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313 General, AA & Peptides Notes

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CHEM 313
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Chem 313 Advanced Organic Chemistry for the Life Sciences 2010 Prof. John Sherman, Office: A243, 822-2305, [email protected] Office Hours: open (call, e-mail, see me after class, make an appointment, or just pop by) TA: Jon Freeman will run two tutorials per week. Day/time to be announced. Text: No formal text. Use any introductory organic text. Handouts: Some notes will be posted on the web and/or handed out. These will be mostly definitions and other data. The majority of notes will be in-class. Outline A. Amino acids and peptides (3-4 weeks) 1. Amino acids-structure, acid/base properties 2. Reactions of amino acids 3. Peptides-structure determination, synthesis of B. Carbohydrates (3-4 weeks) 1. Monosaccharides-structure, conformations (review) 2. Reactions of monosaccharides 3. Protecting groups 4. Glycoside couplings 5. Oligosaccharide structures C. Organophosphate esters (1-2 weeks) 1. Acid/base properties, hydrolysis mechanisms, synthesis of 2. DNA/RNA synthesis D. Pyridoxal phosphate and Thiamin (1-2 weeks) 1. Structure, function 2. Mechanisms of catalytic reactions Breakdown: midterm 1: 10%; midterm 2: 15%; final: 50%; lab: 25%. Note: It is necessary to pass both the lecture and the laboratory parts of the course independently to pass the overall course! ! ! Goals: Survey the organic chemistry that is relevant to biomolecules. Delve into mechanisms to understand why/how reactions go the way they do. Compare and contrast chemists’ and nature’s ways of puttting molecules together. Reviews: For any topic, you should review as needed. Read the relevant chapter or sections from your old textbook. Any organic textbook will do; you can use textbooks from the library or google a topic. For starters, review amino acids, stereochemistry, and acid/base chemistry. 1 Amino Acids and Peptides Amino acids are difunctional compounds that contain both an amine and a carboxylic acid functionality. The acid has the priority in naming. The carbons between the functional groups are designated α, β, γ, δ, etc. starting from the carbonyl. Although there are 100s of naturally occurring amino acids, and many more synthetic derivatives, there are 20 common amino acids, those that are found in proteins. The 20 common amino acids are all α-amino acids. The 20 common amino acids differ in their R group or side chain. They have common names, three letter codes, and one letter codes. Glycine, Gly, or G, has no side chain (R = H). The 20 common amino acids are given on the next page. You do not need to memorize specific amino acids. 2 20 Common Amino Acids and their 3- and 1-letter Codes H 2 CO 2 Lysine Lys K H 2 CO 2 Glycine Gly G H2N CO2H Alanine Ala A NH 2 H2N CO2H Methionine Met M H2N CO2H Valine Val V S H N CO H Leucine Leu L 2 2 H2N CO2H Cysteine Cys C SH H2N CO 2 Serine Ser S H 2 CO 2 Isoleucine Ile I OH H 2 CO 2 Threonine Thr T H 2 CO 2 OH Arginine Arg R H2N CO 2 Phenylalanine Phe F NH HN NH 2 H 2 CO 2 Tyrosine Tyr Y H N CO H 2 2 Aspartic Acid Asp D CO 2 OH H N CO H H 2 CO 2 Histidine His H 2 2 Asparagine Asn N N CONH 2 NH H 2 CO 2 Glutamic acid Glu E H2N CO 2 Tryptophan Trp W CO H 2 N H H 2 CO 2 Glutamine Gln Q CO 2 Proline Pro P HN CONH 2 3 Chirality: Glycine is achiral. The other 19 common amino acids are chiral. Each contains a stereogenic center at the alpha carbon. The 19 common amino acids are of the “L” configuration. This means when drawn in a Fischer Projection (review if needed) with the most oxidiz2d carbon (CO H) at the top, R groups on the bottom, and 2he NH /H are drawn horizontally, 2hen the NH is on the left it is “L”. On the right would be “D”. D/L are conventions. Fischer Projections: Vertical lines go into the page (dashes); horizontal lines come out of the page (wedges). A switch of two substituents leads to the opposite configuration (R goes to S, S to R); two switches and you are back to where you started. Careful rotation about a second-to top or bottom vertical bond also yields the same configuration for that center. See below. CHO CO H CO H 2 2 HO H H2N H