Degree of Unsaturation
Count rings + double bonds from formula
Understanding Degree of Unsaturation
The degree of unsaturation (DU), also called the index of hydrogen deficiency (IHD) or double bond equivalents (DBE), indicates the number of rings and multiple bonds in an organic molecule. Each ring or π bond (double bond) counts as one degree of unsaturation, while each triple bond counts as two.
This calculation is essential in structure elucidation, particularly when combined with spectroscopic data (NMR, IR, MS). By knowing the molecular formula, chemists can immediately determine how many unsaturations exist, guiding interpretation of spectra and narrowing possible structures.
The formula accounts for different element valencies: carbon and silicon contribute positively (valence 4), hydrogen and halogens contribute negatively (valence 1), nitrogen contributes positively (valence 3), while oxygen and sulfur are ignored (valence 2 doesn't affect hydrogen deficiency).
Formula
DU = (2C + 2 + N - H - X) / 2
- DU = degree of unsaturation
- C = number of carbons
- N = number of nitrogens
- H = number of hydrogens
- X = number of halogens (F, Cl, Br, I)
- O and S ignored (valence = 2)
Alternative Formula:
For compounds with Si, P, or other elements: DU = (2C + 2 + N - H - X + P) / 2
Detailed Example: C₆H₆ (Benzene)
Given: benzene, C₆H₆.
Step 1: Identify element counts: C=6, H=6, N=0, X=0
Step 2: Apply formula: DU = (2×6 + 2 + 0 - 6 - 0) / 2
Step 3: Calculate: DU = (12 + 2 - 6) / 2 = 8 / 2 = 4
Step 4: Interpret: Benzene has 3 double bonds + 1 ring = 4 degrees of unsaturation
Answer: DU = 4
This matches benzene's structure: resonance-stabilized ring with three π bonds.
Interpreting DU Values
DU = 0
Saturated hydrocarbon with single bonds only (e.g., alkanes, cycloalkanes counting ring cancels unsaturation)
DU = 1
One double bond OR one ring (e.g., cyclohexane, ethene)
DU = 2
Two double bonds, one triple bond, two rings, or one ring + one double bond
DU = 4
Typically indicates benzene ring (3 double bonds + 1 ring)
DU ≥ 5
Multiple aromatic rings or extensive unsaturation
Structure Elucidation Strategy
Step 1: Calculate DU
Use molecular formula to determine total unsaturations.
Step 2: Check IR Spectrum
Look for C=O (1700 cm⁻¹), C=C (1650 cm⁻¹), C≡C (2100-2260 cm⁻¹), or aromatic C=C (1450-1600 cm⁻¹).
Step 3: Analyze ¹H NMR
Aromatic signals (6.5-8 ppm) indicate benzene ring (DU=4). Vinyl signals (4.5-6 ppm) suggest C=C.
Step 4: Use ¹³C NMR
Count sp² carbons (100-220 ppm) vs sp³ carbons (0-100 ppm) to confirm unsaturations.
Common Examples
| Compound | Formula | DU | Structure Features |
|---|---|---|---|
| Ethene | C₂H₄ | 1 | One C=C double bond |
| Acetylene | C₂H₂ | 2 | One C≡C triple bond |
| Cyclohexane | C₆H₁₂ | 1 | One ring |
| Toluene | C₇H₈ | 4 | Benzene ring |
| Naphthalene | C₁₀H₈ | 7 | Two fused benzene rings |
Important Notes
- Each ring or double bond counts as 1 DU; each triple bond counts as 2 DU.
- Useful in structure elucidation from molecular formula + NMR/IR data.
- Always round to nearest integer; non-integer suggests formula error.
- Radicals and carbenes change the calculation—consult advanced references for these cases.
- Halogens are treated like hydrogens in the formula (same valence contribution).
- DU cannot be negative; if calculation yields negative value, recheck molecular formula.