Major products A and B formed in the following reaction sequence are:

1. ${\text{PBr}}_3$ converts secondary alcohols to alkyl bromides. ${\text{CH}}_3{\text{CH(OH)CH}}_2{\text{CH}}_3\to {\text{CH}}_3{\text{CH(Br)CH}}_2{\text{CH}}_3$ (2-bromobutane). (Product A).2. Alcoholic KOH causes dehydrohalogenation (Elimination). According to Saytzeff's rule, the more substituted alkene is the major product.${\text{CH}}_3{\text{CH(Br)CH}}_2{\text{CH}}_3\stackrel{-\text{HBr}}{\to }{\text{CH}}_3{\text{CH=CHCH}}_3$ (But-2-ene). (Product B).