MCAT Organic Chemistry Flashcards

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

C Bulky groups on either the acid or alcohol slow the reaction due to steric hindrance.

Answer 10

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.

Answer 11

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.

Answer 12

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.

Answer 13

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.

Answer 14

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.

Answer 15

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.

Answer 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.

Answer 17

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.

Answer 18

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.

Answer 19

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.

Answer 20

C Because they have the same molecular formula but different connectivity, 3-methylpentane and hexane are constitutional isomers.

Answer 21

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.

Answer 22

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.

Answer 23

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.

Answer 24

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 ```

Answer 25

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.

Answer 26

A The energies of the subshells within a principle quantum number are as follows: s < p < d < f.

Answer 27

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.

Answer 28

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.

Answer 29

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.

Answer 30

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.

Answer 31

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.

Answer 32

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.

Answer 33

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.

Answer 34

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.

Answer 35

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.

Answer 36

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.

Answer 37

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.

Answer 38

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.

Answer 39

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.

Answer 40

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.

Answer 41

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.

Answer 42

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.

Answer 43

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.

Answer 44

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.

Answer 45

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.

Answer 46

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.

Answer 47

C Because 2-butanol is a secondary alcohol, oxidation by a strong oxidizing agent like dichromate will result in a ketone, butanone.

Answer 48

C Pyridinium chlorochromate is a weak oxidizing agent, and will oxidize an alcohol to an aldehyde. Stronger oxidizing agents are required to convert a primary alcohol to a carboxylic acid.

Answer 49

B An Sn1 reaction is a first-order nucleophilic substitution reaction. It is called first-order because the rate-limiting step involves only one molecule. Choice A is true, but does not explain why Sn1 reactions have first-order kinetics; the rate-limiting step of an Sn2 reaction is also the first (and only) step of that reaction, but Sn2 reactions have second-order kinetics, not first-order. Choice C is a true statement as well, but again does not explain why the reaction is first-order. Finally, Choice D is incorrect because it is the rate-limiting step, not the reaction overall, that involves only one molecule.

Answer 50

D In a protic solvent, the protons in solution can attach to the nucleophile, decreasing nucleophilicity. The larger the nucleophile, and the stronger its conjugate acid, the stronger the nucleophile will be. Of the options, I- will therefore be the strongest nucleophile because it is least likely to associate with the protons in solution.

Answer 51

B In this reaction, there has been an inversion of stereochemistry. The most likely explanation for this is that the reaction proceeded by an Sn2 mechanism. Inversion of stereochemistry is a hallmark of Sn2 reactions, whereas racemization is a hallmark of Sn1.

Answer 52

D To carry out a nucleophile-electrophile reaction, the nucleophile must be able to dissolve in the solvent. Nucleophiles are nearly always polar, and often carry a charge. Polar solvents are therefore preferred for these reactions. Hexane is a nonpolar solvent and will not be useful for a nucleophile-electrophile reaction.

Answer 53

A Aldehydes have one alkyl group connected to the carbonyl carbon, whereas ketones have two. This creates more steric hindrance in ketones, which lowers their reactivity to nucleophiles. Ketones are also less reactive because their carbonyl carbon has less positive charge character; the additional alkyl group can donate electron density - the opposite of Choice D - which decreases the electrophilicity of the compound.

Answer 54

D Remember there is a hierarchy to the reactivity of carboxylic acid derivatives that dictates how reactive they are toward nucleophilic attack. In order from highest to lowest, this hierarchy is anhydrides > carboxylic acids and esters > amides. In practical terms, this means that derivatives of higher reactivity can form derivatives of lower reactivity but not vice versa. Nucleophilic attack of an ester cannot result in a corresponding anhydride because anhydrides are more reactive than esters.

Answer 55

B Alcohols have higher boiling points than their analogous hydrocarbons as a result of their polarized O-H bonds, in which oxygen is partially negative and hydrogen is partially positive. This enables the oxygen atoms of other alcohol molecules to be attracted to the hydrogen, forming a hydrogen bond. Heat is required to overcome these hydrogen bonds, thereby increasing the boiling point. The analogous hydrocarbons do not form hydrogen bonds and, therefore, vaporize at lower temperatures. Choice A is irrelevant; oxygen's bond length is not a factor in determining a substance's boiling point. Choices C and D are true statements, but are also irrelevant to boiling point determination.

