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Learning Objectives
When you have completed Chapter 26, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. use the information provided by an amino acid analysis, an Edman degradation and a carboxypeptidase hydrolysis to determine the structure of an unknown polypeptide.
3. outline the approach that you would use to synthesize a given peptide, providing appropriate mechanistic details if requested to do so.
4. define, and use in context, the key terms introduced in this chapter.
Amino acids are important biochemicals, as they are the building blocks from which proteins and polypeptides are assembled. We begin this chapter with an examination of some of the fundamental chemistry of amino acids: their structures, stereochemistry and synthesis. We then discuss the nature of peptides and of the peptide bond, and present the complex issue of determining the order in which the various amino‑acid residues occur in a given peptide. Once a chemist knows the exact order the of the residues in a given peptide, the next challenge is to determine a method by which the same peptide can be prepared in the laboratory. Thus, two sections are devoted to the problem of protein synthesis. The final sections in the chapter deal with the classification, overall structure and denaturation of proteins.
26: Biomolecules- Amino Acids Peptides and Proteins
Chapter Contents
26.1 Structures of Amino Acids
26.2 Amino Acids and the Henderson–Hasselbalch Equation: Isoelectric Points 26.3 Synthesis of Amino Acids 26.4 Peptides and Proteins 26.5 Amino Acid Analysis of Peptides 26.6 Peptide Sequencing: The Edman Degradation 26.7 Peptide Synthesis 26.8 Automated Peptide Synthesis: The Merrifield Solid-Phase Method 26.9 Protein Structure 26.10 Enzymes and Coenzymes 26.11 How Do Enzymes Work? Citrate Synthase
Continuing our look at the main classes of biomolecules, we’ll focus in this chapter on amino acids, the fundamental building blocks from which up to 150,000 or so different proteins in our bodies are made. We’ll then see how amino acids are incorporated into proteins and examine the structures of those proteins. Any understanding of biological chemistry would be impossible without this knowledge.
Proteins occur in every living organism, are of many different types, and have many different biological functions. The keratin of skin and fingernails, the fibroin of silk and spider webs, and the estimated 50,000 or so enzymes that catalyze the biological reactions in our bodies are all proteins. Regardless of their function, all proteins have a fundamentally similar structure and are made up of many amino acids linked together in a long chain.
Amino acids, as their name implies, are difunctional. They contain both a basic amino group and an acidic carboxyl group.
Their value as building blocks to make proteins stems from the fact that amino acids can join together into long chains by forming amide bonds between the –NH2 of one amino acid and the –CO2H of another. For classification purposes, chains with fewer than 50 amino acids are often called peptides, while the term protein is generally used for larger chains.
26.02: Structures of Amino Acids
We saw in Section 20.3 and Section 24.5 that a carboxyl group is deprotonated and exists as the carboxylate anion at a physiological pH of 7.3, while an amino group is protonated and exists as the ammonium cation. Thus, amino acids exist in aqueous solution primarily in the form of a dipolar ion, or zwitterion (from the German zwitter, meaning “hybrid”).
Amino acid zwitterions are internal salts and therefore have many of the physical properties associated with salts. They have large dipole moments, are relatively soluble in water but insoluble in hydrocarbons, and are crystalline with relatively high melting points. In addition, amino acids are amphiprotic; they can react either as acids or as bases, depending on the circumstances. In aqueous acid solution, an amino acid zwitterion is a base that accepts a proton onto its –CO2 group to yield a cation. In aqueous base solution, the zwitterion is an acid that loses a proton from its –NH3+ group to form an anion.
The structures, abbreviations (both three- and one-letter), and pKa values of the 20 amino acids commonly found in proteins are shown in Table 26.1. All are α-amino acids, meaning that the amino group in each is a substituent on the α carbon—the one next to the carbonyl group. Nineteen of the twenty amino acids are primary amines, RNH2, and differ only in the nature of their side chain—the substituent attached to the α carbon. Proline is a secondary amine whose nitrogen and α carbon atoms are part of a five-membered pyrrolidine ring.
Table 26.1 The 20 Common Amino Acids in Proteins
Name Abbreviations MW Structure pKa α-CO2H pKa α-NH3+ pKa side chain pI
Neutral Amino Acids
Alanine Ala A 89 2.34 9.69 6.01
Asparagine Asn N 132 2.02 8.80 5.41
Cysteine Cys C 121 1.96 10.28 8.18 5.07
Glutamine Gln Q 146 2.17 9.13 5.65
Glycine Gly G 75 2.34 9.60 5.97
Isoleucine Ile I 131 2.36 9.60 6.02
Leucine Leu L 131 2.36 9.60 5.98
Methionine Met M 149 2.28 9.21 5.74
Phenylalanine Phe F 165 1.83 9.13 5.48
Proline Pro P 115 1.99 10.60 6.30
Serine Ser S 105 2.21 9.15 5.68
Threonine Thr T 119 2.09 9.10 5.60
Tryptophan Trp W 204 2.83 9.39 5.89
Tyrosine Tyr Y 181 2.20 9.11 10.07 5.66
Valine Val V 117 2.32 9.62 5.96
Acidic Amino Acids
Aspartic acid Asp D 133 1.88 9.60 3.65 2.77
Glutamic acid Glu E 147 2.19 9.67 4.25 3.22
Basic Amino Acids
Arginine Arg R 174 2.17 9.04 12.48 10.76
Histidine His H 155 1.82 9.17 6.00 7.59
Lysine Lys K 146 2.18 8.95 10.53 9.74
In addition to the 20 amino acids commonly found in proteins, 2 others—selenocysteine and pyrrolysine—are found in some organisms, and more than 700 nonprotein amino acids are also found in nature. γ-Aminobutyric acid (GABA), for instance, is found in the brain and acts as a neurotransmitter; homocysteine is found in blood and is linked to coronary heart disease; and thyroxine is found in the thyroid gland, where it acts as a hormone.
Except for glycine, H2NCH2CO2H, the α carbons of amino acids are chirality centers. Two enantiomers of each are therefore possible, but nature uses only one to build proteins. In Fischer projections, naturally occurring amino acids are represented by placing the –CO2 group at the top and pointing the side chain downwards, as if drawing a carbohydrate (Section 25.2) and then placing the –NH3+ group on the left. Because of their stereochemical similarity to L sugars (Section 25.3), the naturally occurring α-amino acids are often referred to as L amino acids. The nonnaturally occurring enantiomers are called D amino acids.
The 20 common amino acids can be further classified as neutral, acidic, or basic, depending on the structure of their side chains. Fifteen of the twenty have neutral side chains, two (aspartic acid and glutamic acid) have an extra carboxylic acid function in their side chains, and three (lysine, arginine, and histidine) have basic amino groups in their side chains. Note that both cysteine (a thiol) and tyrosine (a phenol), although usually classified as neutral amino acids, nevertheless have weakly acidic side chains that can be deprotonated in a sufficiently basic solution.
At the physiological pH of 7.3, the side-chain carboxyl groups of aspartic acid and glutamic acid are deprotonated and the basic side-chain nitrogens of lysine and arginine are protonated. Histidine, however, which contains a heterocyclic imidazole ring in its side chain, is not quite basic enough to be protonated at pH 7.3. Note that only the pyridine-like, doubly bonded nitrogen in histidine is basic. The pyrrole-like singly bonded nitrogen is nonbasic because its lone pair of electrons is part of the six-π-electron aromatic imidazole ring (Section 24.9).
Humans are able to biosynthesize only 11 of the 20 protein amino acids, called nonessential amino acids. The other 9, called essential amino acids, are biosynthesized only in plants and microorganisms and must be obtained in our diet. The division between essential and nonessential amino acids is not clear-cut, however. Tyrosine, for instance, is sometimes considered nonessential because humans can produce it from phenylalanine, but phenylalanine itself is essential and must be obtained in the diet. Arginine can be synthesized by humans, but much of the arginine we need also comes from our diet.
Problem 26-1
How many of the α-amino acids shown in Table 26.1 contain aromatic rings? How many contain sulfur? How many contain alcohol groups? How many contain hydrocarbon side chains?
Problem 26-2
Of the 19 L amino acids, 18 have the S configuration at the α carbon. Cysteine is the only L amino acid that has an R configuration. Explain.
Problem 26-3
The amino acid threonine, (2S,3R)-2-amino-3-hydroxybutanoic acid, has two chirality centers.
(a) Draw threonine, using normal, wedged, and dashed lines to show dimensionality.
(b) Draw a diastereomer of threonine, and label its chirality centers as R or S. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.01%3A_Why_This_Chapter.txt |
According to the Henderson–Hasselbalch equation (Section 20.3 and Section 24.5), if we know both the pH of a solution and the pKa of an acid HA, we can calculate the ratio of [A] to [HA] in the solution. Furthermore, when pH = pKa, the two forms A and HA are present in equal amounts because log 1 = 0.
$pH = p K a + log [A – ] [HA] or log [A – ] [HA] = pH – p K a pH = p K a + log [A – ] [HA] or log [A – ] [HA] = pH – p K a$
To apply the Henderson–Hasselbalch equation to an amino acid, let’s find out what species are present in a 1.00 M solution of alanine at pH = 9.00. According to Table 26.1, protonated alanine [+H3NCH(CH3)CO2H] has pKa1 = 2.34 and neutral zwitterionic alanine [+H3NCH(CH3)CO2] has pKa2 = 9.69:
Because the pH of the solution is much closer to pKa2 than to pKa1, we need to use pKa2 for the calculation. From the Henderson–Hasselbalch equation, we have:
$log [A – ] [HA] = pH – p K a = 9.00 – 9.69 = –0.69 log [A – ] [HA] = pH – p K a = 9.00 – 9.69 = –0.69$
So
$[A – ] [HA] = antilog(–0.69) = 0.20 and [A – ] = 0.20[HA] [A – ] [HA] = antilog(–0.69) = 0.20 and [A – ] = 0.20[HA]$
In addition, we know that
$[A – ] + [HA] = 1.00 M [A – ] + [HA] = 1.00 M$
Solving the two simultaneous equations gives [HA] = 0.83 and [A] = 0.17. In other words, at pH = 9.00, 83% of alanine molecules in a 1.00 M solution are neutral (zwitterionic) and 17% are deprotonated. Similar calculations can be done at other pH values and the results plotted to give the titration curve shown in Figure 26.2.
Each leg of the titration curve is calculated separately. The first leg, from pH 1 to 6, corresponds to the dissociation of protonated alanine, H2A+. The second leg, from pH 6 to 11, corresponds to the dissociation of zwitterionic alanine, HA. It’s as if we started with H2A+ at low pH and then titrated with NaOH. When 0.5 equivalent of NaOH is added, the deprotonation of H2A+ is 50% complete; when 1.0 equivalent of NaOH is added, the deprotonation of H2A+ is finished and HA predominates; when 1.5 equivalents of NaOH is added, the deprotonation of HA is 50% complete; and when 2.0 equivalents of NaOH is added, the deprotonation of HA is finished.
Look carefully at the titration curve in Figure 26.2. In acid solution, the amino acid is protonated and exists primarily as a cation. In basic solution, the amino acid is deprotonated and exists primarily as an anion. In between the two is an intermediate pH at which the amino acid is exactly balanced between anionic and cationic forms, existing primarily as the neutral, dipolar zwitterion. This pH is called the amino acid’s isoelectric point (pI) and has a value of 6.01 for alanine.
The isoelectric point of an amino acid depends on its structure, with values for the 20 common amino acids given previously in Table 26.1. The 15 neutral amino acids have isoelectric points near neutrality, in the pH range 5.0 to 6.5. The two acidic amino acids have isoelectric points at lower pH so that deprotonation of the side-chain −CO2H does not occur at their pI, and the three basic amino acids have isoelectric points at higher pH so that protonation of the side-chain amino group does not occur at their pI.
More specifically, the pI of any amino acid is the average of the two acid-dissociation constants that involve the neutral zwitterion. For the 13 amino acids with a neutral side chain, pI is the average of pKa1 and pKa2. For the four amino acids with either a strongly or weakly acidic side chain, pI is the average of the two lowest pKa values. For the three amino acids with a basic side chain, pI is the average of the two highest pKa values.
Just as individual amino acids have isoelectric points, proteins have an overall pI due to the cumulative effect of all the acidic or basic amino acids they may contain. The enzyme lysozyme, for instance, has a preponderance of basic amino acids and thus has a high isoelectric point (pI = 11.0). Pepsin, however, has a preponderance of acidic amino acids and a low isoelectric point (pI ~ 1.0). Not surprisingly, the solubilities and properties of proteins with different pI’s are strongly affected by the pH of the medium. Solubility in water is usually lowest at the isoelectric point, where the protein has no net charge, and is higher both above and below the pI, where the protein is charged.
We can take advantage of the differences in isoelectric points to separate a mixture of proteins into its pure components. Using a technique known as electrophoresis, a mixture of proteins is placed near the center of a strip of paper or gel. The paper or gel is moistened with an aqueous buffer of a given pH, and electrodes are connected to the ends of the strip. When an electric potential is applied, the proteins with negative charges (those that are deprotonated because the pH of the buffer is above their isoelectric point) migrate slowly toward the positive electrode. At the same time, those amino acids with positive charges (those that are protonated because the pH of the buffer is below their isoelectric point) migrate toward the negative electrode.
Different proteins migrate at different rates, depending on their isoelectric points and on the pH of the aqueous buffer, thereby effecting a separation of the mixture into its components. Figure 26.3 illustrates this process for a mixture containing basic, neutral, and acidic components.
Problem 26-4
Hemoglobin has pI = 6.8. Does hemoglobin have a net negative charge or net positive charge at pH = 5.3? At pH = 7.3? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.03%3A_Amino_Acids_the_Henderson-Hasselbalch_Equation_and_Isoelectric_Points.txt |
α-Amino acids can be synthesized in the laboratory using some of the reactions discussed in previous chapters. One of the oldest methods of α-amino acid synthesis begins with α bromination of a carboxylic acid by treatment with Br2 and PBr3 (the Hell–Volhard–Zelinskii reaction; Section 22.4). SN2 substitution of the α-bromo acid with ammonia then yields an α-amino acid.
Phenylalanine
(b) Valine
The Amidomalonate Synthesis
A more general method for preparation of α-amino acids is the amidomalonate synthesis, a straightforward extension of the malonic ester synthesis (Section 22.7). The reaction begins with the conversion of diethyl acetamidomalonate into an enolate ion by treatment with base, followed by SN2 alkylation with a primary alkyl halide. Hydrolysis of both the amide protecting group and the esters occurs when the alkylated product is warmed with aqueous acid, and decarboxylation then takes place to yield an α-amino acid. For example, aspartic acid can be prepared from ethyl bromoacetate, BrCH2CO2Et:
Problem 26-6 What alkyl halides would you use to prepare the following α-amino acids by the amidomalonate method? (a)
Leucine
(b) Histidine (c) Tryptophan (d) Methionine
Reductive Amination of α-Keto Acids
Yet another method for the synthesis of α-amino acids is by reductive amination of an α-keto acid with ammonia and a reducing agent. Alanine, for instance, is prepared by treatment of pyruvic acid with ammonia in the presence of NaBH4. As described in Section 24.6, the reaction proceeds through formation of an intermediate imine which is then reduced.
Enantioselective Synthesis
The synthesis of an α-amino acid from an achiral precursor by any of the methods just described yields a racemic mixture, with equal amounts of S and R enantiomers. To use an amino acid in the laboratory synthesis of a naturally occurring protein, however, the pure S enantiomer must be obtained.
Two methods are used in practice to obtain enantiomerically pure amino acids. One way requires resolving the racemic mixture into its pure enantiomers (Section 5.8). A more direct approach, however, is to use an enantioselective synthesis to prepare only the desired S enantiomer directly. As discussed in the Chapter 19 Chemistry Matters, the idea behind enantioselective synthesis is to find a chiral reaction catalyst that will temporarily hold a substrate molecule in an unsymmetrical, chiral environment. While in that chiral environment, the substrate may be more open to reaction on one side than on another, leading to an excess of one enantiomeric product.
William Knowles at the Monsanto Company discovered in 1968 that α-amino acids can be prepared enantioselectively by hydrogenation of a Z enamido acid with a chiral hydrogenation catalyst. (S)-Phenylalanine, for instance, is prepared at 98.7% purity, contaminated by only 1.3% of the (R) enantiomer, when using a chiral rhodium catalyst. For this discovery, Knowles shared the 2001 Nobel Prize in Chemistry.
The most effective catalysts for enantioselective amino acid synthesis are coordination complexes of rhodium(I) with 1,5-cyclooctadiene (COD) and a chiral diphosphine such as (R,R)-1,2-bis(o-anisylphenylphosphino)ethane, the so-called DiPAMP ligand. This complex owes its chirality to the presence of trisubstituted phosphorus atoms (Section 5.10).
Problem 26-7
Show how you could prepare the following amino acid enantioselectively: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.04%3A_Synthesis_of_Amino_Acids.txt |
Proteins and peptides are amino acid polymers in which the individual amino acids, called residues, are linked together by amide bonds, or peptide bonds. An amino group from one residue forms an amide bond with the carboxyl of a second residue, the amino group of the second forms an amide bond with the carboxyl of a third, and so on. For example, alanylserine is the dipeptide that results when an amide bond forms between the alanine carboxyl and the serine amino group.
Note that two dipeptides can result from reaction between alanine and serine, depending on which carboxyl group reacts with which amino group. If the alanine amino group reacts with the serine carboxyl, serylalanine results.
The long, repetitive sequence of –N–CH–CO– atoms that makes up a continuous chain is called the protein’s backbone. By convention, peptides are written with the N-terminal amino acid (the one with the free –NH3+ group) on the left and the C-terminal amino acid (the one with the free –CO2 group) on the right. The name of a peptide is denoted by the abbreviations listed in Table 26.1 for each amino acid. Thus, alanylserine is abbreviated Ala-Ser or A-S, and serylalanine is abbreviated Ser-Ala or S-A. The one-letter abbreviations are more convenient, though less immediately recognizable, than the three-letter abbreviations.
The amide bond that links amino acids together in peptides is no different from any other amide bond (Section 24.3). An amide nitrogen is nonbasic because its unshared electron pair is delocalized by resonance with the carbonyl group. This overlap of the nitrogen p orbital with the p orbitals of the carbonyl group imparts a certain amount of double-bond character to the C–N bond and restricts rotation around it. The amide bond is therefore planar, and the N–H is oriented 180° to the $C═OC═O$.
A second kind of covalent bonding in peptides occurs when a disulfide linkage, RS–SR, is formed between two cysteine residues. As we saw in Section 18.7, a disulfide is formed by mild oxidation of a thiol, RSH, and is cleaved by mild reduction.
A disulfide bond between cysteine residues in different peptide chains links the otherwise separate chains together, whereas a disulfide bond between cysteine residues in the same chain forms a loop. Insulin, for instance, is composed of two chains that total 51 amino acids and are linked by two cysteine disulfide bridges.
Problem 26-8
Six isomeric tripeptides contain valine, tyrosine, and glycine. Name them using both three- and one-letter abbreviations.
Problem 26-9
Draw the structure of M-P-V-G, and indicate its amide bonds.
26.06: Amino Acid Analysis of Peptides
To determine the structure of a protein or peptide, we need to answer three questions: What amino acids are present? How much of each is present? In what sequence do the amino acids occur in the peptide chain? The answers to the first two questions are provided by an automated instrument called an amino acid analyzer.
An amino acid analyzer is based on techniques worked out in the 1950s by William Stein and Stanford Moore at the Rockefeller University, who shared the 1972 Nobel Prize in Chemistry for their work. In preparation for analysis, the peptide is broken into its constituent amino acids by reducing all disulfide bonds, capping the –SH groups of cysteine residues by SN2 reaction with iodoacetic acid, and hydrolyzing the amide bonds by heating with aqueous 6 M HCl at 110 °C for 24 hours. The resultant amino acid mixture is then separated into its components by a technique called chromatography, either high-pressure liquid chromatography (HPLC) or ion-exchange chromatography.
In both HPLC and ion-exchange chromatography, the mixture to be separated is dissolved in a solvent, called the mobile phase, and passed through a metal tube or glass column that contains an adsorbent material, called the stationary phase. Because different compounds adsorb to the stationary phase to different extents, they migrate through the chromatography column at different rates and are separated as they emerge (elute) from the end.
In the ion-exchange technique, separated amino acids eluting from the chromatography column mix with a solution of a substance called ninhydrin and undergo a rapid reaction that produces an intense purple color. The color is measured by a spectrometer, and a plot of elution time versus spectrometer absorbance is obtained.
Because the time required for a given amino acid to elute from a standard column is reproducible, the identities of the amino acids in a peptide can be determined. The amount of each amino acid in the sample is determined by measuring the intensity of the purple color resulting from its reaction with ninhydrin. Figure 26.4 shows the results of amino acid analysis of a standard equimolar mixture of 17 α-amino acids. Typically, amino acid analysis requires about 100 picomoles (2–3 μg) of sample for a protein containing about 200 residues.
Problem 26-10
Show the structure of the product you would expect to obtain by SN2 reaction of a cysteine residue with iodoacetic acid.
Problem 26-11
Show the structures of the products obtained on reaction of valine with ninhydrin. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.05%3A_Peptides_and_Proteins.txt |
With the identities and relative amounts of amino acids known, a peptide can then be sequenced to find out in what order the amino acids are linked. Much peptide sequencing is now done by mass spectrometry, using either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI) linked to a time-of-flight (TOF) mass analyzer, as described in Section 12.4. Also in common use is a chemical method of peptide sequencing called the Edman degradation.
The general idea of peptide sequencing by Edman degradation is to cleave one amino acid at a time from an end of the peptide chain. That terminal amino acid is then separated and identified, and the cleavage reactions are repeated on the chain-shortened peptide until the entire peptide sequence is known. Automated protein sequencers are available that allow as many as 50 repetitive sequencing cycles to be carried out before a buildup of unwanted by-products interferes with the results. So efficient are these instruments that sequence information can be obtained from as little as 1 to 5 picomoles of sample—less than 0.1 μg.
As shown in Figure 26.5, Edman degradation involves treatment of a peptide with phenyl isothiocyanate (PITC), $C6H5–N═C═SC6H5–N═C═S$, followed by reaction with trifluoroacetic acid. The first step attaches the PITC to the –NH2 group of the N-terminal amino acid, and the second step splits the N-terminal residue from the peptide chain, yielding an anilinothiazolinone (ATZ) derivative plus the chain-shortened peptide. Further acid-catalyzed rearrangement of the ATZ derivative with aqueous acid converts it into a phenylthiohydantoin (PTH), which is identified by comparison of its elution time with the known elution times of PTH derivatives of the 20 common amino acids. The chain-shortened peptide is then automatically resubmitted for another round of Edman degradation.
Figure 26.5 MECHANISM Mechanism of the Edman degradation for N-terminal analysis of peptides.
Complete sequencing of large proteins by Edman degradation is impractical because of the buildup of unwanted by-products. To get around this problem, a large peptide chain is first cleaved by partial hydrolysis into a number of smaller fragments, the sequence of each fragment is determined, and the individual fragments are fitted together by matching their overlapping ends. In this way, protein chains with more than 400 amino acids have been sequenced.
Partial hydrolysis of a peptide can be carried out either chemically with aqueous acid or enzymatically. Acid hydrolysis is unselective and gives a more-or-less random mixture of small fragments, but enzymatic hydrolysis is quite specific. The enzyme trypsin, for instance, catalyzes hydrolysis of peptides only at the carboxyl side of the basic amino acids arginine and lysine; chymotrypsin cleaves only at the carboxyl side of the aryl-substituted amino acids phenylalanine, tyrosine, and tryptophan.
Problem 26-12
The octapeptide angiotensin II has the sequence Asp-Arg-Val-Tyr-Ile-His-Pro-Phe. What fragments would result if angiotensin II were cleaved with trypsin? With chymotrypsin?
Problem 26-13
What is the N-terminal residue on a peptide that gives the following PTH derivative upon Edman degradation?
Problem 26-14
Draw the structure of the PTH derivative that would be formed by Edman degradation of angiotensin II (Problem 26-12).
Problem 26-15 Give the amino acid sequence of hexapeptides that produce the following sets of fragments upon partial acid hydrolysis: (a)
Arg, Gly, Ile, Leu, Pro, Val gives Pro-Leu-Gly, Arg-Pro, Gly-Ile-Val
(b) N, L, M, W, V2 gives V-L, V-M-W, W-N-V | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.07%3A_Peptide_Sequencing-_The_Edman_Degradation.txt |
Once the structure of a peptide is known, its synthesis can be undertaken—perhaps to obtain a larger amount for biological evaluation. A simple amide might be formed by treating an amine and a carboxylic acid with a carbodiimide (either DCC or EDC; Section 21.3), but peptide synthesis is a more difficult problem because many different amide bonds must be formed in a specific order, rather than at random.
