From Molecules to Medicines Workshop 5/4 PDF

PA10306 / PA10311 - From Molecules to Medicines - Workshop 5 Carbonyl compounds as nucleophiles in chemistry and biochemistry In this workshop, you will build on what you have learned from the lectures about carbonyl compounds acting as nucleophiles in chemistry and biochemistry. This comes about th...

PA10306 / PA10311 - From Molecules to Medicines - Workshop 5 Carbonyl compounds as nucleophiles in chemistry and biochemistry In this workshop, you will build on what you have learned from the lectures about carbonyl compounds acting as nucleophiles in chemistry and biochemistry. This comes about through formation of enols and enolates. Feel free to discuss your work with other students during the class. The demonstrator(s) will help you with points of difficulty. If you do not finish the work during the workshop time, you must complete it in private study time before the next workshop. 1. Keto-enol tautomerism In the lectures, you were introduced to how carbonyl compounds can tautomerise to form enols and how they can form enolates. Enols are weak nucleophiles and enolates are more powerful nucleophiles but both can react in some enzyme-catalysed reactions. They are also very useful carbon nucleophiles in the synthesis and manufacture of drugs. Below are examples of C-C bond forming reactions covered in the lectures. Draw the enolate generated and show, using curly arrows, how the products are formed. What are the IUPAC names of these products? 2. The Aldol reaction – how a key sugar-metabolising enzyme works You have already met the Aldol reaction in the lectures, using simple molecules. One of the steps in the metabolism of D-glucose involves the cleavage of fructose-1,6-diphosphate (a hexose, a six-carbon sugar) into two products, each containing three carbons. The enzyme that catalyses this reaction is called aldolase; the name gives us a hint as to how the reaction proceeds. Actually, the enzyme can work reversibly and it is the reverse reaction (joining the two three-carbon units together to form fructose-1,6-bisphosphate) which is an Aldol reaction. This is the Aldol reaction which is catalysed by aldolase. Draw “curly arrows” mechanisms for both the Aldol reaction catalysed by aldolase and for the spontaneous cyclisation step which forms a hemiacetal / hemiketal (this second step does not involve enol chemistry). Why is the spontaneous cyclisation so favoured? O P OH O O O HO - OH O P HO O O H O - Dihydroxyacetone phosphate Glyceraldehyde- 3-phosphate OH O P OH O O OH HO O P HO O O O - - D-Fructose-1,6- bisphosphate keto form O O P OH O O OH HO O P HO O O OH - - D-Fructose-1,6- bisphosphate cyclic ketal form Aldol reaction catalysed by aldolase Spontaneous cyclisation OH O P OH O O OH HO O P HO O O O - - O O P OH O O OH HO O P HO O O OH - - Formation of enol O P OH O O O HO - OH O P HO O O H O - O P OH O O HO - O H : Formation of hemiketal / hemiacetal favoured by formation of 5- membered ring Aldol reaction 3. The Claisen condensation – how fatty acids are extended in the cell You have met the Claisen condensation in the lectures as a reaction in which enolates derived from esters react with the carbonyl groups of non-enolised esters to form -oxo esters (-keto esters). This process adds a two-carbon unit to the length of the carbon chain. A very similar reaction is used in mammalian cells to build up fatty acids by two carbons at a time. An acetyl group is attached to Acyl Carrier Protein through a thioester in the complex of enzymes. A base in the enzyme then removed one of the protons to form an enolate (with difficulty – why?) and this enolate reacts with carbon dioxide to make malonyl-Acyl Carrier Protein. The enzyme finds it much easier to remove a proton from the malonyl unit to form another enolate (why is it easier?). This enolate then reacts with a molecule of fatty acyl-Acyl Carrier Protein, as shown below. a. Draw full “curly arrows” reaction mechanisms for both of these steps. Notice that, in the second (Claisen) step, the intermediate has to lose a molecule of carbon dioxide; how does this happen, mechanistically? b. Explain why the malonyl-ACP forms an enolate more readily than does the acetyl-ACP. S O O Acyl carrier protein S O Acyl carrier protein H3C S O Acyl carrier protein S O O O Acyl carrier protein - O C O S O O Acyl carrier protein S O O O Acyl carrier protein - H3C S O Acyl carrier protein O C O S Acyl carrier protein H2C O - S O Acyl carrier protein S O O Acyl carrier protein - O - S O O Acyl carrier protein O O - Decarboxylation, expelling stabilised enolate as leaving group, followed by tautomerisation Claisen condensation It is easier to form the enolate from the malonyl-ACP because the enolate has conjugation 4. Synthesis of drugs and biomolecules through the Mannich reaction In many reactions of carbonyl compounds, the nucleophile is an enolate. Remember that enols are much less nucleophilic than are enolates. If we want to use an enol as a nucleophile in a carbon-carbon bond forming reaction, then the reactivity of the electrophile must be increased. This is what happens in the Mannich reaction. The Mannich reaction (or Mannich condensation) is the reaction of an enolisable ketone, an aldehyde (often formaldehyde) and an amine (usually a secondary amine) to form a new carbon-carbon bond and to attach the amine. This process is catalysed by weak acids and is used in the synthetic manufacture of some drugs and by some enzymes in cells. To understand how it works, we must consider what molecules are actually in solution in the reaction mixture. The ketone will be in equilibrium with a small amount of the corresponding enol. Neither the formaldehyde nor the ketone is sufficiently electrophilic to react with the weakly nucleophilic enol. We must then classify which are nucleophiles and which are electro- philes and which are more reactive and which are less reactive. In the first step, the strongest electrophile (formaldehyde) reacts with the strongest nucleophile (secondary amine) to form an iminium ion. You have learned about the reactions of amines with aldehydes in the lectures. Draw the “curly arrows” mechanisms for this step, including drawing the structures of the tetrahedral intermediate(s) and showing the acid-catalysis. The first curly arrows are already shown for you. The iminium is a much more powerful electrophile than the formaldehyde or the ketone and can react with the weakly nucleophilic enol to form the Mannich product. H2C R O H R" NH R' H3C R O H H O Enolisable ketone Enol Formaldehyde Secondary amine Electrophiles Nucleophiles Weak Strong H3C R O Enolisable ketone H H O Formaldehyde R" NH R' Secondary amine R O N R' R" Weak acid catalysis Mannich product R" NH R' H H O : H H N R’ R” + Iminium H H OH N R' R" H H OH2 N R' R" + The scheme below shows two key steps in the manufacture of the Selective Serotonin Re- uptake Inhibitor (SSRI) drug fluoxetine, which is used to treat depression. One of the trade names for medicines containing this drug is Prozac. Draw “curly arrows” reaction mechanisms for the steps involved in the Mannich reaction step and suggest a suitable reducing agent for the Reduction step. NaBH4 or LiAlH4 would be suitable reducing agents to reduce the ketone to the alcohol. Enol Iminium Mannich product H2C R O H H H N R’ R” + R O N R' R" Me O NH Me O H H O N Me : H H OH N Me PhCH2 H H OH2 N Me PhCH2 + H H N PhCH2 Me + H2C O H In a synthetic process, acetylbenzene (12.0 g) was subjected to the Mannich reaction with formaldehyde (6.00 g) and N-methylbenzylamine (18.2 g). This process produced Inter- mediate A (21.5 g). What was the percentage chemical yield for this process? You need to use mole calculations to work this out: Yield = 85% Intermediate A was then converted into fluoxetine in 75% overall yield through the remaining steps. How much fluoxetine was produced? 75% of 0.085 mol = 0.064 mol of product (MW 309), so mass of fluoxetine produced = 19.7 g PA10306 / PA10311 - From Molecules to Medicines - Workshop 4 The carbonyl group as an electrophile in chemistry and biochemistry In this workshop, you will build on what you have learned from the lectures about the carbonyl group as an electrophile in chemistry and biochemistry. Feel free to discuss your work with other students during the class. The demonstrator(s) will help you with points of difficulty. If you do not finish the work during the workshop time, you must complete it in private study time before the next workshop. 1. Different types of carbonyl groups have different electrophilic reactivity The electrophilic reactivity of a carbonyl group will depend on the electron-density at the carbon atom of the carbonyl. The more + the carbonyl carbon, the more reactive it is as an electrophile. Different types of carbonyl can be ranked according to +; the rank order is shown below. Give explanations (using diagrams where necessary) for the positions of amides, esters and acid chlorides in this rank order, relative to ketones. Me Cl O Me O O Me O H H O Me H O Me Me O Me OH O Me OEt O Me NH2 O Me O O - Acid chloride / Acyl chloride Acid anhydride Methanal / Formaldehyde Aldehyde Ketone Carboxylic acid Amide Carboxylate Ester Cl is strong inductive EWG but only very weakly mesomerically EDG, making the carbonyl very + The central O is inductively EWG. It is sp2 hybridised but the lone pair in the p-orbital has to be shared between two carbonyls, so only weak mesomeric EDG. but only very weakly mesomerically EDG, making the carbonyl very + Hyperconjugation makes the Me a weak EDG, making the carbonyl less + than for formaldehyde. Hyperconjugation makes both Me into weak EDGs. The OH and OEt oxygens are sp2 hybridised, allowing mesomeric donation of electron density from the lone pair into the carbonyl, more than overcoming the inductive electron-withdrawing effect. The N is sp2 hybridised, allowing mesomeric donation of electron density from the lone pair into the carbonyl, more than overcoming the weaker inductive electron-withdrawing effect. An anion! Nucleophiles are unlikely to approach (electrostatics). 2. Reaction of ketones and aldehydes with hydrogen nucleophiles In the lectures, you have already seen carbonyl groups acting as electrophiles in their reactions with oxygen and nitrogen nucleophiles in chemical and biochemical systems. Here, you extend your understanding of the mechanisms of these reactions into putting forward mechanisms and predicting the products of the reactions of various carbonyl compounds with hydrogen nucleophiles. Reactions of carbonyl electrophiles with hydrogen nucleophiles are important in many biosynthetic processes, including the biosynthesis of fatty acids, and in the synthesis and manufacture of drugs. Hydrogen can exist in two ionic forms, the proton (H+) and the hydride ion (H-). The proton is the ion involved in Brønsted acids, such as hydrochloric acid and acetic acid. The hydride ion is present in sodium hydride (Na+ H-), a solid that reacts violently with water and with alcohols. However, although H- is a very strong base, it is completely unreactive as a nucleo-phile. Therefore, more complex equivalents of H- must be used to provide a suitable hydrogen nucleophile. These include lithium aluminium hydride (LiAlH4) and sodium borohydride (NaBH4). The reaction of a hydrogen nucleophile with an electrophilic carbonyl is a reduction. LiAlH4 is a very strong base that reacts violently with alcohols and with water. Therefore, it must be used in an unreactive solvent such as diethyl ether. It is a very powerful nucleophile that will react with any type of carbonyl group, anywhere in the rank order of reactivity. It will reduce aldehydes to primary alcohols and ketones to secondary alcohols. NaBH4 is a much weaker nucleo- phile that is only strong enough to react with ketones and other carbonyl groups which are more polarised. Draw the mechanism of the reaction of NaBH4 with benzaldehyde. R H O H Al H H H R H O H R H HO H R R O H Al H H H R H O R' R H HO R' Addition of water Addition of water R H O H Al H H H R H O H R H HO H R R O H Al H H H R H O R' R H HO R' Addition of water Addition of water Addition of water Addition of water Addition of water Addition of water OH H H H O H B H H H - The two most common biochemical reducing agents are NADH (Nicotinamide Adenine Dinucleotide – reduced form) and NADPH (Nicotinamide Adenine Dinucleotide Phosphate – reduced form). These act like borohydride ion as a source of nucleophilic H- which attacks and reduces electrophilic carbonyl groups. One example is the reduction of the ketone in -keto- fatty acyl units by NADPH in the biosynthesis of fatty acids, shown below. Add curly arrows to the scheme below to give a mechanism for this biochemical reduction. 