Strengthen your MCAT Organic Chemistry skills using curated practice problems and expert guidance on key mechanisms and reaction types.
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Introduction
Organic chemistry is often the most difficult class that premeds must defeat on their way to medical school. But, you’ve now finished organic chemistry and are beginning to study for the MCAT. How much of this dreaded subject will come back to haunt you on the most important exam you’ll take as a premed?
Thankfully, organic chemistry won’t be a large portion of your exam. In fact, the MCAT will ask you only about 6 to 12 questions on organic chemistry out of 230 total questions! As a result, your study schedule should include a balance between content and practice. For organic chemistry questions, you’ll be able to increase your performance significantly by understanding how the test writers at the AAMC like to ask questions.
So, enjoy the practice passages we’ve designed for you here! We’ll test your chemistry knowledge using three MCAT-style passages and five standalone questions. Each explanation for the passage-based questions will have suggestions for what you should review if you miss a question. Good luck!
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MCAT Organic Chemistry Practice Passage #1
In a novel approach, researchers eluted [18F]F− from small (10–12 mg) anion-exchange cartridges with solutions of tetraethylammonium bicarbonate, perchlorate, or tosylate in polar aprotic solvents containing 10–50% water. After dilution with additional aprotic solvent, these solutions were used directly in nucleophilic aromatic and aliphatic 18F-fluorination reactions, which prevents the need for azeotropic drying.
As proof-of-principle, NAMB chemistry was utilized for the synthesis of the dopamine D2/D3 antagonist [18F]fallypride. The standard synthesis of [18F]fallypride is known to be base-sensitive, because of the tendency of the tosyl-fallypride precursor 6 to undergo hydrolysis and elimination side reactions, making this compound a good candidate with which to evaluate this minimally basic synthesis.
Figure 1 NAMB chemistry synthesis of [18F]fallypride.
1. In a novel experiment, researchers find a new [18F]fluoride that can carry out NAMB chemistry upon electron capture. Which of the following best describes the element after positron emission?
A) [18F]fluoride
B) [17F]fluoride
C) [18O]oxygen
D) [17Ne]neon
A) Protic solvents will hydrogen bond with nucleophiles, thereby decreasing nucleophilicity.
B) Protic solvents will hydrogen bond with electrophiles, thereby decreasing electrophilicity.
C) Protic solvents will form Van der Waals interactions with aromatic rings, thereby decreasing nucleophilicity.
D) Protic solvent will form dipole-dipole interactions with electrophiles, thereby decreasing electrophilicity.
A) Ether
B) Ester
C) Amide
D) Amine
A) Anion-exchange because the positive charge on 6 will bind the negatively charged resin.
B) Anion-exchange because the negative charge on 6 will bind the positively charged resin.
C) Cation-exchange because the positive charge on 6 will bind the negatively charged resin.
D) Cation-exchange because the negative charge on 6 will bind the positively charged resin.
Answers and Explanations for MCAT Organic Chemistry Practice Passage #1
1. Answer choice C is correct. The balanced equation would appear as follows:
To solve radioactive decay problems, you must ensure that the top line (18 + 0 = 18) and the bottom line (8 + 1 = 9) are equal to one another (choice C is correct; choices A, B, and D are incorrect).
Review radioactive decay.
2. Answer choice A is correct. Protic solvents hydrogen bond with nucleophiles, thereby decreasing nucleophilicity and preventing the reaction of interest from occurring (choice A is correct; choice C is incorrect). Electrophilicity is not affected by protic solvents in this case (choices B and D are incorrect).
Review nucleophilicity and electrophilicity.
3. Answer choice C is correct. An amide group can undergo hydrolysis through nucleophilic acyl substitution (choice C is correct). An ether (R-OR’) is present on the molecule, but it is unlikely to undergo hydrolysis (choice A is incorrect). An amine is also unlikely to undergo hydrolysis (choice D is incorrect). Though an ester group will be more susceptible to hydrolysis than an amide, this functional group is not present in Compound 6 (choice B is incorrect).
Review functional groups and nucleophilic acyl substitution.
