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## Organic chemistry

### Course: Organic chemistry > Unit 14

Lesson 3: Proton NMR- Introduction to proton NMR
- Nuclear shielding
- Chemical equivalence
- Chemical shift
- Electronegativity and chemical shift
- Diamagnetic anisotropy
- Integration
- Spin-spin splitting (coupling)
- Multiplicity: n + 1 rule
- Coupling constant
- Complex splitting
- Hydrogen deficiency index
- Proton NMR practice 1
- Proton NMR practice 2
- Proton NMR practice 3

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# Hydrogen deficiency index

How to relate molecular structure with the molecular formula using the hydrogen deficiency index (or number of degrees of unsaturation). Created by Jay.

## Want to join the conversation?

- Why do u need to account for the nitrogen (minus #of nitrogen) when dealing with a Benzene ring with chlorine?(14 votes)
- He's just being thorough with the calculation and applying the generic formula in that context. Notice he uses 0 for N when applying the formula to ensure it's accuracy.(15 votes)

- why do some people refer to this as degrees of saturation rather than index of H deficiency? ...(8 votes)
- A compound with no double bonds or rings is called 'saturated' because it has the maximum number of hydrogens possible for a given number of carbons. As the number of double bonds or rings increases for the same number of carbons, the compounds become unsaturated.(11 votes)

- Why is it + for nitrogen but - for chloride?(4 votes)
- HDI = ½ (2C + 2 + N – H – X)

The formula for an alkane is C_nH_(2n+2). For a cycloalkane or an alkene, the formula is C_nH_2n.

Each time you insert a double bond or a ring, you lose two H atoms.

So, a double bond or ring (HDI = 1) means a deficiency of 2 H atoms.

That's where the factor of ½ in the formula comes from.

2C+2 is the number of hydrogen atoms in the saturated alkane.

So, 2C is the number of bonds that each carbon forms to other atoms (i.e., 2 bonds per carbon atom).

Each N atom forms three bonds. That's one more than each C atom. So you must**add**an extra H for each N atom (compare the H count for ethane, CH₃CH₃ and methylamine, CH₃NH₂. You have to add an extra H to get the count back to 6).

Like H, each Cl forms only one bond. We can treat them as if they are hydrogens, so we**subtract**them as well.(8 votes)

- What happens if you have a triple bond? does the 1/2 (2C+2-H) formula still apply?(5 votes)
- Yes, it still works. Think of 3-hexyne. Molecular formula is C6H10.

Applying the rule: 1/2[2(6)+2-10] = 1/2[4] = 2 = HDI

3-hexyne has two extra carbon-carbon bonds relative to hexane, so this matches the HDI(3 votes)

- What if the substituent of a benzene ring is a carbonyl group (=O) ? Does it affect the IHD?(3 votes)
- No the IHD would remain the same you would still have 6 carbons and 6 hydrogens, an oxygen atom has no effect on the IHD. The IHD is 4 and you would still have 2 double bonds inside the ring counting for 2. The one double bond on the carbonyl counting for another 1 making that a total of 3, plus the ring itself equals 4.(2 votes)

- But isn't the Hydrogen Deficiency Index the same as Double Bond Equivalence (DBE)? I mean, at the end of the day, we are going to get the same numbers, right?(2 votes)
- Why do we we doing addition for N and not subtraction?(1 vote)
- Nitrogen forms 3 bonds. So if you replace a hydrogen with a nitrogen, it will also form 2 more bonds to hydrogens. That's a net change of +1 hydrogen, which is why it is +N.

Also if you notice, oxygen is not in the formula, because if you replace a hydrogen with an oxygen, it's still going to bond with 1 hydrogen too, so the net change is 0 hydrogens.(3 votes)

- does it matter which proton we are considering first when drawing he tree to determine the number of peaks of a specific signal?(1 vote)
- No, but it is usually easier and it gives a neater diagram if you start with the largest splittings and work down to the smaller ones.

The order doesn't matter, so you will end up with the same final pattern.(2 votes)

- What is the simplest way?(1 vote)
- So the HDI tells us the number of pi bonds or rings. Does that include double or triple bonds to elements that aren't carbon (like C=O for example)? Or is the HDI only affected by double or triple bonds between carbons (C=C) ?(1 vote)
- AFAiK it covers all forms of desaturation including carbonyls – it definitely is not limited to carbon-carbon bonds.

For carbonyls you can see this by looking at the HDI for formaldehyde (methanal). The fully saturated single carbon hydrocarbon (methane) has 4 hydrogens and formaldehyde has 2 — so the HDI is 2, which is due to the π bond in the carbonyl.(1 vote)

