The reaction of propene with HOCl ($$Cl_2 + H_2O$$) proceeds through the intermediate:

We have propene, whose formula is $$CH_3 - CH = CH_2$$. When chlorine is passed through water, the well-known reaction $$Cl_2 + H_2O \rightarrow HOCl + HCl$$ takes place, so the real reagent that finally adds to the alkene is hypochlorous acid $$HOCl$$.

For the addition to begin, we recall the general rule that $$HOCl$$ polarises to give an electrophile $$Cl^+$$ and a nucleophile $$OH^-$$. The electrophile is the species that first attacks the electron-rich carbon-carbon double bond.

So, in the first step, propene donates the $$\pi$$ electrons of its double bond to the electrophile $$Cl^+$$. Because an electrophilic addition always proceeds through the more stable carbocation, we have to decide to which carbon the $$Cl^+$$ should attach.

Let us examine the two possibilities one by one:

1. Suppose $$Cl^+$$ adds to the middle carbon (C-2). The skeleton after this attack would be $$CH_3 - CHCl - CH_2^+.$$ The positive charge now resides on C-3, which is a primary carbocation and therefore relatively unstable.

2. Suppose instead that $$Cl^+$$ adds to the terminal carbon (C-3). The skeleton produced in that case is $$CH_3 - CH^+ - CH_2Cl.$$ Here the positive charge is on C-2. This centre is attached to two carbon atoms (C-1 and C-3), so it is a secondary carbocation. A secondary carbocation is significantly more stable than a primary one.

Because the mechanism always proceeds through the more stable intermediate, the second alternative is preferred. Hence the reaction passes through the carbocation

$$CH_3 - CH^+ - CH_2 - Cl.$$

This structure corresponds exactly to Option D in the list supplied.

Hence, the correct answer is Option 4.