Introduction

These errors have been discovered by students and are listed here for your convenience. Sometimes errors are corrected between subsequent printings of the same edition of a text.  Because of this you may not see some of the errors listed below. 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.

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.

• Please make sure you have the most recent version of a web site by using the 'reload' or 'refresh' button on your browser.  (The error may have already been reported and repaired.)

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

• When submitting an error please include your first, last name, student ID number, and course in which you are enrolled. A bounty point cannot be awarded if you are partly or fully anonymous. Also, you must also give a correction for the error. A missing or inaccurate correction voids the error bounty point.
  

• You may earn at most six error bounty extra credit points per quarter.
  

• Only the first student who reports an error can get the error bounty extra credit point.
  

• 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.

• The instructor's decision on the viability of any error is final.

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

• Chemistry 30A: The on-line documents will be updated as errors are reported. If you think you've found an error, check the posted version first.
  
• Chemistry 14C Thinkbook, 7^th edition
  
• Chemistry 14C Lecture Supplement, 3^rd edition
  
• Chemistry 14D Thinkbook, 7^th edition

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

• All courses

• Chapter 1
• Problem 29(c), page 47: Some printings of the text use BrF₄^- instead of SF₄. The geometry is pyramidal in both cases.
• 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 696, Activation of Nitric Acid by Sulfuric Acid: HONO₂ is protonated by H₃O⁺, not H₂SO₄.
  
• 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.
  

• 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 29(c), page 11: Some printings of the text use BrF₄^- instead of SF₄. The geometry is pyramidal in both cases.
  
• 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.
  

• Chapter 1
• Page 3, Rule 3: The ground state configuration for carbon is 1s² 2s² 2p_x¹ 2p_y¹ 2p_z⁰.
• Page 23, first paragraph: the predicted H-O-H bond angle is 109.5^o.
• Page 36, sp hybridization diagram, middle of the page: change sp² to sp.
• Page 50, Table 1.11: The data for acetylene is absent. Acetylene is the simplest alkyne, consisting of a carbon-carbon triple bond, with each carbon having a C-H sigma bond. The two C-C pi bonds are constructed by overlapping a pair of 2p orbitals for each pi bond. The C-C sigma bond comes from the overlap of two carbon sp orbitals (one from each carbon), and the C-H sigma bonds from the overlap of a carbon sp orbital with the H 1s orbital. The C-C bond length is 120.3 pm and the C-H bond length is 106.0 pm. The C-C bond strength is 966 kJ mol^-1 (231 kcal mol^-1) whereas the C-H bond strength is 556 kJ mol^-1 (133 kcal mol^-1).
• Chapter 5
• Solution 5.6(c), page 190: The correct name is 1-methyl-4-(1-methylethyl)-cyclohexene.
  
• Chapter 6
• Page 221: The very first mechanism step on this page is missing a curved arrow which shows carbon-bromine bond scission.
• Page 230: In step 3, CH₃CH=CH₂ + 2 H⁺ + 2 e^- ---> CH₃CH₂CH₃.
  

• Chapter 6
• Solution 6.34(a): The alkene is protonated by H₃O⁺, not by H₂SO₄.
• Chapter 9
• Solution 9.22(a), page 187: The by-product of this reaction is HCl, not HBr.

• 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 6, page 7: The hydrogen atom on the solid wedge at the ring junction next to the OH group should be a methyl group. Compare with the correct structure of the same molecule on the previous page.
• Question 5, page 9: Change 'bond-line structure' to 'Lewis structure.'
• Question 7, page 9: Change "and" to "or."
• 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 5, page 18: The given answer is a Kekule structure, not a bond-line structure. In most cases the two structure types are interchangeable.
  
• Solution 7, page 18: Change the last sentence to "CH₃HO implies bonding between the hydrogen atoms of CH₃ and the hydrigen atom of OH."
  
• Solution 15(b), page 21: Change CH₃N to CH₃NO.
• Solution 21, page 22: VSEPR is an acronym for Valence Shell Electron Pair Repulsion.
• Solution 23, page 23: In the H-O-C angle discussion change -CH=CH₂ to -CH₂CH=CH₂.
• Solution 26, page 23: Change "...angle is be a bit larger..." to "...angle is a bit larger...".
  
• Solution 31, page 25: Bond rotation occurs when one atom of a bond is rotated... Decreasing orbital overlap weakens the bond, making the structure less stable.
• Solution 43(a), page 29: The C-C-C bond angles of cyclobutane are 90^o when the molecule is planar...  Also in the last sentence, ...we conclude the greater number of C-C-C bond angles...
  
• Resonance
• Question 1(g), page 40: Draw the resonance contributors, etc., for HONO₂.
  
• Conjugated Molecules
• Question 7, page 54: Allene is a bit unusual among molecules with two pi bonds...
• Solution 1(g), page 61: The resonance hybrid should have a delta+ charge on carbons 2, 4, and 6 of the benzene ring.
• 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.
• Solution 10, page 64: In this case we use 7 C + 6 H₂ + O₂, whose H is zero.
  
