Test Bank for Organic Chemistry 9th Edition by McMurry All Chapters 1-30

Chapter 01 - Structure and Bonding 1. Give the ground-state electron configuration for carbon (atomic number 6).

ANSWER: 1s22s22px12py1 or 1s22s22p2 POINTS: 1 2. Give the ground-state electron configuration for fluorine (atomic number 9).

ANSWER: 1s22s22px2 2py2 2pz1 or 1s22s22p5 POINTS: 1 3. Give the ground-state electron configuration for magnesium (atomic number 12).

ANSWER: 1s22s22p63s2 POINTS: 1 4. How many electrons does silicon have in its valence shell?

ANSWER: four POINTS: 1 Exhibit 1-1 Write valid Lewis (electron-dot) structures for each formula below. Show all electrons as dots and show all non-bonding electrons.

5. C2Cl4 tetrachloroethylene ANSWER:

POINTS: 1 6. CO2 carbon dioxide ANSWER: POINTS: 1 7. CH4O methanol ANSWER:

POINTS: 1 Exhibit 1-2 Consider the structure of urea, shown below, to answer the following question(s).

8. Refer to Exhibit 1-2. Fill in any non-bonding valence electrons that are missing from the line-bond structure.

ANSWER:

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Chapter 01 - Structure and Bonding POINTS: 1 9. Refer to Exhibit 1-2. The carbon atom in urea is: a. sp3 hybridized b. sp2 hybridized c. sp hybridized d. not hybridized ANSWER: b POINTS: 1 10. Refer to Exhibit 1-2. The predicted NH2−C=O bond angle in urea is: a. 109.5° b. 120° c. 180° d. not predictable ANSWER: b POINTS: 1 Exhibit 1-3 Determine the hybridization for the indicated atoms in each structure below.

11. Refer to Exhibit 1-3. The hybridization of this oxygen atom (A) is ______.

ANSWER: sp2 POINTS: 1 12. Refer to Exhibit 1-3. The hybridization of this oxygen atom (B) is ______.

ANSWER: sp3 POINTS: 1 13. Refer to Exhibit 1-3. The hybridization of this carbon atom (C) is ______.

ANSWER: sp3 POINTS: 1 14. Refer to Exhibit 1-3. The hybridization of this carbon atom (D) is ______.

ANSWER: sp POINTS: 1 15. The molecular formula C2H4O can be converted into three-line bond (Kekulé) structures that are consistent with valence rules.

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Chapter 01 - Structure and Bonding

a.

b.

c.

d.

a. Which one of the Kekulé structures is not consistent with valence rules? b. Explain why the structure you chose in part a is not consistent with valence rules.

ANSWER: a. d

The carbon bonded to the oxygen atom in structure d is pentavalent; it has 10 valence electrons. Carbon can only have eight valence electrons. In addition, b. the other carbon has only six valence electrons when it would prefer to have eight.

POINTS: 1 16. The original question was combined with #15. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 17. Convert the following structure to a skeletal drawing and give its molecular formula.

ANSWER:

Molecular formula: C5H7Br POINTS: 1 18. Draw an orbital picture for acetylene, C2H2. Clearly label each bond type and indicate the type of orbitals involved in each bond.

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Chapter 01 - Structure and Bonding ANSWER:

POINTS: 1 Exhibit 1-4 Propose possible structures for a molecule that meets each of the following descriptions.

19. Refer to Exhibit 1-4. Contains two sp3 hybridized carbons and two sp hybridized carbons.

ANSWER:

POINTS: 1 20. Refer to Exhibit 1-4. Contains one sp3 hybridized carbon and two sp2 hybridized carbons.

ANSWER:

POINTS: 1 21. Convert the following molecular model into a condensed structure and a skeletal structure.

ANSWER: POINTS: 1 22. Convert the following molecular model into a condensed structure and a skeletal structure.

ANSWER:

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Chapter 01 - Structure and Bonding

POINTS: 1 23. Convert the following molecular model into a condensed structure and a skeletal structure.

ANSWER: or more specifically;

POINTS: 1 24. Indicate the hybridization on each of the carbon atoms indicated with a number in the molecular model shown.

ANSWER: Carbon 1: sp3 Carbon 2: sp2 POINTS: 1 25. Draw the orbital diagram showing the ground-state electron configuration of sulfur.

ANSWER:

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Chapter 01 - Structure and Bonding POINTS: 1 26. Fill in any nonbonding valence electrons that are missing from the following structural representation of the amino acid cysteine.

ANSWER:

POINTS: 1 27. There are two substances with the molecular formula C2H7N. Draw them and describe how they differ.

ANSWER: The two structures differ in the number of hydrogen atoms and carbon atoms bonded to the nitrogen atom. The first structure contains the carbon atoms bonded in a two carbon chain while in the second structure the two carbons atoms are isolated from each other by the nitrogen atom.

POINTS: 1 28. Overlap of the two atomic orbitals shown could result in a:

a. σ bond b. π bond c. σ or π depending on the direction of the overlap.

ANSWER: c POINTS: 1 29. Hybridization of the atomic orbitals shown would result in:

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Chapter 01 - Structure and Bonding

a. sp3 hybridization b. sp2 hybridization c. sp hybridization ANSWER: b POINTS: 1 30. Which of the following represents a hybrid orbital? a. b.

c.

d.

ANSWER: d POINTS: 1 31. What type of hybridization is exhibited by carbon in the following substance?:

a. sp3 hybridization b. sp2 hybridization c. sp hybridization ANSWER: b Cengage Learning Testing, Powered by Cognero

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Chapter 01 - Structure and Bonding POINTS: 1 32. What type of hybridization is exhibited by carbon in the following substance?:

a. sp3 hybridization b. sp2 hybridization c. sp hybridization ANSWER: c POINTS: 1 33. What type of hybridization is exhibited by the nitrogen atom in the following substance and how many lone pairs are present on the nitrogen?:

a. sp3 hybridization and 1 lone pair b. sp2 hybridization and l lone pair c. sp hybridization and 1 lone pair d. sp3 hybridization and 2 lone pairs e. sp2 hybridization and 2 lone pairs f. sp hybridization and 2 lone pairs ANSWER: a POINTS: 1 34. If all the missing bonds in the following structure are sigma bonds to hydrogen atoms, how many hydrogen atoms are missing from this structure? Atoms other than carbon are labeled.

a. 7 b. 10 c. 12 d. 14 e. None of these is the correct number.

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Chapter 01 - Structure and Bonding ANSWER: e POINTS: 1

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Chapter 02 - Polar Covalent Bonds; Acids and Bases Exhibit 2-1 Give the corresponding letter of the term that best matches the given definition. a. b. c. d. e.

Brønsted-Lowry Acid Brønsted-Lowry Base Lewis Acid Lewis Base Electronegativity

f. g. h. i. j.

Ionic Bond Covalent Bond Polar-Covalent Bond Hydrophobic Hydrophilic

1. _____ Any species that accepts electrons.

ANSWER: c POINTS: 1 2. _____ A bond between two atoms differing in electronegativity by 0.5 − 2.

ANSWER: h POINTS: 1 3. _____ A term used to describe a "water loving" species.

ANSWER: j POINTS: 1 4. _____ A compound that can donate a proton.

ANSWER: a POINTS: 1 5. _____ The ability of an atom to attract the shared electrons in a covalent bond.

ANSWER: e POINTS: 1 6. _____ A term used to describe a "water fearing" species.

ANSWER: i POINTS: 1 7. _____ Any species that donates electrons.

ANSWER: d POINTS: 1 8. _____ A bond between two atoms differing in electronegativity by < 0.5.

ANSWER: g POINTS: 1 9. _____ A compound that can accept a proton.

ANSWER: b POINTS: 1 10. _____ A bond between two atoms differing in electronegativity by > 2.

ANSWER: f POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 02 - Polar Covalent Bonds; Acids and Bases Exhibit 2-2 Calculate the formal charges on the indicated atoms in each compound below.

11. Refer to Exhibit 2-2. The formal charge on phosphorous (A) is ______.

ANSWER: +1 POINTS: 1 12. Refer to Exhibit 2-2. The formal charge on oxygen (B) is ______.

ANSWER: −1 POINTS: 1 13. Refer to Exhibit 2-2. The formal charge on carbon (C) is ______.

ANSWER: −1 POINTS: 1 14. Refer to Exhibit 2-2. The formal charge on oxygen (D) is ______.

ANSWER: +1 POINTS: 1 Exhibit 2-3 Phenylalanine is an amino acid that is essential to human nutrition. The representation below shows the structure of phenylalanine at physiological pH. Consider this structure to answer the following question(s).

15. Refer to Exhibit 2-3. Assign any formal charges to atoms in this representation of phenylalanine.

ANSWER:

POINTS: 1 16. Refer to Exhibit 2-3. The oxygen atom labeled A. has ______ non-bonding electrons.

ANSWER: four POINTS: 1 17. Refer to Exhibit 2-3. The oxygen atom labeled B. has _____ bonding electrons.

ANSWER: two Cengage Learning Testing, Powered by Cognero

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Chapter 02 - Polar Covalent Bonds; Acids and Bases POINTS: 1 Exhibit 2-4 Use the convention and the crossed arrow bonds in the following compounds.

to show the direction of the expected polarity of the indicated

18. Refer to Exhibit 2-4. The C−F bond in fluorobenzene, ANSWER:

POINTS: 1 19. Refer to Exhibit 2-4. The C−Si bond in tetramethylsilane, (CH3)4Si ANSWER:

POINTS: 1

20. Refer to Exhibit 2-4. The C−O bond in furan, ANSWER:

POINTS: 1 Exhibit 2-5 Label the acid, conjugate acid, base, and conjugate base in each reaction below.

21.

ANSWER:

POINTS: 1

22.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases ANSWER:

POINTS: 1 Exhibit 2-6 Refer to the following equation to answer the question(s) below. Place the letter corresponding to the correct answer in the blank.

23. Refer to Exhibit 2-6. The strongest Brønsted-Lowry acid in the equation is ______.

ANSWER: D POINTS: 1 24. Refer to Exhibit 2-6. The strongest Brønsted-Lowry base in the equation is ______.

ANSWER: C POINTS: 1 25. Refer to Exhibit 2-6. Will this reaction take place as written? Explain.

ANSWER: No, the reaction will not take place as written because the strongest acid reacts with the strongest base to give the weakest conjugate acid and the weakest conjugate base. D (pKa = 15.7) is a stronger acid than A (pKa = 18).

POINTS: 1 26. An acid with a low pKa: a. is a weak acid b. is a strong acid c. has a weak conjugate base d. both b and c ANSWER: d POINTS: 1 27. Circle all the Lewis bases in the group of compounds below.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases ANSWER:

POINTS: 1 28. Put a box around all the Lewis acids in the group of compounds below.

ANSWER:

POINTS: 1 29. Draw two resonance structures for the species below.

ANSWER:

POINTS: 1 30. Draw two resonance structures for the species below.

ANSWER:

POINTS: 1 31. Draw two resonance structures for the species below.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases

ANSWER:

POINTS: 1 Exhibit 2-7 Consider the acidity constants below to answer the following question(s).

ACID

STRUCTURE

phenol

pKa

10.00

ethanol

CH3CH2OH

16.00

water

HOH

15.74

32. Refer to Exhibit 2-7. Which acid will be almost completely deprotonated by NaOH? ANSWER: phenol POINTS: 1 33. Refer to Exhibit 2-7. Which acid has the strongest conjugate base?

ANSWER: Ethanol is the weakest acid (largest pKa) so its conjugate base, ethoxide, CH3CH2O−, will be the strongest base.

POINTS: 1 34. Refer to Exhibit 2-7. Explain why phenol has a much lower pKa than ethanol.

ANSWER: Phenol is more acidic (has a lower pKa) than ethanol because the phenoxide anion is resonance stabilized by the pi electrons in the ring.

Ethoxide anion has no resonance stabilization. The negative charge is borne fully by oxygen.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 02 - Polar Covalent Bonds; Acids and Bases Exhibit 2-8 Consider the reaction below to answer the following question(s).

35. Refer to Exhibit 2-8. Using the curved arrow formalism, show the flow of electrons for this reaction.

ANSWER:

POINTS: 1 36. Refer to Exhibit 2-8. Label the acid and the base in the reaction.

ANSWER:

POINTS: 1 Exhibit 2-9 Indole is pleasant smelling in highly dilute solutions and has been used in perfumery. Use the structure of indole, below, to answer the following question(s).

37. Refer to Exhibit 2-9. Indole can function as a Brønsted-Lowry acid in the presence of strong bases. Formulate a reaction, showing electron flow with arrows, that demonstrates this reactivity of indole.

ANSWER:

POINTS: 1 38. Refer to Exhibit 2-9. Indole can function as a Lewis base in the presence of strong acid. Formulate a reaction, showing electron flow with arrows, that demonstrates this reactivity of indole.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases ANSWER:

POINTS: 1 39. The condensed structure for dimethyl ether looks symmetrical. However, dimethyl ether has a dipole moment. Draw a structure that explains this and indicate the expected direction of the molecular dipole moment.

ANSWER:

POINTS: 1 Exhibit 2-10 Consider the acid-base reaction below to answer the following question(s).

40. Refer to Exhibit 2-10. Using the curved arrow formalism, show the flow of electrons for this reaction.

ANSWER:

POINTS: 1 41. Refer to Exhibit 2-10. Write the products of this Lewis acid - base reaction.

ANSWER:

POINTS: 1 42. The following is a representation of the pain reliever, acetaminophen, the active ingredient in Tylenol®. Indicate the positions of any multiple bonds. Atoms other than carbon and hydrogen are labeled.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases

ANSWER:

POINTS: 1 43. Use the curved arrow formalism to show the electron flow in the reaction of ammonia with water.

ANSWER:

POINTS: 1 44. In which series are the elements listed in order of increasing electronegativity? a. P < S < Cl b. Ge < C < P c. As < S < F d. P < Br < N ANSWER: c POINTS: 1 45. Based on electronegativity values, in which of the following is the bond, represented by —, the most polar? a. H3C—I b. H3C—Na c. H3C—Cl

d. H3C—OH

ANSWER: b POINTS: 1 46. Which of the following substances has a zero dipole moment? a. CO2 b. Cl2C=CCl2 c. HOCH2CH2OH d. HCl2CCHCl2 e. All have zero dipole moments.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases ANSWER: e POINTS: 1 47. The following shows a gray-scale image of an electrostatic potential map with the atoms labeled. Which of the numbered regions would appear reddest in a color image?

a. 1 b. 2 c. 3 d. 4 e. 1, 2, and 3 would be the same shade of red. f. 2, 3, and 4 would be the same shade of red.

ANSWER: d POINTS: 1 48. How many resonance forms can be drawn for the NO3– ion? a. 1 b. 2 c. 3 d. 4 e. None, the nitrate ion does not exhibit resonance.

ANSWER: c POINTS: 1 49. Which of the following would represent the strongest acid? a. b. c.

d.

ANSWER: d POINTS: 1 50. Which of the following substances would be expected to have the largest pKa? a. b. c.

d.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases

ANSWER: a POINTS: 1 51. Which of the following does not characterize the curved arrow formalism? a. The arrow shows the movement of electrons not atoms. b. The atom at the head of the arrow is the electron pair acceptor. c. The atom at the tail of the arrow is a Lewis acid. d. The species containing the atom at the head of the arrow will have the smaller pKa. e. All of these correctly describe the curved arrow formalism.

ANSWER: c POINTS: 1 52. Which of the following substances would exhibit hydrogen bonding a. b. c.

d.

e. a, b, and c f. All would exhibit hydrogen bonding.

ANSWER: e POINTS: 1 53. The structure for Vitamin K which is involved in blood clotting is shown below.

This vitamin would be: a. classified as hydrophilic. c. fat-soluble.

ANSWER: c POINTS: 1

b. water-soluble. d. both hydrophilic and hydrophobic.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases 54. The following is generic depiction of a reaction using the curve arrow formalism.

Which of these statements is not correct for this reaction? a. Electrons move from C to B. b. Electrons move from B to A. c. In the products, a bond forms between C and B. d. In the products, A would have a positive charge.

ANSWER: d POINTS: 1 55. Draw the resonance forms of 3,5-heptanedione anion.

ANSWER:

POINTS: 1 56. Draw the resonance forms of methyl acetate anion formed during the protonation of methyl acetate in the presence of sulfuric acid.

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Chapter 02 - Polar Covalent Bonds; Acids and Bases

ANSWER:

POINTS: 1 57. What are the products of the following reaction?

ANSWER:

POINTS: 1

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry Exhibit 3-1 MATCH a structure below to each of the following descriptions and place the letter corresponding to the structure in the blank.

1. _____ is an amino aldehyde.

ANSWER: B POINTS: 1 2. _____ is an aromatic ketone.

ANSWER: E POINTS: 1 3. _____ is a tertiary chloride.

ANSWER: C POINTS: 1 4. _____ is a cyclic alkane with two cis methyl groups.

ANSWER: C POINTS: 1 5. Circle and name each functional group in the following structure.

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry ANSWER:

POINTS: 1 Exhibit 3-2 Label the indicated atoms in the structure below as 1°, 2°, 3°, or 4°.

6. Refer to Exhibit 3-2. The atom at A is _____.

ANSWER: 4° POINTS: 1 7. Refer to Exhibit 3-2. The atom at B is _____.

ANSWER: 3° POINTS: 1 8. Refer to Exhibit 3-2. The atom at C is _____.

ANSWER: 3° POINTS: 1 9. Refer to Exhibit 3-2. The atom at D is _____.

ANSWER: 1° POINTS: 1 Exhibit 3-3 Label the following pairs of compounds as: a. b. c.

identical constitutional isomers neither

Place the letter of the correct answer in the blank.

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry

10. _____ ANSWER: b POINTS: 1

11. _____

ANSWER: c POINTS: 1

12. _____ ANSWER: a POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

13. Name: (CH3)2CHCH2CH2CH(CH2CH3)CH2C(CH3)3 ANSWER: 4-ethyl-2,2,7-trimethyloctane POINTS: 1 14. Name:

ANSWER: 7-bromo-3-ethyl-2,2,5,5-tetramethyloctane or 2-bromo-6-tert-butyl-4,4-dimethyloctane POINTS: 1 15. Name:

ANSWER: 5-isopropyl-3-methyloctane Cengage Learning Testing, Powered by Cognero

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry POINTS: 1 16. Name:

ANSWER: 5-ethyl-3,4-dimethyloctane POINTS: 1 Drawing Instructions: Draw skeletal structures corresponding to each of the given names.

17. Draw: 6-ethyl-4-isopropyldecane ANSWER:

POINTS: 1 18. Draw: 2-fluoro-3-methylpentane ANSWER:

POINTS: 1 19. Draw: 4-(2,2-dibromoethyl)-3,5-dichloroheptane ANSWER:

POINTS: 1 Exhibit 3-4 Experiments have shown that for 1,2-dichloroethane, ClCH2CH2Cl, in carbon tetrachloride solution at 25 °C, 70% of the molecules are in the anti and 30% are in the gauche conformation.

20. Refer to Exhibit 3-4. Draw a Newman projection of the anti conformation of 1,2-dichloroethane.

ANSWER:

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry POINTS: 1 21. Refer to Exhibit 3-4. Draw a Newman projection of the gauche conformation of 1,2-dichloroethane. Explain why the majority of the molecules are not in the gauche conformation.

ANSWER:

The gauche conformation is a higher energy state than the anti conformation due to the torsional strain caused by the close proximity of the two chlorine atoms.

POINTS: 1 Exhibit 3-5 Cipro® (Ciprofloxacin) is a synthetic broad spectrum antibacterial agent. It was most recently in the news as the antibiotic of choice for the treatment of anthrax. The structure of Cipro® is shown below.

22. Refer to Exhibit 3-5. Circle the functional groups in the Cipro® representation above.

ANSWER:

POINTS: 1 Exhibit 3-6 Predict the hybridization of the indicated atoms in Cipro®.

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry 23. Refer to Exhibit 3-6. The hybridization of this nitrogen atom (A) is __________.

ANSWER: sp3 POINTS: 1 24. Refer to Exhibit 3-6. The hybridization of this carbon atom (B) is __________.

ANSWER: sp2 POINTS: 1 25. Refer to Exhibit 3-6. The hybridization of this carbon atom (C) is __________.

ANSWER: sp3 POINTS: 1 26. Put a box around the most polar bond in Cipro® based on electronegativity values.

ANSWER:

POINTS: 1 27. Name the following hydrocarbon.

ANSWER: 2,3-dimethylbutane POINTS: 1 Sight along the C2-C3 bond of 2-methylbutane.

28. How many eclipsed conformations can be drawn?

ANSWER: 3 Cengage Learning Testing, Powered by Cognero

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry POINTS: 1 29. Draw the staggered conformations.

ANSWER:

POINTS: 1 30. Arrange the following conformation from lowest to highest energy.

ANSWER:

POINTS: 1 31. Explain your energy order from lowest to highest.

ANSWER: The lowest energy conformation of those shown is the one staggered conformation. Of the two eclipsed conformations given, the one in which there is one kJ/mol) and one

eclipsed interaction (energy cost: 11

eclipsed interaction (energy cost: 6.0 kJ/mol) and one

eclipsed

interaction(energy cost: 4 kJ/mol, for a total energy cost of 11 kJ/mol + 6 kJ/mol + 4 kJ/mol = 21 kJ/mol) will be higher than the one in which there are three

eclipsed interactions (energy cost: 6.0 kJ/mol x

3 = 18 kJ/mol).

POINTS: 1 32. Draw at least four isomers with molecular formula given below.

C4H10O ANSWER: There are seven total isomers, four alcohols and three ethers.

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry

POINTS: 1 33. Which of the following functional groups if bonded to a three-carbon chain would have the largest δ– charge? a. alcohol b. ether c. phosphate d. sulfide ANSWER: c POINTS: 1 34. Which of the following functional groups contains a carbonyl group? a. ketone b. ester c. carboxylic acid d. amide e. All contain a carbonyl group.

ANSWER: e POINTS: 1 35. How many isomeric chloroalkanes have the molecular formula shown below?

C4H9Cl a. 2 b. 3 c. 4 d. 5 ANSWER: c POINTS: 1 36. If the following alkyl group were attached to a cyclohexane ring,

the group would be named as: a. butyl b. isobutyl Cengage Learning Testing, Powered by Cognero

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Chapter 03 - Organic Compounds: Alkanes and Their Stereochemistry c. sec-butyl ANSWER: d POINTS: 1

d. tert-butyl

37. Which of the following alkanes would have the highest boiling point? a. heptane b. 2-methylhexane c. 2,3,-dimethylpentane d. 2,2,3-trimethylbutane e. All have the same molar mass and would have about the same boiling point.

ANSWER: a POINTS: 1

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry 1. Draw 7 constitutional isomers of a cycloalkane with the formula C6H12. Name each of the isomers you drew.

ANSWER:

cyclohexane, methylcyclopentane, 1,2-dimethylcyclobutane, 1,1-dimethylcyclobutane, 1,3dimethylcyclobutane, 1,2,3-trimethylcyclopropane, and 1,1,2-trimethylcyclopropane POINTS: 1 2. The original question was combined with #1. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 IUPAC Naming Instructions: Provide IUPAC names for each structure.

3. Name:

ANSWER: trans-1-tert-butyl-4-ethylcyclohexane or trans-1-(1,1-dimethylethyl)-4-ethylcyclohexane POINTS: 1 Drawing Instructions: Draw structures corresponding to each of the following names.

4. Draw: cis-1-sec-butyl-2-ethylcyclopentane ANSWER:

POINTS: 1 5. Draw: 3-cyclobutylpentane ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry 6. Draw: 1-chloro-2-isopropylcyclopentane ANSWER:

POINTS: 1 7. Draw: 3,5-dicyclohexylnonane ANSWER:

POINTS: 1 8. Circle all bridgehead carbons in the following structure.

ANSWER:

POINTS: 1 9. Draw the two stereoisomers of 1,3-dibromocyclobutane.

ANSWER:

POINTS: 1 10. This question was omitted on the printed copy. This placeholder question is here to maintain the numbering system integrity between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 Exhibit 4-1 Refer to the structure below to answer the following question(s):

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry

11. Refer to Exhibit 4-1. Which of the labeled bonds in the structure are equatorial bonds?

ANSWER: a and d POINTS: 1 12. Refer to Exhibit 4-1. Which of the labeled bonds are trans to bond b?

ANSWER: e POINTS: 1 13. Refer to Exhibit 4-1. Which labeled bonds have a 1,3-diaxial interaction with each other?

ANSWER: b and c POINTS: 1 Exhibit 4-2 For each substituted cyclohexane below, draw its ring-flip isomer. Circle the most stable conformation.

14. Refer to Exhibit 4-2.

ANSWER:

POINTS: 1 15. Refer to Exhibit 4-2.

ANSWER:

POINTS: 1 16. Refer to Exhibit 4-2.

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry

ANSWER:

POINTS: 1 Exhibit 4-3 The following question(s) refer to the structure of camphor shown below.

17. Refer to Exhibit 4-3. Circle all bridgehead carbons in camphor.

ANSWER:

POINTS: 1 18. Refer to Exhibit 4-3. Camphor is an example of a: a. fused bicyclic molecule. b. bridged bicyclic molecule. c. fused tricyclic molecule. d. bridged tricyclic molecule.

ANSWER: b POINTS: 1 19. Below are the two chair conformations of a 1,2,4-trimethylcyclohexane. Estimate the amount of 1,3-diaxial strain in each conformer and predict which conformer is most stable.

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry ANSWER:

Conformer A C1−CH3 C2−CH3 C4−CH3 Conformer B C1, C2−CH3 C4−CH3

2 (H−CH3) = 2 × 3.8 kJ/mol 2 (H−CH3) = 2 × 3.8 kJ/mol no diaxial interactions Strain energy in conformer A

= 7.6 kJ/mol = 7.6 kJ/mol = 0 kJ/mol = 15.2 kJ/mol

one gauche interactions

= 3.8 kJ/mol = 7.6 kJ/mol = 11.4 kJ/mol

2 (H−CH3) = 2 × 3.8 kJ/mol Strain energy in conformer B

Conformer B is more stable than conformer A by 3.8 kJ/mol.

POINTS: 1 20. The energy difference of 3.8 kJ/mol between gauche and anti butane corresponds to an equilibrium constant, Keq, of approximately 1.9. Calculate the percentage of each conformer at equilibrium.

ANSWER: Keq = (anti)/(gauche) = 1.9 (anti) = 1.9(gauche) and (anti) + (gauche) = 100% Substitute 1.9(gauche) for (anti) and you have:

1.9(gauche) + (gauche) 2.9(gauche) (gauche) (anti) = 1.9(gauche)

= 100% = 100% = 34.4% = 65.6%

POINTS: 1 21. In general, 5-alkyl substituents in 1,3-dioxane exhibit a smaller equatorial preference than they do in cyclohexane. To what might you attribute this observation?

ANSWER: The 5-alkyl substituted dioxanes do not suffer from 1,3-diaxial interactions since there are no hydrogens that are axial to the 5-substituent. Hence, the axial conformation is more stable in 5-substituted 1,3-dioxanes than it is in substituted cyclohexanes.

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry Exhibit 4-4 Label each pair of compounds below as: a. conformational isomers b. stereoisomers c. constitutional isomers d. identical

22. _____ ANSWER: b POINTS: 1

23. _____

ANSWER: c POINTS: 1

24. _____

ANSWER: d POINTS: 1

25. _____

ANSWER: a POINTS: 1

26. _____ ANSWER: b POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry Exhibit 4-5 For each pair of molecules below, circle the most stable.

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry 30. (-)-Menthol can be isolated from the peppermint plant and is responsible for the characteristic flavor and taste of peppermint. The structure of (-)-menthol is:

On the chair template provided below, draw the two chair conformations that are in equilibrium for (-)-menthol.

ANSWER:

POINTS: 1 31. D-Pinitol is an interesting hexahydroxy cyclohexane, whose structure is shown below.

On the templates provided, draw the two chair conformations that are in equilibrium for D-pinitol. Circle the most stable conformation.

ANSWER:

POINTS: 1 32. The two structures show below represent: Cengage Learning Testing, Powered by Cognero

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry

a. constitutional isomers b. stereoisomers c. cis-trans isomers d. both b and c e. a, b and c ANSWER: d POINTS: 1 33. What relationship exists between the following two structures?

a. identical molecules c. stereoisomers ANSWER: c POINTS: 1

b. constitutional isomers d. different molecules

34. Consider the two methyl groups indicated with letters in the following molecular model.

These two groups are: a. A: axial B: axial c. A: equatorial B: axial ANSWER: b POINTS: 1

b. A: axial B: equatorial d. A: equatorial B: equatorial

35. Which of the following would produce the greatest amount of 1,3-diaxial strain when substituted for Cl in the following structure?

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry

a. CN c. C(CH3)3

b. OH d. CO2H

ANSWER: c POINTS: 1 36. Which of the following can exist as cis-trans isomers? a. cyclohexane b. methylcyclohexane c. 1,1-dimethylcyclohexane d. 1,2-dimethylcyclohexane e. both c and d f. all of the above ANSWER: d POINTS: 1 37. The following substituents are attached to a cyclohexane ring. Which substituted cyclohexane is correctly named as a "cyclohexyl alkane"? a. tert-butyl b. 2-methylpentyl c. cyclopentyl d. octyl e. both b and d f. none of the above ANSWER: d POINTS: 1 38. In methylcyclohexane: a. all carbon atoms are sp3 hybridized. b. ring-carbon atoms are sp2 hybridized and the methyl group is sp3 hybridized. c. all bond angles are approximately 120°. d. ring-bond angles are approximately 120° and the ring-methyl bond angle is approximately 109°.

ANSWER: a POINTS: 1 39. Which of the following would have the smallest strain energy (kJ/mol)? a. cyclobutane b. cyclopentane c. cyclohexane d. cyclooctane ANSWER: c POINTS: 1 40. Consider the following table.

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Chapter 04 - Organic Compounds: Cycloalkanes and Their Stereochemistry Number of Carbon Atoms in Ring 3 4 5 6

Heat of Combustion per CH2 (kJ) 696 686 664 659

Based on the data in the table, which of the following compounds would have the largest strain energy? a. cyclopropane b. cyclobutane c. cyclopentane d. cyclohexane ANSWER: a POINTS: 1 41. Consider the following table.

Number of Carbon Atoms in Ring 3 4 5 6

Heat of Combustion per CH2 (kJ) 696 686 664 659

What is the approximate strain energy per CH2 for cyclopropane? a. 12 kJ b. 37 kJ c. 110 kJ d. 230 ANSWER: b POINTS: 1 42. In cyclopropane, which of the following strain types would be the least important in determining the overall energy? a. angle b. torsional c. steric d. All make a significant contribution.

ANSWER: c POINTS: 1

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Chapter 05 - Stereochemistry At Tetrahedral Centers Exhibit 5-1 For the following question(s) MATCH each definition to a term from the list below. Place the letter of the term in the blank to the left of the definition. a. b. c. d. e.

racemates chirality center chirality diastereomers enantiomers

f. g. h. i. j.

meso compounds optically active prochirality center optically inactive achiral

1. ____ are organic molecules which rotate the plane of polarization of plane-polarized light.

ANSWER: g POINTS: 1 2. _____ is the reason for "handedness" in molecules; the property of an object that causes it to be nonsuperimposable on its mirror image.

ANSWER: c POINTS: 1 3. _____ are stereoisomers that are not mirror images.

ANSWER: d POINTS: 1 4. _____ is an atom in a molecule that is bonded to four different atoms or groups of atoms.

ANSWER: b POINTS: 1 5. _____ are molecules which contain a plane of symmetry and chirality centers, but are achiral.

ANSWER: f POINTS: 1 6. _____ is an sp3-hybridized atom that can become a chirality center by changing one of its attached groups.

ANSWER: h POINTS: 1 7. The specific rotation of a compound is denoted by the symbol: a. b. c.

R S α

d.

ANSWER: d POINTS: 1 Exhibit 5-2 Place asterisks at all the chirality centers in each molecule below.

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Chapter 05 - Stereochemistry At Tetrahedral Centers

8.

ANSWER:

POINTS: 1

9.

ANSWER:

POINTS: 1

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Chapter 05 - Stereochemistry At Tetrahedral Centers

10.

ANSWER:

POINTS: 1

11.

ANSWER:

POINTS: 1

12.

ANSWER:

POINTS: 1

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Chapter 05 - Stereochemistry At Tetrahedral Centers

13.

ANSWER:

POINTS: 1 Exhibit 5-3 Assign R, S configurations to each indicated chirality center in the molecules below.

14. Refer to Exhibit 5-3. The configuration of this carbon atom (A) is _____.

ANSWER: R POINTS: 1 15. Refer to Exhibit 5-3. The configuration of this carbon atom (B) is _____.

ANSWER: S POINTS: 1 16. Refer to Exhibit 5-3. The configuration of this carbon atom (C) is _____.

ANSWER: R POINTS: 1 17. Refer to Exhibit 5-3. The configuration of this carbon atom (D) is _____.

ANSWER: S POINTS: 1 18. Refer to Exhibit 5-3. The configuration of this carbon atom (E) is _____.

ANSWER: S POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 05 - Stereochemistry At Tetrahedral Centers Exhibit 5-4 Consider the structure of streptimidone to answer the following question(s).

19. Refer to Exhibit 5-4. Assign R or S configuration to each chirality center indicated in streptimidone.

ANSWER:

POINTS: 1 20. Refer to Exhibit 5-4. Based on the number of chirality centers, how many stereoisomers of streptimidone are possible?

ANSWER: Streptimidone has two chirality centers so there are 22, or 4, possible stereoisomers.

POINTS: 1 21. Refer to Exhibit 5-4. Will streptimidone have a meso stereoisomer? Explain your answer.

ANSWER: No. Meso compounds are compounds that contain chirality centers, but possess a plane of symmetry. Because the substituents on the chiral carbons are different in streptimidone it is impossible for any stereoisomer to have a plane of symmetry.

POINTS: 1 Exhibit 5-5 Label each pair of stereoisomers below as: a. enantiomers b. diastereomers c. identical Place the letter of the correct answer in the blank to the left of the pair of stereoisomers.

22. _____ ANSWER: a POINTS: 1

23. _____ Cengage Learning Testing, Powered by Cognero

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Chapter 05 - Stereochemistry At Tetrahedral Centers ANSWER: c POINTS: 1

24. _____

ANSWER: b POINTS: 1

25. _____ ANSWER: b POINTS: 1

26. _____ ANSWER: c POINTS: 1

27. _____ ANSWER: a POINTS: 1

28. _____

ANSWER: b POINTS: 1 29. Draw a wedge-dash projection of (2R,3S)-dibromobutane.

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Chapter 05 - Stereochemistry At Tetrahedral Centers ANSWER:

POINTS: 1 30. Draw a Newman projection of the most stable conformation of (2R,3S)-dibromobutane sighting down the C2−C3 bond.

ANSWER:

POINTS: 1 31. (2R,3S)-Dibromobutane is: a. optically active. b. racemic. c. dextrorotatory. d. a meso compound.

ANSWER: d POINTS: 1 32. Estriol, a potent estrogenic hormone, has been isolated from the urine of pregnant women. When 40 mg of estriol is dissolved in 1.0 mL of dioxane and placed in a sample tube with 1 dm path length a rotation of +2.32° is observed.

Calculate the specific rotation for estriol.

ANSWER:

POINTS: 1 Exhibit 5-6 Refer to the structure below to answer the following question(s).

33. Refer to Exhibit 5-6. (S)-(−)-Serine: a. is dextrorotatory b. rotates plane-polarized light in a counterclockwise direction c. rotates plane-polarized light in a clockwise direction d. is racemic ANSWER: b POINTS: 1 34. Refer to Exhibit 5-6. Draw the enantiomer of (S)-(−)-serine in a wedge-dash projection.

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Chapter 05 - Stereochemistry At Tetrahedral Centers ANSWER:

POINTS: 1 35. Refer to Exhibit 5-6. Give the complete name of the enantiomer of (S)-(−)-serine.

ANSWER: (R)-(+)-serine POINTS: 1 Exhibit 5-7 A natural product having [α]D = +40.3° has been isolated and purified.

36. Refer to Exhibit 5-7. This information indicates that the natural product: a. is racemic. b. does not rotate plane-polarized light. c. is levorotatory. d. is dextrorotatory.

ANSWER: d POINTS: 1 37. Refer to Exhibit 5-7. Two structures have been proposed for this natural product. Circle the structure that is consistent with the information presented and briefly explain your choice.

ANSWER:

The information presented indicates that the natural product is optically active. To be optically active molecules must be chiral—that is, they must not have a plane of symmetry. The cyclic structure, although has chirality centers, has a plane of symmetry, indicated by the dashed line on the structure, and can, therefore, not be optically active. The circled structure has four chirality centers, and is not symmetric. We would expect it to be optically active.

POINTS: 1 Exhibit 5-8 Identify the indicated hydrogens in the following molecules as pro-R or pro-S.

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Chapter 05 - Stereochemistry At Tetrahedral Centers

38.

ANSWER:

POINTS: 1

39.

ANSWER:

POINTS: 1 Exhibit 5-9 Identify the indicated faces in the following molecules as re or si.

40.

ANSWER:

POINTS: 1

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Chapter 05 - Stereochemistry At Tetrahedral Centers

41.

ANSWER:

POINTS: 1 42. Acetoacetate synthase catalyzes the addition of pyruvate to β-ketobutyrate to yield α-aceto-α-hydroxybutyrate. If the addition occurs from the si face of β-ketobutyrate, what is the stereochemistry of the product?

ANSWER: Addition of the aceto group to the si face results in the S enantiomer of α-aceto-α-hydroxybutyrate.

POINTS: 1 43. Explain why a work glove is chiral but disposable surgical gloves are not.

ANSWER: Surgical gloves are not made to wear on a left hand or right hand but can be worn on either hand with an equally good fit. This type of glove is symmetrical with respect to the positioning of the covering that goes on the thumb - it has a plane of symmetry. or Work gloves have "front" and "back" sides. The thumb is positioned differently with respect to the digits.

Surgical gloves have five symmetrical digits. They do not have "front" and "back" sides.

POINTS: 1 44. A white solid known to be enantiomerically pure is dissolved in water. This solution produced an observed optical rotation of +45.6°. The solution was allowed to stand and the optical rotation was measured repeatedly producing the data shown in the table.

Time (min) 10 20

Observed Rotation (°) +39.9 +5.6

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Chapter 05 - Stereochemistry At Tetrahedral Centers 30 -1.2 40 -23.4 40 -23.4 Explain the pattern of observed rotation.

ANSWER: Some type of reaction, possibly a hydrolysis, is occurring and the optically active dextrorotatory compound is being converted to the levorotatory enantiomer.

The change may be due to the production of a new compound with a specific rotation of -23.4°. This might be verified by using this value to identify the compound if enough other information is known.

Assuming an equilibrium is established, an alternate explanation could be that it appears from the final value of the observed rotation that an equilibrium mixture of the two forms and that the levorotatory from predominates in the mixture.

POINTS: 1 45. Draw a structure for a singly substituted chiral chloroalkane that contains four carbon atoms. Indicate the chirality center with an asterisk (*).

ANSWER:

The chirality center is indicated by the asterisk (*), POINTS: 1 46. Which of the following objects is not chiral? a. baseball glove b. ballerina slipper c. corkscrew d. comb e. All are chiral.

ANSWER: d POINTS: 1 47. Which of the following would be has the highest priority according to the sequence rules? a. —SH b. —NH2 c. —CH3

d. —OH

ANSWER: a POINTS: 1 48. Which of the following arranges the groups in order of decreasing priority according to the sequence rules? a. CH3 > CH2CH3 > CH2OH > CO2H b. CH3 > CH2CH3 > CO2H >CH2OH c. CH2OH > CO2H > CH2CH3 > CH3

d. CO2H > CH2OH > CH2CH3 > CH3

ANSWER: d POINTS: 1 49. In which of the following groups can the carbon atom be a chiral center?

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Chapter 05 - Stereochemistry At Tetrahedral Centers a. b. c. d. e. The carbon atoms cannot be a chiral center in any of these groups.

ANSWER: e POINTS: 1 50. Which of the following compounds will rotate the plane of polarized light? a. b.

c.

d.

ANSWER: c POINTS: 1 51. For which of the following generic substances is it not possible to isolate the enantiomers? (R1, R2 and R3 represent different groups.) a.

b.

c.

d. It is possible to isolate the enantiomers of all of these. e. It is not possible to isolate the enantiomers of any of these.

ANSWER: a POINTS: 1 52. In muscles during strenuous exercise, under anaerobic conditions lactic acid builds up due to the following reaction.

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Chapter 05 - Stereochemistry At Tetrahedral Centers The carbon atom indicated by the asterisk is a. chiral b. achiral c. prochiral d. both achiral and prochiral ANSWER: d POINTS: 1 53. A compound with specific rotation, [α]D = +35.4° reacts to form a single product. The product of the reaction shows no optical activity. This could mean that the product: a. is a racemate. b. does not have any chiral centers. c. is a meso compound. d. Any of the above are possible explanations.

ANSWER: d POINTS: 1

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Chapter 06 - An Overview of Organic Reactions Exhibit 6-1 Classify each reaction below as a(n): a. addition b. elimination c. substitution d. rearrangement Place the letter corresponding to the correct answer in the blank to the left of the reaction.

1. _____ ANSWER: c POINTS: 1

2. _____

ANSWER: d POINTS: 1

3. _____ ANSWER: a POINTS: 1 Exhibit 6-2 Identify the functional groups present in each compound below and predict the direction of polarity in each.

4. mustard gas Cl−CH2CH2−S−CH2CH2−Cl ANSWER:

POINTS: 1

5.

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Chapter 06 - An Overview of Organic Reactions ANSWER:

POINTS: 1 Exhibit 6-3 Classify each structure below as a nucleophile or electrophile and briefly explain your choice.

6.

ANSWER: Azide is a nucleophile since it has a net negative charge (and lots of electron pairs!).

POINTS: 1

7.

ANSWER: Hydronium ion is an electrophile since it has a positive charge.

POINTS: 1

8.

ANSWER: Phenol can be a nucleophile or an electrophile

POINTS: 1 Exhibit 6-4 Identify the nucleophile and electrophile in each reaction below and label them.

9.

ANSWER:

POINTS: 1

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Chapter 06 - An Overview of Organic Reactions

10.

ANSWER:

POINTS: 1 11.

ANSWER:

POINTS: 1 Exhibit 6-5 Add curved arrows to the following reactions to indicate the flow of electrons in each.

12.

ANSWER:

POINTS: 1

13.

ANSWER:

POINTS: 1 14.

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Chapter 06 - An Overview of Organic Reactions ANSWER:

POINTS: 1 15. Acidic ether cleavages are typical substitution reactions. Use the information in Table 6.3 of the textbook to calculate ΔH° for the reaction of ethyl methyl ether with hydrogen chloride and hydrogen iodide. a. b.

Which acid would you predict to be most effective at ether cleavage?

ANSWER: CH3CH2OH +

a. CH 3CH2OCH3 + HCl Bonds broken

CH3Cl Bonds formed

CH3CH2O−CH3

339 kJ/mol

CH3CH2O−H

H−Cl

432 kJ/mol

CH3−Cl

771 kJ/mol ΔH°

436 kJ/mol 351 kJ/mol 787 kJ/mol

= ΔH° bonds broken −ΔH° bonds formed = 771 − 787 kJ/mol = −16 kJ/mol CH3CH2OH + CH3I Bonds formed

b. CH3CH2OCH3C + HI Bonds broken CH3CH2O−CH3

339 kJ/mol

CH3CH2O−H

H−I

298 kJ/mol

CH3−I

637 kJ/mol

ΔH°

436 kJ/mol 234 kJ/mol 670 kJ/mol = ΔH° bonds broken −ΔH° bonds formed = 637 − 670 kJ/mol = −33 kJ/mol

Since cleavage with HI is the most exothermic we would predict that HI would be most effective at ether cleavage.

POINTS: 1 Exhibit 6-6 MATCH each definition to one of the terms below. Place the letter of the term in the blank to the left of the definition.

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Chapter 06 - An Overview of Organic Reactions a. b. c. d. e.

polarization addition reaction homolytic bond breakage electrophile polar reaction

f. g. h. i. j.

substitution nucleophile radical reaction elimination reaction heterolytic bond breakage

16. _____ A process involving symmetrical bond breaking and bond making.

ANSWER: h POINTS: 1 17. _____ This occurs when both bonding electrons remain with one product fragment.

ANSWER: j POINTS: 1 18. _____ A reaction where two reactants exchange parts to give two new products.

ANSWER: f POINTS: 1 19. _____ A general term for species which have electron rich sites that can form a bond by donating a pair of electrons to an electron poor site.

ANSWER: g POINTS: 1 Exhibit 6-7 Consider this reaction when answering the following question(s):

20. Refer to Exhibit 6-7. This reaction is an example of: a. a substitution reaction. b. a rearrangement reaction. c. an addition reaction. d. an elimination reaction.

ANSWER: b POINTS: 1 21. Refer to Exhibit 6-7. a.

