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What is the stereochemistry resulting from hydroboration of 1-methylcyclopentene?
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Pál Váradi Nagy
What is the stereochemistry resulting from hydroboration of 1-methylcyclopentene?
The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.
The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.
Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.
The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.
The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.
Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.
The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.
The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.
Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.
The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.
The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.
Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.
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