H2N H CH OH 2 CH 3 CH 3 L-Glyceraldehyde L-Alanine (of "S" configuration) L-Cysteine, "R" configuration CO 2 CO H2 H N H HSH C2 NH 2 2 CH 2H H Two switches or rotation about C–CO H 2 D/L configurations do not correlate with d/l, +/–, which are real observables based on optical rotation. The alpha carbon of cysteine has R configuration (review Cahn-Ingold-Prelog rules if needed), while the other 18 common amino acids have “S” configuration; again, all 19 common amino acids are “L” (Gly is achiral). Finally, two common amino acids, Threonine and Isoleucine, have a second stereogenic center within their side chains; the other 17 have no stereogenic centers in their side chains. 4 Acid-Base Properties of Amino Acids. Review of acid/base chemistry: CH 3O H2 CH 3O 2 + H pK = 4.74 a CH 3O 2 H -5 K a = 1.8 x 10 CH CO H 3 2 pK a – log (1.8 x 10 ) = 4.74 K a 10 -4.74 CH C3 2 Ka pK a 4.74 If = 1, then = 1 = = pH 4.74 CH C3 H 2 H pK CH 3O 2 a If pH = 5.74, = 10 = pH CH 3O H2 CH CO pKa 3 2 If pH = 3.74, = 0.1 = pH CH 3O H2 At a pH equal to the aK of an acid, the acid and its conjugate base are present in equal amounts. When the pH is one pH unit above the aK , there is 10 times more conjugate base present than conjugate acid. At a pH two units above theapK , there is 100 times more conjugate base present than conjugate acid. At a pH one unit below theapK , there is 10 times more conjugate acid present than conjugate base, etc. If HA has a pK of 5 and HB has a pK of 10, then HA is more acidic than HB, and B is more basic – a a than A . At a pH of 3, the predominant species are HA and HB. At pH 7, the predominant species are A and HB. At pH 12, the predominant species are A and B . 5 Back to Amino Acids: All amino acids have an amino group (basic) and a carboxyl group (acidic). Some have acidic or basic groups in their side chains. Simple amino acids, where the side chain is neither acidic nor basic, exist as zwitterions in neutral aqueous solution; that is, they contain both positive and negative charges. Tha pK of the carboxylic acid is about 2.3a and the pK the ammonium is about 9.6. Why is tha pK of an amino acid so much lower than that of acetic acid (4.74)? CO 2 CO 2 H3N H H 3 H + H R R the acidic proton pK a 2.3 the acidic proton CO 2 CO 2 H N H H N H + H 3 2 R R pKa= 9.6 Some side chains of amino acids contain acidic or basic graups. pK ’s of these groups are given below (you do not need to memorize these). AA pKaof Side Chain Tyrosine 10.1 Lysine 10.5 Arginine 12.5 Histidine 6.1 Cysteine 8.2 Asp 3.7 Glu 4.3 Factors that might affect a reaction: (1) Sterics, (2) Electronics, (3) Hydrogen bonding, and (4) Solvation. Review these concepts. Within Electronics, we have: induction, resonance (delocalization of charge), hybridization, octets (atoms like to have full octets), and aromaticity (achieving aromaticity accords stabilization). Review these concepts. Charges are not stable; reducing charge, even partially or by spreading it around, will stabilize a compound. The carboxylic acid of a typical α-amino acid is more acidic than a simple carboxylic acid because of the inductive effect: the positive charge on the ammonium a few bonds away inductively withdraws electron density from the carboxylate and thereby stabilizes the negative charge; the conjugate base is thus stablized. Resonance and sterics do not come into play. Hydrogen bonding is possible, but it is negligible in such a small molecule in water due to hydration (solvation by water). 6 Pyridinium is 5 pK unats more acidic than ammonium due to hybridization: the lone pair of pyridine is sp while the lone pair of an amine is sp . An sp lone pair is held more tightly to the nucleus and thus has less affinity for a proton. (The more s-character, the closer the orbital/electrons/charge are to the nucleus. This is stabilizing for a negative charge, but destabilizing for a positive charge. The lone pair of pyridine is held closer to the nucleus than in an amine, so pyridine’s lone pair is more stable, less basic than the lone pair of an amine.) No octet is lost, no aromaticity is lost, there are no induction issues. Resonance? Yes, but delocalization of the charge would break up the aromaticity somewhat and lose octets, so