IChO 2014 · Theory problem 9

Heterocycle preparation

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Tetramethylthiurame disulfide (TMTD) is emerging as a useful reagent to prepare many sulfur-nitrogen functional groups and heterocycles in organic chemistry. The reactions of TMTD with primary amines, as well some corresponding post-transformations of the resulting product(s) are presented in the following schemes:

Figure 1
Figure 2
Figure 3
Figure 4
Figure 5

Similar transformations of benzohydrazides (containing nucleophilic \(\ce{NH2}\) group) and TMTD have been observed.

In the synthetic scheme below, the thiocarbamoylation reaction of an aroyl hydrazine with TMTD produces compound C containing a heterocyclic moiety from p-aminobenzoic acid.

Figure 6

During the formation of C from B, an intermediate B′ was observed. This intermediate tautomerizes to B″. C can be formed from B′ or B″.

Figure 7

9.1 Give the structures of A, B, and C.

9.2 Suggest a structure for the tautomer B″ and give a curly-arrow mechanism for the formation of C.

Compound C was then converted to F by the following pathway:

Figure 8

[The group R remains exactly the same throughout the rest of the question.]

9.3 Draw the structures of E, and F. (You do not need to draw the structure for the R group from this point.)

E was only obtained when D was slowly added to the solution of excess \(\ce{N2H4}\) in dioxane. If \(\ce{N2H4}\) was added to the solution of D in dioxane instead, a major side product D′ (\(\ce{R2C14H12N4S2}\)) was formed.

9.4 Give the structure of D′.

Slightly heating of D with ethanolamine (\(\ce{HOCH2CH2NH2}\)) in dioxane for 2 hours yielded G (\(\ce{R-C9H11N2OS}\)).

9.5 Draw the structural formula of G.

9.6 Heating G in the presence of p-toluenesulfonic acid as the catalyst could form a number of different five-membered heterocyclic products:

Figure 9

i) Draw 2 structures that have different molecular formulae.

ii) Draw 2 structures that are constitutional isomers.

iii) Draw 2 structures that are stereoisomers.

Official English text of the problem.

Answers

For C and F, the structures below use the thiol tautomer; their thione tautomers are given in the solution. In subsequent abbreviated SMILES, each * represents the same unchanged group R.

  • 9.1 — A: Methyl 4-aminobenzoate: COC(=O)c1ccc(N)cc1.

  • 9.1 — B: 4-Aminobenzohydrazide: NNC(=O)c1ccc(N)cc1.

  • 9.1 — C: 5-[4-(3,3-Dimethylthioureido)phenyl]-1,3,4-oxadiazole-2-thiol: CN(C)C(=S)Nc1ccc(-c2nnc(S)o2)cc1.

  • 9.2 — B″: The imidic-acid tautomer of B′, N-isothiocyanato-4-(3,3-dimethylthioureido)benzimidic acid. A geometry suitable for cyclization is: CN(C)C(=S)Nc1ccc(/C(O)=N/N=C=S)cc1. Intramolecular oxygen attack on the isothiocyanate carbon, followed by proton transfer, gives C.

  • 9.3 — E: 4-(4-R-Phenyl)thiosemicarbazide: *c1ccc(NC(=S)NN)cc1.

  • 9.3 — F: 5-[(4-R-Phenyl)amino]-1,3,4-thiadiazole-2-thiol: *c1ccc(Nc2nnc(S)s2)cc1.

  • 9.4 — D′: 1,2-Bis[N-(4-R-phenyl)carbamothioyl]hydrazine: *c1ccc(NC(=S)NNC(=S)Nc2ccc(*)cc2)cc1.

  • 9.5 — G: 1-(2-Hydroxyethyl)-3-(4-R-phenyl)thiourea: *c1ccc(NC(=S)NCCO)cc1.

  • 9.6(i): H₁ and H₃, with different molecular formulae:

  • H₁, 1-(4-R-phenyl)imidazolidine-2-thione: *c1ccc(N2CCNC2=S)cc1; formula \(\ce{R-C9H9N2S}\).

