0
$\begingroup$

What is the best way to measure and transfer a sticky compound in the lab, in particular for air/moisture sensitive reactions? I have a really sticky solid Wittig salt which I'm going to use in a Wittig reaction.

$\endgroup$
  • $\begingroup$ PTFE (Teflon etc., e.g. sigmaaldrich.com/catalog/product/aldrich/z115282) may help if the salt is hygroscopic, since both polar (i.e. salt solution) and non-polar substances do not stick well. Silicone tools might also be of use. $\endgroup$ – DrMoishe Pippik Dec 8 '16 at 1:42
  • $\begingroup$ Could you wrap it in plastic wrap? $\endgroup$ – Joseph Hirsch Dec 8 '16 at 3:21
  • $\begingroup$ Glovebox - you can fill it with a neutral dry gas. I would also suggest you checking the purity of your substrate - sticky solids usually mean impurities or solvent presence. $\endgroup$ – vapid Dec 8 '16 at 13:47
1
$\begingroup$

Ideally, you want your sticky solid to already be in a Schlenk flask or similar. If it is unfortunately still in a normal bottle, take the weight of a Schlenk flask, fill in enough of the sticky solid and re-weigh. Note the difference, this is what you have in your flask.

Assuming that the sticky solid has a low enough vapour pressure (spoilers: practically all solids do), evacuate the Schlenk flask under high vacuum and backfill with argon (or nitrogen, if that is your inert gas of choice. Add a known quantity of your (dry, stored over molecular sieve and under inert atmosphere below a septum) solvent. Ideally, you either want an amount of solvent that gives you an easy-to-handle concentration (e.g. $10~\mathrm{M}$) or an amount so that what you need is in an even volume of liquid (e.g. $1~\mathrm{ml}$).

Finally, transfer the solution into your reaction flask.

If you need the sticky solid in different solvents and it is air-stable, you can evaporate the dry solvent at the rotavap. If not, keep it in solution under argon (or nitrogen). If the latter is the case, do your colleagues a favour and choose a standard concentration, not a good filling amount for exactly your reaction.

$\endgroup$
  • $\begingroup$ I do wonder why this answer has been deemed ‘not useful’ or ‘unclear’. $\endgroup$ – Jan Dec 11 '16 at 0:44

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.