Isopropyl iodide is formed after reaction of propene with HI, this is due to

The correct option is more stable carbocation.

When propene reacts with hydroiodic acid (HI), it undergoes an electrophilic addition reaction. The reaction proceeds through a carbocation intermediate via Markovnikov's rule. Let's break down the step-by-step mechanism to understand why this happens:

Step 1: Protonation of the alkene
Propene (${\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2$) reacts with the proton ($H^{+}$) from the strongly acidic HI. The double bond acts as a nucleophile and attacks the proton. This protonation can theoretically yield two different carbocation intermediates depending on which carbon carbon double bond carbon accepts the proton:
1. If the proton adds to the central carbon (C-2), we obtain a primary carbocation:
${\mathrm{CH}}_3-{\mathrm{CH}}^{+}-{\mathrm{CH}}_3$ (Incorrect, this is the secondary path)
Specifically, adding the proton to the terminal carbon (C-3) yields the secondary carbocation: ${\mathrm{CH}}_3-{\mathrm{CH}}^{+}-{\mathrm{CH}}_3$.
2. Adding the proton to the central carbon (C-2) yields a primary carbocation: ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2^{+}$.

Step 2: Stability of the carbocations
A secondary carbocation (${\mathrm{CH}}_3-{\mathrm{CH}}^{+}-{\mathrm{CH}}_3$) is significantly more stable than a primary carbocation (${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2^{+}$) due to the electron-donating inductive effect (+I) of the two methyl groups and hyperconjugation involving six α-hydrogens. The primary carbocation has only one alkyl group and two α-hydrogens, making it much less stable.

Step 3: Attack of the nucleophile
Because the secondary carbocation is the more stable carbocation, the reaction proceeds almost exclusively through this intermediate. The iodide ion ($I^{-}$) then attacks the positively charged central carbon of the secondary carbocation to form isopropyl iodide (${\mathrm{CH}}_3-\mathrm{CH(I)}-{\mathrm{CH}}_3$) as the major product.