Known Typographical Errors

Maintained by Steven Hardinger at the Department of Chemistry & Biochemistry, UCLA

1317	error bounty points awarded to	712	students since Winter 2000 quarter.
Error bounty point 1300 awarded November 4, 2008 Student 700 awarded October 22, 2008 Error bounty point 1200 awarded December 11, 2007 Student 600 awarded October 4, 2007 Error bounty point 1100 awarded April 17, 2007 Error bounty point 1000 awarded August 22, 2006

Introduction

These errors have been discovered by students and are listed here for your convenience. These errors may have been corrected in a later printing, so you may not find them. If you find more errors, please bring them to my attention by email.
My thanks to the students who have brought these errors to my attention.

The errors on this list are true typographical errors, not differences in style. For example, these texts contain a number of naked protons (H⁺). Although I strongly prefer that you use H₃O⁺ instead of H⁺, any naked protons in these texts are generally not considered errors.

Sometimes errors are corrected between various printings of the same edition of a text. Because of this you may not see some of the errors listed below.

Not all errors are listed here. Errors in web site content (exam keys, OWLS solutions, etc.) are usually handled by posting a revised version of the erroneous document.

Bounty Offered

Offered here is a reward for students who find chemically significant errors in any texts, Thinkbooks (current edition only) or any other ancillary used in any of my chemistry courses, or on any web sites constructed by Dr. Hardinger. Course notes archives are not eligible for error bounty points.

There is no bounty for spelling errors, but I do appreciate you bringing them to my attention.

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For each confirmed and verified error that you submit that is not on this list, you will receive one point.

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Errors may be submitted any time during the quarter, but must be submitted before the final exam begins. I appreciate all errors brought to my attention, but once the course is over (i.e. the final exam is completed) errors cannot be evaluated for error bounty points.

My decision on the viability of any error is final.

Error List

Textbooks

Vollhardt and Schore (5^th edition) and Study Guide
Brown, Foote, and Iverson (4^th edition) and solutions manual
Atkins and Jones (3^rd edition) and solutions manual

Lecture Supplements and Thinkbooks

Chemistry 14C Thinkbook, 6^th edition
Chemistry 14C Lecture Supplement, 2^nd edition
Chemistry 14D Thinkbook, 6^th edition
Chemistry 14D Lecture Handout

Ancillaries

Pushing Electrons, 3^rd edition (Weeks)
Organic Chemistry as a Second Language, 2^nd edition (Klein)

Supplementary Reading Material (OCATSA)

All courses

Organic Chemistry: Structure & Function, 5^th edition (Vollhardt and Schore)

Chapter 2

Table 2-2, page 60: Hydronium ion has a formal positive charge (H₃O⁺) not a formal negative charge.

Chapter 3

Solution 14(a), page 42: The first structure is the 1-methylpropyl radical.

Chapter 6

Page 219, first full paragraph: Table 6-3 depicts methyl, primary, and secondary (but not tertiary) halides.
Page 227, last paragraph: The generally accepted explanation for nucleophilic backside attack and the stereochemistry of the S_N2 reaction is that the nucleophile attacks the sigma* (antibonding) orbital associated with the carbon-leaving group bond, and not the small sigma (bonding) orbital backlobe associated with the carbon-leaving group bond. (The stereochemical outcome of these reaction pathways is identical, but the thermodynamics are different.)
Page 237: At the bottom of the page: water has two lone pairs, not three.

Chapter 7

Table 7-2, page 260: In the entry for S_N2 reaction at a tertiary carbon, change "extremely slow" to "no reaction." Because S_N2 reactions at tertiary carbons are virtually unknown, we might accept either rate, but "no reaction" is more consistent with the discussion of this idea in Chapter 6.
"E2 Reactions Proceed In One Step", page 265: This is a false generalization. In every E2 reaction the C-H and C-LG group bonds are broken simultaneously. However, we can imagine a circumstance in which the base must be deprotonated before the elimination event occurs. In this case, the reaction still follows E2 kinetics, but is not concerted.

