MCAT Strategic Supplemental Tests

Dear Future Doctor, The following Strategic Supplemental should be used to practice and to assess your mastery of specific topical information in test format, and to learn how to attack each type of passage. This is an opportunity to practice the STOP, THINK, PREDICT methodology learned in the Kaplan classroom. There are Discrete questions and Passage-based questions that test your ability to apply your foundation knowledge to MCAT-style questions, using critical thinking. Simply completing the tests is inadequate; a solid understanding of your performance through your Score Reports and the explanations is necessary to diagnose your specific weaknesses and address them before Test Day. All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below. Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license. We offer this material for your practice in your own home as a courtesy and privilege. Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation. Sincerely,

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MCAT Organic Chemistry Strategic Supplemental One of the most difficult aspects of the MCAT is reading the passages. Test-takers tend to waste a lot of time reading passages in the same way they read a textbook, with an eye towards memorizing every fact and definition. This type of reading will not pay off on an MCAT passage. MCAT science passages are full of data and information that will always be there if you need it. When you read a passage, you should take a broad look at what is happening. Keep in mind that the MCAT isn’t asking you to cross new frontiers of science. Everything presented in a passage can be understood with the material you have acquired in the Kaplan MCAT course. Included in your arsenal of Organic Chemistry skills for Test Day should be those listed at the end of this Strategic Supplemental. Excelling on the MCAT depends in part on your mastery of this skills set. One last thing to remember about MCAT science passages in general: Don’t take too much time reading each one. Allow yourself a maximum of 60 seconds to absorb what you need from the passage. In this supplemental, you will practice reading Organic Chemistry passages quickly, going into the questions with a good sense of the passage and answering them in the most efficient and effective manner. Reading the Passage As you may have learned from the Strategic Supplementals in the other science areas, MCAT science passages fall into one of three categories:

• Information • Experiment • Persuasive argument

Information passages resemble passages you might find in a textbook. They consist of paragraphs of text that are descriptive in nature. For Organic Chemistry passages, the paragraphs are often relatively short and are often accompanied by diagrams illustrating reactions or mechanisms. If the passage describes a specific laboratory technique or set-up, it may contain lengthier text with the graphic being one of the relevant apparatuses. An information passage in Organic Chemistry may involve one or more of the following:

• presenting a new reaction: its mechanism, its stereochemistry, factors affecting its rate, etc.

• presenting a series of related reactions: e.g., different ways to synthesize a class of compounds, typical reactions that a class of compounds undergoes

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• describing the characteristics of a class of compounds • describing an experimental technique

The following is an example of an information passage. Aspirin, also known as acetylsalicylic acid, is

one of the most useful and economical drugs available. It belongs to a class of drugs known as nonsteroidal anti-inflammatory drugs, and can be used to treat pain and alleviate inflammation and fever. The mechanism of aspirin’s action is not fully understood, although recent research suggests that it functions by inhibiting cyclooxygenase 2, an enzyme that creates prostaglandin precursors. Prostaglandins contribute to the body’s perception of pain and its inflammatory response. Prostaglandins also aid in the formation of blood clots, so aspirin thins the blood and may consequently help prevent heart disease.

Aspirin can be synthesized from salicylic acid

via the reaction shown in Figure 1.

COOH

OH

CH3COCCH3

O O

+

H+, ∆

COOH

OCCH3

+ CH3COOH

O

salicylic acid

aspirin Figure 1

Both salicylic acid and aspirin are aromatic carboxylic acids. Carboxylic acids are generally weak acids, although they are among the strongest organic acids, with pKa’s usually in the range of 3 to 5, much lower than those of corresponding alcohols. The pKa’s of some compounds are given in Table 1.

Table 1 pKa’s of Compounds

Compound pKa Acetic acid 4.76 Fluoroacetic acid 2.66 Difluoroacetic acid 1.24 Trifluoroacetic acid 0.23 2,2-Dimethylpropanoic acid

5.05

Benzoic acid 4.18 p-Nitrobenzoic acid 3.43 2,4,6-Trinitrobenzoic acid

0.65

p-Methoxybenzoic acid

4.47

Phenol 9.95 Ethanol 15.9 Methanol 15.1 Water 14

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Experiment passages describe one or more experiments and usually include a presentation of the results. In Biology and Physics passages, the results are usually presented in the form of a table or a graph. In Organic Chemistry, however, the data may take on a variety of forms: the percentage yield (of a synthesis), a verbal summary of whether the reaction or isolation procedure is successful, a description of the appearance or spectroscopic properties of the product, etc. As a result, an experiment passage may not immediately be as easily recognizable from the visual layout of the passage. An information passage describing a synthesis, for example, would technically be categorized as an experiment passage if the reaction were framed within the context of “A student carried out the following synthesis….” Fortunately, the concept of passage type is not as crucial in Organic Chemistry as it may be for Biology or Physics, for reasons that will be more evident as Organic Chemistry questions are discussed later. One subcategory of Organic Chemistry experiment passages, however, does merit a special mentioning. These are the passages detailing an extraction or purification procedure. Often the passage is accompanied by a flowchart depicting the steps. (However, there may be instances when the entire procedure is only described verbally.) The following is an example of such an experiment passage.

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Soil contains organic matter called humus. Humus is classified into humic and nonhumic materials. Humic materials are operationally divided into three main fractions: humic acid, fulvic acid, and humin. Because soil is heterogeneous and its composition varies from one locale to another, the actual composition of each fraction is elusive. Figure 1 shows the proposed structure for one possible component of fulvic acid.

HOOC

HOOC

OH

COOH

COOH

OH

OCH2OH

COOH

O

OH

COOH

Figure 1

The traditional procedure for the separation of

the three fractions from humic materials is shown in Figure 2.

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Figure 2

Some important properties of the three fractions of humic materials are listed in Table 1.

Table 1 Properties of humic materials

Humic acid

Fulvic acid

Humin

Solubility in water

soluble in alkaline pHs

soluble at all pHs

insoluble at all pHs

Aromatic content

highly aromatic

highly aromatic

highly aromatic

Carboxylic acid content

some high low

Phenolic content

some some some

Molecular weight range (Daltons)

5,000-10,000

800-4,000 >100,000

4

Note that the previous passage again illustrates how the information/experiment distinction in Organic Chemistry may be artificial at times. No actual experiment is carried out in the passage, so it could technically qualify as an information passage. However, the flowchart illustrating the separation procedure lends a very heavy experiment “flavor” to the passage. Persuasive argument passages are ones that present two (or more) different viewpoints or hypotheses. This type of passage is relatively rare in Organic Chemistry. When they do occur, they usually appear in the form of passages presenting different proposed mechanisms for a reaction. However, it is sometimes difficult to draw the distinction between such a passage and an information passage presenting several mechanisms that are not conflicting hypotheses but are instead different mechanisms thought to operate under different conditions (acidic vs. basic, etc.). Such a passage can also resemble (or in fact appear as) an experiment passage in which the different explanations are presented to account for a piece of data, or in which experimental data are included to support each hypothesis. As the previous discussion suggests, categorization of passage types in Organic Chemistry is not quite as meaningful as it may be in other science areas. Nor is it especially effective in helping you answer the questions. The distinction between passage types can be very artificial and fluid. In fact, regardless of what the passage deals with (and to what category it would technically belong), it will generally consist of combinations of elements, some of which are listed below:

• diagram/chemical equation illustrating a reaction or mechanism • table presenting data on related compounds • flowchart presenting a series of extraction and other experimental

procedures • prose summarizing what is already presented in a diagram • prose presenting new information on some aspect of the reaction(s) or

on the compounds involved • prose detailing the experimental procedure for a reaction shown in a

diagram • description of a putative or proposed mechanism

Often the emphasis on one of these elements over the others is what determines the passage type. Despite the problematic issue of categorizing Organic Chemistry passages, however, the technique that you may have learned in the other Strategic Supplementals on reading a passage still holds. You should always aim to perform two tasks on each passage: map the passage and identify the topic.

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Map the passage The idea of mapping a passage lies in reading for structure. As previously mentioned, you shouldn’t read a passage trying to memorize everything. Instead, your aim is to get the gist of what’s going on in each “structural component” of the passage. This means constantly asking yourself: “What is this paragraph/diagram/table/equation doing? What role is it playing?” The amount of attention you need to pay to the prose depends on the nature of the text. Text introducing a reaction presented in a diagram often serves only to provide a context for the reaction and can be skimmed quickly just for any new terms to register in your mind. (You may want to circle or underline the definition of a new term introduced.) Text detailing experimental protocol tends to be extraneous in Organic Chemistry (in contrast to Physics passages, for example). Description of a serial extraction procedure, on the other hand, merits closer attention, especially if no accompanying flowchart is included as a graphic. But again, keep in mind that you can always go back to the passage if you need to; most students spend way too much time absorbing as much information as they can, when it may not even be relevant to answering the questions. Keep in mind that a science passage may appear on different test forms of the MCAT with slightly different questions, so not every element of the passage may be asked about in the set of questions on your test form. The technique of mapping is especially important now that categorization is less relevant. You need to identify how the components are combined to make a passage before you can effectively answer questions calling on information from the passage, as you shall see shortly. Identify the topic In addition to mapping the passage, you should also aim to identify the topic of the passage. The topic is the one-line answer to the question, “What is the author trying to accomplish in this passage?” Notice that this means more than simply coming up with a scientific term corresponding to the main subject of the passage. It means developing an awareness of what the author is trying to do. Is she presenting different reactions that can be used to synthesize alkenes? Is she describing an experiment in which chemists investigate how different substituents affect the rate of reaction of benzene derivatives? This way of characterizing the topic is much more helpful than simply coming up with “alkene synthesis” or “EAS reactions.” On the following pages, you will see the technique applied to the two sample passages you encountered earlier.

