Addition of HI on double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds through

a more stable carbonium ion

The correct answer is option 1. a more stable carbonium ion.

The addition of hydrogen iodide (HI) to propene yields isopropyl iodide as the major product due to the formation of a more stable carbocation intermediate. This process follows Markovnikov's rule, which states that in the addition of HX to an alkene, the hydrogen atom attaches to the carbon with more hydrogen atoms already attached (the less substituted carbon), and the halide (in this case, iodine) attaches to the carbon with fewer hydrogen atoms (the more substituted carbon).

Mechanism:

Formation of the carbocation intermediate:

Propene (\(CH_2=CH-CH_3\)) reacts with HI. The double bond breaks, and the hydrogen atom from HI attaches to the carbon atom with more hydrogen atoms (the terminal carbon), forming a secondary carbocation at the central carbon.

Stability of the carbocation:

The secondary carbocation is more stable than a primary carbocation due to hyperconjugation and inductive effects. The iodide ion (\(I^-\)) then attaches to the positively charged carbon, resulting in the formation of isopropyl iodide.

Thus, the reaction proceeds through the formation of a more stable secondary carbocation intermediate, which leads to the formation of isopropyl iodide rather than n-propyl iodide.