FL 9 - chem/phys

Question 1

Using the PAD detection technique of

f_D = f_O [V/c_A]cos θ

which expression gives the constant velocity of a particle if the angle between the detector and the velocity of the particle is 60°?

A. 2f_D / c_Af₀

B. 2c_Af_D / f₀

C. f_D / c_Af₀

D. f_D / 2c_Af₀

Answer 1

B. 2c_Af_D / f₀

Use the equation for the PA Doppler frequency shift given in the passage, and substitute 0.5 for cos 60°.

f_D = f₀ (V/c_A) cos 60º = (0.5) f₀ V/c_A

The expression can be rearranged to solve for V, the constant velocity of the particle.

V = 2c_Af_D / f₀

Question 2

Sound speed _______ as temperature increases

Answer 2

Sound speed increases as temperature increases

Question 3

Sphingomyelin is most specifically categorized as what type of biological molecule?

A. Phospholipid

B. Fatty acid

C. Glycerolipid

D. Sterol lipid

Answer 3

A. Phospholipid

Sphingolipids are glycosylceramide lipids containing sphingosine, an 18-carbon amino alcohol, bound to a fatty acid head group through an N-acyl linkage. Sphingolipids lacking additional head groups are referred to as ceramides. Sphingomyelin is a complex phospholipid formed by the ester linkage of ceramide to a molecule of phosphocholine or phosphoethanolamine (choice A). The labeled structure of sphingomyelin is shown below:

Question 4

A student performing experiments on gaseous CO₂ would be LEAST likely to cite which of the following assumptions for ideal behavior of the gas?

A. The average kinetic energy of a sample of CO2 gas depends on the mass of the molecules.

B. Collisions between CO₂ molecules are elastic.

C. There are no attractive and repulsive forces between CO₂ molecules.

D. Gas particles have a volume of zero.

Answer 4

A. The average kinetic energy of a sample of CO2 gas depends on the mass of the molecules.

According to the kinetic molecular theory, the average kinetic energy of a sample of gas depends only on the temperature of the sample. The following formula shows this relationship: KE = (3/2)K_BT, where K_B is the Boltzmann constant.

Question 5

Gases are more likely to behave non-ideally under extremely ____ pressures or at very ____ temperatures.

Answer 5

Gases are more likely to behave non-ideally under extremely high pressures or at very low temperatures.

Question 6

How many sp²-hybridized carbons are in colchicine and compound IA, respectively?

A. 13, 13

B. 14, 13

C. 14, 14

D. 16, 14

Answer 6

B. 14, 13

An sp²-hybridized carbon is one that is surrounded by three regions of electron density. Often, this involves a central atom bound to three substituents. Examining the structures provided, IA has 13 sp²-hybridized carbons, while colchicine has 14. Thus, the correct answer should be 14, 13.

Question 7

Disruption of a protein’s primary structure can be achieved by a(n):

A. oxidoreductase.

B. protease.

C. endonuclease.

D. phosphatase.

Answer 7

B. protease.

A protein’s primary structure refers to its sequence of amino acids. The only way to disrupt this sequence is by cleaning a peptide bond, thereby forming two new peptide chains with two new amino acid sequences. Proteases cleave proteins at peptide bonds, so this is the only option that can disrupt a protein’s primary structure.

Question 8

Describe the role of an endonuclease

Answer 8

An endonuclease is an enzyme that cleaves a phosphodiester bond within a chain of nucleotides. It might disrupt the structure of a polynucleotide, such as a DNA strand.

Question 9

What kind of bond is cleaved by an endonuclease?

Answer 9

An endonuclease is an enzyme that cleaves a phosphodiester bond within a chain of nucleotides. It might disrupt the structure of a polynucleotide, such as a DNA strand.

Question 10

What effect would an endonuclease have on the primary structure of a protein?

Answer 10

An endonuclease is an enzyme that cleaves a phosphodiester bond within a chain of nucleotides. It might disrupt the structure of a polynucleotide, such as a DNA strand, but it would have no effect on proteins, which are proteins and contain peptide bonds.

Question 11

What kind of bond does a phosphatase interact with? Does it add or break a bond?

What effect might this have?

Answer 11

A phosphatase is an enzyme that can cleave a phosphate group from a phosphoester, leaving a phosphate ion and an alcohol.

A phosphatase might activate or inactivate an enzyme that relies on phosphorylation to signal its activity, but it would have no effect on a protein’s primary structure.

Question 12

What effect would an oxidoreductase have on the primary structure of a protein?

Answer 12

An oxidoreductase is an enzyme that catalyzes a redox reaction. However, a redox reaction is unlikely to disrupt a protein’s primary structure, or its sequence of amino acids.

Instead, an oxidoreductase might disrupt a protein’s tertiary or quaternary structure by forming or cleaving disulfide linkages.

