0
$\begingroup$

I'm looking for a way to separate reaction product of iodobenzene with ethyl 2-methyl-3-oxopentanoate in 1,4-dioxane (toluene is an alternative as a solvent). The catalyst being used is $\ce{Pd2(DBA)3}$ with $\ce{Ph3P}$ as a ligand and $\ce{K3PO4}$ as a base.

I was thinking about fractional distillation under partial vacuum because there's a risk of explosion if there's some leftover unreacted iodobenzene heated to high temperatures.

Provided the liquids distil off, the reaction flask should be left with some solid junk containing $\ce{[Ph3P + Pd2(DBA)3]}$ and $\ce{K3PO4}$. Then this mixture could be washed with water to get rid of $\ce{K3PO4}$, $\ce{KI}$, water-soluble side products and dry the catalyst + ligand under vacuum and use it again in another run.

So, is it possible for me to separate the product from the liquids using distillation and to reuse catalyst with the procedure I proposed?

Improve this question
$\endgroup$
7
  • $\begingroup$ I don't see any problem with the procedure you describe for re-isolating the Pd-containing stuff, but I would question how worthwhile it is. I'd imagine by the end of the reaction the majority of the Pd is in the form of palladium black, and no longer catalytically active. $\endgroup$
    – PCK
    Nov 22, 2019 at 13:55
  • $\begingroup$ On what scale is this? $\endgroup$
    – Waylander
    Nov 22, 2019 at 14:07
  • $\begingroup$ using 1kg of iodobenzene would mean 50g of palladium catalyst is needed.. given the price for 1g is 20$, this synthesis gets quite expensive, so is there any way how to recycle the catalyst? $\endgroup$
    – steve d.
    Nov 22, 2019 at 14:37
  • $\begingroup$ Could try using Cu as catalyst as here: onlinelibrary.wiley.com/doi/abs/10.1002/anie.201106719 $\endgroup$
    – Waylander
    Nov 22, 2019 at 15:09
  • $\begingroup$ What is your final target? If cost is a major consideration then starting with ethylphenylacetate (just over £60/kg) may be a better route. $\endgroup$
    – Waylander
    Nov 22, 2019 at 15:56

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.