Synthesis of anti-tumor drug derivatives and NMR analysis
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Synthesis of anti-tumor drug derivatives and NMR analysis
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[ID] => 560280
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[post_date] => 2025-01-12 09:44:15
[post_date_gmt] => 2025-01-12 14:44:15
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Combretastatin A-4 (CA-4) was first extracted from the tree combretum caffrum and has proven to be a potent anti-tumor drug. It shares most of its biosynthetic pathway with colchicine, as well as the capacity to inhibit tubulin polymerization. Colchicine inhibits microtubule polymerization by binding to the β-subunit of tubulin. While cancer cells are significantly more vulnerable to colchicine poisoning than normal cells, the overall cytotoxicity makes the therapeutic value of colchicine very limited.
CA-4 is one of the first promising agents shown to have both anti-tumor and anti-vascular properties. By binding to a site different from the colchicine site, CA-4 not only inhibits microtubule assembly but also causes drastic shape change in endothelial cells. By exploiting the differences between tumor and normal tissue endothelium, CA-4 is able to block pre-existing blood vessels and cause extensive cell death in the tumor core. In hopes of finding a novel derivative with greater bioactivity, a research group decides to undergo the design and synthesis of several CA-4 derivatives. The synthetic route is depicted below:
Figure 1. Synthesis of Combretastatin A-4 (CA-4) and derivatives through Perkins Condensation
An alternative route for synthesis is the Wittig reaction, which is a reaction between an aldehyde or ketone with a triphenyl phosphonium ylide (Wittig reagent) to give an alkene and triphenylphosphine oxide:
Figure 2. Wittig Reaction
During the course of the experiment, one of the collection fractions was mislabeled, and the research assistant performed spectral analysis to determine the identity of the sample.
Figure 3. 1H-NMR spectrum of CA-4 derivative in mislabeled sample
Source: Adapted from a paper by M. Ma, et al. Copyright 2013 by Ma et al.; licensee Chemistry Central Ltd.
[post_title] => Synthesis of anti-tumor drug derivatives and NMR analysis
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[quiz_unique_key] => 578908434
[question] => Which of the Combretastatin A-4 derivatives (Compounds 5-9) is represented by the ¹H-NMR spectrum above in Figure 3?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
There are characteristic peaks that you are expected to know for the MCAT. Know that aryl H’s have peaks around 7 ppm. In this NMR, the 3 peaks that appear around 7 ppm correspond to aryl-H’s.
sp²-hydrogens have peaks above 5 ppm. The alkenyl-H’s on the bridge are sp², so the last peak above 5 correspond to those protons.
sp³-hydrogens have peaks between 0 and 5 ppm. The methoxy-H’s appear around 4 ppm since they are sp³ protons near an electron-withdrawing O, hence deshielded. The peaks between 3.5 and 4 ppm correspond to the methoxy-H’s.
There is a small multiplet just under 5 ppm which correspond to the first carbon off the carbonic ester group. The multiplet indicates that there is a proton attached to a C whose neighbors have several H’s each.
The height of the 3 peaks from 3.5-4.0 ppm indicates many protons in such environs, but the height of the multiplet indicates perhaps only one proton. There is one strong signal at around 1.4 ppm, and the height indicates many protons in this chemical environment.
Only isopropyl would have the multiplet as well as a doublet upfield since the 6 protons on the primary carbons are symmetrical around the secondary carbon. Here is a figure with all the hydrogens labeled with their corresponding peaks:
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[answers] => Array
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[0] => Array
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[each_answer] => A. Compound 8
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[1] => Array
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[each_answer] => B. Compound 9
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[2] => Array
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[each_answer] => C. Compound 7
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[3] => Array
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[each_answer] => D. Compound 6
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => Another research group decides to use the Wittig Condensation instead of the Perkin Condensation to prepare Combretastatin A-4. What were the starting reagents for the research group using the Wittig condensation?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Based on Figure 2, for every Wittig reaction, there are two reactants involved: the ylide and an aldehyde or a ketone.
An ylide is a neutral dipolar molecule containing a formally negatively charged atom (usually a carbanion) directly attached to a heteroatom with a formal positive charge (usually N, P or S). Both atoms have a full octet of electrons.