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Aspirin, also known as acetylsalicylic acid, is one of the most useful and economical drugs available. It belongs to a class of drugs known as nonsteroidal anti-inflammatory drugs, and can be used to treat pain and alleviate inflammation and fever. The mechanism of aspirin’s action is not fully understood, although recent research suggests that it functions by inhibiting cyclooxygenase 2, an enzyme that creates prostaglandin precursors. Prostaglandins contribute to the body’s perception of pain and its inflammatory response. Prostaglandins also aid in the formation of blood clots, so aspirin thins the blood and may consequently help prevent heart disease.

Aspirin can be synthesized from salicylic acid

via the reaction shown in Figure 1.

COOH

OH

CH3COCCH3

O O

+

H+, ∆

COOH

OCCH3

+ CH3COOH

O

salicylic acid

aspirin Figure 1

Both salicylic acid and aspirin are aromatic

carboxylic acids. Carboxylic acids are generally weak acids, although they are among the strongest organic acids, with pKa’s usually in the range of 3 to 5, much lower than those of corresponding alcohols. The pKa’s of some compounds are given in Table 1.

Background information on aspirin. (This kind of prose, heavy on physiology and general observations, is usually not very important. Do NOT let yourself be bogged down by the details.) Reaction illustrating how aspirin can be synthesized. A broadening in scope: discussion of organic acids.

7

Table 1 pKa’s of Compounds Compound pKa Acetic acid 4.76 Fluoroacetic acid 2.66 Difluoroacetic acid 1.24 Trifluoroacetic acid 0.23 2,2-Dimethylpropanoic acid

5.05

Benzoic acid 4.18 p-Nitrobenzoic acid 3.43 2,4,6-Trinitrobenzoic acid

0.65

p-Methoxybenzoic acid

4.47

Phenol 9.95 Ethanol 15.9 Methanol 15.1 Water 14

Table presenting pKa’s of different compounds. Topic: The passage deals with aspirin (esp. its synthesis) and then generalizes into the topic of organic acids.

8

Soil contains organic matter called humus. Humus is classified into humic and nonhumic materials. Humic materials are operationally divided into three main fractions: humic acid, fulvic acid, and humin. Because soil is heterogeneous and its composition varies from one locale to another, the actual composition of each fraction is elusive. Figure 1 shows the proposed structure for one possible component of fulvic acid.

HOOC

HOOC

OH

COOH

COOH

OH

OCH2OH

COOH

O

OH

COOH

Figure 1

The traditional procedure for the separation of

the three fractions from humic materials is shown in Figure 2.

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Figure 2 Some important properties of the three fractions

of humic materials are listed in Table 1.

Composition of soil: some new terms introduced. Structure of something that may be found in fulvic acid. Scheme for separating the three fractions of humic materials in soil.

9

Table 1 Properties of humic materials Humic

acid Fulvic acid

Humin

Solubility in water

soluble in alkaline pHs

soluble at all pHs

insoluble at all pHs

Aromatic content

highly aromatic

highly aromatic

highly aromatic

Carboxylic acid content

some high low

Phenolic content

some some some

Molecular weight range (Daltons)

5,000-10,000

800-4,000 >100,000

Characteristics of the different fractions. (Again, do NOT get bogged down with the content. At this point, a look at the categories should suffice.) Topic: The passage discusses the different components of soil and offers a scheme of separating them.

Summary: Reading MCAT Organic Chemistry Passages • There are three types of passages: information, experiment, and persuasive

argument. However, in Organic Chemistry, a typical passage often contains elements of more than one passage type.

• Map the passage as you read. Make sure you know what the author is doing at each point. Read for the appropriate level of detail, while taking notes in the margins and underlining where necessary.

• Identify the topic of the passage. Answer the question, “What is the author trying to do in this passage?” If you can answer this question, you are ready to hit the questions.

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Answering the Questions Correctly answering MCAT Organic Chemistry questions, as with other science areas, is highly correlated to efficiently identifying the essential question and the relevant science. We recommend that you approach each question using the following four steps: • Read the question stem and answer choices. • Understand what you are being asked. • Figure out where to go to get any information that you may need. • Integrating your science knowledge with any necessary passage

research, determine the correct answer. Following these steps on every question will help you develop habits that lead to a higher score on Test Day. Let’s look closely at each step. Read the question stem and answer choices. The first thing you want to do is read the entire question, including the answer choices. You need not, and should not, evaluate the answer choices at this point: You have yet to go through any reasoning process. But you should notice the nature of answer choices: Are they complete statements, continuations of a statement begun in the question stem, scientific terms, chemical reagents, or (rare in Organic Chemistry) mathematical expressions? Some MCAT questions can seem very confusing as you read them. Reading the answer choices right after reading the question stem will help you focus on how to go about solving the problem. More specifically, it will help you perform the next step more effectively. Understand what you are being asked. In this step, your goal is to pare the question down to its essentials and determine as precisely as possible: “What science concepts am I being asked about?” For Organic Chemistry questions, the wording is often sufficiently straightforward that what the question is asking is relatively transparent. Most questions dealing with nomenclature, isomerism, molecular structure, etc., fall into this category. The following are two typical MCAT Organic Chemistry questions, together with the concepts you are being tested on.

The two organomercurial compounds in Equation 3 are: A. constitutional isomers. B. diastereomers. C. enantiomers. D. optical antipoles.

You are asked to identify the isomeric relationship between two compounds.

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How many of the compounds shown in Figure 1 are aromatic? A. 0 B. 1 C. 2 D. 3

You are asked to determine if each compound in a series satisfies the criteria for aromaticity (Huckel’s rule, planarity, etc.)

The answer choices can be useful in helping you specify your task. This is especially important if your outside knowledge does not enable you to answer the question the way it is phrased. Consider the following question pertaining to the passage on aspirin that you encountered earlier:

The conversion of salicylic acid into aspirin as illustrated in Figure 1 is what type of reaction? A. Nucleophilic acyl addition B. Nucleophilic aromatic addition C. Nucleophilic acyl substitution D. Nucleophilic aromatic substitution

You are asked to identify a given reaction by its name/category.

While the characterization of your task shown above is certainly correct, at some point you may realize that you simply do not have the requisite outside knowledge to arrive at the correct answer. Looking at the answer choices will help you recast the question in a way that is more helpful:

The conversion of salicylic acid into aspirin as illustrated in Figure 1 is what type of reaction? A. Nucleophilic acyl addition B. Nucleophilic aromatic addition C. Nucleophilic acyl substitution D. Nucleophilic aromatic substitution

You are asked to determine two things about a reaction: (a) whether the reaction involves the aromatic or the acyl functional group, and (b) whether the reaction is an addition or a substitution.

In some instances, before you use your outside knowledge to actually answer the question, you may need it just to get at the “question behind the question.” In other words, you need to draw on your knowledge to translate or paraphrase the question stem. The most likely scenario for such a case occurs when the question is designed to test your understanding of the underlying scientific principles behind a phenomenon. Consider the following question:

Which of the following compounds has the highest heat of combustion per CH2 group? A. Cyclopropane B. Cyclobutane C. Cyclohexane D. Cycloheptane

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Again, what you are being asked to do is very clearly conveyed by the question stem as it is worded. However, you should rest assured that the MCAT does not expect you to have memorized the heats of combustion of different compounds. To get at the “question behind the question” in this case relies on your ability to identify the factor determining the heat of combustion of a compound, or, equivalently, what the heat of combustion of a compound indicates. If you recall that it is an indication of the stability of a compound, and that the more strained a molecule, the higher the heat of combustion, a much better way to recast the question is: “Which of the compounds is the most strained?” Or, keeping in mind the answer choices, which are all cyclic compounds, an even more precise rewording is “Which of these compounds has the highest ring strain?” Figure out where to go to get any information that you may need. This is the step in which you locate the information you need from the passage, and is precisely why developing a good map of the passage is such an important technique in reading MCAT passages. This step is accomplished by answering the questions: “Do I need to retrieve a specific piece of information from the passage? If so, from where? Do I need to engage in more extensive reasoning from the passage? If so, from which part?” The answer can be as trivial as “Yes, I need to read off a boiling point from Table 1 in the passage,” or it can be more involved; e.g. “I need to walk though the synthesis scheme shown in Figure 3 and imagine what the product would be if the starting material were X instead of Y.” What the answer is obviously depends on the nature of the question. A surprisingly large number of Organic Chemistry questions on the MCAT are actually conceptually independent from the passages they accompany. I.e, they can either stand alone from the passage completely, or they may require nothing more than very specific data from the passage without requiring any in-depth understanding of the passage. For example, the questions may test you on outside knowledge of certain characteristics of one or more compounds mentioned in the passage. The following question pertaining to the aspirin passage is an example of a question that can stand alone from the passage.

