Alkenes are a highly reactive group of compounds as they undergo a wide variety of reactions like hydrogenation, ozonolysis, halogenation, epoxidation and halohydrin formation. Many of the reactions are regioselective or stereoselective reactions. When bromine is added to a solution of 1 -hexene in methanol, the product formed is (a) \(\mathrm{BrCH}_{2}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) a mixture of \(\mathrm{BrCH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{BrCH}_{2}-\mathrm{CH}\left(\mathrm{OCH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{COCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

1-hexene is reacting with bromine (Br2) in methanol, so this reaction is a halogenation reaction. Specifically, it is an electrophilic addition reaction, where the alkene reacts with an electrophilic halogen species (in this case, bromine) to add both atoms of the halogen onto the double bond.

The mechanism of halogenation of alkenes involves several key steps: 1. The alkene double bond acts as a nucleophile, attacking bromine and forming a bromonium ion (a three-membered cyclic intermediate) with one bromine atom still attached. 2. The solvent methanol will open the bromonium ion by acting as a nucleophile, attacking the more substituted carbon and forming an intermediate (Bridgehead Bromonium Methoxide Intermediate). 3. A second nucleophile, a bromide ion, will displace the methoxide ion to give the final product.

Now, let's analyze the given options to determine which corresponds to the correct product: (a) $\mathrm{BrCH}_{2}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a dibromoalkane, which would not occur in this reaction due to the presence of methanol to give a bromohydrin product instead. (b) $\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a monobrominated product where bromine has been added to the less substituted carbon of the alkene. This would be the correct product if the bromide ion opened the bromonium ion, but as described in the mechanism, methanol first opens the bromonium ion to form an intermediate. (c) a mixture of $\mathrm{BrCH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{BrCH}_{2}-\mathrm{CH}\left(\mathrm{OCH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a mixture of a dibromo compound and a bromohydrin compound where the methoxy group has been added to a primary carbon. However, this is not the correct product since methanol would attack the more substituted carbon instead. (d) $\mathrm{H}_{3} \mathrm{COCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a product where methanol has simply replaced the double bond, which does not occur in this reaction because it involves bromine addition. The correct answer is therefore (c), as it represents the mixture of products that form after methanol first attacks the bromonium ion followed by the bromide ion displacement.