Answer 56

A Tertiary alcohols can be oxidized but only under extreme conditions because their substrate carbons do not have spare hydrogens to give up. Alcohol oxidation involves the removal of such a hydrogen so that carbon can instead make another bond to oxygen. If no hydrogen is present, a carbon-carbon bond must be cleaved, which requires a great deal of energy and will, therefore, occur only under extreme conditions. Choice B is incorrect because alcohols are not carbonyl-containing compounds and would more properly describe a carbonyl-containing compound that is unable to form an enolate. Choice C is incorrect because the hydroxyl group of the tertiary carbon is still polarized. Choice D is a false statement; tertiary alcohols are still involved in other reactions, such as Sn1 reactions.

Answer 57

B Remember, diols are named after the parent alkane, with the position of the alcohols indicated, and ending in the suffix -diol. Here the carbon chain is three carbons, with a hydroxyl group on carbons 1 and 2. Thus, the name is propane-1,2-diol.

Answer 58

C All else being equal, boiling points increase with increasing size of the alkyl chain because of increased van der Waals attractions. Isobutyl alcohol has the largest alkyl chain and will thus have the highest boiling point; methanol has the smallest chain and will thus have the lowest boiling point.

Answer 59

B Phenols have significantly more acidic hydroxyl hydrogens than other alcohols because of resonance stabilization of the conjugate base, so this will be the most acidic hydroxyl hydrogen. The acidity of hexanol and cyclohexanol are close, but the hydroxyl hydrogen of hexanol is slightly more acidic because the ring structure of cyclohexanol is slightly electron-donating, which makes its hydroxyl hydrogen slightly less acidic.

Answer 60

B CH3CH2CH2OH is 1-propanol, a primary alcohol. The desired end product, CH3CH2CHO, is propanal, an aldehyde. Of the available options, the only reactant capable of oxidizing primary alcohols to aldehydes is pyridinium chlorochromate (PCC). Chromic trioxide and dichromate salts will both oxidize primary alcohols to carboxylic acids.

Answer 61

D Cyclohexanol is a secondary alcohol, so any of the oxidizing agents listed will convert it to a ketone.

Answer 62

A Acidic conditions, provided by dilute sulfuric acid, are required to complete the Jones oxidation. This reaction is carried out in aqueous conditions. While heat may speed up the reaction, high temperatures are not required for this reaction.

Answer 63

A Methylsulfonyl chloride serves as a protecting group for alcohols, which are converted into mesylates. Reacting with this reagent before continuing with what would normally be an oxidation reaction keeps the alcohol from reacting; when the protecting group is then removed using strong acid, the resultant product is the same as the initial reactant. Neither of the oxidation products in Choices B or C, nor the reduction product in Choice D, will be formed.

Answer 64

A This reaction will create a ketal. This is the first step of the protection of aldehydes or ketones using dialcohols.

Answer 65

B In order to convert phenols into hydroxyquinones, they must first be converted into quinones through an oxidation step; a second oxidation step is required to further oxidize quinones to hydroxyquinones.

Answer 66

C The reaction that converts ubiquinone into ubiquinol is a reduction reaction in which two ketones are reduced to two hydroxyl groups.

Answer 67

A An acetal can be converted to a carbonyl and a dialcohol by treatment with aqueous acid. This is the final step when using alcohols as protecting groups, called deprotection.

Answer 68

D The reactivity of the carbonyl can be attributed to the difference in electronegativity between the carbon and oxygen atoms. The more electronegative oxygen atom attracts the bonding electrons and is therefore electron-withdrawing. Thus, the carbonyl carbon is electrophilic. One resonance structure of the carbonyl pushes the pi electrons onto the oxygen, resulting in a positively charged carbonyl carbon.

Answer 69

B Assuming the length of the carbon chain remains the same, the alkane consistently has the lowest boiling point. The boiling point of the ketone is elevated by the dipole in the carbonyl. The boiling point of the alcohol is elevated further by hydrogen bonding.

Answer 70

B Mutarotation is the interconversion between anomers of a compound. Enantiomerization and racemization, choices (C) and (D), mean the same thing as each other: the formation of a mirror-image or optically inverted form of a compound. Glycosidation, choice (A), is the addition of a sugar to another compound.