The solution to the specificity problem is protection (Section 17.8). If we want to couple alanine with leucine to synthesize Ala-Leu, for instance, we could protect the –NH2 group of alanine and the –CO2H group of leucine to shield them from reacting, then form the desired Ala-Leu amide bond by reaction with EDC or DCC, and then remove the protecting groups.
A number of different amino- and carboxyl-protecting groups have been devised, but only a few are used in peptide synthesis. Carboxyl groups are often protected simply by converting them into methyl or benzyl esters. Both groups are easily introduced by standard methods of ester formation (Section 21.6) and are easily removed by mild hydrolysis with aqueous NaOH. Benzyl esters can also be cleaved by catalytic hydrogenolysis of the weak benzylic C–O bond ($RCO2–CH2Ph+H2→RCO2H+PhCH3RCO2–CH2Ph+H2→RCO2H+PhCH3$).
Amino groups are sometimes protected as their tert-butyloxycarbonyl amide (Boc) or, more commonly, as their fluorenylmethyloxycarbonyl amide (Fmoc). The Boc protecting group is introduced by reaction of the amino acid with di-tert-butyl dicarbonate in a nucleophilic acyl substitution reaction and is removed by brief treatment with a strong acid such as trifluoroacetic acid, CF3CO2H. The Fmoc protecting group is introduced by reaction with fluorenylmethyloxycarbonyl chloride and is removed by treatment with a 20% solution of the amine piperidine in dimethylformamide as solvent.
Thus, five steps are needed to synthesize a dipeptide such as Ala-Leu:
These steps can be repeated to add one amino acid at a time to the growing chain or to link two peptide chains together. Many remarkable achievements in peptide synthesis have been reported, including a complete synthesis of human insulin. Insulin is composed of two chains totaling 51 amino acids linked by two disulfide bridges. The three-dimensional structure of insulin, shown previously, was determined by Dorothy Crowfoot Hodgkin, a British chemist who received the 1964 Nobel Prize in Chemistry for her work on this and other complex biological molecules.
Problem 26-16
Show the mechanism for formation of a Boc derivative by reaction of an amino acid with di-tert-butyl dicarbonate.
Problem 26-17
Write all five steps required for the synthesis of Leu-Ala from alanine and leucine.
26.09: Automated Peptide Synthesis- The Merrifield Solid-Phase Method
As you might imagine, the synthesis of a large peptide chain by sequential addition of one amino acid at a time is a long and arduous process. An immense simplification is possible, however, using methods introduced by R. Bruce Merrifield, who received the 1984 Nobel Prize in Chemistry. In the Merrifield solid-phase method, peptide synthesis is carried out with the growing amino acid chain covalently bonded to small beads of a polymer resin, rather than in solution.
In the original procedure, polystyrene resin was used, prepared so that one of every hundred or so benzene rings contained a chloromethyl (−CH2Cl) group. A Boc-protected C-terminal amino acid was then attached to the resin through an ester bond formed by SN2 reaction.
With the first amino acid bonded to the resin, a repeating series of four steps is then carried out to build a peptide.
The steps in the solid-phase procedure have been improved substantially over the years, but the fundamental idea remains the same. A common procedure is use of the Wang resin, developed by Su-Sun Wang, with the Fmoc protecting group. Less common is using the PAM (phenylacetamidomethyl) resin with Boc as the N-protecting group.
Robotic peptide synthesizers are now used to automatically repeat the coupling, washing, and deprotection steps with different amino acids. Each step occurs in high yield, and mechanical losses are minimized because the peptide intermediates are never removed from the insoluble polymer until the final step. Peptides with 20 amino acids can be routinely prepared in a few hours. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.08%3A_Peptide_Synthesis.txt |
Proteins are usually classified as either fibrous or globular, according to their three-dimensional shape. Fibrous proteins, such as the collagen in tendons and connective tissue and the myosin in muscle tissue, consist of polypeptide chains arranged side by side in long filaments. Because these proteins are tough and insoluble in water, they are used in nature for structural materials. Globular proteins, by contrast, are usually coiled into compact, roughly spherical shapes. These proteins are generally soluble in water and are mobile within cells. Most of the more than 3000 or so enzymes that have been characterized are globular proteins, including the 1300 different enzymes in the human body.
Proteins are so large that the word structure takes on a broader meaning than with simpler organic compounds. In fact, chemists speak of four different levels of structure when describing proteins.
• The primary structure of a protein is simply the amino acid sequence.
• The secondary structure of a protein describes how segments of the peptide backbone orient into a regular pattern.
• The tertiary structure describes how the entire protein molecule coils into an overall three-dimensional shape.
• The quaternary structure describes how different protein molecules come together to yield large aggregate structures.
Primary structure is determined, as we’ve seen, by sequencing the protein. Secondary, tertiary, and quaternary structures are determined either by NMR or by X-ray crystallography (Chapter 12 Chemistry Matters).
The most common secondary structures are the α helix and the β-pleated sheet. An α helix is a right-handed coil of the protein backbone, much like the coil of a spiral staircase (Figure 26.6a). Each turn of the helix contains 3.6 amino acid residues, with a distance between coils of 540 pm, or 5.4 Å. The structure is stabilized by hydrogen bonds between amide N–H groups and $C═OFigure 26.6b).$
A β-pleated sheet differs from an α helix in that the peptide chain is fully extended rather than coiled and the hydrogen bonds occur between residues in adjacent chains (Figure 26.7a). The neighboring chains can run either in the same direction (parallel) or in opposite directions (antiparallel), although the antiparallel arrangement is more common and somewhat more energetically favorable. Concanavalin A, for instance, consists of two identical chains of 237 residues, with extensive regions of antiparallel β sheets (Figure 26.7b).
What about tertiary structure? Why does a protein adopt the shape it does? The forces that determine the tertiary structure of a protein are the same forces that act on all molecules, regardless of size, to provide maximum stability. Particularly important are the hydrophilic (water-loving; Section 2.12) interactions of the polar side chains on acidic or basic amino acids and the hydrophobic (water-fearing) interactions of nonpolar side chains. These acidic or basic amino acids with charged side chains tend to congregate on the exterior of the protein, where they can be solvated by water. Amino acids with neutral, nonpolar side chains tend to congregate on the hydrocarbon-like interior of a protein molecule, away from the aqueous medium.
Also important for stabilizing a protein’s tertiary structure are the formation of disulfide bridges between cysteine residues, the formation of hydrogen bonds between nearby amino acid residues, and the presence of ionic attractions, called salt bridges, between positively and negatively charged sites on various amino acid side chains within the protein.
Because the tertiary structure of a globular protein is delicately maintained by weak intramolecular attractions, a modest change in temperature or pH is often enough to disrupt that structure and cause the protein to become denatured. Denaturation occurs under such mild conditions that the primary structure remains intact, but the tertiary structure unfolds from a specific globular shape to a randomly looped chain (Figure 26.8).
Denaturation is accompanied by changes in both physical and biological properties. Solubility is drastically decreased, as occurs when egg white is cooked and the albumins unfold and coagulate. Most enzymes lose all catalytic activity when denatured, since a precisely defined tertiary structure is required for their action. Although most denaturation is irreversible, some cases are known where spontaneous renaturation of an unfolded protein to its stable tertiary structure occurs, accompanied by a full recovery of biological activity. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.10%3A_Protein_Structure.txt |
An enzyme is a substance—usually a large protein—that acts as a catalyst for a biological reaction. Like all catalysts, an enzyme doesn’t affect the equilibrium constant of a reaction and can’t bring about a chemical change that is otherwise unfavorable. An enzyme acts only to lower the activation energy for a reaction, thereby making the reaction take place more rapidly than it otherwise would. Sometimes, in fact, the rate acceleration brought about by enzymes is extraordinary. Millionfold rate increases are common, and the glycosidase enzymes that hydrolyze polysaccharides increase the reaction rate by a factor of more than 1017, changing the time required for the reaction from millions of years to milliseconds!
Unlike many of the catalysts that chemists use in the laboratory, enzymes are usually specific in their action. Often, in fact, an enzyme will catalyze only a single reaction of a single compound, called the enzyme’s substrate. For example, the enzyme amylase, found in the human digestive tract, catalyzes only the hydrolysis of starch to yield glucose; cellulose and other polysaccharides are untouched by amylase.
Different enzymes have different specificities. Some, such as amylase, are specific for a single substrate, but others operate on a range of substrates. Papain, for instance, a globular protein of 212 amino acids isolated from papaya fruit, catalyzes the hydrolysis of many kinds of peptide bonds. In fact, it’s this ability to hydrolyze peptide bonds that makes papain useful as a cleaner for contact lenses.
Enzymes function through a pathway that involves initial formation of an enzyme–substrate complex (E · S), followed by a multistep chemical conversion of the enzyme-bound substrate into enzyme-bound product (E · P) and final release of product from the complex.
$E + S ⇄ E ⋅ S ⇄ E ⋅ P ⇄ E + PE + S ⇄ E ⋅ S ⇄ E ⋅ P ⇄ E + P$
The overall rate constant for conversion of the E · S complex to products E · P is called the turnover number because it represents the number of substrate molecules a single enzyme molecule turns over into product per unit time. A value of about 103 per second is typical, although carbonic anhydrase can reach a value of up to 600,000.
The extraordinary rate accelerations achieved by enzymes are due to a combination of several factors. One important factor is simple geometry: an enzyme will adjust its shape to hold the substrate, other reactants, and various catalytic sites on acidic or basic residues in the precise geometry needed for reaction. In addition, the wrapping of the enzyme around the substrate can create specialized microenvironments that protect the substrate from the aqueous medium and can dramatically change the behavior of acid–base catalytic residues in the active site. But perhaps most important is that the enzyme stabilizes and thus lowers the energy of the rate-limiting transition state for reaction. That is, it’s not the ability of the enzyme to bind the substrate that matters but rather its ability to bind and stabilize the transition state. Often, in fact, the enzyme binds the transition structure as much as 1012 times more tightly than it binds the substrate or products. An energy diagram for an enzyme-catalyzed process might resemble that in Figure 26.9.
and enzyme-catalyzed processes. The enzyme makes available an alternative, lower-energy pathway. Rate enhancement is due to the ability of the enzyme to bind to the transition state for product formation, thereby lowering its energy.
Enzymes are classified into six categories depending on the kind of reaction they catalyze, as shown in Table 26.2. Oxidoreductases catalyze oxidations and reductions; transferases catalyze the transfer of a group from one substrate to another; hydrolases catalyze hydrolysis reactions of esters, amides, and related substrates; lyases catalyze the elimination or addition of a small molecule such as H2O from or to a substrate; isomerases catalyze isomerizations; and ligases catalyze the bonding of two molecules, often coupled with the hydrolysis of ATP. The systematic name of an enzyme has two parts, ending with -ase. The first part identifies the enzyme’s substrate, and the second part identifies its class. Hexose kinase, for example, is a transferase that catalyzes the transfer of a phosphate group from ATP to a hexose sugar.
Table 26.2 Classification of Enzymes
Class Some subclasses Function
Oxidoreductases
Dehydrogenases
Oxidases
Reductases
Introduction of double bond
Oxidation
Reduction
Transferases
Kinases
Transaminases
Transfer of phosphate group
Transfer of amino group
Hydrolases
Lipases
Nucleases
Proteases
Hydrolysis of ester
Hydrolysis of phosphate
Hydrolysis of amide
Lyases
Decarboxylases
Dehydrases
Loss of CO2
Loss of H2O
Isomerases Epimerases Isomerization of chirality center
Ligases
Carboxylases
Synthetases
Addition of CO2
Formation of a new bond
In addition to their protein part, most enzymes also contain a small non-protein part called a cofactor. A cofactor can be either an inorganic ion, such as Zn2+, or a small organic molecule, called a coenzyme. A coenzyme is not a catalyst but is a reactant that undergoes chemical change during the reaction and requires an additional step to return to its initial state.
Many coenzymes are derived from vitamins—substances that an organism requires in small amounts for growth but is unable to synthesize and must receive in its diet (Chapter 20 Chemistry Matters). Coenzyme A from pantothenate (vitamin B3), NAD+ from niacin, FAD from riboflavin (vitamin B2), tetrahydrofolate from folic acid, pyridoxal phosphate from pyridoxine (vitamin B6), and thiamin diphosphate from thiamin (vitamin B1) are examples. Table 26.3 shows the structures of some common coenzymes.
Problem 26-18 To what classes do the following enzymes belong? (a)
Pyruvate decarboxylase
(b) Chymotrypsin (c) Alcohol dehydrogenase | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.11%3A_Enzymes_and_Coenzymes.txt |
As we saw in the previous section, enzymes work by bringing substrate and other reactant molecules together, holding them in the orientation necessary for reaction, providing any necessary acidic or basic sites to catalyze specific steps, and stabilizing the transition state for reaction. As an example, let’s look at citrate synthase, an enzyme that catalyzes the aldol-like addition of acetyl CoA to oxaloacetate to give citrate. This reaction is the first step in the citric acid cycle, in which acetyl groups produced by degradation of food molecules are metabolized to yield CO2 and H2O. We’ll look at the details of the citric acid cycle in Section 29.7.
Citrate synthase is a globular protein of 433 amino acids with a deep cleft lined by an array of functional groups that can bind to the substrate, oxaloacetate. On binding oxaloacetate, the original cleft closes and another opens up nearby to bind acetyl CoA. This second cleft is also lined by appropriate functional groups, including a histidine at position 274 and an aspartic acid at position 375. The two reactants are now held by the enzyme in close proximity and with a suitable orientation for reaction. Figure 26.10 shows the structure of citrate synthase as determined by X-ray crystallography, along with a close-up of the active site.
Table 26.3 Structures and Functions of Some Common Coenzymes
As shown in Figure 26.11, the first step in the aldol reaction of acetyl CoA and oxaloacetate is generation of the enol of acetyl CoA. The side-chain carboxyl of an aspartate residue acts as base to abstract an acidic α proton, while at the same time the side-chain imidazole ring of a histidine donates H+ to the carbonyl oxygen. The enol thus produced then performs a nucleophilic addition to the ketone carbonyl group of oxaloacetate. The first histidine acts as a base to remove the –OH hydrogen from the enol, while a second histidine residue simultaneously donates a proton to the oxaloacetate carbonyl group, giving citryl CoA. Water then hydrolyzes the thiol ester group in citryl CoA in a nucleophilic acyl substitution reaction, releasing citrate and coenzyme A as the final products.
Figure 26.11 MECHANISM Mechanism of the addition of acetyl CoA to oxaloacetate to give (S)-citryl CoA, catalyzed by citrate synthase.
26.13: Chemistry MattersThe Protein Data Bank
26 • Chemistry Matters 26 • Chemistry Matters
Enzymes are so large, so structurally complex, and so numerous that the use of computer databases and molecular visualization programs has become an essential tool for studying biological chemistry. Of the various databases available online, the Kyoto Encyclopedia of Genes and Genomes (KEGG) database (http://www.genome.jp/kegg/pathway.html), maintained by the Kanehisa Laboratory of Kyoto University Bioinformatics Center, is useful for obtaining information on biosynthetic pathways of the sort we’ll be describing in Chapter 29. For obtaining information on a specific enzyme, the BRENDA database (http://www.brenda-enzymes.org), maintained by the Institute of Biochemistry at the University of Cologne, Germany, is particularly valuable.
Perhaps the most useful of all biological databases is the Protein Data Bank (PDB), operated by the Research Collaboratory for Structural Bioinformatics (RCSB). The PDB is a worldwide repository of X-ray and NMR structural data for biological macromolecules. In mid-2022, data for more than 192,888 structures were available, and more than 7000 new ones were being added yearly. To access the Protein Data Bank, go to www.rcsb.org and a home page like that shown in Figure 26.12 will appear. As with much that is available online, however, the PDB site is changing rapidly, so you may not see quite the same thing.
To learn how to use the PDB, begin by running the short tutorial listed under Learn on the left side column. After that introduction, start exploring. Let’s say you want to view citrate synthase, the enzyme that catalyzes the addition of acetyl CoA to oxaloacetate to give citrate. Type “citrate synthase” (with quotation marks) into the small search box on the top line, click on “Search,” and a list of 2,400 or so structures will appear. Scroll down near the end of the list until you find the entry with a PDB code of 5CTS and the title “Proposed Mechanism for the Condensation Reaction of Citrate Synthase: 1.9-Angstroms Structure of the Ternary Complex with Oxaloacetate and Carboxymethyl Coenzyme A.” Alternatively, if you know the code of the enzyme you want, you can enter it directly into the search box. Click on the PDB code of entry 5CTS, and a new page containing information about the enzyme will open.
If you choose, you can download the structure file to your computer and open it with any of numerous molecular graphics programs to see an image like that in Figure 26.13. The biologically active molecule is a dimer of two identical subunits consisting primarily of α-helical regions displayed as coiled ribbons. For now, just select “View in Jmol” under the enzyme image on the right side of the screen to see some of the options for visualizing and further exploring the enzyme. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.12%3A_How_do_Enzymes_Work_Citrate_Synthase.txt |
26 • Key Terms 26 • Key Terms
• α-amino acids
• α helix
• backbone
• β-pleated sheet
• C-terminal amino acid
• coenzyme
• cofactor
• denatured
• Edman degradation
• enzyme
• fibrous proteins
• globular proteins
• isoelectric point (pI)
• N-terminal amino acid
• peptides
• primary structure
• proteins
• quaternary structure
• residues
• secondary structure
• side chain
• tertiary structure
• turnover number
• zwitterion
26.15: Summary
26 • Summary 26 • Summary
Proteins and peptides are large biomolecules made of α-amino acid residues linked together by amide, or peptide, bonds. Twenty amino acids are commonly found in proteins, and all except glycine have stereochemistry similar to that of L sugars. In neutral solution, amino acids exist as dipolar zwitterions.
Amino acids can be synthesized in racemic form by several methods, including ammonolysis of an α-bromo acid, alkylation of diethyl acetamidomalonate, and reductive amination of an α-keto acid. Alternatively, an enantioselective synthesis of amino acids can be carried out using a chiral hydrogenation catalyst.
Determining the structure of a peptide or protein begins with amino acid analysis. The peptide is hydrolyzed to its constituent α-amino acids, which are separated and identified. Next, the peptide is sequenced. Edman degradation by treatment with phenyl isothiocyanate (PITC) cleaves one residue from the N terminus of the peptide and forms an easily identifiable phenylthiohydantoin (PTH) derivative of the N-terminal amino acid. An automated series of Edman degradations can sequence peptide chains up to 50 residues in length.
Peptide synthesis involves the use of protecting groups. An N-protected amino acid with a free –CO2H group is coupled using DCC or EDC to an O-protected amino acid with a free –NH2 group. Amide formation occurs, the protecting groups are removed, and the sequence is repeated. Amines are usually protected as their tert-butyloxycarbonyl (Boc) or fluorenylmethyloxycarbonyl (Fmoc) derivatives; acids are usually protected as esters. The synthesis is often carried out by the Merrifield solid-phase method, in which the peptide is bonded to insoluble polymer beads.
Proteins have four levels of structure. Primary structure describes a protein’s amino acid sequence; secondary structure describes how segments of the protein chain orient into regular patterns—either α helix or β-pleated sheet; tertiary structure describes how the entire protein molecule coils into an overall three-dimensional shape; and quaternary structure describes how individual protein molecules aggregate into larger structures.
Proteins are classified as either globular or fibrous. Fibrous proteins such as α-keratin are tough, rigid, and water-insoluble; globular proteins such as myoglobin are water-soluble and roughly spherical in shape. Many globular proteins are enzymes—substances that act as catalysts for biological reactions. Enzymes are grouped into six classes according to the kind of reaction they catalyze. In addition to their protein part, many enzymes contain cofactors, which can be either metal ions or small organic molecules called coenzymes.
26.16: Summary of Reactions
26 • Summary of Reactions 26 • Summary of Reactions
1. Amino acid synthesis (Section 26.3)
1. From α-bromo acids
• Diethyl acetamidomalonate synthesis
• Reductive amination of an α-keto acid
• Enantioselective synthesis
• Peptide sequencing by Edman degradation (Section 26.6)
• Peptide synthesis (Section 26.7)
1. Amine protection
2. Carboxyl protection | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.14%3A_Key_Terms.txt |
26 • Additional Problems 26 • Additional Problems
Visualizing Chemistry
Problem 26-19
Identify the following amino acids: (a)
(b)
(c)
Problem 26-20 Give the sequence of the following tetrapeptide (yellow = S): Problem 26-21 Isoleucine and threonine are the only two amino acids with two chirality centers. Assign R or S configuration to the methyl-bearing carbon atom of isoleucine. Problem 26-22 Is the following structure a D amino acid or an L amino acid? Identify it. Problem 26-23 Give the sequence of the following tetrapeptide: Mechanism Problems Problem 26-24 The reaction of ninhydrin with an α-amino acid occurs in several steps. (a) The first step is formation of an imine by reaction of the amino acid with ninhydrin. Show its structure and the mechanism of its formation. (b)
The second step is a decarboxylation. Show the structure of the product and the mechanism of the decarboxylation reaction.
(c) The third step is hydrolysis of an imine to yield an amine and an aldehyde. Show the structures of both products and the mechanism of the hydrolysis reaction. (d)
The final step is formation of the purple anion. Show the mechanism of the reaction.
Problem 26-25
The chloromethylated polystyrene resin originally used for Merrifield solid-phase peptide synthesis was prepared by treatment of polystyrene with chloromethyl methyl ether and a Lewis acid catalyst. Propose a mechanism for the reaction.
Problem 26-26
An Fmoc protecting group can be removed from an amino acid by treatment with the amine base piperidine. Propose a mechanism.
Problem 26-27 Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues by reaction with cyanogen bromide, BrC≡N. The reaction occurs in several steps: (a)
The first step is a nucleophilic substitution reaction of the sulfur on the methionine side chain with BrCN to give a cyanosulfonium ion, [R2SCN]+. Show the structure of the product, and propose a mechanism for the reaction.
(b)
The second step is an internal SN2 reaction, with the carbonyl oxygen of the methionine residue displacing the positively charged sulfur leaving group and forming a five-membered ring product. Show the structure of the product and the mechanism of its formation.
(c)
The third step is a hydrolysis reaction to split the peptide chain. The carboxyl group of the former methionine residue is now part of a lactone (cyclic ester) ring. Show the structure of the lactone product and the mechanism of its formation.
(d)
The final step is a hydrolysis of the lactone to give the product shown. Show the mechanism of the reaction.
Problem 26-28
A clever recent method of peptide synthesis involves formation of an amide bond by reaction of an α-keto acid with an N-alkylhydroxylamine:
The reaction is thought to occur by nucleophilic addition of the N-alkylhydroxylamine to the keto acid as if forming an oxime (Section 19.8), followed by decarboxylation and elimination of water. Show the mechanism.
Amino Acid Structures and Chirality
Problem 26-29
Except for cysteine, only S amino acids occur in proteins. Several R amino acids are also found in nature, however. (R)-Serine is found in earthworms, and (R)-alanine is found in insect larvae. Draw Fischer projections of (R)-serine and (R)-alanine. Are these D or L amino acids?
Problem 26-30
Cysteine is the only amino acid that has L stereochemistry but an R configuration. Make up a structure for another L amino acid of your own creation that also has an R configuration.
Problem 26-31
Draw a Fischer projection of (S)-proline.
Problem 26-32 Show the structures of the following amino acids in their zwitterionic forms: (a)
Trp
(b) Ile (c) Cys (d) His
Problem 26-33
Proline has pKa1 = 1.99 and pKa2 = 10.60. Use the Henderson–Hasselbalch equation to calculate the ratio of protonated and neutral forms at pH = 2.50. Calculate the ratio of neutral and deprotonated forms at pH = 9.70.
Problem 26-34 Using both three- and one-letter codes for amino acids, write the structures of all possible peptides that contain the following amino acids: (a)
Val, Ser, Leu
(b) Ser, Leu2, Pro
Problem 26-35
Look at the side chains of the 20 amino acids in Table 26.1, and then think about what is not present. None of the 20 contain either an aldehyde or a ketone carbonyl group, for instance. Is this just one of nature’s oversights, or is there a likely chemical reason? What complications might an aldehyde or ketone carbonyl group cause?
Amino Acid Synthesis and Reactions
Leucine
(b) Tryptophan
Methionine
(b) Isoleucine
Pro
(b) Val
Problem 26-39
Serine can be synthesized by a simple variation of the amidomalonate method using formaldehyde rather than an alkyl halide. How might this be done?
Problem 26-40 Predict the product of the reaction of valine with the following reagents: (a)
CH3CH2OH, acid
(b) Di-tert-butyl dicarbonate (c) KOH, H2O (d) CH3COCl, pyridine; then H2O
Problem 26-41
Draw resonance forms for the purple anion obtained by reaction of ninhydrin with an α-amino acid (see Problem 26-24).
