I tried to develop a synthesis strategy for 2-[(E)-2-(3-bromophenyl)ethenyl]-1,3-benzoxazole below. I was able to find the synthesis plan shown on the picture below, however my target molecule is one without a fluorine on the benzoxazole. Hence, my questions is: would this reaction go the same way if I were to react the benzoxazole without the halogen (F)? If not, is there a way to remove it from the target molecule?

Here's a link to the synthesis plan I currently have:

enter image description here

And here's my target molecule:

enter image description here


Here's the reagents and the procedure as stated on reaxys:

With N-benzyl-N,N,N-triethylammonium chloride; sodium hydroxide in >dichloromethane; water Inert atmosphere;

2.1. Synthesis General procedure: The SBOs were prepared by the base-catalysed >condensation of the appropriate 5-halogeno-2-methylbenzoxazole with the >requisite aromatic aldehyde under phase transfer conditions. In a typical >experiment, equimolar quantities (5 mmol) of the starting materials were >dissolved in dichloromethane (20–50 ml) in the presence of >benzyltriethylammonium chloride (3 mmol) and stirred magnetically under a >nitrogen atmosphere as an aqueous solution of sodium hydroxide (50percent, >w/v, 5 ml) was added dropwise over a period of 10 min. After being stirred >for 2–36 h until analytical thin layer chromatography indicated that the >reaction was complete, the mixture was diluted with water (50 ml) and the >SBO was extracted with dichloromethane (3×20 ml), dried (MgSO4), filtered, >evaporated under reduced pressure and recrystallized from aqueous methanol >or ethanol.

  • 1
    $\begingroup$ Something that would remove $\ce{F}$ but not $\ce{Br}$? I doubt it. $\endgroup$
    – ringo
    Oct 31, 2016 at 22:51
  • $\begingroup$ @ringo You might be right. That's why I'm asking whether this reaction could be done with benzoxazole that has no halo grp. $\endgroup$
    – monozuki
    Oct 31, 2016 at 22:57
  • $\begingroup$ I guess it's follow up of chemistry.stackexchange.com/questions/61854/… $\endgroup$
    – Mithoron
    Oct 31, 2016 at 23:04
  • $\begingroup$ @Mithoron Indeed it is. $\endgroup$
    – monozuki
    Oct 31, 2016 at 23:10
  • 3
    $\begingroup$ Oh, so it's basically an aldol condensation (the base deprotonates the methyl group on the benzoxazole, you delocalise the negative charge onto nitrogen, and then that anion attacks the aldehyde. The quaternary ammonium salt is a phase transfer catalyst to bring the NaOH into the organic phase). Well, I don't think the C-F should affect anything really. Pretty smart reaction. $\endgroup$ Oct 31, 2016 at 23:36

1 Answer 1


How to remove fluorine from this compound?

There's no need to try that!

Just use the commercially available 2-methyl-1,3-benzoxazole and try to react it with 3-bromobenzaldehyde in N,N-dimethylformamide under reflux for 2-4 h using self-prepared $\ce{KF-Al2O3}$ as a catalyst.

The reaction has been described (DOI) for several benzaldehydes, such as $\ce{4-H, 4-Cl, 4-Me, 4-OCH3, 4-N(CH3)2}$, with yields > 80%.

  • 1
    $\begingroup$ Thank you! This is exactly what I wanted to know :) Oh and this would be a KF/AL2O3 catalysed condensation reaction, right? And if so, would you mind showing me what would the reaction mechanism for this look like? Anywho. Thank you again! $\endgroup$
    – monozuki
    Nov 3, 2016 at 10:28
  • $\begingroup$ I've done some research and it seems that I can make Potassium fluoride on alumina simply by mixing KF with Al2O3. Is that correct? $\endgroup$
    – monozuki
    Nov 3, 2016 at 10:52

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