Chapter 24. The Chemistry of Life: Organic and Biological Chemistry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Transcription

1 Lecture Presentation Chapter 24 The of Life: James F. Kirby Quinnipiac University Hamden, CT

2 Organic and Biochemistry Chapter focus: the molecules that bridge chemistry & biology Most common elements: C, H, O, N Organic chemistry: study of compounds containing carbon Biochemistry: the study of chemistry of living systems

3 24.1 General Characteristics of Organic Molecules 1080 General Characteristics of Organic Molecules Carbon makes four bonds. All single bonds: tetrahedral; sp 3 hybridized One double bond: trigonal planar; sp 2 hybridized One triple bond: linear; sp hybridized C H are most common. C forms stable (strong) bonds with many elements, including C, H, O, N, and the halogens. Groups of atoms that determine how an organic molecule reacts are called functional groups.

4 Solubility Most prevalent bonds are C C and C H, which are nonpolar; solubility in water is low for many organic compounds. Organic molecules, such as glucose, that have polar groups are soluble in polar solvents. Molecules with long nonpolar regions and polar regions act as surfactants, bringing polar material into aqueous solution (used in detergents and soaps).

5 Acid Base Properties Many organic molecules contain acidic or basic functional groups. Carboxylic acids ( COOH) are the most common acids. Amines ( NH 2, NHR, or NR 2, where R is an organic group made up of C and H atoms) are the most common bases.

6 24.2 Introduction to Hydrocarbons 1082 Hydrocarbons Hydrocarbons consist of ONLY carbon and hydrogen. They are grouped based on the number of bonds between carbon atoms. There are four basic types of hydrocarbons: Alkanes Alkenes Alkynes Aromatic hydrocarbons

7 Properties Common to Hydrocarbons Since they are nonpolar, they are insoluble in water but soluble in nonpolar solvents. Melting points and boiling points are determined by dispersion forces (low molar mass hydrocarbons are gases; moderate molar mass hydrocarbons are liquids; high molar mass hydrocarbons are solids).

8 Uses of Some Simple Alkanes Methane (CH 4 ): in natural gas (heating fuel) Propane (C 3 H 8 ): in bottled gas (heating and cooking fuel) Butane (C 4 H 10 ): in disposable lighters and fuel canisters for camping Alkanes with 5 to 12 C atoms: gasoline

9 Methods for Writing Formulas There are a few common methods for writing the structures in organic chemistry. Structural formulas show how atoms are bonded to each other. Condensed structural formulas don t show all C H; they condense them to groupings, like CH 3.

10 Structure of Alkanes Carbons in alkanes are sp 3 hybridized, tetrahedral, and have bond angles. In the straight chain form, all carbons connect in a continuous chain. C can make four bonds, so it is possible for a carbon atom to bond to three or four C atoms, making a branched alkane. Compounds with the same molecular formula but different connections of atoms are called structural isomers, as seen on the next slide.

11 Structural Isomers

12 Systematic Nomenclature of Organic Compounds There are three parts to a compound name: Base: This tells how many carbons are in the longest continuous chain. Suffix: This tells what type of compound it is. Prefix: This tells what groups are attached to the chain.

13 How to Name a Compound 1. Find the longest continuous chain of C atoms, and use this as the base name. 2. Number the chain from the end nearest the first substituent encountered. 3. Name each substituent. (Side chains that are based on alkanes are called alkyl groups.)

14 How to Name a Compound 4. Begin the name with the number(s) on the C atom(s) to which each substituent is bonded. 5. When two or more substituents are present, list them alphabetically.

15 Cycloalkanes Alkanes that form rings or cycles Possible with at least three C atoms, but sp 3 hybridization requires angles not a very stable molecule. Four-C ring is also not very stable. Five-C and more have room for proper bond angle. Naming: add cyclo- as a prefix to alkane name.

