IChO 1991 · Theory problem 2

Structure of a hydrocarbon polymer and its degradation products

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A polymer X contains 88.25 % \(\ce{C}\) and 11.75 % \(\ce{H}\). In dilute solutions it reacts with bromine and ozone. The thermal degradation of X produces a volatile liquid Y in 58 % yield. Y starts boiling at 34 °C and contains 88.25 % \(\ce{C}\) and 11.72 % \(\ce{H}\). Some higher boiling products are formed both from the cracking of X and from Diels-Alder type cyclization of Y. The vapour of Y was 34 times as dense as hydrogen.

The product of bromination of Y yields a compound containing 82.5 % bromine by mass. Ozonolysis of Y followed by mild reduction gave A and B in a molar ratio A : B = 2 : 1. Only compound B gave a positive result for the iodoform reaction.

2.1 Determine the molecular formula and the molar mass of Y.

2.2 Show the structures of compounds Y, A, and B.

2.3 Give an equation for the reaction of Y with bromine.

In the catalytic hydrogenation of 13.6 g of X, 0.2 mole of hydrogen was absorbed. Ozonolysis of X followed by mild reduction yielded compound Z (60.0 % \(\ce{C}\), 8.0 % \(\ce{H}\)).

2.4 Give the molecular formula of Z and the degree of unsaturation of X.

Compound Z gave a positive reaction with Fehling's solution. Mild oxidation of Z gave an acid C. A solution of acid C was titrated with aqueous \(\ce{KOH}\) solution (phenolphthalein as indicator). 0.001 mol \(\ce{KOH}\) was necessary for neutralization of 0.116 g of C.

In the iodoform reaction a probe 2.90 g of C yielded 9.85 g of iodoform. In addition, the alkaline filtrate yielded compound D upon acidification.

2.5 What is the molar mass of C and what functional groups are present in Z?

When heated, D loses water to form E. Both D and E react with an excess of acidified ethanol to F (\(\ce{C8H14O4}\)).

2.6 Sketch the structures of C, D, E, F and Z.

2.7 X exists in isomeric forms which are stereoregular. Show the structure (containing at least 3 monomer units) for two possible stereoisomers of X.

Official English text of the problem.

Answers

  • 2.1 Y: \(\boxed{\ce{C5H8}}\), with \(\boxed{M=68\ \mathrm{g\,mol^{-1}}}\).

  • 2.2 Y is isoprene (2-methylbuta-1,3-diene), C=C(C)C=C. A is methanal (formaldehyde), C=O. B is 2-oxopropanal (methylglyoxal), CC(=O)C=O.

  • 2.3 \[ \ce{CH2=C(CH3)-CH=CH2 + 2Br2 -> BrCH2-C(Br)(CH3)-CHBr-CH2Br} \] The product is 1,2,3,4-tetrabromo-2-methylbutane, BrCC(Br)(C)C(Br)CBr; its stereochemistry is not specified by the data.

  • 2.4 Z: \(\boxed{\ce{C5H8O2}}\). X contains \(\boxed{\text{one C=C bond per }\ce{C5H8}\text{ repeat unit}}\).

  • 2.5 \(\boxed{M(\mathbf C)=116\ \mathrm{g\,mol^{-1}}}\). Z contains one aldehyde group and one methyl ketone group.

  • 2.6

Compound Name SMILES
C 4-Oxopentanoic acid, or levulinic acid CC(=O)CCC(=O)O
D Butanedioic acid, or succinic acid O=C(O)CCC(=O)O
E Succinic anhydride O=C1CCC(=O)O1
F Diethyl butanedioate, or diethyl succinate CCOC(=O)CCC(=O)OCC
Z 4-Oxopentanal CC(=O)CCC=O
  • 2.7 X is 1,4-polyisoprene, with repeat unit \[ \boxed{\left[\ce{-CH2-C(CH3)=CH-CH2-}\right]_n}. \] Three-repeat fragments of the two stereoregular forms are:

cis-1,4-Polyisoprene: *C/C(C)=C\CC/C(C)=C\CC/C(C)=C\C*

trans-1,4-Polyisoprene: *C/C(C)=C/CC/C(C)=C/CC/C(C)=C/C*

Here * denotes continuation of the polymer chain. The backbone substituents are on the same side of every double bond in the cis form and on opposite sides in the trans form.

Solution

Calculations use the conventional rounded atomic masses \(12,\ 1,\ 16,\ 80,\ 127\) for carbon, hydrogen, oxygen, bromine and iodine, respectively.

