**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)/2**

Where:

**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 = 6**

This 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 = 8**

## Calculate the Degree of Unsaturation for C₉H₁₁N

For C₉H₁₁N:

**DU = (2×9 + 1 + 2) - (11 + 0 + 0)/2
= 21 - 5.5 = 5**

### References:

- 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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