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ORGANIC chemistry review handout

ORGANIC chemistry review handout - Hug a7 03 01:55p To the...

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Unformatted text preview: Hug a7 03 01:55p To the Biochemistry Student A useful definition of biochemistry is “the tools of chemistry and physics applied to the problems of biology.” You can interpret this definition in two ways. First, much of the progress in biochemistry is driven by the ex- perimental tools of chemistry and physics, including structural analysis, ki- netics, thermodynamics, radioisotope labeling, spectroscopy, crystallography, and molecular modeling. Understanding biochemistry as a dynamic, evolv- ing field means seeing how currently accepted models have been devised by interpreting data obtained using chemical tools. Second, and of the most im- mediate importance as you begin your study of biochemistry, you will need to use your own tecls of chemical understanding, which you developed in gen- eral and organic chemistry. This may mean that you will need to review ma- terial from previous courses, especially organic chemistry. This chapter pro- vides a brief review of concepts from organic chemistry that you are most likely to need in biochemistry. I hope this review is adequate to get you started, but if any subject I raise here is unfamiliar, look it up in the index of your organic chemistry text to find further information. NOTE: Terms in italics are biochemical terms, and they may be entirely new to you, but I will explain them where they first appear. Terms in bold are review terms, and they should be somewhat familiar. Each review term will appear in one or more biochemical examples designed to help you recall the term and its meaning. ' Overview To begin this review, study Figure 1, which shows a sequence of ten chemical reactions that constitute the metabolic pathway called glycolysis. This sequence of reactions occurs in almost all living organisms. One of its Prelude to Biochemistry: ' . A Review of Important Concepts from Organic Chemistry / Hug a? 03 01:57p 2 Prelude to Biochemistry: A Review of Importont Concepts from Organic Chemistry main functions is to derive energy, in the form of the chemically active sub- stance ATP, from the nutrient glucose. Each compound in the pathway is a product of the previous reaction and a reactant in the succeeding reaction, and each is therefore called a metabolic intermediate. Other reactants and products are shown entering and leaving the pathway by arrows above or below the main reaction arrows You will study this pathway in more detail later. For now, examining it will serve to remind you of, and to illustrate, many areas of basic organic chemistry as they apply to biochemistry. For con- venience, the reactions of the pathway are numbered, and each compound is labeled with its full and abbreviated name. I will use only the abbreviated names in the text. Don’t worry about remembering the names and structures of all these compounds. Focus instead on the review terms and try to under- stand how they apply in these new surroundings. Dihydroxyacetone Phosphate (DHAP) , . _ . . . $H20P0329 Ono/H One/H (IJHQOH ongopoa26 c=o H—clz—OH ADP H—Cll—OH ‘f—O ADP til-0 CH20H A __ _ ATP HO-Clz—H T9: 2 HO—(IS—H ._ “0 ‘f H H0—‘f—H I .— — H—(lz—OH 1 H—Cli—OH 2 H (I: OH 3 H—Cli—OH 4 5 H—rls—OH H—c—oa H"?"°H H“?—°H \ o H . \ CHZOH CHzopoaze cnzopoa26 CH20PO329 \l/ 6 , . H——-G—OH n—Glucose D-GIUCOSB 5-Phosphate D-Fructose 6-phosphate o-Fructose 1.6- bisphosphate C|:H OP 0 29 (G6P) (Fer) _ (FBP) Z a D-Glyceraldehyde 3~phusphatc (G3?) NADl-l p0 26 ATP 6 e . e ' ATP 9 0% /o a o\\ /o 0% {o o.‘ ’0. CW ’0 W, 2 ? MU ‘f t H o ‘f mg) ‘1’ H—CII—OH H—tIS—OH -——- H—CII—OP0329___..2_ Cll_opoaze ——> c=o i 6 CH20P0329 7 CH20P032‘3 8 CH20H 9 CH2' ,. 10 0H3 HP0429 . _ V D-l,3— Bisphosphoglycerate n-S-Phosphoglycerate D-Z-Phosphoglycerate Phosphoenolpyruvate Pymvate (BPG) (3PG) (ZPG) (PEP) Figure 1. The Ten Reactions of Glycolysis Before I describe the pathway in detail, here’s a sampling of the con- cepts I will consider. I will draw your attention first to the structures of the intermediates, and second to the 3.3quth depicted. First, look at structures. Try giving systematic names to a few of them, using oxyphosphory! as the substituent name of the OPnge group. Your names will not be the same as the commonly used names; for example, the oxyphosphoryl group is usually called a phosphate group, as in glucose 6-phosphate, the product of the first reaction. Among the intermediates, can you find pairs of structural isomers? What functional groups can you find and name? What functional groups are stabilized by resonance? Do you remember the meaning of resonance, and how to draw the most important structures that contribute to a resonance Hug a7 03 01:57p I Prelude to Biochemistry: A Review of Important hybrid? Which intermediates are chiral compounds and how many stereoisomers are possible for each structure? Biochemists use D- and L- to designate