The structures below show the step-wise bond making and bond breaking in this reaction.

Draw curved arrows to show the electron flow that has occurred in each step.

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Chapter 06 - An Overview of Organic Reactions

b.

Calculate Keq for the reaction in part a.

c. Calculate ΔG° for the reaction in part a.

ANSWER: a.

b.

Keq = [products]/[reactants] = 76/24 = 3.17

c.

ΔG°= –RT ln Keq = – 8.314 J/(K • mol)(298) ln 3.17 = –2.86 x 103 J/mol = –2.86 kJ/mol

POINTS: 1 22. The original question has been combined with #21 as part b. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 23. Refer to Exhibit 6-7. Draw a qualitative energy diagram for the reaction (assume that the first step is slower than the second step). Label fully.

ANSWER:

POINTS: 1 Exhibit 6-8 MATCH each definition to one of the terms below. Place the letter corresponding to the term in the blank to the left of the Cengage Learning Testing, Powered by Cognero

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Chapter 06 - An Overview of Organic Reactions definition. a. b. c.

transition state endergonic reaction activation energy

d. e. f.

standard Gibbs free energy change exergonic reaction reaction intermediate

24. _____ A species that lies at an energy minimum between steps on a reaction.

ANSWER: f POINTS: 1 25. _____ The energy needed by reactants to reach the transition state.

ANSWER: c POINTS: 1 26. _____ A reaction where ΔG° is negative.

ANSWER: e POINTS: 1 27. _____ ΔG° = −RT ln Keq ANSWER: d POINTS: 1 Exhibit 6-9 Use the reaction energy diagram below to answer the following question(s).

28. Refer to Exhibit 6-9. The reaction depicted in this reaction energy diagram can best be described as: a. a slow exothermic reaction b. a fast exothermic reaction c. a slow endothermic reaction d. a fast endothermic reaction ANSWER: b POINTS: 1 29. Refer to Exhibit 6-9. The transition state is found at point _____ on the diagram.

ANSWER: B POINTS: 1 30. Refer to Exhibit 6-9. The products are found at point _____ on the diagram.

ANSWER: D POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 06 - An Overview of Organic Reactions 31. Refer to Exhibit 6-9. The free-energy change for the reaction is indicated at point _____ on the diagram.

ANSWER: C POINTS: 1 32. Refer to Exhibit 6-9. The reactants are found at point _____ on the diagram.

ANSWER: A POINTS: 1 Exhibit 6-10 Consider the reaction of 2-bromo-2-methylpropane with water, shown below, to answer the following question(s).

Diagram 1: The first step of this reaction is shown below.

Diagram 2: The second and third steps of the reaction are shown below.

33. Refer to Exhibit 6-10. This reaction is an example of: a. a substitution reaction. b. a rearrangement reaction. c. an elimination reaction. d. an addition reaction.

ANSWER: a POINTS: 1 34. Refer to Exhibit 6-10. In Diagram 1, Species B is: a. a carbene b. a carbanion c. a carbocation d. a radical ANSWER: c POINTS: 1 35. Refer to Exhibit 6-10. In Diagram 1, add curved arrows to indicate electron flow.

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Chapter 06 - An Overview of Organic Reactions ANSWER:

POINTS: 1 36. Refer to Exhibit 6-10. In Diagram 2, label the nucleophile, Nu, and the electrophile, E+, in the blanks provided under the structures.

ANSWER:

POINTS: 1 37. Refer to Exhibit 6-10. In Diagram 2, draw arrows on the structures showing electron flow.

ANSWER:

POINTS: 1 38. Refer to Exhibit 6-10. The reaction in Diagram 2 is an example of: a. a pericyclic reaction. b. a radical reaction. c. a concerted reaction. d. a stepwise reaction.

ANSWER: d POINTS: 1 39. Refer to Exhibit 6-10. Using the bond dissociations values in the table below, calculate the ΔH° for the reaction in Diagram 2. Show your calculations for full credit.

Bond (CH3)3C−Br (CH3)3C−OH HO−H H−Br ANSWER: ΔH°

D (kJ/mol) 263 380 498 366 = ΔH° bonds broken − ΔH° bonds formed = 761 kJ/mol − 746 kJ/mol = +15 kJ/mol Bonds Broken 263 kJ/mol (CH3)3C−Br HO−H 498 kJ/mol 761 kJ/mol POINTS: 1 Cengage Learning Testing, Powered by Cognero

Bonds Formed (CH3)3C−OH H−Br

380 kJ/mol 366 kJ/mol 746 kJ/mol

Page 9

Chapter 06 - An Overview of Organic Reactions 40. Refer to Exhibit 6-10. Based on all of the data presented in Exhibit 6-10, draw a qualitative reaction energy diagram for the reaction of 2-bromo-2-methylpropane with water. The first step is the slowest step of the reaction. Fully label the diagram, including the coordinates.

ANSWER:

POINTS: 1 Exhibit 6-11 In each reaction below: a. b.

Label the nucleophile (Nu) and the electrophile (E).

Draw arrows on the structures showing electron flow in the reaction.

41.

ANSWER:

POINTS: 1

42.

ANSWER:

POINTS: 1 43. The following represents the carbocation intermediate in the reaction of an alkene with HBr.

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Chapter 06 - An Overview of Organic Reactions

Draw the skeletal structure of the possible products and reactants.

ANSWER: Two possible reactants:

One product:

POINTS: 1 44. What are the major differences between a reaction occurring in flask in the laboratory compared to a reaction occurring in a biological system?

ANSWER: Solvent: Laboratory reactions occur in nonpolar solvents whereas biological reactions occur in an aqueous environment.

Temperature: Laboratory reactions can be carried out at very low temperatures or very high temperatures or anywhere in between whereas biological reactions are limited to the organism’s body temperature which for warm-blooded species is a very narrow range.

Catalysis: Laboratory reactions can occur with or without catalysts and the catalysts can be acids, bases, metal surfaces, etc. Biological reactions are catalyzed by large protein molecules called enzymes, that often require coenzymes or cofactors to function. Laboratory catalysts in general are nonspecific, an acid or Pt surface can catalyze many different reactants or reactions. Enzymes are highly specific - a given enzyme may only be able to catalyze the reaction of one reactant (substrate) or one class of reactions. pH: Laboratory reactions can occur at a variety of pHs whereas biological reactions must function within narrow physiological pH limits.

POINTS: 1 45. The alkane formed by hydrogenation of (S)-4-methyl-1-hexene is optically active while the one formed by hydrogenation of (S)-3-methyl-1-pentene is not. Explain.

ANSWER: Each starting alkene is optically active because it contains a chiral center⎯a carbon with four different groups bonded to it. The alkane formed by the hydrogenation of (S)-4-methyl-1-hexene is optically because carbon four is still chiral. The product of hydrogenation of (S)-3-methyl-1-pentene is not optically active because two of the groups on carbon three are the same now⎯they are ethyl groups.

POINTS: 1 46. The enzyme aconitase catalyzes the hydration of the alkene functional group of aconitic acid to give two products, citric acid and isocitric acid. Isocitric acid is optically active; citric acid is not optically active. Based on your knowledge of alkene hydration and optical activity, the structure of citric acid is:

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Chapter 06 - An Overview of Organic Reactions

a.

b.

c.

d.

cannot be determined

ANSWER: b POINTS: 1 47. Which of the following could act as an electrophile? a. CN– b. NH2– c. NH3

d. H3O+

ANSWER: d POINTS: 1 48. Which of the following could act as a nucleophile? a. HCl b. CH3NH2 c. BF3

d. CH3Br

ANSWER: b POINTS: 1 49. Consider the following grayscale electrostatic potential map. The regions are labeled as to color.

Which atom is the most electron poor? a. 1 b. 2 c. 3 d. 4 Cengage Learning Testing, Powered by Cognero

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Chapter 06 - An Overview of Organic Reactions ANSWER: b POINTS: 1 50. Consider the following grayscale electrostatic potential map. The regions are labeled as to color.

Which atom is the most likely to be attacked by a nucleophile? a. 1 b. 2 c. 3 d. 4 ANSWER: b POINTS: 1 51. In a polar reaction mechanism, the atom that gives away electrons in a neutral nucleophile will end up as a(n): a. cation b. anion c. radical d. neutral molecule ANSWER: a POINTS: 1 52. Consider the following.

This process would result in the formation of: a. a positive ion and negative ion. b. two neutral species. c. two radicals. d. two negative ions. e. a positive ion and a neutral species. f. a negative ion and a neutral species.

ANSWER: c POINTS: 1 53. Consider the following process.

Which of the following correctly applies? a. termination step b. unsymmetrical bond formation Cengage Learning Testing, Powered by Cognero

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Chapter 06 - An Overview of Organic Reactions c. polar reaction ANSWER: a POINTS: 1

d. produces a radical

54. A reaction that establishes equilibrium with almost no reactants present: a. is exergonic. b. has a negative ΔG°. c. has a large Keq. d. all of the above.

ANSWER: d POINTS: 1 55. A reaction has ΔH° = –14.7 kJ and ΔS° of 35.7 J/(K • mol), will a. be exergonic. b. have a positive ΔG°. c. have a very small Keq. d. will proceed rapidly.

ANSWER: a POINTS: 1 56. Which of the following compounds would produce a single product when treated with Cl2 in the presence of ultraviolet radiation? a. butane b. 2-methylpropane c. 2,2,-dimethylpropane d. All would produce multiple products.

ANSWER: c POINTS: 1 57. Consider a reaction with the following thermodynamic properties. ΔH° = 77.7 kJ ΔS° = –35.7 J/(K • mol) ΔG°= 88.4 kJ This reaction: a. will be spontaneous at low temperatures. b. has bonds in the products that are weaker than the reactants. c. may have fewer and more complicated molecules in the product. d. will proceed very slowly.

ANSWER: c POINTS: 1 58. Consider the following energy diagram for an enzyme-catalyzed reaction.

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Chapter 06 - An Overview of Organic Reactions

How many transition states are involved? a. 2 b. 3 c. 4 d. 5 ANSWER: c POINTS: 1 59. Consider the following energy diagram.

Which step has the least endergonic ΔG±? a. A - C b. C - E c. E - G d. G - I ANSWER: a POINTS: 1 60. Consider the following energy diagram for an enzyme-catalyzed reaction.

Which step is probably the slowest?

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Chapter 06 - An Overview of Organic Reactions a. A - C c. E - G ANSWER: c POINTS: 1

b. C - E d. G - I

61. Consider the following energy diagram for an enzyme-catalyzed reaction.

Which step has the least energetic transition state? a. A - C b. C - E c. E - G d. G - I ANSWER: d POINTS: 1 62. Write the mechanism of the reaction of trans-2-butene with hydrogen bromide.

ANSWER:

POINTS: 1 63. Write the mechanism of the reaction between water and ammonia.

ANSWER:

POINTS: 1 64. Write the mechanism of the reaction between hydroxide ion and methyl chloride.

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Chapter 06 - An Overview of Organic Reactions ANSWER:

POINTS: 1

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Chapter 07 - Alkenes: Structure and Reactivity Exhibit 7-1 Calculate the degree of unsaturation in each formula below. Show your calculations.

1.

ANSWER: The saturated ten carbon compound would have 22 hydrogens. So the degree of unsaturation is: (22 − 18) ÷ 2 = 4 ÷ 2 = 2.

POINTS: 1 2. eicosapentaenoic acid, C20H30O2, a constituent of fish oil ANSWER: Oxygen does not affect the base formula which in this case is C20H30. The saturated 20 carbon compound would have 42 hydrogens. So the degree of unsaturation for eiscosapentaenoic acid is: (42 − 30) ÷ 2 = 12 ÷ 2 = 6.

POINTS: 1 3. diazepam (Valium), C16H13N2OCl ANSWER: Oxygen does not affect the base formula. A hydrogen is added to the base formula for each halogen and subtracted for each nitrogen so the base formula for diazepam is C16H12. The saturated 16 carbon compound would have 34 hydrogens. So the degree of unsaturation for diazepam is: (34 − 12) ÷ 2 = 22 ÷ 2 = 11.

POINTS: 1 Exhibit 7-2 Dieldrin, C12H8Cl6O, is a pentacyclic compound formerly used as an insecticide.

4. Refer to Exhibit 7-2. Calculate the degree of unsaturation for Dieldrin. Show calculations for credit.

ANSWER: Oxygen does not affect the base formula. A hydrogen is added to the base formula for each halogen so the base formula for dieldrin is C12H14. The saturated 12 carbon compound would have 26 hydrogens. So the degree of unsaturation for dieldrin is (26 − 14) ÷ 2 = 12 ÷ 2 = 6.

POINTS: 1 5. Refer to Exhibit 7-2. How many double bonds does dieldrin have?

ANSWER: Dieldrin has 6 degrees of unsaturation and it is pentacyclic, meaning it has 5 rings. Therefore, dieldrin must have one double bond.

POINTS: 1 Drawing Instructions: Draw structures corresponding to each name below. Either condensed or line structures may be used.

6. Draw: (3E)-3,7-dimethyl-1,3,6-octatriene ANSWER:

POINTS: 1 7. Draw: 3,6-dimethyl-1,4-cyclohexadiene

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Chapter 07 - Alkenes: Structure and Reactivity ANSWER:

POINTS: 1 8. Draw: cis-2-hexene ANSWER:

POINTS: 1 9. Draw: trans-4,4-dimethyl-2-pentene ANSWER:

POINTS: 1 Exhibit 7-3 2-Pentene is an example of a disubstituted alkene.

CH3CH=CHCH2CH3 10. Refer to Exhibit 7-3. a. Draw the cis and trans isomers of 2-pentene and label them. b. Circle the isomer which is most stable.

ANSWER: a.

b.

POINTS: 1 11. The original question was combined with #10. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 07 - Alkenes: Structure and Reactivity Exhibit 7-4 Provide names for each structure below. Be sure to include the cis, trans or E, Z designations where applicable.

12. Name:

ANSWER: trans-2-methyl-3-hexene or (E)-2-methyl-3-hexene POINTS: 1 13. Name:

ANSWER: 3-methylcyclopentene POINTS: 1 14. Name:

ANSWER: (2E,4E)-5-ethyl-6-methyl-2,4-heptadiene POINTS: 1 15. Name:

ANSWER: cis-3,3-dimethyl-4-propyl-1,5-octadiene or (Z)-3,3-dimethyl-4-propyl-1,5-octadiene POINTS: 1 Exhibit 7-5 Rank each set of substituents using the Cahn-Ingold-Prelog sequence rules by numbering the highest priority substituent 1 and numbering the lowest priority substituent 4. Place the number in the blank below the substituent.

16.

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Page 3

Chapter 07 - Alkenes: Structure and Reactivity ANSWER:

POINTS: 1

17.

ANSWER:

POINTS: 1

18.

ANSWER:

POINTS: 1

19.

ANSWER: POINTS: 1 Exhibit 7-6 MATCH each alkene below with the appropriate heat of hydrogenation. Place the letter of the correct answer in the blank to the left of the alkene. a. b. c.

125.9 kJ/mol 118.4 kJ/mol 115.5 kJ/mol

20. ______ ANSWER: c POINTS: 1

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Chapter 07 - Alkenes: Structure and Reactivity 21. ______ ANSWER: a POINTS: 1

22. ______ ANSWER: b POINTS: 1 23. Arrange the following bicyclic alkenes in order of increasing stability (least stable to most stable).

a. III < II < I b. I < II < III c. I < III < II d. II < III < I ANSWER: d POINTS: 1 Exhibit 7-7 Assign E or Z configurations to each alkene below.

24.

ANSWER: Z POINTS: 1

25.

ANSWER: E POINTS: 1

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Chapter 07 - Alkenes: Structure and Reactivity

26.

ANSWER: E POINTS: 1 Exhibit 7-8 MATCH each definition to the term it describes. Place the letter corresponding to the term in the blank to the left of the definition. a. b. c.

Hammond Postulate Cahn-Ingold-Prelog Rules Markovnikov's Rule

27. ______ predicts that the more stable carbocation intermediate is formed in electrophilic additions to alkenes.

ANSWER: c POINTS: 1 28. ______ predicts that the transition state of an exergonic reaction step structurally resembles the reactant.

ANSWER: a POINTS: 1 29. ______ assigns priorities to the substituent groups on a carbon.

ANSWER: b POINTS: 1 30. ______ predicts that the transition state of an endergonic reaction step structurally resembles the product.

ANSWER: a POINTS: 1 31. _____ predicts that in additions of HX to alkenes, the H adds to the less substituted alkene carbon and the X adds to the more substituted alkene carbon.

ANSWER: c POINTS: 1 32. Rank the carbocations below in order of increasing stability (least stable = 1; most stable = 3). Place the number corresponding to the carbocation's relative stability in the blank below the structure.

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Chapter 07 - Alkenes: Structure and Reactivity ANSWER:

POINTS: 1 Exhibit 7-9 Consider the following reaction:

33. Refer to Exhibit 7-9. Write the complete stepwise mechanism for this reaction. Show all intermediate structures and show all electron flow using the curved arrow convention.

ANSWER:

POINTS: 1 34. Refer to Exhibit 7-9. Draw a qualitative reaction energy diagram for this reaction. Label the positions of all reactants, intermediates and products.

ANSWER:

POINTS: 1 Exhibit 7-10 Consider the following reaction:

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Chapter 07 - Alkenes: Structure and Reactivity

35. Refer to Exhibit 7-10. Below are all the chemical structures and intermediates involved in this reaction. On the structures provided, show all electron flow using the arrow formalism for the complete stepwise mechanism for this reaction.

ANSWER:

POINTS: 1 36. Refer to Exhibit 7-10. Why does the rearrangement that results in the formation of product B occur?

ANSWER: The 1,2-alkyl shift is the rearrangement of a secondary carbocation to a more stable tertiary carbocation.

POINTS: 1 37. Write the complete stepwise mechanism for the following reaction. Show all intermediate structures and all electron flow with arrows.

ANSWER:

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Chapter 07 - Alkenes: Structure and Reactivity POINTS: 1 Exhibit 7-11 Predict the major organic product for each reaction below.

38.

ANSWER:

POINTS: 1

39.

ANSWER:

POINTS: 1 40.

ANSWER:

POINTS: 1

41.

ANSWER:

POINTS: 1 Exhibit 7-12 Predict the structure of the alkene you would use to prepare each alkyl halide below.

42.

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Page 9

Chapter 07 - Alkenes: Structure and Reactivity ANSWER:

POINTS: 1

43.

ANSWER: POINTS: 1 Exhibit 7-13 The reaction of isobutene with HBr in ether gives one of the two products below as the major product. Answer the following question(s) about this reaction.

44. Refer to Exhibit 7-13. Which product would be the major product?

ANSWER: Product B POINTS: 1 45. Refer to Exhibit 7-13. Which product would be formed via a primary carbocation?

ANSWER: Product A POINTS: 1 46. Refer to Exhibit 7-13. Which product would have a higher energy transition state for the formation of the intermediate leading to it?

ANSWER: Product A POINTS: 1 47. Refer to Exhibit 7-13. According to Hammond's Postulate, which product would have a transition state structure most closely resembling the carbocation intermediate?

ANSWER: Product A POINTS: 1 48. Refer to Exhibit 7-13. Which product would be formed by a carbocation experiencing the greatest degree of hyperconjugative stabilization?

ANSWER: Product B POINTS: 1 49. Refer to Exhibit 7-13. Increasing the temperature at which the reaction is run would lead to an increase in the percentage of which product relative to the other product?

ANSWER: Product A POINTS: 1

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Chapter 07 - Alkenes: Structure and Reactivity Name the following alkenes and draw their skeletal formulas.

50.

ANSWER: trans-3-heptene

POINTS: 1 51.

ANSWER: 1,3,3-trimethylcyclokhexene POINTS: 1 52. Cyclohexanol (—OH bonded to cyclohexane) produces cyclohexene when heated to 70 °C in the presence of a sulfuric acid catalyst. Write the equation for the reaction showing only the reactant of interest and the conditions of the reaction ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 07 - Alkenes: Structure and Reactivity 53. Identify the isoprene units in limonene, an essential oil found in oil of lemon and orange.

ANSWER: There are two isoprene units. Bolded in the structure below.

POINTS: 1 54. The following shows the connectivity of atoms only. Atoms other than carbon and hydrogen are labeled.

How many degrees of unsaturation are present in the compound? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c POINTS: 1 55. If the following compound were dissolved in ether and treated with HCl, which of the following describes the product(s) of the reaction?

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Chapter 07 - Alkenes: Structure and Reactivity

a. 3-bromoheptane b. 4-bromoheptane c. a mixture of 3-bromoheptane and 4-bromoheptane ANSWER: c POINTS: 1 56. If the following compound was dissolved in ether and treated with HBr, which of the following describes the major product(s) of the reaction?

a.

b.

c.

d. a mixture of and

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Chapter 07 - Alkenes: Structure and Reactivity

ANSWER: c POINTS: 1 57. Consider the following molecular model.

I Which of the following could be used to synthesize this compound? a.

b.

c.

d. Either or

e. Any of the above ANSWER: e POINTS: 1 58. Which of the following does not characterize a more highly substituted carbocation? a. forms more rapidly than less highly substituted carbocations b. forms product more rapidly than less highly substituted carbocations c. has a larger ΔG± than less highly substituted carbocations d. is more stable than less highly substituted carbocations Cengage Learning Testing, Powered by Cognero

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Chapter 07 - Alkenes: Structure and Reactivity e. All of these characterize a highly substituted carbocation.

ANSWER: c POINTS: 1 59. Examine the following reaction.

Which of the following is not indicated by this representation? a. KI is the solvent b. Cyclohexene is the reactant of interest. c. The reaction is carried out at room temperature. d. The reaction occurs in solution. e. All indicate the conditions of the reaction.

ANSWER: a POINTS: 1 60. Examine the following biological reaction.

In this reaction, which of the following is not indicated? a. The reaction produces three products. b. Isocitrate is the primary reactant. c. Isocitrate dehydrogenase is the catalyst for the reaction, d. NAD+ is an abbreviation for a required reactant. e. All indicate the conditions of the reaction.

ANSWER: e POINTS: 1 61. Draw the mechanism of the electrophilic addition of HCl to 2-propene.

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Chapter 07 - Alkenes: Structure and Reactivity ANSWER:

POINTS: 1 62. Write the reaction mechanism of the electrophilic reaction of HI to 2-pentene.

ANSWER:

POINTS: 1 63. Draw the mechanism of hydride shift in the reaction between 4-methyl-2-pentene and hydrochloric acid.

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Chapter 07 - Alkenes: Structure and Reactivity ANSWER:

POINTS: 1

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Chapter 08 - Alkenes: Reactions and Synthesis Exhibit 8-1 To answer the question(s) below consider the following reaction: When cyclohexene reacts with chlorine in carbon tetrachloride the trans-dihalide is formed.

1. Refer to Exhibit 8-1. Write the complete stepwise mechanism for this reaction. Be sure to show all intermediate structures and all electron flow using arrows.

ANSWER:

POINTS: 1 2. Refer to Exhibit 8-1. The observed stereochemistry of addition of chlorine to cyclohexene is explained by the intermediacy of a: a. cyclonium ion b. carbocation c. carbene d. chloronium ion ANSWER: d POINTS: 1 3. Refer to Exhibit 8-1. Since the two chlorine atoms add to opposite faces of the cyclohexene double bond, we say that the reaction occurs with: a. syn stereochemistry b. cis stereochemistry c. anti stereochemistry d. retention of stereochemistry ANSWER: c POINTS: 1 4. Refer to Exhibit 8-1. Provide the IUPAC name for the product of the reaction of cyclohexene with chlorine.

ANSWER: trans-1,2-dichlorocyclohexane POINTS: 1 5. Draw both chair conformations of trans-1,2-dichlorocyclohexane on the templates provided below. Circle the least stable conformation.

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Page 1

Chapter 08 - Alkenes: Reactions and Synthesis ANSWER:

POINTS: 1 Exhibit 8-2 Consider the reaction sequence below to answer the following question(s):

6. Refer to Exhibit 8-2. Write the complete reaction mechanism for the first step of this reaction sequence. Show all electron flow with arrows and show all intermediate structures.

ANSWER:

POINTS: 1 7. Refer to Exhibit 8-2. The intermediate in the first step of this reaction sequence is called a: a. carbocation b. cyclonium ion c. mercurinium ion d. mercapto species.

ANSWER: c POINTS: 1 8. Refer to Exhibit 8-2. In the second step of this reaction sequence, the organomercury compound is treated with sodium borohydride, NaBH4, to yield the alcohol product. This replacement of a carbon-mercury bond with a carbon-hydrogen bond is termed: a.

an oxidation

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Chapter 08 - Alkenes: Reactions and Synthesis b. a reduction c. a hydroxylation d. a cycloaddition ANSWER: b POINTS: 1 Exhibit 8-3 Consider the reaction below to answer the following question(s).

Alkenes may be hydrated by the hydroboration/oxidation procedure shown.

9. Refer to Exhibit 8-3. The intermediate formed in the first step of this reaction is:

a.

b.

c.

d.

ANSWER: b POINTS: 1 10. Refer to Exhibit 8-3. Hydroboration of alkenes is an example of: a. a rearrangement reaction. b. a substitution reaction. c. an elimination reaction. d. an addition reaction.

ANSWER: d POINTS: 1 11. Refer to Exhibit 8-3. Hydroboration/oxidation of alkenes occurs with: a. anti stereochemistry. b. trans stereochemistry. c. syn stereochemistry. d. unpredictable stereochemistry.

ANSWER: c POINTS: 1 12. Refer to Exhibit 8-3. The regiochemistry of hydroboration/oxidation of alkenes is: Cengage Learning Testing, Powered by Cognero

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Chapter 08 - Alkenes: Reactions and Synthesis a. Markovnikov. b. non-Markovnikov c. subject to solvent effects. d. unrelated to alkene structure.

ANSWER: b POINTS: 1 Exhibit 8-4 Consider the reaction below to answer the following question(s).

When dichlorocarbene is generated in the presence of an alkene, a dichlorocyclopropane is formed.

13. Refer to Exhibit 8-4. Write the complete stepwise mechanism for the formation of dichlorocarbene, CCl2. Show all intermediate structures and show all electron flow with arrows.

ANSWER:

POINTS: 1 14. Refer to Exhibit 8-4. Draw the complete Lewis electron dot structure for dichlorocarbene, CCl2.

ANSWER: POINTS: 1 15. Refer to Exhibit 8-4. In the reaction of an alkene with dichlorocarbene, the dichlorocarbene is the: a. electrophile. b. Lewis base. c. nucleophile. d. both b and c.

ANSWER: a POINTS: 1 16. Refer to Exhibit 8-4. The reaction of an alkene with dichlorocarbene is: a. regiospecific. b. Markovnikov. c. stereospecific. d. non-Markovnikov.

ANSWER: c POINTS: 1 Exhibit 8-5 The sequence of (1) alkene hydroxylation followed by (2) diol cleavage is often an excellent alternative to direct alkene Cengage Learning Testing, Powered by Cognero

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Chapter 08 - Alkenes: Reactions and Synthesis cleavage with ozone. For this sequence below, answer the following question(s).

17. Refer to Exhibit 8-5. Draw the structure of A. ANSWER:

POINTS: 1 18. Refer to Exhibit 8-5. Give the formula for reagent B. ANSWER: HIO4 POINTS: 1 Exhibit 8-6 Predict the products of each reaction below. Indicate regiochemistry and stereochemistry when relevant.

19.

ANSWER:

POINTS: 1

20.

ANSWER:

POINTS: 1

21.

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Chapter 08 - Alkenes: Reactions and Synthesis ANSWER:

POINTS: 1

22.

ANSWER:

POINTS: 1

23.

ANSWER:

POINTS: 1

24.

ANSWER:

POINTS: 1

25.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 08 - Alkenes: Reactions and Synthesis

26.

ANSWER:

POINTS: 1

27.

ANSWER: POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1 Exhibit 8-7 Choose the best reagent from the list below for carrying out each transformation. Place the letter of the reagent in the blank to the left of the reaction. a. 1.

2.

O3 Zn, H3O+

e. 1.

2.

b. 1.

2.

BH3, THF H2O2, NaOH, H2O

f. KMnO4, acid

c. CHCl3, KOH

g.

CH2I2, Zn(Cu)

d. H2O, H2SO4, heat

h.

1. Hg(OAc)2, H2O

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OsO4 NaHSO3, H2O

Page 7

Chapter 08 - Alkenes: Reactions and Synthesis 2. NaBH4

30. _____ ANSWER: b POINTS: 1

31. _____ ANSWER: a POINTS: 1

32. _____ ANSWER: g POINTS: 1

33. _____ ANSWER: h POINTS: 1 Exhibit 8-8 For each reaction below suggest structures for alkenes that give the indicated reaction products. There may be more than one answer in some cases.

34.

ANSWER: POINTS: 1

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Chapter 08 - Alkenes: Reactions and Synthesis

35.

ANSWER:

POINTS: 1

36.

ANSWER: POINTS: 1

37.

ANSWER: CH2=CHC6C5 POINTS: 1 Exhibit 8-9 To answer the question(s) below consider the following information: In an abandoned laboratory has been found a flammable liquid, A, in a bottle bearing only the label "Compound A: C7H12." Government agents have offered you a considerable sum to determine the structure of this compound. After verifying the molecular formula by elemental analysis, you find that Compound A reacts with 1 mol equiv of hydrogen; and, after treatment with acidic KMnO4, Compound A gives the dicarboxylic acid C (see below). Another bottle from the same laboratory is labeled "Compound B (isomer of A)." Compound B also reacts with 1 mol equiv of hydrogen, but yields cyclohexanone after treatment with acidic KMnO4.

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Chapter 08 - Alkenes: Reactions and Synthesis

38. Refer to Exhibit 8-9. How many degrees of unsaturation does Compound A possess?

ANSWER: A saturated seven carbon compound should have the formula C7H16 so compound A has (16 − 12) ÷ 2 = 2 degrees of unsaturation.

POINTS: 1 39. Refer to Exhibit 8-9. Suggest structures for A and B. ANSWER:

POINTS: 1 40. Refer to Exhibit 8-9. What was the other product formed in the KMnO4 oxidation of B? ANSWER: CO2 POINTS: 1 41. Lyapolate Sodium, whose structure is shown below, is used as an anticoagulant. Identify the monomer unit(s) in lyapolate sodium.

ANSWER: H2C=CH−SO3Na POINTS: 1 42. Povidone is produced commercially as a series of products having mean molecular weights ranging from about 10,000 to 700,000. Complexed with iodine, povidone yields an iodophor, marketed under the tradename Betadine, which is used as a topical anti-infective.

Identify the monomer unit(s) in povidone.

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Chapter 08 - Alkenes: Reactions and Synthesis ANSWER:

POINTS: 1 43. Draw the skeletal structure(s) and name the product(s) of the reaction when the following substance is treated with sulfuric acid and warmed to approximately 50 °C. Atoms other than hydrogen and carbon are labeled.

ANSWER: Two products are possible.

Names respectively: 3-isopropyl-6-methylcyclohexene 1-isopropyl-4-methylcyclohexene POINTS: 1 44. Draw the skeletal structure(s) and name the product(s) of the reaction when the following substance is treated with ethanoic KOH. Atoms other than hydrogen and carbon are labeled.

ANSWER: Two products are possible.

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Chapter 08 - Alkenes: Reactions and Synthesis Names, respectively: 4,4-dimethylcyclopentene 3,3-dimethylcyclopentene POINTS: 1 45. Draw a condensed formula for a polymer containing five repeating units derived from the following monomer. Name the polymer.

ANSWER: polypropylene POINTS: 1 46. Consider the polymeric structure below for amylose (a type of starch). Draw a box around the monomer of amylose.

Extra: Propose a structure for the monomer. Hint: This polymer is formed in a dehydration reaction.

ANSWER:

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Chapter 08 - Alkenes: Reactions and Synthesis

POINTS: 1 47. Consider the following polymer.

Draw the condensed formula for its monomer.

ANSWER:

POINTS: 1 48. Hydration of 4-methyl-1-pentene in the presence of an acid catalyst produces a product that is optically inactive.

Briefly explain why.

ANSWER: While hydration produces alcohols with a chiral center, the product is a racemic mixture. The first step of the hydration reaction involves Markovnikov protonation of 4-methyl-1-pentene to yield a planar, secondary carbocation. In the second step, water attacks the planar carbocation from either the Re or Si face with equal probability to give a racemic mixture of protonated alcohols. Deprotonation of the protonated alcohols produces a racemic, optically inactive mixture of (R)- and (S)-4-methyl-2 pentanol.

POINTS: 1 49. Hydration of an alkene during enzyme catalysis in a biological system produces an optically active product even through there is a potential to form a racemic mixture. Explain why this occurs.

ANSWER: Chemical catalysts do not provide a chiral environment in which a reaction can occur, enzymes provide such an environment. During enzyme catalysis, the planar carbocation is held in the chiral environment of the enzyme’s active site such that the Re and Si faces of the cation are distinguishable. The attacking water molecule can approach only one face of the cation, resulting in one transition state and one enantiomer in the product. The latter producing the observed optical activity.

POINTS: 1 50. Draw the product(s) of the reaction of (R)-3-chlorobutene with HBr. Indicate stereochemistry.

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Chapter 08 - Alkenes: Reactions and Synthesis ANSWER:

POINTS: 1 51. Which of the following will produce only one product upon dehydration? a.

b.

c.

d.

e. b and d f. a, b, and c g. None produce a single product.

ANSWER: e POINTS: 1 52. Consider the following intermediate.

Which is applicable to this structure? a. results in a product which has anti stereochemistry c. is called a bromohydrin ANSWER: a POINTS: 1

b. produces only the cis isomer d. occurs during the addition of HBr

53. Which of the following is a difference between hydroboration/oxidation and oxymercuration?

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Chapter 08 - Alkenes: Reactions and Synthesis a. conversion of an alkene to an alcohol c. hydration ANSWER: b POINTS: 1

b. Markovnikov regiochemistry d. heteroatom-containing intermediate

54. Which of the following is a carbene? a. R3C+ b. R:– c. R2:

d. R.

ANSWER: c POINTS: 1 55. What product forms when the following substance is treated with chloroform in the presence of KOH?

a.

b.

c.

d.

ANSWER: d POINTS: 1 56. Which of the following will not undergo catalytic reduction in the presence of H2 and metal catalyst? a. b.

c.

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d.

Page 15

Chapter 08 - Alkenes: Reactions and Synthesis

ANSWER: b POINTS: 1 57. Hydrogenation reactions use a metal catalyst as depicted below.

Which of the following does not occur during such a reaction? a. The metal provides a surface upon molecules can adsorb. b. π bonds are involved in the interaction between the metal and the unsaturated hydrocarbon. c. Molecular hydrogen attacks the topside of the adsorbed hydrocarbon. d. Hydrogen atoms are transferred to the carbon atoms of the multiple bond one at a time.

ANSWER: c POINTS: 1 58. In the formation of an addition polymer such as polyvinylchloride, a. the monomers must have at least one degree of unsaturation. b. the formation occurs via a polar reaction. c. the products have the same degree of complexity as biopolymers. d. the resulting polymer generally contains two types of monomer.

ANSWER: a POINTS: 1 59. Consider partial hydrogenation of the following substance.

Which of the following describes this reaction? a. 4 moles of hydrogen would required. b. The product would contain trans double bond(s). c. The product would be saturated.

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Chapter 08 - Alkenes: Reactions and Synthesis d. The product would lower blood cholesterol. e. All of the above would occur.

ANSWER: b POINTS: 1 60. Draw the mechanism of the bromination of 3-hexene in the presence of water via bromohydrin formation.

ANSWER:

POINTS: 1 61. Draw the mechanism of the hydration of 2-methyl-propene.

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Chapter 08 - Alkenes: Reactions and Synthesis ANSWER:

POINTS: 1 62. Draw the mechanism of the epoxidation of 2-butene using a halohydrin.

ANSWER:

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Chapter 08 - Alkenes: Reactions and Synthesis POINTS: 1

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis Drawing Instructions: Draw structures corresponding to each of the given names.

1. Draw: ethynylcyclopropane ANSWER: POINTS: 1 2. Draw: 3,10-dimethyl-6-sec-butylcyclodecyne ANSWER:

POINTS: 1 3. Draw: 4-bromo-3,3-dimethyl-1-hexen-5-yne ANSWER:

POINTS: 1 4. Draw: acetylene ANSWER: POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

5. Name:

ANSWER: 4-methyl-2-heptyne POINTS: 1 6. Name:

ANSWER: 1-ethynyl-2-methylcyclopentane POINTS: 1 Exhibit 9-1 The compound below has been isolated from the safflower plant. Consider its structure to answer the following question(s).

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis

7. Refer to Exhibit 9-1. What is the molecular formula for this natural product?

ANSWER: C13H10 POINTS: 1 8. Refer to Exhibit 9-1. What is the degree of unsaturation for this compound?

ANSWER: We can arrive at the degree of unsaturation for a structure in two ways. Since we know that the degree of unsaturation is the number of rings and/or multiple bonds in a compound, we can simply count them. There are three double bonds (3 degrees) and three triple bonds (six degrees), so the degree of unsaturation is 9.

We can verify this by using the molecular formula, C13H10, to calculate a degree of unsaturation. The saturated 13-carbon compound should have the base formula C13H28, so (28 − 10) ÷ 2 = 18 ÷ 2 = 9.

POINTS: 1 9. Refer to Exhibit 9-1. Assign E or Z configuration to each of the double bonds in the compound.

ANSWER:

POINTS: 1 10. Refer to Exhibit 9-1. Provide the name for this unusual natural product.

ANSWER: (3E, 11E)-1,3,11-tridecatrien-5,7,9-triyne or (3E,11E)-trideca-1,3,11-trien-5,7,9-triyne POINTS: 1 Exhibit 9-2 To answer the following question(s), consider the information below: The heat of hydrogenation of 1,4-pentadiene is 254.4 kJ/mol. The heat of hydrogenation of 1-pentyne is approximately 292.9 kJ/mol.

11. Refer to Exhibit 9-2. Which structure is more stable, 1,4-pentadiene or 1-pentyne? Briefly explain your answer.

ANSWER: 1,4-pentadiene The heat of hydrogenation is the amount of energy released when hydrogen adds to the multiple carbon-tocarbon bonds in a compound and converts it to the corresponding saturated compound. The larger the value, the more energy that is released during the hydrogenation. Both of these substances, end up as pentane upon complete hydrogenation. That is, they end up at the same final energy level. Less energy is released by 1,4pentadiene, therefore, this reactant has less energy to start with or is more stable than 1-pentyne.

POINTS: 1 12. Refer to Exhibit 9-2. Show the energy differences between the hydrogenation of 1,4-pentadiene and 1-pentyne on a reaction energy diagram.

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis ANSWER:

POINTS: 1 Exhibit 9-3 Consider an unknown with the molecular formula C4H6.

13. Refer to Exhibit 9-3. How many degrees of unsaturation are present?

ANSWER: A saturated four carbon compound has the formula C4H10. Therefore this compound has (10 − 6) ÷ 2 = 2 degrees of unsaturation.

POINTS: 1 14. Refer to Exhibit 9-3. Draw six of the nine possible structures with this formula.

ANSWER:

POINTS: 1 Exhibit 9-4 Predict the product of each reaction below. Be sure to indicate stereochemistry when appropriate.

15.

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis ANSWER:

POINTS: 1 16.

ANSWER:

POINTS: 1

17.

ANSWER:

POINTS: 1

18.

ANSWER:

POINTS: 1

19.

ANSWER:

POINTS: 1 20.

ANSWER:

POINTS: 1

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis 21.

ANSWER:

POINTS: 1 Exhibit 9-5 To answer the following question(s) consider this reaction:

22. Refer to Exhibit 9-5. On the structures provided, draw arrows showing the electron flow for the reaction mechanism for the electrophilic addition of hydrogen bromide to 1-hexyne.

ANSWER:

POINTS: 1 23. Refer to Exhibit 9-5. The key intermediate in the electrophilic addition reaction shown above is called: a. an acetylenic carbocation b. a vinylic bromide c. a secondary carbocation d. a vinylic carbocation ANSWER: d POINTS: 1 Exhibit 9-6 Select the best reagent or sequence of reagents from the list provided which would best accomplish each transformation below. Place the letter corresponding to the reagent(s) in the blank to the left of each reaction. Each transformation requires only one answer. a. H2SO4, H2O, HgSO4 b. H2, Lindlar Cengage Learning Testing, Powered by Cognero

e.

1. Br2, CCl4 2. 2 NaNH2, NH3 Page 5

Chapter 09 - Alkynes: An Introduction to Organic Synthesis c. 1.

2. d. 1.

2.

BH3, THF H2O2 NaNH2, NH3 CH3CH2I

f.

1. HCl 2. NaNH2, NH3 g. Li/NH3 h. 1. NaNH2, NH3 2. CH3CH2CH2I

24. _____ ANSWER: g POINTS: 1

25. _____

ANSWER: a POINTS: 1 26. _____ ANSWER: h POINTS: 1

27. _____ ANSWER: e POINTS: 1 Exhibit 9-7 Show all reagents and intermediates necessary to carry out the following conversions.

28.

ANSWER:

POINTS: 1

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis 29.

ANSWER:

POINTS: 1 Exhibit 9-8 Using acetylene and any alkyl halides as starting materials, synthesize the following compounds. More than one step may be required. Show all reagents and all intermediate compounds in your synthetic scheme.

30.

ANSWER:

POINTS: 1

31.

ANSWER:

POINTS: 1

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis 32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1

34.

ANSWER:

POINTS: 1 35. Draw the structure(s) of the product(s) formed when the following substance is treated with NaNH2/NH3 and 1bromo-3-methylhexane.

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis

ANSWER: No reaction occurs since this species is an internal alkyne. A terminal alkyne is needed to react under the given conditions.

POINTS: 1 36. Consider the three structures shown below.

Explain why Structure C is unstable ANSWER: The hybridization on the carbon atoms of the triple bond is sp and produces a bond angle of 180°. This dictates a linear geometry about both carbon atoms of the triple bond. In Structure C the internal bond angles between the sp hybridized carbon atoms and the adjacent sp3 carbon atoms of the ring is much less than 180° (approximately 120°). This compression of the bond angle produces too much strain for the ring to be stable.

Puckering of the ring cannot relieve the strain to the degree that is does for Structures A and B. The linear geometry is required to allow for the appropriate orbital overlap to form the sigma bonds between the sp carbon atoms and the sp3 carbon atoms.

POINTS: 1 37. Draw the skeletal formula for the reactant and product that correspond to the enol formed in a hydroboration/oxidation reaction. Atoms other than carbon and hydrogen are labeled.

ANSWER:

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis POINTS: 1 38. Specify the hybridization on the carbons shown in the following structure.

a. 1: sp3 2: sp2

b. 1: sp2 2: sp

c. 1: sp3 2: sp

d. 1: sp2 2: sp

ANSWER: c POINTS: 1 39. Consider the following compound.

This compound a. is weakly acidic. b. could produce a single carboxylic acid upon oxidation. c. would produce a high yield of product upon oxidative cleavage. d. yields an alkene upon treatment with hydrogen in the presence of a Lindlar catalyst. e. None of the above characterize this compound ANSWER: d POINTS: 1 40. What is the term given to the following species?

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis a. carbene c. acetylide anion ANSWER: c POINTS: 1

b. carbanion d. carbocation

41. The following species

a. formed when ethyne is treated with a strong base. b. is the conjugate base of acetylene. c. could react with bromoethane to form 1-butyne. d. could react with bromocyclopentane to form an internal alkyne. e. a and b f. a, b, and c g. a, b, and d h. all of the above ANSWER: f POINTS: 1 42. The term retrosynthesis refers to syntheses developed: a. starting with simple molecules. b. devised by examining the product. c. working forward and backward from intermediates. d. beginning by examining the reactant available.

ANSWER: b POINTS: 1 43. An acetylide anion is more stable than either a vinylic or alkyl anion due to a. the stabilization of the negative charge in orbitals with higher s character. b. the greater distance of the negative charge from the nucleus of the carbon atom. c. the greater stability of an sp hybrid orbital. d. the larger number of π electrons.

ANSWER: a POINTS: 1 44. What is the skeletal formula for the product of the reaction when the following is treated with BH3 in THF followed

by H2O2?

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis

a.

b.

c.

d.

ANSWER: d POINTS: 1 45. Which of the following alkynes will produce a single product upon mercury(II)-catalyzed hydration? a. 2-methyl-5-ethyl-3-heptyne b. 3,3-dimethyl-1-heptyne c. 3-hexyne d. 2-hexyne e. Both 3,3-dimethyl-1-heptyne and 3-hexyne ANSWER: c POINTS: 1 46. When the following substance is treated with excess HBr,

The product of the reaction is a.

b.

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis

c.

d. A mixture of a and b would be produced. e. A mixture of a, b, and c would be produced.

ANSWER: c POINTS: 1 47. When the following substance is treated with 1 equivalent of Br2,

The product of the reaction is a.

b.

c.

d. A mixture of a and b would be produced.

ANSWER: a POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis 48. Draw the mechanism of the bromination of acetylene.

ANSWER:

POINTS: 1 49. Draw the mechanism of the lithium/ammonia reduction of 2-butyne to produce trans-2-butene.

