# PCAT Chemical Processes Practice Test

Below you will find our free PCAT Chemical Processes practice test. The PCAT exam includes 48 Chemistry questions that must be completed within 35 minutes. The topics covered are general chemistry, organic chemistry, and basic biochemistry processes. Our practice questions are designed to be very challenging. Instant scoring is provided along with detailed explanations. Continue your test prep now with our PCAT chemistry practice questions.

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 Question 1

### Which molecule is nonpolar?

 A carbon monoxide B acetone C carbon tetrachloride D ethanol
Question 1 Explanation:
The correct answer is (C). Carbon tetrachloride is the only nonpolar compound because it is the only compound listed in which all of the electrons are shared equally across the four identical sp3-hybridized C-Cl bonds; each chlorine atom bonds with the carbon atom with the same force. Consequently, carbon tetrachloride exhibits a tetrahedral, centrosymmetric structure.
 Question 2

### A 50mL solution of HCl is diluted to 250mL at 10M. What was the initial concentration of the HCl?

 A 500M B 50M C 5M D 0.5M
Question 2 Explanation:
The correct answer is (B). The product of the initial concentration and the initial volume is equal to the product of the final concentration and the final volume of the solution. This can be remembered as:

c1 × v1 = c2 × v2

Substitute the known quantities and solve for the unknown:

c1 = c2 × v2/v1 = 10M × (250mL/50mL) = 50M
 Question 3

### What product results from treating ethanol with a solution of pyridinium chlorochromate (PCC) in dichloromethane?

 A acetaldehyde B formaldehyde C acetone D acetic acid
Question 3 Explanation:
The correct answer is (A). PCC is a mild oxidizing reagent used to oxidize primary alcohols to aldehydes. Reaction of ethanol, a primary alcohol, with PCC results in a 2 carbon aldehyde (because of the 2 carbons initially present in ethanol) called ethanal, or acetaldehyde. Formaldehyde (B), acetone (C) and acetic acid (D) are not oxidized by PCC in CH2Cl2.
 Question 4

### IV.  CH3CO2H

 A II > I > IV > III B IV > II > III > I C III > IV > I > II D I > III > II > IV
Question 4 Explanation:
The correct answer is (A). Categorize the options as strong acids, weak acids, neutral, weak bases, or strong bases and rank. (I) is neutral; (II) is basic; (III) is strongly acidic; (IV) is weakly acidic. Low pKa values correspond to higher acidity levels, so (III) will have the lowest pKa and (II) will have the highest pKa value.
 Question 5

### What is the oxidation number of nitrogen in NF3 ?

 A −3 B 0 C +2 D +3
Question 5 Explanation:
The correct answer is (D). The sum of oxidation numbers of a neutral compound is always 0. The oxidation number of fluorine is always −1. Let x represent the oxidation number of N, set up the equation, and solve for x:

x + (−1) × 3 = 0 → x = 3
 Question 6

### Which formula corresponds to a conjugate base of sulfuric acid?

 A $\text{H}_2 \text{SO}_3$ B $\text{HSO}_4^−$ C $\text{SO}_4^{2−}$ D $\text{H}_3 \text{SO}_4^+$
Question 6 Explanation:
The correct answer is (B). In a solution, sulfuric acid will donate an H+ to water to form hydronium (H3O+). The loss of the single H+ results in H2SO4 becoming:
$\text{HSO}_4^−$
 Question 7

### What would be the result of increasing the number of nucleophiles in a SN1 reaction?

 A The rate would increase exponentially. B The rate would increase linearly. C The rate would decrease. D The rate would not be affected
Question 7 Explanation:
The correct answer is (D). SN1 reactions are substitution reactions that depend on reagents being attracted to electrons (electrophiles). The rate of an SN1 reaction is determined by the product of its first order rate constant and the concentration of substrate but does not depend on the number of nucleophiles involved in the reaction. Consequently, altering the concentration of nucleophiles does not alter the rate of reaction.
 Question 8

### Which of the following species cannot form intermolecular hydrogen bonds with itself?

 A ethanol B acetic acid C ammonia D acetone
Question 8 Explanation:
The correct answer is (D). Hydrogen bonds occur between relatively acidic O-H and/or N-H hydrogen atoms of one molecule with an electronegative atom (oxygen, nitrogen, or fluorine) of another atom. The O-H hydrogen in ethanol (A) and in acetic acid (B) can form intermolecular hydrogen bonds with an oxygen atom in another molecule. Similarly, the N-H hydrogen in ammonia [which is less acidic than the O-H hydrogens in (A) and (B)], can hydrogen-bond to the nitrogen atom in another ammonia molecule. However, the C-H bonds in acetone are not sufficiently acidic to participate in hydrogen bonding; hence, acetone is unable to hydrogen bond with itself.
 Question 9

### Which of the following best describes the result of replacing one mole of H2 by one mole of N2 in a closed container?

