2-Bromo-2-methylbutane elimination

Tertiary alkyl halides are so sterically hindered to nucleophilic attack that the pres ence of any anionic Lewis base favors elimination Usually substitution predominates over elimination m tertiary alkyl halides only when anionic Lewis bases are absent In the solvolysis of the tertiary bromide 2 bromo 2 methylbutane for example the ratio of substitution to elimination is 64 36 m pure ethanol but falls to 1 99 m the presence of 2 M sodium ethoxide... 

The ratio of elimination to substitution is exactly the same (26% elimination) for 2 bromo 2 methylbutane and 2 lodo 2 methylbutane in 80% ethanol/20% water at 25°C... 

One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methylbutane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures ... 

When 2-bromo-2-methylbutane is heated in a mixture of ethanol and water, it gives a 64% yield of substitution products and a 36% yield of elimination products. 

The alkyl halide, in this case 2-bromo-2-methylbutane, ionizes to a carbocation and a halide anion by a heterolytic cleavage of the carbon-halogen bond. Like the dissociation of an alkyloxonium ion to a carbocation, this step is rate-determining. Because the rate-determining step is unimolecular—it involves only the alkyl halide and not the base—this mechanism is known by the symbol El, standing for elimination unimolecular. It exhibits first-order kinetics. 

There is a strong similarity between the mechanism shown in Figure 5.12 and the one shown for alcohol dehydration in Figure 5.6. Indeed, we can describe the acid-catalyzed dehydration of alcohols as an El elimination of their conjugate acids. The main difference between the dehydration of 2-methyl-2-butanol and the dehydrohalogenation of 2-bromo-2-methylbutane is the source of the carbocation. When the alcohol is the substrate, it is the corresponding alkyloxonium ion that dissociates to form the carbocation. The alkyl halide ionizes directly to the carbocation. 

To synthesize 2-methyl-2-butene from 2-bromo-2-methylbutane, you would use Sn2/E2 conditions (a high concentration of HO in an aprotic polar solvent) because a tertiary alkyl halide gives only the elimination product under those conditions. If Sn1/E1 conditions were used (a low concentration of HO in water), both elimination and substitution products would be obtained. 

Stereoelectronic effects are also important in the dehydrohalogenation of acyclic alkyl halides by an E2 pathway. Again, the most favorable arrangement for the hydrogen and the halide being lost is anti coplanar. In the formation of 2-methyl-2-butene from 2-bromo-2-methylbutane shown on page 208, the elimination of HBr occurs readily from the conformation on the left but not from the one on the right. 

Based on the El process shown for it in Mechanism 5.5, would you expect elimination in 2-bromo-2-methylbutane to exhibit a kinetic isotope effect ... 

Ethoxide-promoted E2 elimination from 2-iodo-3-methylbutane produces 18% of the 1-alkene and 82% of the 2-alkene (equation 10.78), while elimination from 2-bromo-2-methylbutane produces 29% of the 1-alkene and 71% of the 2-alkene (equation 10.79). Similarly, E2 elimination of (CH3)2S from dimethyl-sec-butylsulfonium ion produces 74% of the 1-alkene and 26% of the 2-alkenes (equation 10.80), while the bimolecular elimination of (CH3)2S from dimethyl-f-amylsulfonium produces 86% of the 1-alkene and 14% of the 2-alkene (equation 10.81). In each case, an additional methyl substituent on the a-carbon atom leads to a greater proportion of the less substituted alkene. Explain the product distributions from all four reactions in terms of the factors that determine the distribution of products from E2 reactions. 

Dehydrohalogenation may give a mixture of products if the halogen is unsymmetrically located on the carbon skeleton. Eor example, 2-bromo-2-methylbutane (6), the substrate you will use in this experiment, yields both 2-methyl-2-butene (7) and 2-methyl-l-butene (8) on reaction with strong base (Eq. 10.5). Because such elimination reactions are normally irreversible under these experimental conditions, the alkenes 7 and 8 do not undergo equilibration subsequent to their production. Consequently, the ratio of 7 and 8 obtained is defined by the relative rates of their formation. These rates, in turn, are determined by the relative free energies of the two transition states, 9 and 10, respectively, rather than by the relative free energies of the alkenes 7 and 8 themselves. 

Typical GLC traces of the products of elimination of 2-bromo-2-methylbutane. Assignments and peak areas (in parentheses) peak 1 2-methyl-l-butene peak 2 2-methyl-2-butene. (a) Elimination with KOH peak 1 (82210), peak 2 (120706). (b) Elimination with KOClCHi) peak 1 (94260), peak 2 (85140) column and conditions 50-mm X 30-m, 0.25-pfilm of SE-54 80 °C, 40 mL/min. 

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