Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: batty2279 on January 21, 2015, 06:30:42 PM
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I shall preface this post by saying I'm a biochemist and synthesis is not my specialty! haha So this may seem extremely straight-forward, but I'm having problems with the final step of this synth protocol. It is as follows (I found it in 2 separate papers, both very similar):
1) Synthesis of methylacetylphosphonate. (MAP) Dimethylacetylphosphonate (0.5 g, 3.3 mmol) was added to a solution of NaI (5.42 g, 36.2 mmol) in acetone. The reaction was stirred at room
temperature overnight. The resulting solid was filtered and triturated with acetone to remove
remaining NaI.
2) General Procedure fbr the Synthesis of Methyl Acylphosphoncrte Monosalts (2).
Dimethyl acylphosphonate (0.05 mol) was dissolved in dry acetonitrile (or dry acetone) (50 ml), and the solution was added to a solution of lithium bromide (0.55 mol) in dry acetonitrile (30 ml) (or sodium iodide in acetone). The reaction mixture was stirred for 12 h at room temperature after which the precipitate was filtered off, washed with dry acetone, and dried.
This seems super straight forward. I'm assuming the charged Dimethylacetylphosphonate is coupling to free sodium ions to form the salt. However, after the overnight rxn, nothing appears different. Is my product mixed in with the excess of NaI? I have tried to triturate the excess NaI off with cold acetone, but I end up with nothing. I guess my product is highly soluble in it as well. I have no idea how they claim a 93% yield. I have some crystals in my buchner funnel, but when I ran a MS, I got no product, so I'm assuming that was just NaI crashing out on the filter paper. Now I have a large flask full of acetone with all of the salt in solution. Is there anyway to get the MAP (product) out now? If I rotovap it, it will still be a mixture of MAP and NaI correct? UGH! any help would be greatly appreciated!
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This seems super straight forward. I'm assuming the charged Dimethylacetylphosphonate is coupling to free sodium ions to form the salt. However, after the overnight rxn, nothing appears different. Is my product mixed in with the excess of NaI? I have tried to triturate the excess NaI off with cold acetone, but I end up with nothing. I guess my product is highly soluble in it as well. I have no idea how they claim a 93% yield. I have some crystals in my buchner funnel, but when I ran a MS, I got no product, so I'm assuming that was just NaI crashing out on the filter paper. Now I have a large flask full of acetone with all of the salt in solution. Is there anyway to get the MAP (product) out now? If I rotovap it, it will still be a mixture of MAP and NaI correct? UGH! any help would be greatly appreciated!
What are the literature references? The second sentence in the paragraph above is confusing. Do you mean charged (mono)methylacetylphosphonate? One of two things might be true. Either the demethylation occurred but the product stayed in solution, or the demethylation never occurred. Differentiating between these two possibilities might be done by following the reaction with TLC or monitoring the reaction by P-31 NMR.
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What you are doing here is hydrolysing one of the phosphonate esters to the mono phosphonic acid which is then isolated as the mono sodium salt. As NaI is acetone soluble and you are using 10 equivalents the excess dissolves in the acetone and you end up with, hopefully, pure product.
Same thing for the other procedure, except this time you will get either the Li salt or the Na salt. Using 1 equivalent of should deliver the salt of the phosphonic acid without all the excess inorganics hindering the isolation.
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@discodermolide
Reading through the OP's description, I got the impression that little or no precipitate actually formed, although I could be wrong. Assuming that little or no precipitate formed, what do you suppose might have gone wrong?
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that was another thing that confused me. i have to add 10-11 eqs of the NaI to the starting material (it says to add 0.55 mol of NaI). This worked out to be like 67 grams, which was not fully solubilized in the amount of acetone instructed to use. So the solution was already saturated with NaI. Should I have added enough acetone to dissolve all of the NaI, and then look for the precipitated MAP solid after the overnight reaction? I figured with such an excess of salt, the starting material would have no problem forming the product. However, when I went to to triturate the solution, everything dissolved.
I'm fairly certain the starting material is ok. The MS and NMR look good. So somehow I'm screwing up this simple reaction. Should I just try one eq of the NaI as someone mentioned?
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I would use the second method, using LiBr. It uses only equivalent amounts of inorganic materials making the isolation easier!
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how so? it still uses 0.05 mol of starting material and 0.55 mol of LiBr (11 eqs). That's 47 grams of LiBr in 80 total mls of ACN... that too will likely be a super-saturated solution without all of the salt being solubilized. I'd face the same dilemma. Now if I added 200 ml of ACN, then maybe both starting materials would be dissolved, and it would be very easy to discern precipitated product after 24 hours....
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Sorry I mis-read it.
In that case stick to the NaI method. Perhaps you need to heat the reaction mixture.
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From whose laboratory does this work originate? You may have to monitor the reaction yourself, as I indicated before. Fortunately, compounds with phosphorus allow one to use P-31 NMR as one method.