Chapter 15

Page 695, Bromobenzene Formation: The curved arrow which begins at a bromine lone pair should begin at the iron-bromine bond.
Page 697, Mechanism of Aromatic Sulfonation: A proton (H⁺) does not fall off of a molecule by itself, even if its loss from an arenium ion restores aromaticity. A base of some sort removes the proton. In this mechanism the base is either SO₃ or PhSO₃^-.
Page 700, Mechanism of Friedel-Crafts Alkylation: In step 3, the carbon at the six o'clock position of the ring is missing its positive formal charge. In addition, the curved arrow which starts at a one pair on X should begin at the X-Al bond.

Chapter 20

Exercise 20-1: Phenyl chloroformate has C=O not Cl=O.

Chapter 24

Page 1123, "Inversion of Sucrose" scheme: The structure labeled as "32% alpha-D-Glucopyranose" is actually beta-D-glucopyranose.

Chapter 26

Page 1224, first structure: The cytidine structure has the wrong base.

Appendix: Answers to Exercises

Solution 6-8, page A-9: The correct units are mol^-1 L^-1 s^-1.

Study Guice for Organic Chemistry: Structure & Function, 5^th edition (Vollhardt and Schore)

Chapter 1

Solution 25(b), page 7: The middle structure in the second set of structures is missing a curved arrow that depicts an oxygen lone pair becoming a carbon-oxygen pi bond.
Solution 26, page 7: Molecule 24(f) also has resonance.

Solution 27(f), page 8: The most significant resonance contributor for ClO₂^- (chlorite ion) is O=Cl=O, with two lone pairs on each atom and a formal negative charge on chlorine. It has a total of 20 valence electrons.
Solution 39(c): Each oxygen atom also has two lone pairs.
Solution 44: The shortest bond in the molecule is the alkyne triple bond.

Chapter 2

Solution 44(b), part (2): The temperature is 173K not 1.73K.

Chapter 4

Solution 36(e): The ester is not labeled.

Chapter 8

Solution 39(e), page 142: The carbonyl oxygen has an extraneous lone pair.

Chapter 9

Solution 30, page 157: Bromide ion is missing its fourth lone pair.
Solution 409b), page 161: CH₃CN=CHCH₂CH₃ should be 2-pentene.

Organic Chemistry, 4^th Edition (Brown, Foote, and Iverson)

Look also at this list compiled by one of the text authors. Errors on the Iverson list are not eligible for error bounty extra credit points.

Answers that are wrong in the back of the text but right in the Study Guide are not eligible for an error bounty extra credit point. When in doubt, compare your answer with both sources.

Chapter 1

Page 4, third line: The correct electron configuration for carbon is 1s²2s²2p_x¹2p_y¹2p_z⁰.
Page 23: An amide is a derivative of a carboxylic acid in which the carboxylic acid hydroxyl group has been replaced by an amino group.

Chapter 2

Page 88: In the figure in the middle of the page, the labels for axial and equatorial are switched on the right side of the equilibrium.
Page 96, Example 2.15(b) solution: The molecule with bp 126^oC is octane not decane.

Chapter 3

Page 124, structures at the top of page: Change each C-H into C-CH₃.
Problem 3.5, page 124: In structure 4, the H and OH on the front carbon should be switched.

Chapter 6

Page 207, upper right box in table: True when TΔS > ΔH.
Page 219, mechanism box, step 3: The final product does not have a positive charge.
Page 222, second reaction: Ignore the extraneous CH₂ reactant.
Page 223, second reaction: The mechanism step showing Br^- attack on the bromonium ion is missing a curved arrow that shows C-Br bond scission.
Page 227, first sentence below oxymercuration scheme, middle of page: The alkene reactant is 3,3-dimethyl-1-butene.

Chapter 7

Page 271, Example 7.3: The second reagent for the conversion of vicinal dibromide into an alkyne is water, not NH₂O.
Page 275, second reaction: The reactant is 2-methyl-2-butene, not 2-methyl-2-butane.
Page 285, Key Reaction 7: Hydroboration is a syn addition, so the intermediate is (in this case) the E-enol not the Z-enol.

Chapter 8

Page 313, mechanism box, step 6: The Br₂ necessary for step 3 is formed by the reaction of HBr with NBS.
Page 317, mechanism, top of page: The peroxy radical (middle structure) has an extraneous electron (blue dot).
Page 320, mechanism box, chain initiation: Homolytic cleavage of a dialkyl peroxide...

