Paul Bracher - Proposed Errata in Organic Chemistry, 4th Edition, by Janice Smith



	Main Textbook\|Solutions Manual

	This page records proposed errata and personal commentary on Organic Chemistry, 4th edition, by Janice Gorzynski Smith (published by McGraw-Hill). This page was inspired by a smiliar effort for the third edition by Bob Hanson at St. Olaf College. If you have encountered relevant items that are not already included on this list, please contact me. I'll add them with or without attribution—your choice.

Textbook

	Page	Section	Proposed Erratum & Comment

	26 39	Section 1.7B Section 1.10B	The book reads "any atom surrounded by three groups is trigonal planar and has bond angles of 120°." While the bond angles of a symmetrical compound like boron trifluoride are 120°, the bond angles of ethylene are not all 120° (as labeled in the figure). In fact, the H–C–C angle in ethylene is greater than 121°, and the H–C–H angle is closer to 117°. (Source)

	45	Section 1.12 see fIgure	C–H bonds are not nonpolar. This point may seem trivial, but it is extraordinarily important for things like IR spectroscopy. C–H bonds are weakly polar.

	65	Section 2.4	I dislike statements such as "because Table 2.1 is arranged from low to high pKa, an acid can be deprotonated by the conjugate base of any acid below it in the table". It is important to remember that pKa values correspond to equilibria, and just because an equilibrium lies relatively far to the reactants side doesn't mean that no product will form. For instance, hydroxide (pKa of conjugate acid = 15.7) can deprotonate acetone (pKa = 19.2). While the equilibrium does favor hydroxide remaining deprotonated, there will be some small amount of deprotonated acetone present at equilibrium. If this anion is reactive, then it can be consumed and Le Chatelier's Principle will come into play such that more acetone is deprotonated to help reestablish the equilibrium. This is exactly what happens in some aldol reactions. So, refrain from saying a base can't deprotonate an acid unless there is an overwhelming kinetic barrier for doing so. The author is generally careful about this issue, using good wording like "favors deprotonation" or "...strong enough to deprotonate a given acid so that the equilibrium lies to the right." In contrast, the online Connect problems are horrible about making this distinction.

	66	Section 2.4	As mentioned above, one must be careful when making statements like "The base ethoxide is not strong enough to deprotonate acetylene" based purely on differences in pKa values. I will bet if you put hydroxide in a solution of acetylene in THF/D2O and heat it, you will observe some proton exchange.

	87	Section 3.1	The author writes "Ethane, for example, has only C–C and C–H σ bonds, so it has no functional group. Ethane has no polar bonds, no lone pairs, and no π bonds, so it has no reactive sites. Because of this, ethane and molecules like it are very unreactive." First off, the C–H bonds in ethane are polar, although the molecule has no net dipole moment. Furthermore, I would argue that C–H bonds are a functional group, given that we can observe reactions that target them and these groups have characteristic spectroscopic signatures. Finally, ask a fireman if he would consider ethane "very unreactive". Terms like "unreactive" beg for a standard of comparison. Would you consider alkanes "very unreactive" relative to something like carbon tetrachloride?

	92	Section 3.3	The author misuses the term "van der Waals forces" as a synonym for "London forces" (a.k.a. "dispersion forces"). Unfortunately, this is a common error among chemists. According to IUPAC, van der Waals forces comprise all "attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules." Thus, van der Waals forces include not just instantaneous induced dipole–induced dipole interactions (as the author believes), but also dipole–induced dipole and dipole–dipole interactions. Note that while, all induced dipole–induced dipole interactions are van der Waals forces, not all van der Waals forces are induced dipole–induced dipole interactions. Unofrtunately, this error is pervasive throughout chapter 3 and the entire remainder of the textbook.

