MCAT Organic Chemistry Flashcards
Why is the alpha-anomer of d-glucose less likely to form than the beta-anomer?
A) The beta-anomer is preferred for metabolism
B) The beta-anomer undergoes less electron repulsion.
C) The alpha-anomer is the more stable anomer.
D) The alpha-anomer forms more in L-glucose.
B
The hydroxyl group on the anomeric carbon of the beta-anomer is equatorial, thereby creating less steric hinderance than the alpha-anomer, which has the hydroxyl group of anomeric carbon in the axial position.
Ethanol has been popular with humans for more than 10,000 years. Moreover, many animals are known to seek out rotten fruits that have fermented even moderate levels of alcohol. Which of the following agents would oxidize a primary alcohol to an aldehyde?
A) Pyridinium chlorochromate B) Na2Cr2O7 C) K2CrO7 D) CrO3, H2SO4 E) None of the above
A)
Pyridinium Chlorochromate (PCC) converts a primary alcohol to an aldehyde because it lacks the water necessary to hydrate the otherwise easily hydrated aldehyde. Both Na2Cr2O7 and K2Cr2O7 will produce a carboxylic acid. CrO3, H2SO4 will also produce a carboxylic acid.
The benzene ring typically undergoes:
A) Nucleophilic addition
B) Nucleophilic substitution
C) Electrophilic addition
D) Electrophilic substitution
D
Substitution reactions are preferred because they preserve the very stable aromatic structure. Electrophiles can join the ring because of the resonance stabilization of the intermediate ion.
Reaction of alcohols with nucleophiles often requires the presence of acid because:
A) Alcohols only react with hydrogen halides
B) -OH is a poor leaving group and therefore must be pronated to form a better leaving group, -OH2+
C) Acids prevent rearrangement of the carbocation
D) Acids prevent bond cleavage
B
The hydroxide group is a strong base and therefore a poor leaving group. It must be pronated to form a weak base, H2O, so that it can leave the molecule.
What term accurately describes the structural relationship between (1R,2S)-1,2-dibromocyclopentane and (1R,2R)-1,2-dibromocyclopentane?
A) Conformers
B) Enantiomers
C) Diastereomers
D) Constitutional isomers
C
When drawing out the structures, one can see that the basic skeletal arrangements are identical, i.e. stereoisomers (not constitutional isomers). We can dismiss conformers, since the structures cannot be interconverted exclusively by rotations about formally single bonds. The compounds are diastereomers, a class of stereoisomers that are non-superimposable, nonmirror images of one another. More specifically, the compounds are epimers: in chemistry, epimers are diastereomers that differ in configuration of only one stereogenic center.
(Difference between molecules: both have pentane ring, one has Br towards and away, the other has both Br atoms toward).
What is the major product of the reaction below, and what is the mechanism by which it is produced?
(CH3CH2)3CBr + F– –> ; temperature: 500C; solvent: water
A) (CH3CH2)3CF; SN1
B) (CH3CH2)3CF SN2
C) CH3CH=C(CH2CH3)2; E2
D) CH3CH=C(CH2CH3)2; E1
A
This is an example of a SN1 reaction where first step is Br (good leaving group) dissociates to produce a carbocation (stabilized by the polar protic solvent, water). The tertiary carbocation intermediate is very stable, especially in the highly polar protic solvent water. The carbocation then combines with the strong nucleophile, the F- ion. The original substituent (Br) must be a better leaving group than the nucleophile (F-) in order for the reaction to occur). C and D are wrong because this is not an elimination reaction. B is wrong because this is not an SN2 reaction, due to the highly hindered substrate, and the type of solvent used in the reaction (SN2 reactions favor polar aprotic solvents, not polar protic solvents).
Heterocyclic amines are found in the base of DNA. Their most important property includes their basicity due to the presence of nitrogen. Which of the following statements is true about the extra pair of electrons in pyrrole and pyridine? (Google search these amines to see structures.)
A) In pyrrole, an sp2 orbital contains the pair of electrons, while in pyridine, they are involved in the pi cloud
B) In both pyrrole and pyridine, an sp2 orbital contains the pair of electrons
C) In both pyrrole and pyridine, they are invovled in the pi cloud
D) In pyrrole, they are invovled in the pi cloud, while in pyridine, an sp2 orbital contains the pair of electrons
D
In pyrrole, the sigma bonds around nitrogen are formed by sp2 hybrid orbitals, not sp3. A nitrogen electron participates in each of the sigma bonds. The remaining nitrogen electrons are in a nitrogen 2p orbital which overlaps with the carbon 2p orbitals in the ring to form pi bonds and the cyclic compound, delocalizing the lone pair. In pyridine, the nitrogen atom donates one electron to each of the two sigma bonds by use of sp2 orbitals. The third sp2 orbital contains two electrons, which are for acids. One electron of nitrogen is present in a p orbital, which interacts with the carbon p orbitals to form the pi bond.
Heterocyclic amines are a very important type of amino group. One use of them includes being the base of DNA. The most important property however is their basicity due to nitrogen.
