1. Organic and Biological Chemistry Chapter 25 Organic and Biological Chemistry Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

2. Organic and Biological Chemistry Organic Chemistry The chemistry of carbon compounds. Carbon has the ability to form long chains. Without this property, large biomolecules such as proteins, lipids, carbohydrates, and nucleic acids could not form.

3. Organic and Biological Chemistry Structure of Carbon Compounds There are three hybridization states and geometries found in organic compounds:  sp 3 Tetrahedral  sp 2 Trigonal planar  sp Linear

4. Organic and Biological Chemistry Hydrocarbons Four basic types:  Alkanes  Alkenes  Alkynes  Aromatic hydrocarbons

5. Organic and Biological Chemistry Alkanes Only single bonds. Saturated hydrocarbons.  “Saturated” with hydrogens.

6. Organic and Biological Chemistry Formulas Lewis structures of alkanes look like this. Also called structural formulas. Often not convenient, though…

7. Organic and Biological Chemistry Formulas …so more often condensed formulas are used.

8. Organic and Biological Chemistry Properties of Alkanes Only van der Waals force: London force. Boiling point increases with length of chain.

9. Organic and Biological Chemistry Structure of Alkanes Carbons in alkanes sp 3 hybrids. Tetrahedral geometry. 109.5° bond angles.

10. Organic and Biological Chemistry Structure of Alkanes Only  -bonds in alkanes Free rotation about C—C bonds.

11. Organic and Biological Chemistry Isomers Have same molecular formulas, but atoms are bonded in different order.

12. Organic and Biological Chemistry Organic Nomenclature Three parts to a compound name:  Base: Tells how many carbons are in the longest continuous chain.

13. Organic and Biological Chemistry Organic Nomenclature Three parts to a compound name:  Base: Tells how many carbons are in the longest continuous chain.  Suffix: Tells what type of compound it is.

14. Organic and Biological Chemistry Organic Nomenclature Three parts to a compound name:  Base: Tells how many carbons are in the longest continuous chain.  Suffix: Tells what type of compound it is.  Prefix: Tells what groups are attached to chain.

15. Organic and Biological Chemistry To Name a Compound… 1.Find the longest chain in the molecule. 2.Number the chain from the end nearest the first substituent encountered. 3.List the substituents as a prefix along with the number(s) of the carbon(s) to which they are attached.

16. Organic and Biological Chemistry To Name a Compound… If there is more than one type of substituent in the molecule, list them alphabetically.

17. Organic and Biological Chemistry SAMPLE EXERCISE 25.1 Naming Alkanes The parent compound is thus heptane. There are two CH 3 (methyl) groups that branch 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 lowest number possible to the carbons bearing side chains. This means that we should start numbering with the upper left carbon. There is a methyl group on carbon 3, and one on carbon 4. The compound is thus 3,4-dimethylheptane. Solution Analyze: We are given the 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, as summarized in Table 25.1. 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. Give the systematic name for the following alkane: 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. Organic and Biological Chemistry SAMPLE EXERCISE 25.1 continued Answer: 2,4-dimethylpentane PRACTICE EXERCISE Name the following alkane:

19. Organic and Biological Chemistry SAMPLE EXERCISE 25.2 Writing Condensed Structural Formulas Write the condensed structural formula for 3-ethyl-2-methylpentane. Solve: We begin by writing a string of five C atoms attached to each other by single bonds. These represent the backbone of the parent pentane chain: C––C––C––C––C Solution Analyze: We are given the systematic name for a hydrocarbon and asked to write its structural formula. Plan: Because the compound’s name ends in -ane, it is an alkane, meaning that all the carbon–carbon bonds are single bonds. The parent hydrocarbon is pentane, indicating five C atoms (Table 25.1). 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 ethyl group will be attached to the third C atom and the methyl group will be attached to the second C atom. We next place a methyl group on the second C and an ethyl group on the third C atom of the chain. Hydrogens are then added to all the other C atoms to make the four bonds to each carbon, giving the following condensed structure:

20. Organic and Biological Chemistry SAMPLE EXERCISE 25.2 continued The formula can be written even more concisely in the following style: CH 3 CH(CH 3 ) CH(C 2 H 5 ) CH 2 CH 3 In this formula the branching alkyl groups are indicated in parentheses. PRACTICE EXERCISE Write the condensed structural formula for 2,3-dimethylhexane. Answer:

21. Organic and Biological Chemistry Cycloalkanes Carbon can also form ringed structures. Five- and six-membered rings are most stable.  Can take on conformation in which angles are very close to tetrahedral angle.  Smaller rings are quite strained.