Answer 71

B Hemiacetals and hemiketals are usually short-lived because the -OH group will rapidly become protonated in acidic conditions and is lost as water, leaving behind a carbocation that is very susceptible to attack by an alcohol. Once the alcohol has been added, the acetal or ketal becomes more stable because the newly added group is less likely to become protonated and leave as compared to -OH.

Answer 72

A A hemiacetal is a molecule in which one equivalent of alcohol has been added to a carbonyl (-OR) and the carbonyl oxygen has been protonated (-OH). Otherwise, there is the same alkyl group (-R) and hydrogen atom (-H) as the parent aldehyde. Choice B describes an acetal, C a hemiketal, and D a ketal.

Answer 73

A Although both the aldehyde and ketone listed will be reactive with the strongly nucleophilic hydrogen cyanide, aldehydes are slightly more reactive toward nucleophiles than ketones for steric reasons, so the aldehyde and HCN will form major product (which will be a cyanohydrin).

Answer 74

C PCC is a mild anhydrous oxidant that can oxidize primary alcohols to aldehydes, and secondary alcohols to ketones. It is not strong enough to oxidize alcohols or aldehydes to carboxylic acids.

Answer 75

B In a hydration reaction, water adds to a carbonyl, forming a geminal diol - a compound with two hydroxyl groups on the same carbon. Hydrogen peroxide and potassium dichromate are oxidizing agents that can convert an aldehyde to a carboxylic acid. Ethanol will react with a carbonyl compound to form an acetal or a ketal, if excess ethanol is available.

Answer 76

A Ammonia, or NH3, will react with an aldehyde like glutaraldehyde to form an imine. This is a condensation and a substitution reaction, as the C = O of the carbonyl will be replaced with a C ≡ N bond.

Answer 77

C Hydrides like LiAlH4 and NaBH4 are reducing agents; as such, they will reduce aldehydes and ketones to alcohols. The other reagents listed are oxidizing agents, which will not act on a ketone.

Answer 78

D During tautomerization, the double bond between the carbon and nitrogen in an imine is moved to lie between two carbons. This results in an enamine - a compound with a double bond and an amine.

Answer 79

B Keto <---> enol Tautomerization is the interconversion of two isomers in which a hydrogen and a double bond are moved. The keto and enol tautomers of aldehydes and ketones are common examples of tautomers seen on Test Day. Note that the equilibrium lies to the left because the keto form is more stable. Esterification is the formation of esters from carboxylic acids and alcohols. Elimination is a reaction in which a part of a reactant is removed and a new multiple bond is introduced. Dehydration is a reaction in which a molecule of water is eliminated.

Answer 80

A The keto-enol equilibrium lies far to the keto side because the keto form is significantly more thermodynamically stable than the enol form. This thermodynamic stability stems from the fact that the oxygen is more electronegative than the carbon, and the keto tautomer puts more electron density around the oxygen than the enol tautomer. If the enol tautomer is less thermodynamically stable, it is also higher energy than the keto tautomer.

Answer 81

B The aldol condensation is both a dehydration reaction because a molecule of water is lost, and a nucleophilic addition reaction because the nucleophilic enolate attacks and bonds to the carbonyl carbon.

Answer 82

B At high temperatures and with a weak base like NH3, the thermodynamic enolate will be favored. The reaction proceeds slowly with the weak base, giving the kinetic enolate time to interconvert to the more stable thermodynamic enolate.

Answer 83

A Aldehydes are generally more reactive than ketones because the additional alkyl group of a ketone is sterically hindering; this alkyl group is also electron-donating, destabilizing the carbanion intermediate. The carbonyl carbon is highly electrophilic; alkanes lack any significant electrophilicity.

Answer 84

B When ɑ-carbons are deprotonated, the negative charge is resonance stabilized in part by the electronegative carbonyl oxygen, which is electron-withdrawing. Alkyl groups are actually electron-donating, which destabilizes carbanion intermediates; this invalidates statement II.

Answer 85

C All of the answer choices are nitrogen-containing functional groups, but only enamines are tautomers of imines. Imines contain a double bond between a carbon and a nitrogen; enamines contain a double bond between two carbons as well as an amine.

Answer 86

C When succinaldehyde (or any aldehyde or ketone with ɑ-hydrogens) is treated with a strong base like lithium diisopropylamide (LDA), it forms the more nucleophilic enolate carbanion.

