Question 1

Hydroxyl substituents in alcohols are poor leaving groups, and are not readily eliminated during substitution reactions.  Which of the following modifications would make the hydroxyl substituent of an alcohol a better leaving group?

Accepted Answer

11ecba2d_128e_dd44_a3bf_6d22b9bd1ea4_MD0008_00 Substitution reactions require a nucleophile and an electrophile, as well as a good leaving group. A bond is formed between the nucleophile and electrophile, and the bond between the electrophile and leaving group is broken. The leaving group gains a pair of electrons when its bond with the electrophile is broken, and the leaving group is more likely to depart if the electrons gained can form a stable arrangement. Weak bases make better leaving groups than strong bases because they are often neutral molecules or have delocalized charge. Neutral species, such as water, formed after leaving make good leaving groups because uncharged species are more stable than charged species. Alcohols must be modified to form a good leaving group for substitution reactions because hydroxide (⁻OH) is a strong base. The addition of acid to an alcohol causes the hydroxyl group to become protonated, forming bound water with a positive charge on oxygen. Water is a weak base, and becomes neutral once it leaves. Therefore, protonation of an alcohol forms a good leaving group. (Choices A and C) The addition of a strong base will cause deprotonation of the alcohol, creating a negative charge on the oxygen atom of the alcohol. Atoms with a negative charge tend to donate electrons and act as a nucleophile rather than accept electrons and act as a leaving group. (Choice D) Acetals are carbonyl protecting groups formed by the addition of an alcohol to a ketone or aldehyde in acidic conditions. Protecting groups render a functional group unreactive, and therefore would not make alcohols better leaving groups. Educational objective: Hydroxyl groups in alcohols are poor leaving groups because hydroxide is a strong base. Hydroxide must be converted to a good leaving group to participate in substitution reactions. The addition of acid to an alcohol protonates the hydroxyl group, forming a bound, positively charged water molecule. Water is a weak base and forms a neutral molecule upon leaving, which makes it a good leaving group.

Question 2

The reaction below was performed under thermodynamic conditions to yield Compound 1 and Compound 2 as a mixture of structural isomers. If Compound 1 is more stable than Compound 2, what can be inferred about the relative amounts of Compound 1 to Compound 2 in the mixture?

Accepted Answer

11ecba2d_1289_5ee6_a3bf_bd259b3955e2_MD0008_00 Isomers are compounds with the same molecular formula that have a distinguishing structural difference that makes them different compounds with different properties. There are several classifications of isomers, including constitutional isomers, diastereomers, enantiomers, and cis/trans isomers. When a reaction is performed, a mixture of isomers can occur with one isomer predominating over the other. In such a case, the reaction conditions dictate which isomer is the major product (such as in the formation of kinetic and thermodynamic enolates). The kinetic product in a mixture requires a lower activation energy to form; therefore, it forms faster than the thermodynamic product but is also higher in energy (less stable). When a reaction is performed at low temperatures (kinetic conditions), insufficient energy is present to supply the larger activation energy required to form the thermodynamic product in high yield. Although the kinetic product still forms faster at higher temperatures (thermodynamic conditions), the increased thermal energy enables the kinetic product reaction to occur in reverse (back to the intermediate) and supplies the additional activation energy needed to form the lower energy (more stable) thermodynamic product, which then predominates. Compound 1 and Compound 2 are constitutional isomers formed as a mixture in the reaction. Because the reaction was performed under thermodynamic conditions, the more stable isomer will be the major product. Compound 1 is more stable because it contains a more substituted double bond than Compound 2. Therefore, a greater amount of Compound 1 will be in the mixture. (Choices B and C) Because Compound 2 is less stable than Compound 1 and the reaction was performed under thermodynamic conditions, there will be less of Compound 2 than Compound 1. (Choice D) Thermodynamic stability of isomers is related to the amount of each isomer in the mixture. Therefore, the relative amounts of each compound can be determined. Educational objective: Isomers are different compounds with the same molecular formula, and some reactions produce a mixture of isomers. The major product of a reaction is dependent on the reaction conditions: The more stable product will predominate under thermodynamic conditions, and the less stable product will predominate under kinetic conditions.

Question 3

Passage Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds. The organic compounds provided to the students are given in Table 1.Table 1 Organic Compounds Tested First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl). This reagent is used to test for secondary and tertiary alcohols. A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1). Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds. The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test. Compound A was found to be an example of a positive result, as shown in Reaction 2. Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine. This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols. A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3). 3Table 2 Selected Observations from the Classification Tests -Based on the information in the passage, which compound will react most rapidly with the Lucas reagent?

Accepted Answer

11ecba2d_128f_798a_a3bf_bd54d3361d62_MD0008_00 Substitution reactions can occur in one of two ways: an S_N1 or S_N2 mechanism. An S_N1 reaction occurs in two steps whereas an S_N2 reaction occurs in a single step. The first step of an S_N1 reaction involves formation of a carbocation intermediate when the bond between the leaving group and the adjoining carbon is broken. In the second step, the nucleophile adds to the carbocation, forming the product of the reaction. The first step is the slowest step (rate-determining step). Substrates that form a more stable carbocation react faster in an S_N1reaction than those that form a less stable carbocation. The passage states that the Lucas test occurs through an S_N1 mechanism. Because Compound C is a tertiary alcohol, it can form a more stable carbocation than Compounds A and B, which are a secondary and primary alcohol, respectively. Therefore, Compound C will react the most rapidly with the Lucas reagent. (Choice A) Compound A forms a secondary carbocation, which is less stable than the tertiary carbocation formed by a tertiary alcohol. Therefore, Compound A reacts more slowly than Compound C in the Lucas test. (Choice B) Compound B is a primary alcohol and would form a primary carbocation in the S_N1 reaction if it reacted with the Lucas reagent. Because primary carbocations are not stable, it is unfavorable for Compound B to undergo an S_N1 reaction. (Choice D) Compounds A and B react with the Lucas reagent at different rates due to carbocation stability. Compound B does not react with the Lucas reagent because it would form an unstable primary carbocation. Educational objective: An S_N1 reaction occurs in two steps. First, a carbocation intermediate is formed when the bond between the leaving group and adjoining carbon atom is broken; next, the nucleophile adds to the carbocation to form the reaction product. Substrates that form a more stable carbocation react faster than substrates that form a less stable carbocation.

