2-chloro-2-methyl butane, on reaction with aq. KOH gives X as the major product. X is(A) 2-butene(B) 2-methyl-1-butene(C) 2-methyl-2-butene(D) 2-methyl-2-butanol

The reaction of 2-chloro-2-methylbutane (also known as tert-butyl chloride) with aqueous potassium hydroxide (KOH) typically undergoes an elimination reaction known as an E2 mechanism. In this reaction, a base (in this case, OH¯ from KOH) removes a proton adjacent to the carbon bearing the leaving group (Cl¯), while the leaving group leaves, resulting in the formation of a double bond.

The major product X in this reaction is usually the most substituted alkene, which follows Zaitsev's rule. Zaitsev's rule states that in elimination reactions, the most substituted alkene is the major product.

In the case of 2-chloro-2-methylbutane, the major product X would be 2-methyl-2-butene (choice C) because it is the most substituted alkene among the options provided. Therefore, the correct answer is (C) 2-methyl-2-butene.