3.3.7 - Optical Isomerism

Spotting the chiral centre

Optical isomerism is a type of stereoisomerism. That means the structural formula stays the same, but the arrangement of atoms in space changes.

Here, optical isomerism is treated for molecules with a single chiral centre. The key test is not "does this carbon look unusual?" but "are all four groups attached to it different?"

A carbon atom with four bonds is very common; a carbon atom with four different groups is much rarer. For example, in $\ce{CH3CH(OH)COOH}$ the middle carbon is attached to -H, -$\ce{OH}$, -$\ce{CH3}$, and -$\ce{COOH}$, so it is chiral. In $\ce{CH3CH(OH)CH3}$ the middle carbon is attached to -H, -$\ce{OH}$, -$\ce{CH3}$, and another -$\ce{CH3}$, so it is not chiral.

The diagram below compares a chiral tetrahedral carbon with a similar carbon that is not chiral.

[DIAGRAM: asset_name: 3.7 - Optical Isomerism - Diagram 1; asset_slug: 3.7 - Optical Isomerism - Diagram 1; recommended_method: retained_png; description: Side-by-side tetrahedral comparison on a white background. Left: CH3CH(OH)COOH drawn around a starred central carbon with OH above, H below, CH3 on a solid wedge, and COOH on a dashed bond, plus the note four different groups -> chiral. Right: CH3CH(OH)CH3 drawn with OH above, H below, CH3 on a solid wedge, and a second CH3 on a dashed bond, with no star on the central carbon and the note two groups the same -> not chiral.]

A quick reliable method is to find the candidate carbon, then list its four attached groups carefully. Count the whole group, not just the first atom, so -$\ce{CH3}$ and -$\ce{COOH}$ are different even though both begin with carbon.

Once one chiral centre is present, two optical isomers are possible. The next step is being able to show them clearly in a drawing.

Drawing enantiomers correctly

The two optical isomers formed from one chiral centre are called enantiomers. They are mirror images, but you cannot rotate one in space and make it fit exactly on top of the other.

The structural formula of both enantiomers is the same. For example, both optical isomers of 2-hydroxypropanoic acid can be written as $\ce{CH3CH(OH)COOH}$. To show the difference, you need a displayed formula that makes the tetrahedral arrangement around the chiral carbon visible.

A reliable drawing method is:

1. Draw the same groups attached to the same central carbon in both molecules.
2. Show two bonds in the plane of the page, one solid wedge coming out towards you, and one dashed bond going away from you.
3. Draw the mirror image by reversing the 3D arrangement around the chiral centre, not by changing which atoms are connected.

One valid construction keeps $\ce{COOH}$ above the central carbon and $\ce{CH3}$ to its left, both in the plane. Draw $\ce{OH}$ on a solid wedge and H on a dashed bond for one isomer. Keep the two planar groups fixed and swap only $\ce{OH}$ and H between the wedge and dashed bond for the other. Swapping one pair reverses the configuration; swapping two pairs would give another view of the same configuration.

[DIAGRAM: asset_name: 3.7 - Optical Isomerism - Diagram 2; asset_slug: 3.7 - Optical Isomerism - Diagram 2; recommended_method: deterministic_code; description: Correct tetrahedral mirror pair for lactic acid, solid wedges towards the viewer and hashed wedges away, with four distinct groups.]

The two ordinary bonds in a tetrahedral projection should not be drawn as a straight 180° line: the real bond angles are approximately 109.5°. Models or careful wedge drawings help distinguish an actual reflection from a rotation.

A common mistake is to draw the second molecule as just another view of the first. If a simple rotation would make the two drawings match perfectly, you have not drawn enantiomers. Mirror image and non-superimposable must both be true.