16 Reactions of Alkanes Alkanes are relatively unreactive due to the lack of polarity and presence of only C C and C H σ bonds, which are very stable. They do not react with acids, bases, or oxidizing agents. However, the most important reaction observed is combustion: 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(l) (exothermic! ΔH = 2885 kj)

17 Sample Exercise 24.1 Naming Alkanes Give the systematic name for the following alkane: Solution Analyze We are given the condensed structural formula of an alkane and asked to give its name. Plan Because the hydrocarbon is an alkane, its name ends in -ane. The name of the parent hydrocarbon is based on the longest continuous chain of carbon atoms. Branches are alkyl groups, named after the number of C atoms in the branch and located by counting C atoms along the longest continuous chain. Solve The longest continuous chain of C atoms extends from the upper left CH 3 group to the lower left CH 3 group and is seven C atoms long:

18 Sample Exercise 24.1 Naming Alkanes Continued The parent compound is thus heptane. There are two methyl groups branching off the main chain. Hence, this compound is a dimethylheptane. To specify the location of the two methyl groups, we must number the C atoms from the end that gives the lower two numbers to the carbons bearing side chains. This means that we should start numbering at the upper left carbon. There is a methyl group on C3 and one on C4. The compound is thus 3,4-dimethylheptane. Practice Exercise 1 What is the proper name of this compound? (a) 3-ethyl-3-methylbutane, (b) 2-ethyl-2-methylbutane, (c) 3,3-dimethylpentane, (d) isoheptane, (e) 1,2-dimethyl-neopentane.

19 Sample Exercise 24.1 Naming Alkanes Continued Practice Exercise 2 Name the following alkane:

20 Sample Exercise 24.2 Writing Condensed Structural Formulas Write the condensed structural formula for 3-ethyl-2-methylpentane. Solution Analyze We are given the systematic name for a hydrocarbon and asked to write its condensed structural formula. Plan Because the name ends in -ane, the compound is an alkane, meaning that all the carbon carbon bonds are single bonds. The parent hydrocarbon is pentane, indicating five C atoms (Table 24.2). There are two alkyl groups specified, an ethyl group (two carbon atoms, C 2 H 5 ) and a methyl group (one carbon atom, CH 3 ). Counting from left to right along the five-carbon chain, the name tells us that the ethyl group is attached to C3 and the methyl group is attached to C2.

21 Sample Exercise 24.2 Writing Condensed Structural Formulas Continued Solve We begin by writing five C atoms attached by single bonds. These represent the backbone of the parent pentane chain: C C C C C We next place a methyl group on the second C and an ethyl group on the third C of the chain. We then add hydrogens to all the other C atoms to make four bonds to each carbon: The formula can be written more concisely as CH 3 CH(CH 3 )CH(C 2 H 5 )CH 2 CH 3 where the branching alkyl groups are indicated in parentheses.

22 Sample Exercise 24.2 Writing Condensed Structural Formulas Continued Practice Exercise 1 How many hydrogen atoms are in 2,2-dimethylhexane? (a) 6, (b) 8, (c) 16, (d) 18, (e) 20. Practice Exercise 2 Write the condensed structural formula for 2,3-dimethylhexane.

23 24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons 1088 Saturated vs. Unsaturated Hydrocarbons with single bonds only are called saturated hydrocarbons. These are the alkanes. Alkenes, alkynes, and aromatic hydrocarbons have fewer hydrogen atoms than alkanes with the same number of carbon atoms. They are called unsaturated hydrocarbons. Unsaturated hydrocarbons are more reactive than saturated hydrocarbons.

24 Alkenes Contain at least one C C bond No free rotation about the double bond Naming: longest chain must include BOTH carbon atoms that share the double bond; end name in -ene; lowest number possible given to double-bond carbon atoms; isomers also indicated.

25 Geometric Isomers Since there is no free rotation around the double bond, the direction of the longest chain can differ for four or more C atoms. Compounds that have all atoms connected to the same atoms but differ in three-dimensional arrangement are geometric isomers. Alkenes have cis (same side of the double bond) or trans (opposite side of the double bond) isomers.

26 Alkynes Contain at least one C C Unsaturated Naming: longest continuous chain containing both carbon atoms in the triple bond; name ends in -yne (instead of -ane or -ene); give C atoms in triple bond lowest number.

