IChO 1984 · Theory problem 5

Stereochemistry of the bromination-dehydrobromination of a monodeuterated alkene

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One of the diastereotopic methylene protons at the double bond of A was selectively substituted by deuterium. Bromination and subsequent dehydrobromation yields the

Figure 1

deuteriated product B and the non-deuteriated product C.

5.1 Which configuration follows for the monodeuteriated A from the given reaction products?

5.2 The solution of this question requires the formulation of the reaction and a short argumentation why only B and C are formed.

Official English text of the problem.

Answers

  • 5.1 Monodeuteriated A is (3Z)-4-deuterio-3-methylbut-3-en-1-ol, OCC/C(C)=C\[2H]. In the orientation shown, D replaces the lower-right hydrogen: D and the hydroxyethyl group lie on the same side of the double bond.
  • 5.2 Anti bromination gives a racemic vicinal dibromide. Subsequent anti elimination gives:
  • By loss of \(\ce{HBr}\): B, (3Z)-4-bromo-4-deuterio-3-methylbut-3-en-1-ol, OCC/C(C)=C(\Br)[2H].
  • By loss of \(\ce{DBr}\): C, (3E)-4-bromo-3-methylbut-3-en-1-ol, OCC/C(C)=C/Br.

Solution

5.1 Configuration of monodeuteriated A

The unlabelled starting material is 3-methylbut-3-en-1-ol, OCCC(C)=C. Number its chain from the alcohol end, so the double bond is between \(\ce{C3}\) and \(\ce{C4}\).

The stereochemical sequence established in part 5.2 requires the following arrangement in the labelled starting material:

Position in the supplied drawing Substituent
Upper left, on \(\ce{C3}\) \(\ce{CH3}\)
Upper right, on \(\ce{C3}\) \(\ce{CH2CH2OH}\)
Lower left, on \(\ce{C4}\) \(\ce{H}\)
Lower right, on \(\ce{C4}\) \(\ce{D}\)

For the \(E/Z\) assignment:

  • At \(\ce{C3}\), \(\ce{CH2CH2OH}\) has higher priority than \(\ce{CH3}\): after the initial carbon tie, the attached-atom lists are \((\ce{C},\ce{H},\ce{H})\) and \((\ce{H},\ce{H},\ce{H})\).
  • At \(\ce{C4}\), \(\ce{D}\) has higher priority than \(\ce{H}\), because the heavier isotope takes precedence.

The higher-priority groups are on the same side, so monodeuteriated A has the \(Z\) configuration.

5.2 Reaction sequence and stereochemical explanation

Bromination is anti addition, whereas the subsequent E2 elimination requires the departing bromine and the removed hydrogen isotope to be antiperiplanar.

Bromine, BrBr, adds according to the balanced equation

\[ \ce{HOCH2CH2C(CH3)=CHD + Br2 -> HOCH2CH2C(Br)(CH3)CH(Br)D}. \]

Addition from either face gives racemic 3,4-dibromo-4-deuterio-3-methylbutan-1-ol, consisting of:

Enantiomer Stereochemical SMILES
\((3R,4R)\) OCC[C@@](C)(Br)[C@H](Br)[2H]
\((3S,4S)\) OCC[C@](C)(Br)[C@@H](Br)[2H]

The bromine attached to \(\ce{C3}\) is eliminated together with either H or D from \(\ce{C4}\). The following equations explicitly display the eliminated acid for atom balance; under basic conditions, the base captures that acid.

Loss of hydrogen bromide, Br, gives B:

\[ \ce{HOCH2CH2C(Br)(CH3)CH(Br)D -> HOCH2CH2C(CH3)=C(Br)D + HBr}. \]

Loss of deuterium bromide, [2H]Br, gives C:

\[ \ce{HOCH2CH2C(Br)(CH3)CH(Br)D -> HOCH2CH2C(CH3)=CHBr + DBr}. \]

To follow the stereochemistry precisely, use a Newman projection looking from \(\ce{C3}\) toward \(\ce{C4}\). For one enantiomer, an accessible conformation has:

Carbon Substituents at clock-face positions
Front, \(\ce{C3}\) Br at 12 o’clock; methyl at 8; hydroxyethyl at 4
Rear, \(\ce{C4}\) Br at 6 o’clock; H at 10; D at 2

These positions describe a conformation; the SMILES above encode configuration, not a particular conformation.

Eliminating H: Rotate the rear group until H is at 6 o’clock, anti to the front bromine. The rear D then lies at 10 o’clock and the rear Br at 2. Elimination therefore gives, in the paper’s orientation, D on the lower left and Br on the lower right: exactly B. Its higher-priority groups, hydroxyethyl and Br, are on the same side, so B is \(Z\).

Eliminating D: Instead, rotate the rear group until D is at 6 o’clock. The rear Br then lies at 10 o’clock and H at 2. Elimination gives Br on the lower left and H on the lower right: exactly C. Hydroxyethyl and Br are on opposite sides, so C is \(E\).

In each elimination, the base removes the antiperiplanar H or D; the corresponding bond electrons form the new \(\ce{C3=C4}\) π bond while the \(\ce{C3-Br}\) bond breaks, releasing bromide ion, [Br-].

For elimination restoring the original double bond, there are two choices—remove H or remove D—and each fixes one product geometry. The mirror-image dibromide gives the same two achiral alkene products. Thus the sequence accounts for B and C. Putting D on the opposite side in the starting alkene would reverse both predicted product geometries, contradicting the supplied drawings.

The solution was checked against the official solution.