this will be minor. (Six π electrons in a closed loop affords aromaticity, which gains a benzene ring 36 kcal/mol of stabilization. For a protonated pyridine, there are six pi-electrons and six p-orbitals in a closed loop for all resonance structures.) Pyrrolium is 14 pK unias more acidic than ammonium due to a gain of aromaticity upon deprotonation to become pyrrole. If pyrrole is protonated on its nitrogen, it loses aromaticity and has no resonance structures. If it is protonated on its α-carbon, it loses aromaticity and has three resonance structures, all of which lack an octet. It does protonate on its α-carbon over the N (and over the β-carbon), but pyrrolium is very acidic. Guanidiniums (e.g., Arg) have higher pK ’s thaa ammoniums due to delocalization of the positive charge (resonance); the conjugate acid is stablized and is thus less acidic. There is about a 1/3 + charge on each nitrogen. There will only be a small charge on the carbon due to a lost octet on the carbon when it is charged: normally, it is preferable to put a + charge on the less electronegative atom. In guanidinium, it is better to put a + charge on the more electronegative nitrogen (and maintain all octets) than on the less electronegative carbon since the carbon will lose its octet. The hybridization on carbon will be sp , but the nitrogens are each sp in one of three resonance structures and sp in two of3 three resonance structures: you can consider them sp , or in between sp and sp , or just consider 2 them effectively sp since the three nitrogens and the carbon are coplanar. In any case, the higher basicity of guanidine compared to an amine is due to delocalization of the + charge in the conjugate acid. If the hybridization is sp , that would reduce the basicity. Delocalization more than compensates for any effect of hybridization on the pK in tais example. Although the hybridization of guanidinium is an interesting question to consider, it is a good example of a concept that is too grey to be likely to appear on an exam. Phenols (e.g., Tyr) have markedly lower pK ’s thaa alcohols due to the delocalization of the negative charge around the ring (resonance); the conjugate base is stablized. This breaks up the aromaticity somewhat, but no octets are lost, so there is a net gain in stability of the conjugate base versus that of a simple alkoxide. Hybridization: this is a bit complex, but there doesn’t seem to be a way to have much net gain in stability for the conjugate base. Imidazole (His side chain) is aromatic. Like pyridine, protonation does not disturb the aromaticity. Also like pyridine, the lone pair on the nitrogen(s) is sp . An imidazolium should be a bit more stable than a pyridinium from the standpoint of resonance between the nitrogens of the immidazolium, but the extra nitrogen also inductively withdraws, and this likely leads to the overall very similar pK ’s of a imidazolium and pyridinium. Thiols (cysteine side chain): pK areamuch lower than alcohols due to lower charge density on anion. The “hard” alkoxide anion has higher affinity for a “hard” proton. 7 Isoelectric point (pI): The pH at which the amino acid exists in solution predominantly as a neutral species. For simple amino acids (those with no acidic or basic groups in their side chains), the pI = (pKa1 + pKa2)/2. This is the average of tae two pK s. CO 2 CO 2 CO 2 H N H H N H H N H 3 3 2 R R R pK = 2.3 pKa= 9.7 a "Neutral" "Base" "Acid" 2.3 + 9.7 pI = = 6.0 2 pH B 9.7 N B 6.0 N A N 2.3 A [HO ] The acidic form, A, predominates below pH 2.3. The basic form, B, predominates above pH 9.7. The neutral form, N, predominates between pH 2.3 and 9.7, and is maximized at pH 6. A and N are in a 1:1 ratio at pH 2.3. N and B are in a 1:1 ratio at pH 9.7. Thus, the equilibrium arrows above imply species present in equal amounts; all species are in equilibrium at all pH’s. If there is a potential charge in the side chain, the pI is the average of the two pK s that yield a neutral a species. Techniques for characterization and separation such as Electrophoresis take advantage of the different charges of the amino acids (and peptides and proteins) by effecting different mobility toward 8 the positive versus negative ends of