# Proton NMR practice 1 | Spectroscopy | Organic chemistry | Khan Academy

– [Voiceover] Let’s say we’re

given this molecular formula. C5 H10 O and this Proton NMR spectrum. And we’re asked to determine

the structure of the molecule. The first thing you could do is calculate the Hydrogen Deficiency Index. And so if we have five

Carbons here, the maximum number of Hydrogens we could

have is two N plus two, where N is equal to five. So two times five plus two is equal to 12. So 12 is the maximum number of Hydrogens. Here we have only 10 Hydrogens so we’re missing two Hydrogens. We’re missing one pair of Hydrogens. So therefore, the Hydrogen

Deficiency Index is equal to one. Immediately, that makes me

think about a double bond might be present in this

molecule or one ring. Let’s go down to the integration value. For this signal, there’s

an integration value of 27. For this signal, the

integration value is 40.2. For this signal, it’s 28.4. And for this signal, it’s 42.2. Remember what you do. You divide all four integration

values by the lowest one. So the lowest one is, of course, 27. So 27 divided by 27 is, of course, one. 40.2 divided by 27 is pretty close to 1.5. 28.4 divided by 27, that’s

pretty close to one. And 42.2 divided by 27 is once again pretty close to 1.5. Remember, these are just

the relative numbers of protons giving you these signals but you can’t have 1.5

protons giving you a signal. We need a whole number. And we need to account

for 10 total protons here. So if we multiply one by two, then we get two protons. So this signal represents two protons. If we multiply 1.5 by two, we get three. So this signal represents three protons. If we multiply one by two, we get two. And if we multiply 1.5

by two, we get three. So if we add all those up, two plus three plus two plus three, that’s, of course, 10 protons. So now we have accounted

for all 10 protons using our integration values. Next, let’s look at

each signal one by one. We’ll start with this signal. So we have a CH2 since

we have two protons. So this signal represents a CH2. How many neighboring protons

do those CH2 protons have? All right, so we can figure that out by the number of peaks on the signal. So this signal here has

one, two, three peaks. Now if you think about

the N plus one Rule, if you have N neighbors,

you get N plus one peaks. So if we have three peaks, all

we have to do is subtract one to find out how many neighboring protons. So three minus one is equal to two. These CH2 protons have

two neighboring protons. Let’s think about the chemical

shift for this signal. So the chemical shift for this signal is between two parts per million and 2.5. And that’s in the region for

a proton next to a carbonyl. Now it makes a lot of

sense because we calculated an HDI of one indicating their (mumbles) double bond present and we need to account for an oxygen in our molecular formula. Let’s go ahead and draw in a carbonyl. These CH2 protons are

next to the carbonyl. We know that because of the shift, right? The carbonyl, the oxygen deshields these two protons a little bit and gives us a higher value

for the chemical shift compared to something

in Alkene type region. All right, let’s color these

protons in here magenta. So we’re saying this signal

is due to those two protons. Let’s move on to the next signal. So we have three protons. That would be a Metal group, so CH3. How many neighboring protons

for those Metal protons? Well, there’s only one peak here. So one minus one is zero. So zero neighboring protons. What about the chemical shift? The chemical shift for this signal is once again past two parts per million. So this signal is deshielded, all right? So these protons must be

deshielded a little bit. Those must be next to our carbonyl. We’ll draw in our methyl protons, right? Being deshielded because

they’re next to the carbonyl. Let’s make these blue. So these protons in blue here, these three protons are

giving us this signal. And we should expect zero

neighboring protons, all right? So those protons are on this carbon and the next door carbon

has no protons on it. So zero neighboring protons makes sense. Move on to the next signals. This signal represents two

protons, so that’s a CH2. How many neighboring protons? All right, let’s count

how many peaks we have. One, two, three, four, five, six. So six peaks. We subtract one from that. Six minus one is five so we will expect five

neighboring protons. Let’s move on to the next signal. Three protons, all right? That’s a Metal group, so CH3. How many neighbors? Well, we have one, two, three peaks. Three minus one is two. So we will expect two neighboring protons. Now these two signals, this signal here and this signal here, we’re talking about under two parts per million now. So these must be the furthest away from the carbonyl, all right? They’re not being deshielded as much as the two signals in this direction. So these two neighboring