Answer 87

C Aldol condensations contain two main steps. In the first step, the ɑ-carbon of an aldehyde or ketone is deprotonated, generating the enolate carbanion. This carbanion can then attack another aldehyde or ketone, generating the aldol. In the second step, the aldol is dehydrated, forming a double bond. This double bond is between the ɑ- and β-carbons, so the molecule is an ɑ,β-unsaturated carbonyl.

Answer 88

D Because benzaldehyde lacks a ɑ-proton, it cannot be reacted with base to form the nucleophilic enolate carbanion. Therefore, acetone will act as as our nucleophile. In order to perform this reaction, which is an aldol condensation, acetone will be reacted with a strong base - not a strong acid - in order to extract the ɑ-hydrogen and form the enolate anion, which will act as a nucleophile.

Answer 89

B This is an example of an aldol condensation, but stopped after aldol formation (before dehydration). After the aldol is formed using strong base, the reaction may be halted by the addition of acid. Butanal in strong acid would be likely to deprotonate without gaining the hydroxyl group. Methanal in diethyl ether would not be reactive because diethyl ether is not a strong enough base to abstract the ɑ-hydrogen. Reaction of the two aldehydes methanal and ethanal in catalytic base would form 3-hydroxypropanal (which would dehydrate to form propanal), not 3-hydroxybutanal.

Answer 90

B The nomenclature in this question is well above what one needs to be able to draw on the MCAT; however, we can discern that we are forming a ketone and an aldehyde from a single molecule. The hallmark of a reverse aldol reaction is the breakage of a carbon-carbon bond, forming two aldehydes, two ketones, or one of each. In an aldol condensation, we would expect to form a single product by combining two aldehydes, two ketones, or one of each. A dehydration reaction should release a water molecule, rather than breaking apart a large organic molecule into two small molecules. A nucleophilic attack should feature the formation of a bond between a nucleophile and an electrophile; again, we would not expect to break apart a large organic molecule into two smaller molecules. Note that simply noting how many reactants and products are present in the reaction is sufficient to determine the answer.

Answer 91

D The boiling points of compounds depend on the strength of the attractive forces between molecules. In both alcohols and carboxylic acids, the major form of intermolecular attraction is hydrogen bonding; however, hydrogen bonding is much stronger in carboxylic acids as compared to alcohols because carboxylic acids are more polar and the carbonyl also contributes to hydrogen bonding in addition to the hydroxyl group. The stronger hydrogen bonds elevate the boiling points of carboxylic acids compared to alcohols. Boiling points also depend on molecular weight, but in this case, the difference in molecular weight is insignificant compared to the effect of hydrogen bonding. Choices B and C are both true but do not explain the difference in boiling points.

Answer 92

B The acidity of carboxylic acids is significantly increased by the presence of highly electronegative functional groups. Their electron-withdrawing effect increases the stability of the carboxylate anion, favoring proton dissociation. This effect increases as the number of electronegative groups on the chain increases, and it also increases as the distance between the acid functionality and electronegative group decreases. This answer has two halogens bonded to it at a smaller distance from the carbonyl group compared to the other answers.

Answer 93

C Soap is a salt of a carboxylate anion with a long hydrocarbon tail. Choices A and B are not salts of anionic compounds. Choice D is sodium acetate, which is a salt but does not contain the long hydrocarbon tail needed to be considered a soap.

Answer 94

C Carboxylic acids cannot be converted into alkenes in one step. Esters are formed in nucleophilic acyl substitution reactions with alcohols. Amides are formed by nucleophilic acyl substitution reactions with ammonia. Alcohols may be formed using a variety of reducing agents. To form alkenes, carboxylic acids may be reduced to alcohols, which can then be transformed into alkenes by elimination in a second step.

Answer 95

D Lithium aluminum hydride (LiAlH4 or LAH) is a strong reducing agent. LAH can completely reduce carboxylic acids to primary alcohols. Aldehydes are intermediate products of this reaction. The other compounds are not created through the reduction of a carboxylic acid.

Answer 96

D Micelles are self-assembled aggregates of soap in which the interior is composed of long hydrocarbon (fatty) tails, which can dissolve nonpolar molecules. The outer surface is covered with carboxylate groups, which makes the overall structure water-soluble. Soaps, in general, are salts of long-chain hydrocarbons with carboxylate head groups.

