IChO 1983 · Theory problem 6

Identification of compounds A-G from a base-catalysed reaction sequence

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One mole of compound A reacts successively with 3 moles of compound B in aqueous solution in the presence of a basic catalyst (such as \(\ce{Ca(OH)2}\)):

Figure 1

Hydrogenation of compound E yields compound F:

Figure 2

F has the composition: \(\ce{C} = 44.18\,\%,\quad \ce{H} = 8.82\,\%,\quad \ce{O} = 47.00\,\%.\)

Its molar mass: \(M = 136\ \mathrm{g\,mol^{-1}}\)

Knowing that 13.6 g of F reacts with 40.8 g acetic anhydride to form product G and acetic acid write down all chemical equations and assign the letters A, B, C, D, E, F, and G to particular formulas of compounds.

Official English text of the problem.

Answers

  • Problem 6 — no numbered subparts or point values are printed.
Compound Name and condensed structural formula SMILES
A Ethanal (acetaldehyde), \(\ce{CH3CHO}\) CC=O
B Methanal (formaldehyde), \(\ce{HCHO}\) C=O
C 3-Hydroxypropanal, \(\ce{HOCH2CH2CHO}\) O=CCCO
D 3-Hydroxy-2-(hydroxymethyl)propanal, \(\ce{(HOCH2)2CHCHO}\) O=CC(CO)CO
E 3-Hydroxy-2,2-bis(hydroxymethyl)propanal (pentaerythrose), \(\ce{(HOCH2)3CCHO}\) O=CC(CO)(CO)CO
F Pentaerythritol, \(\ce{C(CH2OH)4}\), molecular formula \(\ce{C5H12O4}\) C(CO)(CO)(CO)CO
G Pentaerythritol tetraacetate, \(\ce{C(CH2OCOCH3)4}\), molecular formula \(\ce{C13H20O8}\) C(COC(C)=O)(COC(C)=O)(COC(C)=O)COC(C)=O

All seven compounds are achiral. The balanced equations are developed below in the order of the reaction sequence.

Solution

Determining the molecular formula and hydroxyl content of F.

Write the molecular formula as \(\ce{C_xH_yO_z}\). Using atomic molar masses of \(12\), \(1\), and \(16\ \mathrm{g\,mol^{-1}}\), respectively,

\[ \begin{aligned} x&=\frac{0.4418(136\ \mathrm{g\,mol^{-1}})} {12\ \mathrm{g\,mol^{-1}}}=5.007\approx5,\\ y&=\frac{0.0882(136\ \mathrm{g\,mol^{-1}})} {1\ \mathrm{g\,mol^{-1}}}=11.995\approx12,\\ z&=\frac{0.4700(136\ \mathrm{g\,mol^{-1}})} {16\ \mathrm{g\,mol^{-1}}}=3.995\approx4. \end{aligned} \]

Thus F has molecular formula \(\boxed{\ce{C5H12O4}}\), whose calculated molar mass is \(136\ \mathrm{g\,mol^{-1}}\).

Acetic anhydride, CC(=O)OC(C)=O, has formula \(\ce{C4H6O3}\) and molar mass

\[ M=4(12)+6(1)+3(16)=102\ \mathrm{g\,mol^{-1}}. \]

The reacting amounts are therefore

\[ n(\mathbf F)=\frac{13.6\ \mathrm g}{136\ \mathrm{g\,mol^{-1}}} =0.100\ \mathrm{mol}, \]

\[ n(\text{acetic anhydride}) =\frac{40.8\ \mathrm g}{102\ \mathrm{g\,mol^{-1}}} =0.400\ \mathrm{mol}. \]

Hence one mole of F consumes four moles of acetic anhydride. Each alcohol hydroxyl group consumes one anhydride molecule during acetylation, producing one ester group and one molecule of acetic acid, CC(=O)O. Consequently, F contains four hydroxyl groups, accounting for all four oxygen atoms.

Identifying A and B.

The aqueous basic conditions and successive additions indicate crossed aldol additions. In these reactions, methanal supplies a hydroxymethyl group, \(\ce{-CH2OH}\), while the original aldehyde group remains available for the final hydrogenation.

Three additions of one-carbon methanal to a five-carbon final skeleton leave two carbons in A. The required two-carbon aldehyde is ethanal. Its methyl group has exactly three hydrogens on the carbon adjacent to the aldehyde group, allowing three successive hydroxymethylations.

Thus A is ethanal and B is methanal. The first three equations below are base-catalysed; calcium hydroxide, [Ca+2].[OH-].[OH-], is the example given in the problem.

First addition: A + B → C.

\[ \boxed{\ce{CH3CHO + HCHO -> HOCH2CH2CHO}} \]

One hydrogen on ethanal’s methyl carbon is replaced by a hydroxymethyl group. The product is C, 3-hydroxypropanal.

Second addition: C + B → D.

\[ \boxed{\ce{HOCH2CH2CHO + HCHO -> (HOCH2)2CHCHO}} \]

The carbon adjacent to the aldehyde group still has two hydrogens. A second hydroxymethylation gives D, 3-hydroxy-2-(hydroxymethyl)propanal.

Third addition: D + B → E.

\[ \boxed{\ce{(HOCH2)2CHCHO + HCHO -> (HOCH2)3CCHO}} \]

The last hydrogen on that carbon is replaced, giving E, pentaerythrose, with molecular formula \(\ce{C5H10O4}\). Its central carbon now bears three hydroxymethyl groups and one aldehyde group. It has no remaining hydrogen adjacent to the aldehyde group, so this sequence of aldol additions stops.

These are addition reactions: no small molecule is eliminated.

Hydrogenation: E → F.

Hydrogen, [H][H], reduces the remaining aldehyde group to a fourth hydroxymethyl group:

\[ \boxed{\ce{(HOCH2)3CCHO + H2 -> C(CH2OH)4}} \]

The product F is therefore pentaerythritol, a central carbon bonded to four identical \(\ce{-CH2OH}\) groups. This agrees with both its molecular formula and its four acetylatable hydroxyl groups.

The overall synthesis through hydrogenation is

\[ \ce{CH3CHO + 3 HCHO + H2 -> C(CH2OH)4}. \]

Acetylation: F → G.

All four hydroxyl groups undergo acetylation:

\[ \boxed{ \ce{C(CH2OH)4 + 4 (CH3CO)2O -> C(CH2OCOCH3)4 + 4 CH3COOH} } \]

Thus G is pentaerythritol tetraacetate. Each hydroxyl hydrogen is replaced by an acetyl group, giving a net addition of \(\ce{C2H2O}\) per hydroxyl group. Its molecular formula is consequently

\[ \ce{C5H12O4} + 4(\ce{C2H2O}) \quad\Rightarrow\quad \boxed{\ce{C13H20O8}}. \]

The solution was checked against the official solution.