# Boron enolates double bond geometry

The usage of dialkylboron triflates produces Z enolates from ketones predominately, while the usage of dialkylboron chlorides changes the selectivity favoring the E enolate: How is this trend explained?

## 1 Answer

The image below (Taken from the Evans' lectures, which you might like to consult as there is a whole lot more detail in them) sums up the situation.

Essentially you're interested in how good a leaving group is attached to the boron. The triflate is excellent and as such fully dissociates, whilst the chloride remains (at least partially) in association.

Depending on whether the group on boron dissociates, the 'boron' is significantly more/less bulky, changing the conformation from which enolisation is preferred, which in turn leads to E/Z enolates.

The choice of base is also important. With boron triflates, DIPEA is often used as it is just that bit bulkier and so biases even more towards the Z enolate. In the boron chloride aldols, the bulky DIPEA is a bit too much, and so Et3N is commonly used.

If you study the TS's below, it might be apparent why the boron chloride aldols are typically a little less selective than the boron trifoliate aldols.

Source: Dave Evans' Harvard CHEM207 Lecture notes.

• Yes, his lectures are very enlightening. So, to sum up, regarding the triflate, BL2OTf-Me steric repulsion is much less becase the boron is not tetrahedral and the R-Me repulsion determines the reactive conformation. In the case of the chloride, boron is tetrahedral and the repulsion BL2Cl-Me is significant, so the reactive conformation is the one suffering Me-R repulsion which is why the lower stereoselectivity in this case. Did I get it right?
– EJC
Aug 13 '16 at 22:37