Which of the following best accounts for the higher acidity of phenol compared to ethanol? A. Phenol is hydrophobic. B. Phenol has a higher molecular weight. C. The benzene ring of phenol stabilizes

negative charge. D. The oxygen atom in phenol is sp2

hybridized.

Step 2: You are asked to identify the one statement from among the answer choices that explains why phenol is a stronger acid than ethanol. Step 3: Nothing is needed from the passage. It is based on pure outside knowledge on acidity.

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The relationship of the question to the aspirin passage is rather tenuous: It lies in the appearance of phenol and ethanol (and their pKa’s) in Table 1 of the passage. But again, notice that the wording is entirely “self-contained” and the question could easily have appeared as a discrete question not accompanying any passage. More frequently, however, you will need to go back to the passage, but only to retrieve a very specific piece of information. Consider the following questions, which could accompany the passage on soil composition.

How many stereogenic centers does the structure shown in Figure 1 contain? A. 3 B. 6 C. 9 D. 12 All of the following functional groups are present in the structure shown in Figure 1 EXCEPT: A. Alcohol B. Carboxylic acid C. Ester D. Ketone

Step 2: Identify the number of stereogenic centers (or chiral centers) contained in the molecule. (Or, applying our outside knowledge to more precisely characterize our task: Identify the number of atoms bonded to four different groups arranged in a tetrahedral geometry.) Step 3: Need to examine Figure 1 in the passage. Step 2: Take each answer choice and match it against the structure to see if it appears. Step 3: Need to examine Figure 1 in the passage.

For either question you need to examine the structure shown in Figure 1 (the proposed structure for a component of fulvic acid). However, for the purpose of the questions, you do not really care what role the diagram is playing in the passage. The fact that the structure comes up during a discussion on soil composition is absolutely irrelevant to answering the question. With a slight change in wording, the structure can be included as part of the question stem and the question would be completely independent of the passage. To answer these questions, in other words, you only need specific data from the passage. Just about every Organic Chemistry passage on the MCAT will contain questions of this nature. Whether you need to go back to the passage or not, the important component is our outside knowledge, which again can come into play at an earlier stage in helping you take apart the question. However, certain questions do require an in-depth understanding of (part of) the passage. Sometimes, the question stem makes it easy to identify these questions by beginning with “According to the passage…” or “Based on

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information contained in the passage…” Other times, you’ll have to rely on your having absorbed the passage critically. In Organic Chemistry, questions calling for an understanding of the passage often involve applying a reaction presented in the passage to a new set of compounds, identifying trends in data, interpreting the result of an experiment, etc. In short, they are much more reliant on critical reading of the passage and on the application of scientific reasoning in unfamiliar situations. However, they usually still call for some outside knowledge. Consider the following question accompanying the aspirin passage:

Based on the data in Table 1, what is the expected pKa of the following compound?

COOH

NO2

NO2 A. pKa < 0.65 B. 0.65 < pKa< 3.43 C. 3.43 < pKa< 4.18 D. pKa > 4.18

Step 2: Determine how the structure shown is related to the ones appearing in Table 1 (which involves outside knowledge of nomenclature), and determine how its pKa compares to that of other relevant compounds. (The answer choices are certainly helpful in helping you decide what’s demanded of you: You don’t need to come up with a specific value!) Step 3: Look at the relevant compounds in Table 1 to establish a trend. Use their pKa values as benchmarks to establish the range.

In some instances the purpose of the passage can reveal a likely line of questioning when it comes to questions calling for an understanding of the passage. For example, with a passage dealing with a multistep synthesis, you are likely to be called upon to perform one or more of the following:

• determine what the product would be if the starting compound were a similar but distinct molecule; conversely, determine what the starting compound must have been to synthesize a similar but distinct product using the same method.

• be able to rationalize why each step is necessary (what it accomplishes), and predict what would have been the result if the step were modified or omitted.

• be able to predict where an isotopic label in one of the reactants ends up as a result of the reaction.

Similarly, an experiment passage presenting an extraction procedure, such as the passage on soil composition, may very likely be followed by questions testing your understanding of the rationale behind the scheme, and determine the nature of the compound isolated at specific steps. The following question represents a likely question for this kind of passage.

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In Figure 2, Fraction B is most likely: A. humic acid. B. fulvic acid. C. humin. D. Cannot be determined from the

information given.

Step 2: Apply the solubility data in Figure 2 and, combining it with our knowledge of the properties of the different fractions from Table 1, determine the identity of Fraction B. Step 3: Consult Table 1 and the flowchart in Figure 2.

This last example illustrates one point: Your evaluation of what you are being asked about (Step 2) may undergo some refinement after you have determined exactly what in the passage is going to help you answer the question (Step 3). For example, after having studied Figure 2, a more efficient characterization of your task may be: “Which of the three fractions is most likely to be insoluble in HCl but soluble in NaOH?” This, in turn, helps you narrow down what data is relevant in Table 1. Indeed, as you gain experience in reading MCAT passages effectively and tackling Organic Chemistry questions, you may often find yourself accomplishing Steps 2 and 3 in parallel. Integrating your science knowledge with any necessary passage research, determine the correct answer. This is the step where you essentially “pull everything together” and actually accomplish the ultimate goal of arriving at the correct answer. If you have managed to avoid having to draw on your outside knowledge so far, you will undoubtedly need to now. This is the stage where you need to identify that the structure for the fulvic acid component shown in the passage has 3 chiral centers, and that it contains the alcohol, carboxylic acid and ketone functional groups, but NOT the ester functional group. This is the stage where you need to recall that phenols are more acidic than aliphatic alcohols because the benzene ring can stabilize the negative charge on the conjugate base. This is the stage where you recognize that cyclopropane has the highest ring strain. Let us see in more detail how this step applies to the two questions above:

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Based on the data in Table 1, what is the expected pKa of the following compound?

COOH

NO2

NO2 A. pKa < 0.65 B. 0.65 < pKa< 3.43 C. 3.43 < pKa< 4.18 D. pKa > 4.18

In Figure 2, Fraction B is most likely: A. humic acid. B. fulvic acid. C. humin. D. Cannot be determined from the

information given.

Step 4: The compound is 2,4-dinitrobenzoic acid. It makes most sense to compare this to p-nitrobenzoic acid and 2,4,6-trinitrobenzoic acid in Table 1. p-Nitrobenzoic acid is monosubstituted, while the structure shown is the disubstituted analog, and 2,4,6-trinitrobenzoic acid is the trisubstituted analog. You therefore expect the given structure’s pKa to be between that of p-nitrobenzoic acid and 2,4,6-trinitrobenzoic acid. Choice B is correct.

With our last rephrasing of the question (“Which of the three fractions is most likely to be insoluble in HCl but soluble in NaOH?”), you consult the solubility properties in Table 1 and realize that humic acid, soluble in alkaline pH, fits the bill. Fraction A is fulvic acid while Fraction C is humin. Choice A is correct.

Summary: Answering MCAT Organic Chemistry Questions • Read the question stem and answer choices. • Understand what you are being asked. Be able to specify the task you are

required to perform. Reword the question in your own mind if necessary; let the answer choices help you.

• Figure out where to go to get any information that you may need. For some questions you need not go back to the passage at all. Use your passage map to retrieve any information or data you need efficiently.

• Integrating your science knowledge with any necessary passage research, determine the correct answer.

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Applying What You’ve Learned Starting on the next page you will find three MCAT Organic Chemistry passages with accompanying questions. Spend no more than 6-8 minutes mapping each passage and answering the questions, applying the strategies discussed above. Sample maps and complete explanations to the questions follow.

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Passage I (Questions 1–5) The application of heat often causes

hydrocarbons to undergo decomposition or rearrangement to yield new compounds. At very high temperatures, for example, ethane yields ethylene and hydrogen via a series of radical reactions. Cyclopropane is converted to propylene when heated, as shown below in Figure 1.