  • H₃, N-(4-R-phenyl)-4,5-dihydro-1,3-oxazol-2-amine: *c1ccc(NC2=NCCO2)cc1; formula \(\ce{R-C9H9N2O}\).

  • 9.6(ii): H₁ and H₂, constitutional isomers:

  • H₁ is specified above.
  • H₂, N-(4-R-phenyl)-4,5-dihydro-1,3-thiazol-2-amine: *c1ccc(NC2=NCCS2)cc1.

  • Both have formula \(\ce{R-C9H9N2S}\).

  • 9.6(iii): H₅ and H₇, geometrical stereoisomers:

  • H₅, (Z)-N-(4-R-phenyl)-1,3-thiazolidin-2-imine: *c1ccc(/N=C2\SCCN2)cc1.
  • H₇, (E)-N-(4-R-phenyl)-1,3-thiazolidin-2-imine: *c1ccc(/N=C2/SCCN2)cc1.

Solution

9.1 — Identification of A, B, and C

The starting material, 4-aminobenzoic acid, is Nc1ccc(C(=O)O)cc1. Methanol, CO, and sulfuric acid, OS(=O)(=O)O, effect Fischer esterification:

\[ \ce{H2N-C6H4-CO2H + CH3OH <=>[H2SO4] H2N-C6H4-CO2CH3 + H2O}. \]

The amino and carboxyl-derived substituents remain para throughout. Water is O.

Potassium carbonate, [K+].[K+].[O-]C(=O)[O-], neutralizes the acid during work-up, giving the free amino ester A. The acid-neutralization balance is

\[ \ce{H2SO4 + K2CO3 -> K2SO4 + CO2 + H2O}. \]

Here potassium sulfate is [K+].[K+].[O-]S(=O)(=O)[O-], and carbon dioxide is O=C=O.

Hydrazine, NN, then attacks the ester carbonyl. Nucleophilic acyl substitution replaces methoxy by hydrazino, producing B:

\[ \ce{H2N-C6H4-CO2CH3 + N2H4 -> H2N-C6H4-C(=O)-NH-NH2 + CH3OH}. \]

Thus B contains two primary amino groups: the aromatic amino group and the terminal hydrazide amino group.

TMTD, tetramethylthiuram disulfide, has SMILES CN(C)C(=S)SSC(=S)N(C)C. The supplied transformations account for the two different modifications:

  • The aromatic amino group becomes \(\ce{-NH-C(=S)-N(CH3)2}\).
  • The terminal hydrazide amino group becomes an isothiocyanate group, giving \(\ce{-C(=O)-NH-N=C=S}\).

The resulting B′ is N-isothiocyanato-4-(3,3-dimethylthioureido)benzamide, CN(C)C(=S)Nc1ccc(C(=O)NN=C=S)cc1, exactly matching the supplied intermediate.

For completeness, the supplied reactions give an atom-balanced route to B′, and hence C. At the hydrazide end, reactions (1) and (3) generate the isothiocyanate and two molecules of dimethyldithiocarbamic acid, CN(C)C(=S)S. One acid molecule decomposes by reaction (2), supplying the carbon disulfide needed in reaction (3); the other converts the aromatic amino group into the thiourea by reaction (5).

After cancellation of these intermediates, a balanced net equation is

\[ \ce{C7H9N3O + C6H12N2S4 -> C11H12N4OS2 + (CH3)2NH + H2S + S}. \]

The first reactant is B, and the first product is C. The “2 equiv.” in the experimental scheme specifies reagent loading; the equation above expresses the minimum net consumption along this route.

The small products are dimethylamine, CNC, hydrogen sulfide, S, and elemental sulfur. Carbon disulfide is S=C=S. As in the problem, \(\ce{S}\) denotes an atomic equivalent of elemental sulfur; its usual molecular form, octasulfur, is S1SSSSSSS1.

Cyclization of B′ produces a five-membered ring containing the original carbonyl oxygen, the two hydrazide nitrogens, and the two relevant carbon atoms. This is the 1,3,4-oxadiazole ring in C.