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis ANSWER:

POINTS: 1 50. Draw the mechanism of the alkylation of acetylide anion with chloroethane to give 1-butyne.

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Chapter 09 - Alkynes: An Introduction to Organic Synthesis ANSWER:

POINTS: 1

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Page 16

Chapter 10 - Organohalides Exhibit 10-1 Give the name for each of the following alkyl halides.

1. Name:

ANSWER: 1,1,2-trichloroethane POINTS: 1 2. Name: CHI3 ANSWER: triiodomethane POINTS: 1 3. Name:

, Halothane ANSWER: 2-bromo-2-chloro-1,1,1-trifluoroethane POINTS: 1 4. Name: CH2Cl2 ANSWER: dichloromethane POINTS: 1 Drawing Instructions: Draw structures corresponding to each of the given names.

5. Draw: 3-iodopropene ANSWER: POINTS: 1 6. Draw: trans-1-chloro-3-sec-butylcyclohexane ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 10 - Organohalides 7. Draw: 1,2-dichloro-1,1,2,2-tetrafluoroethane (Cryofluorane) ANSWER:

POINTS: 1 8. Draw: (S)-2-bromobutane ANSWER:

POINTS: 1 Exhibit 10-2 To answer the following question(s) consider the reaction below:

9. Refer to Exhibit 10-2.

When propylbenzene reacts with tert-butylhypochlorite three monochlorinated products are a. formed in the ratios indicated. Calculate a reactivity order for each type of hydrogen atom in propylbenzene.

The reaction of propylbenzene with tert-butylhypochlorite proceeds by a radical substitution b. pathway. Draw the structure of the radical intermediate leading to each product.

Based on your answers to the two questions above explain why (1-chloropropyl)benzene is the c. major product of this reaction.

ANSWER: a. Ar − CH2− = 65% ÷ 2 = 32.5% product ÷ 3.3 = 9.85

−CH2- = 25% ÷ 2 = 12.5% product ÷ 3.3 = 3.8 −Ch3- = 10% ÷ 3 = 3.3% product ÷ 3.3 = 1 Therefore the Relative Reactivity for CH3:CH2:ArCH2 is 1 : 3.8 : 9.85.

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Chapter 10 - Organohalides

b.

In part a, we calculated that the hydrogens next to the aryl group, the benzylic hydrogens, are about 2.5 times more reactive than the methylene hydrogens and about 10 times more reactive than the methyl hydrogens. This means that c. the benzylic radical is more stable than a 2° or 1° radical. The reason the benzylic radical is more stable is because it is resonance stabilized. The resonance forms for the benzylic radical are:

The 2° or 1° radicals have no other resonance forms.

POINTS: 1 10. The original question has been combined with question #9 as part b. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 11. The original question has been combined with question #9 as part c. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 12. Refer to Exhibit 10-2. Are any of the products of this reaction chiral? If so draw them and label the chirality center with an asterisk.

ANSWER: Yes!

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Chapter 10 - Organohalides

POINTS: 1 13. Refer to Exhibit 10-2. Will the product mixture of this reaction display optical activity?

ANSWER: No, the radical intermediates for the chiral products are symmetrical and the transition states leading to chiral products are of equal energy so racemic mixtures are formed.

POINTS: 1 14. In the discussion on relative reactivity of alkane hydrogens towards radical chlorination, we showed that the relative rate of 2° to 1° hydrogen atom abstraction is 3.5 : 1 for butane. Calculate the relative amounts of 1-chloropropane and 2chloropropane obtained by the radical chlorination of propane, using this relative rate of reactivity.

ANSWER: Propane has six 1° hydrogens, so 6 × 1 = 6; and two 2° hydrogens, so 2 × 3.5 = 7. The percentage of 1chloropropane is therefore 6/13 or 46% and the percentage of 2-chloropropane is 7/13 or 54%.

POINTS: 1 Exhibit 10-3 Consider the reaction below to answer the following question(s).

15. Refer to Exhibit 10-3. Draw all the monochlorination products of methylcyclopentane (ignore stereoisomers).

ANSWER:

POINTS: 1 16. Refer to Exhibit 10-3. Label the chirality centers in the monochlorination products of methylcyclopentane with an asterisk.

ANSWER:

POINTS: 1 17. Refer to Exhibit 10-3. Tell whether radical chlorination of methylcyclopentane is an oxidation or a reduction process.

ANSWER: In the radical chlorination of methylcyclopentane a C−H bond is broken and a C−Cl bond is formed, so the process is an organic oxidation.

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Chapter 10 - Organohalides Exhibit 10-4 Consider the reaction below to answer the following question(s).

18. Refer to Exhibit 10-4. Place asterisks(*) at all allylic positions in compound A. ANSWER:

POINTS: 1 19. Refer to Exhibit 10-4. Draw the resonance forms of the allylic radical intermediate that accounts for the formation of B and C. ANSWER: POINTS: 1 20. Refer to Exhibit 10-4. D and E, below, are minor products in this reaction. Explain why.

ANSWER: Products D and E are formed from the other allylic radical, whose resonance forms are shown below.

This allylic radical forms more slowly than the radical for products A and B because it has some 1° character, and thus, is less stable. The allylic radical intermediate for products A and B is 2° allylic.

POINTS: 1 Exhibit 10-5 Choose the best reagent or sequence of reagents from the list provided below for carrying out the following transformations. Place the letter of your response to the left of the reaction. a. PBr3 b. HCl (gas), ether c. 1.

2.

d. e.

Mg, ether D2O

SOCl2, pyridine HBr (gas), ether f. 1.

2.

Mg, ether NBS

21. _____ ANSWER: e POINTS: 1

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Chapter 10 - Organohalides

22. _____ ANSWER: c POINTS: 1

23. _____ ANSWER: d POINTS: 1 Exhibit 10-6 Provide structures for the reactants, intermediates, or products, as indicated, in the following reactions. Draw the structures in the boxes provided.

24.

ANSWER:

POINTS: 1

25.

ANSWER:

POINTS: 1

26.

ANSWER:

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Page 6

Chapter 10 - Organohalides POINTS: 1

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1

30.

ANSWER:

POINTS: 1

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Chapter 10 - Organohalides

31.

ANSWER:

POINTS: 1 Exhibit 10-7 Propose a synthesis of each of the following compounds from the given starting material and any inorganic reagents necessary.

32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1 34. Identify the reagents a and b in the following scheme.

ANSWER: a = NBS, hv b = KOH, ethanol POINTS: 1 Exhibit 10-8 Label each transformation below as: a. b. c.

an oxidation a reduction neither an oxidation nor a reduction

Place the letter corresponding to the correct answer in the blank to the left of the transformation.

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Chapter 10 - Organohalides

35. _____ ANSWER: a POINTS: 1

36. _____

ANSWER: c POINTS: 1

37. _____ ANSWER: b POINTS: 1

38. _____ ANSWER: a POINTS: 1 39. Rank the following compounds in order of increasing oxidation level. Place the number rank (1 = lowest; 4 = highest) in the blank below the structure.

ANSWER:

POINTS: 1 40. The following compound has been used as a less-ozone-depleting replacement for Freons. Name this compound.

CHF2CF3 ANSWER: 1,1,1,2,2-pentafluoroethane POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 9

Chapter 10 - Organohalides 41. Draw the skeletal structure of the product if the following alcohol were treated with HF in pyridine. Atoms other than carbon and hydrogen are labeled.

ANSWER:

with stereochemistry

POINTS: 1 42. Predict the product of the following reaction.

ANSWER:

POINTS: 1 43. The following product is a simplified penicillin analog. Draw the structure of the missing reactant represented by the “?” when the reaction is carried out in the presence of a palladium catalyst.

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Page 10

Chapter 10 - Organohalides ANSWER:

POINTS: 1 44. Draw the condensed formula for the product formed when the following substance is treated with NBS using carbon tetrachloride as the solvent.

ANSWER: Three allylic positions produce:

POINTS: 1 45. Which of the following is the least likely to form a Grignard reagent? a.

b.

c.

d.

e. All will form Grignard reagents.

ANSWER: a POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Organohalides 46. Which of the following pairs are not alternative reagents to prepare the indicated products? a. HCl, SOCl2 - preparation of alkyl b. HBr, NBS - preparation of alkyl bromides from alkenes chlorides from alcohols c. Br2 and HBr - preparation of alkyl d. diorganocopper reagents and organopalladium compounds forming new carbon-to-carbon bonds bromides from alkanes ANSWER: c POINTS: 1 47. Which of the following is a valid comparison of organocopper and organopalladium reagents? a. Both work equally well with alkyl compounds. b. Palladium compounds are more toxic than copper compounds. c. Organopalladium reactions require a larger amount of the metal. d. Organopalladium compounds are particularly useful in from biaryl compounds.

ANSWER: d POINTS: 1 48. Consider the following reaction.

This reaction is classified as: a. an oxidation. b. a reduction. c. neither an oxidation nor a reduction.

ANSWER: a POINTS: 1 49. Consider the following reaction.

This reaction is classified as: a. an oxidation. b. a reduction. c. neither an oxidation nor a reduction.

ANSWER: c POINTS: 1 50. Which of the following correctly orders oxidation levels? a. b. c.

d.

ANSWER: c Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Organohalides POINTS: 1 51. Draw the mechanism of the allylic bromination of 1-cyclopentene.

ANSWER:

POINTS: 1 52. Draw the mechanism of the coupling reaction between lithium diethylcopper and cis-1-chloro-1-octene.

ANSWER:

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Chapter 10 - Organohalides POINTS: 1 53. Draw the mechanism of production of trans-3-decene from trans-1-fluoro-1-heptene.

ANSWER:

POINTS: 1

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations Exhibit 11-1 Circle your response in each set below.

1. Circle the least reactive compound in an SN2 reaction.

ANSWER:

POINTS: 1 2. Circle the best leaving group in an elimination reaction.

ANSWER:

POINTS: 1 3. Circle the best nucleophile in a substitution reaction at a primary carbon.

ANSWER: POINTS: 1 4. Circle the least reactive compound in an SN1 reaction.

ANSWER:

POINTS: 1 5. Circle the best solvent for an SN2 reaction.

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Page 1

Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1 Exhibit 11-2 Consider the pair of reactions below to answer the following question(s). a. or b.

6. Refer to Exhibit 11-2. a. Which reaction would be predicted to be faster? b. Explain your answer to the question in part a.

ANSWER: a. b −SH is a better nucleophile than −OH because nucleophilicity usually increases in going b. down a column of the periodic table and sulfur is below oxygen in group 6A. POINTS: 1 7. The original question has been combined with question #6 as part b. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 8. Refer to Exhibit 11-2. The mechanism for these reactions is: a. SN1 b. SN2 c. E1 d. E2 ANSWER: b POINTS: 1 Exhibit 11-3 Consider the pair of reactions below to answer the following question(s). a. or

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations b.

9. Refer to Exhibit 11-3. a. Which reaction above is faster? b. Explain your answer to the question in part a.

ANSWER: a. a

The conversion of tert-butyl chloride to tert-butyl alcohol proceeds faster in 70% H2O/30% CH3OH than in 50% H2O/50% acetone because SN1 reactions b. are faster in more polar solvents. Acetone (dielectric polarization, P = 20.7) is less polar than CH3OH (P = 33.6), and the percentage of H2O is higher in the water/methanol system.

POINTS: 1 10. The original question has been combined with question #9 as part b. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 11. Refer to Exhibit 11-3. The kinetics of these reactions are: a. second-order b. first-order in nucleophile c. not measurable d. first-order in alkyl halide ANSWER: d POINTS: 1 Exhibit 11-4 Consider the pair of reactions below to answer the following question(s). a. or b.

12. Refer to Exhibit 11-4. The alkyl bromide starting materials in these reactions are classified as: a. 3° b. 2° c. 1° d. 4° ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations 13. Refer to Exhibit 11-4. The solvent in these reactions is: a. nonpolar aprotic b. polar aprotic c. polar protic d. nonpolar protic ANSWER: b POINTS: 1 14. Refer to Exhibit 11-4. The nucleophile in these reactions is: a. K+ b. alkyl group c. Br− d. I− ANSWER: d POINTS: 1 15. Refer to Exhibit 11-4. Which reaction is faster?

ANSWER: b POINTS: 1 16. Refer to Exhibit 11-4. The mechanism for these reactions is: a. SN2 b. E2 c. SN1 d. E1 ANSWER: a POINTS: 1 Exhibit 11-5 Consider the pair of reactions below to answer the following question(s). a.

or

b.

17. Refer to Exhibit 11-5. a. Which reaction above is faster? b. Explain your answer to the question in part a.

ANSWER: a. b Tosylate anion is a better leaving group than amide ion; tosylate is the weak conjugate base of b. a strong acid, while amide is the strong conjugate base of a weak acid.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations POINTS: 1 18. The original question has been combined with question #17 as part b. This placeholder question is here to maintain the integrity of the numbering system between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 19. Refer to Exhibit 11-5. Doubling the concentration of potassium hydroxide in these reactions: a. causes the reaction mechanism to change b. halves the rate of reaction c. has no effect on the rate of reaction d. doubles the rate of reaction ANSWER: d POINTS: 1 20. Refer to Exhibit 11-5. The mechanism for these reactions is: a. SN2 b. E2 c. SN1 d. E1 ANSWER: b POINTS: 1 Exhibit 11-6 Consider the reaction below to answer the following question(s).

21. Refer to Exhibit 11-6. The substrate in the reaction is: ANSWER: A POINTS: 1 22. Refer to Exhibit 11-6. Compound B is the: a. SN2 product b. SN1 product c. E2 product d. E1 product ANSWER: b POINTS: 1 23. Refer to Exhibit 11-6. Compound C is the: Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations a. SN2 product b. SN1 product c. E2 product d. E1 product ANSWER: d POINTS: 1 24. Refer to Exhibit 11-6. Write the complete reaction mechanism for the formation of Compound C in this reaction.

ANSWER:

POINTS: 1 Exhibit 11-7 To answer the following question(s) consider the data below: Reaction of bromomethane with sodium hydroxide in water forms methanol. If sodium iodide is added to the reaction mixture, the rate of methanol formation is dramatically increased (i.e. sodium iodide is a catalyst).

25. Refer to Exhibit 11-7. The mechanism involved in the reaction of CH3Br and NaOH is: a. SN1 b. SN2 c. E1 d. E2 ANSWER: b POINTS: 1 26. Refer to Exhibit 11-7. Write a reaction pathway that accounts for the effect of added NaI. ANSWER: POINTS: 1 27. Refer to Exhibit 11-7. Draw a reaction energy diagram showing the two different reaction pathways (i.e. catalyzed and uncatalyzed). Indicate structures for all energy minima in the diagram.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1 28. Refer to Exhibit 11-7. Explain why adding NaI increases the reaction rate.

ANSWER: Iodide is a better nucleophile than hydroxide so we expect reaction between CH3Br and iodide to proceed faster than the reaction between CH3Br and hydroxide. CH3I is a more reactive alkyl halide than CH3Br because iodide is a better leaving group than bromide so when CH3I forms it reacts with hydroxide faster than CH3Br to form CH3OH. The overall result is faster formation of CH3OH from CH3Br when NaI is added.

POINTS: 1 29. Refer to Exhibit 11-7. Would you expect the same catalytic effect on this reaction if you added NaCl instead? Explain your answer.

ANSWER: No. Chloride ion is a poorer nucleophile as well as a poorer leaving group than bromide ion.

POINTS: 1 Exhibit 11-8 Consider the reaction below to answer the following question(s):

30. Refer to Exhibit 11-8. Write the product that results from the electron flow in the reaction, clearly indicating any stereochemistry.

ANSWER:

POINTS: 1 31. Refer to Exhibit 11-8. Draw a Newman projection of the reactive conformation of the starting material.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1 32. Refer to Exhibit 11-8. The mechanism of this reaction is: a. SN1 b. SN2 c. E1 d. E2 ANSWER: d POINTS: 1 Exhibit 11-9 Consider the reaction below to answer the following question(s).

33. Refer to Exhibit 11-9. Write the complete stepwise mechanism for this reaction. Clearly show the formation of both products. Show all electron flow with arrows and draw all intermediate structures.

ANSWER:

POINTS: 1 34. Refer to Exhibit 11-9. This reaction obeys a rate law of the form: a.

rate = k[RCl][CH3CH2OH]

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations b. rate = k[RCl][H2O][CH3CH2OH] c. rate = k[RCl] d. rate = k[RCl][H2O] ANSWER: c POINTS: 1 Exhibit 11-10 Draw the structure of the major organic product(s) for each of the following reactions. Indicate the stereochemistry for each reaction when appropriate.

35.

ANSWER:

POINTS: 1

36.

ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1

38.

ANSWER:

POINTS: 1

39.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1

40.

ANSWER:

POINTS: 1

41.

ANSWER: POINTS: 1

42.

ANSWER: POINTS: 1

43.

ANSWER:

POINTS: 1

44.

ANSWER:

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations POINTS: 1 Exhibit 11-11 Consider the reaction below to answer the following question(s).

45. Refer to Exhibit 11-11. List all chirality centers in the starting material by number.

ANSWER: Carbon 2 is a chirality center.

POINTS: 1 46. Refer to Exhibit 11-11. As indicated by the carbon numbers, during the course of this reaction the bond between the nitrogen atom and carbon 3 is broken and a bond between the nitrogen atom and carbon 2 is formed. On the structures provided below, draw arrows showing electron flow for the mechanism that accounts for these bonding changes.

ANSWER:

POINTS: 1 47. Refer to Exhibit 11-11. If the absolute configuration at carbon 2 in the starting material is R, what is the absolute configuration at carbon 2 in the product?

ANSWER: The absolute configuration at carbon 2 in the product is S because an intramolecular SN2 reaction occurred in the first step of the reaction to invert the configuration at that carbon.

POINTS: 1 Exhibit 11-12 For each substrate below, choose which reaction type is favored. Place the letter of the reaction type in the blank to the left of the substrate. a. b. c. d.

SN1 SN2 E1 E2

48. ______ ANSWER: b POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

49. ______ ANSWER: d POINTS: 1 50. For the substitution reaction:

Draw the structure present at the point indicated by the asterisk (*) on the following energy diagram. Identify the reactant as R or S or neither.

ANSWER:

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations The reactant is the R enantiomer.

POINTS: 1 51. For the substitution reaction:

Draw the structure present at the point indicated by the asterisk (*) on the following energy diagram.

ANSWER:

POINTS: 1 52. For the elimination reaction:

Draw the structure present at the point indicated by the asterisk (*) on the following energy diagram.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

ANSWER:

POINTS: 1 53. For the elimination reaction:

Draw the structure present at the point indicated by the asterisk (*) on the following energy diagram

ANSWER:

POINTS: 1 54. Consider the two lines shown on the energy diagram below.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

In an SN2 reaction, these compare the effect of the: a. substrate. b. nucleophile. c. leaving group. d. solvent. e. nucleophile or solvent f. substrate or leaving group ANSWER: f POINTS: 1 55. Consider the two lines shown on the energy diagram below.

In an SN2 reaction, a. the upper left line could represent Cl– and the lower left line I–. b. the upper left line could represent OH– and the lower left line CH3COO– . c. the upper left line could represent H2O and the lower left line H2S. d. the upper left line could represent (CH3)2NH and the lower left line (CH3)2N–.

ANSWER: b POINTS: 1 56. Consider the two energy diagrams below for an SN2 reaction.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations AB

Which of the following is correct?: a. The change in the variable shown in A affects the transition state while that of B affects the reactant. b. The change in the variable shown in A affects the reactant while that of B affects the transition state. c. The changes in the variables shown in both A and B affect the transition state. d. The changes in the variables shown in both A and B affect the reactants.

ANSWER: a POINTS: 1 57. Consider the two lines shown on the energy diagram below.

In an SN2 reaction, these lines could compare the effect of which of the following solvents? a. CH3OH and CH3CH2OH b. CH3OH and H2O c. CH3OH and CH3CN

d. DMSO and CH3CN

ANSWER: c POINTS: 1 58. Consider the two energy diagrams labeled A and B given below.

AB

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

Which of the following is correct with respect to these diagrams? a. A represents an SN2 reaction. b. B represents an SN1 reaction. c. B represents an SN2 reaction. d. A represents an SN1 reaction. e. both a and b f. both c and d ANSWER: f POINTS: 1 59. Consider the substitution reaction:

The product(s) of the reaction is(are): a. R enantiomer

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b. S enantiomer

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

c. An equimolar mixture of a and b. e. S enantiomer

d. A mixture of the major product a with the minor product b. f. R enantiomer

g. An equimolar mixture of e and f ANSWER: c POINTS: 1

h. A mixture of the major product f with the minor product e.

60. Consider the substitution reaction:

The product(s) of the reaction is(are): a. S enantiomer

c.

A equimolar mixture of a and b. e. R enantiomer

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b. R enantiomer

d.

A mixture of the major product a with the minor product b. f. S enantiomer

Page 18

Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations g.

A equimolar mixture of e and f.

ANSWER: e POINTS: 1

h.

A mixture of the major product f with the minor product e.

61. Consider the two energy diagrams given below.

AB

Which of the following is correct with respect to these diagrams? a. A could represent an E2 reaction. b. B could represent an E1 reaction. c. A could represent an E1cb reaction.

ANSWER: c POINTS: 1 62. Consider the elimination reaction:

The product(s) of the reaction is(are): a.

b.

c. An equimolar mixture of a and b. d. A mixture of the major product a with the minor product b. e. A mixture of the major product b with the minor product a.

ANSWER: d POINTS: 1 63. Consider the elimination reaction: Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

The product(s) of the reaction is(are): a. b. c. An equimolar mixture of a and b. d. A mixture of the major product a with the minor product b. e. A mixture of the major product b with the minor product a.

ANSWER: d POINTS: 1 64. Nucleophilic substitution and elimination reactions often compete with one another. In general, which of the following would enhance the rate of the E2 reaction over the SN2 reaction? a. use of a weak nucleophile b. branching at the α or β carbon c. use of a weak base d. All of the above would enhance the E2 rate.

ANSWER: b POINTS: 1 65. Draw the mechanism of the SN2 reaction involving (R)-2-chloro-butane, to produce (S)-2-butanol.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1 66. Draw the mechanism of the SN1 reaction of 2-chloro-2-methyl-butane with H2O.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1 67. Draw the E2 mechanism of the reaction involving the reduction of 2-chloro-butane to 2-butene, in the presence of KOH in ethanol.

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Chapter 11 - Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations ANSWER:

POINTS: 1

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy Exhibit 12-1 Select the most reasonable formula for the compounds with the following mass spectral data.

1. Refer to Exhibit 12-1. M+ at m/z = 101 with a minor M+1 peak a. C5H6Cl b. C5H12N2 c. C6H15N d. C9H12O ANSWER: c POINTS: 1 2. Refer to Exhibit 12-1. M+ m/z = 136 and M+ at m/z = 138 of approximately equal intensity a. C6H13OCl b. C4H9Br c. C10H16 d. C9H12O ANSWER: b POINTS: 1 3. The sugar glucose contains C, H, and O and has a mass of 180.0634 amu as determined by high-resolution mass spectrometry. Glucose contains an equal number of carbon and oxygen atoms. What is the molecular formula of glucose?

(1H = 1.00783 amu, 12C = 12.00000 amu, 16O = 15.99491 amu) ANSWER: Since the number of carbons and oxygens is equal, we can add the masses of C and O and then divide 180.0634 by that number. The remainder should be the number of hydrogens.

12.0000 + 15.99491 = 27.99491 180.0634 ÷ 27.99491 = 6 carbons and 6 oxygens with ˜12 remaining So, the formula for glucose should be C6H12O6. We can confirm this by doing the math.

(6 × 12.0000) + (12 × 1.00783) + (6 × 15.99491) = 180.0634 POINTS: 1 Exhibit 12-2 Use the data below to answer the following question(s).

Loratidine is the active ingredient in the antihistamine Claritin®. Mass spectral analysis of loratidine shows M+ at m/z = 382 and M+ at m/z = 384 in an approximate ratio of 3:1 in intensity.

4. Refer to Exhibit 12-2. The mass spectral data indicates that loratidine contains: a. fluorine b. chlorine c. bromine d. iodine ANSWER: b POINTS: 1 5. Refer to Exhibit 12-2. Loratidine is known to contain nitrogen. What is the minimum number of nitrogens in loratidine?

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy ANSWER: Since the molecular ions are even numbers, loratidine must have an even number of nitrogen atoms.

Therefore, the minimum number of nitrogen atoms in loratidine is two.

POINTS: 1 6. Which of the following statements best describes the base peak in a mass spectrum? a. The peak from the most stable radical. b. The peak from the species that has the isotope with the highest atomic number. c. The peak of highest intensity. d. The peak from the molecule minus an electron.

ANSWER: c POINTS: 1 Exhibit 12-3 For each compound below, what are the masses of the charged fragments produced by the indicated cleavage pathways?

7. Refer to Exhibit 12-3. Alpha cleavage of ANSWER: This compound, phenylacetone, has the formula C9H10O and a nominal mass of 134. Alpha cleavage will result in charged fragments at m/z = 43 and m/z = 119.

POINTS: 1

8. Refer to Exhibit 12-3. Alpha cleavage and dehydration of ANSWER: This compound, 4-methyl-3-penten-1-ol, has the formula C6H12O and a nominal mass of 100. Alpha cleavage will result in a charged fragment at m/z = 31. Dehydration (M+ - 18) will result in a charged fragment at m/z = 82.

POINTS: 1 Exhibit 12-4 The following question(s) refer to the mass spectrum shown below.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 9. Refer to Exhibit 12-4. This compound contains C, H, and one other atom. Identify the other atom from the mass spectrum and explain your reasoning.

ANSWER: There are two molecular ions identified in this spectrum of approximately equal intensity. This is characteristic for the presence of a bromine atom (79Br = 50.7% and 81Br = 49.3%).

POINTS: 1 10. Refer to Exhibit 12-4. Calculate a possible molecular formula for this compound.

ANSWER: The M+ = 156 is due to the 79Br, so the carbon and hydrogen portion of the molecule is 156 − 79 = 77. This mass corresponds to C6H5. A possible formula for this compound is C6H5Br. POINTS: 1 11. Refer to Exhibit 12-4. Propose a structure for this compound.

ANSWER: C6H5Br contains four degrees of unsaturation, which is consistent with a benzene ring. A compatible structure is bromobenzene.

POINTS: 1 Exhibit 12-5 Refer to the mass spectrum of 2-methylbutane shown below to answer the following question(s).

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 12. Refer to Exhibit 12-5. What peak represents M+?

ANSWER: The peak at m/z = 72 represents M+.

POINTS: 1 13. Refer to Exhibit 12-5. What peak represents the base peak?

ANSWER: The peak at m/z = 43 is the base peak.

POINTS: 1 14. Refer to Exhibit 12-5. Propose structures for fragment ions at m/z = 57, 43, and 29.

ANSWER: m/z structure

57 43 29

+CH2CH3

POINTS: 1 15. Refer to Exhibit 12-5. Which peak represents the parent peak?

ANSWER: The peak at m/z = 72 represents the parent peak or the molecular ion, M+.

POINTS: 1 16. Below is the mass spectrum of an unknown hydrocarbon. In addition, this hydrocarbon shows characteristic absorption at 2100 cm−1 in its IR spectrum. Give the structure of this unknown.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ ANSWER: The formula weight of 54 corresponds to a molecular formula of C4H6, which has two degrees of unsaturation.

Possible structures for this formula are:

The IR data is consistent with an internal alkyne, so the compound is 2-butyne.

POINTS: 1 17. Could you distinguish 2,2-dimethylbutane from 2-methylpentane using mass spectrometry?

ANSWER: Yes, although M+ at m/z = 86 for both compounds is identical, the fragmentation patterns of the two compounds should show two major differences. We would expect 2,2-dimethylbutane to have significant peaks at m/z = 57, corresponding to C4H9, and m/z = 71, corresponding to C5H11. Fragmentation of 2methylpentane should produce significant peaks at m/z = 43, corresponding to C3H7, and m/z = 57, corresponding to C4H9.

POINTS: 1 18. True or false, both mass spectrometry and infrared spectroscopy involve the interaction of molecules with electromagnetic energy?

ANSWER: False POINTS: 1 19. The amount of energy in electromagnetic radiation is related to the frequency and wavelength of the radiation. High energy radiation like gamma rays is of: Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy a. low frequency and short wavelength b. low frequency and long wavelength c. high frequency and short wavelength d. high frequency and long wavelength ANSWER: c POINTS: 1 20. The amount of energy in infrared light corresponds to: the amount of energy needed to promote one electron from a bonding to an antibonding molecular orbital b. the amount of energy needed to "flip" the spin of a 13C or 1H nucleus c. the amount of energy needed to strip a molecule of one electron to generate a cation radical the amount of energy needed to increase certain molecular motions, such as bond vibrations, in d. organic molecules ANSWER: d POINTS: 1 a.

21. Examining the infrared spectrum of a compound allows us to: a. determine the types of functional groups present in the compound b. determine the carbon-hydrogen framework of the compound c. determine the molecular weight of the compound d. determine the nature of the conjugated pi electron system in the compound ANSWER: a POINTS: 1 Exhibit 12-6 MATCH each of the following groups of bond-types to the region of the infrared spectrum in which their absorptions occur. Place the letter of the region in the blank to the left of the bond-type. a. b. c. d.

4000 to 2500 cm−1 2500 to 2000 cm−1 2000 to 1500 cm−1 below 1500 cm−1

22. _____ C−C, C−O, C−N, and C−X single-bond vibrations.

ANSWER: d POINTS: 1 23. _____ C=O, C=N, and C=C bond absorptions.

ANSWER: c POINTS: 1 24. _____ N−H, C−H, and O−H stretching and bending motions.

ANSWER: a POINTS: 1 25. _____ triple bond stretching vibrations.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy ANSWER: b POINTS: 1 26. Cyclohexene and 2-hexyne both have the molecular formula C6H10. How would you use infrared spectroscopy to distinguish between the two compounds?

ANSWER: To use IR spectroscopy to distinguish between isomers, find a strong IR absorption present in one isomer that is absent in the other isomer.

alkene peak at 3020-3100 cm−1 and at 1650-1670 cm−1

no alkene peaks

no alkyne peaks

alkyne peak of medium intensity at 2100-2260 cm−1

POINTS: 1 27. When 2-bromopropane reacts with ethoxide ion, two products are formed; one is the product of SN2 substitution and the other is the product of E2 elimination. Write the structures of both products, and tell how they could be distinguished using IR spectroscopy.

ANSWER: SN2 product E2 product

CH3CH2OCH(CH3)

CH3CH = CH2

no alkene bands

alkene bands at 3020−3100 cm−1 and 1650−1670 cm−1

C−O stretch near 1100 cm−1

no C−O stretch

POINTS: 1 28. Assume you are carrying out the conversion of 1-bromobutane to 1-butanol. How could you use IR spectroscopy to determine when the reaction is complete?

ANSWER: The infrared spectrum of the starting bromide contains only alkyl C−H absorptions between 2850−2960 cm−1 in the 4000−2500 cm−1 region. As the reaction progresses, a broad absorption at 3400−3640 cm−1, due to the O−H stretch, should develop. When the relative intensity of this band no longer increases, the reaction is complete.

POINTS: 1 Exhibit 12-7 At what approximate positions might the compounds below show IR absorptions?

29.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy ANSWER: This compound should show a medium absorption at 2210−2260 cm−1 for the nitrile, a strong absorption at about 1735 cm−1 for the ester carbonyl and a strong absorption at about 1100 cm−1 for the C−O of the ester.

POINTS: 1

30.

ANSWER: This compound should show a characteristic broad absorption at 3500 cm−1 due to the −OH group and a strong absorption at about 1200 cm−1 for the C−O, as well as the usual 1500 and 1600 cm−1 aromatic bands.

POINTS: 1 Exhibit 12-8 MATCH a structure from the list below to the following IR spectra. Place the letter of the structure in the blank provided.

31. _____

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ ANSWER: F POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy 32. _____

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ ANSWER: A POINTS: 1 33. _____

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ ANSWER: D POINTS: 1 34. Predict the m/z position of the parent peak, the peak due to dehydration of the alcohol and α-cleavage of the carbonyl in the mass spectrum of the compound shown below.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy

ANSWER: Parent peak: 116 Dehydration: 98 α-cleavage: 1.) 101 and 2.) 43 (two positions)

POINTS: 1 35. Mass spectrometry and infrared spectroscopy are complementary techniques because: a. Mass spectrometry provides information about the molar mass and formula while infrared spectroscopy helps identify the functional groups in the formula. b. Infrared spectroscopy provides information about the molar mass and formula while mass spectrometry helps identify the functional groups in the formula. c. Mass spectrometry provides information about the carbon-hydrogen framework while infrared spectroscopy helps identify the functional groups in the framework. d. Infrared spectroscopy provides information about the carbon-hydrogen framework while mass spectrometry helps identify the functional groups in the framework.

ANSWER: a POINTS: 1 36. Which of the following provides information about the functional groups present in a sample? a.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy b.

c.

d.

ANSWER: b POINTS: 1 37. Which of the following fragments would produce a signal in the detector of a mass spectrometer? R represents an organic fragment. a. b. c. d. e. both c and d f. b, c, d g. a, b, c, and d ANSWER: e POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 11

Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy 38. In which of the following ranges of the IR region of the spectrum would the following molecule absorb energy?

a. 3400 - 3650 cm-1 b. 2500 - 3100 cm-1 c. 2850 - 2960 cm-1 d. 1050 - 1150 cm-1 e. all except b f. all of the above ANSWER: e POINTS: 1 39. What m/z ratio would expect to find in the mass spectrum of the following compound due to dehydration?

a. 102 b. 87 c. 85 d. 84 e. 59 ANSWER: d POINTS: 1 40. Consider the following spectrum.

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy

The compound used to generate this spectrum might have which of the following molecular formulas? a. C7H16 b. C7H16O c. C7H14 d. C7H12 e. either c or d ANSWER: e POINTS: 1 41. Which of the following regions of the IR spectrum would be significantly different for the following two substances?

a. 4000 - 2850 cm-1

b. 2850 - 2000 cm-1

c. 2000 - 1500 cm-1

d. 1500 - 400 cm-1

ANSWER: d POINTS: 1 42. Consider the following mass spectrum

Which of the following compounds might have been used to produce this spectrum? a. C6H14 b. C6H12 c. C6H10

d. C5H12O

ANSWER: b Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy POINTS: 1 43. In which of the following regions of the electromagnetic spectrum is the frequency lower than in the IR region? a. radio wave b. visible c. ultraviolet d. X ray ANSWER: a POINTS: 1 44. Which of the following regions of the electromagnetic spectrum would have insufficient energy to produce the following energy transition?

a. radio wave c. ultraviolet ANSWER: a POINTS: 1

b. visible d. X ray

45. When the molecular ion of 2,2-dimethylpentane fragments, the positive charge is most likely to stay on: a. the methyl group. b. the ethyl group. c. the propyl group. d. the tert-butyl group.

ANSWER: d POINTS: 1 46. What are the charged fragments produced if 4,5-dimethylhex-1-en-3-ol undergoes alpha cleavage?

ANSWER: The nominal mass of the compound is 120. Alpha cleavage will result in two charged fragments at m/z=49 and m/z=71.

POINTS: 1 47. Ethyl acetate and 2-butene-1,4-diol both have the molecular formula C4H8O2. How would you use infrared spectroscopy to distinguish between the two?

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Chapter 12 - Structure Determination: Mass Spectrometry and Infrared Spectroscopy ANSWER:

POINTS: 1 48. The peak in the IR absorption spectra for the alkyne group in acetylene would be observed in the range of: a. 2100-2260 cm-1 b. 2370-2650 cm-1 c. 2700-2830 cm-1 d. 2900-3210 cm-1 ANSWER: a POINTS: 1 49. Both methylcycloheptane and ethylcyclohexane have the same molecular mass. Explain how it is possible to distinguish between the two compounds by studying and comparing both their mass spectra.

ANSWER: Molecular ions are shown at M+=113 in both mass spectra, which corresponds to C8H18. However, both of the compounds will have base peaks at different positions in the mass spectra. Ethylcyclohexane will have its base peak at m/z=84, corresponding to the loss of an ethyl group (29 mass units), whereas methylcycloheptane has a very small peak at m/z=84. Methylcycloheptane will instead have its base peak at m/z=98, corresponding to the loss of a methyl group (15 mass units), whereas ethylcyclohexane will have a very small peak at m/z=98.

POINTS: 1

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy 1. For a nucleus to exhibit the nuclear magnetic resonance phenomenon, it must be magnetic. Magnetic nuclei include: a. all nuclei with even numbers of neutrons and protons b. all nuclei with odd numbers of protons c. all nuclei with odd numbers of neutrons d. both b and c ANSWER: d POINTS: 1 2. Nuclear magnetic resonance spectroscopy provides information about a molecule's: a. conjugated pi electron system. b. size and formula. c. carbon-hydrogen framework. d. functional groups.

ANSWER: c POINTS: 1 3. Explain why all protons in a molecule do not absorb rf energy at the same frequency.

ANSWER: All nuclei in molecules are surrounded by electron clouds. When a uniform external magnetic field is applied to a molecule, the circulating electron clouds set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field, so that the effective field actually felt by a nucleus is a bit smaller than the applied field.

Beffective = Bapplied − Blocal This effect is termed shielding. Each nucleus is shielded to a slightly different extent, so each unique kind of proton in a molecule resonates at a slightly different frequency and gives rise to a unique NMR signal.

POINTS: 1 Exhibit 13-1 The following question(s) pertain to the charting of NMR spectra. MATCH a term to each description below. Place the letter of the term in the blank to the left of the description. a. b. c. d.

TMS high-field or upfield side MHz delta (δ)

e. f. g.

low-field or downfield side chemical shift specific absorption

4. _____ When looking at an NMR chart the right-hand part of the chart is the _____.

ANSWER: b POINTS: 1 5. _____ The exact place on the chart at which a nucleus absorbs is called its _____.

ANSWER: f POINTS: 1 6. _____ The calibration standard for 1H and 13C NMR is _____.

ANSWER: a POINTS: 1 7. _____ The NMR charts are calibrated using an arbitrary scale that is divided into _____ units.

ANSWER: d Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy POINTS: 1 Exhibit 13-2 For each of the compounds below tell how many signals you would expect the molecule to have in its normal, broadband decoupled 13C NMR spectra.

8.

ANSWER: three

POINTS: 1

9.

ANSWER: two

POINTS: 1

10.

ANSWER:

five

POINTS: 1

11.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy ANSWER: six

POINTS: 1

12.

ANSWER: five

POINTS: 1

13.

ANSWER: three

POINTS: 1

14.

ANSWER: five POINTS: 1

15.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy ANSWER: three POINTS: 1 Exhibit 13-3 Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic.

16.

ANSWER: homotopic POINTS: 1

17.

ANSWER: enantiotopic POINTS: 1

18.

ANSWER: diastereotopic POINTS: 1

19.

ANSWER: unrelated POINTS: 1 Exhibit 13-4 For each compound below tell how many types of nonequivalent protons there are.

20.

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Page 4

Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy ANSWER:

four

POINTS: 1

21.

ANSWER: one

POINTS: 1

22.

ANSWER:

four

POINTS: 1

23.

ANSWER:

four

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

24.

ANSWER:

four

POINTS: 1

25.

ANSWER:

two

POINTS: 1

26.

ANSWER: five

POINTS: 1

27.

ANSWER: four

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 6

Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy Exhibit 13-5 Predict the splitting patterns you would expect for each proton in the molecules below.

28.

ANSWER:

No. of No. of adjacent equivalent protons neighboring protons 1 1 2

6

Splitting doublet septet

POINTS: 1

29.

ANSWER: No. of adjacent No. of equivalent Splitting protons neighboring protons 1 0 singlet 2

3

quartet

3

2

triplet

POINTS: 1

30.

ANSWER: No. of No. of adjacent equivalent protons neighboring Cengage Learning Testing, Powered by Cognero

Splitting Page 7

Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy protons 1

0

singlet

2

1

doublet

3

1

doublet

POINTS: 1 31. The 1H NMR spectrum of styrene oxide shows that protons 1, 2, and 3 all have different chemical shift values. Proton 1 is coupled to both proton 2 (J = 5.8 Hz) and proton 3 (J = 2.5 Hz). Draw a tree diagram for the proton 1 signal.

ANSWER:

POINTS: 1 Exhibit 13-6 Refer to the structure of 3-methyl-2-butanone below to answer the following question(s).

32. Refer to Exhibit 13-6. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled a.? a. septet b. quartet c. doublet d. singlet ANSWER: d POINTS: 1 33. Refer to Exhibit 13-6. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled b.?

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy a. septet b. quartet c. doublet d. singlet ANSWER: c POINTS: 1 34. Refer to Exhibit 13-6. The carbonyl-carbon resonance of 3-methyl-2-butanone occurs at 208.7 ppm downfield from TMS. How many hertz downfield from TMS would this carbonyl-carbon absorb if the spectrometer used to measure this absorption were operating at 200 MHz? ANSWER:

POINTS: 1 35. Treatment of tert-butyl alcohol with hydrogen chloride yields a mixture of tert-butyl chloride (SN1 product) and 2methylpropene (E1 product). After chromatographic separation, how would you use 1H NMR to help you decide which was which?

ANSWER: 2-Methylpropene has two kinds of hydrogens. It will have a vinylic absorption (4.5−6.5 δ) representing two hydrogens and an unsplit signal (1.0−1.5 δ) due to the six equivalent methyl hydrogens. tert-Butyl chloride has only one kind of hydrogen, which results in one unsplit signal.

POINTS: 1 Exhibit 13-8 Below are three isomeric chlorobutanes and their 13C NMR spectral data. MATCH the spectral data to the correct structures by placing the letter of the spectrum in the blank to the left of the corresponding structure. a = δ, 55.4, 36.2, 19.3 b = δ, 56.2, 35.3 c = δ, 56.0, 36.0, 24.2, 8.2

36. _____ ANSWER: b POINTS: 1

37. _____ ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

38. _____ ANSWER: a POINTS: 1 Exhibit 13-8 Propose structures for compounds that fit the following 1H NMR data.

39. C8H9Br 3H doublet at 2.0 δ, J = 7 Hz 1H quartet at 5.0 δ, J = 7 Hz 5H singlet at 7.3 δ ANSWER:

POINTS: 1 40. C7H14O 6H triplet at 0.9 δ, J = 7 Hz 4H sextet at 1.6 δ, J = 7 Hz 4H triplet at 2.4 δ, J = 7 Hz ANSWER:

POINTS: 1 41. C3H6Br2 2H quintet at 2.4 δ, J = 6 Hz 4H triplet at 3.5 δ, J = 6 Hz ANSWER: BrCH2CH2CH2Br POINTS: 1 42. C10H14 6H doublet at 1.2 δ, J = 7 Hz 3H singlet at 2.3 δ 1H septet at 2.9 δ, J = 7 Hz 4H singlet at 7.0 δ ANSWER:

POINTS: 1 43. C6H14 12H doublet at 0.8 δ Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy 2H septet at 1.4 δ ANSWER:

POINTS: 1 Exhibit 13-9 To answer the following question(s), consider the data and 1H NMR spectrum below: The mass spectrum of this compound shows a molecular ion at m/z = 113, the IR spectrum has characteristic absorptions at 2270 and 1735 cm−1, and the 13C NMR spectrum has five signals.

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 44. Refer to Exhibit 13-9. Based on the mass spectral data and the IR data, what functional groups are present in this compound?

ANSWER: A molecular ion of mass 113 indicates an odd number of nitrogen atoms in the compound. Coupled with the IR absorption at 2270 cm−1, a nitrile functional group is indicated. The IR absorption at 1735 cm−1 indicates a carbonyl group, possibly an ester.

POINTS: 1 45. Refer to Exhibit 13-9. How many types of nonequivalent protons are there in this molecule?

ANSWER: three POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy 46. Refer to Exhibit 13-9. Describe the signal at 3.5 δ in terms of its integration, splitting pattern and chemical shift.

ANSWER: The integration of the signal at 3.5 δ indicates that it is due to two equivalent hydrogens, probably a −CH2− group. Since the signal is a singlet (not split) there are no nonequivalent hydrogens attached to the atoms adjacent to the carbon to which these two hydrogens are bonded. The chemical shift to 3.5 δ indicates that this −CH2− group has at least one electronegative atom or group bonded to it, possibly a carbonyl group.

POINTS: 1 47. Refer to Exhibit 13-9. Describe the signals at 4.35 δ and 1.3 δ in terms of their integration, splitting and chemical shift.

ANSWER: The signal at 4.35 δ is owing to two equivalent hydrogens split by three adjacent hydrogens or a −CH2− next to a −CH3. It is shifted by attachment to an electronegative atom like oxygen. The signal at 1.3 δ is three equivalent hydrogens split by two adjacent hydrogens of a −CH3 next to a −CH2−. It is shifted slightly downfield by the presence of an electronegative atom bonded to the adjacent −CH2−.

POINTS: 1 48. Refer to Exhibit 13-9. What is the significance of the 13C NMR data?

ANSWER: The 13C NMR has five signals, which means that there are five different kinds of carbon in this compound.

