Organic Chemistry 2 · Alcohol transformations
cis-1-Bromo-4-tert-butylcyclohexane by PBr3 Inversion
Conditions: PBr3, pyridine, Et2O, 0 °C
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The mechanism, step by step
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Oxygen bonds to phosphorus
The tert-butyl group locks the chair, so in the trans alcohol the OH is equatorial. Its oxygen bonds to the electrophilic phosphorus of PBr3 and one bromide leaves; the ring carbon is untouched and keeps its configuration.
The curved arrows
- An oxygen lone pair forms a bond to phosphorus.
- One P–Br bond pair leaves as bromide.
Pyridine removes the O–H proton
Pyridine deprotonates the positively charged oxygen and gives a neutral dibromophosphite ester. That O–PBr2 group is now an excellent leaving group on the ring carbon.
The curved arrows
- The pyridine nitrogen lone pair takes the O–H proton.
- The O–H bond pair returns to oxygen.
Axial backside attack by bromide
The leaving group is equatorial, so the back side of the C–O bond points along the axial direction. Bromide attacks there and the C–O bond breaks in the same step, inverting the ring carbon.
The curved arrows
- A bromide lone pair attacks the ring carbon from the axial face, opposite the leaving group.
- The C–O bond pair leaves with oxygen as the dibromophosphite anion.
cis-1-Bromo-4-tert-butylcyclohexane
Inversion puts the bromine axial while the tert-butyl stays equatorial, which is the cis isomer. The trans alcohol therefore gives the cis bromide, a clean test of SN2 stereochemistry.
What you'll practice
Show how PBr3 and pyridine turn trans-4-tert-butylcyclohexanol into cis-1-bromo-4-tert-butylcyclohexane by axial backside attack on the activated equatorial oxygen.
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