protons for the Metal group must be these two protons right here. Let’s go ahead and draw that in. So we have our Metal protons right here. Let me make these protons red. So these protons right here in red are giving us this signal. From the signal, we know

that these Metal protons are next to two neighbors and so we must have a

CH2 next to that metal. And so let’s go ahead and draw in our CH2. This CH2 must be this signal right here. All right, so let’s think

about how many neighbors. We expected five neighbors

for these CH2 protons. Let’s count them up. One, two, three, four, five. So five neighboring protons, matches what we see on the NMR spectrum. Now, those protons, those

magenta and red protons are actually in different environments. And so the simplified version

of the N plus one Rule isn’t quite true but it

works for this example. It works for this example

so we’re gonna go with it because all we care about is getting the structure of our molecular here. All right, so let’s think

about the red protons again. So the red protons have two neighbors. How many neighbors of

the red protons have? Here’s one and here’s

two, so this makes sense. So that would make sense. This one makes sense right here. We just talked about that. Let’s think about the magenta protons. So the magenta protons were

supposed to have two neighbors. Let’s look at the carbon

next door to this one. So here’s the carbon next door. Here we have a neighbor and

here we have a neighbor. So this makes sense. And then, we already talked

about the blue protons, right? Having zero neighbors. So everything seems to make sense here. To sum everything up, make

sure to count all of your atoms and you will get, of course, five Carbons, ten Hydrogens, and one

Oxygen when you do that. So this is the Dot Structure

that we were trying to find. Let’s take a look at another one here. So we have a molecular formula of C8 H10. Let’s go ahead and calculate

the Hydrogen Deficiency Index. So if we have eight

Carbons, then we can have a maximum of two times

eight plus two Hydrogens. So two times eight is 16, plus two is 18. For eight Carbons, 18

Hydrogens is the max. Here we have 10 Hydrogens. So we’re missing eight Hydrogens or missing four pairs of Hydrogens. The Hydrogen Deficiency

Index is equal to four. Remember, anytime your

HDI is equal to four, you should think about a Benzene Ring. I’m going to go ahead and

draw a Benzene Ring in here. Let’s look at the integration. All right, so sometimes you’ll see the integration given like this. This represents five protons for this very complex

looking signal right here. This signal over here

represents two protons and this signal represents three protons. Let’s go back to these five protons with this complex signal, all right. This is in the Aromatic range, all right? So approximately 6.5 to eight. Those must be five Aromatic protons. We can go ahead and draw in five protons off of our Benzene Ring. Even though those protons

are in slightly different environments, because this

integration value is five here, we know a Benzene Ring is

present, we’re done, all right? We don’t have to worry about the slightly different environments. We know that these five

protons are giving us this complicated signal over here. All right, next, let’s

look at this signal. So two protons, so that must be a CH2. How many neighboring protons for the CH2? Well, there’s one, two, three, four peaks. So if there’s four peaks,

four minus one is three. So these CH2 protons have three neighbors. All right, let’s look at this next signal. So three protons. This must be a CH3. How many neighboring protons

do we have for the CH3 protons? Well, we have one, two, three peaks. So three peaks. Three minus one is two, so two neighbors. Two neighboring protons, that must be these two

neighboring protons. We’ll go ahead and draw

an Ethyl group, all right? So this is an Ethyl

group pattern over here. There’s a CH2 and there’s your CH3. So let’s draw in all of our protons. So we have these protons

and we have these protons. Let’s color coordinate here just to make sure everything makes sense. So these red protons right

here, they must be giving us this signal, so this

Metal protons, all right? We expect two neighbors because we have three peaks on our signal. And here are the two neighbors. One, two, so two neighboring protons. Let’s look at these two protons next. So these two protons are

giving us this signal. We’d expect three neighbors

because we have four peaks. One, two, three, four. So three neighbors, so

here’s the next door carbon. So one, two, three, three protons. And then this is also a next door carbon but there are no protons on this carbon. So we see only these three neighbors. And so everything seems to make sense. Once again, count up all of your atoms and make sure you have

accounted for everything. For example, you have eight

carbons you have to worry about. So on your ring, we know there are six. One, two, three, four, five, six, and then we get seven

and then we get eight. So the eight carbons is correct. And if you count all those up, we have our ten Hydrogens

accounted for too. So this is the NMR

spectrum for Ethylbenzene. All right, this one

was a little bit easier than the previous example.