Question 13

During the synthesis of bile, cholesterol is conjugated by the addition of a carboxylic acid residue. Bile salts are thus able to solubilize lipids in aqueous digestive secretions because bile salts:

A. are exclusively polar and thus dissolve completely in digestive secretions.

B. are exclusively polar and fully dissolve dietary lipids.

C. have a polar region and a nonpolar region, thus increasing the water-solubility of lipids.

D. have a polar region and a nonpolar region, and are thus able to neutralize other organic acids.

Answer 13

C. have a polar region and a nonpolar region, thus increasing the water-solubility of lipids.

Bile salts contain both polar elements and a hydrophobic cholesterol core. Because of this amphipathic structure, they contain both polar and nonpolar regions, acting as a detergent and solubilizing lipids in aqueous digestive contents by interacting with both hydrophobic (lipid) and aqueous phases. Sodium cholate, a common bile salt, is shown below:

Question 14

What happens to the IR peak of O-H when taken from a molecule in the gas phase?

Answer 14

The expected broad O-H vibration is replaced by a sharp, shorter vibration seen around 3600 cm^-1

Question 15

The dichromate used in breath analyzers comes from stock solutions of potassium dichromate. What is the oxidation number of chromium in this solution?

A. +2

B. +3

C. +4

D. +6

Answer 15

D. +6

The formula of potassium dichromate is K₂Cr₂O₇. The total charge of this compound is zero. Potassium is in Group 1 of the periodic table and forms only one ion, with a charge of +1. Oxygen, when not part of a peroxide, has a -2 oxidation state. We can then write the following equation, where x represents the oxidation number of the chromium atoms:

0 = 2(+1) + 2x + 7(-2)

0 = 2 + 2x -14

0 = 2x -12

12 = 2x

6 = x

The same result can be obtained by using the formula of the dichromate ion (Cr₂O₇^2-) given in Reaction 1.

• 2 = 2x + 7(-2)
• 2 = 2x -14

12 = 2x

6 = x

Question 16

Which of the following correctly identifies the fundamentals of vibrations for ethanol in Figure 1?

I. The O-H stretch is the sharp peak of medium intensity at 3700 cm^-1.

II. The C-H stretches are the series of strong intensity peaks centered at 2900 cm^-1.

III. The C-O stretch is the weak peak around 1900 cm^-1.

A. I only

B. II only

C. I and II only

D. I, II, and III

Answer 16

C. I and II only

Since the IR spectrum in Figure 1 is of ethanol in the gas phase, the O-H stretch appears as a sharp peak above 3000 cm^-1, rather than the broad feature typically seen in the liquid phase due to hydrogen bonding (I). The various C-H stretches are typically observed slightly below 3000 cm^-1 (II).

Question 17

Where are single bonded C-O stretches often seen in IR spectra?

Answer 17

Singly-bonded C-O stretches are typically found from 1100 to 1000 cm-1

Question 18

When performing normal-phase HPLC and reverse-phase HPLC, respectively, what should be the properties of the stationary phases?

A. Ionic and uncharged

B. Uncharged and ionic

C. Polar and nonpolar

D. Nonpolar and polar

Answer 18

C. Polar and nonpolar

HPLC separates compounds based on polarity and solubility in the stationary phase. In normal-phase HPLC, the stationary phase is polar, and the mobile phase is nonpolar (similar to typical thin-layer chromatography). In reverse-phase HPLC, the stationary phase is nonpolar, and the mobile phase is polar, as shown below:

Question 19

Would it be possible for the researchers to use gas chromatography (GC) for the isolation of retinol during SRA?

A. No; GC can only be used to analyze compounds that are gases at room temperature.

B. No; it would be too expensive compared to fluorometry.

C. Yes, but HPLC would be more accurate.

D. Yes, provided retinol can be vaporized without degradation.

Answer 19

D. Yes, provided retinol can be vaporized without degradation.

In gas chromatography, samples are heated and vaporized before they are introduced into the separating column. Therefore, they must be able to be vaporized without being broken down or combusting so they can be analyzed properly in the gaseous phase.

A is incorrect because samples can be in any phase at room temperature, as long as they are stable when vaporized.

Question 20

In an experiment in which retinol must be isolated by extraction, in which solvent would one expect it to dissolve most readily?

A. Hexane

B. Acetonitrile

C.Water

D. Ethanol

Answer 20

A. Hexane

Figure 1 shows the structure of retinol, which is a large organic molecule. Though it has a polar alcohol group, the vast majority of the molecule is nonpolar. Therefore, it would be most soluble in a nonpolar (hydrophobic) solvent. Of the choices listed, hexane is the most nonpolar, so choice A is correct. This logic is shown below:

Question 21

Which of the following is NOT an example of denaturation?