 A The temperature of the gas at a given pressure and volume will increase. B The pressure of the gas at a given temperature and volume will increase. C The volume of the gas at a given temperature and pressure will increase. D The rate of diffusion of the gas from a small hole in the container will decrease.
Question 9 Explanation:
The correct answer is (D). Recall that one mole of any ideal gas occupies a volume of 1 liter at standard temperature (298 K) and pressure (1 atm). Thus, neither temperature (A) nor pressure (B) nor volume (C) will be affected by increasing the molar mass of a gas when the other two PVT variables are held constant. The correct answer is (D). Graham’s Law of Diffusion states that the rate of diffusion is inversely proportional to the molecular weight of the molecule. Consider replacing H2 (molar mass 2 g/mol) by N2 (molar mass 28 g/mol); substitute these values into Graham’s equation and compare the results:

R(N2)/R(H2) = [M(H2)/M(N2)]1/2 = [2.0/28]1/2 = 0.27

The rate has decreased with the increase in molar mass.
 Question 10

### III.  Asp

 A I < II < III B III < II < I C II < I < III D III < I < II
Question 10 Explanation:
The correct answer is (B). As noted in the stem, the pI of a molecule is its isoelectric point, or the point at which, in this case, the amino acid will have no overall charge. Categorizing the amino acids: (I) is positively charged and basic; (II) is non polar and neutral; (III) is polar and acidic. Acidic amino acids on average have the lowest pI, nonpolar amino acids have a higher pI, and basic amino acids carry the largest pI.
 Question 11

### What is the result of adding HBr to propene in the presence of hydrogen peroxide?

 A 2-propanol B 1-propanol C 2-bromopropane D 1-bromopropane
Question 11 Explanation:
The correct answer is (D). In the presence of an organic peroxide, the addition of HBr (and peroxide) to propene results in an anti-Markovnikov product in which the hydrogen becomes attached to the carbon with fewer hydrogens connected to it. The product of the addition is CH3CH2CH2Br, or 1-bromopropane.
 Question 12

### Which of the following has the highest boiling point?

 A methanol B n-propanol C isopropanol D ethanol
Question 12 Explanation:
The correct answer is (B). Boiling points primarily depend upon the cohesive forces between molecules.

When dealing with molecules of relatively similar molecular weights, the boiling points will depend upon the functional groups involved, and as the molecular weight of the molecule increases, the boiling point also increases.

Lastly, branching, which results in decreasing attractive intermolecular London/van der Waals interactions, decreases boiling point. With these considerations, n−propanol is expected to exhibit the highest boiling point of the options.
 Question 13

### The rate law for a reaction is of the second order. Which statement is true?

 A The rate must depend on both reactants. B The reaction must depend on the square of one reactant. C The reaction must depend on only k squared. D The reaction must depend on at least one of the reactants.
Question 13 Explanation:
The correct answer is (D). In a second-order reaction the reaction rate is dependent upon either the product of the reactants (i.e. k2[A][B]) or the square of one of the reactants (i.e. k2[A]2). Only answer choice (D) fits this definition.
 Question 14

### The solubility of KCl in seawater would be ____ than the solubility of KCl in tap water.

 A higher B the same as C lower D KCl would be completely insoluble in sea water as KCl cannot dissociate in water.
Question 14 Explanation:
The correct answer is (C). Potassium chloride is a Group I metal halide salt. And, while it dissolves readily in water, its solubility is lowered because of the common ions present in the seawater, e.g., Cl formed via dissociation of dissolved salts, such as NaCl. This is otherwise known as the common ion effect.
 Question 15

### An electron pair donor is best classified as a

 A strong acid B Bronsted-Lowry acid C Lewis acid D Lewis base
Question 15 Explanation:
The correct answer is (D). By definition, a Lewis base (D) is an electron pair donor. A Lewis acid (C), on the other hand, is an electron pair acceptor. A strong acid (A) could be either a Lewis acid or a Bronsted-Lowry acid, neither of which can function as an electron pair donor.
 Question 16

