IChO 1992 · Theory problem 3

Coniferyl alcohol: structure and stereoisomers of its dibromide

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Coniferyl alcohol has the molecular formula \(\ce{C10H12O3}\). It is isolated from pine trees. Coniferyl alcohol is not soluble in water or aqueous \(\ce{NaHCO3}\). A solution of \(\ce{Br2}\) in \(\ce{CCl4}\) is decolorized when added to coniferyl alcohol forming A (\(\ce{C10H12O3Br2}\)). Upon reductive ozonolysis coniferyl alcohol produces vanillin (4-hydroxy-3-methoxybenzaldehyd) and B (\(\ce{C2H4O2}\)). Coniferyl alcohol reacts with benzoyl chloride (\(\ce{C6H5COCl}\)) in the presence of a base to form C (\(\ce{C24H20O5}\)). This product rapidly decolorizes \(\ce{KMnO4}\) (aq) and is insoluble in dilute \(\ce{NaOH}\).

Coniferyl alcohol reacts with cold \(\ce{HBr}\) to form D (\(\ce{C10H11O2Br}\)). Hot \(\ce{HI}\) converts \(\ce{ArOR}\) to \(\ce{ArOH}\) and \(\ce{RI}\). Coniferyl alcohol reacts with excess hot \(\ce{HI}\) to give E (\(\ce{C9H9O2I}\)) and \(\ce{CH3I}\). \(\ce{CH3I}\) in aqueous base reacts with coniferyl alcohol to form F (\(\ce{C11H14O3}\)), which is not soluble in a strong base, but decolorizes \(\ce{Br2}/\ce{CCl4}\)-solution.

3.1 Draw the structures of coniferyl alcohol and compounds A - F.

3.2 There are a number of stereoisomers of compound A. Draw structure of compound A. Label each chiral centre in compound A with an asterisk (*). For all stereoisomers draw Fischer projections and label each chiral center with the proper \(R\) or \(S\) designation giving the absolute configuration about the chiral centre.

Official English text of the problem.

Answers

  • 3.1. The structures are as follows. Alkene geometry is left unspecified because the stated experiments do not determine it.
Compound Name SMILES
Coniferyl alcohol 3-(4-Hydroxy-3-methoxyphenyl)prop-2-en-1-ol OCC=Cc1cc(OC)c(O)cc1
A 2,3-Dibromo-3-(4-hydroxy-3-methoxyphenyl)propan-1-ol OCC(Br)C(Br)c1cc(OC)c(O)cc1
B 2-Hydroxyethanal, also called glycolaldehyde O=CCO
C Coniferyl alcohol dibenzoate: 3-[4-(benzoyloxy)-3-methoxyphenyl]prop-2-en-1-yl benzoate O=C(OCC=Cc1cc(OC)c(OC(=O)c2ccccc2)cc1)c1ccccc1
D 4-(3-Bromoprop-1-en-1-yl)-2-methoxyphenol BrCC=Cc1cc(OC)c(O)cc1
E 4-(3-Iodoprop-1-en-1-yl)benzene-1,2-diol ICC=Cc1cc(O)c(O)cc1
F 3-(3,4-Dimethoxyphenyl)prop-2-en-1-ol OCC=Cc1cc(OC)c(OC)cc1
  • 3.2. Number the side chain from the alcohol end: C1 is the hydroxymethyl carbon, C2 is the middle carbon, and C3 is attached to the aromatic ring. C2 and C3 are the two chiral centres. There are four stereoisomers and no meso form.

The following table precisely specifies their Fischer projections. Put the 4-hydroxy-3-methoxyphenyl group at the top, then C3, then C2, and finally \(\ce{CH2OH}\) at the bottom. Horizontal bonds point toward the viewer; vertical bonds point away.

