# How do single, double, and triple bonds differ?

Nov 8, 2016

$\text{By their degree of unsaturation.}$

#### Explanation:

Compare ethane to ethylene to acetylene,

i.e. ${H}_{3} C - C {H}_{3} , {H}_{2} C = C {H}_{2} , H C \equiv C H$

Ethane is fully saturated, and has a formula of ${C}_{n} {H}_{2 n + 2}$; here $n = 2$, and there are 6 hydrogen atoms.

Ethylene has a so-called ${1}^{\circ}$ degree of unsaturation, with $2$ hydrogens LESS than the equivalent saturated formula. That is ethylene has 4 hydrogens, 2 hydrogens LESS than the saturated formula.

And acetylene has ${2}^{\circ}$ so-called degrees of unsaturation, with $4$ hydrogens LESS than the equivalent saturated formula.

And so for a given formula each $\text{degree of unsaturation}$ corresponds to a double bond OR a ring. When we add oxygen to the formula, we assess the degree of unsaturation directly. Where there is nitrogen we substract $N H$ from the formula before assessment. Halogens count for 1 hydrogen.

And thus cyclohexanone, ${C}_{6} {H}_{10} O$ has ${2}^{\circ}$ of unsaturation. Acetylene, $H - C \equiv C H$, also has ${2}^{\circ}$ of unsaturation. Napthalene, ${C}_{10} {H}_{8}$ has how many degrees of unsaturation?

What is trivial in these simple case, becomes useful in a more complicated formula. If you are given a chemical formula, without knowing ANYTHING else, you can assess the $\text{degree of unsaturation,}$ and accurately forecast the number of olefinic bonds, and ring junctions the molecule is likely to have.