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Química Medicinal · Drug design

Reconhecimento molecular e bioisósteros

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O que você vai aprender

  • Identify hydrogen-bond donors and acceptors, salt bridges, cation–π contacts, π-stacking, halogen bonds and the hydrophobic effect in a ligand–receptor complex.
  • Convert between Kd and ΔG° with ΔG° = RT ln Kd, split ΔG° into ΔH° and −TΔS°, and calculate ligand efficiency and lipophilic efficiency.
  • Explain the entropic cost of binding and how conformational restriction, macrocyclic preorganisation and an added methyl group can lower it.
  • Choose carboxylic acid, amide and phenyl bioisosteres (tetrazole, acylsulfonamide, 1,2,3-triazole, 1,2,4-oxadiazole, fluoroalkene, bicyclo[1.1.1]pentane) and draw the reactions that make them.
  • Predict how fluorine changes the basicity of nearby amines and the conformation of a chain, and explain chiral recognition with the three-point model and the eudismic ratio.
  • Interpret host–guest binding by crown ethers and cyclodextrins and calculate a dissociation constant from titration data.

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Non-covalent contacts

A ligand binds a receptor through many weak contacts. A hydrogen bond links a donor X–H (O–H, N–H) to an acceptor lone pair (carbonyl O, pyridine-type N, ether O). Lone pairs that are part of an aromatic sextet (pyrrole N) or delocalised into a carbonyl (amide N) are poor acceptors. A salt bridge is a hydrogen-bonded pair of opposite charges, such as a carboxylate and an ammonium ion. In a cation–π contact a cation, often a quaternary ammonium ion that cannot donate a hydrogen bond, sits over the electron-rich face of a phenylalanine, tyrosine or tryptophan ring. Aromatic rings stack offset or edge-to-face rather than exactly face-to-face, which would put their electron-rich π faces in electrostatically repulsive contact. Heavier halogens bonded to carbon carry a positive cap, the σ-hole, on the C–X axis opposite the C–X bond, so a C–I or C–Br group binds a carbonyl oxygen with a near-linear C–X···O angle; the order is C–I > C–Br > C–Cl, and C–F does not form halogen bonds. The hydrophobic effect is mainly entropic: ordered water around non-polar surfaces is released when those surfaces meet.

Binding free energy

For L + R ⇌ RL the dissociation constant is Kd = [L][R]/[RL]. With the 1 mol/L standard state, ΔG°bind = RT ln Kd (Kd in mol/L), so a stronger binder has a smaller Kd and a more negative ΔG°. At 298 K, RT = 2.478 kJ/mol and every tenfold change in Kd is 2.303RT = 5.71 kJ/mol; Kd = 10 nM corresponds to −45.6 kJ/mol. Isothermal titration calorimetry measures ΔH° directly and gives −TΔS° = ΔG° − ΔH°. Analogues in a series often trade enthalpy for entropy: a tighter, more ordered complex gains ΔH° but loses TΔS°, so ΔG° barely changes (enthalpy–entropy compensation). Ligand efficiency LE = −ΔG°/(number of heavy atoms) shows how much affinity each non-hydrogen atom buys, and lipophilic efficiency LipE = pIC50 − logP shows whether potency comes from specific contacts or just from greasiness.

The entropic price and conformational restriction

A flexible ligand must freeze its rotatable bonds and lose translational and rotational freedom when it binds. If the bound conformation is built into a ring, a double bond or a macrocycle before binding, that part of the entropy cost is already paid, and affinity rises, but only if the locked shape matches the bound one. A locked wrong shape loses affinity, because the right conformation can no longer be reached. 18-Crown-6 binds K⁺ much more strongly than the open-chain polyether with the same six oxygens for the same reason: the macrocycle is preorganised. A single added methyl group can raise potency tenfold or more (the 'magic methyl' effect), most often because ortho or N-methyl groups twist a biaryl or an amide into the bound conformation and because the methyl fills a small hydrophobic pocket.

