CHAPTER 1
Fundamental Models of Organic Chemistry

1. 1.1. An answer to this question should be stated in terms of macroscopic phenomena. A historical exposition provides a rationale for the basis of contemporary chemistry, and several monographs on the history of chemistry can be used to summarize the ideas and observations that led to contemporary chemistry theory.1,2,3
2. 1.2. See, for example, figure 2D in Tieu, P.; Yan, X.; Xu, M.; et al. Small 2021, 17, 2006482.
   1. Transmission electron microscopy was used in this study.
   2. The eye sees a macroscopic image on a computer screen or on a printed page.

      For another example, see the image of kekulene reported by Pozo, I.; Majzik, Z.; Pavliček, N.; et al. J. Am. Chem. Soc. 2019, 141, 15488.

3. 1.3.
   1. Two alternative geometries and their elimination on the basis of number of isomers are:
      1. Square planar. There would be two isomers of CH2Cl2, one “cis,” in which the Cl−C–Cl bond angle is 90°, and one “trans,” in which the Cl−C−Cl bond angle is 180°.
      2. Square pyramid. Similarly, there would be two isomers of CH2Cl2 with the chlorines in the square base plus another isomer with one chlorine at the apex of the pyramid.
   2. In all answers, a substituent is presumed to occupy a position previously occupied by a hydrogen atom in the parent structure.4
      1. If benzene had the structure we now call fulvene, there should be three different derivatives with the formula C6H5Cl.
      2. If benzene had the structure we now call Dewar benzene, there would be two and only two isomers with the formula C6H5Cl.
      3. If benzene had the structure we now call benzvalene, there would be three possible isomers with the formula C6H5Cl.
      4. If benzene had the structure we now call prismane, there would be only one isomer with the formula C6H5Cl, but there would be four isomers (including a pair of enantiomers) with the formula C6H4Cl2.
      5. If benzene had the structure we now call [3]radialene, there would be one and only one isomer with the formula C6H5Cl, but there would be four possible isomers (shown below) with the formula C6H4Cl2.
      6. There are also acyclic structures with the formula C6H6, such as 2,4‐hexadiyne, and they may be analyzed similarly. For example, if benzene were 2,4‐hexadiyne, then there would be one and only one C6H5Cl, but there could be only two structures with the formula C6H4Cl2.
   3. One can never know that something that has not been tested is like something else to which it seems similar. However, it seems unproductive to dwell on this possibility until there is an experimental result that could be rationalized with a structure for chloromethane that is different from the tetrahedral structure of methane. The spectroscopic results for chloromethane are consistent with a tetrahedral geometry.
4. 1.4. The data and equations are given in Bondi, J. J. Phys. Chem. 1964, 68, 441.

   For n‐pentane,

   These results agree with those given by the general formulas for n‐alkanes:

   For isopentane,

   For neopentane,

5. 1.5. Kiyobayashi, T.; Nagano, Y.; Sakiyama, M.; et al. J. Am. Chem. Soc. 1995, 117, 3270.
6. 1.6. Turner, R. B.; Goebel, P.; Mallon, B. J.; et al. J. Am. Chem. Soc. 1968, 90, 4315. Also see Hautala, R. R.; King, R. B.; Kutal, C. in Hautala, R. R.; King, R. B.; Kutal, C., Eds. Solar Energy: Chemical Conversion and Storage; Humana Press: Clifton, NJ, 1979; p. 333. The difference in heats of hydrogenation indicates that quadricyclane is less stable than norbornadiene by 24 kcal/mol, so this is the potential energy storage density for the photochemical reaction.
7. 1.7. Pilcher, G.; Parchment, O. G.; Hillier, I. H.; et al. J. Phys. Chem. 1993, 97, 243.
8. 1.8. See Davis, H. E.; Allinger, N. L.; Rogers, D. W. J. Org. Chem. 1985, 50, 3601.
9. 1.9.
   1. −632.6 ± 2.2 kJ/mol. Roux, M. V.; Temprado, M.; Jiménez, P.; et al. J. Phys. Chem. A 2006, 110, 12477.
   2. 2‐acetylthiophene is 4.7 kJ/mol more stable than 3‐acetylthiophene in the gas phase. Roux, M. V.; Temprado, M.; Jiménez, P.; et al. J. Phys. Chem. A 2007, 111, 11084.
10. 1.10. Wiberg, K. B.; Hao, S. J. Org. Chem. 1991, 56, 5108.
11. 1.11. Fang, W.; Rogers, D. W. J. Org. Chem. 1992, 57, 2294.
12. 1.12.
    1. Using equation 1.9:
    2. Using equation 1.12:
13. 1.13. See Smyth, C. P. in Weissberger, A.; Rossiter, B. W., Eds. Physical Methods of Chemistry, Vol. 1, Part 4; Wiley‐Interscience: New York, 1972; pp. 397–429.
    1. The gas phase dipole moments for CH3–F, CH3–Cl, CH3–Br, and CH3–I are 1.81, 1.87, 1.80 and 1.64 D, respectively. Using the bond length data in Table 1.1 and rewriting equation 1.23 lead to the following partial charges on F, Cl, Br, and I, respectively: −0.27, −0.22, −0.19, −0.16.
    2. The dipole moments do not show a monotonic trend along the series because a dipole moment is a product of two terms. In the series of methyl halides, one term (the partial charge) goes down and the other term (bond length) goes up. The product of these two terms is a maximum at the second member of the series (X = Cl). Note that the assumption that only the carbon and halogen atoms are charged is an over‐simplification. An Extended Hückel calculation indicates that the three methyl hydrogen atoms also bear some charge.
14. 1.14. Because Pauling electronegativities are computed from the properties of atoms in molecules, they generally cannot be computed for the inert gases. However, krypton and xenon fluorides are known, and electronegativities of krypton and xenon were reported by Meek, T. L. J. Chem. Educ. 1995, 72, 17.
15. 1.15. Using equation 1.47 leads to a value of 2.62 for . Therefore, the hybridization of carbon orbitals used for carbon–carbon bonds is sp2.62. The relationship

