Hi UltimaOnline,
I've 1 question to ask here:
N14/P2/Q6b(ii)
In this case, there are 2 chiral centres, which means there are
4 stereoisomers. At the same time, however, it is also possible to
for the compound to display geometric isomerism (E/Z) owing to the
3-D position of the Br & methyl substituents across the ring.
Why aren't there 8 stereoisomers in all then?
My thinking is that there can't be 8 stereoisomers because there
can only be 4 permutations with which the substituents can be
arranged around C1 and C2. However it seems odd that the number of
stereoisomers is only 4 given that there are already 2 chiral
centres - (and the geometric isomers seemingly don't
contribute/coalesce with the optical isomers).
Thank you! :)
Hi Gohby,
This stereochemistry involved in this question goes beyond the
H2 syllabus, so notice that Cambridge doesn't specifically ask the
candidate how many stereoisomers exist in total, or ask the
candidate to draw all of them out.
The reason for this, is that for cyclic molecules like this,
some of the geometric isomers overlap with the optical isomers (eg.
a trans isomer may also be an R isomer, with the exact same
stereochemistry), and you cannot double count them.
As an analogy, if you have an apple, a (red) pen, and a (round)
pineapple, and you're asked to count how many objects you have,
while it is true that apple is red, and the apple is round, that
doesn't mean you have 4 objects (ie. you cannot double count apple
as "red apple" and also as "round apple"), because you only have 3
objects in total (ie. there is overlap between the "red" and the
"round" properties).
In conclusion, due to overlap between optical and geometrical
isomerism for this molecule (which H2 students needn't worry too
much about, as it's beyond the syllabus), there are only 4
stereoisomers in total, not 8.
No prob, Gohby :)