IChO 2019 · Theory problem 9

Synthesis of levobupivacaine

Part I.

The local anesthetic bupivacaine (marketed as Marcaine) is on the World Health Organization List of Essential Medicines. Although the drug is currently used as a racemic mixture, it was demonstrated that one enantiomer of bupivacaine, levobupivacaine, is less cardiotoxic and, therefore, safer than the racemate. Levobupivacaine can be synthesized from the natural amino acid L-lysine.

Figure 1

L-Lysine hydrochloride

9.1 Assign the absolute configuration of the stereogenic center in L-lysine hydrochloride and justify your answer by classifying the substituents in order of their priority.

Configuration: ☐ \(R\)  ☐ \(S\)

Priority \(1 > 2 > 3 > 4\):

Figure 2

9.2 The prefix L in L-lysine refers to relative configuration. Choose all correct statements:

  • ☐ Natural L-amino acids are levorotatory.
  • ☐ Natural L-amino acids can be levorotatory or dextrorotatory.
  • ☐ All natural L-amino acids are \((S)\).
  • ☐ All natural L-amino acids are \((R)\).

Often, we want only one of the amino groups in L-lysine to react. A \(\ce{Cu^2+}\) salt with excess aqueous hydroxide can selectively mask the reactivity of one of the amino groups. After the complex is formed, only the non-complexed \(\ce{NH2}\) group is available to react.

9.3 Considering that L-lysine acts as a bidentate ligand and that two L-lysines coordinate to one \(\ce{Cu^2+}\) ion in the presence of aqueous hydroxide, draw the structure of the intermediate complex.

Fortunately, in the synthesis of levobupivacaine shown below, the same amino group reacts even without \(\ce{Cu^2+}\) salt.

Figure 3
Figure 4

From this point on, you can use the abbreviations proposed in the scheme above.

9.4 Draw the structure of compound A, including the appropriate stereochemistry.

9.5 Transformation of L-lysine into A is (choose proper answer(s)):

  • ☐ an enantioselective reaction
  • ☐ an enantiospecific reaction
  • ☐ a regioselective reaction

9.6 Draw the structures of compounds B – F, including the appropriate stereochemistry.

9.7 What is the role of DCC in the transformation C → D? Tick relevant boxes.

  • ☐ Protecting group for the amino group.
  • ☐ Protecting group for the hydroxy group.
  • ☐ Activating agent for the amide bond formation.

9.8 TsCl is used in the synthesis to enable (tick relevant boxes):

  • ☐ Nucleophilic substitution of an amino group.
  • ☐ Electrophilic substitution of an amino group.
  • ☐ Nucleophilic substitution of a hydroxy group.
  • ☐ Electrophilic substitution of a hydroxy group.

9.9 Mark all possible reagents which could be used as reagent H:

☐ diluted \(\ce{HCl}\) ☐ \(\ce{Zn/HCl}\)
☐ \(\ce{K2CO3}\) ☐ \(\ce{H2SO4}\)
☐ diluted \(\ce{KMnO4}\) ☐ diluted \(\ce{NaOH}\)
☐ \(\ce{SOCl2}\) ☐ \(\ce{PCl5}\)

9.10 Draw the structure of levobupivacaine, including the appropriate stereochemistry.

Part II.

The synthesis of levobupivacaine requires the use of enantiomerically pure L-lysine. A common method to confirm the enantiomeric purity of aminoacids is their transformation into amide using Mosher's acid (see the structure of the \((S)\) isomer below).

Figure 5

\((S)\)-Mosher's acid

9.11 Draw the structure of the amide formed when the \(\alpha\)-amino group of L-lysine is derivatized with \((S)\)-Mosher's acid. Clearly show the stereochemistry of each chiral center.

9.12 How many products will be formed from racemic lysine and \((S)\)-Mosher's acid (consider that only the \(\alpha\)-amino group of lysine is derivatized)? Tick relevant boxes.