POINTS: 1 49. Refer to Exhibit 13-9. Propose a structure for this compound.

ANSWER: By compiling all the information deduced from the data provided, we note that this compound is a) a nitrile and an ester [two carbons accounted for], and b) has two CH2 groups and one CH3 group [three more carbons accounted for]. One of the CH2 groups is attached to the oxygen of the ester and one is attached to the carbonyl of the ester. The singlet CH2 is also bonded to the nitrile. Based on the chemical shifts, the singlet CH2 must be bonded to the carbonyl and the quartet CH2 must be bonded to the oxygen of the ester. So the compound is ethyl cyanoacetate:

POINTS: 1 50. How would you use 1H and 13C NMR to help you distinguish between these two isomeric structures?

ANSWER:

# peaks

Distinguishing Absorptions

1

H

5

methyl doublet at 1.5 δ (overlaps other methylene signals), no vinyl protons

13

C

6

one carbonyl carbon, no vinyl carbons

H

4

split vinylic peak; rel. area 2 methyl singlet between 3.5−4.0 δ

1

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy 13

C

4

one vinylic carbon, no carbonyl carbon

POINTS: 1 Exhibit 13-10 Answer the question(s) for the compound whose 1H NMR spectra is shown below.

C4H8O

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 51. Refer to Exhibit 13-10. Describe each signal in terms of its integration, splitting and chemical shift.

ANSWER: The signals at 1.05 and 2.45 δ are characteristic, in their integration and splitting (a 2H quartet and a 3H triplet), for an ethyl group. The shift of the CH2 to 2.45 indicates that the ethyl group is attached to a carbonyl group. The signal at 2.1 δ, a 3H singlet, is characteristic for a CH3 attached to a carbonyl group.

POINTS: 1 52. Refer to Exhibit 13-10. Propose a structure for this compound.

ANSWER: 2-butanone, CH3COCH2CH3 POINTS: 1 Exhibit 13-11 Answer the question(s) for the compound whose 1H NMR spectra is shown below.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy C5H12O

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 53. Refer to Exhibit 13-11. Calculate the degree of unsaturation for this compound.

ANSWER: The base formula for this compound is C5H12, which corresponds to a saturated compound, so there is no unsaturation in this compound.

POINTS: 1 54. Refer to Exhibit 13-11. Describe each signal in the 1H NMR in terms of its integration, splitting and chemical shift.

ANSWER: In looking at the spectrum, it is important to note that the relative number of hydrogens for each signal adds up to four. The formula of the compound indicates that there are a total of 12 hydrogens. Therefore, the integration of 3:1 corresponds to 9H:3H. The signal at 1.2 δ is a 9H singlet, which is characteristic for a tertbutyl group. The slight downfield shifts indicates that this group is attached to an electronegative atom, in this case, oxygen. The signal at 3.2 δ is a 3H singlet which is characteristic of a methyl group bonded to an oxygen.

POINTS: 1 55. Refer to Exhibit 13-11. Propose a structure for this compound.

ANSWER: tert-butyl methyl ether, (CH3)3COCH3 POINTS: 1 Exhibit 13-12 Answer the question(s) for the compound whose 1H NMR spectra is shown below.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy C8H7ClO

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 56. Refer to Exhibit 13-12. Calculate the degree of unsaturation in this compound.

ANSWER: The base formula for this compound is C8H8; the saturated formula is C8H18; the degree of unsaturation is five [(18 − 8) ÷ 2].

POINTS: 1 57. Refer to Exhibit 13-12. Describe the signals that occur between 7 and 8 δ in terms of integration, splitting and chemical shift.

ANSWER: The chemical shift of the signals between 7 and 8 δ indicates that these hydrogens are aromatic. The integration of four hydrogens and the splitting pattern of two doublets is characteristic of a para disubstituted aromatic compound.

POINTS: 1 58. Refer to Exhibit 13-12. Describe the signal at 2.6 δ in terms of integration, splitting and chemical shift.

ANSWER: This signal is a 3H singlet shifted by attachment to a carbonyl group; a methyl ketone, probably.

POINTS: 1 59. Refer to Exhibit 13-12. Propose a structure for this compound.

ANSWER: 4-chloroacetophenone POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy 60. Propose a structure for a compound, C6H14O, with the following 13C NMR spectral data: Broadband decoupled 13C NMR: DEPT-90: DEPT-135: ANSWER:

23.0, 68.4 δ 68.4 δ positive peaks at 23.0, 68.4 δ; no negative peaks

POINTS: 1 61. Propose a structure for Compound X, which has M+ = 120 and (M + 2) = 122 of approximately equal intensity in its mass spectrum and has the following 13C NMR spectral data: Broadband decoupled 13C NMR: DEPT-90: DEPT-135: ANSWER:

32.6, 118.8, 134.3 δ 134.3 δ positive peaks at 134.3 δ; negative peaks at 32.6, 118.8 δ

POINTS: 1 62. Propose a structure for Compound Z, which has the following spectroscopic properties: MS: IR: 1

H NMR:

M+ = 88 3380 cm−1 0.85 δ (6H doublet); 1.40 δ (3H multiplet); 2.68 δ (1H singlet); 3.55 δ (2H triplet)

Broadband decoupled 13C NMR: DEPT-90: DEPT-135: ANSWER: (CH3)2CHCH2CH2OH

22.7, 25.0, 41.8, 60.5 δ 25.0 δ positive peaks at 22.7, 25.0 δ; negative peaks at 41.8, 60.5 δ

POINTS: 1 63. Describe how you could differentiate the following compounds using a DEPT NMR experiment.

ANSWER: Both compounds contain two different kinds of carbon, so an ordinary broadband decoupled 13C NMR spectrum of either compound will show two peaks. The DEPT-90 spectrum, however, will show no peaks for 1,4-dichlorobutane while the DEPT-90 spectrum of 2,3-dichlorobutane will show one peak. The DEPT-135 spectrum for 1,4-dichlorobutane will show two negative peaks, while the DEPT-135 spectrum of 2,3dichlorobutane will show two positive peaks.

POINTS: 1 64. Which structure of molecular formula C4H8Cl2 fits both the 1H NMR and 13C NMR spectra shown below?

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

a.

b. c.

d.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Spectra obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ ANSWER: b POINTS: 1 65. Which of the following provides information about the carbon framework of a molecule? a.

b.

c.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

d.

ANSWER: a POINTS: 1 66. Which signal(s) remain(s) when a DEPT-90 spectrum is run? a. CH3 b. CH2 c. CH d. quaternary C e. a, b, c ANSWER: c POINTS: 1 67. Which of the following occurs when a DEPT-135 spectrum is run? a. CH3, CH carbons appear as positive signals b. CH2 carbons appears as a negative signal c. quaternary carbon signals are absent d. a and b e. all of the above ANSWER: e POINTS: 1 68. Consider the following structure. Atoms other than carbon and hydrogen are labeled. The carbon atoms are designated by numbers.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Which carbon atom(s) will produce a signal in a DEPT-90 spectrum? a. 1, 6 and 7 b. 2 and 5 c. 3 d. 4 e. All will produce a signal.

ANSWER: d POINTS: 1 69. Consider the following structure. Atoms other than carbon and hydrogen are labeled. The carbon atoms are designated by numbers.

Which carbon atom(s) will produce a signal in the broadband decoupled, the DEPT-90, and the DEPT-135 spectra? a. 1, 6, and 7 b. 2 and 5 c. 3 d. 4 e. All carbons atoms will appear these spectra. f. None of the carbon atoms appear in all these spectra.

ANSWER: d POINTS: 1 70. Consider the following structure. Atoms other than carbon and hydrogen are labeled. The carbon atoms are designated by numbers.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Which carbon atom will produce a signal the furthest downfield in the broadband decoupled 13C NMR spectrum? a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 ANSWER: c POINTS: 1 71. What type of NMR spectrum is the following?

a. broadband decoupled b. DEPT-90 c. DEPT-135 d. It is not possible to judge given just the spectrum.

ANSWER: c POINTS: 1 72. The following is the tree diagram for the protons on a specific carbon.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Which of the following signals in a proton NMR could correspond to these protons? a.

b.

c.

d.

e. None of these due to the coupling constants.

ANSWER: e POINTS: 1 73. How many signals would appear in the proton NMR for the following substance? Atoms other than carbon and hydrogen are labeled.

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

a. 2 b. 3 c. 4 d. 5 ANSWER: b POINTS: 1 74. Identify the splitting pattern of the numbered protons. Atoms other than hydrogen and carbon are labeled.

a. 1: doublet, 2: triplet c. 1: quartet 2: quartet ANSWER: c POINTS: 1

b. 1: triplet, 2: triplet d. 1: singlet, 2: triplet

75. Identify the two numbered hydrogen atoms as:

a. unrelated c. enantiotopic ANSWER: a POINTS: 1

b. homotopic d. diastereotopic

76. Identify the different types of protons in 1-chloro-2-butene. What would be the chemical shift observed for each type of proton?

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

ANSWER: The –CH3 (alkyl) protons absorb wavelengths around 0.7–1.3 δ, the vinylic protons around 4.5–6.5 δ, and the protons of the alkyl halide group 2.5–4.0δ.

POINTS: 1 77. The molecular formula of a certain compound is C4H9OBr. It has the following 1H NMR data: 1.30 δ (6H, singlet), 3.94 δ (2H, doublet), and 2.51 δ (1H, triplet). Propose a structure for the compound using the information given.

ANSWER: From the molecular formula of the compound, we can say that it is a saturated, open-chain molecule.

The six-proton absorption at 1.30 δ is due to the two equivalent alkane-like methyl groups attached to a carbon without any hydrogens. The two-proton doublt at 3.94 δ is due to a secondary alkyl group (CH2) attached to a hydroxyl or ether group. The one-proton triplet at 2.51 δ is mostly due to an alcohol group as there is no other way to account for it. Since there is no peak corresponding to hydrogens attached to halide groups, the bromide group should be attached to a carbon that does not have any hydrogens attached to it. Hence, the bromide group is attached to a carbon that has three other groups, two of which should be the equivalent methyl groups. Therefore, by collating all the information, we get the structure: 2-bromo-2-methyl-propanol.

POINTS: 1 78. Find all the types of nonequivalent protons in the given compound.

ANSWER: Nine

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy POINTS: 1 79. What would be the approximate positions at which you would expect 2-chloro-3-butyne to show 13C NMR absorptions?

ANSWER: 2-chloro-1-butyne has four chemically distinct carbons: two different C which the likely absorptions are 92 δ, 55 δ, and 20 δ, respectively.

C, one C–Cl, and one alkyl C for

POINTS: 1 80. Propose a structure for an alkyl halide, C4H9Br, that has the following 13C NMR data: Broadband decoupled 13C NMR: 18.3, 38.6, 60.2 δ DEPT-90: 38.6 δ DEPT-135: positive peak at 18.3, 60.2 δ; negative peak at 38.6 δ ANSWER: From the molecular formula, we know that the compound is a saturated, open-chain molecule. The compound is supposed to have four carbon atoms but shows absorption peaks at only three positions in the NMR spectrum. This suggests that two of the carbons must be equivalent. Two of the absorptions (18.3 and 38.6 δ) are in the typical alkane region. The third absorption is in the region of a carbon that is attached to an electronegative atom, which in this case is Br. This tells us that the alkyl carbon at 38.6 δ is tertiary (CH). The DEPT-135 spectrum tells us that the alkyl carbon at 18.3 δ is a methyl group and that the carbon at 60.2 δ, which is attached to the bromide group, is secondary (CH2). The two equivalent carbons are probably two methyls attached to the same tertiary carbon. By putting the pieces together, we can tell that the compound is 1-bromo-2-methyl propane.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy

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Page 26

Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy Drawing Instructions: Draw structures corresponding to each of the given names.

1. Draw: 2-methyl-1,3-butadiene ANSWER:

POINTS: 1 2. Draw: (2Z,4Z)-hexadiene ANSWER: POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

3. Name:

ANSWER: 2,3,4,5-tetramethyl-2,4-hexadiene POINTS: 1 4. Name:

ANSWER: (3E)-2-methyl-1,3-pentadiene POINTS: 1 Exhibit 14-1 Circle any conjugated portions in the molecules below.

5.

ANSWER:

POINTS: 1

6.

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Page 1

Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER:

POINTS: 1 7.

ANSWER: POINTS: 1

8.

ANSWER:

POINTS: 1

9.

ANSWER: not conjugated POINTS: 1 Exhibit 14-2 Consider the reaction below to answer the following question(s):

10. Refer to Exhibit 14-2. The nucleophile in this reaction is ____.

ANSWER: A POINTS: 1 11. Refer to Exhibit 14-2. The electrophile in this reaction is ____.

ANSWER: B POINTS: 1 12. Refer to Exhibit 14-2. The kinetically controlled product in this reaction is _____.

ANSWER: D POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy 13. Refer to Exhibit 14-2. The product that results from 1,4-addition is _____.

ANSWER: C POINTS: 1 14. Refer to Exhibit 14-2. Write a stepwise mechanism that accounts for both of the products shown. Show all intermediate structures and electron flow with arrows.

ANSWER:

POINTS: 1 15. Refer to Exhibit 14-2. This reaction shows the preferred electrophilic addition of HBr to this unsymmetrical alkene.

Draw the products of electrophilic addition to the opposite end of the conjugated diene.

ANSWER:

POINTS: 1 16. Electrophilic addition reaction of conjugated dienes that occur at high temperature and/or long reaction times (reversible conditions) are said to be under: a. 1,2-control b. kinetic control c. 1,4-control d. thermodynamic control ANSWER: d POINTS: 1 Exhibit 14-3 Consider the data given below to answer the following question(s).

Cyclobutadiene is an extremely unstable compound whose isolation is impossible even at the lowest temperatures routinely accessible in the organic laboratory. However, free cyclobutadienes almost surely occur in some reactions as intermediates. We might expect cyclobutadienes to undergo Diels-Alder cycloaddition reactions with itself (dimerization) or with reactive dienophiles. In fact, these extremely facile Diels-Alder reactions have been used to recognize intermediate cyclobutadiene. Write the structures of the Diels-Alder products you would expect to isolate from each of the following reactions.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy 17.

ANSWER:

POINTS: 1

18.

ANSWER:

POINTS: 1 Exhibit 14-4 The following molecules may be prepared using the Diels-Alder reactions. Write the structures of the starting diene and dienophile necessary to prepare each molecule and label them.

19.

ANSWER:

POINTS: 1

20.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER:

POINTS: 1

21.

ANSWER:

POINTS: 1 22. For a diene to undergo a Diels-Alder reaction it must: a. be substituted with electron-withdrawing groups b. be able to adopt an s-trans conformation c. be substituted with electron-donating groups d. be able to adopt an s-cis conformation ANSWER: d POINTS: 1 23. For Diels-Alder cycloaddition reactions to take place most rapidly and in highest yield the dienophile must: a. be substituted with electron-withdrawing groups b. be able to adopt an s-trans conformation c. be substituted with electron-donating groups d. be able to adopt an s-cis conformation ANSWER: a POINTS: 1 24. Which of the following compounds will react as a diene in a Diels-Alder reaction? a.

b.

c.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy d.

ANSWER: c POINTS: 1 Exhibit 14-5 Write the structure(s) of the product(s) for each of the reactions below. Be sure to indicate any relevant stereochemistry.

25.

ANSWER:

POINTS: 1 26.

ANSWER:

POINTS: 1

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy

29.

ANSWER:

POINTS: 1

30.

ANSWER:

POINTS: 1 31. When an organic molecule is irradiated with ultraviolet radiation, the energy absorbed by the molecule corresponds to: a. the amount necessary to increase molecular motions in functional groups b. the amount necessary to excite electrons from one molecular orbital to another c. the amount necessary to "flip" the spin of atomic nuclei d. the amount necessary to strip a molecule of one electron to generate a radical cation ANSWER: b POINTS: 1 32. A physical constant that is the quantitative measure of the amount of UV light absorbed by a compound. a. A b. ε c. λmax d. π* ANSWER: b POINTS: 1 33. The amount of energy in electromagnetic radiation is related to the frequency and wavelength of the radiation. High energy radiation like gamma rays is of: a. b. c.

low frequency and short wavelength low frequency and long wavelength high frequency and short wavelength

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy d. high frequency and long wavelength ANSWER: c POINTS: 1 34. Which of the following compounds would show the longest wavelength λmax in its UV spectrum? a.

b.

c.

d.

ANSWER: c POINTS: 1 Exhibit 14-6 Indicate which spectral technique, NMR, IR, UV or MS, would most readily allow differentiation between compounds in each pair of compounds below. Explain your answer in each case.

35.

ANSWER: NMR; these isomers have the same molecular weight and will show no strong IR and no UV absorptions. The compound on the left will have a single peak in its 1H NMR spectrum while the compound on the right will have three multiplets.

POINTS: 1

36.

ANSWER: UV; these compounds will have similar NMR, IR and mass spectra. The UV spectra will be significantly different since the compound on the left is conjugated.

POINTS: 1

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy 37.

ANSWER: MS; these compounds have different molecular weights but similar NMR, IR and UV spectra.

POINTS: 1

38.

ANSWER: IR; these compounds have the same molecular weight, similar 1H NMR (one singlet) spectra, and no strong UV. The compound on the right will have a strong carbonyl absorption in its IR spectrum that the other compound will not.

POINTS: 1 39. In 1839, Charles Goodyear discovered a process by which both natural and synthetic rubbers are hardened. This process is called: a. diazotization b. vulcanization c. polymerization d. condensation ANSWER: b POINTS: 1 40. 2-Chloro-1,3-butadiene is polymerized to yield an excellent, expensive synthetic rubber with good weather resistance called: a. diprene b. isoprene c. styrene d. neoprene ANSWER: d POINTS: 1 Exhibit 14-7 Consider the reaction below to answer the following question(s).

41. Refer to Exhibit 14-7. This is an example of a(n) ____________ reaction. a.

Woodward-Hoffman

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy b. conjugate addition c. Diels-Alder d. electrophilic addition ANSWER: c POINTS: 1 42. Refer to Exhibit 14-7. The dienophile in the reaction is ____.

ANSWER: B POINTS: 1 43. Refer to Exhibit 14-7. The diene in the reaction is ____.

ANSWER: A POINTS: 1 44. Draw the skeletal formulas for the product of the reaction of the following with HBr.

ANSWER: 1,2-additions and 1,4-addition products (both racemic), respectively.

POINTS: 1 45. Which of the following arises from π to π* transitions? a.

b.

c.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy

d.

e.

ANSWER: e POINTS: 1 46. Based on the following spectrum, what is the energy in kilojoules per mole of photons that corresponds to the λmax?

a. b. c. d.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy e.

ANSWER: d POINTS: 1 47. Consider the following spectrum for a conjugated diene.

If a similar spectrum were taken for a conjugated tetraene, the λmax would be: a. close to 217 nm b. less than 217 nm c. greater than 217 nm ANSWER: c POINTS: 1 48. Based on the following table, which compound might have a yellow color? λmax (nm) 165 305 440 790

Compound A B C D a. A b. B c. C d. D e. Both C and D f. All would be colorless.

ANSWER: c POINTS: 1

49. What is the concentration of a solution of a plant pigment (ε = 37,500) if an Absorbance of 0.751 were obtained in a cell with a 5.00 cm path length? a. b. c.

d.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER: b POINTS: 1 50. Consider the following data for a series of hypothetical plant pigments.

Pigment QT7 XR4 B78 L99

ε 15,300 41,000 30,800 10,600

The Absorbance of a Which pigment was in the solution? a. QT7 b. XR4 c. B78 d. L99 ANSWER: d POINTS: 1

solution containing one of these pigments, measured in a 1.00 cm cell, was 0.345.

51. Based on the following table, in which compound is the energy difference between the HOMO and LUMO the largest?

Compound λmax (nm) W 190 X 340 Y 560 Z 250 a. W b. X c. Y d. Z ANSWER: a POINTS: 1 52. If ΔH°hydrog for a monosubstituted alkene is 126 kJ/mol, which of the following systems is most likely to be a conjugated system? a. 3 double bonds, ΔH°hydrog = 379 kJ/mol b. 2 double bond, ΔH°hydrog = 254 kJ/mol c. 4 double bonds, ΔH°hydrog = 505 kJ/mol d. None of these are likely to be conjugated.

ANSWER: d POINTS: 1 53. Which of the following will not undergo a Diels-Alder reaction? a.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy

b.

c.

d.

e. All but a will not undergo a Diels-Alder reaction.. f. All will undergo a Diels-Alder reaction.

ANSWER: c POINTS: 1 54. Which of the following has the highest degree of conjugation? Atoms other than carbon and hydrogen are labeled.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy a.

b.

c.

d.

ANSWER: c POINTS: 1 55. Draw the mechanism of the bromination of 1,3-pentadiene and indicate the major and minor products.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER:

POINTS: 1 56. Draw the mechanism of the bromination of 2-ethyl-1,3-pentadiene.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER:

POINTS: 1 57. Draw the mechanism of the Diels–Alder reaction between propenal and 2-methyl-1,3-butadiene.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER:

POINTS: 1 58. Which of the following compounds is the most stable? a. penta-2,3-diene b. 2,4-hexadiene c. 1,4-pentadiene d. 3,6-octadiene ANSWER: b POINTS: 1 59. What would be the stereochemistry of the product of the following reaction?

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy

ANSWER:

POINTS: 1 60. Which of the following is the most likely to happen when a conjugated diene is irradiated with ultraviolet light? a. The bonds between different atoms in the molecule undergo symmetric stretching. b. A electron is promoted from the highest occupied molecular orbital to the lowest unoccupied molecular orbital. c. A electron is demoted from the lowest occupied molecular orbital to the highest unoccupied molecular orbital. d. The bonds between different atoms in the molecule undergo asymmetric stretching.

ANSWER: b POINTS: 1 61. The unit for molar absorptivity is: a. L/cm b. (L . mol)/cm c. L/(mol . cm) d. mol/L ANSWER: c POINTS: 1 62. Calculate the concentration of β-carotene in a sample which is placed in a cell with a pathlength of 1.0 cm and the UV absorbance reads 0.56. The molar absorbtivity of β-carotene is found to be 138,000 L mol-1 cm-1.

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Chapter 14 - Conjugated Compounds and Ultraviolet Spectroscopy ANSWER:

POINTS: 1

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Chapter 15 - Benzene and Aromaticity Drawing Instructions: Draw structures corresponding to the given names.

1. Draw: m-fluoronitrobenzene ANSWER:

POINTS: 1 2. Draw: p-bromoaniline ANSWER:

POINTS: 1 3. Draw: o-chlorophenol ANSWER:

POINTS: 1 4. Draw: 3,5-dimethylbenzoic acid ANSWER:

POINTS: 1 5. Draw: p-chlorobenzaldehyde ANSWER:

POINTS: 1 6. Draw: 1-phenyl-3-methylpentane ANSWER:

POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

7. Name: Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 15 - Benzene and Aromaticity

ANSWER: m-nitrotoluene or 1-methyl-3-nitrobenzene POINTS: 1 8. Name:

ANSWER: (1-methylethyl) benzene or isopropylbenzene or cumene POINTS: 1 9. Name:

ANSWER: 1-((1S, 3R)-3-methylcyclohexyl)benzene POINTS: 1 10. Name:

ANSWER: 1-methyl-2,4-benzenediamine or 2,4-diaminotoluene POINTS: 1 11. Name:

ANSWER: (E)-2,3-diphenyl-2-hexene POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 2

Chapter 15 - Benzene and Aromaticity Exhibit 15-1 For each molecule below, predict whether the molecule would be expected to show aromatic character or not. Explain your answer in each case.

12.

ANSWER: To meet the criterion of a cyclic, planar, conjugated molecule, the oxygen and the nitrogen bearing the hydrogen must both be sp2-hybridized. If this occurs, both heteroatoms contribute two electrons each to the pi system. This results in a total of eight pi electrons, which is not aromatic according to the Hückel (4n + 2) Rule.

POINTS: 1

13.

ANSWER: The carbocation is sp2-hybridized, so if the oxygen is also sp2-hybridized then each atom in the cycle has a conjugated p-orbital. Each double bond contributes two pi electrons and the oxygen contributes two pi electrons, for a total of six pi electrons in the conjugated system. Therefore, this species is predicted to show aromatic character.

POINTS: 1

14.

ANSWER: This compound, [16]annulene, appears to be a cyclic, planar, conjugated molecule with 16 pi electrons.

However, 16 is not a Hückel number, so the compound is probably not planar, and is not aromatic.

POINTS: 1

15.

ANSWER: This compound has a pyridine-like nitrogen, which is sp2-hybridized and has its lone pair of electrons in the plane of the ring. If oxygen is also sp2-hybridized, then it contributes two pi electrons to the cyclic, conjugated pi system for a total of six π-electrons. Therefore, this molecule is predicted to show aromatic character.

POINTS: 1 Exhibit 15-2 Cengage Learning Testing, Powered by Cognero

Page 3

Chapter 15 - Benzene and Aromaticity Answer the following question(s) concerning sulfathiazole, below.

16. Refer to Exhibit 15-2. What is the hybridization of the nitrogen atom in sulfathiazole?

ANSWER: The hybridization of the nitrogen atom is sp2 in sulfathiazole.

POINTS: 1 17. Refer to Exhibit 15-2. Assuming that the sulfur atom is sp2-hybridized, how many π-electrons are there in the sulfathiazole ring?

ANSWER: There are six π-electrons in the sulfathiazole ring.

POINTS: 1 18. Refer to Exhibit 15-2. What reactivity do you predict for sulfathiazole?

ANSWER: Sulfathiazole is predicted to have aromatic reactivity since it is cyclic, planar, conjugated and has six πelectrons.

POINTS: 1 Exhibit 15-3 Refer to the data below to answer the following question(s).

The 1H NMR spectrum of [14]annulene at −60°C shows two signals, one at 0 ppm and one at 7.6 ppm, with an area ratio of 5:2.

19. Refer to Exhibit 15-3. Is [14]annulene aromatic?

ANSWER: [14]annulene contains a cyclic, conjugated pi system with 14 pi electrons (14 = 4n + 2; n = 3). It is difficult to predict, however, if the molecule is planar without making a model. The 1H NMR signal at 7.6 ppm is consistent with the presence of a ring current, which is characteristic of aromatic molecules. Thus, [14]annulene is aromatic.

POINTS: 1 20. Refer to Exhibit 15-3. Draw in the protons on the [14]annulene skeleton provided below which are responsible for the 1 H NMR signal at 0 ppm.

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Chapter 15 - Benzene and Aromaticity

ANSWER:

POINTS: 1 21. Refer to Exhibit 15-3. Explain why these protons have absorption at such high fields.

ANSWER: When an aromatic ring is oriented perpendicularly to a strong magnetic field the induced ring current deshields protons outside the ring and shields protons inside the ring. Thus, the protons inside the ring absorb at a field higher than normal (0 ppm versus 4.5−6.5 ppm).

POINTS: 1 Exhibit 15-4 Consider the data below to answer the following question(s).

22. Refer to Exhibit 15-4. Conjugate bases QB and ZB are both resonance stabilized. Draw the indicated number of resonance forms for QB and ZB.

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Chapter 15 - Benzene and Aromaticity

ANSWER:

POINTS: 1 23. Refer to Exhibit 15-4. Which of the compounds above, Q or Z, would you predict to be most acidic? Explain your answer.

ANSWER: The conjugate bases of Compounds Q and Z are both resonance stabilized. Let's examine all resonance structures for both conjugate bases:

For the conjugate base QB the negative charge is delocalized into the ring. This generates a cyclopentadienyl anion that is aromatic. For the conjugate base ZB, the negative charge is delocalized by resonance but no aromatic ions result. Therefore, Compound Q should be the most acidic, since its anion is the most stable.

POINTS: 1 24. Most alkyl halides are nonpolar covalent compounds and, therefore, are soluble in nonpolar solvents and insoluble in water. Cycloheptatrienyl bromide is an unusual alkyl halide in that it is insoluble in nonpolar solvents, but is readily soluble in water! This behavior is consistent with cycloheptatrienyl bromide being an ionic compound. Why does cycloheptatrienyl bromide exist as an ionic compound? Explain by comparing the covalent structure to the ionic structure.

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Chapter 15 - Benzene and Aromaticity

ANSWER: In the covalent structure, the carbon bonded to the bromine is sp3-hybridized, so the ring cannot be aromatic.

In the ionic structure, the carbocation is sp2-hybridized, so the structure is a cyclic, conjugated, six π-electron system and aromatic.

POINTS: 1 Exhibit 15-5 Consider the data below to answer the following question(s).

C9H12; 1H NMR:

7.18 δ (broad singlet, 5H) 2.55 δ (triplet, 2H) 1.70 δ (sextet, 2H) 0.9 δ (triplet, 3H)

25. Refer to Exhibit 15-5. Describe the signal at 7.18 δ in terms of its splitting, integration and chemical shift.

ANSWER: The integration of the signal at 7.18 δ indicates that there are five hydrogens that are more or less equivalent (broad singlet); the chemical shift indicates that these five hydrogens are aromatic. This is typical for a monosubstituted benzene.

POINTS: 1 26. Refer to Exhibit 15-5. Describe the signal at 2.55 δ in terms of its splitting, integration, and chemical shift.

ANSWER: The integration of the signal at 2.55 δ indicates that it is two hydrogens that are split (triplet) by two adjacent hydrogens, or a −CH2− next to a −CH2−. The chemical shift indicates that this −CH2− could be next to an aromatic ring.

POINTS: 1 27. Refer to Exhibit 15-5. Describe the signal at 1.70 δ in terms of its splitting, integration and chemical shift.

ANSWER: The integration of the signal at 1.70 δ indicates that it is two hydrogens split by five adjacent hydrogens (sextet); most probably a −CH2− between a −CH2− and a −CH3. Its chemical shift indicates that there is some electronegative group causing some deshielding to occur.

POINTS: 1 28. Refer to Exhibit 15-5. Describe the signal at 0.9 δ in terms of its splitting, integration and chemical shift.

ANSWER: The integration of the signal at 0.9 δ indicates that it is three hydrogens split by two adjacent hydrogens (triplet) or a −CH3 next to a −CH2−. Its chemical shift is in the normal range for a methyl group.

POINTS: 1 29. Refer to Exhibit 15-5. Propose a structure for this compound.

ANSWER:

POINTS: 1 Exhibit 15-6 Consider the data below to answer the following question(s).

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Chapter 15 - Benzene and Aromaticity C9H12; 1H NMR:

6.65 δ (singlet, 3H) 2.25 δ (singlet, 9H)

30. Refer to Exhibit 15-6. Describe the signal at 6.65 δ in terms of its splitting, integration and chemical shift.

ANSWER: The signal at 6.65 δ indicates that it is 3 equivalent hydrogens in the aromatic region of the spectrum. This could be a 1,3,5-trisubstituted compound.

POINTS: 1 31. Refer to Exhibit 15-6. Describe the signal at 2.25 δ in terms of its splitting, integration and chemical shift.

ANSWER: The signal at 2.25 δ indicates that it is nine equivalent hydrogens attached to an aromatic ring; probably three −CH3 groups.

POINTS: 1 32. Refer to Exhibit 15-6. Propose a structure for this compound.

ANSWER:

POINTS: 1 33. What is the structure of a hydrocarbon that shows a molecular ion at m/z = 182 in the mass spectrum and has the following 1H NMR spectrum?

7.2 δ, singlet, 5H 2.9 δ, singlet, 2H ANSWER: The molecular weight of the hydrocarbon corresponds to the formula C14H14, which indicates eight double bonds and/or rings. Since the ratio of protons in the 1H NMR is 5:2, there must be 10 aromatic and 4 aliphatic protons. There must be two equivalently substituted phenyl rings, which accounts for 12 carbons. Since all aliphatic protons are equivalent, and we need two more carbons to complete the structure, the compound must be 1,2-diphenylethane.

POINTS: 1 Exhibit 15-7 Consider the data below to answer the following question(s).

C7H7ClO; IR absorption at 810 cm−1

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Chapter 15 - Benzene and Aromaticity

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/ 34. Refer to Exhibit 15-7. Calculate the degrees of unsaturation for this compound.

ANSWER: The formula C7H7ClO corresponds to a base formula of C7H8. A saturated formula is C7H16 so there are (16 − 8) ÷ 2 = 4 degrees of unsaturation.

POINTS: 1 35. Refer to Exhibit 15-7. What is the significance of an IR absorption at 810 cm−1?

ANSWER: An IR absorption at 810 cm−1 indicates a p-disubstituted aromatic ring.

POINTS: 1 36. Refer to Exhibit 15-7. Describe the signal at 3.7 δ in terms of its splitting, integration and chemical shift.

ANSWER: The signal at 3.7 δ indicates that there are three hydrogens next to zero hydrogens (singlet) shifted by attachment to an electronegative atom like oxygen; probably an −OCH3.

POINTS: 1 37. Refer to Exhibit 15-7. Propose a structure for this compound.

ANSWER:

POINTS: 1 38. Give the IUPAC name for: Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Benzene and Aromaticity

ANSWER: 2-bromotoluene or o-bromotoluene POINTS: 1 39. Draw a picture of the π orbitals in furan (shown below). Is furan aromatic?

ANSWER:

Furan is a planar ring and is conjugated. Furan has 6 π electrons as shown above. The additional electron pair of the oxygen atom is not part of aromatic π system.

4n + 2 = 4(1) + 2 = 6 Furan is aromatic.

POINTS: 1 40. Draw the MO energy diagram for the cycloheptatrienyl anion and classify the anion as aromatic or antiaromatic.

ANSWER:

4n (8) π electrons: antiaromatic POINTS: 1 41. Draw the resonance structures of cyclooctatetraene.

ANSWER:

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Chapter 15 - Benzene and Aromaticity POINTS: 1 42. Explain in terms of MO theory, why systems with 4n + 2 π electrons are aromatic and with 4n π electrons are antiaromatic.

ANSWER: “Aromatic” compounds exhibit greater than expected stability. “Antiaromatic” compounds are less stable. If the MO energy diagram of any 4n + 2 molecule is examined, the π electrons all occupy low energy bonding molecular orbitals. This has a net stabilizing effect. Antiaromatic compounds have some of π electrons occupying higher energy antibonding molecular orbitals. Conversely, this has a net destabilizing effect.

POINTS: 1 43. Consider the following compound. Atoms other than carbon and hydrogen are labeled.

Which of the following is not valid? a. naming of the parent compound, benzene. b. 1,4-disubstitution is present. c. para prefix applicable. d. classified as aromatic.

ANSWER: a POINTS: 1 44. How many monosubstituted products would you expect for the bromination of naphthalene? a. 2 b. 4 c. 6 d. 8 ANSWER: a POINTS: 1 45. Which region of an IR spectrum provides information about the substitution pattern of aromatic compounds? a. 3030 cm-1 b. 1660 - 2000 cm-1 c. 1450 - 1600 cm-1 d. 690 - 840 cm-1 e. b and d f. All of the above can be used.

ANSWER: d POINTS: 1 46. Which of the following does not correctly characterize this compound represented by the molecular model shown Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Benzene and Aromaticity below? Atoms other than carbon and hydrogen are labeled.

a. slightly acidic b. named as isopropyl o-cresol c. contains aryl, alkyl and benzylic proton d. would produce a signal due to dehydration in a mass spectrum e. All of the above characterize this compound.

ANSWER: d POINTS: 1 47. Which of the following is NSAIDs were developed to selectively block the COX-2 enzyme? a.

b.

c.

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Chapter 15 - Benzene and Aromaticity

d.

e. both a and b f. both c and d g. a, b, c, and d ANSWER: d POINTS: 1 48. Which of the following NSAIDs would not produce a peak for benzylic protons in an 1HNMR spectrum? a.

b.

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Chapter 15 - Benzene and Aromaticity

c.

d.

e.

both a and b f. both c and d g. a, b, c, and d ANSWER: f POINTS: 1 49. Which of the following aromatic compounds will have largest number of signals in a 13CNMR? a. 1,4-dimethylbenzene (p-methyl toluene) b. 1,3,5-trimethylbenzene (mesitylene) c. 1,2,4,5-tetramethylbenzene d. All produce the same number of signals.

ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Benzene and Aromaticity 50. What m/z ratio range might be common to many substituted benzene compounds? a. 72 - 77 ppm b. 26 - 38 ppm c. 81 - 88 ppm d. less than 59 ppm ANSWER: a POINTS: 1 51. Which of the choices arranges numbered protons in the following substance in order of increasing downfield chemical shift in a proton NMR? Atoms other than carbon and hydrogen are labeled.

a. 1 = 2 < 3 < 4 c. 4 = 3 < 2 =1 ANSWER: a POINTS: 1

b. 4 < 3 < 2 = 1 d. 1 = 2 < 4 < 3

52. Draw the different resonance forms of anthracene.

ANSWER:

POINTS: 1 53. Draw the different resonance forms of phenanthrene.

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Chapter 15 - Benzene and Aromaticity ANSWER:

POINTS: 1 54. Draw the resonance forms of indene.

ANSWER:

POINTS: 1

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Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution Exhibit 16-1 MATCH a structure or term from the following list with each description below. Place the letter of the structure or term in the blank to the left of the description. a. b. c. d. i.

benzyne +

NO2 R3C+ electron-donating F-TEDA-BF4

e. f. g. h.

+

NO Meisenheimer complex electron-withdrawing

1. ______ The reactive electrophile in Friedel-Crafts acylation reactions.

ANSWER: g POINTS: 1 2. ______ The electrophile in aromatic nitration.

ANSWER: b POINTS: 1 3. ______ Groups which activate aromatic rings towards electrophilic substitution.

ANSWER: d POINTS: 1 4. ______ Groups which activate aromatic rings towards nucleophilic substitution.

ANSWER: h POINTS: 1 5. ______ Intermediate in the elimination-addition mechanism of nucleophilic aromatic substitution.

ANSWER: a POINTS: 1 6. __________Source of F+ in fluorination reactions.

ANSWER: i POINTS: 1 Exhibit 16-2 Consider the Friedel-Crafts alkylation reaction below to answer the following question(s):

7. Refer to Exhibit 16-2. Draw the structure of the electrophilic intermediate in this reaction.

ANSWER: POINTS: 1 8. Refer to Exhibit 16-2. What is the role of the AlCl3 in the reaction?

ANSWER: The AlCl3 is a Lewis acid catalyst that assists in the ionization of the alkyl halide to give the carbocation Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution electrophile.

POINTS: 1 9. Refer to Exhibit 16-2. Write the complete stepwise mechanism for this reaction. Show all electron flow with arrows and include all intermediate structures.

ANSWER:

POINTS: 1 10. Aniline reacts with nitrous acid, HNO2, to yield a stable diazonium salt. This diazonium salt undergoes electrophilic aromatic substitution on activated aromatic rings to yield brightly colored azo compounds that are widely used as dyes.

The intermediate structures for the mechanism of this reaction are given below. Show all electron flow with arrows for this mechanism on the structures provided.

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Page 2

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1 Exhibit 16-3 Consider the data below to answer the following question(s).

The −NH2 group is listed in our textbook as the strongest o,p-directing activator in electrophilic aromatic substitution reactions. However, when aniline is subjected to standard nitration conditions poor yields of m-nitroaniline result.

11. Refer to Exhibit 16-3. Draw all the resonance forms of aniline showing the electron-donating effect of the −NH2 substituent.

ANSWER:

POINTS: 1 12. Refer to Exhibit 16-3. Clearly, the reaction conditions are influencing the directing effect of the −NH2 group. Explain why this occurs, using both words and structures.

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Page 3

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

When aniline is placed in strong acid the nitrogen atom is protonated. Electrophilic aromatic substitution on the anilinium ion⎯whose aromatic ring is now deactivated by a positively charged substituent⎯occurs primarily at the meta position since this keeps the positive charge of the intermediate carbocation away from the positively charged nitrogen. The intermediates shown above for ortho and para substitution are destabilized more than the intermediate for meta substitution, so m-nitroaniline is the major product.

POINTS: 1 Exhibit 16-4 Consider the reaction below to answer the following question(s).

13. Refer to Exhibit 16-4. Write the complete stepwise mechanism for the formation of the ortho product. Show all intermediate structures and show all electron flow with arrows.

ANSWER:

POINTS: 1 14. Refer to Exhibit 16-4. Draw resonance structures for the intermediate carbocation that explain the directing effect of the −Br. ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 4

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution Exhibit 16-5 Rank the compounds in each group below according to their reactivity toward electrophilic aromatic substitution (most reactive = 1; least reactive = 3). Place the number corresponding to the compounds' relative reactivity in the blank below the compound.

15.

ANSWER:

POINTS: 1

16.

ANSWER:

POINTS: 1 17. At what position, and on what ring, is bromination of phenyl benzoate expected to occur? Explain your answer.

ANSWER:

Attack occurs in the activated ring and yields ortho and para bromination.

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Page 5

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution POINTS: 1 Exhibit 16-6 To answer the following question(s), refer to the data below: Isobutylbenzene is the starting material for the industrial synthesis of the NSAID, ibuprofen.

18. Refer to Exhibit 16-6. Attempts to prepare isobutylbenzene by direct Friedel-Crafts alkylation of benzene result in tert-butylbenzene as the major product. Write the complete stepwise mechanism for this reaction, showing all electron flow with arrows and showing all intermediate structures.

ANSWER:

POINTS: 1 19. Refer to Exhibit 16-6. Propose a synthesis for isobutylbenzene which avoids the problems of direct Friedel-Crafts alkylation.

ANSWER:

POINTS: 1 20. Would you expect (nitromethyl)benzene to be more reactive or less reactive than toluene toward electrophilic substitution? Explain.

ANSWER: (Nitromethyl)benzene should be less reactive than toluene owing to the strong inductive electron withdrawing effect of the nitro group.

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Page 6

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution POINTS: 1 Exhibit 16-7 Consider the reaction below to answer the following question(s).

21. Refer to Exhibit 16-7. The nucleophile in the reaction is: ANSWER: A POINTS: 1 22. Refer to Exhibit 16-7. The Lewis acid catalyst in the reaction is: ANSWER: C POINTS: 1 23. Refer to Exhibit 16-7. This reaction proceeds _____ (faster or slower) than benzene.

ANSWER: slower POINTS: 1 24. Refer to Exhibit 16-7. Draw the structure of product D. ANSWER:

POINTS: 1 25. The following reaction proceeds by an intramolecular nucleophilic aromatic substitution mechanism. Write the complete stepwise mechanism, showing all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 26. Tetracyclone is often used to trap benzynes as Diels-Alder adducts. What is the structure of the Diels-Alder adduct that results when benzyne is trapped by tetracyclone?

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Page 7

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution

ANSWER:

POINTS: 1 27. On the structural intermediates below, show all electron flow with arrows for the nucleophilic aromatic substitution reaction of p-nitrochlorobenzene with KOH.

ANSWER:

POINTS: 1 Exhibit 16-8 Give the major organic product(s) of each of the following reactions. If none is predicted, write "N.R."

28.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 8

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution

29.

ANSWER: N.R. Nucleophilic aromatic substitution requires an electron-withdrawing group in the ortho or para position.

A methyl group is electron donating, and substitution will not occur.

POINTS: 1

30.

ANSWER:

POINTS: 1

31.

ANSWER:

POINTS: 1

32.

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Page 9

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1

34.

ANSWER:

POINTS: 1

35.

ANSWER:

POINTS: 1

36.

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Page 10

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1

38.

ANSWER:

POINTS: 1

39.

ANSWER:

POINTS: 1

40.

ANSWER:

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Page 11

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution (Friedel-Crafts reactions do not proceed in the presence of a deactivating group) POINTS: 1

41.

ANSWER:

POINTS: 1

42.

ANSWER:

POINTS: 1

43.

ANSWER:

POINTS: 1

44.

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Page 12

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1 Exhibit 16-9 Choose the best reagent(s) from the list provided below for carrying out the following conversions. Place the letter of the reagent in the box beside the reaction number over the arrow. There is only one answer for each reaction. a. b. c. d. e.

KMnO4, H3O+ Br2, FeBr3 Cl2, FeCl3 CH3Cl, AlCl3 HNO3, H2SO4

f. g. h. i. j. k.

ClCO(CH2)2CH3, AlCl3 CH3CH2CH2CH2Cl, AlCl3 H2/Pd NBS, peroxides (CH3)3CCH2Cl F-TEDA-BF4

45.

ANSWER:

POINTS: 1

46.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 13

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution

47.

ANSWER:

POINTS: 1

48.

ANSWER:

POINTS: 1

49.

ANSWER:

POINTS: 1 Exhibit 16-10 Propose syntheses to carry out each of the following conversions. Assume ortho and para isomers can be separated.

50.

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Page 14

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1

51.

ANSWER:

POINTS: 1

52.

ANSWER:

POINTS: 1

53.

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Page 15

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1

54.

ANSWER:

POINTS: 1

55.

ANSWER:

POINTS: 1 Exhibit 16-11 Propose syntheses of the following compounds starting with benzene or toluene. Assume ortho and para isomers can be separated.

56.

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Page 16

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER:

POINTS: 1

57.

ANSWER:

POINTS: 1 58. It is possible to synthesize 1-bromo-4-nitrobenzene by reacting bromobenzene with a mixture of nitric and sulfuric acids. However, 1-bromo-4-nitrobenzene is not obtained if nitrobenzene is reacted with Br2/FeBr3. Explain the difference.