3. Reaction of esters and amides with hydrogen nucleophiles Sodium borohydride is not a good enough nucleophile to react with carbonyls at the lower end of the reactivity rank order, such as esters and amides. However, lithium aluminium hydride is powerful enough. These carbonyls have potential leaving groups attached, so the overall reaction comprises several steps. Esters are reduced to primary alcohols and amides are reduced to amines. Draw mechanisms for both of the following reactions: O OMe OH i. LiAlH4 ii. water S O O Acyl carrier protein HO H S O Acyl carrier protein N CONH2 H H O OPO3H HO ADPO - NADPH N CONH2 O OPO3H HO ADPO - + NADP OMe O H Al H H H - H OMe O - OH H H H O H Al H H H - 4. Reaction of ketones and aldehydes with carbon nucleophiles Aldehydes and ketones will react with carbon-centred nucleophiles. The simplest of these is cyanide ion, which reacts as shown in this example to form a cyanohydrin. This reaction, which can be enzyme-catalysed in some plants, is reversible. Draw “curly arrows” mechanisms for both the forward and reverse reactions between benzaldehyde and cyanide ion. O NH2 NH2 i. LiAlH4 ii. water NH2 O H Al H H H - H NH2 O - NH2 H H H NH H Al H H H - H N H H O Al H O C N H H O C N - Al is a good Lewis acid One group of carbon nucleophiles, the Grignard reagents, is formed from bromo and iodo compounds. These are very reactive and must be prepared and used in the absence of water or alcohols. Their nucleophilic reactivity is equivalent to that of LiAlH4, i.e. they will react with all types of electrophilic carbonyl groups, including aldehydes and ketones. An example of the reaction mechanism is shown here: Draw a mechanism for the reaction of the Grignard reagent (ethyl magnesium iodide) derived from iodoethane and magnesium and the carbonyl compound acetylbenzene (acetophenone). Grignard reagents are sufficiently powerful nucleophiles to react with esters and amides. The reaction pathway is very similar to that for the reactions above in section 3. Draw the reaction mechanism and name the product of this reaction: O OMe Me I Me Mg I Mg ? + Et Me O - Et Me HO Me O IMg Et Aqueous work-up OMe Me O - Me Me HO OMe O IMg Me + Aqueous work-up Me O IMg Me 2-Phenylpropan-2-ol PA10306 / PA10311 From Molecules to Medicines - Workshop 3 Further topics in stereochemistry Most biomolecules and some drugs are chiral. In this workshop, you will use what you have learned from the lecture course in stereochemistry and apply this knowledge to problems in measurement of optical activity and in drawing representations of biologically important molecules in appropriate conformations and in a formal representation. You will need to use molecular models to answer many of the questions in this workshop. Feel free to discuss your work with other students during the class. The demonstrator(s) will help you with points of difficulty. If you do not finish the work during the workshop time, you must complete it in private study time before the next workshop. 1. Calculation of specific rotation Rotation of the plane of polarised light by a solution of an optically active compound (or of a mixture of enantiomers) is measured in a polarimeter. Of course, the angle  through which this plane is rotated depends on the structure of the compound but it also depends on the number of molecules that the light interacts with. In other words, the observed rotation measured varies with the concentration of the optically active solute and with the length of the light path through the solution. In contrast, the specific rotation [] is independent of concentration and path length and is an intrinsic property of the substance, just like melting point and dipole moment, for example. It is calculated according to this equation: []20 D =  / l c []20 D Specific rotation of the compound at 20C using monochromatic light of wavelength 589 nm. The units are degrees (). The direction of rotation (+ or -) must always be given.  Observed rotation in degrees (). l Path length of the light through the sample solution in decimetres (dm). c Concentration of the sample compound in the solution in grams per cm3 (g cm-3). Calculate the []20 D of the anthelmintic drug levamisole hydrochloride, using these data. A sample of levamisole hydrochloride (2.00 g) was dissolved in water (50 cm3) at 20C. A portion of this solution was used to fill a cylindrical polarimeter tube which had diameter 1.00 cm and length 5.00 cm. Monochromatic plane polarised light of wavelength corresponding to the sodium D line (589 nm) was passed along the axis of this cylinder and the plane of polarisation was found to be rotated anticlockwise through 2.55. 127.5 2.55 / [(0.5 dm) x (0.04 g cm-3)] = 127.5 This is the structure of one enantiomer of this drug. Does it have R or S stereochemistry? S Is it the enantiomer for which the above rotation was measured? Cannot tell – rotation of plane of PPL has nothing to do with R or S. 2. Calculation of enantiomeric excess The main way of expressing the proportion of each enantiomer in a mixture is enantiomeric excess: Enantiomeric excess = (amount of desired enantiomer - amount of other enantiomer)  100% Total amount of compound or Enantiomeric excess = x-y  100% x+y For example, for a sample of the amino acid leucine (2-amino-4-methylpentanoic acid) which comprises 0.9 mol of the naturally occurring S enantiomer and 0.1 mol of the unnatural R enantiomer, the enantiomeric excess is 80%. In the polarimeter, each component of a mixture causes rotation of the plane of polarised light independently of the other components. Hence, the observed rotation (and specific rotation) of a mixture is simply the sum of the rotations of all the components. In particular, it is possible to calculate the proportion of each enantiomer in a mixture of the two enantiomers of a compound. Therefore the enantiomeric purity and enantiomeric excess of the major component can be calculated, if the rotation of one of the pure enantiomers is known. Calculate the enantiomeric excess of a sample of the sugar derivative ribonolactone which has []24 D = +9. Pure D-ribonolactone has []24 D = +18. 50% ee N N H S H Cl Diastereomeric excess is defined similarly for mixtures of diastereoisomers. 3. Fischer projections of sugars – for reference Sugars have many hydroxy groups attached to a carbon chain and therefore have many chiral centres in proximity to one another. Although full stereochemical diagrams will contain all the stereochemical and conformational information and give you a proper impression of the shape of the structure, these can sometimes be a bit complicated. In some textbooks, a short- hand method of representing the chiral centres of sugars is used. In this Fischer projection, the carbon backbone of the sugar is represented as a vertical straight line and the substituent hydroxy groups are represented as being attached by horizontal lines. The method of converting a stereochemical diagram to a Fischer projection will be demonstrated for the sugar D-ribose. Make a model of the structure to help follow this explanation. Step 1. Draw the stereochemical diagram of the sugar in the open-chain structure. Step 2. Rotate the whole structure so that the carbon skeleton is vertical, with the carbonyl group at or near the top. Step 3. Draw a vertical line for the carbon backbone and one horizontal line for each of the chiral carbon atoms. Add the aldehyde (CHO) and terminal (CH2OH) groups. O HO HO OH OH OH HO HO OH O H HO HO OH OH O H OH HO HO OH O H HO HO OH OH O H OH HO HO OH O H CHO CH2OH Step 4. For the carbon atom at position 2, look at the stereochemical diagram (or model). Make sure that the vertical bonds to the rest of the backbone are vertical and going away from the eye and that the carbonyl is at the top. If not, carry out some rotations of parts of the molecule around C-C single () bonds to change the conformation. If necessary, rotate the whole molecule slightly. Draw what you see for that chiral centre. Step 5. Repeat step 4 for each of the other chiral centres. To draw a full stereochemical diagram, starting from a Fischer projection, just reverse the whole process. Notice that the lowest chiral centre in the Fischer projection of D-ribose has the OH group on the right of the vertical line. This is the chiral centre that is used to relate the structures of sugars to the structure of D-glyceraldehyde to determine which sugars are called D and which are called L. D-sugars all have this lowest OH group on the right and L-sugars all have it on the left in the Fischer projection. HO OHC HO OH OH H OH CHO OH H OH H CH2OH OHC OH OH OH HO OHC HO OH OH H OH CH2OH CHO OH H OH H OH H CH2OH CHO OH OH HO OHC HO OH OH H OH CHO R CHO OH H CH2OH HO OHC HO OH OH H OH CHO R CHO OH H CH2OH Below are the Fischer projections of D-glucose, D-galactose, D-mannose and L-glucose. 4. Conformation, stereochemistry and reactivity Conformation and stereochemistry often affect rates and outcomes of (bio)chemical reactions. Conversely, the mechanism governing a particular chemical process will often determine which stereoisomer is formed as the major or only product. Stereochemistry In this example of a SN2 nucleophilic substitution, acetate ion (the nucleophile) displaces iodide (the leaving group) from iodomethane to give methyl acetate. The curly arrow mech- anism is a way of representing the growing overlap of the main lobe of the sp3 orbital on the nucleophile with the 'tail' lobe of the sp3 orbital which makes up the  bond between carbon and iodine. In the transition state, part way through the reaction, the nucleophile is directly opposite the leaving group. There are no stereochemical consequences in this particular example, since neither iodomethane nor methyl acetate is chiral. O OH HO HO OH OH O OH HO OH OH HO O HO HO OH HO OH O OH HO HO HO OH D-glucose D-galactose D-mannose L-glucose a) Draw a stereochemical diagram of the product of the similar reaction between the acetate ion and R-2-iodobutane. Give a systematic name for this product, including appropriate indication of stereochemistry. b) Rivastigmine is a drug used for the treatment of Alzheimer’s disease. It can be manufactured by the reaction below. AcO H H H I - I H H H AcO - AcO I H H H ‡ I H H H AcO Transition state Orbital overlap SN2 AcO H H H AcO H H H I - I - I H H H AcO - I H H H I H H H AcO - AcO - AcO I H H H ‡ AcO I H H H AcO I H H H H H H ‡ I H H H AcO I H H H AcO Transition state Orbital overlap SN2 (i) Identify the stereogenic centres (chiral centres) in the starting material and in rivastigmine and assign them as either R or S. It is good practice to assign priorities and draw a clockwise/anticlockwise arrow as well as to answer R or S. (ii) Suggest a suitable reagent for this transformation and show, using curly arrows, the mechanism for the SN2 reaction. What factors favour the SN2 mechanism? What happens to the stereochemistry in this reaction? Curly arrows should show attack 180o from leaving group, proceeding via a transition state, to give a product with complete inversion of configuration. Factors that favour SN2 – primary/secondary halides, good nucleophiles, nonpolar solvents. (iii) Draw the product(s) if the reaction were to proceed via an SN1 mechanism. What factors favour the SN1 mechanism? What happens to the stereochemistry in this reaction? The curly arrows should show the leaving group going leaving a secondary benzylic carbocation behind. The carbocation is planar (with a vacant p-orbital) so it is achiral Me2HN Br Me H Ar O N O Et Me Me H Br O N O Et Me Me NMe2 H HNMe2 : Rivastigmine Complete inversion of configuration O N O Et Me Me H Br HNMe2 : Nucleophilic attack from front or back is equally likely (planar carbocation), leading to racemic product RO Me H + O N O Et Me Me NMe2 H Front Back O N O Et Me Me H Br 1 2 3 4 Anticlockwise suggests S