Peptides and Enzymes
Problem 26-42 Write full structures for the following peptides: (a)
C-H-E-M
(b) P-E-P-T-I-D-E
Problem 26-43
Propose two structures for a tripeptide that gives Leu, Ala, and Phe on hydrolysis but does not react with phenyl isothiocyanate.
Problem 26-44
Show the steps involved in a synthesis of Phe-Ala-Val using the Merrifield procedure.
I-L-P-F
(b) D-T-S-G-A
Problem 26-46
Which amide bonds in the following polypeptide are cleaved by trypsin? By chymotrypsin?
Phe-Leu-Met-Lys-Tyr-Asp-Gly-Gly-Arg-Val-Ile-Pro-Tyr
Problem 26-47 What kinds of reactions do the following classes of enzymes catalyze? (a)
Hydrolases
(b) Lyases (c) Transferases
Problem 26-48 Which of the following amino acids are more likely to be found on the exterior of a globular protein, and which on the interior? Explain. (a)
Valine
(b) Aspartic acid (c) Phenylalanine (d) Lysine
Problem 26-49 Leuprolide is a synthetic nonapeptide used to treat both endometriosis in women and prostate cancer in men. (a)
Both C-terminal and N-terminal amino acids in leuprolide have been structurally modified. Identify the modifications.
(b) One of the nine amino acids in leuprolide has D stereochemistry rather than the usual L. Which one? (c) Write the structure of leuprolide using both one- and three-letter abbreviations. (d) What charge would you expect leuprolide to have at neutral pH?
General Problems
Problem 26-50
The α-helical parts of myoglobin and other proteins stop whenever a proline residue is encountered in the chain. Why is proline never present in a protein α helix?
Problem 26-51
Arginine, the most basic of the 20 common amino acids, contains a guanidino functional group in its side chain. Explain, using resonance structures to show how the protonated guanidino group is stabilized.
Problem 26-52
Cytochrome c is an enzyme found in the cells of all aerobic organisms. Elemental analysis of cytochrome c shows that it contains 0.43% iron. What is the minimum molecular weight of this enzyme?
Problem 26-53
Evidence for restricted rotation around amide ${\rm O}\text{═}{\rm C}\text{–}{\rm N}$ bonds comes from NMR studies. At room temperature, the 1H NMR spectrum of N,N-dimethylformamide shows three peaks: 2.9 δ (singlet, 3 H), 3.0 δ (singlet, 3 H), and 8.0 δ (singlet, 1 H). As the temperature is raised, however, the two singlets at 2.9 δ and 3.0 δ slowly merge. At 180 °C, the 1H NMR spectrum shows only two peaks: 2.95 δ (singlet, 6 H) and 8.0 δ (singlet, 1 H). Explain this temperature-dependent behavior.
Problem 26-54
Propose a structure for an octapeptide that shows the composition Asp, Gly2, Leu, Phe, Pro2, Val on amino acid analysis. Edman analysis shows a glycine N-terminal group, and leucine is the C-terminal group. Acidic hydrolysis gives the following fragments:
Val-Pro-Leu, Gly, Gly-Asp-Phe-Pro, Phe-Pro-Val
Problem 26-55
Look at the structure of human insulin, and indicate where in each chain the molecule is cleaved by trypsin and chymotrypsin.
Problem 26-56
What is the structure of a nonapeptide that gives the following fragments when cleaved?
Trypsin cleavage: Val-Val-Pro-Tyr-Leu-Arg, Ser-Ile-Arg
Chymotrypsin cleavage: Leu-Arg, Ser-Ile-Arg-Val-Val-Pro-Tyr
Problem 26-57
Oxytocin, a nonapeptide hormone secreted by the pituitary gland, functions by stimulating uterine contraction and lactation during childbirth. Its sequence was determined from the following evidence:
1. Oxytocin is a cyclic compound containing a disulfide bridge between two cysteine residues.
2. When the disulfide bridge is reduced, oxytocin has the constitution Asn, Cys2, Gln, Gly, Ile, Leu, Pro, Tyr.
3. Partial hydrolysis of reduced oxytocin yields seven fragments: Asp-Cys, Ile-Glu, Cys-Tyr, Leu-Gly, Tyr-Ile-Glu, Glu-Asp-Cys, and Cys-Pro-Leu.
4. Gly is the C-terminal group.
5. Both Glu and Asp are present as their side-chain amides (Gln and Asn) rather than as free side-chain acids.
What is the amino acid sequence of reduced oxytocin? What is the structure of oxytocin itself?
Problem 26-58 Aspartame, a nonnutritive sweetener marketed under such trade names as Equal, NutraSweet, and Canderel, is the methyl ester of a simple dipeptide, Asp-Phe-OCH3. (a)
Draw the structure of aspartame.
(b) The isoelectric point of aspartame is 5.9. Draw the principal structure present in aqueous solution at this pH. (c) Draw the principal form of aspartame present at physiological pH = 7.3.
Problem 26-59
Refer to Figure 26.5 and propose a mechanism for the final step in Edman degradation—the acid-catalyzed rearrangement of the ATZ derivative to the PTH derivative.
Problem 26-60
Amino acids are metabolized by a transamination reaction in which the −NH2 group of the amino acid changes places with the keto group of an α-keto acid. The products are a new amino acid and a new α-keto acid. Show the product from transamination of isoleucine.
Problem 26-61
The first step in the biological degradation of histidine is formation of a 4-methylidene-5-imidazolone (MIO) by cyclization of a segment of the peptide chain in the histidine ammonia lyase enzyme. Propose a mechanism.
Problem 26-62
The first step in the biological degradation of lysine is reductive amination with α-ketoglutarate to give saccharopine. Nicotinamide adenine dinucleotide phosphate (NADPH), a relative of NADH, is the reducing agent. Show the mechanism. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.17%3A_Additional_Problems.txt |
Learning Objectives
• When you have completed Chapter 27, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• distinguish among fats and oils, phospholipids, prostaglandins, terpenes and steroids, and be familiar with the sources and biological roles of these substances.
• define, and use in context, the key terms introduced in this chapter.
27: Biomolecules - Lipids
Chapter Contents
27.1 Waxes, Fats, and Oils
27.2 Soap 27.3 Phospholipids 27.4 Prostaglandins and Other Eicosanoids 27.5 Terpenoids 27.6 Steroids 27.7 Biosynthesis of Steroids
We’ve now covered two of the four major classes of biomolecules—proteins and carbohydrates—and have two remaining. In this chapter, we’ll cover lipids, the largest and most diverse class of biomolecules, looking both at their structure and function and at their metabolism.
Lipids are naturally occurring organic molecules that have limited solubility in water and can be isolated from organisms by extraction with nonpolar organic solvents. Fats, oils, waxes, many vitamins and hormones, and most nonprotein cell-membrane components are some examples. Note that this definition differs from the sort used for carbohydrates and proteins in that lipids are defined by a physical property (solubility) rather than by structure. Of the many kinds of lipids, we’ll be concerned in this chapter with only a few: triacylglycerols, eicosanoids, terpenoids, and steroids.
Lipids are classified into two broad types: those like fats and waxes, which contain ester linkages and can be hydrolyzed, and those like cholesterol and other steroids, which don’t have ester linkages and can’t be hydrolyzed.
27.02: Waxes Fats and Oils
Waxes are mixtures of esters of long-chain carboxylic acids with long-chain alcohols. The carboxylic acid usually has an even number of carbons from 16 to 36, while the alcohol has an even number of carbons from 24 to 36. One of the major components of beeswax, for instance, is triacontyl hexadecanoate, the ester of the C30 alcohol (1-triacontanol) and the C16 acid (hexadecanoic acid). The waxy protective coatings on most fruits, berries, leaves, and animal furs have similar structures.
Animal fats and vegetable oils are the most widely occurring lipids. Although they appear different—animal fats like butter and lard are solids, whereas vegetable oils like corn and peanut oil are liquid—their structures are closely related. Chemically, fats and oils are triglycerides, or triacylglycerols—triesters of glycerol with three long-chain carboxylic acids called fatty acids. Animals use fats for long-term energy storage because they are far less highly oxidized than carbohydrates and provide about six times as much energy as an equal weight of stored, hydrated glycogen.
Hydrolysis of a fat or oil with aqueous NaOH yields glycerol and three fatty acids. The fatty acids are generally unbranched and contain an even number of carbon atoms between 12 and 20. If double bonds are present, they have mostly, although not entirely, Z, or cis, geometry. The three fatty acids of a specific triacylglycerol molecule need not be the same, and the fat or oil from a given source is likely to be a complex mixture of many different triacylglycerols. Table 27.1 lists some of the commonly occurring fatty acids, and Table 27.2 lists the approximate composition of fats and oils from different sources.
Table 27.1 Structures of Some Common Fatty Acids
Name No. of carbons Melting point (°C) Structure
Saturated
Lauric 12 43.2 CH3(CH2)10CO2H
Myristic 14 53.9 CH3(CH2)12CO2H
Palmitic 16 63.1 CH3(CH2)14CO2H
Stearic 18 68.8 CH3(CH2)16CO2H
Arachidic 20 76.5 CH3(CH2)18CO2H
Unsaturated
Palmitoleic 16 –0.1 $(Z)-CH3(CH2)5CH═CH(CH2)7CO2H(Z)-CH3(CH2)5CH═CH(CH2)7CO2H$
Oleic 18 13.4 $(Z)-CH3(CH2)7CH═CH(CH2)7CO2H(Z)-CH3(CH2)7CH═CH(CH2)7CO2H$
Linoleic 18 –12 $(Z,Z)-CH3(CH2)4(CH═CHCH2)2(CH2)6CO2H(Z,Z)-CH3(CH2)4(CH═CHCH2)2(CH2)6CO2H$
Linolenic 18 –11 $(all Z)-CH3CH2(CH═CHCH2)3(CH2)6CO2H(all Z)-CH3CH2(CH═CHCH2)3(CH2)6CO2H$
Arachidonic 20 –49.5 $(all Z)-CH3(CH2)4(CH═CHCH2)4CH2CH2CO2H(all Z)-CH3(CH2)4(CH═CHCH2)4CH2CH2CO2H$
Table 27.2 Composition of Some Fats and Oils
Saturated fatty acids (%) Unsaturated fatty acids (%)
Source C12 lauric C14 myristic C16 palmitic C18 stearic C18 oleic C18 linoleic
Animal fat
Lard 1 25 15 50 6
Butter 2 10 25 10 25 5
Human fat 1 3 25 8 46 10
Whale blubber 8 12 3 35 10
Vegetable oil
Coconut 50 18 8 2 6 1
Corn 1 10 4 35 45
Olive 1 5 5 80 7
Peanut 7 5 60 20
More than 100 different fatty acids are known and about 40 occur widely. Palmitic acid (C16) and stearic acid (C18) are the most abundant saturated fatty acids; oleic and linoleic acids (both C18) are the most abundant unsaturated ones. Oleic acid is monounsaturated because it has only one double bond, whereas linoleic, linolenic, and arachidonic acids are polyunsaturated fatty acids because they have more than one double bond. Linoleic and linolenic acids occur in cream and are essential in the human diet; infants grow poorly and develop skin lesions if fed a diet of nonfat milk for prolonged periods. Linolenic acid, in particular, is an example of an omega-3 fatty acid, which has been found to lower blood triglyceride levels and reduce the risk of heart attack. The name omega-3 means that there is a double bond three carbons in from the noncarboxyl end of the chain.
The data in Table 27.1 show that unsaturated fatty acids generally have lower melting points than their saturated counterparts, a trend that is also true for triacylglycerols. Since vegetable oils generally have a higher proportion of unsaturated to saturated fatty acids than animal fats (Table 27.2), they have lower melting points. The difference is a consequence of structure. Saturated fats have a uniform shape that allows them to pack together efficiently in a crystal lattice. In unsaturated vegetable oils, however, the $C═CC═C$ bonds introduce bends and kinks into the hydrocarbon chains, making crystal formation more difficult. The more double bonds there are, the harder it is for the molecules to crystallize and the lower the melting point of the oil.
The $C═CC═C$ bonds in vegetable oils can be reduced by catalytic hydrogenation, typically carried out at high temperature using a nickel catalyst, to produce saturated solid or semisolid fats. Prior to 2015, industrially produced margarine and shortening were produced by hydrogenating soybean, peanut, or cottonseed oil until the proper consistency is obtained. As of 2015, trans fats are no longer considered safe by the FDA and have since been phased out of industrially produced foods. Unfortunately, the hydrogenation reaction is accompanied by some cis–trans isomerization of the remaining double bonds, producing fats with about 10% to 15% trans unsaturated fatty acids. Dietary intake of trans fatty acids increases cholesterol levels in the blood, thereby increasing the risk of heart problems. The conversion of linoleic acid into elaidic acid is an example.
Problem 27-1
Carnauba wax, used in floor and furniture polishes, contains an ester of a C32 straight-chain alcohol with a C20 straight-chain carboxylic acid. Draw its structure.
Problem 27-2
Draw structures of glyceryl tripalmitate and glyceryl trioleate. Which would you expect to have a higher melting point? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.01%3A_Why_This_Chapter.txt |
Soap has been in use for nearly 5000 years. As early as 2800 BC, the Babylonians boiled fats with ashes to create a soap-like substance. Ancient Egyptian medical papyri dating from 1550 BC reveals that Egyptians bathed regularly with soap made from a mixture of animal fats, vegetable oils, and alkaline salts. Chemically, soap is a mixture of the sodium or potassium salts of the long-chain fatty acids produced by hydrolysis (saponification) of animal fat with alkali. Wood ash was used as a source of alkali until the early 1800s, when the LeBlanc process for making Na2CO3 by heating sodium sulfate with limestone became available.
Crude soap curds contain glycerol and excess alkali as well as soap but can be purified by boiling with water and adding NaCl or KCl to precipitate the pure carboxylate salts. The smooth soap that precipitates is dried, perfumed, and pressed into bars for household use. Dyes are added to make colored soaps, antiseptics are added for medicated soaps, pumice is added for scouring soaps, and air is blown in for soaps that float. Regardless of these extra treatments and regardless of price, though, all soaps are basically the same.
Soaps act as cleansers because the two ends of a soap molecule are so different. The carboxylate end of the long-chain molecule is ionic and therefore hydrophilic (Section 2.12), or attracted to water. The long hydrocarbon portion of the molecule, however, is nonpolar and hydrophobic, avoiding water and therefore more soluble in oils. The net effect of these two opposing tendencies is that soaps are attracted to both oils and water and are therefore useful as cleansers.
When soaps are dispersed in water, the long hydrocarbon tails cluster together on the inside of a tangled, hydrophobic ball, while the ionic heads on the surface of the cluster protrude into the water layer. These spherical clusters, called micelles, are shown schematically in Figure 27.2. Grease and oil droplets are solubilized in water when they are coated by the nonpolar, hydrophobic tails of soap molecules in the center of micelles. Once solubilized, the grease and dirt can be rinsed away.
As useful as they are, soaps also have drawbacks. In hard water, which contains metal ions, soluble sodium carboxylates are converted into insoluble magnesium and calcium salts, leaving the familiar ring of scum around bathtubs and a gray tinge on white clothes. Chemists have circumvented this problem by synthesizing a class of synthetic detergents based on salts of long-chain alkylbenzenesulfonic acids. The mechanism of synthetic detergents is the same as that of soaps: the alkylbenzene end of the molecule is attracted to grease, while the anionic sulfonate end is attracted to water. Unlike soaps, though, sulfonate detergents don’t form insoluble metal salts in hard water and don’t leave an unpleasant scum.
Problem 27-3
Draw the structure of magnesium oleate, a component of bathtub scum.
Problem 27-4
Write the saponification reaction of glyceryl dioleate monopalmitate with aqueous NaOH.
27.04: Phospholipids
Just as waxes, fats, and oils are esters of carboxylic acids, phospholipids are esters of phosphoric acid, H3PO4.
Phospholipids are of two general types: glycerophospholipids and sphingomyelins. Glycerophospholipids are based on phosphatidic acid, which contains a glycerol backbone linked by ester bonds to two fatty acids and one phosphoric acid. Although the fatty-acid residues can be any of the C12–C20 units typically present in fats, the acyl group at C1 is usually saturated and the one at C2 is usually unsaturated. The phosphate group at C3 is also bonded to an amino alcohol such as choline [HOCH2CH2N(CH3)3]+, ethanolamine (HOCH2CH2NH2), or serine [HOCH2CH(NH2)CO2H]. The compounds are chiral and have an L, or R, configuration at C2.
Sphingomyelins are the second major group of phospholipids. These compounds have sphingosine or a related dihydroxyamine as their backbone and are particularly abundant in brain and nerve tissue, where they are a major constituent of the coating around nerve fibers.
Phospholipids are found widely in both plant and animal tissues and make up approximately 50% to 60% of cell membranes. Because they are like soaps in having a long, nonpolar hydrocarbon tail bound to a polar ionic head, phospholipids in the cell membrane organize into a lipid bilayer about 5.0 nm (50 Å) thick. As shown in Figure 27.3, the nonpolar tails aggregate in the center of the bilayer in much the same way that soap tails aggregate in the center of a micelle. This bilayer serves as an effective barrier to the passage of water, ions, and other components into and out of cells.
27.05: Prostaglandins and Other Eicosanoids
Prostaglandins are a group of C20 lipids that contain a five-membered ring with two long side chains. First isolated in the 1930s by Ulf von Euler at the Karolinska Institute in Sweden, much of the structural and chemical work on prostaglandins was carried out by Sune Bergström and Bengt Samuelsson. All three shared Nobel Prizes for their work. The name prostaglandin derives from the fact that the compounds were first isolated from sheep prostate glands, but they have subsequently been found to be present in small amounts in all body tissues and fluids.
The several dozen known prostaglandins have an extraordinarily wide range of biological effects. Among their many properties, they can lower blood pressure, affect blood platelet aggregation during clotting, lower gastric secretions, control inflammation, affect kidney function, affect reproductive systems, and stimulate uterine contractions during childbirth.
Prostaglandins, together with related compounds called thromboxanes and leukotrienes, make up a class of compounds called eicosanoids because they are derived biologically from 5,8,11,14-eicosatetraenoic acid, or arachidonic acid (Figure 27.4). Prostaglandins (PG) have a cyclopentane ring with two long side chains; thromboxanes (TX) have a six-membered, oxygen-containing ring; and leukotrienes (LT) are acyclic.
Eicosanoids are named based on their ring system (PG, TX, or LT), substitution pattern, and number of double bonds. The various substitution patterns on the ring are indicated by letter as in Figure 27.5, and the number of double bonds is indicated by a subscript. Thus, PGE1 is a prostaglandin with the “E” substitution pattern and one double bond. The numbering of the atoms in the various eicosanoids is the same as in arachidonic acid, starting with the –CO2H carbon as C1, continuing around the ring, and ending with the –CH3 carbon at the other end of the chain as C20.
Eicosanoid biosynthesis begins with the conversion of arachidonic acid to PGH2, catalyzed by the multifunctional PGH synthase (PGHS), also called cyclooxygenase (COX). There are two distinct enzymes, PGHS-1 and PGHS-2 (or COX-1 and COX-2), both of which accomplish the same reaction but appear to function independently. COX-1 carries out the normal physiological production of prostaglandins, and COX-2 produces additional prostaglandin in response to arthritis or other inflammatory conditions. Vioxx, Bextra, and several other drugs selectively inhibit the COX-2 enzyme but cause potentially serious heart and gastrointestinal problems in weakened patients. (See the Chapter 15 Chemistry Matters.)
PGHS accomplishes two transformations, an initial reaction of arachidonic acid with O2 to yield PGG2 and a subsequent reduction of the hydroperoxide group (–OOH) to the alcohol PGH2. The sequence of steps involved in these transformations was shown in Figure 8.12.
Further processing of PGH2 leads to other eicosanoids. PGE2, for instance, arises by an isomerization of PGH2 catalyzed by PGE synthase (PGES) (Figure 27.6).
Problem 27-5
Assign R or S configuration to each chirality center in prostaglandin E2 (Figure 27.6), the most abundant and biologically potent of mammalian prostaglandins. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.03%3A_Soap.txt |
We saw in the Chapter 8 Chemistry Matters that terpenoids are a vast and diverse group of lipids found in all living organisms. Despite their apparent structural differences, all terpenoids contain a multiple of five carbons and are derived biosynthetically from the five-carbon precursor isopentenyl diphosphate (Figure 27.7). Although formally a terpenoid contains oxygen, while a hydrocarbon is called a terpene, we’ll use the term terpenoid to refer to both for simplicity.
Figure 27.7 Structures of some representative terpenoids.
You might recall from the chapter on Alkenes: Reactions and Synthesis that terpenoids are classified according to the number of five-carbon multiples they contain. Monoterpenoids contain 10 carbons and are derived from two isopentenyl diphosphates, sesquiterpenoids contain 15 carbons and are derived from three isopentenyl diphosphates, diterpenoids contain 20 carbons and are derived from four isopentenyl diphosphates, and so on, up to triterpenoids (C30) and tetraterpenoids (C40). Lanosterol, for instance, is a triterpenoid from which steroid hormones are made, and β-carotene is a tetraterpenoid that serves as a dietary source of vitamin A (Figure 27.7).
The terpenoid precursor isopentenyl diphosphate, formerly called isopentenyl pyrophosphate and thus abbreviated IPP, is biosynthesized by two different pathways, depending on the organism and the structure of the final product. In animals and higher plants, sesquiterpenoids and triterpenoids arise primarily from the mevalonate pathway, whereas monoterpenoids, diterpenoids, and tetraterpenoids are biosynthesized by the 1-deoxyxylulose 5-phosphate (DXP) pathway, also called the methylerithritol phosphate, or MEP, pathway. In bacteria, both pathways are used. We’ll look only at the mevalonate pathway, which is more common and better understood at present.
The Mevalonate Pathway to Isopentenyl Diphosphate
As shown in Figure 27.8, the mevalonate pathway begins with the conversion of acetate to acetyl CoA, followed by Claisen condensation to yield acetoacetyl CoA. A second carbonyl condensation reaction with a third molecule of acetyl CoA, this one an aldol-like process, then yields the six-carbon compound 3-hydroxy-3-methylglutaryl CoA, which is reduced to give mevalonate. Phosphorylation, followed by loss of CO2 and phosphate ion, completes the process.
Step 1 of Figure 27.8: Claisen Condensation
The first step in mevalonate biosynthesis is a Claisen condensation to yield acetoacetyl CoA, a reaction catalyzed by acetoacetyl-CoA acetyltransferase. An acetyl group is first bound to the enzyme by a nucleophilic acyl substitution reaction with a cysteine –SH group. Formation of an enolate ion from a second molecule of acetyl CoA, followed by Claisen condensation, then yields the product.
Step 2 of Figure 27.8: Aldol Condensation
Acetoacetyl CoA next undergoes an aldol-like addition of an acetyl CoA enolate ion in a reaction catalyzed by 3-hydroxy-3-methylglutaryl-CoA synthase. The reaction occurs by initial binding of the substrate to a cysteine –SH group in the enzyme, followed by enolate-ion addition and subsequent hydrolysis to give (3S)-3-hydroxy-3-methylglutaryl CoA (HMG-CoA).
Figure 27.8 MECHANISM
The mevalonate pathway for the biosynthesis of isopentenyl diphosphate from three molecules of acetyl CoA. Individual steps are explained in the text.
Step 3 of Figure 27.8: Reduction
Reduction of HMG-CoA to give (R)-mevalonate is catalyzed by 3-hydroxy-3-methylglutaryl-CoA reductase and requires 2 equivalents of reduced nicotinamide adenine dinucleotide phosphate (NADPH), a close relative of NADH (Section 19.12). The reaction occurs in two steps and proceeds through an aldehyde intermediate. The first step is a nucleophilic acyl substitution reaction involving hydride transfer from NADPH to the thioester carbonyl group of HMG-CoA. Following expulsion of HSCoA as leaving group, the aldehyde intermediate undergoes a second hydride addition to give mevalonate.
Step 4 of Figure 27.8: Phosphorylation and Decarboxylation
Three additional reactions are needed to convert mevalonate to isopentenyl diphosphate. The first two are straightforward phosphorylations by ATP that occur through nucleophilic substitution reactions on the terminal phosphorus. Mevalonate is first converted to mevalonate 5-phosphate (phosphomevalonate) by reaction with ATP; mevalonate 5-phosphate then reacts with a second ATP to give mevalonate 5-diphosphate (diphosphomevalonate). The third reaction results in phosphorylation of the tertiary hydroxyl group, followed by decarboxylation and loss of phosphate ion.
The final decarboxylation of mevalonate 5-diphosphate seems unusual because decarboxylations of acids do not typically occur except in β-keto acids and malonic acids, in which the carboxylate group is two atoms away from an additional carbonyl group. As discussed in Section 22.7, the function of this second carbonyl group is to act as an electron acceptor and stabilize the charge resulting from loss of CO2. In fact, though, the decarboxylation of a β-keto acid and the decarboxylation of mevalonate 5-diphosphate are closely related.
