This degrees of unsaturation (DU) calculator helps compute the total number of rings and multiple bonds in a molecular using DU = (2C + N + 2) – (H + X + M)/2 formula.
It is also known as the Index of Hydrogen Deficiency (IHD).
A molecule with no rings or multiple bonds has the maximum possible hydrogen atoms and is considered fully saturated.
Converting C₃H₈ (propane) to C₃H₆ (propene) shows one degree of unsaturation due to the formation of a double bond.
Degrees of Unsaturation Calculator
| Molecular Formula | Calculation | Degree of Unsaturation |
|---|---|---|
| CH₄ | (2×1 + 0 + 2) – (4 + 0)/2 = 0 | 0 |
| C₆H₆ | (2×6 + 0 + 2) – (6 + 0)/2 = 4 | 4 |
| C₃H₆O | (2×3 + 0 + 2) – (6 + 0)/2 = 1 | 1 |
| C₄H₉Br | (2×4 + 0 + 2) – (9 + 1)/2 = 0 | 0 |
| C₂H₄ | (2×2 + 0 + 2) – (4 + 0)/2 = 1 | 1 |
| C₈H₁₀ | (2×8 + 0 + 2) – (10 + 0)/2 = 6 | 6 |
| C₁₂H₁₈ | (2×12 + 0 + 2) – (18 + 0)/2 = 0 | 0 |
| C₇H₈ | (2×7 + 0 + 2) – (8 + 0)/2 = 5 | 5 |
| C₉H₁₂O | (2×9 + 1 + 2) – (12 + 0)/2 = 3 | 3 |
| C₅H₅Cl | (2×5 + 0 + 2) – (5 + 1)/2 = 4 | 4 |
| C₁₀H₁₂N₂ | (2×10 + 2 + 2) – (12 + 0)/2 = 6 | 6 |
| C₄H₆O₂ | (2×4 + 0 + 2) – (6 + 0)/2 = 1 | 1 |
| C₃H₇Br | (2×3 + 0 + 2) – (7 + 1)/2 = 0 | 0 |
| C₉H₁₂N | (2×9 + 1 + 2) – (12 + 0)/2 = 5 | 5 |
| C₈H₉O | (2×8 + 1 + 2) – (9 + 0)/2 = 6 | 6 |
Degree of Unsaturation Formula
The general formula for calculating DU is:
DU = (2C + N + 2) - (H + X + M)/2Where:
- C = Number of carbon atoms
- N = Number of nitrogen atoms
- H = Number of hydrogen atoms
- X = Number of halogen atoms
- M = Number of monovalent metals
What is Degree of Unsaturation?
Degree of unsaturation represents the number of pi bonds or rings in a molecule. Each of these structural features reduces the total hydrogen count by two. A saturated hydrocarbon follows the formula CₙH₂ₙ₊₂, and any deviation from this indicates unsaturation.
How Do You Find the Degree of Unsaturation?
To calculate DU:
- Count all atoms in the molecular formula.
- Apply the formula mentioned above.
- The result indicates the combined number of rings and multiple bonds.
Example:
For C₆H₆
DU = (2×6 + 0 + 2) - (6 + 0 + 0)/2
= 14 - 3 = 4 (benzene has three double bonds and one ring)What is the Degree of Unsaturation of C₂H₆O?
Let’s analyze C₂H₆O:
Apply the formula, treating oxygen as neutral.
DU = (2×2 + 0 + 2) – (6 + 0 + 0)/2
= 6 – 3 = 0
Ethanol (C₂H₆O) has zero degrees of unsaturation, meaning it’s fully saturated.
Can the Degree of Unsaturation Be in Fractions?
No, the degree of unsaturation cannot be a fraction. Since each pi bond or ring corresponds to a reduction of two hydrogen atoms, DU must always be a whole number.
If a calculation yields 2.5, it indicates an error in the molecular formula or calculation.
What is the Degree of Unsaturation of C₈H₁₀N₄O₂? (Caffeine)
For caffeine:
DU = (2×8 + 4 + 2) - (10 + 0 + 0)/2
= (16 + 4 + 2) - 5
= 22 - 5 = 6This corresponds to caffeine’s structure with four rings and two double bonds.
Calculate the Degree of Unsaturation for C₁₁H₈ClBrO
For C₁₁H₈ClBrO:
DU = (2×11 + 0 + 2) - (8 + 1 + 1)/2
= 24 - 5 = 8Calculate the Degree of Unsaturation for C₉H₁₁N
For C₉H₁₁N:
DU = (2×9 + 1 + 2) - (11 + 0 + 0)/2
= 21 - 5.5 = 5References:
- Journal of Chemical Education: https://pubs.acs.org/doi/10.1021/ed100283v
- Royal Society of Chemistry: https://www.rsc.org/Education/Teachers/Resources/cfb/
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