Chromatography Techniques (5C)

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MCAT Chemical and Physical Foundations of Biological Systems › Chromatography Techniques (5C)

Questions 1 - 10
1

A researcher compares two packed columns for the same normal-phase separation (silica stationary phase; same solvent and flow rate). Column 1 is 10 cm long; Column 2 is 30 cm long, with identical inner diameter and particle size. Which statement best describes the expected separation efficiency outcome?

Both columns must give identical resolution because stationary phase chemistry is the same

The shorter column is likely to provide better resolution because analytes spend less time diffusing

The longer column is likely to provide better resolution because analytes have more opportunity to partition

Column length only affects retention time, not resolution, in chromatography

Explanation

This question tests understanding of how column length affects chromatographic resolution. Resolution depends on the number of theoretical plates, which increases with column length when all other factors are equal. The longer column (30 cm) provides more opportunities for repeated partitioning between mobile and stationary phases, leading to better separation of compounds with similar retention properties. While longer columns do increase analysis time, they improve resolution by allowing small differences in partitioning to be amplified over the greater distance. Choice B incorrectly suggests that longer residence time leads to worse resolution due to diffusion - while diffusion does occur, the increased partitioning opportunities outweigh this effect. The key principle is that resolution generally improves with column length, though practical considerations like analysis time and pressure limits must be balanced.

2

A mixture is separated by gas chromatography (GC) using an inert carrier gas and a nonpolar stationary phase. Two components have similar polarity but different boiling points: Compound S (bp 80°C) and Compound T (bp 140°C). The oven temperature is held constant at 100°C. Which component is expected to elute first, assuming both are stable and volatile enough for GC?

They elute together because polarity is matched to the stationary phase

Compound T, because higher boiling point increases vapor pressure and speeds elution

Compound S, because lower boiling point generally corresponds to higher volatility and shorter retention

Compound S elutes last because smaller molecules interact more strongly with nonpolar phases

Explanation

This question tests understanding of gas chromatography separation based on volatility and boiling points. In GC, compounds are separated based on their partitioning between the gas phase and the stationary phase, which is strongly influenced by volatility. At a column temperature of 100°C, Compound S (bp 80°C) will be completely vaporized and highly volatile, spending more time in the gas phase and less time interacting with the stationary phase. Compound T (bp 140°C) will be less volatile at 100°C, condensing more on the stationary phase and being retained longer. Choice A incorrectly relates higher boiling point to higher vapor pressure - compounds with higher boiling points have lower vapor pressures at a given temperature. The key principle is that in GC, lower boiling point compounds are more volatile and elute before higher boiling point compounds.

3

A lab uses TLC on silica to monitor an esterification. The starting alcohol is polar; the ester product is less polar. Using a solvent system of 70:30 hexane/ethyl acetate, which observation is most consistent with successful product formation?

The starting alcohol spot moves to the solvent front because it is polar

A new spot appears with higher $R_f$ than the starting alcohol spot

No spots appear because silica quenches UV absorption

A new spot appears with lower $R_f$ than the starting alcohol spot

Explanation

This question tests understanding of TLC monitoring for reaction progress based on polarity changes. In normal-phase TLC on silica, less polar compounds have higher Rf values because they interact less with the polar stationary phase. The esterification converts a polar alcohol (with -OH group) to a less polar ester (with -COOR group). Therefore, successful product formation would show a new spot with higher Rf than the starting alcohol. Choice B incorrectly predicts lower Rf for the less polar product, contradicting normal-phase principles. The key concept is that in normal-phase TLC, decreasing polarity leads to increasing Rf values, making this technique ideal for monitoring reactions that change compound polarity.

4

A purification uses ion-exchange chromatography to isolate a DNA-binding protein (Protein Z). The column is cation-exchange resin (negatively charged groups on the stationary phase) equilibrated in buffer at pH 7.4. Protein Z has an isoelectric point $pI = 9.2$. A mixture also contains Protein Y with $pI = 5.1$. Which outcome is most consistent with loading the mixture at pH 7.4 and then eluting with increasing NaCl concentration?

Protein Y binds strongly to the cation exchanger and elutes last

Both proteins fail to bind because NaCl is required for adsorption

Protein Z elutes first because higher $pI$ means more negative charge at pH 7.4

Protein Z binds to the cation exchanger and elutes at higher salt than Protein Y

Explanation

This question tests understanding of ion-exchange chromatography and the relationship between protein pI and charge. At pH 7.4, proteins with pI > 7.4 carry net positive charge, while those with pI < 7.4 carry net negative charge. Since Protein Z has pI = 9.2, it is positively charged at pH 7.4 and will bind to the negatively charged cation exchanger. Protein Y with pI = 5.1 is negatively charged at pH 7.4 and will not bind, flowing through immediately. When NaCl concentration increases, the salt ions compete with Protein Z for binding sites, eventually eluting it. Choice D incorrectly states that higher pI means more negative charge, which is the opposite of the correct relationship. The key principle is that proteins bind to ion exchangers when their net charge is opposite to the resin's charge.