Pyrrole undergoes electrophilic substitution at position 2 because:
(Google search these amines to see structures.)
A) The carbocation intermediate has more resonance structures than the intermediate formed when the electrophile attacks position 3
B) The inductive effect of nitrogen is greater
C) The aromatic ring is maintained in the carbocation
D) Resonance structures cannot be drawn for the carbocation intermediate formed by the attack at position 3
A
Attack at position 2 gives three resonance structures greatly adding to stability of the molecule (positive charge on nitrogen, then position 3, and then position 5). Position 3 only has two resonance structures.
Which of the following statements about 3-bromo-1,1-dimethylcyclobutane are true?
I. It is saturated
II. It is a planar molecule.
III. It has no ring strain
A) I only
B) II only
C) III only
D) I and III
A
A saturated molecule is one where all the valences of the atoms are satisfied by single covalent bonds. In other words, there are no double bonds. Statement I is true.
A planar molecule is a flat one. In this case, we have two opposite corners in a tetrahedral shape. II is not true.
Ring strain is minimized in a cyclohexane and maximized in a cyclopropane, but it is always present in a cyclic molecule. III is not true. Thus the answer is A.
What is the major product of the reaction below, and what is the mechanism by which it is produced?
(CH3CH2)3CBr + F– –> ; temperature: 500C; solvent: water
A) (CH3CH2)3CF; SN1
B) (CH3CH2)3CF SN2
C) CH3CH=C(CH2CH3)2; E2
D) CH3CH=C(CH2CH3)2; E1
A
This is an example of a SN1 reaction where first step is Br (good leaving group) dissociates to produces a carbocation (stabilized by the polar protic solvent, water). The tertiary carbocation intermediate is very stable, especially in the highly polar protic solvent water. The carbocation then combines with the strong nucleophile, the F- ion. (the original substituent (Br) must be a better leaving group than the nucleophile (F-) in order for the reaction to occur). C and D are wrong because this is not an elimination reaction. B is wrong because this is not an SN2 reaction, due to the highly hindered substrate, and the type of solvent used in the reaction (SN2 reactions favor polar aprotic solvents, not polar protic solvents).
The hybridization state and shape of a carbocation is:
A) sp and linear
B) sp2 and trigonal planar
C) sp3 and tetrahedral
D) sp3 and trigonal planar
B
On the surface: A carbocation is bonded to 3 substituents and so it is in the middle of a flat triangle: trigonal planar.
When carbon has:
4 subtituents: sp3 and tetrahedral (109.5º)
3 subtituents: sp2 and trigonal planar (120º)
2 subtituents: sp and linear (180º)
Going Deeper: There are three bonding electron pairs around the carbon atom, and these can be placed as far apart as possible in an sp2 trigonal planar framework. The empty orbital, which results in the charge, is the 2pz atomic (unhybridized) orbital.
(Fill in the blanks) Benzoic acid can be extracted into water from an organic solvent using _________ and then removed from the water into an organic solvent by the addition of _________.
A) Sodium bicarbonate / sodium hydroxide
B) Hydrochloric acid / sodium hydroxide
C) Hydrochloric acid / sulfuric acid
D) Sodium hydroxide / hydrochloric acid
D
Benzoic acid, an organic compound, is insoluble in water. Extraction into water can be achieved with aqueous sodium hydroxide (deprotonation -> polar sodium benzoate, now soluble in water). Transfer of the benzoate back into organic solution is accomplished by treatment with hydrochloric acid (protonation -> less polar benzoic acid).
A high intensity peak in the mass spectrum of 2,2-dimethylpentane is observed at m/z 57. What chemical species does this peak represent?
A) Methyl cation
B) tert-butyl carbocation
C) n-propyl carbocation
D) Ethyl carbocation
B
The most stable fragment likely formed from 2,2-dimethylpentane fragmentation is t-butyl species.
T-butyl species = 57 m/z on mass spectrum
The addition of an acid and heat to 2-oxocyclopentanecarboxylic acid yields a:
A) cyclopentene
B) cyclopentanol
C) cyclopentanone
D) cyclopentanecarboxylic acid
C
Beta-keto carboxylic acids undergo decarboxylation through ring formations. The carbonyl bond of the ketone will react with the alcohol group of the carboxylic acid to form a ring and release the carboxylic acid as CO2. The remaining enol will tautomerize into a carbonyl group again, leaving cyclopentanone.
The data in Table 1 were collected for Reaction I:
2X + Y –> Z
Reaction I
Exp [X] in M [Y] in M Initial Rate of Rxt
1 0.050 0.100 8.5 x 10^-6
2 0.050 0.200 3.4 x 10^-5
3 0.200 0.100 3.4 x 10^-5
What is the rate law for the reaction?
A) Rate = k[X]^2[Y]
B) Rate = k[X]^2[Y]^2
C) Rate = k[X][Y]^2
D) Rate = k[X][Y]
C
This is a common MCAT-type question. By looking at Table 1, we can see that when the concentration of X is quadrupled (factor of 41) while [Y] is unchanged (Exp. 1 and 3), the rate is increased by a factor of 4 = 41. Thus the order of the reaction with respect to X is 1 (= “first order with respect to X”).
