Organic Chemistry 1 · Elimination and competing pathways
Ethoxide E2 of 2-bromo-2-methyl-1-phenylpropane
Conditions: NaOEt, EtOH, Δ
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The mechanism, step by step
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Ethoxide removes a benzylic hydrogen
The tertiary bromide has β hydrogens on two methyl groups and on the benzylic CH2. Removing a benzylic hydrogen anti to bromine forms a C=C bond that is trisubstituted and also conjugated with the benzene ring, so its transition state is the lowest in energy and small ethoxide reaches it easily.
The curved arrows
- An ethoxide oxygen lone pair bonds to a benzylic hydrogen on C1, anti-periplanar to bromine.
- The benzylic C–H bond pair moves between C1 and C2 and becomes a π bond conjugated with the ring.
- The C2–Br bond pair leaves with bromine as bromide while the new π bond forms.
2-Methyl-1-phenylprop-1-ene forms
2-Methyl-1-phenylprop-1-ene is the major product: its trisubstituted double bond overlaps with the benzene π system, which adds extra stability. Ethanol and sodium bromide are the by-products.
What you'll practice
Draw the E2 step in which ethoxide removes a benzylic hydrogen of 2-bromo-2-methyl-1-phenylpropane, giving the conjugated alkene 2-methyl-1-phenylprop-1-ene.
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