Exam 1 C/P Solutions: Passage 1

1) Despite being a passage-based question, we should be able to answer this question using external knowledge. To answer this question, we can do a quick overview of chromatography, then we’ll identify the principal factor determining the migration of individual components in the sample. Chromatography is a separation technique that takes advantage of the different products’ solubilities and relative affinities for the stationary phase used. All chromatography works by having a mobile phase (the part that moves), and a stationary phase (the part that stays still) which allow for separation of the different fractions of the original mixture. The higher the adsorption to the stationary phase, the slower the molecule will move through the column. We’re told in our question stem, water absorbed on cellulose (a very polar compound) functions as the stationary phase. Nonpolar molecules will not be attracted to the water molecules attached to the cellulose and will travel further along the paper, while polar molecules will have a high attraction for the molecules and not travel as far.

1. Hydrogen bonding. Hydrogen bonding to the stationary phase will determine the migration of individual components in the sample. Nonpolar molecules will not be attracted to the water molecules attached to the cellulose and will travel further along the paper, while polar molecules will have a high attraction for the molecules and not travel as far
2. Solute concentration. Solute concentration does not change the migration of individual components in the sample. The migration is dependent on affinities and hydrogen bonding. The most polar nonpolar molecules will travel furthest. 
3. Stationary phase concentration. This is going to be similar to answer choice B. Stationary phase concentration is not going to affect the relative migration of individual components in the sample. That migration still depends on affinities. The most polar nonpolar molecules will still travel furthest. 
4. Thickness of paper. This answer choice is similar to answer choice C. We expect thickness of paper is not going to affect the relative migration of individual components in the sample. The most polar nonpolar molecules will still travel furthest. We can stick with answer choice A as our best answer. 

2) The author goes through Scheme I in the passage to explain the formation of malic acid and subsequently fumaric/maleic acid.

The author then tells us ammonia reacted with fumaric, maleic, and malic acids to afford a prebiological synthesis of aspartic acid. Given this information, we have to decide what assumptions are being made if scientists conclude that aspartic acid was formed by the prebiological synthesis in the passage. 

1. Aspartic acid is unstable at temperatures below 150°C. This is not an assumption that has to be made for aspartic acid to be formed by the prebiological synthesis in the passage. An increased temperature helps the reaction move forward and that temperature is not necessarily the temperature at which aspartic acid is stable.
2. All of the malic acid underwent the dehydration reaction to form fumaric/maleic acid. We are told dehydration of malic acid results in the formation of fumaric acid and its cis isomer, maleic acid. This does not necessarily mean the entirety of the malic acid underwent the dehydration reaction to form fumaric/maleic acid. We don’t know for sure whether a majority or all of the malic aicd underwent this reaction, but we do know fumaric acid and maleic acid are formed. 
3. Compound A and cyanide were available on primitive Earth. This is an assumption that has to be made. Malic acid is formed from Compound A and cyanide, so it stands to reason both were available. If the two were not available, malic acid would not have been formed, and ultimately aspartic acid wouldn’t form. This is going to be the best answer choice so far.
4. The reaction between ammonia and fumaric acid was catalyzed by the presence of water. There is no evidence in the passage that this reaction is catalyzed by the presence of water. Furthermore, we can see where the ammonia group is found in aspartic acid, and there is no hydroxyl group present in that spot in malic acid. We would expect that to be the case in a hydrolysis reaction (catalyzed by the presence of water). We can stick with answer choice C as the correct answer.

3) Although this question is part of the passage-related question set, we can answer it using only our external knowledge.  The retention factor (Rf) of a component is the distance travelled by the component over the distance traveled by the solvent. In this case, we can see the aspartic acid traveled a distance of two units, while the solvent front traveled 10 units. This is going to be a simple division problem:

Rf= 2 units / 10 units = 0.2

The approximate Rf value of aspartic acid is 0.2, or answer choice A.

4) To answer this question, we can go back to the middle of the passage and Scheme 1.

We’re told, “malic acid is thought to have formed via hydrolysis of a cyanohydrin intermediate… Subsequent dehydration of malic acid results in the formation of fumaric acid and its cis isomer, maleic acid.” We have to be careful with the verbiage here. We want to know which of the four statements does NOT describe the dehydration of malic acid to fumaric acid and maleic acid which we see along the bottom of Scheme 1. Three of the statements will describe the reaction, while our correct answer will not. This can be answered by finding the answer choice that does not correctly describe the dehydration taking place.

1. The reaction occurs most readily with tertiary alcohols. There is nothing in the passage that suggests this is not true. The dehydration of malic acid is an E1 reaction and we have a carbocation intermediate. Tertiary alcohols can be a great leaving group in this situation; there is nothing here that implies this is not a correct description.
2. The reaction involves the loss of a water molecule. This is a dehydration reaction where there is a loss of a water molecule. There is nothing in the statement that implies this is not a correct description.
3. The reaction has a carbocation intermediate. This is going to be consistent with what we said in our reasoning for answer choice A. Once again, there is nothing in the statement that implies this is not a correct description.
4. The reaction is stereospecific. We are told in the passage “Subsequent dehydration of malic acid results in the formation of fumaric acid and its cis isomer, maleic acid.” The formation of two products in this case tells us the reaction is not stereospecific. A stereospecific reaction won’t produce a mixture. We can stick with answer choice D as our best answer.