2.1 — Molecular formula and molar mass of Y

At equal temperature and pressure, gas densities are proportional to molar masses. The reference is molecular hydrogen, \(\ce{H2}\), [H][H], so

\[ M(\mathbf Y)=34M(\ce{H2}) =34(2)=68\ \mathrm{g\,mol^{-1}}. \]

The elemental analysis gives

\[ n(\ce C):n(\ce H) =\frac{88.25}{12}:\frac{11.72}{1} =7.354:11.72 \simeq5:8. \]

Thus the empirical formula is \(\ce{C5H8}\), whose formula mass is

\[ 5(12)+8(1)=68. \]

It is therefore also the molecular formula:

\[ \boxed{\mathbf Y=\ce{C5H8},\qquad M(\mathbf Y)=68\ \mathrm{g\,mol^{-1}}}. \]

The quoted percentages total 99.97%; this small discrepancy does not change the formula.

2.2 — Identification of Y, A and B

The degree of unsaturation of Y is

\[ \mathrm{DBE}=\frac{2(5)+2-8}{2}=2. \]

Let \(x\) be the number of moles of bromine added per mole of Y. The bromine mass fraction of the addition product gives

\[ \frac{160x}{68+160x}=0.825, \]

hence \(x\simeq2\). Together with the Diels–Alder behavior and ozonolysis, this identifies Y as a diene.

Ozonolysis produces two molecules of A and one of B, containing five carbon atoms altogether. Because an iodoform-positive carbonyl compound needs at least two carbon atoms, A must contain one carbon and B three carbons.

The one-carbon carbonyl product is therefore methanal, C=O:

\[ \mathbf A=\ce{HCHO}. \]

Cleavage of two double bonds produces four carbonyl oxygen atoms in total. After accounting for two molecules of methanal, the remaining product has formula \(\ce{C3H4O2}\). Its positive iodoform reaction identifies a methyl ketone group. Consequently, B is 2-oxopropanal, CC(=O)C=O:

\[ \mathbf B=\ce{CH3-C(=O)-CHO}. \]

Reconnecting each of its carbonyl carbons to a terminal methylene group reconstructs isoprene, C=C(C)C=C:

\[ \boxed{\mathbf Y=\ce{CH2=C(CH3)-CH=CH2}}. \]

A balanced overall ozonolysis equation, choosing zinc as the mild reducing reagent, is

\[ \ce{CH2=C(CH3)-CH=CH2 + 2O3 + 2Zn -> 2HCHO + CH3-C(=O)-CHO + 2ZnO}. \]

This gives precisely the observed ratio \(\mathbf A:\mathbf B=2:1\).

2.3 — Bromination of Y

Excess bromine, BrBr, adds across both double bonds:

\[ \boxed{\ce{CH2=C(CH3)-CH=CH2 + 2Br2 -> BrCH2-C(Br)(CH3)-CHBr-CH2Br}}. \]

The product is 1,2,3,4-tetrabromo-2-methylbutane, BrCC(Br)(C)C(Br)CBr, with formula \(\ce{C5H8Br4}\). Its bromine content is

\[ w(\ce{Br}) =\frac{4(80)}{68+4(80)}\times100\% =82.47\% \simeq82.5\%. \]

The reported product therefore corresponds to complete addition of two equivalents of bromine.

2.4 — Formula of Z and unsaturation of X

The composition of X corresponds to \(\ce{C5H8}\) repeat units, each of molar mass \(68\ \mathrm{g\,mol^{-1}}\). Thus 13.6 g contains

\[ n(\text{repeat units}) =\frac{13.6\ \mathrm g}{68\ \mathrm{g\,mol^{-1}}} =0.200\ \mathrm{mol}. \]

This absorbs 0.2 mol of hydrogen. Since each carbon–carbon double bond consumes one molecule of hydrogen,

\[ \boxed{\text{X contains one C=C bond per }\ce{C5H8}\text{ repeat unit}}. \]

For Z, the oxygen percentage is

\[ 100.0-60.0-8.0=32.0\%. \]

Its atomic ratio is

\[ \frac{60.0}{12}:\frac{8.0}{1}:\frac{32.0}{16} =5:8:2, \]

giving empirical formula \(\ce{C5H8O2}\).

Cleavage of successive backbone double bonds produces fragments with two carbonyl ends. There is one such double bond per five-carbon repeat interval, so the fragment contains one empirical formula unit:

\[ \boxed{\mathbf Z=\ce{C5H8O2}},\qquad M(\mathbf Z)=100\ \mathrm{g\,mol^{-1}}. \]

These polymer calculations neglect end groups. The measured unsaturation is a per-repeat quantity; the total whole-chain DBE requires the chain length and end-group structures.

2.5 — Molar mass of C and functional groups in Z

Reductive ozonolysis gives carbonyl groups. The positive Fehling reaction shows that at least one of the two carbonyl groups in Z is an aldehyde.

