1. Identify the reaction type

The alkene is propene:

$CH_3-CH=CH_2$

Reaction with $Cl_2/H_2O$ gives a halohydrin formation reaction.

In water, $Cl_2$ forms:

$Cl_2 + H_2O \rightleftharpoons HOCl + HCl$

So effectively, the alkene adds OH and Cl across the double bond.

2. Mechanism concept

The alkene first forms a chloronium ion intermediate. Then water attacks the more substituted carbon because it bears greater partial positive character.

For propene:

$CH_3-CH=CH_2$

• Middle carbon is more substituted.
• Terminal carbon is less substituted.

Hence:

• $OH$ goes to the more substituted carbon.
• $Cl$ goes to the less substituted carbon.

3. Write the product

Adding $OH$ to carbon-2 and $Cl$ to carbon-3 gives:

$CH_3-CH(OH)-CH_2Cl$

This is 1-chloropropan-2-ol.

4. Major product

Therefore, the major product is:

$\boxed{CH_3-CH(OH)-CH_2Cl}$

5. Comparison with stored answer

The stored correct answer is B. Since the actual structures for options A–D are not shown, I infer that B corresponds to:

$CH_3-CH(OH)-CH_2Cl$

So I agree with the stored answer, provided option B is this halohydrin product.