DAT BC OC PT #1-5

How many signals will be present on the 13C NMR spectrum of the following molecule?

13C NMR gives a signal for each chemically unique carbon in a molecule. To determine how many unique carbons a molecule has, locate any planes of symmetry. This molecule has two planes of symmetry. There are only three unique carbons that produce signals on the NMR spectrum. [D] is the answer.

How many signals will be present on the 13C NMR spectrum of ethyl acetate?

13C NMR produces a signal for each chemically unique carbon in a molecule. Locate any planes of symmetry. Ethyl acetate does not have any planes of symmetry. Each carbon will produce a unique signal: [D] is the answer.

What is the IUPAC name for the following molecule?

3-chloro-3-fluoro-2,2-dimethyl-4-propyloctane

cont 47

47 exp

61 mec

61

70 sr

70

73 mech

73

77 mech

77

80 exp

80

90 exp

90

What are the two Brønsted-Lowry acids in the following equilibrium?

A Brønsted-Lowry acid is a proton donor. HCN acts as an acid on the left side of the equation, donating a proton to H2O. H3O+ acts as an acid on the right side of the equation, donating a proton to CN-: [D] is the answer.

What is the major product of the following reaction?

Adding a Grignard reagent to an ester results in a total of two additions of the reagent. In the first addition, CH3MgBr adds to the electrophilic carbonyl in the ester, forming an alkoxide. In such strongly basic conditions, the -OR group in the ester becomes an acceptable leaving group. After it leaves, a ketone is formed: In the second addition, CH3MgBr adds to the ketone, forming a tertiary alcohol: [A] is the answer.

Which of the following statements best describes the equilibrium below?

Amines are strong bases. Nitrogen is less electronegative than oxygen, so it is more willing to donate its lone pairs to act as a base and de-protonate an acid. (CH3)2N- is a stronger base than OH-, so the products are favored. Stronger bases have weaker conjugate acids. The weaker the acid, the higher the pKA. [C] is the answer. A. The equilibrium favors the reactants because (CH3)2N- has a lower pKB than OH-. (CH3)2N- is a stronger base with a lower pKB, but the products are favored. Eliminate [A]. B. The equilibrium favors the reactants because H2O has a lower pKA than (CH3)2NH. The products are favored. H2O has a lower pKA. Eliminate [B]. D. The equilibrium favors the products because (CH3)2N- has a higher pKB than OH-. The products are favored, but (CH3)2N- has a lower pKB. Eliminate [D]. E. The equilibrium favors the reactants because (CH3)2NH has a lower pKA than H2O. H2O has a lower pKA and the products are favored. Eliminate [E].

Which of the following compounds would be the major product of the E1 reaction of (2S)-2-iodo-3-methylpentane?

An E1 reaction will produce the most thermodynamically stable alkene as the major product. The first step is the leaving group leaving, creating a carbocation. This is a 2° carbocation, which will rearrange to form the more thermodynamically stable 3° carbocation. A β-hydrogen will be removed to form the alkene. Removal of a hydrogen labeled Hb results in the least substituted and least thermodynamically stable alkene. This will be a minor product. Removal of a hydrogen labeled Ha results in either the E or Z alkenes. These are more substituted alkenes and are more thermodynamically stable. The E alkene has less steric hindrance than the Z alkene. The E alkene is more stable and is the major product: [B] is the answer.

Which of the following molecules is aromatic?

Aromatic systems must be in a ring and each atom in the ring must have a p-orbital. If there is a sp3-hybridized carbon in the ring, it cannot be aromatic. Answer choices [A] and [D] have sp3 hybridized carbons and are nonaromatic. Hückel's rule states that a system must have 4n + 2 pi electrons in order to be aromatic (where n = 0, 1, 2, 3...). Answer choice [B] has 4 pi electrons, which does not satisfy Hückel's rule, so it is antiaromatic. [C] is the answer.

Which statement best describes molecules I and II shown below?

Assign E and Z to these geometric isomers. The highest priority substituents are the phenyl groups. Molecule I has the phenyl groups on the same side, so it is the Z isomer. Eliminate [A]. Molecule II has the phenyl groups on opposite sides, so it is the E isomer. Eliminate [D]. Molecule I has the two bulkiest groups on the same side, causing strain and results in a less thermodynamically stable compound. Eliminate [B] and [E]. Molecule II has less strain and is more thermodynamically stable. [C] is the answer.

For the molecule below, how many non-hydrogen atoms are in the same plane?

Atoms that are sp2 hybridized and in a conjugated system will have trigonal planar geometry and, thus, all be in the same plane as one another. Referring to the diagram below, that describes atoms 1-10. Atom 11 will also be in the same plane with the others, despite being sp3 hybridized, because it is directly coming off of an atom that has trigonal planar geometry. The only non-hydrogen atom in this molecule that is not in the same plane with the others is the methyl group since it is able to freely rotate around the single bond.

