Degrees of Unsaturation Practice
The degree-of-unsaturation count is the first structural constraint we extract from a molecular formula: how many rings plus π bonds the molecule must contain. It needs to be automatic in both directions: computed from the formula, and counted off a drawn structure. Pick which side you are given below; the other side appears after you answer, and the two must agree.
Common Questions
How do I calculate degrees of unsaturation from a molecular formula?
DoU = (2C + 2 + N − H − X) / 2, where X is the total halogen count. Each degree is one ring or one π bond; a triple bond counts twice. For C₉H₈O₄: (18 + 2 − 8) / 2 = 6. In aspirin that is the benzene ring's four (one ring + three π bonds) plus two C=O groups.
Why do halogens count like hydrogen and oxygen not at all?
The formula counts how far the molecule falls short of a fully saturated chain. A halogen occupies one bonding position exactly the way an H does, so it substitutes one-for-one. Divalent oxygen and sulfur splice into a chain without changing how many hydrogens fit around it — insert an O into C–C or C–H and no H is gained or lost — so they drop out of the count entirely. Trivalent nitrogen brings one extra bonding position, hence the +N term.
Why is 4 such a common answer for degrees of unsaturation?
A benzene ring alone is 4 (one ring + three π bonds), and an enormous share of real organic compounds carries at least one aryl ring. A computed DoU of 4 or more should always make you sketch a benzene ring first and see what's left over. This bank deliberately balances its answers so you cannot score by guessing 4, but on real unknowns, 4 means "look for the aromatic ring."