Isomerism is a fundamental concept in organic chemistry. While structural isomers have different bonding connectivity and stereoisomers like E/Z isomers have restricted rotation about a double bond, optical isomerism introduces a deeper layer of spatial arrangement. Optical isomers arise when a molecule has no plane of symmetry, resulting in structures that are mirror images of each other but cannot be superimposed.
🔑 Key Principle
Optical isomerism is a form of stereoisomerism. It occurs in molecules containing an asymmetric carbon atom, known as a chiral centre. This leads to two non-superimposable mirror image forms called enantiomers. These enantiomers are chemically identical in most environments but rotate the plane of plane-polarised light in opposite directions.
1. Chiral Centres and Chirality
The origin of optical isomerism lies in the tetrahedral geometry of carbon. Carbon forms four covalent bonds directed towards the corners of a tetrahedron, with bond angles of approximately \( 109.5^\circ \).
If a carbon atom is bonded to four different atoms or groups of atoms, the molecule lacks a plane of symmetry. This carbon atom is described as an asymmetric carbon or a chiral centre. In structural drawings, chiral centres are conventionally marked with an asterisk (\( \text{*} \)).
A carbon atom bonded to four different atoms or groups of atoms. Its presence makes the molecule chiral, meaning it lacks a plane of symmetry.
A pair of optical isomers that are non-superimposable mirror images of each other.
To understand why these mirror images are non-superimposable, consider a simple molecule like 2-bromobutane. The carbon at position 2 is bonded to:
- A hydrogen atom (\( \text{-H} \))
- A bromine atom (\( \text{-Br} \))
- A methyl group (\( \text{-CH}_3 \))
- An ethyl group (\( \text{-CH}_2\text{CH}_3 \))
Because all four groups are different, carbon-2 is a chiral centre. If you draw the 3D structure of 2-bromobutane and its mirror image, no matter how you rotate the molecules in space, you can never line up all four groups simultaneously. They are non-superimposable, much like your left and right hands.
When drawing a pair of enantiomers in an exam:
1. Draw the central carbon first.
2. Draw one isomer using 3D wedge-dash lines (one vertical bond, one in-plane side bond, one wedge forward, and one dash backward).
3. Draw a vertical dashed mirror line.
4. Reflect the structure exactly on the other side of the mirror line. Do not change the groups themselves, only their relative spatial positions.
2. Physical and Chemical Properties
Because enantiomers contain the same atoms connected by the same bonds, they share almost identical properties. It is important to know the similarities and differences:
| Property Type | Comparison between Enantiomers |
|---|---|
| Physical Properties | Identical melting points, boiling points, densities, and solubilities in standard achiral solvents. |
| Optical Activity | They rotate the plane of plane-polarised light by equal angles but in opposite directions (clockwise vs anticlockwise). |
| Chemical Properties | Identical reactions with achiral reagents. However, they react differently with chiral reagents, such as enzymes or biological receptors. |
3. Optical Activity and Polarimetry
Normal light consists of electromagnetic waves oscillating in all possible planes perpendicular to the direction of propagation. When normal light passes through a polarising filter (polaroid), all waves are blocked except those oscillating in a single, specific plane. This is called plane-polarised light.
When plane-polarised light is passed through a solution containing a single enantiomer, the plane of polarisation is rotated. Substances that rotate plane-polarised light are described as optically active.
The ability of a single enantiomer to rotate the plane of plane-polarised light when it passes through a solution of the compound.
This rotation is measured using an instrument called a polarimeter:
- One enantiomer rotates the plane clockwise, designated as the dextrorotatory or (+) enantiomer.
- The other enantiomer rotates the plane anticlockwise by the exact same angle, designated as the laevorotatory or (-) enantiomer.
4. Racemic Mixtures (Racemates)
In many organic reactions, the starting materials are not chiral, but the products contain a chiral centre. In such cases, the reaction usually yields a racemic mixture (or racemate).
An equimolar (50:50) mixture of two enantiomers. It is optically inactive because the equal and opposite rotations of plane-polarised light cancel each other out.
A racemic mixture is represented by the prefix (\(\pm\)). Because it contains equal amounts of the (+) and (-) enantiomers, when plane-polarised light passes through it:
- The molecules of the (+) enantiomer rotate the plane clockwise by angle \( \theta \).
- An equal number of molecules of the (-) enantiomer rotate the plane anticlockwise by the exact same angle \( -\theta \).
- The net rotation of the plane-polarised light is exactly zero.
Thus, racemic mixtures are always optically inactive.
Structural Formula: \( \text{CH}_2\text{=CH-CH(CH}_3\text{)-CH}_2\text{-CH}_2\text{-CH}_3 \)
Step 1: Check each carbon atom for four different groups.
- Carbon-1 (\( \text{=CH}_2 \)) is double bonded and has two identical hydrogen atoms. (Not chiral)
- Carbon-2 (\( \text{=CH-} \)) is double bonded and is only bonded to three groups. (Not chiral)
- Carbon-3 (\( \text{-CH(CH}_3\text{)-} \)) is bonded to:
1. A methyl group (\( \text{-CH}_3 \))
2. A hydrogen atom (\( \text{-H} \))
3. An ethenyl/vinyl group (\( \text{-CH=CH}_2 \))
4. A propyl group (\( \text{-CH}_2\text{CH}_2\text{CH}_3 \))
Since all four groups are different, Carbon-3 is a chiral centre (\( \text{C}^* \)). - Carbon-4, 5, and 6 are bonded to multiple identical hydrogen atoms. (Not chiral)
Step 2: Draw the 3D wedge-dash structure centered on C-3.
Let's draw C-3 in the centre. We place the ethenyl group vertically upwards, the propyl group in the plane to the left, the methyl group as a wedge pointing forward-right, and the hydrogen atom as a dashed line pointing backward-left:
|
CH₃CH₂CH₂-C*-CH₃ (wedge)
/ \
H (dash)
If the solution has a net rotation of \( 0.00^\circ \), calculate the percentage of each enantiomer present. Explain your answer.
Solution:
- A net rotation of \( 0.00^\circ \) indicates that the solution is optically inactive.
- Because lactic acid contains a chiral centre, the pure enantiomers are optically active.
- The optical inactivity of the mixture means that the clockwise rotation caused by the (+) enantiomer is exactly balanced by the anticlockwise rotation of the (-) enantiomer.
- This cancellation only occurs when the two enantiomers are present in equal amounts.
- Therefore, the mixture is a racemic mixture containing exactly 50% (+)-lactic acid and 50% (-)-lactic acid.