To solve this problem, we need to understand the concept of dehydrohalogenation and the rules governing the formation of alkenes.$\newline$Dehydrohalogenation is a chemical reaction that involves the elimination of a hydrogen halide (HX) from an alkyl halide to form an alkene.$\newline$In the case of 2-Bromo pentane, the reaction can be represented as:$\newline$2-Bromo pentane$\xrightarrow{Base}$Pent-2-ene$+$HBr$\newline$The major product formed is Pent-2-ene, which is an internal alkene.$\newline$This product formation is based on Saytzeff's Rule, which states:$\newline$• In an elimination reaction, the more substituted alkene is the major product.$\newline$• This rule predicts that the hydrogen atom is removed from the beta-carbon with the fewest hydrogen atoms.$\newline$• As a result, the more stable, more substituted alkene is formed.$\newline$In this case, Pent-2-ene is more substituted than Pent-1-ene, making it the major product.$\newline$Analyzing the options:$\newline$• Option 1: Saytzeff’s Rule - This is correct as it explains the formation of the more substituted alkene.$\newline$• Option 2: Hund’s Rule - This is related to electron configurations and is not applicable here.$\newline$• Option 3: Hofmann Rule - This predicts the formation of the less substituted alkene, which is not the case here.$\newline$• Option 4: Huckel’s Rule - This is related to aromaticity and is not applicable here.$\newline$Therefore, the correct option is Option 1: Saytzeff’s Rule.