Which is the best resonance structure for the following molecule?

B. Molecule B is the most stable resonance structure. Each atom has a full octet, the negative charge is on the most electronegative atom, and there is less separation of charge. [B] is the answer. A.​ Molecule A has greater separation of charge than the other answer choices, and one carbon does not have a full octet, decreasing stability. Eliminate [A]. C. Molecule C lacks a full octet on one carbon, decreasing stability. Eliminate [C]. D. Molecule D lacks a full octet on one carbon and it has a negative charge on a carbon, which is not the most electronegative atom in the molecule. This decreases stability. Eliminate [D]. E. Molecule E is a tautomer, not a resonance structure. A tautomer is a structural isomer. Eliminate [E].

Which of the following reagents would best accomplish the following chemical transformation?

B. H3O+ , heat This reaction converts a nitrile into a carboxylic acid. This can be accomplished by heating the nitrile in aqueous acid. [B] is the answer. A. mCPBA This is an oxidizing agent and is most commonly used to turn alkenes into epoxides or to perform a Baeyer-Villiger oxidation. Eliminate [A]. C. 1) O3, 2) Zn, H2O This would accomplish ozonolysis on an alkene. Eliminate [C]. D. LiAlH4 This would reduce the nitrile. Eliminate [D]. E. PBr3 This is used to convert alcohols into alkyl bromides or carboxylic acids into acyl bromides. Eliminate [E].

Which of the following molecules would display a sharp stretch at about 1,750 cm-1 on its IR spectrum?

Carbonyl groups absorb near 1,700 cm-1 on IR spectra. Molecule A is the only molecule with a carbonyl. [A] is the answer. A carbon nitrogen triple bond has a sharp peak around 2,200 cm-1. Eliminate [B]. An alcohol group has a large, broad trough between 3,000-3,600 cm-1. Eliminate [C]. Carbon hydrogen bonds have big peaks around 3,000 cm-1. Eliminate [D] and [E].

Predict the major product for the following reaction:

Cyanide is a very strong nucleophile, which indicates a SN2 reaction. This reaction is stereospecific, so it will produce a single product with an inversion of configuration. In order for a substitution reaction to occur, there must be a good leaving group. Bromide is a good leaving group, but fluoride is a bad leaving group. The cyanide will only react at the bromine carbon: [C] is the answer.

Which of the following is associated with an E1 reaction?

D. A carbocation intermediate The rate-determining step in an E1 reaction occurs when the leaving group leaves. A carbocation intermediate is left. [D] is the answer. A. An antiperiplanar conformation Because the rate-determining step is unimolecular, a molecule does not need an antiperiplanar conformation. E2 reactions require an antiperiplanar conformation. Eliminate [A]. B. A pentavalent transition state SN2 reactions involve a backside attack with a pentavalent transition state. Eliminate [B]. C. A bimolecular rate law An E1 reaction involves a unimolecular rate-determining step. Eliminate [C]. E. Stereoselectivity in the product Carbocation intermediates are planar, sp2-hybridized species which can be attacked from either side, so they do not lead to stereoselectivity. Eliminate [E].

The following structure has been proposed to be associated with an SN2 reaction. Which of the statements below best describes this structure?

D. A pentavalent transition state with higher energy than the reactants SN2 reactions form a pentavalent transition state. This is a high-energy transition state, and the incoming nucleophile performs a backside attack while the leaving group leaves. [D] is the answer. A. A pentavalent intermediate with lower energy than the reactants The pentavalent transition state is not an intermediate. Eliminate [A]. B. A pentavalent transition state with lower energy than the reactants Transition states represent the highest energy peak, or the activation energy, on a reaction coordinate diagram. Eliminate [B]. C. A pentavalent intermediate with higher energy than the reactants The pentavalent transition state is not an intermediate. Eliminate [C]. E. The formation of a chiral center upon a backside attack There is not chiral center in this reaction. Eliminate [E].

Which of the following best describes the hybridization of the numbered carbons (C1-C8) in the radical compound below?

D. C1-C7 are all sp2-hybridized, while C8 is sp3-hybridized. When determining hybridization, count the number of attached atoms and lone pairs. A unpaired radical electron is not counted. Carbon 7 is sp2-hybridized. All of the carbons on the benzene ring are attached to three atoms, so they are sp2-hybridized. Carbon 8 doesn't have any pi-bonds and is attached to four atoms, so it is sp3-hybridized. [D] is the answer. A. All carbon atoms (C1-C8) in this molecule are sp2-hybridized. Carbon 8 is sp3-hybridized. Eliminate [A]. B. C1, C2, C3, C4, C5, and C6 are sp2-hybridized while C7 and C8 are sp3-hybridized. Carbon 7 is sp2-hybridized. Eliminate [B]. C. C1 and C4 are sp-hybridized, while C2, C3, C5, C6, C7, and C8 are sp2-hybridized. C1-C7 are sp2-hybridized, and C8 is sp3-hybridized. Eliminate [C]. E. The hybridization of C7 cannot be determined because it has an unpaired electron. Carbon 7 is sp2-hybridized. Eliminate [E].

