1. D The maximum number of stereoisomers will be 2n, where, in this case n (the number of chiral centers) is 6. Therefore, 26 = 64.
2. A According to the n + 1 splitting rule, where n = # of non-equivalent neighboring hydrogens, doublets in the NMR spectrum represent hydrogens with one hydrogen neighbor. The C-15 methyl group and the single vinyl hydrogen on C-1 fit this requirement. The methyl group is shifted slightly downfield by the Cl atom on C-14, but should appear between 2-3 ppm. Vinyl hydrogens typically appear around 5-6 ppm, but this one will be shifted even farther downfield by the Cl atom on C-1. The ratio of upfield to down-field is therefore 3:1.
3. B The –SO3– group shown on malhamensilipin A suggests that the –SO3H group can act as a Brønsted acid and donate a proton. This leaves a stable product (due to resonance). In fact, the –SO3H group on the molecule is reminiscent of H2SO4, a strong Brønsted acid. Choice B is the correct answer. The molecule doesn’t have any easily donated electron pairs, and therefore isn’t a strong Lewis base or Brønsted base (eliminate choices A and D). Also, it has no groups that are easily oxidized, so it cannot be a reducing
agent (eliminate choice C).
4. D The reactant contains no stereocenters, but two new ones are introduced in the reaction. The maximum number of stereoisomers any compound with two stereocenters can have is four (22). All four options are not created, however, due to the nature of the reaction and the symmetry of the reactant. The syn addition of two Cl atoms to the double bond in the reactant will lead to the following meso product, regardless of the face of the double bond that reacts:
Cl Cl Cl Cl
Since the product is meso, it has no enantiomer, and therefore we expect only one compound from the reaction.
5. B In the alkylation reaction, both the hydroxyl and the sulfoxide anion will be methylated. This will de-crease the polarity of the compound (eliminate choices A and C). In reverse phase HPLC the stationary phase is nonpolar, and opposite flash column chromatography in which the stationary phase is polar.
Since the stationary phase is nonpolar, less polar compounds will interact more with the stationary phase thus having a longer retention time (eliminate choice D).
| 91
6. C C-13 and C-14 are the two left most chiral carbons in Figure 1. Numbering starts at the double-bond end of the molecule. Shown below are the individual stereocenters in question with the priority of each substituent labeled.
Cl Cl
3 1
2 HO3S
O
C–14
C–13
Cl
Cl Cl
Cl
3
1 2
HO3S
O Cl
Cl
For each stereocenter, the Cl atom gets first priority because it has the highest atomic number. For C-14, the bulk of the molecule gets second priority due to more substitution compared to the methyl group of C-15. The #4 priority hydrogen, not drawn, is already in the back, and connecting 1, 2, and 3 makes a clockwise circle. As such, C-14 is R (eliminate choices C and D). For assigning the second priority for C-13, the first point of difference in the substituents occurs with the Cl on C-14 and the O on C-12. Be-cause Cl has a higher atomic number, the shorter end of the molecule gets second priority. The #4 priority hydrogen for C-14 comes out of the page. Therefore, connecting the ordered substituents will give the opposite configuration. Connecting 1, 2, and 3 shown gives a clockwise rotation, so the actual rotation is counterclockwise, and as such, C-13 is S (eliminate choice B).
STRUCTURE AND STABILITY
Drill
MCAT COMPLETECHAPTER 34 PRACTICE PASSAGE
Gelsemine (Figure 1) has been widely studied due to its complex structure that includes a total of six rings, and several attempts have been made at its stereoselective synthesis.
O H
N
N O
Figure 1 Gelsemine Figure 1 Gelsemine
Part of one synthesis is shown in Figure 2, illustrating the successful addition of the fourth and fifth rings of the compound.
Figure 2 Partial synthesis of gelsemine (Atarashi, S.; et. al., J. Am. Chem. Soc., 1997, 119, 6226-6241.)
Two key steps along the synthetic route to the tetracyclic Compound 1 are shown below in Figure 3.
O
Figure 3 Early steps of gelsemine synthesis
1. Why is the ketone in Compound 3 protected during the first two steps of the synthesis?
A) The hemiacetal prevents the less reactive ketone from being deprotonated.
B) The acetal prevents Grignard addition to the ketone.
C) The acetal, being equatorial, provides extra stability.
D) The hemiacetal prevents the ketone from forming a tertiary alcohol during the Grignard addition.
2. If the conversion of Compound 1 to Compound 2 was monitored by IR spectroscopy, what would be observed as the reaction progressed?
A) The disappearance of all peaks in the region between 1650 cm–1 and 1750 cm–1.
B) The appearance of a moderately strong band near 2100 cm–1.
C) The elimination of one of the two strong peaks between 1650 cm–1 and 1750 cm–1.
D) The disappearance of a set of three moderately strong peaks between 1450 cm–1 and 1580 cm–1.
Drill
| 93 3. Why is no Grignard addition to the carbonyl of the amide
observed in Step 1 of Figure 1?
A) The carbonyl carbon in the amide is lesselectrophilic than that of the ester.
B) The carbonyl carbon in the amide is more electrophilic than that of the ester, but the leaving group (–NR2) is not as stable.
C) The carbonyl carbon of the ester is less electrophilic than that of the amide, but its leaving group (–OEt) is less stable.
D) Addition to the ester results in a release of steric strain.
4. The equilibrium depicted in the first step in Figure 3, in which a carbonyl is protected as an acetal, can be pushed in the forward direction via which of the following strategies?
A) Protecting the amido-carbonyl prior to treatment with the diol
B) Using a 1,3-propandiol with large, bulky substituents at the 2-position
C) The use of an alcohol solvent
D) Rigorously removing water from the system
5. How many chiral centers does gelsemine have?
A) 5 B) 6 C) 7 D) 8
CARBONYL CHEMISTRY
MCAT COMPLETE