| Qu1 |
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(a) Propene will react to give 1,2-dibromopropane which contains 1
stereocentre at C2, therefore the product will be a racemic mixture of
(R)- and (S)-1,2-dibromopropane. |
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(b) E-2-butene will react to give 2,3-dibromobutane which has 2 stereocenters
at C2 and C3. The reaction is stereospecific in nature in an
anti
fashion but we should review the possibilities:
Since the 2 chiral centers in the product have the same 4 groups attached
(-H, -Br, -CH3, and -CHBrCH3), then we have a meso
compound, and all 4 of theses structures are actually the same. If you
aren't convinced, rotate the CHIME image to look like each of them ! |
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(c) Z-2-butene, as above, will react to give 2,3-dibromobutane which
has 2 stereocenters at C2 and C3. Now the possibilities are:
Since the 2 chiral centers in the product we will get the two compounds
(R,R) and (S,S) compounds that are enantiomers so we have a racemic product
mixture. If you aren't convinced that there are only two different compounds
here then rotate the CHIME images to look like each of them ! |
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| Qu2 |
If the alkene from method A gives a pair of enantiomers, then
it must be the E-alkene. |
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Therefore the addition of H2 must have occured with syn
stereochemistry. |
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If the alkene from method B gives a a single meso compound,
then it must be the Z-alkene.
Therefore the addition of H2 must have occured with anti
stereochemistry. |