ANSWER: Bromine is an ortho-para director. The para isomer preferentially forms on nitration due to the steric hindrance of the bulky Br group. The result is a good yield of 1-bromo-4-nitrobenzene. NO2 is a meta director and preferentially forms the meta isomer on bromination. The resulting product is the meta isomer or 1-bromo-3nitrobenzene.

POINTS: 1 59. Which of the following could successfully undergo a Friedel-Crafts alkylation? Assume an appropriate catalyst. a. chlorobenzene reacting with benzene b. 2-chloroethene reacting with benzene c. 2-chloropropane reacting with benzaldehyde d. 2-chlorobutane reacting with benzene ANSWER: d POINTS: 1 60. Which of the following is an ortho- and para- director?

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Page 17

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution a. —NO2

b. —NH2

c. —CN

d. —CO2CH3

ANSWER: b POINTS: 1 61. Which of the following is a meta- director? a. —OH b. —Br c. —CH2CH2CH3 d. —COCH3 ANSWER: d POINTS: 1 62. Which of the following is a meta-deactivating group? a. —CO2H b. —I c. —OH

d. —CH3

ANSWER: a POINTS: 1 63. Which of the following is an ortho- and para- director and a deactivator? a. —NO2 b. —F c. —CN d. —CH2CH3 e. none of these ANSWER: b POINTS: 1 64. What is(are) the product(s) when the following substance reacts with Br2 in the presence of FeBr3?

a. o-bromotoluene b. m-bromotoluene c. p-bromotoluene d. an equimolar mixture of o-bromotoluene and p-bromotoluene e. a mixture of o-bromotoluene (minor) and p-bromotoluene (major) f. a mixture of o-bromotoluene (major) and p-bromotoluene (minor) Cengage Learning Testing, Powered by Cognero

Page 18

Chapter 16 - Chemistry of Benzene: Electrophilic Aromatic Substitution ANSWER: e POINTS: 1 65. Arrange the following compounds in order of increasing reactivity to electrophilic aromatic substitution. bromobenzene nitrobenzene benzene phenol a. phenol < benzene < bromobenzene < nitrobenzene c. nitrobenzene < bromobenene < benzene < phenol ANSWER: c POINTS: 1

b. bromobenzene < nitrobenzene < benzene < phenol d. nitrobenzene < benzene < bromobenzene < phenol

66. When the following compound is treated with a mixture of nitric and sulfuric acids, which of the following will be produced?

a. 5-chloro-2-nitrotoluene b. 3-chloro-4-nitrotoluene c. 3-chloro-2-nitrotoluene d. 5-chloro-3-nitrotoluene e. a mixture of 3-chloro-4-nitrotoluene (major) and 5-chloro-2-nitrotoluene (minor) f. an equimolar mixture of 3-chloro-2-nitrotoluene, 5-chloro-2-nitrotoluene, and 3-chloro-4-nitrotoluene g. a mixture a, b, c, and d ANSWER: e POINTS: 1

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Page 19

Chapter 17 - Alcohols and Phenols Drawing Instructions: Draw structures corresponding to each of the given names.

1. Draw: cis-4-tert-butylcyclohexanol ANSWER:

POINTS: 1 2. Draw: 3-methyl-2-buten-1-ol ANSWER:

POINTS: 1 3. Draw: 2-phenyl-2-propanol ANSWER:

POINTS: 1 4. Draw: glycerol ANSWER: POINTS: 1 5. Draw: 2, 4, 6-trinitrophenol ANSWER:

POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

6. Name:

ANSWER: (E)-2-ethylbut-2-en-1-ol POINTS: 1 7. Name:

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Page 1

Chapter 17 - Alcohols and Phenols HOCH2CH2OH ANSWER: 1, 2-ethanediol or ethylene glycol POINTS: 1 8. Name:

ANSWER: allyl alcohol or 2-propen-1-ol POINTS: 1 9. Name:

ANSWER: hydroquinone or p-dihydroxybenzene or 1,4-benzenediol POINTS: 1 Exhibit 17-1 Rank the following groups of compounds from most acidic (1) to least acidic (4). Place the number corresponding to the compound's relative rank in the blank below the structure.

10.

ANSWER:

POINTS: 1

11.

ANSWER:

POINTS: 1

12.

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Page 2

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1 13. Explain why nonafluoro-2-methyl-2-propoxide is a much weaker base than tert-butoxide.

ANSWER: Nonafluoro-2-methyl-2-propoxide anion is more stable than tert-butoxide anion because the electronwithdrawing fluoride atoms inductively delocalize the negative charge. Since nonafluoro-2-methyl-2propoxide is more stable than tert-butoxide, it is a weaker base.

POINTS: 1 Exhibit 17-2 Refer to the data below to answer the following question(s).

pKas of Some Phenols

Y

pKa

−H

9.89

m-NO2

8.28

p-NO2

7.17

m-OCH3

9.65

p-OCH3

10.21

14. Refer to Exhibit 17-2. The weakest acid in the table is: ANSWER: p-methoxyphenol POINTS: 1 15. Refer to Exhibit 17-2. Which of the acids in the Table has the weakest conjugate base?

ANSWER: p-nitrophenol POINTS: 1 16. Refer to Exhibit 17-2. How do you account for the difference in acidity between meta and para-nitrophenol?

ANSWER: In m-nitrophenol, the inductive effect of the electron-withdrawing nitro group helps to stabilize the negative charge on oxygen. However, when the nitro group is para to the oxygen, direct conjugation of the negative charge on oxygen with the nitro group can occur. p-Nitrophenolate ion is, thus, more stable than mnitrophenolate ion, and, as a result, forms more readily.

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Page 3

Chapter 17 - Alcohols and Phenols

POINTS: 1 Exhibit 17-3 To answer the following question(s), consider the reaction below:

17. Refer to Exhibit 17-3. The best reagents for accomplishing the above transformation are: a.

1. OsO4, pyridine 2. NaHSO3, H2O b. 1. Hg(OAc)2, H2O 2. NaBH4 c. 1. RCO3H, CH2Cl2 2. H3O+ d. 1. BH3, THF 2. H2O2, −OH ANSWER: d POINTS: 1 18. Refer to Exhibit 17-3. The alcohol product is classified as a: a. 1° alcohol b. 2° alcohol c. 3° alcohol d. 4° alcohol ANSWER: b POINTS: 1 19. Refer to Exhibit 17-3. On the templates provided below, draw both conformations of the alcohol product. Circle the least stable conformation.

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Page 4

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1 20. Refer to Exhibit 17-3. Provide the IUPAC name for the product alcohol.

ANSWER: trans-2-Methylcyclohexanol POINTS: 1 Exhibit 17-4 To answer the following question(s), consider the reaction below:

21. Refer to Exhibit 17-4. On the structures provided below, draw arrows which account for the complete stepwise mechanism for this reaction.

ANSWER:

POINTS: 1 22. Refer to Exhibit 17-4. The dehydration of secondary and tertiary alcohols by reaction with POCl3 in pyridine is an example of: a. an E1 process b. an SN1 process c. an E2 process d. an SN2 process ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 5

Chapter 17 - Alcohols and Phenols 23. Refer to Exhibit 17-4. Why is 3-methylcyclohexene the major product of this reaction instead of 1-methycyclohexene?

ANSWER: In E2 elimination, dehydration proceeds most readily when the two groups to be eliminated have a transdiaxial relationship. In this compound, the only hydrogen with the proper geometric relationship to the −OH group is at C6 so the major product of this reaction is 3-methylcyclohexene.

POINTS: 1 24. Acid-catalyzed dehydration of 2,2-dimethylcyclohexanol yields 1,2-dimethylcyclohexene as one of the major products. Write the complete stepwise mechanism for this reaction. Show all electron flow with arrows and show all intermediate structures.

ANSWER:

POINTS: 1 Exhibit 17-5 To answer the following question(s), consider the reaction below:

25. Refer to Exhibit 17-5. Provide the complete IUPAC name for the starting material in this reaction.

ANSWER: (R)-2-heptanol POINTS: 1 26. Refer to Exhibit 17-5. The starting material can be classified as a: a. 1° alcohol b. 2° alcohol c. 3° alcohol d. 4° alcohol ANSWER: b POINTS: 1 27. Refer to Exhibit 17-5. Write the complete stepwise mechanism for the reaction. Show all intermediates and all electron flow with arrows.

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Page 6

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1 28. Refer to Exhibit 17-5. The conversion of an alcohol into an alkyl chloride by reaction with SOC12 is an example of: a. an E1 process b. an SN1 process c. an E2 process d. an SN2 process ANSWER: d POINTS: 1 Exhibit 17-6 Consider the Grignard reaction below to answer the following question(s).

29. Refer to Exhibit 17-6. The electrophile in this reaction is: ANSWER: B POINTS: 1 30. Refer to Exhibit 17-6. The nucleophile in this reaction is: ANSWER: A POINTS: 1 31. Refer to Exhibit 17-6. The alcohol product can be classified as a: a. 1° alcohol b. 2° alcohol c. 3° alcohol d. 4° alcohol ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 7

Chapter 17 - Alcohols and Phenols Exhibit 17-7 A highly useful and general method for the synthesis of alcohols is the addition of Grignard reagents to carbonyl compounds. Show what Grignard reagent and what carbonyl compound you would start with to prepare each alcohol below. List all possibilities.

32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1

34.

ANSWER:

POINTS: 1

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Page 8

Chapter 17 - Alcohols and Phenols 35.

ANSWER:

POINTS: 1 Exhibit 17-8 Give the major organic product(s) of the following reactions or sequences of reactions. Show all relevant stereochemistry.

36.

ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1

38.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 9

Chapter 17 - Alcohols and Phenols

39.

ANSWER:

POINTS: 1

40.

ANSWER:

POINTS: 1

41.

ANSWER:

POINTS: 1

42.

ANSWER:

POINTS: 1

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Page 10

Chapter 17 - Alcohols and Phenols 43.

ANSWER:

POINTS: 1

44.

ANSWER:

POINTS: 1

45.

ANSWER:

POINTS: 1

46.

ANSWER:

POINTS: 1

47.

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Page 11

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1

48.

ANSWER:

POINTS: 1

49.

ANSWER:

POINTS: 1 Exhibit 17-9 Choose the best reagent(s) for carrying out the following conversions from the list provided below. Place the letter of the best choice in the blank to the left of the conversion. Reagents may be used more than once. a. b.

1.

2.

1.

2.

CH3MgBr, ether H3O+ PBr3 NaOH

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e.

1.

2.

p-TosCl, pyridine NaOH

f. g.

CrO3, acetone, H2O 1. NaBH4, ethanol Page 12

Chapter 17 - Alcohols and Phenols c.

d.

1.

2.

3.

1.

2.

(CH3)3SiCl, (CH3CH2)3N CH3MgBr, ether H3O+ LiAlH4, ether H3O+

H3O+ Dess–Martin periodinane , CH2Cl2 2.

h.

50. _____ ANSWER: f POINTS: 1

51. _____

ANSWER: g POINTS: 1

52. _____ ANSWER: h POINTS: 1 53. Propose a synthesis of Dimestrol starting from p-methoxypropiophenone as the only source of carbon.

ANSWER:

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Page 13

Chapter 17 - Alcohols and Phenols POINTS: 1 54. Outline the synthetic steps necessary to carry out the conversion below. You may use any organic or inorganic reagents you need. Show the structures of all intermediate compounds that would probably be isolated during the course of your synthesis, and show all necessary reagents.

ANSWER:

POINTS: 1 55. Synthesize the following alcohol starting with cyclohexene and bromocylopentane as the only organic starting materials. Show all reagents and all intermediates in your synthesis.

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Page 14

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1 Exhibit 17-10 Propose structures for alcohols that have the following 1H NMR spectra.

56. C3H8O

Spectrum obtained from: SBDSWeb: http://www.aist.go.jp/RIODB/SDBS/ Cengage Learning Testing, Powered by Cognero

Page 15

Chapter 17 - Alcohols and Phenols ANSWER: isopropyl alcohol (2-propanol), (CH3)2CHOH POINTS: 1 57. C7H8O (neat solution; no solvent)

Spectrum obtained from: SBDSWeb: http://www.aist.go.jp/RIODB/SDBS ANSWER: benzyl alcohol

POINTS: 1 58. 2,4,6-trinitrophenol is a stronger acid than phosphoric acid. Another name for this compound is picric acid. Explain the acidity of picric acid.

ANSWER: Picric acid’s acidity is partly due to the inductive effect of the three electron-withdrawing nitro groups. In addition, substitution of a nitro group at the ortho and para positions increases resonance stabilization because the negative charge of the phenoxide ion can be delocalized onto an oxygen of the nitro group. These two effects of the three nitro groups increase the acidity of picric acid substantially.

POINTS: 1 59. Give the IUPAC name for the following compound.

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Page 16

Chapter 17 - Alcohols and Phenols

ANSWER: 1-ethylcyclohexanol POINTS: 1 60. Which of the following would have the highest boiling point? a. butane b. 1- butanol c. chloropropane d. pentane ANSWER: b POINTS: 1 61. Which of the following would be the most acidic? a. p-nitrophenol b. phenol c. p-chlorophenol d. p-methylphenol ANSWER: a POINTS: 1 62. Classify the following alcohol. Atoms other than carbon and hydrogen are labeled.

a. primary b. secondary c. tertiary ANSWER: c POINTS: 1 63. Which of the peak in the IR spectrum shown is due to the C—O stretch in an alcohol?

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Page 17

Chapter 17 - Alcohols and Phenols

a. 1 b. 2 c. 3 d. 4 ANSWER: c POINTS: 1 64. In a proton NMR, the signal for the —OH proton a. will generally be upfield from the corresponding alkyl signal. b. is generally unsplit by adjacent protons. c. is lost due to exchange with acidic impurities. d. covers a broader chemical shift range than a phenol —OH. ANSWER: b POINTS: 1 65. Which signal in the following proton NMR is due to an alcohol proton?

a. 1 b. 2 c. 3 d. 4 e. This cannot be determined from the data presented in the spectrum.

ANSWER: c POINTS: 1 66. What product is formed when the following substance is treated with CrO3 under appropriate conditions?

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Page 18

Chapter 17 - Alcohols and Phenols

a.

b.

c.

d.

e.

No reaction occurs under these conditions.

ANSWER: c POINTS: 1 67. Treatment of the following compound with Dess–Martin periodinane in CH2Cl2

would produce which of the following compounds? a.

b.

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Page 19

Chapter 17 - Alcohols and Phenols c.

d.

e. f.

Either a or d Dess–Martin periodinane will not affect this substance.

ANSWER: a POINTS: 1 68. The mass spectrum for the following compound would have a peak at which m/z ratio due to dehydration? Atoms other than carbon and hydrogen are labeled.

a. 116 b. 98 c. 87 d. 59 ANSWER: b POINTS: 1 69. Draw the resonance forms of 4-nitro-m-cresol.

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Page 20

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1 70. Draw the mechanism for the acid-catalyzed dehydration of 2-methyl-2-pentanol to yield 2-methyl-1-butene.

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Page 21

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1 71. Draw the mechanism of the E2 reaction involving the dehydration of 3-methyl-1-cyclohexanol, when reacting with POCl3 in pyridine.

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Page 22

Chapter 17 - Alcohols and Phenols ANSWER:

POINTS: 1

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Page 23

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides Drawing Instructions: Draw structures corresponding to each of the given names.

1. Draw: tetrahydrofuran ANSWER:

POINTS: 1 2. Draw: 12-crown-4 ANSWER:

POINTS: 1 3. Draw: 1-isopropoxycyclopentene ANSWER:

POINTS: 1 4. Draw: allyl benzyl ether ANSWER:

POINTS: 1 5. Draw: diethyl ether ANSWER: CH3CH2−O−CH2CH3 POINTS: 1 6. Draw: 3-methyl-1-butanethiol ANSWER:

POINTS: 1 7. Draw: cyclopropyl ethyl sulfide ANSWER: POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

8. Name: Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

ANSWER: m-chloroanisole or m-chloromethoxybenzene or (m-chlorophenyl) methyl ether POINTS: 1 9. Name:

ANSWER: 1-ethoxy-2,2-dimethylpropane or ethyl neopentyl ether POINTS: 1 10. Name:

ANSWER: dicyclohexyl sulfide POINTS: 1 Exhibit 18-1 Consider the reaction below to answer the following question(s).

11. Refer to Exhibit 18-1. Write the complete stepwise mechanism for the reaction. Show all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 12. Refer to Exhibit 18-1. Mechanistically, the Williamson ether synthesis outlined above is: a. an E1 process b. an SN1 process c. an E2 process d. an SN2 process ANSWER: d Cengage Learning Testing, Powered by Cognero

Page 2

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides POINTS: 1 13. Refer to Exhibit 18-1. Alternatively, cyclopentyl methyl ether may be synthesized from cyclopentene. Outline a synthesis of cyclopentyl methyl ether from cyclopentene.

ANSWER:

POINTS: 1 14. Diphenyl ether is inert to cleavage by HI or HBr. Explain.

ANSWER: Acidic ether cleavages are typical nucleophilic substitution reactions that take place by either an SN1 or SN2 pathway. Diaryl ethers like diphenyl ether are inert under ordinary acidic conditions because the generation of an aryl cation is an SN1 process that is energetically unfavorable, and "back-side" attack at the ring carbon bound to oxygen is sterically unfavorable.

POINTS: 1 15. Write the complete stepwise mechanism for the reaction below. Show all electron flow with arrows and draw all intermediate structures.

ANSWER:

POINTS: 1 Exhibit 18-2 Consider the data below to answer the following question(s).

Acetals are a special class of ethers which have two ether linkages at a single carbon. Tetrahydropyranyl ethers are acetals that are easily cleaved with mild aqueous acids at room temperature to yield alcohols.

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

16. Refer to Exhibit 18-2. Draw arrows showing the electron flow for this reaction on the structures provided below.

ANSWER:

POINTS: 1 17. Refer to Exhibit 18-2. This ether cleavage occurs so easily under such mild conditions because the carbocation intermediate in step two is very stable. Explain why this carbocation is stabilized.

ANSWER: The carbocation generated in the first step of the SN1 process is resonance-stabilized by the adjacent oxygen lone pairs.

POINTS: 1 18. When 1,2-epoxypropane is treated with sodium ethoxide in ethanol 1-ethoxy-2-propanol is the major product while only a trace amount of 2-ethoxy-1-propanol is produced. Explain these results based on the reaction mechanism.

ANSWER: Basic nucleophiles like ethoxide react with epoxides in a typical SN2-type process, so attack takes place at the less hindered epoxide carbon.

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

POINTS: 1 Exhibit 18-3 Consider the data below to answer the following question(s).

Epoxides are synthesized industrially in one step by silver oxide air oxidation of ethylene and on a laboratory scale in one step by treating an alkene with m-chloroperoxybenzoic acid. An alternative two-step process converts alkenes to halohydrins, which are converted by treatment with base to epoxides.

19. Refer to Exhibit 18-3. Show electron flow with arrows on the structures provided below for each step in the above transformation.

ANSWER:

POINTS: 1 20. Refer to Exhibit 18-3. The synthesis of epoxides by base treatment of halohydrins is an example of an intramolecular: a.

SN1 reaction

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Page 5

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides b. hydrolysis reaction c. dehydration reaction d. Williamson ether synthesis ANSWER: d POINTS: 1 Exhibit 18-4 Propose structure(s) for the starting material(s), reagent(s), or major organic product(s) of the following reactions or sequences of reactions. Show all relevant stereochemistry.

21.

ANSWER:

POINTS: 1

22.

ANSWER:

POINTS: 1

23.

ANSWER:

POINTS: 1

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

24.

ANSWER:

POINTS: 1

25.

ANSWER:

POINTS: 1

26.

ANSWER:

POINTS: 1

27.

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Page 7

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1

30.

ANSWER:

POINTS: 1

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

31.

ANSWER:

POINTS: 1

32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1 34. Propose a synthesis of 1,4-dioxane, starting from 1,2-dibromoethane as the only source of carbon.

ANSWER:

POINTS: 1 35. Choose the best reagent for carrying out the following reactions from the list below. Place the letter of the reagent(s) in the box over the reaction arrow. Use only one letter per box.

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

A. NaH, then CH3I

B. NaOCH3, CH3OH

C. m-ClC6H4CO3H

D. CH3MgBr in ether, then H3O+

E. warm H2SO4/H2O

F.

G. H2/Pd

H. PCC, CH2Cl2

I. Cl2, H2O ANSWER:

J.

Hg(O2CCF3)2, CH3OH

LiAlH4 in ether, then H3O+

POINTS: 1 36. Complete the synthetic sequence below by drawing the structures of the reaction in the boxes provided.

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Page 10

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

ANSWER:

POINTS: 1 37. Show how the ether below could be prepared from toluene and any other necessary reagents. Show all reagents and all intermediate structures.

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Page 11

Chapter 18 - Ethers and Epoxides; Thiols and Sulfides ANSWER:

POINTS: 1 38. Propose a synthesis for the following compound using benzene or toluene and any other reagents necessary. Show all major intermediate compounds that would probably be isolated during the course of your synthesis.

ANSWER:

POINTS: 1 39. Name the compound shown below. Atoms other than carbon and hydrogen are labeled.

ANSWER: Ethyl hexyl ether or 1-ethyoxyhexane POINTS: 1 40. Name the following compound. Atoms other than carbon and hydrogen are labeled.

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

ANSWER: o-mercaptotoluene POINTS: 1 41. Predict the product of the following reaction.

ANSWER:

POINTS: 1 42. Draw the structure of the ether that could be used to produce the following compound.

ANSWER:

POINTS: 1 43. Although not always evident in an IR spectrum, which peak in the IR spectrum below might be due to the C—O stretch of an ether?

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

a. 1 b. 2 c. 3 d. 4 ANSWER: b POINTS: 1 44. The following is the proton NMR of a symmetrical alkyl ether. Which peak is most likely due to the protons adjacent to the ether oxygen atom?

a. 1 b. 2 c. 3 ANSWER: c POINTS: 1 45. How many peaks would appear in the 13CNMR for the following compound? Atoms other than carbon and hydrogen are labeled.

a. 4 c. 6

b. 5 d. 7

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides e. 8 f. 9 ANSWER: e POINTS: 1 46. Consider the following species. Atoms other than carbon and hydrogen are labeled.

Which of the following would be produced upon treatment with I2? a. b.

c.

d.

ANSWER: a POINTS: 1 47. What is the product formed upon treatment of the following substance with hydrogen peroxide at room temperature, followed by treatment with a peroxyacid? Atoms other than carbon and hydrogen are labeled.

a.

b.

c.

d.

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Chapter 18 - Ethers and Epoxides; Thiols and Sulfides

ANSWER: b POINTS: 1 48. Which of the following will undergo rearrangement upon heating? a. 2-butenyl phenyl ether b. 1-propenyl 2-propenyl ether c. 2-pentenyl phenyl ether d. All will undergo rearrangement.

ANSWER: d POINTS: 1

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Page 16

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions Drawing Instructions: Draw structures corresponding to each of the given names.

1. Draw: 2,2,2-trichloroethanal (chloral) ANSWER:

POINTS: 1 2. Draw: benzophenone ANSWER:

POINTS: 1 3. Draw: trans-3-isopropylcyclohexanecarbaldehyde ANSWER:

POINTS: 1 4. Draw: 5,5-dimethyl-1,3-cyclohexanedione (dimedone) ANSWER:

POINTS: 1 5. Draw: 2-cyclohexenone ANSWER:

POINTS: 1 6. What is the correct structure for 2-hydroxyacetophenone?

a.

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b.

Page 1

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

c.

d.

ANSWER: c POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

7. Name:

ANSWER: 3-nitrobenzaldehyde POINTS: 1 8. Name:

ANSWER: 4,8-dimethyl-7-nonen-2-one POINTS: 1 9. Name:

ANSWER: 1-phenyl-2-propanone POINTS: 1 10. Name:

ANSWER: cis-1,3-cyclopentanedicarbaldehyde POINTS: 1 11. Name: Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

ANSWER: 5-oxoheptanal POINTS: 1 Exhibit 19-1 Consider the reaction below to answer the following question(s):

12. Refer to Exhibit 19-1. Write the complete stepwise mechanism for the reaction shown above. Show all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 13. Refer to Exhibit 19-1. The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an: a. vicinal diol b. geminal diol c. acetal d. ketal ANSWER: b POINTS: 1 14. Refer to Exhibit 19-1. The exact position of the equilibrium between ketones/aldehydes and their hydrates depends on the structure of the carbonyl compound. Although the equilibrium favors the carbonyl compound in most cases, cyclopropanone forms a stable hydrate. Explain this phenomenon based on the structures of cyclopropanone and its hydrate.

ANSWER: The carbonyl carbon in cyclopropane is highly strained because the preferred bond angle for an sp2-hybridized carbon is 120° but the real angle is 60°. This angle strain is somewhat relieved when cyclopropanone is hydrated⎯the preferred bond angle for an sp3-hybridized carbon is about 109.5°. Going from sp2 to sp3 hybridization relieves angle strain by about 10°. Thus, the hydrate of cyclopropane is more stable than the carbonyl form, and the equilibrium lies to the right in this reaction.

POINTS: 1 15. Refer to Exhibit 19-1. Many nucleophilic addition reactions of aldehydes and ketones are catalyzed by acid or base.

Bases catalyze hydration by: a. making the carbonyl group more electrophilic b. shifting the equilibrium of the reaction c. making the carbonyl group less electrophilic d. converting the water to hydroxide ion, a much better nucleophile ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 3

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions 16. This question was omitted on the printed copy. This placeholder question is here to maintain the numbering system integrity between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 17. Enamines formed from the cyclic secondary amine pyrrolidine are important intermediates in the synthesis of 1,5diketones.

On the structures provided below, draw arrows showing electron flow for the reaction mechanism for the acetic acidcatalyzed formation of an enamine from cyclohexanone and pyrrolidine.

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Page 4

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER:

POINTS: 1 Exhibit 19-2 Consider the data below to answer the following question(s).

Cyanohydrins are important intermediates in the synthesis of α-hydroxycarboxylic acids from ketones and aldehydes. The nitrile functional group can be hydrolyzed by aqueous acid to yield a carboxylic acid. Nitriles can also be hydrolyzed to carboxylic acids using aqueous base. Unfortunately, when a cyanohydrin is treated with aqueous base the original carbonyl compound is isolated.

18. Refer to Exhibit 19-2. The reaction of an aldehyde with hydrogen cyanide is an example of _____ reaction. a. a nucleophilic substitution b. an electrophilic addition c. an electrophilic substitution d. a nucleophilic addition ANSWER: d POINTS: 1 19. Refer to Exhibit 19-2. Identify the electrophile in the reaction of benzaldehyde with hydrogen cyanide.

ANSWER: The benzaldehyde is the electrophile in this reaction.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 5

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions 20. Refer to Exhibit 19-2. The reaction of benzaldehyde with hydrogen cyanide is catalyzed by the addition of a small amount of KCN. Write the complete reaction mechanism for the KCN-catalyzed reaction. Show all electron flow with arrows and show all intermediate structures.

ANSWER:

POINTS: 1 21. Refer to Exhibit 19-2. Formulate a reasonable mechanism for the reaction of the cyanohydrin of benzaldeyhe, shown above, with aqueous NaOH. Show all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 22. In the Wittig reaction, a phosphorus ylide adds to a ketone or aldehyde to yield an alkene. Write the complete stepwise mechanism for the Wittig reaction shown below. Show all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 Exhibit 19-3 Consider the reaction below to answer the following question(s).

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

23. Refer to Exhibit 19-3. The nucleophile in this reaction is: ANSWER: B POINTS: 1 24. Refer to Exhibit 19-3. The catalyst in this reaction is: ANSWER: C POINTS: 1 25. Refer to Exhibit 19-3. The product of this reaction is called: a. an ylide b. an acetal c. a gem diol d. a hydrate ANSWER: b POINTS: 1 26. Refer to Exhibit 19-3. Write the complete stepwise mechanism for this reaction. Show all electron flow with arrows and show all intermediate structures.

ANSWER:

POINTS: 1 Exhibit 19-4 α,β-Unsaturated aldehydes and ketones can undergo reaction with nucleophiles at the β carbon, as shown below.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

27. Refer to Exhibit 19-4. Draw a resonance form for the unsaturated carbonyl that accounts for this reactivity.

ANSWER:

POINTS: 1 28. Refer to Exhibit 19-4. This reaction is called _____ reaction. a. a 1,2-conjugate addition. b. an electrophilic addition. c. a direct addition d. a 1,2-addition.

ANSWER: a POINTS: 1 Exhibit 19-5 Give the major organic product(s) for each of the following reactions or sequences of reactions. Show all relevant stereochemistry.

29.

ANSWER:

POINTS: 1

30.

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Page 8

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER:

POINTS: 1

31.

ANSWER:

POINTS: 1

32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions 34.

ANSWER:

POINTS: 1

35.

ANSWER:

POINTS: 1

36.

ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions 38.

ANSWER:

POINTS: 1

39.

ANSWER: POINTS: 1

40.

ANSWER:

POINTS: 1

41.

ANSWER:

POINTS: 1 Exhibit 19-6 Choose the BEST reagent for carrying out each of the following conversions.

42.

a.

LiAlH4, THF

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions b. c.

NaBH4, ethanol 1. DIBAH, toluene 2. H3O+ d. All of the above work well ANSWER: c POINTS: 1

43.

a. LiAlH4, ether b. NaBH4, ethanol c. CrO3, pyridine d. H2/Pd ANSWER: b POINTS: 1

44.

a.

1.

2.

1.

2.

PhMgBr, ether H3O+ b.

PhCH2MgBr, ether H3O + c. (C6H5)3P=CHC6H5, THF d. Li(C6H5)2Cu, ether ANSWER: d POINTS: 1

45.

a. b.

1.

2.

1.

2.

PhMgBr, ether H3O+ PhCH2MgBr, ether H3O+

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions c. (C6H5)3P=CHC6H5, THF d. Li(C6H5)2Cu, ether ANSWER: c POINTS: 1

46.

a. NaBH4, ethanol b. CH2PPh3 c. NaOH, H2O d. All of the above ANSWER: a POINTS: 1 Exhibit 19-7 Show how the Wittig reaction might be used to prepare each of the following alkenes. Identify the alkyl halide and the carbonyl components that would be used in each synthesis.

47.

ANSWER:

POINTS: 1

48.

ANSWER:

POINTS: 1 Exhibit 19-8 Show how the following conversions might be accomplished. Show all reagents and all intermediate structures. More than one step may be required.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

49.

ANSWER:

POINTS: 1

50.

ANSWER:

POINTS: 1

51.

ANSWER:

POINTS: 1

52.

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Page 14

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER:

POINTS: 1 Exhibit 19-9 Consider the data below to answer the following question(s).

C7H14O IR: MS: 1

H NMR:

1715 cm−1 M+ at m/z = 114, α-cleavage fragment at m/z = 71, McLafferty rearrangement fragment at m/z = 86.

0.92 δ (6H, triplet), 1.59 δ (4H, multiplet), 2.36 δ (4H, triplet)

53. Refer to Exhibit 19-9. Calculate the degree of unsaturation for this compound.

ANSWER: The base formula for C7H14O is C7H14. The formula for the saturated compound is C7H16, so the compound has one degree of unsaturation.

POINTS: 1 54. Refer to Exhibit 19-9. What functional group is indicated by the IR data?

ANSWER: Absorption at 1715 cm−1 in the infrared spectrum indicates the presence of a carbonyl compound, most probably a ketone.

POINTS: 1 55. Refer to Exhibit 19-9. Interpret the mass spectral data.

ANSWER: A fragment at m/z = 71 indicates a loss of 43, or a propyl group, from α-cleavage. A fragment at m/z = 86 indicates a loss of 28, or ethylene, from McLafferty rearrangement (transfer of a hydrogen atom from the gamma carbon to the carbonyl oxygen with concomitant breaking of the bond between the alpha and beta carbon).

POINTS: 1 56. Refer to Exhibit 19-9. Interpret the 1H NMR data.

ANSWER: The 1H NMR indicates that there are only three different kinds of hydrogen in the molecule. The 6H triplet is due to two equivalent CH3 groups next to two equivalent CH2 groups, the 4H triplet is due to two equivalent CH2 groups next to two other equivalent CH2 groups, shifted by a C=O, and the 4H multiplet is two equivalent CH2 groups between a CH2 and a CH3.

POINTS: 1 57. Refer to Exhibit 19-9. Propose a structure consistent with the spectral data presented above.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER: 4-heptanone, CH3CH2CH2COCH2CH2CH3 POINTS: 1 Exhibit 19-10 Consider the data below to answer the following question(s).

C7H14O IR: MS: 1

1715 cm−1 M+ at m/z = 114, α-cleavage fragment at m/z = 71, no McLafferty rearrangement fragment

H NMR:

Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS 58. Refer to Exhibit 19-10. Interpret the mass spectral data for this compound.

ANSWER: A fragment at m/z = 71 indicates a loss of 43, or a propyl group, from α-cleavage. The absence of a McLafferty rearrangement fragment means that there is no hydrogen atom on a gamma carbon, so the propyl group must be isopropyl, and not n-propyl.

POINTS: 1 59. Refer to Exhibit 19-10. Interpret the 1H NMR spectrum.

ANSWER: The 1H NMR spectrum indicates that there are only two kinds of hydrogen in this compound. The 12H doublet (one adjacent H) is most likely four equivalent methyl groups of two equivalent isopropyl groups. The 2H multiplet (many adjacent H) is consistent with two equivalent hydrogens of two isopropyl groups, too. The multiplet is shifted to about 2.77 by the adjacent C=O. POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 16

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions 60. Refer to Exhibit 19-10. Propose a structure for this compound.

ANSWER: 2,4-dimethyl-3-pentanone, (CH3)2CHCOCH(CH3)2 POINTS: 1 61. Name the following compound. Atoms other than carbon and hydrogen are labeled.

ANSWER: 1-chloro-3-methyl-2-butanone POINTS: 1 62. Which of the labeled peaks would allow the distinction of an aldehyde from a ketone based on this spectrum?

a. 1 b. 2 c. 3 d. 4 ANSWER: b POINTS: 1 63. Which of the labeled peaks would allow the distinction of an aldehyde from a ketone based on this spectrum?

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions a. 1 b. 2 c. 3 d. 4 e. 5 f. This is not the spectrum of an aldehyde.

ANSWER: a POINTS: 1 64. Based only on the spectrum below, what is inferred about the compound used to produce the spectrum?

a. aromatic b. aldehyde c. para disubstituted d. ethyl substituent e. All of the above are evident from the spectrum.

ANSWER: e POINTS: 1 65. Consider the following structure. Atoms other than carbon and hydrogen are labeled.

Oxidation of this compound yields: a.

b.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

c.

d.

No reaction ANSWER: d POINTS: 1 66. Which of the following reagents could be used to convert 3-cyclopentyl-2-methylpropanal to the following substance?

Atoms other than carbon and hydrogen are labeled.

a. CrO3

b. O3

c. DIBAH

d. LiAlH4

ANSWER: a POINTS: 1 67. The following substance is heated in the presence of aqueous NaOH. Atoms other than carbon and hydrogen are labeled.

The product of the reaction is: a.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

b.

c.

d.

both b and c

e.

all of the above ANSWER: d POINTS: 1 68. If the following substance were treated with HCN followed by aqueous acid and heat,

which of the following would be produced? Atoms other than carbon and hydrogen are labeled. a.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions

b.

c.

d.

both b and c

e.

all of the above ANSWER: c POINTS: 1 69. Which of the following functional groups does not contain a carbonyl group? a. ester b. carboxylic acid c. ketone d. phenol ANSWER: d POINTS: 1 70. Which of the following is not a characteristic of a carbonyl group? a. sp2 hybridized carbon atom b. planar geometry about the carbonyl bond c. carbonyl carbon atom is electrophilic d. carbonyl oxygen atom is nucleophilic e. weakly polarized carbon to oxygen bond f. All are characteristics of the carbonyl group Cengage Learning Testing, Powered by Cognero

Page 21

Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER: e POINTS: 1 71. Consider similar R groups in the following types of carbonyl compounds. In which type(s) of compound is a negative charge on the acyl group not stabilized to function as a leaving group in nucleophilic substitution reactions? a. acid chloride b. acid anhydride c. ester d. aldehyde e. a and b f. a, b and c g. a, b, c, and d ANSWER: d POINTS: 1 72. Of the types of reactions that characterize carbonyl groups, how is the following reaction classified?

a. nucleophilic addition c. α substitution ANSWER: c POINTS: 1

b. nucleophilic acyl substitution d. carbonyl condensation

73. Draw the mechanism of the nucleophilic addition reaction of pentanal to give an alcohol.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER:

POINTS: 1 74. Draw the mechanism of the nucleophilic addition reaction involving acetaldehyde and a neutral nucleophile, in acidic conditions.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER:

POINTS: 1 75. Draw the mechanism of the Grignard reaction involving acetone and methyl magnesium chloride.

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Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions ANSWER:

POINTS: 1

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Chapter 20 - Carboxylic Acids and Nitriles Drawing Instructions: Draw structures corresponding to each of the given names.

1. Draw: 2-propylpentanoic acid ANSWER:

POINTS: 1 2. Draw: cis-1,3-cyclopentanedicarboxylic acid ANSWER:

POINTS: 1 3. Draw: cyanoacetic acid ANSWER:

POINTS: 1 4. Draw: m-chlorobenzoic acid ANSWER:

POINTS: 1 5. Draw: 2-propenenitrile ANSWER: POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

6. Name:

ANSWER: 1,5-pentanedioic acid POINTS: 1 7. Name: Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 20 - Carboxylic Acids and Nitriles

ANSWER: 2-hydroxybenzoic acid POINTS: 1 8. Name:

ANSWER: 5,5-dimethyl-1,3,6-cycloheptatriene-1-carboxylic acid POINTS: 1 9. Name:

ANSWER: (3R)-3,5-dihydroxy-3-methylpentanoic acid POINTS: 1 10. Name:

ANSWER: (2E)-2-methyl-2-butenoic acid POINTS: 1 11. Name:

ANSWER: phenylacetonitrile or 2-phenylethanenitrile POINTS: 1 12. What is the correct name for the following structure?

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Page 2

Chapter 20 - Carboxylic Acids and Nitriles

a. 2-oxohexanoic acid b. 5-oxohexanoic acid c. methyl butroxoketone d. 4-ketopentanoic acid ANSWER: b POINTS: 1 Exhibit 20-1 Consider the data in the Table below to answer the following question(s): Acidities of Substituted Benzoic and Acetic Acids pKas at 25°C Y−C6H4COOH Y meta Y−CH2COOH H 4.75 4.19 CN 2.47 3.64 3.57 4.09 OCH3

para 4.19 3.55 4.46

13. Refer to Exhibit 20-1. Draw the structure of the strongest acid in the table above.

ANSWER: NC−CH2−COOH POINTS: 1 14. Refer to Exhibit 20-1. Which of the acids in the table above has the strongest conjugate base?

ANSWER: CH3COOH POINTS: 1 15. Refer to Exhibit 20-1. Explain why cyanoacetic acid and methoxyacetic acid are more acidic than their correspondingly substituted benzoic acid counterparts.

ANSWER: Electron-withdrawing groups, like −CN and −OCH3, inductively withdraw electron density, which stabilizes the resulting carboxylate anion and, thus, increases the acidity of the carboxylic acid. These inductive effects are strongly dependent on distance. The −CN and −OCH3 substituents are closer to the carboxylate in the substituted acetic acids than in the substituted benzoic acids, so their effect is greater.

POINTS: 1 16. Even though the para position is one carbon farther from the carboxy group than the meta position, p-cyanobenzoic acid is more acidic than m-cyanobenzoic acid. Explain the differences in acidity between p-cyanobenzoic acid and mcyanobenzoic acid.

ANSWER: In p-cyanobenzoic acid, the negative charge of the carboxylate can be better stabilized than in the meta isomer because the electron-withdrawing cyano group polarizes the benzene ring and places a positive charge next to the negatively charged carboxyl group. The same polarization occurs in m-cyanobenzoic acid, but the effect is not as great because the charge is one carbon farther away from the carboxyl group.

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Chapter 20 - Carboxylic Acids and Nitriles

POINTS: 1 17. Explain the differences in acidity between p-methoxybenzoic acid and m-methoxybenzoic acid.

ANSWER: p-Methoxybenzoic acid is less acidic than m-methoxybenzoic acid because the carboxyl group is directly conjugated, through the benzene ring, with the electron-donating oxygen lone pairs of the methoxy group.

This destabilizes the p-methoxybenzoate anion relative to the m-methoxybenzoate anion by placing two negative charges on the carboxyl group.

POINTS: 1 Exhibit 20-2 Consider the data below to answer the following question(s).

When CO2 is bubbled through an ether solution of benzylmagnesium bromide, and the resulting mixture is acidified, phenylacetic acid is produced. Any unreacted benzylmagnesium bromide is converted to toluene in the acidification step.

18. Refer to Exhibit 20-2. Write the complete reaction sequence for the process described above.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 4

Chapter 20 - Carboxylic Acids and Nitriles 19. Refer to Exhibit 20-2. How could you separate phenylacetic acid from toluene?

ANSWER: Extract the ether solution containing the mixture of products with a dilute aqueous solution of sodium hydroxide. This converts phenylacetic acid to its carboxylate salt, which will dissolve in the aqueous layer.

The toluene will remain in the ether layer. Separate the aqueous and organic layers, acidify the aqueous layer, and extract with an organic solvent to recover the phenylacetic acid.

POINTS: 1 20. Refer to Exhibit 20-2. This reaction is described as a _____ process. a. carbonylation b. carboxylation c. carbaniolation d. cationation ANSWER: b POINTS: 1 Exhibit 20-3 What is the order of increasing acidity for the following compounds? (least to most)

21.

a. IV, I, III, II b. III, II, I, IV c. II, III, I, IV d. III, II, IV, I ANSWER: b POINTS: 1

22.

a. IV, I, III, II b. IV, II, III, I c. II, III, I, IV d. I, III, II, IV ANSWER: a POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Carboxylic Acids and Nitriles

23.

a. IV, II, I, III b. I, II, IV, III c. III, I, II, IV d. III, IV, II, I ANSWER: c POINTS: 1 Exhibit 20-4 Give the major organic product(s) for each of the following reactions or sequences of reactions. Show all relevant stereochemistry.

24.

ANSWER:

POINTS: 1

25.

ANSWER:

POINTS: 1

26.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Carboxylic Acids and Nitriles

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1

30.

ANSWER:

POINTS: 1 31.

ANSWER: POINTS: 1

32.

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Page 7

Chapter 20 - Carboxylic Acids and Nitriles ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1 Exhibit 20-5 Choose the best reagent(s) for carrying out the following conversions from the list below. Place the letter corresponding to the best choice in the blank to the left of the conversion. a. b. c. d. e. f.

KMnO4, H3O+ CH 3I NaBH4, ethanol 1. BH3 2. H3O+ 1. CH3MgBr, ether 2. H3O+ CrO3, H2SO4, H2O

g.

1. Mg, ether

h.

2. CO2 3. H3O+ 1. NaCN

i.

2. H2SO4, H2, heat O3, then Zn and HOAc

34. _____ CH3CH=CHCH2COOH → CH3COOH + HOOCCH2COOH ANSWER: a POINTS: 1

35. _____ ANSWER: g POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Carboxylic Acids and Nitriles

36. _____

ANSWER: f POINTS: 1

37. _____

ANSWER: e POINTS: 1 Exhibit 20-6 Carboxylic acids are synthesized from alkyl halides via Grignard reagent carboxylation or nitrile hydrolysis. Choose the best method for effecting each of the following conversions. Explain each of your choices. If neither method is appropriate, explain.

38.

ANSWER: Since SN2 displacements do not occur at tertiary alkyl halides the Grignard reagent carboxylation is the best method for carrying out this conversion.

POINTS: 1

39.

ANSWER: Either method would be appropriate for this conversion.

POINTS: 1

40.

ANSWER: Neither method of acid synthesis yields the desired product. Any Grignard reagent formed will react with the carbonyl function group present in the starting material. Reaction with cyanide occurs at the carbonyl functional group as well as at the halogen. However, if the aldehyde is first protected by forming an acetal, either method can be used to synthesize a carboxylic acid.

POINTS: 1 41. Propose a synthesis of the anti-inflammatory drug Ibuprofen from benzene. Show all reagents and all intermediate structures. Assume that ortho and para isomers can be separated.

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Chapter 20 - Carboxylic Acids and Nitriles

ANSWER:

POINTS: 1 Exhibit 20-7 Show how you would accomplish the following transformations. More than one step may be required. Show all reagents and all intermediate structures.

42.

ANSWER:

POINTS: 1

43.

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Page 10

Chapter 20 - Carboxylic Acids and Nitriles ANSWER:

POINTS: 1 44. Name the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER: 4-hydroxy-2,2-dimethylbutanoic acid POINTS: 1 45. Name the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER: 3-methylbutanenitrile POINTS: 1 46. Draw the structure of the product formed when the following is treated with 1) NaOH (aq) and 2) H3O+. Explain, if necessary how the product would be affected if the treatment with acid were omitted.

