When 2-bromobutane is treated with aqueous KOH , the major product...

When 2-bromobutane is treated with aqueous $\mathrm{KOH}$, it undergoes nucleophilic substitution. The hydroxide ion acts as a nucleophile and replaces the bromide ion, yielding butan-2-ol as the major product $P$.

When 2-bromobutane is treated with alcoholic $\mathrm{KOH}$ and heated, it undergoes $\beta$-elimination (dehydrohalogenation). The hydroxide or alkoxide ion acts as a base, removing a proton from the $\beta$-carbon. According to Zaitsev's rule, the more highly substituted alkene is formed as the major product. Thus, but-2-ene is the major product $Q$.

Therefore, $P = \text{Butan-2-ol}$ and $Q = \text{But-2-ene}$.

Answer: $P = \text{Butan-2-ol}, \ Q = \text{But-2-ene}$