To determine the major product formed on the reaction of 1-pentene with bromine in aqueous solution, we need to consider the reaction mechanism.
When 1-pentene reacts with bromine (Br2), it undergoes electrophilic addition, where the bromine molecule adds across the double bond of the alkene. Initially, the reaction produces a cyclic bromonium ion intermediate.
In aqueous solution, water (H2O) can act as a nucleophile and attack the cyclic bromonium ion. This results in the formation of a bromohydrin, which is an alcohol with a bromine substituent. The nucleophilic attack can occur at either of the carbons adjacent to the bromine atom.
In the case of 1-pentene, the bromine atom can be added to either the first carbon (C1) or the second carbon (C2) of the alkene. Therefore, there are two possible products: 1-bromo-1-pentanol (where the bromine is added to C1) and 2-bromo-1-pentanol (where the bromine is added to C2).
To determine which product is the major one, we need to consider the stability of the carbocation intermediate that forms during the reaction. The stability of a carbocation depends on the number of alkyl groups attached to the positively charged carbon. A more stable carbocation is formed when the positive charge is on a more substituted carbon.
In this case, the carbocation intermediate formed when the bromine is added to C2 is more stable because it is attached to two alkyl groups, whereas the one formed when the bromine is added to C1 is only attached to one alkyl group. Therefore, the major product formed on the reaction of 1-pentene with bromine in aqueous solution is 2-bromo-1-pentanol (c).
Therefore, the correct answer is c. 2-Bromo-1-pentanol.