Stereoisomer of A C3: left / right C2: left / right Isomeric SMILES
\((2R,3R)\) \(\ce{Br}\) / \(\ce{H}\) \(\ce{H}\) / \(\ce{Br}\) OC[C@@H](Br)[C@H](Br)c1cc(OC)c(O)cc1
\((2S,3S)\) \(\ce{H}\) / \(\ce{Br}\) \(\ce{Br}\) / \(\ce{H}\) OC[C@H](Br)[C@@H](Br)c1cc(OC)c(O)cc1
\((2R,3S)\) \(\ce{H}\) / \(\ce{Br}\) \(\ce{H}\) / \(\ce{Br}\) OC[C@@H](Br)[C@@H](Br)c1cc(OC)c(O)cc1
\((2S,3R)\) \(\ce{Br}\) / \(\ce{H}\) \(\ce{Br}\) / \(\ce{H}\) OC[C@H](Br)[C@H](Br)c1cc(OC)c(O)cc1

Solution

3.1. Identification of coniferyl alcohol and A–F

The degree of unsaturation of coniferyl alcohol is

\[ \mathrm{DBE}=\frac{2(10)+2-12}{2}=5. \]

A benzene ring accounts for four degrees of unsaturation, leaving one additional double bond.

Vanillin is 4-hydroxy-3-methoxybenzaldehyde, O=Cc1cc(OC)c(O)cc1. Its formation by reductive ozonolysis establishes an aromatic fragment bearing hydroxy and methoxy groups at positions 4 and 3 relative to the side chain. The original side chain therefore contains an \(\ce{Ar-CH=}\) unit, where \(\mathrm{Ar}\) denotes the 4-hydroxy-3-methoxyphenyl group.

Benzoylation provides the other essential constraint. Replacing an \(\ce{O-H}\) hydrogen by a benzoyl group adds the net composition \(\ce{C7H4O}\). The observed formula change is

\[ \Delta=\ce{C14H8O2}=2\times\ce{C7H4O}. \]

Thus two hydroxyl groups are benzoylated. One is the phenolic hydroxyl identified through vanillin; the other must be an alcoholic hydroxyl. Together with the two-carbon ozonolysis fragment, this establishes the parent structure:

\[ \boxed{\ce{Ar-CH=CH-CH2OH}}. \]

The lack of bicarbonate solubility is consistent with a phenol, whose acidity is insufficient for the usual bicarbonate reaction of carboxylic acids.

Formation of A: bromine addition.

Bromine adds across the side-chain double bond:

\[ \ce{Ar-CH=CH-CH2OH + Br2 ->[CCl4] Ar-CHBr-CHBr-CH2OH}. \]

This gives A, retaining all twelve hydrogen atoms and adding two bromine atoms, exactly as required by \(\ce{C10H12O3Br2}\).

Formation of B: reductive ozonolysis.

Cleavage of the double bond gives vanillin from the aromatic end and glycolaldehyde from the alcohol end:

\[ \ce{Ar-CH=CH-CH2OH ->[reductive\ ozonolysis] Ar-CHO + OHC-CH2OH}. \]

For an explicitly balanced net equation, dimethyl sulfide, CSC, can be chosen as the reductive work-up reagent; its oxidation product is dimethyl sulfoxide, CS(C)=O:

\[ \ce{Ar-CH=CH-CH2OH + O3 + (CH3)2S -> Ar-CHO + OHC-CH2OH + (CH3)2SO}. \]

The second organic product is therefore B, glycolaldehyde, \(\ce{C2H4O2}\).

Formation of C: benzoylation of both hydroxyl groups.

The phenolic and primary-alcohol hydroxyl groups both form benzoate esters. Writing sodium hydroxide as the base gives the balanced equation

\[ \ce{C10H12O3 + 2 C6H5COCl + 2 NaOH -> C24H20O5 + 2 NaCl + 2 H2O}. \]

In C, the ring substituent \(\ce{-OH}\) has become \(\ce{-OCOC6H5}\), and the side-chain terminus \(\ce{-CH2OH}\) has become \(\ce{-CH2OCOC6H5}\). The methoxy group and alkene remain intact.

