Chapter 23, Problem 23.41P

Acid-catalyzed bromination of pentan $\text{-2-}$ one $\left({\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)$ forms two products: ${\text{BrCH}}_2{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ $\left(A\right)$ and ${\text{CH}}_3\text{COCH}\left(\text{Br}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ $\left(B\right)$. Explain why the major product is $B$, with the $\text{Br}$ atom on the more substituted side of the carbonyl group.