the configurations of chiral compounds. Recall that each chiral center can also be assigned the (R)- or (S-contiguration, according to the system of Cahn, Ingold, and Prelog. Which molecules might exist in equilib‘ rium with forms other than those shown? For example, which might exist in two or more tautomeric forms, and which contain two functional groups that might combine with each other in an intramolecular reaction to produce an alternative form? . Second, examine each reaction. Try to discern what structural change converts the reactant to the product, and try to classify each change as addi- tion, elimination, substitution, or combinations of these simple reaction types. Can you write mechanisms for these reactions, assuming some simple type of catalysis, like specific-acid catalysis (catalysis by protons), and depict- ing with curved arrows the changing allegiances of electrons as bonds break and form? (This powerful form of writing reaction mechanisms is sometimes called electron pushing.) Do you see any oxidation or reduction reactions? One way to detect such changes is to compute the average oxidation number of carbon for each structure, assuming that the oxidation number of hydro- gen is +1 and that of oxygen is —2. (To simplify detection of oxidation-num- ber changes in carbon, you can ignore complex groups that do not change during the reaction.) Now you have an idea of some of the areas of organic chemistry that are essential to understanding a metabolic pathway, and some of the concepts that you need to review and use in your study of biochemistry. Let’s begin our detailed review by looking at the pathway more closely. The Structures The first structure in Figure 1 could be called 2,3,4,5.6-pentahydroxy- hexanal, but fortunately it has a common name, D-glucose. In fact, many bio- logical compounds have common names that were adopted before their struc- tures were known. This prevalence of common or trivial names is in part a blessing, because the names are often much shorter than systematic names, and in part a curse, because you cannot figure out the structures from the names You must therefore simply memorize many new names. Looking at the struc- ture of glucose, you can probably deduce that this compound is water-soluble. Organic compounds having at least one hydroxyl (—OH) or other hydrogen- bonding group per three or four carbons are usually appreciably soluble in water. Looking over all the intermediates in glycolysis, you can see that all would be very soluble in water. Not all biomoleculcs are water-soluble, how‘- ever. Fats and oils, which are esters of fatty acids such as palmitic acid (CH3(CH2)14COOH), are quite insoluble in water because they are mostly nonpolar. They are quite soluble in organic solvents like hexane. Many water- insoluble biomoleculcs belong to the class called lipids. Some are involved in forming membranes, which provide a barrier that makes the cell and its-inter: nal compartments impermeable to many water-soluble substances. . . The sugar glucose is a monosaccharide. a simple carbohydrate. Carbo- hydrates, one of the four main types or classes of biomoleculcs, include sug- ars and starches (the other biomoleculcs are lipids, proteins, and nucleic acids). Most monosaocharides are polyhydroxycarbonyl compounds, either aldehydes like glucose, or ketones like fructose. Notice that glucose has four stereocenters or chiral centers, at carbons 2, 3, 4, and 5 (the aldehyde carbon is C-l). These carbons form covalent bonds (which means that they share electron pairs) through sf-hybridized orbitals, so they form four bonds with [0.3 Concepts from Organic Chemistry 3 Plug 37 03 01:57P 4 Prelude to Biochemistry: A Review of Important Concepts from Organic Chemistry l “'— H—CIJ—OH H— CIJ—OH CHZOH Keto Form tetrahedral geometry. (C-l, in contrast, is stag-hybridized and trigonal pla- nar.) Because two configurations are possible at each stereocenter, there are 2“ or" 16 different stereoisomers of 2,3,4,S,6-pentahydroxyhexanal. Only the full mirror image, or the enantiomer, of D-glucose, shares its name. The enantiomer shown is D-glucose, designated D- because of its structural kin- ship with the simpler chiral compound D-gl'yceraldehyde (the 3-phosphate derivative of D-giyceraldehyde, called D-glyceraldehye 3-phosphate, is one of the products of Reaction 4 in Figure 1). The enantiomer of D-glucose is called L-glucose. The other stereoisomers of glucose, which are its di- astereoisomers or diastereomers, are chemically distinct from D- and L-glu— cose, and have their own common names. The drawings of glucose and the other intermediates in Figure 1 are called Fischer projections. At each chiral center, the groups attached by horizontal bonds in Fischer projections are understood to be projected outward toward the reader, while the groups attached by vertical bonds are pointing back into the page. Because the chiral carbons are sp3 hybridized, the angle formed by the chiral carbon and any pair of attached atoms is approximately 109.5 degrees. Because glucose is an aldehyde with a hydrogen atom on its a—carbon (02), in aqueous solutions it exists in equilibrium with a tautomeric enol form. The equilibrium reaction is shown in Figure 2. Carbons next to carbonyls are weakly acidic. If a base abstracts a pro- H9 H ‘5' ' H 9") cl} H " \C/ -. \C/H -..