5

A chemist performs normal-phase silica column chromatography on a reaction mixture containing three neutral compounds: M (ester), N (aldehyde), and O (carboxylic acid). The mobile phase starts as 100% hexane and is gradually changed to 70:30 hexane/ethyl acetate over the run. Which outcome is most consistent with this gradient strategy?

More strongly retained (more polar) compounds can be eluted later as the mobile phase becomes more polar

Less polar compounds are delayed because increasing ethyl acetate strengthens adsorption to silica

More polar compounds elute earlier because the mobile phase becomes less polar over time

All compounds elute at the same time because gradient elution eliminates retention differences

Explanation

This question tests understanding of gradient elution in normal-phase chromatography. In normal-phase silica chromatography, polar compounds are retained more strongly than nonpolar compounds. Starting with pure hexane (nonpolar) will elute the least polar compounds first, but very polar compounds may remain strongly bound. By gradually increasing the ethyl acetate content (making the mobile phase more polar), the eluting power increases, allowing more strongly retained polar compounds to be eluted. The carboxylic acid (O) is most polar and will elute last when sufficient ethyl acetate is present. Choice A incorrectly reverses the elution order, while choice B misunderstands how polarity affects silica adsorption. The key strategy in gradient elution is to start with weak eluent for less retained compounds and gradually increase eluent strength for more retained compounds.

6

A chemist uses TLC on silica to monitor a reaction converting a nonpolar starting material (SM) to a more polar product (P). The developing solvent is hexane:ethyl acetate = 9:1 (v/v). After development, the solvent front traveled 8.0 cm. The SM spot is at 6.4 cm and the P spot is at 2.4 cm.

Based on these results, which action is most consistent with improving TLC separation (greater distance between SM and P) if the spots are too close in a different trial?

Assume silica is polar and increasing ethyl acetate increases mobile phase polarity.

Switch to distillation, since boiling point differences directly increase TLC resolution

Decrease the ethyl acetate fraction to reduce overall elution strength and increase retention differences on silica

Add more sample to the baseline, since higher loading increases $R_f$ differences between SM and P

Increase the ethyl acetate fraction to move both spots upward, with a larger shift for the polar product

Explanation

This question tests understanding of how to optimize TLC separation by adjusting mobile phase polarity. The current separation shows the nonpolar starting material (SM) with Rf = 0.80 and the polar product (P) with Rf = 0.30, giving a ΔRf of 0.50. To improve separation (increase the distance between spots), we need to increase the difference in how strongly each compound interacts with the stationary phase. Decreasing the ethyl acetate fraction makes the mobile phase less polar, which reduces its ability to compete with silica for both compounds. This causes both compounds to be retained more strongly, lowering both Rf values, but the effect is more pronounced for the polar product, which already has strong interactions with silica. This increases the relative difference in retention between SM and P, improving their separation on the plate. A common misconception is that increasing mobile phase polarity improves separation (choice A), but this would cause both spots to move up the plate with potentially less differentiation. To optimize TLC separation, remember: decreasing mobile phase polarity in normal-phase TLC increases retention differences between compounds of different polarities.

7

A researcher purifies a small-molecule inhibitor from a crude reaction mixture using silica gel column chromatography (polar stationary phase). The mixture contains four neutral compounds at room temperature: hexane (H), ethyl acetate (EA), acetophenone (AP), and aniline (AN). The column is initially eluted with 100% hexane, then switched to 30% ethyl acetate in hexane (v/v) after 5 column volumes. No acid/base modifiers are used. Which component is expected to elute first under the initial 100% hexane mobile phase conditions?

Assume relative polarity increases in the order: H < EA < AP < AN, and that more polar compounds interact more strongly with silica.

Aniline (AN), because its amine group is strongly solvated by hexane and moves fastest

Ethyl acetate (EA), because it is the least polar solvent and therefore elutes first

Acetophenone (AP), because aromatic rings have low affinity for silica and elute before alkanes

Hexane (H), because it is least polar and interacts weakest with the polar stationary phase

Explanation

This question tests understanding of normal-phase column chromatography principles, specifically how compound polarity affects elution order. In normal-phase chromatography with a polar stationary phase (silica gel) and nonpolar mobile phase (100% hexane), compounds are separated based on their differential interactions with the stationary phase. Since hexane is the least polar compound in the mixture, it has the weakest interaction with the polar silica gel and will spend the least time adsorbed to the stationary phase. The more polar compounds (EA, AP, and AN) will interact more strongly with silica through hydrogen bonding and dipole-dipole interactions, causing them to be retained longer on the column. A common misconception is that the mobile phase composition determines which compound elutes first (choice B), but in reality, it's the relative affinity for the stationary phase that matters. To predict elution order in normal-phase chromatography, remember: less polar compounds elute first with nonpolar mobile phases, while more polar compounds are retained longer due to stronger interactions with the polar stationary phase.

