Final answer:
2-chloro-3-methylbutane reacts with alcoholic potash to produce 2-methylbut-2-ene, mainly due to Zaitsev's rule favoring the formation of the most substituted alkene.
Step-by-step explanation:
When 2-chloro-3-methylbutane is treated with alcoholic potash (potassium hydroxide in alcohol), an elimination reaction occurs resulting in the formation of 2-methylbut-2-ene as the major product. This reaction is an example of dehydrohalogenation, where the chloride (Cl) leaving group is removed along with a hydrogen atom from the adjacent carbon. The double bond forms between the second and third carbon atoms, hence the name 2-methylbut-2-ene.
The reason why this product is favored over others is due to the Zaitsev's rule, which states that the most substituted alkene will be the major product. In this case, the hydrogen atom that leaves is from the carbon with more hydrogen atoms initially, resulting in a more substituted and thus more stable alkene. This process follows regioselectivity giving the major product.