IChO 2009 · Theory problem 5

Epoxy resins

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The synthesis of epoxy resins is a multi-billion dollar industry worldwide. Epoxy resins are high performance adhesives synthesised from the reaction of a bis-epoxide with a diamine. The bis-epoxide is made from H and epichlorohydrin, C. Compounds C and H can be synthesised according to the schemes below.

Figure 1
Figure 2

The synthesis of epichlorohydrin C begins with the reaction of propene with chlorine in the presence of light.

5.1 Draw the structures of A and B.

5.2 Give the formula of a suitable reagent for the conversion of B into epichlorohydrin C.

The synthesis of H commences with the reaction of benzene with propene in the presence of an acid catalyst which gives D as the major product and E and F as minor products.

5.3 Draw the structures of D, E, and F from the following data:

D: Elemental composition: C 89.94 %, H 10.06 %; 6 signals in the \(^{13}\mathrm{C}\) NMR spectrum.

E: Elemental composition: C 88.82 %, H 11.18 %; 4 signals in the \(^{13}\mathrm{C}\) NMR spectrum.

F: Elemental composition: C 88.82 %, H 11.18 %; 5 signals in the \(^{13}\mathrm{C}\) NMR spectrum.

Bubbling oxygen through a hot solution of D gives G which on exposure to acid gives phenol (hydroxybenzene) and acetone (propanone). G turns starch iodide paper from white to dark blue. G has 6 signals in the \(^{13}\mathrm{C}\) NMR spectrum and the following signals in the \(^{1}\mathrm{H}\) NMR spectrum: \(\delta\) 7.78 (1H, s), 7.45 – 7.22 (5H, m), 1.56 (6H, s); addition of \(\ce{D2O}\) results in the disappearance of the signal at \(\delta = 7.78\).

5.4 Draw the structure of G.

Exposure of phenol and acetone to hydrochloric acid gives compound H. The \(^{13}\mathrm{C}\) NMR spectrum for H is shown in Fig. 1. The \(^{1}\mathrm{H}\) NMR spectrum is shown in Fig. 2 together with a four-fold expansion of the region 6.5 – 7.1 ppm. The \(^{1}\mathrm{H}\) NMR spectrum after the addition of a drop of \(\ce{D2O}\), is shown in Fig. 3. Peaks due to the solvent are marked with an asterisk (*).

Figure 3
Figure 4
Figure 5

5.5 Draw the structure of H.

5.6 Draw one resonance structure of phenol which explains the regioselective formation of H.

A second compound, I, is also formed in the reaction of phenol with acetone. The \(^{13}\mathrm{C}\) NMR spectrum of I has 12 signals. The \(^{1}\mathrm{H}\) NMR spectrum has the following signals:

\(\delta\) 7.50 – 6.51 (8H, m), 5.19 (1H, s), 4.45 (1H, s), 1.67 (6H, s); addition of \(\ce{D2O}\) results in the disappearance of the signals at \(\delta = 5.19\) and 4.45

5.7 Draw a structure for I.

Excess phenol reacts with epichlorohydrin C in the presence of base to give compound L which has 6 signals in its \(^{13}\mathrm{C}\) NMR spectrum. If the reaction is stopped before completion compounds J and K can also be isolated. Compound L is formed from compound K and compound K is formed from compound J.

Figure 6

5.8 Draw the structures of J, K and L.

Treatment of H with a large excess of epichlorohydrin C and base gives a monomeric bis-epoxide M. M contains no chlorine atoms or OH groups.

5.9 Draw the structure of M.

Treatment of H with a small excess of epichlorohydrin and base gives N. N has the form: endgroup 1-[repeat unit]\(_n\)-endgroup 2 where \(n\) is approximately 10 – 15. N does not contain chlorine atoms and contains one OH group per repeat unit.

5.10 Draw the structure of N in the form indicated above:

(endgroup 1-[repeat unit]\(_n\)-endgroup 2).

5.11 Draw the repeat unit of the polymeric epoxy resin O formed from the reaction of the bis-epoxide M with ethane-1,2-diamine.

Official English text of the problem.

Answers

In the polymer structures below, [*] denotes an attachment point, not an actual atom. Define:

  • \(Q\): the 4,4′-isopropylidenediphenylene group, [*]c1ccc(C(C)(C)c2ccc([*])cc2)cc1. Each external attachment is para to the central \(\ce{C(CH3)2}\) group.
  • \(g\): the glycidyl (oxiran-2-ylmethyl) group, [*]CC1CO1, attached through its exocyclic methylene carbon.

The conditions do not specify absolute stereochemistry. SMILES for chiral products therefore leave configuration unspecified; racemic products or stereoisomeric mixtures are intended.

  • 5.1 — A: allyl chloride, 3-chloroprop-1-ene, C=CCCl. B: 1,3-dichloropropan-2-ol, OC(CCl)CCl.

  • 5.2 — A suitable reagent is sodium hydroxide, \(\ce{NaOH}\), [Na+].[OH-].

  • 5.3 — D: cumene (propan-2-ylbenzene), CC(C)c1ccccc1. E: 1,4-diisopropylbenzene, CC(C)c1ccc(C(C)C)cc1. F: 1,2-diisopropylbenzene, CC(C)c1ccccc1C(C)C.

  • 5.4 — G: cumene hydroperoxide, CC(C)(OO)c1ccccc1.

  • 5.5 — H: bisphenol A, 2,2-bis(4-hydroxyphenyl)propane, CC(C)(c1ccc(O)cc1)c1ccc(O)cc1.

  • 5.6 — The para-carbanionic, oxonium resonance contributor of phenol, [OH+]=C1C=C[CH-]C=C1: oxygen bears a positive charge, and the carbon para to oxygen bears a negative charge.

  • 5.7 — I: 2-(2-hydroxyphenyl)-2-(4-hydroxyphenyl)propane, the ortho/para isomer of bisphenol A, CC(C)(c1ccccc1O)c1ccc(O)cc1.

  • 5.8 — J: 1-chloro-3-phenoxypropan-2-ol, OC(CCl)COc1ccccc1. K: phenyl glycidyl ether, 2-(phenoxymethyl)oxirane, c1ccc(OCC2CO2)cc1. L: 1,3-diphenoxypropan-2-ol, OC(COc1ccccc1)COc1ccccc1.

  • 5.9 — M: bisphenol A diglycidyl ether (DGEBA), CC(C)(c1ccc(OCC2CO2)cc1)c1ccc(OCC2CO2)cc1.

  • 5.10 — N: the glycidyl-terminated bisphenol A–epichlorohydrin oligomer, \[ \underbrace{g\ce{-O-}Q\ce{-O-}}_{\text{endgroup 1}} \underbrace{\left[\ce{CH2-CH(OH)-CH2-O-}Q\ce{-O-}\right]_n}_{\text{repeat units}} \underbrace{g}_{\text{endgroup 2}}. \] Its repeat-fragment SMILES is [*]CC(O)COc1ccc(C(C)(C)c2ccc(O[*])cc2)cc1.

  • 5.11 — O: the ethane-1,2-diamine-cured DGEBA resin. Its opened-DGEBA bridge is [*]CC(O)COc1ccc(C(C)(C)c2ccc(OCC(O)C[*])cc2)cc1. Fully reacted diamine junctions are [*]N([*])CCN([*])[*], with each nitrogen attached to two such bridges. The initial linear repeat, before further reaction of the remaining N–H bonds, is [*]NCCNCC(O)COc1ccc(C(C)(C)c2ccc(OCC(O)C[*])cc2)cc1.

Solution

5.1 — Allylic chlorination and chlorohydrin formation

Propene, C=CC, undergoes allylic substitution with chlorine, ClCl, under the photochemical conditions intended in the scheme. Replacing an allylic hydrogen gives A, allyl chloride:

\[ \ce{CH2=CHCH3 + Cl2 ->[h\nu] CH2=CHCH2Cl + HCl}. \]

The coproduct is hydrogen chloride, Cl.

Hypochlorous acid, OCl, then adds across the double bond. The expected major chlorohydrin has its hydroxyl group on the more substituted alkene carbon:

\[ \ce{CH2=CHCH2Cl + HOCl -> ClCH2CH(OH)CH2Cl}. \]

Thus B has a central hydroxyl-bearing carbon and two identical chloromethyl groups.

5.2 — Formation of epichlorohydrin

Sodium hydroxide converts B into its alkoxide, 1,3-dichloropropan-2-olate, [O-]C(CCl)CCl. The oxygen attacks one adjacent chloromethyl carbon intramolecularly, displacing chloride, [Cl-], and forming the three-membered epoxide ring.

The product C is epichlorohydrin, 2-(chloromethyl)oxirane, ClCC1CO1. The balanced overall equation is

\[ \ce{C3H6Cl2O + NaOH -> C3H5ClO + NaCl + H2O}. \]

The other products are sodium chloride, [Na+].[Cl-], and water, O. The two equivalent chloromethyl groups of B lead to racemic C.

5.3 — Identification of the alkylbenzenes

Using standard atomic molar masses \(12.01\ \mathrm{g\,mol^{-1}}\) for carbon and \(1.008\ \mathrm{g\,mol^{-1}}\) for hydrogen, a \(100.00\ \mathrm{g}\) sample gives:

Compound Amount of carbon atoms Amount of hydrogen atoms H:C atomic ratio
D \(89.94/12.01=7.489\ \mathrm{mol}\) \(10.06/1.008=9.980\ \mathrm{mol}\) \(1.333\)
E, F \(88.82/12.01=7.396\ \mathrm{mol}\) \(11.18/1.008=11.09\ \mathrm{mol}\) \(1.500\)

Alkylating benzene, c1ccccc1, with \(k\) molecules of propene gives the molecular formula

\[ \ce{C_{6+3k}H_{6+6k}}. \]

Consequently,

\[ \frac{6+6k}{6+3k} = \begin{cases} 4/3 &\Rightarrow k=1,\\ 3/2 &\Rightarrow k=2. \end{cases} \]

Thus D is \(\ce{C9H12}\), whereas E and F are \(\ce{C12H18}\).

The acid catalyst is represented by a hydrogen ion, [H+]. Protonation of propene produces a secondary isopropyl electrophile, represented as the propan-2-yl cation, C[CH+]C. Electrophilic aromatic substitution therefore introduces isopropyl groups:

\[ \ce{C6H6 + CH2=CHCH3 ->[H+] C6H5CH(CH3)2}, \]

\[ \ce{C6H6 + 2 CH2=CHCH3 ->[H+] C6H4[CH(CH3)2]2}. \]

The second equation applies separately to the ortho and para products.

Their carbon equivalences identify the structures:

Compound Aromatic carbon environments Side-chain carbon environments Total
D: cumene 4: ipso, ortho, meta, para 2: methine and equivalent methyls 6
E: para-diisopropylbenzene 2: substituted and proton-bearing carbons 2 4
F: ortho-diisopropylbenzene 3: three equivalent pairs 2 5

The meta isomer would have four aromatic environments and six carbon signals overall, so it does not match E or F. The formulas also reproduce the reported elemental percentages to their stated precision.

5.4 — Cumene hydroperoxide

Oxidation of the benzylic position of cumene by dioxygen, O=O, gives G:

\[ \ce{C6H5CH(CH3)2 + O2 -> C6H5C(CH3)2OOH}. \]

The data support a hydroperoxide:

  • The five aromatic protons identify a monosubstituted phenyl group.
  • The \(6\mathrm{H}\) singlet at \(\delta=1.56\) corresponds to two equivalent methyl groups attached to a carbon without a hydrogen.
  • The exchangeable \(1\mathrm{H}\) singlet at \(\delta=7.78\) is the hydroperoxide proton.
  • Four aromatic carbon environments, one central carbon and one methyl environment give six \(^{13}\mathrm{C}\) signals.

A hydroperoxide oxidizes iodide to iodine, producing the dark-blue starch–iodine response. A balanced equation is

\[ \ce{C6H5C(CH3)2OOH + 2 I- + 2 H+ -> C6H5C(CH3)2OH + I2 + H2O}. \]

Here iodide is [I-], iodine is II, and the organic reduction product is cumyl alcohol, 2-phenylpropan-2-ol, CC(C)(O)c1ccccc1.

Acid-catalysed Hock rearrangement of G gives phenol, Oc1ccccc1, and acetone, CC(=O)C:

\[ \ce{C6H5C(CH3)2OOH ->[H+] C6H5OH + (CH3)2CO}. \]

5.5 — Bisphenol A

Two phenol molecules condense with one acetone molecule, eliminating one water molecule:

\[ \ce{2 C6H5OH + (CH3)2CO -> [HCl] (HOC6H4)2C(CH3)2 + H2O}. \]

For H, each hydroxyl group is para to the central isopropylidene group.

