Degree of Unsaturation Formula

Degree of Unsaturation

Degree of Unsaturation Formula- We also know it by the name of the index of hydrogen deficiency (IHD). Furthermore, it is an easier way to calculate the number of multiple bonds or rings in an unknown chemical structure.

In other words, it is the calculation that determines the total number of rings and bonds. Moreover, it helps in drawing chemical structures. Also, it does not give information about the components separately. Like, the specific number of rings and or about a double bond or a triple bond.

Besides, we can verify its final structure with the use of Nuclear Magnetic Resonance (NMR). We can also do this with mass spectrometry and IR spectroscopy as well as qualitative inspection. In addition, it is based on comparing the actual molecular formula to what be the possible formula if the structure was saturated.

General formula of the degree of unsaturation

To calculate the unsaturation formula, firstly, we have to know the molecular formula of the molecule that has the form of  \(C_{V} H_{W} N_{X} O_{Y} X_{Z}\). In this way, we can apply the general formula is:

\(Degrees of unsaturation = \frac{2 C + 2 + N – H – X}{2}\)

Here C, H, N, X mean the number of carbon, hydrogen, nitrogen, and halogen atoms (halogens such as F, Cl, Br, I) that is present in the molecular formula. In addition, oxygen and other divalent atoms (atoms that have a vacancy of two) do not contribute with the unsaturation numbers.

Most noteworthy, in a molecule that has no double bond or rings, we call it ‘saturated’ and the number of hydrogen and carbon  atoms keeps a relation:

#Hydrogen = (2 x #Carbons) + 2

Furthermore, in this, the number of hydrogen (#H), is reduced 2 units with each multiple bond or ring.

Uses of the degree of unsaturation formula

The most useful and first step that helps us to identify an unknown organic molecule is to calculate its degree of unsaturation formula. In addition, you can do it in a simple and quick way by following the next examples:

Example of the degree of unsaturation

For example, a molecule whose formula is \( C_{4} H_{6} O_{2}\), has 4 carbon atoms, 6 hydrogen atoms, and 2 oxygen atoms, therefore:

\(Degrees of unsaturation = \frac{(2 x 4) + 2 + 0 – 6 – 0}{2}\) = 2

Hence, the molecule can have 1 triple bond, 2 double bonds, and 2 rings or any combination that gives 2 unsaturation, such as:

Moreover, a molecule with the molecular formula \( C_{9} H_{14} ClNO_{2}\), has 9 carbon atoms, 14 hydrogen atoms, 1 chloride atom, and 2 oxygen atoms, therefore:

\(Degrees of unsaturation = \frac{(2 x 9) + 2 + 1 – 14 – 1}{2}\) = 3

Hence, the molecule has any combination of 3 unsaturation, such as:

Consideration

We should duly note that the information that we have extracted from the degree of unsaturation formula is useful for getting a partial identification of the molecule.

Although, it does not permit the degree of unsaturation to transform between the multiple bonds or rings. Hence, a degree of saturation of 3 can be a triple bond, two rings or a ring with a double bond.

Solved Example for You

Question: Compute the degree of unsaturation for benzene?

Solution: the formula of benzene is \(C_{6} H_{14}\). Also, its structure has one ring and three double bonds so the degree of unsaturation is:

\(Degrees of unsaturation = \frac{(2 x 6) + 2 + 0 – 14 – 0}{2}\) = 0

So, the degree of unsaturation of benzene is 0.