Zaitsev products

A final piece of evidence involves the stereochemistry of elimination. (Jnlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the El reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsev s rule) product from El reaction, which is just what w e find. To return to a familiar example, menthyl chloride loses HC1 under El conditions in a polar solvent to give a mixture of alkenes in w hich the Zaitsev product, 3-menthene, predominates (Figure 11.22). 

So if we look back at the reaction above, we find that the two possible prodncts are monosubstituted and disubstitnted donble bonds. Whenever we have an elimination reaction where more than one possible donble bond can be formed, we have names for the different products based on which one is more substitnted and which one is less substituted. The more substituted product is called the Zaitsev product, and the... 

Consider the elimination reaction below, which uses a strong base. The product will be a double bond. This reaction will produce two Zaitsev products. One will be cis and one will be trans. Draw these products, and identify which is cis and which is trans. 

However, there are many exceptions in which the Zaitsev product (the more-substituted alkene) is not the major product. For example, if the reaction above is performed with a stericaUy hindered base (rather than using ethoxide as the base), then the major product will be the less-substituted alkene ... 

ANSWER Let s first consider the expected regiochetnical outcome of the reaction. The reaction does not employ a sterically hindered base, so we expect formation of the more substituted alkene (the Zaitsev product) ... 

This is the Zaitsev product that we expect. The stereoisomer of this alkene is not prodnced, because the E2 process is stereospecific ... 

El processes show a regiochemical preference for the Zaitsev product, just as we saw for E2 reactions. For example ... 

The more-substituted alkene (Zaitsev product) is the major product. However, there is one critical difference between the regiochemical outcomes of El and E2 reactions. Specifically, we have seen that the regiochemical outcome of an E2 reaction can often be controlled by carefully choosing the base (sterically hindered or not sterically hindered), hi contrast, the regiochemical outcome of an El process cannot be controlled. The Zaitsev product will generally be obtained. 

The Zaitsev product is generally favored over the Hofmann product, unless a sterically hindered base is used, in which case the Hofmann product will be favored. 

The Zaitsev product is always favored over the Hofmann product. 

For the regiochemical outcome, we expect the Zaitsev product to be the major product, because the reaction does not utilize a sterically hindered base ... 

When doing this type of sequence, there are a few important things to keep in mind. In the hrst step (elimination), we have a choice regarding which way to eliminate do we form the more substituted alkene (Zaitsev product) or do we form the less substituted alkene (Hofmann product) ... 

In the first step above, we need a Markovnikov addition (Br needs to end up at the more substituted carbon), which we can easily accomphsh by using HBr. The second step requires an elimination to give the Zaitsev product, which we can accomphsh if we choose our base carefully (we will need to use a base that is NOT sterically hindered). Therefore, the overall synthesis is ... 

Take special notice of what we can accomplish when we use this technique it gives us the power to move the position of a double bond. When using this technique, we must carefully consider the regiochemistry of each step. In the first step (addition), we must decide whether we want a Markovnikov addition (HBr), or an anti-Markovnikov addition (HBr with peroxides). Also, in the second step (elimination), we must decide whether we want to form the Zaitsev product or the Hofmann product (which we can control by carefully choosing our base, ethoxide or fcrf-butoxide). Get some practice with this technique in the following problems. 

PROBLEMS For each of the follow ing compounds, predict w hat the product w ould be of an E2 reaction (assume the Zaitsev product and focus on stereochemistry) ... 

Stereochemistry = if there are cis/trans isomers, you will get both Regiochemistry = form the more substituted double bond (Zaitsev product) For example,... 

Stereochemistry = if there are cis/trans isomers, you will get the one determined by the H and LG being antiperiplanar (draw Newman projections) Regiochemistry = form the more substituted double bond (Zaitsev product). If you are using a strong, sterically hindered base, then form the less substituted double bond (Hoffmann product)... 

Orientation of Elimination In most El and E2 eliminations with two or more possible products, the product with the most substituted double bond (the most stable product) predominates. This principle is called Zaitsev s rule, and the most highly substituted product is called the Zaitsev product. 

Formation of the Hofmann Product Bulky bases can also accomplish dehydrohalo-genations that do not follow the Zaitsev rule. Steric hindrance often prevents a bulky base from abstracting the proton that leads to the most highly substituted alkene. In these cases, it abstracts a less hindered proton, often the one that leads to formation of the least highly substituted product, called the Hofmann product. The following reaction gives mostly the Zaitsev product with the relatively unhindered ethoxide ion, but mostly the Hofmann product with the bulky tert-butoxide ion. 

Write a mechanism that explains the formation of the following product. In your mechanism, explain the cause of the rearrangement, and explain the failure to form the Zaitsev product. 

In Chapter 7, we saw that eliminations of alkyl halides usually follow Zaitsev s rule that is, the most substituted product predominates. This rule applies because the most-substituted alkene is usually the most stable. In the Hofmann elimination, however, the product is commonly the /east-substituted alkene. We often classify an elimination as giving mostly the Zaitsev product (the most-substituted alkene) or the Hofmann product (the least-substituted alkene). 

El with rearrangement by an alkyl shift. The Zaitsev product violates Bredt s rule. 

---