electrodes. Thus, familiarity with charge states aids in separation and characterization. Given all aK s of an amino acid, you should be able to calculate the pI. Buffers: If amino acids are present in appreciable amounts, they can be used as buffers. They can buffer at any pH that corresponds to one of theia pK ’s. Given tha pK ’s and amounts of species in solution, the pH can be calculated. Likewise, given theapK ’s and the desired pH, the amounts of species needed to create that buffer can be calculated. You should be able to perform simple buffer calculations. Homework: given a solution of alanine at pH 9.00, draw the structure of the species present in appreciable quantities and calculate the concentration of each of these species. Synthesis of Amino Acids Plants can make all 20 amino acids. Humans can make 10. These 10 are considered nonessential, while the other 10 are essential for our diet. Only a few amino acids are made from scratch biosynthetically; most are derived from a few main amino acids. For example glutamine is derived from glutamic acid. We will look at some ways that nature makes amino acids when we discuss reactions of enzymes that use PLP at the end of the course. Beyond that, we will leave amino acid biosynthesis to biochemistry courses. How do chemists synthesize amino acids? Again, many are derivatives from others, where the original source is often natural (i.e., from microorganisms, gelatin, animal skin, etc.). Synthesis is required for non-natural amino acids or for those that are not readily available in suitable quantities. Keys to synthesizing an amino acid are of course (1) incorporating the amino and carboxyl groups in the appropriate positions (e.g., α to each other), and (2) incorporating the appropriate chirality. There are many ways to synthesize amino acids. Chem. Rev. 2007, 107, 4584-4671 is an 88 page review of just catalytic asymmetric amino acid synthesis. That could be a whole course. We will discuss only two methods. Ester Hydrolysis: Before we get into the Gabriel Synthesis, let’s review a “simple” mechanism is some detail. Esters can be hydrolyzed under basic conditions or under acidic conditions. Under basic conditions (Saponification), the stable species are hydroxide, alkoxide, alcohol, and water; any oxygen atom should be neutral or have a formal charge of –1 for any species, including intermediates. H , CH3OH 2 H 3 will not be present in appreciable amounts. They are not stable. Such species should be avoided in any mechanism under basic conditions. Oxygen atoms should not have a +1 formal charge under basic conditions. The aK ’s of said species are below 0, whereas the pH is around 10 or above. The ultimate product will be R2O , not RC2 H, unless of course there is an acid workup. The shortcut notation below is often used with nucleophilic reactions at cabonyls. It denotes formation of and then collapse of a tetrahedral intermediate. O O O O nuc R nuc R OR' R OR' R OR' nuc: nuc: + R'O 9 Esters are not very electrophilic, nor is water a very strong nucleophile. Under basic conditions, the nucleophile is hydroxide. Upon attack by hydroxide, a tetrahedral intermediate forms. Might this species lose its ROH proton? It could under these conditions, but that would yield a dianion, which would not be very stable. No points would be taken off for such a species under these conditions. Could the ROR be protonated before departure? Not likely under basic conditions; points would be taken off for that. Is the shorthand formation/collapse of the tetrahedral intermediate acceptable? I’ll use it in class to save time. I’d prefer to see details on exams, but it is acceptable. Show all bond- making and bond-breaking, including protonations/deprotonations and use the appropriate proton or base sources. Ester hydrolysis under basic conditions will not be reversible since the carboxylate is essentially non- electrophilic with respect to alkoxide. (A carboxylate can react with a strong reducing agent such as LAH, but that is another matter.) And again, virtually no carboxylic acid will be present. If an ester has hydrogens on the α-carbon it may also undergo Claisen condensations under basic conditions. Many reaction mechanisms entail