To determine whether it is an aldehyde or ketone, look at the product CA-4 and what is around its double bond. Since the double bond is cis and has two H’s on the same side, the starting reagent must be an aldehyde. Therefore, this structure would not be correct:
For the 3′-hydroxy-4′-methoxy ring structure, the position of the aldehyde or the ylide should be at 1′, which is para to the methoxy group and meta to the hydroxy group. Therefore, this structure would not be correct:
When the triphenyl group leaves the Wittig reagent, a double bond will remain in its place joining it with the aldehyde. The group must be in the benzylic and not the phenyl position. Therefore, these structures would not be correct:
Furthermore, the molecule on the left is not a ylide, so it cannot participate in the Wittig.
To produce Combretastatin A4, the following reactants undergo a Wittig reaction:
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[answers] => Array
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[0] => Array
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[each_answer] => A.
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[1] => Array
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[each_answer] => B.
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[2] => Array
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[each_answer] => C.
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[3] => Array
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[each_answer] => D.
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => The Perkin condensation is a reaction to convert an aromatic aldehyde and anhydride to an α,β-unsaturated carboxylic acid using sodium acetate, a base, and an acid work-up. Which of the following statements explains the use of HCl in the Perkin condensation in Figure 1?
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[answer] => 1
[description] => Reason for the Correct Answer:
The purpose of the sodium hydroxide is to hydrolyze the acetate group on the C-3′ position through nucleophilic attack of the carbonyl.
When the acetate leaves, it is under basic conditions. So the HCl could not serve to protonate the acetate functional group to increase its lability or to become a leaving group.
Electrophilic aromatic substitution usually involves the hydrogen atom of an aromatic ring being replaced as a result of an electrophilic attack on the aromatic ring. This is not an example of EAS.
What remains after hydrolysis is the alkali salt of the compound, the alkoxide. The desired compound will remain in solution with the negative charge on the O.
To precipitate the alkoxide, it must be protonated and turned into a neutral compound. Compound 4 is a yellow solid.
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[answers] => Array
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[0] => Array
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[each_answer] => A. To protonate the alkoxide in order to precipitate it out of solution
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[1] => Array
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[each_answer] => B. To acidify the sample before using copper powder in quinoline for decarboxylation
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[2] => Array
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[each_answer] => C. To protonate the acetate functional group to increase its lability
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[each_answer] => D. To catalyze the electrophilic aromatic substitution of the acetate group
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[quiz_unique_key] => 2377279144
[question] => The Perkin Condensation of an anhydride and aromatic aldehyde shares a similar set of reaction steps with the aldol condensation. Which of the following statements accurately describes a step in the aldol condensation?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
The first step of aldol condensation involves a strong base like hydroxide abstracting a proton from the α-carbon (not the β-carbon) of a carbonyl compound (aldehyde or ketone) to form the enolate.
The second step of aldol condensation involves the enolate attacking the aldehyde or ketone through a nucleophilic acyl addition mechanism (not substitution). Only carboxylic acid and its derivative can undergo nucleophilic acyl substitution.
What is formed as a result of the nucleophilic attack is the alkoxide ion previously the carbonyl. Protonation of the alkoxide will form the aldol product (aldehyde and alcohol).
The aldol condensation involves two reaction series, the aldol addition reaction and the condensation reaction. Dehydration occurs through an elimination (technically, E1cB) mechanism to form an α, β-unsaturated aldehyde (enal) or ketone (enone), not the aldol.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Protonation of the alkoxide formed from nucleophilic attack is the aldol addition product.
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[1] => Array
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[each_answer] => B. Dehydration occurs through a E2 mechanism to form the aldol.
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[2] => Array
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[each_answer] => C. The enolate attacks the carbonyl compound through a nucleophilic acyl substitution mechanism.
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[3] => Array
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[each_answer] => D. A strong base abstracts a proton from the β-carbon of a carbonyl compound to form the enolate.
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => In Compound 3 of Figure 1, how many sp² hybridized carbons are in the molecule?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
Compound 3 has two benzene rings with 6 sp² carbons each, totaling 12.
Compound 3 has the alkenyl bridge between the rings with 2 sp² carbons, totaling 14.
Compound 3 has 1 more sp² carbon in the carboxylic acid group, totaling 15.
Compound 3 has 1 final sp² carbon in the acetate (AcO) group, totaling 16.
Here is a diagram indicating all the sp² carbons:
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[answers] => Array
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[0] => Array
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[each_answer] => A. 5
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[1] => Array
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[each_answer] => B. 16
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[2] => Array
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[each_answer] => C. 15
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[3] => Array
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[each_answer] => D. 14
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[560280|1] => C
[560280|2] => D
[560280|3] => A
[560280|4] => A
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