EmmisonJ
Hazard to Self
Posts: 89
Registered: 5-1-2009
Member Is Offline
Mood: No Mood
|
|
general alcohol solubility question
so alcohol is miscible in polar and non-polar but is classified as polar, though less polar than say water. if you have a solution which of 50ml
alcohol (EtOH or MeOH) and 50ml non-polar compound. let's say you want the non-polar compound, what i've read on other posts is to wash out the
alcohol by overwhelming it with water.
i understand detergents as having a polar head and non-polar tail, so they can't be washed out in the sense that alcohol can be, but that's how i sort
of pictured alcohols as being and i know this is incorrect. i was wondering if someone could detail what happens when you add say 500ml of water to
the 50ml alcohol / 50ml non-polar solution? is it because the alcohol is more polar than non-polar that when overwhelmed it has to side with polar
(water)? i guess what i'm really asking is how much water would it take to overwhelm the alcohol and make it insoluble in the non-polar. i know the
more water you'd add the more separation you'd get, but what exactly is happening here with the alcohol being "overwhelmed" and how much water would
it take to make anhydrous EtOH or MeOH separate from the non-polar entirely?
one physical example we can use where this happens is the washing of MeOH out of OTC DCM sources. i would imagine you're still washing out at least
SOME of the DCM, but how much would it take to lose NO DCM in the wash? perfect or near perfect separation pushing the alcohol out of the non-polar
|
|
|
UnintentionalChaos
International Hazard
Posts: 1454
Registered: 9-12-2006
Location: Mars
Member Is Offline
Mood: Nucleophilic
|
|
Dichloromethane is actually somewhat soluble (13g/l) in water, so you'll never lose none of it.
How soluble in water is this nonpolar compound? If it's barely soluble, I'd use several volumes of water.
[Edited on 6-7-09 by UnintentionalChaos]
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
|
|
|
merrlin
Hazard to Others
Posts: 110
Registered: 3-4-2009
Member Is Offline
Mood: No Mood
|
|
I would recommend using a volume of water that is no greater than the volume of the solution from which you are extracting the alcohol. Agitate to
reduce the diffusion distance and increase the interfacial area between the immiscible phases. The alcohol will partition between the nonpolar solvent
and water. If you repeat this a number of times will be able to pull most of the alcohol out of the non-polar solvent. For a given amount of water,
you will be able to remove more alcohol by repetitive washing using a fraction of your available water for each washing.
|
|
|
EmmisonJ
Hazard to Self
Posts: 89
Registered: 5-1-2009
Member Is Offline
Mood: No Mood
|
|
UnintentionalChaos, good point. my mistake, let me re-word this.
merrlin thanks, so multiple washes will pull more and more alcohol from the non-polar. but i'll have to ask one more question before i get the
picture of what's going on. i'll go ahead and spill my beans so that you guys can see the gap in my understanding:
this is all in theory just to help my understanding of how exactly alcohol becomes the "binder" between polar and non-polar, so let's say we're
working with an oil that is completely insoluble with water (so let's drop the DCM idea, my mistake), however it is all one solution due to an alcohol
being the "binder" between the water and oil. i know how detergents do it, a polar head and non-polar tail, so a water molecule on one end and an oil
molecule on the other end (or multiple molecules, depending how long the head/tail of the detergent molecule is). i read that alcohols similarly have
a polar and non-polar end. so if this polar end is attracted to water and the non-polar end is attracted to the water-insoluble oil. then when you
wash the solution with water, does the alcohol lose the attraction to the water-insoluble oil? certainly it still has a non-polar end which would
still be attracted to the oil, so it wouldn't really separate completely even though this theoretical oil is water-insoluble, there could be some
present in the aqueous layer right? because the alcohol molecules would have dragged some over with them due to attraction?
[Edited on 7-6-2009 by EmmisonJ]
|
|
|
merrlin
Hazard to Others
Posts: 110
Registered: 3-4-2009
Member Is Offline
Mood: No Mood
|
|
Think of a neutral organic molecule as a spatial arrangement of positive nuclei surrounded by a cloud of electrons that has a varying spatial density.
The spatial density of the electron cloud is a function of the distribution and charge of the positive nuclei, and various quantum mechanical
constraints. It also oscillates over time. Look up "Van der Waals force" and "London dispersion force" on Wikipedia. For polar substances, the
electron distribution produces a net dipole. Nonpolar molecules have no net dipole, but can develop polarization under the influence of nearby
dipoles.
