EmmisonJ - 7-8-2009 at 12:01
if someone wanted to extract a small amount of a water immiscible liquid (say like 10ml or so) from a solution which also consists of 50ml MeOH and
150ml h2o using a solvent with a low dielectric constant such as hexane, toluene, etc. would one need to be concerned with the azeotrope that hexane
or toluene form with methanol or would the presence of the excess amount of water (polarity) be stronger than the MeOH/nonpolar solvent azeotrope?
i read that heptane/meoh form an azeotrope of 51% heptane and 49% MeOH. i also read that toluene/meoh form an azeotrope of 32% toluene and 68% meoh
so the question comes down to meoh being pulled towards the polar layer or forming an azeotrope with the heptane or toluene? i'm assuming its
polarity would favor it alongside the water, but how much water would one need to make sure as little MeOH as possible forms this azeotrope for a
successful extraction? the extraction would then probably need to be washed with water a few times to, again, aid in MeOH removal if some of it does
form the azeotrope?
[Edited on 7-8-2009 by EmmisonJ]
JohnWW - 7-8-2009 at 14:52
The most common industrial method for preparing reagent-grade absolute ETHANOL at least 99% pure, from grain alcohol (which is about 10.57% water due
to the constant-minimum boiling-point azeotrope (78.15ºC), after fractional distillation from fermented grain or other carbohydrate feedstock) is by
azeotropic redistillation mixed with benzene; see http://en.wikipedia.org/wiki/Ethanol and http://en.wikipedia.org/wiki/Azeotropic_distillation and http://en.wikipedia.org/wiki/Extractive_distillation . This depends on the ethanol favoring mixing with benzene rather than with water, which is
thereby depleted in ethanol; there are ternary water-ethanol-benzene phase diagrams somewhere. However, the product from redistillation with benzene
has traces of benzene in it, being toxic if consumed. Other common methods involve desiccation using adsorbents such as starch, corn grits, or
zeolites, or glycerol, which adsorb water preferentially.
HOWEVER, the same processes do NOT apply to purification of METHANOL. which forms a constant-minimum-boiling-point azeotrope with benzene containing
61.4% methanol (and also similar ones with hexane, cyclohexane, toluene, and heptane), and does NOT form an azeotrope with water. This would be
because methanol would have a greater preference for the aqueous layer than ethanol. I think you may have confused the two alcohols.
EmmisonJ - 7-8-2009 at 15:49
i see EtOH shares an azeotrope with the solvents in question (benzene/heptane/toluene) as well which are fairly close % by weight as they do with
MeOH.
so:
EtOH would actually favor forming its azeotrope instead of favoring polarity and going into the aqueous layer. so EtOH would primarily move away from
the aqueous layer to satisfy its azeotrope. if more EtOH is used than say benzene or heptane, then it would satisfy its azeotrope and the remaining
non-azeotropic EtOH would go into the aqueous layer?
and MeOH is the opposite, it would primarily favor polarity by migrating to the aqueous layer (it has no azeotrope with water) instead of forming its
azeotrope with benzene or heptane or whichever?
so MeOH favors the aqueous layer more in this situation than EtOH. but if using a solution of MeOH/h2o/heptane would you still end up with
small/minor amounts of MeOH that do form an azeotrope with heptane while the majority is in the aqueous layer or would it be attracted so strongly to
the polar/aqueous layer that no MeOH would be found in the heptane?
thanks so much for the reply!
[Edited on 7-8-2009 by EmmisonJ]