When the concentration of Y is doubled (factor of 21) while [X] remains the same (Exp. 1 and 2), the rate of reaction is quadrupled (factor of 4 = 22). Thus, the order of reaction with respect to Y is 2 (= “second order with respect to Y”).
The rate equation is Rate = [X][Y]2. Since the overall rate of reaction is the sum of exponents, we can say that the reaction is (1 + 2) a third order reaction overall.
{Notice that the stoichiometric coefficients are not relevant; the order of reaction being based on data.}
Additional learning point: In Organic Chemistry, an SN2 reaction is typical of a second order reaction where the rate must be equally proportional to the concentration of both reactants; for example, first order with respect to the alkyl halide and first order with respect to the nucleophile, so overall second order.
The anomers of D-(+)-glucose differ in their configuration around which of the following?
A) C-1
B) C-5
C) C-6
D) C-2
A
Glucose anomers have a different orientation of the hydroxyl group on the C-1 carbon.
A radioactive substance A was placed in a lead-lined container and the radiation it emitted was allowed to pass through a small aperture in the container and below a positively charged plate. The path of the particle changed such that it angled up toward the plate. Which of the following could it be?
A) Alpha particle
B) Beta particle
C) Proton
D) Gamma particle
B
Since the radioactive emission angled up toward the positive plate, it must be oppositely charged, that is, negatively charged. The only negatively charged species in the answer choices is the beta particle which is, by definition, an electron or a positron. Note that electromagnetic radiation like gamma rays (particles) and infrared radiation are not deflected in electromagnetic fields.
Heterocyclic amines are a very important type of amino group. One use of them includes being the base of DNA. The most important property however is their basicity due to nitrogen.
Pyridine is a much stronger base than pyrrole because:
A) Pyrrole will lose its aromatic character if nitrogen accepts an acid
B) Pyridine has a pair of electrons in an sp2 orbital that is available for sharing
C) The extra pair of electrons on nitrogen in pyrrole is involved in the pi cloud and therefore less available for sharing
D) All of the above
D
In pyrrole, the sigma bonds around nitrogen are formed by sp2 hybrid orbitals, not sp3. A nitrogen electron participates in each of the sigma bonds. The remaining nitrogen electrons are in a nitrogen 2p orbital which overlaps with the carbon 2p orbitals in the ring to form pi bonds and the cyclic compound, delocalizing the lone pair. In pyridine, the nitrogen atom donates one electron to each of the two sigma bonds by use of sp2 orbitals. The third sp2 orbital contains two electrons, which are for acids. One electron of nitrogen is present in a p orbital, which interacts with the carbon p orbitals to form the pi bond.
Which molecules contain an IR stretching frequency of 1680-1740 cm-1?
I. Alcohols
II. Carboxylic acids
III. Nitriles
IV. Amides
A) I and II
B) II and III
C) II and IV
D) III and IV
C
All carbonyl compounds absorb in the region 1760-1665 cm-1 due to the stretching vibration of the C=O bond. Different carbonyl compounds absorb in narrow ranges within this general region. Carboxylic acids and amides contain carbonyl functionality. Alcohols have broad absorption at approx. 3200-3500 cm-1; nitriles show absorptions at approx. 2250 cm-1.
In the phrase “SN1 Reaction”, what does the “1” indicate?
A) The equilibrium constant
B) The number of reactants involved
C) The stereochemical outcome
D) The rate order
D
“SN” stands for nucleophilic substitution and the “1” represents the fact that the rate-determining step is unimolecular.
In proton NMR spectroscopy, which compounds or functional groups show a characteristic peak in the range 9-10 ppm?
A) Halogenated alkanes
B) Aldehydes
C) Alcohols
D) Aromatic compounds
B
Nuclear magnetic resonance (NMR) is concerned with the magnetic properties of certain nuclei. Anything that removes electron density from around a nucleus deshields it and shifts the absorption to higher frequency. The more electronegative the atoms that are attached to the nucleus, the higher the absorption frequency. Thus, NMR is able to provide information on the types of functional groups present as each has a characteristic chemical shift range. ALDEHYDES HAVE A VERY CHARACTERSTIC SHIFT OF 9-10 PPM.
Choose the molecules containing an IR stretching frequency of 3500-3000 cm-1.
I. Amines
II. Carboxylic Acids
III. Phenols
IV. Alcohols
A) (I) and (II)
B) (II) and (III)
C) (III) and (IV)
D) All of them
D
The range 3500-3000 cm-1 is typically associated with stretches between heteroatoms and a hydrogen atom, for example NH and OH bonds. They may appear broad, intense and jagged (COOH), broad and smooth (OH) or less intense and sharper (NH). The preceding sentence is likely more than you would need to know for the MCAT. In general, for MCAT purposes, it is useful to remember APPROX. 3300 FOR HYDROXYL AND APPROX. 1700 FOR CARBONYL (C=O).