4. Answer choice C is correct. Protonation of the amide nitrogen would lead to an additional positive charge (choices B and D are incorrect). An anion-exchange column binds negatively charged molecules, meaning its resin (which is in the column itself) will be positive (choice A is incorrect). The cation-exchange column binds positively charged molecules, and its resin will therefore be negative (choice C is correct).
Review protonation and column chromatography.
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MCAT Organic Chemistry Practice Passage #2
In a new study, researchers aim to synthesize diindolylmethane (DIM) derivatives based-thiadiazole as a new class of urease inhibitors. Diindolylmethane is a natural product alkaloid used extensively in medicinal chemistry.
Figure 1 DIM-bearing thiadiazole analog synthesis.
Compound 8 was the most potent inhibitor amongst all derivatives (IC50 = 0.50 ± 0.01 µM). In the next step, researchers used x-ray crystallography to determine structure-activity relationships for the compounds. Researchers discovered that compound 8 forms a pi-stacking interaction with a conserved Y87 residue within urease. Mutation of Y87 to alanine decreases compound 8 potency by 3.7-fold. The key binding interactions of most active compounds with urease were confirmed through molecular docking studies.
1. Which of the following is NOT a strong acid?
A) Hydrochloric acid
B) Acetic acid
C) Hydrobromic acid
D) Sulfuric acid
2. During the first step of the reaction in Figure 1, which of the following functional groups present on the reactants is NOT present in the product?
A) Aromatic ring
B) Aldehyde
C) Carboxylic acid
D) Amine
3. Which of the following urease mutations at position 87 is MOST likely to maintain pi-stacking interactions with compound 8?
A) Alanine
B) Proline
C) Threonine
D) Phenylalanine
4. Researchers want to decrease the rate of product formation from the second step of the reaction in Figure 1. Which of the following would NOT decrease the rate of compound II formation according to Le Chatelier’s principle?
A) Decrease the concentration of the bromine-containing initial reactant
B) Decrease the concentration of the aldehyde- and carboxylic acid-containing initial reactant
C) Decrease the concentration of compound II
D) Decrease the concentration of compound I
Answers and Explanations for MCAT Organic Chemistry Practice Passage #2
1. Answer choice B is correct. Acetic acid is a weak acid while the others are strong acids (choice B is correct; choices A, C, and D are incorrect).
Review strong and weak acids.
2. Answer choice B is correct. The aldehyde (CHO) present in the reactant is not present in the product of the first step of the reaction in Figure 1 (choice B is correct). The aromatic rings, carboxylic acid, and amines are maintained (choices A, C, and D are correct).
Review functional groups and practice interpreting Figure 1.
3. Answer choice D is correct. Pi-stacking interactions are formed between aromatic rings. There are several aromatic rings in compound 8. In addition, the original amino acid at position 87 was tyrosine, which is an aromatic amino acid. Of the answer choices, only phenylalanine is aromatic and could maintain the pi-stacking interaction with the compound (choice D is correct; choices A, B, and C are incorrect).
Review amino acid properties and pi-stacking interactions.
4. Answer choice C is correct. Decreasing the concentration of compound II would lead to an increase in the forward reaction and formation of compound II via Le Chatelier’s principle (choice C is correct). The other actions would decrease the rate of formation of compound II (choices A, B, and D are incorrect).
Review Le Chatelier’s principle.
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MCAT Organic Chemistry Practice Passage #3
Researchers have discovered that the concise construction of diene scaffolds is useful in the synthesis of polyenes. Many diene building blocks have been developed based on Suzuki, Still and Hiyama couplings. In a new study, the commercially available compound dienedioic acid was used as a diene building block. Broad substrate scope, good functional group tolerance, and late-stage derivatization of complex drug molecules were achieved.
The researchers found that different moieties could be introduced to both sides. For example, piperine and the methyl ester of azoxymycin C were prepared in three steps. Additionally, one product showed promising anticancer activities in leukemia K562 and MV-4-11 cells. Further mechanistic studies indicated that the reaction proceeds through a Heck-decarboxylate coupling procedure, and the carboxyl group acts as a directing group to promote the reaction and control regioselectivity.