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Chapter 20 - Carboxylic Acids and Nitriles

ANSWER:

The corresponding conjugate base would be produced.

POINTS: 1 47. Draw the structure of the reactant needed to form the following upon treatment with SOCl2 at a moderately elevated temperature. Atoms other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 48. Which of the following is not an acyl derivative of a carboxylic acid? a. nitrile b. ester c. amide d. acid chloride ANSWER: a POINTS: 1 49. Which of the following is not a common feature of both carboxylic acids and nitriles? a. carbon atom with three bonds to an electronegative atom b. presence of π bonds c. can act as electrophiles d. occur frequently in living organisms e. All characterize both functional group classes.

ANSWER: d Cengage Learning Testing, Powered by Cognero

Page 12

Chapter 20 - Carboxylic Acids and Nitriles POINTS: 1 50. A solution of glycolic acid is prepared such that the glycolic acid is 69% dissociated. Calculate the pH of this solution. (

)

a. 3.83 c. 4.18 ANSWER: a POINTS: 1

b. 6.06 d. 3.48

51. In a solution of acetic acid ( substance as: a. CH3CO2H

) at physiological pH of 7.3, it is most accurate to represent this

b. CH3CO2– c. CH3CO2H and CH3CO2– ANSWER: b POINTS: 1 52. What reactant could produce the following substance upon treatment with CrO3 under acidic conditions? Atoms other than carbon and hydrogen are labeled.

a. 2, 2-dimethylbutanal b. 2, 2-dimethyl-1-butanol c. 2, 2-dimethylbutanenitrile d. a and b e. a and c f. a, b, and c ANSWER: d POINTS: 1 53. In order to produce a carboxylic acid as a product of the Grignard reaction, the Grignard reagent reacts with: a. CO2 b. a methyl ketone c. a methyl ester Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Carboxylic Acids and Nitriles d. an aldehyde e. either or CO2 a methyl ketone ANSWER: a POINTS: 1 54. Which labeled peaks in the following spectrum are characteristic of a carboxylic acid?

a. 1 b. 2 c. 3 d. 4 e. 5 f. 1 and 2 g. 2 and 3 h. All five peaks ANSWER: f POINTS: 1 55. Which labeled peaks would disappear if D2O were added to the sample?

a. 1 b. 2 c. 3 d. 4 e. All peaks would disappear.

ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Carboxylic Acids and Nitriles 56. The following spectrum is representative which of the following types of compounds?

a. carboxylic acid b. aldehyde c. nitrile d. alcohol e. halide f. Data presented does not provide enough information.

ANSWER: c POINTS: 1 57. A sample is known to contain either hexanoic acid or 3-methylpentanoic acid. Which of the following spectra would not distinguish between the two isomers? a. 1H NMR b. 13C NMR c. DEPT-90 d. DEPT-135 e. All would distinguish between the two isomers.

ANSWER: b POINTS: 1 58. Draw the mechanism of the basic hydrolysis of methyl nitrile to yield methyl amide, which is then hydrolyzed to yield an acetate ion.

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Chapter 20 - Carboxylic Acids and Nitriles ANSWER:

POINTS: 1 59. Draw the mechanism of the conversion of isopropyl nitrile to an amine.

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Chapter 20 - Carboxylic Acids and Nitriles ANSWER:

POINTS: 1 60. Draw the mechanism for the preparation of nitriles by dehydration of a primary amide, e.g., propanamide, using thionyl chloride.

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Chapter 20 - Carboxylic Acids and Nitriles ANSWER:

POINTS: 1

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Page 18

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions IUPAC Naming Instructions: Provide proper IUPAC names.

1. Name:

ANSWER: Ethyl 4-aminobenzoate POINTS: 1 2. Name:

ANSWER: bis(2-methylpropanoic) anhydride POINTS: 1 3. Name:

ANSWER: methyl phenylacetate POINTS: 1 4. Name:

ANSWER: trans-2-butenoyl chloride POINTS: 1 5. Name:

ANSWER: N-cyclopentyl-3-methylbutanamide POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions Drawing Instructions: Draw structures corresponding to each of the given names.

6. Draw: 2-propenamide ANSWER:

POINTS: 1 7. Draw: 3,4,5-trimethoxybenzoyl chloride ANSWER:

POINTS: 1 8. Draw: acetic formic anhydride ANSWER:

POINTS: 1 9. Draw: N,N-dimethylformamide ANSWER:

POINTS: 1 10. Draw: methyl cis-3-ethylcyclobutanecarboxylate ANSWER:

POINTS: 1 11. Draw: (E)-2,4-dimethyl-2-hexenoyl chloride

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Page 2

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

POINTS: 1 12. What is the correct structure for phenylbenzoate?

a.

b.

c.

d.

ANSWER: a POINTS: 1 Exhibit 21-1 What is the order of decreasing reactivity towards nucleophilic acyl substitution for the carboxylic acid derivatives? (most reactive first)

13.

a. I, II, III, IV b. I, III, IV, II c. II, IV, III, I d. II, I, III, IV ANSWER: b POINTS: 1

14.

a. I, III, II, IV b. II, III, I, IV c. III, II, I, IV d. IV, I, III, II ANSWER: d POINTS: 1

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Page 3

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions 15. The ester ethyl p-nitrobenzoate is more reactive towards nucleophilic acyl substitution than ethyl pmethyoxybenzoate. Explain this reactivity difference using both words and structures.

ANSWER: In general, any factor that makes the carbonyl carbon more polarized will increase the reactivity of the carboxylic acid derivative. In these particular examples, both are aromatic ethyl esters. It is the substituent on the aromatic ring that is affecting the reactivity difference. The electron-withdrawing nitro group increases the polarization of the carbonyl by withdrawing electrons from it through the aromatic ring. The electron donating methoxy group decreases the polarization of the carbonyl by donating electrons to it through the aromatic ring.

POINTS: 1 16. Write the complete stepwise mechanism for the basic hydrolysis of acetamide, shown below. Show all electron flow with arrows and draw all intermediate structures.

ANSWER:

POINTS: 1 Exhibit 21-2 Consider the reaction below to answer the following question(s):

Acid halides react with diazomethane to yield diazoketones. Excess diazomethane is used to prevent the HCl produced in the reaction from reacting with the diazoketone.

17. Refer to Exhibit 21-2. Diazomethane is an example of a dipolar molecule; a molecule which is neutral overall but has charges on individual atoms. One resonance form of diazomethane is drawn below. Draw the Lewis structure of the other resonance form of diazomethane. Be sure to include all formal charges.

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

ANSWER:

POINTS: 1 18. Refer to Exhibit 21-2. The intermediate structures for the mechanism for the reaction of propanyl chloride with diazomethane are provide below. Show all electron flow with arrows on these structures.

ANSWER:

POINTS: 1 Exhibit 21-3 Consider the reaction below to answer the following question(s):

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

19. Refer to Exhibit 21-3. Write the complete stepwise mechanism for this reaction. Show intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 20. Refer to Exhibit 21-3. This reaction is an example of: a. an intermolecular nucleophilic acyl substitution reaction b. an intramolecular nucleophilic acyl substitution reaction c. an intermolecular SN2 reaction d. an intramolecular SN2 reaction ANSWER: b POINTS: 1 21. Refer to Exhibit 21-3. The purpose of the base catalyst in this reaction is: a. to polarize the carbonyl group to make it more electrophilic b. to convert the ester to an intermediate carboxylic acid c. to convert the alcohol group to an alkoxide anion, which is a better nucleophile d. all of the above ANSWER: c POINTS: 1 22. Refer to Exhibit 21-3. The product of this reaction is: a. a lactone b. an anhydride c. a lactam d. an ether ANSWER: a POINTS: 1 23. Methyl butanoate has been isolated from pineapple oil and can be prepared by the Fischer esterification reaction Cengage Learning Testing, Powered by Cognero

Page 6

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions shown below. Write the complete stepwise mechanism for this reaction. Show all electron flow with arrows and include all intermediate structures.

ANSWER:

POINTS: 1 24. Write the complete stepwise mechanism for the acid-catalyzed hydrolysis of the following amide to yield mandelic acid. Show all electron flow with arrows and draw the structures of all intermediate species.

ANSWER:

POINTS: 1 25. The purpose of the acid catalyst in the hydrolysis of an amide is: a. b.

to enhance the nucleophilicity of the water molecule to enhance the electrophilicity of the amide carbonyl carbon

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Page 7

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions c. to enhance the electrophilicity of the water molecule d. to shift the equilibrium of the reaction ANSWER: b POINTS: 1 Exhibit 21-4 Consider the information below to answer the following question(s).

The reaction of a carboxylic acid with an alcohol in the presence of acid is termed Fischer esterification.

26. Refer to Exhibit 21-4. The nucleophile in this reaction is _____.

ANSWER: B POINTS: 1 27. Refer to Exhibit 21-4. Compound C functions as _____ in this reaction. a. a base scavenger b. a solvent c. a catalyst d. a neutralizer ANSWER: c POINTS: 1 28. Refer to Exhibit 21-4. Fischer esterification is an example of: a. nucleophilic acyl addition b. nucleophilic acyl substitution c. nucleophilic acyl elimination d. nucleophilic acyl rearrangement ANSWER: b POINTS: 1 29. Refer to Exhibit 21-4. Write the stepwise mechanism for the Fischer esterification reaction of benzoic acid and methanol given above. Show all electron flow by using curved arrows, and include all intermediate structures.

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Page 8

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

POINTS: 1 Exhibit 21-5 Provide structure(s) for the starting material(s), reagent(s) or the major organic product(s) of each of the following reactions or sequences of reactions. Show all relevant stereochemistry.

30.

ANSWER:

POINTS: 1

31.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1

34.

ANSWER:

POINTS: 1

35.

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Page 10

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

POINTS: 1

36.

ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1

38.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

39.

ANSWER:

POINTS: 1

40.

ANSWER:

POINTS: 1

41.

ANSWER:

POINTS: 1

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

42.

ANSWER:

POINTS: 1

43.

ANSWER:

POINTS: 1

44.

ANSWER:

POINTS: 1 Exhibit 21-6 Choose the best reagent(s) from the list provided below for carrying out the following transformations. Place the letter corresponding to the best choice in the blank to the left of the transformation. a. (Ph)2CuLi, ether Cengage Learning Testing, Powered by Cognero

f.

1. DIBAH, toluene Page 13

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions b. NaBH4, ethanol c. 1.

2. d. 1.

2. e. (CH3)2CuLi, ether

CH3MgBr, ether H3O+ LiAlH4, THF H3O+

2. H3O+ g. 1. (Ph)2CHMgBr 2. H3O+ h. 1. 2 PhMgBr, ether 2. H3O+

45. _____

ANSWER: h POINTS: 1

46. _____

ANSWER: a POINTS: 1

47. _____

ANSWER: d POINTS: 1

48. _____

ANSWER: f POINTS: 1

49. _____ ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions Exhibit 21-7 Show how you would accomplish each of the following transformations. More than one step may be required. Show all reagents and all intermediate structures.

50.

ANSWER:

POINTS: 1

51.

ANSWER:

POINTS: 1

52.

ANSWER:

POINTS: 1

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

53.

ANSWER:

POINTS: 1 54. Aklomide, 2-chloro-4-nitrobenzamide, is an ingredient in veterinary antibacterial preparations. Propose a synthesis of aklomide starting with toluene. Show all reagents and all intermediate structures.

ANSWER:

POINTS: 1 55. Ethyl phenylacetate is a pleasant smelling compound used in perfumery. Draw structures for each of the intermediates in the synthesis of ethyl phenylacetate below.

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Page 16

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

POINTS: 1 Exhibit 21-8 For each pair of compounds below, tell which spectroscopic technique you would use to distinguish between the two members of the pair. Tell what differences you would expect to see.

56.

ANSWER: IR

2250 cm−1 (nitrile) 1730−1760 cm−1 (four-membered lactam)

Although you would not be expected to know the absorption of a four-membered lactam, you should be able to predict that IR spectroscopy would differentiate these two compounds based solely on the nitrile absorption in the first compound and the fact that the lactam would have a carbonyl absorption somewhere in the region of 1660−1750 cm−1.

POINTS: 1

57.

ANSWER: IR

1

H NMR

1735 cm−1 (aliphatic ester)

1720 cm−1 (aromatic ester)

methyl singlet methylene singlet

one triplet one quartet

POINTS: 1

58.

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Page 17

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER: IR

1

H NMR

1820, 1760 cm−1 (anhydride)

2500−3300 cm−1 (acid−OH) 1710−1760 cm−1 (carboxyl) 1715 cm−1 (ketone)

one triplet one quartet

three overlapping triplets one quartet one singlet

POINTS: 1 Exhibit 21-9 Refer to the data below to answer the following question(s): Kodel® is a staple and filament fiber prepared from dimethyl terephthalate and 1,4-cyclohexanedimethanol. Fabric made from Kodel® has good crease resistance.

59. Refer to Exhibit 21-9. Draw the structure of the Kodel® polymer.

ANSWER:

POINTS: 1 60. Refer to Exhibit 21-9. Kodel® is an example of: a. a polyurethane b. a polyester c. a polyamide d. a polycarbonate ANSWER: b POINTS: 1 61. Poly(ethylene terephthalate), PET, is the polymeric material of Mylar® and Dacron®. What are the monomers from which PET is prepared?

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Page 18

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

POINTS: 1 62. Name the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER: 4-methylpentanoyl chloride POINTS: 1 63. Name the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER: ethyl 1-chloropropanoate POINTS: 1 64. Which of the following will take place via nucleophilic acyl substitution? a. thioester producing an acid chloride b. ester producing a thioester c. ester producing an amide d. acid anhydride producing an acid chloride ANSWER: c POINTS: 1 65. The following reaction could be classified as which type of nucleophilic acyl substitution? More than one step might be needed.

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

a. amoniolysis b. alcoholysis. c. reduction. d. hydrolysis. e. Grignard reaction. f. c or e depending on reaction conditions. g. The reaction will not occur via nucleophilic acyl substitution.

ANSWER: d POINTS: 1 66. The following reaction could be classified as which type of nucleophilic acyl substitution? More than one step might be needed.

a. amoniolysis b. alcoholysis. c. reduction. d. hydrolysis. e. Grignard reaction. f. c or e depending on reaction conditions. g. The reaction will not occur via nucleophilic acyl substitution.

ANSWER: b POINTS: 1 67. The following reaction could be classified as which type of nucleophilic acyl substitution? More than one step might be needed.

a. amoniolysis b. alcoholysis. c. reduction. d. hydrolysis. e. Grignard reaction. f. c or e depending on the reaction conditions.

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Page 20

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions g. The reaction will not occur via nucleophilic acyl substitution.

ANSWER: c POINTS: 1 68. The following reaction could be classified as which type of nucleophilic acyl substitution?

a. amoniolysis b. alcoholysis. c. reduction. d. hydrolysis. e. Grignard reaction. f. c or e depending on reaction conditions. g. The reaction will not occur via nucleophilic acyl substitution.

ANSWER: g POINTS: 1 69. What is the product when the following substance reacts with trimethylamine ((CH3)3N)? Atoms other than carbon and hydrogen are labeled.

a.

b.

c.

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

d.

No reaction occurs.

ANSWER: d POINTS: 1 70. What is the name of the product when the following substance reacts with NaCH3COO? Atoms other than carbon and hydrogen are labeled.

a. 4-methylbutanoic acid b. acetic 4-methylbutanoic anhydride c. 4-methylbutyl acetate d. acetyl 4-methylbutanoate e. No reaction occurs.

ANSWER: b POINTS: 1 71. Which line in the following spectrum would indicate the presence of a carbonyl group?

a. 1 b. 2 c. 3 d. 4 e. 5 Cengage Learning Testing, Powered by Cognero

Page 22

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER: e POINTS: 1 72. Based on the following spectrum

the sample used would probably be classified as a(n): a. aldehyde. b. ketone. c. ester. d. The spectrum is not specific enough to classify the substance.

ANSWER: c POINTS: 1 73. The following is an excerpt from a spectrum taken on an unknown sample.

To which of the following functional groups might the unknown belong? a. amide b. acid chloride c. carboxylic acid d. ester e. None of these can be eliminated.

ANSWER: d Cengage Learning Testing, Powered by Cognero

Page 23

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions POINTS: 1 74. Consider the following molecular model. Atoms other than carbon and hydrogen are labeled.

Which of the following reactants could be used to produce this compound? a. 2-chloropropanoic acid and ethanol b. propanoic acid and chloroethanol c. propanoic acid and ethanol ANSWER: a POINTS: 1 75. Consider the structure given below which might be found in a protein.

Which of the following is not applicable to this structure? a. contains three monomers b. contains amide bonds c. formed from a diacid and diamine d. contains three types of monomers e. All of these apply to this structure.

ANSWER: c POINTS: 1 76. Which of the following describes nucleophilic acyl substitution? a. Nucleophile adds to the electrophilic carbon of the carbonyl bond. b. Electrophile replaces a group on the carbon alpha to the carbonyl. c. Nucleophile replaces a group adjacent to the carbon of carbonyl. d. Two carbonyl groups react with each other.

ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 24

Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions 77. Draw the mechanism of the Fischer esterification of butanoic acid, in the presence of ethanol and hydrochloric acid, which is the catalyst.

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions 78. Draw the mechanism of base-induced hydrolysis of ethyl propanoate.

ANSWER:

POINTS: 1 79. Draw the mechanism of acid-catalyzed hydrolysis of propyl ethanoate to yield propanoic acid.

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Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions ANSWER:

POINTS: 1

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Page 28

Chapter 22 - Carbonyl Alpha-Substitution Reactions Exhibit 22-1 Refer to the compounds below to answer the following question(s):

1. Refer to Exhibit 22-1. Indicate all the acidic hydrogens in Compounds I through IV. ANSWER: Acidic hydrogens are underlined.

POINTS: 1 2. Refer to Exhibit 22-1. Indicate which hydrogens in Compound II are the most acidic. Explain your answer.

ANSWER: The methylene hydrogens are the most acidic because they are flanked by two electron-withdrawing groups, −NO2 and C=O. The enolate generated by their removal is highly resonance stabilized; it has three resonance forms.

In contrast, the enolate generated by the abstraction of a methyl proton is stabilized by only one electronwithdrawing group and has two resonance forms:

POINTS: 1 3. Refer to Exhibit 22-1. Choose the most acidic compound from Compounds I - IV. Explain your choice.

ANSWER: Compound III is the most acidic compound because the methine hydrogen is flanked by three carbonyl groups so the enolate ion generated by its removal has four resonance forms.

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Page 1

Chapter 22 - Carbonyl Alpha-Substitution Reactions 4. Refer to Exhibit 22-1. Draw the structures for all enols of Compound I. ANSWER:

POINTS: 1 Exhibit 22-2 Consider the structures below to answer the following question(s).

5. Refer to Exhibit 22-2. Indicate the most acidic hydrogens in each of the molecules.

ANSWER:

The most acidic hydrogens are underlined.

POINTS: 1 6. Refer to Exhibit 22-2. Rank the molecules above in order of increasing acidity (least acidic to most acidic). a. III, II, I b. II, III, I c. I, II, III d. II, I, III ANSWER: b POINTS: 1 7. Nitroethane [CH3CH2NO2, pKa = 8.6] is a much stronger acid than ethane [CH3CH3, pKa ≈ 60]. Explain.

ANSWER: When a proton is removed from the carbon bearing the nitro group in nitroethane, the negative charge is shared by the electron-withdrawing nitro group. Thus, the anion is stabilized by resonance between two forms.

When a proton is abstracted from ethane, the carbon atom must bear the full negative charge.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 2

Chapter 22 - Carbonyl Alpha-Substitution Reactions Exhibit 22-3 Consider the reaction below to answer the following question(s).

8. Refer to Exhibit 22-3. The strongest base in the reaction is: ANSWER: B POINTS: 1 9. Refer to Exhibit 22-3. The weakest acid in the reaction is: ANSWER: D POINTS: 1 10. Refer to Exhibit 22-3. The enolate ion in the reaction is: ANSWER: C POINTS: 1 11. Refer to Exhibit 22-3. On the structures provided above, draw arrows indicating electron flow in the generation of the intermediate C. ANSWER:

POINTS: 1 Exhibit 22-4 Consider the reaction sequence below to answer the following question(s):

12. Refer to Exhibit 22-4. Compound X, diethyl propanedioate, is more commonly known as _____. a. b. c. d.

ethyl acetoacetate acetoacetic ester oxalic ester malonic ester

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER: d POINTS: 1 13. Refer to Exhibit 22-4. Write the complete stepwise mechanism for the conversion of Compound X into Compound Y. Show all electron flow with arrows and draw all intermediate structures.

ANSWER:

POINTS: 1 14. Refer to Exhibit 22-4. Below are the structures and electron flow for the conversion of Compound Y into Compound Z. Draw the structure of Compound Z. (Hint: Compound Z is an alcohol.)

ANSWER:

POINTS: 1 Exhibit 22-5 Consider the reaction sequence below to answer the following question(s):

15. Refer to Exhibit 22-5. The starting material A in this reaction sequence is called _____. a. a β-keto ester b. an α-carboethoxy ketone c. a malonic ester d. an acetoacetic ester ANSWER: a POINTS: 1 16. Refer to Exhibit 22-5. Conversion of A into B is a type of reaction termed _____. a. b.

an acylation an enolation

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Chapter 22 - Carbonyl Alpha-Substitution Reactions c. an alkylation d. a phenylation ANSWER: c POINTS: 1 17. Refer to Exhibit 22-5. Conversion of B into C involves hydrolysis of the ester followed by decarboxylation. On the structures provided below, show the electron flow for the decarboxylation step.

ANSWER:

POINTS: 1 18. Refer to Exhibit 22-5. The initial product formed on decarboxylation is an enol, which rapidly equilibrates to its keto form under the acidic reaction conditions. Write the complete stepwise mechanism for the acid-catalyzed conversion of the enol into its keto form, Compound C.

ANSWER:

POINTS: 1 Exhibit 22-6 Consider the reaction below to answer the following question(s):

19. Refer to Exhibit 22-6. Write the complete stepwise mechanism for the reaction above. Show all intermediate structures and all electron flow with arrows.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER:

POINTS: 1 20. Refer to Exhibit 22-6. Explain the product ratio in this reaction.

ANSWER: Lithium diisoproylamide is a sterically hindered base, so abstraction of a proton is more favorable at the less sterically crowded side of the ketone. Hence, the substitution occurs primarily at position 6 rather than position 2.

POINTS: 1 21. Write the complete stepwise mechanism for the reaction of cyclopentanone with bromine in acetic acid to give 2bromocyclopentanone. Show all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 Exhibit 22-7 Give the major organic product(s) of each of the following reactions or sequences of reactions. Show all relevant stereochemistry.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions

22.

ANSWER:

POINTS: 1

23.

ANSWER:

POINTS: 1

24.

ANSWER:

POINTS: 1

25.

ANSWER:

POINTS: 1

26.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 22 - Carbonyl Alpha-Substitution Reactions

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1 30. Diethyl malonate can be prepared by the following reaction sequence. Draw the structures of each of the missing intermediates in the boxes provided.

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Page 8

Chapter 22 - Carbonyl Alpha-Substitution Reactions

ANSWER:

POINTS: 1 Exhibit 22-8 How would you prepare each of the following compounds using either an acetoacetic ester synthesis or a malonic ester synthesis? Show all intermediate structures and all reagents.

31.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER:

POINTS: 1

32.

ANSWER:

POINTS: 1

33.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER:

POINTS: 1

34.

ANSWER:

POINTS: 1

35.

ANSWER:

POINTS: 1 Exhibit 22-9 Show how you would accomplish each of the following transformations. More than one step may be required. Show all reagents and all intermediate structures.

36.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1 38. Mechanistically like halogenation, acetone can be deuterated by treatment with D3O+. Describe how an H NMR could be used to verify that deuteration actually occurs.

ANSWER: Acetone containing only 1H nuclei would show one singlet in the 1H NMR. Upon deuteration, all of the hydrogen atoms will no longer exist in chemically equivalent environments and splitting of the signal may occur. If the deuteration goes to completion, CD3COCD3, the 1H NMR signal will be lost because 2H (D) is a nonmagnetic nucleus possessing an even number of nucleons.

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Page 12

Chapter 22 - Carbonyl Alpha-Substitution Reactions POINTS: 1 39. Explain why the following reaction does not occur to any significant extent respect when the reactant is treated with bromine in the presence of acetic acid. Atoms other than carbon and hydrogen are labeled.

Reactant Product ANSWER: Carboxylic acids, like esters and amides, do not enolize to any appreciable extent. The enol tautomer is the necessary intermediate in alpha substitution using Br2 in acetic acid.

POINTS: 1 40. Draw the structure of the major product formed when the substance represented by the following molecular model is treated with: 1. Br2/acetic acid, 2. pyridine/heat. Atoms of other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 41. Explain how to use an alkylation reaction to produce:

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Page 13

Chapter 22 - Carbonyl Alpha-Substitution Reactions

Atoms other than carbon and hydrogen are labeled.

ANSWER: Treatment of 3-methylbutanenitrile with LDA/THF followed by the alkylation using iodoethane, CH3CH2I, will yield 2-ethyl-3-methylbutanenitrile via deprotonation and alkylation of the α-carbon.

POINTS: 1 42. Which of the following is not correct? a. Tautomers are constitutional isomers. b. Tautomers rapidly interconvert. c. The enol form is generally more stable. d. Tautomerization is catalyzed by both acids and bases. e. All of the above are correct with respect to tautomers.

ANSWER: c POINTS: 1 43. Examine the following generalized structure.

This represents a. the acid catalyzed intermediate between an keto and enol tautomer. b. the base catalyzed intermediate between an keto and enol tautomer. c. one of the resonance forms of the acid catalyzed intermediate between an keto and enol tautomer. d. one of the resonance forms of the acid catalyzed intermediate between an keto and enol tautomer.

ANSWER: c POINTS: 1 44. Consider the following molecular model. Atoms other than carbon and hydrogen are labeled.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions Bromination of the substance represented would be dependent on the concentration of: a. ketone. b. acid. c. halogen. d. a and b. e. a and c. f. b and c. g. a, b, d.

ANSWER: d POINTS: 1 45. The alkyl halide that should be used to produce octanoic acid via the malonic ester synthesis is: a. 1-bromooctane b. 1-bromohexane c. 1-bromopentane d. 1-bromodecane e. 1-bromoundecane ANSWER: b POINTS: 1 46. Using the acetoacetic ester synthesis, to produce 5-methyl-2-heptanone, the alkyl halide that should be used is: a. 1-bromo-2-methylpentane b. 1-bromo-2-methylnonane c. 1-bromo-2-methylbutane d. 1-bromo-2-methyldecane e. 1-bromo-5-methylheptane ANSWER: c POINTS: 1 47. The following substance is produced using acetoacetic ester synthesis. Atoms other than carbon and hydrogen are labeled.

Which of the following regions of an IR spectrum could be used to monitor the progress of the reaction? a. 2850 - 29060 cm–1 b. 2500 - 3100 cm–1 c. 1670 - 1780 cm–1 d. 500 - 600 cm–1 Cengage Learning Testing, Powered by Cognero

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Chapter 22 - Carbonyl Alpha-Substitution Reactions e. either c or d ANSWER: c POINTS: 1 48. Which of the hydrogen atoms indicated by a number is the most acidic in the structure below? Atoms other than carbon and hydrogen are labeled.

a. 1 b. 2 c. 3 d. 4 ANSWER: a POINTS: 1 49. Draw the mechanism of enol formation from acetone in a reaction catalyzed by hydrochloric acid.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER:

POINTS: 1 50. Draw the mechanism of the acid-catalyzed chlorination of acetophenone in the presence of a base.

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Chapter 22 - Carbonyl Alpha-Substitution Reactions ANSWER:

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Page 18

Chapter 22 - Carbonyl Alpha-Substitution Reactions 51. Draw the general mechanism of the α-substitution reaction between acetone and the electrophilic methyl group.

ANSWER:

POINTS: 1

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Page 19

Chapter 23 - Carbonyl Condensation Reactions Exhibit 23-1 Draw the structure of the aldol self-condensation product for each of the following compounds. If a compound does not undergo aldol self-condensation, explain why it does not.

1.

ANSWER:

POINTS: 1

2.

ANSWER: Benzophenone does not undergo aldol self-condensation because it has no α-hydrogens.

POINTS: 1

3.

ANSWER:

POINTS: 1

4.

ANSWER: 2,2-Dimethylpropanal does not undergo aldol self-condensation because it has no α-hydrogens.

POINTS: 1 Exhibit 23-2 Consider the reaction below to answer the following question(s):

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Page 1

Chapter 23 - Carbonyl Condensation Reactions 5. Refer to Exhibit 23-2. Write the complete stepwise mechanism for the reaction above. Show all intermediate structures and all electron flow with arrows.

ANSWER:

POINTS: 1 6. Refer to Exhibit 23-2. This reaction is an example of: a. an intramolecular Claisen condensation b. an intramolecular aldol condensation c. a Robinson annulation d. a Michael reaction ANSWER: b POINTS: 1 7. Refer to Exhibit 23-2. The product of this reaction is: a. a β, γ-unsaturated aldehyde b. an α, β-unsaturated ketone c. an α, β-unsaturated aldehyde d. an enol ANSWER: c POINTS: 1 Exhibit 23-3 Consider the data below to answer the following question(s): The Friedlander Quinoline Synthesis, first reported in 1882, is the base-catalyzed condensation of 2-aminobenzaldehydes with ketones to form quinoline derivatives.

8. Refer to Exhibit 23-3. The first step of the Friedlander Quinoline Synthesis is a mixed aldol condensation. Write the complete stepwise mechanism for this reaction.

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Chapter 23 - Carbonyl Condensation Reactions

ANSWER:

POINTS: 1 9. Refer to Exhibit 23-3. The second step of the Friedlander Quinoline Synthesis is a nucleophilic addition of a primary amine to a ketone yielding an imine. Write the complete stepwise mechanism for this imine forming reaction. Show all electron flow with arrows and show all intermediate structures.

ANSWER:

POINTS: 1 10. In the alkylation of cyclohexanone, better yields are obtained by first reacting cyclohexanone with an equivalent of lithium diisopropylamide in THF and then adding the alkyl halide, rather than mixing cyclohexanone, alkyl halide, and a catalytic amount of sodium ethoxide in ethanol. Explain this observation by pointing out what the problems with the second reaction conditions might be and how the first set of reaction conditions help alleviate the problems.

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Page 3

Chapter 23 - Carbonyl Condensation Reactions ANSWER: When cyclohexanone is placed in a solution of sodium ethoxide in ethanol, the enolate is generated in small amounts in the presence of large amounts of ketone-ideal conditions for an aldol self-condensation.

Consequently, aldol self-condensation competes with alkylation of the enolate, and product mixtures result.

Conversely, when cyclohexanone is treated first with an equivalent of LDA, the enolate is rapidly generated and no ketone remains. Addition of alkyl halide then yields the desired alkylated product.

POINTS: 1 Exhibit 23-4 Each of the following compounds can be prepared by a mixed aldol condensation reaction. Give the structures of the aldehyde and/or ketone precursors for each aldol product and formulate the reaction.

11.

ANSWER:

POINTS: 1

12.

ANSWER:

POINTS: 1

13.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 23 - Carbonyl Condensation Reactions Exhibit 23-5 Consider the reaction below to answer the following question(s).

14. Refer to Exhibit 23-5. Which carbonyl compound functions as the electrophile in this reaction?

ANSWER: A POINTS: 1 15. Refer to Exhibit 23-5. Draw the structure of the enolate ion that is generated during the course of this reaction.

ANSWER:

POINTS: 1 16. Refer to Exhibit 23-5. This reaction is an example of: a. a mixed Claisen condensation. b. a Dieckman condensation. c. a Michael reaction. d. a mixed aldol reaction.

ANSWER: d POINTS: 1 Exhibit 23-6 Draw the structure of the product you would expect to obtain by Claisen condensation of each of the following esters. If an ester does not undergo Claisen condensation, explain why it does not.

17.

ANSWER: This ester does not undergo Claisen condensation, because it has no α-hydrogens and cannot undergo enolization.

POINTS: 1

18.

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Chapter 23 - Carbonyl Condensation Reactions ANSWER:

POINTS: 1

19.

ANSWER:

POINTS: 1 Exhibit 23-7 Consider the reaction below to answer the following question(s): Acetoacetic ester can be prepared by the Claisen self-condensation reaction of ethyl acetate.

20. Refer to Exhibit 23-7. Write the complete stepwise mechanism for this reaction. Show all electron flow with arrows and draw all intermediate structures.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 23 - Carbonyl Condensation Reactions 21. Ethyl acetate can be prepared from ethanol as the only organic starting material. Show all reagents and structures for all intermediates in this preparation.

ANSWER:

POINTS: 1 22. Give the structures of the ester precursors for the following Claisen condensation product and formulate the reaction.

ANSWER:

POINTS: 1 Exhibit 23-8 Consider the compound 2-methyl-2-carboethoxycyclopentanone, whose structure is shown below, to answer the following question(s).

23. Refer to Exhibit 23-8. Formulate a synthesis of 2-methyl-2-carboethoxycyclopentanone starting with acyclic precursors using a Dieckmann cyclization as a key carbon-carbon bond forming step. Show all reagents and all intermediate structures.

ANSWER:

POINTS: 1 24. Refer to Exhibit 23-8. When 2-methyl-2-carboethoxycyclopentanone is treated with sodium ethoxide in ethanol solution followed by a mild aqueous acid work-up, 5-methyl-2-carboethoxycyclopentanone is isolated as the major product. This reaction proceeds by a reverse Claisen condensation mechanism followed by a recyclization. On the structures provided below, show electron flow with arrows in this interesting reaction.

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Chapter 23 - Carbonyl Condensation Reactions

ANSWER:

POINTS: 1 25. Oxaloacetic acid is an important intermediate in the biosynthesis of citric acid. Synthesize oxaloacetic acid using a mixed Claisen condensation as a key carbon-carbon bond forming reaction.

ANSWER:

POINTS: 1 Exhibit 23-9 Draw the structures of the precursors to each of the following Michael reaction products. Label the Michael donor and the Michael acceptor in each case and formulate the reaction.

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Chapter 23 - Carbonyl Condensation Reactions

26.

ANSWER:

POINTS: 1

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

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Chapter 23 - Carbonyl Condensation Reactions

29.

ANSWER:

POINTS: 1

30.

ANSWER:

POINTS: 1 Exhibit 23-10 Consider the reaction below to answer the following question(s).

The Stork enamine reaction is a variation on the Michael reaction which utilizes an enamine nucleophile.

31. Refer to Exhibit 23-10. On the structures above, draw arrows indicating electron flow in each step of this reaction.

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Chapter 23 - Carbonyl Condensation Reactions ANSWER:

POINTS: 1 32. Refer to Exhibit 23-10. Draw the structures of the Ketone + Amine products of this reaction.

ANSWER:

POINTS: 1 33. Refer to Exhibit 23-10. Show how you might use a Stork enamine reaction to prepare the following compound.

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Chapter 23 - Carbonyl Condensation Reactions ANSWER:

POINTS: 1 34. Provide the indicated starting material and intermediate in the synthetic sequence below involving a Dieckmann cyclization, followed by a Robinson annulation.

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Chapter 23 - Carbonyl Condensation Reactions ANSWER:

POINTS: 1 35. Show how you might use a Robinson annulation reaction to synthesize the following compound. Draw the structures of both reactants and the structure of the intermediate Michael addition product.

ANSWER:

POINTS: 1 Exhibit 23-11 Give the major organic product(s) for each of the following reactions or reaction sequences.

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Chapter 23 - Carbonyl Condensation Reactions

36.

ANSWER:

POINTS: 1 37. This question was omitted on the printed copy. This placeholder question is here to maintain the numbering system integrity between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1

38.

ANSWER:

POINTS: 1

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Chapter 23 - Carbonyl Condensation Reactions

39.

ANSWER:

POINTS: 1

40.

ANSWER:

POINTS: 1

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Chapter 23 - Carbonyl Condensation Reactions

41.

ANSWER:

POINTS: 1 42. A multistep synthesis can be used to prepare 3,3-dimethylindanone. From the list provided below, choose the best reagent(s) for each step in this sequence. Place the letter in the box over the reaction arrow. a. Cl2, HOAc 0.05 equiv. NaOEt, ethanol, b. heat c. 1.

2. d. 1.

2.

3.

e.

I2, NaOH, H 2O

f.

H2SO4, Δ

PhMgBr, ether +

H3O Mg, ether CO2 H3O+

g. 1.

2. h. 1.

2. i. SOCl2, CHCl3

(Ph)2CuLi, ether H3O+ LDA, THF (CH3)2CHBr

ANSWER: step 1 = b; step 2 = g; step 3 = e; step 4 = i POINTS: 1 43. Consider the following molecular model. Atoms other than carbon and hydrogen are labeled.

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Chapter 23 - Carbonyl Condensation Reactions

A synthesis is set-up based upon treatment of this compound with NaOH in ethanol. This synthesis aims to produce the following substance.

What is wrong with this synthesis?

ANSWER: Benzaldehyde and its derivatives, such as p-bromobenzaldehye shown, will not undergo Aldol reactions because they do not possess α-hydrogens.

POINTS: 1 44. Draw the products of the Claisen condensation of the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 45. In carbonyl condensation reactions, a. the carbonyl group acts as an electrophile. b. the carbonyl group acts as a nucleophile. c. the carbonyl group acts as both an electrophile and a nucleophile. d. the carbonyl group acts as neither an electrophile nor a nucleophile.

ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 23 - Carbonyl Condensation Reactions 46. Which of the following does not characterize carbonyl condensation reactions? a. rare in biological systems b. form new carbon-to-carbon bonds c. enolate ion can act as a nucleophile d. an α-hydrogen required for an aldol condensation e. All of these characterize carbonyl condensation reactions.

ANSWER: a POINTS: 1 47. The product of an aldol condensation reaction is generally favored if a. an aldehyde is disubstituted. b. a ketone is disubstituted. c. an aldehyde has no alpha substituents. d. substituents near the carbonyl are large.

ANSWER: c POINTS: 1 48. Which of the following will not undergo an aldol condensation reaction? a. butanal b. 2-methylbutanal c. 2, 2-dimethylbutanal d. 3, 3-dimethyl-2-butanone e. neither c nor d f. All will undergo an aldol condensation.

ANSWER: c POINTS: 1 49. For which of the following will the aldol reaction go the furthest toward completion? a. b.

c.

d.

ANSWER: d POINTS: 1 50. Which of the following when reacted with butanal would be the most likely to produce a single aldol product? a.

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Chapter 23 - Carbonyl Condensation Reactions

b.

c.

d.

ANSWER: d POINTS: 1 51. Experimentally to produce 2-ethylpentanal from pentanal, how should the reaction conditions be adjusted to ensure the alpha substitution product rather than the aldol product result or a mixture of the two products? a. use a catalytic amount of a relatively weak base b. add a full equivalent of a strong base c. add the base and alklylation agent simultaneously d. add the alkylating agent after the addition of the base e. warm the reaction mixture f. a and c g. b and d h. b, d, and e ANSWER: g POINTS: 1 52. Which of the following constitutes a difference between an aldol and a Claisen condensation? a. basic conditions b. tetrahedral intermediate c. enolate ion formation d. protonation of the intermediate ANSWER: d POINTS: 1 53. Which of the following will produce a five-membered ring via a Dieckmann cyclization? a. 1,6-diester b. 1,6-diketone c. 1,7-diester Cengage Learning Testing, Powered by Cognero

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Chapter 23 - Carbonyl Condensation Reactions d. 1,7-diketone e. a and c f. b and d ANSWER: a POINTS: 1 54. Which of the following would be classified as a Michael donor? a. propenamide b. 1-nitropentane c. 3-buten-2-one d. propylpropenoate ANSWER: b POINTS: 1 55. What is the structure of the enamine produced in the reaction of cyclohexanone with piperidine? See structure below.

a.

b.

c.

d.

ANSWER: d POINTS: 1 56. Which of the following is the ester analog of an aldol condensation? a. Robinson annulation reaction b. Michael reaction c. Claisen condensation d. Dieckmann cyclization e. c and d f. All are analogs of an aldol condensation.

ANSWER: e POINTS: 1 57. Which of the following can result in cyclization? a. aldol condensation b. Claisen condensation c. Robinson annulation d. All can result in cyclization.

ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 23 - Carbonyl Condensation Reactions 58. Draw the general mechanism of a carbonyl condensation reaction between two molecules of acetaldehyde.

ANSWER:

POINTS: 1 59. Draw the mechanism of the Claisen condensation reaction between two molecules of methyl acetate in the presence of sodium ethoxide (abbreviated as “EtO-”).

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Chapter 23 - Carbonyl Condensation Reactions ANSWER:

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Page 22

Chapter 23 - Carbonyl Condensation Reactions 60. Draw the mechanism of the Michael reaction between ethyl acetoacetate and hex-5-en-4-one. (The ethyl group is abbreviated as “Et.”) ANSWER:

POINTS: 1

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Page 23

Chapter 24 - Amines and Heterocycles Exhibit 24-1 Classify each of the following nitrogen atoms in the following compounds as primary, secondary, tertiary, or quaternary.

1.

ANSWER: The nitrogen atom in ephedrine is secondary.

POINTS: 1

2.

ANSWER: The nitrogen atom in amphetamine is primary.

POINTS: 1

3.

ANSWER: The nitrogen in mepiquat chloride is a quaternary ammonium salt.

POINTS: 1

4.

ANSWER: The nitrogen atom in fexofenadine is tertiary POINTS: 1 Drawing Instructions: Draw structures corresponding to each of the given names.

5. Draw: N,N-diisopropylamine ANSWER:

POINTS: 1 6. Draw: N,N-dimethylcyclopentanamine Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1 7. Draw: 1,4-butanediamine ANSWER: H2NCH2CH2CH2CH2NH2 POINTS: 1 8. Draw: p-methoxyaniline ANSWER:

POINTS: 1 IUPAC Naming Instructions: Provide proper IUPAC names.

9. Name:

ANSWER: N-isopropyl-N-methylcyclobutylamine POINTS: 1 10. Name:

ANSWER: 2,4-hexanediamine POINTS: 1 11. Name:

ANSWER: methyl 2-aminobutanoate POINTS: 1 12. Name:

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Chapter 24 - Amines and Heterocycles

ANSWER: N-ethylaniline POINTS: 1 13. Name:

ANSWER: m-chloro-N,N-dimethylaniline POINTS: 1 14. Name:

ANSWER: p-nitroaniline POINTS: 1 Exhibit 24-2 Consider the reaction below to answer the following question(s).

Methamphetamine can be synthesized by reacting phenyl-2-propanone with methylamine in the presence of H2/Ni.

15. Refer to Exhibit 24-2. Identify the nucleophile in the initial reaction of phenyl-2-propanone to yield intermediate A. ANSWER: The nucleophile is methylamine.

POINTS: 1 16. Refer to Exhibit 24-2. Intermediate A is an example of: a. an imine b. an enamine c. an iminium ion d. an imide ANSWER: a POINTS: 1

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Chapter 24 - Amines and Heterocycles 17. Refer to Exhibit 24-2. Although the yield of methamphetamine is good, some unreacted phenyl-2-propanone remains after the reaction is complete. Describe how methamphetamine can be separated from phenyl-2-propanone.

ANSWER: Dissolve the reaction mixture containing the ketone and amine in ether and extract with aqueous HCl. The basic amine will dissolve in the aqueous layer as its hydrochloric salt. The neutral ketone will remain in the ether layer. Separate the two layers and neutralize the aqueous layer with NaOH. Extract the neutralized layer with ether. The purified amine will dissolve in the ether.

POINTS: 1 Exhibit 24-3 Refer to the Table of pKas below to answer the following question(s): pKas of Some Arylammonium Ions

Y −H −NH2

pKa 4.63 1.74 6.15

18. Refer to Exhibit 24-3. Based on the pKas for their corresponding ammonium ions, which arylamine above is the strongest base?

ANSWER: p-aminoaniline POINTS: 1 19. Refer to Exhibit 24-3. Explain the difference in acidity between p-cyanoanilinium ion and anilinium ion. Use both words and structures.

ANSWER: p-Cyanoanilinium ion is a stronger acid (pKa = 1.74) than anilinium ion (pKa = 4.63). p-Cyanoanilinium ion gives up a proton on nitrogen more readily than anilinium ion because the electron-withdrawing cyano group polarizes the benzene ring and thus decreases the stability of the positively charged ion.

POINTS: 1 20. Refer to Exhibit 24-3. Explain the difference in acidity between p-aminoanilinium ion and anilinium ion. Use both words and structures.

ANSWER: p-Aminoanilinium ion (pKa = 6.15) is less acidic than anilinium ion (pKa = 4.63) because p-aminoanilinium ion is stabilized by the electron-donating amino group. The p-amino group donates electron density to the benzene ring, which helps stabilize the positive charge on nitrogen.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 24 - Amines and Heterocycles 21. Rank the following compounds in order of increasing basicity. Label the least basic compound "1" and the most basic compound "4". Place the number corresponding to the compound's rank in the blank below the compound.