Chapter 9

Page 330, lower left corner, definition of electrophile: An electrophile loves electrons and not nuclei.
Page 340, generalization #1: Within a row of the periodic table, nucleophilicity increases from right to left (i.e., F^- is a poorer nucleophile than HO^-).
Page 357, E1 mechanism box: The mechanism box illustrates the E2 reaction of 2-bromopropane. In addition, bromide ion is missing a lone pair and formal negative charge.
Page 366, Table 9.14, entry for RCH₂X: Acetate ion is a weak base and good nucleophile (favors S_N2). Tert-butoxide is a strong base and good nuclophile (favors E2). Also, iodide ion is a good nucleophile but a weak base.

Organic Chemistry, 4^th Edition (Brown, Foote, and Iverson) Solutions Manual

Chapter 1

Solution 1.35(a), page 9: The O-H bond is more polar than the C-O bond.

Chapter 2

Solution 2.23(d), page 33: The correct formula for this structure is C₄H₈O.

Chapter 6

Solution 6.13(g), top of page 109: The bond dissociation enthalpy for C=C is 727 (not 627) kJ/mol, and the reaction enthalpy is 5969 (not 5869) kJ/mol.
Solution 6.33, page 121: The alkene in the first mechanism structure has an extra hydrogen atom.
Solution 6.40, page 125: The molecule is bicyclo[2.2.2]-2-octene not bicyclo[2.2.2]-2-heptene.

Chapter 8

Solution 8.16(a), page 156: The reaction is exothermic by 63 kJ/mol (15 kcal/mol), as shown by the calculation.

Chapter 9

Solution 9.22(a), page 176: The last product is HCl not HBr. Better yet, the haloalkane reactant is 2-bromobutane, not 2-chlorobutane.

Chem 14C Thinkbook, 6^th edition (printed August 2008)

Previous editions are outdated and not eligible for error bounty points.

Spectroscopy Data Table, inside front cover: The characteristic ¹H-NMR chemical shift range for an -OCH₂- proton is 3.6-4.6 ppm.
Molecular Structure: Introduction and Review

Question 15(b), page 10: Change CH₃N to CH₃NO.
Question 19: The structure of methyl stessonate can be found in question 13, not question 12.
Question 42, page 13: The structure of Xenical can be found in question 39, not 37.
Question 51, page 14: Change 2-chloro-1-methylcyclohexane to 1-chloro-2-methylcyclohexane.
Solution 15(b), page 21: Change CH₃N to CH₃NO.
Solution 21, page 22: VSEPR is an acronym for Valence Shell Electron Pair Repulsion.

Resonance

Question 1(g), page 40: Draw the resonance contributors, etc., for HONO₂.

Conjugated Molecules

Solution 2(g), page 62: It is not possible to destroy all conjugation in this molecule with the addition of just two hydrogen atoms. Either the benzene ring conjugation or the ester conjugation can be destroyed, but not both.

Introduction to Aromaticity

Solution 11(b), page 78: The resonance contributor using the other Kekulé benzene structure has be omitted.

Stereochemistry

Solution 2, page 81: A portion of the solution appears at the top of page 82.

Mass Spectrometry

Solution 1(e), page 106: Change 'because it one' to 'because it contains one'.
Solution 5, page 110: The relative abundance of the M+1 peak is the sum of all the natural abundances of atoms present in the molecule whose mass is one more than the lowest mass isotope (²H, ¹³C, etc.), times the number of these atoms present.

Infrared Spectroscopy

Solution 3, page 121: A few are due to coupled vibrations, such as the zone 5 benzene ring peaks.
Solution 5, page 128: Molecule B is eliminated because the IR shows a peak due to alcohol O-H stretch at 3350 cm^-1. Molecule A is a reasonable fit.

Proton NMR Spectroscopy

Solution 4(b), page 145: H_a is more shielded than a typical aromatic proton.

Chem 14C Lecture Supplement, 2^nd edition (printed August 2008) Same errors also appear on the PowerPoint CD

Previous editions are outdated and not eligible for error bounty points

Spectroscopy Data Table, inside front cover: The characteristic ¹H-NMR chemical shift range for an -OCH₂- proton is 3.6-4.6 ppm.
Introduction and Review

Page 14, top slide: Change H 1s + C 2p_x + C 2p_y to H 1s + C 2p_x + C 2p_z.

¹H-NMR Spectroscopy

Page 128, bottom slide: Replace the first Newman projection with the Newman projection shown below.