Carboxylic acid bioisosteres

A carboxylic acid (pKa about 4–5) is almost fully ionised at pH 7.4, which helps binding to arginine or lysine but limits membrane permeability. A 5-substituted 1H-tetrazole has a similar pKa (5-phenyl about 4.5) and, as the anion, spreads its charge over four ring nitrogens, so it makes the same salt bridges with a slightly larger, more lipophilic anion. Tetrazoles are made from nitriles and azide ion, with NH₄Cl or ZnBr₂ to activate the nitrile: azide adds to the nitrile carbon to give an imidoyl azide, which closes by 1,5-electrocyclisation. N-Acylsulfonamides, RCONHSO₂R′, are also acidic (pKa about 4–5) because the N⁻ is flanked by a carbonyl and a sulfonyl group. They are made by activating the acid with carbonyldiimidazole (CDI) and treating the acyl imidazole with a sulfonamide and DBU. A plain sulfonamide (pKa about 10) or a hydroxamic acid (about 9) is far less acidic.

Amide and phenyl replacements

Amides can be hydrolysed by peptidases. A 1,4-disubstituted 1,2,3-triazole, made by copper-catalysed azide–alkyne cycloaddition (CuAAC), places its two substituents like the ends of a trans amide; the 1,5-disubstituted isomer, made by ruthenium-catalysed cycloaddition (RuAAC, for example with Cp*RuCl(PPh₃)₂), holds them on adjacent ring atoms like the ends of a cis amide. The uncatalysed (thermal) cycloaddition of a terminal alkyne gives both regioisomers. A 1,2,4-oxadiazole mimics an ester or amide and is not hydrolysed by peptidases; it is made by O-acylation of an amidoxime, itself formed from a nitrile and hydroxylamine, followed by cyclodehydration on heating. A (Z)-fluoroalkene, R–CF=CH–R′, copies the shape and dipole of a trans amide with F in place of O. Bicyclo[1.1.1]pentane (BCP) replaces a phenyl or para-phenylene ring with a saturated cage whose bridgehead bonds point 180° apart; it is shorter (about 1.85 Å between bridgehead carbons against about 2.8 Å for the para carbons of benzene), cannot π-stack, and usually improves solubility.

Fluorine and methyl effects

Fluorine is small (van der Waals radius 1.47 Å against 1.20 Å for H) but the most electronegative element. Each fluorine on the carbon β to an amine nitrogen lowers the pKa of the ammonium ion by about 1.7 units: ethylamine 10.7, 2-fluoroethylamine 9.0, 2,2-difluoroethylamine 7.3, 2,2,2-trifluoroethylamine 5.7. Fluorine also biases conformation: 1,2-difluoroethane prefers the gauche conformer, because each C–H bond donates best into an anti C–F σ* orbital, and in 2-fluoroethylammonium ions the gauche preference is reinforced by the C–F/N⁺ charge–dipole attraction. An aryl C–F bond (about 525 kJ/mol) is much stronger than an aryl C–H bond (about 470 kJ/mol), so fluorine is placed at sites where C–H bonds are oxidised. Organic fluorine is a weak hydrogen-bond acceptor.

Chirality in recognition

Receptors are chiral, so enantiomers bind differently. In the Easson–Stedman three-point model the more active enantiomer (the eutomer) makes three complementary contacts with three sites of the binding pocket; its mirror image (the distomer) can match only two at a time. The eudismic ratio is the ratio of activities, for example Ki(distomer)/Ki(eutomer), and RT ln(eudismic ratio) is the difference in binding free energy between the two enantiomers. If the distomer is inactive, a racemate contains only half as much eutomer, so its IC50 is twice that of the pure eutomer.

Host–guest models

Synthetic hosts show the same rules in a simple form. 18-Crown-6 binds K⁺ about 60 times more strongly than Na⁺ in methanol (log K about 6.1 and 4.3), because K⁺ fits its cavity and all six oxygens contact the ion; the smaller Na⁺ cannot reach all six without strain. β-Cyclodextrin, a cyclic heptamer of glucose, has a hydrophobic cavity about 6–6.5 Å wide that includes a benzene ring or an adamantyl group in water, driven by release of cavity water and by dispersion contacts. For a 1:1 complex, Kd = [H][G]/[HG]; if the fraction of guest bound is measured at known total concentrations, mass balance gives every concentration and hence Kd.