Infrared spectroscopy is often used to determine the presence of a carbonyl group in a compound. It is a strong band appearing at about 1700 cm^-1 in aldehydes, ketones, and carboxylic acids and their derivatives. A carboxylic acid will show another carbon oxygen stretching band at:
A) A higher frequency
B) A lower frequency
C) The same frequency
D) A frequency in the ultraviolet region of the electromagnetic spectrum
B
A C-O single bond is weaker than a double bond. Therefore, less energy is needed to induce the stretching vibration. Remember reciprocal centimeters is a measure of frequency and energy is directly proportional to the frequency.
Which reagent would you choose to convert hexan-1-ol to hexanal?
A) PCC/CH2Cl2
B) DIBAL/diethyl ether
C) KMnO4/aqueous H2SO4/acetone
D) K2Cr2O7/aqueous H2SO4/acetone
A
Pyridinium chlorochromate (PCC) is a mild oxidant, which converts primary and secondary alcohols to aldehydes and ketones respectively. KMnO4 and K2Cr2O7 are harsher oxidants, and would further oxidize primary alcohols to the corresponding carboxylic acids. Di-isobutylaluminum hydride (DIBAL) is a reducing agents.
Which statement below about recrystallization is NOT correct?
A) The method relies on the fact that solids are more soluble in a cold solvent than they are in the same amount of hot solvent
B) Precipitation is initiated by adding a seed crystal of the compound being purified
C) Scratching the side of the vessel at the air/solution interface can induce crystal nucleation
D) Slowly cooling the solution promotes precipitation of the compound
A
This statement is reversed: The process of recrystallization involves dissolution of the solid in appropriate solvent at a high temperature and the subsequent re-formation of the crystals upon cooling, so that any impurities remain in solution.
_____________ refers to the method by which an optically reactive reagent or catalyst is used to transform an optically inactive starting material into an optically active product.
A) Racemization
B) Enantioenrichment
C) Chiral chromatography
D) Asymmetric induction
D
Racemization refers to the conversion of an optically active substance to a racemic form (has equal amounts of left- and right-handed enantiomers of a chiral molecule). Enantioenrichment refers to the act of enriching an already existing mixture of stereoisomers such that it contains more than 50% of one enantiomer. Chiral chromatography involves the separation of stereoisomers using a chiral stationary phase, which does not involve chemical transformations. “Optical reduction” as a technique or phrase does not exist.
Which of the following solutions/compounds is/are optically inactive?
A) A racemic mixture showing no rotation of plane polarized light
B) A meso compound
C) A 50-50 mixture of a D and L form of a substance
D) All of the above
D
All the mentioned forms of compounds in this question show no rotation of plane polarized light. Racemic mixtures showing no rotation of plane polarized light, like a 50-50 mix of D and L compounds, are exact racemic mixtures of two enantiomeric forms of a compound thus cancelling the polarization effect of plane polarized light allowing for optical isomerization. A meso compound is superposable on its mirror image meaning that the molecule is achiral (i.e. optically inactive) though it may contain chiral centers.
The technique of fractional distillation is most effective for:
A) Separating solids on the basis of their different solubilities in a particular liquid
B) Identifying the functional groups in mixtures of more than one liquid
C) Separating liquids based on their different boiling points
D) Determining the boiling points of a variety of liquids
C
Using fractional distillation one can increase the temperature of a liquid mixture and as the different liquids vaporize at their respective boiling point temperatures one can collect the vapor by condensing it and collecting the desired liquid at the corresponding temperature.
Which angle is common to tetrahedral molecules?
A) 109.5°
B) 120°
C) 60°
D) 90°
A
Tetrahedral geometry has three-dimensional placement with atoms positioned at 109.5° to each other. Trigonal planar molecules have angles of 120°, interior angles of 60° are associated with strained cyclopropanes bonding, 90° with cyclobutanes and 180° for alkynes.
Which of the following is the most likely to react with Magnesium (Mg)?
A) O
B) Ca
C) S
D) Cl
D
Electronegativity increases to the right of the periodic table, and also from bottom to top. We note that chlorine is the closest to the upper right of the periodic table, meaning it is the most electronegative of the available choices. Thus, it is the most likely to react with Mg, which normally carries a 2+ charge.
How many orbitals can occupy the 5f subshell?
A) 5
B) 7
C) 10
D) 14
B
There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. Thus, the answer is B. Do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. Each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.
Toluene will show peaks in its:
A) IR spectrum only
B) UV and IR spectrum only
C) IR, UV, and NMR spectrum
D) UV spectrum only
C
Aromatic systems show strong absorptions in the UV region of the system. C-C and C-H bonds will absorb in the IR region. The NMR will show the inequivalent hydrogens.
Acetic acid reacts more rapidly with methanol than with ethanol to form an ester. Most likely this is due to:
A) An inductive electronic effect
B) A resonance effect
C) Steric hindrance
D) An ortho effect
C
Bulky groups on either the acid or alcohol slow the reaction due to steric hindrance.
Which of the following lists the correct common names for ethanal, methanal, and ethanol, respectively?