H2C CH2

H2C ∆

H2C

HC

CH3

Figure 1 The decomposition of hydrocarbons under heat

is the basis for the industrial process of thermal cracking, used to refine the different organic components of oil. In the process, a mixture of petroleum-distilled alkanes is heated with steam to 500-900°C for a fraction of a second and then immediately cooled. As a result, the alkanes are broken down into smaller hydrocarbons such as methane, ethylene, propylene and alkylbenzenes. The compounds obtained are often in turn used to synthesize other commercially available products. Methane can be oxidized to formaldehyde, ethylene can be oxidized to produce ethylene glycol (antifreeze), and propylene is used in the production of phenol, which is important in adhesives. Propylene can also be converted into isopropyl alcohol via the following reaction.

H2C

HC

CH3

+ H2OH3C C CH3

HO HH2SO4

Figure 2

Ethylene and propylene isolated from the

thermal cracking process can be linked to aromatics to form alkyl-substituted aromatic compounds, which can then be used to form polystyrene and various solvents.

+ H2C CH2catalyst

+catalyst

H2C

HC

CH3

Figure 3

1. Which of the following is the most likely catalyst for

the reactions shown in Figure 3? A. H2O B. Pt/H2 C. H2SO4 D. heat

2. The reaction shown in Figure 2 is an example of: A. oxymercuration-demercuration. B. hydroboration-oxidation. C. acid-catalyzed hydrolysis. D. acid-catalyzed hydration.

3. In the reaction shown in Figure 1, the C-C-C bond angle changes from: A. 60° to 109.5°. B. 60° to 120°. C. 120° to 109.5°. D. 120° to 180°.

4. The bond rupture that occurs for alkanes at high temperatures can be described as: A. pyrolysis. B. heterolysis. C. annealing. D. condensation.

5. The reaction shown in Figure 2 is most effectively reversed by: A. heating the alcohol. B. using a base. C. heating the alcohol in the presence of NaOH. D. heating the alcohol in the presence of acid.

19

Passage II (Questions 6–10) Mechanistically, nitration reactions of aromatic

compounds in strongly acidic media appear to proceed via the initial formation of the nitronium ion. This ion presumably then serves as the electrophile which forms the cationic aryl adduct which can subsequently decompose to form the final product. The proposed mechanism for the nitration of benzene in concentrated sulfuric acid is outlined in Figure 1. This reaction is found to be first order with respect to the aromatic substrate. Substitution of the deuterated analog (d6-benzene) for the substrate has no measurable effect on rate.

HNO3 + H2SO4 H2NO3+ + HSO4

−

H2NO3+ H2O + NO2

+

+ NO2+

H

NO2

H

NO2

NO2

+ H+

Figure 1

Nitration in the presence of transition metal

catalysts appears to follow a different mechanism. The accepted mechanism for such reactions is illustrated in Figure 2 for the reaction of benzene with nitric acid in the presence of mercury (II) nitrate. Complexation of the aromatic donor with the electron deficient metal cation is believed to occur to a steady state limit, followed by rate-determining loss of a proton from the cationic adduct. The resulting organometallic cation then reacts rapidly with nitric acid, presumably via a concerted step involving transfer of hydroxide to the metal center. Initial rate studies indicate that the reaction is first order in the aromatic substrate, and first order in the catalyst when it is homogeneous. Further, deuterium exchange studies exhibit a kH/kD ratio of about 6, indicating that the loss of the proton in the second step is rate determining.

+ Hg2+

H

Hg

H

HgHg

+ H+

Hg

+ HNO3

NO2

+ Hg(OH)+

Figure 2

6. Based on the information provided in the passage, what can be concluded about the reaction shown in Figure 1? A. The first, reversible step of the mechanism is

rate-determining. B. The equilibrium constant for the first step is

small, and the fourth step is rate-determining. C. The equilibrium constant for the first step is

large, and the fourth step is rate-determining. D. The third step of the mechanism is rate-

determining.

7. Which of the following is a resonance structure of the

product in Step 3 of the reaction shown in Figure 1?

A. C.

H

NO2

H

NO2

B. D.

H

NO2

H

NO2

20

8. Compared to the nitration of benzene via the scheme shown in Figure 1, the nitration of cyanobenzene under the same conditions is expected to have: A. a smaller rate constant and a slower

rate. B. a smaller rate constant but the same

rate. C. a larger rate constant and a higher rate. D. a larger rate constant but the same rate.

9. Which of the following can be used to

convert nitrobenzene into aniline? A. NaOH B. NH3 C. CF3CO3H D. Sn/HCl

10. With certain aromatic compounds, the rate

of the nitration reaction as shown in Figure 1 is found to be independent of the substrate concentration. For what type of compounds is this most likely to be true? A. Aromatic compounds less reactive than

benzene B. Aromatic compounds more reactive

than benzene C. Aromatic compounds with a higher

molecular weight than benzene D. Aromatic compounds with a lower

molecular weight than benzene

21

Passage III (Questions 11−15) Spectroscopic determination of the structure of

long-chain polyunsaturated carboxylic acids is often difficult because of the relatively large number of functionally similar isomers possible for a given molecular formula. While standard wet analytical techniques, e.g., bromination or hydrogenation, are generally employed to elucidate such facets of an unknown acid's identity as its degree of unsaturation, a combination of chemical and spectral analysis is usually necessary to determine the exact structure of the carboxylic acid.

A researcher attempting to determine the structure

of an unknown carboxylic acid with molecular formula C18H26O2 took a mass spectrum of a small sample of the unknown, then performed a series of chemical tests:

Test 1. Catalytic hydrogenation over palladium on

charcoal (Pd/C) yielded stearic acid, CH3(CH2)16COOH.

Test 2. Hydrogenation over Lindlar's catalyst resulted in the absorption of exactly two equivalents of H2.

Test 3. Oxidation of the unknown in KMnO4 followed by acidification produced four distinct, nonisomeric compounds, three of which were dicarboxylic acids.

Test 4. Ozonolysis of the unknown acid produced four compounds, one of which was formaldehyde.

Test 5. Titration with KHP indicated that the unknown had only one carboxylic acid functionality.

Product mixtures resulting from Tests 3 and 4

were separated by HPLC. Spectral analysis was employed to determine the structures of the various reaction products. Based on the results of the five chemical tests and the interpretation of the spectral data, the researcher concluded that the unknown compound was one of the compounds shown below:

H2C CH(CH2)4C CC C(CH2)7COOH

H2C CH(CH2)4C C(CH2)7C CCOOH

22

11. The major difference between hydrogenation Pd/C and hydrogenation over Lindlar’s catalyst isthat:

over Map of Passage I

A. the use of Lindlar’s catalyst results in thhydogenation of only one pi bond in each tbond, while Pd/C catalyzes the hydrogenatioall the carbon-carbon pi bonds in the compou

e riple n of nd.

e onds

The application of heat often causes hydrocarbons to undergo decomposition or rearrangement to yield new compounds. At very high temperatures, for example, ethane yields ethylene and hydrogen via a series of radical reactions. Cyclopropane is converted to propylene when heated, as shown below in Figure 1.

B. the use of Lindlar’s catalyst results in thhydrogenation of carbon-oxygen double bwhile the use of Pd/C does not.

C. quantitative measurement of the hydrogen absorbed during catalytic hydrogenation is possible only when Lindlar’s catalyst is used.

D. hydrogenation over Pd/C is an oxidative process while that over Lindlar’s catalyst is reductive.

H

12. One of the products of Test 3 is not a dicarboxylicacid. Its most likely identity is:

the

rbon

rans

The decomposition of hydrocarbons under heat is the basis for the industrial process of thermal cracking, used to refine the different organic components of oil. In the process, a mixture of petroleum-distilled alkanes is heated with steam to 500-900°C for a fraction of a second and then immediately cooled. As a result, the alkanes are broken down into smaller hydrocarbons such as methane, ethylene, propylene and alkylbenzenes. The compounds obtained are often in turn used to synthesize other commercially available products. Methane can be oxidized to formaldehyde, ethylene can be oxidized to produce ethylene glycol (antifreeze), and propylene is used in the production of phenol, which is important in adhesives. Propylene can also be converted into isopropyl alcohol via the following reaction.

A. CO2. B. CH3COOH. C. CH3OH. D. CH4.

13. Which of the following can be inferred fromresults of Test 1, but NOT from that of Test 5? A. The unknown contains unsaturated ca

atoms. B. The unknown is not capable of cis/t

isomerism. C. The unknown contains an acidic hydrogen. D. The unknown is unbranched.

14. Which of the following observations would be most

helpful in distinguishing stearic acid from the unknown compound? A. The existence of a 1H-NMR singlet between 10

and 12 ppm B. A strong, broad IR absorption from 2500 to

3400 cm−1

C. The absence of an IR stretch around 2200 cm−1 of

Ethylene and propylene isolated from the thermal cracking process can be linked to aromatics to form alkyl-substituted aromatic compounds, which can then be used to form polystyrene and various solvents.

D. A boiling point lower than that CH3(CH2)17OH

15. What is the relationship between the two proposedstructures for the unknown?

A. They are geometric isomers. B. They are diastereomers. C. They are constitutional isomers. D. They are not isomers of each other.

