IChO 2007 · Theory problem 7

Atherosclerosis and intermediates of cholesterol biosynthesis

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Cholesterol is a lipid wide-spread in living nature. Disruption of its metabolism leads to atherosclerosis and related potentially fatal diseases.

Substances X and Y are two key intermediates of cholesterol biosynthesis in animals.

X is an optically active monocarbonic acid composed of atoms of only three elements. It is formed in organisms from (S)-3-hydroxy-3-methylpentanedioyl-coenzyme A (HMG-CoA). This reaction is catalyzed by enzyme E1 (which catalyses two types of reactions) and does not involve water as a substrate. X is further metabolized into X1 through a three-stage process requiring enzymes E2, E3, E4, which catalyze reactions of one and the same (and only one) type. Finally, X1 spontaneously (non-enzymatically) decomposes to give isopentenyl pyrophosphate (3-methylbut-3-enyl diphosphate, IPP) and inorganic products:

Figure 1

Scheme 1

7.1 In the Answer Sheet, choose the reaction type(s) for E1 and E3.

7.2 Draw the structure of X with stereochemical details and indicate absolute configuration (R or S) of the stereocenter.

Y is an unsaturated acyclic hydrocarbon. Its reductive ozonolysis leads to a mixture of only three organic substances Y1, Y2 and Y3 in a molar ratio of 2:4:1. Y is formed as a result of a number of successive coupling reactions of two isomeric substances: IPP and dimethyl allyl pyrophosphate (3-methylbut-2-enyl diphosphate, DAP) with subsequent reduction of a double bond in the final coupling product Y5. Carbon atoms IPP and DAP involved in the formation of C-C bonds during biosynthesis of Y are marked with asterisks.

Figure 2

7.3 Write down the overall reaction equation for reductive ozonolysis of DAP, if dimethyl sulfide is used as the reducing agent.

The product of the final coupling reaction (hydrocarbon Y5) is formed when two hydrocarbon residues (R) of intermediate Y4 are combined:

Figure 3

Scheme 2

At each coupling stage but that shown in Scheme 2, pyrophosphate is released in a molar ratio of 1 : 1 to the coupling product.

7.4 Determine molecular formula of Y, if it is known that Y2 and Y3 contain 5 and 4 carbon atoms, respectively.

7.5 Calculate the number of IPP and DAP molecules needed to give Y5, if it is known that all carbon atoms of isomeric pyrophosphates are incorporated into Y.

7.6 Draw the product of coupling reaction of one IPP molecule with one DAP molecule (C-C bond can be formed only by carbon atoms marked with asterisks), if it is known that subsequent reductive ozonolysis of the product of the coupling reaction gives Y1, Y2 and one more product, the latter containing phosphorus.

The only double bond reduced in Y5 during its metabolism into Y was formed in the reaction described in Scheme 2. All double bonds in Y and Y4 exist in trans configuration.

7.7 Draw structures of Y and Y4 with stereochemical details.

Official English text of the problem.

Answers

In the condensed structures below, \(\ce{-OPP}\) denotes the diphosphate ester group \(\ce{-O-P(=O)(O^-)-O-P(=O)(O^-)O^-}\). Its charge is included implicitly.

  • 7.1 E1: reduction and elimination of the CoA thiol fragment—nonhydrolytic cleavage of the thioester \(\ce{C-S}\) bond. E3: phosphorylation.

  • 7.2 X is (R)-mevalonic acid, or (3R)-3,5-dihydroxy-3-methylpentanoic acid, \(\ce{C6H12O4}\).
    SMILES: C[C@@](O)(CCO)CC(=O)O

  • 7.3 \[ \ce{(CH3)2C=CHCH2OPP + O3 + (CH3)2S -> (CH3)2CO + OHCCH2OPP + (CH3)2SO} \] The products are acetone, 2-oxoethyl diphosphate, and dimethyl sulfoxide.

  • 7.4 \(\boxed{\ce{Y = C30H50}}\).

  • 7.5 Each molecule of Y5 requires \(\boxed{4\text{ IPP molecules}+2\text{ DAP molecules}}\).

  • 7.6 The product is geranyl diphosphate, (2E)-3,7-dimethylocta-2,6-dien-1-yl diphosphate: \[ \ce{(CH3)2C=CHCH2CH2C(CH3)=CHCH2OPP}. \] SMILES: CC(C)=CCC/C(C)=C/COP(=O)([O-])OP(=O)([O-])[O-]

  • 7.7 Y is all-trans-squalene, with \(E\) configurations at positions \(6,10,14,18\).
    SMILES: CC(C)=CCC/C(C)=C/CC/C(C)=C/CC/C=C(C)/CC/C=C(C)/CCC=C(C)C

Y4 is (2E,6E)-farnesyl diphosphate, or (2E,6E)-3,7,11-trimethyldodeca-2,6,10-trien-1-yl diphosphate.
SMILES: CC(C)=CCC/C(C)=C/CC/C(C)=C/COP(=O)([O-])OP(=O)([O-])[O-]

Solution

7.1 — Reaction types

The thioester end of HMG-CoA is converted into a primary alcohol. The functional-group balance is

\[ \ce{A-C(=O)-SCoA + 4[H] -> A-CH2OH + HSCoA}, \]

where \(A\) represents the unchanged remainder of the substrate and \([H]\) denotes formal hydrogen equivalents.

