The degree of unsaturation, also known as the index of hydrogen deficiency (IHD), tells us how many π bonds and/or rings are present in a molecule. It’s one of the first things we calculate when analyzing a molecular formula, especially in structure determination, spectroscopy, or mass spectrometry problems.

What Does “Unsaturation” Mean?

A molecule is saturated if it has the maximum number of hydrogens possible.
The simplest examples are alkanes, where all the carbons are connected to four other atoms with single bonds, making it impossible to add any more hydrogens.

Unlike alkanes, we can add hydrogens to alkenes and alkynes because of the π bonds present in those structures which makes them unsaturated. The addition of hydrogen breaks the π bonds, and new single bonds are formed:

Therefore, alkanes are called saturated hydrocarbons, while alkenes and alkynes are unsaturated hydrocarbons.

Their general formulas and unsaturation indexes are:

• Alkanes: C_nH_2n+2 (saturated) – For example, C₄H₁₀, C₆H₁₄, C₇H₁₆
• Alkenes: C_nH_2n (1 degree of unsaturation) – For example, C₄H₈, C₆H₁₂, C₇H₁₄
• Alkynes: C_n​H_2n−2​ (2 degrees of unsaturation) – For example, C₄H₆, C₆H₁₀, C₇H₁₂

Alkenes differ from alkanes by the presence of one π bond, and alkynes differ from alkenes by the presence of an additional π bond. Therefore, the pattern is that every π bond introduces one degree of unsaturation, meaning the molecule has two hydrogens fewer than the corresponding saturated analog.

Calculating the Degree of Unsaturation

There are two ways to calculate the degree of unsaturation:

1. By evaluating the molecular structure, and
2. By using a formula.

The first approach is more practical and especially useful when the structure of the molecule is given. Let’s start with that by looking at a few examples.

Degree of Unsaturation from Molecular Structure

If the structure of the molecule is given, determining the degree of unsaturation is much easier.
All you need to remember is that:

• Each double bond or ring contributes one degree of unsaturation,
• Each triple bond contributes two degrees of unsaturation.

This applies to any double or triple bond – not just carbon–carbon bonds.

For example, hexane is a saturated hydrocarbon because it contains only single bonds and no rings or π bonds, so it has zero degrees of unsaturation, i.e., it is a saturated molecule.

Hexene, on the other hand, has one double bond, which contributes one degree of unsaturation.

Hexyne contains a triple bond, which counts as two degrees of unsaturation (each π bond contributes one).

Cyclohexane has a ring but no double or triple bonds, so it has one degree of unsaturation.

Cyclohexene has both a ring and a double bond, giving it two degrees of unsaturation.

The same principle applies to the set of pentane analogs and to any other molecule in general – each ring or π bond contributes one degree of unsaturation.

Carbonyl Compounds Have One Degree of Unsaturation

A quick pattern to recognize is that carbonyl compounds have one degree of unsaturation, due to the C=O double bond. Acids, esters, aldehydes, ketones, amides, and similar compounds all share this feature.

So, if you are solving a spectroscopy problem and determine from the IR data that a carbonyl group is present, you can immediately account for one degree of unsaturation in your analysis before considering additional rings or double/triple bonds.

This, of course, only applies if the molecule doesn’t contain other π bonds or rings.
For example, esters and amides have one degree of unsaturation (from the C=O bond), but their cyclic counterparts, lactones and lactams, have two degrees of unsaturation (one from the ring and one from the C=O bond).

For the examples of lactones and lactams shown above, we did not use any calculation to determine the degree of unsaturation. Instead, we dropped the oxygen in the case of lactones, and for lactams, we removed one hydrogen along with the nitrogen. What we are trying to do is get the ratio of the carbon and hydrogen atoms in the molecule and see if it matches the general formula for alkanes, alkenes, or alkynes, because we know their degrees of unsaturation.

So yes, very often, we can also calculate the degree of unsaturation by applying the common patterns involving oxygen, halogens, and nitrogen. Let’s see what they are.

Oxygen and Degree of Unsaturation

First, what does it mean to assess the role/effect of an element in the degree of unsaturation for the given compound?

Once again, the idea here is to compare how much the presence of each element (for example, a halogen, an oxygen, or a nitrogen) is changing the expected number of hydrogen atoms based on the number of carbons compared to the general formula of the corresponding alkane.

Now, oxygen is the easiest atom to deal with when calculating the degree of unsaturation because it has no effect on the hydrogen count.

So, if the molecular formulas are C₃H₆O or C₄H₁₀O, we can simply treat them as C₃H₆O and C₄H₁₀. C₃H₆ corresponds to the general formula of an alkene, meaning one degree of unsaturation; therefore, C₃H₆O corresponds to a molecule with a one double bond, which can be a C=C or C=O.

C₄H₁₀ corresponds to a saturated compound; therefore, C₄H₁₀O is the formula of an alcohol or an ether, as it also has zero degrees of unsaturation.

This does not tell us the exact structure but rather gives a pool of possible constitutional isomers.

For example, it can be diethyl ether, which is structurally similar to butane in that it has the same number of carbon and hydrogen atoms, differing only by an oxygen atom replacing one of the C–C bonds.

