# How do you calculate the degree of unsaturation?

You convert the molecular formula to its equivalent alkane formula. Then you compare the number of H atoms to those in the alkane with the number of C atoms.
The degree of unsaturation or U-value gives the total number of rings and double bonds in a molecule. Here’s a way to calculate it.
The formula for an alkane is ##C_nH_(2n+2)##. Hexane is C₆H₁₄.
The formulas for hexene and cyclohexane are C₆H₁₂. Thus, the introduction of a double bond or a ring removes 2 H atoms. Each ring or double bond counts as a degree of unsaturation. A triple bond counts as two double bonds.
We could write this as
U = ½(H atoms in alkane – H atoms in compound).
If we have a compound with formula C₃H₄, the saturated alkane would be C₃H₈. Then U = ½(8 – 4) = 2.
This tells that the molecule must have 2 double bonds, 1 ring + 1 double bond, 2 rings, or a triple bond. It must be either propa-1,2-diene, cyclopropene, or propyne.
If there are heteroatoms X, we make the following substitutions:
Group 17: X → H
Group 16: Ignore
Group 15: X → CH
Group 14: X → C
Thus,
C₃H₉N → C₃H₉(CH) = C₄H₁₀; U = ½(10 – 10) = 0. The compound is saturated.
C₂H₆O → C₂H6; U = ½(6 – 6) = 0
For C₆H₆6 (benzene), U = ½(14 – 6) = 4. Any time you get U ≥ 4, you should suspect the presence of an aromatic ring.
C₆H₇NO → C₆H₇(CH) = C₆H₈. The corresponding alkane is C₆H₁₄. U = ½(14 – 6) = 4. Aromatic?
C₆H₁₂O₆ → C₆H₁₂; U = ½(14 – 12) = 1.
C₆H₁₂N₂Br₂ → C₆H₁₂(CH)₂H₂ = C₈H₁₆; U = ½(18 – 16) = 1.

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