What are the two Bronsted-Lowry bases in the following equilibrium? CH3NH2 + H2O <-> CH3NH3+ + OH- A. CH3NH2 and H2O B. CH3NH2 and CH3NH3+ C. H2O and OH- D. CH3NH2 and OH-

D. CH3NH2 and OH-

Which of the following reagents will produce the desired diol shown below from cyclopentene?

D. OsO4 , H2O2 Reacting an alkene with OsO4 with H2O2 results in a syn addition, giving the desired cis diol: [D] is the answer. A. 1. mCPBA 2. H3O+ Anti dihydroxylation results in the formation of a trans diol: B. Na2Cr2O7 , H2SO4 These reagents oxidize alcohols to ketones or carboxylic acids: C. 1. BH3·THF 2. H3O+ These reagents result in an anti-Markovnikov addition of one hydroxy group: E. 1. O3 2. Zn, H2O An ozonolysis reaction forms two aldehydes:

In the following reaction, which observations on an IR spectrum would best indicate the conversion of reactants into products?

D. The appearance of a stretch at 1700 cm-1 The product has a -C=O group that isn't present in the reactant. As the -C=O is formed, a stretch at 1700 cm-1 will appear. [D] is the answer. A. The appearance of a stretch at 3300 cm-1 The reactant and the product have an -OH group, which has a stretch at 3300 cm-1. Eliminate [A]. B. The disappearance of a stretch at 3300 cm-1 The reactant and the product have an -OH group, which has a stretch at 3300 cm-1. Eliminate [B]. C. The appearance of a stretch at 2200 cm-1 Stretches at 2200 cm-1 correspond to triple bonds, which are not present in the product. Eliminate [C]. E. The disappearance of all stretches at 3000 cm-1 Stretches at 3000 cm-1 correspond to -C-H bonds, which are present in the reactant and product. Eliminate [E].

How many peaks would appear on the 13C NMR spectrum of the following molecule?

Determine how many chemically unique carbon atoms are in the molecule. Each chemically unique carbon will give a signal on the spectrum. Find any planes of symmetry. The cyclobutane ring has a plane of symmetry, so carbon atoms d give only one signal. There are five signals: [E] is the answer.

Which of the following would undergo an aldol condensation with 3,3-dimethyl-2-hexanone to yield a single major product?

Draw each compound and examine its alpha hydrogens: When performing an aldol cross-condensation, only one product will be produced if 3,3-dimethyl-2-hexanone reacts with a compound without any alpha hydrogens. If a compound has two different sets of alpha hydrogens, or both reactants have alpha-hydrogens, the reaction will produce a mixture of products. The only answer choice without any alpha-hydrogens is benzaldehyde. [C] is the answer.

Which of the following methods would most effectively separate two volatile compounds? A. Silica gel chromatographyB. Thin layer chromatographyC. ExtractionD. CrystallizationE. Gas-liquid chromatography

E. Gas-liquid chromatography Volatile compounds are best separated by a method that involves the gas phase. [E] is the answer. A. Silica gel chromatography Silica gel chromatography cannot separate volatile compounds. Eliminate [A]. B. Thin layer chromatography Thin layer chromatography uses silica gel, which cannot separate volatile compounds. Eliminate [B]. C. Extraction Extraction separates solid or liquid compounds based on solubility and acid/base properties. Eliminate [C]. D. Crystallization Crystallization involves formation of a pure solid. Eliminate [D].

Which of the following is an enantiomer of D-glucose (shown below)?

Enantiomers are unique pairs of molecules that are non-superimposable mirror images of one another. They have opposite configurations at each chiral center. The enantiomer of D-glucose will have a dash where D-glucose has a wedge, and a wedge where it has a dash, for example. [C] is the answer.

What is the structure of the major product (X) of the following aldol cross-condensation?

First an enolate will be formed. The ketone is the only reactant with an α-carbon that has an acidic hydrogen, so it will become the enolate. The enolate attacks the aldehyde and forms a β-hydroxyketone. This molecule undergoes a dehydration reaction to form the more thermodynamically stable trans-α,β-unsaturated ketone:

What is the correct IUPAC name for the following compound?

First, find the longest continuous carbon chain. The alcohol takes priority in numbering, so it should have the lowest carbon number possible. There is an ethyl group on C5, a methyl on C6, and a chlorine on C7. List the substituents in alphabetical order. The IUPAC name of this compound is 7-chloro-5-ethyl-6-methyloctan-3-ol. [A] is the answer.

The following compounds all have similar molecular weights. Which compound would be isolated FIRST in the process of fractional distillation?