Answer 97

D The reaction described is esterification, in which the nucleophile oxygen atom of 1-pentanol attacks the electrophilic carbonyl carbon of butanoic acid, ultimately displacing water to form pentyl butanoate. The acid catalyst is regenerated from 1-pentanol's released proton. Choice A reverses the carbon chains, considering the butyl tail to be the esterifying group. Ethers do not form under these conditions.

Answer 98

D The ɑ-hydrogen of a carboxylic acid is relatively acidic as far as organic compounds go, due to resonance stabilization. However, the hydroxyl hydrogen is significantly more acidic because it is able to share the negative charge resulting from deprotonation between the electronegative oxygen atoms in the functional group.

Answer 99

B The reaction of formic acid, which is a simple carboxylic acid, with sodium borohydride, which is a mild reducing agent, will result in no reaction, and therefore will result in maintenance of the carboxylic acid. Sodium borohydride is too mild to reduce carboxylic acids, and therefore cannot produce the primary alcohols that lithium aluminum hydride, a strong reducing agent, would.

Answer 100

C 5-aminopentanoic acid contains a carboxylic acid and an amine. If this molecule undergoes intramolecular nucleophilic acyl substitution, it will form a cyclic amide. These molecules are called lactams. Lactones are cyclic esters, not amides.

Answer 101

A Butanoic anhydride is an anhydride with two butane R groups. Anhydrides are produced by the reaction of two carboxylic acids with the loss of a water molecule. Therefore, butanoic anhydride would be produced by the reaction of two molecules of butanoic acid.

Answer 102

C Nucleophilic acyl substitutions are favored in basic solution, which makes the nucleophile more nucleophilic; in acidic solution, which makes the electrophile more electrophilic; and by good leaving groups. However, strong bases do not make good leaving groups; weak bases do.

Answer 103

D Based on its name, caprylic acids must be a carboxylic acid. The reaction between a carboxylic acid and ammonia (NH3) would produce an amide - which is not one of the options listed. Instead, we should take a look at the type of reaction occurring. The production of an amide from a carboxylic acid and ammonia occurs through a condensation reaction in which a molecule of water is removed as a leaving group.

Answer 104

A Methyl propanoate is an ester; it can be synthesized by reacting a carboxylic acid with an alcohol in the presence of acid. Here, the parent chain is propanoate, and the esterifying group is a methyl group. Choice B reverses the nomenclature, and would therefore form propyl methanoate. The other reactions listed would not form esters.

Answer 105

C This question requires knowledge of the nomenclature of cyclic molecules. A δ-lactam has a bond between the nitrogen and the fourth carbon away from the carbonyl carbon. This ring will have six elements: the nitrogen, the carbonyl carbon, and the four carbons in between. Cyclohexane carboxylic acid has cyclohexane, a six-membered cycloalkane. The anhydride formed from pentanedioic acid will have the five carbons in the parent chain and one oxygen atom closing the ring, meaning there are still six elements. γ-butyrolactone will have five elements because it contains a bond between the ester oxygen and the third carbon away from the carbonyl carbon. The five elements will be the oxygen, the carbonyl carbon, and the three carbons in between.

Answer 106

D With the same R groups, steric influence is the same in each case, so we can therefore rely solely on electronic effects. When this is all that is taken into account, reactivity toward nucleophiles is highest for anhydrides, followed by esters and carboxylic acids, then amides.

Answer 107

A Methylamine would react readily to form an amide. The less substituted the nucleophile, the easier it will be for the nucleophile to attack the carbonyl carbon and form the amide. In fact, triethylamine would not be able to form an amide at all because it does not have a hydrogen to lose while attaching to the carbonyl carbon.

Answer 108

A Propanamide is an amide; as such, it is the least reactive of the carboxylic acid derivatives discussed in the chapter. Without strong acid or base, propanamide will not be able to undergo nucleophilic acyl substitution and no reaction will occur.

Answer 109

D β-lactams are amides in the form of four-membered rings; amides are generally the least reactive type of carboxylic acid derivative. β-lactams experience significant ring strain from both eclipsing interactions (torsional strain) and angle strain, and are therefore more susceptible to hydrolysis than the linear form of the same molecule.