2C CH2

H2C ∆

H2C

HC

CH3

Figure 1

H2C

HC

CH3

+ H2OH3C C CH3

HO HH2SO4

Figure 2

23

What can happen when hydrocarbons are heated. Examples of ethane (described in words) and cyclopropane (illustrated in addition). Description of industrial-scale process (thermal cracking). Examples of compounds obtained. Further reactions (synthetic uses) of compounds obtained in thermal cracking: methane and propylene described in words. Further reactions (synthetic uses) of compounds obtained in thermal cracking: propylene illustrated.

+ H2C CH2catalyst

+catalyst

H2C

HC

CH3

Figure 3

24

Further reactions (synthetic uses) of compounds obtained in thermal cracking: ethylene and propylene illustrated. Topic: The author presents a collection of reactions that are either examples of what happens when an alkane is heated to high temperatures, or what these products can be used for.

Map of Passage II Mechanistically, nitration reactions of aromatic

compounds in strongly acidic media appear to proceed via the initial formation of the nitronium ion. This ion presumably then serves as the electrophile which forms the cationic aryl adduct which can subsequently decompose to form the final product. The proposed mechanism for the nitration of benzene in concentrated sulfuric acid is outlined in Figure 1. This reaction is found to be first order with respect to the aromatic substrate. Substitution of the deuterated analog (d6-benzene) for the substrate has no measurable effect on rate.

HNO3 + H2SO4 H2NO3+ + HSO4

−

H2NO3+ H2O + NO2

+

+ NO2+

H

NO2

H

NO2

NO2

+ H+

Figure 1

Nitration in the presence of transition metal

catalysts appears to follow a different mechanism. The accepted mechanism for such reactions is illustrated in Figure 2 for the reaction of benzene with nitric acid in the presence of mercury (II) nitrate. Complexation of the aromatic donor with the electron deficient metal cation is believed to occur to a steady state limit, followed by rate-determining loss of a proton from the cationic adduct. The resulting organometallic cation then reacts rapidly with nitric acid, presumably via a concerted step involving transfer of hydroxide to the metal center. Initial rate studies indicate that the reaction is first order in the aromatic substrate, and first order in the catalyst when it is homogeneous. Further, deuterium exchange studies exhibit a kH/kD ratio of about 6, indicating that the loss of the proton in the second step is rate determining.

25

Verbal description of mechanism of one nitration reaction. Kinetics data of reaction. Results and conclusion of isotope-labeling experiment. Illustration of mechanism described above. Verbal description of mechanism of another nitration reaction. Miscellaneous observations/beliefs about the kinetics. Results and conclusion of isotope-labeling experiment.

+ Hg2+

H

Hg

H

HgHg

+ H+

Hg

+ HNO3

NO2

+ Hg(OH)+

Figure 2

26

Illustration of mechanism described above. Topic: The author presents two alternative ways in which nitration of benzene can occur and discusses some kinetic/mechanistic details of each.

Map of Passage III Spectroscopic determination of the structure of

long-chain polyunsaturated carboxylic acids is often difficult because of the relatively large number of functionally similar isomers possible for a given molecular formula. While standard wet analytical techniques, e.g., bromination or hydrogenation, are generally employed to elucidate such facets of an unknown acid's identity as its degree of unsaturation, a combination of chemical and spectral analysis is usually necessary to determine the exact structure of the carboxylic acid.

A researcher attempting to determine the structure

of an unknown carboxylic acid with molecular formula C18H26O2 took a mass spectrum of a small sample of the unknown, then performed a series of chemical tests:

Test 1. Catalytic hydrogenation over palladium on

charcoal (Pd/C) yielded stearic acid, CH3(CH2)16COOH.

Test 2. Hydrogenation over Lindlar's catalyst resulted in the absorption of exactly two equivalents of H2.

Test 3. Oxidation of the unknown in KMnO4 followed by acidification produced four distinct, nonisomeric compounds, three of which were dicarboxylic acids.

Test 4. Ozonolysis of the unknown acid produced four compounds, one of which was formaldehyde.

Test 5. Titration with KHP indicated that the unknown had only one carboxylic acid functionality.

Product mixtures resulting from Tests 3 and 4

were separated by HPLC. Spectral analysis was employed to determine the structures of the various reaction products. Based on the results of the five chemical tests and the interpretation of the spectral data, the researcher concluded that the unknown compound was one of the compounds shown below:

H2C CH(CH2)4C CC C(CH2)7COOH

H2C CH(CH2)4C C(CH2)7C CCOOH

27

General discussion on methodology of determining structure of long-chain polyunsaturated carboxylic acids. (Background info to set the context.) Specific scenario at hand. (The molecular formula is given here.) Tests conducted and results/deductions. (Underline key terms to quickly isolate the nature of each test.) (Filler prose on minutiae of experimental protocol, leading to the important part…) Two possible structures for the unknown. Topic: The author describes a series of experiments conducted to determine the structure of an unknown.

28

Answers and Explanations Answer Key 1. C 2. D 3. B 4. A 5. D 6. D 7. D 8. A 9. D 10. B 11. A 12. A 13. D 14. C 15. C Explanations 1. C Your task is relatively straightforward from the wording of the question stem: identify the catalyst for the reactions shown in Figure 3 of the passage. If your outside knowledge does not allow you to predict what the species should be, you may conceive of your task as follows: Go through the answer choices in turn and see if the reaction would occur as illustrated. Examining the reactions themselves is a must, but does the text of the passage perhaps give any hints of information that may be useful? The answer, unfortunately, is no. With your map of the passage, you should know exactly where to go look to see if there is any helpful information. But the paragraph immediately preceding the diagram merely recaps the reactants and products and what the products can in turn be used for. The question, then, is one that requires only data from the passage, and you need to rely primarily on your outside knowledge. Since the reaction places an alkyl (more precisely an alkenyl) group on benzene, you may recognize it as a Friedel-Crafts alkylation reaction, catalyzed by the Lewis acid AlCl3 or FeCl3. However, there are two problems with this prediction based on outside knowledge: In such reactions, the

alkyl group to be added is in the form of an alkyl chloride, which is not the case here, and besides, neither AlCl3 nor FeCl3 is among the answer choices. You therefore need to fall back on an examination of each answer choice. Choice A, water, will not catalyze the reaction. You may realize from outside knowledge that water rarely acts as a catalyst. It is difficult to imagine what role water could conceivably play other than adding to a double bond if the conditions are right. (See for example the reaction shown in Figure 2 of the passage.) Choice B is something that causes hydrogenation of double bonds. While the alkenyl functionality has been removed in the product, hydrogenation is not the only thing that occurs: The aliphatic hydrocarbon actually needs to add to the benzene ring. This is not something that Pt/H2 can accomplish. Choice C is an acid. In your study of organic chemistry you have encountered many instances of acid-catalyzed reactions, so at least this choice is promising. Could it catalyze the reaction? At this stage you need to draw on your expertise in organic chemistry mechanisms. What might a proton do to either the benzene or the alkene? Recall that the addition of a proton to an alkene is often the first step in a Markovnikov addition. Let us pursue this with the second reaction shown in Figure 3. The first step then may be as follows:

H2C

HC

CH3

H+

H3C

HC

CH3 Notice that the proton adds to the terminal methylene group so that the carbocation obtained is the more stable secondary one. What might happen next? You have a carbocation, which is electrophilic and will therefore be drawn to the pi electrons in the benzene ring. What follows, then, is the classic electrophilic aromatic substitution reaction. (Recall that in all EAS reactions, one needs to generate a positively charged or positively polarized electrophile to attack the benzene ring.)

29

H

H

− H+

Choice C is therefore the correct answer. Choice D, heat, is by itself insufficient to catalyze the reaction. 2. D From a scan of the question stem and the answer choices, your task is to categorize or label the reaction shown in Figure 2. It would appear that it is primarily your outside knowledge that is being tested, although the reaction itself and the surrounding prose may give you some help. Is this the case? The only reference to the diagram within the passage is in the sentence immediately preceding it, and it only mentions the reactant and the product. If there is anything to help you, therefore, it will be in the reaction itself, something that you'll need to look at anyway. Notice that the result of the reaction is the addition of a water molecule across a double bond. Furthermore, sulfuric acid is used as a catalyst. Armed with these observations, you should be able to arrive at choice D, acid-catalyzed hydration, as the correct answer. Choices A and B both lead to the (net) addition of a water molecule across a double bond. However, they do not refer to the reaction shown. From their names, you could have surmised that oxymercuration-demercuration would involve the element mercury at some point in the reaction, while hydroboration-oxidation would involve some boron-containing compound as a reagent. Neither is present in the reaction. Choice D, acid-catalyzed hydrolysis, refers to the breaking of a compound by water under

acidic conditions. An ester, for example, can undergo acid-catalyzed hydrolysis to yield a carboxylic acid and an alcohol. 3. B Your task is actually two-fold: determine the C-C-C bond angle in the starting material of the reaction in Figure 1, and the C-C-C bond angle in the product. What you need from the passage is nothing more than the identity of the starting material and the product, which you obtain by looking at Figure 1. You would then rely on your outside knowledge to determine the bond angles. The starting material for the reaction is cyclopropane. From the triangular geometry of the molecule, you can see that the C-C-C bond angle is 60°. The product, propylene, has a C-C-C bond angle of 120°: Because the central carbon atom is sp2 hybridized, the three groups (=CH2, –CH3 and –H) are arranged in a trigonal planar geometry around it, making an angle of 120° with each other. 4. A In this question, just as in question #2, you are asked to categorize or characterize a reaction. However, this time the question is truly independent of the passage. No reference is made to any component of the passage. Paraphrasing the question: “When an alkane C-C bond breaks because of heat, what is the process called?” Nothing in the passage is likely to give you the answer, but since the passage is largely concerned with such reactions, its content may prove helpful, although at this stage it is difficult to speculate how. For the moment, you need to take each choice and evaluate it. Choice A, pyrolysis, may not be familiar to you, but maybe you can take a stab at its meaning by recognizing its roots: pyro- having to do with fire and –lysis referring to the breaking of something. This is certainly conceivably applicable to the process. Heterolysis (choice B) may not be a familiar term, but a bond is said to break (or cleave) heterolytically if the bonding pair of electrons ends up on one atom only.

