4
$\begingroup$

I have been using a reaction with an organozincate starting material. This zincate is freshly prepared from zinc chloride and a Grignard reagent. The experimental procedure ‘handed down’ from more senior members of the lab include fusing the zinc chloride before use. ‘Fusing’ means:

  • heating under high vacuum with a heat gun until the salt melted
  • cooling the liquid down rapidly with the cold air stream of another heat gun to receive a glass-like solid

This is then dissolved in THF and the Grignard added.

In a recent group seminar I spoke about the trouble I had in achieving consistency with that reaction. Part of the resulting discussion revolved around the method of fusing and what I had been doing wrong. The PI noted how important it is for the fusing to be performed correctly as it influences the reactivity.

What actually happens to the zinc chloride during this process? My little understanding of inorganic solid-state chemistry leads me to believe that the initial crystals break down but the resolidified glass should essentially have the same structure. I am also at a loss how to explain any differences in reactivity following this method.

While it is not the most easy task to search for such references, the concept of fused zinc chloride does exist in the literature; the following two references use it in their experimental sections:

  1. P. Dotta, A. Magistrato, U. Rothlisberger, P. S. Pregosin, A. Albinati, Organometallics 2002, 21, 3033–3041. DOI: 10.1021/om020314q. Synthesis of the ligand 4 on page 3040.

  2. P. G. Gassman, J. G. Schaffhausen, J. Org. Chem. 1978, 43, 3214–3223. DOI: 10.1021/jo00410a024. Synthesis of compound 28 on page 3220.

$\endgroup$
  • $\begingroup$ Zinc Chloride is highly hygroscopic so what you are doing by melting it under high vac is driving off the water it has absorbed. When it resolidifies I doubt its structure will be greatly different (except for the missing water) $\endgroup$ – Waylander Feb 7 '18 at 14:36
  • $\begingroup$ If there's enough water around to formally form the hydrate of zinc chloride that gives Zn(OH)Cl on heating and could possibly account for the variability you're observing $\endgroup$ – Waylander Feb 7 '18 at 14:44
  • $\begingroup$ @Waylander One thing to note is that our zinc chloride is stored and filled in a glove box. Not sure how much water there can still be remaining. $\endgroup$ – Jan Feb 7 '18 at 15:06
  • $\begingroup$ But if you use it without first fusing it, what happens? $\endgroup$ – Waylander Feb 7 '18 at 15:20
  • $\begingroup$ @Waylander Haven’t tried that; only wrong fusing (i.e. letting it cool down slowly rather than rapidly which was identified as one of the possible points of concern.) $\endgroup$ – Jan Feb 7 '18 at 15:38
4
$\begingroup$

One purpose is to eliminate water as @Waylander has already pointed out.

Another is to have a clean crystal surface. Aging is a common process of crystals. The surface is transformed slowly. The crystalline structure changes since it is not the same to be surrounded with other atoms (molecules, ions) of the same composition than with nothing or a gas.

Finally, another purpose is to have a crystal size as small as possible. The smaller the crystals the bigger the surface area, so more is available and the reaction proceeds faster. The small crystals are obtained in the rapid cooling. The faster, the smaller.

The zinc chloride solidifies in one lump, but there are still crystals. They only have grown together. If it were a real glass, the reactivity would be the highest since there would be no packing. This can only be achieved with extreme rapid cooling. When cooling is fast but still takes some time, crystals grow. When the cooling is slower, the crystals grow bigger. The smaller they are, the more surface the lump has.

Melting must be done for the other reasons, but when you cool it must be done fast because, if not, you will get a less reactive mass. It is not a comparison between molten and non molten product, but between fast or slow cooling of the molten salt.

$\endgroup$

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.