A. the melting of a protein at high temperatures

B. the conversion of a large protein into constituent subunits

C. the digestion of proteins in the stomach by pepsin

D. the disruption of side-chain interactions by a strong acid

Answer 21

C. the digestion of proteins in the stomach by pepsin

The digestion of proteins in the stomach by pepsin is an example of the action of a protease, whereby a protein’s primary structure is disrupted by cleavage at a peptide bond. The hydrolysis of a peptide bond is not an example of protein denaturation, as it changes the protein’s amino acid sequence, not just its three-dimensional structure.

Question 22

In a gas experiment, a student wished to measure the deviation of non-ideal xenon gas at various temperatures and pressures. For one mole of non-ideal xenon gas at a pressure of 1000 atm and a temperature of 100 K, what is the most likely value for the quantity PV/RT?

A. 0.3

B. 0.5

C. 1

D. 3

Answer 22

D. 3

For one mole of an ideal gas (n = 1) the quantity PV/RT equals 1 for all pressures. Deviations from the ideal gas law occur at high pressures and low temperatures, which are the conditions provided in this question. To understand these deviations, we must first remember two assumptions we make about ideal gases: that individual particles have no volume, and that there are no attractive or repulsive forces between gas particles.

Here, the main assumption that is broken is that gas particles have no volume. Remember, as P increases, V (the volume of the container) decreases. At a pressure as high as 1000 atm, the gas must be compressed into a very small volume. Under these conditions, the volumes of the gas particles will begin to be substantial in comparison to the overall volume. This causes the real volume to increase, making V_real > V_ideal. (If this is still confusing, imagine pressure increasing further, approaching infinite pressure. According to PV = nRT, volume would approach zero. However, at a certain point, volume will not be able to decrease, because the particle volume cannot be compressed. As a result, the higher the pressure, the more V_real will deviate from V_ideal, or V_predicted.)

If V_real is significantly higher than expected, the PV term in PV/nRT will also be especially high, and PV/nRT will exceed 1. The only answer choice greater than 1 is choice D.

Question 23

True or False:

Between two DNA double-helix strands A and B, strand A has twice the guanine content of strand B. Compared to strand B, strand A has the double the pyrimidine content

Answer 23

False.

Any length of a double-helix strand of DNA will have 50% pyrimidine content and 50% purine content, as every purine will be base-paired to a pyrimidine on the opposite strand, and every pyrimidine to a purine. Therefore, strand A has equivalent pyrimidine content to strand B: 50%. Alternatively, consider hypothetical values for the content of each base for the two strands. For strand A, a hypothetical makeup of 40% guanine, 40% cytosine, 10% adenine, and 10% thymine (note that the guanine content must match the cytosine content and the adenine content must match the thymine content for proper Watson-Crick base pairing). Consequently, for strand B, a hypothetical makeup of 20% guanine (1/2 that of strand A’s guanine content), 20% cytosine, 30% adenine, and 30% thymine. For both of these hypothetical base compositions, the pyrimidine content (= cytosine content + thymine content) is 50%.

Question 24

Between two DNA double-helix strands A and B, strand A has twice the guanine content of strand B. Compared to strand B, strand A has:

I. half the pyrimidine content.

II. a higher melting point.

III. twice the cytosine content.

A. I only

B. II and III only

C. I and III only

D. I, II, and III

Answer 24

B. II and III only

Strand A has twice the guanine content of strand B, and therefore has twice as many G-C base pairs as strand B. G-C base pairs have more stable interactions than A-T base pairs due to their hydrogen bonding structure (G-C base pairs have 3 hydrogen bonds, while A-T base pairs have only 2). Therefore, strand A’s melting point, defined as the temperature at which the two single strands have lost sufficient hydrogen bonding interactions to denature the DNA molecule, will be higher than that of strand B (II). In addition, if strand A has twice the guanine content of strand B, then it must also have twice the cytosine content, as the guanine and cytosine content within one DNA double-strand must match to achieve proper Watson-Crick base pairing (III).

Question 25

Starch is readily digestible by humans, but cellulose is not. This is due to a difference in:

A. monosaccharides.

B. branching.

C. anomeric carbon stereochemistry.

D. polysaccharide length.

Answer 25

C. anomeric carbon stereochemistry.

Monomers of cellulose are connected by beta-1,4-glycosidic linkages, while monomers of starch are connected primarily by alpha-1,4-glycosidic linkages. The key difference is in the stereochemistry of the anomeric carbon of the monosaccharides along the chain of each polysaccharide. In cellulose, the anomeric carbon of each glucose monomer is oriented in the beta position, while in starch, the anomeric carbon of each glucose monomer is oriented in the alpha position. Humans lack the enzyme to cleave beta-1,4-glycosidic linkages but possess amylase, an enzyme capable of lysing the alpha-1,4-glycosidic linkages in starch.

Question 26

What kind of linkage holds monomers of cellulose together?

