Solve in English

The reactant is 1-bromopropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$), which is a primary alkyl halide. The reagent is potassium hydroxide (KOH) in ethanol with heat. This combination is characteristic of an E2 elimination reaction, where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms to form an alkene.

In 1-bromopropane:

• The carbon atom directly bonded to the bromine is the alpha ($\alpha$) carbon.
• The carbon atom adjacent to the alpha carbon is the beta ($\beta$) carbon.

The beta hydrogens are the hydrogen atoms attached to the beta carbon. In this case, there is only one type of beta carbon with available hydrogens.

The strong base (ethoxide, formed from KOH in ethanol) will abstract a beta-hydrogen. Simultaneously, the electrons from the C-H bond will form a new C=C double bond, and the bromine atom will depart as a bromide ion ($\text{Br}^-$). This occurs in a single, concerted step.

The main organic product formed from the elimination of HBr from 1-bromopropane is propene.

The final product is: