IChO 1977 · Theory problem 4

Hydrolysis of an optically active ester C6H12O2

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An optically active ester (11.6 g) having the summary formula \(\ce{C6H12O2}\), was hydrolysed by heating with an excess of aqueous sodium hydroxide solution. After terminating the hydrolysis the alkaline reaction mixture was several times extracted with ether. The aqueous solution was not optically active. The united ether extracts were dried with anhydrous magnesium sulphate. The ether solution was filtrated, ether was distilled off from it and the residue was redistilled. 7.4 g (100 %) of a liquid boiling at 100 °C was obtained.

Problems:

  1. Write the structural formula of the ester.
  2. What would be the structure of an ester with identical summary formula, i. e. \(\ce{C6H12O2}\), if the aqueous solution after the alkaline hydrolysis obtained in the above mentioned way, were optically active?
  3. Write down equations for the alkaline hydrolysis of both esters with sodium hydroxide solution.

Official English text of the problem.

Answers

The stereochemical SMILES below depict representative enantiomers. Replacing @ by @@ gives the corresponding opposite enantiomers, which are equally consistent with the information supplied.

  1. Butan-2-yl ethanoate (sec-butyl acetate), CC(=O)O[C@H](C)CC: \[ \ce{CH3C(=O)OCH(CH3)CH2CH3}. \]

  2. Methyl 2-methylbutanoate, COC(=O)[C@H](C)CC: \[ \ce{CH3CH2CH(CH3)C(=O)OCH3}. \]

  3. The balanced hydrolysis equations are: \[ \ce{CH3C(=O)OCH(CH3)CH2CH3 + NaOH -> CH3COONa + CH3CH(OH)CH2CH3} \] \[ \ce{CH3CH2CH(CH3)C(=O)OCH3 + NaOH -> CH3CH2CH(CH3)COONa + CH3OH}. \]

The first reaction produces sodium ethanoate, [Na+].CC(=O)[O-], and butan-2-ol, O[C@H](C)CC. The second produces sodium 2-methylbutanoate, [Na+].[O-]C(=O)[C@H](C)CC, and methanol, CO. The reagent is sodium hydroxide, [Na+].[OH-].

Solution

1. Identification of the original ester

Alkaline hydrolysis of an ester produces one mole of alcohol and one mole of sodium carboxylate per mole of ester. The neutral alcohol is recovered from the organic extracts, while the sodium carboxylate remains in the aqueous solution.

Using the usual integer atomic masses, the ester’s molar mass is \[ M_{\text{ester}} =\bigl[6(12)+12(1)+2(16)\bigr]\ \mathrm{g\,mol^{-1}} =116\ \mathrm{g\,mol^{-1}}. \] Thus, \[ n_{\text{ester}} =\frac{11.6\ \mathrm{g}}{116\ \mathrm{g\,mol^{-1}}} =0.100\ \mathrm{mol}. \]

Hydrolysis and alcohol recovery are quantitative, so \[ n_{\text{alcohol}}=0.100\ \mathrm{mol}, \qquad M_{\text{alcohol}} =\frac{7.4\ \mathrm{g}}{0.100\ \mathrm{mol}} =74\ \mathrm{g\,mol^{-1}}. \]

The ester has one degree of unsaturation: \[ \mathrm{DBE}=\frac{2(6)+2-12}{2}=1. \] Its carbonyl group accounts for this completely; consequently, the alcohol fragment is saturated and acyclic. A saturated monohydric alcohol has formula \(\ce{C_{k}H_{2k+2}O}\), giving \[ M_{\text{alcohol}}=(14k+18)\ \mathrm{g\,mol^{-1}}. \] Therefore, \[ k=\frac{74-18}{14}=4, \] so the alcohol has molecular formula \(\ce{C4H10O}\).

Four of the ester’s six carbon atoms belong to the alcohol fragment, leaving two in the acid fragment. The latter must therefore be the ethanoate group, which is achiral. The ester’s optical activity must arise from the four-carbon alcohol fragment.

Among the saturated four-carbon alcohols, only butan-2-ol, O[C@H](C)CC, is chiral. Its hydroxyl-bearing carbon has four different substituents: hydroxyl, hydrogen, methyl and ethyl. Its boiling point is consistent with the stated \(100\,^\circ\mathrm{C}\).

The ester is therefore butan-2-yl ethanoate, CC(=O)O[C@H](C)CC. Hydrolysis leaves achiral sodium ethanoate in the aqueous phase, explaining its optical inactivity.

2. Identification when the aqueous solution is optically active

After extraction of the alcohol, optical activity in the aqueous solution requires a chiral, nonracemic carboxylate.

The smallest saturated, acyclic monocarboxylic acid capable of having a stereogenic carbon is 2-methylbutanoic acid, O=C(O)[C@H](C)CC: \[ \ce{CH3CH2CH(CH3)COOH}. \] Its stereogenic carbon bears hydrogen, methyl, ethyl and carboxyl groups, all different. This acid fragment contains five carbon atoms.

Since the ester contains six carbon atoms altogether, only one remains for the alcohol-derived group. That group must be methyl. The required ester is consequently methyl 2-methylbutanoate, COC(=O)[C@H](C)CC.

Its hydrolysis produces optically active sodium 2-methylbutanoate in the aqueous phase and achiral methanol as the alcohol.

3. Hydrolysis equations and stereochemical outcome

On heating with aqueous sodium hydroxide, the first ester undergoes: \[ \boxed{\ce{ CH3C(=O)OCH(CH3)CH2CH3 + NaOH -> CH3COONa + CH3CH(OH)CH2CH3 }} \]

The ester’s acyl–oxygen bond is cleaved; the bond between oxygen and the stereogenic carbon remains intact. Thus, the stereogenic centre is retained in butan-2-ol.

For the second ester: \[ \boxed{\ce{ CH3CH2CH(CH3)C(=O)OCH3 + NaOH -> CH3CH2CH(CH3)COONa + CH3OH }} \]

Here the stereogenic centre remains in the acid-derived product, sodium 2-methylbutanoate. Both equations consume one mole of sodium hydroxide per mole of ester and are balanced in atoms and charge.

The solution was checked against the official solution.