Fractional distillation separates compounds based on boiling point. The compound with the lowest boiling point will be isolated first. Compounds with the weakest intermolecular forces have the lowest boiling points. Hydrogen bonding is the strongest intermolecular force. Eliminate [B] and [D]. The remaining answer choices are non-polar hydrocarbons with the same molecular weight. The only intermolecular force present is London dispersion forces. When molecular weights are equal, a higher degree of branching leads to a lower degree of intermolecular forces, and a lower boiling point. The most highly branched alkane will have the lowest boiling point. [E] is the answer.

Which of the following statements best explains why Grignard reactions must be performed in aprotic solvents?

Grignard reactions must be carried out in an aprotic solvent. Aprotic solvents are not capable of hydrogen bonding. Protic solvents have a slightly acidic hydrogen attached to an oxygen or a nitrogen. Grignard reagents are extremely strong bases, so they will react quickly and exothermically with a protic solvent. [C] is the answer.

IR

IR

Which of the following is a meso compound?

In order for a molecule to be meso, it must have chiral centers and a plane of symmetry. Molecule A has two chiral centers and a plane of symmetry: [A] is the answer. Molecule B has two chiral centers but no plane of symmetry: Molecules C, D and E do not have chiral centers:

All of the following are aromatic compounds EXCEPT one. Which one is the EXCEPTION?

In order to be aromatic a molecule must be cyclic or polycyclic, all atoms must be sp2 or sp hybridized, and it must satisfy Hückel's rule. Each answer choice satisfies the first two rules. Hückel's rule states that a molecule must have 4n + 2 π electrons, where n = 0, 1, 2, 3... Answer choices [A], [B], [C], and [D] have 6 π electrons. They satisfy Hückel's rule: 4n + 2 = 6 4n = 4 n = 1 Answer choice [E] has 4 π electrons. This molecule is anti-aromatic because it satisfies the first two rules, but does not satisfy Hückel's rule: 4n + 2 = 4 4n = 2 n is not a whole number [E] is the answer.

Which of the following represents an initiation step in the mechanism of the free radical bromination of toluene with NBS shown below?

In radical halogenation reactions, initiation steps involve the homolytic cleavage of a covalent bond to form two radicals. Eliminate [B], [C], and [E]. In this reaction NBS is the brominating agent, not Br2. Eliminate [A]. [D] is the answer.

Which of the following reactions represents a propagation step in the free radical bromination of propane?

In radical halogenation reactions, the propagation step has one radical on the reactant side and a different radical on the product side: [C] is the answer. Answer choice [A] is an initiation step with zero radicals on the reactant side and two radicals on the product side. Answer choices [B], [D], and [E] are termination steps with two radicals on the reactant side and zero radicals on the product side.

In thin layer chromatography, which of the following molecules would have the highest Rf?

In thin layer chromatography (TLC), the plate is made of polar silica gel. More polar molecules are more attracted to the plate, and travel a shorter distance. Less polar molecules are less attracted to the plate, and travel a farther distance. The Rf is the distance traveled by the compound divided by the total distance traveled by the solvent. The less polar the molecule, the higher the Rf. Answer choice D is the only non-polar molecule, so it will have the largest Rf. [D] is the answer.

What is the major product of the Markovnikov addition of H-Br to the following molecule?

Normally the Markovnikov addition of H-Br to an alkene results in the addition of H to the least substituted carbon and the Br to the most substituted carbon. However, this reaction involves a carbocation. Carbocations are capable of rearrangement into a more stable intermediate. The addition of H+ to the alkene forms a secondary carbocation. This carbocation rearranges to a more stable tertiary carbocation. Once this more stable carbocation reacts with Br-, molecule A is formed: [A] is the answer.

Which of the following statements best describes compound I with respect to compound II?

On compound I, the two highest ranking substituents are on the same side, so this is the Z configuration. Eliminate answer choices [A], [C], and [D]. The stability of the alkene depends on the steric bulk of its substituents. The phenyl ring is bulkier than the methyl group. Because the phenyl group and the tert-butyl group are on the same side, compound I is less stable than compound II. In general, Z isomers are less stable than E isomers. Eliminate [E]. [B] is the answer.

Which of the following reactants would undergo an addition-elimination reaction with water most rapidly?

Reactants that are more reactive will react more rapidly. All of these compounds are carboxylic acid derivatives. The electrophilicity of the carbonyl carbon as well as the leaving group ability affect reactivity. Amides, answer choice [A], are the least reactive, followed by slightly more reactive esters, answer choices [D] and [E], and anhydrides, answer choice [C]. The most reactive are acid halides, answer choice [B]. [B] is the answer.

Which statement is true with regard to the following addition reactions (I and II)?

