HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Haloalkanes and HaloarenesConceptualNumerical
2-Bromo-3-methylbutane is heated in water, with no added base…
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2-Bromo-3-methylbutane is heated in water, with no added base. Only and E1 reactions take place, with water as the nucleophile and the solvent. The first carbocation is secondary. It can capture water or lose a proton directly, and it can also change by a 1,2-hydride shift into a more stable carbocation, which can again capture water or lose a proton. List all the different organic compounds that can form, counting a pair of enantiomers as two compounds. Give the number of alcohols, the number of alkenes and the total.
Need a reminder? The key ideaShow
is first order, , with half-life . A planar carbocation gives a racemic or partly racemic product. The enantiomeric excess is the difference of the two shares, and the observed rotation is the specific rotation times tube length times concentration times the excess. A carbocation can rearrange to a more stable one by a hydride shift.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Follow each cation to its products. Then remove repeats.
Step 1. The secondary cation. Water gives 3-methylbutan-2-ol. This carbon holds four different groups (OH, H, and an isopropyl group), so the product is a racemic mixture of two enantiomers. Loss of a proton from carbon 1 gives 3-methylbut-1-ene. Loss from carbon 3 gives 2-methylbut-2-ene.
Step 2. The hydride shift. The hydrogen on carbon 3 moves with its electron pair to carbon 2. A tertiary cation forms, .
Step 3. The tertiary cation. Water gives 2-methylbutan-2-ol, which is not chiral. Loss of a proton gives 2-methylbut-2-ene again, from the methyl or the side, and also 2-methylbut-1-ene.
Step 4. Count. Alcohols: the two enantiomers of 3-methylbutan-2-ol and 2-methylbutan-2-ol, so 3. Alkenes: 3-methylbut-1-ene, 2-methylbut-2-ene and 2-methylbut-1-ene, so 3. Total: .
Answer: 3 alcohols, 3 alkenes and 6 compounds in all.
Check. 2-Methylbut-2-ene comes from both cations, but it is one compound, so it is counted once.
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