The reaction pathway / energy diagram is shown below. The first step
in the reaction is between the achiral aldehyde 10.10 and the enantiomerically
pure hydrazine 10.11 to give a single enantiomer of hydrazone 10.12.
Since no new stereocentre (or other stereofeature) is created in this step,
there is only one curve in this part of the pathway. Note that this part
of the reaction pathway has been simplified in the diagram below since
a single transition state and no intermediates are shown. This is not actually
the case but is not important in the context of the question. The next
stage in the reaction is the deprotonation of enantiomerically pure hydrazone
10.12 by the achiral base LDA. Once again, there is only one possible
enantiomer of the anionic product so a single curve is shown in the diagram,
however see the note at the end of this explanation.
The next step is the key step in the reaction pathway since anion 10.13
can react with the achiral hexyl bromide on either diastereotopic face
to create a new stereocentre. Reaction on one face (the back face as drawn
in Scheme 10.2) proceeds through a lower energy transition state
than reaction on the other face due to steric hindrance to reaction on
the front face as drawn in Scheme 10.2. The products obtained by
reaction on the two faces of the anion are diastereomers of one another
since they will differ in configuration at one stereocentre but not at
the other, thus in this step of the reaction both the transition states
and products are at different energies and the observed product (10.14)
will be that obtained from the lower energy transition state as shown by
the red curve in the diagram below. The final step of the reaction is the
hydrolysis of the diastereomeric hydrazones (10.14 and its diastereomer)
by achiral acid. The products of this reaction are 10.11 and
either 10.15 (which will be formed from 10.14) or the
enantiomer of 10.15 (which will be obtained from the diastereomer
of 10.14). Since the products of the reaction are enantiomers of
one another, they will have equal energies. However, the two possible starting
materials are diastereomers rather than enantiomers so the starting materials
and transition states will be at different energies as shown in the reaction
pathway / energy diagrams.
It can be argued that there should be two curves in the region of the
reaction pathway between 10.12 and 10.13 since the two hydrogen
atoms which the base can abstract are diastereotopic as shown in Scheme
10.2. This is a valid argument, and would result in two curves starting
at 10.12 and ending at 10.13 but passing through transition
states of different energies. Which diagram is correct will depend upon
whether the base reacts selectively with one of the two diastereotopic
hydrogen atoms in which case the reaction pathway as drawn above is correct,
or whether the base randomly abstracts either of the two diastereotopic
hydrogen atoms in which case the modified diagram would be appropriate.
Note that in either case, there is a single structure for 10.13,
so this has no relevance to the eventual origin of the new stereocentre.
This reaction pathway / energy diagram will be very similar to that described above for Scheme 10.2. Thus, for the first two steps (formation of the enantiomerically pure imide from 10.16 and the achiral acid chloride and deprotonation of the imide for the lithium enolate), only one reaction pathway is possible since no new stereoisomers are formed. Again, the actual reaction profile has been simplified in the diagram below since only one transition state is shown for each step, and the same comment concerning the diastereotopic nature of the hydrogens in 10.12 discussed in the first example is also relevant to the diastereotopic hydrogens of the imide intermediate in this case.
The next step is the key step in the reaction pathway since the
enolate can react with the achiral alkyl halide on either diastereotopic
face to create a new stereocentre. Reaction on one face (the back face
as drawn in Scheme 10.3) proceeds through a lower energy transition
state than reaction on the other face due to steric hindrance to reaction
on the front face as drawn in Scheme 10.3. The products obtained
by reaction on the two faces of the enolate are diastereomers of one another
since they will differ in configuration at one stereocentre but not at
the other, thus in this step of the reaction both the transition states
and products are at different energies and the observed product will be
that obtained from the lower energy transition state as shown by the red
curve in the ,diagram below. The final step of the reaction is the hydrolysis
of the diastereomeric imides by achiral lithium hydroxide. The products
of this reaction are 10.16 and either enantiomer of the product
carboxylic acid. Since the products of the reaction are enantiomers of
one another, they will have equal energies. However, the two possible starting
materials are diastereomers rather than enantiomers so the starting materials
and transition states will be at different energies as shown in the reaction
pathway / energy diagrams.
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