# Symmetric and asymmetric stretching | Spectroscopy | Organic chemistry | Khan Academy

– [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.

## 7 Replies to “Symmetric and asymmetric stretching | Spectroscopy | Organic chemistry | Khan Academy”

that was very helpful…

Nice presentation. Good to understand.

Thank you!!!

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Thx!

Thank u khan academy for providing such helpful resources.

does plane chnges occur in asymmetric stretching ?