Answer 26

Monomers of cellulose are connected by beta-1,4-glycosidic linkages

Question 27

What is the major difference between starch and cellulose?

Answer 27

Monomers of cellulose are connected by beta-1,4-glycosidic linkages,

while monomers of starch are connected primarily by alpha-1,4-glycosidic linkages.

Humans contain the enzyme amylase, capable of breaking alpha-1,4-glycosidic linkages, this is why we can digest starch but not cellulose.

Question 28

# Define:
Tautomer

Answer 28

Tautomers are isomers of a compound that differ only in the position of the protons and electrons.

Question 29

What is the net charge of creatine in aqueous solution at a pH of 2?

A. -1

B. 0

C. +1

D. +2

Answer 29

D. +2

Remember, a group will be protonated at pH levels below its pK_a, while it will be deprotonated when exposed to a pH higher than its pK_a. In amino acids, the pK_a of a typical carboxylic acid terminal is slightly above 2; we can extend that reasoning and assume that the carboxylic acid group on creatine will be protonated at pH 2, making it neutral. The amino group will gain a proton and become positive, as the pKa of an amino functionality is typically 8 or higher. Finally, Figure 1 shows the imine side-chain protonated even when the other groups are deprotonated. Since the other groups will be protonated at pH 2, the imine certainly will be as well, giving it a +1 charge. +1 + 0 + 1 = +2 net charge.

Question 30

Which of the following must be true for a biochemical reaction at equilibrium?

A. ΔGº’ = 0

B. ΔG = 0

C. K_eq=0

D. Q=0

Answer 30

B. ΔG = 0

The Gibbs free energy change (ΔG) is used to predict the spontaneity of a reaction, with negative values indicating that the reaction will proceed in the forward direction (from reactants to products) and positive values indicating that it will proceed in the reverse direction (from products to reactants). When the reaction has reached equilibrium, ΔG is zero. Therefore, choice B is the best answer.

Question 31

Based on the information in Table 1, what is the ΔG°‘ at 25°C for the hydrolysis of one mole of pyrophosphate into two phosphate groups?

A. -30.5 kJ/mol

B. -33.4 kJ/mol

C. -63.9 kJ/mol

D. -94.4 kJ/mol

Answer 31

B. -33.4 kJ/mol

The reaction for the hydrolysis of pyrophosphate (PP) into two phosphate groups (P) is:

PP → 2 P

Using the concept of Hess’s law, we can combine the reactions in Table 1. This way, they give the desired reaction when added together, and summing the free energies of the reactions gives the desired ΔG° value.

AMP + PP → ATP ΔG° = +30.5 kJ/mol

ADP → AMP + P ΔG° = -33.4 kJ/mol

ATP → ADP + P ΔG° = -30.5 kJ/mol

PP → 2 P ΔG° = -33.4 kJ/mol

Question 32

If the red LED used in this experiment was designed to operate at 30 mW and there was a voltage drop of 3.2 V when the LED was lit, what was the current flowing in the circuit?

A. 9.4 x 10^-3 amps

B. 1.0 x 10^-1 amps

C. 9.4 x 10⁰ amps

D. 1.1 x 10² amps

Answer 32

A. 9.4 x 10^-3 amps

The electrical power, P, is given by the equation P = IV, where V represents the voltage drop and I is the current in amperes. Be sure to convert milliwatts into watts, as follows: 30 mW = 30 x 10^-3 W = 3.0 x 10^-2 W. Note also that 1 watt is the same as 1 volt-amp.

I = P/V = (3.0 x 10^-2 V-amp) / (3.2 V) ~ 1 x 10^-2 amps

Since 3.0/3.2 is slightly less than 1, the answer must be slightly less than 1 x 10^-2 amps, or about 0.9 x 10^-2 amps. This value is equivalent to 9 x 10^-3 amps, which is closest to choice A.

Question 33

What is the chemical formula of Zinc Sulfate

Answer 33

Zinc sulfate has a formula of ZnSO₄, as it is composed of Zn²⁺ cations and SO₄^2- anions.

Question 34

What are the hybridization and molecular geometry of the carbon in formaldehyde?

A. sp², tetrahedral

B. sp², trigonal planar

C. sp³, tetrahedral

D. sp³, trigonal planar

Answer 34

B. sp², trigonal planar

Formaldehyde, the simplest aldehyde, is a good structure to know for the MCAT. Its central atom (and the only carbon in the molecule) is bonded to three different atoms and has no lone pairs, so its hybridization is sp². We expect trigonal planar geometry with sp²-hybridized species unless there are lone pairs of electrons to deflect the bond angle. Thus, choice B is correct. The geometry of formaldehyde is shown below to help clarify this visually:

Question 35

The hydrotrope sodium benzoate is formed by the reaction of NaOH and benzoic acid. In an aqueous solution, this hydrotrope is:

A. weakly acidic.

B. strongly acidic.

C. weakly basic.

D. strongly basic.

Answer 35

C. weakly basic.

The reaction of NaOH (a strong base) and benzoic acid (a weak acid) will deprotonate the benzoic acid and produce sodium benzoate, a salt.