### III.  +∆H, −∆S

 A I only B II only C I and III only D I and II only
Question 16 Explanation:
The correct answer is (D). Under favorable conditions, ∆H < 0, and ∆S > 0, a reaction will be spontaneous at any temperature. If one of the conditions is unfavorable with either ∆H > 0 or ∆S < 0, a reaction can be spontaneous if it satisfies ∆G < 0 using the standard-state free energy of reaction equation: ∆G = ∆H − T∆S. Roman numeral (I) satisfies both favorable conditions and will be spontaneous. Roman numeral (II) can be spontaneous if ∆H is sufficiently low. Roman numeral (III) can never be spontaneous.
 Question 17

### Which of the following alcohols is the most easily dehydrated?

 A (CH3)3C−OH B (CH3)2CH−OH C CH3CH2−OH D CH3−OH
Question 17 Explanation:
The correct answer is (A). Classify the alcohols as primary, secondary, or tertiary: (A) is tertiary because the −OH group is directly attached to 3 alkyl groups; (B) is secondary because the carbon attached to the −OH is connected to 2 carbon atoms; (C) is primary because the carbon connected to the −OH group is connected to 1 carbon atom; and although (D) lacks an attached alkyl group, it is considered primary. As tertiary alcohols form the most stable carbocations, (A) is correct.
 Question 18

### What is the net charge on the heptapeptide Ile-Ala-Lys-Phe-Glu-Asn-Asp at human physiological pH?

 A −1 B 0 C +1 D +2
Question 18 Explanation:
The correct answer is (A). Recall that amino acids are classified as basic, neutral or acidic according to the properties of their side chain. Only those side chains containing a proton-ionizable functionality will be affected by changing pH of the medium.

In the present example, Ile, Ale, Phe, and Asn possess neutral side chains. Lys contains a basic side chain (NH2 group); at human physiological pH (ca. 7.4) the NH2 group in Lys will be protonated (NH3+) and thus bear +1 charge. However, both Asp and Glu contain acidic side chains (CO2H groups), which both will be deprotonated at pH 7.4 to afford CO2- (-1 charge on each group). Hence, the net charge on the heptapeptide at pH 7.4 is expected to be -1.
 Question 19

### Which of the following reagents can best be used to carry out the reaction indicated below?

 A DIBAL-H (Diisobutylaluminum hydride) B NaBH4 C LiAIH4 D Pyridinium chlorochromate (PCC)
Question 19 Explanation:
The question asks examinees to recognize the best reagent that can be used to selectively reduce a ketone C=O group in the presence of an ester group. Note that PCC is an oxidizing agent, not a reducing agent; eliminate (D). DIBAL-H capable of reducing ester groups to carboxaldehydes (HC=O groups); eliminate (A). Sodium borohydride (B) can be used to perform the reduction depicted in the stem. Lithium aluminum hydride is a more powerful reducing agent that will reduce both the ketone and ester functionalities to CH2OH; eliminate C.
 Question 20

### What is the conjugate acid of HCl?

 A Cl‾ B Cl2‾ C H2Cl+ D H2O
Question 20 Explanation:
The correct answer is (C). The conjugate acid of a compound results from the addition of a proton (H+). The theoretical conjugate acid of HCl would be: HCl + H+ → H2Cl+.
 Question 21

### Which of the following are reducing agents?

Question 21 Explanation:
The correct answer is (D). Reducing agents are those that donate electrons in redox chemical reactions. Cellular respiration involves the carrying of electrons by both FADH2 and NADH, which generate 2 ATP and 3 ATP molecules, respectively. Because both must be included in the answer, (D) is correct.
 Question 22

### What is the partial pressure of nitrogen in a gas at 30 atm consisting of equal moles of oxygen, nitrogen and carbon?

 A 3 atm B 10 atm C 15 atm D 30 atm
Question 22 Explanation:
The correct answer is (B). Dalton’s law states that the total pressure of a gas mixture is equal to the sum of the partial pressures. In this case, the total pressure is 30 atm, which is the result of 3 equally combined gases. Dividing the total pressure by the number of gases yields the partial pressure of each gas: 30 ÷ 10 = 10 atm.
 Question 23

### Which statement correctly describes the effect produced when a nonvolatile solute is added to a chemically pure solvent?

