[davidenarb]

Hi all,

Hydroboration-Oxidation is an anti-Makovnikov addition, and for the starting material shown below, the regioselectivity is not relevant because we have the same number of the substituent on each side of the pi bond (one on the right, and one the left). However, if the -OH group goes to carbon 2, we will have chirality (so we will have one pair of enantiomers, S and R), and if it goes to the carbon 3, then we will not have chirality, and thereby it is not the same products.

So, where does the -OH group go and why ?

[davidenarb]

Will we have a mixture of 3 products = The S product + The R product + The product without chirality  ?

[orgopete]

Will we have a mixture of 3 products = The S product + The R product + The product without chirality  ?

That is my expectation, but not in equal amounts. Not greatly different, but the substituents are different so the ratios will also be different, maybe 55:45 favoring the 3-pentanol.

[kriggy]

Why do you think that there will be more of 3-pentanol in the mixture? I would suggest that the main product will be 2-pentanol. Just from my understanding of this reaction I think that the the boron will add to the 2nd carbon

[davidenarb]

 

Why do you think that there will be more of 3-pentanol in the mixture? I would suggest that the main product will be 2-pentanol. Just from my understanding of this reaction I think that the the boron will add to the 2nd carbon

Good question! However, I don't think the boron will play an essential role because the reaction is not regioselective, so even if it's Acid Catalyzed Hydration or Oxomercuration-Demercuration, then we will have the same product. I, nevertheless, still don't know which product will major though !

[orgopete]

Why do you think that there will be more of 3-pentanol in the mixture? I would suggest that the main product will be 2-pentanol. Just from my understanding of this reaction I think that the the boron will add to the 2nd carbon

Even though the alkene carbons are similar, they are not the same. This is not like a racemic mixture of isomers in which the only difference is absolute stereochemistry. A methyl group is different than an ethyl group. However, I only expect a relatively small effect on the regiochemistry. Since boron is inherently electrophilic in this reaction, and I believe this governs the regiochemistry, the a replacing a hydrogen with a more electron donating carbon should increase any partial charges generated at that carbon, though only by a small amount. For this reason I suggested 3-pentanol should be found in a slightly larger quantity.

That was the basis of my speculation, now all we need is some actual data.

[kriggy]

Exactly thats why I think the boron should add to the second carbon

The picture is crappy but it ilustrates my thought that the ethyl group with stronger inductive efect will move the electron density to the 2nd carbon. The differece will not be that big but I would expect more of the pentane-2-ol just from this thought that the 2nd carbon has slightly higher electron density.

[orgopete]

Okay then, we are disagreeing on the mechanism. I think BH3 lacks an octet and reacts with the electrons at the most negatively charged carbon. If it were a terminal alkene, the negative charge would be on the terminal carbon.

[kriggy]

exaclty. I think we think the same about the mechanism - BH₃ reacts with the most negatively charged carbon. Thats why I asked why do you expect the 3-pentanol to be major (even only by small percentage) product. The way I see it, the carboon in position 2 is more electron rich than the one in position 3. Its not that big of a difference between those two so I would expect some mixture of products anyway.

[orgopete]

Oops, I agree. I guess that is what can happen when you don't write anything down.