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    The question is:

    A sample of lactic acid (CH3CH(OH)COOH) was extraced from a natural source and sound to be optically active, it was then subjected to 2 reactions:

    CH3CH(OH)COOH (sample 1) ->(step A)-> CH3COCOOH ->(Step B)-> CH3CH(OH)COOH (sample 2)

    Previous questions arent relevant, just asking what are the reactions, reagents, conditions for Steps A and B which I hope Im correct in saying:

    Step A- Oxidation of an alcohol to make a ketone with H+/Cr2O7 (2-) and reflux and Step B is Reduction with NaBH4.

    Why is sample 2 not optically active? Its the same as sample 1 (well, its not as it isnt optically active but it has the same structure). Im thinking its to do with under oxidation the carbon only has 3 groups attached to it and not the 4 to make it a chiral carbon, and during reduction an additional group is added (artifically?) to make it have 4 atoms/groups attached to it.

    Im stuck, any help would be appreciated, btw its a 3 mark question so I really don't know what to write.

    Thanks again!
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    Lactic acid from a natural source will be chiral, but oxidising the alcohol in step A you remove the chirality. The reduction in step B is a non-chiral reduction and so you will get a mixture of the two enatiomers - in fact a racemic mixture and so you won't see any optical activity
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    (Original post by EierVonSatan)
    Lactic acid from a natural source will be chiral, but oxidising the alcohol in step A you remove the chirality. The reduction in step B is a non-chiral reduction and so you will get a mixture of the two enatiomers - in fact a racemic mixture and so you won't see any optical activity
    Thankyou for your help.

    If the oxidation removed chirality, and reaction B has no chirality so gives sample 2 no chirality (if that makes sense?) how can it be a racemic mixture as there are no optical isomers in there to be in equal amounts since all chirality has been removed in step A?

    Im not sure if my question has been worded how I wanted it to, sorry. Please ask again if you don't understand.

    Thanks again for your first bit of help!
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    Sample 2 will be a 50:50 mixture of both possible optical isomers - the natural one (which you start with) and the unnatural one. When you pass polarised light through a racmic mixture such as this it will give you a zero reading because the two rotate in opposite directions equally

    You are 'creating chirality' in the reduction step but there will be no preference to which optical isomer is prefered - so we call it a non-chiral reduction. It is possible to do chiral reductions i.e. to favour one isomer over the other but the reagents used here won't do that.
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    (Original post by EierVonSatan)
    Sample 2 will be a 50:50 mixture of both possible optical isomers - the natural one (which you start with) and the unnatural one. When you pass polarised light through a racmic mixture such as this it will give you a zero reading because the two rotate in opposite directions equally

    You are 'creating chirality' in the reduction step but there will be no preference to which optical isomer is prefered - so we call it a non-chiral reduction. It is possible to do chiral reductions i.e. to favour one isomer over the other but the reagents used here won't do that.
    Aha, thanks for clearing that up, I understand you now.

    I misinterpreted what you said before, and thought that the reduction didn't make any chiral compounds, not that it makes a mixture of the 2 optical isomers.

    Thanks again, you've really helped me!
 
 
 
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