In general, the salt of an acidic species will be weakly basic when dissolved in solution. Since the other product of the reaction is water (which is neutral), the solution as a whole will be weakly basic.

In this case in particular, because benzoic acid is a weak acid, it relatively firmly holds on to its acidic proton. So after the benzoic acid has been deprotonated, the resulting benzoate grabs the few protons from the dissociated water, leaving only hydroxides. By LeChatelier’s Principle, more water dissociates, resulting in more hydroxides and protons. Benzoates take the protons, which leads to more benzoic acid and more dissociation of water. The process continues until both water’s and benzoic acid’s equilibrium constants are reached. A moderate excess of hydroxides, beyond the number of protons, results.

Question 36

Assuming a constant acceleration, how can you find the average velocity if given time?

Answer 36

acceleration = average velocity / time

rearranging yields average velocity = acceleration*time

Question 37

Complete hydrolysis of every bond in the ethyl chains on the end of the diethyl phthalate molecules would:

A. cause London dispersion forces to manifest themselves.

B. result in hydrogen bonding.

C. result in a more polar molecule, which then would exhibit a smaller attractive force with identical molecules.

D. result in the formation of an ionic solid.

Answer 37

B. result in hydrogen bonding.

Hydrogen bonding between molecules will be present when the molecules contain an O, N, or F atom covalently bonded to an H atom. Here, hydrolyzing the ethyl chains will turn diethyl phthalate into phthalic acid; in other words, it will turn the molecule on the right side of Figure 1 into the molecule on the left side of Figure 1. Phthalic acid contains O atoms covalently bonded to hydrogen; hence, there will be hydrogen bonding between these molecules.

Question 38

Methane gas is nonpolar. What are the possible implications of this?

A. It depends on whether a carbon atom is present, which, if so, will cause hydrogen bonding.

B. It depends on whether a bromine atom is present, which, if so, will cause dipole-dipole bonding.

C. Being nonpolar, dipole-dipole forces will not be present.

D. Being nonpolar, Van der Waals’ forces will not be present.

Answer 38

C. Being nonpolar, dipole-dipole forces will not be present.

Dipole-dipole forces are present only between polar molecules, and methane is nonpolar.

Question 39

What type of bonding can BPA be expected to exhibit with identical molecules?

I. Dipole-dipole bonding

II. Hydrogen bonding

III. Induced-dipole bonding

A. I and II only

B. I and III only

C. I, II, and III

D. Neither I, II, nor III

Answer 39

A. I and II only

Despite having one plane of symmetry, BPA has a polar structure overall. As such, it will experience dipole-dipole forces with other polar molecules (I). The -OH groups on BPA, as shown in Figure 2, will exhibit hydrogen bonding with the -OH groups on other molecules (II).

Induced-dipole bonding occurs only when a polar molecule induces a nonpolar molecule to become temporarily polar. Since BPA is already polar, it cannot be induced from a nonpolar into a polar state (III).

Question 40

When do induced-dipole forces occur?

Answer 40

Induced-dipole bonding occurs only when a polar molecule induces a nonpolar molecule to become temporarily polar.

Question 41

When do dipole-dipole interactions occur?

Answer 41

Dipole-dipole interactions only occur between polar molecules, not between polar and nonpolar molecules.

Question 42

A solution was formed using dipropyl ether as a solvent and diethyl phthalate as a solute. Dipropyl ether is nonpolar. Which of the following interactions will be strongest, assuming that the diethyl phthalate dissolved as intended?

A. Hydrogen bonding between dipropyl ether and diethyl phthalate

B. Induced dipole interactions between dipropyl ether and diethyl phthalate

C. Dipole-dipole interactions between dipropyl ether and diethyl phthalate

D. London dispersion force interactions between dipropyl ether and diethyl phthalate

Answer 42

B. Induced dipole interactions between dipropyl ether and diethyl phthalate

Diethyl phthalate is a polar molecule and will induce a dipole in neighboring nonpolar molecules. As such, a permanent dipole will exist next to a temporary dipole, and the two will then attract each other due to electrostatic force (the attractive force between positively-charged and negatively-charged atoms). The name for this phenomenon is “induced-dipole” interactions.

Question 43

The two molecules shown below are:

A. diastereomers.

B. optical isomers.

C. conformational isomers.

D. constitutional isomers.

Answer 43

A. diastereomers.

Examining the figure, we can see that these two molecules have identical molecular formulas, as well as identical stereochemistry except for the carbon attached to the carboxyl group, which switches its chirality. This means that these monosaccharides represent a type of diastereomer. Diastereomers are stereoisomers that differ at some, but not all, chiral centers.