but lowest priority group is at front, so it’s actually R R-configuration Lowest priority group is at BACK; anticlockwise = S O N O Et Me Me NMe2 H 4 2 3 1 S-configuration and it is equally likely that the nucleophile can attack from above or below, which will give a 50:50 mixture of enantiomers = racemic mixture. Factors that favour SN1: secondary/tertiary/benzylic halides (to make lower-energy carbocation), very good leaving group, poor nucleophiles, polar protic solvents. Conformations In lectures you will have seen the chair conformation of cyclohexane, which has axial and equatorial positions. Chair conformations undergo “ring flipping” rapidly at room temperature, where an axial bond becomes equatorial and vice versa, without breaking any bonds. Below is an example of a substituted cyclohexane, which forms two chair conformations of different energy. Use this background to answer the question below. Lomustine is a highly lipophilic drug used to treat a range of cancers, in particular brain tumours. It forms various metabolites in vivo including trans-4-hydroxylomustine and cis-4-hydroxylomustine. Draw the metabolites trans-4-hydroxylomustine and cis-4-hydroxylomustine using chair diagrams with appropriate labels. Which metabolite is lower in energy and why? Are these compounds chiral? Why do you think that lomustine is particularly effective towards brain tumours? These molecules are NOT chiral, as they all have a plane of symmetry. These drugs must pass the Blood-Brain Barrier to enter the brain. Most drugs cross the BBB by transmembrane diffusion, which is favoured for small lipophilic molecules like lomustine. HN OH N O N O Cl HN OH N O N O Cl NHR HO HO NHR NHR HO HO NHR cis-4-Hydroxylomustine trans-4-Hydroxylomustine Both these conformations have one axial and one equatorial substituent; on the upper face; they have similar energies These conformations have substituents on opposite faces. The diequatorial conformer is of lower energy than the diaxial conformer PA10306 / PA10311 From Molecules to Medicines - Tutorial 5 Functional group interconversions 1. Draw the structure of an aldehyde of your choice which contains five carbon atoms; give its systematic name. Lecture 7. Draw the structure of and name the product of oxidation of your aldehyde. What reagent could you use for this oxidation? Lecture 21. KMnO4, or K2Cr2O7 or Na2Cr2O7 with catalytic acid are strong oxidising agents and will oxidise an aldehyde (+2) to a carboxylic acid (+3). Pyridinium chlorochromate (PCC) is a mild oxidant and will only be able to oxidise a primary alcohol (+1) to an aldehyde (+2), but not usually to a carboxylic acid (+3). Draw the structure of and name the product of reduction of your aldehyde. What reagent could you use for this reduction? Lecture 22. You learned about NaBH4/EtOH (mild reducing agent) and LiAlH4/Et2O (strong reducing agent) in the lectures. Either will do, although best to go for the milder agent. 2. What are the structures of the reducing agents LiAlH4 and NaBH4? What types of functional group are they used to reduce? Lecture 22. NaBH4 will reduce acid chlorides and acid anhydrides (even though they are both +3) because although it is a mild reducing agent, these carbonyls are reactive (remember your rank order of carbonyls!). How many molar equivalents of NaBH4 are required to reduce acetone to propan-2-ol? 0.25 Remember that, in NaBH4, there are four B—H bonds to deliver a nucleophilic hydrogen How many molar equivalents of LiAlH4 are required to reduce ethyl benzoate to ethanol and benzyl alcohol? 0.5 Remember that, in LiAlH4, there are four Al—H bonds to deliver a nucleophilic hydrogen What use can you think of for NaB2H4? The 2 refers to the hydrogen! Remember your lecture on isotopes (Lecture 4), so you should know of 1H (protium), 2H (D, deuterium) and 3H (T tritium). Isotopically labelling drugs to identify metabolites. 3. Below is a synthesis of paracetamol. Suggest reagents and using curly arrows show the mechanism of reaction, explaining any selectivity. Lectures 18/19, 22, 15 and 23. 4. Biosynthetic reactions use different forms of nicotinamide adenine dinucleotide to perform oxidation/reduction reactions. Nicotinamide adenine dinucleotide phosphate (NADPH) is a reducing agent, whilst nicotinamide adenine dinucleotide (NAD+) can oxidise. What are the differences in the structures of these two forms? See below OH NH2 OH NH O Me O NH O Me O Me OH NO2 :OH HNO3 NO2 + HO- is o- and p- directing Some 2-nitrophenol also formed – need to separate H2 / Pd or Fe / HCl or Sn / HCl, etc. Me OH O Me O O Me O Me Cl O or BUT NOT The OH is also nucleophilic, so some ester (above) mayalso be formed. This ester is easily hydrolysed to make paracetamol (esters much more electrophilic than amides) The NH2 is more nucleophilic than the OH 5. NADPH can be represented as shown below. Using curly arrows, show how this could reduce butenoyl-ACP (Acyl Carrier Protein), a key step in lipid biosynthesis. See mechanism below – note conjugate addition. Lectures 20 & 22. 6. NAD+ can be represented as shown below. Using curly arrows, show how this could oxidise 3S-hydroxyhexadecanoyl-CoA (co-enzyme A), a key step in metabolism. Lecture 21 & 22. SACP O A H SACP H H O N H H H2NOC R : N H2NOC R + SACP O H H - H27C13 SCoA O O H H H27C13 SCoA O O N H H H2NOC R N H2NOC R + PA10306 / PA30311 From Molecules to Medicines - Tutorial 4 Alkenes and aromatics 1. Using curly arrows to show your working out, draw the product(s) formed by addition of HCl to indene. What factors stabilise a carbocation? Which product, if any, is favoured? Lecture 17 - In order to answer the question properly, you must have drawn curly arrows and considered the relative stabilities of the different carbocation intermediates generated. You are expected to show your working out! Factors which lower the energy of a carbocation include being more substituted (e.g. tertiary carbocation is more stable than a secondary carbocation) due to i) more stabilising inductive effects and ii) hyperconjugation. Benzylic and allylic carbocations are also have their energy lowered further by additional resonance forms (mesomeric delocalisation), spreading out the carbocation charge. Remember that carbocations are sp2. H Cl H H Cl Cl H H H + H H H + Cl - Cl - p Carbocation is secondary but not benzylic – higher in energy Carbocation is secondary and benzylic – lower in energy ONLY PRODUCT NOT FORMED 2. Using curly arrows and chair diagrams, draw the product(s) formed by addition of Br2 to cyclohexene. Show the different chair conformations of the product and indicate which is lower in energy. As above, you are expected to show your working out, using curly arrows and adding comments where relevant (Lectures 8 & 17). Br Br Br Br Br Br Br + Br Br p Br - SN2 Trans product only Diaxial Diequatorial Higher in energy, owing to unfavourable 1,3-diaxial interactions Lower in energy, owing to less steric clash 3. Atrolactamide is a very potent anesthetic but was found to display toxicity due to the formation of the by-product 2-phenylacrylamide. Show, using curly arrows, how this by- product could be formed and explain why it is toxic. Draw the structure of the tripeptide glutathione (reduced form; GSH). What are its functions in the body? Show, using curly arrows, how glutathione could detoxify the toxic metabolite 2-phenylacrylamide. Glutathione has important roles in protecting cells against oxidants and against electrophilic toxins. Key molecules in the cell, including enzymes and DNA, have many nucleophilic sites and glutathione acts as a sacrificial strong nucleophile to trap electrophiles before they can react with these key nucleophiles. Most cells have ca. 1 mM glutathione but hepatocytes have ca. 7 mM glutathione, as more electrophiles are generated there by metabolism. The sacrificial trapping of the electrophilic 2- phenylacrylamide metabolite of atrolactamide is an example of this protection. Glutathione is also involved in the transport of amino-acids and in maintain the redox status of the cell. Lecture 20 – conjugate addition. Base elimination to afford the 2-phenyl¬propenamide is shown above, but could also be obtained under acidic conditions, as shown below (Lecture 12 – Elimination). NH2 O NH2 O S R NH2 O S H R - + NH2 O HO H Base SH N H O O H3N CO2 H N CO2 - - + S H R ≡ Atrolactamide Glutathione 2-Phenylacrylamide Potent Michael acceptor enone – reacts with key cellular nucleophiles, therefore toxic E2 Elimination Congugate addition Loss of H+ and enolate – carbonyl tautomerism S: H R Metabolite / glutathione conjugate 4. L-Thyroxine is a synthetic drug used to treat thyroid hormone deficiency. It is derived from L-tyrosine, which can be selectively iodinated, as shown below. Show, using curly arrows, the reaction mechanism and explain why the major isomer obtained is the one shown. What is the role of NEt3 in this reaction? Lecture 19. This question applies the knowledge that you gained in the Aromatics – directing influences lecture. The OH group on phenol is strongly electron-donating (strong mesomeric effects), so it is ortho- and para-directing. Since the para position is already substituted, it first iodinates at one ortho position, and then the other, to give the observed product. The triethylamine is a base. O CO2 H3N I I I OH I + - OH CO2 H3N I I + - OH CO2 H3N + - I I : OH I H + H3N + CO2 - :NEt3 L-Tyr Iodinated product OH CO2 H3N I + - Electrophilic substitution L-thyroxine OH I H + H3N + CO2 - + OH I H H3N + CO2 - OH I H + H3N + CO2 - OH I H + H3N + CO2 - ≡ OH is activating (no need for Lewis acid) and is ortho / para-directing Repeat Stabilisation of cation by resonance / mesomerism PA10306 / PA10311 From Molecules to Medicines - Tutorial 2 Stereochemistry 1. Methamphetamine (MA) is a highly active central nervous system (CNS) stimulant and is chiral. A sample obtained by police was found to be a mixture of 60% (+)-MA and 40% (–)-MA and has []20 D = +2. What are the enantiomeric excess of this mixture? From Lecture 9 and in Workshop 3: ee = (x-y)/(x+y) = (60-40)/(60+40) = 20% ee of (+)-MA What is the []20 D of pure (+)- MA? X (see diagram) = +10 What is the []20 D of a mixture comprising 25% (+)-MA and 75% (–)-MA? Y (See scheme for answers) = 5. Draw both enantiomers of methamphetamine, assign priorities and label which is R and S. Can you tell which is (+) and (–)? 2. All amino acids in proteins (except glycine) are chiral and have L-configuration. Using the Cahn, Ingold, Prelog rules assign R- or S-stereochemistry for L-serine and L-cysteine. 3. What is meant by d, D, R and (+)? d – Small d signifies that the plane of Plane Polarised Light (PPL) is rotated clockwise. It exactly the same as (+). D – Big D is the old convention for assigning stereochemical configuration relative to D- glyceraldehyde. R – This is from using CIP rules for assigning stereochemistry, and signifies clockwise rotation of the assigned groups 123, with 4 at the back. It has nothing to do with PPL! 4. Ibuprofen is a nonsteroidal anti-inflammatory drug (NSAID) for relieving pain. It is often administered as a racemic mixture, but only the S-enantiomer is active. Draw the S- and the R-enantiomers of ibuprofen. See below In the body, the enantiomers of ibuprofen are converted into their CoA thioesters. These CoA thioesters can racemise – show mechanistically how epimerisation can occur. [Hint: thioesters are similar to ketones in the rank order of reactivity of carbonyls] See below and Lecture 11. Me Me CO2H Me R-Ibuprofen Me Me CO2H Me S-Ibuprofen 1 2 3 1 2 3 Me Me Me SCoA O Me Me Me SCoA O Me Me Me SCoA O - Formation of enolate -Carbon now sp2, so planar – loss of stereogenic / chiral centre HS N H O N H O OH Me Me O P O O O P O O O O N O OH N N N H2N P O OH O - - - Structure of coenzyme A (for information) Department of Pharmacy & Pharmacology The Chemistry of Drugs Aromatic Lecture: Directing influences Dr Lorenzo Caggiano SL12009/SL12101 Directing influences Directing influences R=Strongly e- donating – Activating stabilise ortho and para so direct ortho and para R=Weakly e- donating – Activating stabilise ortho and para so direct ortho and para Directing influences R=Strongly e- withdrawing – Deactivating destabilise ortho and para so direct meta R=Weakly e- withdrawing – Deactivating stabilise ortho and para so direct ortho and para PA10306 / PA10311 - From Molecules to Medicines - Workshop 1 Structure, electronic effects, bonding and moles In this workshop, you will build on your previous knowledge of chemical structure and bonding and on what you have learned from the lectures. Feel free to discuss your work with other students during the class. The demonstrator(s) will help you with points of difficulty. If you do not finish the work during the workshop time, you must complete it in private study time before the next workshop. 