IChO 2015 · Theory problem 6

Bread is the stuff of life

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When you pass by the bakery, you are stopped by the smell of freshly baked bread. The hero of one of the novels said on a similar occasion: "If you tell me that this is not perfect, you are my enemy forever." The principle bread flavour component was identified in 1969 as compound X which occurs in equilibrium with its tautomer Y in a 2 : 1 ratio. Unfortunately, both forms are labile, and after some hours bread has no the same nice smell.

Figure 1

This tautomeric mixture of X and Y was synthesized in 1993 from piperidine by the reaction sequence given in Scheme 1. It is noteworthy that the initial ratio of X and Y was 1:4; on standing this ratio gradually changed to an equilibrium one.

Figure 2

Scheme 1.

Compound B which is characterized by 3-fold axis of symmetry (i.e., rotation by \(120^\circ\) results in a molecule indistinguishable from the original) occurs in equilibrium with its diastereomer C. The interconversion of these two forms proceeds via intermediate A which is also intermediate in B and C formation as well as their transformation to D. Compounds A, B, and C have the same elemental composition: \(w_{\mathrm C}=72.24\%\), \(w_{\mathrm H}=10.91\%\), \(w_{\mathrm N}=16.85\%\).

6.1 Write down the structural formulae of compounds A - E, X, Y.

Treatment of compound E with \(\ce{CH3Li . LiBr}\) complex in \(\ce{(C2H5)2O}\) at \(0\,^\circ\mathrm C\) failed to produce the target products X and Y. Instead, a yellow precipitate F was initially formed. Aqueous workup of this precipitate led to the mixture of compound E and its tautomer G.

6.2 Write down the structural formulae of compounds F and G.

Another approach to compound X is based on the use of pipecolinic acid derivative H. It was shown that X can be synthesized by reaction sequence presented in Scheme 2.

Figure 3

Scheme 2

6.3 Write down the structural formulae of compounds I and J.

Official English text of the problem.

Answers

  • 6.1
Compound Name and structure
A 1-Piperideine, or 3,4,5,6-tetrahydropyridine: N1=CCCCC1
B All-cis tripiperideine, the threefold-symmetric cyclic trimer of A: N12[C@H]3CCCCN3[C@H]4CCCCN4[C@H]1CCCC2
C Tripiperideine with one ring-junction hydrogen reversed, the diastereomer of B: N12[C@H]3CCCCN3[C@H]4CCCCN4[C@@H]1CCCC2
D Piperidine-2-carbonitrile: N#CC1CCCCN1
E 3,4,5,6-Tetrahydropyridine-2-carbonitrile: N#CC1=NCCCC1
X 2-Acetyl-1,4,5,6-tetrahydropyridine, the enamine tautomer: CC(=O)C1=CCCCN1
Y 2-Acetyl-3,4,5,6-tetrahydropyridine, the imine tautomer: CC(=O)C1=NCCCC1
  • 6.2
Compound Name and structure
F Lithium salt of 2-cyano-1,4,5,6-tetrahydropyridine, shown in its N-anion resonance form: [Li+].[N-]1C(C#N)=CCCC1
G 1,4,5,6-Tetrahydropyridine-2-carbonitrile: N#CC1=CCCCN1
  • 6.3
Compound Name and structure
I tert-Butyl 2-(1-ethoxyethenyl)piperidine-1-carboxylate: C=C(OCC)C1CCCCN1C(=O)OC(C)(C)C
J 2-Acetylpiperidine, or 1-(piperidin-2-yl)ethan-1-one: CC(=O)C1CCCCN1

For B, the three ring-junction carbon-bound hydrogens are on the same face of the central ring. For C, two are on one face and the third is on the opposite face. The SMILES specify representative relative configurations.

D is obtained as a racemate. No absolute configuration is specified for the starting material H, so none is assigned to I or J.

Solution

6.1 — Identification of A–E, X and Y

First determine the empirical formula shared by A, B, and C. Using a \(100\,\mathrm g\) sample and conventional atomic molar masses,

\[ n_{\mathrm C}:n_{\mathrm H}:n_{\mathrm N} = \frac{72.24}{12.011}: \frac{10.91}{1.008}: \frac{16.85}{14.007} \approx 6.015:10.82:1.203 \approx 5:9:1. \]

Thus their common empirical formula is \(\ce{C5H9N}\).

Piperidine, N1CCCCC1, has formula \(\ce{C5H11N}\). Treatment with tert-butyl hypochlorite, CC(C)(C)OCl, first gives N-chloropiperidine, ClN1CCCCC1. Base-induced elimination then gives the cyclic imine A, with loss of two hydrogen atoms overall. A formal net equation is

\[ \ce{C5H11N + (CH3)3COCl -> C5H9N + (CH3)3COH + HCl}. \]

The base neutralizes the hydrogen chloride represented in this equation. Therefore A is 1-piperideine.

Its degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(5)+2+1-9}{2}=2, \]

accounted for by one ring and one imine double bond.

Formation and stereochemistry of B and C. Three molecules of A undergo reversible cyclic trimerization:

\[ \ce{3 C5H9N <=> C15H27N3}. \]

The three imine double bonds become single bonds, and three new intermolecular carbon–nitrogen bonds close a central six-membered ring. Its atoms alternate between nitrogen and carbon. Each alternating N–CH edge is also part of a piperidine ring, completed by a four-methylene bridge.

This addition process changes the molecular formula but preserves the elemental percentages.