Constitutional isomers differ in the order in which the atoms are connected, so they can contain different functional groups and/or bonding patterns.

Question 44

The structure of sucrose, a dimer of glucose and fructose, is shown in the following figure.

The two monomeric subunits of sucrose are best described as:

A. structural isomers.

B. diastereomers.

C. enantiomers.

D. epimers.

Answer 44

A. structural isomers.

Glucose and fructose are structural, or configurational, isomers. Remember that glucose is a hexose, meaning that it contains a six-membered ring. In contrast, fructose (a pentose) includes a five-membered ring. Both monosaccharides contain the same number of carbon, oxygen, and hydrogen atoms.

Question 45

Which of these solvents would best facilitate a bimolecular nucleophilic substitution reaction?

A. Ethanol, due to its ability to hydrogen bond

B. Water, due to its polar protic nature

C. Acetone, due to its lack of an –OH or –NH bond

D. Ammonia, due to its basic tendencies

Answer 45

C. Acetone, due to its lack of an –OH or –NH bond

Acetone is a polar aprotic solvent, meaning that it cannot donate hydrogen bonds. S_N2 reactions proceed best in the presence of polar aprotic solvents.

Question 46

Though relatively slow, nucleophilic substitution reactions feature in a number of reaction schemes commonly employed in a laboratory setting. One such process, the Finkelstein reaction, involves the treatment of a primary alcohol with tosyl chloride, followed by addition of a halide salt. This series of steps permits the synthesis of a primary alkyl halide. Which of the following conditions would best optimize the speed and yield of the Finkelstein reaction?

A. Low pH with acetone as the solvent

B. Low pH with water as the solvent

C. Neutral pH with ethanol as the solvent

D. Neutral pH with dimethylformamide as the solvent

Answer 46

D. Neutral pH with dimethylformamide as the solvent

Since the initial substrate is primary, it cannot form a carbocation; as a result, we know that the reaction described must proceed according to an SN2 mechanism. SN2 reactions progress most rapidly in polar aprotic solvents, of which the only ones listed are acetone and dimethylformamide. However, an acidic environment would protonate and reduce the available concentration of the halide anion in solution, slowing down the substitution. Optimally, the reaction should instead be conducted at neutral pH, as choice D mentions.

Question 47

When an alkene is exposed to HBr, the anti-Markovnikov product:

A. involves the bromine atom adding to the more substituted end of the alkene, and occurs during radical reactions.

B. involves the bromine atom adding to the more substituted end of the alkene, and occurs during the simple addition of HBr.

C. involves the bromine atom adding to the less substituted end of the alkene, and occurs during radical reactions.

D. involves the bromine atom adding to the more substituted end of the alkene, and occurs during the simple addition of HBr.

Answer 47

C. involves the bromine atom adding to the less substituted end of the alkene, and occurs during radical reactions.

When an alkene is reacted with hydrobromic acid, the Markovnikov product typically forms. Since this product includes the halogen bound to the more substituted end of the bond, it tends to be especially stable. Radical reactions, particularly those initiated by peroxides, form an exception to this rule. In these processes, hydrogen adds to the less substituted end, while the bromine ion adds to the other position.

Question 48

p-chlorophenol can most easily be synthesized by reacting phenol with:

A. chlorine in water.

B. chlorine and chromium trioxide in water.

C. acyl chloride in water.

D. chlorine in carbon disulfide.

Answer 48

D. chlorine in carbon disulfide.

The process described is the halogenation of a phenol. When deciding which solvent to use for such a procedure, make sure to distinguish between monohalogenation and polyhalogenation. Here, since we are attempting monohalogenation, we should use a nonpolar solvent, such as carbon disulfide.

Water, which is polar, is a better solvent for polysubstitution than for monosubstitution. The polar nature of this solvent favors the existence of charged species and transition states.

Question 49

A sample of nonanoic acid is analyzed using nuclear magnetic resonance (NMR) spectroscopy. Which of this molecule’s protons will exhibit a peak that is farthest upfield?

A. The proton(s) on carbon 9, which will display a shift of around 1.0 ppm

B. The proton(s) on carbon 9, which will display a shift of around 11.5 ppm

C. The proton on the –COOH group, which will display a shift of around 1.0 ppm

D. The proton on the –COOH group, which will display a shift of around 11.5 ppm

Answer 49

A. The proton(s) on carbon 9, which will display a shift of around 1.0 ppm

With regard to NMR, upfield refers to the right-hand side of a spectrum. Here, signals appear if they correspond to nuclei that are shielded, or physically distant from electronegative atoms. The most shielded nuclei belong to those of the hydrogen atoms that are far from the –COOH terminal. As such, they will possess very low ppm values.