Catalyzed by mevalonate-5-diphosphate decarboxylase, the substrate is first phosphorylated on the free –OH group by reaction with ATP to give a tertiary phosphate, which undergoes spontaneous SN1-like dissociation to give a tertiary carbocation. The positive charge then acts as an electron acceptor to facilitate decarboxylation in the same way a β carbonyl group does, giving isopentenyl diphosphate. (In the following structures, the diphosphate group is abbreviated OPP.)
Problem 27-6
The conversion of mevalonate 5-phosphate to isopentenyl diphosphate occurs with the following result. Which hydrogen, pro-R or pro-S, ends up cis to the methyl group, and which ends up trans?
Conversion of Isopentenyl Diphosphate to Terpenoids
The conversion of isopentenyl diphosphate (IPP) to terpenoids begins with its isomerization to dimethylallyl diphosphate, abbreviated DMAPP and formerly called dimethylallyl pyrophosphate. These two C5 building blocks then combine to give the C10 unit geranyl diphosphate (GPP). The corresponding alcohol, geraniol, is itself a fragrant terpenoid that occurs in rose oil.
Further combination of GPP with another IPP gives the C15 unit farnesyl diphosphate (FPP), and so on, up to C25. Terpenoids with more than 25 carbons—that is, triterpenoids (C30) and tetraterpenoids (C40)—are synthesized by dimerization of C15 and C20 units, respectively (Figure 27.9). Triterpenoids and steroids, in particular, arise from dimerization of farnesyl diphosphate to give squalene.
Figure 27.9 An overview of terpenoid biosynthesis from isopentenyl diphosphate.
The isomerization of isopentenyl diphosphate to dimethylallyl diphosphate is catalyzed by IPP isomerase and occurs through a carbocation pathway. Protonation of the IPP double bond by a hydrogen-bonded cysteine residue in the enzyme gives a tertiary carbocation intermediate, which is deprotonated by a glutamate residue as base to yield DMAPP. X-ray structural studies on the enzyme show that it holds the substrate in an unusually deep, well-protected pocket to shield the highly reactive carbocation from reaction with solvent or other external substances.
Both the initial coupling of DMAPP with IPP to give geranyl diphosphate and the subsequent coupling of GPP with a second molecule of IPP to give farnesyl diphosphate are catalyzed by farnesyl diphosphate synthase. The process requires Mg2+ ion, and the key step is a nucleophilic substitution reaction in which the double bond of IPP behaves as a nucleophile in displacing diphosphate ion leaving group (PPi) or DMAPP. Evidence suggests that the DMAPP develops a considerable cationic character and that spontaneous dissociation of the allylic diphosphate ion in an SN1-like pathway probably occurs (Figure 27.10).
Figure 27.10 Mechanism of the coupling reaction of dimethylallyl diphosphate (DMAPP) and isopentenyl diphosphate (IPP), to give geranyl diphosphate (GPP).
Further conversion of geranyl diphosphate into monoterpenoids typically involves carbocation intermediates and multistep reaction pathways that are catalyzed by terpene cyclases. Monoterpene cyclases function by first isomerizing geranyl diphosphate to its allylic isomer linalyl diphosphate (LPP), a process that occurs by spontaneous SN1-like dissociation to an allylic carbocation, followed by recombination. The effect of this isomerization is to convert the C2–C3 double bond of GPP into a single bond, thereby making cyclization possible and allowing E/Z isomerization of the double bond.
Further dissociation and cyclization by electrophilic addition of the cationic carbon to the terminal double bond then gives a cyclic cation, which might either rearrange, undergo a hydride shift, be captured by a nucleophile, or be deprotonated to give any of the several hundred known monoterpenoids. As just one example, limonene, a monoterpenoid found in many citrus oils, arises by the biosynthetic pathway shown in Figure 27.11.
Figure 27.11 Mechanism for the formation of the monoterpenoid limonene from geranyl diphosphate.
Worked Example 27.1
Proposing a Terpenoid Biosynthesis Pathway
Propose a mechanistic pathway for the biosynthesis of α-terpineol from geranyl diphosphate.
Strategy
α-Terpineol, a monoterpenoid, must be derived biologically from geranyl diphosphate through its isomer linalyl diphosphate. Draw the precursor in a conformation that approximates the structure of the target molecule, and then carry out a cationic cyclization, using the appropriate double bond to displace the diphosphate leaving group. Since the target is an alcohol, the carbocation resulting from cyclization evidently reacts with water.
Solution
Problem 27-7 Propose mechanistic pathways for the biosynthetic formation of the following terpenoids:
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.06%3A_Terpenoids.txt |
In addition to fats, phospholipids, eicosanoids, and terpenoids, the lipid extracts of plants and animals also contain steroids, molecules that are derived from the triterpenoid lanosterol (Figure 27.7) and whose structures are based on a tetracyclic ring system. The four rings are designated A, B, C, and D, beginning at the lower left. The three 6-membered rings (A, B, and C) adopt chair conformations but are prevented by their rigid geometry from undergoing the usual cyclohexane ring-flips (Section 4.6).
Two cyclohexane rings can be joined in either a cis or a trans manner. With cis fusion to give cis-decalin, both groups at the ring-junction positions (the angular groups) are on the same side of the two rings. With trans fusion to give trans-decalin, the groups at the ring junctions are on opposite sides.
As shown in Figure 27.12, steroids can have either a cis or a trans fusion of the A and B rings, but the other ring fusions (B–C and C–D) are usually trans.
Substituent groups on the steroid ring system can be either axial or equatorial. As with simple cyclohexanes, equatorial substitution is generally more favorable than axial substitution for steric reasons. The hydroxyl group of cholesterol, for example, has the more stable equatorial orientation.
Problem 27-8 Draw the following molecules in chair conformations, and tell whether the ring substituents are axial or equatorial: (a)
(b)
Problem 27-9
Lithocholic acid is an A–B cis steroid found in human bile. Draw lithocholic acid showing chair conformations, as in Figure 27.12, and tell whether the hydroxyl group at C3 is axial or equatorial.
Steroid Hormones
In humans, most steroids function as hormones, chemical messengers that are secreted by endocrine glands and carried through the bloodstream to target tissues. There are two main classes of steroid hormones: the sex hormones, which control maturation, tissue growth, and reproduction, and the adrenocortical hormones, which regulate a variety of metabolic processes.
Sex Hormones
Testosterone and androsterone are the two most important male sex hormones, or androgens. Androgens are responsible for the development of male secondary sex characteristics during puberty and for promoting tissue and muscle growth. Both are synthesized in the testes from cholesterol. Androstenedione is another minor hormone that has received particular attention because of its use by prominent athletes.
Estrone and estradiol are the two most important female sex hormones, or estrogens. Synthesized in the ovaries from testosterone, estrogenic hormones are responsible for the development of female secondary sex characteristics and for regulation of the menstrual cycle. Note that both have a benzene-like aromatic A ring. In addition, another kind of sex hormone called a progestin is essential in preparing the uterus for implantation of a fertilized ovum during pregnancy. Progesterone is the most important progestin.
Adrenocortical Hormones
Adrenocortical steroids are secreted by the adrenal glands, small organs located near the upper end of each kidney. There are two types of adrenocortical steroids, called mineralocorticoids and glucocorticoids. Mineralocorticoids, such as aldosterone, control tissue swelling by regulating cellular salt balance between Na+ and K+. Glucocorticoids, such as hydrocortisone, are involved in the regulation of glucose metabolism and in the control of inflammation. Glucocorticoid ointments are widely used to bring down the swelling from exposure to poison oak or poison ivy.
Synthetic Steroids
In addition to the hundreds of steroids isolated from plants and animals, many more have been synthesized in pharmaceutical laboratories in the search for new drugs. Among the best-known synthetic steroids are oral contraceptives and anabolic agents. Most birth control pills are a mixture of two compounds, a synthetic estrogen, such as ethynylestradiol, and a synthetic progestin, such as norethindrone. Anabolic steroids, such as methandrostenolone (Dianabol), are synthetic androgens that mimic the tissue-building effects of natural testosterone. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.07%3A_Steroids.txt |
Steroids are heavily modified triterpenoids that are biosynthesized in living organisms from farnesyl diphosphate (C15). A reductive dimerization first converts farnesyl diphosphate to the acyclic hydrocarbon squalene (C30), which is converted into lanosterol (Figure 27.13). Further rearrangements and degradations then take place to yield various steroids. The conversion of squalene to lanosterol is among the most intensively studied of biosynthetic transformations. Starting from an achiral, open-chain polyene, the entire process requires only two enzymes and results in the formation of six carbon–carbon bonds, four rings, and seven chirality centers.
Lanosterol biosynthesis begins with the selective epoxidation of squalene to give (3S)-2,3-oxidosqualene, catalyzed by squalene epoxidase. Molecular O2 provides the epoxide oxygen atom, and NADPH is required, along with a flavin coenzyme. The proposed mechanism involves reaction of FADH2 with O2 to produce a flavin hydroperoxide intermediate (ROOH), which transfers an oxygen to squalene in a pathway initiated by nucleophilic attack of the squalene double bond on the terminal hydroperoxide oxygen (Figure 27.14). The flavin alcohol formed as a by-product loses H2O to give FAD, which is reduced back to FADH2 by NADPH. As noted in Section 8.7, this biological epoxidation mechanism is closely analogous to the mechanism by which peroxyacids (RCO3H) react with alkenes to give epoxides in the laboratory.
The second part of lanosterol biosynthesis is catalyzed by oxidosqualene-lanosterol cyclase and occurs as shown in Figure 27.15. Squalene is folded by the enzyme into a conformation that aligns the various double bonds for a cascade of successive intramolecular electrophilic additions, followed by a series of hydride and methyl migrations. Except for the initial epoxide protonation/cyclization, the process is probably stepwise and appears to involve discrete carbocation intermediates that are stabilized by electrostatic interactions with electron-rich aromatic amino acids in the enzyme.
Steps 1, 2 of Figure 27.15: Epoxide Opening and Initial Cyclizations
Cyclization begins in step 1 with protonation of the epoxide ring by an aspartic acid residue in the enzyme. Nucleophilic opening of the protonated epoxide by the nearby 5,10 double bond (steroid numbering; Section 27.6) then yields a tertiary carbocation at C10. Further addition of C10 to the 8,9 double bond in step 2 next gives a bicyclic tertiary cation at C8.
Step 3 of Figure 27.15: Third Cyclization
The third cationic cyclization is somewhat unusual because it occurs with non-Markovnikov regiochemistry and gives a secondary cation at C13 rather than the alternative tertiary cation at C14. There is growing evidence, however, that the tertiary carbocation may in fact be formed initially and that the secondary cation arises by subsequent rearrangement. The secondary cation is probably stabilized in the enzyme pocket by the proximity of an electron-rich aromatic ring.
Figure 27.15 MECHANISM Mechanism of the conversion of 2,3-oxidosqualene to lanosterol. Four cationic cyclizations are followed by four rearrangements and a final loss of H+ from C9. The steroid numbering system is used for referring to specific positions in the intermediates (Section 27.6). Individual steps are explained in the text.
Step 4 of Figure 27.15: Final Cyclization
The fourth and last cyclization occurs in step 4 by addition of the cationic center at C13 to the 17,20 double bond, giving what is known as the protosteryl cation. The side-chain alkyl group at C17 has β (up) stereochemistry, although this stereochemistry is lost in step 5 and then reset in step 6.
Steps 5–9 of Figure 27.15: Carbocation Rearrangements
Once the tetracyclic carbon skeleton of lanosterol has been formed, a series of carbocation rearrangements occur (Section 7.11). The first rearrangement, hydride migration from C17 to C20, occurs in step 5 and results in establishment of R stereochemistry at C20 in the side chain. In step 6, a second hydride migration occurs from C13 to C17 on the α (bottom) face of the ring and reestablishes the 17β orientation of the side chain. Finally, two methyl migrations, the first from C14 to C13 on the top (β) face and the second from C8 to C14 on the bottom (α) face, place the positive charge at C8. A basic histidine residue in the enzyme then removes the neighboring β proton from C9 to give lanosterol.
From lanosterol, the pathway for steroid biosynthesis continues on to yield cholesterol. Cholesterol then becomes a branch point, serving as the common precursor from which all other steroids are derived.
Problem 27-10
Compare the structures of lanosterol and cholesterol, and catalog the changes needed for the transformation shown. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.08%3A_Biosynthesis_of_Steroids.txt |
27 • Chemistry Matters 27 • Chemistry Matters
We hear a lot these days about the relationship between saturated fats, cholesterol, and heart disease. What are the facts? It’s well established that a diet rich in saturated animal fats often leads to an increase in blood serum cholesterol, particularly in sedentary, overweight individuals. Conversely, a diet lower in saturated fats and higher in polyunsaturated fats leads to a lower serum cholesterol level. Studies have shown that a serum cholesterol level greater than 240 mg/dL (a desirable value is < 200 mg/dL) is correlated with an increased incidence of coronary artery disease, in which cholesterol deposits build up on the inner walls of coronary arteries, blocking the flow of blood to the heart muscles.
A better indication of a person’s risk of heart disease comes from a measurement of blood lipoprotein levels. Lipoproteins are complex molecules with both lipid and protein components that transport lipids through the body. They can be divided into three types according to density, as shown in Table 27.3. Very-low-density lipoproteins (VLDLs) act primarily as carriers of triglycerides from the intestines to peripheral tissues, whereas low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs) act as carriers of cholesterol to and from the liver.
Evidence suggests that LDLs transport cholesterol as its fatty-acid ester to peripheral tissues, whereas HDLs remove cholesterol as its stearate ester from dying cells. If LDLs deliver more cholesterol than is needed, and if insufficient HDLs are present to remove it, the excess is deposited in arteries. Thus, a low level of low-density lipoproteins is good because it means that less cholesterol is being transported, and a high level of high-density lipoproteins is good because it means that more cholesterol is being removed. In addition, HDL contains an enzyme that has antioxidant properties, offering further protection against heart disease.
As a rule of thumb, a person’s risk drops about 25% for each increase of 5 mg/dL in HDL concentration. Normal values are >40 mg/dL for men and <50 mg/dL for women, perhaps explaining why premenopausal women appear to be somewhat less susceptible than men to heart disease.
Table 27.3 Serum Lipoproteins
Name Density (g/mL) % Lipid % Protein Optimal (mg/dL) Poor (mg/dL)
VLDL 0.930–1.006 90 10
LDL 1.019–1.063 75 25 <100 >130
HDL 1.063–1.210 67 33 >60 <40
Not surprisingly, the most important factor in gaining high HDL levels is a generally healthful lifestyle. Obesity, smoking, and lack of exercise lead to low HDL levels, whereas regular exercise and a sensible diet lead to high HDL levels. Distance runners and other endurance athletes have HDL levels nearly 50% higher than the general population. Failing that—some of us, but not everyone, wants to run 30 miles or bike 80 miles per week—diet is also important. Diets high in cold-water fish, like salmon and whitefish, raise HDL and lower blood cholesterol because these fish contain almost entirely polyunsaturated fat, including a large percentage of omega-3 fatty acids. Animal fat from red meat and cooking fats should be minimized because saturated fats and monounsaturated trans fats raise blood cholesterol.
27.10: Key Terms
27 • Key Terms 27 • Key Terms
• anabolic steroid
• androgen
• eicosanoid
• estrogen
• fatty acid
• hormone
• lipid
• lipid bilayer
• lipoprotein
• micelle
• mineralocorticoid
• omega-3 fatty acid
• phospholipid
• polyunsaturated fatty acid
• progestin
• prostaglandin
• saponification
• steroid
• terpenoid
• triacylglycerol
• wax
27.11: Summary
27 • Summary 27 • Summary
Lipids are the naturally occurring materials isolated from plants and animals by extraction with a nonpolar organic solvent. Animal fats and vegetable oils are the most widely occurring lipids. Both are triacylglycerols—triesters of glycerol with long-chain fatty acids. Animal fats are usually saturated, whereas vegetable oils usually have unsaturated fatty acid residues.
Phospholipids are important constituents of cell membranes and are of two kinds. Glycerophospholipids, such as phosphatidylcholine and phosphatidylethanolamine, are closely related to fats in that they have a glycerol backbone esterified to two fatty acids (one saturated and one unsaturated) and to one phosphate ester. Sphingomyelins have the amino alcohol sphingosine for their backbone.
Eicosanoids and terpenoids are still other classes of lipids. Eicosanoids, of which prostaglandins are the most abundant kind, are derived biosynthetically from arachidonic acid, are found in all body tissues, and have a wide range of physiological activity. Terpenoids are often isolated from the essential oils of plants, have an immense diversity of structure, and are produced biosynthetically from the five-carbon precursor isopentenyl diphosphate (IPP). Isopentenyl diphosphate is itself biosynthesized from 3 equivalents of acetate in the mevalonate pathway.
Steroids are plant and animal lipids with a characteristic tetracyclic carbon skeleton. Like the eicosanoids, steroids occur widely in body tissues and have a large variety of physiological activities. Steroids are closely related to terpenoids and arise biosynthetically from the triterpenoid lanosterol. Lanosterol, in turn, arises from cationic cyclization of the acyclic hydrocarbon squalene. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.09%3A_Chemistry_MattersSaturated_Fats_Cholesterol_and_Heart_Disease.txt |
27 • Additional Problems 27 • Additional Problems
Visualizing Chemistry
Problem 27-11
The following model is that of cholic acid, a constituent of human bile. Locate the three hydroxyl groups, and identify each as axial or equatorial. Is cholic acid an A–B trans steroid or an A–B cis steroid?
Problem 27-12
Propose a biosynthetic pathway for the sesquiterpenoid helminthogermacrene from farnesyl diphosphate.
Problem 27-13
Identify the following fatty acid, and tell whether it is more likely to be found in peanut oil or in red meat:
Mechanism Problems
Problem 27-14
Propose a mechanistic pathway for the biosynthesis of caryophyllene, a substance found in clove oil.
Problem 27-15
Suggest a mechanism by which ψ-ionone is transformed into β-ionone on treatment with acid.
Problem 27-16
Isoborneol is converted into camphene on treatment with dilute sulfuric acid. Propose a mechanism for the reaction, which involves a carbocation rearrangement.
Fats, Oils, and Related Lipids
Problem 27-17
Fatty fish like salmon and albacore are rich in omega-3 fatty acids, which have a double bond three carbons in from the noncarboxyl end of the chain and have been shown to lower blood cholesterol levels. Draw the structure of 5,8,11,14,17-eicosapentaenoic acid, a common example. (Eicosane = C20H42.)
Problem 27-18
Fats can be either optically active or optically inactive, depending on their structure. Draw the structure of an optically active fat that yields 2 equivalents of stearic acid and 1 equivalent of oleic acid on hydrolysis. Draw the structure of an optically inactive fat that yields the same products.
Problem 27-19
Spermaceti, a fragrant substance from sperm whales, was widely used in cosmetics until it was banned in 1976 to protect whales from extinction. Chemically, spermaceti is cetyl palmitate, the ester of cetyl alcohol (n-C16H33OH) with palmitic acid. Draw its structure.
Problem 27-20 Show the products you would expect to obtain from reaction of glyceryl trioleate with the following reagents: (a)
Excess Br2 in CH2Cl2
(b) H2/Pd (c) NaOH/H2O (d) O3, then Zn/CH3CO2H (e) LiAlH4, then H3O+ (f) CH3MgBr, then H3O+
Problem 27-21 How would you convert oleic acid into the following substances? (a)
Methyl oleate
(b) Methyl stearate (c) Nonanal (d) Nonanedioic acid (e) 9-Octadecynoic acid (stearolic acid) (f) 2-Bromostearic acid (g) 18-Pentatriacontanone, CH3(CH2)16CO(CH2)16CH3
Problem 27-22
Plasmalogens are a group of lipids found in nerve and muscle cells. How do plasmalogens differ from fats?
Problem 27-23
What products would you obtain from hydrolysis of a plasmalogen (Problem 27-22) with aqueous NaOH? With H3O+?
Problem 27-24
Cardiolipins are a group of lipids found in heart muscles. What products would be formed if all ester bonds, including phosphates, were saponified by treatment with aqueous NaOH?
Problem 27-25
Stearolic acid, C18H32O2, yields stearic acid on catalytic hydrogenation and undergoes oxidative cleavage with ozone to yield nonanoic acid and nonanedioic acid. What is the structure of stearolic acid?
Problem 27-26
How would you synthesize stearolic acid (Problem 27-25) from 1-decyne and 1-chloro-7-iodoheptane?
Terpenoids and Steroids
Problem 27-27 Without proposing an entire biosynthetic pathway, draw the appropriate precursor, either geranyl diphosphate or farnesyl diphosphate, in a conformation that shows a likeness to each of the following terpenoids: (a) (b) Problem 27-28
Indicate by asterisks the chirality centers present in each of the terpenoids shown in Problem 27-27. What is the maximum possible number of stereoisomers for each?
Problem 27-29
Assume that the three terpenoids in Problem 27-27 are derived biosynthetically from isopentenyl diphosphate and dimethylallyl diphosphate, each of which was isotopically labeled at the diphosphate-bearing carbon atom (C1). At what positions would the terpenoids be isotopically labeled?
Problem 27-30
Assume that acetyl CoA containing a 14C isotopic label in the carboxyl carbon atom is used as starting material for the biosynthesis of mevalonate, as shown in Figure 27.8. At what positions in mevalonate would the isotopic label appear?
Problem 27-31
Assume that acetyl CoA containing a 14C isotopic label in the carboxyl carbon atom is used as starting material and that the mevalonate pathway is followed. Identify the positions in α-cadinol where the label would appear.
Problem 27-32
Assume that acetyl CoA containing a 14C isotopic label in the carboxyl carbon atom is used as starting material and that the mevalonate pathway is followed. Identify the positions in squalene where the label would appear.
Problem 27-33
Assume that acetyl CoA containing a 14C isotopic label in the carboxyl carbon atom is used as starting material and that the mevalonate pathway is followed. Identify the positions in lanosterol where the label would appear.
General Problems
Problem 27-34
Flexibilene, a compound isolated from marine coral, is the first known terpenoid to contain a 15-membered ring. What is the structure of the acyclic biosynthetic precursor of flexibilene? Show the mechanistic pathway for the biosynthesis.
Problem 27-35
Draw the most stable chair conformation of dihydrocarvone.
Problem 27-36
Draw the most stable chair conformation of menthol, and label each substituent as axial or equatorial.
Problem 27-37 As a general rule, equatorial alcohols are esterified more readily than axial alcohols. What product would you expect to obtain from reaction of the following two compounds with 1 equivalent of acetic anhydride? (a) (b) Problem 27-38
Propose a mechanistic pathway for the biosynthesis of isoborneol. A carbocation rearrangement is needed at one point in the scheme.
Problem 27-39
Digitoxigenin is a heart stimulant obtained from the purple foxglove Digitalis purpurea and used in the treatment of heart disease. Draw the three-dimensional conformation of digitoxigenin, and identify the two –OH groups as axial or equatorial.
Problem 27-40
What product would you obtain by reduction of digitoxigenin (Problem 27-39) with LiAlH4? By oxidation with the Dess–Martin periodinane?
Problem 27-41
Vaccenic acid, C18H34O2, is a rare fatty acid that gives heptanal and 11-oxoundecanoic acid [OHC(CH2)9CO2H] on ozonolysis followed by zinc treatment. When allowed to react with CH2I2/Zn(Cu), vaccenic acid is converted into lactobacillic acid. What are the structures of vaccenic and lactobacillic acids?
Problem 27-42
Eleostearic acid, C18H30O2, is a rare fatty acid found in the tung oil used for finishing furniture. On ozonolysis followed by treatment with zinc, eleostearic acid furnishes one part pentanal, two parts glyoxal (OHC–CHO), and one part 9-oxononanoic acid [OHC(CH2)7CO2H]. What is the structure of eleostearic acid?
Problem 27-43
Diterpenoids are derived biosynthetically from geranylgeranyl diphosphate (GGPP), which is itself biosynthesized by reaction of farnesyl diphosphate with isopentenyl diphosphate. Show the structure of GGPP, and propose a mechanism for its biosynthesis from FPP and IPP.
Problem 27-44
Diethylstilbestrol (DES) has estrogenic activity even though it is structurally unrelated to steroids. Once used as an additive in animal feed, DES has been implicated as a causative agent in several types of cancer. Show how DES can be drawn so that it is sterically similar to estradiol.
Problem 27-45
Cembrene, C20H32, a diterpenoid hydrocarbon isolated from pine resin, has a UV absorption at 245 nm, but dihydrocembrene (C20H34), the product of hydrogenation with 1 equivalent of H2, has no UV absorption. On exhaustive hydrogenation, 4 equivalents of H2 react, and octahydrocembrene, C20H40, is produced. On ozonolysis of cembrene, followed by treatment of the ozonide with zinc, four carbonyl-containing products are obtained:
Propose a structure for cembrene that is consistent with its formation from geranylgeranyl diphosphate.
