Hi myodyne :)
Are you talking about the freebase or a salt? The freebase seems to be a liquid as only the boiling points at various pressures are given in Beilstein. No mention is made of any salt derivative. The boiling points are:
82oC at 20 torr
67-68oC at 7 torr
55-57oC at 5 torr
40oC at 2 torr
I assume you're following one of the methods from 'Synthesis of diphenyl-2-pyrrolidinyl-methanol and diphenyl-2-pyrrolidinyl-methane' (https://www.thevespiary.org/rhodium/Rhodium/chemistry/pyrrolidinyl.html)
(https://www.thevespiary.org/rhodium/Rhodium/chemistry/pyrrolidinyl.html)?
For the sake of others, the bioassays are in Post 395847 (https://www.thevespiary.org/talk/index.php?topic=7712.msg39584700#msg39584700)
(Nemo_Tenetur: "Bioassay not too pleasant ...", Chemistry Discourse) and the reports by Fastandbulbous in the thread beginning with Post 483951 (missing)
(blunts: "(R)-(+)-alpha,alpha-Diphenyl-2-pyrrolidinemethanol", General Discourse). Nemo's not-too-pleasant bioassay appears to fit Fastandbulbous' experience with the S-isomer, whereas the more positive bioassay of the R-isomer indicates this gives the better experience. The S-isomer doesn't seem worth bothering with, at least on it's own. Nobody seems to have tried the racemic drug.
Unfortunately your end product will be the S-isomer if starting from L-proline. It may be wise to wait until you get hold of some D- (or racemic) proline for a better end result.
Have you considered the reduction of the compound to diphenyl-2-pyrrolidinyl-methane? The reduction of the tertiary alcohol with it's two attached phenyl groups is extremely easy. A standard HI/P reduction is over in minutes (according to the above link). Though the potency decreases the entire synthesis is so cheap and easy it would still be worthwhile if the product turned out to be interesting.
I haven't made either diphenyl-2-pyrrolidinyl-methanol or diphenyl-2-pyrrolidinyl-methane but I do plan on making both sometime in the future. If you start with the proline ester you only need to perform one step before you can bioassay your compound; another step, and you have the next compound to try.
Hello again :)
As I mentioned in my initial post, the boiling points are taken from Beilstein, and correspond to the boiling points of the proline freebase ethyl ester. I don't see why they should be considered strange; what is it about them which makes you think this?
As Nicodem said, it is impossible to accurately answer your question without more information. It seems likely that proline ethyl ester freebase is a liquid. Are your crystals the freebase or a salt?
The melting point of plain proline is suspiciously similar to that of your product. Beilstein reports melting points ranging from 203oC up to 220-222oC. Recrystallising proline from ethanol will give a product with a melting point of 208-210oC. If you are talking about the freebase having such a high melting point, then there is a possibility you have proline instead of the ester.
If you want to try a HCl catalysed esterification, you may like to take a look at the following articles which deal with this:
J. Amer. Chem. Soc., 71, 1949, 3100.
Justus Liebigs Ann. Chem., 326?, 1903, 99.
Tetrahedron, 52 (1), 1996, 225-232. [Using HCl gas to give a quantitative yield]
An alternative is to first form the acyl chloride with SOCl2 then react this with ethanol. The highest yielding and/or most accessible references are:
J. Org. Chem, 55 (7), 1990, 2254-2256. [97% yield]
Tetrahedron Asymmetry, 11 (6), 2000, 1367-1374.
Can. J. Chem. 80, 2002, 1662-1667. [99% yield]
Edit: Here is the experimental from Tetrahedron, 52 (1), 1996, 225-232. The title refers to the N-protected amino ester. I suppose you'll have to be the judge of what is an 'enough amount' of HCl gas:
Preparation of Ethyl N-(2,2-Dimethylethoxycarbonyl)-(L)-prolinate (6a):
To a solution of (L)-proline (10 g, 87 mmol) in EtOH (150 mL) was introduced enough amount of hydrogen chloride gas and the resulting solution was heated at reflux for 1 h. Evaporation of the solvent gave ethyl (L)-prolinate hydrochloride (15.6 g, 100%) as a colorless oil.