Reaction of methacrylic acid with BH3 If methacrylic acid (2-methylprop-2-enoic acid) is reacted with first $\ce{BH_3/THF}$ and second  $\ce{H_2O_2/HO^{-}}$ then $\ce{OH}$ group is attached at less subtituted carbon. 

But if it is treated with $\ce{CH_3COOH}$ in the second step, what is the end product?
It's simply a protonolysis:


Except in the picture D is used instead of H. Hydroboration based reactions always proceed via this boron-ate complex (negatively charged) then migration and loss of leaving group. Search google for carbonylation, amination, cyanidation using boron.