27 Addition Reactions of Alkenes and Alkynes Add atoms to the double or triple bond, making it saturated or more saturated π bonds are broken and electrons form σ bonds to added atoms. Work with H 2 (hydrogenation), HX (hydrogen halides or water), or X 2 (halogenation)

28 Aromatic Hydrocarbons Aromatic hydrocarbons have six-membered rings containing localized and delocalized electrons. The π ring is much more stable than a π bond. So, aromatic hydrocarbons are much less reactive than alkenes and alkynes. They undergo substitution reactions rather than addition reactions: groups replace H on a ring (e.g., nitration, halogenation, alkylation).

29 Aromatic Nomenclature Many aromatic hydrocarbons are known by their common names. Others are named as derivatives of benzene. Substitution positions for two substituents: 1,2 = ortho-; 1,3 = meta-; 1,4 = para-

30 Sample Exercise 24.3 Drawing Isomers Draw all the structural and geometric isomers of pentene, C 5 H 10, that have an unbranched hydrocarbon chain. Solution Analyze We are asked to draw all the isomers (both structural and geometric) for an alkene with a five-carbon chain. Plan Because the compound is named pentene and not pentadiene or pentatriene, we know that the five-carbon chain contains only one carbon carbon double bond. Thus, we begin by placing the double bond in various locations along the chain, remembering that the chain can be numbered from either end. After finding the different unique locations for the double bond, we consider whether the molecule can have cis and trans isomers. Solve There can be a double bond after either the first carbon (1-pentene) or second carbon (2-pentene). These are the only two possibilities because the chain can be numbered from either end. Thus, what we might erroneously call 3-pentene is actually 2-pentene, as seen by numbering the carbon chain from the other end:

31 Sample Exercise 24.3 Drawing Isomers Continued Because the first C atom in 1-pentene is bonded to two H atoms, there are no cis trans isomers. There are cis and trans isomers for 2-pentene, however. Thus, the three isomers for pentene are (You should convince yourself that cis-3-pentene is identical to cis-2-pentene and trans-3-pentene is identical to trans-2-pentene. However, cis-2-pentene and trans-2-pentene are the correct names because they have smaller numbered prefixes.)

32 Sample Exercise 24.3 Drawing Isomers Continued Practice Exercise 1 Which compound does not exist? (a) 1,2,3,4,5,6,7-octaheptaene, (b) cis-2-butane, (c) trans-3-hexene, (d) 1-propene, (e) cis-4-decene. Practice Exercise 2 How many straight-chain isomers are there of hexene, C 6 H 12?

33 Sample Exercise 24.4 Naming Unsaturated Hydrocarbons Name the following compounds: Solution Analyze We are given the condensed structural formulas for an alkene and an alkyne and asked to name the compounds. Plan In each case, the name is based on the number of carbon atoms in the longest continuous carbon chain that contains the multiple bond. In the alkene, care must be taken to indicate whether cis trans isomerism is possible and, if so, which isomer is given. Solve (a) The longest continuous chain of carbons that contains the double bond is seven carbons long, so the parent hydrocarbon is heptene. Because the double bond begins at carbon 2 (numbering from the end closer to the double bond), we have 2-heptene. With a methyl group at carbon atom 4, we have 4-methyl-2-heptene. The geometrical configuration at the double bond is cis (that is, the alkyl groups are bonded to the double bond on the same side). Thus, the full name is 4-methyl-cis-2-heptene.

34 Sample Exercise 24.4 Naming Unsaturated Hydrocarbons Continued (b) The longest continuous chain containing the triple bond has six carbons, so this compound is a derivative of hexyne. The triple bond comes after the first carbon (numbering from the right), making it 1-hexyne. The branch from the hexyne chain contains three carbon atoms, making it a propyl group. Because this substituent is located on C3 of the hexyne chain, the molecule is 3-propyl-1-hexyne. Practice Exercise 1 If a compound has two carbon carbon triple bonds and one carbon carbon double bond, what class of compound is it? (a) an eneyne, (b) a dieneyne, (c) a trieneyne, (d) an enediyne, (e) an enetriyne. Practice Exercise 2 Draw the condensed structural formula for 4-methyl-2-pentyne.