What to carry forward

Affinity is a free energy: ΔG° = RT ln Kd, 5.7 kJ/mol for every tenfold change at 298 K, built from many weak contacts and paid for in lost freedom. Judge leads by efficiency (LE, LipE), not by raw potency. Bioisosteres keep the contacts and fix a liability: tetrazoles and acylsulfonamides keep the acidity of a carboxylic acid, triazoles, oxadiazoles and fluoroalkenes keep the geometry of an amide without its hydrolysis, BCP keeps the exit vectors of a phenyl ring without its flatness, and fluorine tunes basicity and conformation. Chiral receptors distinguish enantiomers, and crown ethers and cyclodextrins show the same size, fit and preorganisation rules in simple host–guest systems.

Problemas de prática (40)

Praticar é grátis com uma conta.

Desenhe o mecanismo 9

Acid and amide bioisosteres

  • N-Acylsulfonamide from an acyl imidazole, methanesulfonamide and DBUDraw the acylation of methanesulfonamide by 1-benzoylimidazole with DBU as the base, explain why the imidazole leaves only after protonation, and show why the product is formed as its DBU salt.
  • Benzamidoxime from benzonitrile and hydroxylamineDraw the addition of hydroxylamine nitrogen to benzonitrile and the proton transfers that give (Z)-benzamidoxime, the precursor of 1,2,4-oxadiazoles.
  • O-Acylation of benzamidoxime by acetyl chlorideDraw the O-acylation of benzamidoxime by acetyl chloride in pyridine and explain why the oxime oxygen, not the NH₂ group, is acylated.
  • Thermal azide–alkyne cycloaddition: benzyl azide and DMADDraw the concerted [3+2] cycloaddition of benzyl azide with dimethyl acetylenedicarboxylate to a 1,2,3-triazole, and explain why the same reaction of a terminal alkyne needs copper to give a single regioisomer.
  • Acylating bicyclo[1.1.1]pentan-1-amine: a saturated stand-in for anilineDraw the addition–elimination of bicyclo[1.1.1]pentan-1-amine with benzoyl chloride and the deprotonation by triethylamine, and explain why the bicyclo[1.1.1]pentyl group is used in place of a phenyl ring.
  • Cyclodehydration of an O-acylamidoxime to a 1,2,4-oxadiazoleDraw the ring closure of O-acetyl benzamidoxime onto its ester carbonyl, the proton transfers through pyridine, and the loss of water that gives aromatic 5-methyl-3-phenyl-1,2,4-oxadiazole.
  • Deprotonating 5-phenyl-1H-tetrazole: a delocalised anion like carboxylateDraw the proton transfer from 5-phenyl-1H-tetrazole to hydroxide and a resonance form of the tetrazolide anion that shows its charge spread over the ring nitrogens.
  • Making a tetrazole: benzonitrile, sodium azide and ammonium chlorideDraw how azide ion and an ammonium proton convert benzonitrile into an imidoyl azide, how that intermediate closes to 5-phenyl-1H-tetrazole, and why the tetrazole ends up as its ammonium salt before work-up.

Non-covalent interactions

  • Making the ions of a salt bridge: acetic acid and methylamineDraw the proton transfer from acetic acid to methylamine that gives the carboxylate and ammonium ions of a salt bridge, and decide from pKa values which way it goes.

Preveja o produto 1

  • Locking a flexible ligandPredict how locking a flexible ligand into the bound or a non-bound conformation changes its affinity.

Classificar 3

Acid and amide bioisosteres

  • A fluoroalkene for a trans amideClassify fluoroalkene stereoisomers as mimics of a trans amide bond.

Non-covalent interactions

  • Which nitrogen accepts a hydrogen bond?Decide from where its lone pair sits whether a nitrogen atom can act as a hydrogen-bond acceptor.
  • How do aromatic rings stack?Classify arrangements of two aromatic rings as favourable or unfavourable π-stacking geometries.

