# Absorption Band and Wavenumbers in Infrared Spectroscopy

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This online lecture. Were going to discuss the relationship between wave numbers and the the types of bonds. Giving rise to those wave numbers and heres what we to get out of this number one the stronger the bond.
The higher the frequency or wave. Number we also want to know number two the lighter the atoms the higher the frequency or wave number and by the way what i mean here by higher frequency means higher frequency of bond vibration. Which would therefore give a higher wave number on the ir spectrum.
And were also going to see number three here the broadening of an oha absorption band is due to hydrogen bonding. Were also going to see number four electron delocalization that is resonance can affect the wave number of a functional group and lastly. Were going to see that electron withdrawing or donating can affect the wave number of a functional group as well so lets go back to our wave number chart here all were doing here is trying to explain why all of these bonds peak roughly at these corresponding wave.
Numbers notice. They all vary one of the ways. We can explain it is using an equation you might have learned in physics called hookes law.
If you do remember this from physics. You might remember that it was related to a spring and think about it its kind of like how were conceptualizing bonds in organic chemistry. Were thinking of them as springs vibrating and stretching and bending and wagging so thats why this spring equation is appropriate.
But in our particular case we want our equation to equal wave number if you learn this equation in physics you probably saw it equal to frequency. But remember frequency and wavelength are proportional as one goes up so does the other now hopefully if you did take physics. Your professor derived this equation for you for organic chemistry.
Its way beyond the scope of this lecture. Were just going to accept this equation as true. But what i wanted here is point out some of the variables in the equation and connect them to organic chemistry.
Like for instance. In this equation. You have an f in this case f is equal to the force constant or in other words the bond strength other items in this equation you have our mass you have both m 1 and m 2.
For us these are the masses of the atoms and remember every bond. Technically connects two atoms so thats why theres an m one and an m two. Now with this understood we can see from our equation.
That number one the stronger. The bond that means the higher the frequency or wave. Number again.
What were saying. Here is that the bigger the f value is the bigger. The wave number lets also look at the relationship between mass and wave number if you study this equation youre going to see that the lighter the atoms also the higher the frequency or wave number notice.
We can see that right here in the equation. We have mass one being multiplied by mass to notice that would give a bigger number than what we see in the numerator. Which is when were adding mass one to mass two this means that the smaller m 1.
And m 2. Are the greater the wave number or we can say theyre inversely proportional to each. Other lets look at some examples of what were trying to see here notice here.
We have c triple bonded to cc double bonded to c. And then carbon singly bonded to carbon. If you remember from a previous online lecture.
We noted that single bonds are longer than double bonds. Which are longer than triple bonds. So the trend is as we move up we go from a longer bond to a shorter bond when we first learn this we also learned this relationship as well that is the shorter the bond.
The stronger. It is which means again the trend here would be weakest on the bottom two strongest bond on the top and what were trying to learn in this online lecture is the relationship between bond strength and wave number and we saw through hookes law. That this would be the relationship here that the sea singly bonded to another carbon would have the lower wave.
Number. And the c triply bonded to another carbon would have a higher wave. Number another way you could think about this is remember if triple bonds are shorter.
We could think of them as tighter or stiffer springs. Whereas if single bonds are longer. Theyre longer less stiff springs and a smaller tighter spring will vibrate faster than a longer lets say lankier spring.
This is how you could think about this concept without using hookes law and notice. Its this exact trend that we see on our wave number chart notice. The c.
Triple bond has a wave number at 2122 60. Heres the c double bond. It has a lower wave number however not on this chart is the c singly bonded to another c.
But we can rest assure that it would be even lower than the c double bond c. Now lets look at this example right. Here could we predict the relative wave numbers for a co bond compared to a ch bond well in the co bond.
We can say the carbon is bonded to oxygen. Which is a heavier atom than hydrogen. So.
The ch bond has a lighter atom bonded to carbon remember we saw in hookes law. That mass is inversely proportional to wave number therefore the heavier the atom. The lower the wave number the lighter the atom the higher the wave number so we should see the ch absorption being at a higher wave.
Number and sure enough going back to our wave. Number chart here here is the ch bond and here is the co bond. Notice.
The ch bond has a much greater wave number however while were looking at this lets talk about this. Oh a ch bond. Notice his intensity its not only strong.
But its also considered broad. What does that mean well this is the ir spectrum of a molecule that has a no h. We would see this band right here broad in this case means wide.
Why does this band appear broad well. Its simply due to hydrogen bonding and the more hydrogen bonding. You have the wider the intensity therefore.
Lets say if you have a sample that you put in the ir that has a very high concentration of o h. Then we can expect a nice broad peak between thirty two hundred and thirty five fifty if you run the same molecule but at a lower concentration you can see that range narrow. Now its only thirty five 92 3650 the lower concentration sample would have less hydrogen bonding.
Now lets take what weve learned here so far and apply to some problems here here sample problem one. It says in which molecule below would we observe a higher wave number due to the carbonyl group. So were focusing on the c double bonded to the oxygen.
But notice the difference here in these two molecules. This one on the right happens to have a double bond right here. Which means that this molecule on the right has some resonance.
Lets look at that resonance. Remember weve saw before that were allowed to move. Pi.