	170	Section 5.3	It is a relatively minor and common mistake to use the word "superimposable" instead of "superposable". To superimpose an object is to place it over another. Any two molecules are superimposable. Two molecules are superposable only if their atoms and bonds coincide exactly when the molecules are placed against each other. If you don't care to listen to me, perhaps you care to listen to IUPAC?

	291	Section 8.2A	The H–C–H bond angle in ethylene is not 120° (see erratum for page 26).

	311	Problem 8.22	I would like to see a reference for the reactions in "a" and "c". I believe these double dehydrohalogenation products have a tendency to favor terminal alkynes (vs. internal alkynes) due to the alkyne zipper reaction. This isomerization is not a concern in "b" (the solutions manual shows the terminal alkyne as the exclusive product) and "d" (where isomerization is impossible).

	318	Problem 8.35	This might be the first example of a recurring and frustrating error in the book: the use of tert-butoxide to produce the more-substituted alkene from an alkyl halide (i.e., the Zeitsev product in higher yield than the Hoffman product). The author uses tert-butoxide as a non-nucleophilic, strong base that will favor E2 dehydrohalogenation, but indicates that Zaitsev's Rule will be followed to generate the more substituted alkene when presented with a choice. The literature reveals that selection of a base as bulky as tert-butoxide will favor the less substituted product (see Brown, et al. J. Am. Chem. Soc. 1956, 78, 2193 and Chapter 10 in Carroll's Perspectives on Structure and Mechanism in Organic Chemistry). Note that the author has correctly applied the literature in her initial discussion of Zeitsev's Rule in Section 8.5 by using smaller (non-bulky) bases like hydroxide and ethoxide.

	373	Section 10.1	The H–C–H bond angle in ethylene is not 120° (see erratum for page 26).

	385	Section 10.7	The use of tert-butoxide to eliminate HBr from 1-bromo-1-methylcyclohexane to produce the more-substituted alkene is problematic (see erratum for page 318).

	554	Problem 14.64	I would be very surprised if this reaction occurred in anything close to reasonable yield. I would love to see a reference for it.

	615	Section 16.13D	The green box offering a bulleted summary of endo vs. exo products gets it wrong by stating, "a substituent on one bridge is endo if it is closer to the longer bridge that joins the two carbons common to both rings." Determining endo vs. exo can be done by examining the competing transition states shown on page 616 in Figure 16.11. The underlying mechanism has nothing to do with the length of any bridge, but rather the secondary orbital interactions present between the internal atoms of the diene and the atoms on the substituents of the dieneophile. While the rule stated in the green box holds for cyclopentadiene, it is ambiguous for 1,3-cyclohexadiene and fails for 1,3-cycloheptadiene. Also, you needn't form a bicyclic product to apply the endo rule.

	884	Mechanism 22.7	I do not like the order of events drawn in this mechanism. I think it is more reasonable for the proton transfer from the acid to the DCC to occur prior to the attack by oxygen because (i) imines are much stronger bases than carboxylates, (ii) proton transfers are among the fastest reactions in organic chemistry, (iii) the proton transfer will produce a more active nucleophile (carboxylate vs. carboxylic acid) and electrophile (a protonated diimide vs. a non-protonated diimide). For similar reasons, in part [2] of the mechanism, I would argue that the proton transfer will probably occur before loss of the leaving group.

	929	Mechanism 23.3	In the final step of the mechanism, bromide is shown as the base used to deprotonate the protonated carbonyl group. I do not like mechanisms that invoke a base that is weaker than another base also present in higher concentration. In this case, acetic acid (even though it's "already" protonated) is both a stronger Brønsted base and present in higher concentration than bromide. The pKa of protonated acetic acid is −6.4 (ref) and the pKa of bromide is −9 (ref). It is far more likely that a molecule of acetic acid will serve as the base used to deprotonate the intermediate than bromide, which is an extraordinarily weak Brønsted base.