\C/ ,Ie II II .f—OH (it—OH H9 lc—OH ‘ + HO—CI:'—'H I i H0_?_H HO_?'—H H—Cl>_OH H—‘f" OH H—(IIJ—OH H——rl3—0H H_C__.0H H—(IJ-OH CHZOH (ISHZOH CH20H Enolate Ion Enol Form (Resonance Hybrid) Figure 2. Tautomerism in Glucose ton from the tit-carbon of the aldehyde (which is called the keto tautomer), a resonance-stabilized enolate ion (shown in square brackets) results. The enolate may be reprotonated (almost certainly by a different proton from the solvent) at 0-2 or at 0-1 (the oxygen of C-1). Protonation at 0-1 (shown) produces the enol form of glucose. This process is reversible, and because the reactant, glucose, is less acidic than the enol, it is more stable and predomi- nates in the tautomeric equilibrium. The fraction of enol-glucose at equilib- rium is so small that we always write glucose as shown in Figure 1, but tau- tomerism is important in explaining both spectroscopic and chemical properties of aldehydes and ketones, as I will show later. Hug a7 03 01:57p [0.5 Prelude to Biochemistry: A Review of importont Concepts from Organic Chemistry 5 Now examine the structures shown within the square brackets in Fig- ure 2. These structures, taken together, depict the enolate ion as a reso- nance hybrid of the structures shown, which are called resonance contrib- utors. Organic chemists depict molecules or ions as resonance hybrids when one Lewis diagram is inadequate to depict them fully. In this enolate ion, the negative charge is distributed between C—2 and the oxygen of C-1, so neither of the two drawings alone fully depicts the properties of the ion. This terminology does not imply that an enolatc is an equilibrium mixture of the resonance contributors It is better to say that the enolate ion is a W of the contributors, which means that it has some of the proper~ ties of both. A telling analogy from an organic text of the late sixties helped me un- derstand the meaning of resonance. It goes like this: A rhinoceros is a resonance hybrid of a dragon and a unicorn. This does not imply that the rhino spends part of its time as a dragon and part of its time as a unicorn. Instead, it means that the rhino has some dragon attributes (like a tough, thick coat), and some unicorn attributes (like its horn). In addition the rhino, like the resonance hybrid, is a real animal, but the dragon and unicorn, like the contributors, are fictional. So actually, the resonance contributors do not exist at all. They are fic- tional characters whose structures suggest the properties we want to depict for the real enolate ion. We view the hybrid as being most like the contribu- tor or contributors .we judge to be the most stable, based on criteria such as charge distribution and number of covalent bonds. Contributors with less separation of opposite charge and more shared electrons are more stable. And the more equivalent forms we can write, the greater is the stabilization by resonance. (While I think of the contributors as useful fictions, I leave for philosophers the question of whether molecules and ions themselves are real. Like the vivid characters in good fiction, they certainly help us to see the real world more clearly.) - Looking again at the enolate ion, you should conclude that, if the two contributors were’distinct substances, the second contributor would be more stable because the negative charge resides on oxygen, which is more elec- tronegative than carbon. Chemists assume, therefore, that the enolate ion is more like the more stable contributor, so you would expect the enolate ion to protonate faster on oxygen because a greater portion of the negative charge resides there. If so, why is the keto rather than the enol form favored at equi- librium? It must be that the lifetime of the enol form is shorter than that of the keto form, or to put it another way, the enol form is more acidic. The faster protonation of the enolate ion at oxygen is a kinetic effect, while the predominance of the keto form at equilibrium is a thermodynamic effect. In chemical reactions that can produce two alternative products, the one that forms faster is called the kinetic product, while the more stable one is called the thermodynamic product. In this example, the enol form is the kinetic product of protonation, but the keto form is the thermodynamic product. Given enough time, the keto form will predominate. Chemists can often con- trol reaction conditions to obtain a desired product. For instance, short reac- tion times favor kinetic products, while long reaction times at high tempera- tures favor the thermodynamic product by providing plenty of opportunity for the reaction to come to equilibrium. ' ‘ ' Plug 37 03 01:56P 6 Prelude to Biochemistry: A Review of Importont Concepts from Organic