1. Inductive effects on pKa pKa is a measure of the strength of an acid. It refers to the equilibrium constant for a proto- tropic reaction at equilibrium: The pKa value is the negative log10 of Ka. The smaller the value of pKa, the stronger (more likely to lose H+) is the conjugate acid in the equilibrium and the weaker (less likely to gain H+) is the conjugate base. As many drugs and biomolecules have several ionisable groups, we must be careful to define clearly the equilibrium to which a particular pKa value refers. Remember that a pKa value refers to an equilibrium, not a substance. Effects which lower the energy of the conjugate acid or conjugate base in an equilibrium will affect the equilibrium and, therefore, the pKa, through Le Chatelier’s Principle. Please ask the Demonstrator if you do not know about Le Chatelier’s Principle. For acidic compounds, including drugs, the pKa for dissociation into a proton and the conjugate base will depend on factors which lower the energy of the anionic conjugate base, i.e. factors which “spread out” the negative charge over several atoms. Inductively electron-withdrawing atoms can do this. H A H + A - Conjugate acid Conjugate base Proton Ka Equilibrium constant Ka = [H+] [A-] [HA] For the first part of this workshop, draw the equilibria (including conjugate acids and conjugate bases) for the ionisation of these acids and explain the differences between the pKa values. Now draw the equilibria (including conjugate acids and conjugate bases) for the protonation of these amine bases and explain the differences between the pKa values. O CH3 O H O CH3 O - H + pKa = 4.76 O O H Me OH O O Me OH - H + pKa = 3.86 O O H F F F O O F F F - H + pKa = 0.23 The energy of the carboxylate anion is lowered by delocalisation of the negative charge across both oxygens. The energy of the carboxylate anion is lowered by delocalisation of the charge across both oxygens and by the presence of an inductively electron- withdrawing oxygen in the OH. The energy of the carboxylate anion is lowered by delocalisation of the charge across both oxygens and by the presence of three inductively electron- withdrawing fluorine atoms. H2N Me H3N Me + H + pKa = 10.7 H2N F F F H3N F F F + pKa = 8.3 H + N H O N H2 O + pKa = 8.3 H + The energy of the cation (the conjugate acid) is raised by the inductive effect of the three fluorines withdrawing electron-density from the cationic nitrogen. The energy of the cation (the conjugate acid) is raised by the inductive effect of the oxygen withdrawing electron-density from the cationic nitrogen. 2. Mesomeric effects on pKa You have met the mesomeric effect in the lectures. It is also known as resonance. It happens when a lone pair of electrons in a p-orbital (usually on a nitrogen or oxygen) overlaps with an adjacent -orbital (usually on a C=C double bond or an aromatic ring). It usually results in the electron density of the lone pair being pushed into the -orbital. Mesomeric electron- donation like this is called a +M electron-donating mesomeric effect. Sometimes there is overlap between a C=C or aromatic -orbital and an adjacent p-orbital which has low electron-density. This results in a –M electron-withdrawing mesomeric effect. Mesomeric effects can only happen when p- and -orbitals overlap. They are usually stronger than inductive effects. Draw the prototropic equilibria (including conjugate acids and conjugate bases) and explain the mesomeric effects responsible for the different pKa values of these phenols. OH OH O Me pKa = 9.95 H + O - O - O - O - These are the resonance forms or mesomers showing how the negative charge can be delocalised, lowering the energy of the anion Overlap of the oxygen p-orbital containing the charge with the -orbital of the aromatic ring – students should be thinking in this way pKa = 10.2 O - sp2 H + O - O - O - O - O Me O Me O Me O Me : : : : O - sp2 O : sp2 Me The same the resonance forms or mesomers are possible here. However, the methoxy oxygen is sp2-hybridised, putting a lone pair into a p-orbital, which pushes electron- density into the aromatic ring, disfavouring the delocalisation and raising the energy of the anion Here the lone pair of the +M methoxy group also shares electron-density into the -orbital of the aromatic ring O N O O - - + OH N O O - + pKa = 7.2 H + O - O - O - O - N O O - + N O O - + N O O - + N O O - + The –M nitro group gives more opportunity for the negative charge to be delocalised (blue arrows), lowering the energy of the anion. Delocalisation onto oxygen is particularly favoured The mesomeric effect also affects the protonation of aromatic amine bases (anilines / aminobenzenes). Draw the prototropic equilibria (including conjugate acids and conjugate bases) and explain the mesomeric effects responsible for the different pKa values of the conjugate acids of these anilines. pKa = 4.6 H + NH2 Mesomeric overlap can occur in the neutral conjugate base but not in the cationic conjugate acid. This raises the energy of the protonated amine. H2N sp2 NH3 + H3N sp3 + pKa = 5.3 H + The MeO oxygen is sp2 hybridised. Electron-density from the lone pair in its p-orbital is shared into the aromatic ring, lowering the energy of the cation O : sp2 Me H2N sp2 NH3 + O Me NH2 O Me H3N sp3 + O : sp2 Me pKa = 1.0 H + As above, the –M nitro group is strongly mesomerically electron-withdrawing and raises the energy of the cation NH3 + NO2 NH2 NO2 3. Non-covalent bonds and interactions The conformations (folded shapes) of proteins are held mostly by non-covalent bonds and interactions. Enzymes bind to their substrates mostly by non-covalent bonds and interactions. Drugs bind to their targets (usually proteins) mostly by non-covalent bonds and interactions. You have met several different types of non-covalent bonds and interactions in the lectures. They include hydrogen bonds (H-bonds), ionic interactions (attraction and repulsion), van der Waals forces, hydrophobic interactions and -stacking. Suggest the likely types of non-covalent bonding of olaparib to the target enzyme PARP-1. Mark the part of the molecule involved in each type of bonding on a diagram. For potentially hydrogen-bonding groups, you should identify whether they are H-bond Donors or H-bond Acceptors. H-bond acceptors. Students should consider which heteroatoms have lone pairs available in sp2 or sp3 orbitals. Lone pairs in p-orbitals are involved in mesomeric overlap (e.g. in amides) and cannot accept an H in a H-bond. H-bond donor. Stacking. Stacking can occur onto both of the aromatic rings. There can also be -stacking onto the amides. Hydrophobic interactions and van der Waals forces can also help bind this drug into the binding site. Additional information, if you are interested: This new anti-cancer drug binds into the nicotinamide-binding site of PARP-1 through -stacks from the top aromatic ring and two tyrosine side-chains (forming a sandwich), through a H-bond from the N-H onto a Gly carbonyl and H-bonds from a Ser OH and a Gly N-H onto the top carbonyl oxygen. This was established by a crystal structure. N N O H F N O N O Tyr907 Gly863 Ser904 H-bonds H-bond -stack 4. Molecular weights and concentrations It is important for pharmacists and pharmacologists to be able to be able to make up solutions of drugs accurately. Concentrations of solutions can be expressed as mass-per-volume concentrations (e.g. mg mL-1, % w/v, g mL-1) or in molar units. 4a. Mass-per-volume concentrations What volume of solvent should be added to 20 mL of a 5.0% (w/v) solution of the anxiolytic drug diazepam in buffer to make a 0.2% (w/v) solution? 480 mL What volume of solvent should be added to 5.0 mL of a 12 mg mL-1 solution of the antihistamine drug diphenhydramine to make a 0.1% (w/v) solution? 55 mL 4b. Molar concentrations A 1.00 molar (M) solution is defined as a solution in which 1.00 mol of the solute is dissolved to make 1.00 L of solution. In pharmacy and pharmacology, solutions are often more dilute and are expressed in millimolar (mM) ( 10-3), micromolar (M) ( 10-6) or nanomolar (nM) ( 10-9). Calculate the mass of cinnarizine (an antihistamine drug used to treat sea- sickness) required to make 100 mL of a 2.0 mM solution in water. Mass of cinnarizine required: 73.6 mg Calculate the mass of silodosin (an - adrenergic drug) required to make 1.00 L of a 2.0 M solution in water. Mass of silodosin required: 0.992 mg 5. Drawing chemical structures To help you to visualise the three-dimensional structures of drug molecules, it is important that you draw good stereochemically and conformationally accurate structural diagrams. In particular, you should be careful to draw appropriate bond angles at all the atoms in a diagram; it is best to draw ‘stick’ structures where you only label heteroatoms and omit all unimportant hydrogens: Now draw good diagrams of: a. 2-Acetoxybenzoic acid (Aspirin). b. Trans-butenedioic acid (Fumaric acid). H C C H H H H C C H H C H H OH C C C C H C OH H H H H H H H H Et OH WRONG! Better Best SL12009 / SL12101 – The Chemistry of Drugs - Tutorial 3 Carbonyl chemistry in manufacture and metabolism / cleavage of drugs 1. Draw the structures of these compounds and place them in rank order of rate of reaction with nucleophiles: acetic acid, acetyl chloride, acetamide, acetaldehyde, acetone. Lecture 13 2. Draw the structures of these compounds and of their enols. Place them in rank order of extent of enolisation at equilibrium: ethyl 3-oxobutanoate, ethyl acetate, pentane-2,4- dione, acetone, N,N-dimethylacetamide. Lecture 13 - Students should know that the major factors affecting enolisation are: Conjugation in enol and Intramolecular H-bonding in enol. 3. Procaine is a local anaesthetic drug, often used in dentistry. It is metabolised in the plasma by the enzyme pseudocholinesterase through hydrolysis to 4-aminobenzoic acid (PABA), which is then excreted by the kidneys. Draw separate chemical mechanisms (with curly arrows) for (a) the acid-catalysed hydrolysis and (b) the base-catalysed hydrolysis of procaine. Lecture 15 Me Cl O Me OH O Me H O Me Me O Me NH2 O Acetamide Acetic acid Acetone Acetaldehyde Acetyl chloride >>> > >> >> Me O O Me Me O Me O H Me O O OEt Me O OEt O H H3C O OEt H2C OEt O H H3C O NMe2 H2C NMe2 O H H3C O Me H2C Me O H >>> > > = H2N O O NEt2 H3N O OH NHEt2 + + + O OH2 H H3N O NHEt2 + + + H O OH H H3N O NHEt2 + + + HO NHEt2 + H3N OH O + H + Prototropic equilibrium OH2 : Acid-catalysed hydrolysis For the acid-catalysed hydrolysis, the red arrows show the forward (hydrolysis) reaction and the blue arrows show the reverse process. The water is a weak nucleophile and cannot attack the ester carbonyl until the carbonyl is protonated. Students should notice that all the steps are equilibria and that the overall reaction is under the control of Le Chatelier’s Principle. For the base-catalysed hydrolysis, the red arrows show the forward (hydrolysis) reaction and the blue arrows show the reverse process. Hydroxide is a strong nucleophile and can attack the ester carbonyl. The leaving-group abilities of the alkoxide (forward reaction) and hydroxide (reverse reaction from the tetrahedral intermediate) are about the same. Students should notice that the first step is an equilibrium and that the overall reaction is under the control of Le Chatelier’s Principle, through removing one component from the equilibrium by deprotonation (making a non-electrophilic carboxylate). 