 A The boiling point of the resulting solution is expected to be lower than that of pure solvent. B The melting point of the resulting solution is expected to be lower than that of pure solvent. C The freezing point of the resulting solution is expected to be higher than that of pure solvent. D The vapor pressure of the resulting solution is expected to be higher than that of pure solvent.
Question 23 Explanation:
The correct answer is (B). All four choices describe colligative properties, which depend upon the number but not the nature of solute particles in solution.

Recall that addition of solutes increases the boiling point [eliminate (A)] but lowers the freezing/melting point of pure solvent. Eliminate (C).

Recall that the vapor pressure of a solution is determined by applying Raoult’s law, i.e., the partial pressure exerted by individual components of an ideal liquid mixture is equal to the product of the vapor pressure of each pure component and its mole fraction present in the mixture.

Addition of a nonvolatile solute decreases the mol fraction of solvent in solution, thereby decreasing the vapor pressure of the resulting solution relative to that of pure solvent. Eliminate (D). The correct response is (B).
 Question 24

### An element whose atomic number is 16 would have how many unpaired electrons?

 A 0 B 1 C 2 D 4
Question 24 Explanation:
The correct answer is (C). The atomic number indicates the number of protons (and electrons). Writing the electron configuration: 1s2 2s2 2p6 3s2 3p4 (2 + 2 + 6 + 2 + 4 = 16). The 3p orbital will contain 1 filled orbital, and 2 half-filled orbitals (recall the Hund Principle of Maximum Multiplicity), which requires two unpaired electrons.

 Question 25

### If a substance has a constant half life of 10 weeks, how many grams of a 12 g sample would remain after 30 weeks?

 A 8 g B 6 g C 3 g D less than 2 g
Question 25 Explanation:
The correct answer is (D). Thirty weeks corresponds to 3 half-lives. After the first half life (10 weeks), there will be 12 ÷ 2 = 6 g remaining. After another 10 weeks, there will be 3 g remaining. And after the final 10 weeks, there will be 1.5 g remaining.
 Question 26

### IV.  CH3CH2CO2H

 A II, III, IV, I B II, IV, I, III C II, III, I, IV D IV, I, III, II
Question 26 Explanation:
The correct answer is (C). There are no strong acids present in the answer choices. There is only one weak acid present in the answer choices, CH3CH2CO2H. The presence of the carboxyl group indicates a weak acid. Because there are no other weak acids, (IV) must be the most acidic compound included, so answer choice (C) is correct.
 Question 27

### Which of the following would show the greatest number of peaks at different chemical shift values in its proton NMR spectrum?

 A CH4 B CH3CH3 C CH3CH2CH2CH3 D CH3CH2CH2OH
Question 27 Explanation:
The correct answer is (D). Chemical shift values of protons in a molecule are determined by their relative magnetic environment. Since the four protons in methane (A) are chemically and magnetically equivalent, the proton NMR spectrum of methane is expected to display a single absorption signal (peak).

Similarly, all protons in (B) are magnetically equivalent due to rapid rotation about the C-C bond at ambient temperature and thus give rise to a single peak.

The two CH3 groups in (C) are equivalent but occupy a magnetic environment that differs from that of the two equivalent CH2 groups in n-butane. Hence, the proton NMR spectrum of (C) is expected to display two peaks.

The correct answer is (D), whose proton NMR spectrum is expected to display four peaks.
 Question 28

### Which of the following statements best describes how resonance interactions contribute to the molecular stability of acetate anion, CH3CO2-?

 A Resonance decreases the charge per unit volume in acetate anion. B Resonance decreases the bond order of the carbon-carbon CH3-CO2- bond in acetate anion. C Resonance strengthens the carbonyl C=O dipole moment in acetate anion. D Resonance delocalizes the negative charge in acetate anion over both carbon atoms.
Question 28 Explanation:
The correct answer is (A). The two equivalent resonance forms for acetate anion along with the corresponding resonance hybrid are depicted below:

Therein it can be seen that resonance serves to delocalize the negative charge on the acetate anion over the two oxygen atoms. The resonance hybrid bears a single negative charge spread over a larger volume (i.e., over two electronegative oxygen atoms instead of one as depicted in each canonical form). As a result the charge per unit volume in the resonance hybrid is lower than that in either canonical form, thereby resulting in net stabilization of the acetate anion. The correct response is (A).

The carbon-carbon CH3-CO2- bond in acetate anion is not part of the delocalized electronic system; hence its bond order is not affected by resonance [eliminate (B) and (D)].