Answer 110

C As far as we can tell, we are converting one ester to another in this reaction. The fact that this reaction is acid-catalyzed should confirm the suspicion that this is transesterification.

Answer 111

A In order to prepare butyl acetate from butanol, we need to perform a nucleophilic acyl substitution reaction. If the product is an ester, we need to start with a reactant that is more reactive than the ester itself, or the reaction will not proceed. Anhydrides are more reactive than esters, but amides are less reactive. Reaction with propanoic anhydride would result in butyl propanoate.

Answer 112

D The presence of water in an esterification reaction would likely revert some of the desired esters back into carboxylic acids. Small carboxylic acids, like formic or acetic acid, are easily dissolved in water. The polarity of water plays little role in affecting the leaving group; if anything, water can be used to increase electrophilicity of the carbonyl carbon by protonating the carbonyl oxygen. Finally, this is a nucleophilic substitution mechanism, not a nucleophilic addition mechanism. Further, hydrogen bonding would likely augment the reaction.

Answer 113

D Glycine's R group is a hydrogen atom; this amino acid is therefore achiral because the central carbon is not bonded to four different substituents. The other amino acids are all chiral and therefore have both L- and D-enantiomers.

Answer 114

B This reaction is similar to the Gabriel synthesis. Phthalimide acts as nucleophile, the methyl carbon acts as an electrophile, and bromide acts as the leaving group. Therefore, the reaction between methyl bromide and phthalimide results in the formation of methyl phthalimide. Subsequent hydrolysis then yields methylamine.

Answer 115

B Cysteine is well known for containing a sulfur atom because it is able to form disulfide bridges; however, methionine also contains a sulfur atom in its R group.

Answer 116

B An amide is formed from an amine and a carboxyl group or its acyl derivatives. In this question, an amine is already given; the compound to be identified must be an acyl compound. The only acyl compound among the choices given is a carboxylic acid.

Answer 117

C During the Strecker synthesis, ammonia attacks a carbonyl, forming an imine. This imine is attacked by cyanide, forming a amine and a nitrile. Amide bonds are formed between amino acids, but do not appear during the Strecker synthesis.

Answer 118

D One resonance structure of a C - N bond in an amide has the double bond between the C and N, not between the C and O. Thus, the C - N bond of an amide has some sp2 character, and sp2-hybridized atoms exhibit planar geometry.

Answer 119

D Both the Strecker and Gabriel syntheses contain planar intermediates, which can be attacked from either side by a nucleophile. This results in a racemic mixture of enantiomers, and the solution will therefore be optically inactive.

Answer 120

A During the Gabriel synthesis, phthalimide attacks a secondary carbon in diethyl bromomalonate. The secondary carbon is the electrophile, and bromide is the leaving group.

Answer 121

C The Gabriel synthesis includes two nucleophilic substitution steps, followed by hydrolysis and decarboxylation. Dehydration - the loss of a water molecule - is not part of this reaction.

Answer 122

B The pka2 of phosphoric acid is close to physiological pH; therefore, [H2PO4-] ≈ [HPO4^2-] at this pH.

Answer 123

C Pyrophosphate is unstable in aqueous solution and will degrade to form two equivalents of inorganic phosphate. The solvent is water, which should retain its polarity regardless of the presence of solutes. Pyrophosphate and inorganic phosphate are small, charged molecules which are relatively soluble.

Answer 124

B The amino acid in question is aspartic acid, which is an acidic amino acid because it contains an extra carbonyl group. At neutral pH, both of the carbonyl groups are ionized, so there are two negative charges on the molecule. Only one of the charges is neutralized by the positive charge on the amino group, so the molecule has an overall negative charge.

Answer 125

A As DNA is synthesized, it forms phosphodiester bonds, releasing pyrophosphate, PPi. Pyrophosphate is an inorganic phosphate-containing molecule, but it is not the single phosphate group commonly referred to as inorganic phosphate. The DNA molecule itself is referred to as an organic phosphate.

Answer 126

B Phosphoric acid has three hydrogens with pKa values spread across the pH range. This allows some degree of buffering over almost the entire standard pH range from 0 to 14.

