DoU measures how far a molecule sits from the saturated alkane — each ring or π bond removes two hydrogens, so counting missing H tells you how many are present before you draw anything.
Same six carbons, three different levels of unsaturation — read directly from each formula. Structures drawn live.
1. Each Ring or π Bond Is Exactly One Degree of Unsaturation
From the alkane CnH2n+2, closing a ring or adding one double bond each removes an H₂, so both cyclohexane and 1-hexene (C₆H₁₂) are 1 DoU — the formula can't tell them apart.
2. DoU = (2C + 2 + N − H − X) / 2
The formula gives the saturated max hydrogen count (2C + 2), subtracts the H you have, and halves the difference; benzene C₆H₆ works out to (12 + 2 − 6)/2 = 4.
3. Oxygen and Sulfur Are Free; Nitrogen Adds One; Halogens Count Like Hydrogen
Valence explains it: divalent O and S add no H (dropped), trivalent N adds one to the numerator, monovalent halogens count like H — so acetone is 1, acetic acid 1, and acetonitrile 2.
4. Triple Bonds Count as Two, and a Benzene Ring Counts as Four
A triple bond stacks two π bonds (hex-1-yne = 2) and a benzene ring bundles four (three π bonds + one ring), so fused naphthalene reaches 7 — recognizing these bundles consumes big chunks of a total at once.
5. Work Backward: DoU Turns a Formula Into a Short List of Structures
DoU narrows the candidates: C₆H₈ = 3 points to 1,3-cyclohexadiene (ring + two double bonds), while aniline's 4 again forces an aromatic ring, now bearing nitrogen.
6. DoU Is Your First Move in Spectroscopy Problems
Compute DoU from the formula (often the mass spec molecular ion) before reading spectra: 4+ says hunt for an aromatic ring in NMR and IR, as with styrene's 5 (ring 4 + vinyl 1).
7. Summary
DoU = (2C + 2 + N − H − X) / 2 · O and S ignored, N adds one, halogens count like H · each degree = one ring or π bond · triple bond = 2, benzene = 4 · 0 = saturated chain, 4+ = aromatic.
Worked example
- DoU = (2C + 2 + N − H − X) / 2.
- Here = (2·6 + 2 − 10) / 2 = (14 − 10) / 2 = 2.
- Two degrees = any mix of rings and/or π bonds: a ring + a double bond, two double bonds, or one triple bond.
Answer. 2 degrees of unsaturation (e.g. cyclohexene, a diene, or an alkyne).
Quiz yourself
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DoU = (2·4 + 2 − 6)/2 = 4/2 = 2. Two degrees could be two double bonds (1,3-butadiene), one ring plus one double bond (cyclobutene), two rings (bicyclobutane), or a single triple bond (but-1-yne or but-2-yne).
O and S are divalent — inserting an –O– or –S– into a saturated skeleton neither adds nor removes hydrogens, so they have no effect on the count. Nitrogen is trivalent, giving an extra bonding site, so each N adds one to the numerator (the "+ N" term).
Halogens count like hydrogen (the "− X" term): DoU = (2·6 + 2 − 11 − 1)/2 = 2/2 = 1. One ring or one double bond — for example chlorocyclohexane, or a chlorinated hexene.
DoU = (18 + 2 − 10)/2 = 5. Four of those degrees almost certainly belong to a benzene ring (the aromatic bundle), leaving one more — likely a C=O or a C=C. The oxygen plus a leftover degree points toward an aromatic aldehyde, ketone, or a phenyl vinyl ether. Confirm the carbonyl with an IR band near 1700 cm⁻¹.
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.