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What is the major product formed in the elimination reaction between 2-bromo-2-methylpropane and KOH in ethanol? Draw the mechanism for the reaction and explain why this product is favored.

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The major product formed in the elimination reaction between 2-bromo-2-methylpropane  also known as tert-butyl bromide  and KOH in ethanol is 2-methylpropene  also known as isobutylene . This reaction follows the E2 mechanism.Mechanism:1. The KOH acts as a strong base, and the hydroxide ion  OH-  abstracts a -hydrogen from the 2-bromo-2-methylpropane molecule. In this case, the -hydrogen is the one adjacent to the carbon with the bromine atom.2. Simultaneously, the carbon-bromine bond breaks, and the electrons from this bond form a new double bond between the -carbon  the one with the bromine atom  and the -carbon.3. The bromide ion  Br-  leaves as a leaving group, resulting in the formation of 2-methylpropene.The reason why this product is favored is due to Zaitsev's rule, which states that the more substituted alkene  the one with more alkyl groups attached to the double bond  will be the major product in an elimination reaction. In this case, 2-methylpropene is more substituted than the alternative product, propene, because it has two methyl groups and one hydrogen attached to the double bond, while propene has only one methyl group and two hydrogens attached to the double bond. The increased substitution leads to a more stable alkene, making it the major product.
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