A) Acetaldehyde, formaldehyde, ethyl alcohol
B) Ethyl alcohol, propionaldehyde, isopropyl alcohol
C) Ethyl alcohol, formaldehyde, acetaldehyde
D) Isopropyl alcohol, ethyl alcohol, formaldehyde
A
The common name of ethanal is acetaldehyde, the common name of methanal is formaldehyde, and the common name of ethanol is ethyl alcohol. Isopropyl alcohol is the common name of 2-propanol. Propionaldehyde is the common name of propanal.
Which of the following are considered terminal functional groups?
I. Aldehydes
II. Ketones
III. Carboxylic acids
A) I only
B) III only
C) I and III only
D) I, II, and III
C
Aldehydes and carboxylic acids are characterized by their positions at the ends of carbon backbones and are thus considered terminal groups. As a result, the carbons to which they are attached are usually designated carbon 1. Ketones are internal by definition because there must be a carbon on either side of the carbonyl.
NADH is a coenzyme that releases high-energy electrons into the electron transport chain. It is known as nicotinamide adenine dinucleotide or diphosphopyridine nucleotide. What functional groups exist in this molcule?
I. Phosphate
II. Amide
III. Anhydride
A) I only
B) II only
C) I and II only
D) I, II, and III
C
The suffix -amide in nicotinamide indicates that this compound contains an amide functional group. The prefix diphospho- indicates that there are two phosphate groups, as well. Even if we did not know the prefix phospho- from this chapter, we should recognize that nucleotides, mentioned in the name, contain a sugar, a phosphate group, and a nitrogenous base.
Which of the following are common names for carboxylic acid derivatives?
I. Acetic anhydride
II. Formic acid
III. Methyl formate
A) I and II only
B) I and III only
C) II and III only
D) I, II, and III
B
Acetic anhydride is the common name for ethanoic anhydride. Methyl formate is the common name for methyl methanoate; we can infer this from the common root form - and the ester suffix -oate (which is sometimes shortened to -ate for pronunciation purposes). Formic acid is the common name for methanoic acid, but this is a carboxylic acid - not a derivative.
Consider the name 2,3-diethylpentane. Based on the structure implied by this name, the correct IUPAC name for this molecule is:
A) 2,3-diethylpentane
B) 1,2-diethylbutane
C) 3-ethyl-4-methylhexane
D) 3-methyl-4-ethylhexane
C
Draw out the molecule, and you will see that the longest carbon chain with the substituents at the lowest possible carbon numbers is actually different from the one chosen in the original name. The correct IUPAC name for this molecule is 3-ethyl-methylhexane.
Which of the following does NOT show optical activity?
A) (R)-2-butanol
B) (S)-2-butanol
C) A solution containing 1 M (R)-2-butanol and 2 M (S)-2-butanol
D) A solution containing 2 M (R)-2-butanol and 2 M (S)-2-butanol
D
This is a racemic mixture of 2-butanol because it consists of equimolar amounts of (R)-2-butanol and (S)-2-butanol. The (R)-2-butanol molecule rotates the plan of polarized light in one direction, and the (S)-2-butanol rotates it by the same angle but in the opposite direct; as a result, no net rotation of polarized light is observed.
How many stereoisomers exist for the following aldehyde?
O
||
C - H
|
HO - C - H
|
HO - C - H
|
HO - C - H
|
HO - C - H
|
H
A) 2
B) 8
C) 9
D) 16
B
The maximum number of stereoisomers of a compound equals 2^n, where n is the number of chiral carbons in the compound. In this molecule, C-1 (the aldehydic carbon) is not chiral, nor is C-5 (because it is attached to two hydrogen atoms). Therefore, with three chiral centers, there are 2^3 = 8 stereoisomers.
Which of the following compounds is optically inactive?
A) CH3
|
H - | - Cl
|
Cl - | - H
|
CH3
B) CH3
|
Cl - | - H
|
H - | - Cl
|
CH3
C) CH3
|
H - | - Cl
|
H - | - Cl
|
CH3
D) CH2Cl
|
H - | - Cl
|
H - | - H
|
CH3
C
This answer choice is an example of a meso compound - a compound that contains chiral centers but has an internal plane of symmetry:
CH3
|
H - | - Cl
-----------|------------- Plane of
H - | - Cl Symmetry
|
CH3
Owing to this internal plane of symmetry, the molecule is achiral and, hence, optically inactive. Choices A and B are enantiomers of each other and will certainly show optical activity on their own. Choice D is incorrect because it contains a chiral carbon and no internal plane of symmetry, thus, it is optically active as well.
(+)-Glyceraldehyde and (-)-glyceraldehyde refer to the (R) and (S) forms of 2,3-dihydroxypropanal, respectively. These molecules are considered:
A) Enantiomers
B) Diastereomers
C) Meso compounds
D) Structural isomers
A
(+)-Glyceraldehyde and (-)-glyceraldehyde, or (R)- and (S)-2,3-dihydroxypropanal, are enantiomers. Enantiomers are nonsuperimposable mirror images. Each has only one chiral center (C-2), which has the opposite configuration in these two molecules.
Consider (E)-2-butene and (Z)-2-butene. This is a pair of what type(s) of isomers?