The absence of a free phenolic hydroxyl explains insolubility in dilute sodium hydroxide. The retained alkene explains rapid permanganate decolorization. Under the usual mild alkaline test conditions, a balanced representative reaction is

\[ \ce{3 C24H20O5 + 2 KMnO4 + 4 H2O -> 3 C24H22O7 + 2 MnO2 + 2 KOH}. \]

Here the organic product is 2,3-dihydroxy-3-[4-(benzoyloxy)-3-methoxyphenyl]propyl benzoate, O=C(OCC(O)C(O)c1cc(OC)c(OC(=O)c2ccccc2)cc1)c1ccccc1. The problem gives only the decolorization observation, so a particular isolated oxidation product is not needed for identifying C.

Formation of D: replacement of the alcoholic hydroxyl by bromine.

The formula change corresponds to loss of \(\ce{OH}\) and introduction of \(\ce{Br}\):

\[ \ce{Ar-CH=CH-CH2OH + HBr ->[cold] Ar-CH=CH-CH2Br + H2O}. \]

Thus D is the allylic bromide. Simple addition of hydrogen bromide across the alkene would retain all three oxygen atoms and would not match the stated formula.

Formation of E: ether cleavage and alcohol substitution.

Excess hot hydrogen iodide causes two transformations:

  1. The aromatic methoxy group becomes a hydroxyl group, releasing iodomethane (methyl iodide), CI.
  2. The side-chain alcoholic hydroxyl is replaced by iodine.

The balanced overall equation is

\[ \ce{C10H12O3 + 2 HI -> C9H9O2I + CH3I + H2O}. \]

Hence E has adjacent phenolic hydroxyl groups and the side chain \(\ce{-CH=CH-CH2I}\). Its formula requires retention of the alkene.

Formation of F: methylation of the phenolic hydroxyl.

In aqueous base, the phenolic hydroxyl is preferentially deprotonated and methylated. The balanced net equation is

\[ \ce{C10H12O3 + CH3I + NaOH -> C11H14O3 + NaI + H2O}. \]

The product F therefore has two aromatic methoxy groups and retains its primary-alcohol group. With no free phenolic hydroxyl, it does not acquire the usual phenolate-mediated solubility in aqueous strong base.

Its remaining double bond consumes bromine:

\[ \ce{C11H14O3 + Br2 -> C11H14O3Br2}. \]

The addition product is 2,3-dibromo-3-(3,4-dimethoxyphenyl)propan-1-ol, OCC(Br)C(Br)c1cc(OC)c(OC)cc1.

3.2. Chiral centres, Fischer projections, and absolute configurations

The starred carbon atoms in A are

\[ \ce{HOCH2-} \underset{\mathrm{C2}^{*}}{\ce{CH(Br)}} \ce{-} \underset{\mathrm{C3}^{*}}{\ce{CH(Br)}} \ce{-Ar}. \]

Each has four different substituents. The hydroxymethyl carbon is not chiral because it bears two hydrogen atoms.

The Cahn–Ingold–Prelog priorities are:

  • At C2: \[ \ce{Br} > \ce{CH(Br)-Ar} > \ce{CH2OH} > \ce{H}. \] The two carbon ligands are distinguished by their next attached atoms: \((\ce{Br},\ce{C},\ce{H})\) outranks \((\ce{O},\ce{H},\ce{H})\).

  • At C3: \[ \ce{Br} > \ce{CH(Br)-CH2OH} > \ce{Ar} > \ce{H}. \] The side-chain carbon has bromine in its next comparison sphere and therefore outranks the aromatic carbon.

These priorities give the Fischer projections specified in the Answers table. In every listed projection, hydrogen lies on a horizontal bond and points toward the viewer, so the apparent clockwise or counterclockwise \(1\rightarrow2\rightarrow3\) order must be reversed.

There are

\[ 2^2=\boxed{4} \]

configurational stereoisomers. The different aromatic and hydroxymethyl ends prevent an internal symmetry element from producing a meso form. The enantiomeric pairs are

\[ (2R,3R)\longleftrightarrow(2S,3S), \qquad (2R,3S)\longleftrightarrow(2S,3R). \]

Members of different pairs are diastereomers. Anti bromination of a single alkene geometrical isomer ordinarily produces one enantiomeric pair; the question asks for all stereoisomers of the dibromide.

The solution was checked against the official solution.