Solving Spectroscopy Problems

Page 161, bottom slide: Typical -OCH₂- chemical shift = 3.6-4.6 ppm.
Page 162, top slide: C₅H₁₃N also fits the MS data, but is rejected because it does not fit the ¹H-NMR integration.

Chem 14D Thinkbook, 6^th edition (printed August 2008)

Previous editions are outdated and not eligible for error bounty points

Lecture Supplement: Elimination Reactions

Page 14: The C-H bonds in the two 2-butene transition states are partial bonds (C-----H-----OH).

Ionic Substiution Reactions: S_N1

Question 13, page 116: Change 2-bromobutane to 2-bromopropane.
Solution 17(b), page 126: The curved arrow in the second mechanism step is imprecise. Here is a better rendition:

Solutions 20(c) and (d), page 128: These solutions are switched.

Elimination Reactions

Solution 10, page 148: In the carbocation structure on the lower right hand corner of the page replace the broken wedge with a 'normal' line. (The carbocation carbon has trigonal planar geometry, not tetrahedral geometry.)
Solution 23(e), page 157: Change 'least' to 'most'.

Addition Reactions of Carbon-Carbon Pi Bonds

Solution 4(b), page 178: The syn and anti labels are switched.
Solution 8(c), page 181: The rds is formation of the carbocation.

Electrophilic Aromatic Substitution

Question 3, page 201: Delete this question.

Solution 3, page 206: Question 3 was deleted, so the solution numbering is correct.

Radicals

Solution 31, page 230: The correct equation is 2 GS-H + 2 HO radical ---> GS-SG + 2 H₂O
Solution 30, page 242: For the 'atom or group transfer' fate change O₂ (molecule oxygen) to superoxide.

Pushing Electrons, 3^rd Edition (Weeks)

Chapter 1

Page 11: The very first structure on this page has an oxygen atom with too many electrons. This oxygen atom should have two (not three) lone pairs.
Page 16: Under "is much easier than" the bromine atom on the far right is missing a lone pair.
Solution 37, page 44: In the second structure, the upper oxygen atom should have two (not three) lone pairs.
Solution 57, page 46: The first structure should be (CH₃CH₂)₂NH₂. The lower left methyl group is missing a hydrogen atom.

Chapter 2

Question 41 (b): Change "into the right ring" to "within the right ring."
Question 45, structure 9: This structure should have two lone pairs on the oxygen atom, not three.
Solution 47: Structure 20 has an extra positive charge.
Solution 54, page 85: The resonance contributors with the alternate position for the benzene ring double bonds should also be included.
Solution 57, page 85: There should be a C=N double bond in the right hand structure.
Question 62: Each oxygen should have two lone pairs of electrons.
Solution 63: The last double-headed arrow is extraneous.
Solution 64: The carbon atoms is missing its formal charge of -1.

Chapter 3

Question 31: The water molecule is missing a proton.
Question 38: The structure on the far left is missing a methyl group. The correct structure is CH₃CH₂O₂CCH₃.
Question 72: The last sentence of the statement should be "The positively charged carbon is now neutral."
Question 74: The chlorine atom of the left-hand structure is missing a lone pair. (If lone pairs of any given atom are specified, then all lone pairs must be shown for that atom.)

Solution 58: The answer has been cut off. The correct answer is:

Solution 68: The oxygen attached to the carbon bearing the nitrile group is missing its formal negative charge.
Solution 91: The answer has an extra double bond between the carbon bearing the fluorine and the carbon bearing the negative charge.
Solution 101: Chloride ion has four lone pairs, not three.
Solution 108: The correct answer is CH₃CH₂CH₂⁺. The given answer has an erroneous double bond.
Solution 126: The carbon that is receiving the migrating phenyl group should start out with a positive charge.

Chapter 4

Exercise 2, page 164, second line should read: "Taking CO₂ as a landmark, the only carbon in the starting material that has two oxygens attached to it is C1."

Organic Chemistry as a Second Language (Klein; 2^nd edition)

Chapter 1

Page 5, first condensed formula: The correct formula is (CH₃)₃CHCH=COCH₃.

Chemical Principles, Atkins and Jones (3^rd edition)

Chapter 2

Last paragraph, page 68: The electron configuration of a generic noble gas is ns² np⁶.
Example 2.7, page 69: The sulfur atom of SF₄ has a lone pair.
Problem 2.45(c), page 80: The second structure has an extraneous lone pair on nitrogen.

