Química Orgânica 1 · Energia de reação e cinética
Locate the rate-determining step of an SN1 reaction
Condições: 2-Bromo-2-methylbutane in warm ethanol (ethanolysis). One ethanol is the nucleophile and a second ethanol removes the proton. Follow the substitution branch to 2-ethoxy-2-methylbutane; E1 elimination competes but is not requested. The energy profile is schematic: its values are illustrative, not measured.
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Ionize the substrate
Heterolysis of C–Br is uphill: it separates charge and leaves carbon with only six electrons. It has the highest barrier on the profile, so it is the rate-determining step. Ethanol takes no part in it, so rate = k[2-bromo-2-methylbutane], first order overall.
As setas curvas
- The C–Br bond pair leaves with bromine, forming bromide and a tertiary carbocation.
Apply the Hammond postulate
Ionization is strongly endergonic, so by the Hammond postulate its transition state (‡1) lies close in energy, and therefore in structure, to the carbocation: the C–Br bond is largely broken and C2 already carries substantial positive charge. Anything that stabilizes the cation (three alkyl groups, a polar protic solvent) also stabilizes ‡1 and speeds the reaction. The cation itself sits in a shallow well, so it is an intermediate, not a transition state.
Capture the carbocation
This fast step starts from a high-energy intermediate and has only a small barrier (‡2). Bromide can also recombine with the cation, but ethanol, the solvent, is present in huge excess.
As setas curvas
- An ethanol oxygen lone pair forms the new C–O bond to the empty p orbital.
Deprotonate the oxonium ion
The oxonium intermediate (oxygen with three bonds, one lone pair, +1) loses its proton to a second solvent molecule through a small barrier (‡3).
As setas curvas
- A lone pair on the second ethanol accepts the oxonium proton.
- The oxonium O–H bond pair returns to oxygen, which becomes a neutral ether oxygen.
2-Ethoxy-2-methylbutane
Three elementary steps give three transition states (‡1–‡3) and two intermediates (the carbocation and the oxonium ion). Only ‡1, the highest point reached from the starting material, controls the rate; the later steps are fast and do not appear in the rate law. The ether oxygen ends neutral with two bonds and two lone pairs; the second ethanol is now protonated (+1), balanced by bromide (−1): the reaction mixture becomes acidic (HBr).
O que você vai praticar
Map each elementary step of an SN1 solvolysis onto its energy profile and apply the Hammond postulate to the ionization transition state.
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