5) This is another question in this same problem set that is going to rely predominantly on our external knowledge. Each amino acid is attached to its neighboring amino acid by a covalent bond, known as a peptide or amide bond. During the formation of this amide bond, the carboxyl (-COOH) group of one amino acid and the amino (-NH2) group of the other amino acid combine and release a molecule of water. The functional group formed is an amide group.

The only answer choice that is consistent with our breakdown here is going to be answer choice C, an amide group. 

 

Exam 1 C/P Solutions: Passage 2

6) This question relies on us revisiting the passage (specifically Figure 3) and using external knowledge about the rate of substitution of different reactions. 

Figure 3 is the S_N2 Reaction of 1-pentanol and we have a 70%/30% product distribution as shown above. The author tells us later in the passage, “The rate of the reaction shown in Figure 3 increases as the concentration of alcohol, halide, or H+ increases.” We’re replacing 1-pentanol with 2-pentanol which is a secondary alcohol. We can visualize pentane with a hydroxyl group at position 2:

1. the C–O bond in 2-pentanol is stronger than the C–O bond in 1-pentanol. The C-O bond is not going to be stronger in the different alcohols. Therefore, the rate of substitution is not going to differ because of this reason. 
2. there is a competing elimination reaction that slows the rate of substitution. We’re told in the question stem we’re looking at the S_N2 reaction in Figure 3, only with 2-pentanol. S_N2 reactions favor primary alcohols more than they do secondary alcohols, so the rate of substitution will not slow.
3. there is more steric hindrance at the oxygen atom in 2-pentanol than in 1-pentanol, making protonation less likely. There is not more steric hindrance at the oxygen atom itself in 2-pentanol; the 2-position of 2-pentanol has more steric hindrance, but not at the oxygen atom. We expect the likelihood of protonation will not change.
4. there is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-pentanol. This is going to tie into our reasoning for answer choice C. A secondary alcohol has been substituted for a primary alcohol. There is more steric hindrance at the 2-position, which we can see comparing Figure 3 and our 2-pentanol visual. Steric hindrance ultimately decreases the rate of substitution. Answer choice D is going to be our best option. 

7) To answer this question, we can pull up Reaction 2 from the passage. 

We’re told the reaction is repeated with HCl and the compound from the question stem. We want to go through our four answer choices and find the one that is NOT a direct product (without rearrangement). That means three of the answer choices will be direct products, while our correct answer will not.

1. This answer choice is not going to be a direct product. I want you to pay attention to the verbiage in the question stem. We’re explicitly told “without arrangement.” How is answer choice A made? The carbocation shifts, or in other words, a rearrangement. This is going to be a strong answer choice.
2. The way to get answer choice B is through an elimination reaction. Our hydroxyl group is protonated and chlorine ion deprotonates an adjacent hydrogen. 
3. This is possible through an E1 and SN1 reaction. We stick with answer choice A as our best answer.
4. This is possible through an E1 and SN1 reaction. The only answer choice that is not a direct product is going to be answer choice A. 

8) To answer this question, we can do an overview of gas chromatography, then we can decide the compound responsible for the first peak observed in the gc trace.

Gas chromatography involves samples being vaporized and passed through a liquid or solid stationary phase using a gaseous mobile phase. The stationary phase lets polar molecules elute more slowly. The molecules with the lowest boiling points come out of the column first. The molecules with the higher boiling points come out of the column last. The test-maker asks, “the first peak observed in the gc trace is attributable to which compound?” In other words, which substance will migrate the fastest?

1. 2-Methyl-2-butanol. This is going to be the reactant in our reaction above. We have the ability to hydrogen bond given the hydroxyl group, and the polarity of the molecule compared to the alkene in the products is going to make this a weak answer choice. 
2. 2-Methyl-2-butene. This is going to be the only alkene in the reaction above, and it is also nonpolar; alkenes have only carbon-carbon and carbon-hydrogen bonds. I mentioned the stationary phase lets polar molecules elute more slowly, so the nonpolar alkene would elute the fastest. This is looking like a strong answer choice. 
3. 2-Chloro-2-methylbutane. Adding a chlorine to the hydrocarbon increases molecular weight and polarity as well. That rules out answer choice C.
4. 2-Bromo-2-methylbutane. Reasoning here is going to be the same as answer choice C. Adding that bromine to the hydrocarbon increases molecular weight and polarity as well. We can stick with answer choice B as our best answer. 

9) To answer this question, we will once again reference the passage, and specifically Figure 3.

First thing we can do is figure out the configuration on Compound 1 by assigning priority to the different substituents. Hydroxyl group will be first as oxygen has the highest atomic number. The carbon chain will be second as carbon has the second highest atomic number. Third is the deuterium (more neutrons than hydrogen), and lastly hydrogen is priority 4. The original compound is in (R) configuration. We have an SN2 reaction and an inversion of this configuration; our answer is going to be (S) configuration. The bromine from HBr replaces our hydroxyl group on the first carbon. We get (S)-1-bromo-1-deuteriopentane, or answer choice C.

 

To see the remaining Advanced Solutions, please download our Chrome Extension by clicking here. Once downloaded, while viewing the official AAMC products in ‘review’ mode, the Jack Westin AAMC Chrome Extension will overlay our Advanced Solutions to your screen.

This tutorial video will walk you through everything you need to know about using the Jack Westin Chrome Extension.

Please email us at support@jackwestin.com if you need assistance with using this free tool.