The resonance hybrid in acetate anion no longer contains a carbonyl functionality [eliminate (C)].
 Question 29

### Which of the following would have the highest pH in a solution?

 A hydrofluoric acid B water C ammonia D hydrochloric acid
Question 29 Explanation:
The correct answer is (C). Ammonia is both a Bronsted base and a Lewis base and thus can accept protons from water molecules. Water is considered neutral, whereas HF and HCl are Bronsted acids that display low pH values due to ionic dissociation in aqueous solutions. Ammonia, a base, has the highest pH among the answer choices.
 Question 30

### Which compound is expected to be the major product formed by heating an equimolar solution of CH3CH2-CHBr-CH3 and sodium ethoxide (NaOCH2CH3) in ethanol at 80°C?

 A 1-Butene B Z-2-butene C E-2-butene D 2-Ethoxybutane
Question 30 Explanation:
Reaction of a secondary alkyl bromide with a relatively strong base (NaOCH2CH3), when performed in a polar, protic solvent (ethanol) at elevated temperature (80 °C), is expected to favor E2 elimination vis-à-vis SN2 substitution [eliminate (A)].

Note that bimolecular E2 elimination in 2-bromobutane can proceed to afford alkenes with either (or both) of two possible orientations. Recall that Zaitsev-rule E2 elimination would afford the more highly substituted (and hence thermodynamically favored) product, i.e., 2-butene. In contrast, Hofmann-rule elimination is expected to produce the corresponding terminal alkene, i.e., 1-butene.

The use of a relatively small base (ethoxide ion) favors the product alkene that results via application of Zaitzev-rule [eliminate (D)].

Note that E-2-butene is less sterically crowded than the corresponding Z-isomer; hence, the former alkene (C) is expected to be the major product of the reaction described in the stem. Eliminate (B).
 Question 31

### Glyceraldehyde-3-phosphate (G3P) --------→ Pyruvate

 A cytoplasm B nucleus C mitochondria D endoplasmic reticulum
Question 31 Explanation:
The correct answer is (A). The oxidation of G3P into pyruvate occurs during the glycolysis stage of cellular respiration.

(C) Glycolysis occurs in the cytoplasm, outside of the mitochondria. Other steps involved in glycolysis, e.g., the citric acid cycle, occur in mitochondria (C), but glycolysis does not. Eliminate (C).

(B) and (D) The endoplamic reticulum (ER) is a membrane-bound organelle in eukaryotic cells. Several essential cellular functions take place in the ER, e.g., protein and lipid synthesis along with calcium ion storage and release. Neither the nucleus nor the ER are involved in glycolysis or the citric acid cycle. Eliminate (B) and (D).
 Question 32

### Which would yield an endothermic reaction?

 A ∆G = 298 kcal-mol-1 T = 298 K ∆S = −1 kcal-mol-1K-1 B ∆G = −298 kcal-mol-1 T = 298 K ∆S= −2 kcal-mol-1K-1 C ∆G = −298 kcal-mol-1 T = 0 K ∆S = 1 kcal-mol-1K-1 D ∆G = −298 kcal-mol-1 T = 298 K ∆S = +2 kcal-mol-1K-1
Question 32 Explanation:
The correct answer is (D). The free energy change of a reaction, ∆G, is equal to the difference between ∆H, the change in enthalpy, and the product of the temperature, T, with the change in entropy, ∆S, of the system: ∆G = ∆H − T∆S. An endothermic reaction requires a ∆H > 0. Solve for ∆H in each case. (A): 0, (B): −894, (C): −298, (D): +298.
 Question 33

### What is the total vapor pressure above a solution containing 39 g of benzene and 50 g of n-heptane at 300 K? The vapor pressure of each pure liquid at 300 K is 102 torr (benzene) and 48 torr (n-heptane).

 A 11 torr B 54 torr C 75 torr D 150 torr
Question 33 Explanation:
Raoult’s law states that the partial pressure exerted by individual components of an ideal liquid mixture is equal to the product of the vapor pressure of each pure component and its mole fraction present in the mixture, i.e.,

$P_{\text{total}} = \sum_{i} p_{i} \chi_{i}$

The molecular mass of benzene (C6H6) is 78 g/mol, whereas that of n-heptane (C7H16) is 100 g/mol. The number of moles of each component in solution is found by dividing the number of grams of the pure component by its molecular mass: benzene = 39/78 = 0.50 mol; n-heptane = 50/100 = 0.50 mol. Applying Raoult’s law:

$P_{\text{total}} = \sum p_{i} \chi_{i} =$
$(p_{\text{benzene}})(\chi_{\text{benzene}}) + (p_{n-\text{heptane}})(\chi_{n-\text{heptane}})$

$P_{\text{total}}$ $= (102 \text{ torr} × 0.50) + (48 \text{ torr} × 0.50)$ $= (51 + 24 \text{ torr})$ $= 75 \text{ torr}$

 Question 34

### What effect would doubling the volume and temperature of an ideal gas have on its pressure?

 A double the pressure B half the pressure C make the pressure zero D no effect
Question 34 Explanation:
The correct answer is (D). The ideal gas law states: PV = nRT, where P is pressure, V is volume, n is the amount, R is the ideal gas constant, and T is the gas’s temperature. Doubling the volume, V = 2V, and doubling the temperature, T = 2T, will have not alter the pressure because the 2’s on both sides of the equation will cancel, yielding the original pressure.
 Question 35

### Which one of the following reactions is not expected to afford acetic acid?

 A Diethyl malonate + aqueous H2SO4, heat B Acetone + KMnO4, reflux C (1) Iodomethane + Mg/dry ether; (2) CO2, (3)H3O+ D Acetonitrile + aqueous H2SO4, prolonged reflux
Question 35 Explanation:
(A) Diethyl malonate, (EtO2C-CH2-CO2Et) when heated with aqueous acid is expected to be hydrolyzed to malonic acid + ethanol. Subsequently, since malonic acid is a β-dicarboxylic acid, this compound will undergo decarboxylation to afford acetic acid + CO2. Elilminate (A).

(B) Vigorous oxidation of acetone results in rupture of carbon-carbon bonds. Under forcing conditions the oxidation products will be CO2 and H2O. The correct response is (B).

(C) Iodimethane + Mg/dry ether produces the corresponding Grignard reagent (CH3MgBr), which subsequently reacts with CO2 to afford CH3C(O)OMgBr. Aqueous acidic workup then is expected to produce acetic acid. Eliminate (C).

(D) Acetonitrile is hydrolyzed by aqueous acid, initially affording acetamide [CH3C(O)NH2], which upon continued heating with aqueous acid suffers further hydrolysis to yield acetic acid. Eliminate (D).
 Question 36

### The hybridization of an oxygen atom in CO2 is:

 A s B sp C sp2 D sp3
Question 36 Explanation:
The correct answer is (C). A molecular orbital diagram and a Lewis structure for CO2 that includes all chemical bonds and unshared electron pairs are shown below.

Therein it can be seen that the bond between carbon and each oxygen atoms is a C=O double bond. Accordingly, both oxygen atoms are sp2 hybridized. Note that CO2 is a linear molecule; the central carbon atom is sp hybridized.
 Question 37

### An enzyme reaction reaches _____ at _____.

 A Vmax, Km B Vmax, ½Km C ½Vmax, Km D ½Vmax, ½Km
Question 37 Explanation:
The correct answer is (C). The kinetic behavior of many enzymes is described by the Michaelis-Menten equation

$V = \dfrac{V_{\text{max}}[\text{S}]}{\text{K}_{\text{m}} + [\text{S}]}$

in which Vmax is the maximum rate at which an enzymatic reaction can occur. Km is the substrate concentration yielding a reaction rate that is half of the maximum reaction rate.
 Question 38

### The infrared spectrum of which of the following is expected to display an absorption signal at 2190 cm−1?

 A alkane B alkene C alkyne D ketone
Question 38 Explanation:
The correct answer is (C). Alkynes possess a carbon to carbon triple bond, the IR stretching frequency of which ranges between 2260 cm−1 and 2100 cm−1. Alkanes, alkenes, and ketones possess C-H, C=C, and C=O bonds, respectively. None of these other bonds are expected to display IR absorption signals in the region 2260-2100 cm−1.
 Question 39

### Ozonolysis of Compound X, C7H12, followed by reductive workup of the reaction mixture with dimethyl sulfide (DMS) produced two aldehydes. Which one of the following compounds is most likely to correspond to Compound X?

 A 4-Methylhexa-1,4-diene B 5-Methylhexa-1,4-diene C 1-Methylcyclohexene D 1,6-Heptadiene
Question 39 Explanation:
The correct answer is (D). Inspection of the molecular formula CnH2n-2 indicates that Compound X contains two degrees of unsaturation. Given the four compounds (A)–(D), Compound X can either be a diene or a cyclic monoalkene.

