Products Hofmann

In conformation I the ethyl group is between Br and Me, while in J it is between Br and H. This means that J is more stable, and most of the elimination should occur from this conformation. This is indeed what happens, and 51% of the trans isomer is formed (with KOEt) compared to 18% of the cis (the rest is the Hofmann product). These effects become larger with increasing size of A, B, and E. 

For an elimination reaction where there is more than one possible aUcene that can be formed, we have names for the different products based on which alkene is more substituted and which is less substituted. The more substituted alkene is called the Zaitsev product, and the less substituted alkene is called the Hofmann product. Usually, the Zaitsev product is the major product ... 

In this case, the Hofmann product is the major product, because a sterically hindered base was used. This case illustrates an important concept The regiochemical outcome of an E2 reaction can often be controlled by carefully choosing the base. Below are two examples of sterically hindered bases that will be encountered frequently throughout your organic chemistry course ... 

PROBLEMS Draw the Zaitsev and Hofmann products that are expected when each of the following compounds is treated with a strong base to give an E2 reaction. For the following problems, don t worry about identifying which product is major and which is minor, since the identity of the base has not been indicated. Just draw both possible products ... 

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. 

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) ... 

However, we must be careful to control the regiochemistry properly in each of these two steps. In the elimination step, we need to form the less substituted double bond (i.e., the Hofmann product), and therefore, we must use a stericaUy hindered base. Then, in the addition step, we need to place the Br on the less substimted carbon (anh-Markovnikov addition), so we must use HBr with peroxides. This gives us the following overall synthesis ... 

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. 

Y is -t-vely charged or not) also leads to increasing formation of the Hofmann product. (3-Substituents that would help stabilise a developing -ve charge promote formation of the Hofmann product,... 

It is a general observation that, where different alkene products can arise through E2 elimination, the more-substituted alkene predominates. 2-Menthene contains a double bond with two alkyl substituents, whereas the double bond in 3-menthene has three substituents. The more-substituted alkene is termed the Saytzeff product the less-substituted alkene is termed the Hofmann product. We recommend you disregard the proper names, and think of the products in terms of more-substituted alkene and less-substituted alkene . 

Problem 7.33 Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with C2H3O K to give mainly 2-pentene (the Sayzteff product), with MCjCO K it gives mainly 1-pentene (the anti-Sayzteff, Hofmann, product). 

This E2 elimination (Table 7-3) give the less substituted alkene (Hofmann product) rather than the more substituted alkene (Saytzeff product Section 6.3). 

In some cases, especially with cyclic compounds, the more stable olefin forms and Zaitsev s rule applies. For example, menthyl acetate gives 35% of the Hofmann product and 65% of the Zaitsev, even though a cis (3 hydrogen is present on both sides and the statistical distribution is the other way. A similar result was found for the pyrolysis of menthyl chloride.138... 

H. C. Brown has suggested that steric factors are of primary and almost sole importance in determining the position of the double bond. According to Brown, Hofmann product predominates when a large leaving group makes it even more difficult for the base to abstract the more hindered protons.96 He has asserted that data similar to those of Table 7.12, which seem at first glance to be contrary to his theory, support it further He says that fluorine takes up more space in the transition state than iodine because fluorine is more solvated.97 However, the entropies of activation for Reaction 7.37 with X = F, Cl, Br, or I are all very similar therefore increased solvation of fluorine seems not to be the proper explanation for the preponderance of Hofmann product when X = F.98... 

Notice that the LEAST stable alkene is the major product in the Hofmann elimination, called the Hofmann product. 

Formation of Regioisomeric Alkenes by /3-Elimination Saytzeff and Hofmann Product(s)... 

Fig. 4.7. Regioselectivity of the elimination of H/Het from Rsec—Het. When C/5 has fewer alkyl substituents than CI3 the constitutionally isomeric products are Hofmann product (A) and Saytzeff product (B).

If the /1-elimination of H/Het from R —Het can, in principle, afford regioisomeric alkenes whose C=C double bonds (Figure 4.7) contain a different number of alkyl substituents, they are differentiated as Hofmann and Saytzeff products the Hofmann product is the alkene with the less alkylated double bond, and the Saytzeff product is the alkene with the more alkylated double bond. Because C=C double bonds are stabilized by alkyl substituents, a Hofmann product is, in general, less stable than its Saytzeff isomer. Accordingly, eliminations of H/Het from Rv(,f —Het, which exhibit product development control, furnish a Saytzeff product with some regioselectivity. 

H. If a Ctert—H were present, it would not react at all. As a consequence, the Hofmann product is produced predominantly, but not exclusively. 

Fig. 4. 22. Saytzeff product is produced preferentially although not exclusively. The reaction of the sterically hindered base tBuO is directed preferentially toward the more readily accessible H atoms at the primary C atom. Therefore, it mainly provides the Hofmann product.

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