Reaction I, the addition of water to an alkene in acid, is a Markovnikov addition. In this case there is a carbocation rearrangement, which results in the 3° alcohol major product. This is still Markovnikov addition because the -OH is on the more substituted carbon. Reaction II, the oxymercuration-demercuration reaction, is a Markovnikov addition. This intermediate cannot undergo a rearrangement. The -OH group is placed on the more substituted carbon. [B] is the answer.

Carb Acid Derivative Rxns

Rxns

Which of the following statements best describes the stereochemistry at the reactive site of an SN2 reaction?

SN2 reactions are stereospecific. They produce one product with a specific stereochemistry. This stereospecificty results from the pentavalent transition state that occurs during backside attack of the nucleophile. After the leaving group leaves, the product has an inversion of configuration. [B] is the answer.

Which of the following molecules would be collected first in the eluent from a silica gel chromatography column?

Silica gel chromatography separates compounds based on polarity. Silica gel has a polar surface, so it interacts more with polar molecules. Polar molecules stick to the column and take longer to elute. Nonpolar molecules elute first. The most nonpolar molecule is the butane, which has no polar bonds. [E] is the answer.

Which structure below best represents a reasonable resonance structure of the following molecule?

Structure B is the only resonance structure that does not violate the octet rule. Each atom has a full octet: [B] is the answer. Structure A has a carbon bonded to 5 atoms: Structure C has a negative charge on a nitrogen with two lone pairs and 3 bonds: Structure D has a sulfur with only two lone pairs and a single bond: Structure E has an extra hydrogen as well as an illogical electronic structure:

Of the following, which is the most stable resonance form?

Structures 1, 2, 4 and 5 are valid resonance forms of a benzylic carbocation. Structure 3 is an impossible Lewis structure. Eliminate [C]. Structures 1, 2, and 4 lose aromaticity when cyclic conjugation of the benzene ring disrupted. Structure 5's aromaticity is intact, so it is the most stable. [E] is the answer.

The following functional group is the predominant linkage found in nucleic acids. What is this functional group called?

The backbone of RNA and DNA is made up of phosphodiester linkages. This is a phosphorous double-bonded to an oxygen atom and two other groups consisting of oxygen bound to a carbon (a diester). [D] is the answer.

Which of the following is an intermediate in the bromination of toluene?

The bromination of toluene is an electrophilic aromatic substitution reaction. The methyl group on toluene is o,p-directing. The electron-rich aromatic ring attacks the electrophile, Br, in the para position. This forms of a positively-charged arenium ion. This carbocation is unstable from the positive charge on the molecule and the temporary loss of aromaticity. It is partially stabilized by resonance, which allows the positive charge to be distributed: [B] is the answer.

The overlap of which orbitals forms the double bond in the following molecule?

The carbon atoms in this molecule are attached to three atoms, so they are sp2-hybridized. The carbon atoms are bonded through a sigma-bond formed by the overlap of sp2 orbitals. Eliminate answer choices [A], [C], and [E]. The second bond in the double bond is formed by the overlap of two p orbitals, forming a pi-bond. Eliminate [B]. [D] is the answer.

What is a major organic product in the following reaction sequence?

The first step is oxidation using the Jones reagent. This oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid. The resulting β-ketoacid undergoes a decarboxylation reaction in acid and heat. This decarboxylation produces a ketone and carbon dioxide: [A] is the answer.

What is the final major organic product of the following reaction series?

The first step of this reaction is an acid-catalyzed Fisher esterification, producing a methyl ester. The next step is a Grignard reaction. When a Grignard reacts with an ester, it adds twice, producing in tertiary alkoxide. Finally, the alkoxide is protonated in aqueous acid to yield a tertiary alcohol: [A] is the answer.

What is a product of the following reaction sequence

The first step of this reaction is the nitration of the toluene. The methyl substituent is slightly electron donating, so it is activating and ortho/para directing. The major product is para-nitrotoluene. The second step is the reduction of the nitro group to an amine. The final step is an acetyl protection of the amine. The amine acts as the nucleophile and the acetic anhydride is the electrophile. The pyridine deprotonates the molecule: [C] is the answer.

What is the final product (2) in the following reaction sequence?

The first step of this reaction sequence is the reduction of an aldehyde to a primary alcohol, using NaBH4. The second step converts the alcohol into an alkyl bromide, using PBr3: [E] is the answer.

What is the IUPAC name of the following molecule?

The longest continuous chain of carbon atoms is seven carbons long, called heptane. Eliminate [E]. The alcohol has higher priority in naming than the alkyne. Eliminate [A] and [C]. The triple bond is an alkyne, so it should be denoted as -yn, not -en. Eliminate [D]. [B] is the answer.

Which of the following compounds has the lowest pKA?

The lower the pKA, the more acidic the acid. The best way to assess the acidity of a molecule is to look at the stability of the conjugate base. The more stable the conjugate base, the more acidic the acid. Carboxylic acids are much stronger than alkanes and alcohols. Eliminate [C], [D], and [E]. Compound A is a conjugated diene, which is very stable due to resonance. The conjugate base of compound A has more resonance than that of compound B. Compound A is the most acidic and has the lowest pKA. [A] is the answer.