Recall that ozonolysis is employed to cleave C=C double bonds in alkenes and carbon to carbon triple bonds in alkynes. Oxidative workup of the reaction mixture obtained via ozonlysis of alkenes produces carboxylic acids, whereas the corresponding reductive workup affords aldehydes. Ozonolysis of internal alkynes followed by aqueous workup affords carboxylic acids.

(A) Ozonolysis of H3C-CH=C(CH3)-CH2CH=CH2 is expected to afford three products: H3C-CH=O + H3C-C(O)-CH2CH=O + H2C=O. Eliminate (A).

(B) Ozonolysis of (H3C)2C=CH-CH2CH=CH2 is expected to afford three products: (H3C)2C=O + O=CH-CH2-CH=O + H2C=O. Eliminate (B).

(C) Ozonolysis of 1-methylcyclohexene is expected to proceed with concomitant ring opening to afford only one product, i.e., H3C-C(O)-(CH2)4CH=O. Eliminate (C).

(D) Ozonolysis of H2C=CH(CH2)3CH=CH2 is expected to afford two products: O=CH(CH2)3CH=O + H2C=O (2 equivalents). The correct response to this question is (D).
 Question 40

### (b) Which cofactor is required to carry out this step?

 A a) Enzyme-mediated oxidation of isocitrate; (b) NAD+ B (a) Enzyme-mediated oxidation of isocitrate; (b) NADH C (a) Enzyme-mediated reduction of isocitrate; (b) NAD+ D (a) Enzyme-mediated reduction of isocitrate; (b) NADH
Question 40 Explanation:
A basic amino acid residue in the enzyme (α-ketoglutarate dehydrogenase) abstracts a proton from the substrate as indicated below with concomitant transfer of hydride ion from OH to NAD+ as illustrated below. Thus, the C-OH carbon atom in isocitrate is oxidized with concomitant reduction of the nicotinamide cofactor to NADH. The correct answer is (A).

 Question 41

### Which of the following has the largest bond order?

 A $\text{O}_2$ B $\text{NO}_3^−$ C $\text{NO}_2^+$ D $\text{CN}^−$
Question 41 Explanation:
The correct answer is (D). Bond order for diatomic molecules is found by determining the type of bonds between the 2 atoms.

In cases where there are more than two atoms involved, count the number of bonds, then count the number of bond groups between atoms, then calculate the ratio of the number of bonds to the total number of bond groups.

Lewis structures for the four choices appear below.

The Lewis structure for (A) shows a double bond between oxygen atoms, and its bond order is 2.

(B) Shows a double bond between N and O, and a single bond between N and each O.

$\frac{\text{total number of bonds}}{\text{number of groups}}$ $= \frac{4}{3} = 1.33$

(C) Shows 2 double bonds between N and O: 4 ÷ 2 = 2.

(D) Shows 1 triple bond between C and N and a bond order of 3. It follows that (D) has the largest bond order.
 Question 42

### What is the osmotic pressure of a 2.00 M NaCl solution at 273 K?

 A 44.8 atm B 45.5 atm C 0 atm D 97 atm
Question 42 Explanation:
The correct answer is (A). Calculate osmotic pressure by using the van't Hoff expression π = MRT, where π is pressure, M is the molar concentration of the dissolved solution, R is the ideal gas constant (0.08206 L atm mol−1 K−1), and T is temperature in Kelvin. Substituting and solving:

π = MRT = (2.0 mol-L−1) × (0.08206 L-atm-mol−1-K−1) × (273 K) = 44.8 atm
 Question 43

### Which compound forms when acetyl-CoA enters the TCA cycle?

 A oxaloacetate B acetic acid C citrate D FADH2
Question 43 Explanation:
The correct answer is (C). Prior to the TCA cycle, glycolysis functions to break down glucose into pyruvate, which is then decarboxylated to afford acetyl-CoA. Acetyl-CoA combines with oxaloacetate to produce citrate, which marks the start of the TCA cycle.
 Question 44

### Which compound has the lowest boiling point?

 A CH3(CH2)2CH3 B CH3CH(CH3)2 C CH3(CH2)2CH2OH D CH3CH(CH3)OH
Question 44 Explanation:
The correct answer is (B). When comparing boiling points among compounds, consider these three generalities: (i) boiling point increases relative to the strength of the intermolecular forces present in the compound, which decrease in the order: hydrogen bonding > dipole-dipole interactions > van der Waals forces. (ii) Boiling point also increases as the molecular weight of the compound increases. (iii) Increasing the degree of branching within the compound decreases the boiling point by mitigating van der Waals attractive forces. Answer choice (B) is correct.
 Question 45

### Which statement offers the best rationale to account for the fact that CO2 fails to behave as an ideal gas at STP (298 K, 1 atm)?

