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# Hydrogen deficiency index

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

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• Why do u need to account for the nitrogen (minus #of nitrogen) when dealing with a Benzene ring with chlorine?
• 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.
• why do some people refer to this as degrees of saturation rather than index of H deficiency? ...
• 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.
• Why is it + for nitrogen but - for chloride?
• 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.
• What happens if you have a triple bond? does the 1/2 (2C+2-H) formula still apply?
• 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
• What if the substituent of a benzene ring is a carbonyl group (=O) ? Does it affect the IHD?
• 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.
• 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?
• 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.
• 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.
• how can I calculate the HDI in case of benzoic acid or alkyl benzoate ?
(1 vote)
• First you we need the chemical formula of your benzene derivative (And if you were doing this in a lab you would have done mass spec. so you should know that early on). For benzoic acid, a benzene with a carboxylic acid, the formula would be C7H6O2. So for the HDI formula, C = 7, H = 6, N =0, X = O. And oxygen doesn't affect the HDI so we don't need them to calculate it. (2(7) + 2 - 6)/2 = 5, so benzoic acid has an HDI of 5.

For an alkyl benzoate, which is an ester with a benzene bonded to the carbonyl carbon and an alkyl group coming off the single bonded oxygen, it will have a similar HDI. So I'll use the simplest benzoate, methyl benzoate, which has the chemical formula C8H8O2. So the formula would look like, (2(8)+2-8)/2 = 5. So methyl benzoate has an HDI of 5, but even if I had used ethyl, propyl, butyl, etc. benzoate it would have also been 5 too.

Which makes sense because what HDI ultimately tells you is how many pi bonds and rings you have in your molecule. Both benzoic acid and alkyl benzoates have 4 pi bonds and 1 ring. You can look at the bond line structure of a molecule and tell its HDI just by counting the pi bonds and rings. So benzene has 3 pi bonds and 1 ring, so it has an HDI of 4, which I can figure out just by looking at the structure of benzene without having to calculate it using the formula. Hope that helps.