Question 4

The amino acids Val, Leu, and Ile are all degraded outside the liver by an aminotransferase and an α-keto acid dehydrogenase complex.  These amino acids can all be classified as:

Accepted Answer

11ecba2d_128c_ba5e_a3bf_5189b902d4d0_MD0008_00 Amino acids have three components: an amino group, a carboxyl group, and a side chain, which are each bonded to the same carbon atom (known as the α-carbon). The various amino acid structures differ only by the composition of the substituent side chains, which contain different functional groups that give amino acids specific characteristics. This permits classification of amino acids based on the different side-chain characteristics. Major classifications include charged (acidic or basic), hydrophobic (nonpolar), and hydrophilic (polar and charged). Some amino acids fall into more than one of these categories. Val, Leu, and Ile are the three-letter abbreviations for valine, leucine, and isoleucine, respectively. The side chains of these amino acids are similar in functionality. Each is a hydrocarbon chain of varying length with branching at some point in the side chain. Because the side chains of these amino acids are composed of nonpolar hydrocarbon chains, they can be classified as hydrophobic. (Choices A and B) Acidic and basic amino acids contain negatively and positively charged side chains, respectively, at physiological pH, and have a nonzero net charge. Valine, leucine, and isoleucine contain uncharged side chains, and have a net charge of zero at physiological pH. (Choice D) The side chains of valine, leucine, and isoleucine do not contain any polar functional groups. The hydrocarbon side chains are nonpolar, making these amino acids hydrophobic rather than hydrophilic. Educational objective: Amino acids are made up of an amino group, a carboxyl group, and a side chain, all bonded to the same carbon atom. The side chains contain different functional groups that give amino acids specific characteristics, which allow the classification of amino acids based on their properties. Four main classifications of amino acids include acidic, basic, hydrophilic, and hydrophobic.

Question 5

What functional group forms in an acid-catalyzed reaction between a secondary amine and a ketone?

Accepted Answer

11ecba2d_128d_2f8f_a3bf_e114c2e9c5b5_MD0008_00 Ketones and aldehydes react with primary (1°) and secondary (2°) amines to form a carbon-nitrogen bond. The reaction begins with an acid-catalyzed addition of the amine to the carbonyl, which involves protonation of the carbonyl, nucleophilic attack of the carbonyl by the amine, and deprotonation of the amine. This forms an intermediate known as an α-aminoalcohol. The second part of the reaction is an acid-catalyzed dehydration involving protonation of the -OH group, loss of H₂O, and deprotonation. The product formed is dependent on the type of amine used (1° or 2°). If a 1° amine is used, an imine (nitrogen analogue to ketones and aldehydes) is formed because the nitrogen has a proton available for removal in the last step, permitting C=N bond formation. If a 2° amine is used, an enamine (nitrogen analogue to an enol) is formed because the nitrogen does not have a proton available to be removed in the last step. Therefore, a proton is removed from the α-carbon, forming a C=C bond. (Choice B) A carbon-nitrogen bond is formed in the reaction, but the functional group is not a nitrile because the bond between carbon and nitrogen is a single bond rather than a triple bond. (Choice C) If a ketone reacts with a 1° amine rather than a 2° amine, an imine would form. (Choice D) Although a carbon-carbon double bond is formed in the reaction, an alkene indicates that the functional group contains only a double bond. Because a nitrogen atom is bonded to the sp² carbon, the functional group would be classified as an enamine rather than an alkene. Educational objective: Ketones and aldehydes undergo acid-catalyzed addition of 1° and 2° amines to form imines or enamines, respectively. Like ketones and aldehydes, imines contain a carbon-heteroatom double bond (nitrogen in imines, oxygen in ketones and aldehydes). Enamines are the nitrogen analogue to enols, containing a carbon-carbon double bond.

Question 6

Passage Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds. The organic compounds provided to the students are given in Table 1.Table 1 Organic Compounds Tested First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl). This reagent is used to test for secondary and tertiary alcohols. A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1). Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds. The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test. Compound A was found to be an example of a positive result, as shown in Reaction 2. Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine. This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols. A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3). 3Table 2 Selected Observations from the Classification Tests -An unknown compound containing both a secondary alcohol and an aldehyde in its structure was given to a student. Suppose the Lucas test confirmed there was an alcohol present. If a protecting group were added to the alcohol, would a positive result from the chromic acid test be indicative of only the presence of the aldehyde?