  • ☐ Two diastereoisomers.
  • ☐ Four diastereoisomers.
  • ☐ A racemic mixture of two enantiomers.
  • ☐ Four compounds: two enantiomers and two diastereoisomers.

9.13 Choose the method(s) which can be used to quantitatively determine the enantiomeric purity of lysine after its derivatization with \((S)\)-Mosher's acid. Tick the relevant boxes:

  • ☐ NMR spectroscopy
  • ☐ Liquid chromatography
  • ☐ NMR spectroscopy
  • ☐ Liquid chromatography

Official English text of the problem.

Answers

Cbz denotes benzyloxycarbonyl; Ts denotes \(p\)-toluenesulfonyl. The stereochemical SMILES below specify the configurations independently of any drawing orientation.

  • 9.1: \(S\). Priorities: \[ \ce{-NH3+}>\ce{-COO^-}>\ce{-(CH2)4NH3+}>\ce{-H}. \] Fill the four boxes, from left to right, with 3, 1, 2, 4.

  • 9.2: Tick only “Natural L-amino acids can be levorotatory or dextrorotatory.”

  • 9.3: Bis(L-lysinato)copper(II), with each ligand coordinated through its \(\alpha\)-amino nitrogen and one carboxylate oxygen. Both terminal amino groups remain free.

Dative-bond SMILES: [Cu+2]12(<-N[C@@H](CCCCN)C(=O)[O-]->1)<-N[C@@H](CCCCN)C(=O)[O-]->2

Draw a square-planar \(\ce{CuN2O2}\) core with two five-membered chelate rings; an arrangement with the two nitrogen donors opposite each other is suitable. The SMILES encodes the coordination connectivity, not this geometry.

  • 9.4: A: \(N^\varepsilon\)-benzylidene-L-lysine, shown as its zwitterion; the original stereocenter remains \(S\).

[NH3+][C@@H](CCCCN=Cc1ccccc1)C(=O)[O-]

  • 9.5: Tick “a regioselective reaction.”

  • 9.6: All five compounds retain the \(2S\) configuration.

Compound Name Stereochemical SMILES
B \(N^\alpha\)-Cbz-L-lysine, zwitterionic form O=C(OCc1ccccc1)N[C@@H](CCCC[NH3+])C(=O)[O-]
C (2S)-6-Acetoxy-2-(Cbz-amino)hexanoic acid O=C(OCc1ccccc1)N[C@@H](CCCCOC(C)=O)C(=O)O
D (2S)-6-Acetoxy-2-(Cbz-amino)-N-(2,6-dimethylphenyl)hexanamide O=C(OCc1ccccc1)N[C@@H](CCCCOC(C)=O)C(=O)Nc1c(C)cccc1C
E (2S)-2-(Cbz-amino)-N-(2,6-dimethylphenyl)-6-(tosyloxy)hexanamide O=C(OCc1ccccc1)N[C@@H](CCCCOS(=O)(=O)c1ccc(C)cc1)C(=O)Nc1c(C)cccc1C
F (2S)-2-Amino-N-(2,6-dimethylphenyl)-6-(tosyloxy)hexanamide N[C@@H](CCCCOS(=O)(=O)c1ccc(C)cc1)C(=O)Nc1c(C)cccc1C
  • 9.7: Tick “Activating agent for the amide bond formation.”

DCC is N,N′-dicyclohexylcarbodiimide, C1CCCCC1N=C=NC1CCCCC1.

  • 9.8: Tick “Nucleophilic substitution of a hydroxy group.”

TsCl is \(p\)-toluenesulfonyl chloride, Cc1ccc(S(=O)(=O)Cl)cc1.

  • 9.9: Tick potassium carbonate, \(\ce{K2CO3}\), [K+].[K+].[O-]C(=O)[O-], and diluted sodium hydroxide, \(\ce{NaOH}\), [Na+].[OH-].

