Organic Chemistry 1 · Substitution and elimination: exam review
Anti E2 of (1R,2R)-1-bromo-1,2-diphenylpropane
Conditions: NaOEt, EtOH
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The mechanism, step by step
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Anti-periplanar E2
Ethoxide removes the single hydrogen on C2, and E2 requires that hydrogen to be anti-periplanar to bromine on C1. Rotating the C1–C2 bond into that one allowed conformation fixes where the two phenyl groups end up, so this stereoisomer can only give the (Z) alkene.
The curved arrows
- The ethoxide oxygen lone pair removes the C2 hydrogen that lies anti to the C–Br bond.
- The C–H bonding pair forms the C1=C2 π bond, keeping the groups in the anti-periplanar geometry.
- The C–Br bonding pair leaves with bromide from the face opposite the departing hydrogen.
(Z)-1,2-Diphenylpropene
Only the (Z)-alkene forms, with the two phenyl groups on the same side of the double bond; this is what the anti-periplanar conformation of the (1R,2R) isomer delivers. The (1R,2S) diastereomer gives the (E)-alkene instead, which shows the reaction is stereospecific.
What you'll practice
Draw the anti-periplanar E2 of (1R,2R)-1-bromo-1,2-diphenylpropane with ethoxide and explain why the only alkene formed is (Z)-1,2-diphenylpropene.
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