Which of the following has the lowest pKa?

The lower the pKa the more acidic the molecule is. Answer choices [C] and [E] are alcohols. Alcohols have higher pKa's than carboxylic acids due to less resonance stabilization. Eliminate [C] and [E]. The remaining answer choices are carboxylic acids. Answer choice [D] is the strongest acid because the conjugate base formed after deprotonation is stabilized by both resonance and induction. [D] has two electronegative fluorine atoms, which inductively stabilize the negative charge through their electron withdrawing properties. This spreads the negative charge throughout the molecule. The more stable the conjugate base, the more acidic the acid. [D] is the answer.

Which of the following best represents the most stable conformation of the substituted cyclohexane shown below?

The molecule has its two substituents on the same side of the ring, in the cis form. The isopropyl group has the most steric bulk, so it is most stable in the equatorial position. The two substituents are on carbons 1 and 4. [A] is the answer.

Which of the following is the least acidic?

The more stable the conjugate base, the more acidic the acid. Draw each compound's conjugate base. First, consider atom size. The larger the atom bearing the negative charge, the more stable the compound. Compounds B and D are larger, more stable and more acidic: Eliminate [B] and [D]. Next, consider resonance stabilization. Compound E has extra resonance stabilization due to the para-nitro group, so it is a stronger acid than the other phenols: Eliminate [E]. Finally, in compound A, the fluorine inductively stabilizes the negative charge, increasing acidity: Eliminate [A]. Compound C is the only conjugate base with no extra stabilization, making it the least acidic: [C] is the answer. Topic: Acid Base Chemistry

Predict the major product of the following reaction:

The presence of a phosphonium ylide, (Ph)3P+-CH2-, the Wittig reagent, indicates that an alkene will be formed in between the carbon in the Wittig reagent and the carbonyl carbon: The other major organic product is Ph3P=O. [B] is the answer.

What is the major product of the following reaction?

The reactant is a tertiary haloalkane, ruling out a SN2 reaction. Eliminate answer choices [D] and [E]. The base, NaOCH3, is strong and unhindered, indicating an E2 reaction. The more thermodynamically stable highly substituted alkene will be the major product. Eliminate answer choice [B], a less stable terminal alkene. The leaving group, Br, and the ß-hydrogen must be antiperiplanar. Rotate the carbon carbon bond: After elimination occurs, the product is in the trans configuration: [A] is the answer.

Which of the following statements best describes the reaction sequence below?

The starting material is an aldehyde and it is reduced to an alcohol. The -OH is attached to a carbon that is attached to one other carbon, so it is a 1º alcohol. [C] is the answer.

Which of the following compounds will have the highest melting point?

The stronger the intermolecular forces, the higher the melting point. Hydrogen-bonding is the strongest intermolecular force, so the molecule with the highest capability for H-bonding will have the highest melting point. Choices B, C and D are not capable of H-bonding. Eliminate [B], [C], and [D]. Carboxylic acids can H-bond to a higher degree than alcohols, increasing their melting point. [A] is the answer.

Which of the following best describes the absolute configuration of the molecule below?

There is no chiral center at carbon 4. Eliminate [D] and [E]. Rank the groups around carbon 1 (red) and carbon 2 (blue). The lowest priority group on carbon 1 is hydrogen, labeled with *, which is moving backwards into the plane. Follow the order of the ranking of the substituents. This is a counter-clockwise rotation, and a S configuration. The lowest priority group on carbon 2 is hydrogen, labeled with #, which is moving forwards out of the plane. Reverse the order of the ranking of the substituents. This is a clockwise rotation, and a R configuration. [C] is the answer.

What method would most effectively separate a mixture of the following two compounds?

These molecules are diastereomers, stereoisomers that are not mirror images of one another and are non-superimposable. Diastereomers can be separated by traditional means unlike enantiomers, which are mirror images of one another. D. Distillation Distillation separates compounds based on boiling point. Diastereomers have different physical properties. [D] is the answer. A. Polarimetry Polarimetry is not a method of chemical separation. It measures optical activity. Eliminate [A]. B. Extraction Extraction separates compounds using acid-base properties. These molecules do not have different acid-base properties. Eliminate [B]. C. Mass spectrometry Mass spectrometry is not a method of chemical separation. This calculates a compound's molecular weight. Eliminate [C]. E. Decanting Decanting separates solids and liquids. Eliminate [E].

Which term best describes the relationship between the following two molecules?

These molecules are superimposable mirror images, so they are identical. To determine if a molecule is superimposable on its mirror image, determine if it is meso. A meso compound has two or more chiral centers and a mirror plane: Another way to solve is to rotate one molecule by 180o:

The specific rotations of the molecules X and Y are shown below. Which of the following statements best describes the optical activity of a solution containing equal concentrations of compounds X and Y?

