Instructor Supplemental Solutions to Problems Marc Loudon Joseph G. Stowell to accompany Organic Chemistry 5th Edition

Instructor Supplemental Solutions to Problems Marc Loudon Joseph G. Stowell to accompany Organic Chemistry 5th Edition This manual provides the solutions to the problems that are not provided in the Study Guide and Solutions Manual. These answers are provided as electronic files in Portable Document Format (PDF). Each chapter is provided as a separate file. For the conventions used in these solutions, see the Preface of the Study Guide and Solutions Manual. Permission is given to adopting instructors to post these answers only in electronic form for the benefit of their students. Distribution in print form or resale is a violation of copyright. Copyright © 2010 ROBERTS & COMPANY PUBLISHERS Greenwood Village, Colorado© 2010 Roberts and Company Publishers Chapter 1 Chemical Bonding and Chemical Structure Solutions to In-Text Problems 1.1 (b) The neutral calcium atom has a number of valence electrons equal to its group number, that is, 2. (d) Neutral Br, being in Group 7A, has 7 valence electrons; therefore, Br+ has 6. 1.2 (b) The positive ion isoelectronic with neon must have 10 electrons and 11 protons, and therefore must have an atomic number = 11. This is the sodium ion, Na+. (d) Because Ne has atomic number = 10 and F has atomic number = 9, the neon species that has 9 electrons is Ne+. 1.3 (b) (d) 1.5 The structure of acetonitrile: 1.6 (b) The overall charge is –2. 1.8 (b) Formal charge does not give an accurate picture, because O is more electronegative than H; most of the positive charge is actually on the hydrogens. (d) An analysis of relative electronegativities would suggest that, because C is slightly more electronegative than H, a significant amount of the positive charge resides on the hydrogens. However, carbon does not have its full complement of valence electrons—that is, it is short of the octet by 2 electrons. In fact, both C and H share the positive charge about equally. 1.9 The bond dipole for dimethylmagnesium should indicate that C is at the negative end of the C—Mg bond, because carbon is more electronegative than magnesium. 1.10 (a) Water has bent geometry; that is, the H—O—H bond angle is approximately tetrahedral. Repulsion between the lone pairs and the bonds reduces this bond angle somewhat. (The actual bond angle is 104.5°.)INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 1 2 (c) The formaldehyde molecule has trigonal planar geometry. Thus, both the H—C—H bond angle and the H— CAO bond angle are about 120°. 1.11 (a) Bond angles: aa, ab, bc, bd, cd, de, df, and ef are all about 120°, because all are centered on atoms with trigonal planar geometry; fg is predicted to have the tetrahedral value of 109.5. The bond lengths increase in the order a
g e b d
f c (In Chapter 4, you’ll learn that C—H bonds attached to carbons of double bonds are shorter than C—H bonds attached to carbons of single bonds. For this reason, a g.) 1.13 (a) Because the carbon has trigonal-planar geometry, the H—CAO bond angle is 120°. (b) The two structures are as follows: 1.15 The resonance structures of benzene: Each bond is a single bond in one structure and a double bond in the other. On average, each bond has a bond order of 1.5. 1.17 (b) A 4s orbital is four concentric spheres of electron density, each separated by a node, as shown in “cutaway” diagram (b) in Fig. IS1.1. 1.18 (b) The chloride ion, Cl– (atomic number = 17 and one negative charge, therefore 18 electrons): This ion has the same electronic configuration as argon: (1s)2(2s)2(2p)6(3s)2(3p)6. The valence orbitals are the 3s and 3p orbitals, and the valence electrons are the eight electrons that occupy these orbitals. (d) The sodium atom (atomic number = 11) has 11 electrons. Therefore its electronic configuration is (1s)2(2s)2(2p)6(3s)1. The valence orbital is the 3s orbital, and the valence electron is the one electron that occupies this orbital. Figure IS1.1 Cross section of a 4s orbital in the solution to Problem 1.17b.INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 1 3 1.19 The pictures and energy levels of the molecular orbitals for parts (a), (b), (d), and (e) are essentially the same as they are for the dihydrogen molecular orbitals in Fig. 1.14 of the text. For part (c), the atomic orbitals involved are 2s orbitals, which have a node. However, the various species differ in their electron occupancies. (b) The H2– ion contains three electrons. This can be conceived as the combination of a hydrogen atom with a hydride ion (H–). By the aufbau principle, two occupy the bonding molecular orbital, and one occupies the antibonding molecular orbital. This is shown in the electron-occupancy diagram (b) of Fig. IS1.2. Because the bonding molecular orbital contains a greater number of electrons than the antibonding molecular orbital, this species is stable. Notice that in terms of electron occupancy, H2– and He2+ are identical. (d) The H22– ion can be conceived to result from the combination of two hydride ions (H–). This species contains four electrons; two occupy the bonding molecular orbital, and two occupy the antibonding molecular orbital. This is shown in the electron-occupancy diagram (d) of Fig. IS1.2. In this species, the energetic advantage of the electrons in the bonding molecular orbital is cancelled by the energetic disadvantage of the same number of electrons in the antibonding molecular orbital. This species consequently has no energetic advantage over two dissociated hydride (H–) ions, and therefore it readily dissociates. 1.22 (a) The oxygen of the hydronium ion has approximately tetrahedral geometry and is therefore sp3-hybridized. Two of the sp3 hybrid orbitals contain an unshared electron pair. One of these becomes the lone pair in H3O+; the other overlaps with a proton (H+) to give one of the O—H bonds. The other two sp3 hybrid orbitals contain one electron each; each of these overlaps with the 1s orbital of a hydrogen atom (which contains one electron) to give the three sp3–1s
bonds (the O—H bonds) of H3O+. (b) The H—O—H bond angles in H3O+ should be somewhat larger than those in water, because there is one less lone pair and a smaller associated repulsion. Figure IS1.2 Electron-occupancy diagram for the solution to Problems 1.19b and 1.19d.INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 1 4 Solutions to Additional Problems 1.23 (b) Compound (2), cesium fluoride, would be most likely to exist as an ionic species, because Cs and F come from opposite corners of the periodic table. 1.24 The formal charge on all the hydrogens is 0. For the other atoms: (b) Nitrogen has a complete octet and a formal charge of 0. (d) Boron has a sextet and a formal charge of 0. (f) Boron has an octet and a formal charge of –1. 1.26 The lengths of the C—C bonds in isobutane have no significance; two are longer merely to accommodate the rest of the structure. Remember that these Lewis structures show only connectivity—not geometry. 1.27 (b) (d) (f)