4. Bupivacaine is one of the most commonly used drugs in epidural anaesthesia during labour. Draw a chemical mechanism (with curly arrows) for its manufacture from 2,6- dimethylaniline and the acyl chloride A. Lecture 15 What would happen if 2,6-dimethylaniline were treated with the carboxylic acid B? All that would happen is that they would form a salt. It is important that the students understand that amine + carboxylic acid does NOT give an amide. O OH H2N O NEt2 Base-catalysed hydrolysis H2N O OH NEt2 OH - - H2N OH O HO NEt2 H2N O O - O NEt2 - Me Me N H O N Bu Cl O N Bu Me Me NH2 : N H O Cl Me Me N Bu - enolate 5. The anticoagulant drug warfarin is manufactured by the process shown below. Using curly arrows draw a chemical mechanism for the first manufacturing step only (Hint – two separate steps: base is needed for the Aldol reaction (Lecture 16), then for the elimination of water (Lecture 12). elimination O Me Me Me O O H Me CH2 O - H HO Me O H H OH - base PA10306 / PA10311 - From Molecules to Medicines - Tutorial 1 Bonding, structure and electronic effects in synthetic and natural product drugs The questions in this and subsequent tutorial sheets are intended to help you reinforce your learning and understanding of the importance of chemical structure and reactivity in the design of drugs, in how they interact with their targets and in how they are delivered as medicines. You must prepare answers to all the questions on a sheet before the tutorial. In this way, you will be ready to present your answers to the group and to discuss any points that you do not understand. Your Unit tutorial leader will also then be able to identify and to help you with any points that you find more difficult. Lectures in this Unit will have covered hybridisation and bonding, inductive and mesomeric effects on reactivity and how these can affect chemical, physicochemical and biochemical properties. This knowledge and understanding will help you in later parts of this Unit and in other Units, such as in understanding the stability of drugs, formulation of medicines and issues in absorption, distribution, metabolism and excretion of drugs. 1. For each of these three compounds, give the state of hybridisation of all the carbon atoms (sp, sp2 or sp3). a. Bromochlorofluoroiodomethane sp3 b. 1-Bromo-2-fluoroethene sp2 c. 1,2-Dichlorobenzene sp2 2. Using this structure of an analogue of the local anaesthetic drug butanilicaine, answer the following questions. a. Show where significant inductive effects occur. Inductive effects occur in -bonds only: C→Cl, C→N, C→O b. Show where mesomeric effects occur. Mesomeric effects occur by overlap of lone pairs in p-orbitals with adjacent - orbitals: NH2 into Ar ring, amide NH into Ar ring, amide NH into carbonyl, weak effect Cl into ring. Please consider orbital overlap rather than resonance forms / mesomers. c. Which nitrogen atom is the most basic in this drug and why? Draw the structure of the hydrochloride salt of this compound. The aliphatic amine is the most basic (pKa of the conjugate acid ca. 10). The others are less basic because the N atoms are sp2 hybridised and their lone pairs are in p- orbitals which overlap mesomerically with adjacent -bonds (delocalisation). Please draw a diagram where the Cl- counter- ion is well away from the N+, to avoid any idea that there is a bond between them. 3. Nimbinin is present in neem oil from the tree Azadirachta indica (formerly Melia indica). It is closely related to the main biologically active component, nimbin. a. Identify and name the oxygen- containing functional groups in nimbinin. b. Give the molecular formula of nimbinin. C28H34O6. What is the mass of nimbinin is required to make up 100 mL of a 10.0 M solution in water? 466 g c. Draw an enol tautomer of nimbinin, taking care not to impose excessive strain in any rings. No other enol is possible (either lack of -hydrogens or increase in bond angle strain). ,-unsaturated Nimbinin O O O O O O Furan Ketone Oxirane / epoxide Ester Ketone OH O O O O O O O O O HO O Best (lowest energy - enol C=C conjugated with aromatic furan) OK (unconjugated enol) A neem tree 4. Sulfasalazine is a sulfonamide drug used in the treatment of Crohn’s disease. a. What is Crohn’s disease? An inflammatory auto-immune disease of the GI tract that may affect any part of the gastrointestinal tract, causing a wide variety of symptoms. It primarily causes abdominal pain, diarrhoea, vomiting or weight loss. b. Show where intramolecular hydrogen bonding may occur in sulfasalazine, taking care not to impose excessive strain in any rings. Intramolecular H-bonds between the phenol and the carboxylic acid are possible but those involving the pyridine and sulfonamide are not, owing to excessive ring-strain in 3- and 4-membered rings. c. Rank the three acidic functional groups in order of their pKa and give a rationale for your ranking. Carboxylic acid (pKa ca. 4), sulfonamide (pKa ca. 7), phenol (pKa ca. 10). Rationale – stabilisation of anion in the prototropic equilibrium. d. What is the molecular weight of sulfasalazine? 398 Da. What mass of sulfasalazine is needed to make up 10.0 mL of a 0.5% w/v solution? 50 mg Department of Pharmacy & Pharmacology From Molecules to Medicines 1 Lecture 28: The Modern Drug Development Process Why does it take so long? Why is it so expensive? Dr Andrew Watts 1 The Development Process 2 Statistics on Drug Development 3 The Development Process 4 Why do drugs fail getting to market? 5 A significant cost of drug development comes from the candidates that DON’T make it to market. Most of these reasons are related to the chemistry of the drug! Drug companies have made significant efforts to try improve the success rates during drug design. Modern Drug Design 6 Drug Molecules 7 A drug molecule possesses one or more functional groups positioned in three- dimensional space on a structural framework that holds the functional groups in a defined position that enables the molecule to bind specifically to a targeted biological macromolecule, the receptor. Shape Department of Pharmacy & Pharmacology From Molecules to Medicines 1 Modern Drug Design Techniques 8 Modern Drug Design 9 Target ID and validation 10 We are comparing the diseased state to the healthy state. Disease state is assessed by taking clinical samples from patients, or using established ‘models’. Genomic, proteomic and genetic association methods are used to identify the target. Target Identification 11 Genomic methods seek to identify differential expression of mRNA in the disease state. As mRNA is used to make protein, changes to levels of mRNA can lead to changes in the amount of protein (enzymes) present in the disease state. Target Identification 12 Proteomic methods seek to identify differential expression of proteins in the disease state. Proteomic methods can also identify functional changes to proteins (ie. Specific activity and point mutations). Modern Drug Design 13 High Throughput Screening 14 High throughput screening is a technique where large numbers of chemical compounds (libraries) are rapidly tested for ‘activity’ against our chosen target. Libraries consist of 10,000’s of compounds Assays can evaluate 100,000’s of samples/day Often 384 wells/plate High Throughput Screening 15 Advantage is that the process is highly automated. High Throughput Screening 16 Major pharmaceutical companies now typically maintain compound collections of up to 1 million compounds, carefully selected and characterised (potential resource for new HTS campaigns) Specialist companies also provide “off the shelf” libraries: E.g., Maybridge (UK) offer 53 000 compounds for screening in microplate format also Chembridge, Prestwick Considerations This type of screening is random. The ‘active’ compounds usually provide us with a starting point. ie. identification of our scaffold. Modern Drug Design 17 Virtual Screening 18 A process where we use computational methods to predict: - how a compound will bind to a protein - what compounds might bind to a protein - how strongly a compound might bind to a protein Advantages: - We do not need to physically make the compounds Disadvantages: - We need to know the exact position in 3-dimensional space, of all atoms in the protein !! Protein Crystallography 19 X-ray crystallography can reveal 3D structure of protein and bound compounds Visualization of these “complexes” of proteins and potential drugs can help scientists understand the mechanism of action of the drug and to improve the design of a drug Visualization uses computational “ball and stick” model of atoms and bonds, as well as surfaces Virtual Screening 20 Put a compound in the approximate area where binding occurs Generic algorithm encodes orientation of compound and rotatable bonds Optimize binding to protein Minimize energy Hydrogen bonding Hydrophobic interactions Provides a ‘rational screening’ method for identifying the scaffold. Modern Drug Design 21 Combinatorial Chemistry 22 Techniques for producing large numbers of compounds in a short period of time, using defined reaction routes and a large variety of starting materials and reagents. Combinatorial Chemistry 23 An example of a ‘multi-component’ reaction. Quickly generates a library of 160,000 compounds with different R1, R2, R3 and R4 Combinatorial Chemistry 24 Advantages if achievable: Vast increases in productivity Chemistry ‘bottlenecks’ in drug discovery process can be prevented Companies can patent sooner Cost savings and increased time for revenue generation Requirements: - Highly efficient chemistry - Limited work-up, isolation and purification - Rapid throughput automation Modern Drug Design 25 Molecular Modelling 26 Molecular modelling uses similar methods to virtual screening, but is carried out after we already have our ‘lead’ compound. It is used to try optimise the pharmacophores for improved binding (efficacy). Optimize pharmacophore binding to the protein: Minimize energy Hydrogen bonding Hydrophobic interactions Modern Drug Design 27 ADME Models 28 Traditionally, animals were used for pre-human testing. However, animal tests are expensive, time consuming and ethically undesirable ADME (Absorbtion, Distribution, Metabolism, Excretion) techniques help model how the drug will likely act in the body These methods can be experimental (in vitro) using cellular tissue, or in silico, using computational models 29 Based around real tissue samples, which have similar properties to those in the body Cuts down animal tests, by acting as a “pre-screen” Enables ADME data to be discovered on many more compounds Example is CACO-2 cell line for predicting oral bioavailability. In vitro ADME Models 30 Caco-2 permeability assay Helps to understand the suitability of a compound for oral dosing by predicting human intestinal permeability and drug efflux. Caco-2 permeability assay to investigate intestinal permeability predicts the in vivo absorption of drugs across the gut wall by measuring the rate of transport of a compound across the Caco-2 cell line. The Caco-2 cell line is derived from a human colon carcinoma. The cells have characteristics that resemble intestinal epithelial cells such as the formation of a polarised monolayer, well-defined brush border on the apical surface and intercellular junctions. In vitro ADME Models 31 In silico ADME Models Computational methods can predict compound properties important to ADME, e.g. – LogP, a lipophilicity measure – Solubility – Permeability – Cytochrome p450 metabolism Means estimates can be made for millions of compounds, helping reduce “attrition” – the failure rate of compounds in late stage 32 Questions Department of Pharmacy & Pharmacology From Molecules to Medicines 1 Lecture 26: Introduction to Drug Discovery and Design Dr Andrew Watts 1 Overview Introduction to Drug Discovery and Design A look at the history of drug discovery; sources of drug molecules and classes of drug molecules; types of drug targets. Basic principles of Drug Design & Drug Design pipeline Definition and properties of a drug molecule; Receptor theory; Designing molecules to fit receptors; Rational drug design; What makes a ‘good drug’? Steps involved in taking a new drug to market, from discovery to regulatory approval; what a new drug costs to develop; where do most drug candidates fail? Modern Drug Design techniques Overview of modern techniques used in drug design and the impact they have had on the drug design process. 