– [Voiceover] If you shine

infrared light on a molecule, it’s possible for the molecule to absorb energy from the light. Energy from the light can

cause a bond to stretch. We call that a stretching vibration. You can have other kinds

but we’re only going to focus on stretching here. The stretching vibration of a

bond is like the oscillation of a spring so you can think about a bond as being like a spring. Let’s think about this bond right here. The bond between the

carbon and the hydrogen. We’re going to model

that bond as a spring. We’re going to attempt

to draw a spring in here. So here’s the spring. Then let’s put in the carbon on one side. So we have the carbon on one side and the hydrogen on the other side. So the stretching vibration of the bond is like the oscillation of the spring. So if you had a spring

and you had two masses on either end of the spring,

if you put some energy in, you can stretch that spring. So you could pull the carbon this way and you could pull the hydrogen

that way so it’s like the stretching of this bond here. We know that springs also contract. So then the spring can

pull back in this direction and you get an oscillation

of the spring which is once again how we model the

stretching vibration of a bond. Let’s look at the IR

spectrum for this molecule. So we’re talking about one octyne here. If you shine a range of

infrared frequencies through a sample of this compound, some of the frequencies are

absorbed by the compound. You can tell which

frequencies are absorbed by looking at your infrared spectrum here. For right now, let’s

think about these numbers, like 3,000 or 4,000. Let’s think about those as representing frequencies of light. Over here we have “% transmittance.” If you had 100 percent transmittance, let me go ahead and draw a line up here. So 100 percent transmittance,

let’s say we’re talking about this frequency of light. I look at the frequency

of light, I go up to here, and I can see I have 100

percent transmittance. 100 percent transmittance

means all that light was transmitted through your sample. If all the light went through your sample, nothing was absorbed. So this particular frequency was not absorbed by your compound. If you’re talking about less

than 100 percent transmittance, let’s say for this frequency right here. So for this frequency,

we have the signal here, appearing at this frequency. We don’t have 100 percent

transmittance so that means not all of the light went

through the compound. Some of it was absorbed. So this specific frequency

was absorbed by the molecule. That energy can cause a bond to stretch, and we get a stretching vibration. Actually this signal

corresponds to the bond that we’ve been talking about. So this signal indicates the stretching of this bond right here. Let’s think about wave number next. We just talked about

percent transmittance, let’s think about wave number. I’ve been calling all

these things frequencies, different frequencies of light. Let’s see how wave number

relates to the frequency of light and also the wavelength of light. The definition of wave

number, so a wave number, here’s the symbol for wave number. The definition of wave

number is it’s equal to one over the wavelength in centimeters. So if we had a wavelength of

light of .002 centimeters, so let’s go ahead and plug that in. A wavelength of light of .002 centimeters, what would be the wave number? Let’s get out the calculator

here and let’s do that math. One divided by .002 is equal to 500. So that’s equal to 500. Your units would be one over centimeters, or you could write that,

meaning the same thing. So that’s the wave number. That’s the wave number. If you go over here, a wave number of 500, you can think about

this corresponding to a particular wavelength of light. Of course this also relates

to frequency because we know that wavelength and frequency

are related to each other. So let’s get some more room down here. We know that the wavelength

times the frequency, wavelength times the frequency, so lambda times nu, is

equal to the speed of light. That’s equal to c. So if I wanted to solve for

the frequency, solve for nu, the frequency is equal to

the speed of light divided by lambda divided by the wavelength. That’s the same thing as one over lambda times the speed of light. One over lambda was our

definition for wave number. So you can say that the

frequency of light is equal to the wave number

times the speed of light. So let’s go ahead and do

that calculation here. We talked about this as being

a particular wavelength. We found the wave number. Let’s plug that wave number

into here and see what we get. So the wave number was 500. Units were one over centimeters. Multiply that by the speed of light. We need to have the speed

of light in centimeters. That’s three times approximately, three times 10 to the tenth

centimeters per second is the speed of light. Notice what happens to the units. The centimeters cancel. Let’s do that math. Let’s get some room over here. We take the wave number

and we multiply that by the speed of light in centimeters. So three times 10 to the tenth. We get 1.5 times 10 to the thirteenth. The frequency would be 1.5

times 10 to the thirteenth. The units would be one over seconds or you could use hertz for that. A wave number corresponds

to the wavelength, And you can also get

the frequency from that. So we have, let me just

rewrite this really quickly, so frequency is equal to wave number times the speed of light. Wave number is equal to the frequency divided by the speed of light. We’ll use this in a later video. We’ll come back to this idea. For right now the frequency

of light is directly proportional to the wave number. You can look at an infrared spectrum and, go back up here. You can look at an infrared

