Chromatography Techniques (5C)
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MCAT Chemical and Physical Foundations of Biological Systems › Chromatography Techniques (5C)
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.
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
Aniline (AN), because its amine group is strongly solvated by hexane and moves fastest
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.
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 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
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
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.
A peptide mixture is separated by size-exclusion chromatography (SEC) using an aqueous buffer (pH 7.4, 50 mM phosphate, 150 mM NaCl) on a porous gel filtration column. The mixture contains three globular proteins that do not bind the stationary phase: Protein A (200 kDa), Protein B (66 kDa), and Protein C (12 kDa). The void volume $V_0$ is 8.0 mL; the total column volume $V_t$ is 24.0 mL.
Which protein is expected to elute first (smallest elution volume)?
Protein C (12 kDa), because smaller proteins move faster through the packed bed
Protein A (200 kDa), because the largest protein is excluded from pores and spends less time in the stationary phase volume
Protein B (66 kDa), because intermediate size best matches pore size and minimizes retention
All three elute together at $V_t$ because SEC does not separate proteins without charge differences
Explanation
This question tests understanding of size-exclusion chromatography (SEC) separation principles. In SEC, separation occurs based on the differential ability of molecules to enter the pores of the stationary phase gel, with no chemical interactions involved. Large molecules that cannot enter the pores are excluded from the stationary phase volume and travel only through the mobile phase between gel particles, eluting at the void volume (V0). Smaller molecules can enter the pores and explore more of the total column volume, causing them to elute later. Protein A (200 kDa), being the largest, will be excluded from most or all pores and will elute first, near the void volume. Protein B (66 kDa) will partially enter pores and elute at an intermediate volume, while Protein C (12 kDa) will fully access the pore volume and elute last. A common misconception is that smaller proteins move faster (choice A), which incorrectly applies electrophoresis logic to SEC. To predict SEC elution order, remember: larger molecules elute first because they are excluded from the stationary phase pores, while smaller molecules elute later after exploring the full column volume.
A chemist uses ion-exchange chromatography to separate two peptides at pH 7.0. A cation-exchange resin is used (negatively charged stationary phase). The mobile phase is 20 mM phosphate buffer, pH 7.0, and a linear NaCl gradient from 0 to 500 mM is applied. Peptide P has net charge +2 at pH 7.0; peptide Q has net charge −1 at pH 7.0.
Based on the setup, which outcome is most consistent with the chromatographic separation?
Assume both peptides are soluble and stable across the salt gradient.
Both peptides elute at the same time because a salt gradient affects only size-based separations
Peptide P elutes before peptide Q because higher positive charge increases repulsion from the resin
Peptide Q elutes last because negative peptides bind most strongly to a cation exchanger
Peptide Q elutes in the flow-through (or very early), while peptide P elutes later at higher NaCl
Explanation
This question tests understanding of ion-exchange chromatography principles, specifically cation exchange. In cation-exchange chromatography, the stationary phase carries negative charges that attract and retain positively charged analytes through electrostatic interactions. At pH 7.0, peptide P has a net charge of +2 and will bind strongly to the negatively charged cation-exchange resin, while peptide Q has a net charge of -1 and will be repelled by the negative charges on the resin. Therefore, peptide Q will not bind and will elute in the flow-through or very early in the chromatogram. Peptide P will remain bound until the increasing NaCl concentration provides enough ionic strength to disrupt the electrostatic interactions between the peptide and the resin, causing it to elute at higher salt concentrations. A common misconception is that negative peptides bind to cation exchangers (choice A), which confuses cation and anion exchange principles. To predict ion-exchange behavior, remember: cation exchangers (negative stationary phase) bind positive molecules, while anion exchangers (positive stationary phase) bind negative molecules.
A student performs paper chromatography to compare two inks (Ink 1 and Ink 2) using water as the mobile phase. The paper is cellulose (polar). The solvent front travels 10.0 cm from the origin. Ink 1 produces one dye spot at 6.0 cm. Ink 2 produces one dye spot at 3.0 cm.
Based on the setup, which statement is most consistent with these results?
Assume both dyes are neutral and do not chemically react with the paper.
Ink 1’s dye moved farther because the stationary phase is nonpolar and repels polar dyes
Ink 2’s dye is more soluble in water (mobile phase) than Ink 1’s dye
Ink 2’s dye moved less because it has a lower molecular weight and diffuses more slowly
Ink 1’s dye has a higher $R_f$ and is less strongly retained by the polar cellulose than Ink 2’s dye
Explanation
This question tests understanding of paper chromatography principles and Rf value interpretation. In paper chromatography with cellulose (polar stationary phase) and water (polar mobile phase), separation occurs based on differential partitioning between the stationary and mobile phases. The Rf value is calculated as the distance traveled by the compound divided by the distance traveled by the solvent front. Ink 1's dye traveled 6.0 cm with an Rf of 0.60, while Ink 2's dye traveled 3.0 cm with an Rf of 0.30. A higher Rf value indicates that the compound spends more time in the mobile phase and less time adsorbed to the stationary phase, suggesting weaker interactions with the polar cellulose. Therefore, Ink 1's dye is less strongly retained by the cellulose than Ink 2's dye, which could be due to Ink 1's dye being less polar or having weaker hydrogen bonding capacity. A common misconception is that higher solubility in the mobile phase alone determines movement (choice A), but it's the balance between mobile and stationary phase interactions that matters. To interpret chromatography results, remember: higher Rf values indicate weaker retention by the stationary phase and stronger affinity for the mobile phase.
A researcher optimizes a silica column chromatography purification of a neutral product. Two trial runs use identical columns and sample loads but different mobile phases. Fractions are monitored by TLC, and the product is collected when it elutes.
