IChO 2012 · Theory problem 7

An artificial enzyme for a Diels–Alder reaction

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An artificial enzyme was designed to bind the two substrate molecules shown below (diene and dienophile) and catalyze a Diels-Alder reaction between them.

a. There are eight potential products from a Diels-Alder reaction involving these two molecules in the reaction without any enzyme.

7.1 Draw the structures of any two of the potential products that are regioisomers of each other, in the boxes that are given below. Use wedges (\(\blacktriangleleft\)) and dashes (\(\cdots\)) to show the stereochemistry of each product in your drawings. Use R and R′ shown below to represent the substituents in the molecules that are not directly involved in the reaction.

Figure 1
Figure 2

7.2 Draw the structures of any two of the potential products that are enantiomers of each other, in the boxes that are given below. Use wedges (\(\blacktriangleleft\)) and dashes (\(\cdots\)) to show the stereochemistry of each product in your drawings. Use R and R′ as in part (7.1).

7.3 Draw the structures of any two of the potential products that are diastereomers of each other, in the boxes that are given below. Use wedges (\(\blacktriangleleft\)) and dashes (\(\cdots\)) to show the stereochemistry of each product in your drawings. Use R and R′ as in part (7.1).

b. The rate and regioselectivity of a Diels-Alder reaction depend on the degree of electronic complementarity between the two reactants. The structures of the diene and the dienophile from part a are given below.

7.4 Circle the carbon atom in the diene that has increased electron density and therefore can act as an electron donor during the reaction. Draw one resonance structure of the diene in the box to support your answer. Indicate all non-zero formal charges on the atoms in the resonance structure that you draw.

7.5 Circle the carbon atom in the dienophile that has decreased electron density and therefore can act as an electron acceptor during the reaction. Draw one resonance structure of the dienophile in the box to support your answer. Indicate all non-zero formal charges on the atoms in the resonance structure that you draw.

7.6 Based on your assignments in parts (7.4) and (7.5), predict the regiochemistry of the uncatalyzed Diels-Alder reaction of the diene and dienophile by drawing the major product. You need not show the stereochemistry of the product in your drawing.

c. The figure below shows the Diels-Alder reactants as they are bound before they enter the transition state for product formation in the active site of the artificial enzyme. The gray area represents a cross-section through the enzyme. The dienophile is below the cross-section plane whereas the diene is above the cross-section plane, when the two molecules are bound in the active site that is shown.

7.7 Draw the structure of the product of the enzyme-catalyzed reaction in the box given below. Show the stereochemistry of the product in your drawing and use R and R′ as you did for question a.

Figure 3

d. Consider the following statements about enzymes (artificial or natural). For each statement, indicate whether that statement is True or False (draw a circle around “True” or “False”).

7.8 Enzymes bind more tightly to the transition state than to the reactants or products of the reaction.

True  False

7.9 Enzymes alter the equilibrium constant of the reaction to favour the product.

True  False

7.10 Enzymatic catalysis always increases the entropy of activation of the reaction compared to the uncatalyzed reaction.

True  False

e. Modified versions of the artificial enzymes with different catalytic activities were prepared (enzymes I, II, III, and IV, shown in the figure below).

Figure 4

Two amino acid residues that differ among the different enzymes are shown. Assume that the enzyme functional groups shown are located in close proximity to the matching fragments of the reagents when they form the transition state in the enzyme active site.

7.11 Of these four enzymes which one would cause the greatest increase in the rate of the Diels-Alder reaction compared to the uncatalyzed reaction?

f. The substrate specificity of the artificial enzymes V and VI (see below) was tested by using the dienophile reactants 1 - 6, shown below.

Figure 5

7.12 Dienophile #1 reacted most rapidly in the reaction catalyzed by artificial enzyme V (see below). However, artificial enzyme VI catalyzed the reaction most rapidly with a different dienophile. Of the six dienophiles shown above, which one would react most rapidly in the Diels-Alder reaction catalyzed by enzyme VI?

Figure 6

Official English text of the problem.