Let \(q\) be the number of aldehyde groups oxidized to carboxyl groups. Each oxidation adds one oxygen atom without changing the carbon or hydrogen count, so

\[ M(\mathbf C)=100+16q. \]

The titration gives the acid’s equivalent mass:

\[ \frac{M(\mathbf C)}{q} =\frac{0.116\ \mathrm g}{0.001\ \mathrm{mol}} =116\ \mathrm{g\,mol^{-1}} \]

per mole of neutralized carboxyl groups. Therefore,

\[ 100+16q=116q, \qquad q=1. \]

Hence Z contains one aldehyde group and one ketone group, and

\[ \boxed{M(\mathbf C)=116\ \mathrm{g\,mol^{-1}}}, \qquad \mathbf C=\ce{C5H8O3}. \]

Iodoform, or triiodomethane, IC(I)I, has molar mass

\[ M(\ce{CHI3})=12+1+3(127)=394\ \mathrm{g\,mol^{-1}}. \]

The experimental amounts are

\[ n(\mathbf C) =\frac{2.90}{116} =0.0250\ \mathrm{mol}, \]

\[ n(\ce{CHI3}) =\frac{9.85}{394} =0.0250\ \mathrm{mol}. \]

Thus each molecule of C produces one molecule of iodoform, establishing one methyl ketone group. Mild oxidation preserves this ketone group, so Z contains

\[ \boxed{\text{one }\ce{-CHO}\text{ group and one }\ce{-C(=O)CH3}\text{ group}}. \]

2.6 — Structures of C, D, E, F and Z

The haloform reaction converts the methyl ketone group of C into a carboxyl group while removing its methyl carbon as iodoform. After acidification, D is therefore a four-carbon dicarboxylic acid.

Formation of the diethyl ester F, \(\ce{C8H14O4}\), is consistent with

\[ \mathbf D=\ce{C4H6O4}, \]

because replacing the two acidic hydrogens by ethyl groups adds \(\ce{C4H8}\) overall.

The loss of water on heating identifies D as succinic acid, O=C(O)CCC(=O)O:

\[ \mathbf D=\ce{HOOC-CH2-CH2-COOH}. \]

Its two carboxyl groups form a stable five-membered cyclic anhydride. The alternative four-carbon diacid, methylmalonic acid, CC(C(=O)O)C(=O)O, characteristically decarboxylates on heating instead.

Working backward from succinic acid gives:

\[ \boxed{\mathbf C=\ce{CH3-C(=O)-CH2-CH2-COOH}} \]

and

\[ \boxed{\mathbf Z=\ce{CH3-C(=O)-CH2-CH2-CHO}}. \]

Thus C is levulinic acid, CC(=O)CCC(=O)O, and Z is 4-oxopentanal, CC(=O)CCC=O.

The balanced haloform reaction in alkaline solution is

\[ \ce{CH3COCH2CH2COOH + 3I2 + 5OH- -> C4H4O4^2- + CHI3 + 3I- + 4H2O}. \]

The product \(\ce{C4H4O4^2-}\) is the succinate ion, O=C([O-])CCC(=O)[O-]. Acidification gives D:

\[ \ce{C4H4O4^2- + 2H+ -> HOOC-CH2-CH2-COOH}. \]

Heating gives succinic anhydride E, O=C1CCC(=O)O1:

\[ \ce{C4H6O4 ->[\Delta] C4H4O3 + H2O}. \]

To sketch E, draw a five-membered ring containing, consecutively, one oxygen atom, a carbonyl carbon, two methylene carbons and a second carbonyl carbon, closing back to the ring oxygen.

Both D and E give diethyl succinate F, CCOC(=O)CCC(=O)OCC, with excess acidified ethanol, CCO:

\[ \ce{C4H6O4 + 2C2H5OH <=>[H+] C8H14O4 + 2H2O}, \]

\[ \ce{C4H4O3 + 2C2H5OH <=>[H+] C8H14O4 + H2O}. \]

The structure of F is

\[ \boxed{\ce{C2H5O-C(=O)-CH2-CH2-C(=O)-OC2H5}}. \]

2.7 — Two stereoregular structures of X

The ozonolysis product

\[ \ce{OHC-CH2-CH2-C(=O)-CH3} \]

shows that successive double bonds in X are separated by two methylene groups. One alkene carbon bears hydrogen, and the other bears methyl. Reconnecting successive carbonyl ends therefore gives 1,4-polyisoprene:

\[ n\,\ce{CH2=C(CH3)-CH=CH2} \longrightarrow \left[\ce{-CH2-C(CH3)=CH-CH2-}\right]_n. \]

Its two stereoregular forms are:

  • cis-1,4-Polyisoprene, the principal polymer of natural rubber:
    *C/C(C)=C\CC/C(C)=C\CC/C(C)=C\C*

At every double bond, draw the two backbone methylene substituents on the same side. These double bonds have \(Z\) configuration in the continuing polymer chain.

  • trans-1,4-Polyisoprene, the principal polymer of gutta-percha:
    *C/C(C)=C/CC/C(C)=C/CC/C(C)=C/C*

At every double bond, draw the two backbone methylene substituents on opposite sides. These double bonds have \(E\) configuration.

Each SMILES contains three complete five-carbon repeat units. The asterisks denote chain continuation; the SMILES specify double-bond configuration, without imposing a particular three-dimensional chain conformation.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.