I. Cis-trans isomers
II. Diastereomers
III. Enantiomers
A) I only
B) II only
C) I and II only
D) I and III only
C
(E)-2-butene can also be called trans-2-butene; (Z)-2-butene can also be called cis-2-butene. As such, they are cis-trans isomers. Remember that cis-trans isomers are a subtype of diastereomers in which the position of substituents differs about an immovable bond. Diastereomers are molecules that are non-mirror-image stereoisomers (molecules with the same atomic connectivity). These are not enantiomers because they are not mirror images of each other.
3-methylpentant and hexane are related in that they are:
A) Enantiomers
B) Diastereomers
C) Constitutional isomers
D) Conformational isomers
C
Because they have the same molecular formula but different connectivity, 3-methylpentane and hexane are constitutional isomers.
(R)-2-chloro-(S)-3-bromobutane and (S)-2-chloro-(S)-3-bromobutane are:
A) Enantiomers
B) Diastereomers
C) Meso compounds
D) The same molecule
B
These two molecules are stereoisomers of one another, but are NOT nonsuperimposable mirror images. Therefore, they are diastereomers. Note that these molecules differ by at least one, but not all, chiral carbons.
A scientist takes a 0.5g/mol solution of an unknown pure dextrorotatory organic molecule and places it in a test tube with a diameter of 1cm. He observes that a plane of polarized light is rotated 12 degrees under these conditions. What is the specific rotation of this molecule?
A) -240 degrees
B) -24 degrees
C) +24 degrees
D) +240 degrees
D
Remember that the equation for specific rotation is [alpha] = a(obs)/(c x l). In this example a(obs) is +12 degrees (remember that dextrorotatory, or clockwise, rotation is positive), c = 0.5g/ml, l = 1cm = 0.1dm. Remember that path length is ALWAYS measured in decimeters when calculating specific rotation. Therefore, the specific rotation can be calculated as:
[alpha] = +12 / (0.5g/ml x 0.1 dm) = +240 degrees.
Omeprazole is a proton pump inhibitor commonly used in gastroesophageal reflux disease. When omeprazole, a racemic mixture, went off-patent, pharmaceutical companies began to manufacture esomeprazole, the (S)-enantiomer of omeprazole, by itself. Given 1 M solution of omeprazole and esomeprazole, which solution(s) would likely exhibit optical activity?
A) Omeprazole only
B) Esomeprazole
C) Both omeprazole and esomeprazole
D) Neither omeprazole nor esomeprazole
B
Racemic mixtures like omeprazole contain equimolar amounts of two enantiomers and thus have no observed optical activity. Each of the two enantiomers causes rotation in opposite directions, so their effects cancel out. Esomeprazole only contains one of the two enantiomers and thus should cause rotation of plane-polarized light.
(2R,3S)-2,3-dihydroxybutanedioic acid and (2S,3R)-2,3-dihydroxybutanedioic acid are:
I. Meso compounds
II. The same molecule
III. Enantiomers
A) I only
B) III only
C) I and II only
D) I and III only
C
Draw out these structures. The two names describe the same molecule, which also happens to be a meso compound because it contains a plane of symmetry. These compounds are not enantiomers because they are superimposable mirror images of one another, NOT nonsuperimposable mirror images. These compounds are better termed meso-2,3-dihydroxybutanedioic acid:
COOH
|
H - | - OH
------------|--------------
H - | - OH
|
COOH
If the methyl groups of butane are 120 degrees apart, as seen in a Newman projection, this molecule is in its:
A) Highest-energy gauche form
B) Lowest-energy staggered form
C) Middle-energy eclipsed form
D) Highest-energy eclipsed form
C
In butane, the position at which the two methyl groups are 120 degrees apart is an eclipsed conformation. This has a moderate amount of energy, although not as high as a totally eclipsed conformation in which the two methyl groups are 0 degrees apart.
Within one principle energy level, which subshell has the least energy?
A) s
B) p
C) d
D) f
A
The energies of the subshells within a principle quantum number are as follows: s < p < d < f.
Which of the following compounds possesses at least one sigma bond?
A) CH4
B) C2H2
C) C2H4
D) All of the above contain at least one sigma bond
D
All single bonds are sigma bonds; double and triple bonds each contain one sigma bond and one or two pi bonds, respectively. The compounds CH4, C2H2, and C2H4 all contain at least one single bond and therefore contain at least one sigma bond.
A carbon atom participates in one double bond. As such, this carbon contains orbitals with:
A) Hybridization between the s-orbital and the one p-orbital
B) Hybridization between the s-orbital and two p-orbitals
C) Hybridization between the s-orbital and three p-orbitals
D) Unhybridized s character
B
In a carbon with one double bond, sp2 hybridization occurs - that is, one s-orbital hybridizes with two p-orbitals to form three sp2-hybridized orbitals. The third p-orbital of the carbon atom remains unhybridized and takes part in the formation of the pi bond of the double bond. Although there is an unhybridized p-orbital, there are no unhybridized s-orbitals.
