3-methyl-2-pentene on reaction with HOCl gives:A) 3-chloro-3-methyl pentanol-2B) 2,3-dichloro-3-methylpentaneC) 2-chloro-3-methyl pentanol-3D) 2,3-dimethyl butanol-2

The reaction of 3-methyl-2-pentene with HOCl (hypochlorous acid) is an example of halogenation, specifically chlorination. In this reaction, the double bond in the alkene is replaced by a chlorine atom. Let's go through the reaction step by step.

The structure of 3-methyl-2-pentene is:

CH3 - CH = C(CH3)2 - CH3

When this alkene reacts with HOCl, it undergoes an addition reaction, and the Cl atom will be added to one of the carbon atoms in the double bond. The product will have a chlorine atom bonded to one of the carbons, and the other carbon will have the original methyl group. The product can be represented as:

CH3 - CH(Cl) - C(CH3)2 - CH3

So, the correct product is:

A) 3-chloro-3-methyl pentanol-2