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Degrees of unsaturation

How to calculate degrees of unsaturation (the index of hydrogen deficiency) from a molecular formula, with rules for heteroatoms.

Quick answer DoU = (2C + 2 + N − H − X) / 2. Every degree of unsaturation = one ring OR one double bond. Two = one triple bond, or two rings, or a benzene-like ring with three π bonds plus one ring = 4.

The formula

For a molecule with C carbons, H hydrogens, N nitrogens, X halogens, and any number of oxygens or sulfurs: DoU = (2C + 2 + N − H − X) / 2. Oxygen and sulfur don't appear in the formula because they're divalent — they don't change the H count of a saturated structure.

Worked example: C₆H₆

Benzene. DoU = (2·6 + 2 − 6) / 2 = 8/2 = 4. That accounts for one ring + three double bonds = 4. Or one ring + a triple bond + a double bond. Or two rings + two double bonds. The formula tells you the total — figuring out the arrangement is the next step.

Worked example: C₅H₁₁N

DoU = (10 + 2 + 1 − 11) / 2 = 2/2 = 1. One ring OR one double bond. With an N in the structure, candidates include piperidine (a 6-membered saturated ring with N) or pent-2-enylamine (a chain with one double bond).

Why it's useful

When you have a molecular formula and need to guess at structures (from a mass spec problem, a synthesis exercise, or a natural product), DoU narrows the field hugely. A high DoU strongly suggests rings or aromatic systems; DoU = 0 means a fully saturated acyclic structure.

Draw this on the whiteboard

Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.

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