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instructor_supplement_18 - Instructor Supplemental...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 18 The Chemistry of Aryl Halides, Vinylic Halides, and Phenols. Transition-Metal Catalysis Solutions to In-Text Problems 18.1 (b) 1-Bromocyclohexene, a vinylic halide, does not react by the S N 2 mechanism; 1-(bromomethyl)cyclohexene, an allylic halide, reacts most rapidly. (See text Sec. 17.4, text p. 802.) 18.3 (b) The reactivity order is B << C < A . The reaction of compound B is slowest because vinylic halides are virtually inert in S N 1 reactions; and the reaction of compound A is fastest because its ionization gives a resonance-stabilized allylic carbocation. 18.4 (b) The product results from nucleophilic aromatic substitution by the thiolate group: 18.5 (b) The second compound, p -fluoronitrobenzene, reacts most rapidly because only in the reaction of this compound is the intermediate Meisenheimer complex stabilized by resonance interaction of an unshared electron pair with the nitro substituent. 18.7 (b) The PPh 3 ligands are L-type ligands; hence, there are no X-type ligands, and, because the charge on Pd is 0, the oxidation state of Pd is 0. You might be wondering about the prefix tetrakis in the name of this complex. The prefixes bis , tris , and tetrakis are used as numerical prefixes instead of di , tri , and tetra when the group that is enumerated itself contains multiple substituents. Thus, the ligand triphenylphosphine has three phenyl groups on the phosphorus (thus the prefix tri in the name of this ligand). There are four triphenylphosphine ligands—thus the prefix tetrakis . 18.9 (b) Pd has ten valence electrons in the neutral atom. There are no charges and no X-type ligands in the complex; hence, this is a d 10 complex. Using Eq. 18.24, text p. 836, with an oxidation state of 0, we get the same answer. 18.11 Neutral iron (Fe) has 8 electrons. Because CO is an L-type ligand, it is counted twice in the electron count. We simply solve for x in 8 + 2 x = 18 and obtain x = 5. Fe(CO) 5 , or pentacarbonyliron(0), is in fact a stable complex that can be purchased commercially.
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 18 2 18.13 (b) Triphenylphosphine (PPh 3 ) is an L-type ligand. If we strip the four PPh 3 ligands from the Pd (palladium), a Pd (0) atom remains. From Fig. 18.3 on text p. 832, Pd has 10 valence electrons. This is exactly the number needed to fill all of the 4 d orbitals with two electrons each. This leaves four valence orbitals—the 5 s and the 5 p orbitals—empty. These are hybridized to form more directed orbitals. (This situation is exactly like carbon hybridization in methane, except that we are using orbitals from period 5.) Hybridize of one 5 s and three 5 p orbitals gives four sp 3 hybrid orbitals, which, as we know from methane, are directed to the corners of a regular tetrahedron. Each of these empty orbitals accepts a pair of electrons from a PPh 3 ligand. Thus, the Pd(PPh 3 ) 4 complex is tetrahedral.
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