The formula follows directly by adding its structural fragments:

\[ \underbrace{\ce{C6H4}}_{\text{phenylene}} + \underbrace{\ce{C3H7N2S}}_{\text{dimethylthioureido group}} + \underbrace{\ce{C2HN2OS}}_{\text{oxadiazole-thiol group}} = \ce{C11H12N4OS2}. \]

9.2 — Tautomer B″ and electron-flow mechanism

Let \(\ce{X}\) denote the 4-(3,3-dimethylthioureido)phenyl group. The relevant tautomerization is

\[ \ce{X-C(=O)-NH-N=C=S <=> X-C(OH)=N-N=C=S}. \]

This is amide–imidic-acid tautomerism: the acyl nitrogen loses its proton, oxygen gains that proton, and the carbonyl double bond is replaced by a carbon–nitrogen double bond. Acid/base-assisted proton transfer can accomplish this change.

For the cyclization, label the original carbonyl carbon \(C_a\), its adjacent nitrogen \(N_a\), the terminal nitrogen \(N_b\), and the isothiocyanate carbon \(C_b\). The chain in B″ is therefore

\[ \ce{X-C_a(OH)=N_a-N_b=C_b=S}. \]

The precise curly-arrow sequence from B″ is:

  1. Ring closure: Draw an arrow from a lone pair on the hydroxyl oxygen to \(C_b\). Simultaneously, draw an arrow from the \(\ce{C_b=S}\) π bond onto sulfur. This forms the five-membered ring with atom order \[ O-C_a-N_a-N_b-C_b. \]

  2. The resulting cyclic oxonium–thiolate zwitterion has SMILES CN(C)C(=S)Nc1ccc(C2=NN=C([S-])[OH+]2)cc1. Oxygen bears a positive charge and still carries hydrogen; the exocyclic sulfur bears a negative charge.

  3. Proton transfer: Draw an arrow from the negatively charged sulfur to the oxygen-bound hydrogen, and an arrow from the \(\ce{O-H}\) bond back to oxygen. A proton shuttle may mediate this transfer. The product is the neutral oxadiazole-2-thiol C.

The configuration specified for B″ places the hydroxyl oxygen and the continuing nitrogen chain on the same side of the imidic \(\ce{C=N}\) bond, allowing this five-membered closure.

Direct formation from B′ is also possible. The carbonyl oxygen attacks \(C_b\), while the \(\ce{C_b=S}\) π electrons move to sulfur. The resulting carbonyl-derived cyclic oxonium–thiolate is CN(C)C(=S)Nc1ccc(C2=[O+]C([S-])=NN2)cc1.

Transfer of the acyl \(\ce{N-H}\) proton to sulfur, followed by donation of the nitrogen lone pair into the \(C_a-N_a\) bond and movement of the \(C_a=O\) π electrons onto oxygen, gives the same C.

The thiol form of C is tautomeric with 5-[4-(3,3-dimethylthioureido)phenyl]-1,3,4-oxadiazole-2(3H)-thione, CN(C)C(=S)Nc1ccc(C2=NNC(=S)O2)cc1. In that representation, the proton is on ring nitrogen and the exocyclic sulfur is double-bonded to carbon.

9.3 — Identification of E and F

From this point, write \(\ce{Ar}\) for the 4-R-phenyl group. In the thiol representation used above, R is the 5-sulfanyl-1,3,4-oxadiazol-2-yl substituent.

Hydrogen chloride, Cl, promotes cleavage of the dimethylthiourea group in C. Protonation traps the departing dimethylamine as dimethylammonium chloride, C[NH2+]C.[Cl-]:

\[ \ce{ArNH-C(=S)-N(CH3)2 + HCl -> Ar-N=C=S + [(CH3)2NH2]Cl}. \]

Thus D is 5-(4-isothiocyanatophenyl)-1,3,4-oxadiazole-2-thiol, S=C=Nc1ccc(-c2nnc(S)o2)cc1.

Hydrazine attacks the electrophilic carbon of the isothiocyanate. Proton transfer gives the thiosemicarbazide E, without elimination of a small molecule:

\[ \ce{Ar-N=C=S + N2H4 -> ArNH-C(=S)-NH-NH2}. \]

Treatment of E with TMTD converts its terminal \(\ce{-NH2}\) group into \(\ce{-N=C=S}\). The intermediate is N-isothiocyanato-N′-(4-R-phenyl)thiourea, *c1ccc(NC(=S)NN=C=S)cc1.