2 Drug Design and Development Drug design covers the identification and optimisation of the active pharmaceutical ingredient (API) of a medicine. Focus of the rest of M2M1 and continued throughout your course. Development of the API into a Medicine continues in M2M2 (Semester 2) 3 Sources of Drug Molecules 4 1. Natural Products Source of the earliest drugs 2. Semi-synthetic Derivatives of natural products Receptor selectivity 3. Synthetic Small molecule drugs Large biological drugs So where does the API come from? Natural Products 5 The Ebers Papyrus (c. 1550 BC) Origins of Drug Discovery The Ebers Papyrus is one of the oldest compendiums of treatments for disease, dated to around 1550 BC. It documents around 800 prescriptions written in hieroglyphics for over 700 remedies. Includes mention of mental disorders, and remedies for conditions such as depression and dementia. Aloe vera - worms, relieves headaches, soothes chest pains, burns, ulcers and for skin disease and allergies Juniper tree (Juniperis phonecia; Juniperus drupacea)- digestive, soothes chest pains, soothes stomach cramps Henna (Lawsonia inermis) - astringent, stops diarrhea, close open wounds (and used as a dye). Parsley (Apium petroselinum) – diuretic Sesame (Sesamum indicum)- soothes asthma Tamarind (Tamarindus indica)- laxative Natural Products 6 Sources of Natural Products Natural products may be extracted from cells, tissues and secretions. These may come from -Microorganisms Penicillin -Plants Tannins Alkaloids Paclitaxel Morphine -Animals Conotoxins Natural Products 7 Example: Morphine Derived from the seedpods of the poppy, Papaver somniferum (opium) has been used for thousands of years by early civilizations By the 1800’s in Europe, opium was the painkiller of choice for physicians, but its effects were hard to predict. Friedrich Sertürner (1783-1841) sought to isolate the active ingredient in opium, from which predictable and reliable doses could then be produced. Friedrich named his new drug for the Greek god of dreams, Morpheus, but to keep it consistent with standard naming conventions, labeled it morphine. By the mid-1820s, pharmaceutical companies, including that founded by Heinrich Emanuel Merck, were producing standardized doses of the drug. In addition to being sold as an analgesic, initially morphine was also marketed as a non-addictive cure for addiction to both alcohol and opium. Natural Products 8 Other Examples Despite being a ‘traditional’ source, natural products still remain an important source of current drug molecules Examples of some current antibiotics or cancer drugs isolated from, or derived from, natural sources. Semi-Synthetic Drugs 9 Definition sem·i·syn·thet·ic (sem'ē-sin-thet'ik), Describing the process of synthesizing a particular chemical using a naturally occurring chemical as a starting material, thus obviating part of a total synthesis, for example, the conversion of cholesterol (obtained from a natural source) into a corticosteroid. What are the advantages of semi-synthesis? 1. Components of the molecule are ‘complex’ and too difficult to synthesise chemically. 2. Limited availability from natural sources 3. Modification of a natural compound can result in altered receptor selectivity or improved specificity, leading to additional/alternative uses. Semi-Synthetic Drugs 10 Examples: Taxol Components of the molecule are ‘complex’ and too difficult to synthesise chemically. Taxol Taxus brevifolia (Pacific Yew) In the early 1960's, Jonathan Hartwell of the US National Cancer Institute organized collection of plants from the U.S. for evaluation as potential sources of anticancer drugs. One plant collected as part of the NCI program was the Pacific yew, Taxus brevifolia, and the extract was confirmed as active in an animal test system. Component identified as Taxol. Taxol interferes with cell division by binding to the protein tubulin, which is a key factor in mitosis. Unlike some other cancer drugs which prevented tubulin from assembling into microtubules, taxol bound to assembled microtubules and blocked them from disassembling. Semi-Synthetic Drugs 11 Examples: Taxol The NCI signed an agreement with the pharmaceutical company Bristol Myers to fund and carry out further work, including securing a larger supply of drug. Taxol is found in the bark, and the collection of yew bark from Oregon and Washington was expanded so that more trials could be run. At their peak, Pacific yew bark collections were several hundred thousand pounds of bark per year. The bark from a single tree only yielded enough taxol for about one dose of the drug. Semi-Synthetic Drugs 12 Examples: Taxol Taxus baccata (Common Yew) Bristol Myers addressed the supply problem by finding an alternative source of the compound. The common yew, Taxus baccata, which is widely grown as an ornamental tree, contains a taxol relative, 10-deacetyl baccatin III, which is relatively more abundant, and which can be converted to taxol by chemical manipulation. 10-DAB Taxol Semi-synthetic synthesis of Taxol Semi-Synthetic Drugs 13 Example: Morphine analogues Modification of a natural compound can result in alternative receptors selectivity or improved specificity, leading to additional/alternative uses. Modification of morphine results in opioid analogues with varying potency and receptor specificity. Semi-Synthetic Drugs 14 Example: Morphine analogues There are three types of opioid receptors. Chemical modifications to morphine can change which receptor type (  ) the compound binds to and the effect it causes (agonist, antagonist). Synthetic Drugs 15 Total synthesis of drugs using chemical methods, usually starting from readily available materials that are relatively cheap. Advantages: Reliability of production, cost effective, quality control Disadvantages: Limited chemical space, limited number of steps, IDENTITY OF TARGET? Drug Design Drug Design 16 In the case of natural products (and many semi-synthetic drugs), Nature has evolved and produced the API for us. In drug design, we need to rationally discover the identity of the API using a series of techniques and methods. This is referred to as Rational Drug Design. Department of Pharmacy & Pharmacology From Molecules to Medicines 1 Basic Principles of Drug Design Dr Andrew Watts 17 The Development Process 18 Druggable Targets 19 Enzymes Receptors Ion channels Substrates, Metabolites and Proteins Transport systems DNA/RNA and the ribosome Physicochemical mechanism When designing a drug, there are a number of targets we can focus on: Druggable Targets 20 There are a number of key factors which make a target ‘druggable’. Target is disease-modifying and/or has a proven function in the pathophysiology of a disease. Modulation of the target is less important under physiological conditions or in other diseases. Target expression is not uniformly distributed throughout the body. Druggable Targets 21 There are a number of key factors which make a target ‘druggable’. Properties of an ideal drug target: If the druggability is not obvious (e.g. as for kinases) a 3D- structure for the target protein or a close homolog should be available for a druggability assessment. Target has a favorable ‘assayability’ enabling high throughput screening. A target/disease-specific biomarker exists to monitor therapeutic efficacy. Favorable prediction of potential side effects according to phenotype data (e.g. in k.o. mice or genetic mutation databases). Target has a favorable IP situation (no competitors on target, freedom to operate). Drug Molecules 22 “Every year, millions of new molecules are prepared, but only a small fraction of these are ever considered as possible drug candidates. A chemical compound must possess certain characteristics if it is to cross the hurdle from being an organic molecule to becoming a drug molecule.” What is a drug molecule? What is a drug like molecule? Drug Molecules 23 Chemical Properties Lipinski’s Rule of Five does a good job of quantifying drug like properties. According to this rule a drug should have: molecular weight < 500 a log P < 5 < 5 hydrogen bond donors < 10 hydrogen bond acceptors Drug Molecules 24 A drug molecule possesses one or more functional groups positioned in three- dimensional space on a structural framework that holds the functional groups in a defined position that enables the molecule to bind specifically to a targeted biological macromolecule, the receptor. Shape Drug Molecules 25 In vivo properties These structurally distinct molecular fragments are referred to as bioisosteres Bioisosteres need to be compatible with a variety of conditions in vivo. ie. solubility, absorption, stability, toxicity Drug Molecules 26 Types of Drug Molecules 27 As molecular weight increases, so does specificity. Size Aspirin 21 atoms Growth hormone 3,000 atoms Monoclonal Antibody 25,000 atoms 1º 1º, 2º, 3º, 4º Structure Small molecule Biological Low MW Low specificity High MW High specificity Department of Life Sciences Dr Charlotte Dodson Fundamentals of Pharmacy / Pharmacology: The Chemistry of Drugs Lecture 25 – Making and breaking biological outcomes SL12101 SL12009 Intended learning outcomes (L24 & L25) At the end of these lectures you should be able to: Describe the chemical properties of amino acids (backbone and side-chain atoms and chemical structures) and apply this knowledge in a biological context. Describe the factors important in achieving specificity in protein interactions in a biological context and the effect of this specificity on the biological outcome. Use your knowledge of the chemical properties of amino acids to predict the effect of novel protein mutations on protein function. The importance of chemistry in biology Much of biology relies on chemistry …to generate energy and transfer it into a useful form (usually ATP, but also ion or metabolite gradients) …to partition molecules into different compartments …to catalyse reactions so that they occur on a meaningful timescale …to maintain molecular structure (proteins, DNA packaging, membranes) …to regulate the activity of other molecules …to transmit and store information The pharmaceutical industry also exploits chemistry in the treatment of disease, through developing small molecule drugs and biological therapies (eg antibodies, nucleic acid therapies) Some of this chemistry involves making and breaking covalent bonds (catalysis, energy generation in the form of ATP, storage of information) Other chemistry (molecular structure, molecular partition, ion / metabolite gradients) relies on the same weak forces that stabilise protein tertiary structure (see SL12100 / SL12010). Irrespective of whether covalent or non-covalent chemistry is involved, life as we know it relies on most molecular interactions being specific. The importance of chemistry in biology Much of biology relies on chemistry …to generate energy and transfer it into a useful form (usually ATP, but also ion or metabolite gradients) …to partition molecules into different compartments …to catalyse reactions so that they occur on a meaningful timescale …to maintain molecular structure (proteins, DNA packaging, membranes) …to regulate the activity of other molecules …to transmit and store information The pharmaceutical industry also exploits chemistry in the treatment of disease, through developing small molecule drugs and biological therapies (eg antibodies, nucleic acid therapies) Some of this chemistry involves making and breaking covalent bonds (catalysis, energy generation in the form of ATP, storage of information) Other chemistry (molecular structure, molecular partition, ion / metabolite gradients) relies on the same weak forces that stabilise protein tertiary structure (see SL12100 / SL12010). Irrespective of whether covalent or non-covalent chemistry is involved, life as we know it relies on most molecular interactions being specific. The importance of chemistry in biology Much of biology relies on chemistry …to generate energy and transfer it into a useful form (usually ATP, but also ion or metabolite gradients) …to partition molecules into different compartments …to catalyse reactions so that they occur on a meaningful timescale …to maintain molecular structure (proteins, DNA packaging, membranes) …to regulate the activity of other molecules …to transmit and store information The pharmaceutical industry also exploits chemistry in the treatment of disease, through developing small molecule drugs and biological therapies (eg antibodies, nucleic acid therapies) Some of this chemistry involves making and breaking covalent bonds (catalysis, energy