Accepted Answer

11ecba2d_1291_c382_a3bf_c3470f4acf79_MD0008_00 Organic compounds often contain multiple functional groups that can both react under the same conditions. In cases where a reaction is desired in one functional group but not another functional group, protecting groups are utilized. Protecting groups are organic molecules added to a functional group to prevent it from reacting and forming an undesired product. Protecting groups tend to be stable under oxidizing and reducing conditions. After the desired reaction is complete, the protecting group can be removed to provide the original functional group. The chromic acid test is an oxidation reaction, and the alcohol and aldehyde would both react with chromic acid to give a positive result (blue-green suspension). Primary alcohols and aldehydes will be oxidized to carboxylic acids, and secondary alcohols will be oxidized to ketones. To confirm there is an aldehyde present in the unknown compound, the alcohol can be protected. Because protecting groups are unreactive under oxidizing conditions, only the aldehyde will be oxidized by chromic acid to form a carboxylic acid. (Choices A and B) An aldehyde is reactive under oxidizing conditions, and because the chromic acid test is an oxidation reaction, an aldehyde would give a positive result. Because the protected alcohol cannot react and there are no other reactive functional groups present, a positive result is solely indicative of an aldehyde in this compound. (Choice D) A protected alcohol is unreactive under oxidizing conditions, and therefore would not react in the chromic acid test. Educational objective: Chromic acid is an oxidizing agent that can oxidize primary alcohols and aldehydes to carboxylic acids and secondary alcohols to ketones. Protecting groups are organic molecules added to a functional group to prevent them from reacting under a given set of reaction conditions. Protecting groups tend to be stable against oxidation and reduction.

Question 7

A wax is made up of a long-chain fatty acid and a long-chain alcohol joined by:

Accepted Answer

11ecba2d_1293_bf55_a3bf_03c0d106d97f_MD0008_00 Lipids are hydrophobic molecules that are classified based on characteristics such as length, linkage type, and solubility. These molecules are broadly classified as hydrolyzable (lipid contains hydrolyzable ester or amide linkages) or nonhydrolyzable (lipid does not contain hydrolyzable ester or amide linkages). The hydrolyzable lipids, such as phospholipids, glycolipids, and waxes, are further classified based on the structure of their backbone (glycerophospholipids, glycosphingolipids, etc.). Waxes are an example of a hydrolyzable lipid that serves as a form of protection in plants and animals. Waxes are made up of a long-chain fatty acid and a long-chain alcohol. The wax forms when the acid and the alcohol combine through dehydration to form an ester bond. (Choice A) Amide linkages are present in sphingolipids but not in waxes. For an amide to form, the carboxyl group of the fatty acid would need to combine with an amine rather than an alcohol. (Choice C) Like esters, anhydrides are carboxylic acid derivatives. However, for an anhydride to form, the carboxyl group from the fatty acid would need to combine with an acid chloride rather than an alcohol. (Choice D) A glycosidic bond joins a sugar to another molecule. Because waxes are composed of a fatty acid and an alcohol and do not contain sugars, a glycosidic bond cannot be present. Educational objective: Lipids are hydrophobic molecules broadly classified as hydrolyzable or nonhydrolyzable, and more specifically classified based on their backbone structure. Waxes are hydrolyzable lipids that contain an ester bond formed by the linkage between a long-chain fatty acid and a long-chain alcohol.

Question 8

Passage Fatty acids (FAs) are lipids that play important roles in cellular function, including energy storage and membrane structure. Most polyunsaturated fatty acids (PUFAs) are made in the body; however, omega-3 alpha-linolenic acid and omega-6 linoleic acid must be obtained from the diet. These two PUFAs are precursors to other omega-3 and omega-6 long-chain FAs that are necessary for physiological processes such as neurological development, normal growth, and immune response. Insufficient omega-3 and omega-6 long-chain FAs can lead to an increase in omega-9 fatty acids such as mead acid, and to essential fatty acid deficiency (EFAD), which can cause infertility, decreased growth, and increased susceptibility to infection.Total FA levels in plasma and serum can be used to diagnose EFAD. Plasma and serum samples from several individuals were hydrolyzed to release FAs from triacylglycerols and phospholipids. Once isolated, the free FAs were converted to pentafluorobenzyl (PFB) derivatives and separated by gas chromatography (GC). Total levels of each type of fatty acid were quantified by negative chemical ionization-mass spectrometry (NCI-MS) immediately after GC. NCI is an ionization technique that creates negatively charged ions with little to no fragmentation. The NCI-MS revealed that two of the peaks from GC had the same m/z ratio (Figure 1). They were both confirmed to be forms of an unsaturated, unbranched 22-carbon FA called omega-9 docosenoic acid. Figure 1 Selected ion chromatogram of docosenoic acidResearchers have proposed that an elevated ratio of the triene omega-9 mead acid to the tetraene omega-6 arachidonic acid, known as the triene/tetraene (TT) ratio, can signify EFAD. Using GC and NCI-MS, they measured the TT ratio, concentration of mead acid, and individual concentrations of omega-3 and omega-6 FAs in plasma and serum samples from three patients and compared their results to normal ranges. An increased TT ratio and mead acid concentration and decreased omega-3 and omega-6 FA concentrations were observed in the patient samples compared to normal ranges. Adapted from E. Kish-Trier et al., "Quantitation of total fatty acids in plasma and serum by GC-NCI-MS." Clinical Mass Spectrometry. ©2016 Elsevier. -Based on the passage, which compound was most likely analyzed by GC and NCI-MS in the experiment?

