3
$\begingroup$

I'm watching MIT chemistry by Donald Sadoway. In one of his lectures devoted to solutions and phase separation, he performs experiments with absinthe. First he mixes absinthe with 5 $\times$ water which turns into milky louche.

Then he adds some cognac and the mixture becomes transparent again. I don't grasp his explanation of this.

Here is a timestamped link to the video.
I'll also briefly summarize his argument below:

... and why did he add cognac? ... it's cause if you have a fat phase here and you've got an aqueous phase here and if you add alcohol, you got

$(fat)\:\textrm{CH}_3 \textrm{CH}_2 \textrm{COOH} \: \dot{}\dot{}\:\textrm{OH}_2$

... this can bond to the water by hydrogen bond and this aliphatic tail can stab the fat and bring them into solution. This is why you have these recipes...

What does "stabbing the fat" mean?

Does it mean that the ethanol somehow makes the surface between fat and water disintegrate thus turning the mixture into a single phase solution? Or maybe ethanol + water + fat somehow combine to form a single molecule so that, again, there is a single phase solution now?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
6
$\begingroup$

Absinthe is a strongly alcoholic beverage containing anethole, which is insoluble in water and very soluble in ethyl alcohol. When water is added to the absinthe, the alcohol becomes too dilute to keep the anethole dissolved, and it appears milky. When more ethyl alcohol is added in the form of cognac, the solution again becomes strong enough in alcohol to dissolve the anethole and the solution becomes clear again.

The phrase "stab the fat" is a bit odd, but I'll do my best to explain. Anethole is a fat-like or fat-loving oil, meaning it dissolves well in other fatty sorts of compounds, but not in water (which is lipohpobic, or fat-avoiding/fat-fearing). Ethyl alcohol has two parts to it. One end of it (labeled (fat) in the chemical formula in the question) is lipophilic (attracted to fat) and the other end is hydrophilic (attracted to water). If there is enough alcohol in the mixture, the lipophilic end of the alcohol molecules can surround the anethole ("stabbing the fat") which leaves the other end of the alcohol free to be dissolved in the water (and also any excess hydrophilic portions of the alcohol).

I hope the "stab the fat" explanation was clear. Don't hesitate to ask for specific clarifications in the comments below.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for the answer! I'd like to clarify a bit about dissolving. As I understand, the milky colour of the original mixture is caused by the light scattered from the fat-phase/aqueous-phase boundary. So after more cognac is introduced into the solution does this boundary disappear? and we get a single phase without separation? $\endgroup$
    – xaxa
    Commented May 5, 2017 at 8:54
  • $\begingroup$ To make a kind of illustration to my question: in the original mixture we have [... w w w |f f f| w w w w w |f f f f| ...] After the cognac is added to the solution we get [... wef wef wef wef ...] here w stands for water molecule, f stands for fats (whichever are there), e for ethanol and | represents a phase boundary. Is this correct? $\endgroup$
    – xaxa
    Commented May 5, 2017 at 8:59
  • $\begingroup$ First off, of course you are welcome ;) You're correct about the cause of the milky color, but we need to clarify what the milk is. In this case, it is a suspension of macrosopic droplets of anethole. As anethole happens to have nearly the exact density of water, it's tendency to settle out is very small. As the solution as a whole (water + alcohol) has a density that is less than anethole, I suspect that eventually there would be a layer of anethole on the bottom, with the alcohol/water layer on top. My point is, the anethole is not dissolved at all when you see the milky stuff (to be cont.). $\endgroup$
    – airhuff
    Commented May 5, 2017 at 19:18
  • $\begingroup$ So, when there is not enough alcohol to surround the anethole with it's lipophilic tails (stabbing the fat) there is truly a phase boudary. Once there is enough alcohol to surround each anethole molecule, it becomes dissolved, there is just a single phase, and it appears clear. I think the point of confusion might be one of scale. When the anethole is dissolved, we're talking about each individual molecule being surrounded by a layer of the alcohol. When it's milky (undissolved) the droplets are trillions of trillions of molecules in size, just holding onto each other to escape the solvent. $\endgroup$
    – airhuff
    Commented May 5, 2017 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.