Org Prep Daily

September 24, 2006


Filed under: procedures — milkshake @ 12:50 am




The chlorobenzodioxole acid 5.00g (24.93mmol) and triethylamine 4.9mL solution in anh. THF (220mL) was cooled to 0C and ethyl chloroformate 2.9mL (30mmol) was added dropwise over 5 min. The mixture was stirred at 0C for 90 min, ice-cooled solution of sodium azide 8.2g in water 80mL was added, followed by ice-cold water 120mL. The cooling bath was removed and the mixture was stirred at ambient temperature for 10 hours (overnight). The reaction mixture was extracted twice with ether (2x250mL), the extracts were washed with brine (250mL), combined, dried (MgSO4) and evaporated. The crude product (yellow solid, 3.89g) was applied onto a column of silica (80g) in hexane, and eluted with EtOAc gradient, 0 to 30% (in hexane). Y= 3.278g (76.5%) of a pale-yellow crystalline solid.


1H(d6-DMSO, 400MHz): 6.739(d, 8.4Hz, 1H), 6.241(d, 8.4Hz, 1H), 5.976(s, 2H), 5.117(very br s, 2H); 13C(d6-DMSO, 100MHz): 146.03, 133.79, 129.28, 121.60, 111.80, 106.96, 101.04. 97.79




  1. Hi, might be a silly question, but I was wondering why you get the acyl azide formed on the benzodioxole acid as opposed to the azide attacking the ethyl chloroformate derived carbonate carbon. Surely steric factors would favour this, so what other factor/s cause the formation of the observed product.

    Comment by Q — September 25, 2006 @ 4:48 am

  2. You get acyl azide. Azide eliminates N2 to form acyl nitrene that immediately rearranges to isocyanate (carbonyl migrates – look up Curtius rearrangement in the book.) I could not put all intermediates in the scheme – the structures must be large (at least 0.3 median bond) or they will not display correctly here. If the scheme is too long, the right side of the reaction scheme will not fit printout and 17 inch monitor.

    Comment by milkshake — September 25, 2006 @ 9:57 am

  3. Yeh i know what the curtius rearrangement is, Probably my question wasn’t clear but why do you get the desired benzodioxoleacyl azide as opposed to the ethyl formylazide, and the benzodioxolate? I can understand why it would attack at the benzodioxole “end” if a more hinder mixed anhydride partner was chosen, i.e. isobutyl chloroformate… but ethyl…??

    Comment by Q — September 25, 2006 @ 11:10 am

  4. That is good question – I don’t know. But it seems quite general that in these mixed anhydrides it is the ROCO2- that works as a leaving group. Perhaps it has to do something with electron density but I am not sure.
    There is another similar example: mixed anhydride HCO-O-COCH3 made by mixing equal volumes of acetic anhydride and 96% formic acid (and letting the mix sit under nitrogen for 4 hours at RT.) This is an excellent reagent for formylating alcohols and very unreactive amines. Contrary what one would expect, there is no acetylated product in most cases, only the formyl gets transfered.

    Comment by milkshake — September 25, 2006 @ 11:22 am

  5. Entropically you would expect elimination of the EtCOO- with decomposition to EtO- and CO2. I think you are on the right track Milkshake, I do not have a Gaussian package handy, but I recall that “carbonates ” are much less electrophilic than “esters” (like in your substrate).

    If you use the isopropyl chloroformate, you introduce a steric factor as well as an electronic one.

    Nonetheless, it is a nifty transformation.

    Comment by milo — September 25, 2006 @ 12:18 pm

  6. Milo you are right, carbonates C=O is much less reactive than the corresponding ester. Q, the easiest way to rationalize this decreased electrophilicity is to say that the non-carbonyl oxygen acts as a nucleophile toward the partially positive carbonyl C. The presence of two rather than one “internal” nucleophiles in carbonates as opposed to esters makes it much harder for an “external” nucleophiles, such as azide, to attack that center. That to me is the simplest explanation, but it makes sense.

    Comment by aa — September 25, 2006 @ 6:43 pm

  7. The carbonate carbonyl has more even distribution of the electron density, and therefore less dipole movement. Electronically speaking, with every electronegative atom that you add to the carbonyl system (N>O>F>Cl>Br) you are significantly raising the energy of the corresponding LUMO. This clearly explains the reactivity:

    Aldehyde > Acyl Chloride > Ester >>> Amide > Carbonate

    Comment by Ryan K. — September 26, 2006 @ 4:20 am

  8. Hi Milkshake,

    I am going through your old stuff. Still find them helpful. Here can you help me understand why in this case the Curtius rearrangement happens at low temperature? I remember that you need to heat up to 80 degree to make the isocyanate rearrange. Is this condition general?

    Comment by diketenes — February 3, 2007 @ 10:01 pm

  9. I don’t know – I was suprised also, because what I thought was acyl azide after isolation turned out to be the final aniline actualy (and quite unpolar one). I have seen this again with another acyl chloride (Curtius done with TMS-N3, at 0 to RT). In both cases the acyl azide was o,o-disubstituted so I think a possible explanation is that steric influences make the cabronyl out of plane with the aryl which makes it less stable – so it decomposes around room temp. But I don’t know for sure.

    Comment by milkshake — February 3, 2007 @ 11:53 pm

  10. Thanks Milkshake, I was looking for a procedure to make aniline from aromatic acid or phenol.

    Comment by pmgb — May 17, 2008 @ 2:17 pm

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