Answer 127

A Infrared spectroscopy is most useful for distinguishing between different functional groups. Almost all organic compounds have C - H bonds so except for fingerprinting a compound, these absorptions are not useful. Little information about the optical properties of a compound can be obtained by IR spectroscopy.

Answer 128

C Because molecular oxygen is homonuclear (composed of only one element) and diatomic, there is no net charge in its dipole moment during vibration or rotation; in other words, the compound does not absorb in a measurable way in the infrared region. IR spectroscopy is based on the principle that, when the molecule vibrates or rotates, there is a change in dipole moment. Choice A is incorrect because oxygen does have molecular motions; they are just not detectable in IR spectroscopy. Choice B is incorrect because it is possible to record the IR of a gaseous molecule as long as it shows a change in its dipole moment when it vibrates. Choice D is incorrect because lone pairs do not have an effect on the ability to generate an IR spectrum of a compound.

Answer 129

D In this reaction, the functional group is changing from a hydroxyl to an aldehyde. This means that a sharp peak will appear around 1750 cm^-1, which corresponds to the carbonyl functionality. Choice C is the opposite of what occurs; the reaction will be characterized by the disappearance of the O-H peak at 3100 to 3500 cm^1, not its appearance. Comparing the fingerprint regions will provide evidence that a reaction is occurring, but it is not as useful for knowing that the reaction that occurred was indeed the one that was desired.

Answer 130

A The peak at 9.5 ppm corresponds to an aldehydic proton. This signal lies downfield because the carbonyl oxygen is electron-withdrawing and deshields the proton. Choice C corresponds to a carboxyl proton and is even further downfield because the acidic proton is deshielded to a greater degree than the aldehydic proton. Choice B corresponds to aromatic protons. Choice D is characteristic of an alkyl proton on an sp3-hybridized carbon.

Answer 131

C This isotope has no magnetic moment and will therefore not exhibit resonance with an applied magnetic field. Nuclei with odd mass numbers (1H, 11B, 13C, 15N, 19F, and so on) or those with an even mass number but an odd atomic number (2H, 10B) will have a nonzero magnetic moment.

Answer 132

D Spin-spin coupling (splitting) is due to influence on the magnetic environment of one proton by protons on the adjacent atom. These protons are three bonds away from each other. Splitting in other NMR spectra can include coupling with carbon atoms, but not in 1H-NMR.

Answer 133

C Most conjugated alkenes have an intense ultraviolet absorption. Aldehydes, ketones, acids, and amines all absorb in the ultraviolet range. However, other forms of spectroscopy (mainly IR and NMR) are more use for precise identification. Isolated alkenes can rarely be identified by UV spectroscopy.

Answer 134

B The region in question often gives information about the types of alkyl groups present. Specifically, ethanol will give a characteristic triplet for the methyl group (which is coupled to -CH2-) and a quartet for -CH2- (which is coupled to the methyl group). Isopropanol will have a septet for the -CH- group (which is coupled to both methyl groups combined) and a doublet for the two methyl groups (which are coupled to -CH-). In both cases, the proton in the alcohol does not participate in coupling. The alcohol hydrogen likely lies downfield for both compounds because it is bonded to such an electronegative element.

Answer 135

A The HOMO is the highest occupied molecular orbital. Only after absorbing ultraviolet light is an electron excited from the HOMO to the LUMO, the lowest unoccupied molecular orbital.

Answer 136

A Carbonyl groups (C=O) in conjugation with double bonds tend to absorb at lower wavenumbers because the delocalization of pi electrons causes the C=O bond to lose double-bond character, shifting the stretching frequency closer to C-O stretches. Remember that higher-order bonds tend to have higher absorption frequencies, so loss of double-bond character would decrease the absorption frequency of the group.

Answer 137

B Wavenumber (1/λ) is directly proportional to frequency (c/λ). It is inversely proportional to wavelength and has no proportionality to percent transmittance or absorbance.

Answer 138

A Enantiomers will have identical IR spectra because they have the same functional groups and will therefore have the exact same absorption frequencies. Enantiomers have opposite specific rotations, but specific rotation actually has no effect on the IR spectrum.

Answer 139

A The oxygen of the hydroxyl group will deshield the hydroxyl hydrogen, shifting it downfield, or left. Hydrogens in carboxylic acids can have some of the most downfield absorbances, around 10.5 to 12 ppm.