These two molecules are enantiomers. They are non-superimposable mirror images of one another and they have equal and opposite specific rotations. If there is a 50:50 mixture of two enantiomers, it is a racemic mixture. Racemic mixtures have an optical rotation of 0. [A] is the answer.

What is the major organic product of the following reaction?

This is a Diels-Alder reaction. The furan acts as the diene, and the di-substituted trans-alkene acts as the dienophile: [A] is the answer. Diels-Alder reactions form 6-membered rings. Molecules D and E contain 5- and 7-membered rings. Eliminate [D] and [E]. Because one of the reactants is a ring, the product will be a fused ring system. Eliminate [B]. The alkene is trans, so the product should also be trans. Molecule C is a cis-ester. Eliminate [C].

Which statement best describes the conformation of the following molecule?

This is a Newman projection looking down from C2 to C3 in this deuterated butane. All groups are staggered, not eclipsed. The two bulkiest substituent groups are the methyl groups. The methyl groups are 180 degrees from each other which is anti, reducing steric strain. This is the most stable conformation because it is staggered and anti. [E] is the answer.

The following reaction yields two isomeric substitution products. What is their relationship?

This is a SN1 reaction because water is not a strong nucleophile. The bromine leaves, forming a benzylic carbocation. Carbocation carbons are sp2-hybridized in a trigonal planar shape. Water can nucleophilically attack from either side of the plane, forming a racemic mixture of stereocenters: These compounds are non-superimposable mirror images of each and, therefore, enantiomers. [B] is the answer. DAT Pro-tip: Diastereomers contain at least two stereocenters. The carbon on the left of the molecule is not chiral; it is bonded to two methyl groups. Therefore, diastereomers is not the correct answer.

What is the major product of the following reaction?

This is a Williamson Ether Synthesis. The strongly basic NaH deprotonates the alcohol, creating a nucleophilic alkoxide. This alkoxide goes performs a SN2 reaction with the alkyl halide: [C] is the answer.

What is the final major organic product (2) of the following reaction series?

This is a carbon-carbon bond formation reaction. The terminal hydrogen of the alkyne is slightly acidic, so adding a strong base, NaNH2, deprotonates the alkyne. The alkyne anion act as a nucleophile and attacks the electrophilic carbonyl carbon. The final step quenches the alkoxide, resulting in the a tertiary alcohol: [A] is the answer.

Which solvent would best dissolve the following compound?

This is a chlorine salt of a protonated aniline. The ionic nature increases solubility in polar solvents. The best polar solvent option is water. [A] is the answer. Nonpolar solvents (Et2O, benzene, and toluene) will not be effective at dissolving charge/ionic species. Eliminate [B], [C], and [D]. Tetrahydrofuran, or THF, is an aprotic polar solvent. However, it is not nearly as polar as water. Eliminate [E].

Which of the following alkenes would produce a secondary alkyl halide upon reacting with HBr and H2O2?

This is a radical addition reaction using peroxide. The reaction is an Anti-Markovnikov addition because the reaction uses HBr and H2O2. Br adds to the least substituted carbon:

Which of the following reagents will undergo the reaction shown below the most rapidly?

This is a substitution reaction in which the stereocenter is inverted. This occurs in a SN2 reaction, which occurs most rapidly with strong nucleophiles. Charge partially determines nucleophile strength. Stronger nucleophiles have a negative charge, like CN-. [C] is the answer.

What is the major product in the following reaction?

This is a typical Diels-Alder reaction. The diene is cyclopentadiene. The dienophile is the alkene with two cis ester groups. In Diels-Alder reactions, groups that begin cis will remain cis in the product. Groups with pi-bonds will be in the endo position on the ring, pointing down: [A] is the answer.

What is the major product in the following reaction?

This is an electrophilic aromatic substitution reaction. The starting material has an ester, which is an electron withdrawing group (EWG). EWGs are deactivating and are meta-directors. The chlorine will be placed in the meta-position on the ring: [C] is the answer.

Predict the major product of the following reaction:

This is an organolithium reaction used to form carbon-carbon bonds. The nucleophilic carbon in the ethyl lithium reagent attacks the electrophilic carbonyl carbon. This adds an ethyl group at C1 and forms an alkoxide. This alkoxide is protonated by the H3O+ to produce an alcohol: [C] is the answer.

What functional group will be present in the major organic product of the following reaction?

This is the oxidation reaction of a primary alcohol, using the Jones reagent. In strongly acidic conditions, the alcohol is oxidized fully into a carboxylic acid: [E] is the answer.

Which set of reagents could accomplish the following chemical transformation?