spectrum and call this down here, you can call this wave number, you can refer to it as a frequency, you can call it whatever

you want as long as you understand what’s going on here. Let’s look more in detail

at this infrared spectrum. Let’s draw a line at approximately 1,500 wave numbers right here. The left side, the left side of that line, so we’ve divided our

spectrum into two regions. The region on the left is

called the diagnostic region. So this is called, this is called the diagnostic region of our spectrum. It’s because a signal in

this region can be diagnostic for a certain functional group. For example, this signal right here, if we go down here to the wave number, that signal is at approximately 2,100 for this wave number here. That’s corresponding to

the triple bond here. This tells us a functional group. This tells us that a

functional group is present. This triple bond is present. It’s diagnostic. It helps you figure out the

structure of the molecule. You can figure out

different functional groups present in molecules using IR spectra. So the right side, the

right side of this line is called the fingerprint region. So this is the fingerprint region. It’s harder to interpret

the fingerprint region. It’s much more complicated. It’s not as easy to see different signals. It’s extremely complicated, but it is unique to each molecule. So it’s like a fingerprint

for the molecule and so you can match up IR spectra. If you have an unknown you can look at the fingerprint region, and again it’s unique. All these different lines

are unique to that molecule. So we have the diagnostic region

and the fingerprint region. We’re not gonna deal much

with the fingerprint region. Maybe a little bit. We’re going to focus in

on the diagnostic region. We’re gonna focus where

the signals appear. All right, so I look at the signal, I go down to here, and I

get a specific wave number. So the location of the

signal is pretty important. I did want to point out that,

if you look at what I used for the wave number here, I

changed how I did everything. So the spacing is different. It doesn’t really matter. I only did this to fit

this video and to make it work for this video. I’m also gonna hand

draw all my IR spectra, so it’s really not gonna be perfect. The idea is not worry

too much about what they give you here for the

scaling for the wave number. Think about where the signals appear. So the location of this

signal, approximately 2,100 wave numbers. You also want to think about

the intensity of the signal, and the shape of the signal. We’ll talk much more about

those in later videos. On the next video we need

to develop this idea of bonds as springs a little bit more. So we’ll think a little bit

about some classical physics.

Is speed ofd light not 3 x 10^8? why did you use 3 x 10 ^10 instead at 5:15

Wavenumber is also proportional to the energy of the photons that the substance is absorbing.

Very lucid lecture! Thanks for the useful videos!

sorry! what is the kind of the software that you use for writting in your lectures

How do you know what regions of the IR spectrum correspond to different bonds, such as double and triple?

Even the first 14 seconds was very informative. Thank you

good video but i don't likee the convention of using v for frequency as it can be confused with speed for v = f lamda

Fantastic lecture

u r doing a great job

sooooooooooooooooooooooooooooooooooooo stupid:(:(

great. if you could use f, instead of v for frequency, it would be a lot less confusing. but amazing video

Thank you

Is there a subject you aren't good at?! Great video!

how do you calculate number of infrared stretches for a metal carbonyl complex?

what software did you use to make this video?

Spectrometer/spectroscopy cannot detect atom star and telling how much water, air etc from far away. Stupid scientists still believe this false. I wonder how many years they are going to realize that they are wrong.

My god.

I thought speed of light is 3.00×10^8…

thanks a lot 💜💛💚💘❤💓💕💞💝💟💗💙

very nice

why transmittance would be more than 100%?

why is 1500 the boundary between diagnostic and fingerprint

What separates the "fingerprint" area from the "diagnostic" one? Why is that wave number of 1500 not some arbitrary line? There are 3 dips in the curve to its left, and two to its right. But why the need for two zones. The teacher leaves this unclear.

Thanks a lot¡¡¡

audio could be better

Good

IR made easy, can't thank you enough !

wheres salmon khan at? no one wants to hear ur dead voice

I thought v is wave energy and f is wave frequency?

superb…

Thanks!

sounds exactly like jackfilms at 2.12 and 2.35

seriously the calculation in the middle was a big waste of time …. when people click on this video they want to know the general basics of everything dealing with IR spec…… like show us examples of how to identify groups and all that not some stupid calculation

I wasted 10 min of my day on this video and i still know the same amount of info on IR that I knew before lol not much ….. cut the crap and get to the point ….not trying to be mean but i dont like wasting my time

Your videos are so detailed and helpful! I wished I found these videos sooner. 😩Thank you so much!! ♥️

kyr sp33dy?

Thank you so much sir for such a nice initiative taken by you🙂🙂 🙏. Your videos make learning very simple 🙏🙏