What we see is that oxygen has no effect on the expected number of hydrogen atoms per carbon. The molecule still has 10 hydrogens for 4 carbons, which corresponds to the general formula of alkanes, C_nH_2n+2.

So, whenever calculating the HDI for a molecule containing an oxygen atom(s), simply ignore it.

As an example, the degree of unsaturation for C₇H₁₄O₂ is the same as C₇H_14, which corresponds to the C_nH_2n general formula a with one degree of unsaturation.

Halogens and Degree of Unsaturation

Putting a halogen in place of a hydrogen did not create any degree of unsaturation and did not affect the number of expected hydrogen atoms. Therefore, when calculating the degree of unsaturation of a halogenated compound, simply replace the halogen with H and compare the resulting formula to the C_nH_2n+2 for that number of carbons.

For example, for the formula C₄H₉Br, the expected number of hydrogens is 10 (C₄H₁₀, an alkane with the CnH_2n+2 general formula). The number of hydrogen atoms is, however, 9 for 4 carbons, even though there is no unsaturation occurring because of the bromine.

In this case, C₄H₉Br should have the same HDI as C₄H_10, and we know that C₄H₁₀ is, in fact, a saturated compound.

Another example. What is the degree of unsaturation in the following compound: C₅H₉Cl?

Replacing the Cl with H gives us C₅H_10, which corresponds to the C_nH_2n general formula. C_nH_2n has two hydrogens less than its corresponding alkane (C_nH_2n+2), and this means the compound has one degree of unsaturation (HDI = 1). This compound has the same HDI as C₅H₁₀.

So, remember: for calculating the HDI, treat the halogen as if it were a hydrogen atom.

Nitrogen and Degree of Unsaturation

Comparing the structures of the following alkane and the amine, we can see that nitrogen is changing the number of expected hydrogen atoms.

The nitrogen brought an extra H, and it is now 9 instead of 8 hydrogens for three carbons. Therefore, when there is a nitrogen atom present in the molecular formula, we must subtract one hydrogen atom from the molecular formula. This means that C₃H₉N should have the same HDI as C₃H₈.

As an example: The HDI for compounds with molecular formula C₃H₅N is the same as for C₃H_4, which belongs to the general formula of alkynes (CnH_2n-2), which is four hydrogens less than the corresponding alkane; therefore, the HDI here is 2.

Now that we have covered the effect of N, O, and halogens on the HDI, let’s determine the degree of unsaturation for C₆H₁₀ClNO_2:

1. Replace the Cl with an H. We get C₆H₁₁NO₂
2. Ignore the Oxygens – C₆H₁₁N
3. Subtract one H for the N – C₆H₁₀

This corresponds to the CnH_2n-2 formula, which means the molecule has a 2-degree unsaturation. And this is the structure of the molecule that I had in mind when writing this example:

The suggested molecule has 2 π bonds, thus the HDI = 2. Of course, this is not the only structure that would work for this molecular formula. Any constitutional isomer is possible.

Formulas for Calculating the Degree of Unsaturation

For a hydrocarbon containing only carbon and hydrogen, the general formula is:

where C is the number of carbons and H is the number of hydrogens. Each double bond or ring contributes one degree of unsaturation, and a triple bond contributes two.

For example, let’s calculate the degree of unsaturation for a molecule with the chemical formula C₉H₁₆.

Here, C = 9 and H = 16. Plugging in the values:

This means the molecule has two degrees of unsaturation, which could correspond to two double bonds, one ring and one double bond, or a triple bond.

Degree of Unsaturation for Molecules Containing Heteroatoms

When a molecule contains heteroatoms such as oxygen, nitrogen, or halogens, the basic formula for calculating the degree of unsaturation (DU) is slightly adjusted:

The general formula then becomes:

For example, what is the degree of unsaturation of a compound with the general formula C₁₇H₂₅BrN₂O₄?

Six degrees of unsaturation is quite high, and generally suggests the presence of aromaticity – a benzene ring alone accounts for four degrees of unsaturation (three π bonds plus one ring). The formula is fairly large with 17 carbons, so I drew a hypothetical structure to match it. Can you evaluate the structure to see if it has six degrees of unsaturation?

Summarizing Degree of Unsaturation

You can determine the degree of unsaturation (DU) either by evaluation or by using a formula. For a faster approach, the evaluation method follows these rules:

• Oxygen (O): Ignore oxygen atoms, as they do not affect the hydrogen count needed for a saturated hydrocarbon.
• Halogens (F, Cl, Br, I): Count each halogen as a hydrogen. For example, HBr is treated as H.
• Nitrogen (N): Drop one hydrogen for each nitrogen and check if the formula matches an alkane, alkene, or an alkyne.

Alternatively, for complex chemical formulas, you can use the formula-based approach to calculate the degree of unsaturation.

Check this 60-question, Multiple-Choice Quiz with a 2-hour Video Solution covering Lewis Structures, Resonance structures, Localized and Delocalized Lone Pairs, Bond-line structures, Functional Groups, Formal Charges, Curved Arrows, Constitutional Isomers, and Hydrogen Deficiency Index.

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