 A Intermolecular dipole-dipole interactions render the volume occupied by CO2 at STP less that occupied by an ideal gas. B Intermolecular van der Waals interactions render the volume occupied by CO2 STP less that occupied by an ideal gas. C Intermolecular interactions between relatively large CO2 molecules render the volume occupied by the gas at STP larger that occupied by an ideal gas. D Intermolecular interactions between highly polarizable CO2 molecules render the volume occupied by the gas at STP larger that occupied by an ideal gas.
Question 45 Explanation:
Deviation from ideal gas behavior results for two main reasons: (i) The assumption that molecules of a gas are point-sized and thus occupy negligible volume within their container becomes invalid. (ii) The assumption that there are no attractive intermolecular forces between gas molecules (i.e., that molecular collisions are perfectly elastic) breaks down.

(A): Due to molecular symmetry the C=O dipoles in CO2 cancel one another; hence, CO2 possesses no net dipole moment. Eliminate (A).

(B): Intermolecular van der Waals forces result in attractive interactions, which reduce the volume occupied by CO2 at STP relative to that occupied by an ideal gas. The correct response is (B).

(C): The net result of intermolecular interactions between CO2 molecules at STP is that they are attractive and thus reduce the volume occupied by CO2 relative to that occupied by an ideal gas. Eliminate (C).

(D): “Polarizability” offers a measure of the ease with which electron clouds can be distorted. Electrons in CO2 π-orbitals are more polarizable than those in, e.g., H2, which are tightly held in the 1s orbital. Attractive intermolecular London dispersion forces are expected to be more important for CO2 than for H2 and thus reduce the volume occupied by CO2 relative to that occupied by an ideal gas. Eliminate (D).
 Question 46

### How many sigma bonds does acetylene (C2H2) have?

 A 1 B 2 C 3 D 4
Question 46 Explanation:
The correct answer is (C). Recall that single bonds are always sigma bonds, double bonds consist of a sigma bond and a pi bond, and triple bonds contain one sigma bond and two pi bonds. HCCH contains two C-H bonds and one C-C triple bond. So, 2 C-H bonds contribute two sigma bonds, along with another sigma bond from the triple bond thereby affording a total of three sigma bonds.
 Question 47

### Which one of the following statements is most consistent with these experimental results?

 A Thermolysis of Compound X proceeds via an E1 elimination reaction mechanism. B Thermolysis of Compound X proceeds via an E2 elimination reaction mechanism. C Thermolysis of Compound X proceeds via a cyclic transition state. D 1,3-Dimethylcyclohezene is formed from Compound X via anti elimination.
Question 47 Explanation:
(A) Recall that E1 eliminations proceed via formation of a carbocationic intermediate. The intermediate carbocation derived from either Compound X or Y is expected to lose a proton to afford 1,3-dimethylcyclohexene along with O=C-(SH)SCH3. However, these products were not obtained upon thermolysis of Y. Eliminate (A).

(B) Recall that E2 eliminations proceed in a single step with concomitant stereospecific anti elimination. The fact that thermolysis of Compound Y fails to afford an alkene is inconsistent with a mechanism that proceeds via anti elimination. Eliminate (B).

(C) Thermolysis of Compound X proceeds via a cyclic transition state with concomitant syn eliminaton. The all-cis isomer (Compound Y) does not possess a syn β-hydrogen and thus fails to afford an alkene upon thermolysis. A reasonable mechanism that accounts for the course of thermolysis of Compound X is shown below. The correct answer is (C).

(D) As indicated above, thermolysis of Compound X results in formation of 1,3-Dimethylcyclohezene via a syn rather than anti elimination mechanism. Eliminate (D).
 Question 48

### The viscosity of a liquid:

 A increases with decreasing temperature B increases with increasing temperature C decreases with decreasing temperature D is independent of temperature
Question 48 Explanation:
The correct answer is (A). A liquid’s viscosity, or its resistance to flow, changes with a change in temperature. As a liquid’s temperature increases, intermolecular attractions within the liquid are reduced, thereby permitting the liquid to move more freely and easily and thus consequently decreasing its viscosity.
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