Accepted Answer

11ecba2d_1294_82ab_a3bf_957a9502fb1f_MD0008_00 Fatty acids (FAs) consist of a nonpolar hydrocarbon chain with a polar carboxyl head group, and can be classified as either saturated or unsaturated. Saturated FAs do not contain any double bonds in the hydrocarbon chain whereas the hydrocarbon chain of unsaturated FAs contains at least one double bond, usually in the cis conformation . Free FAs are usually not found in blood plasma but rather exist as a derivative, such as a triglyceride or phospholipid. Hydrolysis of triacylglycerides and phospholipids releases free FAs. In the experiment, plasma and serum samples were hydrolyzed to give free FAs, which were then converted to a pentafluorobenzyl (PFB) derivative. The PFB FA derivatives were analyzed by GC and NCI-MS to determine total FA levels. Therefore, the structure with a hydrocarbon tail and a carboxyl group attached to a PFB substituent was most likely analyzed in the experiment. (Choices A and D) These structures were derived from a triacylglyceride and the phospholipid phosphatidylcholine, respectively, both of which are present in plasma and serum samples. However, triacylglycerides and phospholipids were not analyzed directly; they were hydrolyzed to give free FAs, which were then converted to PFB derivatives and analyzed by GC and NCI-MS. (Choice C) This structure is derived from cholesterol. Although cholesterol is a lipid, it contains a steroid backbone and is made up of isoprene units rather than FAs. Educational objective: Fatty acids (FAs) are made up of a nonpolar hydrocarbon chain with a polar carboxyl head group, and can be classified as either saturated or unsaturated. Free FAs are usually not found in blood plasma but rather exist as derivatives, such as a triacylglyceride or phospholipid. Hydrolysis of triacylglycerides and phospholipids releases free FAs.

Question 9

Compound 1 shown below is converted to Compound 2 by a base-catalyzed aldol condensation. If the carbonyl labeled with an asterisk is the carbonyl in the final product, what is the structure of Compound 2?

Accepted Answer

11ecba2d_128b_5ac7_a3bf_cb80bc575253_MD0008_00 An aldol condensation requires two carbonyl groups (from ketones and/or aldehydes), and these may be from separate molecules or within the same molecule. The reaction begins with deprotonation of an α-carbon (carbon adjacent to a carbonyl) to form an enolate. Next, the nucleophilic enolate attacks another carbonyl, forming a new carbon-carbon bond (the aldol product), in which the carbonyl adjacent to the nucleophile remains intact. Finally, -OH elimination takes place to give an α,β-unsaturated enone. A reaction between two separate compounds results in the addition of one compound to the other, whereas an intramolecular reaction within the same molecule (as in a diketone) results in a cyclization. Compound 1 is a 1,4-diketone and forms a five-membered ring in an aldol condensation. Because the carbonyl in the α,β-unsaturated product is on carbon 5, the nucleophile must be either carbon 4 or 6 (α-carbons). It is energetically unfavorable for carbon 4 to be the nucleophile because a strained three-membered ring would form, whereas a favorable five-membered ring forms if carbon 6 is the nucleophile. Subsequent deprotonation of carbon 6 results in -OH elimination from carbon 2 and the formation of a double bond between carbons 2 and 6 to yield an α,β-unsaturated ketone. (Choices A and B) These structures form if carbon 1 acts as the nucleophile rather than carbon 6, leaving carbon 2, rather than carbon 5, with the carbonyl. (Choice D) The alkyl substituents are on the correct carbon atoms in this structure; however, the double bond in the ring is on the incorrect side of the carbonyl. Because the double bond is a result of -OH elimination, the double bond should be between the carbon with acidic protons (carbon 6) and the carbon with the -OH leaving group (carbon 2). Educational objective: An aldol condensation results in a new carbon-carbon bond and an α,β-unsaturated enone. The reaction requires two carbonyl groups (from ketones and/or aldehydes) and occurs by the nucleophilic attack of an enolate on a carbonyl followed by elimination. A diketone substrate will result in cyclization and formation of a five- or six-membered ring.

Question 10

Compound 1 and PBr₃ react to form Compound 2 as shown below. Which position in Compound 2 corresponds to the hydrogen(s) that appear as a doublet at approximately 4.0 ppm in the ¹H NMR spectrum?

Accepted Answer

11ecba2d_1288_e9b4_a3bf_415dcbd8fbf9_MD0008_00 Proton nuclear magnetic resonance spectroscopy (¹H NMR) detects hydrogen atoms (¹H isotope) in a molecule as an external magnetic field and radio frequency pulse are applied to the sample. The hydrogen atoms in different magnetic environments (resulting from molecular structure) are nonequivalent, and neighboring nonequivalent hydrogens within three bonds of each other cause spin-spin splitting of the peaks in the NMR spectrum. The splitting pattern can be determined by n+ 1 rule, where n is the number of nonequivalent hydrogen atoms on adjacent carbons. Position IV in Compound 2 contains two hydrogen atoms that are adjacent to a single H atom bonded to an sp² carbon. According to the n + 1 rule, n = 1 because there is only one coupled hydrogen to the hydrogen atoms at position IV. Therefore, because n + 1 = 2, the peak will be split in two, making the splitting pattern a doublet. (Choice A) Position I contains two hydrogen atoms that are split by three adjacent hydrogen atoms. Although a quartet might be expected, complex, secondary splitting causes overlapping peaks to appear as a multiplet at 2.1 ppm. (Choice B) Position II consists of a methyl group (three hydrogen atoms) that is adjacent to a 4° carbon (no bonds to hydrogen). Because there are no neighboring hydrogen atoms coupled to the methyl hydrogens, the peak is not split and appears as a singlet at 1.8 ppm. (Choice C) Position III contains an alkene hydrogen atom that is adjacent to an alkyl (-CH₂-) group. The two alkyl hydrogen atoms are coupled to the alkene hydrogen, causing the peak to be split into a triplet. Educational objective: ¹H NMR detects hydrogen atoms in a molecule as an external magnetic field and radio frequency pulse are applied to the sample. Hydrogen atoms in different magnetic environments within three bonds of each other cause spin-spin splitting of the peaks in the spectrum. The splitting pattern can be determined by the n + 1 rule (n = number of nonequivalent hydrogen atoms on adjacent carbons).

