Degrees of Unsaturation Explained

degrees of unsaturation are important for identifying candidates

Core Concepts

In this article, we take a close look at “Degrees of Unsaturation,” including what they are, how to calculate them, and some important caveats with the concept.

Topics Covered in Other Articles

What are Degrees of Unsaturation?

“Degree of Unsaturation”, also known as “Hydrogen Deficiency Index”, is an important tool chemists use when they’re trying to identify an unknown compound. In organic chemistry, the raw molecular formula doesn’t give away all that much information about the structure of a compound. For instance, we can imagine a multitude of structures that fit the formula C7H9O3N.

multiple options for the same formula. degrees of unsaturation

However, just by looking at the formula, chemists can calculate the compound’s “degrees of unsaturation,” which provides a little more insight. Put simply, a structure’s degrees of unsaturation are the total sum of its pi bonds and ring structures

Importantly, “unsaturation” doesn’t refer to saturation in the solvent-solute sense, but rather the degree to which carbon is “saturated” with hydrogens. You may recall from biochemistry that saturated fatty acids have more hydrogens than unsaturated fatty acids.

these fatty acids have different degrees of unsaturation

In addition, unsaturated fatty acids have more pi bonds, in the form of double bonds, than saturated fatty acids. From this, we can observe that there exists some inverse relationship between pi bonds and hydrogens. Also, ring structures tend to have fewer hydrogens than noncyclic structures; for example, take hexane (C6H14) versus cyclohexane (C6H12).

hexane and cyclohexane have different degrees of unsaturation

These relationships between hydrogens, pi bonds, and rings all affect how we calculate degrees of unsaturation.

How To Calculate Degrees of Unsaturation

For a hydrocarbon without heteroatoms, we can use the following formula to calculate degrees of unsaturation:

DoU = (2C + 2 – H)/2

C = Number of carbons in formula

H = Number of hydrogens in formula

As mentioned before, the total degrees of unsaturation indicate the combined pi bonds and rings in the structure. This formula comes from the fact that the maximum number of hydrogens in a hydrocarbon is twice the number of carbons plus two. We can see this phenomenon in the totally-saturated compounds propane, hexane, and dodecane.

completely saturated substances have 2c+2 hydrogens. zero degrees of unsaturation

If a hydrocarbon has fewer hydrogens, it must have pi bonds or ring structures. This becomes incredibly useful, as it limits the possible structures of a formula to only a few plausible options.

degrees of unsaturation narrows the options

However, even if a compound has zero degrees of unsaturation and is therefore completely saturated, there are still multiple structures the formula could indicate. For instance, C6H14 most often indicates hexane, but 3-methylpentane has the same formula.

these both have zero degrees of unsaturation hexane and 3-methylpentane

Also, as a useful tip, if a relatively small structure (less than 10 carbons) has an unsaturation number of 4 or more, the chances are high that it has a phenyl group. Though alternate structures exist in these cases, they likely don’t have the same structural stability and simplicity as an aromatic molecule.

Degrees of unsaturation seem useful for hydrocarbons, but what does it tell us when our unknown has oxygen, nitrogen, etc?

Degrees of Unsaturation with Heteroatoms

When assessing the effect of a heteroatom (non-carbon-or-hydrogen) on unsaturation, we need to ask ourselves one important question: does this atom add or take away possible locations for hydrogen?


To start, let’s look at halogens. Notice, when chlorine or bromine exists in a structure, it essentially replaces hydrogen. 

halogen replacement of hydrogen

For instance, for a structure with two carbons and one chlorine, the structure can only have a maximum number of five hydrogens. However, since the structure has no pi bonds or rings, it must still be saturated and have zero units of unsaturation. 

chloroethane, saturated

Thus, when calculating degrees of unsaturation, we must add the number of halogens to the hydrogen total. This makes sense when you consider that each halogen in a compound indicates a position where hydrogen could occupy.


Next, let’s consider nitrogen. Notice, when nitrogen occupies a space that hydrogen could occupy, it provides two additional spaces for hydrogen. 

amino replacementt for hydrogen

That’s a net gain of one spot for hydrogen to bond. Thus, for a structure to be fully saturated, it must have an additional hydrogen per nitrogen. Therefore, we need to subtract the number of nitrogens from the hydrogens when calculating unsaturation.


Oxygen and Sulfur

Finally, let’s look at oxygen and sulfur. Notice, when either replaces hydrogen, it provides one space for hydrogen to bind.

methyl to alcohol

That means the presence of oxygens and sulfur has a net-zero impact on unsaturation. We can see this in ethanol and ethanethiol, both of which having completely saturated structures. 

saturated ethanol and ethanethiol

Thus, we can leave out oxygens and sulfur completely when calculating degrees of unsaturation.

Deluxe Degrees of Unsaturation Formula

Incorporating what we know about heteroatoms, we can thus use the following formula to calculate unsaturation for heterocompounds:

Degrees of Unsaturation = (2C + 2 – (H + X – N))/2

C = Number of carbons in formula

H = Number of hydrogens in formula

X = Number of halogens in formula

N = Number of nitrogens in formula

Note: if your unknown compound has a heteroatom different than those we’ve talked about, such as a transition metal, it would have an unpredictable effect on degrees of unsaturation because it likely has a range of possible oxidation states. However, the above formula applies to the vast majority of compounds encountered in organic chemistry courses.

As mentioned before, though calculating degrees of unsaturation narrows your options, you’ll still have multiple possible compounds for a given formula. Thus, degrees of unsaturation are best used in concert with Mass Spec, IR, and NMR techniques, which provide their own pieces to the puzzle. With each analysis technique, you can often derive the exact structure of an organic compound. To see this in action, check out this article on structural analysis.