22. Organic and Biological Chemistry Reactions of Alkanes Rather unreactive due to presence of only C—C and C—H  -bonds. Therefore, great nonpolar solvents.

23. Organic and Biological Chemistry Alkenes Contain at least one carbon–carbon double bond. Unsaturated.  Have fewer than maximum number of hydrogens.

24. Organic and Biological Chemistry Structure of Alkenes Unlike alkanes, alkenes cannot rotate freely about the double bond.  Side-to-side overlap makes this impossible without breaking  -bond.

25. Organic and Biological Chemistry Structure of Alkenes This creates geometric isomers, which differ from each other in the spatial arrangement of groups about the double bond.

26. Organic and Biological Chemistry Properties of Alkenes Structure also affects physical properties of alkenes.

27. Organic and Biological Chemistry Nomenclature of Alkenes Chain numbered so double bond gets smallest possible number. cis- alkenes have carbons in chain on same side of molecule. trans- alkenes have carbons in chain on opposite side of molecule.

28. Organic and Biological Chemistry Reactions of Alkenes Addition Reactions  Two atoms (e.g., bromine) add across the double bond.  One  -bond and one  -bond are replaced by two  -bonds; therefore,  H is negative.

29. Organic and Biological Chemistry Mechanism of Addition Reactions Two-step mechanism:  First step is slow, rate-determining step.  Second step is fast.

30. Organic and Biological Chemistry Mechanism of Addition Reactions In first step,  -bond breaks and new C—H bond and cation form.

31. Organic and Biological Chemistry Mechanism of Addition Reactions In second step, new bond forms between negative bromide ion and positive carbon.

32. Organic and Biological Chemistry SAMPLE EXERCISE 25.5 Identifying the Product of a Hydrogenation Reaction Write the 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 ). Plan: To determine the structural formula of the reaction product, we must first write the structural formula or Lewis structure of the reactant. In the hydrogenation of the alkene, H 2 adds to the double bond, producing an alkane. (That is, each carbon atom of the double bond forms a bond to an H atom, and the double bond is converted to a single bond.) 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 the third C from that end (position 3): Comment: The longest chain in the product alkane has five carbon atoms; its name is therefore 3-methylpentane. Hydrogenation—the addition of two H atoms to the carbons of the double bond—leads to the following alkane: PRACTICE EXERCISE Addition of HCl to an alkene forms 2-chloropropane. What is the alkene? Answer: propene

33. Organic and Biological Chemistry Alkynes Contain at least one carbon–carbon triple bond. Carbons in triple bond sp-hybridized and have linear geometry. Also unsaturated.

34. Organic and Biological Chemistry Nomenclature of Alkynes Analogous to naming of alkenes. Suffix is -yne rather than –ene. 4-methyl-2-pentyne

35. Organic and Biological Chemistry Reactions of Alkynes Undergo many of the same reactions alkenes do. As with alkenes, impetus for reaction is replacement of  -bonds with  -bonds.

36. Organic and Biological Chemistry SAMPLE EXERCISE 25.3 Drawing Isomers Draw all the structural and geometric isomers of pentene, C 5 H 10, that have an unbranched hydrocarbon chain. 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 4-pentene is actually 1-pentene, as seen by numbering the carbon chain from the other end.) 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 can begin by first placing the double bond in various locations along the chain, remembering that the chain can be numbered from either end. After finding the different distinct locations for the double bond, we can consider whether the molecule can have cis and trans isomers.

37. Organic and Biological Chemistry SAMPLE EXERCISE 25.3 continued Because the first C atom in 1-pentene is bonded to two H atoms, there are no cis-trans isomers. On the other hand, there are cis and trans isomers for 2-pentene. Thus, the three isomers for pentene are Answer: five (1-hexene, cis-2-hexene, trans-2-hexene, cis-3-hexene, trans-3-hexene) (You should convince yourself that cis- or trans-3-pentene is identical to cis- or trans-2-pentene, respectively.) PRACTICE EXERCISE How many straight-chain isomers are there of hexene, C 6 H 12 ?

