Calculating Degrees of Unsaturation
Because of the presence of a double bond, alkenes have two fewer hydrogens than the corresponding parent hydrocarbon. For example, ethene (H2C=CH2) has the molecular formula C2H4 and ethane (CH3CH3) has the formula C2H6 (following the 2n + 2 rule). Cycloalkanes, likewise, have two fewer hydrogens than the parent hydrocarbon since two of the valences are utilized to close the ring: cyclohexane is C6H12, while hexane is C6H14.
Knowing this relationship, it is possible to take a molecular formula and calculate the degree of unsaturation; that is, the total number of multiple bonds or rings in a molecule. This information can then be utilized in the conversion of analytical data into structural possibilities.
For hydrocarbons, the process is simple:
- take the parent hydrocarbon and calculate the number of hydrogens using the 2n + 2 rule,
- every two hydrogens that are "missing" in the analysis of the unknown represents one degree of unsaturation.
For compounds containing elements other than carbon and hydrogen, degrees of unsaturation can be calculated as follows:
- Organohalogen compounds: since a halogen is simply a replacement for a hydrogen in an organic molecule (a valence of one), you simply add the total number of halogens to the carbon-hydrogen analysis, and calculate the unsaturation number as described above.
- Organooxygen compounds: since oxygen is divalent, it has no effect on the calculation for the degree of unsaturation, and can be simply ignored. This can be demonstrated by considering ethanol (CH3CH2OH); removing the oxygen produces ethane (CH3CH2-H). The calculation for ethanol (ignoring the oxygen) therefore gives no degrees of unsaturation. For a carbonyl, (i.e., acetone, CH3COCH3), ignoring the oxygen gives C3H6, two hydrogens short of (2n + 2), and one degree of unsaturation. The carbonyl is therefore equivalent to one degree of unsaturation.
- Organonitrogen compounds: since nitrogen is trivalent, an organonitrogen compound has one more hydrogen than an equivalent hydrocarbon has, and therefore you should subtract the number of nitrogens from the total number of hydrogens and calculate as described above.
Some examples:
|
C6H10O |
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"C6H10" which is 4 short of C6H14 |
|
2 DU |
|
C5H9N |
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"C5H8" which is 4 short of C5H12 |
|
2 DU |
|
C6H1?N |
|
"C6H12" which is 2 short of C6H14 |
|
1 DU |
|
C6H13N4 |
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"C6H8" which is 6 short of C6H14 |
|
3 DU |
|
C8H8O |
|
"C8H8" which is 10 short of C8H18 |
|
5 DU |

Each of the six rings in this molecule is shown in a separate color. Note that, when counting rings in a polycyclic compound, you are not allowed to retrace any path, only to connect atoms following new paths.
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