The hybridization of the carbon and nitrogen atoms in CN- are:
A) sp3 and sp3, respectively
B) sp3 and sp, respectively
C) sp and sp3, respectively
D) sp and sp, respectively
D
The carbon and nitrogen atoms are connected by a triple bond in CN- (:C≡N:-). A triple-bonded atom is sp hybridized; one s-orbital hybridizes with one p-orbital to form two sp-hybridized orbitals. The two remaining unhybridized p-orbitals take part in the formation of two pi bonds.
Which of the following hybridizations does the Be atom in BeH2 assume?
A) sp
B) sp2
C) sp3
D) sp3d
A
Beryllium has only two electrons in its valence shell. When it bonds to two hydrogens, it requires two hybridized orbitals, meaning that its hybridization must be sp. Note that the presence of only single bonds does not mean that the hybridization must be sp3; this is a useful assumption for carbon, but does not apply to beryllium because of its smaller number of valence electrons. The two unhybridized p-orbitals around beryllium are empty in BeH2, which takes on the linear geometry characteristic of sp-hybridized orbitals.
Two atomic orbitals may combine to form:
I. A bonding molecular orbital
II. An antibonding molecular orbital
III. Hybridized orbitals
A) I only
B) III only
C) I and II only
D) I, II and III
D
When atomic orbitals combine, they form molecular orbitals. When two atomic orbitals with the same sign are added head-to-head or tail-to-tail, they form bonding molecular orbitals. When two atomic orbitals with opposite signs are added head-to-head or tail-to-tail, they form antibonding molecular orbitals. Atomic orbitals can also hybridize, forming sp3, sp2 or sp orbitals.
Molecular orbitals can contain a maximum of:
A) One electron
B) Two electrons
C) Four electrons
D) 2n^2 electrons, where n is the principle quantum number of the combining atomic orbitals
B
Like atomic orbitals, molecular orbitals each can contain a maximum of two electrons with opposite spins. The 2n^2 rule in Choice D refers to the total number of electrons that can exist in a given energy shell, not in a molecular orbital.
Pi bonds are formed by which of the following orbitals?
A) Two s-orbitals
B) Two p-orbitals
C) One s-orbital and one p-orbital
D) Two sp2-hybridized orbitals
B
Pi bonds are formed by the parallel overlap of unhybridized p-orbitals. The electron density is concentrated above and below the bonding axis. A sigma bond, on the other hand, can be formed by the head-to-head overlap of two s-orbitals or hybridized orbitals. In a sigma bond, the density of the electrons is concentrated between the two nuclei of the bonding atoms.
How many sigma bonds and pi bonds are present in the following compound?
H O
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H - C - C
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H H
A) Six sigma bonds and one pi bond
B) Six sigma bonds and two pi bonds
C) Five sigma bonds and one pi bond
D) Five sigma bonds and two pi bonds
A
Each single bond has one sigma bond, and each double bond has one sigma and one pi bond. In this question, there are five single bonds (five sigma bonds) and one double bond (one sigma bond and one pi bond), which gives a total of six sigma bonds and one pi bond.
The four C-H bonds of CH4 point toward the vertices of a tetrahedron. This indicates that the hybridization of the carbon atom in methane is:
A) sp
B) sp2
C) sp3
D) sp3d
C
The four bonds point to the vertices of a tetrahedron, which means that the angle between two bonds is 109.5 degrees, a characteristic of sp3 orbitals. Hence, the carbon atom of CH4 is sp3-hybridized.
Why is a single bond stronger than a pi bond?
I. Pi bonds have greater orbital overlap
II. s-orbitals have more overlap than p-orbitals
III. sp3 hybridization is always unstable
A) I only
B) II only
C) I and III only
D) II and III only
B
Bond strength is determined by the degree of orbital overlap; the greater the overlap, the greater the bond strength. A pi bond is weaker than a single bond because there is significantly less overlap between the unhybridized p-orbitals of a pi bond (due to their parallel orientation) than between the s-orbitals or hybrid orbitals of a sigma bond. sp3-hybridized orbitals can be quite stable, as evidenced by the number of carbon atoms with this hybridization forming stable compounds.
The p character of the bonds formed by the carbon atom in HCN is:
A) 25%
B) 50%
C) 67%
D) 75%
B
The carbon bond in hydrogen cyanide (H-C≡N:) is triple-bonded, and because triple bonds require two unhybridized p-orbitals, the carbon must be sp-hybridized; sp-hybridized orbitals have 50% s character and 50% p character.
A resonance structure describes:
I. The hybrid of all possible structures that contribute to electron distribution
II. A potential arrangement of electrons in a molecule
III. The single form that the molecule most often takes
A) I only
B) II only
C) I and II only
D) I, II, and III
B
A resonance structure describes an arrangement of electrons in a molecule. Different resonance structures can be derived by moving electrons in unhybridized p-orbitals throughout a molecule containing conjugated bonds. In molecules that contain multiple resonance structures, some are usually more stable than others; however, each resonance structure is not necessarily the most common form a molecule takes, eliminating statement III. Statement I has reversed the terminology for resonance structures: the electron density in a molecule is the weighted average of all possible resonance structures, not the other way around.
