Main content

## 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Symmetric and asymmetric stretching

How symmetric and asymmetric stretching of two identical groups can lead to two distinct signals in IR spectroscopy. Created by Jay.

## Want to join the conversation?

- Why does the C-H bond have a higher intensity in the IR than the N-H bond, if the dipole moment for N-H bond is greater?(14 votes)
- My best guess is statistically the IR has a higher chance to interact with the C-H bonds (18 times more of them) so the absorbance goes way up (transmission goes down)(6 votes)

- why can't the secondary amine stretch asymmetrically ?(6 votes)
- Because there is only one hydrogen that gives only one possibility of stretching (out and in). With primary amines, there are two hydrogen, giving two possiblities of stretching: the same direction (both out/both in) or opposite directions (one out, one in).(9 votes)

- Hi guys,

At3:34, the peak at 3300 cm-1 could also be C-H (sp hybridized), how do we know which is which?

Thank ;)(7 votes)- the reason is that: if there is a sp hybridized C-H bond occur, there usually would be a C(triple bond)C occur around 2150 on the spectrum, which there is no such peak in that spectrum.(4 votes)

- As there is nitrogen in the amine there must be hydrogen bonding between any two amine molecules. So we must get broad signals, right?(4 votes)
- i think the answer might be that the hydrogen bonding in amine is weaker than the hydrogen bonding of hydroxyl therefore we donot get a very broad signal as compared to that of alcohol.(1 vote)

- In the last example, at9:10, why are we observing the symmetric and asymettric stretch for the carbonyl group. We have two different carbonyl groups on two different carbon atoms.So from my understanding, both of these double bonds should stretch individually, giving us a single peak.(3 votes)
- Note: That is a really good question, I am not sure if my answer is correct, but I think its generally the right idea....

Answer: The electron withdrawing/donating effect of functional groups can be seen throughout a molecule. They are strongest on directly bound atoms, but continue to have an effect on atoms adjacent to those they are directly bonded to, and beyond. Thus in the case of an anhydride, the bond characteristics of each carbonyl bond are somewhat impacted by the stretching of the other carbonyl bond.

This results in a differential energy of stretching for symmetric vs asymmetric stretching.

I hope this was helpful.(3 votes)

- we see the C-O bond in 1100, but why this signal is so broad?(3 votes)
- It's simply a property of the bond. Different bonds will have different stretching with different intensities and broadness. (Luckily, the exact reason is beyond undergraduate university level)(2 votes)

- In general, if you have a double-tipped peak like that, is it indicative of the asymmetric and symmetric streching?(2 votes)
- I would say it really depends on the structure of your molecule, if you have only one functional group like the acid anhydride in the video, than its obvious the signal belongs to the symm. and asymm. stretch vibration.

But you could also have two carbonyl groups in your molecule with different environment, which results in two different signals. Then it would be not obvious if two signals belong to a sym. and asym. stretch or two different carbonyl groups. Maybe take a look at the next video.(4 votes)

- At6:28why asymmetric stretch is taking more energy than symmetric stretch ?(4 votes)
- Here's my guess:

Consider a CH₂ group.

In a symmetric stretch, the two H atoms are going in the same direction. The group dipole moment changes considerably because both bonds are going in and out at the same time.

In an asymmetric stretch, they are going in opposite directions. One bond gets longer as the other gets shorter, so the change in dipole moment is much less.

It takes energy to separate positive and negative charges from each other, so the symmetric vibration has a higher frequency (energy).(0 votes)

- So it seems in and out with one N-H makes one signal. With two N-Bonds he says the signals are not because of the two N-Hs but because of the combination where they could both be going out symmetrically. I see three possibilities for energy level in that case. one and and one out which averages to the same in/out of the signal carbon. Out/Out would be a symmetric stretch and create a signal. But I am having a hard time wrapping my head around in/in not creating a third signal. If there is a difference between in/out and out/out then it seems there must be a energy difference between in/out, out/out, and in/in. How does it boil down to two things?(1 vote)
- In/in and out/out are opposite extremes of the same symmetric vibration. They are not different vibrations.

In the same way, in/out and out/in are opposite extremes of the same asymmetric vibration.

Hence there are only two types of vibrations.(2 votes)

- In trying to combine all of the concepts, how would I relate the Kinetic Energy of the electrons? Is it correct to say that a stronger bond that requires more energy (from photon of light) will have an increase in KE of the electrons, thus an increase in vibrations?