Question 50

Which solvent would be most suitable for the recrystallization of molecule 1, a solid collected after a synthesis procedure? Note that the known melting point for molecule 1 is 168 °C and the sample collected is displaying a melting point of 164-167°C.

A. Solvent A, in which molecule 1 is very soluble at 80°C but far less soluble at 19.5°C

B. Solvent B, in which molecule 1 is highly (and equally) soluble at all temperatures

C. Solvent C, in which both molecule 1 and its common impurities are fairly soluble at 80°C but insoluble at 19.5°C

D. None of the above; recrystallization either would not work or is not necessary for the purification of this sample.

Answer 50

A. Solvent A, in which molecule 1 is very soluble at 80°C but far less soluble at 19.5°C

Recrystallization is a method used to remove impurities from solid samples. For these procedures, we must choose a solvent in which our compound is relatively insoluble at cold temperatures but relatively soluble at warm ones. This way, we can heat our compound / solvent mixture to dissolve the sample, then cool the vessel to recrystallize it and exclude impurities from its new structure.

Question 51

Methyl acetate is:

A. a three-carbon ether.

B. a three-carbon ester.

C. a four-carbon ester.

D. a four-carbon ether.

Answer 51

B. a three-carbon ester.

Typically, esters end in the suffix “-oate,” but since this molecule is a derivative of acetic acid, it simply ends in “-ate.” In any case, this species cannot possibly be an ether, which are designated by the full word “ether” in their nomenclature. Acetate substituents contain two carbons, while methyl contains one, yielding three in total.

Question 52

An organic chemistry professor wants to demonstrate a substitution reaction to his students. If he needs this process to occur in a single step and invert the stereochemistry of the original reactant, what starting material should he use?

A. A quaternary alkyl halide, because S_N1 reactions are favored by substituted reagents

B. A tertiary alkyl halide, because the electron-donating nature of its substituents will promote the formation of a carbocation

C. A primary alkyl halide, because S_N2 reactions require such a substrate

D. A primary alkyl halide, because it lacks steric hindrance

Answer 52

D. A primary alkyl halide, because it lacks steric hindrance

The question stem describes an S_N2 procedure. These processes, formally known as bimolecular nucleophilic substitution reactions, happen in one step and involve a “backside attack” by a strong nucleophile. Since the attacking atom must bind at the same time as the leaving group is removed, these reactions require an unhindered substrate.

Question 53

Consider the structure below.

The bond labeled C is part of which functional group?

A. An anhydride

B. An imine

C. An enamine

D. An amide

Answer 53

B. An imine

An imine involves a nitrogen atom that is connected to a carbon atom with a double bond.

Question 54

An unsaturated free fatty acid must contain:

I. an alkene or alkyne.

II. an ester.

III. a carboxylic acid.

A. I only

B. III only

C. I and II

D. I and III

Answer 54

D. I and III

The term “unsaturated” indicates that there is at least one multiple bond, generally an alkene, in the molecule. Additionally, any fatty acid, as its name implies, must have a carboxylic acid at one terminal.

Question 55

A Ph. D. student conducts a TLC procedure to comparatively analyze three different compounds, labeled below as X, Y, and Z. He uses a silica stationary phase and a toluene solvent.

According to the diagram above, the compound with the lowest Rf is:

A. Compound X, because it is the smallest distance from the solvent front.

B. Compound X, because it interacts the most favorably with the stationary phase.

C. Compound Y, because it is the least polar.

D. Compound Y, because it interacts least favorably with the solvent.

Answer 55

D. Compound Y, because it interacts least favorably with the solvent.

Rf is a measure of the distance traveled by a compound in comparison to the distance covered by the solvent front. In other words, the lowest Rf will be attained by the molecule that traveled the shortest distance along the plate. Here, that is Compound Y, which did so because it interacted more readily with the plate than with the solvent.

Question 56

Consider the figure below:

Which of the following carbohydrates is NOT an epimer of the molecule on the right?

A. Fructose

B. Galactose

C. Mannose

D. Allose

Answer 56

A. Fructose

A is correct. As a ketose, fructose cannot possibly be an epimer of glucose, an aldose. Epimers differ only in the stereoconfiguration at a single chiral center.

Question 57

D-glucose and L-glucose have identical IR spectra because:

A. they have the same functional groups.

B. they have the same specific rotation.

C. they have the same boiling point.

D. they have the same molecular weight.

Answer 57

A. they have the same functional groups.

IR tests for functional groups by measuring the vibration states of the bonds involved. D-glucose and L-glucose are stereoisomers and possess identical functional groups.

Question 58

The following diagram shows several reactants and intermediates involved in a multi-step reaction.

If a laboratory researcher is attempting to conduct this reaction, which conditions should he use to facilitate the first step?

