IChO 1993 · Theory problem 5

Asymmetric hydroboration of 1-methylcyclopentene

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By treatment with a chiral optically active hydroborating agent (\(\ce{R2^{*}BH}\)) and subsequent oxidative work-up, 1-methylcyclopentene affords a mixture of alcohols that rotates the plane of the polarized light.

Questions:

5.1 Write the structures of the alcohols.

5.2 Assign the configurations (\(R, S\)) to the chiral centres.

5.3 Explain in a sentence why the obtained mixture of alcohols exhibits optical activity.

Official English text of the problem.

Answers

  • 5.1 The two enantiomers of trans-2-methylcyclopentan-1-ol: \((1R,2R)\)-2-methylcyclopentan-1-ol, O[C@H]1[C@H](C)CCC1, and \((1S,2S)\)-2-methylcyclopentan-1-ol, O[C@@H]1[C@@H](C)CCC1.
  • 5.2 Numbering the hydroxyl-bearing carbon C1 and the methyl-bearing carbon C2, the configurations are \((1R,2R)\) and \((1S,2S)\).
  • 5.3 The chiral hydroborating agent preferentially attacks one enantiotopic face of the alkene, producing unequal amounts of the two enantiomers, so their opposite optical rotations do not cancel.

Solution

5.1

The starting alkene is 1-methylcyclopentene, CC1=CCCC1.

Hydroboration follows anti-Markovnikov regiochemistry: boron attaches to the less substituted alkene carbon, while hydrogen attaches to the carbon bearing the methyl group. The hydrogen and boron add syn, from the same face of the double bond.

Oxidative work-up replaces the carbon–boron bond with a carbon–oxygen bond with retention of spatial configuration at carbon. Therefore, the resulting hydroxyl group and the newly introduced hydrogen on the adjacent carbon occupy the same ring face. The methyl group occupies the opposite face, making the hydroxyl and methyl groups trans.

Addition from the two faces gives the two enantiomers listed in Answer 5.1.

To depict their structures precisely, draw a pentagon with the hydroxyl-bearing C1 at the top and number the remaining ring carbons clockwise, placing the methyl group on C2:

  • In one structure, draw the C1–\(\ce{OH}\) bond as a solid wedge toward the viewer and the C2–\(\ce{CH3}\) bond as a hashed wedge away.
  • In the other structure, reverse both wedges.

These drawings specify configuration; they do not imply that the cyclopentane ring is actually planar.

5.2

The CIP priorities are:

Stereocentre Priority 1 Priority 2 Priority 3 Priority 4
C1 Hydroxyl oxygen Ring path through C2 Ring path through C5 Hydrogen
C2 Ring path through C1 Ring path through C3 Methyl carbon Hydrogen

At C1, the two directly attached ring carbons are distinguished by their next attached atoms: C2 has \((\ce{C},\ce{C},\ce{H})\), whereas C5 has \((\ce{C},\ce{H},\ce{H})\). Thus, the path through C2 has higher priority.

At C2, the corresponding comparisons are:

\[ (\ce{O},\ce{C},\ce{H}) > (\ce{C},\ce{H},\ce{H}) > (\ce{H},\ce{H},\ce{H}), \]

giving the order C1 path \(>\) C3 path \(>\) methyl.

Applying these priorities to the specified drawings, the structure with hydroxyl toward the viewer and methyl away is \((1R,2R)\); reversing both wedges gives \((1S,2S)\).

The reagent’s detailed structure and absolute configuration are unspecified, so neither the predominant enantiomer nor the sign of optical rotation can be assigned.

5.3

The chiral reagent makes attack on the two enantiotopic alkene faces proceed through diastereomeric transition states of different energies, yielding unequal amounts of the two alcohol enantiomers whose optical rotations therefore do not cancel.

The solution was checked against the official solution.