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Q: Dear Erowid, I'm a chemist and I recently calculated that the molecular weight for LSD tartrate on your LSD Chemistry page is incorrect. You list it as 398.475 but it should be 473.519. The LSD base (C20H25N3O) by itself is 323.44 and the Tartaric acid (C4H6O6) by itself is 150.087.

While it is true that tartaric acid is a diprotic, the acid dissociation constants are not identical: the first pKa is 2.98 and the second pKa is 4.34. What that means is that it is much harder to remove the second proton from the same molecule of tartaric acid than it is to remove the first proton from another, undissociated tartaric acid molecule. In a solution of a weak base with tartaric acid, *all* the tartaric acid molecules will lose one proton before any of them loses two protons; the first dissociation will go to completion before any of the second dissociation occurs.

So if one adds a one-to-one molar ratio of a weak base such as LSD to tartaric acid, each LSD molecule will associate with one molecule of tartaric acid and the LSD-tartrate will precipitate as a 1-to-1 ratio.

I know that all the synthesis protocols, even TIHKAL, use a 2-to-1 molar ratio of LSD base to tartaric acid, but I have not yet seen an analysis that demonstrates that the actual association is 2-to-1. The pKa of LSD is only 7.8, making it a very weak base and I have doubts that it is strong enough to pull off that second proton from tartaric acid.

A: Thanks for writing in to ask us to verify the molecular weight of LSD. Great question. However, we are pretty sure we have the correct value.

First, checking the existing literature, both the Merck Index and TIHKAL give a 2-to-1 ratio for LSD to tartaric acid, so you're climbing uphill to argue against the canon.

Second, let's walk through what's happening with the diprotic tartaric acid. Because each tartaric acid molecule has two protons (hydrogens) that it can share with the LSD molecules, the LSD tartrate form is essentially [LSD-tartaric acid-LSD]. The total molecular weight of that 2-to-1 complex is 796.95, but each of those composite molecules contains two fully active LSD molecules, each of which will be available when the LSD tartrate is dissolved in solution. Therefore the molecular weight of LSD tartrate is considered half that, or 398.475.

Contrary to what you argue, tartaric acid's second pKa (acid dissocation constant) of 4.34 is plenty acidic enough to protonate LSD base. There are chemistry pH / equilibrium calculators available (like CuTiPot), into which one can enter the pKa values for tartaric acid and LSD and see that the vast majority of the tartaric acid will become doubly ionized and the 2-to-1 [LSD-tartaric acid-LSD] species will be the most common.

Third, while it is true that some diprotic acids only use one of their acidic functions in the salt, forming 1-to-1 complexes, which salt forms is governed more by the solubility of the 1-to-1 versus the 2-to-1 complex than by the pKa values of the two acids. Even if you add 1-to-1 stoichiometry, if the 2-to-1 salt is much less soluble, that's what will crystallize out. This is the case with LSD tartrate.

Finally, we checked with a senior LSD chemist on this issue and he agreed with Merck, TIHKAL, and Erowid:
"The ratio of LSD to tartaric acid is 2-to-1. The only ergoline salts that are commonly formed as 1-to-1 are maleate salts. As far as the pKa argument goes, it works for maleic acid (1.91 vs 6.33) but tartaric and fumaric acid 2-to-1 salts form readily. I prefer to form 2-to-1 salts because they are more stable at room temperature. We have made 1-to-1 tryptamine salts but they lacked stability at room temperature. When we started making LSD for research, we added enough tartaric acid to form 1-to-1 complexes, but the crystals that formed were 2-to-1. You can know what the composition is of the crystalline solid either because the elemental analysis is very different, or in the NMR you only see half the tartartic acid protons that you'd see if it was a 1-to-1 complex. Further, I determine my salt forms experimentally by NMR analysis and by the careful addition of one molar equivalent of acid to two molar equivalents of base." -- S
Thanks for getting us to nail this down,


[Note: This answer was finalized after several back-and-forth emails with the individual who asked the question. In the end, s/he agreed that this answer is correct.]

Asked By : A PhD Pharmacologist and Chemist
Answered By : earth
Edited By : fire
Published Date : 2 / 3 / 2010
Last Edited Date : 2 / 3 / 2010
Question ID : 3141

Categories: [ Chemicals ] [ LSD ] [ Chemistry ]

Ask Erowid v1.7 - Jul, 2005

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