ANSWER:

POINTS: 1 22. When a THF solution of phenyl-2-propanone and bromobutane is treated with 50% aqueous NaOH, poor yields of 3phenyl-2-heptanone result. However, when a small amount of benzyltriethylammonium chloride is added to the reaction mixture, the yield of 3-phenyl-2-heptanone increases to 90%. Explain these results.

ANSWER: The reaction that is occurring is a carbonyl α-substitution reaction. Hydroxide ion abstracts a proton from C-l on phenyl-2-propanone to generate an enolate ion, which then undergoes SN2 attack at the primary carbon of bromobutane. However, since the organic layer and water layer are immiscible, the hydroxide is unable to come into contact with the ketone in the organic phase, so there is little or no reaction.

When the quaternary ammonium salt is added, it dissolves in both the aqueous and organic layers. When it moves into the organic layer, it takes hydroxide ion with it, to preserve the charge neutrality. Once in the organic layer, hydroxide ion reacts with the ketone to generate the enolate, which then undergoes SN2 displacement of bromide to give the carbonyl α-substitution product.

POINTS: 1 Exhibit 24-4 Refer to the reaction below to answer the following question(s):

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Chapter 24 - Amines and Heterocycles

23. Refer to Exhibit 24-4. Draw arrows on the structures provided below which show electron flow in the complete stepwise mechanism for the reaction above.

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Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1 24. Refer to Exhibit 24-4. This reaction is an example of: a. a Curtius rearrangement b. a Hofmann elimination reaction c. a Gabriel synthesis d. a Hofmann rearrangement ANSWER: d POINTS: 1 Exhibit 24-5 Give the major organic product(s) of each of the following reactions or sequences of reactions. Show all relevant stereochemistry.

25.

ANSWER:

POINTS: 1

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Chapter 24 - Amines and Heterocycles 26.

ANSWER:

POINTS: 1

27.

ANSWER:

POINTS: 1

28.

ANSWER:

POINTS: 1

29.

ANSWER:

POINTS: 1

30.

ANSWER:

POINTS: 1

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Chapter 24 - Amines and Heterocycles 31.

ANSWER:

POINTS: 1

32.

ANSWER:

POINTS: 1

33.

ANSWER:

POINTS: 1

34.

ANSWER:

POINTS: 1

35.

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Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1

36.

ANSWER:

POINTS: 1

37.

ANSWER:

POINTS: 1

38.

ANSWER:

POINTS: 1

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Chapter 24 - Amines and Heterocycles 39.

ANSWER:

POINTS: 1

40.

ANSWER:

POINTS: 1

41.

ANSWER:

POINTS: 1

42.

ANSWER:

POINTS: 1

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Chapter 24 - Amines and Heterocycles 43.

ANSWER:

POINTS: 1

44.

ANSWER:

POINTS: 1

45.

ANSWER:

POINTS: 1 Exhibit 24-6 Show how each of the following transformations might be best accomplished. More than one step may be required. Show all reagents and all intermediate structures.

46.

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Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1

47.

ANSWER:

POINTS: 1

48.

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Page 13

Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1

49.

ANSWER:

POINTS: 1

50.

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Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1 Exhibit 24-7 Choose the best series of reactions for accomplishing each conversion below.

51.

a. b. c. d.

ANSWER: c POINTS: 1

52.

a. b.

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Chapter 24 - Amines and Heterocycles c.

d.

ANSWER: c POINTS: 1

53.

a. b.

c. d.

ANSWER: a POINTS: 1 54. Mescaline is a hallucinogenic alkaloid isolated from peyote cactus. Synthesize mescaline from 3,4,5trimethoxytoluene. Show all reagents and all intermediate structures.

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Chapter 24 - Amines and Heterocycles ANSWER:

POINTS: 1 Exhibit 24-8 Refer to the data below to answer the following question(s): Triclocarban is a disinfectant prepared from 3,4-dichloroaniline and 4-chlorophenylisocyanate.

55. Refer to Exhibit 24-8. Prepare 4-chlorophenylisocyanate from toluene.

ANSWER:

POINTS: 1 56. Refer to Exhibit 24-8. Prepare 3,4-dichloroaniline from benzene.

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Chapter 24 - Amines and Heterocycles

ANSWER:

POINTS: 1 57. Refer to Exhibit 24-8. Propose a mechanism for the reaction of 3,4-dichloroaniline and 4-chlorophenylisocyanate to yield Triclocarban.

ANSWER:

POINTS: 1 58. From the list provided below, choose the best reagent(s) for each step in the following synthesis. There is only one answer for each reaction. a. NaBH4, ethanol b. KCN, acetone c. 1.

2.

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d.

LiAlH4, THF H2O

1. HNO2, H2SO4 2. CuCN, KCN e. 1. SnCl2, H3O+ 2. NaOH, H2O f. HNO3, H2SO4 Page 18

Chapter 24 - Amines and Heterocycles

ANSWER: step 1 = e; step 2 = d; step 3 = c POINTS: 1 59. Draw the structures for each of the intermediates in the boxes provided for the synthesis below.

ANSWER:

POINTS: 1 60. Propose a structure that is consistent with the following spectral data: Cengage Learning Testing, Powered by Cognero

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Chapter 24 - Amines and Heterocycles MS: IR: 1

M+ at 73 3350 cm−1 (weak, single band) δ1.05 (br s, 1H), δ1.15 (t, 6H), δ2.65 (q, 4H)

H NMR: ANSWER: diethylamine, (CH3CH2)2NH POINTS: 1

61. Name the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER: N,N-dimethylbutylamine POINTS: 1 62. Draw the product of the reaction of the following substance with 1) LiAlH4 in ether, 2) H2O. From what alkyl bromide could this substance be produced when reacted with NaCN? Atoms other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 63. Write the equation for the reaction represented by the pKa of methyl amine.

ANSWER:

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Chapter 24 - Amines and Heterocycles POINTS: 1 64. Describe the importance of the formation of arenediazonium salts.

ANSWER: Arenediazonium salts are stable and readily form from primary aryl amines upon treatment with HNO2/H2SO4. These salts form an intermediate that readily undergo nucleophilic substitution with a wide variety of nucleophiles. This allows for the preparation of a wide variety of substituted aromatic compounds: nitriles, halides, carboxylic acids, phenols, etc. Diazonium salts also undergo coupling reactions to produce azo compounds.

POINTS: 1 65. Classify the following amine. Atoms other than carbon and hydrogen are labeled.

a. primary c. tertiary ANSWER: c POINTS: 1

b. secondary d. quaternary

66. Compared to pentylamine, pentane is a. more soluble in water. b. has a higher boiling point. c. is less basic. d. more offensive in terms of odor.

ANSWER: c POINTS: 1 67. The pKa of diethylammonium ion is 10.98. What is the pKb of diethylamine? a. 3.02 b. 14.00 c. 24.98 d. 7.00 ANSWER: a POINTS: 1 68. What is the percentage of neutral species in a 0.00500 M solution of pyridine at pH 9.50? pKa of pyridine is 5.25. a. 25% b. 75% c. 50% d. nearly 100% e. close to 0% ANSWER: d Cengage Learning Testing, Powered by Cognero

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Chapter 24 - Amines and Heterocycles POINTS: 1 69. At which pH will a solution of a 0.00050 M solution of aniline (pKa = 4.63) contain the greatest amount of the neutral species? a. 2.74 b. 4.63 c. 7.31 d. 9.90 ANSWER: d POINTS: 1 70. Which of the following would best represent the form of ethylamine (pKa) at physiological pH? a. CH3CH2NH3+ b. CH3CH2NH2 c. CH3CH2NH–

d. a combination of a and b

ANSWER: a POINTS: 1 71. The amino acid phenylalanine has the structure below.

Which of the following is the best representation of this acid at pH 7.3? a. b.

c.

d.

ANSWER: c POINTS: 1 72. Which of the following methods of preparation of amines results in an amine with one less carbon atom than the starting material?

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Chapter 24 - Amines and Heterocycles a. reduction of a nitrile c. reductive amination of a ketone ANSWER: d POINTS: 1

b. Gabriel synthesis from an alkyl halide d. Hofman rearrangement of an amide

73. Which of the following methods of preparation of amines can be used to prepare primary, secondary, and tertiary amines? a. reduction of a nitrile b. Gabriel synthesis from an alkyl halide c. reduction amination of a ketone d. Hofman rearrangement of an amide e. None of these methods is applicable to all types of amines.

ANSWER: c POINTS: 1 74. Which of the following is not classified as a heterocycle? a.

b.

c.

d.

e.

c and d ANSWER: d POINTS: 1 75. An unknown substance produced the following spectrum.

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Chapter 24 - Amines and Heterocycles

This unknown is likely to be a: a. primary amine b. secondary amine c. tertiary amine d. amide ANSWER: a POINTS: 1 76. Examine the following spectrum.

Which of the following would best describe this compound? a. The compound may contain zero nitrogen atoms. b. The compound may contain one nitrogen atoms. c. The compound may contain two nitrogen atoms. d. The compound may contain three nitrogen atoms. e. a or c f. b or d g. The possible number of nitrogen atoms cannot be determined from the spectrum.

ANSWER: f POINTS: 1 77. Which of the following is not a valid comparison of the spectral data for alkyl amines and alkyl alcohols? a. IR: strong absorptions > 3000 cm–1. b. MS: peaks due α-cleavage c. 13C NMR: adjacent carbon signal shifted downfield relative to an alkane d. 1H NMR: proton bonded to the heteroatom appears over a wide range e. All are valid comparisons between the spectral data of amines and alcohols.

ANSWER: e Cengage Learning Testing, Powered by Cognero

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Chapter 24 - Amines and Heterocycles POINTS: 1 78. Draw the mechanism of the reductive amination of 2-butanone in the presence of ammonia and hydrochloric acid.

ANSWER:

POINTS: 1 79. Draw the mechanism for the Hofmann rearrangement of acetamide to acetamine.

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Chapter 24 - Amines and Heterocycles ANSWER:

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Chapter 24 - Amines and Heterocycles 80. Draw the mechanism of the Sandmeyer reaction that converts o-methyl aniline to o-bromotoluene.

ANSWER:

POINTS: 1

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Chapter 25 - Biomolecules: Carbohydrates Exhibit 25-1 Convert the following Fischer projections into tetrahedral representations, and assign R or S stereochemistry to each.

1.

ANSWER:

POINTS: 1

2.

ANSWER:

POINTS: 1

3.

ANSWER:

POINTS: 1 Exhibit 25-2 Label each pair of stereoisomers below as: a. b. c.

enantiomers diastereomers identical

Place the letter of the correct answer in the blank to the left of the pair of stereoisomers.

4. _____ ANSWER: b POINTS: 1

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Chapter 25 - Biomolecules: Carbohydrates

5. _____

ANSWER: b POINTS: 1

6. _____ ANSWER: a POINTS: 1 7. Draw a Fischer projection of (2R,3S)-dibromobutane.

ANSWER:

POINTS: 1 Exhibit 25-3 Refer to the monosaccharides below to answer each of the following question(s):

Classify each sugar by type; for example, glucose is an aldohexose.

8. Refer to Exhibit 25-3. Xylulose is _____.

ANSWER: a ketopentose POINTS: 1 9. Refer to Exhibit 25-3. Erythrulose is _____.

ANSWER: a ketotetrose POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 25 - Biomolecules: Carbohydrates 10. Refer to Exhibit 25-3. Sorbose is _____.

ANSWER: a ketohexose POINTS: 1 Exhibit 25-4 From the sugars below, choose the one that best fits each description below. Place the letter of the sugar in the blank to the left of the description. There is only one correct answer for each question.

11. _____ a D-sugar.

ANSWER: d POINTS: 1 12. _____ a deoxy sugar.

ANSWER: b POINTS: 1 13. _____ a sugar with three chirality centers.

ANSWER: a POINTS: 1 14. _____ a sugar which yields a single alditol upon reduction.

ANSWER: b POINTS: 1 15. Refer to Exhibit 25-4. Assign R or S configuration to each chirality center in sorbose.

ANSWER:

POINTS: 1 16. Refer to Exhibit 25-4. Name sorbose systematically.

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Chapter 25 - Biomolecules: Carbohydrates ANSWER: (3S, 4R, 5S)-1,3,4,5,6-pentahydroxy-2-hexanone POINTS: 1 17. Refer to Exhibit 25-4. Draw both chair conformations of the α-anomer of rhamnose in its pyranose form. Circle the more stable conformation.

ANSWER:

POINTS: 1 18. Refer to Exhibit 25-4. Provide the complete name for the α-anomer of rhamnose in its pyranose form.

ANSWER: α-L-Rhamnopyranose POINTS: 1 Exhibit 25-5 Refer to the equilibrium below to answer the following question(s).

19. Refer to Exhibit 25-5. Classify ribose by carbonyl type and number of carbons.

ANSWER: Ribose is an aldopentose.

POINTS: 1 20. Refer to Exhibit 25-5. Which enantiomer of ribose is drawn, D or L? ANSWER: D POINTS: 1 21. Refer to Exhibit 25-5. The correct name for the cyclic structure is _______. a. α-L-ribofuranose b. β-D-ribofuranose c. α-L-ribopyranose d. β-D-ribopyranose ANSWER: b POINTS: 1 Exhibit 25-6 Refer to the sugars below to answer the following question(s). Choose the sugar that best fits each description and place Cengage Learning Testing, Powered by Cognero

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Chapter 25 - Biomolecules: Carbohydrates the letter of the sugar in the blank to the left of the description. There is only one correct answer for each question, but sugars may be used more than once.

22. _____ a D-ketohexose ANSWER: a POINTS: 1 23. _____ oxidizes to an optically inactive aldaric acid ANSWER: c POINTS: 1 24. _____ a dextrorotary hexose ANSWER: b POINTS: 1 25. _____ a ketose with two chirality centers ANSWER: d POINTS: 1 26. The Fischer projection for D-iodose corresponds to which pyranose below?

a.

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b.

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Chapter 25 - Biomolecules: Carbohydrates

c.

d.

ANSWER: d POINTS: 1 27. What is the correct structure for α-D-glucopyranose?

a.

b.

c.

d.

ANSWER: a POINTS: 1 Exhibit 25-7 What is the relationship between the following compounds?

28.

a. anomers b. meso compounds c. enantiomers d. equivalent structures ANSWER: c POINTS: 1

29.

a. b. c.

anomers meso compounds enantiomers

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Chapter 25 - Biomolecules: Carbohydrates d. equivalent structures ANSWER: a POINTS: 1 Exhibit 25-8 Draw structures for the products you would expect to obtain from reaction of β-D-galactopyranose with each of the following reagents. Be sure to include all relevant stereochemistry.

β-D-galactopyranose

30. CH3I, Ag2O ANSWER:

POINTS: 1 31. warm dilute HNO3 ANSWER:

POINTS: 1 32. (CH3CO)2O, pryridine ANSWER:

POINTS: 1 33. NaBH4 in H2O

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Chapter 25 - Biomolecules: Carbohydrates ANSWER:

POINTS: 1 34. CH3OH, HCl ANSWER:

POINTS: 1 35.

1. HCN 2. H2, Pd/BaSO4 3. H3O+ ANSWER:

POINTS: 1 36. Br2, H2O ANSWER:

POINTS: 1 37.

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Chapter 25 - Biomolecules: Carbohydrates 1. H2NOH 2. (CH3CO)2O, CH3CO2Na 3. NaOCH3 ANSWER:

POINTS: 1 Exhibit 25-9 Choose a structure from the list provided below that best fits each of the following descriptions. Place the letter of the structure in the blank to the left of the description. There is only one correct answer for each question.

A.

B.

C.

D.

E.

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Chapter 25 - Biomolecules: Carbohydrates

F.

38. _____ a reducing monosaccharide ANSWER: C POINTS: 1 39. _____ a non-reducing disaccharide ANSWER: D POINTS: 1 40. _____ amylopectin ANSWER: F POINTS: 1 41. _____ a 1,4' β-glycoside ANSWER: E POINTS: 1 42. _____ cellulose ANSWER: E POINTS: 1 Exhibit 25-10 Choose a structure from the list provided below that best fits each of the following descriptions. Place the letter of the structure in the blank to the left of the description. There is only one correct answer for each question. a.

starch

c.

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b.

cellulose

d.

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Chapter 25 - Biomolecules: Carbohydrates

e.

f.

sucrose

43. _____ a monosaccharide that gives a negative Benedict's Test.

ANSWER: e POINTS: 1 44. _____ a β-1,4'-glycoside ANSWER: b POINTS: 1 45. _____ a disaccharide ANSWER: f POINTS: 1 46. Fructose reduces Tollens' reagent even though it contains no aldehyde group. This occurs because fructose is readily isomerized to an aldose in basic solution. Write the complete stepwise mechanism for the base-catalyzed isomerization of fructose to an aldohexose. Show all intermediate structures and all electron flow with arrows.

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Chapter 25 - Biomolecules: Carbohydrates ANSWER:

POINTS: 1 47. Melibiose is a reducing disaccharide that forms D-glucose and D-galactose on hydrolysis. Reaction of melibiose with iodomethane and silver oxide yields an octamethyl derivative, which can be hydrolyzed with aqueous acid to give one equivalent of 2,3,4,6-tetra-O-methyl-D-galactopyranose and one equivalent of 2,3,4,-tri-O-methyl-D-glucopyranose. If melibiose contains an α-glycosidic bond, what is its structure?

ANSWER:

POINTS: 1 48. This question was omitted on the printed copy. This placeholder question is here to maintain the numbering system integrity between the printed copy and ExamView. Therefore, it has been marked "do not use on test" in ExamView's Cengage Learning Testing, Powered by Cognero

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Chapter 25 - Biomolecules: Carbohydrates question information dialog. As a result, this placeholder question is automatically prevented from being chosen as a test question.

ANSWER: Answer not provided.

POINTS: 1 49. Compound U is an aldotetrose that can be oxidized to an optically active aldaric acid, compound V. Upon Wohl degradation, compound U is converted into D-glyceraldehyde. Upon Kiliani-Fischer chain extension, compound U is converted into compounds W and X. Compound W can be oxidized to an optically inactive aldaric acid, compound Y, but compound X is oxidized to an optically active aldaric acid Z. What are the structures of compounds U-Z? ANSWER:

POINTS: 1 Exhibit 25-11 Consider the reaction below to answer the following question(s).

50. Refer to Exhibit 25-11. Place a triangle around the anomeric carbon in compound Q.

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Chapter 25 - Biomolecules: Carbohydrates ANSWER:

POINTS: 1 51. Refer to Exhibit 25-11. Compound Z is: a. the D-anomer. b. the α-anomer. c. the β-anomer. d. the L-anomer.

ANSWER: b POINTS: 1 52. Refer to Exhibit 25-11. Which anomer is the LEAST stable?

Q or Z ANSWER: Z POINTS: 1 53. Refer to Exhibit 25-11. The process by which anomer Q is converted into a mixture of both anomers is called: a. anomeric rotation. b. glycosidation. c. pyranorotation. d. mutarotation.

ANSWER: d POINTS: 1 54. Refer to Exhibit 25-11. Q and Z are cyclic examples of: a. acetals b. hemiacetals c. alditols d. hemialditols ANSWER: b POINTS: 1 55. This cyclization is an example of an intramolecular nucleophilic addition of an alcohol with an aldehyde. This process is catalyzed by mild acid. On the structures provided below draw arrows indicating electron flow in this reaction mechanism.

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Chapter 25 - Biomolecules: Carbohydrates

ANSWER:

POINTS: 1 Exhibit 25-12 Below is the structure of the antigenic determinant for the blood group B. Use this structure to answer the following question(s).

56. Refer to Exhibit 25-12. Put a BOX around the 1,3'-glycosidic linkage.

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Chapter 25 - Biomolecules: Carbohydrates ANSWER:

POINTS: 1 57. Refer to Exhibit 25-12. One of the monosaccharides is an unusual deoxy sugar called L-fucose. CIRCLE the L-fucose moiety.

ANSWER:

POINTS: 1 58. Consider the following structure. Atoms other than carbon and hydrogen are labeled.

Which of the following is not a correct classification for this substance? a. Aldose b. Pentose Cengage Learning Testing, Powered by Cognero

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Chapter 25 - Biomolecules: Carbohydrates c. Nonreducing sugar d. Monosaccharide e. All are correct classifications.

ANSWER: c POINTS: 1 59. Consider the following structure. Atoms other than carbon and hydrogen are labeled.

How many chirality centers are present? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c POINTS: 1 60. In comparing Fischer projections, which of the following are allowed rotations in the plane of the page? a. 90° b. 180° c. 270° d. a and c e. a, b and c ANSWER: b POINTS: 1 61. Examine the following molecular models. Atoms other than carbon and hydrogen are labeled.

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Chapter 25 - Biomolecules: Carbohydrates A

B

Which of the following correctly describes these substances? a. A is a D sugar and B is an L sugar b. A is an L sugar and B is a D sugar c. Both are D sugars. d. Both are L sugars.

ANSWER: a POINTS: 1 62. For a hypothetical aldoheptose with 5 chirality centers, how many stereoisomers are possible? a. 4 b. 8 c. 16 d. 32 e. 64 ANSWER: d POINTS: 1 63. Examine the following molecular models. Atoms other than carbon and hydrogen are labeled.

Which of the following correctly describes these substances? a. Equivalent structures b. Enantiomers c. Diastereomers d. Constitutional isomers ANSWER: c POINTS: 1 64. What relationships exists between the following two substances? Atoms other than carbon and hydrogen are labeled.

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Chapter 25 - Biomolecules: Carbohydrates

a. Identical c. Anomers ANSWER: c POINTS: 1

b. Enantiomers d. Meso compounds

65. Examine the following structure. Atoms other than carbon and hydrogen are labeled.

What letter labels the anomeric center? a. A b. B c. C d. D e. E f. F ANSWER: a POINTS: 1 66. Examine the following structure. Atoms other than carbon and hydrogen are labeled.

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Chapter 25 - Biomolecules: Carbohydrates

What letter labels carbon 6? a. A b. B c. C d. D e. E f. F ANSWER: f POINTS: 1 67. Classify the following substance. Atoms other than carbon and hydrogen are labeled.

a.

b.

c.

d.

ANSWER: b POINTS: 1 68. Consider the following reaction sequence.

1.) HCN 2.) H2/Pd 3.) H3O+ Cengage Learning Testing, Powered by Cognero

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Chapter 25 - Biomolecules: Carbohydrates

The result of this treatment to a monosaccharides would be: a. oxidation of the carbonyl group b. lengthening the carbon chain by one carbon atom c. shortening the carbon chain by one carbon atom d. formation of a glycoconjugate ANSWER: b POINTS: 1 69. The intermediate common to both chain lengthening and shortening of a simple sugar is a(n): a. imine. b. oxime. c. enamine. d. cyanohydrin.

ANSWER: d POINTS: 1 70. Which of the following is correct as applies to the essential carbohydrates? a. Cannot be synthesized by metabolic pathways b. Simple aldohexoses c. Arise from glucose or galactose d. Synthesis is highly endergonic ANSWER: d POINTS: 1 71. Which of the following is a difference between amylopectin and glycogen? a. b. c. Energy storage function d. Glucose monomers e. Molar mass f. All characteristics listed are shared by amylopectin and glycogen.

ANSWER: e POINTS: 1

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins Exhibit 26-1 Refer to the data below to answer the following question(s): Leucine is an essential amino acid with the systematic name 2-amino-3-methylpentanoic acid. It has pKa1 = 2.36 and pKa2 = 9.60.

1. Refer to Exhibit 26-1. Draw the condensed structure for leucine, and label all chirality centers with an asterisk.

ANSWER:

POINTS: 1 2. Refer to Exhibit 26-1. How many possible stereoisomers of leucine are there?

ANSWER: Since leucine has one chirality center, there are 21, or 2 possible leucine stereoisomers.

POINTS: 1 3. Refer to Exhibit 26-1. Draw a Fischer projection of L-leucine and label the chirality center(s) as R or S. ANSWER:

POINTS: 1 4. Refer to Exhibit 26-1. What is the pI of leucine?

ANSWER: The pI is the isoelectric point and is the average of pKa1 and pKa2. For leucine:

POINTS: 1 5. Refer to Exhibit 26-1. Draw the structure of the predominant form of leucine at pH = 10.00.

ANSWER:

POINTS: 1 6. Refer to Exhibit 26-1. Draw the structure of the predominant form of leucine at pH = 1.50.

ANSWER:

POINTS: 1 7. Refer to Exhibit 26-1. Leucine is described as an essential amino acid. What does this mean?

ANSWER: Humans are able to synthesize only 10 of the 20 amino acids necessary for protein synthesis. The remaining 10 are called essential amino acids since they must be obtained from dietary sources. Failure to include an Cengage Learning Testing, Powered by Cognero

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins adequate dietary supply of these essential amino acids can lead to severe deficiency diseases.

POINTS: 1 8. Refer to Exhibit 26-1. Show how leucine might be synthesized using the Knowles enantioselective synthesis.

ANSWER:

POINTS: 1 9. Refer to Exhibit 26-1. Show the alkyl halide you would use to prepare leucine by the amidomalonate method.

ANSWER: POINTS: 1 Exhibit 26-2 MATCH a structure from the list below to each of the following terms. Place the letter of the structure in the blank to the left of the term which it describes.

A.

B.

C.

Val−Phe−Leu−Met−Tyr−Pro−Gly−Trp−Cys−Glu

D.

E.

F.

Asp−Tyr−Ile−His−Pro−Phe−Arg−Val

G.

apoenzyme

H.

I.

J.

K.

L. Val−Lys−Phe−Gly−Arg−Met−Arg−Phe

M. vitamins

10. _____ an octapeptide with a C-terminal valine.

ANSWER: F POINTS: 1 11. _____ an amino acid in its zwitterionic form.

ANSWER: B POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins 12. _____ a polypeptide which gives four fragments on treatment with chymotrypsin.

ANSWER: C POINTS: 1 13. _____ the product of an Edman degradation.

ANSWER: J POINTS: 1 14. _____ a carboxyl-protected amino acid.

ANSWER: D POINTS: 1 15. _____ a peptide coupling reagent.

ANSWER: H POINTS: 1 16. _____ small organic molecules which act as coenzymes.

ANSWER: M POINTS: 1 Exhibit 26-3 Refer to the data below to answer the following question(s): Amino Acid Arginine Glutamic Acid Tryptophan

Isoelectric point 10.76 3.22 5.89

17. Refer to Exhibit 26-3. At what pH would you carry out an electrophoresis experiment if you wanted to separate a mixture of lysine, aspartic acid and phenylalanine? Explain.

ANSWER: The optimum pH for the electrophoresis of three amino acids occurs at the isoelectric point of the amino acid intermediate in acidity. At this pH, one amino acid migrates toward the cathode (the least acidic), one migrates toward the anode (the most acidic), and the amino acid intermediate in acidity does not migrate. In this example, electrophoresis at 5.89 allows the maximum separation of the three amino acids.

POINTS: 1 18. Define isoelectric point.

ANSWER: The isoelectric point is the pH at which an amino acid is exactly balanced between anionic and cationic forms and exists primarily as the neutral, dipolar zwitterion.

POINTS: 1 19. The most basic amino acid is _____.

ANSWER: Arginine POINTS: 1 20. The most acidic amino acid is ____.

ANSWER: Aspartic acid POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins Exhibit 26-4 Refer to the data below to answer the following question(s): Porcine dynorphin is a neuropeptide having 17 amino acid residues. Its structure is shown below.

Tyr−Gly−Gly−Phe−Leu−Arg−Arg−Ile−Arg−Pro−Lys−Leu−Lys−Trp−Asp−Asn−Gln 21. Refer to Exhibit 26-4. Leu5-enkephalin is a pentapeptide contained as the N-terminal sequence of dynorphin. Write the structure of Leu5-enkephalin using both the three letter and one letter abbreviations for the amino acids.

ANSWER: Tyr−Gly−Gly−Phe−Leu and Y-G-G-F-L POINTS: 1 22. Refer to Exhibit 26-4. What fragments would result if dynorphin were cleaved by trypsin?

ANSWER: Trypsin catalyzes the hydrolysis of peptides as the carboxyl side of arginine and lysine. Consequently, the fragments resulting from trypsin hydrolysis of dynorphin are: Tyr−Gly−Gly−Phe−Leu−Arg Arg Ile−Arg Pro−Lys Leu−Lys Trp−Asp−Asn−Gln POINTS: 1 23. Refer to Exhibit 26-4. What fragments would result if dynorphin were cleaved by chymotropsin?

ANSWER: Chymotropsin catalyzes the hydrolysis of peptides at the carboxyl side of the aryl-substituted amino acids phenylalanine, tyrosine, and tryptophan. Thus, the fragments resulting from chymotropsin hydrolysis of dynorphin are: Tyr Gly−Gly−Phe Leu−Arg−Arg−Ile−Arg−Pro−Lys−Leu−Lys−Trp Asp−Asn−Gln POINTS: 1 Exhibit 26-5 Refer to the data below to answer the following question(s): The octapeptide saralasin is a specific antagonist of angiotensin II. A derivative of saralasin is used therapeutically as an antihypertensive. Amino acid analysis of saralasin show the presence of the following amino acids: Ala, Arg, His, Pro, Sar, Tyr, Val, Val 24. Refer to Exhibit 26-5. Sar is the abbreviation for sarcosine, N-methyl aminoethanoic acid. Draw the structure of sarcosine.

ANSWER: POINTS: 1 25. Refer to Exhibit 26-5. N-Terminal analysis by the Edman method shows saralasin contains sarcosine at the Nterminus. Partial hydrolysis of saralasin with dilute hydrochloric acid yields the following fragments: Cengage Learning Testing, Powered by Cognero

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins Tyr−Val−His Sar−Arg−Val His−Pro−Ala Val−Tyr−Val Arg−Val−Tyr What is the structure of saralasin?

ANSWER: Sar−Arg−Val Arg−Val−Tyr Val−Tyr−Val Tyr−Val−His His−Pro−Ala The complete sequence: Sar−Arg−Val−Tyr−Val−His−Pro−Ala POINTS: 1 26. Show the steps involved in a synthesis of the peptide "F-G-I" using the Merrifield procedure.

ANSWER: 1.

2.

3.

4.

5.

6.

7.

POINTS: 1 Exhibit 26-6 During peptide synthesis, the carboxylic acid group may be protected as the methyl ester. Aqueous base hydrolysis is utilized to remove the ester protecting group.

27. Refer to Exhibit 26-6. Write the complete stepwise mechanism for the hydrolysis of the methyl ester of BOC−Val−Gly−OCH3, above. Show all intermediate structures and all electron flow with arrows.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins ANSWER:

POINTS: 1 28. Refer to Exhibit 26-6. The reaction above is an example of: a. nucleophilic addition b. electrophilic addition c. nucleophilic acyl substitution d. electrophilic substitution ANSWER: c POINTS: 1 Exhibit 26-6 MATCH a term from the list below to each definition. Place the letter of the term in the blank to the left of the definition. a. b. c. d.

Ligases Fibrous proteins Conjugated protein Hydrolases

e. f. g. h.

Simple protein Globular proteins Lyases Transferases

29. _____ Proteins that are tough and insoluble in water.

ANSWER: b POINTS: 1 30. _____ Enzymes that catalyze the breaking away of a small molecule such as from a substrate.

ANSWER: g POINTS: 1 31. _____ Enzymes that catalyze the bonding together of two substrates.

ANSWER: a POINTS: 1 32. Assign an S or R configuration and a D or L configuration to the following amino acid. Atoms other than carbon and hydrogen are labeled.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins

ANSWER: S and L POINTS: 1 33. On the titration curve for an amino acid. In terms of the letters on the graph, label the isoelectric point, the pKa of the amino group and the pKa of the carboxyl group.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins ANSWER:

POINTS: 1 34. Show how you could prepare the following amino acid starting with the appropriate carboxylic acid. Atoms other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 35. What alkyl halide should be used to prepare the following amino acid by the amidomalonate method? Atoms other than carbon and hydrogen are labeled.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins

ANSWER: 2-bromobutane

POINTS: 1 36. Draw and name the structure of the tripeptide formed from the following amino acids. Use B as the N-terminal acid and A as the C-terminal end. Atoms other than carbon and hydrogen are labeled.

ABC ANSWER:

isoleucylleucylhistidine POINTS: 1 37. Classify the following amino acid. Atoms other than carbon and hydrogen are labeled.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins

a. acidic b. basic c. neutral ANSWER: c POINTS: 1 38. Which of the following is a correct representation of the amino acid below? Atoms other than carbon and hydrogen are labeled.

a. Isoleucine b. ILe c. I d. None of these present this amino acid. e. All of these represent this amino acid.

ANSWER: d POINTS: 1 39. The following amino acid: (Atoms other than carbon and hydrogen are labeled.)

a. has two chirality centers. b. is classified as secondary α-amino acid. c. is characterized by three pKas d. would have a net +1 charge as shown.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins ANSWER: c POINTS: 1 40. Consider the following titration curve for an amino acid.

What is the pI of this amino acid? a. A b. B c. D d. F ANSWER: c POINTS: 1 41. Consider the following titration curve for an amino acid.

At which of the following “pH” values would the amino acid be fully deprotonated?

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins a. A b. B c. D d. F ANSWER: d POINTS: 1 42. Consider the following titration curve for an amino acid.

At which of the following “pH” values would the amino acid have a net charge of zero? a. A b. B c. D d. F ANSWER: c POINTS: 1 43. Consider the following data for a hypothetical basic amino acid. pKa 3.15 8.22 10.93 What is the pI of this substance? a. 5.68 b. 7.04 c. 8.22 d. 9.58 ANSWER: d POINTS: 1 44. How many different peptides could be produced using one each of the following amino acids?

Ala, Gly, His, Cys a. 6 b. 16 c. 24 d. 64 Cengage Learning Testing, Powered by Cognero

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins ANSWER: c POINTS: 1 45. Consider the following image.

Which level of protein structure is shown here? a. Primary b. Secondary c. Tertiary d. Quaternary ANSWER: b POINTS: 1 46. Consider the following image.

Which level of protein structure is shown here? a. Primary b. Secondary c. Tertiary d. Quaternary ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins 47. Consider the following image.

Which level of protein structure is shown in the box? a. Primary b. Secondary c. Tertiary d. Quaternary ANSWER: a POINTS: 1 48. Which level of protein structure is disrupted upon mild heating or alteration in pH? a. Primary b. Secondary c. Tertiary ANSWER: c POINTS: 1 49. The enzyme that would catalyze the following reaction would belong in which classification? a. kinase b. oxidase c. protease d. transaminase ANSWER: a POINTS: 1 50. Consider the image below.

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Chapter 26 - Biomolecules: Amino Acids, Peptides, and Proteins

What type of interaction is indicated by the arrow? a. Salt bridges b. Disulfide bonds c. Hydrogen bonds d. Dispersion forces ANSWER: c POINTS: 1 51. The coenzymes NAD+ and FAD are involved in: a. acyl transfer. b. methyl transfer. c. decarboxylation. d. oxidation/reduction.

ANSWER: d POINTS: 1 52. Which of the following contributes most substantially to rate acceleration produced by an enzyme? a. Ability to bind the substrate b. Ability to bind the transition state c. Ability to bind the product d. All of the above contribute equally.

ANSWER: b POINTS: 1

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Chapter 27 - Biomolecules: Lipids Exhibit 27-1 MATCH each of the following terms to a structure from the list below. There is only one correct structure for each term, and structures may be used more than once. Place the letter of the structure in the blank to the left of the corresponding term.

A.

B.

C.

D.

E.

F.

G.

Malonyl ACP

I.

H.

J.

1. _____ starting material for fatty acid biosynthesis.

ANSWER: I POINTS: 1 2. _____ a phosphoglyceride ANSWER: F POINTS: 1 3. _____ a prostaglandin ANSWER: D POINTS: 1 4. _____ monoterpene precursor ANSWER: J POINTS: 1 Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 27 - Biomolecules: Lipids 5. _____ a triglyceride ANSWER: C POINTS: 1 6. _____ a sphingomyelin ANSWER: B POINTS: 1 7. _____ a soap ANSWER: H POINTS: 1 8. _____ a glycerophospholipid ANSWER: F POINTS: 1 9. _____ a steroid ANSWER: A POINTS: 1 10. _____ a sesquiterpene ANSWER: E POINTS: 1 11. Bolecic acid, C18H26O2, yields stearic acid, CH3(CH2)16COOH, on catalytic hydrogenation. Ozonolysis of bolecic acid yields the following products: CH2O HOOC−COOH, HOOC(CH2)7COOH, and OHC(CH2)4COOH Propose a structure for bolecic acid.

ANSWER: There are two possible structures for bolecic acid:

POINTS: 1 Exhibit 27-2 Refer to the structures of the terpenes below to answer the following question(s).

Show the positions of the isoprene units in each of the indicated terpenes.

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Chapter 27 - Biomolecules: Lipids

12. grandisol ANSWER:

POINTS: 1 13. α-santonin ANSWER:

POINTS: 1 14. abietic acid ANSWER:

POINTS: 1 15. linalool

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Chapter 27 - Biomolecules: Lipids ANSWER:

POINTS: 1 16. ocimene ANSWER:

POINTS: 1 17. lineatin ANSWER:

POINTS: 1 Exhibit 27-3 Indicate by asterisks the stereogenic centers present in each of the indicated terpenes.

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Chapter 27 - Biomolecules: Lipids ANSWER:

POINTS: 1 19. α-santonin ANSWER:

POINTS: 1 20. abietic acid ANSWER:

POINTS: 1 21. linalool ANSWER:

POINTS: 1 22. lineatin ANSWER:

POINTS: 1 23. Abietic acid is an example of: a. a monoterpene b. a sesquiterpene c. a diterpene d. a sesterterpene ANSWER: c Cengage Learning Testing, Powered by Cognero

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Chapter 27 - Biomolecules: Lipids POINTS: 1 24. Suggest a mechanism by which caryophyllene is transformed into humulene by treatment with acid.

ANSWER:

POINTS: 1 25. Draw the most stable chair conformation of terpin hydrate.

ANSWER:

POINTS: 1 Exhibit 27-4 Draw the structures of the major organic product(s) you would expect to obtain from reaction of androsterone with each of Cengage Learning Testing, Powered by Cognero

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Chapter 27 - Biomolecules: Lipids the following reagents.

26. PCC ANSWER:

POINTS: 1 27. NaBH4, ethanol ANSWER:

POINTS: 1 28. CH3COCl, pyridine ANSWER:

POINTS: 1 29. NaH, then CH3I

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Chapter 27 - Biomolecules: Lipids ANSWER:

POINTS: 1 30. Mixtures of esters of long-chain carboxylic acids with long-chain alcohols are called: a. waxes b. fats and oils c. fatty acids d. soaps ANSWER: a POINTS: 1 31. Chemically, these are triacylglycerols, triesters of glycerol with three long-chain carboxylic acids. a. waxes b. fats and oils c. fatty acids d. soaps ANSWER: b POINTS: 1 32. The major lipid components of cell membranes are: a. sphingolipids b. prostaglandins c. phosphoglycerides d. terpenoids ANSWER: c POINTS: 1 33. A group of C20 lipids that contain a five-membered ring with two long side chains. a. sphingolipids b. prostaglandins c. phosphoglycerides d. terpenoids ANSWER: b POINTS: 1 34. Plant extracts, called essential oils, are largely a mixture of these lipids. a. b. c.

sphingolipids prostaglandins phosphoglycerides

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Chapter 27 - Biomolecules: Lipids d. terpenoids ANSWER: d POINTS: 1 35. When soaps disperse in water, the long hydrocarbon tails cluster together in a hydrophobic ball, while the ionic heads on the surface of the cluster stick out into the water. These spherical clusters are called: a. fatty acid esters b. triacylglycerols c. lipid bilayers d. micelles ANSWER: d POINTS: 1 36. Cell membranes, which form an effective barrier to the passage of ions and other components into and out of the cell, are: a. fatty acid esters b. tricacylglycerols c. lipid bilayers d. micelles ANSWER: c POINTS: 1 37. Male sex hormones like testosterone are called: a. androgens b. estrogens c. progestins d. glucocorticoids ANSWER: a POINTS: 1 38. In humans, most steroids function as: a. enzymes b. hormones c. nucleic acids d. proteins ANSWER: b POINTS: 1 39. Draw and name the product of the complete reaction of the following with glycerol. Atoms other than hydrogen and oxygen are labeled.

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Chapter 27 - Biomolecules: Lipids ANSWER: glyceryl tristearate

POINTS: 1 40. Explain why soaps produce a scum when used in hard water (water containing divalent ions such as magnesium and calcium).

ANSWER: The salts formed between these cations and the anions from the soap are insoluble in water.

POINTS: 1 41. Write the saponification reaction of glyceryl monooleate distearate with aqueous NaOH. ANSWER:

POINTS: 1 42. Describe how a micelle and cell membrane are similar.

ANSWER: Both are composed of molecular species that have polar head and a nonpolar tail. The polar head in both Cengage Learning Testing, Powered by Cognero

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Chapter 27 - Biomolecules: Lipids orients toward an aqueous environment while the nonpolar tail orients toward a nonpolar environment. In a micelle a spherical structure results with the polar heads on the surface toward the aqueous environment and the nonpolar tails inward toward the oil or grease (nonpolar environment). In a cell membrane a bilayer structure results with the polar head on the cell surface in contact with the aqueous environments inside and outside the cell while the nonpolar tails orient inward toward each other forming a nonpolar environment.

POINTS: 1 43. Using a condensed formula, draw the product of the complete reaction of the following with the 30-carbon saturated straight chain alcohol, triacontanol. Atoms other than hydrogen and oxygen are labeled.

ANSWER:

POINTS: 1 44. Examine the following structure. Atoms other than carbon and hydrogen are labeled.

This substance: a. is a saturated fatty acid. b. would be found in animal fat. c. could react with glycerol to form a triacylglycerol. d. could react with a long chain alcohol to form a wax. e. All of the above characterize this structure.

ANSWER: e POINTS: 1 45. Examine the following structure. Atoms other than carbon and hydrogen are labeled.

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Chapter 27 - Biomolecules: Lipids

Which of the following does not characterize this structure? a. Polyunsaturated fatty acid b. Cis fatty acid c. Omega-6 fatty acid d. Essential in the human diet e. All of the above characterize this structure.

ANSWER: c POINTS: 1 46. Which feature is common to all lipids? a. Contain carbonyl groups b. Fused rings c. Polar character d. Solubility in nonpolar solvents e. Ester linkages ANSWER: d POINTS: 1 47. What product besides glycerol would be produced when glyceryl dipalmitate monostearate is treated with KOH? a. Palmitic acid b. Potassium palmitate c. Stearic acid d. Potassium stearate e. a and c f. b and d g. a, b, c, d ANSWER: f POINTS: 1 48. Which of the following would be the general structure of a phospholipid? a.

b.

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Chapter 27 - Biomolecules: Lipids

c.

d.

All of these could be a phospholipid depending on the identity of the R groups.

ANSWER: b POINTS: 1 49. Which of the following correctly does not correctly describe the composition of a phosphatidylethanolamine? a. Glycerol residue b. Phosphate group c. One fatty acid residue d. Ethanolamine residue e. All of the above ANSWER: c POINTS: 1 50. Which of the following is common to both glycerophospholipids and sphingomyelins? a. Phosphatidic acid b. Sphingosine c. Phosphate d. Choline ANSWER: c POINTS: 1 51. Classify the following structure.

a. wax c. oil ANSWER: a POINTS: 1

b. fat d. soap

52. The following compound could be classified as all of the following except which one?

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Chapter 27 - Biomolecules: Lipids

a. Phospholipid c. Sphingomyelin ANSWER: c POINTS: 1

b. Glycerophosopholipid d. Phosphodiester

53. Which of the following types/classes of compounds are derived from the following structure?

a. Thromboxanes b. Leukotrienes c. Prostaglandins d. Eicosanoids e. All are derived from this structure.

ANSWER: e POINTS: 1 54. The following generalized structure is that of a:

a. prostaglandin b. thromboxane c. leukotriene ANSWER: c POINTS: 1 55. Which of the following would have the highest melting point? Atoms other than carbon and hydrogen are labeled. a.

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Chapter 27 - Biomolecules: Lipids

b.

c.

d.

All have about the same melting point.

ANSWER: a POINTS: 1

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Chapter 28 - Biomolecules: Nucleic Acids Exhibit 28-1 MATCH each of the following terms to a structure from the list below. There is only one correct structure for each term and structures may be used more than once. Place the letter of the structure in the blank to the left of the corresponding term.

A. C.

Sanger dideoxy method Watson-Crick

E.

G. I.

GAUCGUAAA translation

F.

transcription AUGGCUGAG

K.

M.

B. D.

H. J.

Maxam-Gilbert method replication

L.

AGATCGCTC

1. _____ a pyrimidine nucleoside ANSWER: E POINTS: 1 2. _____ RNA base sequence with guanine at the 3' end.

ANSWER: I POINTS: 1 3. _____ DNA base sequence with cytosine at the 3' end.