Answers

The substituents retain their definitions from the problem:

  • \(R\): [(4-carboxylatobenzyl)oxycarbonyl]amino, fragment SMILES [*]NC(=O)OCc1ccc(C(=O)[O-])cc1.
  • \(R'\): N,N-dimethylcarbamoyl, fragment SMILES [*]C(=O)N(C)C.

Here [*] denotes attachment to the rest of the structure. Product names below use these fully specified substituents; all product SMILES expand them explicitly. Number the cyclohexene ring with its double bond at C1=C2 and \(R\) at C3.

  • 7.1: One suitable pair of regioisomers is:

A: cis-(3R,4S)-3-\(R\)-4-\(R'\)-cyclohex-1-ene:
C1=C[C@@H](NC(=O)OCc2ccc(C(=O)[O-])cc2)[C@@H](C(=O)N(C)C)CC1

B: cis-(3S,5S)-3-\(R\)-5-\(R'\)-cyclohex-1-ene:
C1=C[C@@H](NC(=O)OCc2ccc(C(=O)[O-])cc2)C[C@@H](C(=O)N(C)C)C1

  • 7.2: A and its enantiomer Ā, cis-(3S,4R)-3-\(R\)-4-\(R'\)-cyclohex-1-ene:
    C1=C[C@H](NC(=O)OCc2ccc(C(=O)[O-])cc2)[C@H](C(=O)N(C)C)CC1

  • 7.3: A and T, trans-(3R,4R)-3-\(R\)-4-\(R'\)-cyclohex-1-ene:
    C1=C[C@@H](NC(=O)OCc2ccc(C(=O)[O-])cc2)[C@H](C(=O)N(C)C)CC1

  • 7.4: Circle the diene’s terminal \(\ce{CH2}\) carbon farthest from nitrogen. A supporting charge-separated resonance contributor of the diene is:
    [CH2-]C=CC=[NH+]C(=O)OCc1ccc(C(=O)[O-])cc1

  • 7.5: Circle the dienophile’s terminal \(\ce{CH2}\) carbon, the carbon \(\beta\) to its carbonyl. A supporting charge-separated resonance contributor of N,N-dimethylprop-2-enamide is:
    [CH2+]C=C([O-])N(C)C

  • 7.6: The major connectivity is 3-\(R\)-4-\(R'\)-cyclohex-1-ene, with adjacent substituents; stereochemistry is unspecified:
    C1=CC(NC(=O)OCc2ccc(C(=O)[O-])cc2)C(C(=O)N(C)C)CC1

  • 7.7: T, trans-(3R,4R)-3-\(R\)-4-\(R'\)-cyclohex-1-ene:
    C1=C[C@@H](NC(=O)OCc2ccc(C(=O)[O-])cc2)[C@H](C(=O)N(C)C)CC1

  • 7.8: True.

  • 7.9: False.
  • 7.10: False.
  • 7.11: Enzyme II.
  • 7.12: Dienophile 5, N-ethyl-N-(2-hydroxyethyl)prop-2-enamide:
    C=CC(=O)N(CC)CCO

Solution

The starting diene is (4-carboxylatobenzyl) N-[(1E)-buta-1,3-dien-1-yl]carbamate, C=C/C=C/NC(=O)OCc1ccc(C(=O)[O-])cc1. The dienophile is N,N-dimethylprop-2-enamide, also called N,N-dimethylacrylamide, C=CC(=O)N(C)C.

Label the diene carbons D1–D4, starting at the end bonded to nitrogen. Thus its backbone is \(R\)–D1=D2–D3=D4. Label the dienophile’s carbonyl-adjacent alkene carbon \(\alpha\), and its terminal alkene carbon \(\beta\).

Cycloaddition forms two carbon–carbon single bonds, leaves D2=D3 as the ring double bond, and eliminates no atoms. Its balanced molecular equation is

\[ \ce{C13H12NO4^- + C5H9NO -> C18H21N2O5^-}. \]

For all product drawing instructions below, draw the hexagon with C1 upper left, C2 left, C3 lower left, C4 lower right, C5 right, and C6 upper right. Put the double bond between C1 and C2. A solid wedge points toward the viewer; a hashed wedge points away.

