Need help preparing for the Organic Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach everything you need to know about extraction for Separations and Purifications. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!
Extractions are a laboratory technique that is used to separate compounds based on their solubility in organic or aqueous solvents. Recall that solubility generally follows the ‘like dissolves like’ principle, that is polar compounds are dissolved in polar solvents and non-polar compounds are dissolved in non-polar solvents.
Toluene is almost perfectly non-polar, and it dissolves readily in organic solvents. Glucose on the other hand has notable hydroxyl groups that make it amenable to solvation aqueous solvents. Benzoic acid is a bit of a more difficult call to make - the benzene ring is non-polar, and the carboxylic acid group clearly polar. Since the benzene ring is far more massive, it acts largely as a non-polar molecule and dissolves in the organic layer. But amusingly enough, the conjugate base of benzoic acid, benzoate, dissolves readily in the aqueous layer and very poorly in the organic layer. This is due to the negative charge on the carboxylate group. Charged molecules will almost always be found in the aqueous layer, unless they sport an impressively large non-polar group!
These stark differences in solubility are something we can take advantage of in extraction procedures. Our knowledge of acid-base chemistry allows us to introduce positive and negative charges to bases and acids, respectively.
As a quick reminder: Strong acids, and some of the stronger organic acids, can be deprotonated with both weak and strong bases. Weaker acids, such as the majority of organic acids, can only be deprotonated by strong bases. For this reason, a solution of carboxylic acid and phenol can be separated by introducing a weak base such as sodium bicarbonate first, which will turn the carboxylic acid overwhelmingly into a carboxylate and move it to the organic layer. After separating the two solutions, the phenol could then be moved into the aqueous layer for further extraction as well, by simply introducing a strong base such as NaOH.
Bases, such as amine groups, can be selectively moved to the aqueous layer as well. To do so, we would simply use an acid - thereby introducing a positive charge to the amine, and forcing it into the aqueous layer. Strong acids such as hydrochloric acid are the most common choice for this procedure.
Picture a sample of how 4 different molecules might be extracted from a single organic layer. The molecules in this case are intentionally left innominate and simply designated with numbers. All we care about are their most relevant functional groups.
Before we commence our procedure, all 4 molecules are in the organic layer. This is to be expected, as they all have large non-polar groups and only moderate dipole moments on their polar head-groups. Additionally they all sport a alkane chain as an R-group, contributing further to their favorable solubility in our organic layer.
First, we're going to add hydrochloric acid in an aqueous solution. This creates two immiscible layers, an aqueous layer and an organic layer. A little bit of perturbation of our mixture will encourage molecules to move around (rather than waiting entirely on Brownian motion), and the mixture will re-separate readily into two separate layers along with the relevant soluble molecules in each layer. We can see that molecule #2 sports an amine group, which should have been protonated by our hydrochloric acid - suddenly making it soluble in the aqueous layer. If we decant (pour off) the aqueous layer now, we’ve already extracted one of our 3 target molecules.
Next we will add sodium bicarbonate, a weak base. Neither molecule #1 and #3 have a carboxylic acid functional group, but #4 does. As a result, only #4 will be deprotonated in the presence of our weak base, and we can again expect to find it in the aqueous layer after a short while. Following another round of decanting, we can finally add a strong base to separate molecule 1 from molecule 3 - since the proton on molecule 3 is far more labile, and will actually dissociate in the presence of a strong base, while the proton in molecule 1 will not. In case it is not entirely clear why, resonance stabilization by the aromatic ring makes the phenolate anion energetically more favorable than the simple alkoxide of molecule #1.
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