Organic Chemistry 2 · Ethers and epoxide opening
1-Bromo-2-methylbutan-2-ol from 2-methylbut-1-ene and NBS
Conditions: NBS, H2O/DMSO
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The mechanism, step by step
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Bromonium ion forms
The alkene π pair attacks the electrophilic bromine of NBS while a bromine lone pair bonds back to the other alkene carbon, giving a three-membered bromonium ion. The N–Br bond breaks and the succinimide anion leaves, stabilized by its two carbonyl groups.
The curved arrows
- The alkene π pair forms a bond from the CH2 carbon to bromine.
- A bromine lone pair bonds to the tertiary alkene carbon, closing the bromonium ring.
- The N–Br bond pair moves onto nitrogen, releasing the succinimide anion.
Water opens the ring at the tertiary carbon
Water attacks the bromonium ion at the tertiary carbon, which carries more of the positive charge, from the face opposite the bromine. The C–Br bond to that carbon breaks and bromine stays on the CH2 carbon (Markovnikov orientation).
The curved arrows
- A water lone pair forms the new C–O bond to the tertiary carbon.
- The C–Br bond pair moves onto bromine, which becomes neutral.
Succinimide anion takes the proton
The succinimide anion, the strongest base present, removes a proton from the oxonium ion. This gives the neutral bromohydrin and succinimide.
The curved arrows
- The nitrogen lone pair of the succinimide anion takes the oxonium proton.
- The O–H bond pair stays on oxygen, giving the neutral alcohol.
1-Bromo-2-methylbutan-2-ol
1-Bromo-2-methylbutan-2-ol forms with OH on the tertiary carbon and Br on the primary carbon. The new stereocentre is formed from either face of the alkene, so the bromohydrin is racemic; base would close it to 2-ethyl-2-methyloxirane.
What you'll practice
Draw NBS transferring bromine to 2-methylbut-1-ene as a bromonium ion, water opening it at the tertiary carbon, and the succinimide anion removing the proton to give 1-bromo-2-methylbutan-2-ol.
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