Comparar 4

Acid and amide bioisosteres

  • Rank carboxylic acid replacements by acidityRank benzoic acid and candidate acid bioisosteres by acidity from their pKa values.

Fluorine, chirality and host–guest binding

  • Fluorine and amine basicityRank fluorinated ethylamines by basicity from the number of β-fluorines.

Non-covalent interactions

  • Rank aryl halides as halogen-bond donorsRank aryl halides by the strength of the halogen bond they form to a carbonyl oxygen.

Binding energy and efficiency

  • Rank a fragment, a hit and a lead by efficiencyRank compounds of different size and potency by ligand efficiency computed from Kd.

Calcular 11

Fluorine, chirality and host–guest binding

  • How much better does 18-crown-6 bind K⁺?Convert the difference in log K for two cations binding one crown ether into a free-energy difference.
  • Kd of a cyclodextrin complexCalculate the dissociation constant of a 1:1 host–guest complex from total concentrations and the fraction of guest bound.
  • How much difluoroethylamine is neutral?Calculate the neutral percentage of a β,β-difluorinated amine at pH 7.4 and compare it with the non-fluorinated amine.
  • Energy gap between enantiomersCalculate the eudismic ratio of two enantiomers and convert it into a difference in binding free energy.
  • IC50 of a racemateCalculate the IC50 of a racemate from the IC50 of its eutomer when the distomer is inactive.

Binding energy and efficiency

  • Binding free energy from KdCalculate a standard binding free energy from a dissociation constant at 298 K.
  • Energy of a hundredfold gainCalculate the extra binding free energy that corresponds to a hundredfold drop in Ki.
  • The entropy term from ITCSplit a binding free energy into its enthalpy and entropy terms from isothermal titration calorimetry data.
  • Kd from a binding free energyConvert a standard binding free energy into a dissociation constant in nanomolar units.
  • Ligand efficiency of a leadCalculate ligand efficiency in kJ/mol per heavy atom from Kd and the heavy-atom count.
  • Lipophilic efficiency of an inhibitorCalculate lipophilic efficiency from an IC50 and a logP value.

Encontre o erro 1

  • A salt bridge that cannot formFind the error in a proposed salt bridge by comparing the ligand amine's pKa with the pH.

Interprete as evidências 3

Binding energy and efficiency

  • Three analogues, one free energyInterpret calorimetry data for an analogue series that shows enthalpy–entropy compensation.
  • Potent and greasy, or weaker and lean?Use lipophilic efficiency to choose between a potent lipophilic analogue and a less potent polar one.

Non-covalent interactions

  • What drives the extra binding?Recognise the hydrophobic effect from the enthalpy and entropy changes measured by isothermal titration calorimetry.

Trabalhe de trás para frente 1

  • Disconnect a 1,2,4-oxadiazoleWork back from a 3,5-disubstituted 1,2,4-oxadiazole to the amidoxime and acyl chloride that give it.

Planeje uma síntese 2

Acid and amide bioisosteres

  • Replace a cis amide bondChoose the triazole regioisomer and catalyst that mimic a cis amide bond.

Fluorine, chirality and host–guest binding

  • Block para-hydroxylationChoose the substituent that blocks oxidative metabolism at an aromatic C–H with the smallest change in size.

Explique por quê 5

Non-covalent interactions

  • Acetylcholine and a tryptophan ringExplain why a quaternary ammonium ion binds the face of an aromatic side chain rather than a hydrogen-bond acceptor.

Fluorine, chirality and host–guest binding

  • Why 18-crown-6 prefers K⁺Explain the K⁺/Na⁺ selectivity of 18-crown-6 from the match between ion size and cavity size.
  • Why is 1,2-difluoroethane gauche?Explain the gauche preference of vicinal C–F bonds by hyperconjugation.
  • Why one enantiomer binds betterExplain with the three-point model why a chiral receptor distinguishes two enantiomers.

Binding energy and efficiency

  • A methyl group worth fiftyfoldExplain how an ortho-methyl group on a biaryl can raise potency by changing the preferred conformation.

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