35 Sample Exercise 24.5 Predicting the Product of an Addition Reaction Write the condensed structural formula for the product of the hydrogenation of 3-methyl-1-pentene. Solution Analyze We are asked to predict the compound formed when a particular alkene undergoes hydrogenation (reaction with H 2 ) and to write the condensed structural formula of the product. Plan To determine the condensed structural formula of the product, we must first write the condensed structural formula or Lewis structure of the reactant. In the hydrogenation of the alkene, H 2 adds to the double bond, producing an alkane. Solve The name of the starting compound tells us that we have a chain of five C atoms with a double bond at one end (position 1) and a methyl group on C3: Hydrogenation the addition of two H atoms to the carbons of the double bond leads to the following alkane: Comment The longest chain in this alkane has five carbon atoms; the product is therefore 3-methylpentane.

36 Sample Exercise 24.5 Predicting the Product of an Addition Reaction Continued Practice Exercise 1 What product is formed from the hydrogenation of 2-methylpropene? (a) propane, (b) butane, (c) 2-methylbutane, (d) 2-methylpropane, (e) 2-methylpropyne. Practice Exercise 2 Addition of HCl to an alkene forms 2-chloropropane. What is the alkene?

37 24.4 Organic Functional Groups 1096 Functional Groups The chemistry of an organic molecule is largely determined by the functional groups it contains. R represents the alkyl portion (C,H) of an organic molecule.

38 Alcohols Alcohols contain one or more OH group (the alcohol group or the hydroxyl group). They are named from the parent hydrocarbon; the suffix is changed to -ol and a number designates the carbon to which the OH group is attached.

39 Properties and Uses of Alcohols Polar molecules: lead to water solubility and higher boiling points Methanol: used as a fuel additive Ethanol: in alcoholic beverages Ethylene glycol: in antifreeze Glycerol: cosmetic skin softener and food moisturizer Phenol: making plastics and dyes; topical anesthetic in throat sprays Cholesterol: important biomolecule in membranes, but can precipitate and form gallstones or block blood vessels

40 Ethers R O R Formed by dehydration between alcohol molecules Not very reactive (except combustion) Used as solvents for organic reactions.

41 Functional Groups Containing the Carbonyl Group The carbonyl group is C O. Functional groups containing C O: Aldehydes Ketones Carboxylic acids Esters Amides

42 Aldehydes and Ketones Aldehydes have at least one hydrogen atom attached to the carbonyl carbon atom. Ketones have two R groups attached to the carbonyl carbon atom.

43 Important Aldehydes and Ketones Many aldehydes are natural flavorings: vanilla, cinnamon, spearmint, and caraway are from aldehydes. Ketones are used extensively as solvents; the most important solvent other than water is acetone, which dissolves in water and dissolves many organic compounds.

44 Carboxylic Acids Structure: hydroxyl group bonded to the carbonyl group H on the hydroxyl group is weakly acidic. Important in manufacturing polymers for films, fibers, and paints Oxidation product of alcohols (some make aldehydes)

45 Esters Esters are the products of reactions between carboxylic acids and alcohols. They are found in many fruits and perfumes. Naming: name the alcohol part as an alkyl name; separate word is the acid part as an -ate anion.

46 Decomposition of Esters Heating an ester in the presence of an acid catalyst and water can decompose the ester. (This is the reverse reaction of the preparation of an ester; it is an equilibrium.) Heating an ester in the presence of a base results in saponification (making soap).

47 Nitrogen Containing Organic Compounds Amines are organic derivatives of ammonia (NH 3 ). One, two, or all three H atoms can be replaced by R groups (the same or different R groups). If H in NH 3 or an amine is replaced by a carbonyl group (N directly attached to C O), an amide is formed.

