What is the major product formed on addition of Hydrogen Iodide to Pent-2-ene?
I would say the reason that the iodine ends up in the 2-position is because at the 2 and 3 positions the charge of the carbo-cation is on a secondary position and has no thermodynamic driving force. The adjacent secondary hydrides may shift without much energy loss leaving the major product to be kinetically determined. Since there are two positions for an addition to the 2-position carbon and one for the 3-position carbon, your 2-position addition has a statistical advantage. Combine this with the fact that two adjacent ethyl groups would be more stericly hindered to attack than a methyl and propyl group and it makes reasonable sense that the addition occurs on the 2-position.