	935	Section 23.8B	The conditions given for generation of the thermodynamic enolate are far from optimal. Begin by constrasting the pKa values for a ketone (~19–20) and protonated tert-butoxide (~17). If we let the system reach equilibrium, as the figure suggests, only ~1% of the ketone will have been converted to the enolate. So, when we add the methyl iodide, we're going to have a lot of tert-butoxide competing with a little enolate to attack the alkyl halide. Even though the tert-butoxide is bulky, methyl iodide is a great, unhindered electrophile and we have put this reaction to good use before (in the Williamson ether synthesis). But, fear not! There is a better way! Sutiable conditions can be found by asking why it is important to use LDA at −78 °C to make the less-substituted enolate. We use low temperature because once LDA has deprotonated the less-hindered carbon, we don't want to reprotonate it (with a proton from the other side of the carbonyl group) and establish an equilibrium that will favor the more-substituted (thermodynamic) enolate. So, if we actually want the thermodynamic enolate, all we have to do is increase the temperature. LDA at 0 °C is a good, general value. Also, it helps to use a slight excess (~5%) of the ketone (relative to LDA) to serve as a weak source of protons to favor reaching equilibrium faster.

	1019	Figure 25.11	The author uses tert-butoxide as a non-nucleophilic, strong base that will favor E2 dehydrohalogenation, but indicates that Zaitsev's Rule will be followed to generate the more substituted alkene when presented with a choice. The literature reveals that selection of a base as bulky as tert-butoxide will favor the less substituted product (see Brown, et al. J. Am. Chem. Soc. 1956, 78, 2193 and Chapter 10 in Carroll's Perspectives on Structure and Mechanism in Organic Chemistry).




Solutions Manual

		Problem 1.18	Some of the bond angles are close, but wrong for the number of significant digits provided. (See note for Problem 1.59, below).

		Problem 1.59	I really don't understand why the bond angles in these answers are given as they are. In "a", the bond angles in chloromethane are not all 109.5° because the four groups on carbon are not all the same. The H–C–Cl angles are 110.8° and the H–C–H angles are 108.2°, see NIST). Why provide four significant figures when you are only accurate to two? Why not just say "~110°"? Also, in part "d" the author does explicitly use a "~" to denote an approximate value, which implies the other angles are exact. Odd. Anyway, most of the bond angles given as answers to this problem are wrong, although close.

		Problem 1.78	For part "d", the statement that "all C–H bonds are nonpolar σ bonds" is incorrect. These bonds are polar, albeit weakly polar. (See discussion for page 45, above.)

		Problem 2.39	In the answer given for part "c", I would be very hesitant to write the deprotonation of acetylene by hydroxide with a single left-to-right arrow as opposed to a set of equilibrium arrows that indicate this equilibrium lies strongly to the side of the reactants.

		Problem 16.41	As written, the answer is ambiguous. If a diene is more stable, less energy should be released upon hydrogenation. If the heat of hydrogenation is defined as ΔH for the reaction, then the least stable diene should have the most negative (and hence, lowest/smallest) value.

		Problem 23.10	It should be noted that LDA will produce the enolate in quantitative yield, while treatment with methoxide will produce a small amount of enolate (<1%) at equilibrium.

		Problem 23.19	For "b" and "d", it is much better to use 0.95 eq. LDA at 0°C than tert-butoxide in tert-butanol to generate the thermodynamic enolate (see note for page 935, above).

		Problem 23.51	For "d", it is much better to use 0.95 eq. LDA at 0°C than tert-butoxide in tert-butanol to generate the thermodynamic enolate (see note for page 935, above).

		Problem 23.66	For "f", it is much better to use 0.95 eq. LDA at 0°C than tert-butoxide in tert-butanol to generate the thermodynamic enolate (see note for page 935, above).

		Problem 25.31	The author shows the Zaitsev elimination products as major, when the Hoffman products should predominate for a bulky base like tert-butoxide (see discussion for page 1019, above). If you want to generate the more-substituted alkenes, use a smaller base, like methoxide.