ANSWER: M POINTS: 1 4. _____ uridine ANSWER: E POINTS: 1 5. _____ a purine nucleoside ANSWER: L POINTS: 1 6. _____ DNA sequencing method for the human genome Cengage Learning Testing, Powered by Cognero

Page 1

Chapter 28 - Biomolecules: Nucleic Acids ANSWER: A POINTS: 1 7. _____ RNA base sequence complementary to the DNA base sequence which codes for Leu−Ala−Phe: CTAGCATTT ANSWER: B POINTS: 1 8. _____ 2'-deoxyadenosine 5'-phosphate ANSWER: K POINTS: 1 9. _____ process by which mRNA directs protein synthesis ANSWER: D POINTS: 1 10. The process invented in 1986 which allows multiple copies of a given DNA sequence to be produced is: a. the polymerase chain reaction. b. the Sanger dideoxy method. c. the Maxam-Gilbert method. d. the restriction endonuclease reaction.

ANSWER: a POINTS: 1 11. Describe, in general terms, the steps involved in the automated synthesis of the DNA sequence GCT. ANSWER: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6

Attachment of 5'-DMT-N-benzoyl-2'-deoxyaguanosine to the polymer support by a 3' ester linkage.

Deprotection of the 5'-hydroxyl by cleavage of the DMT ether with dichloroacetic acid in CH2Cl2.

Phosphite formation between the free 5'-hydroxyl of the polymer-bound N-protected 2'deoxyguanosine and the phosphoramidite protected 5'-DMT-2'-deoxycytidine in the presence of tetrazole followed by oxidation to the phosphate triester with I2.

Repeat step 2 to deprotect the 5'-hydroxyl of the polymer−GC. Repeat step 3 to form the phosphate triester between 5'-hydroxyl of 2'-deoxycytidine residue and phosphoramidite protected 5'-DMT-N-benzoyl-2'-deoxythymidine.

Cleave all protecting groups from the heterocyclic bases from the phosphates, as well as cleave the ester bond holding the DNA to the silica by treatment with aqueous ammonia.

POINTS: 1 12. Draw and label the four DNA nucleosides.

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Chapter 28 - Biomolecules: Nucleic Acids ANSWER:

POINTS: 1 13. Draw and label a G-C base pair. Indicate all hydrogen bonds with a dashed line.

ANSWER:

POINTS: 1 14. Draw and label an A-T base pair. Indicate all hydrogen bonds with a dashed line.

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Chapter 28 - Biomolecules: Nucleic Acids ANSWER:

POINTS: 1 15. Of the two sets of DNA base pairs, which would you expect to be stronger and why?

ANSWER: The G-C base pair is the stronger of the two as it has three hydrogen bonds whereas the A-T base pair has only two.

POINTS: 1 16. Draw and label the four RNA nucleosides.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 28 - Biomolecules: Nucleic Acids 17. Draw and label an A-U base pair. Indicate all hydrogen bonds with a dashed line.

ANSWER:

POINTS: 1 18. What is the primary structural difference between RNA nucleotides and DNA nucleotides?

ANSWER: As their names suggest, RNA bases are attached to ribose sugars while DNA bases are attached to deoxyribose sugars. The difference is a lack of the 2' hydroxyl group in the sugar backbone of the DNA nucleotides.

POINTS: 1 19. Because of the structure of the RNA sugar-phosphate backbone, RNA is subject to base-promoted hydrolysis, as shown below. Suggest a mechanism for base-promoted RNA hydrolysis.

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Chapter 28 - Biomolecules: Nucleic Acids ANSWER:

POINTS: 1 20. While the Watson-Crick G-C and A-T base pairs are the most common, other non-standard base pairs are encountered.

Draw and label a G-U base pair. Indicate all hydrogen bonds with a dashed line.

ANSWER:

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Chapter 28 - Biomolecules: Nucleic Acids POINTS: 1 21. When DNA is replicated, the new chain is synthesized in what direction?

ANSWER: As DNA is replicated, the new chain is synthesized in the 5' to 3' direction.

POINTS: 1 22. Give a one sentence definition of replication.

ANSWER: Replication is the process by which identical copies of DNA are made.

POINTS: 1 23. Give a one sentence definition of transcription.

ANSWER: Transcription is the process by which genetic messages are read and carried out of the cell to ribosomes, where protein synthesis occurs.

POINTS: 1 24. Give a one sentence definition of translation.

ANSWER: Translation is the process by which genetic messages are decoded and used to synthesize proteins.

POINTS: 1 25. Give the single letter amino acid sequence that is coded for by the following segment of DNA coding strand (sense strand)?

(5') GCA—TTT—CTA—ACT—TGA—GGG (3') ANSWER: The mRNA strand is: (Remembering U replaces T in RNA.) (5') GCA—UUU—CUA—ACU—UGA—GGG (3') The amino acid squence is: (UGA is a stop codon.)

Three-letter code: Ala—Phe—Leu—Thr One-letter code (See Table 26.1.)

A—F—L—T POINTS: 1 26. Describe the difference between a 2',3'-dideoxyribonucleoside triphosphate and 2'-deoxyribonucleoside triphosphate.

ANSWER: A 2'-deoxyribonucleoside triphosphate is one in which the 2' OH group is missing from ribose. A 2',3'dideoxyribonucleoside triphosphate is one in which both 2' and 3' OH groups are missing from ribose. In DNA sequencing the significance of this removal is that DNA replication stops at this nucleotide since the 3’ —OH needed for the continued addition of nucleotides.

POINTS: 1 27. On the diagram of DNA, identify: 1.) major groove 2.) minor groove 3.) position of the sugar-phosphate backbone 4.) site of possible intercalation

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Chapter 28 - Biomolecules: Nucleic Acids

ANSWER:

POINTS: 1 28. Identify any errors in the figure shown below.

ANSWER: Two errors: 1.) In the left lower nucleotide, the base is indicated as U (uracil) which only appears in RNA. 2.) In the left upper nucleotide, the sugar is ribose which only occurs in RNA. Cengage Learning Testing, Powered by Cognero

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Chapter 28 - Biomolecules: Nucleic Acids POINTS: 1 29. Consider the following diagram.

1.) Label the anticodon 2.) Label the position of attachment of the amino acid 3.) Draw the structure of the amino acid that would be attached 4.) Indicate the corresponding mRNA codon

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Chapter 28 - Biomolecules: Nucleic Acids ANSWER:

Amino acid: Valine

mRNA codon: GUA POINTS: 1 30. In the diagram below, fill in the terms in the appropriate places indicated by a letter.

a. A - replication, B - transcription, C - translation b. A - replication, B - translation, C - transcription c. A - transcription, B - replication, C - translation d. A - translation, B - transcription, C - replication Cengage Learning Testing, Powered by Cognero

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Chapter 28 - Biomolecules: Nucleic Acids ANSWER: a POINTS: 1 31. Consider the following molecular model. Atoms other than carbon and hydrogen are labeled.

Which of the following is a characteristic of this compound? a. Found in DNA b. Found in both DNA and RNA c. Pyrimidine base d. Pairs with thymine ANSWER: c POINTS: 1 32. Consider the following molecular model. Atoms other than carbon and hydrogen are labeled.

Which of the following is a characteristic of this structure? a. Nucleoside b. Contains ribose c. Contains a purine base d. Found only in DNA ANSWER: d POINTS: 1 33. Consider the following diagram.

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Chapter 28 - Biomolecules: Nucleic Acids

Which letter should be filled in on the bottom line? a. GCU b. UCG c. GCT d. TGC ANSWER: a POINTS: 1 34. Consider the following diagram.

What DNA sequence (5' to 3') would correspond to the third codon? a. GCU b. TCG c. UGC d. GCT ANSWER: b POINTS: 1 35. Consider the following diagram.

What amino acid corresponds to the second codon from the left? a. Tyrosine b. Asparagine c. Isoleucine d. None, the mRNA codon is a stop signal ANSWER: d POINTS: 1 36. Consider DNA sequencing experiments using 2',3'-dideoxyribonucleoside triphosphates. Which of the following is not correct? a. 2',3'-dideoxyribonucleoside triphosphates are added to the replicating strand by DNA polymerase.

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Chapter 28 - Biomolecules: Nucleic Acids b. There are four types of 2',3'-dideoxyribonucleoside triphosphates used. c. The sequence of 2',3'-dideoxyribonucleoside triphosphates in the replicated DNA yields the DNA structure. d. DNA replication ceases when a 2',3'-dideoxyribonucleoside triphosphates is added to the replicating strand.

ANSWER: c POINTS: 1 37. In determining DNA sequencing using the Sanger dideoxy method, how many colors will appear in the resulting data display? a. 2 b. 4 c. 6 d. 8 ANSWER: b POINTS: 1 38. During DNA synthesis, how is the 5' —OH of deoxyribose protected? a. With a p-dimethoxytrityl group b. With a benzoyl group c. With an isobutryl group d. With a β-cyanoethyl group e. This group does not need to be protected.

ANSWER: a POINTS: 1 39. During DNA synthesis, the free amino group does not need to be protected for which base? a. Adenine b. Guanine c. Thymine d. Cytosine ANSWER: c POINTS: 1 40. What is the purpose of the addition of aqueous ammonia during the synthesis of DNA? a. Cleavage of the ester bond to the silica support b. Removal of the groups protecting the free amino group of the bases c. Removal of the protecting group of the 5’ —OH of the sugar d. Ammonia accomplishes all of the above.

ANSWER: d POINTS: 1 41. Consider the DNA polymerase chain reaction. After 5 cycles have been completed, how many copies of the original DNA are present? a. 10 b. 32 c. 64 d. 128 ANSWER: b POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 28 - Biomolecules: Nucleic Acids 42. Which of the following summarizes the steps in DNA synthesis using the PCR method? a. Anneal-denature-synthesize b. Synthesize-denature-anneal c. Denature-anneal-synthesize d. Denature-synthesize-anneal ANSWER: c POINTS: 1 43. When the β-cyanoethyl group is removed during DNA synthesis, H2CCHCN is produced as a product. This reaction is likely to be: a. E1 b. E2 c. E1cB d. SN1 e. SN2 ANSWER: b POINTS: 1 44. Using the PCR method for synthesizing DNA, why don't the two DNA strands recombine during the cooling phase? a. The temperature used b. The concentration of the primers 2+ c. Presence of Mg d. Presence of Taq polymerase ANSWER: b POINTS: 1 45. Which of the following statement correctly describes DNA replication? a.

Replication occurs in the direction. b. Ligases link the Okazaki fragments to form the lagging strand. c. The leading strand is replicated discontinuously. d.

One strand is replicated while the other strand is replicated

.

ANSWER: b POINTS: 1

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Chapter 29 - The Organic Chemistry of Metabolic Pathways Exhibit 29-1 MATCH a term or structure from the list below to each of the following definitions or names. Place the letter of the term or structure in the blank to the left of the definition or name which it describes.

A.

electron-transport chain

B.

C.

citric acid cycle

D.

E.

glycolysis

F.

G.

Hans Krebs

H.

I.

ATP

J.

K.

fatty acid spiral

L.

M.

acetyl CoA

N.

O.

Embden-Meyerhoff

catabolism

anabolism

P.

gluconeogenesis

Q.

metabolism

1. _____ Pathways that synthesize larger biomolecules from smaller ones.

ANSWER: L POINTS: 1 2. _____ Acetyl groups are oxidized to yield carbon dioxide and water.

ANSWER: C POINTS: 1 3. _____ The "energy currency" of the cell.

ANSWER: I Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways POINTS: 1 4. _____ Coenzyme NAD+ required in the β-oxidation of fatty acids, glycolysis, and the citric acid cycle.

ANSWER: B POINTS: 1 5. _____ Discoverers of the glycolysis pathway.

ANSWER: D POINTS: 1 6. _____ Pyridoxal phosphate, the cofactor in transaminations.

ANSWER: J POINTS: 1 7. _____ Uses the energy produced in the citric acid cycle to make ATP. ANSWER: A POINTS: 1 8. _____ The product of the citric acid cycle, which is a reactant in the first step.

ANSWER: O POINTS: 1 9. _____ The pathway by which organisms make glucose from simple precursors.

ANSWER: P POINTS: 1 10. _____ Common fatty acids have an even number of carbons because they are biosynthesized from _____.

ANSWER: M POINTS: 1 11. Olive oil is comprised of 80% oleic acid, CH3(CH2)7CH=CH(CH2)7COOH. How many molecules of acetyl CoA are produced by catabolism of one molecule of oleic acid and how many passages of the β-oxidation pathway are needed for its complete catabolism?

ANSWER: Since oleic acid is a C18 fatty acid, nine molecules of acetyl CoA will be produced by catabolism of one molecule of oleic acid. This will require eight passages of the β-oxidation pathway.

POINTS: 1 Exhibit 29-2 Use the data below to answer the following question(s): Under certain metabolic conditions associated with a high rate of fatty acid oxidation, the liver produces considerable quantities of acetoacetate and β-hydroxybutyrate. Acetoacetate continually undergoes spontaneous decarboxylation to yield acetone. The interrelationship of these compounds is shown below.

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Chapter 29 - The Organic Chemistry of Metabolic Pathways

12. Refer to Exhibit 29-2. Write a mechanism for the decarboxylation of acetoacetate to yield acetone.

ANSWER:

POINTS: 1 13. Refer to Exhibit 29-2. The reduction of acetoacetate to β-hydroxybutyrate is analogous to the hydride reductions of ketones studied in Chapter 9. Write the mechanism of this reduction.

ANSWER:

POINTS: 1 14. Refer to Exhibit 29-2. The reduction of acetoacetate to β-hydroxybutyrate is an example of a: a. nucleophilic substitution reaction. b. nucleophilic acyl substitution reaction. c. nucleophilic addition reaction. d. nucleophilic exchange reaction.

ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways Exhibit 29-3 Use the data below to answer the following question(s).

Mevalonate is an important intermediate in the biosynthesis of cholesterol. This six-carbon compound can be formed from acetyl CoA utilizing two separate pathways. The most significant one involves the intermediate β-hydroxy-βmethylglutaryl-CoA.

15. Refer to Exhibit 29-3. What is the by-product represented as a question mark (?) in each step of this reaction sequence?

ANSWER: HS−CoA POINTS: 1 16. Refer to Exhibit 29-3. In step 1 of the mevalonate biosynthesis, acetyl CoA is converted into acetoacetyl CoA. This reaction is analogous to: a. a Claisen condensation b. an aldol reaction c. the malonic ester synthesis d. a carbonyl reduction reaction ANSWER: a POINTS: 1 17. Refer to Exhibit 29-3. In step 2 of the mevalonate biosynthesis, acetoacetyl CoA is converted into β-hydroxy-βmethylglutaryl-CoA. This reaction is analogous to: a. b. c.

a Claisen condensation an aldol reaction the malonic ester synthesis

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Chapter 29 - The Organic Chemistry of Metabolic Pathways d. a carbonyl reduction reaction ANSWER: b POINTS: 1 18. Refer to Exhibit 29-3. Propose a mechanism for step 2 of the mevalonate biosynthesis.

ANSWER:

POINTS: 1 19. What is the structure of the α-keto acid formed by transamination of the amino acid isoleucine?

ANSWER:

POINTS: 1 20. Which of the following statements about catabolism is false? a. The ultimate products of food catabolism are CO2, H2O, and ATP. b. The citric acid cycle is a stage of catabolism. c. Gluconeogenesis is the catabolism of glucose. d. Transamination is the catabolism of proteins.

ANSWER: c POINTS: 1 21. Which of the following statements about anabolism is false? a.

Anabolic reactions "spend" ATP by transferring a phosphate group to another molecule.

As a rule, the anabolic pathway by which an organism makes a substance is the exact reverse b. of the catabolic pathway. c. Pathways that synthesize larger biomolecules from smaller ones are known as anabolism. d. Gluconeogenesis is the anabolic pathway by which organisms make glucose from pyruvate.

ANSWER: b POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways 22. Which of the following statements about ATP is false? a. ADP is synthesized from ATP. b. ATP is produced in the fourth stage of catabolism, the electron-transport chain. c. Catabolic reactions "pay off" in ATP. d. ATP is a phosphoric acid anhydride.

ANSWER: a POINTS: 1 Exhibit 29-4 Consider the reaction below to answer the following question(s):

23. Refer to Exhibit 29-4. This reaction proceeds through a common enol structure. Draw the structure of the glucose/fructose enol.

ANSWER:

POINTS: 1 24. Refer to Exhibit 29-4. Write the mechanism for this isomerization.

ANSWER:

POINTS: 1 25. Refer to Exhibit 29-4. The isomerization of glucose 6-phosphate to fructose 6-phosphate is an example of: a. an elimination reaction b. an oxidation reaction c. keto-enol tautomerism d. a conjugate addition reaction ANSWER: c POINTS: 1 Exhibit 29-5 Consider the reaction below to answer the following question(s):

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Chapter 29 - The Organic Chemistry of Metabolic Pathways

26. Refer to Exhibit 29-5. This reaction is an example of: a. an aldol reaction. b. a Claisen reaction. c. a reverse Claisen reaction. d. a retro aldol reaction.

ANSWER: d POINTS: 1 27. Refer to Exhibit 29-5. Draw arrows on the structure below showing electron flow in this reaction.

ANSWER:

POINTS: 1 28. Phosphoenolpyruvate could undergo hydrolysis to form pyruvate and hydrogen phosphate. This reaction could be represented as shown below. (ΔG°' = -61.9 kJ/mol)

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Chapter 29 - The Organic Chemistry of Metabolic Pathways This reaction can be used to phosphorylate ADP. Write the net reaction when this reaction is coupled with ADP phosphorylation and calculate ΔG°' for the reaction.

ANSWER:

POINTS: 1 29. Draw the structure of the amino acid that is the precursor to the substance shown below. Atoms other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 30. Phosphoenolpyruvate could undergo hydrolysis to form pyruvate and hydrogen phosphate. This reaction could be represented as shown below. This reaction can be coupled to the phosphorylation of ADP.

The product exists as tautomers. Draw their structures.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways 31. During β-oxidation the following reaction occurs. What substances do the two question marks on the curve arrow represent?

ANSWER: 1. NAD+ 2. NADH POINTS: 1 32. Fatty acids always have an even number of carbon atoms. Explain.

ANSWER: Biosynthesis of fatty acids occurs by sequential addition of two-carbon units to acetyl CoA (also a two carbon unit).

POINTS: 1 33. During the synthesis of fatty acids, the following sequence occurs. Draw the products presented by the question marks and identify the type of reaction.

ANSWER:

POINTS: 1 34. During the synthesis of fatty acids, the following substance is formed by two pathways.

Draw the structure(s) of the starting material(s) for each path?

ANSWER: The ubiquitous acetyl CoA:

POINTS: 1 35. Draw the structure of the product formed on the transamination of the following substance. Atoms other than carbon and hydrogen are labeled.

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Chapter 29 - The Organic Chemistry of Metabolic Pathways

ANSWER: POINTS: 1 36. The following reaction occurs in the citric acid cycle.

Draw the structure of the product.

ANSWER:

POINTS: 1 37. Which of the following does not represent a correct relationship in catabolism? a. b. c.

d.

ANSWER: b POINTS: 1 38. What type of high energy bonds are found in ATP and ADP? a. carboxylic anhydride b. phosphodiester c. thioester d. phosphoanhydride ANSWER: d POINTS: 1 39. Which of the following does not correctly describe a stage of metabolism in humans? a. first stage: ester and peptide bond broken b. first stage: fatty acids and simple sugars degraded in cytoplasm c. second stage: produces acetyl groups Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways d. third stage: involves the citric acid cycle ANSWER: b POINTS: 1 40. In coupled reactions, the net reaction has: a. ΔG = 0 b. ΔG > 0 c. ΔG < 0 ANSWER: c POINTS: 1 41. Consider the following reaction:

Which of the following reactions per mole could not be driven by the hydrolysis of ATP? a. Reaction 1: ΔG°' = 12.6 kJ/mol b. Reaction 1: ΔG°' = 22.5 kJ/mol c. Reaction 1: ΔG°' = 29.6 kJ/mol d. Reaction 1: ΔG°' = 40.6 kJ/mol ANSWER: d POINTS: 1 42. Consider the following reaction:

Hydrolysis of which of the following biomolecules could drive the formation of ATP? a. glycerol-1-phosphate ΔG°' = -9.2 kJ/mol b. glucose-6-phosphate ΔG°' = -13.8 kJ/mol c. fructose-6-phosphate ΔG°' = -15.8 kJ/mol d. phosphocreatine ΔG°' = -49.5 kJ/mol e. All could drive the formation of ATP. ANSWER: d POINTS: 1 43. Which of the following is an anabolic pathway? a. glycolysis b. citric acid cycle c. β-oxidation d. gluconeogenesis ANSWER: d POINTS: 1 44. The transformation of FAD to FADH2 accompanies which of the following types of reactions? a. isomerization b. double bond formation c. thioester formation d. alcohol oxidation ANSWER: b Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways POINTS: 1 45. During β-oxidation the following reaction occurs. What type of reaction is this?

a. Claisen condensation c. aldol condensation ANSWER: b POINTS: 1

b. retro-Claisen condensation d. retro-aldol condensation

46. During β-oxidation the following reaction occurs.

How many times will this reaction occur for the following acid? Atoms other than carbon and hydrogen are labeled.

a. 8 b. 9 c. 16 d. 18 ANSWER: a POINTS: 1 47. In which of the following steps of glycolysis is ATP produced? a. Steps 1 and 3 b. Steps 7 and 10 c. Steps 1, 3 and 10 d. Steps 1, 3, and 7 e. Steps 1, 3, 7, and 10 ANSWER: b POINTS: 1 48. Which of the following correctly describes glycolysis? a. produces a net of 4 ATP b. intermediate path in glucose catabolism c. produces two equivalents of pyruvate ions d. involves one dephosphorylation and two phophorylations.

ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 29 - The Organic Chemistry of Metabolic Pathways 49. Which of the following does not apply to the citric acid cycle? a. The products of one step are the reactants of the next step. b. The starting material is derived from the pyruvate formed during glycolysis. c. Most of the energy output is derived from reduced coenzymes. d. The first step is a Claisen condensation that converts acetyl CoA and oxaloacetate to citrate. e. The last step involves an oxidation that produces (S)-malate. f. All of these do characterize the citric acid cycle.

ANSWER: e POINTS: 1 50. Consider the following conversion that occurs in the citric acid cycle.

What formula should be placed at the tail of the arrow? a. FADH2 b. FAD c. NAD+

d. NADH

ANSWER: b POINTS: 1

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions 1. There are three major classes of pericyclic reactions. Which of the following types of reactions is not an example of a pericyclic process? a. sigmatropic rearrangements b. cycloaddition reactions c. annulation reactions d. electrocyclic reactions ANSWER: c POINTS: 1 2. The mnemonic phrase "The Electrons Circle Around", TECA, assists in predicting stereochemistry of pericyclic reactions. Describe application of this mnemonic.

ANSWER: The mnemonic TECA assists us in remembering that "Thermal pericyclic reactions with Even numbers of electrons will occur with Conrotatory or Antarafacial stereochemistry". As a result, the following statements also apply: Photochemical pericyclic reactions with even numbers of electrons will occur with disrotatory or suprafacial stereochemistry. Thermal pericyclic reactions with odd numbers of electrons will occur with disrotatory or suprafacial stereochemistry. Photochemical pericyclic reactions with odd numbers of electrons will occur with conrotatory or antarafacial stereochemistry.

POINTS: 1 Exhibit 30-1 To answer the following question(s), consider the π molecular orbitals of a conjugated diene shown below:

3. Refer to Exhibit 30-1. Which molecular orbital of a conjugated diene contains two nodes between nuclei? a. ψ1 b. ψ2 c. ψ3* d. ψ4* ANSWER: c POINTS: 1 4. Refer to Exhibit 30-1. Which molecular orbital of a ground-state conjugated diene is the highest occupied molecular orbital (HOMO)? a. b.

ψ1 ψ2

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions c. ψ3* d. ψ4* ANSWER: b POINTS: 1 5. Refer to Exhibit 30-1. Which molecular orbital of an excited-state conjugated diene is the lowest unoccupied molecular orbital (LUMO)? a. ψ1 b. ψ2 c. ψ3* d. ψ4* ANSWER: d POINTS: 1 6. Draw the product(s) you would expect to obtain from photochemical cyclization of (2Z, 4E)-hexadiene.

ANSWER:

POINTS: 1 Exhibit 30-2 Consider the reaction below to answer the following question(s):

7. Refer to Exhibit 30-2. Has this reaction taken place in a conrotatory manner or in a disrotatory manner?

ANSWER: In order for the hydrogens to be trans in the product, rotation must occur in a disrotatory manner.

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

POINTS: 1 8. Refer to Exhibit 30-2. Under what conditions, thermal or photochemical, would you carry out this reaction? Explain.

ANSWER: For a disrotatory cyclization to occur the HOMO must have the symmetry pictured. For a 4 π electron system this HOMO must arise from photochemical excitation of a pi electron (see questions 3-5 in printed test bank).

To obtain a product with the correct stereochemistry, the reaction must be carried out under photochemical conditions.

POINTS: 1 Exhibit 30-3 To answer the following question(s), consider the reaction below:

9. Refer to Exhibit 30-3. Two pericyclic reactions are involved in this synthesis of compound Y. Draw the structure of intermediate X and name it using IUPAC nomenclature.

ANSWER:

POINTS: 1 10. Refer to Exhibit 30-3. The thermal reaction which generates X is an example of: a. an electrocyclic reaction b. a sigmatropic rearrangement c. a reverse cycloaddition reaction d. a cycloaddition reaction ANSWER: a POINTS: 1 11. Refer to Exhibit 30-3. Explain the stereochemistry of compound Y. Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions ANSWER: Thermal electrocyclic opening of the cis-substituted cyclobutene yields (1E, 3Z, 5Z)-cyclodecatriene (X).

Symmetry-allowed thermal cyclization of the cyclodecatriene occurs in a disrotatory fashion to yield the observed compound Y.

POINTS: 1 Exhibit 30-4 Consider the reaction below to answer the following question(s):

12. Refer to Exhibit 30-4. What type of pericyclic reaction is involved in this transformation? a. sigmatropic rearrangement b. reverse cycloaddition reaction c. electrocyclic reaction d. cycloaddition reaction ANSWER: d POINTS: 1 13. Refer to Exhibit 30-4. How many pairs of electrons are involved in this pericyclic reaction? a. two b. four c. eight d. sixteen ANSWER: c POINTS: 1 14. Refer to Exhibit 30-4. Explain the stereochemistry of this reaction.

ANSWER: The reaction of heptafulvene with tetracyanoethylene is a [14 + 2] cycloaddition that has occurred with antarafacial stereochemistry.

POINTS: 1 15. Refer to Exhibit 30-4. Would you expect this reaction to occur under photochemical or thermal conditions?

ANSWER: Since this reaction is a [14 + 2] cycloaddition with eight electron pairs, we would expect the thermal reaction to occur with antarafacial stereochemistry and the photochemical reaction to occur with suprafacial stereochemistry. The product has antarafacial stereochemistry resulting from a thermal reaction.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions Exhibit 30-5 To answer the following question(s), consider the reaction below:

16. Refer to Exhibit 30-5. What type of pericyclic reaction is involved in this transformation? a. sigmatropic rearrangement b. reverse cycloaddition reaction c. electrocyclic reaction d. cycloaddition reaction ANSWER: d POINTS: 1 17. Refer to Exhibit 30-5. How many pairs of electrons are involved in this pericyclic reaction? a. two b. four c. eight d. sixteen ANSWER: a POINTS: 1 18. Refer to Exhibit 30-5. Propose a mechanism for the reaction that fully accounts for the formation of both products.

ANSWER:

This reaction occurs with suprafacial stereochemistry that would be expected for a photochemical [2 + 2] cycloaddition. However, suprafacial cycloaddition can occur in two different ways⎯with the methyl groups of both alkenes on the same side or on opposite sides.

POINTS: 1 Exhibit 30-6 Consider the reaction sequence below to answer the following question(s): Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

19. Refer to Exhibit 30-6. Step 1 is an example of: a. an electrocyclic reaction b. a cycloaddition reaction c. a sigmatropic rearrangement d. a reverse cycloaddition reaction ANSWER: a POINTS: 1 20. Refer to Exhibit 30-6. Explain the stereochemistry of product B. ANSWER: Step 1 is the thermal electrocyclic reaction of a conjugated triene (three electron pairs) that occurs with disrotatory stereochemistry to yield a cis-fused cyclopropane.

POINTS: 1 21. Refer to Exhibit 30-6. Step 2 is an example of: a. an electrocyclic reaction. b. a cycloaddition reaction. c. a sigmatropic rearrangement. d. a reverse cycloaddition reaction.

ANSWER: c POINTS: 1 22. Refer to Exhibit 30-6. Explain the stereochemistry of product C. ANSWER: Step 2 is a thermal [1, 5] sigmatropic rearrangement (three electron pairs) that occurs with suprafacial stereochemistry.

POINTS: 1 23. Refer to Exhibit 30-6. Show how C is formed from B. ANSWER:

POINTS: 1 Exhibit 30-7 Consider the Sommelet-Hauser rearrangement below to answer the following question(s): Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

24. Refer to Exhibit 30-7. The Sommelet-Hauser rearrangement is an example of a [2, 3] sigmatropic rearrangement. How many pairs of electrons are involved in this reaction? a. two b. three c. four d. five ANSWER: b POINTS: 1 25. Refer to Exhibit 30-7. Would you expect the Sommelet-Hauser rearrangement to be antarafacial or suprafacial?

Explain.

ANSWER: Since three pairs of electrons are involved in the rearrangement, the Sommelet-Hauser rearrangement should occur with suprafacial geometry.

POINTS: 1 Exhibit 30-8 Classify each of the following sigmatropic reactions by order [x, y].

26.

ANSWER: [3, 3]

POINTS: 1

27.

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions ANSWER:

[1, 7]

POINTS: 1

28.

ANSWER: [1, 5]

POINTS: 1 Exhibit 30-9 Vitamin D3, cholecalciferol, is synthesized under the skin by the photochemical pericyclic process shown below. Answer the following question(s) about these reactions.

29. Refer to Exhibit 30-9. The first step in this process is a(n): a. sigmatropic rearrangement b. cycloaddition reaction c. annulation reaction d. electrocyclic reaction ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions 30. Refer to Exhibit 30-9. The second step in this process is a(n): a. sigmatropic rearrangement b. cycloaddition reaction c. annulation reaction d. electrocyclic reaction ANSWER: a POINTS: 1 31. Predict the product (including stereochemistry) when the following substance reacts under thermal conditions.

ANSWER:

POINTS: 1 32. Predict the product of the following reaction.

ANSWER:

POINTS: 1 33. Define a pericyclic reaction.

ANSWER: A pericyclic reaction is a concerted reaction that occurs through a cyclic transition state. It occurs in one step without the formation of intermediates.

POINTS: 1 34. Draw the structure of the product when the following substance reacts with maleic anhydride. Classify the reaction according to the number of π electrons that interact. Atoms other than carbon and hydrogen are labeled.

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

ANSWER: [4+2]

POINTS: 1 35. The photochemical cycloaddition of the structure below is related to DNA mutations causing skin cancer. Draw the structure of this dimer. Atoms other than carbon and hydrogen are labeled.

ANSWER: [2+2] cycloaddition occurs due to photochemical conditions.

POINTS: 1 36. Which of the following is not an electrocyclic reaction? a.

b.

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

c.

d.

e. All are electrocyclic reactions.

ANSWER: e POINTS: 1 37. For an even number of electron pairs (double bonds), which of the following rules is correct?: a. thermal conditions - controtatory, photochemical conditions - disrotatory b. thermal conditions - disrotatory, photochemical conditions - conrotatory ANSWER: a POINTS: 1 38. Consider the following image which shows the two outermost lobes a polyene.

For a bond to form which of the following is correct? a. Disrotatory motion is required. b. Both must rotate clockwise. c. One must rotate clockwise and the other counterclockwise. d. Conrotatory motion is required. e. a and c f. b and d ANSWER: d POINTS: 1 39. Under thermal conditions, the following substance will undergo: Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

a. disrotatory cyclization. b. conrotatory cyclization. c. no reaction.

ANSWER: a POINTS: 1 40. Under photochemical conditions, the following substance will undergo:

a. disrotatory cyclization. b. conrotatory cyclization. c. no reaction.

ANSWER: a POINTS: 1 41. How many π MO’s does the following compound have?

a. 2 b. 4 c. 6 d. 8 ANSWER: c POINTS: 1 42. Consider the following MO diagram for a conjugated triene.

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions

When irradiated with light of the appropriate wavelength, how many electrons will be in bonding MO’s? a. 6 b. 5 c. 4 d. 3 ANSWER: b POINTS: 1 43. For an odd number of electron pairs (double bonds), for cycloadditions reactions which of the following rules is correct?: a. thermal conditions - antarafacial, photochemical conditions - suprafacial b. thermal conditions - suprafacial, photochemical conditions - antarafacial ANSWER: b POINTS: 1 44. In which type of pericyclic reaction is a σ bond broken and a new σ bond formed? a. cycloaddition b. photochemical electrocyclic c. thermal electrocyclic d. sigmatropic rearrangement. e. all of the above ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions 45. Consider the reaction given below.

Which of the following is correct? a. [1,5] antarafacial b. [3,3] antarafacial c. [1,5] suprafacial d. [3,5] antarafacial ANSWER: b POINTS: 1 46. Which of the following reaction types are pericyclic reactions? a. Diels-Alder reaction b. Cope rearrangement c. Claisen rearrangement d. Stork reaction e. All except d are pericyclic reactions. f. All are pericyclic reactions.

ANSWER: e POINTS: 1 47. Classify the following reaction.

a. electrocyclic reaction b. cycloaddition c. sigmatropic rearrangement ANSWER: c POINTS: 1 48. Classify the following reaction.

a. electrocyclic reaction b. cycloaddition c. sigmatropic rearrangement ANSWER: a Cengage Learning Testing, Powered by Cognero

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Chapter 30 - Orbitals and Organic Chemistry: Pericyclic Reactions POINTS: 1 49. Classify the following reaction.

a. electrocyclic reaction b. cycloaddition c. sigmatropic rearrangement ANSWER: b POINTS: 1

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Chapter 31 - Synthetic Polymers Exhibit 31-1 MATCH a term or structure from the list below to each of the following definitions or names. Place the letter of the term or structure in the blank to the left of the definition or name which it describes.

A. C.

BF3 + H2O chain-growth polymer

E.

G. I. K. M.

B. D.

elastomer AL(CH2CH3)3 + TiCl4

F.

thermoplastic plasticizers copolymers Grubb’s catalyst

H. J. L. N.

homopolymer thermosetting resins step-growth polymer metallacycle

O.

P. 1. ____ Produced by reactions in which each bond in the polymer is formed independently of the others.

ANSWER: L POINTS: 1 2. ____ Isotactic propylene ANSWER: E POINTS: 1 3. ____ Polymers obtained when two or more different monomers are allowed to polymerize together.

ANSWER: K POINTS: 1 4. ____ a polyurethane ANSWER: O POINTS: 1 5. ____ Small organic molecules that act as lubricants between polymer chains.

ANSWER: I POINTS: 1 6. ____ Amorphous polymers that have the ability to stretch out and spring back to their original shape.

ANSWER: B POINTS: 1 7. ____ Bakelite is this type of polymer.

ANSWER: J Cengage Learning Testing, Powered by Cognero

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Chapter 31 - Synthetic Polymers POINTS: 1 8. ____ a Ziegler-Natta catalyst ANSWER: D POINTS: 1 9. _____ General structure: M=CHR ANSWER: M POINTS: 1 10. ______ four-membered metal containing intermediate ANSWER: N POINTS: 1 11. Rank the following monomers in order of increasing reactivity toward cationic polymerization (least reactive to most reactive).

a. III, IV, I, II b. II, I, IV, III c. I, II, IV, III d. IV, III, I, II ANSWER: b POINTS: 1 12. Rank the following monomers in order of increasing reactivity toward anionic polymerization (least reactive to most reactive).

a. III, II, IV, I b. II, III, I, II c. I, IV, III, II d. IV, III, II, I ANSWER: c POINTS: 1 13. Do you expect nitroethylene to be more or less reactive than ethylene toward cationic polymerization? Explain.

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Chapter 31 - Synthetic Polymers

ANSWER: Cationic polymerization is favored when the vinyl monomer contains an electron-donating group (EDG) capable of stabilizing the chain-carrying carbocation intermediate.

Since the nitro group is an electron-withdrawing group (EWG) it should destabilize the chain-carrying carbocation intermediate and, thus, be less reactive than ethylene toward cationic polymerization.

POINTS: 1 14. Do you expect nitroethylene to be more or less reactive than ethylene toward anionic polymerization? Explain.

ANSWER: Vinyl monomers with electron-withdrawing groups (EWG) can be polymerized by anionic catalysts in Michael-type reactions. The EWG stabilizes the negative charge of the chain-carrying anionic intermediate.

Since the nitro group is a strong EWG it should stabilize the chain-carrying carbanion intermediate in anionic polymerizations, and thus, be more reactive than ethylene.

POINTS: 1 15. Monomer A is copolymerized with Monomer B and is found to have the following structure:

The type of copolymer formed is: a. a random copolymer b. an alternating copolymer c. a block copolymer d. a graft copolymer ANSWER: c POINTS: 1 Exhibit 31-2 Draw the structure(s) of the monomer(s) used to make each of the following polymers.

16. Draw: ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 31 - Synthetic Polymers

17. Draw:

ANSWER:

POINTS: 1

18. Draw: ANSWER:

POINTS: 1

19. Draw: ANSWER:

POINTS: 1 Exhibit 31-3 Classify each polymer below as step-growth or chain-growth.

20.

ANSWER: chain-growth POINTS: 1

21.

ANSWER: chain-growth POINTS: 1

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Chapter 31 - Synthetic Polymers 22.

ANSWER: step-growth POINTS: 1

23.

ANSWER: step-growth POINTS: 1 Exhibit 31-4 Draw the structure of the polymers obtained from the following reactions.

24.

ANSWER:

POINTS: 1

25.

ANSWER:

POINTS: 1

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Chapter 31 - Synthetic Polymers 26.

ANSWER:

POINTS: 1

27.

ANSWER:

POINTS: 1 28. Monomer A is copolymerized with Monomer B and is found to have the following structure:

The type of copolymer formed is: a. a random copolymer b. an alternating copolymer c. a block copolymer d. a graft copolymer ANSWER: d POINTS: 1 29. Monomer A is copolymerized with Monomer B and is found to have the following structure: -(—A—B—B—B—A—B—A—B—A—A—B—A—A—)The type of copolymer formed is: a.

a random copolymer

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Chapter 31 - Synthetic Polymers b. an alternating copolymer c. a block copolymer d. a graft copolymer ANSWER: a POINTS: 1 30. Draw two repeating units of the polymer formed by the reaction of 1,3-butadiene with the following substance. Atoms other than carbon and hydrogen are labeled.

ANSWER:

POINTS: 1 31. Draw the structure of the product produced by the acyclic diene metathesis (ADMET) of 1,6-heptadiene.

ANSWER: POINTS: 1 32. Draw the structure of the monomer used to produce the following polymer segment using the ROMP method.

ANSWER:

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 31 - Synthetic Polymers 33. Classify the following polymer.

a. isotactic b. syndiotactic c. atactic ANSWER: a POINTS: 1 34. Classify the following polymer.

a. isotactic b. syndiotactic c. atactic ANSWER: c POINTS: 1 35. Which of the following represents the general structure of a urethane? a. b.

c.

d.

ANSWER: b POINTS: 1 36. Within a crystallite, the polymer chains are held together by: a. disuflide bridges b. hydrogen bonds c. van der Walls forces d. covalent bonds ANSWER: c POINTS: 1 37. Which of the following is not a property of a crystalline polymer? a. unaffected by steric properties of substituent groups b. generally hard and durable c. crystalline regions become amorphous on heating d. contain highly organized regions Cengage Learning Testing, Powered by Cognero

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Chapter 31 - Synthetic Polymers ANSWER: a POINTS: 1 38. Which type of polymer has a high glass transition temperature (Tg)? a. thermoplastic b. fiber c. elastomer d. thermosetting resins ANSWER: a POINTS: 1 39. Which type of polymer is characterized by irregular chain length, low Tg, and only a small degree of cross-linking? a. thermoplastic b. fiber c. elastomer d. thermosetting resins ANSWER: c POINTS: 1 40. Consider the following diagram representing crystallites.

Which type of polymer is represented? a. thermoplastic b. fiber c. elastomer d. thermosetting resins ANSWER: b POINTS: 1 41. In general, which of the following types of polymers would probably be the least crystalline? a. isotactic b. atactic c. syndiotactic ANSWER: b POINTS: 1 42. Which of the following monomers could produce atactic, isotactic, and syndiotactic polymers? a. H2C=CH2 b. H2C=CClH c. H2C=CCl2

d. Cl2C=CCl2

ANSWER: b POINTS: 1 43. Which of the following is not an advantage of olefinic metathesis polymerization?

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Chapter 31 - Synthetic Polymers a. It can be used in the presence of many functional groups. b. It results in an unsaturated polymer. c. The resulting polymer is susceptible to further chemical changes. d. ROMP reactions run with high efficiency but ADMET reactions do not. e. All characterize olefinic metathesis polymerization procedures.

ANSWER: d POINTS: 1

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PAR 1 (appears after Chapter 6) 1. R and S enantiomers of thalidomide have different effects on living systems because _____. a. they have different physical and chemical properties b. they can metabolize only the R enantiomer of any compound c. they can metabolize only the S enantiomer of any compound d. each enantiomer is metabolized by different parts of any living system ANSWER: a POINTS: 1 2. The carbon atoms in the phenyl group of thalidomide are _____. a. sp-hybridized b. sp2-hybridized c. sp3-hybridized d. not hybridized ANSWER: b POINTS: 1

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PAR 2 (appears after Chapter 11) 1. Find the number of chiral centers in melphalan and mark them on the structure below.

ANSWER: One

POINTS:

1

2. In the reaction where the aziridinium ion interacts with guanine, guanine acts as a: a. Lewis acid. b. free radical. c. nucleophile. d. Brønsted–Lowry base.

ANSWER: c POINTS: 1

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PAR 3 (appears after Chapter 14) 1. If the molar absorptivity of phthalocyanine is 162,000 L/( mol · cm), calculate the concentration of the compound in a solution placed in a cell of length 1-cm length and with a UV absorbance of 0.89. a. 4.49 x 10-6 mol/L b. 5.49 x 10-6 mol/L c. 6.49 x 10-6 mol/L d. 7.49 x 10-6 mol/L ANSWER: b POINTS: 1 2. Nucleic acids absorb wavelengths between 250 and 300 nm, and hemoglobin absorbs wavelengths around 600 nm. This means that: a. nucleic acids cannot absorb longer wavelengths as they are smaller than hemoglobin molecules are. b. nucleic acids are more conjugated than hemoglobin is. c. hemoglobin is more conjugated than nucleic acids are. d. hemoglobin molecules can absorb higher wavelengths because they are bigger than nucleic acids are.

ANSWER: c POINTS: 1

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PAR 4 (appears after Chapter 19) 1. If D-phenylalanine is prepared from an aldehyde compound, the chemicals that would be used are: a. KMnO4 and hot HNO3. b. NaBH4. c. CH3MgCl in ether. d. NH2OH. ANSWER: a POINTS: 1 2. Aldehydes and ketones cannot undergo nucleophilic addition–elimination reactions because: a. they are extremely stable molecules and do not react chemically in most cases. b. they do not have an electronegative atom attached to the carbonyl group to act as a good leaving group. c. they do not have an electronegative atom attached to the α-carbon to act as a good leaving group. d. the bond angles in these molecules are not conducive for simultaneous nucleophilic addition and elimination.

ANSWER: b POINTS: 1

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PAR 5 (appears after Chapter 23) 1. Which of the following reagents would be used if dUMP was to be converted to dTMP artificially in a laboratory? a. Na+ -OEt in ethanol b. NaOH in ethanol c. NaH/THF d. LDA/THF ANSWER: d POINTS: 1 2. What would be the IR absorption of dUMP? a. 1705 cm-1 b. 1730 cm-1 c. 1685 cm-1 d. 1650 cm-1 ANSWER: c POINTS: 1

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PAR 6 (appears after Chapter 17) There are no printable questions.

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PAR 7 (appears after Chapter 27) 1. How many π molecular orbitals are present in in 5-hydroxytryptophan? a. 12 b. 5 c. 9 d. 18 ANSWER: b POINTS: 1 2. How many prochiral centers are present in the structure of serotonin? a. 5 b. 4 c. 3 d. 2 ANSWER: d POINTS: 1

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PAR 8 (appears after Chapter 30) 1. According to the frontier orbital theory, the symmetry of electrocyclic reactions is determined by: a. the total number of double bonds present in the molecule. b. the symmetry of the HOMO. c. the symmetry of the LUMO. d. proper orientation of the HOMO of one atom with the LUMO of the other atom.

ANSWER: b POINTS: 1 2. If lobes of like sign are on opposite sides of a molecule, and both orbitals must rotate in the same direction (either both clockwise or both counterclockwise), the kind of motion observed is: a. suprafacial. b. antarafacial. c. disrotatory. d. conrotatory.

ANSWER: d POINTS: 1

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