This intermediate cyclizes analogously to B′, but the original thiocarbonyl sulfur now becomes the ring heteroatom. The product therefore contains a 1,3,4-thiadiazole ring, giving F.

An atom-balanced net equation, with the generated dimethyldithiocarbamic acid written in its decomposed form, is

\[ \ce{R-C7H8N3S + C6H12N2S4 -> R-C8H6N3S2 + 2(CH3)2NH + CS2 + S}. \]

The formula of F, excluding unchanged R, is

\[ \ce{C6H4}+\ce{NH}+\ce{C2HN2S2} = \ce{C8H6N3S2}, \]

as required.

The corresponding thione tautomer is 5-[(4-R-phenyl)amino]-1,3,4-thiadiazole-2(3H)-thione, *c1ccc(NC2=NNC(=S)S2)cc1.

9.4 — Identification of D′ and the addition-order effect

Both nitrogen atoms of hydrazine can react with isothiocyanate molecules. Addition of two equivalents of D gives

\[ \ce{2Ar-N=C=S + N2H4 -> ArNH-C(=S)-NH-NH-C(=S)-NHAr}. \]

This is D′, the bis-adduct specified in the answers.

Its formula contains two para-disubstituted phenyl groups, two thiocarbonyl carbons, and four \(\ce{NH}\) groups:

\[ \ce{R2}+\ce{C12H8}+\ce{C2S2}+\ce{N4H4} = \ce{R2C14H12N4S2}. \]

When D is added slowly to excess hydrazine, an incoming molecule of D preferentially encounters free hydrazine, forming E. When hydrazine is added to a solution rich in D, the initially formed E can react again through its remaining terminal amino group, producing D′.

9.5 — Identification of G

Ethanolamine, 2-aminoethan-1-ol, has SMILES NCCO. Its amino group attacks the isothiocyanate carbon more readily than its hydroxyl group. Addition and proton transfer give G:

\[ \ce{Ar-N=C=S + H2N-CH2-CH2-OH -> ArNH-C(=S)-NH-CH2-CH2-OH}. \]

No atoms are lost. The formula, excluding R, is consequently

\[ \ce{C7H4NS}+\ce{C2H7NO} = \ce{C9H11N2OS}. \]

9.6 — Classification of the heterocyclic products

The catalyst, p-toluenesulfonic acid, has SMILES Cc1ccc(S(=O)(=O)O)cc1.

Formation of the selected sulfur-containing products can be balanced as dehydration:

\[ \ce{R-C9H11N2OS -> R-C9H9N2S + H2O}. \]

Formation of the oxygen-containing product H₃ instead corresponds to loss of hydrogen sulfide:

\[ \ce{R-C9H11N2OS -> R-C9H9N2O + H2S}. \]

9.6(i): Different molecular formulae — H₁ and H₃.

H₁ contains sulfur and has formula \(\ce{R-C9H9N2S}\). H₃ contains oxygen in its place and has formula \(\ce{R-C9H9N2O}\). Their molecular formulae therefore differ.

9.6(ii): Constitutional isomers — H₁ and H₂.

Both have formula \(\ce{R-C9H9N2S}\), but their atom connectivity differs:

  • In H₁, both nitrogen atoms are ring members, and sulfur is an exocyclic thione sulfur.
  • In H₂, sulfur and one nitrogen are ring members, while the aryl-bearing nitrogen is exocyclic.

They are therefore constitutional isomers.

9.6(iii): Stereoisomers — H₅ and H₇.

Both have the same thiazolidine-ring connectivity and an exocyclic \(\ce{C=N-Ar}\) bond. They differ in configuration about that double bond.

At the imine carbon, sulfur has higher CIP priority than nitrogen. At the imine nitrogen, the aryl group has higher priority than the lone-pair position. In the supplied drawings:

  • H₅: The aryl group and ring sulfur are on the same side of the exocyclic double bond, giving the Z configuration.
  • H₇: They are on opposite sides, giving the E configuration.

The directional bonds in their answer SMILES encode this distinction.

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