The spectra establish the symmetry and substitution pattern:

Approximate \(^{1}\mathrm{H}\) shift Integral Assignment
\(9.1\ \mathrm{ppm}\) 2H Two equivalent phenolic OH groups
\(7.0\ \mathrm{ppm}\) 4H One set of aromatic protons
\(6.65\ \mathrm{ppm}\) 4H The other set of aromatic protons
\(1.55\ \mathrm{ppm}\) 6H Two equivalent methyl groups

The two aromatic sets give the characteristic pattern of equivalent para-disubstituted rings. Addition of deuterium oxide, [2H]O[2H], removes the OH resonance through hydrogen–deuterium exchange.

Excluding the starred solvent peaks, the \(^{13}\mathrm{C}\) spectrum contains six signals: four aromatic environments, the central quaternary carbon, and the equivalent methyl carbons. This identifies bisphenol A.

5.6 — Resonance and regioselectivity

Use the contributor

[OH+]=C1C=C[CH-]C=C1.

To draw it precisely, number the hydroxyl-bearing ring carbon \(1\):

  • Draw a double bond from carbon \(1\) to oxygen; oxygen retains its hydrogen and bears a positive charge.
  • Draw ring double bonds between carbons \(2\)–\(3\) and \(5\)–\(6\).
  • Place a negative charge on carbon \(4\), para to oxygen; that carbon retains its hydrogen.

The electron flow is donation of an oxygen lone pair into the oxygen–carbon bond, accompanied by movement of ring \(\pi\) electrons toward the para carbon. This is a resonance contributor of phenol, not a separate reaction product.

It shows enhanced electron density at the para position. Related contributors enhance the ortho positions. Phenol therefore directs electrophilic substitution ortho and para; the less crowded para positions favour formation of H.

5.7 — The ortho/para bisphenol isomer

Compound I has two nonequivalent phenolic OH groups: their separate \(1\mathrm{H}\) singlets disappear after deuterium oxide addition. Its eight aromatic protons and equivalent six methyl protons are consistent with another bisphenol A isomer.

The carbon-signal count distinguishes the ortho/para structure:

\[ \underbrace{6}_{\text{ortho-disubstituted ring}} + \underbrace{4}_{\text{para-disubstituted ring}} + \underbrace{1}_{\text{central carbon}} + \underbrace{1}_{\text{equivalent methyls}} =12. \]

Thus draw one hydroxyl group ortho to the central carbon substituent and the other para to it.

The balanced condensation is

\[ \ce{2 C6H5OH + (CH3)2CO ->[HCl] C15H16O2 + H2O}, \]

where the \(\ce{C15H16O2}\) product has the connectivity specified for I.

5.8 — Chlorohydrin, glycidyl ether and diether

Base produces the phenoxide ion, [O-]c1ccccc1, which attacks the less substituted carbon of the epoxide ring of C. Ring opening and proton transfer give J:

\[ \ce{C6H5OH + C3H5ClO ->[base] C6H5OCH2CH(OH)CH2Cl}. \]

This product has formula \(\ce{C9H11ClO2}\), as required.

Base then promotes intramolecular displacement of chloride, forming the epoxide K:

\[ \ce{C9H11ClO2 + NaOH -> C9H10O2 + NaCl + H2O}. \]

Further phenoxide attack opens K at its terminal epoxide carbon:

\[ \ce{C9H10O2 + C6H5OH ->[base] C6H5OCH2CH(OH)CH2OC6H5}. \]

The resulting L has formula \(\ce{C15H16O3}\). Its two ends are identical, so its six carbon signals comprise:

\[ 4\ \text{phenyl environments} +1\ \text{equivalent methylene environment} +1\ \text{central hydroxymethine environment}. \]

5.9 — The monomeric bis-epoxide

Each phenolic group of H undergoes the chlorohydrin-formation/ring-closure sequence established in 5.8. With a large excess of epichlorohydrin, both ends become glycidyl ethers, giving M.

Draw a bisphenol A framework bearing an \(\ce{-O-CH2-}\) linkage to an oxirane ring at each para position. The oxirane oxygen bridges its adjacent methine and terminal methylene carbons.

The balanced overall reaction is

\[ \ce{C15H16O2 + 2 C3H5ClO + 2 NaOH -> C21H24O4 + 2 NaCl + 2 H2O}. \]

The four oxygen atoms in M are two aryl ether oxygens and two epoxide oxygens. There are no OH groups or chlorine atoms.

5.10 — The oligomer and its end groups

With a smaller excess of epichlorohydrin, phenolic groups also open glycidyl epoxide groups on other molecules. Each such chain-extension step creates the linkage

\[ \ce{-O-CH2-CH(OH)-CH2-O-}, \]

which contains exactly one secondary alcohol group.

The required structure is

\[ \boxed{ \underbrace{g\ce{-O-}Q\ce{-O-}}_{\text{endgroup 1}} \left[ \underbrace{\ce{CH2-CH(OH)-CH2-O-}Q\ce{-O-}}_{\text{repeat unit}} \right]_n \underbrace{g}_{\text{endgroup 2}} }. \]

The fragment SMILES, with attachment points, are:

Fragment SMILES
Endgroup 1: glycidylated bisphenol A ether end CC(C)(c1ccc(OCC2CO2)cc1)c1ccc(O[*])cc1
Repeat unit [*]CC(O)COc1ccc(C(C)(C)c2ccc(O[*])cc2)cc1
Endgroup 2: glycidyl group [*]CC1CO1

Join the atoms adjacent to matching attachment points, removing the dummy atoms. Both termini are epoxides. Each repeat contributes one OH group, and no chlorine remains. Setting \(n=0\) recovers M, which checks the end-group assignment.

The repeat formula is \(\ce{C18H20O3}\), and the complete oligomer formula is

\[ \ce{C_{18n+21}H_{20n+24}O_{3n+4}}. \]

Its balanced preparation can be written

\[ \begin{aligned} &(n+1)\ce{C15H16O2}+(n+2)\ce{C3H5ClO}+(n+2)\ce{NaOH}\\ &\quad\ce{->} \ce{C_{18n+21}H_{20n+24}O_{3n+4}} +(n+2)\ce{NaCl}+(n+2)\ce{H2O}. \end{aligned} \]

For \(n=10\)–15, the ideal incorporated epichlorohydrin:bisphenol A ratio is

\[ \frac{n+2}{n+1}=1.063\text{–}1.091\ \mathrm{mol\,mol^{-1}}, \]

consistent with a small molar excess of epichlorohydrin.

5.11 — Amine opening and crosslinking

Ethane-1,2-diamine, NCCN, attacks the less substituted carbon of each epoxide. The epoxide oxygen becomes an OH group, while the attacked terminal carbon forms a carbon–nitrogen bond. No small molecule is eliminated.

Define the doubly opened M bridge \(P\) as

\[ P= \ce{-CH2-CH(OH)-CH2-O-}Q \ce{-O-CH2-CH(OH)-CH2-}, \]

with fragment SMILES

[*]CC(O)COc1ccc(C(C)(C)c2ccc(OCC(O)C[*])cc2)cc1.

If each amino nitrogen reacts once, the linear addition repeat is

\[ \boxed{\left[\ce{-NH-CH2-CH2-NH-}P\right]_m}. \]

Its fragment SMILES is

[*]NCCNCC(O)COc1ccc(C(C)(C)c2ccc(OCC(O)C[*])cc2)cc1.

The corresponding repeat composition is obtained by direct addition:

\[ \ce{C21H24O4 + C2H8N2 -> C23H32N2O4} \qquad\text{(repeat-unit composition)}. \]

Each nitrogen still has one N–H bond. These secondary amines can open further epoxide rings, producing crosslinks. In the fully reacted resin, each nitrogen has three carbon bonds: one to the ethylene bridge and two to opened-epoxide methylene groups.

To draw that network motif precisely:

  1. Draw two nitrogen atoms joined by \(\ce{-CH2-CH2-}\).
  2. Give each nitrogen two additional bonds to \(\ce{-CH2-CH(OH)-CH2-O-}\) arms.
  3. Continue each arm through \(Q\), then through another \(\ce{-O-CH2-CH(OH)-CH2-}\) arm to a nitrogen elsewhere in the resin.

A representative four-attachment structural cell is

[*:1]N(CCN([*:2])CC(O)COc1ccc(C(C)(C)c2ccc(OCC(O)C[*:3])cc2)cc1)CC(O)COc3ccc(C(C)(C)c4ccc(OCC(O)C[*:4])cc4)cc3.

Attachments 1 and 2 terminate at nitrogens; attachments 3 and 4 terminate at methylene carbons. Connect nitrogen ends to methylene ends of other cells. This specifies the local covalent connectivity; the long-range crosslink arrangement need not be periodic.

A fully reacted cell contains two M molecules per diamine, because one diamine supplies four reactive N–H bonds:

\[ \ce{2 C21H24O4 + C2H8N2 -> C44H56N2O8} \qquad\text{(network-cell composition)}. \]

It contains four newly formed OH groups and no residual N–H bonds.

The answers were checked against the official answers published by the organisers.