Three possible reaction pathways could be involved in the reaction: (1) “Suzuki type”, in which decarboxylation occurred first to form C–M (M = Cu, Ag, and Pd) bond followed by transmetallation and reductive elimination similar to Suzuki coupling (Fig. 1, path a); (2) “Decarboxylation-Heck type”, in which decarboxylation occurred first to form alkene followed by Heck coupling (Fig. 1, path b); and (3) “Heck-decarboxylation type”, in which Heck coupling took place first followed by decarboxylation, and the carboxylic acid played as a directing group to control the α-selectivity (Fig. 1, path c). These three different possible reaction mechanisms are shown in Figure 1.
Figure 1 Three possible reaction pathways.
1. Which of the following stereochemical designations for compound II is correct?
A) R
B) S
C) E
D) Z
2. What best describes the reaction occurring in the formation of compound II from compound I through “path a” (Figure 1)?
A) Oxidation
B) Reduction
C) Carboxylation
D) Circularization
3. In a new experiment, researchers use radioactively labeled propanol to carry out nucleophilic acyl substitution on Compound 1. Which of the following carbon positions on Compound 1 will be closest to the radioactively labeled propanol?
A) Alpha
B) Beta
C) Gamma
D) Delta
Answers and Explanations for MCAT Organic Chemistry Practice Passage #3
1. Answer choice C is correct. A chiral carbon is not present, so R and S cannot be assigned (choices A and B are incorrect). E indicates trans stereochemistry across a double bond, and both double bonds have an E configuration in compound II (choice C is correct). Z indicates cis stereochemistry across a double bond, which is not observed in compound II (choice D is incorrect).
Review double bond stereochemistry.
2. Answer choice B is correct. Three bonds to oxygen are removed in order to derive compound II from compound I, and decreasing the number of bonds to oxygen is known as reduction (choice B is correct). Oxidation would increase the number of bonds to oxygen (choice A is incorrect). Compound II undergoes decarboxylation, not carboxylation (choice C is incorrect). Circularization does not occur (choice D is incorrect).
Review oxidation and reduction.
3. Answer choice A is correct. This question tests your knowledge of nucleophilic acyl substitution. The radioactively labeled propanol would replace the OH on Compound 1. As a result, the alpha carbon would be the closest (choice A is correct; choices B, C, and D are incorrect).
Review nucleophilic acyl substitution.
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MCAT Organic Chemistry Practice Questions (Standalone)
1. A researcher performs an organic reaction in which a benzene with an alcohol substituent turns into a benzene with an aldehyde substituent. Which of the following peaks may appear for the product in IR spectroscopy?
A) Broad, 3500 cm-1
B) Sharp, 3500 cm-1
C) Broad, 1700 cm-1
D) Sharp, 1700 cm-1
2. A molecule has two chiral centers. Which of the following pairs represents enantiomers?
A) R,R and R,S
B) R,S and R,R
C) R,S and S,R
D) S,S, and S,R
3. Which of the following molecule pairs can MOST likely be separated based on distinct boiling points?
A) R,R and R,R
B) R,S and S,R
C) S,R and R,S
D) S,S, and S,R
A) Reactant
B) Product
C) They will both have the same Rf
D) Thin layer chromatography cannot be used to measure the progress of this specific reaction
A) 28P
B) 30P
C) 28Al
D) 30Al
1. Answer choice D is correct. In the product, a carbonyl group is forming (C=O), and this is seen on an IR spec at 1700 cm-1 as a sharp peak (choice D is correct). A broad peak around 3500 cm-1 is characteristic of an OH, which we would see for the reactant, not the product (choice A is incorrect).
2. Answer choice C is correct. Enantiomers are non-superimposable mirror images, and they must have the reverse stereochemical designation at each chiral center (choice C is correct). Choices A, B, and D represent diastereomers.
3. Answer choice D is correct. Diastereomers have differing stereochemical designations that are not completely flipped (examples of completely flipped are seen in answer choices B and C, which are enantiomers). Diastereomers have different chemical reactivities and properties (such as boiling point), which allows them to be separated based on those properties (choice D is correct). Choice A shows the same molecules, which would have the same boiling points (choice A is incorrect).
5. Answer choice C is correct. An alpha particle is written as 4He, and by losing it, the mass number of 32P decreases from 32 to 28 (top number). Then, the atomic number of 15 decreases by 2 to 13 (implied bottom number), which is the atomic number for aluminum (choice C is correct; choices A, B, and D are incorrect).
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