1 (70)

Prévia do material em texto

56 Chapter 3 REACTIONS OF ALKANES Now C3: Relative reactivity 90% of a hydrogen on / 2 chlorination = 90 Relative reactivity 1.5% C3 of a hydrogen on C3 / 3 C3 chlorination, Recalling the results from propane (Section 3-6), the hydrogens at C2 (secondary) were about 4 times more reactive than those at either or C3 (primary). The reason was that secondary alkyl radicals are more stable than primary alkyl radicals. In the molecule in this problem, 1-bromopropane, the most reactive hydrogens are those on the C1, where the Br is attached. Evidently, the bromine atom strongly stabilizes a radical at C1. It is reasonable to speculate that this stabilization may have something to do with the lone pairs on the Br, because we learned that radicals are electron deficient: They are stabilized in alkanes by hyperconjugation, which allows electrons from neighboring bonds to delocalize towards the half-empty radical p orbital. In 1-bromopropane, we can imagine that a p orbital containing a lone pair on the Br could align and overlap with the half-empty p orbital on carbon, an overlap that can also be represented by resonance: C Br C Br 39. All necessary DH° values are in Table 3-1, except for X-X, which are given in Table 3-4. (a) The answers (kcal mol using = DH° (bond broken) - DH° (bond formed), are Reaction for X = F CI Br I (1) + CH₄ + H. -5 +20 +35 +48 (2) + X₂ -98 -45 -41 -35 (1) + (2) + X₂ + HX = 103 -25 -6 +13 (b) In every case the for the hypothetical first propagation step above is much less favorable than the for the generally accepted, correct step (Table 3-5). Therefore, the Eₐ values for the first to steps the above will, in all probability, be much larger than the Eₐ values for the correct steps. Relative correct propagation steps, then, the reaction + CH₄ very, very slow, and unlikely to compete kinetically. CH₃X + H. will probably be species 40. Inhibition in usually comes about by reaction of the inhibitor with one of the reactive with inhibitors. a propagation step. In the case of radical halogenation, the alkyl radical is "chain-carrying" reaction The products of the inhibition process are not reactive enough to continue susceptible on to the propagation step, chain so the propagation "chain" is broken in much the same way that termination to steps another break propagation process. We have the following: Cl₂ 2 Initiation + CH₄ HCI + Propagation step 1: = +2 kcal mol⁻¹ Then, however, in the presence of + + 1. Inhibition step: = -21 kcal mol⁻¹ The kcal chain started by propagation step 1 is now broken because cannot reaction +34 system. from Table 3-5). A chain carrying radical has been permanently react with CH₄ lost from = the