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What products will be formed during an E1 elimination reaction of 2-bromo-2-methylpropane in the presence of ethanol as a solvent? Please provide the step by step mechanism for the reaction.

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An E1 elimination reaction involves the formation of a carbocation intermediate, followed by the removal of a proton from an adjacent carbon to form a double bond. In this case, the starting material is 2-bromo-2-methylpropane, and the solvent is ethanol. The products formed will be an alkene and a hydrogen bromide  HBr . Here's the step-by-step mechanism for the reaction:1. Ionization: The 2-bromo-2-methylpropane molecule ionizes, with the bromide ion  Br-  leaving as a leaving group. This results in the formation of a carbocation intermediate, specifically a tertiary carbocation, at the 2-position of the propane molecule.2-bromo-2-methylpropane  [2-methylpropyl]+ + Br-2. Proton removal: A molecule of ethanol  acting as a weak base  abstracts a proton from an adjacent carbon to the carbocation center. In this case, the only available carbon is the one next to the carbocation center  the 1-position . This results in the formation of a double bond between the 1- and 2-positions of the propane molecule, creating an alkene.[2-methylpropyl]+ + EtOH  2-methylpropene + EtOH2+3. Deprotonation: The ethanol molecule that gained a proton in the previous step loses it to a nearby ethanol molecule, regenerating the ethanol solvent and forming a hydrogen bromide  HBr  molecule.EtOH2+ + EtOH  2 EtOH + HBrThe final products of the E1 elimination reaction are 2-methylpropene  an alkene  and hydrogen bromide  HBr .
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