Erythro enantiomers

For example, ephedrine erythro-(S) and pseudoephedrinethreo-(5)have the same substitution pattern and two asymmetric centers, so there are four possible stereoisomers. The drug ephedrine is a mixture of the erythro enantiomers (li ,2S) and (1S,2R) the threo pair... 

If addition of hydrogen to an alkene forms a product with two asymmetric carbons, only two of the four possible stereoisomers are obtained because only syn addition can occur. (The other two stereoisomers would have to come from anti addition.) One stereoisomer results from addition of both hydrogens from above the plane of the double bond, and the other stereoisomer results from addition of both hydrogens from below the plane. The particular pair of stereoisomers that is formed depends on whether the reactant is a c/5-alkene or a fran -alkene. Syn addition of H2 to a cis-alkene forms only the erythro enantiomers. (In Section 5.8, we saw that the erythro enantiomers are the ones with identical groups on the same side of the carbon chain in the eclisped conformers.)... 

If each of the two asymmetric carbons is bonded to the same four substituents, a meso compound will be obtained instead of the erythro enantiomers. 

Similarly, the addition of Br2 to the trans alkene forms only the erythro enantiomers. Because the cis and trans isomers form different products, the reaction is stereospecific as well as stereoselective. 

Because the addition of Br2 to the cis alkene forms the threo enantiomers, we know that addition of Br2 is an example of anti addition because if syn addition had occurred to the cis alkene, the erythro enantiomers would have been formed. The addition of Bt2 is anti because the reaction intermediate is a cyclic bromonium ion (Section 4.7). Once the bromonium ion is formed, the bridged bromine atom blocks that side of the ion. As a result, the negatively charged bromide ion must approach from the opposite side (following either the green arrows or the red arrows). Thus, the two bromine atoms add to opposite sides of the double bond. Because only anti addition of Br2 can occur, only two of the four possible stereoisomers are obtained. 

Addition of H2 Addition of borane syn cis > erythro enantiomers trans > threo enantiomers... 

Addition of Br2 anti cis > threo enantiomers trans > erythro enantiomers ... 

Two asymmetric carbons have been created in the product. Because the reactant is trans and addition of Br2 is anti, the erythro enantiomers are formed. 

Two asymmetric carbons have been created in the product. Because the reactant is trans and addition of Br2 is anti, one would expect the erythro enantiomers. However, the two asymmetric carbons are bonded to the same four groups, so the erythro product is a meso compound. Thus, only one stereoisomer is formed. 

For example, ephedrine and pseudoephedrine have the same substitution pattern, but substitution of both carbons 1 and 2 means four stereoisomers are possible. Racemic ( )-ephedrine is a mixture of the erythro enantiomers 1R,2S and 1S,2R, whereas the threo pair of enantiomers. 1R.2R and 1S.2S. are known as racemic pseudoephedrine ( )-ephedrine). As discussed for a-methyinorepinephrine, (-)-ephedrine is the naturally occurring stereoisomer and has the 1R.2S absolute configuration with a mixed direct activity on both a- and p-receptors and some indirect activity. Its 1S.2R-(+)-enantiomer exhibits primarily indirect activity. 1S,2S-(+)-Pseudoephedrine has virtually no direct receptor activity and is mostly indirect acting. 

When Fischer projections are drawn for stereoisomers with two adjacent asymmetric centers (such as those for 3-chloro-2-butanol), the enantiomers with the hydrogens on the same side of the carbon chain are called the erythro enantiomers (see Problem 48), whereas those with the hydrogens on opposite sides are called the threo enantiomers. Therefore, 1 and 2 are the erythro enantiomers of 3-chloro-2-butanol (the hydrogens are on the same side), whereas 3 and 4 are the threo enantiomers. 

When perspective formulas are drawn to show the stereoisomers in their less stable eclipsed conformations (those shown next), we can easily see that the erythro enantiomers have similar groups on the same side. We will use both perspective formulas and Fischer projections to depict the arrangement of groups bonded to an asymmetric center. 

The particular pair of stereoisomers that is formed depends on whether the reactant is a cis alkene or a trans alkene. Syn addition of H2 to a cis alkene forms only the erythro enantiomers. (In Section 4.11 we saw that the erythro enantiomers are the ones with the hydrogens on the same side of the carbon chain in the eclipsed conformers.)... 

Because the addition of Br2 to the cis alkene forms the threo enantiomers, we know that anti addition must have occurred, since we have just seen that syn addition would have formed the erythro enantiomers. The addition of Br2 is anti because the two bromine atoms add to opposite sides of the double bond (Figure 6.9). 

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