This transformation involves reduction of the acyl carbon and cleavage of its bond to sulfur. Mechanistically, reduction of the thioester produces an aldehyde with elimination of the CoA thiol fragment; further reduction gives the primary alcohol. The overall thioester cleavage is a hydrogenolysis, and no water is consumed.

Thus E1 performs reduction and elimination/nonhydrolytic C–S cleavage.

The three subsequent steps are phosphorylations:

  1. E2 phosphorylates the terminal primary alcohol.
  2. E3 adds a second phosphoryl group to form the diphosphate ester.
  3. E4 phosphorylates the tertiary hydroxyl group.

The last phosphorylation makes that oxygen substituent a suitable leaving group. Subsequent loss of phosphate and decarboxylation produces IPP. Therefore E3 catalyzes phosphorylation.

7.2 — Identification and absolute configuration of X

Reduction of the thioester group gives

\[ \ce{HOOC-CH2-C(OH)(CH3)-CH2-CH2OH}, \]

which is mevalonic acid. Its stereocenter is C3.

The CIP priorities in the product are

\[ \ce{OH} > \ce{CH2COOH} > \ce{CH2CH2OH} > \ce{CH3}. \]

In the starting HMG-CoA, the thioester-containing side chain has higher priority than the carboxylic-acid-containing side chain: the comparison ultimately encounters sulfur versus oxygen. After reduction, the carboxylic-acid-containing chain has higher priority than the hydroxyethyl chain.

Consequently, priorities 2 and 3 exchange places while the spatial arrangement at C3 is retained. The starting \(S\) descriptor therefore becomes \(R\).

The required drawing is (R)-mevalonic acid, SMILES C[C@@](O)(CCO)CC(=O)O. An unambiguous Fischer projection has:

  • \(\ce{CH2COOH}\) above the stereocenter;
  • \(\ce{CH2CH2OH}\) below;
  • \(\ce{CH3}\) on the left;
  • \(\ce{OH}\) on the right.

As usual for a Fischer projection, the horizontal bonds point toward the viewer and the vertical bonds point away.

7.3 — Reductive ozonolysis of DAP

DAP is 3-methylbut-2-en-1-yl diphosphate:

CC(C)=CCOP(=O)([O-])OP(=O)([O-])[O-]

Its alkene has two methyl groups on one carbon and one hydrogen plus a diphosphate-bearing side chain on the other:

\[ \ce{(CH3)2C=CH-CH2-OPP}. \]

Reductive cleavage therefore produces a ketone at the dimethyl-substituted carbon and an aldehyde at the other carbon. Dimethyl sulfide accepts the remaining oxygen atom from ozone:

\[ \boxed{ \ce{(CH3)2C=CHCH2OPP + O3 + (CH3)2S -> (CH3)2CO + OHCCH2OPP + (CH3)2SO} } \]

The named species and their SMILES are:

Species Name SMILES
\(\ce{O3}\) Ozone [O-][O+]=O
\(\ce{(CH3)2S}\) Dimethyl sulfide CSC
\(\ce{(CH3)2CO}\) Acetone, or propanone CC(=O)C
\(\ce{OHCCH2OPP}\) 2-Oxoethyl diphosphate O=CCOP(=O)([O-])OP(=O)([O-])[O-]
\(\ce{(CH3)2SO}\) Dimethyl sulfoxide CS(=O)C

The equation conserves every atom and the diphosphate ester’s charge. Each alkene consumes one molecule each of ozone and dimethyl sulfide.

7.4 — Molecular formula of Y

The DAP-derived terminal \(\ce{(CH3)2C=}\) groups yield acetone on ozonolysis. Hence Y1 is acetone, containing three carbon atoms.

The final coupling joins two prenyl chains, each with one such terminal group. One molecule of Y therefore gives two acetone molecules, so the stated ratio corresponds to

\[ 2\,\mathrm{Y1}+4\,\mathrm{Y2}+1\,\mathrm{Y3} \]

per molecule of Y.

