[Nescafe]

Hello,

I was wondering if someone could help me understand the mechanism. Imagine we start with R-OH, reacting it with the reagents specified would mean that the alcohol attacks the BF3.OEt2 (strong lewis acid) yielding a good leaving group (R-OH-BF3). Next the hydride transfer (source: Et3Si-H) which kicks of HO-B(-)F3 forming R-H as the product and releases Et3Si+? Or is it more of a SN1 route where HO-BF3 leaves giving a R(+) to which the hyride transfers to?

Nescafe.

[Nescafe]

Any advice on this?

[orgopete]

I don't know the answer. If the alcohol is tertiary, I think the reaction is  SN1. I don't think triethylsilane is very nucleophilic, so I am not thinking it is going to be SN2. Does the reaction work with a primary alcohol? If not, not SN2.

[Nescafe]

I see, the alcohol is secondary, it is cyclobutanol.

[orgopete]

I'm still guessing SN1.

[Nescafe]

I'm still guessing SN1.

Its done in ether though, I thought SN1s are typically favored in polar protic solvents to stabilize the charge of the T.S. Hmmm, but I see your point it makes sense, and SN1 does not have to be done in a polar protic solvent, i would imagine ether is sufficient to dissolve the reagents.