# What are the similarities and differences between structural, geometric, and optical isomers?

Jun 7, 2016

Structural isomers differ in their connectivity; geometric isomers differ in their geometry.

#### Explanation:

Take a simple organic formula, ${C}_{4} {H}_{8}$. This has one degree of unsaturation. What does this mean? A saturated molecule has a formula of ${C}_{n} {H}_{2 n + 2}$. Each degree of unsaturation, here a double bond, OR a ring, reduces the hydrogen count by 2.

Butene has 2 structural isomers: $\text{1-butene}$; and $\text{2-butene}$. For these isomers the connectivity with respect to the double bond is different.

Now consider $\text{2-butene}$; because of the position of the double bond, the pendant methyl groups can be on the same side of the double bond, to generate the $\text{cis}$ isomer, versus opposite sides, to generate the $\text{trans}$ isomer. For both isomers, the connectivity is the same: ${C}_{1}$ connects to ${C}_{2}$......to ${C}_{4}$. Nevertheless the geometry is different.

Now consider optical isomerism (a subset of geometric isomerism). For a pair of optical isomers, here the connectivity is precisely the same. Nevertheless they are handed; i.e. non-superposable mirror images. For instance, if you had an identical twin, can your left hand shake his right hand, or ice versa? The answer is no, even though their right hand is the mirror image of the left hand. (Interestingly, I am told that identical twins do not possess identical sets of fingerprints).

Here is a tip with regard to chiral centres. If you have correctly represented a chiral centre, $C {R}_{1} {R}_{2} {R}_{3} {R}_{4}$, on the printed page, or by constructing a model (which you certainly should do if time permits), the interchange of any 2 substituents (i.e. ${R}_{1}$ for ${R}_{2}$ or ${R}_{1}$ for ${R}_{3}$) results in depiction of the enantiomer. Interchange again (and it could be original pair OR the other two) you get back to the original stereoisomer. Capisce?