IChO 2018 · Theory problem 6

Let’s go mushrooming

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Mushrooming belongs to Czech and Slovak traditional pastimes. While some of our mushroom species are edible, some are inedible or even poisonous.

Inky cap (Coprinopsis atramentaria) is considered edible and delicious. It contains a natural compound called coprine (E), which can be easily synthesized from ethyl 3-chloropropanoate (1).

Figure 1

6.1 Draw the formulae of compounds A–E including stereochemistry when necessary.

Hint: The first reaction affording compound A proceeds via an organometallic compound which then cyclizes.

In the human body, coprine undergoes hydrolysis to L-glutamic acid (3) and compounds C and 4, which are responsible for the coprine adverse side-effects. They inhibit the enzyme acetaldehyde dehydrogenase, which is involved in the metabolism of alcohol. When the enzyme is inhibited, acetaldehyde formed by alcohol dehydrogenase accumulates in the body, causing strong symptoms of hangover (so called antabuse effect). The active site of the enzyme contains a cysteine \(\ce{SH}\) group, which is blocked either by compound C or 4.

Figure 2

Enzyme = acetaldehyde dehydrogenase

6.2 Using the pictogram for acetaldehyde dehydrogenase above, draw the structure F of the enzyme inhibited by compound 4.

The antabuse effect got its name after antabuse (5), the most known drug used in alcohol-addiction treatment. This drug can be synthesized according to the following scheme.

Figure 3

6.3 Draw the formulae of compounds G and H.

Hint: Compound H contains five carbon atoms.

6.4 Mark all possible reagents which could be used for I from the following list.

☐ m-chloroperbenzoic acid (mCPBA) ☐ diluted \(\ce{H2O2}\)
☐ \(\ce{Zn / CH3COOH}\) ☐ \(\ce{NaBH4}\)
☐ \(\ce{I2}\) ☐ hot concentrated \(\ce{H2SO4}\)
☐ \(\ce{K2CO3}, \ce{H2O}\) ☐ \(\ce{AlCl3}\)

The way antabuse inhibits acetaldehyde dehydrogenase is similar to the effect of compounds C and 4.

Figure 4

Enzyme = acetaldehyde dehydrogenase

6.5 Using the pictogram for acetaldehyde dehydrogenase above, draw the structure J of the enzyme inhibited by antabuse (5). Hint: Three sulfur atoms should be in the structure.

False morel (Gyromitra esculenta) is another interesting mushroom. Although it was considered edible in the past (esculentus means edible in Latin), there is clear evidence that this mushroom is poisonous due to the content of gyromitrin (M). This natural compound can be prepared from N-methylhydrazine (6):

Figure 5

6.6 Draw the formulae of compounds K – M.

In human body, gyromitrin (M) hydrolyzes and provides N-methylhydrazine (6), which is strongly hepatotoxic. Gyromitrin (M) hydrolysis occurs as soon as it enters the acidic environment in human stomach where both its amide and imine groups are hydrolyzed.

Let us focus on the hydrolysis of the amide moiety within the gyromitrin molecule. The vibrational wavenumber of the stretching mode of the relevant \(\ce{C-N}\) bond amounts to \(1\,293.0\ \mathrm{cm^{-1}}\) and the potential energy surface does not significantly alter its shape with isotope substitution effect.

6.7 Calculate the highest possible hypothetical kinetic isotope effect at the temperature of human body, \(37\,^\circ\mathrm{C}\), for the given hydrolysis reaction assuming that both relevant nitrogen and carbon atoms were simultaneously substituted, \(\ce{^{14}N}\) with the \(\ce{^{15}N}\) isotope and \(\ce{^{12}C}\) with the \(\ce{^{13}C}\) isotope. Consider that only the zero point vibrational energy affects the rate constants. Assume that the molar masses of all isotopes are integers. In all steps consider five significant digits.

6.8 After making these isotopic changes, the rates of hydrolysis are not significantly different. Which of the following is the most likely the rate determining step?

  • ☐ Nucleophilic attack of water on a protonated amidic moiety
  • ☐ \(\ce{C-N}\) bond cleavage
  • ☐ Protonation of the gyromitrin molecule

Official English text of the problem.

Answers

In the enzyme-adduct SMILES, * represents the rest of the enzyme attached to the active-site cysteine sulfur.

  • 6.1
Compound Identification SMILES
A 1-Ethoxy-1-(trimethylsilyloxy)cyclopropane CCOC1(O[Si](C)(C)C)CC1
B 1-Ethoxycyclopropan-1-ol CCOC1(O)CC1
C 1-Aminocyclopropan-1-ol NC1(O)CC1
D \(N^\alpha\)-Boc-protected coprine α-tert-butyl ester; amino-acid centre \(S\) CC(C)(C)OC(=O)N[C@@H](CCC(=O)NC1(O)CC1)C(=O)OC(C)(C)C
E Coprine, \(N^5\)-(1-hydroxycyclopropyl)-L-glutamine; amino-acid centre \(S\) N[C@@H](CCC(=O)NC1(O)CC1)C(=O)O

  • 6.2 F is the enzyme-bound cyclopropanone hemithioketal, *SC1(O)CC1. The same cyclopropane carbon bears \(\ce{OH}\) and \(\ce{S-Enzyme}\).

  • 6.3 G is carbon disulfide, S=C=S. H is sodium \(N,N\)-diethyldithiocarbamate, CCN(CC)C(=S)[S-].[Na+].

  • 6.4 Select:

  • m-chloroperbenzoic acid, O=C(OO)c1cccc(Cl)c1;
  • diluted hydrogen peroxide, OO;
  • iodine, II.

Do not select hot concentrated sulfuric acid: it promotes decomposition of the dithiocarbamate.

  • 6.5 J is the enzyme–diethyldithiocarbamate mixed disulfide, *SSC(=S)N(CC)CC. Draw \(\ce{Enzyme-S-S-C(=S)-N(Et)2}\).

  • 6.6

Compound Identification SMILES
K Acetic formic anhydride, the mixed anhydride of acetic and formic acids CC(=O)OC=O
L \(N\)-Amino-\(N\)-methylformamide CN(N)C=O
M Gyromitrin, acetaldehyde \(N\)-methyl-\(N\)-formylhydrazone CC=NN(C)C=O

The supplied scheme does not specify the hydrazone’s \(E/Z\) geometry.

  • 6.7 \[ \boxed{\left(\frac{k_{\mathrm{light}}}{k_{\mathrm{heavy}}}\right)_{\max} \approx 1.12} \] Five-significant-digit intermediate rounding gives 1.1165; retaining guard digits gives 1.1166.

  • 6.8 Select nucleophilic attack of water on a protonated amidic moiety.

Solution

Throughout the equations, \(\mathrm{Et}\) denotes ethyl and \(\mathrm{Enz}\) denotes the enzyme residue bonded to its active-site sulfur. Neutral organic structures are used for the hydrolysis and deprotection balances; protonation in acidic solution does not change the identified connectivity.

6.1 — Synthesis of coprine

Formation of A. Starting material 1 is ethyl 3-chloropropanoate, CCOC(=O)CCCl. Sodium generates its organosodium derivative:

\[ \ce{ClCH2CH2COOEt + 2 Na -> NaCH2CH2COOEt + NaCl}. \]

The intermediate, ethyl 3-sodiopropanoate, has the ionic SMILES CCOC(=O)C[CH2-].[Na+]. Its terminal carbanion attacks the ester carbonyl intramolecularly. The carbonyl \(\pi\) electrons move onto oxygen, producing a three-membered ring: sodium 1-ethoxycyclopropan-1-olate, CCOC1([O-])CC1.[Na+].

Chlorotrimethylsilane, C[Si](C)(C)Cl, traps this alkoxide at oxygen, giving A. The overall balance is

\[ \ce{ClCH2CH2COOEt + 2 Na + (CH3)3SiCl -> C8H18O2Si + 2 NaCl}. \]

Sodium chloride is [Na+].[Cl-]. The structure of A contains the required eight carbons: three in the ring, two in ethoxy, and three attached to silicon.

Formation of B. Fluoride, [F-], attacks silicon and removes the trimethylsilyl protecting group. Methanol, CO, supplies the proton for the liberated oxygen. A balanced net ionic representation is

\[ \ce{C8H18O2Si + F^- + CH3OH -> C5H10O2 + (CH3)3SiF + CH3O^-}. \]

The silicon-containing product is fluorotrimethylsilane, C[Si](C)(C)F; methoxide is C[O-]. Thus B is the ethyl hemiketal of cyclopropanone, with \(\ce{OH}\) and \(\ce{OEt}\) on the same ring carbon.

Formation of C. Sodium amide, [Na+].[NH2-], in liquid ammonia, N, replaces the ethoxy group by an amino group. The formal overall balance is

\[ \ce{C5H10O2 + NaNH2 -> C3H7NO + NaOC2H5}. \]

Sodium ethoxide is CC[O-].[Na+]. The transformation can proceed through cyclopropanone, O=C1CC1, followed by nitrogen addition and proton transfer. Consequently, C is the corresponding hemiaminal, 1-aminocyclopropan-1-ol.

Formation of D. Compound 2 is \(N\)-Boc-L-glutamic acid α-tert-butyl ester:

CC(C)(C)OC(=O)N[C@@H](CCC(=O)O)C(=O)OC(C)(C)C.

Its side-chain carboxyl group is free, whereas its α-amino and α-carboxyl groups are protected. Coupling therefore joins the side-chain carboxyl group to the amino group of C, giving D.

Dicyclohexylcarbodiimide, DCC, C1CCCCC1N=C=NC1CCCCC1, becomes 1,3-dicyclohexylurea, O=C(NC1CCCCC1)NC1CCCCC1. HOBt, 1-hydroxybenzotriazole, On1nnc2ccccc21, assists acyl activation; tetrahydrofuran, C1CCOC1, is the solvent.

The balanced overall coupling, in molecular formulas, is

\[ \ce{C14H25NO6 + C3H7NO + C13H22N2 ->[HOBt] C17H30N2O6 + C13H24N2O}. \]

The reactants are 2, C, and DCC; the products are D and dicyclohexylurea. DCC formally accepts the elements of water released during amide formation.

Formation of E. Aqueous hydrogen chloride, Cl, removes both acid-labile protecting groups. The Boc group releases carbon dioxide, O=C=O; tert-butyl cleavage can be balanced using 2-methylpropene, C=C(C)C:

\[ \ce{C17H30N2O6 ->[HCl(aq)] C8H14N2O4 + CO2 + 2 C4H8}. \]

This gives coprine E, with a free α-amino group and α-carboxyl group.

The amino-acid stereocentre retains the \(S\) configuration supplied in 2. None of these steps requires breaking a bond at that stereocentre. The substituted cyclopropane carbon is not stereogenic because its two ring paths are identical.

6.2 — Inhibition by compound 4

The stated hydrolyses are balanced as follows:

\[ \ce{C8H14N2O4 + H2O -> C5H9NO4 + C3H7NO}, \]

\[ \ce{C3H7NO + H2O -> C3H6O2 + NH3}. \]

The products of the first equation are L-glutamic acid 3, N[C@@H](CCC(=O)O)C(=O)O, and C. The second produces cyclopropane-1,1-diol 4, OC1(O)CC1, and ammonia. Water is O.

Compound 4 is the hydrate of cyclopropanone:

\[ \ce{C3H6O2 <=> C3H4O + H2O}. \]

The enzyme thiol, represented by *S, adds to cyclopropanone. Sulfur forms a bond to the carbonyl carbon, the carbonyl \(\pi\) bond opens toward oxygen, and proton transfer gives the hemithioketal F.

The net balance starting from the hydrate is

\[ \ce{Enz-SH + C3H6O2 -> Enz-S-C3H5O + H2O}. \]

Here the product formula denotes F, *SC1(O)CC1. For the requested drawing, retain the enzyme oval and connect it through sulfur to one vertex of a cyclopropane triangle. Attach an \(\ce{OH}\) group to that same vertex. The cyclopropane ring remains intact.

6.3 — Identification of G and H

Diethylamine, CCNCC, already contains four carbons. Carbon disulfide G supplies the fifth carbon required in H.

The amine attacks the electrophilic carbon of carbon disulfide, and sodium hydroxide, [Na+].[OH-], removes the nitrogen-bound proton:

\[ \ce{Et2NH + CS2 + NaOH -> Et2NC(=S)SNa + H2O}. \]

Thus H is sodium \(N,N\)-diethyldithiocarbamate. Its four ethyl-group carbons and one thiocarbonyl carbon account for the five-carbon hint.

6.4 — Suitable oxidants

Antabuse 5 is disulfiram, tetraethylthiuram disulfide:

CCN(CC)C(=S)SSC(=S)N(CC)CC.

Let \(X^-\) denote the \(N,N\)-diethyldithiocarbamate anion, CCN(CC)C(=S)[S-], and let \(X_2\) denote disulfiram, whose two units are joined by an \(\ce{S-S}\) bond. The required oxidation half-reaction is

\[ \ce{2 X^- -> X2 + 2 e^-}. \]

The three suitable reagents give the following balanced net ionic equations:

\[ \ce{2 X^- + I2 -> X2 + 2 I^-}, \]

\[ \ce{2 X^- + H2O2 -> X2 + 2 OH^-}, \]

\[ \ce{2 X^- + ArC(=O)OOH -> X2 + ArCOO^- + OH^-}. \]

In the last equation, \(\mathrm{Ar}\) is 3-chlorophenyl, so the oxidant is mCPBA. The ionic by-products are iodide, [I-]; hydroxide, [OH-]; and 3-chlorobenzoate, O=C([O-])c1cccc(Cl)c1.

Hot concentrated sulfuric acid is unsuitable. Sulfuric acid, OS(=O)(=O)O, strongly acidifies the dithiocarbamate. The resulting \(N,N\)-diethyldithiocarbamic acid, CCN(CC)C(=S)S, is susceptible to cleavage into diethylamine and carbon disulfide; heating promotes this destruction. Including protonation of the liberated amine gives

\[ \ce{Et2NC(=S)S^- + 2 H+ -> Et2NH2+ + CS2}. \]

The protonated amine is diethylammonium, CC[NH2+]CC. Therefore, sulfuric acid’s oxidizing character does not make hot concentrated sulfuric acid a suitable reagent for preserving and coupling this substrate.

The other unselected reagents also do not accomplish the required oxidation:

Reagent Name and SMILES Reason
\(\ce{Zn / CH3COOH}\) Zinc [Zn]; acetic acid CC(=O)O A reducing system
\(\ce{NaBH4}\) Sodium borohydride [Na+].[BH4-] A reducing agent
\(\ce{K2CO3},\ce{H2O}\) Potassium carbonate [K+].[K+].[O-]C(=O)[O-]; water O Supplies basic conditions, not the required oxidation
\(\ce{AlCl3}\) Aluminium chloride Cl[Al](Cl)Cl A Lewis acid, not the required oxidant

6.5 — Inhibition by disulfiram

The enzyme’s cysteine sulfur attacks one sulfur of disulfiram’s disulfide bond. Cleavage of that bond releases a diethyldithiocarbamate unit and leaves the enzyme in a mixed disulfide.

A balanced neutral-form equation is

\[ \ce{Enz-SH + X2 -> Enz-S-SC(=S)NEt2 + Et2NC(=S)SH}. \]

The second product is the dithiocarbamic acid identified above.

For J, draw the enzyme oval followed by

\[ \ce{-S-S-C(=S)-N(Et)2}. \]

Its three sulfur atoms are the original cysteine sulfur, the sulfur joining it to the thiocarbonyl carbon, and the thiocarbonyl sulfur. This matches *SSC(=S)N(CC)CC.

6.6 — Synthesis of gyromitrin

Formation of K. Formic acid, O=CO, reacts with acetyl chloride, CC(=O)Cl, to produce the mixed anhydride K:

\[ \ce{HCOOH + CH3COCl -> HC(=O)OC(=O)CH3 + HCl}. \]

This mixed anhydride provides an activated formyl group.

Formation of L. \(N\)-Methylhydrazine 6, CNN, is formylated at the methyl-bearing nitrogen, giving L. The terminal \(\ce{NH2}\) group must remain available for the subsequent aldehyde condensation.

Triethylamine, CCN(CC)CC, captures the acid formed. A combined balance is

\[ \ce{CH3NHNH2 + HC(=O)OC(=O)CH3 + Et3N -> HC(=O)N(CH3)NH2 + Et3NH+ + CH3COO^-}. \]

The salt by-product is triethylammonium acetate, CC[NH+](CC)CC.CC(=O)[O-].

Formation of M. The remaining terminal amino group condenses with acetaldehyde, also called ethanal, CC=O, eliminating water:

\[ \ce{HC(=O)N(CH3)NH2 + CH3CHO -> HC(=O)N(CH3)N=CHCH3 + H2O}. \]

The product is gyromitrin M. It contains both the formamide group and the hydrazone \(\ce{C=N}\) group described in the text. The reaction scheme specifies its connectivity but does not establish its \(E/Z\) configuration.

Hydrolysis of both groups has the overall balance

\[ \ce{HC(=O)N(CH3)N=CHCH3 + 2 H2O -> HCOOH + CH3NHNH2 + CH3CHO}. \]

Thus the products are formic acid, \(N\)-methylhydrazine, and acetaldehyde.

6.7 — Maximum hypothetical kinetic isotope effect

For a harmonic bond-stretching vibration,

\[ \widetilde{\nu}=\frac{1}{2\pi c}\sqrt{\frac{f}{\mu}}, \]

where \(f\) is the force constant and \(\mu\) is the reduced mass. The unchanged potential-energy surface means that \(f\) is unchanged by isotopic substitution.

The reduced masses are

\[ \mu_{\mathrm{light}} =\frac{12\times14}{12+14}\,\mathrm{u} =6.4615\ \mathrm{u}, \]

\[ \mu_{\mathrm{heavy}} =\frac{13\times15}{13+15}\,\mathrm{u} =6.9643\ \mathrm{u}. \]

Therefore,

\[ \frac{\widetilde{\nu}_{\mathrm{heavy}}} {\widetilde{\nu}_{\mathrm{light}}} = \sqrt{\frac{\mu_{\mathrm{light}}}{\mu_{\mathrm{heavy}}}} =0.96323, \]

and

\[ \widetilde{\nu}_{\mathrm{heavy}} =(1293.0)(0.96323) =1245.5\ \mathrm{cm^{-1}}. \]

The molar zero-point energy of this mode is

\[ E_0=\frac12N_{\mathrm A}hc\widetilde{\nu}. \]

The maximum effect occurs when the relevant stretching vibration is completely lost at the transition state. Its isotope-dependent zero-point-energy difference then contributes fully to the difference in activation energies:

\[ \Delta E^\ddagger_{\mathrm{heavy}} -\Delta E^\ddagger_{\mathrm{light}} = E_{0,\mathrm{light}}-E_{0,\mathrm{heavy}}. \]

Using standard constants to five significant digits,

\[ \begin{aligned} h&=6.6261\times10^{-34}\ \mathrm{J\,s},\\ c&=2.9979\times10^{10}\ \mathrm{cm\,s^{-1}},\\ N_{\mathrm A}&=6.0221\times10^{23}\ \mathrm{mol^{-1}},\\ R&=8.3145\ \mathrm{J\,mol^{-1}\,K^{-1}}, \end{aligned} \]

the difference, using the rounded intermediate wavenumber, is

\[ \begin{aligned} \Delta E_0 &=\frac12N_{\mathrm A}hc(1293.0-1245.5)\ \mathrm{cm^{-1}}\\ &=284.11\ \mathrm{J\,mol^{-1}}. \end{aligned} \]

The temperature is

\[ T=37+273.15=310.15\ \mathrm K, \]

so

\[ RT=2578.7\ \mathrm{J\,mol^{-1}}. \]

Since only zero-point energies are to affect the rate constants,

\[ \begin{aligned} \left(\frac{k_{\mathrm{light}}}{k_{\mathrm{heavy}}}\right)_{\max} &=\exp\!\left(\frac{\Delta E_0}{RT}\right)\\ &=\exp(0.11018)\\ &=\boxed{1.1165}. \end{aligned} \]

Thus the light molecule can hydrolyze approximately 12% faster in this limiting model.

Retaining guard digits throughout instead gives \(\Delta E_0=284.38\ \mathrm{J\,mol^{-1}}\) and a five-significant-digit ratio of \(1.1166\). The difference is solely intermediate rounding.

6.8 — Most likely rate-determining step

The best choice is nucleophilic attack of water on the protonated amide.

Carbonyl protonation is normally a rapid pre-equilibrium. The relevant protonated species is O-protonated gyromitrin, CC=NN(C)C=[OH+]:

\[ \ce{HC(=O)N(CH3)N=CHCH3 + H+ <=> HC(=OH+)N(CH3)N=CHCH3}. \]

Water then attacks the activated carbonyl carbon:

\[ \ce{HC(=OH+)N(CH3)N=CHCH3 + H2O -> HC(OH)(OH2+)N(CH3)N=CHCH3}. \]

The product of this addition is the protonated tetrahedral intermediate, CC=NN(C)C(O)[OH2+].

During this addition, the amide \(\ce{C-N}\) bond remains intact. A small isotope effect is therefore consistent with addition being rate determining. Substantial \(\ce{C-N}\) cleavage in the rate-determining step would be expected to produce a more pronounced primary isotope effect. Among the choices, water addition is consequently the most likely slow step.

The solution was checked against the official solution.