  • 9.10: Levobupivacaine: (2S)-1-butyl-N-(2,6-dimethylphenyl)piperidine-2-carboxamide.

CCCCN1[C@@H](CCCC1)C(=O)Nc1c(C)cccc1C

  • 9.11: The \(\alpha\)-N-[(S)-Mosher acyl] derivative of L-lysine. Both the Mosher center and the lysine center are \(S\).

CO[C@@](C(F)(F)F)(c1ccccc1)C(=O)N[C@@H](CCCCN)C(=O)O

  • 9.12: Tick “Two diastereoisomers.” Their configurations, ordered as (Mosher center, lysine center), are \((S,S)\) and \((S,R)\).

  • 9.13: Both NMR spectroscopy and liquid chromatography. Since the printed problem repeats both choices without qualification, all four printed entries describe applicable methods.

Solution

9.1 — Absolute configuration of L-lysine

The depicted L-lysine hydrochloride is (S)-2,6-diaminohexanoic acid hydrochloride:

[NH3+][C@@H](CCCC[NH3+])C(=O)[O-].[Cl-]

For the stereogenic carbon, compare the directly attached atoms. Nitrogen has higher atomic number than carbon, while hydrogen has the lowest:

\[ \ce{N}>\ce{C}>\ce{H}. \]

To distinguish the two carbon substituents, compare the atoms attached to those carbons, counting a double bond according to the CIP duplicate-atom convention:

\[ \begin{aligned} \text{carboxylate carbon:}&\quad (\ce{O},\ce{O},\ce{O}),\\ \text{side-chain carbon:}&\quad (\ce{C},\ce{H},\ce{H}). \end{aligned} \]

Therefore,

\[ 1=\ce{-NH3+},\qquad 2=\ce{-COO^-},\qquad 3=\ce{-(CH2)4NH3+},\qquad 4=\ce{-H}. \]

In the supplied drawing, the amino group points away and the implicit hydrogen points toward the observer. The apparent clockwise \(1\to2\to3\) sequence must therefore be reversed, giving \(S\). The chloride counterion does not enter the priority comparison.

9.2 — Relative configuration and optical rotation

The designation L describes a relative stereochemical relationship; it does not specify the sign of optical rotation. Consequently, an L-amino acid can be levorotatory or dextrorotatory.

Nor does L always mean \(S\). For example, L-cysteine, N[C@@H](CS)C(=O)O, has the \(R\) configuration: its sulfur-containing side chain outranks its carboxyl group in the CIP comparison.

Thus only the second statement is correct.

9.3 — Copper complex

In excess hydroxide, the carboxyl group is deprotonated and both amino groups can be present as neutral amines. Each lysinate ligand binds through:

  1. the lone pair of the \(\alpha\)-amino nitrogen;
  2. one carboxylate oxygen.

This gives a favorable five-membered chelate ring with the sequence

\[ \ce{Cu-N-C_{\alpha}-C-O}. \]

Two such ligands give bis(L-lysinato)copper(II), whose coordination SMILES is given in the Answers. Each ligand has charge \(-1\), so the complex with \(\ce{Cu^2+}\) is neutral.

A balanced equation starting with the hydrochloride shown in the question is

\[ \ce{ 2C6H15ClN2O2 + Cu^2+ + 4OH^- -> [Cu(C6H13N2O2)2] + 2Cl^- + 4H2O }. \]

The drawing should show two five-membered chelate rings around copper. Both stereocenters remain \(S\), and both terminal \(\varepsilon\)-amino groups project outward without coordinating. Those terminal amino groups remain available for reaction.

9.4 — Identification of A

Benzaldehyde, O=Cc1ccccc1, condenses selectively with the terminal \(\varepsilon\)-amino group to form a Schiff base, or imine.

The sequence is:

  • lithium hydroxide neutralizes one acid equivalent;
  • the terminal amino group reacts with benzaldehyde;
  • loss of water produces the terminal benzylidene imine.

The overall balanced equation is

\[ \ce{ C6H15ClN2O2 + LiOH + C7H6O -> C13H18N2O2 + LiCl + 2H2O }. \]

Thus A is \(N^\varepsilon\)-benzylidene-L-lysine. Its terminal group is \(\ce{-N=CHPh}\), while the \(\alpha\)-amino group remains available for subsequent protection.

No bond at the stereogenic carbon is broken. Draw its substituents in the same spatial arrangement as in the starting lysine: \(S\). The SMILES leaves the imine’s \(E/Z\) geometry unspecified because the scheme does not establish it.

9.5 — Classification of the transformation

The reaction distinguishes between two amino groups at different positions and forms the imine at the terminal \(\varepsilon\)-amino group. It is therefore regioselective.

No new stereogenic center is generated, and the reaction does not select between enantiomers. The existing \(S\) center is simply preserved. Accordingly, “enantioselective” is not appropriate.

9.6 — Identification of B–F

The sequence uses the terminal imine as a temporary protecting group, allowing the \(\alpha\)-amino group to be protected independently.

A → B: protect the \(\alpha\)-amino group, then remove the imine.

Benzyl chloroformate, or Cbz–Cl, O=C(Cl)OCc1ccccc1, converts the free \(\alpha\)-amino group into a Cbz carbamate. Diluted hydrochloric acid then hydrolyzes the terminal imine, releasing benzaldehyde and restoring the terminal amino group.

A balanced net equation for protection followed by imine hydrolysis is

\[ \ce{ C13H18N2O2 + C8H7ClO2 + NaOH -> C14H20N2O4 + C7H6O + NaCl }. \]

The acid/base work-up establishes the final protonation state. At pH 6.2, B, \(N^\alpha\)-Cbz-L-lysine, is predominantly represented with a carboxylate and a terminal ammonium group, as shown in its SMILES.

Its molecular formula is the supplied \(\ce{C14H20N2O4}\).

B → C: replace the terminal amino group with acetate.

Acidified nitrite generates nitrous acid, ON=O. The unprotected terminal primary amine undergoes nitrosation and deamination, releasing nitrogen. In the acetic acid/acetate medium, the terminal carbon acquires an acetoxy group.

Acetic acid is CC(=O)O; an acetoxy substituent has connectivity \(\ce{-O-C(=O)-CH3}\).

A balanced net equation consistent with the reagents is

\[ \ce{ C14H20N2O4 + NaNO2 + 2CH3COOH -> C16H21NO6 + N2 + CH3COONa + 2H2O }. \]

The protected \(\alpha\)-nitrogen remains intact. Thus C is the terminal acetate, with formula \(\ce{C16H21NO6}\). This formula distinguishes it from the corresponding terminal alcohol.

C → D: form the aromatic amide.

The carboxylic acid reacts with 2,6-dimethylaniline, Nc1c(C)cccc1C, using DCC as the coupling reagent. The terminal acetate and the Cbz group remain intact.

DCC becomes N,N′-dicyclohexylurea, O=C(NC1CCCCC1)NC1CCCCC1.

The balanced coupling equation is

\[ \ce{ C16H21NO6 + C8H11N + C13H22N2 -> C24H30N2O5 + C13H24N2O }. \]

Therefore D is the \(N\)-(2,6-dimethylphenyl) carboxamide, with formula \(\ce{C24H30N2O5}\).

D → E: hydrolyze the acetate, then form the tosylate.

Aqueous potassium carbonate removes the terminal acetyl protecting group. The intermediate is (2S)-2-(Cbz-amino)-N-(2,6-dimethylphenyl)-6-hydroxyhexanamide:

O=C(OCc1ccccc1)N[C@@H](CCCCO)C(=O)Nc1c(C)cccc1C

Its formula is \(\ce{C22H28N2O4}\). The hydrolysis can be balanced in ionic form:

\[ \ce{ C24H30N2O5 + OH^- -> C22H28N2O4 + CH3COO^- }. \]

The resulting alcohol reacts with TsCl and triethylamine, CCN(CC)CC, to give the terminal tosylate E:

\[ \ce{ C22H28N2O4 + C7H7ClO2S + C6H15N -> C29H34N2O6S + C6H16ClN }. \]

The product formula agrees with the supplied \(\ce{C29H34N2O6S}\).

E → F: remove the Cbz group by hydrogenolysis.

Hydrogen over palladium on carbon removes the Cbz protecting group. The products accompanying the free amine are toluene, Cc1ccccc1, and carbon dioxide, O=C=O:

\[ \ce{ C29H34N2O6S + H2 -> C21H28N2O4S + C7H8 + CO2 }. \]

Thus F contains a free \(\alpha\)-amino group and a terminal tosylate. Its formula is the supplied \(\ce{C21H28N2O4S}\).

Throughout B–F, the bonds at the original stereogenic carbon retain their spatial arrangement. Nitrogen remains its highest-priority substituent, the carboxyl-derived group remains second, and the side chain remains third. Every compound therefore remains \(2S\).

9.7 — Role of DCC

DCC activates the carboxyl group of C for amide formation.

The carboxyl group reacts with the carbodiimide to form an activated acyl intermediate, conventionally described as an O-acylisourea. The aniline nitrogen attacks the acyl carbon, and collapse of the intermediate gives the amide D and dicyclohexylurea.

DCC is therefore an activating agent for amide bond formation, not an amino- or hydroxy-protecting group.

9.8 — Role of TsCl

An unmodified hydroxy group is a poor leaving group. Tosylation converts it into a sulfonate ester, \(\ce{-OTs}\).

During the subsequent ring closure, the amino nitrogen attacks the carbon bearing that group and expels the \(p\)-toluenesulfonate anion, Cc1ccc(S(=O)(=O)[O-])cc1.

Consequently, TsCl enables nucleophilic substitution of the hydroxy group, after its conversion into a tosylate.

9.9 — Reagent H and ring closure

The free \(\alpha\)-amino group in F attacks the terminal carbon bearing the tosylate through an intramolecular \(\mathrm{S_N2}\) reaction. The resulting ring contains:

  • the amino nitrogen;
  • the original stereogenic carbon;
  • the four intervening methylene carbons.

It is therefore a six-membered piperidine ring.

The immediate cyclization product is an ammonium species; base removes a proton to give the neutral secondary amine. A base also keeps the starting amine available in its unprotonated, nucleophilic form. Suitable choices are potassium carbonate and diluted sodium hydroxide.

The product G is (2S)-N-(2,6-dimethylphenyl)piperidine-2-carboxamide:

N1[C@@H](CCCC1)C(=O)Nc1c(C)cccc1C

Balanced equations for the two acceptable bases are

\[ \ce{ C21H28N2O4S + NaOH -> C14H20N2O + C7H7NaO3S + H2O } \]

and

\[ \ce{ 2C21H28N2O4S + K2CO3 -> 2C14H20N2O + 2C7H7KO3S + CO2 + H2O }. \]

Acids protonate the amino group and suppress its nucleophilicity. The oxidizing, reducing, and chlorinating alternatives listed do not provide the required basic cyclization conditions.

The substitution occurs at a terminal methylene carbon, not at the original stereocenter. Its \(S\) configuration is retained.

9.10 — Levobupivacaine

The secondary ring nitrogen of G reacts with 1-bromobutane, CCCCBr, by nucleophilic substitution, installing an \(n\)-butyl group. Triethylamine accepts the proton released in forming the neutral tertiary amine.

The balanced equation is

\[ \ce{ C14H20N2O + C4H9Br + C6H15N -> C18H28N2O + C6H16BrN }. \]

The resulting levobupivacaine is (2S)-1-butyl-N-(2,6-dimethylphenyl)piperidine-2-carboxamide, with the SMILES given in the Answers.