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No one's lab in particular. I got the protocol from these papers:
1) 1-DEOXY-D-XYLULOSE 5-PHOSPHATE SYNTHASE: A NOVEL RANDOM SEQUENTIAL
MECHANISM IN THIAMINE DIPHOSPHATE-DEPENDENT ENZYMOLOGY
Leighanne A. Brammer,§a Jessica M. Smith,§a Herschel Wade,b and Caren Freel Meyersa
2) ACYLPHOSPHONIC ACIDS AND METHYL HYDROGEN ACYLPHOSPHONATES: PHYSICAL AND CHEMICAL PROPERTIES AND THEORETICAL CALCULATIONS
Karaman
I don't think I'll be able to use NMR to monitor the reaction unfortunately. I still feel like I'm just over-looking something simple here. I have no experience synthesizing salts however, so I'm not sure what I'm looking for.
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Rafik Karaman, Amiram Goldblum, Eli Breuer and Haim Leader
J. Chem. Soc., Perkin Trans. 1, 1989, 765-774
DOI: 10.1039/P19890000765
Until he retired, Dr. Breuer was an important figure in acylphosphonate chemistry. I might try contacting him for advice. I think it will be fairly easy to follow this reaction by TLC. The product will hardly move under conditions where the reactant will have significant mobility.
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I'm a little confused by this. Is this the exact procedure that gave a 93% yield or were the concentrations changed to adjust to how you thought the procedure should be run?
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This is the verbatim procedure:
General Procedure fbr the Synthesis of Methyl Acylphosphoncrte Monosalts (2).
Dimethyl acylphosphonate (0.05 mol) was dissolved in dry acetonitrile (or dry acetone) (50 ml), and the solution was added to a solution of lithium bromide (0.55 mol) in dry acetonitrile (30 ml) (or sodium iodide in acetone). The reaction mixture was stirred for 12 h at room temperature after which the precipitate was filtered off, washed with dry acetone, and dried. Methyl soiliutn ac.et?ilpliosphonate (2a). Yield 93%, m.p. 1 9 1 - 192 "C; v,,,(KBr) 1670s, 1210s, 1090s, and 1025s cm-'; 6,(D20) 2.50 (3 H, d, J4.8 Hz) and 3.69 (3 H, d, J 10.63 Hz); 6, -0.83 p.p.m. (q) (Found: C, 22.5; H, 4.15. Calc. for C,H,Na0,*0.5H20: C. 21.30; H, 4.14%).
I did not recover 50 mmol of Dimethyl acylphosphonate... only 39.9 mmol. Thus I adjusted the volume of acetone to 39.9 mL. Also I do not have LiBr, so I used NaI instead. The amount I needed was 39.9 mmol x 11 eqs = 438.9 mmol NaI, which works out to 65.7 g, which was "dissolved" in 42 mL of acetone... I used the quotes because that isn't nearly enough solvent to completely dissolve that large mass of NaI. Therefore, the entire time there is a huge excess of salt in the reaction. This is one of the things I'm confused about. Should I add a minimum amount of acetone until all of the NaI is solubilized, then add the starting material, then run overnight and look for a precipitate? The way the protocol is stated, there is no way all that LiBr (or NaI in my case) would be disssolved. Half a mole is a lot of salt!
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Given that the formation of precipitate is one way to follow the reaction, as well as to purify the produce, I am tempted to say that one has to start with everything in solution. The way the procedure is worded, I think that the salt was dissolved in a separate portion of solvent, not the same portion as the starting material. The 30 mL value for acetonitrile sounds low.* BTW did you make or buy the starting material?
*ETA or as Dan pointed out, the amount of salt was too high due to a typographical error. I have to admit upon reflection, that the large excess of salt should have suggested to me that something was not right.
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I think I may have found the problem. I think there is a typo in the paper:
The reference you provided [J. Chem Soc., Perkin Trans. 1 1989, 765] suggests the use of 11 equiv NaI in a quite clearly unrealistic solution volume of acetone.
Now, if you read the text, the authors say that this is a literature method, citing (in endnote 3c) Biochim. Phys. Acta 1980, 10 (Link (http://www.sciencedirect.com/science/article/pii/0005274480901862)) in which 1.1 equiv. NaI is used for the same reaction:
Preparation of sodium methyl acetylphosphonate. 120.3 g (0.79 mol)
dimethyl acetylphosphonate [13], were dissolved in 1.5 1 acetone (dried over
MgSO4) with 135g (0.90mol) NaI, and the solution stirred at ambient
temperature for 3 h. The resulting precipitate was filtered and dried in vacuo
yielding 122 g (0.76 mol, 96%). The precipitate was recrystallized from
ethanol/water (15 : 1) in 73% yield.
So I think they meant 0.055 mol (not 0.55 mol) in the perkin trans paper...
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oh wow. that is a huge difference. i didn't have access to the full paper. thank you so much! i'll try this next week!!