Answer:
Yes, that is correct. When 1-bromobutane is heated with a strong base like potassium tert-butoxide, it undergoes dehydrohalogenation via an E2 (elimination bimolecular) mechanism, in which the bromine atom is removed from the molecule along with a proton from a neighboring carbon atom. This results in the formation of an alkene (butene) and a molecule of hydrogen bromide. The reaction follows second-order kinetics because it involves two species (1-bromobutane and the base) in the rate-determining step.
Step-by-step explanation: