Question 1

Ubiquinone (shown below) is an important molecule in the electron transport chain, where it receives electrons in complex I originating from NADH.  What is the structure of the product formed when electrons are added to ubiquinone?

Accepted Answer

A)   
B)   
C)   
D)

Question 2

Passage
High blood pressure, or hypertension, is a prevalent condition in the United States.  One in three adults has hypertension, and approximately 20% do not know that they have it.  If left untreated, high blood pressure can lead to major health issues, including stroke, heart attack, and kidney disease or failure.  A number of drugs can help control hypertension, including those shown in Table 1.  These drugs can be classified into different groups based on their mechanism of action.Table 1  Drugs Used to Treat High Blood Pressure
  Labetalol (Figure 1) exists as four stereoisomers and is administered as a mixture of all four.  This drug can be classified as a pseudohybrid drug because two of the isomers, (R,R) and (S,R), exhibit biological activity; the other two isomers, (S,S) and (R,S), are not active.  The (R,R) isomer acts as a nonselective β-adrenergic receptor blocker, and the (S,R) isomer acts as a selective α-adrenergic receptor blocker.
  Figure 1  Structure of (±)-labetalol and l-epinephrinel-epinephrine (Figure 1) is the agonist of the α- and β-adrenergic receptors.  The structural motifs that allow l-epinephrine to bind to the adrenergic receptors include an amine separated from an aromatic ring by two carbon units, a hydroxyl group at a chiral center beta to the amine, and two hydroxyl groups on the aromatic ring in the meta and para positions.  With a few structural modifications, the agonist l-epinephrine can be converted into the antagonist labetalol.  Substitution of a hydroxyl on the aromatic ring with an amide group and extension of the methyl group on the amine transforms l-epinephrine into labetalol.  These substitutions keep a majority of the structural motifs necessary for labetalol to mimic l-epinephrine and bind to the adrenergic receptors.
-The number of stereoisomers possible for a given molecule is 2n.  The variable n denotes the number of:

Accepted Answer

A) enantiomers. 
B) stereocenters. 
C) diastereomers. 
D) epimers.

Question 3

Passage
Fluorescent amino acids are useful for labeling proteins because they can be visualized spectroscopically, facilitating the study of protein structure and interactions.  The noncanonical amino acid dansylalanine, a fluorescent amino acid that is sensitive to the hydrophobicity of its environment, is a good candidate to be incorporated into proteins because of its small size and fluorescent capabilities.  Dansyl chloride and Compound 1 were used to synthesize dansylalanine, as shown in Reaction 1.  Dansylalanine absorbs maximally at 340 nm and emits at approximately 540 nm.
  Reaction 1  Synthesis of dansylalanineA leucyl-tRNA synthetase from Escherichia coli was engineered to aminoacylate bacterial tRNAs containing a CUA anticodon with dansylalanine.  After verifying that the tRNA/synthetase pair does not interact with endogenous amino acids or tRNAs, it was transfected into the yeast Saccharomyces cerevisiae along with a plasmid encoding a mutated human superoxide dismutase (hSOD).  Using this system, dansylalanine was genetically incorporated into hSOD in place of either Gln-16 or Trp-33, and expressed in S. cerevisiae.Position 16 is located on the surface of hSOD at the N-terminus of a β-strand, and position 33 lies in the center of a strand within a β-barrel.  To probe changes in the structure of hSOD, researchers exposed the purified protein to guanidinium chloride (GdmCl) to denature the protein at concentrations ranging 0-4.5 M.  The experiment was conducted in a 35 mM sodium phosphate buffer at pH 7.2.  The fluorescence of hSOD with dansylalanine at either position 16 or 33 was monitored at each GdmCl concentration, and the results of the unfolding experiment are shown in Figure 1.
  Figure 1  Plot of dansylalanine fluorescence at different concentrations of GdmCl
Adapted from Summerer D, Chen S, Wu N, Deiters A, Chin JW, Schultz PG. A genetically encoded fluorescent amino acid. Proc Natl Acad Sci USA. 2006.
-What is the charge of the N-terminus during the protein unfolding experiment?

Accepted Answer

A) +1, because the amine group is not protonated 
B) +1, because the amine group is in the conjugate acid form 
C) 0, because the amine group is protonated 
D) 0, because the amine group is in the conjugate base form

Question 4

Compound 1 is shown below.  All of the following are equivalent to Compound 1 EXCEPT:

Accepted Answer

A)   
B)   
C)   
D)

Question 5

Passage
The β-lactam scaffold is an important feature in a class of broad-spectrum antibiotics that includes the penicillins, cephalosporins, and monobactams.  These antibiotics are used to treat a variety of diseases caused by bacteria.  The penicillins work by inhibiting a step in the synthesis of peptidoglycan.  However, the development of antibiotic resistance in the form of enzymes such as β-lactamase is an ongoing problem.An aromatic ring linked to the β-lactam ring has been reported to participate in hydrophobic interactions with β-lactamase active sites, and an aromatic ring bonded to the nitrogen of the β-lactam has shown to be beneficial to biological activity.  With this in mind, researchers synthesized a group of penicillanic acid analogues with substituted aromatic rings bonded to the β-lactam ring in an effort to overcome the challenge of antibiotic resistance.  The substituents on the aromatic rings were varied, including -OCHS1U1B13S1U1B0 and -NOS1U1B12S1U1B0, to make several analogues.  A key step in the synthesis of these analogues was the coupling of Compounds 1 and 2 to form a compound with two β-lactam rings (Reaction 1).
  Reaction 1The researchers studied the analogues' structure-activity relationship and antimicrobial activity against the gram-negative bacteria Escherichia coli and the gram-positive bacteria Staphylococcus aureus.  Ampicillin, a penicillin derivative used to treat illnesses brought about by gram-positive and gram-negative bacteria, was used as a control.  The minimal inhibitory concentrations are shown in Table 1.Table 1  Minimal Inhibitory Concentration (MIC)
  The penicillanic acid analogues were tested for their effect on mammalian cell viability because toxicity to cells is also an important factor to consider for potential drug candidates.  The general acceptable level of cell viability for a drug candidate is 70%.  These results are shown in Figure 1.
  Figure 1  Cell viability assay results
Adapted from De rosa M, Vigliotta G, Palma G, Saturnino C, Soriente A. Novel Penicillin-Type Analogues Bearing a Variable Substituted 2-Azetidinone Ring at Position 6: Synthesis and Biological Evaluation. Molecules. 2015.
-The synthesis of Compound 1 requires hydrolysis of an ester with LiOH and produces a carboxylate ion.  The reaction is quenched in the next step by the addition of HCl.  What is the expected result if blue litmus paper is immersed in the fully quenched solution?  The blue litmus paper will:

Accepted Answer

A) turn red as the carboxylate becomes protonated. 
B) turn red as the carboxylate becomes deprotonated. 
C) remain blue as the solution becomes acidic. 
D) remain blue as the solution becomes basic.

Question 6

Phospholipids contain all of the following structural features EXCEPT:

Accepted Answer

A) a phosphate. 
B) a phenol. 
C) a glycerol backbone. 
D) fatty acid chains.

Question 7

Passage
High blood pressure, or hypertension, is a prevalent condition in the United States.  One in three adults has hypertension, and approximately 20% do not know that they have it.  If left untreated, high blood pressure can lead to major health issues, including stroke, heart attack, and kidney disease or failure.  A number of drugs can help control hypertension, including those shown in Table 1.  These drugs can be classified into different groups based on their mechanism of action.Table 1  Drugs Used to Treat High Blood Pressure
  Labetalol (Figure 1) exists as four stereoisomers and is administered as a mixture of all four.  This drug can be classified as a pseudohybrid drug because two of the isomers, (R,R) and (S,R), exhibit biological activity; the other two isomers, (S,S) and (R,S), are not active.  The (R,R) isomer acts as a nonselective β-adrenergic receptor blocker, and the (S,R) isomer acts as a selective α-adrenergic receptor blocker.
  Figure 1  Structure of (±)-labetalol and l-epinephrinel-epinephrine (Figure 1) is the agonist of the α- and β-adrenergic receptors.  The structural motifs that allow l-epinephrine to bind to the adrenergic receptors include an amine separated from an aromatic ring by two carbon units, a hydroxyl group at a chiral center beta to the amine, and two hydroxyl groups on the aromatic ring in the meta and para positions.  With a few structural modifications, the agonist l-epinephrine can be converted into the antagonist labetalol.  Substitution of a hydroxyl on the aromatic ring with an amide group and extension of the methyl group on the amine transforms l-epinephrine into labetalol.  These substitutions keep a majority of the structural motifs necessary for labetalol to mimic l-epinephrine and bind to the adrenergic receptors.
- What position(s) on the analog of labetalol would be considered stereocenters?

Accepted Answer

A) Positions 1 and 2 only 
B) Position 2 only 
C) Position 3 only 
D) Positions 3 and 4 only

Question 8

Which structure is a tautomer of guanine?

Accepted Answer

A)   
B)   
C)   
D)

Question 9

Passage
The drug paracetamol, also known as acetaminophen, is used to relieve pain and fever, and is a metabolite of the antipyretic drug phenacetin (Figure 1).  These two drugs have similar medicinal properties, but phenacetin has been shown to be carcinogenic and cause kidney damage.  Acetaminophen is a safer alternative to phenacetin when taken in therapeutic doses.
  Figure 1  Structure of phenacetinA student synthesizes acetaminophen in a laboratory from phenol according to the reactions shown.  The synthesis begins with the nitration of phenol by sodium nitrate (NaNOS1U1B13S1U1B0) to produce the mixture of the isomers 2-nitrophenol and 4-nitrophenol (Reaction 1).  These isomers can be easily separated by steam distillation to isolate the desired product, 4-nitrophenol.  Reduction of the nitro group with sodium borohydride (NaBHS1U1B14S1U1B0) (Reaction 2) is followed by acetylation of the amine with acetic anhydride (Reaction 3) to afford the final product, acetaminophen.
  Reaction 1
  Reaction 2
  Reaction 3
-Based on the structure of acetaminophen shown in the passage, in which region of the infrared spectrum will it most likely show a strong absorption?

Accepted Answer

A) 1690-1650 cm−1 
B) 2150-2100 cm−1 
C) 2260-2240 cm−1 
D) 2850-2750 cm−1

Question 10

Passage
Two key ingredients found in many soaps are PEG 150 distearate and ethylhexyl glycerin.  Both can be synthesized from glyceryl tristearate (Figure 1).
  Figure 1  Formation of PEG 150 distearate and ethylhexyl glycerin from glyceryl tristearateThe first step in this process requires the saponification of glyceryl tristearate.  The resulting stearate molecules can then react with PEG 150 in an acid-catalyzed Fischer esterification to form PEG 150 distearate.  The glycerol produced during saponification is used as a nucleophile to form ethylhexyl glycerin through a base-catalyzed SS1U1B1NS1U1B02 reaction with an ethylhexyl halide.Studies have shown that hand washing with soap can substantially reduce the number of germs present on hands and that washing hands before preparing food is particularly effective in reducing illness.  Researchers examined the effect of washing for 20 seconds with different amounts of soap containing PEG 150 distearate and ethylhexyl glycerin and found an inverse correlation between the volume of soap used and the amount of bacteria recovered from the hands.  The data are shown in Figure 2.
  Figure 2  Bacteria recovered from hands after washing with varying amounts of soapThe researchers then investigated the effect of hand washing with different kinds of soap on the transfer of bacteria from hands to objects.  Volunteers washed with a fixed volume of soap for 30 seconds, then handled sterilized plastic spheres for 30 seconds.  Bacteria were then recovered from the spheres.  The results are shown in Table 1.Table 1 Bacteria Recovered from Objects
  Adapted from  Fuls JL, Rodgers ND, Fischler GE, et al. Alternative hand contamination technique to compare the activities of antimicrobial and nonantimicrobial soaps under different test conditions. Appl Environ Microbiol. 2008.
-Which layer in the B&D and GC methods in Figure 2 is expected to contain sterols isolated from the Y. lipolytica extract?

Accepted Answer

A) L1 in the B&D method and L2 in the GC method 
B) L3 in the B&D method and L1 in the GC method 
C) L3 in the GC method and L2 in the B&D method 
D) L3 in the GC method and L1 in the B&D method

Question 11

Hydroxyl substituents in alcohols are poor leaving groups, and are not readily eliminated during substitution reactions.  Which of the following modifications would make the hydroxyl substituent of an alcohol a better leaving group?

Accepted Answer

A) Deprotonation 
B) Addition of an acid 
C) Addition of a strong base 
D) Conversion to an acetal

Question 12

A researcher needs to use an amino acid that is a zwitterion at physiological pH.  Which of the following amino acids could NOT be used in the experiment?

Accepted Answer

A) D 
B) W 
C) F 
D) T

Question 13

Passage
The cryptophycins, isolated from cyanobacteria, are a family of macrocyclic molecules.  Researchers have studied the biological activity of these compounds and found that many of them are active against multi-drug-resistant tumor cells.Cryptophycin-52, an analogue of the natural cryptophycins, has been tested in clinical trials for its efficacy against cancer cells, but the studies were suspended due to neurotoxic side effects.  Nevertheless, the synthesis of similar analogues continues to be of interest in cancer research.  A key step in the synthesis of cryptophycin-52 is the stereospecific formation of an epoxide from an alkene and oxone (Figure 1), which can be facilitated by the Shi catalyst.  The Shi catalyst (Figure 2) is derived from the carbohydrate D-fructose and is known among organic chemists for its ability to form epoxides in hydrocarbons in a stereospecific manner.
  Figure 1  Shi epoxidation of compound 1
  Figure 2  Structure of the Shi catalystThe epoxidation of compound 1 was monitored by thin-layer chromatography (TLC).  When the reaction was completed, high-performance liquid chromatography (HPLC) was used to analyze the epoxide product.  A chiral column that had an affinity for the desired epoxide was used.  A standard of the desired epoxide was obtained to optimize the conditions and determine the retention time of the epoxide.  The HPLC chromatogram of the epoxide standard is shown in Figure 3.
  Figure 3  HPLC of epoxide standard
Adapted from Weiß C, Bogner T, Sammet B, Sewald N. Total synthesis and biological evaluation of fluorinated cryptophycins. Beilstein J Org Chem. 2012.
-The Shi catalyst is derived from D-fructose based on the reaction below.  The structure of D-fructose in the form shown below can be classified as:  pyranose.furanose.ketohexose.aldohexose.

Accepted Answer

A) I and III only 
B) I and IV only 
C) II and III only 
D) II and IV only

Question 14

Passage
Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds.  The organic compounds provided to the students are given in Table 1.Table 1  Organic Compounds Tested
  First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl).  This reagent is used to test for secondary and tertiary alcohols.  A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1).
  Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds.  The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test.  Compound A was found to be an example of a positive result, as shown in Reaction 2.
  Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine.  This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols.  A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3).
  3Table 2  Selected Observations from the Classification Tests
 
-Based on the information in the passage, which compound will react most rapidly with the Lucas reagent?

Accepted Answer

A) Compound A 
B) Compound B 
C) Compound C 
D) Each compound will react at the same rate

Question 15

Passage
Hemophilia B is a blood clotting disorder caused by a factor IX (FIX) deficiency.  FIX is a 57-kDa, vitamin K-dependent protease that activates factor X, leading to the conversion of prothrombin to thrombin for propagation of the clotting cascade.  Activated FIX has two major domains:  a γ-carboxyglutamic acid domain and a serine protease domain.  The γ-carboxyglutamic acid domain participates in the oxidation of vitamin K using metallic cofactors, as shown in Figure 1.
  Figure 1  Oxidation of vitamin KTo further analyze the γ-carboxyglutamic acid-rich domain of FIX, an analogous synthetic peptide composed of matching residues 1 through 49 on FIX was evaluated by proton nuclear magnetic resonance (NMR) spectroscopy.  Analysis of the proton chemical shift before the addition of metal ions suggested that the synthetic peptide contained normal structural elements.  Large chemical shifts were observed after the addition of calcium and beryllium, as shown in Figure 2.
  Figure 2  Results of NMR spectroscopy (tetramethylsilane [TMS] peak has been removed)The synthetic analog was then placed in solution with vitamin K hydroquinone and cofactors required for vitamin K oxidation.  The oxidation products of vitamin K in Reactions 1 and 2 were collected and evaluated under high-performance liquid chromatography (HPLC) using hexane as the mobile phase.  Analysis demonstrated cis and trans isomers of vitamin KS1U1B11S1U1B0 and a trans-epoxy vitamin KS1U1B11S1U1B0, as shown in Figure 3.
  Figure 3  Results of HPLC separation
Adapted from Freedman SJ, Furie BC, Furie B, Baleja JD. Structure of the metal-free gamma-carboxyglutamic acid-rich membrane binding region of factor IX by two-dimensional NMR spectroscopy. J Biol Chem. 1995.
-The addition of metallic ions to FIX causes a conformational change that leads to induction of FIX activity, as shown in Figure 2.  The 1 ppm (parts per million) proton peak in the inactivated FIX most likely represents:

Accepted Answer

A) a comparatively large number of deshielded protons. 
B) a large amount of hydrogen bonding. 
C) a nearby atom with high electronegativity. 
D) a portion of FIX with comparatively shielded protons.

Question 16

A student needs to separate a mixture of chloroform (bp 61°C) and benzene (bp 80°C).  What type(s) of distillation would be expected to give the best separation of the two compounds?Simple distillationFractional distillationVacuum distillation

Accepted Answer

A) I only 
B) II only 
C) I and III only 
D) II and III only

Question 17

A mixture of 1-butanol, 1-hexanol, 1-octanol, and 1-propanol is separated by gas-liquid chromatography.  What compound elutes last?

Accepted Answer

A) 1-butanol 
B) 1-hexanol 
C) 1-octanol 
D) 1-propanol

Question 18

Passage
Fatty acids (FAs) are lipids that play important roles in cellular function, including energy storage and membrane structure.  Most polyunsaturated fatty acids (PUFAs) are made in the body; however, omega-3 alpha-linolenic acid and omega-6 linoleic acid must be obtained from the diet.  These two PUFAs are precursors to other omega-3 and omega-6 long-chain FAs that are necessary for physiological processes such as neurological development, normal growth, and immune response.  Insufficient omega-3 and omega-6 long-chain FAs can lead to an increase in omega-9 fatty acids such as mead acid, and to essential fatty acid deficiency (EFAD), which can cause infertility, decreased growth, and increased susceptibility to infection.Total FA levels in plasma and serum can be used to diagnose EFAD.  Plasma and serum samples from several individuals were hydrolyzed to release FAs from triacylglycerols and phospholipids.  Once isolated, the free FAs were converted to pentafluorobenzyl (PFB) derivatives and separated by gas chromatography (GC).  Total levels of each type of fatty acid were quantified by negative chemical ionization-mass spectrometry (NCI-MS) immediately after GC.  NCI is an ionization technique that creates negatively charged ions with little to no fragmentation.  The NCI-MS revealed that two of the peaks from GC had the same m/z ratio (Figure 1).  They were both confirmed to be forms of an unsaturated, unbranched 22-carbon FA called omega-9 docosenoic acid.
  Figure 1  Selected ion chromatogram of docosenoic acidResearchers have proposed that an elevated ratio of the triene omega-9 mead acid to the tetraene omega-6 arachidonic acid, known as the triene/tetraene (TT) ratio, can signify EFAD.  Using GC and NCI-MS, they measured the TT ratio, concentration of mead acid, and individual concentrations of omega-3 and omega-6 FAs in plasma and serum samples from three patients and compared their results to normal ranges.  An increased TT ratio and mead acid concentration and decreased omega-3 and omega-6 FA concentrations were observed in the patient samples compared to normal ranges.
Adapted from E. Kish-Trier et al., "Quantitation of total fatty acids in plasma and serum by GC-NCI-MS." Clinical Mass Spectrometry. ©2016 Elsevier.
-Based on the passage, which compound was most likely analyzed by GC and NCI-MS in the experiment?

Accepted Answer

A)   
B)   
C)   
D)

Question 19

Passage
Electrospray ionization mass spectrometry (ESI-MS) employs a soft ionization technique that, unlike traditional ionization via electron bombardment, results in little to no fragmentation of target molecules.  During electron bombardment, electrons collide with the analyte and cause the molecule to lose an electron whereas during ESI-MS, ions are produced by passing the analyte solution through an electrospray needle that has a potential difference applied with respect to a counterelectrode.  This leads to the formation of charged droplets, which are repelled from the needle toward the counterelectrode and mass spectrometer, as shown in Figure 1.
  Figure 1  ESI-MS ionization mechanismDuring the droplet's flight, the solvent is evaporated off analyte molecules.  Ionization occurs through the protonation of basic sites on target molecules.  Reaction 1 shows the number of protons n added to molecule M.M → (M + nH)S1U1P1nS1S1P0S1U1P1+S1S1P0Reaction 1Researchers used ESI-MS to quantify the formation of four pyridyloxobutyl-DNA (POB-DNA) adducts, shown in Figure 2.  POB-DNA adduct formation involves highly mutagenic pyridyloxobutylating agents, which are formed from cytochrome P450-mediated hydroxylation of 4-(methylnitrosamino)-1-(3-pyridyl)-1-butanone (NNK), an abundant N-nitrosamine carcinogen found in cigarette smoke.  Adducts were produced in vitro by treating samples of calf thymus DNA with NNK.  The DNA adducts were then separated by reversed-phase high-performance liquid chromatography (RP-HPLC) and analyzed via ESI-MS.
  Figure 2  Molecular structure of POB-DNA adducts
Adapted from Lao Y, Villalta PW, Sturla SJ, Wang M, Hecht SS. Quantitation of pyridyloxobutyl DNA adducts of tobacco-specific nitrosamines in rat tissue DNA by high-performance liquid chromatography-electrospray ionization-tandem mass spectrometry. Chem Res Toxicol. 2006.
-Based on the procedure described in the passage, which of the following spectra most likely corresponds to the detection of the singly charged POB Adduct 1 ion?

Accepted Answer

A)   
B)   
C)   
D)

Question 20

Passage
The N-terminus of an amino acid must be protected with 9-fluorenylmethoxycarbonyl (Fmoc) during solid state peptide synthesis to avoid side reactions (Scheme I).
  Scheme IThe Fmoc protecting group is typically removed with piperidine (Scheme II), a heterocyclic secondary amine.
  Scheme IIBecause the use of piperidine is restricted, structurally similar amines were examined as possible replacements.  To determine whether piperidine analogues were suitable alternatives for Fmoc removal, deprotection of Compound 2 was carried out with several amine bases (Scheme III).
  Scheme IIICompound 2 was prepared from L-tryptophan and Fmoc-Cl, and the reaction was monitored by thin-layer chromatography (TLC) and visualized by ultraviolet light.  The deprotections were done with piperidine and each analogue using a microwave set to 155 watts at 75 °C for 15 seconds.  The products of the deprotections were analyzed by high-performance liquid chromatography (HPLC) and mass spectrometry; the retention times of the products were compared to tryptophan and Compound 2 standards.
Adapted from O.P. Luna et al, "Deprotection Reagents in Fmoc Solid Phase Peptide Synthesis: Moving Away from Piperidine?." Molecules. ©2016 MDPI.
-Which reaction would best give data to support the need of a cyclic amine for removal of the Fmoc group?

Accepted Answer

A) React tryptophan with a secondary noncyclic amine and determine whether the Fmoc group was removed. 
B) React tryptophan with a tertiary noncyclic amine and determine whether the Fmoc group was removed. 
C) React Compound 2 with a secondary noncyclic amine and determine whether the Fmoc group was removed. 
D) React Compound 2 with a tertiary noncyclic amine and determine whether the Fmoc group was removed.

Exam 6: Organic Chemistry

Ubiquinone (shown below) is an important molecule in the electron transport chain, where it receives electrons in complex I originating from NADH. What is the structure of the product formed when electrons are added to ubiquinone?

Compound 1 is shown below. All of the following are equivalent to Compound 1 EXCEPT:

Phospholipids contain all of the following structural features EXCEPT:

Which structure is a tautomer of guanine?

Hydroxyl substituents in alcohols are poor leaving groups, and are not readily eliminated during substitution reactions. Which of the following modifications would make the hydroxyl substituent of an alcohol a better leaving group?

A researcher needs to use an amino acid that is a zwitterion at physiological pH. Which of the following amino acids could NOT be used in the experiment?

A student needs to separate a mixture of chloroform (bp 61°C) and benzene (bp 80°C) . What type(s) of distillation would be expected to give the best separation of the two compounds?Simple distillationFractional distillationVacuum distillation

A mixture of 1-butanol, 1-hexanol, 1-octanol, and 1-propanol is separated by gas-liquid chromatography. What compound elutes last?