30

X Y X + Y+

heterolytic bond cleavage This, however, is not an accurate description of the process at hand. In order for radicals to be generated (as the passage says would occur in the first paragraph), the bond must cleave homolytically: The bonding pair should be split between the two atoms as the bond breaks. So choice B is incorrect. Choice C, annealing, refers to the heating and subsequent slow cooling of a metal or glass for the purpose of strengthening it. Choice D, condensation, refers either to the physical process of gaseous vapor liquefying or to a reaction in which two molecules come together. So choice A is the correct answer. 5. D You are asked to evaluate what is most effective in causing the reaction shown in Figure 2 to proceed in the reverse direction. As with question #2, the prose immediately preceding the diagram is of limited utility as it only refers to the starting material and product. So you have to go with the reaction itself as illustrated in the diagram, plus your outside knowledge. The reaction as illustrated is an acid-catalyzed hydration reaction. The reverse reaction is an elimination reaction of an alcohol to form an alkene. In other words, once you have actually examined the reaction in Figure 2, you could redefine the “question behind the question” as follows: “What would cause an alcohol to undergo elimination to form an alkene?” Alcohol elimination proceeds in an E1 manner in the presence of heat and acid, so choice D is correct. Choice A, heating the alcohol, will not accomplish the task: The alcohol as is does not possess a good leaving group. Choices B and C, using a base, may be tempting, since an acid is used in the forward (hydration) reaction. However, keep in mind that the acid functions as a catalyst, and one characteristic of catalysts in general is that they catalyze both the forward and reverse reactions. (Only then will they not affect the

thermodynamics of the reaction.) In other words, it is not correct to assume that if an acid is present in the forward reaction, then a base will automatically cause the reverse reaction to occur. Indeed, in this case, a base is not going to be very effective. While a base can take part in an E2 reaction, again there is no good leaving group on the alcohol. 6. D The question stem itself is rather vaguely worded. You need to look at the answer choices to specify your task: determine which step in the mechanism shown in Figure 1 is rate-limiting, and maybe something about the first step in addition. (Compare choices B and C.) As the question stem indicates, you will need to go back to the passage for information or clues. This is an example of a question requiring an understanding of the passage, as you will need to engage in some critical assessment of information contained in the passage. Since the question refers to Figure 1, where is helpful information most likely located? With your map of the passage, you know that the second half of paragraph one conveys information on the kinetics of the mechanism. You are told that the reaction is first order in the aromatic substrate. The fact that the concentration of the aromatic substrate affects the reaction rate means that the rate-limiting step cannot be earlier than the third step, when the substrate makes its first appearance. (If, for example, the second step were rate-limiting, then the rate of the reaction would be unaffected by how fast the third step occurs, i.e., it would be unaffected by the concentration of benzene.) Your decision, then, is to decide whether the third or the fourth step is rate-limiting. The end of paragraph 1 tells you that using C6D6 has no effect on the rate. What is the significance of this information? If using the deuterated analog has no effect on the rate, then the rate-limiting step cannot involve the breaking of the C-D (or C-H) bond. Because of the isotope effect, the C-D bond is harder to break than the C-H bond. If the breaking of that bond is rate-limiting (or is part of what goes on in the rate-limiting step), then the deuterated compound would exhibit a slower

31

reaction rate. Hence, the fourth step cannot be rate-limiting, and choice D is correct. 7. D Your task here is to identify a resonance structure of one shown in the passage. This is another classic example of a question requiring nothing more than data from the passage. You need to look at Step 3 of the reaction shown in Figure 1 for the structure, but you are not interested in what role the compound actually plays conceptually in the passage. Whenever you are asked to identify a resonance structure, you can first and foremost eliminate any choice that does not carry the same net charge as the given structure. Unfortunately, in this case, this criterion is satisfied by every choice (although see the discussion below on the nitro group). You therefore have to actually see if, by shifting the electrons, you can obtain any one of the four choices (and you should be able to obtain one, and only one). Luckily, you do not have far to go, and you can even identify the correct choice immediately if your knowledge of the electrophilic aromatic substitution mechanism is strong:

HNO2

HNO2

The structure you obtain is choice D. You can also draw a third resonance structure in which the formal charge is placed on the “top” carbon (the other carbon atom ortho to the nitro group). Choice A is incorrect because you cannot shift the electrons in such a way as to place the formal charge on the meta carbon. Choices B and C are incorrect because the carbon bearing the nitro group forms five bonds, which is not allowed. On the surface the two structures do look as if they have the right charge, but the nitrogen in the nitro group always has a formal positive charge:

NO O

NOO

This formal positive charge is balanced by the negative charge spread out over the two oxygen atoms, so in the other structures these charges are omitted and the nitro group is simply represented as an overall neutral group. 8. A After having read the answer choices, you can see that there are two things you need to determine about the nitration of cyanobenzene: its rate constant and its rate, as compared to the same reaction with unsubstituted benzene. Your outside knowledge should allow you to get at the “question behind the question”: Is the cyano group an activator or a deactivator? Furthermore, once you have determined the answer, you have to translate that into how exactly the nature of the group affects the rate law. Both in wording and conceptually, this question is completely independent of the passage. The cyano group has the structure –C≡N. You may either know from pure memory that it is a deactivator, or you nay deduce this from its structure: Nitrogen is highly electronegative and withdraws electron density from carbon. The positively polarized carbon, in turn, becomes electron deficient and withdraws electron density from the benzene ring. This has the effect of destabilizing the intermediate. Now that you know that the cyano group is a deactivator, you have to get at how the rate law is affected. The rate law of a reaction can be expressed as rate = k[reactant A]x[reactant B]y… The presence of a cyano group is not going to affect the concentration of the reactants. (You are told that the nitration occurs under the same conditions, so the concentration of the substrate should be kept constant.) So what does being a deactivator mean? It means that the rate of reaction is slower, and since you just concluded that the reactant concentrations remain constant, this must occur via a

32

reduced rate constant k. Choice A is therefore correct. 9. D You are asked to identify the correct chemical reagent that will effect the conversion given in the question stem, from nitrobenzene to aniline. Your nomenclature knowledge is a must in this question, which is independent of the passage. Specifically, you have to know what the structure of aniline is. Once you do, an effective pictorial representation of the question is as follows:

NO2 NH2

?

The process can be recognized as a reduction. Among the reagents, only choice D, Sn/HCl, can accomplish a reduction. Choices A and B, sodium hydroxide and ammonia, can act as a base or a nucleophile, but not as a reducing agent. Choice C is a peroxyacid. 10. B The first sentence in the question stem gives you a piece of information on the kinetics of a subclass of aromatic compounds as they undergo the nitration illustrated in Figure 1. You are then asked to identify what this subclass is. Essentially you need to figure out the following: How are these compounds special and what makes them special? To answer the first question, you need to contrast the information given in the first sentence of the question stem to the information presented in the passage. The question stem tells you that the rate of nitration is independent on substrate concentration. How does this jibe with what you are told in the passage? With a map of the passage, you would know to go to the second half of the first paragraph, where the “usually observed” kinetics is presented. There, you are told that the reaction is first-order with respect to the substrate; i.e., the rate increases as the substrate concentration increases. With this

comparison, you can further rephrase the question as follows: For what kind of compounds will the reaction rate not depend on the concentration? This is a question calling for a relatively high level of critical thinking based on the passage. The answer choices offers two “lines of attack” (and thus a further refinement of your task): Is this phenomenon more likely due to the reactivity of the compound (choices A and B), or to its molecular weight (choices C and D)? Your outside knowledge (or perhaps more accurately your “chemical intuition” since it is hardly something that textbooks spell out explicitly) may allow you to reject choices C and D because molecular weight is expected to influence kinetics only through the isotope effect or indirectly through steric interactions. A difference in molecular weight in and of itself (say, between anthracene and benzene, or between ethylbenzene and pentylbenzene) is not able to cause or account for differences in kinetics. You should therefore focus on reasoning along the lines of reactivity. If the rate is dependent on the concentration of the substrate, then the rate-determining step can occur no earlier than the step where the substrate first appears. In this case this is the third step as shown in Figure 1. The more reactive the substrate, the faster the rate. However, if the substrate were reactive enough, it may be that one of the first two steps becomes the rate-limiting step, in which case the rate will be found to be independent on the substrate concentration. In other words, the scenario described in the question stem most likely arises when the substrate is very reactive compared to benzene, choice B. Choice A is incorrect: If the substrate is less reactive than benzene, the third step is still the rate-limiting step (and in fact becomes “more so”); the dependence on substrate concentration will remain the same but the rate of reaction will be slower. Choices C and D are rejected based on the reasoning discussed above. 11. A The question is asking you to identify a difference between two different

