(NOTE: this is a somewhat long read, but I promise, I made it an easy and perhaps enjoyable read, with an intreaguing technical question at the end)
Ok, this is along similar lines to my last thread (the sepiation of ammonium nitrate and urea thread).
Suppose I have two esters: estradiol benzoate ("E") and trenbolone acetate ("T") (both esters, as you can see), and I desire to separate the two. The
positive to this is in boiling points: E melts at about 180*C, while T melts at a comparatively cool 98*C. Before considering a solution to this
chemical puzzle, here are the facts:
1. Both esters are inert. They couldn't chemically react with Jack. It's as if they're noble compounds in that regard.
2. These esters have identical solubility profiles. You can consider any solvent you wish; what dissolves one will dissolve the other. Conversely, if
one is insoluble in a solvent, there is a 99.99% chance the other is likewise insoluble (in the course of my research, I found a rather obscure
solvent called Digitonin may dissolve estradiol, but remain insoluble to all others. The reliability of this info isn't 100%, however).
3. Oily solvents can't be used, because of the improbably of separating the oily solvent from the steroid ester after separation is effected.
You aren't permitted to de-esterify by means of the addition of NaOh, as NaOh destroys the parent along with the ester.
Now to finally get to my question: would it be possible to effect a separation of E from T by putting them in the Buchner with filter paper, then
introducing it to the oven? Turn the oven to a carefully controlled 150*C or so (give or take like 30* due to the widely differing melting points),
allow sufficient time for the two crystalline powders to form a slurry, then crank on the dual-vane to begin the vacuum filtration of the two?
Will this separate the two?
One of the following two conditions will result from ovenizing (it's a word) the powders. Which of the two will it be?:
1. The now molten T (T melts at 98*C, remember) will act as a sort of liquid solvent to the E (E melts at 180*C, remember), causing the E to dissolve
into the T, making separation by filtration impossible. Or...
2. Or will the now molten T act as a 'super-saturated' liquid to E, preventing any dissolution of powdered E into the (now) liquid T (in much the same
way that a 'super-saturated' salt solution prevents any further dissolution of salt in water), meaning we can now proceed with the vacuum filtration,
and a successful separation of the two?
Discuss!Oxy - 23-7-2021 at 18:37
Personally I would try to separate them chromatographically.
Find a system by means of TLC which gives a good separation and run the column. That will be the easiest I suppose and will give cleanest product.
Trying to separate by melting point may be not so easy as the melting point of mixture of components is highly affected by it's purity which is really
visible when m.p is taken during purification phase of synthesis. That means you may get a mixture with single melting point or the difference might
be much less than for pure compounds. A compound having higher melting point might be soluble in liquified compound which melts in lower.
AvBaeyer - 23-7-2021 at 19:13
Your Buchner funnel proposal is pure folly. If you want to do the separation thermally like you desire you need to look into the concept of zone
refining. Complicated and tedious but it does work if you can find the equipment.
AvBzed - 24-7-2021 at 21:48
Perhaps the Ketone will form an insoluble oxime. Easy-Peasey.
[Edited on 26-7-21 by unionised]walruslover69 - 26-7-2021 at 08:48
Could you provide some of the solubility data regarding the two steroids? Looking at the structure recrystallization might be a potential option. Even
though you said the compounds were soluble in most organic compounds, I imagine there could be a significant difference in their relative solubility
for some solvents. For example since estradiol benzoate ("E") has an OH group while trenbolone acetate ("T") has only the ester and ketone groups. (E)
is a hydrogen bond donor and acceptor while T is only an acceptor as it is lacking any liable protons.
Polar Aprotic solvents like Acetone or Acetonitrile would only be able to hydrogen bond with (E) and not (T), so while both compounds might be
soluble, the solubility of (E) is probably higher in one of these solvents than (T). It also might be the case that their solubility's are similar at
room temp, but the hydrogen bonding (E) would be much more soluble at higher temps due to the favorable entropic interactions of the hydrogen bonds.
This is just a guess though.
Recrystalizations work great, but finding the solvents and solvent mixtures can be tedious.