generation in the form of ATP, storage of information) Other chemistry (molecular structure, molecular partition, ion / metabolite gradients) relies on the same weak forces that stabilise protein tertiary structure (see SL12100 / SL12010). Irrespective of whether covalent or non-covalent chemistry is involved, life as we know it relies on most molecular interactions being specific. How do we obtain specificity in biology? Specific interactions involving proteins can be with large molecules (eg other proteins) or small molecules (eg neurotransmitters, substrates). Interactions can occur at the protein surface or in an active site. Specific protein-protein interaction on protein surface Specific protein-substrate interaction at active site Specificity in molecular interactions relies on: Molecular shape: different molecules fitting together like a 3D jigsaw Chemical complementarity: positive charges interacting with negative charges, hydrogen bonding potential fulfilled, clusters of hydrophobic moieties Spatiotemporal overlap: molecules being in the same place at the same time Interactions in an active site: Class A -lactamases in antimicrobial resistance -lactam antibiotics are the single most prescribed antibiotic class and include penicillins, cephalosporins, carbapenems and monobactams -lactam antibiotics target enzymes known as penicillin-binding proteins (PBPs). PBPs are involved in the biosynthesis of the cell wall of Gram-negative bacteria and are irreversibly inhibited by -lactam antibiotics. -lactamases are a class of enzymes found in some Gram-negative bacteria -lactamases hydrolyse the amide bond of the four- membered -lactam ring found in many antibiotics -lactam hydrolysis means that the antibiotic can no longer bind to its target, leading to loss of activity If an antibiotic has no activity in its target bacteria, this means that the bacteria is resistant to the antibiotic. -lactam hydrolysis is therefore a mechanism of antimicrobial resistance Penicillin Cephalosporin Carbapenem Molecular shape of benzylpenicillin binding to PC1 − lactamase from S. aureus Penicillin binds in a pocket on the surface of PC1 (the enzyme) Penicillin fits well within its binding pocket (there are no big gaps) Bound drug displaced so that binding site visible Chemical complementarity of benzylpenicillin binding to PC1 -lactamase from S. aureus Hydrogen bonding to Gln237 backbone Benzylpenicillin (active antibiotic) Benzylpenicillin after hydrolysis (inactivated molecule) Interactions of benzylpenicillin (hydrolysed) bound to the active site the of -lactamase PC1 (inactivating enzyme) Hydrogen bonds to solvent (water) Hydrogen bond to Asn132 sidechain Hydrophobic residues around benzine ring Ring stacking between benzine ring and Tyr105 sidechain Chemical complementarity of benzylpenicillin binding to PC1 -lactamase from S. aureus Tyr105 Asn132 Gln237 Ile239 H2O Comparing -lactamase binding pocket with that in a PBP -lactamase (breaks down antibiotic, causes AMR) PBP (antibiotic action / inhibited by antibiotic) Both pockets form hydrogen bonds to same chemical groups PBP additionally forms a covalent bond to the sidechain of Ser434 (forms an acyl-enzyme) You are investigating a mutant PBP enzyme in which Asn489 is mutated to Asp. What do you predict the effect will be on antibiotic binding? Aspartic acid Covalent chemistry in -lactam hydrolysis Deprotonated serine acts as a nucelophile Reaction stops here in PBP enzyme! (Acyl-enzyme is covalently inhibited) -lactamase has hydrolysed -lactam. - lactam is now inactive. Summary Chemistry is essential for biology to take place Biology relies on both covalent and non-covalent chemistry Specificity in biological interactions is obtained by Molecular shape Chemical complementarity Spatiotemporal overlap An example of complementary molecular shape and chemical complementarity is the interaction between -lactam antibiotics and - lactamase / PBP enzymes Similar interactions in both target and off-target enzymes makes it difficult to design new antibiotics Fundamentals of Pharmacy / Pharmacology: The Chemistry of Drugs SL12101 / SL12009 Structural, physical and chemical properties of amino acids: workbook (Lecture 24) Dr Charlotte Dodson Name………………………………. Department of Life Sciences 2 Introduction Throughout this workbook I have included a number of tasks for you to complete (highlighted in blue). These are of varying difficulty, but I hope that none of them is too onerous. The point of these tasks is to encourage to you to engage in active learning and to think about the material as you work your way through the workbook. This may feel like hard work now, but there is lots of evidence that active learning makes us learn the material more deeply (better) and – in due course – makes revision much easier. On this page there is a table with a reminder of the first few letters of the Greek alphabet. At the back I have included the structures of the proteogenic amino acids for reference. Please note that you are expected to learn the structures of these amino acids for this unit. Reminder of the first few letters of the Greek alphabet Letter Pronounciation  alpha b beta  gamma  delta  epsilon  zeta 3 Chemistry of amino acids Revision of amino acids and carbon naming In SL12100 / SL12010 and at school you covered the general structure of an amino acid. At school level, an amino acid consists of an amine group, a carboxylic acid, a central carbon linking these two moieties, and a variable sidechain or R-group (Figure 1). Figure 1: General structure of an amino acid. Since school you have learnt that the carbon adjacent to a carboxylic acid is called the alpha carbon (C). This rule holds in amino acids, and so we can also label the C in Figure 1 above. This convention of carbon numbering continues, moving away from the carboxylic acid. Therefore, depending on the amino acid side chain, we may also have a C and a C etc. Alanine (Figure 2) has a C and a C only. Figure 2: Structure of alanine with alpha and beta carbons labelled. In SL12100 / SL12010 we briefly mentioned the structure of the different amino acid sidechains. You will need to learn these chemical structures for this unit. As a reminder, a copy of the chemical structures can be found inside the back cover of this booklet. 4 Task: Draw the structures of aspartic acid and isoleucine below. Label the amine, the carboxylic acids and the different carbons in the sidechain (C, C etc). Table I: Labelling the atoms in an amino acid Aspartic acid Isoleucine In proteins, the carboxylic acid group of one amino acid is covalently bonded to the amine group of another. This results in the nitrogen of a NH group being covalently bound to the carbon of a C=O, ie in formation of an amide. This explains why the bond between two amino acids is called an amide bond. Importance of the chemistry of amino acids Biology – life itself – is a beautiful phenomenon. At a fundamental level, the way in which molecules interact with each other and in which information is stored and transferred in a biological cell underpins much of what we call life. The chemical properties of amino acids are intrinsic to the mechanisms by which much of this occurs – and to our ability to manipulate this in living organisms when the organism becomes sick. The chemical properties of amino acids are important in: i) Maintaining protein structure ii) Protein-protein interactions iii) Protein location within the cell iv) Channel specificity v) Information transfer (cellular signalling, gene regulation) vi) Catalysis vii) Small molecule drug binding and mechanism of action 5 Stereochemistry of amino acids All  amino acids are chiral (except glycine). Task: Look up the structures of L- and D- amino acids in a textbook or online. Is the amino acid in Figure 1 an L- or D- isomer? Copy this amino acid into the correct box in Table II below, and draw the structure of the other isomer. Table II: L- and D-isomers of  amino acids L-amino acid D-amino acid Only L-amino acids (and glycine) are incorporated into peptides and proteins in biology. D-amino acids are synthesised by prokaryotes and found in peptidoglycan in the bacterial cell wall. D-amino acids (eg D-Asp and D-Ser) have recently been found in humans – these are thought to be a consequence of ageing (racemization during life). NB all proteogenic amino acid are L-amino acids. Due to the way that naming conventions work, they are all S isomers with the exception of cysteine (which is R). L-alanine L-cysteine S-alanine R-cysteine Figure 3: Stereochemical naming of alanine and cysteine. Protonation state of amino acids The amine and carboxylic acid groups of free amino acids (ie monomeric amino acids not in a protein or peptide) can both titrate. Earlier in the semester Dr Caggiano talked about the pKa of common chemical groups: the pKa of a primary amine is around 10.6 and that of a carboxylic acid around 2.1. This means that at pH 7 an amino acid has a positive charge on one group and a negative charge on the other. It is therefore a zwitterion (it is zwitterionic). Question: What is the net charge on glycine at pH 7? 6 Task: Using the pKa values for an amine and carboxylic acid given below (and on the previous page), complete Figure 4 by drawing the structure of a general amino acid at pH 1 and pH 13. Which one is a cation? Which one is an anion? Figure 4: The protonation equilibria for a general amino acid Remember, even though one form is predominant at each pH value, all three forms exist to some extent at all pH values. Properties of the sidechains of amino acids In SL12100 / SL12010 we listed the weak non-covalent interactions which stabilise protein tertiary structure. These interactions provide the free energy (G) necessary to stabilise the folded protein structure. They also contribute free energy and – most importantly – the specificity which ensures that a small molecule drug binds to the correct target protein. The two most relevant interactions for this unit are salt bridges (interactions between permanent positive and negative charges; ie an interaction between a + (plus) and a – (minus)), and hydrogen bonds. Protonation state of amino acid sidechains Like the amine and carboxylic acid groups of free amino acids (and the N- and C-termini of peptides), some amino acid sidechains will titrate. Reminder: pKa is related to the log of the acid constant Ka. Ka is the dissociation constant of the protonation equilibrium (think back to equilibria at school). It can also be thought about as the pH at which half of the sidechain molecules will be protonated and half will be deprotonated. a K HA H A + − + [ ][ ] [ ] a H A K HA + − = 10 log ( ) a a pK K = − When the pH of a solution is below the pKa of a titrating group, the group will be protonated. When the pH of a solution is above the pKa of a titrating group, the group will be deprotonated. The protonation state of an amino acid will determine its charge, and thus its ability to form salt bridges. Task: Complete Table III (on the next page) by drawing the amino acid sidechains at the indicated pH values. Remember that each titrating group has its own pKa, so you will need to consider the protonation state of amine, carboxylic acid and sidechain in turn. pKa values for amino acid sidechains are given on p15. 