Test: Separations and Purifications - MCAT MCQ

Cresol is like phenol, but with an extra methyl group at the para position. Due to its alcohol functional group, it is considered a weak acid and would only be extracted with the stronger base. With base extraction, consider the strength of the acid before picking the base.

Methoxyethane is an ether. While ethers can get protonated, the carbocation is only an intermediate and is of transient existence. It will remain in the organic layer once all the other 3 compounds are extracted.

N-methylethanamine is a base due to its amine group, and amines are best extracted by a strong acid.

Since we are retaining the top layer of each extraction, after the first wash, we have benzoic acid in the aqueous layer, which is the top layer. If we wash this layer with NaOH, we would have benzoate ion in the aqueous layer. No further separation is possible.

Benzoic acid can be extracted and isolated with either a weak or strong base, but if in a mixture with an alcohol, it should be extracted first with a weak base like LiHCO₃ and the alcohol extracted a strong base like Na₂s

Column chromatography is based around the choice of the adsorbent, whether it be polar or non-polar. Ion-exchange, affinity, and size-exclusion are some of the additional uses of column chromatography that takes advantage of a solute’s charge, ability to bind antibodies and to navigate through small porous beads.

Gas chromatography can be used to separate a mixture by boiling points.

Size-exclusion chromatography can be used to separate a mixture by molecular weight, but size, of course, foremost.

Column chromatography separates mostly by differential adsorption or how well a solute adheres to the adsorbent in the column.

Azeotropes are mixtures that cannot be separated by simple distillation techniques. Generally additional separation techniques must be applied.

Adding another agent to create a ternary azeotrope coupled with further distillation would result in greater purification of ethanol.

Adding calcium oxide in water would create insoluble calcium hydroxide. Calcium oxide is essentially a dessicant, and what is left is pure ethanol.

Dissolving a salt will increase the boiling point or decreases the volatility of the solvent, namely water. Further purification of ethanol can be achieved.

Nitrogen is extremely inert, and distillation conducted in an all-nitrogen gas environment would not allow for greater purification of the mixture.

Like dissolves like. Sugars are polyhydroxy aldehydes or ketones, and salt is an ionic compound.

Water can dissolve both sugar through hydrogen bonding and salt through solvation. Water is not effective in recrystallization for either.

Benzene and petroleum ether are both very non-polar. It would not dissolve salt to any appreciable degree.

Petroleum ether is a mixture of alkanes (pentane, hexane, heptane, etc.) and should not be confused with the ether functional group.

Ethanol would dissolve sugar and allow for its separation from table salt through recrystallization.

In thin-layer chromatography, there is a mobile phase or solvent which can vary in its eluting power. The more polar a solvent, the greater the eluting power. This eluent usually moves by capillary action up the plate, not by gravity.

The stationary phase consists of a solid polar, not non-polar, adsorbent, usually silica or alumina gel.

R_f  values are only constant from one experiment to another if all the following conditions are also constant: solvent system, adsorbent, thickness of the adsorbent, amount of material spotted, and temperature

The correct answer is that a solvent with a greater eluting power than petroleum ether is acetic acid. Here is the order of eluting power from least to greatest: petroleum ether (hexane; pentane), cyclohexane, carbon tetrachloride, benzene, dichloromethane, chloroform, ether (anhydrous), ethyl acetate (anhydrous), acetone (anhydrous), ethanol, methanol, water, pyridine, and finally organic acids.

Here we have a liquid-vapor phase diagram for an azeotrope. More specifically, this is a positive azeotrope, which is pressure-maximum and temperature-minimum, and will boil at a lower temperature than any other ratio of its constituents.

This is the distillation procedure. Heating a mixture with a composition B will boil at a temperature Bb’, and the vapor will have a composition b’ and will condense with a liquid composition C. By heating this condensed liquid it will boil at temperature Cc’.

