Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Treatment of bromine water with propene yields 1-bromopropan-2-ol.
Reason (R): Attack of water on bromonium ion follows Markovnikov rule and results in 1-bromopropan-2-ol.
In the light of the above statements, choose the most appropriate answer from the options given below:

We have the alkene propene, whose condensed formula is $$\mathrm{CH_3-CH=CH_2}$$.

When bromine is added in the presence of excess water, the actual reagent in aqueous medium is written as $$\mathrm{Br_2/H_2O}$$. The overall transformation is the formation of a bromohydrin, that is, simultaneous addition of $$\mathrm{Br}$$ and $$\mathrm{OH}$$ across the C=C double bond.

The first elementary step is electrophilic attack of $$\mathrm{Br_2}$$ on the double bond. The $$\pi$$ electrons of the C=C bond act as a nucleophile and displace $$\mathrm{Br^-}$$ from $$\mathrm{Br_2}$$, producing a cyclic bromonium ion. This step may be written as

$$\mathrm{CH_3-CH=CH_2 \;+\; Br_2 \;\longrightarrow\; [CH_3-CH-CH_2]^+}$$
with $$\mathrm{Br}$$ bridging the two carbons, and a free $$\mathrm{Br^-}$$ in the medium.

Now water, which is abundant and a better nucleophile than $$\mathrm{Br^-}$$ because of its higher concentration, attacks this three-membered bromonium ring. A key point is that the attack is Markovnikov. Before using that rule, we state it explicitly:

Markovnikov rule: In an electrophilic addition to an alkene, the nucleophilic part of the reagent attaches to the more substituted carbon atom of the double bond, while the electrophilic part attaches to the less substituted carbon.

In the bromonium ion derived from propene, carbon 2 (the middle carbon) is secondary, whereas carbon 1 (the terminal carbon) is primary. According to Markovnikov rule, the nucleophile $$\mathrm{H_2O}$$ must therefore attack carbon 2. We write the attack step as

$$\mathrm{[CH_3-CH-CH_2]^+ + H_2O \;\longrightarrow\; CH_3-CH(OH)-CH_2Br + H^+}$$

The proton released is immediately taken up by $$\mathrm{Br^-}$$ or by another water molecule, regenerating neutral medium; the essential organic product remains:

$$\mathrm{CH_3-CH(OH)-CH_2Br}$$

Let us name this compound systematically. We number the longest carbon chain from either end, but we keep the lowest possible locant for the functional group with priority, which here is the hydroxyl group. Thus carbon 2 bears the $$\mathrm{OH}$$ group and carbon 1 bears the $$\mathrm{Br}$$ atom:

$$\mathrm{HO-CH(CH_3)-CH_2Br}$$ = 1-bromopropan-2-ol

So, the experimental observation given in the Assertion is completely correct.

Now we examine the Reason. It says that the attack of water on the bromonium ion follows Markovnikov rule and therefore gives the same product, 1-bromopropan-2-ol. We have just followed the mechanism and indeed found that application of Markovnikov orientation is precisely why $$\mathrm{OH}$$ ends up on C 2 and $$\mathrm{Br}$$ on C 1. Thus the Reason is also true and it is the exact explanation of the Assertion.

In the list of options, the statement “Both (A) and (R) are true and (R) is the correct explanation of (A)” corresponds to Option C (the third option).

Hence, the correct answer is Option C.