E2" reactions

An illustration of the effect of group size is provided by the elimination of HBr from the esters 10.14 and 10.15 (both are made similarly by a reaction that we will meet in a later chapter). Elimination from the methyl ester, which has low steric demand, gives mainly the 2-alkene, with 10 %-15 % of the 1-alkene. However, using a bulkier ester and a bulkier base, the 1-alkene is produced with 95 % selectivity. In the reaction of 10.16, there is only one possible elimination product, but 5 2 substitution predominates in this process. 

Elimination of an amino group, NRj, is a difficult problem, whatever mechanism is involved. The typical pK of an amine, RjNH, is 35, so expulsion of RjN is unrealistic. If we need to do this, we must first make the amine into a quaternary ammonium salt, by reaction with methyl iodide. 

The product is useful in Diels-Alder reactions that we discuss in Chapter 11 

Note that the more stable, conjugated product is formed 

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Effects that arise because one spatial arrangement of electrons (or orbitals or bonds) IS more stable than another are called stereoelectronic effects There is a stereoelec tromc preference for the anti coplanar arrangement of proton and leaving group in E2 reactions Although coplanarity of the p orbitals is the best geometry for the E2 process modest deviations from this ideal can be tolerated In such cases the terms used are syn periplanar and anti periplanar... 

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... 

The first example of a stereo electronic effect in this text concerned anti elimination in E2 reactions of alkyl halides (Section 5 16)... 

The molecule below has four stereoisomeric forms exoO exoCH2Br, exoO endoCH2Br, and so on. Examine electrostatic potential maps of the four ions and identify the most nucleophilic (electron-rich) atom in each. Examine the electron-acceptor orbital (the lowest-unoccuped molecular orbital or LUMO) in each and identify electrophilic sites that are in close proximity to the nucleophilic. Which isomers can undergo an intramolecular E2 reaction Draw the expected 8 2 and E2 products. Which isomers should not readily undergo intramolecular reactions Why are these inert ... 

E2 Reaction C-H and OX bonds break simultaneously, giving the alkene in a single step without intermediates. 

The E2 reaction (for elimination, bimolecular) occurs when an alkyl halide is treated with a strong base, such as hydroxide ion or alkoxide ion (RO-). It is the most commonly occurring pathway for elimination and can be formulated as shown in Figure 11.17. 

Mechanism of the E2 reaction of an alkyl halide. The reaction takes place in a single step through a transition state in which the double bond begins to form at the same time the H and X groups are leaving. 

Figure 11.18 The transition state for the E2 reaction of an alkyl halide with base. Overlap of the developing p orbitals in the transition state requires periplanar geometry of the reactant.

Predicting the Double-Bond Stereochemistry of the Product in an E2 Reaction... 

Anti periplanar geometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms (Section 4.8). As pointed out by Derek Barton in a landmark 1950 paper, much of the chemical reactivity of substituted cyclohexanes is controlled by their conformation. Let s look at the E2 dehydro-halogenation of chlorocyclohexanes to see an example. 

The anti periplanar requirement for E2 reactions overrides Zaitsev s rule and can be met in cyclohexanes only if the hydrogen and the leaving group are trans diaxial (Figure 11.19). If either the leaving group or the hydrogen is equatorial, E2 elimination can t occur. 

Figure 11.19 The geometric requirement for E2 reaction in a substituted cyclohexane. The leaving group and the hydrogen must both be axial for anti peri-planar elimination to occur.

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). 

Draw the structure and assign Z or stereochemistry to the product von expect from E2 reaction of the following molecule with NaOH (yellow-green = Cl) ... 

In light of your answer to Problem 11.49, which alkene, E or Z, would you expect from an E2 reaction on the tosylate of (2/ ,3R)-3-phenyl-2-butanol Which alkene would result from E2 reaction on the (25,3 P) and (25,35) tosy-lates Explain. 

Although anti periplanar geometry is preferred for E2 reactions, it isn t absolutely necessary. The deuterated bromo compound shown here reacts with strong base to yield an undeuterated alkene. Clearly, a svn elimination has occurred. Make a molecular model of the reactant, and explain the result. 

In light of your answer to Problem 11.61, explain why one of the following isomers undergoes E2 reaction approximately 100 times as fast as the other. Which isomer is more reactive, and why ... 

There are eight diastereomers of 1,2,3,4,5,6-hexachlorocyclohexane. Draw each in its more stable chair conformation. One isomer loses HC1 in an E2 reaction nearly 1000 times more slowly than the others. Which isomer reacts so slowly, and why ... 

Amines are converted into alkenes by a two-step process called the Iiofnunn elimination. SN2 reaction of the amine with an excess of CH3I in the first stej yields an intermediate that undergoes E2 reaction when treated with silvei oxide as base. Pentylamine, for example, yields 1-pentene. Propose a structim for the intermediate, and explain why it undergoes ready elimination. 

Figure 13.11 The 13C NMR spectrum of 1-methylcyclohexene, the E2 reaction product from treatment of 1-chloro-1-methylcyclohexane with base.

Benzyl chloride can be converted into benzaldebyde by treatment with nitromethane and base. The reaction involves initial conversion of nilro-methane into its anion, followed by SN2 reaction of the anion with benzyl chloride and subsequent E2 reaction. Write the mechanism in detail, using curved arrows to indicate the electron flow in each step. 

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