Structural Biochemistry/Degree of Unsaturation

General Info
In addition to NMR/IR spectroscopy and mass spectroscopy, the ‘Degree of Unsaturation’ analysis is a tool helpful in identifying molecular structure. To utilize this method, one must understand the meaning of unsaturated and saturated compounds. In the case of acyclic alkanes, a saturated compound is one that has the maximum number of single bonds, following the formula, CnH2n+2. In contrast, an example of an unsaturated compound includes an acyclic alkene that follows the formula CnH2n (two hydrogens less, result of double bond). Similarly, cycloalkanes follow the same formula: CnH2n. From these examples, it can be seen that hydrocarbons with any extra bonds (double or triple bonds) or rings will deviate the saturated formula CnH2n+2. This leads to the concept of the degree of unsaturation, which can defined as the sum of numbers of rings and pi bonds present in a molecule.

Each degree of unsaturation refers to a decrease in two hydrogens in the molecule, as a result of the presence of a pi bond or a ring. A degree of unsaturation of 0 means that the molecule follows the formula for an acyclic alkane (CnH2n+2). A degree of unsaturation of 1 means that there is a decrease of two hydrogens in the molecular formula and that the resulting formula will CnH2n. Following the same logic, a degree of unsaturation value of 2 will give a molecular formula of CnH2n-2. In this situation (degree of unsaturation = 2), there are three possible scenarios. There can be two pi bonds present, one pi bond and one ring present, or just two rings present. Similarly, with a degree of unsaturation value of 3, the scenarios are as follows: three pi bonds present, two pi bonds and one ring present, or one pi bond and two rings present.

Degrees of unsaturation can help us determine how an alkane molecule will act. The more saturated a chain is the less fluidity it has. This can have many implications for structural biochemistry because it changes the way a molecule behaves or its function by changing it's shape and fluidity. Degree of unsaturation can illuminate structure by showing double or triple bonds that exist. This phenomena can also be described as hydrogen deficiency. The more rings and double bonds a molecule has the less saturated the molecule. It is important to remember what these double bonds and rings can do to function and behavior of a molecule. Usually when determining degrees of unsaturation molecules that are not carbons and hydrogens are ignored. These can be molecules such as oxygen, nitrogen or any others.

The examples given above apply to simple situations where there are no heteroatoms present. When these atoms are involved, the degree of unsaturation calculations require a little more work. The same process is involved; however, the presence of halogens reduces the number of hydrogens required for saturation, the presence of nitrogen increases the number of hydrogens required, and the number of oxygen doesn’t affect the hydrogens needed. A general equation can be used to determine the degree of unsaturation from a molecular formula:

Degree of Unsaturation = (2Nc + 2 + Nn – Nx – Nh)/2

Nc = number of carbons, Nn = number of nitrogens, Nx = number of halogens, Nh = number hydrogens

This formula can be verified by drawing out various structures with an assortment of rings, pi bonds, halogens, nitrogens, and oxygens.

Understanding the Equation
The presence of nitrogen increases the number of hydrogens required to reach saturation because nitrogen is trivalent. Adding an amine to the compound results in one extra hydrogen, since the nitrogen component is essentially taking away a C-H bond and adding two N-H bonds and a C-N bond. This can be seen in the comparison between ethane (C2H6) and ethanamine (C2H7N). For halogens, there is essentially one C-H bond being replaced with a C-X bond; therefore, the presence of halogens reducs the number of hydrogens required for saturation: compare ethane (C2H6) and chloroethane (C2H5)Cl. Oxygen is divalent, and so it will have no effect on the calculation of degree of unsaturation. To verify, compare ethane (C2H6) and ethanol (C2H6O).

Examples
A simple example is finding the degree of unsaturation of the molecular formula C4H8. Utilizing the formula above, the degree of unsaturation will be (2*4 + 2 + 0 - 0 - 8)/2 = 1. Therefore, the structure of the molecular formula of C4H8 has either an alkene present (1 pi bond), or it is a cyclobutane. A slightly more challenging example is a molecular formula of C5H9N. In this scenario, the degree of unsaturation will be (2*5 + 2 + 1 – 0 – 9)/2 = 2. Thus, there can be two pi bonds present, one pi bond and one ring, or two rings present in the molecule. A third example is the molecular formula C10H13ClO, which will have a degree of unsaturation of (2*10 + 2 + 0 – 1 – 13)/2 = 4. From this value, it can be seen that there are a variety of combinations of pi bonds/rings (carbonyl group included) that would correspond to the calculated value.