7 Table III: Protonation state of different amino acid side chains at indicated pH Aspartic acid at pH 7 Lysine at pH 7 Glutamic acid at pH 3 Histidine at pH 5 Hydrogen bonding between amino acids A hydrogen bond occurs when a hydrogen atom covalently attached to one electronegative atom (often N or O) is shared with another electronegative atom that possesses a lone pair of electrons (Figure 5 below): Figure 5: A hydrogen bond (shown in green) between a protein backbone carbonyl and a lysine sidechain We encountered hydrogen bonds between the backbone atoms of amino acids when we looked at protein secondary structure in SL12100 / SL12010. Alpha helices and beta sheets are stabilised by hydrogen bonds between the oxygen of a backbone carbonyl and the hydrogen of a backbone amine. Task: In the space below, draw a backbone hydrogen bond that might be found in a beta sheet: 8 Hydrogen bonding of amino acid sidechains In addition to backbone-backbone hydrogen bonds, hydrogen bonds can occur between protein sidechains or between protein sidechains and the protein backbone (eg Figure 5 above). These are the hydrogen bonds which make up protein tertiary structure. Hydrogen bonds can also occur between protein sidechains and small molecule drugs. The hydrogen bonding potential of functional groups on a small molecule drug with amino acid sidechains (or backbone) in the binding site gives rise to much of the specificity of a drug-target interaction. Task: The hydrogen bonding potential of different amino acid sidechains is shown in Figure 6 below. Have a look at the different side chains and circle the groups could donate or accept a hydrogen bond. Hydrogen bond donors Hydrogen bond acceptors Hydrogen bond donors or acceptors Figure 6: Hydrogen bonding properties of amino acid sidechains at physiological pH (pH 7). -stacking of aromatic sidechains Aromatic sidechains (with -bonds) often stack on top of each other in protein structures due to attractive - interactions between the aromatic rings. The same interactions are used in drug design to increase the binding energy between a small molecule drug and its protein binding site (Figure 7). 9 Figure 7: -stacking between the aromatic ring of a drug molecule and a phenylalanine residue in a protein binding site. Note the two leucine residues creating an hydrophobic environment in the binding pocket. The protein is shown in green (most sidechains not drawn). Carbon atoms of the drug are shown in pink, nitrogen atoms in blue and sulphur in yellow. Note: − interactions are usually less specific than hydrogen bonding interactions. In drug design, - interactions are often used to increase affinity (ie make the drug bind the target more tightly) and hydrogen bonds to increase specificity (ie make the drug bind to the right target in the right place). Task: One amino acid has been omitted from Figure 8 below. Which one do you think it is? Draw your answer in the space provided. Figure 8: The structure of four amino acids which can engage in -stacking Amino acid sidechains and protein solubility So far we have focussed on the interactions which stabilise the internal structure of a protein (secondary, tertiary etc) or which stabilise drug-protein binding. However, the chemical properties of the amino acids on a protein surface are also important! Cytosolic proteins (proteins found in the cytosol of a cell) are in a hydrophilic environment. In order to remain stably folded in solution, cytosolic proteins must be able to interact with the charged environment created by water molecules. Task: Name two amino acids whose sidechains would increase the ability of the surface of a protein to interact with an aqueous environment. 10 Other proteins (membrane proteins) are found in the hydrophobic environment of the lipid bilayer which makes up the membranes of a cell. The surface of these proteins must be able to interact with their environment. Task: Name two amino acids which would increase the ability the surface of a protein to interact with the lipid bilayer. Some transmembrane proteins form channels or pumps to enable charged species such as Na+ or K+ to cross the cellular membrane. Name an amino acid which might help a positively charge ion such as Na+ to cross a lipid bilayer. Amino acid sidechains as chemical reactants The chemistry of amino acid sidechains is not only important in drug binding and in stabilising protein structure. Amino acid sidechains can themselves react: i) in biosynthesis ii) by forming covalent drug-target interactions (covalent inhibitors have a very high affinity as they are covalently attached to their target) iii) in enzyme catalysis iv) to form modified amino acids (for cellular signalling, to regulate gene expression, in response to O2 levels etc) Amino acids as nucleophiles Four amino acids can act as nucleophiles in chemical reactions at or near physiological pH (Figure 9). The first three (cysteine, aspartic acid and glutamic acid) are frequently found, particularly on the surface of proteins. The fourth (deprotonated serine) is not normally found on the surface of proteins, but is found in the active site of serine proteases. Note the protonation state of the sidechains of all four amino acids drawn. Figure 9: Amino acids which can act as nucleophiles in chemical reactions The relative order of reactivity of the different nucleophilic functional groups is: 2 R S R NH R COO R O − − − −  −  −  − 11 Modifications of amino acids Amino acid sidechains may be post-translationally modified (be chemically modified by enzymes once they are fully folded and released from the ribosome) as part of normal cellular function. Common modifications are phosphorylation, acetylation, methylation and hydroxylation. Post- translational modifications can usually be reversed by a different enzyme to give a dynamic situation in which the chemistry of an amino acid reflects current conditions in the cell. Protein phosphorylation Amino acids may be phosphorylated on a sidechain hydroxyl. Phosphorylation is a chemical signal used to transfer and integrate information in the cell and is used in cell signalling. Threonine, serine and tyrosine all possess hydroxyl groups on their sidechains, and phosphorylation of these amino acids in proteins is common in protein kinase signalling. Task: Complete Table IV by drawing the phosphorylated serine and tyrosine residues in the spaces provided Table IV: Amino acid phosphorylation Phosphothreonine Phosphoserine Phosphotyrosine Amino acid acetylation and methylation Lysine residues in proteins can also be modified – either by acetylation or by methylation. Acetyllysine and methyllysine are important in the regulation of gene expression. While acetylation is the addition of a single chemical group, lysine can undergo mono-, di- and tri-methylation, each of which communicates different biological information. Task: Complete Table V on the next page 12 Table V: Lysine modifications Acetyllysine mono-methyllysine di-methyllysine tri-methyllysine Proline hydroxylation Proline residues can be hydroxylated. This may be to signal cellular oxygen levels, or to increase protein structural stability (eg in collagen). The structure of hydroxyproline is given in Figure 10 below. Figure 10: The structure of hydroxyproline 13 Summary The chemical properties of amino acids give rise to many of the macroscopic properties of biology – and life as we know it. Amino acids have a stereocentre and may be L- or D- isomers. Depending on their chemistry, amino acid sidechains may protonate / deprotonate or form - stacking interactions. If the pH of the environment is known, the protonation state of an amino acid sidechain can be determined from its pKa. The chemical properties of amino acid sidechains determine the interactions made within and by a protein molecule. They also determine its reactivity, its ability to signal, and the specificity of its interaction with a small molecule drug. The material in this booklet is © Charlotte Dodson and is published under a CC-BY-SA 4.0 licence https://creativecommons.org/licenses/by-sa/4.0/. 14 Reminder of amino acid structures and sidechain pKa values Amino acids with aliphatic sidechains Amino acids with aromatic sidechains 15 Amino acids with acidic sidechains sidechain pKa = 3.65 sidechain pKa = 4.25 sidechain pKa = 8.18 Amino acids with basic sidechains sidechain pKa = 12.48 sidechain pKa = 6.00 sidechain pKa = 8.95 Amino acids with neutral polar sidechains Lecture 30: New & Emerging Drug Technologies Dr Paul De Bank [email protected] SL12009/12101 Department of Life Sciences Fundamentals of Pharmacy / Pharmacology: The Chemistry of Drugs After completing this lecture, you should be better able to: 1. Explain how genes, oligonucleotides and the manipulation of genetic material can be used to treat disease 2. Describe strategies for cell-based therapies 3. Outline the rationale for controlled and targeted drug delivery and how these can be achieved Learning outcomes Nucleic acid-based medicines 1. A mutation in DNA results in a defective protein required for normal cell function, causing a disease 2. Some diseases are exacerbated by increased levels of a particular protein 3. Sub-physiological levels of protein can be overcome by delivering exogenous protein DNA mRNA Protein Nucleic acids can be used as a treatment in all three scenarios DNA mRNA Therapeutic Protein Gene therapy → therapeutic protein Overcomes the difficulties of producing & purifying recombinant proteins by getting the patient’s cells to do it! Transient, so repeated injections of gene/vector as required Adapted from www.nature.com/nrg/journal/v4/n10/box/nrg1183_BX2.html RNA interference - siRNA Small interfering RNA (siRNA) 21-25 bp long Can “knock down” gene expression if sequence is known Readily designed Readily synthesised  Micro RNA (miRNA) is ~22 nucleotide ssRNA Derived from genes that specifically code miRNAs, not proteins Important in regulating gene expression (~1000 miRNAs regulate about one third of our genes) Don’t usually cause mRNA destruction, just repress or destabilize target mRNA Strong evidence of involvement in a number of disease states, including cancer RNA interference - miRNA RNA interference - miRNA Nature (2005), 435, 834-8 High Expression Low Expression DNA coding underexpressed or missing miRNAs could be delivered to cells to replace the missing miRNAs Overexpressed miRNAs can be inhibited by anti-miRNAs, complementary nucleotides (~22 bases) that bind to the problem miRNA RNA vaccines https://investors.modernatx.com/static-files/34f97bb2-d89a-45e4-a770-cae0591fa807 YouTube video: How do mRNA vaccines work? Cell therapies Delivery of cells to the patient to correct disease or damage May be combined with genetic strategies, e.g. delivering cells which secrete a therapeutic protein, or reprogramming immune cells http://stemcells.nih.gov N.B. bone marrow transplants are a well-known type of stem cell therapy Strimvelis - first ex vivo stem cell gene therapy Treats Severe Combined Immunodeficiency caused by adenosine deaminase deficiency Autologous CD34+ cells expressing functional ADA Rare: ~15 children per year in Europe http://www.ema.europa.eu/ema/index.jsp?curl=pages/medicines/human/medicines/003854/human_med_001985.jsp&mid=WC0b01a c058001d124 https://www.cancer.gov/publications/dictionaries/cancer-terms/def/car-t-cell-therapy Cell therapies – CAR T cells Cell therapies – CAR T cells Larson, R.C. and Maus, M.V. Nature Reviews Cancer, 21, 145–161 (2021) Approved CAR T cell therapies: Kymriah Yescarta Tecartus Breyanzi ABECMA Carvykti Related emerging cell therapies: TCR-T cells CAR-M CAR-NK cells  Combination cell-gene therapy trialed in 56 patients Mesenchymal stem cells transfected with a gene called TRAIL An “off the shelf” treatment Scaling up cell culture is key Funded through the Biomedical Catalyst (MRC and Innovate UK) http://www.mrc.ac.uk/news-events/news/combination-cell- gene-therapy-for-lung-cancer-to-be-tested-in-uk-patients/ Cell therapies – future approaches  Current approaches to nucleic acid therapy are often transient and have varying degrees of success Permanently editing the genome is an attractive prospect - e.g. correcting a defect, knocking down a specific gene or activating a gene Gene editing dsDNA Double strand break Donor DNA? No Yes Yes Gene disruption Gene correction Gene insertion Science (2013), 341, 833–6 CRISPR/Cas9 is a system from bacteria which can be used to edit DNA In the last decade, it has been used to modify genes in human cells in vitro and in vivo - huge potential Can be used to inactivate, correct, and insert genes Gene editing T cell membrane CCR5 Viral coat protein Virus CD4 Gene editing: a cure for HIV? Gene editing: a cure for HIV? A number of other clinical trials are underway Stem cell therapies https://en.wikiversity.org/wiki/WikiJournal_of_Medicine/Medical_galler y_of_Mikael_H%C3%A4ggstr%C3%B6m_2014 https://opentextbc.ca/anatomyandphysiology/chapter/3-6- cellular-differentiation/ Limbal stem cell transplantation To cure blindness due to corneal burning Began work on the concept in 1990, first 2 patient trial in 1997 Holoclar approved by the EMA in December 2014 http://www.nature.com/news/behind-the-scenes-of-the-world-s- first-commercial-stem-cell-therapy- 1.17022?WT.mc_id=TWT_NatureNews S