2
$\begingroup$

I'm wondering if any reaction pathway exists (not involving enzymes) to do the following in aqueous solvent at temperatures < ~100°C:

To clarify, it's important however that the reaction is concerted, and that the W - X and Y - Z bonds are not broken until both the W-Y and X-Z bonds are formed. Bonds here must be covalent, but otherwise can be of any type. Any kind of catalyst can be used. Of course, it would be nice to know of an efficient reaction with minimally exotic conditions or reagents (i.e. it would be nice to avoid heavy metals for oxidative-addition type reactions), but I won't push my luck.

Does something like this exist?

$\endgroup$

closed as too broad by hBy2Py, airhuff, ron, M.A.R., Zhe Apr 8 '17 at 19:14

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ (Let's get unnecessarily technical) What exactly defines a 'covalent bond'? $\endgroup$ – LordStryker May 12 '14 at 14:45
  • 1
    $\begingroup$ @LordStryker I hardly see why getting technical in this manner is unnecessary. I'm alright here with any bond that will be stable @ 50 C in aqueous solvent for, on order, a few days. By "covalent" perhaps I should have said "not hydrogen or Van der Waals bonded". $\endgroup$ – user5475 May 12 '14 at 14:57
  • $\begingroup$ 'Covalent' is a widely used term to generally describe a bond. I describe my question as being 'unnecessary' because, while it serves as a good reminder that nobody really knows what a covalent bond truly is (not a line drawn between two atoms, not an 'equal' sharing of a pair of electrons, etc.), it is rather silly to launch into a debate/discussion about it anytime someone wants to use 'covalent bonding' as a general description of something (whether here on SE or in the literature). That is all. $\endgroup$ – LordStryker May 12 '14 at 15:18
  • $\begingroup$ @LordStryker Oh ok I see. Just saying though, I wouldn't want to rule out pseudo-ionic bonds between, for example, carbon and heavy metals (which are sort of a mix between covalent and ionic bonds). $\endgroup$ – user5475 May 12 '14 at 15:23
4
$\begingroup$

$$\ce{W=X + Y=Z ->[\mathrm{cat}] W=Y + X=Z}$$ is known as olefin metathesis, achieved using Grubbs/Hoyveda/Fürstner catalysts.

Water-soluble Ru catalysts have been prepared in the Grubbs group (DOI).

$\endgroup$
  • 1
    $\begingroup$ And we can learn a lot about that for free from Nobelprize.org. $\endgroup$ – Martin - マーチン May 12 '14 at 14:19
  • $\begingroup$ @Klaus Warzecha Thanks for the additional link! So, I keep seeing the reaction: CH3=C-R1 + R2-C=CH3 --> R1-C=C-R2 (releasing H2C=CH2 gas). Can those terminal methane groups be anything? $\endgroup$ – user5475 May 12 '14 at 15:14
  • $\begingroup$ @Klaus Warzecha I should specify that by "anything" I mean, can those CH3 groups be CH2 - CH3 / etc. groups. $\endgroup$ – user5475 May 12 '14 at 15:20
  • $\begingroup$ @user5475 I rather doubt that it will work with anything but at least one $\ce{R1R2C=CH2}$. $\endgroup$ – Klaus-Dieter Warzecha May 12 '14 at 15:27
  • $\begingroup$ @KlausWarzecha Sorry, it's my fault but I can't understand your comment? I'm kind of wondering if R1R2 C=C R3R4 will be able to react with R5R6 C=C R7R8 to give R1R2 C=C RR5R6, R1R2 C=C RR7R8, etc. (where the Ri groups are carbons). Why is methane special? $\endgroup$ – user5475 May 12 '14 at 15:31
3
$\begingroup$

And the Wittig reaction works in a quite similar fashion:

Wikipedia Wittig_reaction_scheme

Note that the Phosphorous Oxygen Ylide Bond may has mesomeric forms: $$\ce{-{}^{+}P-C^{-}-<->-P=C-{}}$$

And we can again learn a lot for free at Nobelprize.org.

$\endgroup$
  • $\begingroup$ To nitpick, the Wittig reaction isn't concerted (a condition specified by OP). $\endgroup$ – orthocresol Apr 8 '17 at 17:36

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