Question 11

The number of different stereoisomers that exist for the compound shown below is:

Accepted Answer

11ecba2d_128e_4103_a3bf_238a65aab51f_MD0008_00 Stereoisomers are compounds that contain the same molecular formula and same connectivity but differ in spatial arrangement. The number of stereoisomers for a molecule is related to the number of stereocenters in the compound. The maximum number of stereoisomers possible for a compound is determined by the expression 2ⁿ, where n is the number of stereocenters. The actual number of stereoisomers for a particular compound could be less than 2ⁿ if two of the stereoisomers have an internal mirror plane of symmetry (a meso compound), such that the symmetry causes the stereoisomers to be conformations of the same compound. This compound has two stereocenters (on carbons 1 and 3) because each carbon has four different substituents. Using the expression 2ⁿ, the maximum number of stereoisomers is 2² = 4. The (1S, 3S) and (1R, 3R) stereoisomers are nonsuperimposable mirror images, as are the (1S, 3R) and (1R, 3S) stereoisomers, indicating that each of these compound pairs are enantiomers. None of the four stereoisomers has an internal plane of symmetry. Therefore, the compound in this question has four different stereoisomers. (Choice A) The number of stereocenters in this compound is two, but this value does not provide the number of different stereoisomers. (Choice B) The maximum possible number of stereoisomers is four, but the actual number of stereoisomers could be three only if one of the stereocenter configuration combinations for this compound were a meso compound (with an internal plane of symmetry). In such a case, two of the possible four stereochemical configurations would be the same compound, which would decrease the number of different stereoisomers by one. (Choice D) There would be eight maximum possible stereoisomers if this compound had three stereocenters (2³ = 8). However, the central carbon is not chiral, so there are only two stereocenters. Educational objective: The maximum number of stereoisomers possible for a compound is determined by the expression 2ⁿ, where n is the number of stereocenters. However, the actual number of stereoisomers for a particular compound could be less than 2ⁿ if two of the possible stereochemical configurations have an internal mirror plane of symmetry (meso compound), making them conformations of the same compound.

Question 12

Compounds 1 and 2 can be described as:

Accepted Answer

11ecba2d_128d_a4c1_a3bf_3384cef2518f_MD0008_00 Isomers are compounds that have the same molecular formula but a different arrangement of atoms. Conformational isomers are structures that have the same connectivity and can be interconverted by the rotation of σ bonds. Because conformational isomers are identical except for bond rotations, they are the same compound. Six-membered rings can be depicted in two different ways: as a wedge-dash projection or a chair conformation. In wedge-dash projections, wedges indicate the substituent is up and above the ring, and dashes indicate the substituent is down and below the ring. Substituents can be either axial or equatorial in a chair conformation. However, axial and equatorial do not dictate whether the substituent is up and above the ring or down and below the ring. The wedge-dash projection of Compound 1 shows the -OH and the -CH₃ both on a dash (down). Likewise, the chair conformation of Compound 2 shows the -OH and -CH₃ as equatorial down. To represent this wedge-dash projection as a chair conformation, both substituents can be shown as axial down. A chair flip puts the substituents in an equatorial down position (still on the same side of the ring), yielding an identical structure to Compound 2. Because these compounds have the same connectivity and can be interconverted from one to the other through bond rotations (via chair flip), they are conformational isomers. (Choice A) Enantiomers are nonsuperimposable mirror images in which all stereocenters are inverted. The stereocenter configurations in Compounds 1 and 2 are identical, and therefore cannot be enantiomers. (Choice B) Diastereomers are stereoisomers in which at least one, but not all, stereocenters are inverted. Because none of the stereocenters in Compounds 1 and 2 are inverted, they cannot be diastereomers. (Choice D) Compounds 1 and 2 have the same molecular formula and the same connectivity, so they are not constitutional isomers. Educational objective: Conformational isomers are structures that have the same formula and connectivity, and can be interconverted by the rotation of σ bonds. Because conformational isomers are identical except for structural bond rotations, they are the same compound.

Question 13

Passage Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds. The organic compounds provided to the students are given in Table 1.Table 1 Organic Compounds Tested First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl). This reagent is used to test for secondary and tertiary alcohols. A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1). Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds. The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test. Compound A was found to be an example of a positive result, as shown in Reaction 2. Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine. This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols. A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3). 3Table 2 Selected Observations from the Classification Tests -Compound D gives a positive result for the chromic acid test. What is the structure of the product formed by the reaction of Compound D with the chromic acid reagent?

Accepted Answer

11ecba2d_1290_b210_a3bf_d1d7323c023e_MD0008_00 Oxidation of an organic molecule often involves a decrease in the number of C-H bonds and an increase in the number of C-O bonds. Oxidizing agents will convert primary alcohols to carboxylic acids (except PCC, which converts primary alcohols to aldehydes), secondary alcohols to ketones, and aldehydes to carboxylic acids. The passage demonstrates that when a secondary alcohol gives a positive result in the chromic acid test, it is converted to a ketone in the process. The increase in the number of C-O bonds as a result of reaction indicates that the chromic acid test is an oxidation. Compound D is a four-carbon aldehyde (butanal), and the question states that it gives a positive result for the chromic acid test (meaning it is oxidized). Aldehydes already have two C-O bonds (as a C=O group), but when aldehydes are oxidized, they form a compound with an additional C-O bond (a carboxylic acid). Therefore, a four-carbon carboxylic acid will be produced when Compound D reacts with the chromic acid reagent. (Choice B) Formation of 1-butanol would result from a reduction rather than an oxidation. (Choice C) This compound is a ketone, which would result from the oxidation of a secondary alcohol rather than an aldehyde. (Choice D) This compound is the aldehyde butanal (the same as Compound D). Because Compound D gives a positive result for the chromic acid test, it must be transformed by oxidation into a new compound rather than remain unchanged. Educational objective: Oxidation of organic molecules often increases the number of C-O bonds and decreases the number of C-H bonds in the molecule. Oxidizing agents, such as chromic acid, will convert primary alcohols to aldehydes, secondary alcohols to ketones, and aldehydes to carboxylic acids.