Or am I just trying to combine way too many concepts?(1 vote)- I'm pretty sure that you must have things backwards — a strong bond is more stable and that should mean it has lower energy — so I would guess that if kinetic energy is relevant that the electrons would have lower kinetic energy in a stronger bond ...

However, to be honest, I'm not really sure if talking about the KE of bonding electrons is helpful ...

As for the vibrations, I believe that most of the movement is of the nuclei rather than the electrons.(1 vote)

## Video transcript

- [Voiceover] Before we get into symmetric and asymmetric stretching, let's look at the IR
spectrum for dibutylamines. So here's our dot structure
for a dibutylamine. And we start by drawing
a line about 3,000, and we know to the right
of that, we expect to find the signal for the
carbon-hydrogen bond stretch, where we're talking about
an SP3 hybridized carbon. So this is the bond to hydrogen
region on the IR spectrum. And notice we have
another signal right here, so a signal that is at
a higher wave number than the carbon-hydrogen bond stretch. So if we drop down here, we
can estimate the wave number. It's approximately, this'll be 31, 32, 33. So at approximately 3,300 wave numbers, we get another signal, so another bond of an atom to hydrogen. This is the bond stretch
for nitrogen-hydrogen, so that's this bond
stretching right there. So in magenta, this is the
nitrogen-hydrogen bond stretch. Let's compare the strength of that bond to a carbon-hydrogen bond where the carbon is SP3 hybridized. We know that the wave
number is dependent on two things from an earlier video. We know it's dependent
upon the force constant, k, or the spring constant
k, and the reduced mass. Well, the reduced mass for these two bonds is approximately the same. So if you did a calculation
for the reduced mass for nitrogen-hydrogen
and for carbon-hydrogen, you're gonna get
approximately the same value for the reduced mass. And so that's not what's
affecting the different wave numbers for these signals here. So it must be the force constant. It must be k. And since the nitrogen-hydrogen
bond has a greater wave, the signal shows up at
a higher wave number, that must mean it's a stronger bond, because if you increase
the force constant, increase the strength of the bond, you increase the wave number. You increase the frequency. And so the nitrogen-hydrogen
bond is stronger than the carbon-hydrogen bond, where the carbon is SP3 hybridized. And if it's stronger, it takes more energy to cause that bond to stretch. So let's really quickly talk about energy. Energy is equal to h,
which is Planck's constant, times the frequency, so
when you're talking about the energy of a photon,
it's equal to h nu. And nu is our frequency, and we know that relates to wave number. So the frequency is equal to the wave number times the speed of light. We talked about this in an earlier video. So if you take this and plug it into here, you can see the energy is
directly proportional to the wave number, so this
would be E is equal to h times the wave number
times the speed of light. And this is one of the reasons why you see IR spectrum done in wave numbers, because you can also
think about energy, right? So if you increase the wave number, if you're talking about
increased wave number, you're talking about increased energy. So as you go this way, as
you increase in wave number, you're also talking about
increasing in energy. So you can think about
it takes more energy, more energy is needed to
stretch a stronger bond. And so it takes more energy to stretch this nitrogen-hydrogen bond. So again, think about a bond as a spring. If you have a really
stiff or strong spring, it takes more energy to
stretch that spring out as compared to a looser spring. And so that's thinking about
energy, and also looking at a typical IR spectrum here
for a secondary amine. So this nitrogen here is
bonded to two carbons, so this is a secondary amine. And in a secondary amine,
you're going to get one signal approximately 3,300 here. So let's compare this IR
spectrum of a secondary amine with another amine, so
this is a primary amine. Let's compare it to butylamine. So over here, this is a primary amine. The nitrogen is bonded to one carbon, so we're talking about
a primary amine now. And let's analyze the IR spectrum. So once again, we're gonna
draw a line around 3,000, and we know that this
in here is talking about the carbon-hydrogen bond stretch for an SP3 hybridized carbon. Alright, once again, let's
look at just past that, right in the bond to hydrogen region, and we get two signals this time, right? So if we look over here,
there are two signals. This signal, let's drop down,
this is approximately 3,300, so we have one signal approximately 3,300. And then we have another signal. Let me go ahead and make that green here. So we've got another signal right here, which is a little bit higher
in terms of the wave number. So we drop down, this signal
is approximately 3,400. So we know this is where
we would expect to find the nitrogen-hydrogen bond stretch. We get two signals, and
we need to figure out what's going on here. Well, this has to do with symmetric and asymmetric stretching. So let's look at two generic amines here, and let's talk about