The procedure progresses from point B to point d’ until reaching the azeotropic point. When starting right of the azeotrope point, the experiment progresses from points g’ to e’ closing in on the azeotrope point from the other direction. Remember that at the azeotrope point, the two compounds cannot be separated by simple distillation methods.

At a composition D, the distillate will boil at a lower temperature because it has a lower percentage of Y than the original mixture.

In terms of solvents, cyclohexane, benzene, petroleum ether, and carbon tetrachloride have the least eluting power, while acetone, methanol, and ethanol have greater eluting power.

If plate A has alumina gel as adsorbent and ethanol as eluent, plate B could not have a greater R_f with the same adsorbent and petroleum ether as eluent. The R_f's would be switched.

Since cellulose is the least adsorptive, when plate A has cellulose as adsorbent and cyclohexane as eluent, the R_f values should be greater.

Two answer choices have that plate A using silica gel as the adsorbent and solvent with a low eluting power, i.e. cyclohexane and benzene. Let’s evaluate plate B. In order for the R_f to increase, the adsorbent must have less adsorptive power or the eluent must have greater eluting power.

Only when plate B has silica gel as adsorbent and acetone as eluent does the chromatograph match. Here, the eluent (acetone) has a greater eluting power.

As final confirmation, look at plate C. It has an adsorbent of alumina gel, which has a greater adsorptive power, so R_f values would decrease. The answer is that Plate A: silica gel, cyclohexane, Plate B: silica gel, acetone, and Plate C: alumina gel, petroleum ether.

The four products are separated by boiling point, and product D elutes out of the column first.

Products A and B form one peak together because of their nearly identical boiling points.

The ratio of the areas under the peaks is 22:33:45 or approximately 2:3:4. Here, it is ambiguous whether the lowest yield product is D since product A and B falls under the highest peak. In fact, product A is the lowest yield at 15%, and product B comes in at 30%.

The smallest peak or peak D does correspond to 2-chloro-2-methylbutane, meaning it has the lowest boiling point. It is the most compact and branched of all the chlorinated methylbutanes, so it has the least van der Waals force, hence lowest boiling point.

To address the last point, it elutes the quickest because it has the lowest boiling point. Since it is further to the right, it elutes first. It was also mentioned earlier that the products are separated by boiling point.

The tetrapeptide has two basic amino acids: lysine and arginine. The other two, proline and threonine, do not have R groups to consider.

Lysine and arginine have two amine groups with pKa’s around 10 and above. Arginine has an exceptionally high pK_α of 12.48 due to the resonance of the guadinium.

Threonine starts the sequence, so there is a terminal amino group to consider. Generally, amino groups have pK_a's from about below 9 to above 10, so basically 9-10.

Arginine ends the sequence, so we have to consider its terminal carboxy group. Generally, carboxy groups have pKa’s of under 2 to 3, so 2-3.

At pH = 6, only the carboxy group has relinquished its proton to form the carboxylate ion and will bear a negative charge of -1. The 3 amino groups will all still be protonated and bear a positive charge of +1. The net total charge is +3 plus -1 for +2.

Now to consider how it travels on the gel electrophoresis. It is clearly positively charged, since pH < pI definitely, the tetrapeptide travels to the cathode.

Here we have a mechanism for chiral resolution for the drug Duloxetine. Generally, a racemic mixture is resolved into its enantiomer through a chiral resolving agent.

The chiral resolving agent here is mandelic acid without having assigned stereochemistry. The molecule that is shown is (S)-mandelic acid. It will protonate the amine group in the Duloxetine.

Upon protonation, the ammonium ion is formed, and mandelic acid becomes the mandalate ion. There are two diastereomeric salts formed, (R)-Duloxetine-(S)-mandalate and (S)-Duloxetine-(S)-mandalate.

Compound 5 needs to have the stereochemistry assigned, which is (S)-mandelic acid.

Compound 4 is the (S,S) diastereometric salt and can be separated by precipitation, leaving (R)-Duloxetine behind.