38. Organic and Biological Chemistry SAMPLE EXERCISE 25.4 Naming Unsaturated Hydrocarbons Name the following compounds: Solve: (a) The longest continuous chain of carbons that contains the double bond is seven in length. The parent compound is therefore heptene. Because the double bond begins at carbon 2 (numbering from the end closest to the double bond), the parent hydrocarbon chain is named 2-heptene. A methyl group is found at carbon atom 4. Thus, the compound is 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. Solution Analyze: We are given the structural formulas for two compounds, the first an alkene and the second 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 case of the alkene, care must be taken to indicate whether cis-trans isomerism is possible and, if so, which isomer is given. (b) The longest continuous chain of carbon atoms containing the triple bond is six, so this compound is a derivative of hexyne. The triple bond comes after the first carbon (numbering from the right), making it a derivative of 1-hexyne. The branch from the hexyne chain contains three carbon atoms, making it a propyl group. Because it is located on the third carbon atom of the hexyne chain, the molecule is 3-propyl-1-hexyne.

39. Organic and Biological Chemistry SAMPLE EXERCISE 25.4 continued PRACTICE EXERCISE Draw the condensed structural formula for 4-methyl-2-pentyne. Answer:

40. Organic and Biological Chemistry Aromatic Hydrocarbons Cyclic hydrocarbons. p-Orbital on each atom.  Molecule is planar. Odd number of electron pairs in  -system.

41. Organic and Biological Chemistry Aromatic Nomenclature Many aromatic hydrocarbons are known by their common names.

42. Organic and Biological Chemistry Reactions of Aromatic Compounds Unlike in alkenes and alkynes,  - electrons do not sit between two atoms. Electrons are delocalized; this stabilizes aromatic compounds.

43. Organic and Biological Chemistry Reactions of Aromatic Compounds Due to stabilization, aromatic compounds do not undergo addition reactions; they undergo substitution. Hydrogen is replaced by substituent.

44. Organic and Biological Chemistry Structure of Aromatic Compounds Two substituents on a benzene ring could have three possible relationships  ortho-: On adjacent carbons.  meta-: One carbon between them.  para-: On opposite sides of ring.

45. Organic and Biological Chemistry Reactions of Aromatic Compounds Reactions of aromatic compounds often require a catalyst. Halogenation Friedel-Crafts Reaction

46. Organic and Biological Chemistry Functional Groups Term used to refer to parts of organic molecules where reactions tend to occur.

47. Organic and Biological Chemistry Alcohols Contain one or more hydroxyl groups, —OH Named from parent hydrocarbon; suffix changed to -ol and number designates carbon to which hydroxyl is attached.

48. Organic and Biological Chemistry Alcohols Much more acidic than hydrocarbons.  pK a ~15 for most alcohols.  Aromatic alcohols have pK a ~10.

49. Organic and Biological Chemistry Ethers Tend to be quite unreactive. Therefore, they are good polar solvents.

50. Organic and Biological Chemistry Carbonyl Compounds Contain C—O double bond. Include many classes of compounds.

51. Organic and Biological Chemistry Aldehydes At least one hydrogen attached to carbonyl carbon.

52. Organic and Biological Chemistry Ketones Two carbons bonded to carbonyl carbon.

53. Organic and Biological Chemistry Carboxylic Acids Have hydroxyl group bonded to carbonyl group. Tart tasting. Carboxylic acids are weak acids. CH 3 COOH

54. Organic and Biological Chemistry Carboxylic Acids

55. Organic and Biological Chemistry Esters Products of reaction between carboxylic acids and alcohols. Found in many fruits and perfumes.

56. Organic and Biological Chemistry SAMPLE EXERCISE 25.6 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 group to the O atom of the carboxyl group. We’ve 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.

57. Organic and Biological Chemistry SAMPLE EXERCISE 25.6 continued The products are benzoate ion and ethanol. The products are butyrate ion and phenol. (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. The net ionic equation for the reaction of phenyl butyrate with hydroxide ion is 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

58. Organic and Biological Chemistry SAMPLE EXERCISE 25.6 continued PRACTICE EXERCISE Write the structural formula for the ester formed from propyl alcohol and propionic acid. Answer:

59. Organic and Biological Chemistry Amides Formed by reaction of carboxylic acids with amines.

60. Organic and Biological Chemistry Amines Organic bases. Generally have strong, unpleasant odors.

61. Organic and Biological Chemistry Chirality Carbons with four different groups attached to them are handed, or chiral. Optical isomers or stereoisomers If one stereoisomer is “right-handed,” its enantiomer is “left-handed.”