Chapter 3

Figure 3.33, page 109: The bonding orbital is shown on the right and the antibonding orbital on the left.
Self-Test 3.10B solution page 112: The second pi 2p* electron is a duplicate, and therefore should be deleted.
Exercise 3.9(d), page 119: Change NO₂ (nitric oxide) to NO₂^- (nitrite anion).
Exercise 3.19(a), page 119: Pyridine is C₅H₅N.

Chapter 16

Figure 16.4, page 609: The symbol for technetium is Tc (not Te, which is tellurium). The symbol for rhodium is Rh (not Rb, which is rubidium).

Chemical Principles: Solutions Manual Atkins and Jones (3^rd edition)

Chapter 2

Solution 2.41, page SM-66: Each nitrogen has a +1 formal charge, and each oxygen with thre lone pairs has a -1 formal charge.
Solution 2.45(a), page SM-67: The chlorine atom of the "lower energy" structure is missing a lone pair.
Solution 2.77(d), page SM-73: Acetylide dianion features a carbon-carbon triple bond, and a lone pair and negative formal charge on each carbon atom.

Chapter 3

Solution 3.27(c), page SM-85: The best Lewis structure for POCl₃ has a P-O double bond, three P-Cl single bonds, and no formal charges. In this structure oxygen has two lone pairs, each chlorine has three lone pairs and phosphorus has no lone pairs.

Chapter 5

Solution 5.1: Hydrogen bonding is also at work.

Supplementary Reading Material

Chapter 11: Ionic Substitution Reactions

Page 2, section 11.2, first paragraph, fourth line: "In other words, the base replaces the conjugate base."
Page 10, Caution box: The reaction shown converts trans into cis (not cis into trans). The "inverted" and "not inverted" labels are switched.
Page 21, Concept Focus Question 11.24: Change CH₃CH₂Cl to CH₃I.
Page 36, last paragraph: The polar bonds in ethanol are C-O and O-H.
Page 43, end of second paragraph: The equilibrium has been revered. The last line in this paragraph should read: "Taken together, these two competing reactions cause the CH₃Cl + HO^- ---> CH₃OH + Cl^- equilibrium lie to the right."
Page 47, CFQ 50(d): Ignore the extra methyl group.

Page 48, first paragraph: The reactant is tert-butyl chloride, not tert-butyl iodide.
Page 54, Think Ahead Question 11.54: The second product is a tertiary bromide, not a tertiary iodide.
Page 59, Figure 11.21: The starting material is missing a carbon atom.
Page 61: There are three S_N1 reaction variables (not four).
Page 71, section 11.7D: An S_N1 reaction is fastest in a polar protic solvent.

Chapter 11: Ionic Substitution Reactions - Solutions

CFQ solution 9(a): The nucleophile should be iodide ion (not azide ion), and the propyl group should be an ethyl group.
CFQ solution 9(c): Change CH₂=CHCH₂- to CH₂=CH-.
CFQ solution 10(d): The transition state structure has a pentavalent carbon. Change CH₂=CH₂- to CH₂=CH-. Also switch the position of Br and OCH₃ in the transition state.
CFQ solution 11.4: The problem requires you to also draw the leaving groups, but these were not included in the answers. The leaving groups are: (a) Br^-, (b) I^-, (c) I^- and (d) Cl^-.
CFQ solution 36: Of the unlabeled methyl groups, only the CH₃O and (CH₃)₂N are methyl carbons. The others are all primary carbons.
CFQ solution 52(f): The trans, trans isomer is the major product.
CFQ solution 64(b): The carbocation is primary, not secondary.
CFQ solution 65(d): Change "low polarity solvent" to "polar solvent."
CFQ solution 66: The solution for part (a) belong to question (b), and vice versa.
CRQ solution 15, page 35: The change from methanol to ethanol makes the reaction slower.

Chapter 30: Electrophilic Aromatic Substitution

Page 12: Cl₂ and Fe converts benzene into chlorobenzene, not bromobenzene.
page 29, Caution box: However, these atoms usually have one or more lone pairs...
Page 40, Think Ahead Question 30.21: The reactant is toluene not benzene.
Page 42: "Incipient carbocation" paragraph, middle of the page: The leaving group is on a methyl carbon, not a secondary carbon.