33

hydrogenation techniques. Unfortunately, the answer choices do not suggest an angle from which the problem can be attacked. The statements deal with many different aspects. If you approach this as a question independent from the passage (which the wording does allow), you will have to have very strong outside knowledge, possibly beyond the scope of what the MCAT requires. But the question can be approached as one that is passage-based, one requiring an understanding and synthesis of different components of the passage. Where in the passage does the topic of hydrogenation come up? In the first two tests described in the passage. It is imperative to pay close attention to the different results in the two cases. In the first test, you are told hydrogenation over Pd/C yielded stearic acid. From the condensed structural formula given, you should be able to tell that the alkyl portion of the acid is completely saturated; i.e., there are no carbon-carbon double or triple bonds. In test 2, with the use of Lindlar’s catalyst, only two molar equivalents of hydrogen were absorbed. In other words, only two pi bonds were saturated. The results of these two tests by themselves do not suggest any immediately obvious difference, and in fact they may have led to the exact same compound. The question to ask, then, is do the two tests actually yield the same compound? Is stearic acid obtained by saturating two pi bonds? To answer this you need to turn to the possible structures of the compounds given at the end of the passage. Both structures contain two carbon-carbon triple bonds and one carbon-carbon double bond. To convert the unknown into stearic acid requires the saturation of every carbon-carbon pi bond in the compound and would therefore lead to the absorption of five molar equivalents of hydrogen. (Recall that a triple bond contains two pi bonds and a double bond contains one pi bond.) Absorption of two molar equivalents of hydrogen (using Lindlar’s catalyst) would still leave some pi bonds intact. What functionality do you have two of in the unknown? Triple bonds. It is therefore reasonable to assume that each triple bond is converted into a double bond, leaving a product containing three C=C bonds. Choice A is therefore correct.

Choice B is not supported by evidence in the passage and can also be rejected because of its implausibility: Saturating a carbon-oxygen double bond (i.e., converting a carbonyl functionality into a hydroxy group) requires only one molar equivalent of hydrogen. Choice C is also unsupported and implausible: If you could determine the identity of the end product and if you knew the degree of unsaturation of the unknown, then there is no reason why quantitative measurement of hydrogen absorbed is not possible. Choice D is incorrect because hydrogenation reactions are by definition reduction reactions. 12. A The question asks you to identify the one compound obtained in Test 3 that is not a dicarboxylic acid. The nature and results of Test 3 are therefore worth revisiting in close detail. The unknown, you are told, is oxidized in KMnO4. If your outside knowledge is strong, you may recall that except in cold, dilute conditions, KMnO4 will cleave an carbon-carbon double and triple bond and generally converting each carbon on the end into a carboxylic acid functionality (after acid work-up). Even if you have forgotten the exact outcome, the results given to you in the passage, together with an awareness of the possible structures of the unknown, should lead you to the same conclusion. For example, if the unknown had the first of the two possible structures, the three dicarboxylic acids would have been obtained in the following manner:

H2C CH(CH2)4C CC C(CH2)7COOH

HOOC(CH2)4COOH HOOCCOOH HOOC(CH2)7COOH If you have gotten this far in the question, you will now realize that the question has become: What happens to the terminal methylene group upon oxidation? That product is the one species that is not a dicarboxylic acid, whose identity you are trying to identify. Again, if your outside

34

knowledge is strong, you can arrive at carbon dioxide as the correct answer immediately. But even if you are not aware of this “exception,” you may still have been able to eliminate the other choices. Your first intuition may be to treat the methylene group no differently: The carbon atom would become a carboxylic acid functional group:

H2C[O]

HCOOH

However, formic acid, HCOOH, is not among the answer choices. Choice B can be rejected because potassium permanganate cannot generate an extra carbon. The number of carbon atoms in each species is determined by the structure of the original compound (where the double and triple bonds occur). Choice C may be harder to reject, but you may surmise that if the reagent is capable of oxidizing all the other functionalities “all the way” to a carboxylic acid (with three C-O) bonds, it is strange to halt the oxidation of the terminal methylene group at the alcohol stage. In other words, you would expect the species to be more oxidized. Choice D can be rejected because it is not an oxidation product but a reduction product. After you have tentatively rejected the other answer choices, you may be able to justify to yourself why carbon dioxide is a reasonable choice. It is as oxidized as one can get (consistent with the production of dicarboxylic acids), and stoichiometrically it works out since the hydrogen can escape as a gas:

H2C[O]

CO2 + H2

13. D The important point in reading the question stem for this question is not to read too much into it. The question is asking you to identify the one piece of information that can be deduced based on Test 1, but cannot be deduced based on Test 5. The results of Test 5 need not contradict the information, and indeed, it should not, since the tests are conducted on the same compound. So, the results of Test 5 should be unrelated to the property established by Test 1.

Test 1 indicates that upon hydrogenation, the unknown was converted into stearic acid, whose condensed structural formula is then given. Test 5 indicates that the unknown only has one –COOH group. The results of the two tests certainly do not contradict each other (and as mentioned above, they had better not!). What remains to be done now is to evaluate each answer choice in turn and see if it satisfies two criteria: (a) it is a natural conclusion based on the fact that it yields stearic acid upon hydrogenation, and (b) it is consistent with, but not a necessary consequence of, the molecule having one –COOH group. This question, as you can see, calls for quite sophisticated reasoning both based on the passage and based on your science knowledge. Choice A states that the unknown contains unsaturated carbon atoms. This is an incorrect choice because it is necessarily true given the result of Test 5: The carbonyl carbon of a carboxylic acid functional group is unsaturated. Choice B states that the unknown is not capable of cis/trans isomerism. Cis/trans isomerism can occur if the unknown contains a C=C bond, AND if each carbon of the double bond is attached two different groups. Test 1 tells you that the compound is unsaturated, but by itself it doesn’t tell you if the compound contains C=C bonds (as opposed to the unsaturation arising solely from triple bonds), nor does it tell you what groups are attached to the double-bonded carbon atoms if they exist. Choice B is therefore not inferable from Test 1. Choice C states that the unknown contains an acidic hydrogen. This is inferable from BOTH tests, so cannot be the correct choice. Choice D, by the process of elimination, must be correct, but you should still make sure you can convince yourself why it is true. From the condensed structural formula of stearic acid, you can tell that the alkyl portion is straight-chained (seventeen carbons in a row followed by a –COOH group). The unknown must therefore also not be branched, since hydrogenation by itself does not rearrange the carbon skeleton. The fact that the unknown is unbranched is therefore inferable from Test 1. However, it is not inferable from Test 5 which has nothing to do with the alkyl portion. Just because it contains one

