Optical Isomerism?

Hi,

We're currently looking at multi-stage synthesis and optical isomerism and have been given a few sheets to do on it this week. On the isomerism sheet, a similar question asking 'why both optical isomers are formed in the mixture' e.g. "The radical above [CH₃C•HCO₂H] then reacts with Cl₂ to form both optical isomers of CH₃CHClCO₂H. Suggest why both optical isomers are formed.

Thanks for any help,

Dan

EDIT: I realised that wasn't a clear representation of the radical in the example so heres a picture of the structure:

The product formed has a chiral carbon centre, it has 4 different groups on a carbon and so it is optically active, there would be 2 enantiomers both mirror images of each other

Testing for optical isomerism?

Hi, sorry I have another question but thought it would be best to post it here.

Another question on the sheet is talking about the need for safety tests on optically-active drugs- the question asks how they test for it. I've read the spread on chirality in the OCR book and looked at the Chemguide article but can't find an answer to this...

Thanks,

Dan

EDIT: is the question just asking me to explain that BOTH isomers must be tested?