CHEM 2444 Chapter Notes - Chapter 13: Steric Effects, Vinyl Chloride, Sodium Borohydride

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9 Sep 2016

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Or is always on the more highly substituted carbon because you always form the more highly substituted carbocation product: oxymercuration. Akene reacts as a lewis base, donating electrons to mercury, creating a carbon-mercury bond and breaking one of the oxygen-mercury bonds, giving a carbocation (more stable carbocation) If you do this reaction in the presence of water, the water attacks the carbocation, forms an oxonium ion, ending up with an alcohol product. Sodium borohydride (nabh4) reducing agent, introduced to mercuric alcohol product to get the markovnikov alcohol by replacing the mercury with a hydrogen. Mercury stabilizes carbocation; prevents rearrangement: hydroboration. Reacts via 4-center transition state that places b on the carbon with the least steric hindrance. In the second chemical step, borane is treated with sodium hydroxide and hydrogen peroxide to generate hydrogen peroxide anion. Boron goes on less-substituted carbon to minimize steric hindrance in the 4- centered-transition state. Enol goes under tautomerization to generate a ketone.

Related Questions

1) There are several reagents that can be used to effect addition to a double bond, including: acid and water, oxymercuration–demercuration reagents, and hydroboration–oxidation reagents. The transformation shown utilizes oxymercuration–demecuration conditions. The starting material is a 5 carbon chain where there is a double bond between carbons 1 and 2 and a methyl substituent on carbon 3. The first step reacts with mercury diacetate in THF and water. The second step reacts with sodium borohydride and hydroxide. The product is a 5 carbon chain where carbon 2 has a hydroxy group and carbon 3 has a methyl group. Inspect the final product and select all the reasons why oxymercuration–demercuration was chosen to effect the following transformation instead of the other reagents. Select the correct reasons.

a. Hydroboration–oxidation reagents yield the anti‑Markovnikov product of addition.

b. Oxymercuration–demercuration reagents prevent carbocation rearrangements.

c. The reaction requires the Markovnikov product without carbocation rearrangement.

d. Addition with acid and water as reagents avoids carbocation rearrangements.

e. Addition with acid and water as reagents allows carbocation rearrangements.

f. Hydroboration–oxidation reagents yield the Markovnikov product of addition.

g. The reaction requires the anti‑Markovnikov product with carbocation rearrangement.

h. Oxymercuration–demercuration reagents favor carbocation rearrangements.

2) Complete the mechanism for the reaction by adding curved arrows. Note, the acetate ion is a spectator throughout. Step 1: Draw curved arrows to form the mercurinium ion. Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase C H O Hg