Since reductive ozonolysis preserves the carbon atoms in these organic fragments,

\[ N_{\mathrm C} =2(3)+4(5)+1(4) =30. \]

There are seven fragments. Cleaving each double bond in an acyclic carbon skeleton increases the number of separate fragments by one, so

\[ N_{\text{double bonds}}=7-1=6. \]

An acyclic hydrocarbon with 30 carbon atoms and six double bonds has

\[ N_{\mathrm H}=2(30)+2-2(6)=50. \]

Thus

\[ \boxed{\ce{Y = C30H50}}. \]

7.5 — Numbers of IPP and DAP molecules

Both IPP and DAP contain five carbon atoms. No carbon is lost during assembly, and the final reduction changes only the hydrogen count. Therefore six five-carbon building blocks are required:

\[ N_{\mathrm{IPP}}+N_{\mathrm{DAP}}=\frac{30}{5}=6. \]

The two identical residues in Scheme 2 each contain 15 carbon atoms, hence three five-carbon units.

Each residue has one DAP-derived terminal dimethyl-substituted alkene. The remaining two units extend that chain and come from IPP. Thus each Y4 is assembled from

\[ 1\text{ DAP}+2\text{ IPP}. \]

For two molecules of Y4,

\[ \boxed{N_{\mathrm{DAP}}=2,\qquad N_{\mathrm{IPP}}=4}. \]

This also satisfies the diphosphate balance: four ordinary coupling steps release four diphosphates, and the final coupling releases two more, accounting for all six starting diphosphate groups.

7.6 — Coupling one IPP with one DAP

IPP is 3-methylbut-3-en-1-yl diphosphate:

C=C(C)CCOP(=O)([O-])OP(=O)([O-])[O-]

Numbering from the carbon bearing the diphosphate ester, the required new bond joins C1 of DAP to C4 of IPP, the terminal methylene carbon.

The DAP diphosphate group leaves, while the IPP diphosphate group is retained. Formation of the new bond and loss of a proton relocate the IPP double bond from C3–C4 to C2–C3. The product is

\[ \ce{PP-O-CH2-CH=C(CH3)-CH2-CH2-CH=C(CH3)2}. \]

This is geranyl diphosphate, with the stereochemical SMILES given in the Answers.

Its reductive ozonolysis gives

\[ \ce{GPP + 2O3 + 2(CH3)2S -> (CH3)2CO + CH3COCH2CH2CHO + OHCCH2OPP + 2(CH3)2SO}. \]

The three organic fragments are:

  • Y1: acetone, CC(=O)C;
  • Y2: 4-oxopentanal, \(\ce{CH3COCH2CH2CHO}\), CC(=O)CCC=O;
  • 2-oxoethyl diphosphate, O=CCOP(=O)([O-])OP(=O)([O-])[O-].

Their carbon counts are \(3+5+2=10\), as required.

Ozonolysis determines the connectivity but does not distinguish alkene geometries. The later trans requirement for Y4, whose existing double bonds are retained during chain extension, selects the \(2E\) geranyl structure.

7.7 — Structures and stereochemistry of Y and Y4

Adding a second IPP unit to geranyl diphosphate gives the 15-carbon intermediate farnesyl diphosphate, Y4:

\[ \ce{PP-O-CH2-CH=C(CH3)-CH2-CH2-CH=C(CH3)-CH2-CH2-CH=C(CH3)2}. \]

For the required drawing, place the double bonds at C2–C3 and C6–C7 in the \(E\) configuration. The terminal C10–C11 double bond has two identical methyl substituents at C11 and therefore has no \(E/Z\) designation.

Thus Y4 is (2E,6E)-farnesyl diphosphate:

CC(C)=CCC/C(C)=C/CC/C(C)=C/COP(=O)([O-])OP(=O)([O-])[O-]

In Scheme 2, the two farnesyl residues join through the carbons that originally bore diphosphate. The newly formed double bond is subsequently reduced. Therefore, to draw Y, replace the two terminal \(\ce{C1-O}\) bonds of two Y4 molecules with a C1–C1′ single bond, retaining all pre-existing alkene configurations.

The central segment is consequently

\[ \ce{-C(CH3)=CH-CH2-CH2-CH=C(CH3)-}. \]

The resulting hydrocarbon is squalene, specifically

(6E,10E,14E,18E)-2,6,10,15,19,23-hexamethyltetracosa-2,6,10,14,18,22-hexaene:

CC(C)=CCC/C(C)=C/CC/C(C)=C/CC/C=C(C)/CC/C=C(C)/CCC=C(C)C

Its ozonolysis provides a direct structural check:

Product Origin in squalene Molecules per molecule of Y SMILES
Y1: acetone Two terminal dimethyl-substituted alkene carbons 2 CC(=O)C
Y2: 4-oxopentanal Four intervening five-carbon segments 4 CC(=O)CCC=O
Y3: butane-1,4-dial, or succinaldehyde Central four-carbon segment 1 O=CCCC=O

The balanced overall check is

\[ \ce{C30H50 + 6O3 + 6(CH3)2S -> 2(CH3)2CO + 4CH3COCH2CH2CHO + OHCCH2CH2CHO + 6(CH3)2SO}. \]

This reproduces the required \(2:4:1\) ratio and the specified carbon counts.

The solution was checked against the official solution.