Optical Isomerism and Non-racemic Mixture

In this JC2 webinar we want to discuss the optical activity of different combinations of optical isomers.

Let's have the following example 1-bromoethanol.

Assuming the optical activity of pure 1-bromoethanol to be 10 degrees, we will have the following combinations:

optical isomers and optical activity

Pure 100% (+)-1-bromoethanol will have optical activity of 10 degree clockwise.

Pure 100% (-)-1-bromoethanol will have optical activity of 10 degree counter-clockwise.

Racemic mixture of 50% (+)-1-bromoethanol and 50% (-)-1-bromoethanol will be optically inactive since the optical activity of equal amounts of both enantiomers will cancel out exactly.

There is also different combinations of both enantiomers where the amount is not 50%-50% and we can refer to them as non-racemic mixtures.

The optical activity of non-racemic mixtures can be easily determined since the major species will determine whether plane polarised light is rotated clockwise or counter-clockwise.

If there is majority (+)-isomer, the optical activity will be overall clockwise, but the magnitude will be less than the pure enantiomer.

For example, 80% (+)-1-bromoethanol and 20% (-)-1-bromoethanol will have optical activity of less than 10 degree clockwise.

Conversely, if there is majority (-)-isomer, the optical activity will be overall counter-clockwise and magnitude will be less than pure enantiomer.

For example, 30% (+)-1-bromoethanol and 70% (-)-1-bromoethanol will have optical activity of less than 10 degree counter-clockwise.

Application to Nucleophilic Substitution

We learnt from Nucleophilic substitution of halogenoalkanes that the optical activity of the product depends on whether the mechanism is via SN1 or SN2.

If you wish to know more details about SN1 and SN2 mechanism please check out this video about nucleophilic substitution of halogenoalkanes.

If the mechanism is via SN1 only, the product will be racemic and optically inactive.

If the mechanism is via SN2 only, the product will be one of the pure enantiomers and optically active.

It's also possible for the mechanism to be a combination of SN1 and SN2, then the product will be a non-racemic mixture.

Let's try to use a simple example of 50% SN1 and 50% SN2 to show how a non-racemic mixture is formed.

how SN1 and SN2 gives non racemic mixture

Since half of the reactants goes via SN1, a racemic mixture of 25% (+)-isomer product B and 25% (-)-isomer product B will be formed.

The remaining 50% of reactants goes via SN2 and forms 50% of either (+)-isomer B or (-)-isomer B. Let's say 50% of (+)-isomer B is formed in this case.

The resultant solution will be a mixture of 75% (+)-B and 25% (-)-B which is our non-racemic mixture.

We will expect this mixture to be able to rotate plane polarised light clockwise since there is majority (+)-isomer, and the magnitude of rotation will be less than that of pure enantiomer.

Topic: Halogenoalkanes, Organic Chemistry, A Level Chemistry, Singapore

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