Ultimately, solubility is dependent upon a solvent accomodating a solute molecule by achieving a lower energy configuration for the system as whole.
The electron clouds surrounding a solute molecule and those of adjacent solvent molecules adapt to each other. In terms of free energy, the addition
of a single impurity molecule to a pure solvent provides a very large change in entropy. This is why trying to get very pure anything is difficult.
Some finite quantity of a nonpolar solvent will mix with water. Remember benzene in Perrier?
When you add water to the alcohol/nonpolar solvent mixture, there is initially a very large driving force for alcohol molecules to shift to the water.
The overall system energy is reduced with each molecule that moves to the water. At some point a dynamic equilibrium will be reached, and the net
change in energy is zero for an alcohol molecule moving to the water from the nonpolar solvent (or vice versa). If the replacement of water is
repeated enough times, the last few alcohol molecules will eventually be extracted, since pure water will win out over pure oil as a home for the last
few alcohol molecules.
|
|
|
UnintentionalChaos
International Hazard
Posts: 1454
Registered: 9-12-2006
Location: Mars
Member Is Offline
Mood: Nucleophilic
|
|
Not quite. The bonding that keeps the nonpolar tails attracted are london dispersion forces. Compared to the hydrogen bonding of the alcohol groups
and waters, these nonpolar forces are pitifully weak. This is especially true when the nonpolar tail is as short as ethanol's. The London dispersion
forces are stronger when the tails are longer, and when the tails are straight or have long, straight regions that can overlap and form lots of
individual "bonds." A good example are butanol isomers. n-butanol has limited water solubility, while tert-butanol is miscible with water.
When working with miscible mixes of solvents, you can consider the overall polarity as an intermediate. In your case, the more water you add, the more
polar the mix becomes. This is the exact principle used in any of the liquid mobile-phase chromatography types, which rely on various blends of lower
alcohols, water, acetonitrile, acetic acid, ethyl acetate, etc. Using multiple washes will actually decrease yield since the first wash or two
(containing a reasonable amount of alcohol) will probably solvate some of your target compound. One larger wash and a few smaller ones with DI water
is probably the way to go.
IIRC, surfactacants work by making micelle formation easier, not by dissolving the target compounds. Detergent is a much broader definition and I
would avoid it.
[Edited on 6-7-09 by UnintentionalChaos]
[Edited on 6-7-09 by UnintentionalChaos]
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
|
|
|
Magpie
lab constructor
Posts: 5939
Registered: 1-11-2003
Location: USA
Member Is Offline
Mood: Chemistry: the subtle science.
|
|
I think merrlin has give you a good answer and I will just supplement it a little.
If you have a 3 component system, say alcohol, water, and oil, each component is going to be soluble in the others to some degree, and that degree all
depends on what gives the lowest energy for the system as a whole.
What you should take a look at is an equilibrium diagram for such a 3 component system. This will very graphically show you how the distribution
changes with the changing relative amounts of each component. Also, there will be a different diagram for each temperature.
These diagrams are used to design liquid-liquid extraction equipment on an industrial scale. For example, one of my chemical engineering text books
has such a ternary diagram for the system water-isopropyl ether-acetic acid. Another is for the system propane-oleic acid-cottonseed oil.
I'll see if I can find one of these diagrams and post it.
Edit 1: Here's a site that might be helpful:
http://www.ddpsinc.com/ProcessProfiles/pp50.html
The "envelope" is the area enclosed by the curved line. The region above that is where all 3 components are mutually soluble. Within the envelope
two phases will form, and they lie on the curved line. Their compositions are determined by using the "tie lines" shown within the envelope.
Edit 2: Here's information on the "distribution coefficient" way of looking at liquid-liquid extraction. This is probably more commonly useful in
the laboratory:
http://en.wikipedia.org/wiki/Liquid-liquid_extraction
[Edited on 7-6-2009 by Magpie]
[Edited on 7-6-2009 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
|
|
|
EmmisonJ
Hazard to Self
Posts: 89
Registered: 5-1-2009
Member Is Offline
Mood: No Mood
|
|
thank you everybody for the posts, they were very informative and helped incredibly. your replies answered my question in its entirety but you guys
also gave me some good info to research for further understanding of what's going on.
thanks
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
