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- Aug 6, 2006
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these are from part I of BR ochem
page 227, #37. how can you explain a 0 optical rotation after a reaction?
a. SN1
d. E1
the answer is d, because any nucleophilic substituion reaction that proceeds with inversion or retention of a chiral center remains chiral...
but then on page 232, #52, they contradict this by saying the answer says if a reactions proceeds by pure Sn1, you get 50% R and 50% S.
I thought Sn1 gives you a racemic mixture so you get 0 degree optical rotation?
with Sn2, you get all R or all S, there is no other possibilites.
with Sn1, do you get exactly 50% R 50% S? or its possible to get 40% R 60% S?
on page 158, the 2 NMR spectra are supposed to be different because of coupling. the answer says "in spectrum I, the distance between the peaks in the alkene region (coupling constant) is greater than in spectrum II). what are they talking about? to me, they look exactly the same. and why does coupling change the distance between peaks?
page 227, #37. how can you explain a 0 optical rotation after a reaction?
a. SN1
d. E1
the answer is d, because any nucleophilic substituion reaction that proceeds with inversion or retention of a chiral center remains chiral...
but then on page 232, #52, they contradict this by saying the answer says if a reactions proceeds by pure Sn1, you get 50% R and 50% S.
I thought Sn1 gives you a racemic mixture so you get 0 degree optical rotation?
with Sn2, you get all R or all S, there is no other possibilites.
with Sn1, do you get exactly 50% R 50% S? or its possible to get 40% R 60% S?
on page 158, the 2 NMR spectra are supposed to be different because of coupling. the answer says "in spectrum I, the distance between the peaks in the alkene region (coupling constant) is greater than in spectrum II). what are they talking about? to me, they look exactly the same. and why does coupling change the distance between peaks?