one or more proton transfers. We will write these out at first, but + eventually, we may just say “H transfer.” It is usually not the same proton that is actually transferred (thus the expression “proton transfer” is misleading), although this is possible, particularly in an enzyme’s active site. The transfer may be internal, especially if 6-membered transition state is possible, but the “transfer” is often inter-molecular. The solvent involved is usually not the same molecule that removes and delivers the proton. There are typically heaps of protons and solvent molecules around. "H Transfer" H2O: H H O OH H O OH H O OH R OR' R O R' R OR' H H H H O O H H Under acidic conditions, ester hydrolysis is reversible. The ratio of ester to acid is dependent on the ratio of water to alcohol (and on the relative thermodynamic stabilities of the ester and acid). The nucleophile is water, as no hydroxide, alkoxide, or oxygen with a formal charge of –1 will be present in appreciable amounts (pH is around 3 or below). The ester must be protonated first to make it electrophilic enough to react with water. Does it matter which lone pair of the ester carbonyl is protonated? Well, they are non-equivalent, but no it doesn’t matter. You should challenge yourself with such questions. Can the leaving group be an alkoxide? No, not under acidic conditions. For proton transfers, I’d like to see details. A proton rarely actually transfers between atoms of a molecule; it is usually a two step process. Back to Amino Acid Synthesis: The Gabriel Synthesis involves addition of potassium phthalimide to diethyl-2-bromomalonate. The ensuing diester is hydrolyzed to the diacid, then decarboxylated, then further hydrolyzed: the imide is hydrolyzed to yield an amine (amino acid) and the diacid phthalic acid. You should be comfortable with such a mechanism, which involves tautomerization. Use of the 10 Gabriel Synthesis yields a racemic product: the tautomerization following the decarboxylation entails protonation at the α-carbon, which will occur 50:50 from each side. There are methods to resolve (separate enantiomers) amino acids such as formation of diastereomers. Alternatively, homochiral (one enantiomer only) amino acids are usually obtained from natural sources or from enantioselective syntheses. Gabriel Synthesis Overview O O O O O – KBr OEt N K + EtO OEt N Br SN2 OEt O O O potassium 1,2-benzenedicarboxylicimide + (potassium phthalimide) Ester hydrolysisH3O , heat 2 times CO 2 O O O hydrolysis decarboxylation OH CO 2 N N + OH – CO2 OH O O HO 2 NH 3 O O There are many methods to create unusual homochiral amino acids. We will look at only one, and very briefly at that. Chiral catalysts have been used to synthesize homochiral amino acids. Industrial syntheses include those to make L-DOPA (used to treat Parkinson’s disease) and Aspartame (artificial sweetener). The 2001 Nobel Prize in Chemistry was awarded in part for such work. Enantiospecific amino acid synthesis using a chiral auxiliary. An enantiospecific process is one where a starting enantiomer gives predominantly one enantiomeric product; the other starting enantiomer will give the other enantiomeric product. A chiral auxiliary is a homochiral (one enantiomer) helper molecule that is incorporated into an achiral molecule of interest, helps it achieve a stereoselective process, and is then removed, thus leaving a new chiral molecule that is not racemic. We’ll look at (1R,2S)-2-amino-1,2-diphenylethanol acting as a chiral auxiliary to effect enantiospecific amino acid synthesis. In the first step, an SN2 reaction between the amine (which is more nucleophilic than the alcohol) and the α-bromoester yields an intermediate that quickly cyclizes to a six-membered ring; the low nucleophilicity of the alcohol is compensated by the intramolecularity of the reaction. In the second step, a weak base mops up the acid produced and thus helps convert the remaining starting material from an ammonium to an amine. The second step protects the amine as a carbamate (t- butoxycarbonyl, t-Boc); the carbamate is not nucleophilic. Third step: sodium hexamethylsilazide is a very strong base that is sterically hinde
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