Run 1 mobile phase: 95:5 hexane:ethyl acetate (v/v); product elutes after 18 column volumes.
Run 2 mobile phase: 70:30 hexane:ethyl acetate (v/v); product elutes after 6 column volumes.
Which interpretation is most consistent with these observations?
Assume silica is polar and increasing ethyl acetate increases mobile phase polarity.
Run 2 elutes faster because a more polar mobile phase competes more effectively with silica for the product
Run 2 elutes faster because silica becomes less polar in the presence of ethyl acetate
Run 1 elutes slower because higher hexane content increases the product’s volatility and delays elution
The difference is most consistent with a higher column temperature in Run 2
Explanation
This question tests understanding of how mobile phase composition affects elution in normal-phase column chromatography. In silica gel chromatography, increasing the proportion of the polar component (ethyl acetate) in the mobile phase increases its elution strength, causing compounds to elute faster. Run 1 used 5% ethyl acetate and required 18 column volumes for product elution, while Run 2 used 30% ethyl acetate and required only 6 column volumes. The more polar mobile phase in Run 2 competes more effectively with the silica gel for interaction with the product, reducing the product's retention time. This is because polar solvents can better solvate compounds and disrupt their interactions with the polar stationary phase through competitive hydrogen bonding and dipole interactions. A common misconception is that silica becomes less polar in the presence of ethyl acetate (choice B), but the stationary phase properties remain constant. To optimize separations, remember: in normal-phase chromatography, increasing mobile phase polarity decreases retention times by competing with the stationary phase for analyte binding.
A biochemist separates a mixture of nucleotides by reverse-phase HPLC using a C18 column (nonpolar stationary phase). The mobile phase starts as 95% water/5% acetonitrile (v/v) and is ramped to 60% water/40% acetonitrile over 10 minutes. Two analytes are injected: compound M (more hydrophobic) and compound N (more hydrophilic). Detection is by UV absorbance.
Which outcome is most consistent with this method?
Assume both compounds are stable and have similar UV response factors.
Compound N elutes after compound M because hydrophilic compounds bind more strongly to C18
Elution order depends only on molecular weight, so M and N cannot be ranked by hydrophobicity
Compound M elutes after compound N, and increasing acetonitrile promotes elution of M
Both compounds elute earlier as the mobile phase becomes more aqueous
Explanation
This question tests understanding of reverse-phase HPLC separation principles. In reverse-phase chromatography with a C18 column (nonpolar stationary phase), compounds are separated based on their hydrophobicity, with more hydrophobic compounds having stronger interactions with the nonpolar stationary phase. The mobile phase starts highly aqueous (95% water) and becomes more organic (40% acetonitrile), which increases its elution strength for hydrophobic compounds. Compound N, being more hydrophilic, has weaker interactions with the C18 stationary phase and will elute first when the mobile phase is still highly aqueous. Compound M, being more hydrophobic, binds more strongly to the C18 phase and requires a higher proportion of organic solvent (acetonitrile) to disrupt these hydrophobic interactions and promote elution. A common misconception is that hydrophilic compounds bind more strongly to C18 (choice A), which reverses the fundamental principle of reverse-phase chromatography. To predict reverse-phase elution order, remember: hydrophilic compounds elute first, hydrophobic compounds elute later, and increasing organic solvent promotes elution of hydrophobic compounds.
A crude reaction mixture is purified by silica column chromatography. The product contains a carboxylic acid; the major impurity is a nonpolar alkane. The mobile phase is 90:10 hexanes:ethyl acetate. Which observation is most consistent with this separation?
Assume the acid is not ionized and can hydrogen bond strongly to silica.
The carboxylic acid elutes first because it is most attracted to the polar mobile phase
The alkane elutes in early fractions; the carboxylic acid elutes later (or may require more polar solvent)
Elution order depends primarily on melting point, so the higher-melting compound elutes last
Both elute together because silica cannot retain acids
Explanation
This question evaluates elution order in normal-phase silica chromatography for compounds of differing polarity. Silica retains polar compounds like carboxylic acids via hydrogen bonding, while nonpolar alkanes elute quickly in nonpolar solvents. With 90:10 hexanes:ethyl acetate, the alkane impurity elutes early, but the acid requires more polar conditions. This follows because the acid's strong silica interactions delay elution. Choice B is incorrect as it claims the acid elutes first due to mobile phase attraction, but polar compounds are retained more on polar stationary phases. In similar purifications, rank compounds by polarity and H-bonding. A useful check is to expect nonpolar first in normal-phase with nonpolar mobile phases.
A mixture of two nucleotides is separated by anion-exchange chromatography (positively charged stationary phase) at pH 8.0. Nucleotide U has one phosphate group; nucleotide V has three phosphate groups. Which outcome is most consistent with these conditions?
Assume both are fully deprotonated at pH 8.0 and elution is by increasing salt concentration.
U and V cannot be separated because ion-exchange works only for proteins
U elutes later because fewer phosphates increase ionic binding strength
V elutes first because higher charge increases mobility through the column
V elutes later because it carries more negative charge and binds more strongly to the positively charged resin
Explanation
This question assesses charge-based separation in anion-exchange chromatography. Anion-exchange resins (positive) bind negatively charged species, with stronger binding for higher charge; elution uses salt gradients. At pH 8, nucleotide V (three phosphates) has more negative charge than U (one), binding more strongly. Thus, V elutes later, requiring higher salt. Choice D fails because higher charge increases binding, not mobility, in ion-exchange. In similar problems, count deprotonated groups for net charge. A useful strategy is: more charge matching resin = stronger binding = later elution.
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
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
Both compounds will have $R_f \approx 1$ on C18 because nonpolar stationary phases do not retain analytes
D will always have lower $R_f$ than E on both plates because polarity always decreases $R_f$
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.