For the requested drawing, use a piperidine ring bearing an \(n\)-butyl group on nitrogen and a \(\ce{-C(=O)-NH}\)-(2,6-dimethylphenyl) substituent at C2. Show C2 as \(S\). No particular chair conformation is required.

9.11 — Mosher amide of L-lysine

(S)-Mosher’s acid, or (S)-3,3,3-trifluoro-2-methoxy-2-phenylpropanoic acid, is

CO[C@@](C(F)(F)F)(c1ccccc1)C(=O)O.

The neutral connectivity representation of L-lysine is

N[C@@H](CCCCN)C(=O)O.

The Mosher carboxyl group forms an amide with the \(\alpha\)-amino nitrogen of lysine. The lysine carboxyl group and terminal amino group remain unchanged.

The formal condensation balance is

\[ \ce{ C6H14N2O2 + C10H9F3O3 -> C16H21F3N2O4 + H2O }. \]

This represents the net amidation; experimental formation of the amide requires suitable carboxyl activation.

Neither stereogenic carbon undergoes substitution. Both remain \(S\). At the Mosher center, the priority order in the amide is

\[ \text{methoxy oxygen} > \text{trifluoromethyl carbon} > \text{amide carbonyl carbon} > \text{phenyl carbon}. \]

Thus amidation also preserves the Mosher center’s \(S\) descriptor.

For the drawing, connect the Mosher carbonyl carbon to lysine’s \(\alpha\)-nitrogen, and label both stereogenic centers \(S\). The complete product SMILES is given in Answer 9.11; it uses the neutral acid/amine representation.

9.12 — Products from racemic lysine

Racemic lysine contains both configurations at its \(\alpha\)-carbon. Reaction with a single enantiomer of Mosher’s acid gives:

Product Configuration: (Mosher, lysine)
(S)-Mosher amide of L-lysine \((S,S)\)
(S)-Mosher amide of D-lysine \((S,R)\)

The second product, the \(\alpha\)-N-[(S)-Mosher acyl] derivative of D-lysine, has SMILES

CO[C@@](C(F)(F)F)(c1ccccc1)C(=O)N[C@H](CCCCN)C(=O)O.

These products differ at only one of their two stereogenic centers, so they are diastereoisomers. They are not mirror images: the Mosher center has the same configuration in both.

Therefore, two diastereoisomers are formed.

9.13 — Quantitative determination of enantiomeric purity

Derivatization converts the original enantiomers into diastereoisomers. These can have different properties even in an achiral analytical environment.

NMR spectroscopy: The two derivatives can give distinct chemical shifts. Integrating suitable resolved signals under quantitative conditions gives their molar ratio. The \(\ce{^{19}F}\) resonances of the trifluoromethyl groups are particularly useful because each derivative contributes three equivalent fluorine nuclei.

For integrals \(I_L,I_D\) representing \(p_L,p_D\) nuclei per molecule,

\[ \frac{n_L}{n_D} = \frac{I_L/p_L}{I_D/p_D}. \]

Liquid chromatography: The diastereoisomers can have different retention times, including on a suitable achiral stationary phase. Resolved peak areas give their molar ratio after allowing for detector response factors:

\[ \frac{n_L}{n_D} = \frac{A_L/r_L}{A_D/r_D}, \]

where \(A_i\) is peak area and \(r_i\) is response per amount of substance.

The original lysine composition follows from

\[ \%L=100\frac{n_L}{n_L+n_D}, \]

and the magnitude of the enantiomeric excess is

\[ ee(\%)= 100\frac{|n_L-n_D|}{n_L+n_D}. \]

This requires derivatization without racemization and without unequal, uncorrected conversion of the two lysine enantiomers. Both listed analytical methods are therefore suitable.

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