- 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
How dipole moment determines the intensity of a signal in an IR spectrum. Created by Jay.
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- Can you please elaborate more on the reason why changes in dipole moments correlates with the absorbed IR intensity?(9 votes)
- The light has an oscillating electrical component. If the frequency of this electrical vibration matches the vibration frequency of the bond dipole, the two systems can exchange energy.
The greater the dipole moment, the greater the amount of energy exchanged.
It's like two runners in a relay race. It's easier to pass the baton when they are both going at the same speed.(11 votes)
- If you don't see a signal in the double bond region for symmetrical molecules like the last example in the video, do you still see some peaks above 3000 to indicate a =C-H ? The drawing looks right on the border of 3,000 but its hand drawn so could go either way. Thanks!(5 votes)
- Yes you would, because the bonds surrounding the C-H bonds are not symmetrical and therefore have a dipole moment.(1 vote)
- why does a molecule need to have a dipole moment for IR spectroscopy to work?(2 votes)
- IR works because the oscillating electric vector of the infrared radiation must match the frequency of the oscillating dipole moment in the molecule for energy exchange to occur.
No dipole moment means no energy exchange.(5 votes)
- Around time5:13you show the last example. I noticed that there are both sp3 and sp2 hybridized C-H bonds shown in the spectrum (around 3000 1/cm). But, how there can there be sp2 hybridized C-H bond signals when there are not any hydrogens bonded to the sp2 carbons?(2 votes)
- The peak looks like it might be an alkene C-H stretch, but it is really an alkane C-H stretch.
When you assign a peak to some type of vibration, you should look elsewhere in the spectrum for confirmation.
In this case, there should be a fairly strong alkene C-H bending vibration somewhere in the 1000-650 cm⁻¹ region.
That peak is missing, so there is no alkene C-H.(4 votes)
- How come C=O has a greater frequency then C=C when O is heavier than C?(3 votes)
- Great question.
C=O has a larger dipole moment than C=C, meaning C=O has a stronger bond strength. This is because of the difference in electronegativity values of C and O.
As you already know, a greater mass of the atoms of a bond means a lower frequency. But C=O has a greater bond strength, which apparently compensates for the difference in mass.
Simply, the increase in bond strength is larger than the increase in mass.(2 votes)
- If the signal isn't present for a symmetrical alkene then how do we know it's there?(2 votes)
- Why are only the vibration that produce a change in the dipole moment observed on the spectra? Why not all vibrations?(2 votes)
- The change in the dipole moment it's just a way for the equipment to detect how much radiation of a certain wavenumber the molecule absorbed. But not all IR techniques measure the change in dipole moment.(2 votes)
- How a composite signal be decomposed in multiple sine waves?(1 vote)
- Is there a way to figure out what the spike around 3000 wavenumbers in the first example was caused by? Thanks in advance!(1 vote)
- [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.