    then gives a value of 3.47 for the carbon orbitals used for the carbon–hydrogen bonds.

16. 1.16. Mastryukov, V. S.; Schaefer, H. F., III.; Boggs, J. E. Acc. Chem. Res. 1994, 27, 242. Also see the discussion in Gilardi, R.; Maggini, M.; Eaton, P. E. J. Am. Chem. Soc. 1988, 110, 7232.
    1. As the bond angle increases, the C–C bond length decreases. Conversely, as the bond angle decreases, the C–C bond length increases.
    2. The larger the α, the greater the contribution of p character to the orbital of C2 used for the C2–C3 bond. This means greater s character in the orbital of C2 used for the C1–C2 bond, which results in a shorter C1–C2 bond. The same result can be rationalized using the VSEPR approach. As the angle α increases, there is less repulsion of the electrons comprising the C1–C2 bond with the electrons in the C2−C3 bond. This allows the electrons in the C1–C2 bond to move closer to C2, thus decreasing the bond length.
17. 1.17. Maksić, Z. B.; Randić, M. J. Am. Chem. Soc. 1970, 92, 424. The bond lengths are a function of the hybridization of the carbon atoms.
    1. ethyne, ethene, cyclopropane, cyclobutane, ethane.
    2. 1,3‐butadiyne, 1‐butene‐3‐yne, 1,3‐butadiene, propene, 2‐methylpropene, 2‐methylpropane, ethane.
18. 1.18.
    1. According to the bent bond formulation, the electrons in the bent C–C bonds are pulled in toward the other olefinic carbon atom, so the electrons in these bonds repel the electrons in the carbon–hydrogen bonds less than they would in propane. Therefore, the H–C–H bond angle opens to a larger value.
    2. The electrons in formaldehyde should be pulled even more strongly away from the carbon atom than is the case in ethene. Therefore, the repulsion of electrons in either C–O bond with electrons in a C–H bond is even less than the repulsion of electrons in the C–C bonds with electrons in a C–H bond in ethene. Therefore, the H–C–H bond angle in formaldehyde should be greater than that in ethene.
19. 1.19. Based on an H–C–H angle of 116.2° for ethene, Robinson, E. A.; Gillespie, R. J. J. Chem. Educ. 1980, 57, 329 (appendix, p. 333) reported sp2.26 or 30.6% s character for the C–H bond. Using 117° for the H–C–H angle5 leads to sp2.20, or 31.2% s character. For formaldehyde, using an H–C–H angle of 125.8° similarly leads to 36.9% s character for the carbon orbital used for carbon–hydrogen bonding.6
20. 1.20.
    1. The formula is given by Newton, M. D.; Schulman, J. M.; Manus, M. M. J. Am. Chem. Soc. 1974, 96, 17. Rewrite equation 1.52 as J = 5.7 × (% s) − 18.4 Hz. Then 500/(1 + λ2) =...