7.1 — Regioisomers

There are two possible orientations for joining the unsymmetrical reactants:

New bonds Positions of substituents in the product
D1–\(\alpha\) and D4–\(\beta\) \(R\) at C3 and \(R'\) at C4
D1–\(\beta\) and D4–\(\alpha\) \(R\) at C3 and \(R'\) at C5

These give different connectivities and therefore regioisomers.

Draw A with \(R\) at C3 and \(R'\) at C4, both on hashed wedges. Draw B with \(R\) at C3 and \(R'\) at C5, again both on hashed wedges. At each stereocenter, the implicit hydrogen points toward the viewer.

In full, the first product’s structural name is (3R,4S)-3-{[(4-carboxylatobenzyl)oxycarbonyl]amino}-4-(N,N-dimethylcarbamoyl)cyclohex-1-ene; the second is the corresponding (3S,5S)-5-(N,N-dimethylcarbamoyl) regioisomer. Their full SMILES are given in the answers.

Each connectivity has two stereogenic centers and four configurations. There is no meso form, so the total number is

\[ 2\text{ connectivities}\times 4\text{ stereoisomers}=8. \]

7.2 — Enantiomers

Enantiomers have identical connectivity and opposite configurations at every stereogenic center.

Use A, with both substituents on hashed wedges, and Ā, with both substituents on solid wedges:

\[ (3R,4S)\longleftrightarrow(3S,4R). \]

These structures are nonsuperimposable mirror images. The implicit hydrogen at each center is opposite its displayed substituent.

7.3 — Diastereomers

Use A and T. Both have \(R\) at C3 and \(R'\) at C4.

Keep the C3–\(R\) bond on a hashed wedge in both drawings. Change only the C4–\(R'\) bond from a hashed wedge in A to a solid wedge in T:

\[ (3R,4S)\longrightarrow(3R,4R). \]

Thus A is cis and T is trans. They have the same connectivity but are not mirror images, so they are diastereomers.

7.4 — Electron-rich carbon of the diene

The nitrogen lone pair can donate into the conjugated diene. Draw the supporting resonance contributor by making these changes simultaneously:

  1. Convert the N–D1 bond into an N=D1 double bond.
  2. Move the D1=D2 \(\pi\) electrons to form D2=D3.
  3. Move the D3=D4 \(\pi\) electrons onto D4.

The resulting bond sequence is N=D1–D2=D3–D4. Nitrogen has formal charge \(+1\), and terminal D4 has formal charge \(-1\).

The complete resonance contributor is:

[CH2-]C=CC=[NH+]C(=O)OCc1ccc(C(=O)[O-])cc1

The pre-existing carboxylate oxygen must also retain its formal charge \(-1\). Thus the complete structure contains three nonzero formal charges, and its net charge remains \(-1\).

This contributor identifies D4, the terminal carbon remote from nitrogen, as the electron-rich reacting terminus.

7.5 — Electron-poor carbon of the dienophile

Conjugation with the carbonyl withdraws electron density from the terminal alkene carbon:

\[ \ce{CH2=CH-C(=O)N(CH3)2 <-> ^+CH2-CH=C(O^-)N(CH3)2}. \]

To draw this contributor, move the alkene \(\pi\) electrons toward the bond between the \(\alpha\) carbon and the carbonyl carbon, and move the carbonyl \(\pi\) electrons onto oxygen.

The resulting contributor, [CH2+]C=C([O-])N(C)C, has formal charge \(+1\) on the terminal \(\beta\) carbon and \(-1\) on oxygen. Its overall charge is zero.

Therefore, circle the terminal \(\beta\)-\(\ce{CH2}\) carbon.

7.6 — Uncatalyzed regioselectivity

The electron-rich diene terminus D4 preferentially bonds to the electron-poor dienophile terminus \(\beta\). Ring closure then makes the second bond between D1 and \(\alpha\).

Since \(R\) is attached to D1 and \(R'\) to \(\alpha\), the two substituents become adjacent.

