Final answer:
Bromine (Br₂) reacts with polar protic solvents to form a variety of products. In the presence of alkenes, bromine forms bromohydrins. The rate of formation of Br₂ from the disappearance of Br⁻ can be calculated using the stoichiometry from the balanced chemical reaction.
The corrcet answer is 4).
Step-by-step explanation:
When Bromine (Br₂) reacts with polar protic solvents, a variety of reactions can occur depending on the specific conditions and solvents involved. However, in the context of organic reactions, bromine can react with alkenes in the presence of a polar protic solvent to form bromohydrins. In this reaction, the addition of Br₂ across the double bond occurs through a cyclic bromonium ion intermediate, which is then attacked by the solvent, leading to the formation of a bromohydrin (a molecule containing both a hydroxyl and a bromine).
Regarding the reaction provided, the rate of disappearance of Br⁻(aq) is 3.5 × 10⁻⁴ mol L⁻¹ s⁻¹ and the balanced chemical equation shows a stoichiometry of 5 Br⁻ to 3 Br₂. To find the rate of formation of Br₂, we would divide the rate of disappearance of Br⁻ by 5 (the stoichiometric coefficient) and then multiply by 3 (the stoichiometric coefficient of Br₂), as shown in the equation:
Rate of formation of Br₂ = (3.5 × 10⁻⁴ mol L⁻¹ s⁻¹ / 5) × 3 = 2.1 × 10⁻⁴ mol L⁻¹ s⁻¹.
Therefore, the correct option is: 4) Formation of bromoalkane