To draw the two products precisely:

  • Draw the central ring with alternating N and CH atoms, and attach the three four-methylene bridges across alternating N–CH edges.
  • For B, draw all three central C–H bonds as wedges. The three identical units are related by rotation through \(120^\circ\).
  • For C, reverse one of those C–H bonds to a dashed bond. This gives the required diastereomer and removes the threefold symmetry.

The stereochemical SMILES in the answers encode these relative configurations; they do not prescribe a particular chair conformation.

Formation of D. Cyanide adds to the electrophilic carbon of the protonated imine A. Net addition of hydrogen cyanide across its imine bond gives piperidine-2-carbonitrile, D:

\[ \ce{C5H9N + HCN -> C6H10N2}. \]

Attack on either face of the planar imine gives racemic D. Consumption of A also draws B and C through their equilibria toward D.

Formation of E. Repeating the N-chlorination–elimination sequence on D removes the N–H hydrogen and the hydrogen on the carbon bearing the cyano group. The result is E, with an imine bond between N1 and C2:

\[ \ce{C6H10N2 + (CH3)3COCl -> C6H8N2 + (CH3)3COH + HCl}. \]

The formula \(\ce{C6H8N2}\) gives

\[ M_{\mathrm E} = 6(12.011)+8(1.008)+2(14.007) = 108.144\,\mathrm{g\,mol^{-1}}, \]

and therefore

\[ w_{\mathrm N} = \frac{28.014\,\mathrm{g\,mol^{-1}}} {108.144\,\mathrm{g\,mol^{-1}}} \times100\% = 25.90\%, \]

in agreement with Scheme 1.

Formation and assignment of X and Y. One equivalent of methylmagnesium iodide adds a methyl group to the nitrile carbon. Acidic hydrolysis converts the resulting nitrogen-containing group into an acetyl group. A balanced net equation for this addition and acidic workup is

\[ \ce{C6H8N2 + CH3MgI + H2O + 2H+ -> C7H11NO + NH4+ + Mg^2+ + I-}. \]

The immediate organic product retains the ring imine bond and is therefore Y, 2-acetyl-3,4,5,6-tetrahydropyridine.

Transfer of a proton from C3 to nitrogen, accompanied by movement of the double bond from N1–C2 to C2–C3, gives X, 2-acetyl-1,4,5,6-tetrahydropyridine. Both have formula \(\ce{C7H11NO}\).

The initially greater abundance of Y is consistent with its direct formation from E:

\[ (X:Y)_{\text{initial}}=1:4. \]

Subsequent imine–enamine equilibration increases the amount of X until

\[ (X:Y)_{\text{equilibrium}}=2:1. \]

6.2 — Identification of F and G

Methyllithium acts as a base toward E, removing a proton from C3, adjacent to the imine carbon:

\[ \ce{C6H8N2 + CH3Li -> LiC6H7N2 + CH4}. \]

The resulting anion is delocalized. Its C3-carbanion resonance contributor can be written as

[Li+].N1=C(C#N)[CH-]CCC1.

Moving the C3 lone pair into the C2–C3 bond and moving the imine \(\pi\)-electron pair onto nitrogen gives the N-anion representation of F:

[Li+].[N-]1C(C#N)=CCCC1.

Thus F is the lithium salt of the enamine tautomer G. The supplied \(\ce{LiBr}\) does not change this organic connectivity.

During aqueous workup, protonation can occur at either end of the delocalized anion:

  • Protonation at C3 regenerates the imine E.
  • Protonation at N1 gives the enamine G, 1,4,5,6-tetrahydropyridine-2-carbonitrile.

For either organic tautomer, the balanced formula-level workup is

\[ \ce{LiC6H7N2 + H2O -> C6H8N2 + LiOH}. \]

The ionic SMILES specifies the lithium salt’s composition and organic bonding. It does not specify lithium coordination or aggregation in the precipitate.

6.3 — Identification of I and J

The starting material H is ethyl N-Boc-pipecolinate:

CCOC(=O)C1CCCCN1C(=O)OC(C)(C)C.

The titanocene reagent in Scheme 2 performs Petasis methylenation of the side-chain ester carbonyl. It replaces that carbonyl oxygen with a terminal methylene group while retaining the ethoxy group on the same carbon.

Consequently, I contains the substituent

\[ \ce{-C(OC2H5)=CH2} \]

at C2 of the N-Boc-protected piperidine ring. It is tert-butyl 2-(1-ethoxyethenyl)piperidine-1-carboxylate.

The formula provides an independent check: replacing one oxygen of H, \(\ce{C13H23NO4}\), by \(\ce{CH2}\) gives exactly the stated formula

\[ \boxed{\ce{C14H25NO3}}. \]

Acid treatment has two consequences:

  1. Hydrolysis of the enol ether produces the methyl ketone.
  2. Removal of the Boc protecting group exposes the secondary amine.

After the stated pH adjustment, the resulting structure is J, 2-acetylpiperidine.

A balanced net equation for hydrolysis and Boc removal, representing the final organic product as its free base, is

\[ \ce{C14H25NO3 + H2O -> C7H13NO + C2H5OH + CO2 + C4H8}. \]

The carbon-containing by-products are ethanol, CCO; carbon dioxide, O=C=O; and 2-methylpropene, C=C(C)C.

Finally, oxygen oxidizes the secondary amino ketone J to the ring imine Y by removing the N–H hydrogen and the C2 hydrogen. The balanced substrate oxidation half-reaction is

\[ \ce{C7H13NO -> C7H11NO + 2H+ + 2e-}. \]

The resulting Y undergoes the imine–enamine tautomerization established in part 6.1, giving the mixture of X and Y.

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