A. Acidic conditions only

B. Basic conditions only

C. Either acidic or basic conditions

D. Neutral conditions only

Answer 58

C. Either acidic or basic conditions

The first step in this reaction is the formation of the hemiacetal. Since the oxygen atom in methanol is a fairly good nucleophile, and since nucleophiles readily attack the carbons of carbonyl groups, this step can occur under either acidic or basic conditions. Note that the formation of the acetal, the second step, is only possible under acidic conditions.

Question 59

Tertiary alkyl halides are less susceptible to S_N2 reaction than primary alkyl halides due to:

A. leaving group stability.

B. steric hindrance.

C. carbocation instability.

D. solvation effects.

Answer 59

B. steric hindrance.
In an S_N2 reaction, no carbocation intermediate is formed, and the nucleophile must attack the molecule in its original state. Tertiary alkyl halides are too hindered for this to occur effectively.

The degree of substitution of an alkyl halide has no effect on the leaving group stability, as that is inherent of the leaving group itself.

Question 60

A biochemist wishes to convert the molecule below into 4,5-dimethyl-2-heptanol.He could best accomplish this task by:

A. reacting the molecule with one equivalent of K₂Cr₂O₇.

B. reacting the molecule with one equivalent of NaBH₄.

C. reacting the molecule with H₂ and a platinum catalyst.

D. none of the above.

Answer 60

D. none of the above.

To convert the eight-carbon structure shown into a nine-carbon molecule, we must somehow establish a new carbon-carbon bond. Interestingly, the Grignard reaction could accomplish this, but that would require reaction with CH₃MgBr. This process is not listed, and none of the answer choices could do the same.

Question 61

Several researchers test out a new mass spectrometry apparatus using a known sample of 3-pentanone. They note several peaks, one of which corresponds to an m/z value of 15. How can this reading be interpreted?

A. One of the fragments has a mass of 15 amu, likely a methyl group.

B. One of the fragments has a mass of 15 amu, likely the carbonyl functionality.

C. One of the fragments has a mass-to-charge ratio of 15 amu/charge, likely a methyl group.

D. This scenario is impossible, since 3-pentanone is not an ion.

Answer 61

C. One of the fragments has a mass-to-charge ratio of 15 amu/charge, likely a methyl group.

Mass spectrometry peaks represent mass-to-charge ratios. More specifically, mass spec involves the fragmentation and ionization of the molecule in question. Typically, this ionization simply entails the removal of one electron, leaving the mass of the fragment virtually unchanged. Here, the molecule was likely broken so as to remove a methyl group, and that group was likely made to carry a charge of +1. As a result, we see an m/z reading of (12 + 1 + 1 + 1) / (1) or 15.

Question 62

Consider the structures of trans-3-pentene and cis-3-pentene, two geometrically isomeric alkenes.

Which of these compounds is predicted to have the lower melting point?

A. Trans-2-pentene, because its molecules cannot stack as efficiently

B. Trans-2-pentene, because its molecules can stack more efficiently

C. Cis-2-pentene, because its molecules cannot stack as efficiently

D. Cis-2-pentene, because its molecules can stack more efficiently

Answer 62

C. Cis-2-pentene, because its molecules cannot stack as efficiently

The cis isomer of this alkene should melt at a lower temperature than the trans isomer. In general, this trend is true of alkenes, as the trans form (pictured on the left in this case) is better able to stack on top of like molecules. Better stacking equates to more surface area over which intermolecular forces may be exerted, making the substance more difficult to melt.

Question 63

Which of the following carboxylic acids will be least acidic?

A. CH₃CH₂CHClCH₂COOH

B. CH₃CCl₂CHClCH₂COOH

C. CH₃CH₂CFClCH₂COOH

D. CBrH₂CH₂CH₂CH₂COOH

Answer 63

D. CBrH₂CH₂CH₂CH₂COOH

Halogens are electron-withdrawing and help stabilize the negative charge on the conjugate base. This choice only has 1 halogen atom, and it is located the farther from the carboxylic acid group. The farther the halogen is from the carboxylic acid group, the more difficult it is for the halogen to stabilize the negative charge on the conjugate base. Bromine is also less electron-withdrawing than fluorine or chlorine and would stabilize the negative charge on the anionic base the least.

Question 64

Which of the following statements is/are true?

I. The carbonyl carbon in an aldehyde is less sterically hindered than that of a ketone.

II. The carbonyl carbon in a ketone is more positively charged than in that of an aldehyde.

III. Aldehydes are more reactive than ketones.

IV. A C=O bond is shorter than a C=C bond.

A. I only

B. I and III

C. I, II, and IV

D. I, III, and IV

Answer 64

D. I, III, and IV

For statement I, remember that a ketone’s carbonyl carbon is hindered by two methyl groups, while the same carbon in an aldehyde has only one. As a result, aldehydes can be more easily approached by nucleophiles, making these compounds more reactive as well. Finally, in comparison to C=C bonds, C=O bonds are shortened slightly due to the electronegativity of the oxygen atom.