Enantiomers anti-addition

If a reaction can go through only an anti addition, then the two groups must be added across opposite sides of the double bond. So we will get only the set of enantiomers where the groups are on opposite sides ... 

Sometimes, the reaction is not stereoselective. In other words, we get both syn and anti addition. So we get all fonr prodncts (both sets of enantiomers). 

So let s go back to our example above with the addition of Br and Br across a double bond. This reaction is an anti addition, so we get only the set of enantiomers that has the two Br groups on opposite sides of the ring ... 

In total, there are four possible products (two pairs of enantiomers). The two prodncts of a syn addition represent one pair of enantiomers. And the two products from an anti addition represent the other pair of enantiomers. 

In cases like this, the stereochemistry is still irrelevant (as long as the compound does not possess any other stereocenters). Why With only one stereocenter, there will only be two possible products (not four). These two products will represent a pair of enantiomers (one will be R and the other will be S). You will get both of these products whether the reaction proceeds through a syn addition or through an anti addition. If the reaction is a syn addition, the OH group can come from above the plane or from below the plane of the double bond, giving both possible products. Similarly, if the reaction is an anti addition, the OH group can come from above the plane or from below the plane of the donble bond, giving both possible products. Either way, we get the two possible products. 

Therefore, the fact that the reaction proceeds through a syn addition is not important for predicting the products. If the reaction had been an anti addition, we would have obtained the same products. In fact, if the reaction had not been stereospecific at all, we still would have obtained the same two products (the pair of enantiomers above). 

Answer (a) We are adding Br and Br, so the regiochemistry is irrelevant. But what about the stereochemistry We look to see whether we are creating two new stereocenters. In this case, we are. So, the stereochemistry is relevant. We have explored the mechanism and justified why the reaction must be an anti addition. So, we must draw the pair of enantiomers that we would get from an anti addition. To do this properly, it will be helpful to redraw the alkene, as we have done many times before ... 

Answer (a) We are adding OH and OH, and therefore, regiochemistry will be irrelevant. What abont stereochemistry In this case, we are creating two new stereocenters, so we mnst carefully consider the stereochemistry of this reaction in order to draw the correct pair of enantiomers. This two-step synthesis gives an anti addition of OH and OH. Therefore,... 

Thus, (Z)-but-2-ene will react to give 2,3-dibromo-butane as a pair of enantiomers, R,R and S,S, a result of the anti addition. A racemic product will thus be formed, because there is equal probability of... 

Two facts demonstrate that halogenation follows a different mechanism from that of hydroha-logenation or hydration. First, no rearrangements occur, and second, only anti addition of X2 is observed. For example, treatment of cyclohexene with Br2 yields two trans enantiomers formed by anti addition. 

The reagent (Br2 + HpO) adds the elements of Br and OH to a double bond in an anti fashion— that is, from opposite sides. To draw two products of anti addition add Br from above and OH from below in one product then add Br from below and OH from above in the other product. In this example, the two products are nonsuperimposable mirror images—enantiomers. 

This reaction proceeds via a cyclic bromonium ion. The anti-addition of the bromide anion may take place at either of the two carbon atoms that form part of the ring, and so two products are formed in a 1 1 ratio. These are the threo d,l pair of stereoisomers. The resultant mixture is not optically active, because each enantiomer is produced in equal amounts. 

The two olefins carry a total of two enantiomeric pairs of diastereotopic faces. When a tartrate-titanium epoxidation system is allowed to react with this substrate, approach to only one of these four faces simultaneously satisfies the requirements of both the catalyst (which prefers the Re face) and the substrate (which prefers 1,2-anti addition). To the extent that the rate of epoxidation at this face exceeds that of the others (ki in Scheme 8.10), one product predominates. Minor diastereomers result from pathways k2 and k4. However, note that the pathway with a rate of ky (mismatched 1,2-anti diastereoselectivity combined with disfavored Si enantiofacial attack) affords the enantiomer of the major isomer. 

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.)... 

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. 

Because only anti addition occurs, addition of Br2 to a cyclohexene forms only the enantiomers that have the added bromine atoms on opposite sides of the ring. 

Again, since the starting material is an alkane, the first reaction must be a radical substitution. Bromination leads to selective substitution of the tertiary hydrogen. Under E2 conditions, tertiary alkyl halides undergo only elimination, so there will be no competing substitution product in the next reaction. Because addition of Br2 involves only anti addition, the target molecule (as well as its enantiomer) is obtained. 

The transformation proceeds through anti addition. The possible products arising from addition to the trans and the cis isomers of the starting compound are shown below. Addition to the trans isomer gives the required 2S,3S isomer, but as a racemic mixture with its 2R,3R enantiomer, because initial attack by bromine on the double bond has an equal chance of occurring from the top and bottom faces of the double bond. 

Notice that the stereochemistry of the product of the osmium tetroxide oxidation of lrans-2-butene is opposite that formed on the addition of bromine to lra s-2-butene. Osmium tetroxide oxidation gives the glycol as a pair of enantiomers forming a racemic mixture. Addition of bromine to trcms-2-butene gives the dibromoalkane as a meso compound. A similar difference is observed between the stereochemical outcomes of these reactions with ds-2-butene. The difference in outcomes occurs because bromination of an alkene involves anti addition, whereas oxidation by osmiinn tetroxide involves syn addition. 

Anti addition across r-2-butene leads to a pair of enantiomers, while anti addition across trans-2-butene leads to a meso compound. These examples illustrate that the configuration of the starting alkene determines the configuration of the product for halogenation reactions. 

The next step is to identify the stereochemical outcome. In this case, two new chirality centers are formed, so we expect only the pair of enantiomers that would result from anti addition. That is, the OH and Br will be installed on opposite sides of the it bond ... 

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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