2-Bromo-2-methylpropane elimination reaction

Mechanism of the SnI substitution and El elimination reactions of tert-butyl bromide (2-bromo-2-methylpropane). 

Consider another elimination reaction involving the dehydrohalogenation of 2-bromo-2-methylpropane with sodium ethoxide to give 2-methylpropene. 

Now let s draw the forward scheme. The 3° alcohol is converted to 2-methylpropene using strong acid. Anti-Markovnikov addition of HBr (with peroxides) produces l-bromo-2-methylpropane. Subsequent reaction with sodium acetylide (produced from the 1° alcohol by dehydration, bromination and double elimation/deprotonation as shown) produces 4-methyl-1-pentyne. Deprotonation with sodium amide followed by reaction with 1-bromopentane (made from the 2° alcohol by tosylation, elimination and anfi -Markovnikov addition) yields 2-methyl-4-decyne. Reduction using sodium in liquid ammonia produces the E alkene. Ozonolysis followed by treatment with dimethylsulfide produces an equimolar ratio of the two products, 3-methylbutanal and hexanal. 

Although nucleophilic substitution with acetylide anions is a very valuable carbon-carbon bondforming reaction, it has the same limitations as any Sn2 reaction. Steric hindrance around the leaving group causes 2° and 3° alkyl halides to undergo elimination by an E2 mechanism, as shown with 2-bromo-2-methylpropane. Thus, nucleophilic substitution with acetylide anions forms new carbon-carbon bonds in high yield only with unhindered CH3X and 1 ° alkyl halides. 

Tertiary alkyl halides E2 elimination occurs when a base such a OH" or RO" is used. For example, 2-bromo-2-methylpropane gives 97% elimination product when treated with ethoxide ion in ethanol- By contrast, reaction under neutral conditions (heating in pure I ethanol) leads to a mixture of products resulting from both SnI sub-1 stitution and El elimination. "... 

At one extreme, breaking of the C—Lv bond to give a carbocation is complete before any reaction occurs with the base to lose a hydrogen and form the carbon-carbon double bond. This mechanism is designated an El reaction, where E stands for Elimination and 1 stands for unimolecular. One species (in this case, the haloalkane) is involved in the rate-determining step. The mechanism of an El reaction is illustrated here by the reaction of 2-bromo-2-methylpropane to form 2-methylpropene. 

When 2-bromo-2-methylpropane is dissolved in methanol, it disappears rapidly. As expected, the major product, 2-methoxy-2-methylpropane, arises by solvolysis. However, there is also a signihcant amount of another compound, 2-methylpropene, the product of elimination of HBr from the original substrate. Thus, in competition with the SnI process, which leads to displacement of the leaving group, another mechanism transforms the tertiary halide, giving rise to the alkene. What is this mechanism Is it related to the SnI reaction ... 

Which nucleophile in each of the following pairs will give a higher elimination substitution product ratio in reaction with l-bromo-2-methylpropane ... 

In all cases where substitution and elimination compete, higher reaction temperatures lead to greater proportions of elimination products. Thus, the amount of elimination accompanying hydrolysis of 2-bromo-2-methylpropane doubles as the tenpCTature is raised from 25 to 65"C, and that from reaction of 2-bromopropane with ethoxide rises from 80% at 25"C to nearly 100% at 55"C. Explain. 

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