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Orginal question:

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(1) One of the carbocations formed (CC on the right) is only slightly more stable than the other (both are 2°).

(2) Iodine ion ($\ce{I-}$) is a slightly bigger ion and it faces more steric hinderance while attaching to one of the carbocation (CC on the right).

These two factors are contradictory hence how do i compare them and arrive at a major product?

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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.

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  • $\begingroup$ Thanks for the answer, i just have a little bit trouble understanding this part "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". $\endgroup$
    – Sujith Sizon
    Commented Dec 7, 2015 at 16:25
  • $\begingroup$ This is because if the positive charge is on the 3-position, then hydride shift can come from either of the ajacent carbons (2) $\endgroup$
    – A.K.
    Commented Dec 7, 2015 at 16:28
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    $\begingroup$ Also your answer reminded me that Hyper-Conjugation (obviously >inductive effect) is making stability of 1st carbocation > 2nd carbocation. I would be pleased if you could mention that too in your answer. $\endgroup$
    – Sujith Sizon
    Commented Dec 7, 2015 at 16:29

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