Question 14

Passage Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds. The organic compounds provided to the students are given in Table 1.Table 1 Organic Compounds Tested First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl). This reagent is used to test for secondary and tertiary alcohols. A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1). Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds. The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test. Compound A was found to be an example of a positive result, as shown in Reaction 2. Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine. This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols. A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3). 3Table 2 Selected Observations from the Classification Tests -Compound E does not give a positive result for the Lucas test. However, after reaction with LiAlH₄, a positive result for the Lucas test is observed. Which compound is the product of the reaction of Compound E with LiAlH₄?

Accepted Answer

11ecba2d_1291_2741_a3bf_3b4b0e059a4e_MD0008_00 Reduction of a carbonyl (C=O) decreases its oxidation state, decreases the number of C-O bonds, and increases the number of C-H bonds, yielding an alcohol. The metal hydride LiAlH₄ is a reducing agent and contains four hydride ions (H:⁻). The hydride ion acts as a nucleophile and attacks the electrophilic carbonyl carbon, forming a new C-H bond and decreasing the number of C-O bonds (a reduction). In this manner, aldehydes and ketones are reduced to primary (1°) and secondary (2°) alcohols, respectively. Accordingly, if Compound E (a ketone) is reacted with LiAlH₄, Compound E will be reduced to a secondary alcohol, which gives a positive result for the Lucas test. Compounds are named based on the number of carbons in the longest continuous carbon chain (prefix) and the highest-priority functional group (suffix). The carbons are numbered to give the highest-priority functional group the smallest number, which is added to the beginning of the parent chain name. Applying the nomenclature rules, Compound E is a four-carbon (prefix but-) ketone (suffix -one) with the name 2-butanone. The name of the 2° alcohol formed from the reduction of 2-butanone is obtained by changing the ketone suffix -one to the alcohol suffix -ol to give butanol. The hydroxyl group for this compound is on carbon 2, so the name of the 2° alcohol product is 2-butanol. (Choices A and B) Reduction of a ketone results in a 2° alcohol rather than an aldehyde or a 1° alcohol. (Choice D) Compound E is 2-butanone. Because it is what reacts with LiAlH₄, it cannot be the product of the reaction. Educational objective: Reduction of an aldehyde or ketone with a reducing agent, such as LiAlH₄, reduces the number of C-O bonds and increases the number of C-H bonds to form a 1° and 2° alcohol, respectively. Ketones are named using the suffix -one; alcohols are named using the suffix -ol.

Question 15

Passage Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds. The organic compounds provided to the students are given in Table 1.Table 1 Organic Compounds Tested First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl). This reagent is used to test for secondary and tertiary alcohols. A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1). Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds. The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test. Compound A was found to be an example of a positive result, as shown in Reaction 2. Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine. This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols. A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3). 3Table 2 Selected Observations from the Classification Tests -Based on the results presented in Table 2, which compounds would give a positive result for the iodoform test?CH₃CH₂CHOCH₃(CH₂)₄OHCH₃(CH₂)₂C(O)CH₃CH₃(CH₂)₂CH(OH)CH₃

Accepted Answer

11ecba2d_1292_fc03_a3bf_21dec0749f02_MD0008_00 A general expression of the iodoform test is given by Reaction 3: 11ecba2d_1292_fc04_a3bf_c182af8c9d4e_MD0008_00 Table 2 shows that Compounds A and E give a positive result (yellow precipitate forms) whereas Compound B gives a negative result (no precipitate) for the iodoform test. Compound A is a secondary (2°) alcohol, and Compound E is a methyl ketone. These functional groups have an internal C-O bond adjacent to a -CH₃ group, allowing for the reaction with I₂ and NaOH to occur and form the yellow precipitate. However, Compound B is a primary (1°) alcohol, which has a terminal C-O bond and therefore cannot be adjacent to a -CH₃, which prevents its reaction with I₂ and NaOH, resulting in no precipitate formation. Among the new compounds presented in this question, CH₃(CH₂)₂C(O)CH₃ and CH₃(CH₂)₂CH(OH)CH₃ are a methyl ketone and a 2° alcohol, respectively, and each has an internal C-O bond adjacent to a -CH₃. Based on the results from the model compounds shown in Table 2, a 2° alcohol (Compound A) and a methyl ketone (Compound E) should give a positive result in the iodoform test. Because CH₃(CH₂)₂C(O)CH₃ and CH₃(CH₂)₂CH(OH)CH₃ contain the same functional groups as the model Compounds A and E, CH₃(CH₂)₂C(O)CH₃ and CH₃(CH₂)₂CH(OH)CH₃ (Numbers III and IV) will react in a similar fashion with I₂ and NaOH to give a positive result for the iodoform test. (Numbers I and II) The oxygen atom in aldehydes and 1° alcohols is bonded to a terminal carbon, so these compounds are adjacent to a -CH₂- rather than a -CH₃. Because the iodoform test requires a -CH₃ adjacent to the C-O bond to form CH₃I, aldehydes and 1° alcohols do not give a positive result for the iodoform test. Educational objective: Analysis of the results from an organic reaction with respect to structural features of the reacting compounds is important for drawing conclusions about what types of functional groups will react with the particular reagents. Such an analysis helps predict what compounds will or will not react with the same reagent under similar conditions.