what the difference is between symmetric and
asymmetric stretching. If you have symmetric
stretching, so these bonds are stretching in phase, if you will. You can think about the
hydrogens stretching away from the nitrogen at the same time. So this is called symmetric stretching. This is symmetric stretching. And this one over here,
let me go ahead and draw what's happening over here. So this time these two
nitrogen-hydrogen bonds are stretching out of phase. So if that hydrogen is
stretching this way, this hydrogen might be contracting here, so that's an asymmetric stretch. Let me go ahead and write that. So we're talking about an
asymmetric stretch here. So this is what's happening. This is why we get these
two different signals. It turns out it takes less energy to do the symmetric stretching. So if it takes less energy to
do the symmetric stretching, this is the one that we
find at a lower wave number. Remember, wave numbers
correspond to energy. So it takes less energy to
do a symmetric stretching, and so that's this signal. It takes a little more energy
to do asymmetric stretch. And so that's this signal, right up here. So we get two different signals
here for our primary amine. Two signals, right? And it's tempting to say,
"Oh, we get two signals "because we have two
nitrogen-hydrogen bonds. "So here's a nitrogen-hydrogen bond, "and here's a nitrogen-hydrogen bond." But that's not really what's happening. Some of the molecules are
having a symmetric stretch, and some of the molecules are
having an asymmetric stretch. And so that's why you see
these two different signals. Once again, let's just
really quickly compare these two different amines. A secondary amine is going
to give you only one signal on your IR spectrum,
whereas a primary amine is going to give you two signals, right? These two different signals, here. That's something to look out for. Also, we can think about the carbon-hydrogen SP3 stretching here. For example, if you have a CH2 here, so a CH2, let me go ahead and draw two
different situations here, so a CH2 in a molecule, you
can have the same thing. You can have a symmetric stretch
and an asymmetric stretch. So if these hydrogens are both
stretching at the same time, that's a symmetric stretch. You could also have an
asymmetric stretch like this. And, once again, you'll
find the asymmetric stretch, so this one, actually takes
a little bit more energy. So you're gonna find this signal at a slightly higher wave number. It's a pretty small difference, but it is a slightly higher
wave number for this stretch. And that's one of the
reasons why you get such a hard to interpret signal
in here, so less than 3,000 there's a lot of stuff going on. It's too difficult to worry
about in great detail. Just understand that that's
where you would expect to find your carbon-hydrogen bond stretch, where I'm talking about
an SP3 hybridized carbon. Finally, let's look at one more example of a symmetric and asymmetric stretch, and that's an acid and hydride. So let's look at an IR spectrum, just a generic IR spectrum
for an acid and hydride. Let's draw a line around 1,500. To divide our two regions, we
draw a line around 3,000 here. So we know this is our
carbon-hydrogen in here. And then we get these two
very intense signals, right? So let's figure out where
these are, approximately. This signal right in here, let me use a different color for right here. We drop down, so where
is that, approximately? Well, if this is 1,500, 1,600, 1,700, so that's pretty close to, let's say, 1,760 here for this signal. So, 1,760 wave numbers for that signal. And then this other
signal, we drop down here. This is just a tiny bit past 1,800. We'll say approximately 1,800 here. So we get these two
different, strong signals for an acid and hydride, and once again, we're talking about symmetric
and asymmetric stretching. So a symmetric stretch,
let me go down here, so this carbonyl, of course,
is what we're talking about. We're talking about this
really strong absorbence in the double bond region, right? So in here is the double bond
region on our IR spectrum. We get these two strong
signals, and the signal at a lower wave number is
due to symmetric stretching. So this carbonyl could
be stretching in phase with this carbonyl, so that's
our symmetric stretch signal. So that's this one, down here. And then once again, we could
get an asymmetric stretch. So we could have one of
these, we could have this one stretching and this one
contracting for our spring. It takes more energy to
do an asymmetric stretch, and so that's this higher signal here. So if you see an IR spectrum,
and you see these two really intense signals, we're
talking about the carbonyl. So right in here, this is
our double bond region. We're talking about the carbonyl stretch, with a large dipole
moment, so a large change in dipole moment when it stretches. So that's why we see such
an intense signal here. And then we see two of
them, because we have symmetric and asymmetric stretching. There are other examples of symmetric and asymmetric stretching,
but hopefully this gives you an idea about what to look
for on your IR spectrum.