An E2 Reaction Is Regioselective

An alkyl halide such as 2-bromopropane has two 8-carbons from which a proton can be removed in an E2 reaction. Because the two 8-carbons are identical, the proton can be removed equally easily from either one. 

Alkyl Chlorides Alkyl Bromides, and Alkyl Iodides Preferentially Form the More Stable Product 

The major product of the E2 reaction of 2-bromobutane and methoxide ion is 2-butene (indicated by the blue line), because the transition state leading to its formation is more stable than the transition state leading to formation of 1-butene (indicated by the red line). 

In the transition state leading to the alkene, the C—H and C—Br bonds are partially broken and the double bond is partially formed (the partially broken and partially formed bonds are indicated by dashed lines), giving the transition state an alkene-like structure. Therefore, factors that stabilize the alkene will also stabilize the alkene-like transition state, allowing the more stable alkene to be formed faster. 

The following reaction also forms two elimination products. Because 2-methyl-2-butene is the more substituted alkene (it has a greater number of alkyl substituents 

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In Section 11.1, we saw that an E2 reaction is regioselective, which means that more of one constitutional isomer is formed than the other. For example, the major product formed from the E2 elimination of 2-bromopentane is 2-pentene. 

An E2 reaction is regioselective. The major product is the more stable alkene, unless the reactants are sterically hindered or the leaving group is poor. The more stable alkene is... 

For alkyl halides that can undergo both E2 and El reactions, the E2 reaction is favored by the same factors that favor an Sn2 reaction (namely, a high concentration of a strong base) and the El reaction is favored by the same factors that favor an SnI reaction (namely, a weak base). An E2 reaction is regioselective the major product is the more stable alkene, unless the reactants are sterically hindered or the leaving group is poor. 

Zaitsev s rule is If more than one alkene can be formed by an elimination reaction, the more stable alkene is the major product. This means that the E2 reaction is regioselective because of two alkene products one is formed as the major product. Note that this says major product— not the only product. Although small amounts of 8 may be formed in the E2 reaction of 7, the m or product will be 9. The percentage of each remains unknown without doing the experiment, so it is only possible to make predictions about major and minor products. 

What is the regioselectivity of an E2 reaction In other words, what are the factors that dictate which of the two elimination products will be formed in greater yield ... 

The reason for the two different regioselectivities is a change in mechanism. As we have already discussed, acid-catalysed elimination of water from tertiary alcohols is usually El, and you already know the reason why the more substituted alkene forms faster in El reactions (p. 394). It should come as no surprise to you now that the second elimination, with a strong, hindered base, is an E2 reaction. But why does E2 give the less substituted product This time, there is no problem getting C-H bonds anti-periplanar to the leaving group in the conformation... 

Dehydrohalogenation of alkyl halides (Sections 5 14-5 16) Strong bases cause a proton and a halide to be lost from adjacent carbons of an alkyl halide to yield an alkene Regioselectivity is in accord with the Zaitsev rule The order of halide reactivity is I > Br > Cl > F A concerted E2 reaction pathway is followed carbocations are not involved and rearrangements do not occur An anti coplanar arrangement of the proton being removed and the halide being lost characterizes the transition state... 

Although El reactions show some stereo and regioselectivity, the level of selectivity in E2 reactions can be much higher because of the more stringent demands on the transition state for E2 elimination. In an E2 elimination, the new n bond is formed by overlap of the C—H o bond with the C—X a antibonding orbital. The two orbitals have to lie in the same plane for best overlap, and now there are two conformations that allow this. One has H and X syn-periplanar, the other anti-periplanar. The anti-periplanar conformation is more stable because it is staggered (the syn-periplanar conformation is eclipsed) but, more importantly, only in the anti-periplanar conformation are the bonds (and therefore the orbitals) truly parallel. 

SgZ fashion (labeled in structure 22) leads to the chelated rr-allylpalladium intermediate 22, which can potentially lose either of two protons (labeled H], and in structure 22) to afford either 18 or 19, respectively. From recent work in the literature, notably that of Krafift and co-workers,it is recognized that chelated rr-allylpalladium intermediates can show unusual regioselectivity in addition reactions. Perhaps related factors govern the regioselective loss of to afford intermediate 23 and, ultimately, the enediene product 19 there is evidence suggesting that loss of proton H, occurs via an E2 rather than /S-hydride elimination pathway. 

Regiochemistry is an issue in E2 reactions, which are said to be regioselective, because the more substituted alkene, called the Zaitsev product, is generally the major product. 

The regioselectivity also results from steric effects in the transition state for the E2 ehmination. A (CH3)3N group has approximately the same van der Waals radius as a tert-huvy group. A trans periplanar arrangement of the proton to be eliminated and the (CH3)3N group is required for an E2 elimination reaction of the quaternary ammonium salt to take place. 

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