Question 16

Compound 1 is shown below. All of the following are equivalent to Compound 1 EXCEPT:

Accepted Answer

11ecba2d_128a_223c_a3bf_0dcf0bb887fb_MD0008_00 Compounds can be illustrated in different ways and orientations. Molecules must have all substituents in the same relative position and stereochemical orientation to be considered the same compound. The cyclohexane ring, for example, can be depicted using the wedge-dash projection or a chair conformation. In the wedge-dash projection, the wedges indicate that the substituent is above or in front of the plane of the ring, and the dashes indicate that the substituent is below or behind the plane. In a chair conformation, the wedges and dashes can be shown as axial or equatorial positions. The axial positions are perpendicular to the plane of the ring and alternate pointing above or below the ring plane. The equatorial positions are oriented radially with the ring plane and alternate positions slightly above or below the plane. Compound 1 is a six-membered ring shown in the wedge-dash projection with three substituents: −OH, −Br, and −CH₃. The −OH group is bonded to carbon shown by a wedge, and −Br and −CH₃ are bonded to carbon atoms shown by dashed lines. If the carbon atoms are numbered 1-6, beginning at the carbon bonded to −OH and continuing clockwise, −Br is on carbon 3 and −CH₃ is on carbon 5. The answer choices are chair conformations and must maintain the positions and stereochemical orientations of Compound 1 to be considered the same molecule. The chair conformations that depict −OH above the plane of the ring ("up") and −Br and −CH₃ below the plane ("down") and on carbons 3 and 5 clockwise from −OH, respectively, are the same molecule as Compound 1 (Choices A, B, and C). The −OH substituent is equatorial "down" and the −Br and −CH₃ substituents are axial "up" in this chair conformation, causing the stereochemical orientation in this structure to be the complete opposite of Compound 1. Therefore, these structures are different compounds and considered enantiomers. Educational objective: Compounds can be illustrated in different orientations and forms, including wedge-dash projections and chair conformations. For two depictions of a molecule to be considered the same compound, the atoms must have the same relative position and stereochemical orientation.

Question 17

If Staphylococcus aureus V8 protease is specific for acidic amino acids, which of the following residues would be cleaved by this protease?

Accepted Answer

11ecba2d_128c_6c3d_a3bf_85fc931eb0a0_MD0008_00 Amino acids are the fundamental components of proteins and contain three main constituents: An amino group, a carboxyl group, and a side chain, which are all bonded to one carbon atom (the α-carbon). Amino acids differ only in the structure of the side chain, which contains different functional groups with varying characteristics. Therefore, amino acids can be grouped together and classified based on their side chain functional groups. Four common amino acid classifications include hydrophilic, hydrophobic, acidic, and basic. Acidic amino acids have a side chain with a proton-donating (acidic) functional group that is negatively charged at physiological pH. The amino group is protonated (positively charged) and the carboxyl group is deprotonated (negatively charged) at physiological pH, giving amino acids with uncharged side chains a zero net charge. Therefore, acidic amino acids have a −1 net charge at physiological pH because they have charged side chains. Staphylococcus aureus V8 protease specifically cleaves peptide bonds at the carboxyl terminal of acidic amino acids. The distinguishing feature of acidic amino acids is the carboxyl group at the end of the side chain. The only answer choice with a carboxyl group as part of the side chain is aspartic acid. (Choices A, C, and D) These amino acids are methionine, arginine, and threonine, which can be classified as hydrophobic, basic, and hydrophilic, respectively. Educational objective: Amino acids are the fundamental components of proteins and are made of an amino group, a carboxyl group, and a side chain. The structure of amino acids differs only in the side chains, which are made of different functional groups with varying characteristics. Amino acids can therefore be grouped based on their side-chain classification, including hydrophilic, hydrophobic, acidic, or basic.

Question 18

Passage Students used three classification tests to study the reactivity of alcohol, aldehyde, and ketone functional groups in organic compounds. The organic compounds provided to the students are given in Table 1.Table 1 Organic Compounds Tested First, the students performed the Lucas test by mixing a few drops of each compound with 2 mL of the Lucas reagent (ZnClS1U1B12S1U1B0 and HCl). This reagent is used to test for secondary and tertiary alcohols. A positive test result is indicated by a cloudy reaction mixture upon addition of the Lucas reagent due to the formation of an insoluble compound by an SS1U1B1NS1U1B01 reaction with a secondary or tertiary alcohol (Reaction 1). Reaction 1Next, a test for alcohols and aldehydes was performed by the addition of a drop of chromic acid reagent (CrOS1U1B13S1U1B0 dissolved in aqueous HS1U1B12S1U1B0SOS1U1B14S1U1B0) into samples of the organic compounds. The disappearance of the orange color from chromic acid and formation of a blue-green suspension is indicative of a positive result for this test. Compound A was found to be an example of a positive result, as shown in Reaction 2. Reaction 2Finally, samples of the compounds were also dissolved in water, followed by the addition of NaOH and iodine. This test is known as the iodoform test, which is used to identify methyl ketones and secondary alcohols. A positive result is indicated by the formation of iodoform (CHIS1U1B13S1U1B0), a yellow precipitate (Reaction 3). 3Table 2 Selected Observations from the Classification Tests -If the chromic acid reagent is added to Compound C, no reaction occurs because:

Accepted Answer

11ecba2d_1290_15cb_a3bf_b1003135ce32_MD0008_00 Oxidation of an organic molecule requires a decrease in the number of C-H bonds and an increase in the number of C-O bonds. Primary and secondary alcohols can be oxidized to form aldehydes and ketones, respectively, using various oxidizing agents such as chromic acid. However, the tertiary carbon of a tertiary alcohol does not have any C-H bonds to lose; therefore, its number of bonds to oxygen cannot be increased. If the number of bonds to oxygen were increased without losing a bond to hydrogen, the carbon atom would have five bonds exceeding the maximum number possible (four bonds). The passage indicates that a secondary alcohol gives a positive result for the chromic acid test and that it is converted into a ketone in the process. As a result, an increase in the number of C-O bonds is achieved in the reaction indicating that the chromic acid test is an oxidation. Because (CH₃)₃COH (Compound C) is a tertiary alcohol, it cannot be oxidized. (Choice A) A tertiary alcohol can be reduced to a hydrocarbon; however, the chromic acid test is an oxidation rather than a reduction. (Choices C and D) An aldehyde can be both oxidized and reduced, but Compound C is an alcohol rather than an aldehyde. Educational objective: Oxidation of an organic molecule requires a decrease in the number of C-H bonds and an increase in the number of C-O bonds. Primary and secondary alcohols can be oxidized to aldehydes and ketones, respectively, by various oxidizing agents. Tertiary alcohols cannot be oxidized because the tertiary carbon does not have any C-H bonds to lose.

Question 19

The enzyme phenylalanine hydroxylase catalyzes the conversion of phenylalanine to tyrosine.  The disease phenylketonuria (PKU) is caused by an inactivating mutation in the gene encoding phenylalanine hydroxylase.  Which structure corresponds to the amino acid that will NOT form as a result of PKU?

Accepted Answer

11ecba2d_128a_be81_a3bf_2fb37d49303e_MD0008_00 Amino acids are important biological molecules that make up proteins and are a source of metabolic energy through oxidative degradation. Amino acids are made up of an amino group, a carboxyl group, and a side chain all bonded to the same carbon atom (known as the α-carbon). Because amino acid structures differ only by the attached side chain, amino acids can be grouped together by their side-chain classification, which relates to the side-chain functional group. Accordingly, amino acid classifications include aromatic, hydrophilic or hydrophobic, acidic or basic, and charged (positive or negative) or uncharged. Phenylketonuria is characterized by a lack of activity of phenylalanine hydroxylase, which catalyzes the hydroxylation of phenylalanine to tyrosine. Because the enzyme is not functioning properly, tyrosine, which is characterized as an aromatic amino acid with a phenol functional group, is not formed. (Choice A) This aromatic amino acid is phenylalanine; it builds up in the blood because it cannot be hydroxylated to form tyrosine as a result of phenylketonuria. (Choices C and D) Tryptophan and histidine are also aromatic amino acids, but they are not involved in the hydroxylation catalyzed by phenylalanine hydroxylase; therefore, formation of these amino acids is not affected by phenylketonuria. Educational objective: Amino acids are important biological molecules that make up proteins. Amino acids contain an amino group, a carboxyl group, and a side chain all bonded to an α-carbon. The amino acids can be grouped together based on the classification of the side chains and functional group. Classifications include aromatic, hydrophilic or hydrophobic, acidic or basic, and charged (positive or negative) or uncharged.

Question 20

Phospholipids contain all of the following structural features EXCEPT:

Accepted Answer

11ecba2d_128b_cff8_a3bf_9feb168d3b0a_MD0008_00 Phospholipids are a main component of the cell membrane and are responsible for the cell membrane's structure. They are amphiphilic molecules made of a hydrophobic (nonpolar) tail and a hydrophilic (polar) head group. This characteristic causes the phospholipid molecules to arrange into a bilayer in aqueous environments. The hydrophobic tails form the inner portion of the bilayer to keep away from water, and the hydrophilic heads encompass the outer portion of the bilayer toward water because the polar head group molecules can form favorable interactions with water, such as hydrogen bonds. The general structure of a phospholipid includes a polar head group and nonpolar tail joined together by a molecule referred to as the backbone. The hydrophilic head contains a substituted phosphate, and the hydrophobic tail is made of two fatty acid chains (Choices A and D). The hydrophilic head and the hydrophobic tail are linked together by a glycerol backbone (Choice C). Two adjacent carbons of glycerol are bonded to the fatty acid chains through an ester bond, and the third glycerol carbon is bonded to phosphate through a phosphodiester bond. A variety of phospholipids can be made with different head groups on phosphate, a different backbone, and different fatty acids (saturated or unsaturated). The fatty acid double bond configuration causes the hydrophobic tail to take on different shapes and affects the properties of the phospholipid, such as membrane fluidity. Phenols are present in biological molecules, such as the amino acid tyrosine; however, they are not a component of the phospholipid structure. Educational objective: Phospholipids are the main component in the cell membrane and provide structure to the cell membrane. These molecules are made of a hydrophilic head containing a substituted phosphate group and a fatty acid hydrophobic tail. The head and tail are linked together by a glycerol backbone.

Exam 6: Organic Chemistry

Hydroxyl substituents in alcohols are poor leaving groups, and are not readily eliminated during substitution reactions. Which of the following modifications would make the hydroxyl substituent of an alcohol a better leaving group?