IChO 1992 · Theory problem 4

Geraniol from rose oil

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Rose oil is an essential oil obtained from the steam distillation of plant material from roses. It contains a number of terpenes, one of which is geraniol, \(\ce{C10H18O}\) (A). Upon oxidation, geraniol can either give a ten-carbon aldehyde or a ten-carbon carboxylic acid. Reaction with two moles of bromine gives a tetrabromide (\(\ce{C10H18OBr4}\)) (B). Geraniol reacts with \(\ce{HBr}\) to give two bromides of formula \(\ce{C10H17Br}\).

When geraniol is vigorously oxidized, three products are obtained:

Figure 1

4.1 Give the structure of geraniol A.

4.2 Give the structure of compound B.

4.3 Give the structures of the two bromides of formula \(\ce{C10H17Br}\).

4.4 Indicate which of the two bromides is formed in greater proportions.

Official English text of the problem.

Answers

The following are the intended answers for geraniol. The middle oxidation-product drawing contains a carbon-count discrepancy, explained below.

  • 4.1 Geraniol, \((2E)\)-3,7-dimethylocta-2,6-dien-1-ol:
    OC/C=C(C)/CCC=C(C)C \[ \ce{HOCH2-CH=C(CH3)-CH2-CH2-CH=C(CH3)2} \]

  • 4.2 B, 2,3,6,7-tetrabromo-3,7-dimethyloctan-1-ol:
    OCC(Br)C(Br)(C)CCC(Br)C(Br)(C)C
    This connectivity represents a mixture of stereoisomers. \[ \ce{HOCH2-CHBr-C(Br)(CH3)-CH2-CH2-CHBr-C(Br)(CH3)2} \]

  • 4.3 The two constitutional substitution products are:

  • Geranyl bromide, \((2E)\)-1-bromo-3,7-dimethylocta-2,6-diene:
    BrC/C=C(C)/CCC=C(C)C

  • Linalyl bromide, racemic 3-bromo-3,7-dimethylocta-1,6-diene:
    C=CC(Br)(C)CCC=C(C)C

  • 4.4 Linalyl bromide is the expected major product under the allylic-carbocation model intended by the question.

Solution

4.1 — Identification of geraniol

The degree of unsaturation is \[ \mathrm{DBE}=\frac{2(10)+2-18}{2}=2. \] Addition of two equivalents of bromine accounts for two carbon–carbon double bonds, using both degrees of unsaturation.

Oxidation to either an aldehyde or a carboxylic acid, retaining all ten carbons, identifies a primary alcohol, \(\ce{-CH2OH}\).

There is a problem with the printed degradation diagram:

Position in the figure Product actually drawn SMILES Carbon atoms
Left Propan-2-one, acetone CC(=O)C 3
Middle 3-Oxobutanoic acid, acetoacetic acid CC(=O)CC(=O)O 4
Right Ethanedioic acid, oxalic acid O=C(O)C(=O)O 2

These fragments contain \(3+4+2=9\) carbons. For the ordinary oxidative cleavage of ten-carbon geraniol, the middle product must instead be 4-oxopentanoic acid, levulinic acid, CC(=O)CCC(=O)O. Its structure contains an additional \(\ce{CH2}\) group.

Using that explicit correction, the fragments reconstruct the molecule:

  • Acetone identifies an alkene end of the form \(\ce{=C(CH3)2}\): an alkene carbon without hydrogen gives a ketone.
  • Oxalic acid identifies the alcohol-containing end \(\ce{HOCH2-CH=}\). The primary alcohol becomes one carboxyl group, while oxidative cleavage of the alkene carbon bearing hydrogen produces the other.
  • Levulinic acid, \(\ce{CH3COCH2CH2COOH}\), identifies the intervening portion \(\ce{=C(CH3)-CH2-CH2-CH=}\).