48 Sample Exercise 24.6 Naming Esters and Predicting Hydrolysis Products In a basic aqueous solution, esters react with hydroxide ion to form the salt of the carboxylic acid and the alcohol from which the ester is constituted. Name each of the following esters, and indicate the products of their reaction with aqueous base. Solution Analyze We are given two esters and asked to name them and to predict the products formed when they undergo hydrolysis (split into an alcohol and carboxylate ion) in basic solution. Plan Esters are formed by the condensation reaction between an alcohol and a carboxylic acid. To name an ester, we must analyze its structure and determine the identities of the alcohol and acid from which it is formed. We can identify the alcohol by adding an OH to the alkyl group attached to the O atom of the carboxyl (COO) group. We can identify the acid by adding an H to the O atom of the carboxyl group. We have learned that the first part of an ester name indicates the alcohol portion and the second indicates the acid portion. The name conforms to how the ester undergoes hydrolysis in base, reacting with base to form an alcohol and a carboxylate anion.

49 Sample Exercise 24.6 Naming Esters and Predicting Hydrolysis Products Continued Solve (a) This ester is derived from ethanol (CH 3 CH 2 OH) and benzoic acid (C 6 H 5 COOH). Its name is therefore ethyl benzoate. The net ionic equation for reaction of ethyl benzoate with hydroxide ion is Using the rate law and the data from experiment 1, we have The products are benzoate ion and ethanol.

50 Sample Exercise 24.6 Naming Esters and Predicting Hydrolysis Products Continued (b) This ester is derived from phenol (C 6 H 5 OH) and butanoic acid (commonly called butyric acid) (CH 3 CH 2 CH 2 COOH). The residue from the phenol is called the phenyl group. The ester is therefore named phenyl butyrate or phenyl butanoate. The net ionic equation for the reaction of phenyl butyrate with hydroxide ion is The products are butyrate ion and phenol.

51 Sample Exercise 24.6 Naming Esters and Predicting Hydrolysis Products Continued Practice Exercise 1 For the generic ester RC(O)OR, which bond will hydrolyze under basic conditions? (a) the R (b) the C O bond (c) the C O bond (d) the O R bond (e) more than one of the above C bond Practice Exercise 2 Write the condensed structural formula for the ester formed from propyl alcohol and propionic acid.

52 24.5 Chirality in Organic 1105 Chirality Carbons with four different groups attached to them are chiral. These are optical isomers, or enantiomers. Enantiomers have the same physical and chemical properties when they react with nonchiral reagents. Enantiomers rotate plane-polarized light in opposite directions.

53 Chirality and Pharmaceuticals Many drugs are chiral compounds. Equal mixtures of enantiomers is called a racemic mixture. Often only one enantiomer is clinically active; the other can be inert OR harmful.

54 24.6 Introduction to Biochemistry 1106 Biomolecules Biopolymers (large biological molecules built from small molecules) Proteins Polysaccharides (carbohydrates) Nucleic acids Lipids are large biomolecules, but they are NOT polymers.

55 24.7 Proteins 1106 Amino Acids and Proteins Amino acids have amine and carboxylic acid functional groups. Proteins are polymers of α-amino acids. A condensation reaction between the amine end of one amino acid and the acid end of another produces a peptide bond, which is an amide linkage.

56 The Natural α-amino Acids

57 Protein Structure Primary structure: the sequence of amino acids in the polypeptide/protein chain Secondary structure: interactions between the chain atoms (C O and N H atoms) that give structure to proteins Tertiary structure: intermolecular forces between side-chain atoms that give structure to proteins Quaternary structure: arrangement of multiple units and/or incorporation of non amino acid portions of proteins

58 Protein Structure

59 Primary Structure Formation of the amide bond between amino acids Repeats MANY times for a polypeptide/protein

60 Secondary Structure Two common types: α-helix: looks like a corkscrew or spiral staircase; the C O forms H bonds with the N H from another amino acid in the chain. β-sheets: two or more pleated regions (looking like the shape of a pleated skirt or corrugated cardboard) are held together by the same H bonds as in an α-helix; one major difference is how far apart from each other the amino acids are in the amino acid sequence (α-helix atoms are much closer together).

61 Tertiary Structure Side chains interact with each other and the surrounding environment (usually aqueous environment). The forces we called intermolecular are mostly what drives tertiary structure. The side chains have polar, nonpolar, and charged groups; these give rise to ion ion, ion dipole, dipole dipole, and dispersion force interactions.

