over to here. So now it's going to compress the spring. And now there's a force of the spring back towards this direction, like that. Of course the spring is now compressed, all the energy is stored in
the compression of the spring, it pushes on the box
and the box, therefore, goes back towards the
equilibrium position, the center. But it's going to keep
on going until it reaches this position where we started,
so the original position. That's like one oscillation. And the time it takes for one oscillation is called the period. So that would be the period. The period is measured in seconds. So the period is measured in seconds, the time it takes for one oscillation. Right, so one over the period ... One over the period is
equal to the frequency. So you could write frequency like that, or you could write frequency like this. And so the frequency would be equal ... The units would be one over seconds. And this is talking about the number of oscillations per second. So frequency is number of oscillations, number of oscillations per second. So what affects the frequency? Well, once again, the spring constant affects the frequency, so k. If you increase the value for k, you're going to increase the frequency. If you have a strong spring, it's going to cause that mass to oscillate faster. Let me write that down. Increase k, all right, increase
the strength of the spring, you increase the frequency. What about the mass? If you increase the mass,
what happens to the frequency? If we increase this number, just think about it mathematically, right? That would decrease this number. So an increase in the mass, right, would decrease the frequency. You wouldn't get as many
oscillations per second, you get a slower oscillation. And so this is what we're thinking about when we're treating this bond as a spring. So let's go back up here
to this diagram, right, where we have the bond as a spring. So let's think about keeping the carbon stationary for now. So we keep the carbon stationary,
and we're going to pull, we're going to pull on the hydrogen here. So we're going to pull to the right. Just let me go ahead and draw a line. So we're going to pull to the right and stretch the spring out, right? So we're going to pull the
hydrogen here to the right. So the force that we apply
is in this direction. So the hydrogen feels a restoring
force in this direction. That's the force of the spring. And so, once again, if you
have a really strong bond, right, so a really strong bond, that means you have an
increased value for k, right? So a stronger bond means a stronger spring constant, if you will. So you're going to have an increased frequency of oscillation. So after you release this hydrogen, it's going to oscillate in a way analogous to this spring-mass system here. Once again, if you increase
the strength of the spring, you increase the frequency. What happens if you
change the mass, right? What happens if you ... Let me use a different color here. What happens if you change the mass? Instead of hydrogen, if
you change it to carbon or oxygen, or something maybe
has more mass than hydrogen? Right, what would happen to
the frequency of oscillation? If you increase the mass, you decrease the frequency of oscillation. This is a pretty good model,
if you think about it this way. However, we're talking about only the hydrogen moving this time. But we know that, when we're talking about a stretching vibration, both of these, both of these are moving. So let's get some more space and let's deal with that next. We have this situation
where we have two masses. I'm just going to generalize it now. So we have m one, and then
we have m two over here and the bond between them. So m one and m two are
the masses of the nuclei of the atoms that we're talking about. So both masses are actually moving in our situation, so we have to amend the equation for frequency slightly. So let's go back up here and let's look at this equation again. The frequency of oscillation is equal to one over two pi square root of k over m. But this is assuming
only one mass is moving. We have both masses moving when we're talking about
a stretching vibration. We have to use something different for m. We use what's called the reduced mass. Let me go ahead and write it down here. The frequency is equal to one over two pi square root of k over m, but we can't write m any more because of the slightly
different situation. We're going to use that
symbol to represent the reduced mass. The reduced mass is equal to the m one times m two
divided by m one plus m two. And we're talking about
the mass of the nuclei in AMUs, we'll worry about
units in the next video. And you can use the atomic mass to get an approximate value
for the mass of the nuclei. Let's think about a carbon-hydrogen bond. So a carbon-hydrogen bond, that would be like m
one is equal to carbon, and m two is equal to hydrogen. What is the reduced mass equal to? The reduced mass is equal to ... Well, the atomic mass of carbon is 12, so it'd be 12 times the
atomic mass of hydrogen, which is one, all right? Divide that by 12 plus
one, so let's do that math. Let's go ahead and do that really quickly. 12 times one is 12,
divided by 12 plus one, which is 13, and so we get .923. We get .923, so that's the
reduced mass of our system. Let's do another one. Let's do carbon-carbon this time. So carbon-carbon, the
reduced mass is equal to 12 times 12 over 12 plus 12. So what is the reduced mass equal to now? 12 times 12 is 144, divide that
by 12 plus 12, which is 24, so we get six. So the reduced mass is equal to six here. Let's think about what that does to the frequency of vibration, all right? So if we increase, if we
increase the reduced mass, so if we're going from
a reduced mass of .923 to a reduced mass of six,
we're increasing this. We're increasing, we're
increasing the reduced mass. What happens to the
frequency of vibration? Well, if we increase this value, of course we're going to decrease the frequency of vibration. So we would expect a
carbon-carbon single bond to have a lower frequency of vibration than a carbon-hydrogen single bond. Let's do this idea one more time, except, let's think about ... Let's think about a double bond. This was a carbon-carbon
single bond, right? What about a carbon-carbon double bond? Well, the reduced mass
would be the same, right? It would still be six. But what happens to the spring constant? What happens to the force constant? Well, for a carbon-carbon double bond, we can pretend like
this is twice as strong as a single bond. So if the value for the single bond ... If the spring constant
was k for the single bond, for the double bond, it would be two k. It would be twice that. So we're increasing the
value of the spring constant. We're increasing k because a double bond, we're assuming a double
bond is twice as strong as a single bond. So what happens to that
frequency of vibration if you increase k? If you increase this value, you're going to increase this value. So if you increase k,
you're going to increase the frequency of vibration
because you have a stronger bond. So the bottom line is, the
things you have to remember, are, stronger bonds vibrate faster. All right, so that's what
this is saying right here. A stronger bond, increased
k, you get an increased, increased frequency of vibration. So a stronger bond vibrates faster. And what about a lighter atom here? So if we're thinking about
a hydrogen, all right ... So a hydrogen, a lighter atom, that has a smaller value
for the reduced mass. So that's going to vibrate
faster than a heavier atom. So it's, stronger bonds vibrate faster, and lighter atoms vibrate faster, too. And this is what we
gather from thinking about bonds and springs. These are two very important
points to think about. What affects the frequency of vibration would be the strength of the bond and also the mass that
we're thinking about here, the reduced mass.