Degree of Unsaturation Calculator

The degree of unsaturation — also called the index of hydrogen deficiency or the double bond equivalent — is one number squeezed out of a molecular formula. It counts the rings and pi bonds that formula can carry: a ring costs one unit, a double bond costs one, and a triple bond costs two.

You do not need the structure to get it, and that is the whole point. Count the carbon, hydrogen, nitrogen and halogen atoms in the formula and the arithmetic is finished. Benzene, $\mathrm{C_6H_6}$, comes out at 4 — exactly what its structure spends on three double bonds and one ring.

That makes it a checking number rather than a descriptive one. If a structure has been proposed for a formula, the two have to agree: a formula worth 4 units of unsaturation cannot belong to a molecule drawn with a single double bond and no ring anywhere. Students use it to test their own answers; anyone reading a formula from a mass spectrum, a combustion analysis or a database uses it to see how much unsaturation the formula is carrying before a pencil touches paper. If all you have is a percentage composition rather than a formula, turn that into a formula first with the empirical formula calculator.

The whole page is one line of algebra, solved in whichever direction you leave open:

The numerator counts hydrogens that are missing from a reference skeleton, and the 2 in front, together with the $2\,\mathrm{C}$, is that skeleton's capacity. A chain of $\mathrm{C}$ carbons, joined by single bonds only and closed into no ring, holds $2\,\mathrm{C} + 2$ hydrogens. Close one ring, or replace one single bond with a double bond, and the same carbons can carry two hydrogens fewer. Dividing by two turns "two hydrogens missing" into "one unit of unsaturation".

The same equation, written as missing hydrogens

Written this way, each box has an obvious job in building the reference: no memorising required, only valence.

The halogen line is the one people trip over. A halogen does not create unsaturation on its own — swap a hydrogen for a chlorine and the degree of unsaturation does not move — but it occupies a position a hydrogen could have occupied, so it has to sit on the same side of the subtraction as hydrogen. What the numerator really measures is $\mathrm{H} + \mathrm{X}$, the monovalent atoms standing on the skeleton, against the reference ceiling.

One unit, several possible structures

A degree of unsaturation of 4 describes benzene, but it does not prove benzene. The same 4 is satisfied by four separate double bonds, by two triple bonds, by two rings with two double bonds, or by one ring with three double bonds. The arithmetic does not prefer a ring to a pi bond, and $\mathrm{C_4H_6}$ is the classroom illustration: buta-1,3-diene (two double bonds), but-1-yne (one triple bond) and cyclobutene (one ring and one double bond) all give 2.

Reading the number in practice: 0 means no rings and no pi bonds at all, as in hexane $\mathrm{C_6H_{14}}$ or ethanol $\mathrm{C_2H_6O}$; 1 is a single ring or a single double bond, as in cyclohexane $\mathrm{C_6H_{12}}$ or propene $\mathrm{C_3H_6}$; 2 covers anything totalling two, including two rings in a fused or bridged system; and a larger number is simply a larger total, to be allocated by evidence rather than by the count.

Why a half can appear

Doubling the formula gives $2\,\mathrm{DoU} = 2 + 2\,\mathrm{C} + \mathrm{N} - \mathrm{H} - \mathrm{X}$. The first two terms are even, so the right-hand side is even exactly when $\mathrm{N} + \mathrm{H} + \mathrm{X}$ is even. When that sum is odd the degree of unsaturation is a half-integer, and for a neutral molecule that is always informative: either the species has an unpaired electron, as the methyl radical $\mathrm{CH_3^{\bullet}}$ does with its 0.5, or one of the counts is wrong — most often a halogen left out. The page prints the half rather than rounding it, because 1.5 is the clue and 2 would be the mistake.