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picture above describes two ways of forming epoxides from alkene.
(a) peroxy acid
(b) Br2 and H2O / NaOH
it says R groups stay in the same position.
and I want to ask some about (b).
is it possible R groups that were on the same side before reaction occurs can be opposite side of where the double bond used to be after product is formed?
I've drawn this in the picture below.
The problem with NK Yu's diagram is that the bond rotation is unnecessary to obtain the correct conformation for epoxide formation. The epoxidation and the bromination/base give the same product and both processes are stereospecific. (Z)-Butene (1) gives the cis-oxirane 2 while (E)-butene affords the trans-oxirane (not shown).
Peracid epoxidation adds oxygen to either face of the π-system of (Z)-butene to provide meso-oxirane 2. Bromination of 1 affords an intermediate meso-bromonium ion 3 by the same type of addition as was seen in the peroxidation. When hydroxide effects an SN2 (blue arrow) inversion of stereochemistry occurs (R $\rightarrow$ S), the S,S-bromohydrin 3 is formed. With equal probability the path of the red arrow leads to the R,R-enantiomer 4. Deprotonation of the racemic bromohydrin leads to a second inversion of stereochemistry to form meso-oxirane 2, i.e., R,R $\rightarrow$ R,S and S,S $\rightarrow$ R,S. The commonality here is that direct, peracid epoxidation involves no inversions of stereochemistry while the bromination/base route involves two inversions of stereochemistry. The bottom line is that an even number of inversions, zero being considered even, produces the same stereochemistry.
