Química Orgânica 1 · Reações radicalares e seletividade
Brominate isobutane at its single tertiary hydrogen
Condições: 2-Methylpropane (isobutane) with Br₂ and light (hν). Split the upper-left Br₂ molecule, then draw one propagation cycle: a bromine radical removes the tertiary hydrogen and the tert-butyl radical reacts with the remaining Br₂. The unused radical from initiation remains; its later fate is outside this task. Monobromination only.
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Initiate: split Br₂ with light
Light supplies enough energy to break the weak Br–Br bond evenly (homolysis). Each bromine keeps its three lone pairs and gains one unpaired electron; no ions form.
As setas curvas
- One bonding electron of the upper-left Br₂ stays on one of its two bromines.
- The other bonding electron stays on the other bromine of that same molecule.
Remove the tertiary hydrogen
This is the selectivity-determining step. One bromine radical reacts; the other is still present. Abstraction by Br• is endothermic, so its transition state resembles the radical being formed, and the tertiary radical (tertiary C–H about 400 kJ/mol) is reached far more easily than a primary one (about 420 kJ/mol).
Nota: Isobutane has nine primary hydrogens and one tertiary hydrogen; follow the tertiary branch that gives the major product.
As setas curvas
- The bromine radical nearer the tertiary hydrogen supplies one electron of the new H–Br bond.
- One C–H bonding electron completes H–Br.
- The other C–H electron stays on the tertiary carbon.
Transfer bromine from a second Br₂
The planar tert-butyl radical forms a bond to bromine. The new Br• can now start another cycle, so one initiation event drives many propagation cycles.
As setas curvas
- The tert-butyl unpaired electron supplies one electron of the new C–Br bond.
- One Br–Br bonding electron completes the C–Br bond.
- The other Br–Br electron regenerates Br•.
tert-Butyl bromide, HBr and two Br radicals
The central carbon now bonds to three methyl groups and bromine: 2-bromo-2-methylpropane. Despite nine primary hydrogens and only one tertiary, bromination gives tert-butyl bromide almost exclusively (over 99%). The unused bromine radical from initiation and the new one from propagation are both neutral radicals; the radical count stays two because propagation neither makes nor destroys radicals.
O que você vai praticar
Draw a complete photobromination sequence from Br–Br homolysis through both propagation steps and connect the tertiary radical to the nearly exclusive tert-butyl bromide product.
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