Problem 27-46
α-Fenchone is a pleasant-smelling terpenoid isolated from oil of lavender. Propose a pathway for the formation of α-fenchone from geranyl diphosphate. A carbocation rearrangement is required.
Problem 27-47
Fatty acids are synthesized by a multistep route that starts with acetate. The first step is a reaction between protein-bound acetyl and malonyl units to give a protein-bound 3-ketobutyryl unit. Show the mechanism, and tell what kind of reaction is occurring.
Problem 27-48
Propose a mechanism for the biosynthesis of the sesquiterpenoid trichodiene from farnesyl diphosphate. The process involves cyclization to give an intermediate secondary carbocation, followed by several carbocation rearrangements. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids/27.12%3A_Additional_Problems.txt |
Learning Objectives
When you have completed Chapter 28, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. draw the structure of a given nucleotide.
3. discuss the structure of DNA and RNA.
4. describe the processes involved in DNA replication, transcription, translation, and protein synthesis.
5. define, and use in context, the key terms introduced in this chapter.
Two types of nucleic acids are found in cells—deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). These highly complex substances are built up from a number of simpler units, called nucleotides. Each nucleotide consists of three parts: a phosphoric acid residue, a sugar and a nitrogen‑containing heterocyclic base. Thus, in order to understand the biochemistry of the nucleic acids, you must first study the chemistry of the sugars (see Chapter 25) and simple heterocyclic systems. We have already discussed certain aspects of the structure of heterocyclic ring systems during our study of aromaticity (Sections 15.5–15.6). You may find it helpful to review this chapter.
Chapter 28 examines the structure and replication of DNA and then describes the structure and synthesis of RNA. The chapter closes with a brief study of the role played by RNA in the biosynthesis of proteins.
28: Biomolecules - Nucleic Acids
Chapter Contents
28.1 Nucleotides and Nucleic Acids
28.2 Base Pairing in DNA 28.3 Replication of DNA 28.4 Transcription of DNA 28.5 Translation of RNA: Protein Biosynthesis 28.6 DNA Sequencing 28.7 DNA Synthesis 28.8 The Polymerase Chain Reaction
Nucleic acids are the last of the four major classes of biomolecules we’ll consider. So much has been written and spoken about DNA in the media that the basics of DNA replication and transcription are probably known to you. Thus, we’ll move fairly quickly through the fundamentals and then look more closely at the chemical details of DNA sequencing, synthesis, and metabolism. This field is moving rapidly, and there’s a lot you may not be familiar with.
The nucleic acids, deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), are the chemical carriers of a cell’s genetic information. Coded in a cell’s DNA is the information that determines the nature of the cell, controls its growth and division, and directs biosynthesis of the enzymes and other proteins required for cellular functions.
In addition to nucleic acids themselves, nucleic acid derivatives such as ATP are involved as phosphorylating agents in many biochemical pathways, and several important coenzymes, including NAD+, FAD, and coenzyme A, have nucleic acid components. See Table 26.3 for their structures.
28.02: Nucleotides and Nucleic Acids
Just as proteins are biopolymers made of amino acids, the nucleic acids are biopolymers made of nucleotides, joined together to form a long chain. Each nucleotide is composed of a nucleoside bonded to a phosphate group, and each nucleoside is composed of an aldopentose sugar linked through its anomeric carbon to the nitrogen atom of a heterocyclic purine or pyrimidine base.
The sugar component in RNA is ribose, and the sugar in DNA is 2′-deoxyribose. (In naming and numbering nucleotides, numbers with a prime superscript refer to positions on the sugar and numbers without a prime superscript refer to positions on the heterocyclic base. Thus, the prefix 2′-deoxy indicates that oxygen is missing from C2′ of ribose.) DNA contains four different amine bases: two substituted purines (A, adenine, and G, guanine) and two substituted pyrimidines (C, cytosine, and T, thymine). Adenine, guanine, and cytosine also occur in RNA, but thymine is replaced in RNA by a closely related pyrimidine base called U, uracil.
The structures of the four deoxyribonucleotides and the four ribonucleotides are shown in Figure 28.2. Although similar chemically, DNA and RNA differ dramatically in size. Molecules of DNA are enormous, containing as many as 245 million nucleotides and having molecular weights as high as 75 billion. Molecules of RNA, by contrast, are much smaller, containing as few as 21 nucleotides and having molecular weights as low as 7000.
Nucleotides are linked together in DNA and RNA by phosphodiester bonds [RO–(PO2)–OR′] between phosphate, the 5′-hydroxyl group on one nucleoside, and the 3′-hydroxyl group on another nucleoside. One end of the nucleic acid polymer has a free hydroxyl at C3′ (the 3′ end), and the other end has a phosphate at C5′ (the 5′ end). The sequence of nucleotides in a chain is described by starting at the 5′ end and identifying the bases in order of occurrence, using the abbreviations G, C, A, T (or U in RNA). Thus, a typical DNA sequence might be written as TAGGCT.
Problem 28-1
Draw the full structure of the DNA dinucleotide AG.
Problem 28-2
Draw the full structure of the RNA dinucleotide UA. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.01%3A_Why_This_Chapter.txt |
Samples of DNA isolated from different tissues of the same species have the same proportions of heterocyclic bases, but samples from different species often have greatly differing proportions of bases. Human DNA, for example, contains about 30% each of adenine and thymine and about 20% each of guanine and cytosine. The bacterium Clostridium perfringens, however, contains about 37% each of adenine and thymine and only 13% each of guanine and cytosine. Note that in both examples the bases occur in pairs. Adenine and thymine are present in equal amounts, as are cytosine and guanine. Why?
In 1953, Rosalind Franklin, Maurice Wilkins, James Watson, and Francis Crick published scientific reports describing the secondary structure of DNA. According to their model, DNA under physiological conditions consists of two polynucleotide strands, running in opposite directions and coiled around each other in a double helix like the handrails on a spiral staircase. The two strands are complementary rather than identical and are held together by hydrogen bonds between specific pairs of bases, A with T and C with G. That is, whenever an A base occurs in one strand, a T base occurs opposite it in the other strand; when a C base occurs in one, a G occurs in the other (Figure 28.3). This complementary base-pairing thus explains why A and T are always found in equal amounts, as are G and C.
, while the edges have positive and negative regions. Pairing G with C and A with T brings together oppositely charged regions.
A full turn of a DNA double helix is shown in Figure 28.4. The helix is 20 Å wide, there are 10 base pairs per turn, and each turn is 34 Å in length. Notice in Figure 28.4 that the two strands of the double helix coil in such a way that two kinds of “grooves” result, a major groove 12 Å wide and a minor groove 6 Å wide. The major groove is slightly deeper than the minor groove, and both are lined by flat heterocyclic bases. As a result, a variety of other polycyclic aromatic molecules are able to slip sideways, or intercalate, between the stacked bases. Many cancer-causing and cancer-preventing agents function by interacting with DNA in this way.
An organism’s genetic information is stored as a sequence of deoxyribonucleotides strung together in the DNA chain. For the information to be preserved and passed on to future generations, a mechanism must exist for copying DNA. For the information to be used, a mechanism must exist for decoding the DNA message and implementing the instructions it contains.
What Crick called the “central dogma of molecular genetics” says that the function of DNA is to store information and pass it on to RNA. The function of RNA, in turn, is to read, decode, and use the information received from DNA to make proteins. This view is greatly oversimplified but is nevertheless a good place to start. Three fundamental processes take place.
• Replication is the process by which identical copies of DNA are made so that information can be preserved and handed down to offspring.
• Transcription is the process by which genetic messages are read and carried out of the cell nucleus to ribosomes, where protein synthesis occurs.
• Translation is the process by which the genetic messages are decoded and used to synthesize proteins.
Worked Example 28.1
Predicting the Complementary Base Sequence in Double-Stranded DNA
What sequence of bases on one strand of DNA is complementary to the sequence TATGCAT on another strand?
Strategy
Remember that A and G form complementary pairs with T and C, respectively, and then go through the sequence replacing A by T, G by C, T by A, and C by G. Remember also that the 5′ end is on the left and the 3′ end is on the right in the original strand.
Solution
Original: (5′) TATGCAT (3′)
Complement: (3′) ATACGTA (5′) or (5′) ATGCATA (3′)
Problem 28-3
What sequence of bases on one strand of DNA is complementary to the following sequence on another strand?
(5′) GGCTAATCCGT (3′) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.03%3A_Base_Pairing_in_DNA.txt |
DNA replication is an enzyme-catalyzed process that begins with a partial unwinding of the double helix at various points along the chain, brought about by enzymes called helicases. Hydrogen bonds are broken, the two strands separate to form a “bubble,” and bases are exposed. New nucleotides then line up on each strand in a complementary manner, A to T and G to C, and two new strands begin to grow from the ends of the bubble, called the replication forks. Each new strand is complementary to its old template strand, so two identical DNA double helices are produced (Figure 28.5). Because each of the new DNA molecules contains one old strand and one new strand, the process is described as semiconservative replication.
Addition of nucleotides to the growing chain takes place in the 5′ → 3′ direction and is catalyzed by the DNA polymerase enzyme. The key step is the addition of a nucleoside 5′-triphosphate to the free 3′-hydroxyl group of the growing chain with the loss of a diphosphate leaving group.
Because both new DNA strands are synthesized in the 5′ → 3′ direction, they can’t be made in exactly the same way. One new strand must have its 3′ end nearer a replication fork, while the other new strand has its 5′ end nearer the replication fork. What happens is that the complement of the original 5′ → 3′ strand is synthesized continuously in a single piece to give a newly synthesized copy called the leading strand, while the complement of the original 3′ → 5′ strand is synthesized discontinuously in small pieces called Okazaki fragments (named for Tsuneko Okazaki, in recognition of her discovery of them) that are subsequently linked by DNA ligases to form the lagging strand.
The magnitude of the replication process is staggering. The nucleus of every human cell contains 2 copies of 22 chromosomes plus an additional 2 sex chromosomes, for a total of 46. Each chromosome consists of one very large DNA molecule, and the sum of the DNA in each of the two sets of chromosomes is estimated to be 3.0 billion base pairs, or 6.0 billion nucleotides. Despite the size of these enormous molecules, their base sequence is faithfully copied during replication. The entire copying process takes only a few hours and, after proofreading and repair, an error gets through only about once per 10 to 100 billion bases. In fact, only about 60 of these random mutations are passed on from parent to child per human generation.
28.05: Transcription of DNA
As noted previously, RNA is structurally similar to DNA but contains ribose rather than deoxyribose and uracil rather than thymine. RNA has three major types, each of which serves a specific purpose. In addition, there are a number of small RNAs that appear to control a wide variety of important cellular functions. All RNA molecules are much smaller than DNA, and all remain single-stranded rather than double-stranded.
• Messenger RNA (mRNA) carries genetic messages from DNA to ribosomes, small granular particles in the cytoplasm of a cell where protein synthesis takes place.
• Ribosomal RNA (rRNA) complexed with protein provides the physical makeup of the ribosomes.
• Transfer RNA (tRNA) transports amino acids to the ribosomes, where they are joined together to make proteins.
• Small RNAs, also called functional RNAs, have a variety of functions within the cell, including silencing transcription and catalyzing chemical modifications of other RNA molecules.
The genetic information in DNA is contained in segments called genes, each of which consists of a specific nucleotide sequence that encodes a specific protein. The conversion of that information from DNA into proteins begins in the nucleus of cells with the synthesis of mRNA by transcription of DNA. In bacteria, the process begins when RNA polymerase recognizes and binds to a promoter sequence on DNA, typically consisting of around 40 base pairs located upstream (5′) of the transcription start site. Within the promoter are two hexameric consensus sequences, one located 10 base pairs upstream of the start and the second located 35 base pairs upstream.
Following formation of the polymerase–promoter complex, several turns of the DNA double helix untwist, forming a bubble and exposing 14 or so base pairs of the two strands. Appropriate ribonucleotides then line up by hydrogen-bonding to their complementary bases on DNA, bond formation occurs in the 5′ → 3′ direction, the RNA polymerase moves along the DNA chain, and the growing RNA molecule unwinds from DNA (Figure 28.6). At any one time, about 12 base pairs of the growing RNA remain hydrogen-bonded to the DNA template.
Unlike what happens in DNA replication, where both strands are copied, only one of the two DNA strands is transcribed into mRNA. The DNA strand that contains the gene is often called the sense strand, or coding strand, and the DNA strand that gets transcribed to give RNA is called the antisense strand, or noncoding strand. Because the sense strand and the antisense strand in DNA are complementary, and because the DNA antisense strand and the newly formed RNA strand are also complementary, the RNA molecule produced during transcription is a copy of the DNA sense strand. That is, the complement of the complement is the same as the original. The only difference is that the RNA molecule has a U everywhere that the DNA sense strand has a T.
Another part of the picture in vertebrates and flowering plants is that genes are often not continuous segments of the DNA chain. Instead, a gene will begin in one small section of DNA called an exon, then be interrupted by a noncoding section called an intron, and then take up again farther down the chain in another exon. The final mRNA molecule results only after the noncoded sections are cut out of the transcribed mRNA and the remaining pieces are joined together by spliceosome enzymes. The gene for triose phosphate isomerase in maize, for instance, contains eight noncoding introns accounting for approximately 70% of the DNA base pairs and nine coding exons accounting for only 30% of the base pairs.
Problem 28-4
Show how uracil can form strong hydrogen bonds to adenine.
Problem 28-5
What RNA base sequence is complementary to the following DNA base sequence?
(5′) GATTACCGTA (3′)
Problem 28-6
From what DNA base sequence was the following RNA sequence transcribed?
(5′) UUCGCAGAGU (3′) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.04%3A_Replication_of_DNA.txt |
The primary cellular function of mRNA is to direct the biosynthesis of the thousands of diverse peptides and proteins required by an organism—as many as 150,000 in a human. The mechanics of protein biosynthesis take place on ribosomes, small granular particles in the cytoplasm of a cell that consist of about 60% ribosomal RNA and 40% protein.
The specific ribonucleotide sequence in mRNA forms a message that determines the order in which amino acid residues are to be joined. Each “word,” or codon, along the mRNA chain consists of a sequence of three ribonucleotides that is specific for a given amino acid. For example, the series UUC on mRNA is a codon directing incorporation of the amino acid phenylalanine into the growing protein. Of the 43 = 64 possible triplets of the four bases in RNA, 61 code for specific amino acids and 3 code for chain termination. Table 28.1 shows the meaning of each codon.
Table 28.1 Codon Assignments of Base Triplets
Third base (3′ end)
First base (5′ end) Second base U C A G
U U Phe Phe Leu Leu
C Ser Ser Ser Ser
A Tyr Tyr Stop Stop
G Cys Cys Stop Trp
C U Leu Leu Leu Leu
C Pro Pro Pro Pro
A His His Gln Gln
G Arg Arg Arg Arg
A U Ile Ile Ile Met
C Thr Thr Thr Thr
A Asn Asn Lys Lys
G Ser Ser Arg Arg
G U Val Val Val Val
C Ala Ala Ala Ala
A Asp Asp Glu Glu
G Gly Gly Gly Gly
The message embedded in mRNA is read by transfer RNA (tRNA) in a process called translation. There are 61 different tRNAs, one for each of the 61 codons that specify an amino acid. A typical tRNA is single-stranded and has roughly the shape of a cloverleaf, as shown in Figure 28.7. It consists of about 70 to 100 ribonucleotides and is bonded to a specific amino acid by an ester linkage through the 3′ hydroxyl on ribose at the 3′ end of the tRNA. Each tRNA also contains on its middle leaf a segment called an anticodon, a sequence of three ribonucleotides complementary to the codon sequence. For example, the codon sequence UUC present on mRNA is read by a phenylalanine-bearing tRNA having the complementary anticodon base sequence GAA. [Remember that nucleotide sequences are written in the 5′ → 3′ direction, so the sequence in an anticodon must be reversed. That is, the complement to (5′)-UUC-(3′) is (3′)-AAG-(5′), which is written as (5′)-GAA-(3′).]
As each successive codon on mRNA is read, different tRNAs bring the correct amino acids into position for enzyme-mediated transfer to the growing peptide. When synthesis of the proper protein is completed, a “stop” codon signals the end, and the protein is released from the ribosome. This process is illustrated in Figure 28.8.
Worked Example 28.2
Predicting the Amino Acid Sequence Transcribed from DNA
What amino acid sequence is coded by the following segment of a DNA coding strand (sense strand)?
(5′) CTA-ACT-AGC-GGG-TCG-CCG (3′)
Strategy
The mRNA produced during translation is a copy of the DNA coding strand, with each T replaced by U. Thus, the mRNA has the sequence
(5′) CUA-ACU-AGC-GGG-UCG-CCG (3′)
Each set of three bases forms a codon, whose meaning can be found in Table 28.1.
Solution
Leu-Thr-Ser-Gly-Ser-Pro.
Problem 28-7 List codon sequences for the following amino acids: (a)
Ala
(b) Phe (c) Leu (d) Tyr
Problem 28-8 List anticodon sequences on the tRNAs carrying the following amino acids. (a)
Ala
(b) Phe (c) Leu (d) Tyr
Problem 28-9
What amino acid sequence is coded by the following mRNA base sequence?
CUU-AUG-GCU-UGG-CCC-UAA
Problem 28-10
What is the base sequence in the original DNA strand on which the mRNA sequence in Problem 28-9 was made? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.06%3A_Translation_of_RNA_-_Protein_Biosynthesis.txt |
A scientific revolution is now under way in molecular biology, as scientists learn how to manipulate and harness the genetic machinery of organisms. None of the extraordinary advances of the past several decades would have been possible, however, were it not for the discovery in 1977 of methods for sequencing immense DNA chains.
The first step in DNA sequencing is to cleave the enormous chain at known points to produce smaller, more manageable pieces, a task accomplished by the use of restriction endonucleases, or restriction enzymes. Each different restriction enzyme, of which more than 4000 are known and approximately 600 are commercially available, cleaves a DNA molecule at a point in the chain where a specific base sequence occurs. For example, the restriction enzyme AluI cleaves between G and C in the four-base sequence AG-CT. Note that the sequence is a palindrome, meaning that the sequence (5′)-AGCT-(3′) is the same as its complement (3′)-TCGA-(5′) when both are read in the same 5′ → 3′ direction. The same is true for other restriction endonucleases.
If the original DNA molecule is cut with another restriction enzyme having a different specificity for cleavage, still other segments are produced whose sequences partially overlap those produced by the first enzyme. Sequencing all the segments, followed by identification of the overlapping regions, allows for complete DNA sequencing.
A dozen or so different methods of DNA sequencing are now available, and many others are under development. The Sanger dideoxy method is among the most frequently used and was the method responsible for first sequencing the entire human genome of 3.0 billion base pairs. In commercial sequencing instruments, the dideoxy method begins with a mixture of the following:
• The restriction fragment to be sequenced
• A small piece of DNA called a primer, whose sequence is complementary to that on the 3′ end of the restriction fragment
• The four 2′-deoxyribonucleoside triphosphates (dNTPs)
• Very small amounts of the four 2′,3′-dideoxyribonucleoside triphosphates (ddNTPs), each of which is labeled with a fluorescent dye of a different color (A 2′,3′-dideoxyribonucleoside triphosphate is one in which both 2′ and 3′ –OH groups are missing from ribose.)
DNA polymerase is added to the mixture, and a strand of DNA complementary to the restriction fragment begins to grow from the end of the primer. Most of the time, only normal deoxyribonucleotides are incorporated into the growing chain because of their much higher concentration in the mixture, but every so often, a dideoxyribonucleotide is incorporated. When that happens, DNA synthesis stops because the chain end no longer has a 3′-hydroxyl group for adding further nucleotides.
When reaction is complete, the product consists of a mixture of DNA fragments of all possible lengths, each terminated by one of the four dye-labeled dideoxyribonucleotides. This product mixture is then separated according to the size of the pieces by gel electrophoresis (Section 26.2), and the identity of the terminal dideoxyribonucleotide in each piece—and thus the sequence of the restriction fragment—is determined by noting the color with which the attached dye fluoresces. Figure 28.9 shows a typical result.
So efficient is the automated dideoxy method that sequences up to 1100 nucleotides in length, with a throughput of up to 19,000 bases per hour, can be sequenced with 98% accuracy. After a decade of work and a cost of about \$500 million, preliminary sequence information for the entire human genome of 3.0 billion base pairs was announced early in 2001 and complete information was released in 2003. More recently, the genome sequencing of individuals, including that of James Watson, one of the discoverers of the double helix, has been accomplished. The sequencing price per genome is dropping rapidly and is currently less than \$1,000, meaning that the routine sequencing of individuals is within reach.
Remarkably, our genome appears to contain only about 21,000 genes, less than one-fourth the previously predicted number and only about twice the number found in the common roundworm. It’s also interesting to note that the number of genes in a human (21,000) is much smaller than the number of kinds of proteins (perhaps 500,000). This discrepancy arises because most proteins are modified in various ways after translation (posttranslational modifications), so a single gene can ultimately give many different proteins. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.07%3A_DNA_Sequencing.txt |
The ongoing revolution in molecular biology has brought with it an increased demand for the efficient chemical synthesis of short DNA segments, called oligonucleotides, or simply oligos. The problems of DNA synthesis are similar to those of peptide synthesis (Section 26.7) but are more difficult because of the complexity of the nucleotide monomers. Each nucleotide has multiple reactive sites that must be selectively protected and deprotected at specific times, and coupling of the four nucleotides must be carried out in the proper sequence. Automated DNA synthesizers are available, however, that allow the fast and reliable synthesis of DNA segments up to 200 nucleotides in length.
DNA synthesizers operate on a principle similar to that of the Merrifield solid-phase peptide synthesizer (Section 26.8). In essence, a protected nucleotide is covalently bonded to a solid support, and one nucleotide at a time is added to the growing chain by the use of a coupling reagent. After the final nucleotide has been added, all the protecting groups are removed and the synthetic DNA is cleaved from the solid support. Five steps are needed:
STEP 1
The first step in DNA synthesis is to attach a protected deoxynucleoside to a silica (SiO2) support by an ester linkage to the 3′ –OH group of the deoxynucleoside. Both the 5′ –OH group on the sugar and free –NH2 groups on the heterocyclic bases must be protected. Adenine and cytosine bases are protected by benzoyl groups, guanine is protected by an isobutyryl group, and thymine requires no protection. The deoxyribose 5′ –OH is protected as its p-dimethoxytrityl (DMT) ether.
STEP 2
The second step is removal of the DMT protecting group by treatment with dichloroacetic acid in CH2Cl2. The reaction occurs by an SN1 mechanism and proceeds rapidly because of the stability of the tertiary, benzylic dimethoxytrityl cation.
STEP 3
The third step is the coupling of the polymer-bonded deoxynucleoside with a protected deoxynucleoside containing a phosphoramidite group [R2NP(OR)2] at its 3′ position. The coupling reaction takes place in the polar aprotic solvent acetonitrile, requires catalysis by the heterocyclic amine tetrazole, and yields a phosphite, P(OR)3, as product. Note that one of the phosphorus oxygen atoms is protected by a β-cyanoethyl group, $–OCH2CH2C≡N–OCH2CH2C≡N$. The coupling step takes place with better than 99% yield.
STEP 4
With the coupling accomplished, the phosphite product is oxidized to a phosphate by treatment with iodine in aqueous tetrahydrofuran in the presence of 2,6-dimethylpyridine. The cycle of (1) deprotection, (2) coupling, and (3) oxidation is then repeated until an oligonucleotide chain of the desired sequence has been constructed.
STEP 5
The final step is removal of all protecting groups and cleavage of the ester bond holding the DNA to the silica. All these reactions are done at the same time by treatment with aqueous NH3. Purification by electrophoresis then yields the synthetic DNA.
Problem 28-11
p-Dimethoxytrityl (DMT) ethers are easily cleaved by mild acid treatment. Show the mechanism of the cleavage reaction.
Problem 28-12
Propose a mechanism to account for cleavage of the β-cyanoethyl protecting group from the phosphate groups on treatment with aqueous ammonia. (Acrylonitrile, H2C$\text{═}$CHCN, is a by-product.) What kind of reaction is occurring?
28.09: The Polymerase Chain Reaction
It often happens that only a tiny amount of DNA can be obtained directly, as might occur at a crime scene, so methods for obtaining larger amounts are sometimes needed to carry out sequencing and characterization. The invention of the polymerase chain reaction (PCR) by Kary Mullis in 1986 has been described as being to genes what Gutenberg’s invention of the printing press was to the written word. Just as the printing press produces multiple copies of a book, PCR produces multiple copies of a given DNA sequence. Starting from less than 1 picogram (pg) of DNA with a chain length of 10,000 nucleotides (1 pg = 10–12 g; about 105 molecules), PCR makes it possible to obtain several micrograms (1 μg = 10–6 g; about 1011 molecules) in just a few hours.