62. Organic and Biological Chemistry Chirality Many pharmaceuticals are chiral. Often only one enantiomer is clinically active. S-ibuprofen

63. Organic and Biological Chemistry SAMPLE EXERCISE 25.8 Identifying Chiral Centers How many chiral carbon atoms are there in the open-chain form of glucose (Figure 25.28)? 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 25.7). We need to identify those carbon atoms in glucose. Carbon atoms 1 and 6 have only three different substituents on them. Thus, there are four chiral carbon atoms in the glucose molecule. Solve: The carbon atoms numbered 2, 3, 4, and 5 each have four different groups attached to them, as indicated here:

64. Organic and Biological Chemistry SAMPLE EXERCISE 25.8 continued PRACTICE EXERCISE How many chiral carbon atoms are there in the open-chain form of fructose (Figure 25.28)? Answer: three

65. Organic and Biological Chemistry Amino Acids and Proteins Proteins are polymers of  -amino acids. A condensation reaction between the amine end of one amino acid and the acid end of another produces a peptide bond.

66. Organic and Biological Chemistry Amino Acids and Proteins Hydrogen bonding in peptide chains causes coils and helices in the chain. Kinking and folding of the coiled chain gives proteins a characteristic shape.

67. Organic and Biological Chemistry Amino Acids and Proteins Most enzymes are proteins. The shape of the active site complements the shape of the substrate on which the enzyme acts  hence, the “lock- and-key” model.

68. Organic and Biological Chemistry SAMPLE EXERCISE 25.7 Drawing the Structural Formula of a Tripeptide Draw the full structural formula for alanylglycylserine. Solution Analyze: We are given the name of a substance with peptide bonds and asked to write its full 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 first-named amino acid (alanine, in this case) has a free amino group and the last-named one (serine) has a free carboxyl group. Thus, we can construct the structural formula of the tripeptide from its amino acid building blocks (Figure 25.23). We can abbreviate this tripeptide as Ala-Gly-Ser. 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 group. The resulting tripeptide consists of three “building blocks” connected by peptide bonds:

69. Organic and Biological Chemistry SAMPLE EXERCISE 25.7 continued Answer: serylaspartic acid; Ser-Asp PRACTICE EXERCISE Name the dipeptide that has the following structure, and give its abbreviation:

70. Organic and Biological Chemistry Carbohydrates Simple sugars are polyhydroxy aldehydes or ketones.

71. Organic and Biological Chemistry Carbohydrates In solution they form cyclic structures. These can form chains of sugars that form structural molecules such as starch and cellulose.

72. Organic and Biological Chemistry Nucleic Acids Two of the building blocks of RNA and DNA are sugars (ribose or deoxyribose) and cyclic bases (adenine, guanine, cytosine, and thymine or uracil).

73. Organic and Biological Chemistry Nucleic Acids These combine with a phosphate to form a nucleotide.

74. Organic and Biological Chemistry Nucleic Acids Nucleotides combine to form the familiar double-helix form of the nucleic acids.

75. Organic and Biological Chemistry SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together It is formed in the body from carbohydrate metabolism. In the muscle it is reduced to lactic acid in the course of exertion. The acid-dissociation constant for pyruvic acid is 3.2  10 –3. (a) Why does pyruvic acid have a higher 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 about 7.4 and an acid concentration of 2  10 –4 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 25.17). (Although H atoms don’t exist as such in biochemical systems, biochemical reducing agents deliver hydrogen for such reductions.) Pyruvic acid has the following structure: Solution (a) The acid ionization constant for pyruvic acid should be somewhat greater than that of acetic acid because the carbonyl function on the  -carbon atom exerts an electron-withdrawing effect on the carboxylic acid group. In the C––O––H bond system the electrons are shifted from hydrogen, facilitating loss of the hydrogen 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

76. Organic and Biological Chemistry SAMPLE INTEGRATIVE EXERCISE continued The second term in the brackets is negligible compared to the first, so x = [Pv – ] = 6.4  10 –7 /3.2  10 –3 = 2  10 –4 M. 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. Let [Pv – ] = x. Then the concentration of undissociated acid is 2  10 –4 –x. The concentration of [H + ] is fixed at 4.0  10 –8 (the antilog of the pH value). Substituting, we obtain Solving for x, we obtain x[3.2  10 –3 + 4.0  10 –8 ] = 6.4  10 –7. (c) Pyruvic acid should be quite soluble in water because it has polar functional groups and a small hydrocarbon component. It is miscible with water, ethanol, and diethyl ether. (d) The methyl group carbon has sp 3 hybridization. The carbon carrying the carbonyl group has sp 2 hybridization because of the double bond to oxygen. Similarly, the carboxylic acid carbon is sp 2 hybridized. Essentially, the ketonic functional group has been reduced to an alcohol. (e) The balanced chemical equation for this reaction is