This case is exactly analogous to the propanal used in Scheme 10.2. The only difference is that the methyl group of propanal has been replaced by a phenyl group. Thus, this aldehyde will react in exactly the same way as propanal and will give an enantiomerically pure product for the reasons discussed in the text.
b)
This case is also exactly analogous to the propanal used in Scheme 10.2. The only difference is that the hydrogen atom attached to the carbonyl group of propanal has been replaced by a phenyl group. Ketones like aldehydes, react with hydrazines to give hydrazones, thus this ketone will react in exactly the same way as propanal and will give an enantiomerically pure product for the reasons discussed in the text.
c)
This aldehyde will undergo the chemistry shown in Scheme 10.2 in exactly the same way as propanal. However, the product will not contain a stereocentre since there are two identical methyl groups attached to the carbon atom adjacent to the carbonyl. Thus, in this case an enantiomerically pure product will not be obtained.
d)
Esters do not react with hydrazines to give hydrazones, so this compound will not undergo the chemistry shown in Scheme 10.2
e) and f)
In both of these cases, the aldehyde will react with hydrazine 10.11
to produce an enantiomerically pure hydrazone analogous to 10.12.
However, there are no hydrogen atoms attached to the carbon atom adjacent
to the carbonyl and so the reaction will not be able to progress any further.
Only hydrogen atoms attached to the carbon atom adjacent to a carbonyl
group are acidic enough to be removed by LDA. Thus, in both case no enantiomerically
pure product will be obtained.
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