Organic Chemistry 1 · Carbocation rearrangements and ring expansions
Ethanolysis of 2-bromo-3-methylbutane with a hydride shift
Conditions: EtOH, Δ
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The mechanism, step by step
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Bromide leaves
In polar protic ethanol the C–Br bond ionizes on its own, giving a secondary carbocation and bromide. This is the rate-determining step.
The curved arrows
- The C–Br bonding pair leaves with bromine.
1,2-Hydride shift
The hydrogen on the neighbouring CH carbon moves with its bonding pair to the cationic carbon, giving the tertiary 2-methylbutan-2-yl cation.
The curved arrows
- The C–H bond pair carries the hydrogen to the cationic carbon.
Driving force
The shift trades a secondary cation for a tertiary one, which is stabilized by more hyperconjugation.
Ethanol captures the cation
An ethanol oxygen lone pair bonds to the tertiary cationic carbon, giving an oxonium ion.
The curved arrows
- An oxygen lone pair forms the new C–O bond.
Deprotonate the oxonium ion
A second ethanol removes the proton from the positively charged oxygen, leaving the neutral ether.
The curved arrows
- An ethanol lone pair takes the proton.
- The O–H pair stays on the ether oxygen.
2-Ethoxy-2-methylbutane
The ethoxy group sits on the tertiary carbon of the rearranged cation; protonated ethanol and bromide amount to HBr.
What you'll practice
Draw the solvolysis of 2-bromo-3-methylbutane in ethanol: ionization, 1,2-hydride shift, ethanol capture and deprotonation to 2-ethoxy-2-methylbutane.
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