An electron is known to be in the n = 4 shell and the l = 2 subshell. How many possible combinations of quantum numbers could this electron have?
A) 1
B) 2
C) 5
D) 10
D
An electron in the n = 4 shell and the l = 2 subshell can have five different values for ml; -2, -1, 0, 1, or 2. In each of these orbitals, electrons can have positive or negative spin. Thus, there are 5 x 2 = 10 possible combinations of quantum numbers for this electron.
Compared to single bonds, triple bonds are:
A) Weaker
B) Longer
C) Made up of fewer sigma bonds
D) More rigid
D
Pi bonds do not permit free rotation, unlike sigma bonds; this makes triple bonds more rigid than single bonds. Triple bonds are stronger and shorter bonds than single bonds. Both single and triple bonds contain one sigma bond.
Which of the following are Lewis bases?
I. Ag+
II. H2O
III. NH3
A) I only
B) I and II only
C) II and III only
D) I, II, and III
C
NH3 and H2O are Lewis bases because nitrogen and oxygen can donate lone pairs. Ag+ is a Lewis acid because it can accept a lone pair into an unoccupied orbital.
Rank the following in order of decreasing nucleophilicity in an aprotic solvent: RO-, RCOOH, ROH, HO-
A) RCOOH > ROH > RO- > HO-
B) HO- > ROH > RO- > RCOOH
C) RO- > HO- > ROH > RCOOH
D) RCOOH > RO- > HO- > ROH
C
Remember, good nucleophiles tend to have lone pairs or pi bonds and are negatively charged or polarized. Alkoxide (OR-) and hydroxide (OH-) anions are strong nucleophiles. Alcohol (ROH) and carboxylic acids (RCOOH) are weak nucleophiles. The alkyl group of an alkoxide anion donates additional electron density, making it more reactive than the hydroxide ion. The carboxylic acid contains more electron-withdrawing oxygen atoms than the alcohol, making it less nucleophilic.
Rank the following in order of decreasing electrophilicity; CR3+, CH3OH, CH3OCH3, CH3Cl
A) CH3OCH3 > CR3+ > CH3OH > CH3Cl
B) CR3+ > CH3OH > CH3OCH3 > CH3Cl
C) CH3OCH3 > CH3Cl > CR3+ > CH3OH
D) CR3+ > CH3Cl > CH3OH > CH3OCH3
D
Good electrophiles are positively charged or polarized. CH3+ is a tertiary carbocation; it has a positive charge, which makes it very electrophilic. CH3Cl and CH3OH are both polarized; however, the leaving groups differ between these two. Cl- is a weaker base than OH- (HCl is a stronger acid than H2O). As such, Cl- will be more stable in solution than OH-, which increases the electrophilic reactivity of CH3Cl above CH3OH. CH3OCH3 has a much less stable leaving group, CH3O-, and is therefore significantly less electrophilic.
Rank the following in order of decreasing leaving group ability: H2O, HO-, Br-, H-
A) H2O > Br- > HO- > H-
B) H2O > HO- > Br- > H-
C) HO- > Br- > H2O > H-
D) HO- > H- > H2O > Br-
A
Good leaving groups are weak bases, which are the conjugates of strong acids. Leaving groups must also be stable once they leave the molecule. H2O is, by far, the most stable leaving group and will be extremely unreactive once it leaves the molecule through heterolysis. Br- is the conjugate base of HBr; HO- is the conjugate base of water. HBr is a much stronger acid than water, so Br- is a better leaving group than HO-. Finally, hydride (H-) is a very poor leaving group because it is extremely unstable in solution.
Rank the following in order of decreasing oxidation state: amine, carboxylic acid, aldehyde, alkane
A) Aldehyde, amine, alkane, carboxylic acid
B) Carboxylic acid, aldehyde, amine, alkane
C) Carboxylic acid, amine, aldehyde, alkane
D) Alkane, amine, aldehyde, carboxylic acid
B
Carboxylic acids are the second most oxidized form of carbon (only carbon dioxide is more oxidized). In carboxylic acids, the carbon atom has three bonds to oxygen. In aldehydes, the carbon atom has two bonds to oxygen. In amines, the carbon atom has one bond to nitrogen. In an alkane, the carbon only has bonds to other carbons and hydrogens.
If cinnamaldehyde was treated with LiAlH4, what reaction would occur?
A) Reduction, resulting in a primary alcohol
B) Oxidation, resulting in a carboxylic acid
C) An acid-base reaction, resulting in a diol
D) No reaction would occur
A
All that we need to know about cinnamaldehyde is that is is an aldehyde, and therefore will be reduced by a strong reducing agent like LiAlH4 to a primary alcohol.
If 2-butanol was treated with dichromate, what reaction would occur?
A) Reduction, resulting in the formation of butene
B) Oxidation, resulting in the formation of butanoic acid
C) Oxidation, resulting in the formation of butanone
D) No reaction would occur
C
Because 2-butanol is a secondary alcohol, oxidation by a strong oxidizing agent like dichromate will result in a ketone, butanone.