Joining these pieces gives \[ \boxed{\ce{HOCH2-CH=C(CH3)-CH2-CH2-CH=C(CH3)2}}. \]

The name geraniol specifies the \(E\) configuration at the C2=C3 double bond. Thus the \(\ce{CH2OH}\) group and the continuing carbon chain lie on opposite sides of that double bond. The C6=C7 bond has no \(E/Z\) configuration because C7 bears two identical methyl groups.

The mild oxidation products are:

  • Geranial, \((2E)\)-3,7-dimethylocta-2,6-dienal:
    O=C/C=C(C)/CCC=C(C)C
  • Geranic acid, \((2E)\)-3,7-dimethylocta-2,6-dienoic acid:
    O=C(O)/C=C(C)/CCC=C(C)C

Writing \(\ce{[O]}\) for one oxygen-atom equivalent supplied by the unspecified oxidant, the balanced equations are \[ \ce{C10H18O + [O] -> C10H16O + H2O}, \] \[ \ce{C10H18O + 2[O] -> C10H16O2 + H2O}. \] Here water has SMILES O.

The balanced vigorous-oxidation equation, with the corrected middle product, is \[ \ce{C10H18O + 8[O] -> CH3COCH3 + CH3COCH2CH2COOH + HOOC-COOH + H2O}. \]

4.2 — Formation of the tetrabromide

Bromine, BrBr, adds across each double bond, placing one bromine atom on each former alkene carbon: \[ \ce{C10H18O + 2Br2 -> C10H18OBr4}. \]

Consequently, \[ \boxed{\ce{B = HOCH2-CHBr-C(Br)(CH3)-CH2-CH2-CHBr-C(Br)(CH3)2}}. \]

Ordinary electrophilic bromination proceeds through bromonium ions, followed by attack of bromide ion, [Br-], from the opposite face. Addition is therefore anti.

The product has stereogenic centers at C2, C3 and C6; C7 is not stereogenic because it bears two methyl groups. One permitted stereoisomer of B is represented by
OC[C@H](Br)[C@@](Br)(C)CC[C@H](Br)C(Br)(C)C.

Its mirror image and the corresponding C6 epimers give four possible stereoisomers, arranged in two enantiomeric pairs. The problem does not specify their relative proportions.

4.3 — Formation of the two bromides

The product formula indicates substitution of the hydroxyl group by bromine, with loss of water. For either constitutional product, the balanced net equation with hydrogen bromide, [H]Br, is \[ \ce{C10H18O + HBr -> C10H17Br + H2O}. \]

Protonation gives protonated geraniol,
[OH2+]C/C=C(C)/CCC=C(C)C.

The C–O bonding electron pair then leaves with water, generating an allylic carbocation. Its two resonance contributors are represented by:

  • Geranyl-cation contributor, with charge at C1:
    [CH2+]/C=C(C)/CCC=C(C)C
  • Linalyl-cation contributor, with charge at C3:
    C=C[C+](C)CCC=C(C)C

These are two representations of the same delocalized ion. Moving the C2=C3 \(\pi\)-electron pair toward C1 forms C1=C2 and shifts the positive charge to C3.

A lone pair on bromide can attack either end of this allylic system:

  1. Attack at C1 gives geranyl bromide: \[ \ce{BrCH2-CH=C(CH3)-CH2-CH2-CH=C(CH3)2}. \]

  2. Attack at C3 gives linalyl bromide, with the first double bond shifted to C1=C2: \[ \ce{CH2=CH-C(Br)(CH3)-CH2-CH2-CH=C(CH3)2}. \]

Attack at the planar C3 center can occur from either face. In an achiral medium, linalyl bromide is therefore racemic; its two enantiomeric SMILES are
C=C[C@](Br)(C)CCC=C(C)C and
C=C[C@@](Br)(C)CCC=C(C)C.

4.4 — Major bromide

The contributor with positive charge at C3 places the charge on a tertiary carbon; the C1 contributor places it on a primary carbon. The tertiary contributor is better stabilized by the surrounding alkyl groups.

The intended qualitative argument therefore predicts preferential bromide capture at C3, giving linalyl bromide as the major product. No numerical product ratio can be obtained from the supplied information.

The solution was checked against the official solution.