1559T_ch22_389-40811/7/0515:40Page403 EQA Solutions to Problems.403 The Sl pathwa earlier (Section 13-9)that alkenyl halides has maximum density in the vicinity of the atomic nucleus.where attraction between the positive t.In contrast.p s have a node at the nuc 。 n since Chapter 7 have had unoccupied p orbitals,derived from cleavage of a bond with an sporbital,which is 14s and 34p in character: An sporbital is 13s and 23 p in nature.mores and less p.Departure of a leaving group to leave it charged e fact that 3δ÷d It is an open guestion whether a phenyl cation is actually involved in reactions such as this.One alternative is addition-elimination (Section 22-4).but this option is not supported by kinetic evidence.which indicates that This mechanism is more likely because it has a unimolecular rate-determining prop tion step.loss ofN from the phe a much sier spe phenyl cation is not an intermediate.It is still considered to be the best candidate for the The SN1 pathway requires formation of a carbocation. We saw earlier (Section 13-9) that alkenyl halides are unreactive toward displacement reactions. The SN2 pathway is poor for the same reason we stated above for benzene compounds. In addition, alkenyl cations are high-energy species because the placement of a positive charge on an sp2 -hybridized carbon is very unfavorable. The same issue arises in the case of benzene: The phenyl cation is a high-energy species and forms only with difficulty because it contains an sp2 -hybridized, positively charged carbon atom. Why is an sp2 -hybridized, positively charged carbon atom a bad thing? Recall the basics of hybridization. Hybrid orbitals combine the characteristics of the contributing simple atomic orbitals. An occupied s orbital has maximum electron density in the vicinity of the atomic nucleus, where attraction between the positive nucleus and the negative electrons is the greatest. In contrast, p orbitals have a node at the nucleus; electrons in p orbitals therefore are less strongly held and more easily lost. So given a choice between having a vacant p orbital or a vacant s orbital, a cationic atom will invariably choose to put electrons in the s orbital and leave the p orbital vacant. All carbocations we have seen since Chapter 7 have had unoccupied p orbitals, derived from cleavage of a bond with an sp3 orbital, which is 1/4 s and 3/4 p in character: An sp2 orbital is 1/3 s and 2/3 p in nature, more s and less p. Departure of a leaving group to leave it unoccupied and positively charged is disfavored by its greater partial s character, and also by the fact that in benzenes and alkenes no geometrical change to a more favorable orbital arrangement can occur. 57. It is an open question whether a phenyl cation is actually involved in reactions such as this. One alternative is addition–elimination (Section 22-4), but this option is not supported by kinetic evidence, which indicates that the displacement process is not bimolecular but unimolecular. Another possibility is a radical process, initiated by reduction of the phenyldiazonium cation to a radical by the iodide ion, which is a good reducing agent. This mechanism is more likely because it has a unimolecular rate-determining propagation step, loss of N2 from the phenyldiazo radical to give a phenyl radical, a much easier species to form than the phenyl cation. Although the exact mechanism for iodobenzene formation from the diazonium cation is unclear, the phenyl cation is not an impossible intermediate. It is still considered to be the best candidate for the intermediate in phenol formation from arenediazonium salts and hot water (Section 22-4). In addition, it is known that diazonium cations exchange their bonded N2 with gaseous N2 (detected in isotope studies),
H, H2O  
N2  I
N N   N (One possibility) N I H OH2
H NH2 N N H O N O    H H N N H OH X H H H sp3 hybridized sp2 hybridized p orbital vacant H X
H  H Solutions to Problems • 403 1559T_ch22_389-408 11/7/05 15:40 Page 403