8

A lab compares thin-layer chromatography (TLC) of a mixture of two analgesics on silica plates (polar stationary phase) using two different mobile phases. Spots are visualized under UV, and $R_f$ values are recorded.

Mobile phase 1: hexane:ethyl acetate = 8:2 (v/v)

Mobile phase 2: hexane:ethyl acetate = 2:8 (v/v)

Compound X is more polar than compound Y. Based on the setup, which outcome is most consistent with switching from mobile phase 1 to mobile phase 2?

Assume both compounds are neutral and remain chemically unchanged during development.

Both $R_f$ values remain unchanged because $R_f$ depends only on the stationary phase

Only compound Y’s $R_f$ increases because nonpolar compounds respond more to polar solvents

Both $R_f$ values increase, with a larger increase for the more polar compound X

Both $R_f$ values decrease because a more polar mobile phase increases adsorption to silica

Explanation

This question tests understanding of how mobile phase polarity affects Rf values in thin-layer chromatography. In TLC with a polar stationary phase (silica), the Rf value represents the ratio of compound migration distance to solvent front distance, which depends on the balance between compound interactions with the stationary and mobile phases. When switching from a less polar mobile phase (8:2 hexane:ethyl acetate) to a more polar one (2:8 hexane:ethyl acetate), the increased polarity of the mobile phase competes more effectively with the silica for interaction with both compounds. This increased competition causes both compounds to spend less time adsorbed to the stationary phase and more time dissolved in the mobile phase, resulting in higher Rf values for both compounds. The more polar compound X will show a larger increase because the polar mobile phase provides stronger solvation for polar compounds. A common misconception is that increasing mobile phase polarity would decrease Rf values (choice A), but this confuses the effect with increasing stationary phase interactions. To predict Rf changes, remember: increasing mobile phase polarity in normal-phase TLC increases Rf values, with larger increases for more polar compounds.

9

A scientist compares normal-phase TLC on silica versus reverse-phase TLC on C18 for the same pair of neutral compounds: D (more polar) and E (less polar). Both plates are developed with the same moderately polar solvent mixture. Which outcome is most consistent with switching from silica to C18?

Assume the solvent is compatible with both plates and development conditions are identical.

The relative migration order must remain the same because $R_f$ depends only on molecular weight

D will always have lower $R_f$ than E on both plates because polarity always decreases $R_f$

Both compounds will have $R_f \approx 1$ on C18 because nonpolar stationary phases do not retain analytes

The relative migration order is expected to invert: D tends to have lower $R_f$ than E on silica but higher $R_f$ than E on C18

Explanation

This question compares migration order in normal- vs. reverse-phase TLC. Normal-phase on silica retains polar compounds more (lower Rf for D), while reverse-phase on C18 retains nonpolar more (higher Rf for polar D). Switching to C18 in the same solvent inverts the order: D (polar) has higher Rf than E. This reflects reversed polarity affinities. Choice B is incorrect as Rf depends on polarity interactions, not just weight. For similar comparisons, predict order reversal. A key strategy is to note phase type dictates whether polar or nonpolar elutes faster.

10

A mixture is purified by silica column chromatography. The chemist increases the mobile-phase polarity (more ethyl acetate in hexanes) and observes that all components elute in fewer fractions, but two peaks become less resolved (more overlapping). Which statement best accounts for the decreased resolution?

Assume the column packing and flow rate are unchanged.

Resolution decreases because ethyl acetate increases boiling points, causing co-elution

A stronger eluent can reduce differences in retention factors between analytes, compressing elution times and lowering resolution

Increasing polarity increases stationary-phase binding, so peaks should spread out and resolve better

Resolution decreases only when the stationary phase is nonpolar, so this cannot occur on silica

Explanation

This question evaluates how mobile phase strength affects resolution in normal-phase chromatography. In silica columns, a stronger (more polar) eluent reduces retention factors and compresses elution ranges, potentially decreasing resolution by minimizing retention differences. Increasing ethyl acetate causes faster elution but overlaps peaks. This occurs as the eluent overpowers selectivity. Choice B fails because higher polarity weakens, not strengthens, binding to silica, leading to compression. In similar optimizations, balance eluent strength. A strategy is to use weaker eluents for better resolution when peaks are close.

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