Electrons. And you could actually move these pi electrons. This way when you make that move these pi electrons.
Right here jump up on top of this oxygen notice what this does to the c double bond o. If we look at the resulting. Resonance structure.
It turns. It into a single bond. Now think about it would we get the same resonance for this molecule.
Over here. The answer is no therefore. We would say that this c double bond o.
Has more double bond character. Whereas. The carbonyl group in the molecule on the right has less double bond character.
Remember we discussed before in resonance. That if one resonance structure has a double bond. And the other resonance structure has a single bond.
Then the bond is neither single nor double. Its somewhere in between so. This molecule on the right has a c double bond o.
Thats partially single partially double. Which means if the molecule on the left has more double bond character it means that its a shorter bond and if the molecule on the right has some partial single bond character. It should be a relatively longer bond and remember we learned the relationship between bond length and bond strength and that is shorter bonds are stronger longer bonds are weaker and weve learned in this online lecture that the stronger the bond.
The higher the wave number the weaker the bond. The lower the wave number. So.
The answer is the carbonyl group on the left molecule would have a higher wave. Number which means this if we looked at the spectrum for each one of these molecules. Notice the top molecule right here his carbonyl peaks right about here notice.
It is peaking at a higher wave number than the molecule below. So notice sometimes you got a break out your resonance moves to compare wave numbers. Lets look at another example.
Here. Sample problem. Two in which molecule below would we observe a higher wave number due to the carbonyl group.
So same question here. But again. Lets look at resonance for the molecule on the left.
We can actually remember move. These. Lone pair electrons this way and that would force the pi electrons in the carbonyl group to jump up on top of this oxygen.
The resulting. Resonance structure would be this right here. So right away.
We can see our carbonyl group for the molecule on the left has some partial double and single bond character. But notice we get the same resonance for the molecule on the right again these electrons can jump up here. The pi electrons can move up on the oxygen.
And we end up with this resulting resonance structure. So at first glance. We would probably think well the wave numbers should be roughly the same.
But careful here these resonance structures. All rely on the fact that either the nitrogen or the oxygen has to give up electrons and move in that direction. And the question is which atom is more likely to give up the electrons or which atom is less likely to give up the electrons well remember think of electronegativity nitrogen is less electronegative and oxygen is more electronegative.
If nitrogen is less electronegative. Hes more likely to donate the electrons to the left whereas oxygen being more electronegative. Hes more likely to hold on to those electrons more tightly.
Therefore. Hes less electron donating that means the molecule on the left will therefore have a greater resonance and the molecule on the right would have less resonance. Which means.
The left molecule would have more single bond character for its carbonyl group. And the molecule on the right would have less single bond character for its carbonyl group. Unless single bond character remember means shorter bond and therefore higher wave.
Number. So. Our answer is the molecule on the right would have a greater wave number for its carbonyl group in fact again comparing them side by side notice.
You would see this same relationship. Right here. Lets look at another sample problem here in which molecule below would we observe a higher wave number this time due to the c single bonded to o bond.
So for the molecule on the left. Were focusing on this bond right here and for the molecule on the right were focusing on this co bond right here notice here molecule on the left has resonance. We can say that the electrons on that oxygen can fall this way and boot.
The pi electrons up the resulting resonance structure would look like this which means. Our co bond. In this case.
Is partially double bond partially single bond whereas in the other molecule. We wouldnt get that analogous resonance. The electrons on that o would not jump to the left.
Therefore. The molecule on the left has less single bond character giving the molecule on the right more single bond character. Which means.
The co bond in the left molecule would have a higher wave. Number and the co bond for the right molecule would have a lower wave. Number.
So. The answer is the molecule on the left would have the corresponding higher way ape number four its ceo bond. So what have we learned here a lot of points here.
Number one we saw the stronger the bond. The higher the frequency or wave. Number we saw the lighter the atoms the higher the frequency or wave.
Number. We also saw that oh h. Absorption bands are broad due to hydrogen bonding.
We also saw number four electron delocalization can affect the wave number of a functional group and lastly we saw five electron withdrawing or donating can affect the wave number of a functional group you .

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