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 C?H??.

The formulas for hexene and cyclohexane are C?H??. 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 C?H?, the saturated alkane would be C?H?. 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:

1)Group 17: X → H

2)Group 16: Ignore

3)Group 15: X → CH

4)Group 14: X → C

Thus,

C?H?N → C?H?(CH) = C?H??; U = ½(10 – 10) = 0. The compound is saturated.

C?H?O → C?H6; U = ½(6 – 6) = 0

For C?H?6 (benzene), U = ½(14 – 6) = 4. Any time you get U ≥ 4, you should suspect the presence of an aromatic ring.

C?H?NO → C?H?(CH) = C?H?. The corresponding alkane is C?H??. U = ½(14 – 6) = 4. Aromatic?

C?H??O? → C?H??; U = ½(14 – 12) = 1.

C?H??N?Br? → C?H??(CH)?H? = C?H??; U = ½(18 – 16) = 1.

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