Degree of Unsaturation Calculator (Double Bond Equivalents)
Number of rings plus double bonds (and twice the triple bonds) in an organic molecule from its molecular formula.
Results
Degrees of unsaturation (rings + π bonds)
4.0
Possible for a neutral molecule?
Yes
What this tool does
Each ring or double bond removes two hydrogens from the fully saturated chain CnH2n+2, so counting the missing hydrogens tells you how many there are before you draw a single structure — the first step in solving a spectrum. Halogens count like hydrogen, each nitrogen adds one place, and oxygen and sulfur change nothing. Benzene, C6H6, gives 4: one ring and three double bonds. A fraction or a negative number means the formula cannot belong to a neutral molecule.
Formula
DBE = C − H/2 − X/2 + N/2 + 1 (oxygen and sulfur do not count)
Variables
| Symbol | Meaning | Unit |
|---|---|---|
c | Carbon atoms (C) | — |
h | Hydrogen atoms (H) | — |
n | Nitrogen atoms (N) | — |
x | Halogen atoms (F, Cl, Br, I) | — |
D | Degrees of unsaturation (rings + π bonds) | — |
OK | Possible for a neutral molecule? | — |
Worked example
- Carbon atoms (C)6
- Hydrogen atoms (H)6
- Nitrogen atoms (N)0
- Halogen atoms (F, Cl, Br, I)0
- Degrees of unsaturation (rings + π bonds)4.0
- Possible for a neutral molecule?Yes
Limitations
- The default values are typical reference figures, not measurements of your situation. Replace them with your own data whenever you have it.
- The calculation runs at full precision and only the display is rounded. If you copy an intermediate value and retype it, small differences can appear.