• Introduction to Aromaticity
• Question 10(b), page 73: ...the hydrogenation of benzene was interrupted before all of the benzene had reacted.
• Solution 6, page 76: The first molecule is the cycloheptatrienyl anion.
• Solution 8, page 77: The molecule in question has only four pi electrons...
  
• Solution 11(b), page 78: The resonance contributor using the other Kekulé benzene structure has be omitted.
  
• Stereochemistry
• Question 5, page 80: The correct structure of carvone is:

• Solution 2, page 81: A portion of the solution appears at the top of page 82.
• Question 1(a), page 84: The first structure shown is gulose, not glucose. This error does not change the answer.
• Question 12(a), page 86: Change to "Draw all enantiomers of thalidomide".
• Solution 12(a), page 91: Add the R enantiomer of thalidomide.
  
• 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.
• Solution 6(a), page 147: Add (CH₂)₂CHCH to 2.45 ppm signal implications.
• Solution 6(e), page 151: Add 2 x CH in CH₂CH(CH)₂ to the 1.60 ppm implications. Add 2 x CH in (CH)₂CHCH₃ to the 1.39 ppm implications.
• Solving Spectroscopy Problems
• Solution 9, page 175: In IR zone 1, amide/amine N-H is absent due to no nitrogen in the formula.
  
• ¹³C-NMR, 2D-NMR, and MRI
• Question 11(a), page 185: Draw eight isomers.
• Solution, page 193: Change 2,2-dimethyl-2-propanol to 2-methyl-2-propanol. (The structure is correct but the name is in error.) Also include isopropyl methyl ether, which has two quartets and one doublet in its ¹³C-NMR spectrum.
  
• Solution 16, page 198: DBE = 16 - (18/2) + (0/2) + 1 = 8
• Solution 17, page 200: For IR zone 4, delete 'Not enough DBE for both pi-conjugated carbonyl and benzene'.
  
• Amino Acids, Peptides, and Proteins
• Solution 11, page 274: Oxytocin is Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly-NH₂, with a line connecting the two cysteines indicating the disulfide linkage.
  

• Conjugated Molecules
• Page 38, bottom slide: The "these atoms lie in a plane" box should appear as follows:

• Solving Spectroscopy Problems
• Page 158, top slide: zone 5 is 1680-1450 cm^-1.
  

• Ionic Substitution Reactions: S_N2
• Solution 9, page 101: "...a molecule bearing a negative charge is more nucleophilic..."
• Solution 31(b), page 110: The nitrile group in the product should be equatorial down not axial up.
• Ionic Substitution reactions: S_N1
• Question 3(c), page 118: The product of each reaction should be 3-methyl-1-buten-3-ol.
• Question 4(a), page 118: The product's alkene has been cut off.
• Solution 23(b), page 135: "...high electronegativity of oxygen and charge stabilization by resonance decrease nucleophilicity..."
• Elimination Reactions
• Question 24(e), page 151: "What would you use in place of NaOCH₃ to favor formation of the product with the greatest number of vinylic hydrogens?"
• Solution 11, page 157: The 'be deprotonated; form pi bond' carbocation fate should give two alkenes:

• Solution 27(a), page 167: The dielectric constant of ethanol is 25.
• Solution 28(b), page 169: In the mechanism at the bottom of the page, the middle structure should not have a wedged phenyl group, because the carbocation is planar. In addition the caption 'Two enantiomers' belongs under the last structure in the mechanism.
• Electrophilic Aromatic Substitution
• Solution 19, page 235: The tetrachloroethyl group is not a carnivore. It causes meta attack.
• Radicals
• Solution 30, page 260: Under 'Atom transfer' the OOR portion of the product is cut off.
  

• Chapter 1
• Page 5, first condensed formula: The correct formula is (CH₃)₃CHCH=COCH₃.
• Page 16, first line: If oxygen has a positive charge, then it must have three bonds and one lone pair (except in the rare circumstance where oxygen has one bond, two lone piars, and an open octet).
  
• Chapter 4
• Solution 4.12, page 340: The carbonyl oxygen atom is sp².
• Solution 4.13, page 340: The carbonyl oxygen atom is sp².
• Solution 4.15, page 340: The carbonyl oxygen atom is sp².
• Solution 4.17, page 340: The carbonyl oxygen atom is sp².
• Chapter 5
• Solution 5.57, page 341: Some (but not all) copies of the book give the answer as trans-5-ethyl-4-methyloct-2-ene. The correct answer is trans-4-ethyl-5-methyloct-2-ene.
• Chapter 11
• Solution 11.36, page 349: The first mechanism step is missing a curved arrow which depicts scission of the H-Br bond.
  

• 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).
  

• 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.
  

• Chapter 9: Introduction to Structure and Reactivity: Acids and Bases
• Page 39, section 9.4: A proton transfer equilibrium equation favors the weakest acid (highest pK_a).
  
• Chapter 11: Ionic Substitution Reactions
• None so far.
• Chapter 16 : Infrared Spectroscopy
• Page 24, IR Spectrum 16.32: A primary amine gives two N-H stretch peaks in zone 1, and a secondary amine gives one.
• 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.