Draw a cyclohex-1-ene ring with \(R\) at C3 and \(R'\) at C4, using ordinary bonds without specifying stereochemistry. This is the constitutional product named and encoded in answer 7.6.

7.7 — Product selected by the enzyme

The binding diagram aligns D1 with \(\alpha\) and D4 with \(\beta\), giving the same adjacent-substituent connectivity.

The specified depth ordering also fixes the reacting faces: the dienophile approaches the diene from below its plane. Preserving the substituent orientations shown in the figure gives, in the product drawing convention defined above:

  • C3–\(R\): hashed wedge, away from the viewer.
  • C4–\(R'\): solid wedge, toward the viewer.

Consequently, the substituents are trans, and the product is T, trans-(3R,4R)-3-\(R\)-4-\(R'\)-cyclohex-1-ene.

The CIP priorities are:

\[ \begin{aligned} \text{At C3:}\quad& \text{N of }R > \text{C4 path} > \text{C2 path} > \ce{H},\\ \text{At C4:}\quad& \text{carbonyl C of }R' > \text{C3 path} > \text{C5 path} > \ce{H}. \end{aligned} \]

These priorities give the stated \((3R,4R)\) configuration and the stereochemical SMILES in answer 7.7.

7.8 — Transition-state binding

True. Preferential stabilization of the transition state relative to the reactants lowers the activation free energy. This is the basis of the transition-state binding description of enzyme catalysis.

7.9 — Equilibrium constant

False. An enzyme changes reaction rates, but it does not change the standard reaction Gibbs energy or the equilibrium constant:

\[ K=\exp\!\left(-\frac{\Delta_{\mathrm r}G^\circ}{RT}\right). \]

It accelerates equilibration in both directions without shifting the equilibrium composition.

7.10 — Activation entropy

False. Catalysis requires a lower activation free energy, not necessarily a larger activation entropy:

\[ \Delta G^\ddagger=\Delta H^\ddagger-T\Delta S^\ddagger. \]

Preorganization can reduce an entropic penalty, but favorable enthalpic interactions can also account for acceleration. There is no requirement that \(\Delta S^\ddagger\) always increase.

7.11 — Most effective enzyme

Enzyme II supplies both complementary hydrogen-bonding interactions.

Its glutamine residue, side-chain fragment [*]CCC(=O)N, presents a carbonyl oxygen that accepts a hydrogen bond from the diene’s N–H group. This interaction supports the nitrogen’s electron-donating role and stabilizes the developing positive character represented by the diene resonance contributor.

Its tyrosine residue, side-chain fragment [*]Cc1ccc(O)cc1, donates a hydrogen bond from its phenolic O–H group to the dienophile’s carbonyl oxygen. This interaction increases dienophile polarization and stabilizes developing negative character on oxygen.

Here the side-chain wildcard denotes attachment to the protein backbone.

Enzyme II therefore combines activation of both reaction partners. Enzyme III supplies the tyrosine interaction, enzyme IV supplies the glutamine interaction, and enzyme I supplies neither of these two polar contacts.

7.12 — Substrate preferred by enzyme VI

The preferred substrate is dienophile 5, N-ethyl-N-(2-hydroxyethyl)prop-2-enamide, C=CC(=O)N(CC)CCO.

Two changes in the binding pocket explain this choice:

  • Isoleucine → valine: the side-chain fragments change from [*]C(C)CC to [*]C(C)C. The smaller valine side chain provides additional space for a larger substrate substituent.
  • Alanine → asparagine: the side-chain fragments change from [*]C to [*]CC(=O)N. The pocket-facing asparagine carbonyl oxygen provides a hydrogen-bond acceptor.

Dienophile 5 combines larger N-substituents with a terminal hydroxyl group positioned to donate a hydrogen bond to the asparagine carbonyl oxygen.

The distinction from dienophile 6 is significant: N-acryloylmorpholine, C=CC(=O)N1CCOCC1, has an ether oxygen but no hydroxyl hydrogen. It cannot donate the same hydrogen bond to the pocket-facing carbonyl oxygen.

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