The products AB and B are

It is haloform test

$-{\mathrm{CH}}_3$ group of $-\mathrm{CO}-{\mathrm{CH}}_3$ gets halogenated first to form $-\mathrm{CO}-{\mathrm{CX}}_3.$

$-\mathrm{CO}-{\mathrm{CX}}_3$ undergoes alkaline hydrolysis to give haloform and a carboxylic acid with one carbon lesser than the parent is formed during which C = C is not oxidized.

${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{CO}-{\mathrm{CH}}_3 \stackrel{\mathrm{NaOCl}}{\to } {\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{COONa} + {\mathrm{CHCl}}_3$