Question
Asked Dec 24, 2019
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Elimination of HBr from 2-bromobutane affords a mixture of but-1-ene and but-2-ene. With sodium ethoxide as base, but2-ene constitutes 81% of the alkene products, but with potassium tert-butoxide, but-2-ene constitutes only 67% of the alkene products. Offer an explanation for this difference.

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Expert Answer

Step 1

According to Zeitsev rule the more alkylated alkene in the major product.  When HBr is elimination from 2-bromobutane, 2-butane is formed as the major product.

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Br Base but-1-ene (E)-but-2-ene 2-bromobutane (less stable) (less substituted alkene) (more stable) (more substituted alkene)

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Step 2

Potassium tert-butoxide is a sterically hindered base and will be not much efficient in proton abstraction from 2-bromobutane.  Hence, it forms but-2-ene only to the extent of 67%.

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Br CH3 Н-с Ċ-C- ннн H H н на Sterically free protons Sterically hindered protons :Br: CHз Н2 H2C Н. CH3 H. н Н but-1-ene Н, H3Ç CH3 Насснз H3C :0 :Br: CH3 н H3C. Н c CH3 Н `H. steric repulsions Н н (E)-but-2-ene 67% Нас CHз to Нассн, H3C-

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Step 3

Sodium ethoxide is a less hindered base and will abstract protons more ...

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:Br: Н CH3 Нас `CH3 Н Н н н Н (E)-but-2-ene H. 81 % :0: Нас-

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