62 Quaternary Structure Some proteins are made up of more than one polypeptide chain. Some proteins contain portions that are NOT amino acid in nature. The combination of subunits is quaternary structure.

63 Classifying Proteins One method to classify proteins is based on solubility: Globular proteins are roughly spherical and dissolve in aqueous environments. Fibrous proteins are usually long fibers that are insoluble in water and are used as structural materials.

64 Sample Exercise 24.7 Drawing the Structural Formula of a Tripeptide Draw the structural formula for alanylglycylserine. Solution Analyze We are given the name of a substance with peptide bonds and asked to write its structural formula. Plan The name of this substance suggests that three amino acids alanine, glycine, and serine have been linked together, forming a tripeptide. Note that the ending -yl has been added to each amino acid except for the last one, serine. By convention, the sequence of amino acids in peptides and proteins is written from the nitrogen end to the carbon end: The first-named amino acid (alanine, in this case) has a free amino group and the last-named one (serine) has a free carboxyl group. Solve We first combine the carboxyl group of alanine with the amino group of glycine to form a peptide bond and then the carboxyl group of glycine with the amino group of serine to form another peptide bond: We can abbreviate this tripeptide as either Ala-Gly-Ser or AGS.

65 Sample Exercise 24.7 Drawing the Structural Formula of a Tripeptide Continued Practice Exercise 1 How many nitrogen atoms are in the tripeptide Arg-Asp-Gly? (a) 3, (b) 4, (c) 5, (d) 6, (e) 7. Practice Exercise 2 Name the dipeptide and give the two ways of writing its abbreviation.

66 24.8 Carbohydrates 1111 Carbohydrates The name comes from an empirical formula for sugars: C x (H 2 O) y for the simplest sugars, x = y. Simple sugars (monosacchharides) are polyhydroxy aldehydes or ketones. They are often drawn as chains but most frequently exist as rings in solution.

67 Monosaccharides The two most common monosaccharides are glucose and fructose. Glucose is an aldehyde; fructose is a ketone.

68 Disaccharides Dehydration between two monosaccharides forms a disaccharide. Sucrose and lactose are both disaccharides. Disaccharides are often referred to as sugars.

69 Polysaccharides Dehydration can create long-chain polysaccharides. The three most common are starch, glycogen, and cellulose. Starch consists of many different-sized and various branching chains of glucose prepared by plants to store energy. Glycogen is often called animal starch it is for temporary energy storage in animals (which use fats for long-term energy storage). Cellulose is a structural polysaccharide in plants, making up cell walls; it is unbranched.

70 Polysaccharides

71 Sample Exercise 24.8 Identifying Functional Groups and Chiral Centers in Carbohydrate How many chiral carbon atoms are there in the open-chain form of glucose (Figure 24.20)? Solution Analyze We are given the structure of glucose and asked to determine the number of chiral carbons in the molecule. Plan A chiral carbon has four different groups attached (Section 24.5). We need to identify those carbon atoms in glucose.

72 Sample Exercise 24.8 Identifying Functional Groups and Chiral Centers in Carbohydrate Continued Solve Carbons 2, 3, 4, and 5 each have four different groups attached to them: Thus, there are four chiral carbon atoms in the glucose molecule.

73 Sample Exercise 24.8 Identifying Functional Groups and Chiral Centers in Carbohydrate Continued Practice Exercise 1 How many chiral carbon atoms are there in the open-chain form of fructose (Figure 24.20)? (a) 0, (b) 1, (c) 2, (d) 3, (e) 4 Practice Exercise 2 Name the functional groups present in the beta form of glucose.

74 24.9 Lipids 1114 Lipids Lipids are a broad class of nonpolar bioorganic molecules. They are grouped because they are insoluble in water. They are used biologically to store energy (fats, oils) and for biological structure (phospholipids in cell membranes).

75 Fats and Oils Fats and oils are made from long-chain carboxylic acids and glycerol. Fats have only saturated carboxylic acids. They are solids. These are often called the bad fats in your diet. Oils have at least one unsaturated carboxylic acid. They are liquids. The more unsaturated, the better for you (polyunsaturated versus monounsaturated fats). Essential fatty acids (with double bonds) must be included in our diet. Our bodies can t produce them. They are often called omega-3 and omega-6 fatty acids.