This is the preparation of a Grignard reagent. First, the alcohol is converted into an alkyl halide, using PBr3. The final step is a metal insertion reaction with magnesium: [D] is the answer. Mg and CH3MgBr react violently with alcohols, resulting in a highly exothermic acid-base reaction. Eliminate [A] and [B]. Br2 and MgBr2 cannot convert an alcohol into an alkyl bromide. Eliminate [C] and [E].

The following molecule contains which functional groups?

This molecule contains a cyclic ester and an alcohol: [B] is the answer.

How many signals will be present on the 1HNMR spectrum in the following molecule?

This molecule has a vertical plane of symmetry. There are four chemically unique hydrogen on this molecule: [D] is the answer.

What is a major product of the following reaction sequence?

This reaction begins with an epoxidation of the alkene using mCPBA. This epoxide is opened with NaOH, forming a trans-diol. Finally, the Jones reagent oxidizes the secondary alcohols into ketones: [E] is the answer.

Which reagent will best accomplish the following chemical transformation?

This reaction converts a carboxylic acid into an acyl halide. The most effective way to convert a carboxylic acid into an acyl chloride is to use thionyl chloride (SOCl2). [C] is the answer. Answer choices [A], [B], [D], and [E] do not react with carboxylic acids. No product would be formed.

Which set of reagents will accomplish the following chemical transformation?

This reaction is the reduction of an alkyne to an alkene. Na in NH3 selects for the trans alkene. [C] is the answer. DAT Pro-tip: The Lindlar catalyst, answer choice [B], selects for the the cis alkene. In the answer, the hydrogens are on opposite sides (E) of the alkene bond, because of the orientation.

Rank the following compounds from LEAST to MOST acidic.

To assess the acidity of a molecule look at the stability of its conjugate base. The more stable the conjugate base, the more acidic the acid. Draw each compound's conjugate base: All four compounds have the same amount of resonance. Electronegative fluorine helps stabilize the negative charge through induction. The most stable conjugate base and the strongest acid is compound I, with two fluorine atoms. Eliminate [C] and [E]. Compound II does not have any fluorine atoms, so its conjugate base is the least stable and is the weakest acid. Eliminate [B]. Compounds III and IV each have one fluorine atom. The closer fluorine is to the negative charge, the greater its stabilizing effect. Compound III's fluorine atom is closer to the electronegative oxygen anion, stabilizing its conjugate base and increasing acidity. From least to most acidic: II < IV < III < I. [A] is the answer.

Select the molecule with the lowest overall pKa value.

To rank acids, we must compare conjugate base stability. The more stable the conjugate base, the more acidic the acid. To determine which conjugate base is most stable, we will use CARDIO: Charge is the same. All of the conjugate bases have a -1 charge. Atom is the same. Every conjugate base places the charge on the nitrogen atom. Resonance is a major point of difference. Choice [D] doesn't have resonance throughout the entire ring, it is non-aromatic. Two of its carbons are sp3-hybridized, thus the ring is not planar nor conjugated: The rest of the molecules have resonance and are aromatic. Dipole Induction - Because N has some degree of electronegativity, the aromatic ring with the most nitrogens will be the most acidic. More nitrogens = more dipole induction = more stable conjugate base = more acidic molecule = lower pKa . Thus, Choice [E] will have the lowest pKa overall.

What is the major product of the following reaction?

Under basic conditions, the ethyl lithium reagent attacks the least sterically-hindered side of the epoxide. This occurs in one step, SN2-like. The ethyl lithium attacks the carbon on the right side of the epoxide, and the stereochemistry on the left carbon is retained: [A] is the answer. DAT Pro-tip: The reaction is under basic conditions when the epoxide attack occurs. The H3O+ is added in a separate second step to protonate the O-. In acidic conditions, the nucleophile attacks the more substituted carbon.

All of the following molecules are capable of hydrogen bonding with water EXCEPT one. Which one is the EXCEPTION?

Water can hydrogen bond with compounds containing fluorine, oxygen or nitrogen, with at least one lone pair on the atom. The only answer choice without fluorine, nitrogen or oxygen is [D], which is a haloalkane and not capable of hydroge bonding with water. [D] is the answer.

Which of the following compounds best corresponds to the 1H NMR spectrum shown below?

[A] is the answer. B. An -OH signal is a broad singlet, which is not present. Eliminate [B]. C. Aromatic hydrogen atoms have signals in the 7-8 ppm range. Eliminate [C]. D. The given spectrum must have an electronegative atom that deshields enough to have a signal above 3.5 ppm. Eliminate [D]. E. Alkene hydrogen atoms have a signal near 5-6 ppm. Eliminate [E].

Which of the following reactants could produce the product below upon reaction with mCPBA?

mCPBA reacts with alkenes to form epoxides:[B] is the answer. Eliminate answer choices without alkenes. Eliminate [A], [C], and [D]. To place the epoxide on the ring, the alkene must be in the ring. Eliminate [E].

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