Question

I have a reaction in which I perform a deprotonation of an acidic proton at a phosphate and later do a P-C-P coupling. The coupling partner is diethylchlorophosphate. Deprotonation is performed with LDA in dry THF. Deprotonation is performed at -78 Degrees Celsius and argon. Unfortunately I also observe triethylphosphate (TEP) as a side product that can not easily be separated of from the desired product by silica flash coloumn chromatography. I am wondering what the mechanism could be, how diethylchlorophosphate forms triethylphosphate and how this formation can be avoided completely? Can LDA abstract an ethanol/ ethoxy unit that subsequently reacts with diethylchlorophosphate? Than less base should have an effect? Might the slower addition of the diethylchlorophosphate to the carben ion also reduce the formation?

Answer 1

Triethylphosphate (TEP) forms as a side product due to the reaction between diethylchlorophosphate and an ethanol/ethoxy unit. To avoid TEP formation, you can reduce the amount of base used and add diethylchlorophosphate slowly.

Step-by-step explanation:

In the reaction you described, the formation of triethylphosphate (TEP) as a side product can be explained by the reaction between diethylchlorophosphate and LDA-abstracted ethanol/ethoxy unit.

In this mechanism, the LDA deprotonates an acidic proton at the phosphate group, forming a carbanion. This carbanion then reacts with diethylchlorophosphate, resulting in the formation of triethylphosphate as a side product.

To avoid the formation of TEP completely, you can try reducing the amount of base used, as using less base may reduce the likelihood of LDA abstracting an ethanol/ethoxy unit. The slower addition of diethylchlorophosphate to the carbanion may also help reduce the formation of TEP.