76 Phospholipids How their structure is similar to fats: glycerol with ester linkage to two fatty acids (instead of three) How it differs: the third site has a phosphate ester linkage connected to a charged or polar group, such as choline. They cluster together in water: polar regions pointing toward water; nonpolar fatty acid regions pointing toward each other forming a lipid bilayer the start of a cell membrane.

77 Comparing Phospholipids to Fats

78 24.10 Nucleic Acids 1115 Nucleic Acids Class of biopolymers that are chemical carriers of genetic information Two types: Deoxyribonucleic acid (DNA) Ribonucleic acid (RNA)

79 Comparing Nucleic Acid Types DNA Huge molecules (6 to 16 million amu) Primarily in nucleus of the cell Stores genetic information Specifies which proteins the cell can synthesize RNA Smaller molecules (20,000 40,000 amu) Mostly outside of nucleus (in cytoplasm) Carries stored info from DNA into cytoplasm Information is used in protein synthesis outside of nucleus

80 Structure of Nucleic Acids Nucleic acids consist of a five-c sugar (ribose or deoxyribose); a phosphate group; a N-containing base (adenine, guanine, cytosine, and thymine or uracil). Polynucleotides form by condensation reactions between a phosphate and a sugar OH.

81 The Double Helix Nucleotides combine to form the familiar double-helix form of the nucleic acids. H-bonding, dipole dipole interactions, and dispersion forces hold the double helix together. Complementary base pairs form ideal H-bonding partners: A T; C G. (Note: number of H-bonds, NOT a double/triple bond!)

82 Sample Integrative Exercise Putting Concepts Together Pyruvic acid, is formed in the body from carbohydrate metabolism. In muscles, it is reduced to lactic acid inthe course of exertion. The acid-dissociation constant for pyruvic acid is (a) Why does pyruvic acid have a larger acid-dissociation constant than acetic acid? (b) Would you expect pyruvic acid to exist primarily as the neutral acid or as dissociated ions in muscle tissue, assuming a ph of 7.4 and an initial acid concentration of M? (c) What would you predict for the solubility properties of pyruvic acid? Explain. (d) What is the hybridization of each carbon atom in pyruvic acid? (e) Assuming H atoms as the reducing agent, write a balanced chemical equation for the reduction of pyruvic acid to lactic acid (Figure 24.13). (Although H atoms do not exist as such in biochemical systems, biochemical reducing agents deliver hydrogen for such reductions.)

83 Sample Integrative Exercise Putting Concepts Together Continued Solution (a) The acid-dissociation constant for pyruvic acid should be somewhat greater than that of acetic acid because the carbonyl function on the α-carbon atom of pyruvic acid exerts an electron-withdrawing effect on the carboxylic acid group. In the C O H bond system, the electrons are shifted from H, facilitating loss of the H as a proton. (Section 16.10) (b) To determine the extent of ionization, we first set up the ionization equilibrium and equilibrium-constant expression. Using HPv as the symbol for the acid, we have Let [Pv ] = x. Then the concentration of undissociated acid is x. The concentration of [H + ] is fixed at (the antilog of the ph value). Substituting, we obtain Solving for x, we obtain x = [Pv ] = M

84 Sample Integrative Exercise Putting Concepts Together Continued This is the initial concentration of acid, which means that essentially all the acid has dissociated. We might have expected this result because the acid is quite dilute and the acid-dissociation constant is fairly high. (c) Pyruvic acid should be quite soluble in water because it has polar functional groups and a small hydrocarbon component. We would predict it would be soluble in polar organic solvents, especially ones that contain oxygen. In fact, pyruvic acid dissolves in water, ethanol, and diethyl ether. (d) The methyl group carbon has sp 3 hybridization. The carbon of the carbonyl group has sp 2 hybridization because of the double bond to oxygen. Similarly, the carboxylic acid carbon is sp 2 hybridized. (e) The balanced chemical equation for this reaction is Essentially, the ketonic functional group has been reduced to an alcohol.