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This reaction is a dehydrohalogenation of an alkyl halide, specifically 1-bromopropane, using a strong base (KOH) in an alcoholic solvent (ethanol) with heat. These conditions favor an E2 (Elimination, bimolecular) reaction. Step 1: Identify the reactants and reaction type. The reactant is 1-bromopropane (CH_3CH_2CH_2Br), 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. Step 2: Identify the alpha () and beta () carbons and hydrogens. In 1-bromopropane: The carbon atom directly bonded to the bromine is the alpha () carbon. The carbon atom adjacent to the alpha carbon is the beta () carbon. CH_3 - CH_2 - CH_2 - Br 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. Step 3: Perform the E2 elimination. 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 (Br^-). This occurs in a single, concerted step. CH_3CH_2CH_2Br + KOH Ethanol, Heat CH_3CH=CH_2 + KBr + H_2O Step 4: Identify the final product. The main organic product formed from the elimination of HBr from 1-bromopropane is propene. The final product is: Propene 3 done, 2 left today. You're making progress.