Calculating the Degree of Unsaturation.
The degree of unsaturation is valuable because it tells you how many
rings and/or multiple bonds a structure has. This can save a lot
of time, because if the structure doesn't have the right number of
unsaturations it CANNOT be correct! It is easier and faster than
counting hydrogens.
The degree of unsaturation is simply the number of hydrogens expected
on a saturated moleclue, minus the number of hydrogens in the molecule
of interest, divided by two.
We have to divide by two because to make a ring or double bond we
must remove two hydrogens: one on each atom that will be connected.
Degree of Unsaturation = (Number of H in saturated molecule) - (Number of H on this molecule)
2
1. Determine the number of hydrogens expected in the molecule.
Lots of texts give a fancy formula to memorize, but it is easier
just to remember what you know about organic moleclues. The basic
rule is that there are 2n+2 hydrogens, where 'n' is the number of
carbon atoms. Where did this come from? Picture a linear
organic hydrocarbon: butane, pentane, dodecane, it doesn't matter:
H-CH2-CH2-CH2-CH2-H
Note that every carbon has two hydrogens, and then there
is one extra hydrogen at each end to complete the valence: 2n+2.
How many hydrogens would you
expect for a saturated molecule with six carbons?
(choose one)
10 12 14 16
Two hydrogens for each carbon, plus one for each end.
Heteroatoms can get a bit messy, and it is easy to get messed up if you
are trying to memorize those fancy formulas. The easier way is to
picture how you would have to change any old molecule if you added the
atom of interest.
For oxygen (and sulfur), imagine slipping an oxygen between a C-H bond.
H3C-H ---> H3C-O-H
Since we didn't have to add any hydrogens to make the molecule
stable, we can always ignore oxygen when calculating the degree of
unsaturation!
Ignore Oxygen atoms.
For nitrogen (and phosphorous), again imagine slipping a nitrogen between a C-H bond:
H3C-H ---> H3C-N-H
This time
we are not finished, though. Nitrogen needs three bonds, but only
has two. So we must add another hydrogen to make the molecule
stable.
Add one hydrogen for each Nitrogen.
For Halogens (F, Cl, Br, I) you
cannot simply insert the halogen between a C-H bond, because that would
give the halogen two bonds but they can only have one. So we
must replace a hydrogen with a halogen:
H3C-H ---> H3C-Cl
Subtract one hydrogen for each halogens.
Practice on the following molecules:
Molecular formula
Number of rings and/or multiple
bonds (choose one)
C5H10
0 1 2 3 4
C6H6O
0 1 2 3 4
C8H13NO2
0 1 2 3 4
C9H12Cl3NO
0 1 2 3 4