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I want the following conversion to take place$$\ce{Ph-R-CH2-BH2 -> Ph-R-CH3}$$ Does acetic acid react with the given alkyl borane to get this alkane? If yes, what is the correct mechanism? Please explain the mechanism in detail.

I am a newbie to organic chemistry mechanisms(spent 3 weeks,hard work)and I have studied quite a bit of acid-base reactions,electrophilic and nucleophlic substitutions; but cant make this one out.

I have a tried a reaction similar to oxidation and hydrolysis of alkylboranes using hydrogen peroxide (refer to this):

There we have an $\ce{H-O-O-}$ ion,and after it gets attached to the boron by donating an electron pair to form a $\ce{Ph-R-B^{-}-H2-O-O-H}$, thereafter Boron donates a proton to the Oxygen atom attached to it and simultaneously the $\ce{-OH}$ group leaves.That is how the mechanism proceeds in that case.
But here,the problem is in the ion $\ce{CH3COO-}$; even if the boron donates a proton to the bonded $\ce{O}$ atom,$\ce{CH3C^{+}=O}$ is a VERY bad leaving group, hence it wont be correct. I tried thinking about a simple electrophilic substitution, but couldn't understand the nature of $\ce{BH2+}$ as a leaving group.

Any correct way in which acetic acid reacts?

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It would be better if you showed us what you have tried. –  ManishEarth Dec 25 '12 at 6:11
    
I am actually solving a theoretical question.I am a newbie to organic chemistry mechanisms(spent3 weeks,hard work)and I want to know whether a reaction is possible using acetic acid,as i have studied quite a bit of acid--base reactions,electrophilic&nucleophlic substitutions;but cant make this one out. –  scienceauror Dec 25 '12 at 6:14
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OK, but if you "can't make this one out", you must have tried a few mechanisms and failed to get them to fit well. You ought to edit those into your post, as well as your current level of understanding. –  ManishEarth Dec 25 '12 at 6:15
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Now my question is edited correctly –  scienceauror Dec 25 '12 at 11:30
    
@aditya-sriram can you help –  scienceauror Dec 28 '12 at 9:06
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1 Answer

Yes, but you'll need to really beat on it. See the heading under "protonolysis" in this paper.

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