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Unit 12 Stereochemistry II Stereoselective Reactions

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Overview

We are now beginning to see how stereochemistry can give us a three dimensional picture of a reaction. This may include such factors as the direction of attack and the shape of the transition state. The C=C double bond is highly reactive, in part because its planar faces are open to attack. Reactivity in the SN2 reaction is largely determined by crowding about pentavalent carbon in the transition state.

We have seen that the SN2 reaction proceeds with complete configurational inversion. This not only shows that attack is from the back side, but is strong evidence that the reaction involves one single step, with bond-breaking and bond-breaking occurring in concert with one another. Free radical chlorination of an optically active alkane, however, proceeds with complete racemization, indicating that the C-H bond to the chiral center is broken before the C-Cl bond is formed. Between these two extremes of stereochemical behavior, we have encountered a third kind, partial racemization, in the SN1 reaction, which again gives essential information about the reaction mechanism.   

Our next step is to utilize this understanding of the reaction mechanism in order to control the stereochemistry of the reaction.  This includes the selection of the proper reagent, conditions and catalyst so that we obtain our product in precisely the stereochemical form that we desire.

We have previously established in Unit 6 that if no bond to a chiral center is broken, then the configuration about this center must be retained. Furthermore, optically inactive reactants in an achiral medium can only give optically inactive products. What we must now consider is the fact that whenever a chiral center or a double bond is generated in a molecule, both configurations about that chiral center or double bond may result.   

In this Unit, we shall examine the concepts of stereospecificity and stereoselectivity. We shall proceed from what can or cannot happen in a given reaction to what does actually happen. In so doing, we shall study the stereochemistry of both addition and elimination.

Addition of Halogens to Alkenes

Let's begin with the stereochemistry of the addition reaction of halogens to alkenes. E.G. The addition of bromine to 2-butene yields 2,3-dibromobutane.  

Two chiral centers are generated in the reaction. Thus, the product can exist as a pair of enantiomers (I and II) as well as a meso compound (III).

The reactants, too, exist as stereoisomers; a pair of geometric isomers cis and trans.

If we start with, say, cis 2-butene, which of the stereoisomeric products do we get ? The products will definitely NOT include all of them. The cis alkene yields only racemic 2,3-dibromobutane (50/50 mixture) . None of the meso compound is obtained. 

A reaction that yields predominantly one stereoisomer (or one pair of enantiomers) of several possible diasteroemers is called a stereoselective reaction.

Now, suppose we start with trans 2-butene. The trans compound yields only the meso compound. 

The product obtained depends upon which stereoisomers we start with. This concept is known as stereospecificity.  This term is used in a broader sense, to indicate any kind of discrimination on a stereochemical basis between different reactants.

The addition of bromine to alkenes is both stereoselective and stereospecific. It is stereoselective since, from a given alkene, we obtain either 1) only one diastereomer, or 2) one pair of enantiomers. It is stereospecific since stereoisomeric alkenes react differently to give stereochemically different products.

In syn-addition, the added groups become attached to the same face of the double bond.

In anti-addition, the added groups become attached to opposite faces of the double bond.

                                   Addition of bromine (or chlorine) to the 2-butenes involves anti-addition.

Thus, If we start with cis 2-butene, we can attach the bromine atoms to opposite faces of the double bond in two different ways. Attachment as in reaction path (a) or (b) gives enantiomers I and II respectively. Since a and b are equally likely (regardless of the reaction mechanism) we obtain the racemic modification.

Starting with trans 2-butene, we can again attach the bromine atoms to opposite faces fo the double bond in two ways. Attachment as in reaction path (c) or (d) both give the same structural isomer of this compound: the meso dibromide. 

It was to account for the observed stereospecificity that the proposed halonium ion mechanism was given.

So how does the halonium ion mechanism account for the anti-addition reaction ? Let's consider first the addition of bromine to cis 2-butene. In the first step, positive bromine becomes attached to either the top or bottom face of the alkene. What if bromine were to become attached to the top face ? Then the C atoms of the double bond tend to become tetrahedral, and the H atoms and methyl groups are displaced downward. However, the methyl groups are still located across from each other, as they were in the alkene. In this way, bromonium ion VI is formed.