The key to the polymerase chain reaction is Taq DNA polymerase, a heat-stable enzyme isolated from the thermophilic bacterium Thermus aquaticus found in a hot spring in Yellowstone National Park. Taq polymerase is able to take a single strand of DNA having a short, primer segment of complementary chain at one end and then finish constructing the entire complementary strand. The overall process takes three steps, as shown in Figure 28.10. More recently, improved heat-stable DNA polymerases have become available, including Vent polymerase and Pfu polymerase, both isolated from bacteria growing near geothermal vents in the ocean floor. The error rate of both enzymes is substantially less than that of Taq.
STEP 1
The double-stranded DNA to be amplified is heated in the presence of Taq polymerase, Mg2+ ion, the four deoxynucleotide triphosphate monomers (dNTPs), and a large excess of two short oligonucleotide primers of about 20 bases each. Each primer is complementary to the sequence at the end of one of the target DNA segments. At a temperature of 95 °C, double-stranded DNA denatures, spontaneously breaking apart into two single strands.
STEP 2
The temperature is lowered to between 37 and 50 °C, allowing the primers, because of their relatively high concentration, to anneal by hydrogen-bonding to their complementary sequence at the end of each target strand.
STEP 3
The temperature is then raised to 72 °C, and Taq polymerase catalyzes the addition of further nucleotides to the two primed DNA strands. When replication of each strand is complete, two copies of the original DNA now exist. Repeating the denature–anneal–synthesize cycle a second time yields four DNA copies, repeating a third time yields eight copies, and so on, in an exponential series.
PCR has been automated, and 30 or so cycles can be carried out in an hour, resulting in a theoretical amplification factor of 230 (∼109). In practice, however, the efficiency of each cycle is less than 100%, and an experimental amplification of about 106 to 108 is routinely achieved for 30 cycles. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.08%3A_DNA_Synthesis.txt |
28 • Chemistry Matters 28 • Chemistry Matters
The invention of DNA sequencing has affected society in many ways, few more dramatic than those stemming from the development of DNA fingerprinting. DNA fingerprinting arose from the discovery in 1984 that human genes contain short, repeating sequences of noncoding DNA, called short tandem repeat (STR) loci. Furthermore, the STR loci are slightly different for everyone except identical twins. By sequencing these loci, a pattern unique to each person can be obtained.
Perhaps the most common and well-publicized use of DNA fingerprinting is that carried out by crime laboratories to link suspects to biological evidence—blood, hair follicles, skin, or semen—found at a crime scene. Many thousands of court cases have been decided based on DNA evidence.
For use in criminal cases, forensic laboratories in the United States have agreed on 13 core STR loci that are most accurate for the identification of an individual. Based on these 13 loci, a Combined DNA Index System (CODIS) has been established to serve as a registry of convicted offenders. When a DNA sample is obtained from a crime scene, the sample is subjected to cleavage with restriction endonucleases to cut out fragments containing the STR loci, the fragments are amplified using the polymerase chain reaction, and the sequences of the fragments are determined.
If the profile of sequences from a known individual and the profile from DNA obtained at a crime scene match, the probability is approximately 82 billion to 1 that the DNA is from the same individual. In paternity cases, where the DNA of male parent and offspring are related but not fully identical, the identity of the male parent can be established with a probability of around 100,000 to 1. Even after several generations, paternity can still be inferred from DNA analysis of the Y chromosome of direct male-line descendants. The most well-known such case is that of Thomas Jefferson, one of the nation's founders, who likely impregnated enslaved person Sally Hemings. Although Jefferson himself has no male-line descendants, DNA analysis of the male-line descendants of Jefferson’s paternal uncle contained the same Y chromosome as a male-line descendant of Eston Hemings, the youngest son of Sally Hemings. Thus, a mixing of the two genomes is clear, although the male individual responsible for that mixing can’t be identified with 100% certainty.
Among its many other applications, DNA fingerprinting is widely used for the diagnosis of genetic disorders, both prenatally and in newborns. Cystic fibrosis, hemophilia, Huntington’s disease, Tay–Sachs disease, sickle cell anemia, and thalassemia are among the many diseases that can be detected, enabling early treatment of an affected child. Furthermore, by studying the DNA fingerprints of relatives with a history of a particular disorder, it’s possible to identify DNA patterns associated with the disease and perhaps obtain clues for an eventual cure. In addition, the U.S. Department of Defense requires blood and saliva samples from all military personnel. The samples are stored, and DNA is extracted if the need for identification of a casualty arises.
28.11: Key Terms
28 • Key Terms 28 • Key Terms
• anticodon
• antisense strand
• codon
• deoxyribonucleic acid (DNA)
• double helix
• 3′ end
• 5′ end
• messenger RNA (mRNA)
• nucleoside
• nucleotides
• polymerase chain reaction (PCR)
• replication
• ribonucleic acid (RNA)
• ribosomal RNA (rRNA)
• Sanger dideoxy method
• sense strand
• small RNAs
• transcription
• transfer RNA (tRNA)
• translation
28.12: Summary
28 • Summary 28 • Summary
DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) are biological polymers that act as chemical carriers of an organism’s genetic information. Enzyme-catalyzed hydrolysis of nucleic acids yields nucleotides, the monomer units from which RNA and DNA are constructed. Further enzyme-catalyzed hydrolysis of the nucleotides yields nucleosides plus phosphate. Nucleosides, in turn, consist of a purine or pyrimidine base linked to the C1 of an aldopentose sugar—ribose in RNA and 2-deoxyribose in DNA. The nucleotides are joined by phosphate links between the 5′ phosphate of one nucleotide and the 3′ hydroxyl on the sugar of another nucleotide.
Molecules of DNA consist of two complementary polynucleotide strands held together by hydrogen bonds between heterocyclic bases on the different strands and coiled into a double helix. Adenine and thymine form hydrogen bonds to each other, as do cytosine and guanine.
Three processes take place in deciphering the genetic information of DNA:
• Replication of DNA is the process by which identical DNA copies are made. The DNA double helix unwinds, complementary deoxyribonucleotides line up in order, and two new DNA molecules are produced.
• Transcription is the process by which RNA is produced to carry genetic information from the nucleus to the ribosomes. A short segment of the DNA double helix unwinds, and complementary ribonucleotides line up to produce messenger RNA (mRNA).
• Translation is the process by which mRNA directs protein synthesis. Each mRNA is divided into codons, ribonucleotide triplets that are recognized by small amino acid–carrying molecules of transfer RNA (tRNA), which deliver the appropriate amino acids needed for protein synthesis.
Sequencing of DNA is carried out by the Sanger dideoxy method, and small DNA segments can be synthesized in the laboratory by automated instruments. Small amounts of DNA can be amplified by factors of 106 using the polymerase chain reaction (PCR). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.10%3A_Chemistry_MattersDNA_Fingerprinting.txt |
Visualizing Chemistry
Problem 28-13
Identify the following bases, and tell whether each is found in DNA, RNA, or both:
(a)
(b)
(c)
Problem 28-14
Identify the following nucleotide, and tell how it is used:
Problem 28-15
Amine bases in nucleic acids can react with alkylating agents in typical SN2 reactions. Look at the following electrostatic potential maps, and tell which is the better nucleophile, guanine or adenine. The reactive positions in each are indicated.
Mechanism Problems
Problem 28-16
The final step in DNA synthesis is deprotection by treatment with aqueous ammonia. Show the mechanisms by which deprotection occurs at the points indicated in the following structure:
Problem 28-17
The final step in the metabolic degradation of uracil is the oxidation of malonic semialdehyde to give malonyl CoA. Propose a mechanism.
Problem 28-18
One of the steps in the biosynthesis of a nucleotide called inosine monophosphate is the formation of aminoimidazole ribonucleotide from formylglycinamidine ribonucleotide. Propose a mechanism.
Problem 28-19
One of the steps in the metabolic degradation of guanine is hydrolysis to give xanthine. Propose a mechanism.
Problem 28-20
One of the steps in the biosynthesis of uridine monophosphate is the reaction of aspartate with carbamoyl phosphate to give carbamoyl aspartate followed by cyclization to form dihydroorotate. Propose mechanisms for both steps.
General Problems
Problem 28-21
Human brain natriuretic peptide (BNP) is a small peptide of 32 amino acids used in the treatment of congestive heart failure. How many nitrogen bases are present in the DNA that codes for BNP?
Problem 28-22
Human and horse insulin both have two polypeptide chains, with one chain containing 21 amino acids and the other containing 30 amino acids. They differ in primary structure at two places. At position 9 in one chain, human insulin has Ser and horse insulin has Gly; at position 30 in the other chain, human insulin has Thr and horse insulin has Ala. How must the DNA for the two insulins differ?
Problem 28-23
The DNA of sea urchins contains about 32% A. What percentages of the other three bases would you expect in sea urchin DNA? Explain.
Problem 28-24
The codon UAA stops protein synthesis. Why does the sequence UAA in the following stretch of mRNA not cause any problems?
-GCA-UUC-GAG-GUA-ACG-CCC-
(a) GAATTC
(b) GATTACA
(c) CTCGAG
Problem 28-26 For what amino acids do the following ribonucleotide triplets code?
(a) AAU
(b) GAG (c) UCC (d) CAU
Problem 28-27
From what DNA sequences were each of the mRNA codons in Problem 28-26 transcribed?
Problem 28-28
What anticodon sequences of tRNAs are coded for by the codons in Problem 28-26?
Problem 28-29
Draw the complete structure of the ribonucleotide codon UAC. For what amino acid does this sequence code?
Problem 28-30
Draw the complete structure of the deoxyribonucleotide sequence from which the mRNA codon in Problem 28-29 was transcribed.
Problem 28-31
Give an mRNA sequence that will code for the synthesis of metenkephalin.
Tyr-Gly-Gly-Phe-Met
Problem 28-32
Give an mRNA sequence that will code for the synthesis of angiotensin II.
Asp-Arg-Val-Tyr-Ile-His-Pro-Phe
Problem 28-33
What amino acid sequence is coded for by the following DNA coding strand (sense strand)?
(5′) CTT-CGA-CCA-GAC-AGC-TTT (3′)
Problem 28-34
What amino acid sequence is coded for by the following mRNA base sequence?
(5′) CUA-GAC-CGU-UCC-AAG-UGA (3′)
Problem 28-35
If the DNA coding sequence -CAA-CCG-GAT- were miscopied during replication and became -CGA-CCG-GAT-, what effect would there be on the sequence of the protein produced?
Problem 28-36
Show the steps involved in a laboratory synthesis of the DNA fragment with the sequence CTAG.
Problem 28-37
Draw the structure of cyclic adenosine monophosphate (cAMP), a messenger involved in the regulation of glucose production in the body. Cyclic AMP has a phosphate ring connecting the 3′- and 5′-hydroxyl groups on adenosine.
Problem 28-38
Valganciclovir, marketed as Valcyte, is an antiviral agent used for the treatment of cytomegalovirus. Called a prodrug, valganciclovir is inactive by itself but is rapidly converted in the intestine by hydrolysis of its ester bond to produce an active drug, called ganciclovir, along with an amino acid.
(a) What amino acid is produced by hydrolysis of the ester bond in valganciclovir?
(b) What is the structure of ganciclovir? (c) What atoms present in the nucleotide deoxyguanine are missing from ganciclovir? (d) What role do the atoms missing from deoxyguanine play in DNA replication? (e) How might valganciclovir interfere with DNA synthesis? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/28%3A_Biomolecules_-_Nucleic_Acids/28.13%3A_Additional_Problems.txt |
Chapter Contents
29.1 An Overview of Metabolism and Biochemical Energy 29 • Why This Chapter?
In this chapter, we’ll look at some of the pathways by which organisms carry out their chemistry, focusing primarily on how they metabolize fats and carbohydrates. The treatment will be far from complete, but it should give you an idea of the kinds of processes that occur.
Anyone who wants to understand or contribute to the revolution now taking place in the biological sciences must first understand life processes at the molecular level. This understanding, in turn, must be based on a detailed knowledge of the chemical reactions and pathways used by living organisms. Just knowing what occurs is not enough; it’s also necessary to understand how and why organisms use the chemistry they do.
Biochemical reactions are not mysterious. Even though the biological reactions that take place in living organisms often appear complicated, they follow the same rules of reactivity as laboratory reactions and they operate by the same mechanisms.
Some of the molecules we’ll be encountering are substantially larger and more complex than those we’ve been dealing with thus far. But don’t be intimidated; keep your focus on the parts of the molecules where changes occur, and ignore the parts where nothing changes. The reactions themselves are exactly the same additions, eliminations, substitutions, carbonyl condensations, and so forth, that we’ve been dealing with all along. By the end of this chapter, it should be clear that the chemistry of living organisms is organic chemistry.
29.02: An Overview of Metabolism and Biochemical Energy
29.1 • An Overview of Metabolism and Biochemical Energy
The many reactions that occur in the cells of living organisms are collectively called metabolism. The pathways that break down larger molecules into smaller ones are called catabolism, and the pathways that synthesize larger biomolecules from smaller ones are known as anabolism. Catabolic reaction pathways are usually exergonic and release energy, while anabolic pathways are often endergonic and absorb energy. Catabolism can be divided into the four stages shown in Figure 29.2.
In the first catabolic stage, commonly called digestion, food is broken down in the mouth, stomach, and small intestine by hydrolysis of ester, acetal (glycoside), and amide (peptide) bonds to yield fatty acids, simple sugars, and amino acids. These smaller molecules are then absorbed and further degraded in the second stage of catabolism to yield acetyl groups attached by a thioester bond to the large carrier molecule, coenzyme A. The resultant compound, acetyl coenzyme A (acetyl CoA), is a key substance in the metabolism of food molecules and in many other biological pathways. As noted in Section 21.8, the acetyl group in acetyl CoA is linked to the sulfur atom of phosphopantetheine, which is itself linked to adenosine 3',5'-bisphosphate.
Acetyl groups are oxidized inside cellular mitochondria in the third stage of catabolism, the citric acid cycle, to yield CO2. (We’ll see the details of the process in Section 29.7.) Like most oxidations, this stage releases a large amount of energy, which is used in the fourth stage, the electron-transport chain, to accomplish the endergonic phosphorylation of adenosine diphosphate (ADP) with hydrogen phosphate ion (HOPO32, abbreviated Pi) to give adenosine triphosphate (ATP).
As the final result of food catabolism, ATP has been called the “energy currency” of the cell. Catabolic reactions “buy” ATP with the energy they release to synthesize it from ADP and hydrogen phosphate ion. Anabolic reactions then spend the ATP by transferring a phosphate group to another molecule, thereby regenerating ADP. Energy production and use in living organisms thus revolves around the ATP $⇄⇄$ ADP interconversion.
ADP and ATP are both phosphoric acid anhydrides, which contain linkages analogous to the linkage in carboxylic acid anhydrides. Just as carboxylic acid anhydrides react with alcohols by breaking a C–O bond and forming a carboxylic ester, ROCOR' (Section 21.5), phosphoric acid anhydrides react with alcohols by breaking a P–O bond and forming a phosphate ester, ROPO32. The reaction is, in effect, a nucleophilic acyl substitution at phosphorus. Note that phosphorylation reactions with ATP generally require the presence of a divalent metal cation in the enzyme, usually Mg2+, to form a Lewis acid–base complex with the phosphate oxygen atoms and to neutralize negative charge.
How does the body use ATP? Recall from Section 6.7 that the free-energy change ΔG must be negative and energy must be released for a reaction to be energetically favorable and occur spontaneously. If ΔG is positive, the reaction is energetically unfavorable and the process can’t occur spontaneously.
For an energetically unfavorable reaction to occur, it must be “coupled” to an energetically favorable reaction so that the overall free-energy change for the two reactions together is favorable. To understand what it means for reactions to be coupled, imagine that reaction 1 does not occur to any reasonable extent because it has a small equilibrium constant and is energetically unfavorable; that is, the reaction has ΔG > 0.
where A and B are the biochemically “important” substances while m and n are enzyme cofactors, H2O, or other small molecules.
Imagine also that product n can react with substance o to yield p and q in a second, highly favorable reaction that has a large equilibrium constant and ΔG << 0.
Taking the two reactions together, they share, or are coupled through, the common intermediate n, which is a product in the first reaction and a reactant in the second. When even a tiny amount of n is formed in reaction 1, it undergoes essentially complete conversion in reaction 2, thereby removing it from the first equilibrium and forcing reaction 1 to continually replenish n until reactant A is gone. That is, the two reactions added together have a favorable ΔG < 0, and we say that the favorable reaction 2 “drives” the unfavorable reaction 1. Because the two reactions are coupled through n, the transformation of A to B becomes favorable.
For an example of two reactions that are coupled, look at the phosphorylation reaction of glucose to yield glucose 6-phosphate plus water, an important step in the breakdown of dietary carbohydrates.
The reaction of glucose with HOPO32– does not occur spontaneously because it is energetically unfavorable, with ΔG°' = 13.8 kJ/mol. (The standard free-energy change for a biological reaction is denoted ΔG°' and refers to a process in which reactants and products have a concentration of 1.0 M in a solution with pH = 7.) At the same time, however, the reaction of water with ATP to yield ADP plus HOPO32– is strongly favorable, with ΔG°' = –30.5 kJ/mol. When the two reactions are coupled, glucose reacts with ATP to yield glucose 6-phosphate plus ADP in a reaction that is favorable by about 16.7 kJ/mol (4.0 kcal/mol). That is, ATP drives the phosphorylation reaction of glucose.
It’s this ability to drive otherwise unfavorable phosphorylation reactions that makes ATP so useful. The resultant phosphates are much more reactive as leaving groups in nucleophilic substitutions and eliminations than the alcohols they’re derived from and are therefore more chemically useful.
Problem 29-1
One of the steps in fat metabolism is the reaction of glycerol (1,2,3-propanetriol) with ATP to yield glycerol 1-phosphate. Write the reaction, and draw the structure of glycerol 1-phosphate. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.01%3A_Why_This_Chapter.txt |
29.2 • Catabolism of Triacylglycerols: The Fate of Glycerol
The metabolic breakdown of triacylglycerols begins with their hydrolysis in the stomach and small intestine to yield glycerol plus fatty acids. The reaction is catalyzed by a lipase, whose mechanism is shown in Figure 29.3. The active site of the enzyme contains a catalytic triad of aspartic acid, histidine, and serine residues, which act cooperatively to provide the necessary acid and base catalysis for the individual steps. Hydrolysis is accomplished by two sequential nucleophilic acyl substitution reactions, one that covalently binds an acyl group to the side chain –OH of a serine residue on the enzyme and a second that frees the fatty acid from the enzyme.
Figure 29.3 MECHANISM
Mechanism for the action of lipase. The active site of the enzyme contains a catalytic triad of aspartic acid, histidine, and serine, which react cooperatively to carry out two nucleophilic acyl substitution reactions. Individual steps are explained in the text.
Steps 1–2 of Figure 29.3: Acyl Enzyme Formation
The first nucleophilic acyl substitution step—reaction of the triacylglycerol with the active-site serine to give an acyl enzyme—begins with deprotonation of the serine alcohol by histidine to form the more strongly nucleophilic alkoxide ion. This proton transfer is facilitated by a nearby side-chain carboxylate anion of aspartic acid, which makes the histidine more basic and stabilizes the resultant histidine cation by electrostatic interactions. The deprotonated serine adds to a carbonyl group of a triacylglycerol to give a tetrahedral intermediate, which expels a diacylglycerol as a leaving group and produces an acyl enzyme. This step is catalyzed by a proton transfer from histidine to make the leaving group a neutral alcohol.
Steps 3–4 of Figure 29.3: Hydrolysis
The second nucleophilic acyl substitution step hydrolyzes the acyl enzyme and gives the free fatty acid by a mechanism analogous to that of the first two steps. Water is deprotonated by histidine to give hydroxide ion, which adds to the enzyme-bound acyl group. The tetrahedral intermediate then expels the neutral serine residue as the leaving group, freeing the fatty acid and returning the enzyme to its active form.
The fatty acids released on triacylglycerol hydrolysis are transported to mitochondria and degraded to acetyl CoA, while the glycerol is carried to the liver for further metabolism. In the liver, glycerol is first phosphorylated by reaction with ATP and then oxidized by NAD+. The resulting dihydroxyacetone phosphate (DHAP) enters the carbohydrate glycolysis pathway, which we’ll discuss in Section 29.5.
You might note that C2 of glycerol is a prochiral center (Section 5.11) with two identical “arms.” As is typical for enzyme-catalyzed reactions, the phosphorylation of glycerol is selective. Only the pro-R arm undergoes reaction, although this can’t be predicted in advance.
Note also that the phosphorylation product is named sn-glycerol 3-phosphate, where the sn- prefix means “stereospecific numbering.” In this convention, the molecule is drawn as a Fischer projection with the − OH group at C2 pointing to the left and the glycerol carbon atoms numbered from the top.
29.04: Catabolism of Triacylglycerols- -Oxidation
29.3 • Catabolism of Triacylglycerols: β-Oxidation
The fatty acids that result from triacylglycerol hydrolysis are converted into thioesters with coenzyme A and then catabolized by a repetitive four-step sequence of reactions called the β-oxidation pathway, shown in Figure 29.4. Each passage along the pathway results in the cleavage of an acetyl group from the end of the fatty-acid chain, until the entire molecule is ultimately degraded. As each acetyl group is produced, it enters the citric acid cycle and is further degraded to CO2, as we’ll see in Section 29.7.
Figure 29.4 MECHANISM
The four steps of the β-oxidation pathway, resulting in the cleavage of an acetyl group from the end of the fatty-acid chain. The key chain-shortening step is a retro-Claisen reaction of a β-keto thioester. Individual steps are explained in the text.
Step 1 of Figure 29.4: Introduction of a Double Bond
The β-oxidation pathway begins when two hydrogen atoms are removed from C2 and C3 of the fatty acyl CoA by one of a family of acyl-CoA dehydrogenases to yield an α,β-unsaturated acyl CoA. This kind of oxidation—the introduction of a conjugated double bond into a carbonyl compound—occurs frequently in biochemical pathways and usually involves the coenzyme flavin adenine dinucleotide (FAD). Reduced FADH2 is the by-product.
The mechanisms of FAD-catalyzed reactions are often difficult to establish because flavin coenzymes can follow both two-electron (polar) and one-electron (radical) pathways. As a result, extensive studies of the family of acyl-CoA dehydrogenases have not yet provided a clear picture of how these enzymes function. What is known is that: (1) The first step is abstraction of the pro-R hydrogen from the acidic α position of the acyl CoA to give a thioester enolate ion. Hydrogen-bonding between the acyl carbonyl group and the ribitol hydroxyls of FAD increases the acidity of the acyl group. (2) The pro-R hydrogen at the β position is transferred to FAD. (3) The α,β-unsaturated acyl CoA that results has a trans double bond.
One suggested mechanism has the reaction taking place by a conjugate nucleophilic addition of hydride, analogous to what occurs during alcohol oxidations with NAD+. Electrons on the enolate ion might expel a β hydride ion, which could add to the doubly bonded N5 nitrogen on FAD. Protonation of the intermediate at N1 would give the product.
Step 2 of Figure 29.4: Conjugate Addition of Water
The α,β-unsaturated acyl CoA produced in step 1 reacts with water by a conjugate addition pathway (Section 19.13) to yield a β-hydroxyacyl CoA in a process catalyzed by enoyl CoA hydratase. Water as nucleophile adds to the β carbon of the double bond, yielding an intermediate thioester enolate ion that is protonated on the α position.
Step 3 of Figure 29.4: Alcohol Oxidation
The β-hydroxyacyl CoA from step 2 is oxidized to a β-ketoacyl CoA in a reaction catalyzed by one of a family of L-3-hydroxyacyl-CoA dehydrogenases, which differ in substrate specificity according to the chain length of the acyl group. As in the oxidation of sn-glycerol 3-phosphate to dihydroxyacetone phosphate mentioned at the end of Section 29.2, this alcohol oxidation requires NAD+ as a coenzyme and yields reduced NADH/H+ as by-product. Deprotonation of the hydroxyl group is carried out by a histidine residue at the active site.
Step 4 of Figure 29.4: Chain Cleavage
Acetyl CoA is split off from the chain in the final step of β-oxidation, leaving an acyl CoA that is two carbon atoms shorter than the original. The reaction is catalyzed by β-ketoacyl-CoA thiolase and is mechanistically the reverse of a Claisen condensation reaction (Section 23.7). In the forward direction, a Claisen condensation joins two esters together to form a β-keto ester product. In the reverse direction, a retro-Claisen reaction splits apart a β-keto ester (or β-keto thioester in this case) to form two esters (or two thioesters).
