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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?

methacrylic acid

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    $\begingroup$ Are there any other conditions for the second reaction, like heating or acid catalysis? If that were the case then I would suggest an ester as the product, after a reduction to the primary alcohol by borane in the first step. $\endgroup$ – bon Jul 20 '16 at 18:46
  • $\begingroup$ Last step requires a nucleophile. It can be water, base or acetate (from acetic acid. Nucleophile will attach to terminal CH2 group. In your case it will be acetate. $\endgroup$ – sixtytrees Jul 20 '16 at 19:17
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    $\begingroup$ A nucleophile is required, yes. But to form the boron-ate complex shown in my answer. Also, @sixtytrees notice that the nucleophile in the first step of the question is not water it's hydrogen peroxide with base to give a peroxide ion. This forms an ate complex and there is a migration followed by hydrolysis to give the alcohol. NOT a substitution. $\endgroup$ – RobChem Jul 20 '16 at 20:50
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It's simply a protonolysis: enter image description here

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.

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  • $\begingroup$ Can you show me the mechanism of my reaction $\endgroup$ – Koolman Jul 21 '16 at 9:30
  • $\begingroup$ It's exactly the same. The double bond is the thing that matters. Perhaps I should have made it clear that there is and initial syn addition of BH3 across the double bond. This is called hydroboration. The reaction in my answer starts with 1-methyl-cyclopent-1-ene. $\endgroup$ – RobChem Jul 21 '16 at 9:31
  • $\begingroup$ so what is the role of compound which I have shown in image $\endgroup$ – Koolman Jul 21 '16 at 9:34
  • $\begingroup$ It contains the double bond which is what reacts with bh3. $\endgroup$ – RobChem Jul 21 '16 at 10:46
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The acid would be reduced by the borane to the primary alcohol first. Subsequent hydroboration of the product allylic alcohol would probably give you the diol.

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    $\begingroup$ Can you please draw the mechanism as i am not able to understand like this proparly $\endgroup$ – Koolman Jul 20 '16 at 18:25
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    $\begingroup$ Whilst I think this is probably true, it doesn't answer the second part of the question about what happens with 1. $\ce{BH3/THF}$, 2. $\ce{CH3COOH}$. $\endgroup$ – bon Jul 20 '16 at 18:44
  • $\begingroup$ @bon yes I also think you are correct $\endgroup$ – Koolman Jul 21 '16 at 9:28

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