Now bromonium ion VI is attacked by a bromide ion. A new C-Br bond is formed, and an old C-Br bond is broken. This is a familiar reaction: nucleophilic substitution, either on one end (C2) or the other (C3) of the bromonium ion. In this case, the negative bromide ion Br- is the nucleophile, and the positive bromine ion Br+ is the leaving group. Note, however, that although Br+ it is the leaving group, it does not go far. In fact, the Br+ remains attached to the opposite side of the molecule, which has been cleaved, and no longer constitutes a cyclic ring halonium ion intermediate.   

Attack by the bromide ion is from the backside (along the bottom face of VI) so that the bond being formed is on the opposite side of the carbon form the bond being broken. There is an inversion of the configuration about the C atom being attacked.

Attack on VI can occur by reaction path (a) to yield structure VII or by reaction path (b) to yield structure VIII. We recognize VII and VIII as enantiomers. Since attack by either (a) or (b) is equally likely, the enantiomers are formed in equal amounts, yielding a racemic modification. The same results are obtained if the Br+ attache initially to the bottom face of cis 2-butene. 

Let us now consider trans 2-butene. This time, bromonium ion IX is formed. Attack on it by path (c) yields X. Attack on it by (d) yields XI, which is identical in all ways to X. It is the meso compound. The same results are obtained if the Br+ is initially attached to the bottom face of trans 2-butene. 

That such cyclic ring intermediates can give rise to anti-addition is demonstrated clearly by hydroxylation of alkenes with peroxy acids. In that case, analogous intermediates in the form of epoxides can be isolated and clearly characterized w/r/to structure.

The idea of a halonium ion may appear strange to us, in contrast to the already familiar hydronium (H30+) and ammonium (NH3+) ions. The tendency of a halogen to share two pairs of electrons and acquire a positive charge should be weak due to the high electronegativity of the halogens themselves. Yet the evidence shows that this tendency is appreciable. In halogen addition, what we need to ask ourselves is: Which is more stable ?

1) An open carbocation in which carbon has a sextet of electrons.

2) A halonium ion in which each atom (except hydrogen) has a complete octet.

It is not a matter of which atom (H or C) can better accommodate a positive charge. It is instead a matter of completeness or incompleteness of octets. In halonium ion formation, we see one more example of what underlies all carbocation behavior: the need to get a pair of electrons to complete the octet of the positively charged carbon.

E2: syn / anti Elimination

Let's consider the stereochemistry of elimination using another familiar reaction: dehydrohalogenation by E2. E.G. the following compound contains two chiral centers.

We can easily show that it can exist as two pairs of enantiomers: I and II (called erythro) and II and IV (called threo). Each pair is diastereomeric with the other pair.

The product, too, exists as stereoisomers: a pair of geometric isomers, Z and E. 

Now if we start with the erythro halide, I and II, we obtain only the Z alkene.

If we start with the threo halide, II and IV, we obtain only the E alkene.

Other studies have shown that these results are typical: E2 elimination is both stereoselective and stereospecific. In order to describe the kinds of stereoselectivity that may be observed in elimination reactions, the concepts of syn-elimination and anti-elimination are utilized. These terms are not the names of specific mechanisms. They simply indicate the stereochemical facts: that the eliminated groups are either lost from the same face (syn) or from opposite faces (anti) of the developing double bond. As this example and many others show,

                                               E2 elimination typically involves anti-elimination.

In the transition state, the hydrogen and the leaving group are located in the anti relationship as contrasted to gauche or eclipsed. 

                                  Diastereomer I (or its enantiomer II) gives the Z alkene:

                                Diastereomer III (or its enantiomer, IV) gives the E alkene.

Previously, we likened E2 to SN2 in that the halide is pushed out of the molecule by a kind of nucleophilic attack. the "nucleophile" is the beta-carbon which, using the electron pair left behind by the departing proton, begins to form a bond -- the pi bond -- to the alpha-carbon. On this basis, the preference for anti elimination indicates that this "nucleophilic attack" occurs preferentially at the face of the alpha-carbon most remote from the departing halide -- the familiar back-side attack of nucleophilic substitution.  

The stereochemistry observed for these elimination reactions is entirely consistent with the E2 mechanism. The high degree of stereoselectivity indicates a strong preference for a particular spatial relationship between the two departing groups. This is quite understandable if they are departing simultaneously.