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–COOH group reveals nothing about what the alkyl portion might be like. 14. C The question asks you to determine the one property that will set stearic acid and the unknown apart. From the answer choices, you can tell that the property can be based on either IR spectrum, NMR spectrum, or boiling point. The correct answer must therefore apply to one of the two compounds. Incorrect answer choices will be true of both compounds or of neither compound. What is needed from the passage is an awareness of the structural differences between stearic acid and the unknown. The condensed structural formula of stearic acid and the possible structures for the unknown are given in the passage. The unknown, unlike stearic acid, has carbon-carbon double and triple bonds. You would suspect that the distinguishing feature you are after hinges on the presence (or absence) of these unsaturated bonds. With this in mind, you can evaluate the answer choices. Your outside knowledge of spectroscopy or physical properties will be called for. Choice A refers to a singlet in the proton NMR spectrum between 10 and 12 ppm. This chemical shift corresponds to the proton of a carboxylic acid group, which is present in both compounds. This feature therefore cannot be used to distinguish between the two. Choice B refers to a strong and broad IR absorption from 2500 to 3400 cm−1. This band is characteristic of an O-H bond, which again is present on both compounds. So this too cannot be used to distinguish between them. Choice C refers to the absence of any IR stretches around 2200 cm−1. This is

where a C≡C bond is expected to absorb, so its absence would signify that the compound is stearic acid and not the unknown, which you know to contain triple bonds. Choice C is therefore the correct answer. Finally, choice D can be rejected because both compounds, being carboxylic acids, will have higher boiling points than the corresponding alcohol. In fact, the alcohol listed in the answer choice has one fewer

carbon atom, making its boiling point even lower. 15. C You are asked to identify what type of isomers the two proposed structures are (if they are isomers at all). This question is therefore a classic example of a question requiring data from the passage: You need only examine the structures given. However, in this case, it helps to keep in mind that since they are proposed structures for the same compound, they have the same molecular formula and are therefore isomers of each other. What needs to be done is apply the definition of the different isomer types. Geometric isomers are isomers arising from the arrangement of groups around a double bond. While the structures do both contain a C=C bond, the fact that one carbon is attached to two hydrogen atoms (two identical groups) precludes the possibility of geometric isomers. So choice A is incorrect. Choice B is also incorrect because diastereomers are either geometric isomers (rejected as a possibility) or stereoisomers with multiple chiral centers that are not enantiomers of each other. The structures proposed do not contain any chiral center, so they cannot be diastereomers. Choice C is correct: They differ in the connectivity of the atoms. In one case, the carboxyl group is attached to a chain of seven saturated carbon atoms; in the other case, the carboxyl group is directly bonded to an sp hybridized carbon. Choice D is incorrect because, as discussed above, the two structures have the same molecular formula and are therefore isomers of each other.

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In preparing for the MCAT, I should …

1. Molecular Structure: Hybridization, Bonding, Polarity, Resonance, Physical Properties

• be able to identify the hybridization of carbon atoms in different molecules

• know the implications of hybridization on molecular geometry, bond order, etc.

• be familiar with the characteristics of sigma vs. pi bonds

• be able to assign formal charges to atoms in a molecule

• be able to identify polar bonds and polar molecules

• be able to draw resonance structures and assess their stability (and hence contribution)

• know how inductive effects and resonance stabilize charge

• be able to identify conjugated systems, and know how conjugation affects the stability of a compound

• know the implications of resonance on hybridization and geometry

• be able to identify compounds capable of hydrogen bonding

• know the trend of physical properties in hydrocarbons

• know the common reactions of hydrocarbons, and how structural factors affect them (strain, stability of radicals, etc.)

• be able to classify a carbon atom as methyl, primary, secondary, tertiary, or quaternary, and know the implications on reactivity

• know how the stability of a molecule can be quantified in terms of heats of combustion, heats of hydrogenation, etc.

• know how polarity, hydrogen bonding, molecular weight, etc., affect physical properties of compounds

2. Functional Groups • be able to identify the functional

groups present in a molecule • know how to interpret condensed

structural formulas • be aware of major characteristics of

functional groups (hydrogen bonding, acidity, solubility, etc.)

• know the fundamental rules of nomenclature for each family of compounds

3. Acidity and Basicity • be able to identify the most acidic

proton in an organic molecule • know the factors affecting acidity and

basicity • be able to account for the acidity of

carboxylic acids, phenols, etc. • be able to predict (or rationalize)

trends in acidity and basicity 4. Isomerism • know the definitions of the different

isomers • be able to identify the isomeric

relationship between two compounds • be able to identify the number of

stereogenic centers in an organic molecule

• be able to classify a compound as chiral or achiral

• know the origins of optical activity (specific rotation)

• be able to identify meso compounds

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• be able to assign the absolute configuration of a chiral center

• be able to interpret different representations of compounds: Fischer and Newman projections, etc.

• know the conformational dynamics of ethane, butane, substituted cyclohexanes, etc.

• know the differences/similarities in chemical and physical properties between two isomers

• be able to identify means of separating different isomer types

5. Reactions and Syntheses: General

Principles

• be able to apply a reaction scheme given in a passage to a new set of reactants

• be able to keep track of how bonds are formed and broken in a given mechanism by “pushing electrons,” be able to apply a given, but previously unknown, mechanism to specific compounds

• be able to visualize the mechanism in three-dimensional space to deduce the stereochemistry of the product

• be able to deduce the structure of transition states and intermediates in a given reaction

• for a multistep synthesis, be able to recognize what happens at each step (what the step accomplishes)

• be able to predict where an isotopic label ends up if one is employed in the reaction

6. SN1, SN2, E1, E2 • know the mechanism of each

reaction • know the rate law of each reaction • be able to identify reaction profiles of

single-step vs. multistep reactions • be able to predict (in simple

situations) which reaction is likely to predominate

• be aware of the synthetic uses of the reactions

• know the factors affecting the rate of each reaction

• be able to predict the relative stability of carbocation intermediates

• be able to identify good nucleophiles and good leaving groups

• know the stereochemical implications of SN1 vs. SN2

• be aware of some specific examples: dehydration, solvolysis, Williamson ether synthesis, etc.

• be able to apply Zaitsev’s rule to determine the most stable elimination product

7. Addition Reactions • know the difference between

Markovnikov and anti-Markovnikov addition

• be able to predict the addition product of each

• know the fundamental concepts of radical chemistry

• know the mechanism of anti addition via a cyclic ion intermediate; be able to predict the product (and its most stable conformation)

• know the stereochemistry of catalytic hydrogenation

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8. Aromaticity and EAS

• be able to identify aromatic compounds by applying Hückel’s rule, etc.

• be aware of the special stability of aromatic compounds

• be familiar with electrophilic aromatic substitution reactions: common reagents and catalysts

• know the rate and orientation effects of different substituents; be able to classify substituents as ortho/para- or meta-directing, activating or deactivating

9. Carbonyl Chemistry

• be able to identify different families

of compounds containing the carbonyl group

• be aware of the characteristics of the carbonyl group and how they affect its chemical reactivity

• know the nomenclature of simple carbonyl compounds

• know the relative position of different carbonyl compounds in the oxidation-reduction scheme

• know the common oxidizing and reducing agents used on (or used to synthesize) carbonyl compounds

• know the mechanism of addition reactions of aldehydes and ketones: (hemi)acetal formation

• know the mechanism of nucleophilic acyl substitution reactions in both acidic and basic conditions, and how these reactions can be used to synthesize carboxylic acid derivatives

• know the terms for specific examples: esterification, hydrolysis, saponification

• know the relative reactivity of different carboxylic acid derivatives

• know how long-chain carboxylic acids form micelles in aqueous solution

• be able to recognize the principles of carbonyl chemistry in the context of the reactions of fatty acids and triacylglycerols

• know the relative order of reactivity of different carboxylic acid derivatives

• be aware of the acidity of alpha protons, and how abstraction of an alpha proton generates a carbonyl-containing nucleophile (leading to aldol condensation)

• be aware of the thermodynamics and the mechanisms of keto-enol tautomerism

10. Amines • know the rules for naming simple

amines • be able to classify an amine as

primary, secondary or tertiary • know the stereochemical and

physical properties of amines • know how amines can form

quaternary salts • know the factors affecting the

basicity of amines, including aromatic amines

• know the key ways to synthesize amines

• know how amines act as nucleophiles in carbonyl reactions

11. Carbohydrate Chemistry • be able to classify monosaccharides

as aldoses or ketoses • be able to identify stereogenic

centers in (straight-chain) monosaccharides

• be able to classify a monosaccharide as a D- or L-sugar

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• be able to characterize the isomeric relationship between monosaccharides

• be aware of the cyclic form of hexoses in aqueous solution

• be able to apply the principles of conformational dynamics to the cyclic form of hexoses

• know what epimers and anomers are • be aware of some of the common

reactions of monosaccharides • know how monosaccharides form

disaccharides, and how the linkage can be broken

12. Protein Chemistry

• be aware of the amphoteric nature of

amino acids • be aware of the presence of a chiral

center in an amino acid • be able to predict the predominant

form of an amino acid in acidic and basic environments

• be able to classify amino acids according to their side chains

• know the characteristics of a peptide bond

• know the different levels of protein structure

• know the special role played by certain amino acids (proline, cystine) in the structure of a protein

• know the importance of the isoelectric point of an amino acid or a protein

13. Separation and Purification

Techniques

• be aware that the presence of impurities is often revealed by broadening and depression of melting point • know the principles at work in

common techniques used to separate and purify compounds

• be able to select the most efficient technique in separating two given compounds

• be able to apply acid-base principles to extraction based on solubility

• be able to apply principles of intermolecular forces to chromatographic techniques

14. Spectroscopy • know the principles at work in

different types of spectroscopy • be able to identify the spectroscopic

technique that would be most effective in distinguishing between given compounds

• know the IR frequency of common functional groups

• know how to interpret a proton NMR spectrum: splitting, chemical shift, etc.

• know the chemical shift of common proton types

• be able to predict general features of the proton NMR spectrum of an organic compound

• be able to identify a compound based on a combination of IR and NMR (and perhaps UV-vis) data

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