The retro-Claisen reaction occurs by nucleophilic addition of a cysteine −SH group on the enzyme to the keto group of the β-ketoacyl CoA to yield an alkoxide ion intermediate. Cleavage of the C2–C3 bond then follows, with the expulsion of an acetyl CoA enolate ion that is immediately protonated. The enzyme-bound acyl group then undergoes nucleophilic acyl substitution by reaction with a molecule of coenzyme A, and the chain-shortened acyl CoA that results re-enters the β-oxidation pathway for further degradation.
Look at the catabolism of myristic acid shown in Figure 29.5 to see the overall results of the β-oxidation pathway. The first passage converts the 14-carbon myristoyl CoA into the 12-carbon lauroyl CoA plus acetyl CoA, the second passage converts lauroyl CoA into the 10-carbon caproyl CoA plus acetyl CoA, the third passage converts caproyl CoA into the 8-carbon capryloyl CoA, and so on. Note that the final passage produces two molecules of acetyl CoA because the precursor has four carbons.
Most fatty acids have an even number of carbon atoms, so none are left over after β-oxidation. Those fatty acids with an odd number of carbon atoms yield the three-carbon propionyl CoA in the final β-oxidation. Propionyl CoA is then converted to succinate by a multistep radical pathway, and succinate enters the citric acid cycle (Section 29.7). Note that the three-carbon propionyl group should technically be called propanoyl, but biochemists generally use the nonsystematic name.
Problem 29-2
Write the equations for the remaining passages of the β-oxidation pathway following those shown in Figure 29.5.
Problem 29-3 How many molecules of acetyl CoA are produced by catabolism of the following fatty acids, and how many passages of the β-oxidation pathway are needed? (a)
Palmitic acid, CH3(CH2)14CO2H
(b) Arachidic acid, CH3(CH2)18CO2H | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.03%3A_Catabolism_of_Triacylglycerols-_The_Fate_of_Glycerol.txt |
29.4 • Biosynthesis of Fatty Acids
One of the most striking features of the common fatty acids is that they have an even number of carbon atoms (Table 27.1). This occurs because all fatty acids are derived biosynthetically from acetyl CoA by sequential addition of two-carbon units to a growing chain. The acetyl CoA, in turn, arises primarily from the metabolic breakdown of carbohydrates in the glycolysis pathway, which we’ll see in Section 29.5. Thus, dietary carbohydrates consumed in excess of immediate energy needs are turned into fats for storage.
As a general rule in biological chemistry, the anabolic pathway by which a substance is made is not the reverse of the catabolic pathway by which the same substance is degraded. The two paths must differ in some respects for both to be energetically favorable. Thus, the β-oxidation pathway for converting fatty acids into acetyl CoA and the biosynthesis of fatty acids from acetyl CoA are related but are not exact opposites. Differences include the identity of the acyl-group carrier, the stereochemistry of the β-hydroxyacyl reaction intermediate, and the identity of the redox coenzyme. FAD is used to introduce a double bond in β-oxidation, while NADPH is used to reduce the double bond in fatty-acid biosynthesis.
In bacteria, each step in fatty-acid synthesis is catalyzed by a separate enzyme. In vertebrates, however, fatty-acid synthesis is catalyzed by an immense, multienzyme complex called a synthase that contains two identical subunits of 2505 amino acids each and catalyzes all steps in the pathway. In fact, for an 18-carbon fatty acid, the synthase catalyzes 42 separate steps! An overview of fatty-acid biosynthesis is shown in Figure 29.6.
Steps 1–2 of Figure 29.6: Acyl Transfers
The starting material for fatty-acid biosynthesis is the thioester acetyl CoA, the final product of carbohydrate breakdown, as we’ll see in Section 29.6. The pathway begins with several priming reactions, which transport acetyl CoA and convert it into more reactive species. The first priming reaction is a nucleophilic acyl substitution reaction that converts acetyl CoA into acetyl ACP (acyl carrier protein).
Notice that the mechanism of the nucleophilic acyl substitution in step 1 can be given in an abbreviated form that saves space by not explicitly showing the tetrahedral reaction intermediate. Instead, electron movement is shown as a heart-shaped path around the carbonyl oxygen to imply the two steps of the full mechanism. Biochemists commonly use this kind of abbreviated format, and we’ll also use it on occasion through the rest of this chapter.
Figure 29.6 MECHANISM
The pathway for fatty-acid biosynthesis from the two-carbon precursor, acetyl CoA. Individual steps are explained in the text.
In bacteria, ACP is a small protein of 77 residues that transports an acyl group from one enzyme to another. In vertebrates, however, ACP appears to be a long arm on a multienzyme synthase complex, whose apparent function is to shepherd an acyl group from site to site within the complex. As in acetyl CoA, the acyl group in acetyl ACP is linked by a thioester bond to the sulfur atom of phosphopantetheine. The phosphopantetheine is in turn linked to ACP through the side-chain –OH group of a serine residue in the enzyme.
Step 2, another priming reaction, involves a further exchange of thioester linkages by another nucleophilic acyl substitution and results in covalent bonding of the acetyl group to a cysteine residue in the synthase complex that catalyzes the upcoming condensation step.
Steps 3–4 of Figure 29.6: Carboxylation and Acyl Transfer
Step 3 is a loading reaction in which acetyl CoA is carboxylated by reaction with HCO3 and ATP to yield malonyl CoA plus ADP. This step requires the coenzyme biotin, which is bonded to the lysine residue of acetyl CoA carboxylase and acts as a carrier of CO2. Biotin first reacts with bicarbonate ion to give N-carboxybiotin, which then reacts with the enolate ion of acetyl CoA and transfers the CO2 group. Thus, biotin acts as a carrier of CO2, binding it in one step and releasing it in another.
The mechanism of the CO2 transfer reaction with acetyl CoA to give malonyl CoA is thought to involve CO2 as the reactive species. One proposal is that the loss of CO2 is favored by hydrogen-bond formation between the N-carboxybiotin carbonyl group and a nearby acidic site in the enzyme. Simultaneous deprotonation of acetyl CoA by a basic site in the enzyme gives a thioester enolate ion that can react with CO2 as it forms (Figure 29.7).
Figure 29.7 MECHANISM Mechanism of step 3 in Figure 29.6, the biotin-dependent carboxylation of acetyl CoA to yield malonyl CoA.
Following the formation of malonyl CoA, another nucleophilic acyl substitution reaction occurs in step 4 to form the more reactive malonyl ACP, thereby binding the malonyl group to an ACP arm of the multienzyme synthase. At this point, both acetyl and malonyl groups are bound to the enzyme, and the stage is set for their condensation.
Step 5 of Figure 29.6: Condensation
The key carbon–carbon bond-forming reaction that builds the fatty-acid chain occurs in step 5. This step is simply a Claisen condensation between acetyl synthase as the electrophilic acceptor and malonyl ACP as the nucleophilic donor. The mechanism of the condensation is thought to involve decarboxylation of malonyl ACP to give an enolate ion, followed by immediate nucleophilic addition of the enolate ion to the carbonyl group of acetyl synthase. Breakdown of the tetrahedral intermediate then gives the four-carbon condensation product acetoacetyl ACP and frees the synthase binding site for attachment of the chain-elongated acyl group at the end of the sequence.
Steps 6–8 of Figure 29.6: Reduction and Dehydration
The ketone carbonyl group in acetoacetyl ACP is next reduced to the alcohol β-hydroxybutyryl ACP by β-keto thioester reductase and NADPH, a reducing coenzyme closely related to NADH. R stereochemistry results at the newly formed chirality center in the β-hydroxy thioester product. (Note that the systematic name of a butyryl group is butanoyl.)
Subsequent dehydration of β-hydroxybutyryl ACP by an E1cB reaction in step 7 yields trans-crotonyl ACP, and the carbon–carbon double bond of crotonyl ACP is reduced by NADPH in step 8 to yield butyryl ACP. The double-bond reduction occurs by conjugate nucleophilic addition of a hydride ion from NADPH to the β carbon of trans-crotonyl ACP. In vertebrates, this reduction occurs by an overall syn addition, but other organisms carry out similar chemistry with different stereochemistry.
The net effect of the eight steps in the fatty-acid biosynthesis pathway is to take two 2-carbon acetyl groups and combine them into a 4-carbon butyryl group. Further condensation of the butyryl group with another malonyl ACP yields a 6-carbon unit, and still further repetitions add two carbon atoms at a time until the 16-carbon palmitoyl ACP is produced.
Further chain elongation of palmitic acid occurs by reactions similar to those just described, but CoA rather than ACP acts as the carrier group, and separate enzymes are needed for each step rather than a multienzyme synthase complex.
Problem 29-4
Write a mechanism for the dehydration reaction of β-hydroxybutyryl ACP to yield crotonyl ACP in step 7 of fatty-acid synthesis.
Problem 29-5
Evidence for the role of acetate in fatty-acid biosynthesis comes from isotope-labeling experiments. If acetate labeled with 13C in the methyl group (13CH3CO2H) were incorporated into fatty acids, at what positions in the fatty-acid chain would you expect the 13C label to appear?
Problem 29-6
Does the reduction of acetoacetyl ACP in step 6 occur on the Re face or the Si face of the molecule? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.05%3A_Biosynthesis_of_Fatty_Acids.txt |
29.5 • Catabolism of Carbohydrates: Glycolysis
Glucose is the body’s primary short-term energy source. Its catabolism begins with glycolysis, a series of ten enzyme-catalyzed reactions that break down glucose into 2 equivalents of pyruvate, CH3COCO2. The steps of glycolysis, also called the Embden–Meyerhoff pathway after its discoverers, are summarized in Figure 29.8.
Figure 29.8 MECHANISM
The ten-step glycolysis pathway for catabolizing glucose to two molecules of pyruvate. Individual steps are described in the text.
: (Continued)
Steps 1–2 of Figure 29.8: Phosphorylation and Isomerization
Glucose, produced by the digestion of dietary carbohydrates, is phosphorylated at the C6 hydroxyl group by reaction with ATP in a process catalyzed by hexokinase. As noted in Section 29.1, the reaction requires Mg2+ as a cofactor to complex with the negatively charged phosphate oxygens. The glucose 6-phosphate that results is then isomerized by glucose 6-phosphate isomerase to give fructose 6-phosphate. This isomerization takes place by initial opening of the glucose hemiacetal ring to its open-chain form, followed by keto–enol tautomerization to a cis enediol, $HO─C═C─OHFigure 29.10).$
in glycolysis, the isomerization of glucose 6-phosphate to fructose 6-phosphate.
Step 3 of Figure 29.8: Phosphorylation
Fructose 6-phosphate is converted in step 3 to fructose 1,6-bisphosphate (FBP) by a phosphofructokinase-catalyzed reaction with ATP (recall that the prefix bis- means two). The mechanism is similar to that in step 1, with Mg2+ ion again required as cofactor. Interestingly, the product of step 2 is the α anomer of fructose 6-phosphate, but it is the β anomer that is phosphorylated in step 3, implying that the two anomers equilibrate rapidly through the open-chain form. The result is a molecule ready to be split into the two three-carbon intermediates that will ultimately become two molecules of pyruvate.
Step 4 of Figure 29.8: Cleavage
Fructose 1,6-bisphosphate is cleaved in step 4 into two 3-carbon pieces, dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP). The bond between C3 and C4 of fructose 1,6-bisphosphate breaks, and a $C═OSection 23.1) and is catalyzed by an aldolase. A forward aldol reaction joins two aldehydes or ketones to give a$β-hydroxy carbonyl compound, while a retro-aldol reaction, as in this case, cleaves a β-hydroxy carbonyl compound into two aldehydes or ketones.
Two classes of aldolases are used by organisms for catalysis of the retro-aldol reaction. In fungi, algae, and some bacteria, the retro-aldol reaction is catalyzed by class II aldolases, which function by coordination of the fructose carbonyl group with Zn2+ as Lewis acid. In plants and animals, the reaction is catalyzed by class I aldolases and does not take place on the free ketone. Instead, fructose 1,6-bisphosphate undergoes reaction with the side-chain –NH2 group of a lysine residue on the aldolase to yield a protonated enzyme-bound imine (Section 19.8), which is often called a Schiff base in biochemistry.
Because of its positive charge, the iminium ion is a better electron acceptor than a ketone carbonyl group. Retro-aldol reaction ensues, giving glyceraldehyde 3-phosphate and an enamine, which is protonated to give another iminium ion that is hydrolyzed to yield dihydroxyacetone phosphate (Figure 29.11).
in Figure 29.8, the cleavage of fructose 1,6-bisphosphate to yield glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. The reaction occurs through an iminium ion formed by reaction with a lysine residue in the enzyme.
Step 5 of Figure 29.8: Isomerization
Dihydroxyacetone phosphate is isomerized in step 5 by triose phosphate isomerase to form a second equivalent of glyceraldehyde 3-phosphate. As in the conversion of glucose 6-phosphate to fructose 6-phosphate in step 2, the isomerization takes place by keto–enol tautomerization through a common enediol intermediate. A base deprotonates C1 and then reprotonates C2 using the same hydrogen. The net result of steps 4 and 5 together is the production of two glyceraldehyde 3-phosphate molecules, both of which pass down the rest of the pathway. Thus, each of the remaining five steps of glycolysis takes place twice for every glucose molecule entering at step 1.
Steps 6–7 of Figure 29.9: Oxidation, Phosphorylation, and Dephosphorylation
Glyceraldehyde 3-phosphate is oxidized and phosphorylated in step 6 to give 1,3-bisphosphoglycerate (Figure 29.12). The reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase and begins by nucleophilic addition of the –SH group of a cysteine residue in the enzyme to the aldehyde carbonyl group to yield a hemithioacetal (Section 19.10), the sulfur analog of a hemiacetal. Oxidation of the hemithioacetal –OH group by NAD+ then yields a thioester, which reacts with phosphate ion in a nucleophilic acyl substitution step to yield 1,3-bisphosphoglycerate, a mixed anhydride derived from a carboxylic acid and a phosphoric acid.
in Figure 29.9, the oxidation and phosphorylation of glyceraldehyde 3-phosphate to give 1,3-bisphosphoglycerate. The process occurs through initial formation of a hemiacetal that is oxidized to a thioester and converted into an acyl phosphate.
Like all anhydrides (Section 21.5), the mixed carboxylic–phosphoric anhydride is a reactive substrate in nucleophilic acyl (or phosphoryl) substitution reactions. Reaction of 1,3-bisphosphoglycerate with ADP occurs in step 7 by substitution on phosphorus, resulting in transfer of a phosphate group to ADP and giving ATP plus 3-phosphoglycerate. The process is catalyzed by phosphoglycerate kinase and requires Mg2+ as cofactor. Together, steps 6 and 7 accomplish the oxidation of an aldehyde to a carboxylic acid.
Step 8 of Figure 29.9: Isomerization
3-Phosphoglycerate isomerizes to 2-phosphoglycerate in a step catalyzed by phosphoglycerate mutase. In plants, 3-phosphoglycerate transfers its phosphoryl group from its C3 oxygen to a histidine residue on the enzyme in one step and then accepts the same phosphoryl group back onto the C2 oxygen in a second step. In animals and yeast, however, the enzyme contains a phosphorylated histidine, which transfers its phosphoryl group to the C2 oxygen of 3-phosphoglycerate and forms 2,3-bisphosphoglycerate as intermediate. The same histidine then accepts a phosphoryl group from the C3 oxygen to yield the isomerized product plus the regenerated enzyme. As explained in Section 29.4, we’ll occasionally use an abbreviated mechanism for nucleophilic acyl substitution reactions to save space.
Steps 9–10 of Figure 29.9: Dehydration and Dephosphorylation
Like most β-hydroxy carbonyl compounds, 2-phosphoglycerate undergoes a ready dehydration in step 9 by an E1cB mechanism (Section 23.3). The process is catalyzed by enolase, and the product is phosphoenolpyruvate, abbreviated PEP. Two Mg2+ ions are associated with the 2-phosphoglycerate to neutralize the negative charges.
Transfer of the phosphoryl group to ADP in step 10 then generates ATP and gives enolpyruvate, which tautomerizes to pyruvate. The reaction is catalyzed by pyruvate kinase and requires that a molecule of fructose 1,6-bisphosphate also be present, as well as 2 equivalents of Mg2+. One Mg2+ ion coordinates to ADP, and the other increases the acidity of a water molecule necessary for protonation of the enolate ion.
The overall result of glycolysis can be summarized by the following equation:
Problem 29-7
Identify the two steps of glycolysis in which ATP is produced.
Problem 29-8
Look at the entire glycolysis pathway, and make a list of the kinds of organic reactions that take place—nucleophilic acyl substitutions, aldol reactions, E1cB reactions, and so forth. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.06%3A_Catabolism_of_Carbohydrates-_Glycolysis.txt |
29.6 • Conversion of Pyruvate to Acetyl CoA
Pyruvate, produced by catabolism of glucose (and by degradation of several amino acids), can undergo several further transformations depending on the conditions and on the organism. In the absence of oxygen, pyruvate can either be reduced by NADH to yield lactate [CH3CH(OH)CO2] or, in yeast, fermented to give ethanol. Under typical aerobic conditions in mammals, however, pyruvate is converted by a process called oxidative decarboxylation to give acetyl CoA plus CO2. (Oxidative because the oxidation state of the carbonyl carbon rises from that of a ketone to that of a thioester.)
The conversion occurs through a multistep sequence of reactions catalyzed by a complex of enzymes and cofactors called the pyruvate dehydrogenase complex. The process occurs in three stages, each catalyzed by one of the enzymes in the complex, as outlined in Figure 29.13. Acetyl CoA, the ultimate product, then acts as fuel for the final stage of catabolism, the citric acid cycle.
Step 1 of Figure 29.13: Addition of Thiamin Diphosphate
The conversion of pyruvate to acetyl CoA begins by reaction of pyruvate with thiamin diphosphate, a derivative of vitamin B1. Formerly called thiamin pyrophosphate, thiamin diphosphate is usually abbreviated as TPP. The spelling thiamine is also correct and frequently used.
The key structural element in thiamin diphosphate is the thiazolium ring—a five-membered, unsaturated heterocycle containing a sulfur atom and a positively charged nitrogen atom. The thiazolium ring is weakly acidic, with a pKa of approximately 18 for the ring hydrogen between N and S. Bases can therefore deprotonate thiamin diphosphate, leading to formation of an ylide much like the phosphonium ylides used in Wittig reactions (Section 19.11). As in the Wittig reaction, the TPP ylide is a nucleophile and adds to the ketone carbonyl group of pyruvate to yield an alcohol addition product.
Figure 29.13 MECHANISM
Mechanism for the conversion of pyruvate to acetyl CoA through a multistep sequence of reactions that requires three different enzymes and four different coenzymes. The individual steps are explained in the text.
Step 2 of Figure 29.13: Decarboxylation
The TPP addition product, which contains an iminium ion β to a carboxylate anion, undergoes decarboxylation in much the same way that a β-keto acid decarboxylates in the acetoacetic ester synthesis (Section 22.7). The $C═N+C═N+$ bond of the pyruvate addition product acts like the $C═OC═O$ bond of a β-keto acid to accept electrons as CO2 leaves, giving hydroxyethylthiamin diphosphate (HETPP).
Step 3 of Figure 29.13: Reaction with Lipoamide
Hydroxyethylthiamin diphosphate is an enamine ($R2N─C═CSection 23.11). It therefore reacts with the enzyme-bound disulfide lipoamide by nucleophilic attack on a sulfur atom, displacing the second sulfur in an$S$N2-like process.
Step 4 of Figure 29.13: Elimination of Thiamin Diphosphate
The product of the HETPP reaction with lipoamide is a hemithioacetal, which eliminates thiamin diphosphate ylide. This elimination is the reverse of the ketone addition in step 1 and generates acetyl dihydrolipoamide.
Step 5 of Figure 29.13: Acyl Transfer
Acetyl dihydrolipoamide, a thioester, undergoes a nucleophilic acyl substitution reaction with coenzyme A to yield acetyl CoA plus dihydrolipoamide. The dihydrolipoamide is then oxidized back to lipoamide by FAD (Section 29.3), and the FADH2 that results is in turn oxidized back to FAD by NAD+, completing the catalytic cycle.
Problem 29-9
Which carbon atoms in glucose end up as –CH3 carbons in acetyl CoA? Which carbons end up as CO2? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.07%3A_Conversion_of_Pyruvate_to_Acetyl_CoA.txt |
29.7 • The Citric Acid Cycle
The initial stages of catabolism result in the conversion of both fats and carbohydrates into acetyl groups that are bonded through a thioester link to coenzyme A. Acetyl CoA then enters the next stage of catabolism—the citric acid cycle, also called the tricarboxylic acid (TCA) cycle, or Krebs cycle, after Hans Krebs, who unraveled its complexities in 1937. The overall result of the cycle is the conversion of an acetyl group into two molecules of CO2 plus reduced coenzymes by the eight-step reaction sequence shown in Figure 29.14.
As its name implies, the citric acid cycle is a closed loop of reactions in which the product of the final step (oxaloacetate) is a reactant in the first step. The intermediates are constantly regenerated, flowing continuously through the cycle, which operates as long as the oxidizing coenzymes NAD+ and FAD are available. To meet this condition, the reduced coenzymes NADH and FADH2 must be reoxidized via the electron-transport chain, which in turn relies on oxygen as the ultimate electron acceptor. Thus, the cycle is dependent on the availability of oxygen and on the operation of the electron-transport chain.
Figure 29.14 MECHANISM
The citric acid cycle is an eight-step series of reactions that results in the conversion of an acetyl group into two molecules of CO2 plus reduced coenzymes. Individual steps are explained in the text.
Step 1 of Figure 29.14: Addition to Oxaloacetate
Acetyl CoA enters the citric acid cycle in step 1 by nucleophilic addition to the oxaloacetate carbonyl group, to give (S)-citryl CoA. This addition is an aldol reaction and is catalyzed by citrate synthase, as discussed in Section 26.11. (S)-Citryl CoA is then hydrolyzed to citrate by a typical nucleophilic acyl substitution reaction with water, catalyzed by the same citrate synthase enzyme.
Note that the hydroxyl-bearing carbon of citrate is a prochirality center and contains two identical arms. Because the initial aldol reaction of acetyl CoA to oxaloacetate occurs specifically from the Si face of the ketone carbonyl group, the pro-S arm of citrate is derived from acetyl CoA and the pro-R arm is derived from oxaloacetate.
Step 2 of Figure 29.14: Isomerization
Citrate, a prochiral tertiary alcohol, is next converted into its isomer, (2R,3S)-isocitrate, a chiral secondary alcohol. The isomerization occurs in two steps, both of which are catalyzed by the same aconitase enzyme. The initial step is an E1cB dehydration of a β-hydroxy acid to give cis-aconitate, the same sort of reaction that occurs in step 9 of glycolysis (Figure 29.8). The second step is a conjugate nucleophilic addition of water to the C═C bond (Section 19.13). The dehydration of citrate takes place specifically on the pro-R arm—the one derived from oxaloacetate—rather than on the pro-S arm derived from acetyl CoA.
Step 3 of Figure 29.14: Oxidation and Decarboxylation
(2R,3S)-Isocitrate, a secondary alcohol, is oxidized by NAD+ in step 3 to give the ketone oxalosuccinate, which loses CO2 to give α-ketoglutarate. Catalyzed by isocitrate dehydrogenase, the decarboxylation is a typical reaction of a β-keto acid, just like that in the acetoacetic ester synthesis (Section 22.7). The enzyme requires a divalent cation as a cofactor to polarize the ketone carbonyl group and make it a better electron acceptor.
Step 4 of Figure 29.14: Oxidative Decarboxylation
The transformation of α-ketoglutarate to succinyl CoA in step 4 is a multistep process just like the transformation of pyruvate to acetyl CoA that we saw in Figure 29.13. In both cases, an α-keto acid loses CO2 and is oxidized to a thioester in a series of steps catalyzed by a multienzyme dehydrogenase complex. As in the conversion of pyruvate to acetyl CoA, the reaction involves an initial nucleophilic addition reaction of thiamin diphosphate ylide to α-ketoglutarate, followed by decarboxylation. Reaction with lipoamide, elimination of TPP ylide, and finally a transesterification of the dihydrolipoamide thioester with coenzyme A yields succinyl CoA.
Step 5 of Figure 29.14: Acyl CoA Cleavage
Succinyl CoA is converted to succinate in step 5. This reaction is catalyzed by succinyl CoA synthetase and is coupled with phosphorylation of guanosine diphosphate (GDP) to give guanosine triphosphate (GTP). The overall transformation is similar to that of steps 6 through 8 in glycolysis (Figure 29.8), in which a thioester is converted into an acyl phosphate and a phosphate group is then transferred to ADP. The overall result is a “hydrolysis” of the thioester group without involvement of water.