Answer 140

D The coupling constant is a measure of the degree of splitting introduced by other atoms in a molecule, and is the frequency of the distance between subpeaks. It is measured in hertz. The coupling constant is independent of the value of n + 1, and is not changed by calibration with tetramethylsilane.

Answer 141

B Amino acids in their fully protonated form contain all three of the peaks that should be memorized for Test Day: C-O, N-H, and O-H. While statements I and II correctly give the peaks for the C=O bond (sharp peak at 1750 cm^-1) and the N-H bond (sharp peak at 3300 cm^-1), the peak for the O-H bond is in the wrong place. In a carboxylic acid, the C=O bond withdraws electron density from the O-H bond, shifting the absorption frequency down to about 3000 cm^1. Statement III is therefore incorrect.

Answer 142

B Fractional distillation is the most effective procedure for separating two liquids that boil within a few degrees of each other. Ethyl acetate and ethanol boil well within 25℃ of each other and thus would be good candidates for fractional distillation. Fractional distillation could also be used for the liquids in Choice C, but would require lower pressures because of their high boiling points.

Answer 143

C By extracting with sodium hydroxide, benzoic acid will be converted to its sodium salt, sodium benzoate. Sodium benzoate, unlike its conjugate acid, will dissolve in an aqueous solution. The aqueous layer simply has to be acidified afterward to retrieve benzoic acid. Choice A is incorrect because diethyl ether and tetrahydrofuran are both nonpolar and are miscible. Hydrochloric acid will not form benzoic acid into a soluble salt. Finally Choice D is incorrect because protonated benzoic acid has limited solubility in water.

Answer 144

D It is more effective to perform four successive extractions with small amounts of ether than to perform one extraction with a large amount of ether.

Answer 145

D Hexane is less polar than ether and, therefore, is less likely to displace polar compounds absorbed to the silica gel. This would decrease the distance these polar compounds would travel, decreasing Rf values.

Answer 146

D In column chromatography, as in TLC, the less polar compound travels most rapidly. This means that 1-naphthalenemethanol, with the highest Rf value, would travel most rapidly and would be the first to elute from the column.

Answer 147

B Because methylene chloride is denser than brine (salt water), the organic layer will settle at the bottom of the funnel. Methylene chloride is nonpolar, so it cannot mix with brine.

Answer 148

B Silica gels are polar. Polarity is used to selectively attract specific solutes within a nonpolar solvent phase. Although silica gels have other properties, this is the most important to TLC.

Answer 149

B In this question, three substances must be separated using a combination of techniques. The first step should be the most obvious: remove the sand by filtration. The remaining compounds - benzoic acid and naphthalene - are still dissolved in ether. If the solution is extracted with aqueous base, the benzoate anion is formed and becomes dissolved in the aqueous layer, while naphthalene, a nonpolar compound, remains in the ether. Finally, evaporation of the ether will yield purified naphthalene.

Answer 150

C This is the only option that would be effectively separated by simply distillation. Choice B would require vacuum distillation because the boiling points are over 150℃. In Choice A and D, the boiling points are within 25℃ of each other and would therefore require fractional distillation in order to be separated.

Answer 151

C Affinity chromatography, using the target for the biological effector or a specific antibody, would work best in this case. It will specifically bind the protein of interest and keep it in the column.

Answer 152

D Because the solution is composed of a much larger molecule and a much smaller molecule, size-exclusion chromatography would effectively remove the smaller insulin molecule into the fraction retained in the column and allow the titin to be eluted. Affinity chromatography could also be used, but comes with a risk of rendering the titin unusable; the eluent run through an affinity chromatography column often binds to the target molecule.

Answer 153

B Each of these is the mobile phase of the system, in which the solutes are dissolved and move. The stationary phase in gas chromatography is usually a crushed metal or polymer; the stationary phase in paper chromatography is paper.

Answer 154

C Warm or hot solvent is generally used in gravity filtration to keep the desired product soluble. This allows the product to remain in the filtrate, which can then be collected. In this case, the student likely used solvent that was too cold, and the product crystallized out. The product should be present in the residue.

Answer 155

B Because the lactoferrin proteins are likely to be charged, as is the resin described in the question, this is an example of ion-exchange chromatography. The charged protein molecules will stick to the column, while the remainder of the milk washes through and can later be washed off of the column and collected.