## Video transcript

- [Voiceover] When you
get a spectroscopy problem on a test, you're often
given the molecular formula and asked to determine the
structure of the molecule. And sometimes the
hydrogen deficiency index can be useful for figuring out the structure of your molecule. Let's start off by looking at hexanes. So here we have hexane with a molecular formula of C six H 14. We say that hexane is completely
saturated with hydrogen. So it has the maximum
number of hydrogen atoms possible for the number of carbons. So if we have six carbons, 14 is the maximum number of
hydrogens that we can have. And remember when you're
talking about alkanes and you have n carbons, you get two n plus two hydrogens. So here n is equal to six,
so we have six carbons. So two times six plus two gives us 14 hydrogens. Let's compare hexane with one hexene. So now we have a double bond present. And that changes the
molecular formula, of course. So now the molecular
formula is C six H 12. We have the same number of carbons, six, but we have 12 hydrogens. And up here we had 14. So we're missing two hydrogens. And so we say that's one
degree of unsaturation, or a hydrogen deficiency
index equal to one. So an HDI equal to one. So it's like we're missing
one pair of hydrogens here. Let's look at cyclohexanes. So here we have cyclohexane. And the molecular formula is C six H 12. Once again we have six carbons, but we have only 12 hydrogens, so we're missing two compared to hexane. We're missing two hydrogens, we're missing one pair of hydrogens, so the hydrogen deficiency index, the HDI, is equal to one. So if you were given this
molecular formula, C six H 12, right here it's the
same molecular formula, and you think about how many hydrogens you're missing from the maximum number. You're missing two hydrogens. That's a hydrogen deficiency
index equal to one. Here we had one double-bond and here we had one ring. So with an HDI equal to one, one possibility is you have one ring, another possibility is
you have one double bond. Alright, let's look at benzene. So here we have benzene, a molecular formula of C six H six. So with six carbons, the maximum number of
hydrogens you can have is 14. And here we have six. So 14 minus 6 is equal to eight. So it's like we're
missing eight hydrogens. So that's four pairs of hydrogens. So the HDI is equal to four. So the hydrogen deficiency
index is equal to four. And I like to think about
why the HDI is equal to four, because, if this is four, we
have three double bonds here. One, two, three. And we have one ring. Right, so we have one ring here. So three plus one is equal to four and we get an HDI equal to four, right? Up here we had one ring and we had an HDI equal to one. Up here we had one double bond and an HDI equal to one. We could also calculate
the HDI using a formula. So we could write that the
hydrogen deficiency index is equal to one-half times what's in the parenthesis. Which is two times the
number of carbons plus two. That's the whole two n plus two idea. Minus the number of hydrogens. So let's go ahead and do it for this molecular formula with six carbons and six hydrogens. Let's plug in those numbers, so we have one-half, this would be two times
the number of carbons, so that's two times six plus 2 minus the number of hydrogens, that's six. So let's do the math. So this will be one-half. Two times six plus two is 14. Minus six. 14 minus six is eight, times one-half. We get an HDI equal to four. So if you have a spectroscopy problem and you calculate the HDI equal to four, think about the possibility of
a benzene ring being present in the structure of your molecule. Let's compare benzene to another molecule where we're adding in a nitrogen. So here's benzene and we
know there are six hydrogens. One, two, three, four, five, six. So six hydrogens. Let's add in a nitrogen. Right, so we are adding in a nitrogen. So now we have aniline
as our compound here. How many hydrogens? So this would be one, two,
three, four, five, six, seven. So we have seven hydrogens here. So we've added one nitrogen and we get one more hydrogen. So we need to modify the formula for HDIs. Let's go ahead and do that. So we would write the
hydrogen deficiency index is equal to one-half times
what's in our parenthesis, which would be two carbons plus... two times the number of carbons plus two. And now we're going to add the number of nitrogens that we have. So add the number of nitrogens, and then subtract the number of hydrogens. So let's go ahead and do this calculation. So one-half times, once again we have six carbons one, two, three, four, five, and six, so two times six plus two plus the number of nitrogens, we have one nitrogen, so we add a one in here, minus the number of hydrogens, we said there were seven, so minus seven, and let's do this, this will be one-half times, this would be 14 plus one is 15, so 15 minus seven is equal to eight times one-half is equal to four. So an HDI of four, remember an HDI of four, think about a benzene ring being present in the
structure of your molecule. Alright, let's do another one. Let's compare benzene to chlorobenzene. So over here on the left we have benzene, once again six hydrogens. And here we have chlorobenzene, so we've added in a halogen. And how many hydrogens do
we have in chlorobenzene? One, two, three, four, five. So only five hydrogens this time. So we've added a halogen
and we've lost a hydrogen. We went from six to five. So we need to modify our formula again. So HDI is equal to one-half times what's in the parenthesis, two times the number of carbons plus two plus the number of nitrogens, so plus number of nitrogens, minus the number of hydrogens, and we would also need to subtract the number of halogens. So minus the number of
halogens that you have. Alright, let's do it for chlorobenzene. So one-half times, how many carbons do we have? Still six. One, two, three, four, five, and six. So two times six plus two, zero nitrogens, alright, so zero nitrogens, how many hydrogens? Five for chlorobenzene, so minus five and we have one halogen. Right we have one halogen here, it's the chlorine, so minus one. So let's do that math. So we have one-half times two times six, plus two is 14. So we have 14 minus six over here. So 14 minus six is eight, times a half is equal to four. So we get an HDI equal to four and once again we think benzene ring. So HDI is equal to four. We think about a benzene
ring being present. Alright, let's think
about one more example. So, if we look on the left, once again we have benzene. So here's benzene with the six hydrogens. And we've added in an oxygen. So for this molecule
we've added in one oxygen. What happened to the number of hydrogens? So we have five on the ring. One, two, three, four, five. And then this one for six. So even though we've added in a hydrogen, there's no change in the number... even though we've added in
an oxygen, I should say, there's no change in
the number of hydrogens. And since there's no change
in the number of hydrogens, we don't need to modify our formula here for the hydrogen deficiency index. So I'm gonna go right
back up here and box this, so this is our final version
for calculating the HDI. And you don't have to use
this formula every time. Sometimes you can just think about how many hydrogens are missing. But sometimes using this formula allows you to calculate the HDI which helps you to think
about what's present in the structure of your molecule. And so I'll show you several examples of how to use the HDI to figure out your dot structure.