The value is proportional to the affinity of the solute to the solvent. The solvent acts as the mobile phase along a polar paper stationary phase. Polar compounds will interact more with the paper, travelling slowly, while nonpolar compounds will interact more with the solvent, travelling more quickly.

A large  value represents a large proclivity for the mobile solvent in the experiment. Because we are using ether, a non-polar solvent, we would expect non-polar compounds to travel the farthest on our plate. Of the answer choices, alkanes are the least polar and would thus travel farthest into the mobile phase of the four functional groups. A polar functional group, like a halide, will interact more in the stationary phase, and will thus has a significantly smaller  value. 

Generally, extraction is the best means of separating two solutes based on polarity. This technique allows separation based on solubility in two different solvents, which separate based on polarity.

Extraction, however, is not offered as an answer. The next best option would be thin layer chromatography. In this process, a polar stationary phase is introduced to a nonpolar solvent. Solutes are placed on the stationary phase. The nonpolar solvent acts as the mobile phase. Nonpolar solvents interact more with the mobile solvent, travelling quickly along the polar stationary phase, while polar solutes are attracted to the stationary phase and travel more slowly. This property allows for separation based on polarity.

Ion exchange chromatography is used to separate compounds with different charges, not necessarily differing polarities. Mass spectroscopy will identify compounds based on mass, and distillation will allow for separation based on differences in boiling point and vapor pressure.

The stationary phase in chromatography is typically attracted to the more polar compounds in a solution, while the mobile phase carried the nonpolar compounds. As a result, more polar compounds will move a shorter distance, resulting in a lower R_f factor. Glucose is a very polar molecule, and would move a shorter distance compared to the other options.

In the normal phase chromatography system described, the most nonpolar compound would elute first and the most polar compound would elute last. The silica stationary phase will interact with more polar molecules, while the hexane mobile phase will carry nonpolar molecules. This would slow the progress of polar molecules as they bond to the silica, and enhance the progress of nonpolar molecules as they interact with the mobile phase.

Compound C is the most nonpolar compound because it contains only hydrogen and carbon. Compounds A and B are more polar because of the presence of oxygen, and hence the presence of polarized carbon-oxygen bonds. The alcohol group of compound B makes this compound the most polar of the three molecules by virtue of hydrogen bonding capabilities as well as the carbon-oxygen dipole. Compound B would thus elute last.

DNA is negatively charged, so it will migrate toward the positive electrode during electrophoresis. As a result, it will migrate from the center line into the green region of the gel.

To isolate the amine, it is first necessary to treat it with acid to form an ammonium salt. This would go into the aqueous phase along with the HCl, while the dialkyl ether byproducts would remain behind in the organic phase of the crude mixture.

After discarding the organic phase, treating the aqueous phase with 1N NaOH would convert the ammonium salt back to the free amine, which could then be extracted with ethyl acetate.

As for the other answers, extracting the organic phase with aqueous NaOH would leave the amine behind in the organic phase, as would washing with water. Also, the protonated amine would have to be neutralized in the aqueous phase with NaOH in order for it to be extracted back into ethyl acetate.

During the reaction, the ester is converted to a carboxylic acid, however, because of the basic conditions the acid would be deprotonated to form a carboxylate salt. This salt would be soluble in the aqueous phase, and would have to be protonated by adding dilute acid to convert it into a neutral molecule. We want to isolate the carboxylic acid product via extraction, forcing it from the aqueous phase to the organic phase. Only when the product is neutral can it be extracted by organic ethyl acetate.

Simply concentrating the reaction mixture, either by itself or after neutralizing any base with acid, would be insufficient because residual inorganic salts would remain with the desired carboxylic acid. Likewise, extracting the reaction with ethyl acetate without adding any acid would not extract the product, as it would remain in the aqueous phase as a carboxylate salt.

Extraction is a useful separation technique when there is a mixture of compounds in a solution that have similar polarities, but different solubilities. The three-step process of extraction can take advantage of different solubilities by introducing the mixture to different acidic and basic conditions.

The correct answer is arrow C. Separation into the aqueous layer during liquid-liquid extraction is most easily accomplished by creating a salt somewhere in the molecule. There are two common targets for creating salts in organic molecules. The first target is to create a carboxylate salt, by the addition of an inorganic base (such as NaOH) to a carboxylic acid or an existing carboxylate group, as shown in the diagram at arrow C.

Another common target for forming a salt would be to protonate the nitrogens of the molecule. Addition of an inorganic acid (such as HCl) would be a good way to protonate the nitrogens and also provide a negatively charged ion that would form the salt. In this case, however, the nitrogen at arrow B is involved in an amide functional group, and cannot be protonated. For the nitrogen to be useful in extraction, it must be part of an amine group.

The main point to take away from here is that during liquid-liquid extraction, the goal is to separate compounds into aqueous and organic layers and isolate the desired product alone in one of these layers. Ions and salts will generally be extracted into the aqueous layer, where nonpolar large structures will generally be extracted into the organic layer. Look for areas on the molecule that are susceptible to acid/base reactions, and you will find a good target for creating an ion for liquid-liquid extraction.