1. B Hemiacetals and acetals have the general structures shown below. These structures illustrate that the functional group in the molecule is an acetal (eliminate choices A and D).
OR C
R R
OR
C
R R
OH
OH
acetal hemiacetal
The acetal has two bonds to the six-membered ring, one of which is axial and the other equatorial (elimi-nate choice C). Ketones are more reactive than esters; therefore, if the ketone had been present during the Grignard addition, the Grignard reagent would have added there instead.
2. C As Compound 1 is converted to Compound 2, one of the two carbonyl moieties is converted into an ether as a result of the addition of two phenyl rings. The C=O stretches of carbonyl compounds appear in the same area, but do not necessarily appear at the same exact wavenumbers. Seeing one of the two initial stretches in the carbonyl region between 1650 cm–1 and 1750 cm–1 disappear would be a clear indication that the reaction had progressed as shown. Elimination of all the stretches in this region would indicate the reaction of all C=O units in the molecule, which does not fit the reaction scheme (eliminate choice A). Stretches near 2100 cm–1 generally derive from carbon-carbon triple bonds, which play no role in the chemistry presented (eliminate choice B). Sets of multiple peaks in the region between 1450 cm–1 and 1580 cm–1 usually come from aromatic C=C bonds; therefore, the appearance of these peaks would be
expected, not their disappearance (eliminate choice D).
3. A Amides are less electrophilic at the carbonyl carbon because the two oxygen atoms of the ester induc-tively withdraw more electron density from the carbon than the nitrogen and oxygen atoms of the amide (choice A is correct; eliminate choice B). When comparing the possible leaving groups (NR2 or –OR), –OR is the better leaving group because oxygen is more electronegative than nitrogen, and it can better stabilize a negative charge (eliminate choice C). Addition to the amide might result in a release of strain from breaking the ring; however, there is no such strain present in the ester (eliminate choice D).
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4. D The formation of the acetal in the presence of H+ results in the elimination of water. This water can re-attack the acetal, causing a reversion to the carbonyl, giving rise to the equilibrium in the equation. If this water is removed from the system by a drying agent, the equilibrium will be pushed in the forward direc-tion. The amido-carbonyl does not need protection, as the amide is already of sufficient stability for the following steps, and its protection has no bearing on the ease of protection of the ketone in the equation (eliminate choice A). A bulky substituent at the 2-position of the 1,3-propandiol will, if anything, inhibit the attack of the diol, pushing the equilibrium to the reactants side of the equations (eliminate choice B).
The use of an alcohol solvent would greatly complicate the equilibrium, as there would be competing reac-tions involving the addition of solvent molecules, rather than the desired diol (eliminate choice C).
5. C Carbons with four different substituents are chiral. The chiral centers of gelsemine are marked in the structure below.
O
N O
* * *
* * *
*
HN CARBONYL CHEMISTRY
Drill
MCAT COMPLETECHAPTER 35 PRACTICE PASSAGE
The Maillard reaction is a nonenzymatic reaction between a reducing sugar and an amino acid, called glycation,
characterized in the browning of food. This multi-step reaction is sensitive to heat, pH, water activity, and the presence of other compounds. The reaction results in numerous products, all of which are responsible for the aroma, flavor, and texture of cooked foods.
When asparagine reacts with a reducing sugar in this way, a Schiff base forms. The Schiff base can be irreversibly isomerized and decarboxylated to form acrylamide, which is a known neurotoxin and potential animal carcinogen. The pathway for the formation of acrylamide is shown in Figure 1.
O N
N-linked glycosylation is the name given the endogenous version of the Maillard reaction. N-linked glycosylation represents an enzymatic co- or post-translational modification whereby an oligosaccharide chain is linked to the side-chain of an asparagine residue in a protein. The monosaccharides that compose these chains are often D-mannose and N-acetyl-D-glucosamine units assembled into various linear and branched structures. These oligosaccharide chains play important biological roles in intercellular interactions with carbohydrate binding proteins, called lectins.
Drill
| 97 1. Lectins are proteins capable of binding specific
carbohydrate moieties. Which of the following would NOT be recognized by a lectin?
A) Lipoproteins B) Glycoproteins C) Glycolipids
D) Soluble monosaccharides or disaccharides
2. Which of the following accurately describes D-mannose and N-acetyl-D-glucosamine?
A) D-Mannose is shown in the α configuration and N-acetyl-D-glucosamine is shown in the β configuration.
B) N-acetyl-D-glucosamine can form β-1,4 glycosidic linkages with D-mannose.
C) The N-acetyl group in N-acetyl-D-glucosamine can form a Schiff base.
D) Mannose is a structural isomer of glucose.
3. How many 1H NMR resonances will be displayed in the
4. Which of the following reagents would perform the isomerization shown in Step 3 of the reaction?
A) H2SO4 B) NaOCH2CH3 C) CH3MgBr D) H3CCOCH3
5. Which of the following statements regarding Schiff base formation is true?
A) Under strongly acidic conditions the rate of the reaction increases.
B) The removal of water from the reaction system decreases the yield of the Schiff base product.
C) The loss of water is the thermodynamic driving force in the reaction.
D) The reaction outcome is under kinetic control, favoring the formation of the Schiff base product.
6. Glycated hemoglobin (hemoglobin-A1c) is an adduct that forms when glucose reacts with an amino group in hemoglobin as in the first three steps shown in Figure 1.
Which of the following statements about hemoglobin-A1c is FALSE?
A) Glucose and mannose can react to form identical hemoglobin-A1c adducts.
B) A sustained period of hyperglycemia (elevated blood glucose concentration), as in diabetes mellitus, will result in elevated hemoglobin-A1c levels.
C) High-performance liquid chromatography is an effective method of separating hemoglobin-A1c from non-glycated hemoglobin.
D) The isomerization process that ultimately forms hemoglobin-A1c is reversible in an aqueous acidic environment.
7. A biologist researching a particular protein is interested in determining its degree of N-linked glycosylation and begins digesting the protein into fragments of various sizes using peptidases. Which chromatographic technique is most appropriate to separate the glycopeptide
fragments from the non-glycosylated peptide fragments in the digested mixture?
MCAT COMPLETE