Chemistry Finally, notice the three curved arrows in Figure 2. These arrows depict the movement of electrons as bonds break and form. In the first step of the mechanism shown, an unspecified base removes a proton (hydrogen ion) from glucose. The curved arrow shows that the bonding electron pair in the 0—H bond becomes an unshared pair on carbon, leaving the proton with no electrons. In the second step, the curved arrow shows that an unshared elec- tron pair on oxygen becomes a bonding pair between oxygen'and a proton. Notice that the arrows depict the movement of electrons, not nuclei — for ex- ample, no arrow follows the lost proton. In writing reaction mechanisms, curved arrows are used only to depict electron movement. Now focus on the various functional groups of the intermediates in Fig- ure 1. Glucose, with its carbonyl ( >C=O) group at C-1, is an aldehyde. It also possesses hydroxyl (—-OI-I) groups, the functional groups found in alcohols. The product of Reaction 2 in Figure 1, fructose 6-phosphate, or F61”, is a ke- tone, because the carbonyl carbon is attached to two carbons Like glucose, it is also an alcohol. Joined to 0-6 of fructose is a phosphate (-—-OPO329) group. F6? is thus also an ester, which is the product of reaction between an alcohol and an acid. You are most familiar with esters of carboxylic acids, which con- tain carboxyl groups (#COOH). F6P is the product of reaction between an alcohol (the C—6 hydroxyl of fructose) and phosphoric acid. Figure 321 com- pares a carboxylate ester with a phosphate ester. Notice the similarity in bonding in the ester linkages (in boxes). if if ' CH CH ex CH20H3 - e P--.5 2 3 H30 0 . 0 co Ethyl Acetate a. Ethyl Phosphate Phosphoric Anhydride b. (Pyrophosphate) Acetyl Phosphate, a Mixed Anhydn'de Figure 3. Analogous Carbon and Phosphorus Compounds. a. Carhoxyiate and Phosphate Esters b. Carboxylate and Phosphate Anhydrides c. A Mixed Anhydride The product of Reaction 6 (Figure 1), BPG, is an anhydride. As with esters, you are most familiar with anhydrides of carboxylic acids. 1,3-Bispho- sphoglyceric acid (BPG) is a mixed anhydride of phosphoric acid and a car- boxylic acid, 3-phosphoglyceric acrd. Figures 3b and 3c compare carboxylic, mixed carboxylic-phosphoric, and phosphoric anhydrides (called phosphoan- Hug a7 03 01:55p [0.7 Prelude to Biochemistry: A Review of Important Concepts from Organic Chemistry 7 hydrides) Notice the similarity in bonding in the anhydride linkages (in boxes). Hydrolysis (literally, cleavage by water) of anhydrides gives acids. Reaction ’7 produces a carboxylic acid (3P6) by transfer of the phos- phate from BPG to adenosine 5 '-diphosphate (ADP). The carboxyl group of 3PG is shown in Figure 1 in its unprotonated, or carhoxylate, form, because this form would predominate at pH values common to the cellular and intra- cellular fluids of living organisms (pH 7 to‘ 8.5). The pK,I values for simple carboxylic acids are well below 7. When the pH of the solution is numerically the same as the pK, of an acid, the acid is 50% ionized; that is, the protonated and unprotonated forms are present in equal quantities. At higher pH (lower concentration of hydrogen ions), the unprotonated form predominates by a factor of 10 for each unit of difference between the pH and the pKa. For ex ample, if the pK,, of an acid is 4.0, then at pH 7.0, the unprotonated groups (carboxylates) outnumber the protonated groups (carboxyls) by 1000 to 1. At pH 4.0, carboxylates and carboxyls are present in equal quantities. The carboxylate group provides another example of resonance stabi- lization, as shown for acetate ion in Figure 4. Curved arrows on the first con- tributor show that you can use electron pushing to help you write resonance .Ie :?: 1?: H 009 c ., H30’ ‘50:. ch/ i:53 Figure 4. Resonance Stabilization of Acetate Ion contributors See if you can use curved arrows to generate the second con- tributor for the enolate ion in Figure 2. The two resonance forms of the carboxylate ion are equivalent; that is, we judge them to be of equal energy, because bonding and charge distribution are identical in the two forms. For this reason, we expect resonance stabiliza- tion to be greater in the carboxylate ion than in the enolate ion, in which the resonance forms are not equivalent (one carries negative charge on carbon, the other on oxygen). This greater resonance stabilization of the carboxylate partially explains the much greater acidity of carboxyl groups (with typical pKa values between '2 and 6) in comparison to carbon acids like ketones or aldehydes having a—hydrogens (with typical pKa values greater than 20). Glucose provides one more opportunity to review a class of organic compounds that is important in carbohydrate chemistry. Glucose contains the aldehyde carbonyl as well as the hydroxyl groups found in alcohols. A1- cohols react with aldehydes (and ketones) to form, first, herniacetals, R- CH(OH)(OR’), and then acetals, R—CH(OR’)2, (with ketones the pr...
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