Step 6 of Figure 29.14: Dehydrogenation
Succinate is dehydrogenated in step 6 by the FAD-dependent succinate dehydrogenase to give fumarate. This process is analogous to what occurs during the β-oxidation pathway of fatty-acid catabolism (Section 29.3). The reaction is stereospecific, removing the pro-S hydrogen from one carbon and the pro-R hydrogen from the other.
Steps 7–8 of Figure 29.14: Hydration and Oxidation
The final two steps in the citric acid cycle are the conjugate nucleophilic addition of water to fumarate to yield (S)-malate and the oxidation of (S)-malate by NAD+ to give oxaloacetate. The addition is catalyzed by fumarase and is mechanistically similar to the addition of water to cis-aconitate in step 2. This reaction occurs through an enolate-ion intermediate, which is protonated on the side opposite the OH, leading to a net anti addition.
The final step is the oxidation of (S)-malate by NAD+ to give oxaloacetate, a reaction catalyzed by malate dehydrogenase. The citric acid cycle has now returned to its starting point, ready to revolve again. The overall result of the cycle is
Problem 29-10
Which of the substances in the citric acid cycle are tricarboxylic acids, thus giving the cycle its alternative name?
Problem 29-11
Write mechanisms for step 2 of the citric acid cycle, the dehydration of citrate and the addition of water to aconitate.
Problem 29-12
Is the pro-R or pro-S hydrogen removed from citrate during the dehydration in step 2 of the citric acid cycle? Does the elimination reaction occur with syn or anti geometry? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.08%3A_The_Citric_Acid_Cycle.txt |
29.8 • Carbohydrate Biosynthesis: Gluconeogenesis
Glucose is the body’s primary fuel when food is plentiful, but in times of fasting or prolonged exercise, glucose stores can become depleted. Most tissues then begin metabolizing fats as their source of acetyl CoA, but the brain is different. The brain relies almost entirely on glucose for fuel and is dependent on receiving a continuous supply in the blood. When the supply of glucose fails, even for a brief time, irreversible damage can occur. Thus, a pathway for synthesizing glucose from simple precursors is crucial.
Higher organisms are not able to synthesize glucose from acetyl CoA but must instead use one of the three-carbon precursors lactate, glycerol, or alanine, all of which are readily converted into pyruvate.
Pyruvate then becomes the starting point for gluconeogenesis, the 11-step biosynthetic pathway by which organisms make glucose (Figure 29.15). The gluconeogenesis pathway is not the reverse of the glycolysis pathway by which glucose is degraded. As with the catabolic and anabolic pathways for fatty acids (Section 29.3 and Section 29.4), the catabolic and anabolic pathways for carbohydrates differ in some details so that both are energetically favorable.
Figure 29.15 MECHANISM
The gluconeogenesis pathway for the biosynthesis of glucose from pyruvate. Individual steps are explained in the text.
: (Continued)
Step 1 of Figure 29.15: Carboxylation
Gluconeogenesis begins with the carboxylation of pyruvate to yield oxaloacetate. The reaction is catalyzed by pyruvate carboxylase and requires ATP, bicarbonate ion, and the coenzyme biotin, which acts as a carrier to transport CO2 to the enzyme active site. The mechanism is analogous to that of step 3 in fatty-acid biosynthesis (Figure 29.6), in which acetyl CoA is carboxylated to yield malonyl CoA.
Step 2 of Figure 29.15: Decarboxylation and Phosphorylation
Decarboxylation of oxaloacetate, a β-keto acid, occurs by the typical retro-aldol mechanism like that in step 3 in the citric acid cycle (Figure 29.14), and phosphorylation of the resultant pyruvate enolate ion by GTP occurs concurrently to give phosphoenolpyruvate. This reaction is catalyzed by phosphoenolpyruvate carboxykinase.
Steps 3–4 of Figure 29.15: Hydration and Isomerization
Conjugate nucleophilic addition of water to the double bond of phosphoenolpyruvate gives 2-phosphoglycerate by a process similar to that of step 7 in the citric acid cycle. Phosphorylation of C3 and dephosphorylation of C2 then yields 3-phosphoglycerate. Mechanistically, these steps are the reverse of steps 9 and 8 in glycolysis (Figure 29.8), which have equilibrium constants near 1 so that substantial amounts of reactant and product are both present.
Steps 5–7 of Figure 29.15: Phosphorylation, Reduction, and Tautomerization
Reaction of 3-phosphoglycerate with ATP generates the corresponding acyl phosphate, 1,3-bisphosphoglycerate, which binds to the glyceraldehyde 3-phosphate dehydrogenase by a thioester bond to a cysteine residue. Reduction of the thioester by NADH/H+ yields the corresponding aldehyde, and keto–enol tautomerization of the aldehyde gives dihydroxyacetone phosphate. All three steps comprise a mechanistic reversal of the corresponding steps 7, 6, and 5 of glycolysis and have equilibrium constants near 1.
Step 8 of Figure 29.16: Aldol Reaction
Dihydroxyacetone phosphate and glyceraldehyde 3-phosphate, the two 3-carbon units produced in step 7, join by an aldol reaction to give fructose 1,6-bisphosphate, the reverse of step 4 in glycolysis (Figure 29.11). As in glycolysis, the reaction is catalyzed in plants and animals by a class I aldolase and takes place on an iminium ion formed by reaction of dihydroxyacetone phosphate with a side-chain lysine –NH2 group on the enzyme. Loss of a proton from the neighboring carbon then generates an enamine, an aldol-like reaction ensues, and the product is hydrolyzed.
Steps 9–10 of Figure 29.16: Hydrolysis and Isomerization
Hydrolysis of the phosphate group at C1 of fructose 1,6-bisphosphate gives fructose 6-phosphate. Although the result of the reaction is the exact opposite of step 3 in glycolysis, the mechanism is not. In glycolysis, phosphorylation is accomplished by reaction of fructose with ATP, with formation of ADP as by-product. The reverse of that process, however—the reaction of fructose 1,6-bisphosphate with ADP to give fructose 6-phosphate and ATP—is energetically unfavorable because ATP is too high in energy. Thus, an alternative pathway is used in which the C1 phosphate group is removed by a direct hydrolysis reaction, catalyzed by fructose 1,6-bisphosphatase.
Following hydrolysis, keto–enol tautomerization of the carbonyl group from C2 to C1 gives glucose 6-phosphate. The isomerization is the reverse of step 2 in glycolysis.
Step 11 of Figure 29.16: Hydrolysis
The final step in gluconeogenesis is the conversion of glucose 6-phosphate to glucose by a second phosphatase-catalyzed hydrolysis reaction. As just discussed for the hydrolysis of fructose 1,6-bisphosphate in step 9, and for the same energetic reasons, the mechanism of the glucose 6-phosphate hydrolysis is not the exact opposite of the corresponding step 1 in glycolysis.
Interestingly, however, the mechanisms of the two phosphate hydrolysis reactions in steps 9 and 11 are not the same. In step 9, water is the nucleophile, but in the glucose 6-phosphate reaction of step 11, a histidine residue on the enzyme attacks phosphorus, giving a phosphoryl enzyme intermediate that subsequently reacts with water.
The overall result of gluconeogenesis is summarized by the following equation:
Problem 29-13
Write a mechanism for step 6 of gluconeogenesis, the reduction of 3-phosphoglyceryl phosphate with NADH/H+ to yield glyceraldehyde 3-phosphate. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.09%3A_Carbohydrate_Biosynthesis-_Gluconeogenesis.txt |
29.9 • Catabolism of Proteins: Deamination
The catabolism of proteins is much more complex than that of fats and carbohydrates because each of the 20 α-amino acids is degraded through its own unique pathway. The general idea, however, is that (1) the α amino group is first removed as ammonia by a deamination process, (2) the ammonia is converted into urea, and (3) the remaining amino acid carbon skeleton (usually an α-keto acid) is converted into a compound that enters the citric acid cycle.
Transamination
Deamination is usually accomplished by a transamination reaction in which the –NH2 group of the amino acid is exchanged with the keto group of α-ketoglutarate, forming a new α-keto acid plus glutamate. The overall process occurs in two parts, is catalyzed by aminotransferases, and involves participation of the coenzyme pyridoxal phosphate, abbreviated PLP, a derivative of pyridoxine (vitamin B6). The aminotransferases differ in their specificity for amino acids, but the mechanism remains the same.
The mechanism of the first part of transamination is shown in Figure 29.17. The process begins with reaction between the α-amino acid and pyridoxal phosphate, which is covalently bonded to the aminotransferase by an imine linkage between the side-chain –NH2 group of a lysine residue in the enzyme and the PLP aldehyde group. Deprotonation/reprotonation of the PLP–amino acid imine effects tautomerization of the imine $C═NC═N$ bond, and hydrolysis of the tautomerized imine gives an α-keto acid plus pyridoxamine phosphate (PMP).
Figure 29.17 MECHANISM Mechanism for the enzyme-catalyzed, PLP-dependent transamination of an α-amino acid to give an α-keto acid. Individual steps are explained in the text.
Step 1 of Figure 29.17: Transimination
The first step in transamination is transimination—the reaction of the PLP–enzyme imine with an α-amino acid to give a PLP–amino acid imine plus expelled enzyme as the leaving group. The reaction occurs by nucleophilic addition of the amino acid –NH2 group to the $C═NSection 19.8). The protonated diamine intermediate undergoes a proton transfer and expels the lysine amino group in the enzyme to complete the step.$
Steps 2–4 of Figure 29.17: Tautomerization and Hydrolysis
Following formation of the PLP–amino acid imine in step 1, a tautomerization of the $C═NC═N$ bond occurs in step 2. The basic lysine residue in the enzyme that was expelled as a leaving group during transimination deprotonates the acidic α position of the amino acid, with the protonated pyridine ring of PLP acting as the electron acceptor. Reprotonation occurs on the carbon atom next to the ring, generating a tautomeric product that is the imine of an α-keto acid with pyridoxamine phosphate, abbreviated PMP.
Hydrolysis of this PMP–α-keto acid imine then completes the first part of the transamination reaction. This is the mechanistic reverse of imine formation and occurs by nucleophilic addition of water to the imine, followed by proton transfer and expulsion of PMP as leaving group.
Regeneration of PLP from PMP
With PLP plus the α-amino acid now converted into PMP plus an α-keto acid, PMP must be transformed back into PLP to complete the catalytic cycle. The conversion occurs by another transamination reaction, this one between PMP and an α-keto acid, usually α-ketoglutarate. The products are PLP plus glutamate, and the mechanism is the exact reverse of that shown in Figure 29.17. That is, PMP and α-ketoglutarate give an imine; the PMP–α-ketoglutarate imine undergoes tautomerization of the $C═NC═N$ bond to give a PLP–glutamate imine; and the PLP–glutamate imine reacts with a lysine residue on the enzyme in a transimination process to yield PLP–enzyme imine plus glutamate.
Problem 29-14
Write all the steps in the transamination reaction of PMP with α-ketoglutarate plus a lysine residue in the enzyme to give the PLP–enzyme imine plus glutamate.
Problem 29-15
What α-keto acid is formed on transamination of leucine?
Problem 29-16
From what amino acid is the following α-keto acid derived? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.10%3A_Catabolism_of_Proteins-_Deamination.txt |
29.10 • Some Conclusions about Biological Chemistry
As promised in the chapter introduction, the past few sections have been a fast-paced tour of a large number of reactions. Following it all undoubtedly required a lot of work and a lot of page-turning to look at earlier sections.
After examining the various metabolic pathways, perhaps the main conclusion about biological chemistry is the remarkable similarity between the mechanisms of biological reactions and the mechanisms of laboratory reactions. If you were to look at the steps of vitamin B12 biosynthesis, you would see the same kinds of reactions we’ve been seeing throughout the text—nucleophilic substitutions, eliminations, aldol reactions, nucleophilic acyl substitutions, and so forth. There are, of course, some complexities, but the fundamental mechanisms of organic chemistry remain the same, whether in the laboratory with smaller molecules or in organisms with larger molecules.
So what is there to be learned from studying metabolism? One good answer is given in the following Chemistry Matters, which relates the story of how knowledge of a biosynthetic pathway led to the design of new drugs that have saved millions of lives.
29.12: Chemistry MattersStatin Drugs
29 • Chemistry Matters 29 • Chemistry Matters
Coronary heart disease—the buildup of cholesterol-containing plaques on the walls of heart arteries—is the leading cause of death for people older than 20 in industrialized countries. It’s estimated that up to one-third of women and one-half of men will develop the disease at some point in their lives.
The onset of coronary heart disease is directly correlated with blood cholesterol levels (see the Chapter 27 Chemistry Matters), and the first step in disease prevention is to lower those levels. It turns out that only about 25% of your blood cholesterol comes from what you eat; the remaining 75%—about 1 gram each day—is biosynthesized in your liver from dietary fats and carbohydrates. Thus, any effective plan for lowering your cholesterol level means limiting the amount that your body makes, which is where a detailed chemical knowledge of cholesterol biosynthesis comes in.
We saw in Section 27.5 and Section 27.7 that all steroids, including cholesterol, are biosynthesized from the triterpenoid lanosterol, which in turn comes from acetyl CoA through isopentenyl diphosphate. If you knew all the mechanisms for all the chemical steps in cholesterol biosynthesis, you might be able to devise a drug that would block one of those steps, thereby short-circuiting the biosynthetic process and controlling the amount of cholesterol produced.
But we do know those mechanisms! Look back at the pathway for the biosynthesis of isopentenyl diphosphate from acetyl CoA, shown in Figure 27.8. It turns out that the rate-limiting step in the pathway is the reduction of 3-hydroxy-3-methylglutaryl CoA (abbreviated HMG-CoA) to mevalonate, brought about by the enzyme HMG-CoA reductase. If that enzyme could be stopped from functioning, cholesterol biosynthesis would also be stopped.
To find a drug that blocks HMG-CoA reductase, chemists did two simultaneous experiments on a large number of potential drug candidates isolated from soil microbes. In one experiment, the drug candidate and mevalonate were added to liver extract; in the second experiment, only the drug candidate was added without mevalonate. If cholesterol was produced only in the presence of added mevalonate but not in the absence of mevalonate, the drug candidate must have blocked the enzyme for mevalonate synthesis.
The drugs that block HMG-CoA reductase, and thus control cholesterol synthesis in the body, are called statins. In just the 10-year period following their introduction in 1994, the death rate from coronary heart disease decreased by 33% in the United States.
Like many drugs, statins don’t come without risks and have some serious side effects that people considering their use should be aware of, but approaching 30 years later, they remain among the most widely prescribed drugs in the world, with estimated annual sales of \$14 billion. Atorvastatin (Lipitor), simvastatin (Zocor), rosuvastatin (Crestor), pravastatin (Pravachol), and lovastatin (Mevacor) are some examples. An X-ray crystal structure of the active site in the HMG-CoA reductase enzyme is shown in the accompanying graphic, along with a molecule of atorvastatin (purple) that is tightly bound in the active site and stops the enzyme from functioning. A good understanding of organic chemistry certainly paid off in this instance.
29.13: Key Terms
29 • Key Terms 29 • Key Terms
• anabolism
• β-oxidation pathway
• catabolism
• citric acid cycle
• deamination
• gluconeogenesis
• glycolysis
• metabolism
• Schiff base
• transamination
29.14: Summary
29 • Summary 29 • Summary
Metabolism is the sum of all chemical reactions in the body. Reactions that break down large molecules into smaller fragments are called catabolism, and those that build up large molecules from small pieces are called anabolism. Although the details of specific biochemical pathways are sometimes complex, all the reactions that occur follow the normal rules of organic chemical reactivity.
The catabolism of fats begins with digestion, in which ester bonds are hydrolyzed to give glycerol and fatty acids. The fatty acids are degraded in the four-step β-oxidation pathway by removal of two carbons at a time, yielding acetyl CoA. Catabolism of carbohydrates begins with the hydrolysis of glycoside bonds to give glucose, which is degraded in the ten-step glycolysis pathway. Pyruvate, the initial product of glycolysis, is then converted into acetyl CoA. Acetyl CoA next enters the eight-step citric acid cycle, where it is further degraded into CO2. The cycle is a closed loop of reactions in which the product of the final step (oxaloacetate) is a reactant in the first step.
Catabolism of proteins is more complex than that of fats or carbohydrates because each of the 20 different amino acids is degraded by its own unique pathway. In general, though, the amino nitrogen atoms are removed and the substances that remain are converted into compounds that enter the citric acid cycle. Most amino acids lose their nitrogen atom by transamination, a reaction in which the –NH2 group of the amino acid trades places with the keto group of an α-keto acid such as α-ketoglutarate. The products are a new α-keto acid and glutamate.
The energy released in catabolic pathways is used in the electron-transport chain to make molecules of adenosine triphosphate, ATP. ATP, the final result of food catabolism, couples to and drives many otherwise unfavorable reactions.
Biomolecules are synthesized as well as degraded, but the pathways for anabolism and catabolism are not the exact reverse of one another. Fatty acids are biosynthesized from acetate by an 8-step pathway, and carbohydrates are made from pyruvate by the 11-step gluconeogenesis pathway. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.11%3A_Some_Conclusions_about_Biological_Chemistry.txt |
29 • Additional Problems 29 • Additional Problems
Visualizing Chemistry
Problem 29-17
Identify the amino acid that is a catabolic precursor of each of the following α-keto acids: (a)
(b)
Problem 29-18 Identify the following intermediate in the citric acid cycle, and tell whether it has R or S stereochemistry: Problem 29-19 The following compound is an intermediate in the biosynthesis of one of the 20 common α-amino acids. Which one is it likely to be, and what kind of chemical change must take place to complete the biosynthesis? Problem 29-20 The following compound is an intermediate in the pentose phosphate pathway, an alternative route for glucose metabolism. Identify the sugar it is derived from. Mechanism Problems Problem 29-21 In the pentose phosphate pathway for degrading sugars, ribulose 5-phosphate is converted to ribose 5-phosphate. Propose a mechanism for the isomerization. Problem 29-22 Another step in the pentose phosphate pathway for degrading sugars (see Problem 29-21) is the conversion of ribose 5-phosphate to glyceraldehyde 3-phosphate. What kind of organic process is occurring? Propose a mechanism for the conversion. Problem 29-23 One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of sedoheptulose 7-phosphate with glyceraldehyde 3-phosphate in the presence of a transaldolase to yield erythrose 4-phosphate and fructose 6-phosphate. (a) The first part of the reaction is the formation of a protonated Schiff base of sedoheptulose 7-phosphate with a lysine residue in the enzyme followed by a retro-aldol cleavage to give an enamine plus erythrose 4-phosphate. Show the structure of the enamine and the mechanism by which it is formed. (b)
The second part of the reaction is a nucleophilic addition of the enamine to glyceraldehyde 3-phosphate followed by hydrolysis of the Schiff base to give fructose 6-phosphate. Show the mechanism.
Problem 29-24 One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of xylulose 5-phosphate with ribose 5-phosphate in the presence of a transketolase to give glyceraldehyde 3-phosphate and sedoheptulose 7-phosphate. (a)
The first part of the reaction is nucleophilic addition of thiamin diphosphate (TPP) ylide to xylulose 5-phosphate, followed by a retro-aldol cleavage to give glyceraldehyde 3-phosphate and a TPP-containing enamine. Show the structure of the enamine and the mechanism by which it is formed.
(b) The second part of the reaction is addition of the enamine to ribose 5-phosphate followed by loss of TPP ylide to give sedoheptulose 7-phosphate. Show the mechanism.
Problem 29-25
The amino acid tyrosine is biologically degraded by a series of steps that include the following transformations:
The double-bond isomerization of maleoylacetoacetate to fumaroylacetoacetate is catalyzed by practically any nucleophile, :Nu. Propose a mechanism.
Problem 29-26
Propose a mechanism for the conversion of fumaroylacetoacetate to fumarate plus acetoacetate (see Problem 29-25).
Problem 29-27
Propose a mechanism for the conversion of acetoacetate to acetyl CoA (see Problem 29-25).
Problem 29-28
Design your own degradative pathway. You know the rules (organic mechanisms), and you’ve seen the kinds of reactions that occur in the biological degradation of fats and carbohydrates into acetyl CoA. If you were Mother Nature, what series of steps would you use to degrade the amino acid serine into acetyl CoA?
Problem 29-29
The amino acid serine is biosynthesized by a route that involves reaction of 3-phosphohydroxypyruvate with glutamate to give 3-phosphoserine. Propose a mechanism.
Problem 29-30
The amino acid leucine is biosynthesized from α-ketoisocaproate, which is itself prepared from α-ketoisovalerate by a multistep route that involves (1) reaction with acetyl CoA, (2) hydrolysis, (3) dehydration, (4) hydration, (5) oxidation, and (6) decarboxylation. Show the steps in the transformation, and propose a mechanism for each.
Problem 29-31 The amino acid cysteine, C3H7NO2S, is biosynthesized from a substance called cystathionine by a multistep pathway. (a)
The first step is a transamination. What is the product?
(b) The second step is an E1cB reaction. Show the products and the mechanism of the reaction. (c) The final step is a double-bond reduction. What organic cofactor is required for this reaction, and what is the product represented by the question mark in the equation?
Enzymes and Coenzymes
Problem 29-32
What chemical events occur during the digestion of food?
Problem 29-33
What is the difference between digestion and metabolism?
Problem 29-34
What is the difference between anabolism and catabolism?
Problem 29-35
Draw the structure of adenosine 5′-monophosphate (AMP), an intermediate in some biochemical pathways.
Problem 29-36
Cyclic adenosine monophosphate (cyclic AMP), a modulator of hormone action, is related to AMP (Problem 29-35) but has its phosphate group linked to two hydroxyl groups at C3′ and C5′ of the sugar. Draw the structure of cyclic AMP.
Problem 29-37
What general kind of reaction does ATP carry out?
Problem 29-38
What general kind of reaction does NAD+ carry out?
Problem 29-39
What general kind of reaction does FAD carry out?
Problem 29-40 What enzyme cofactor is associated with each of the following kinds of reactions? (a)
Transamination
(b) Carboxylation of a ketone (c) Decarboxylation of an α-keto acid
Problem 29-41
Lactate, a product of glucose catabolism in oxygen-starved muscles, can be converted into pyruvate by oxidation. What coenzyme do you think is needed? Write the equation in the normal biochemical format using a curved arrow.
Metabolism
Problem 29-42
Write the equation for the final step in the β-oxidation pathway of any fatty acid with an even number of carbon atoms.
Problem 29-43 Show the products of each of the following reactions: (a) (b) (c) Problem 29-44 Why aren’t the glycolysis and gluconeogenesis pathways the exact reverse of each other? Problem 29-45 How many moles of acetyl CoA are produced by catabolism of the following substances? (a) 1.0 mol of glucose (b)
1.0 mol of palmitic acid
(c) 1.0 mol of maltose
Problem 29-46 How many grams of acetyl CoA (MW = 809.6 amu) are produced by catabolism of the following substances? Which substance is the most efficient precursor of acetyl CoA on a weight basis? (a)
100.0 g of glucose
(b) 100.0 g of palmitic acid (c) 100.0 g of maltose
Problem 29-47 What is the structure of the α-keto acid formed by transamination of each of the following amino acids? (a)
Threonine
(b) Phenylalanine (c) Asparagine
Problem 29-48
The glycolysis pathway shown in Figure 29.8 has a number of intermediates that contain phosphate groups. Why can 3-phosphoglyceryl phosphate and phosphoenolpyruvate transfer a phosphate group to ADP while glucose 6-phosphate cannot?
Problem 29-49
Write a mechanism for the conversion of α-ketoglutarate to succinyl CoA in step 4 of the citric acid cycle (Figure 29.14).
Problem 29-50
In step 2 of the citric acid cycle (Figure 29.14), cis-aconitate reacts with water to give (2R,3S)-isocitrate. Does –OH add from the Re face of the double bond or from the Si face? What about –H? Does the addition of water occur with syn or anti geometry?
General Problems
Problem 29-51
In glycerol metabolism, the oxidation of sn-glycerol 3-phosphate to give dihydroxyacetone phosphate is catalyzed by sn-glycerol-3-phosphate dehydrogenase, with NAD+ as cofactor. The reaction is stereospecific, occurring exclusively on the Re face of the nicotinamide ring.
Which hydrogen in the NADH product comes from sn-glycerol 3-phosphate? Does it have pro-R or pro-S stereochemistry?
Problem 29-52
The primary fate of acetyl CoA under normal metabolic conditions is degradation in the citric acid cycle to yield CO2. When the body is stressed by prolonged starvation, however, acetyl CoA is converted into compounds called ketone bodies, which can be used by the brain as a temporary fuel. Fill in the missing information indicated by the four question marks in the following biochemical pathway for the synthesis of ketone bodies from acetyl CoA:
Problem 29-53
The initial reaction in Problem 29-52, conversion of two molecules of acetyl CoA to one molecule of acetoacetyl CoA, is a Claisen reaction. Assuming that there is a base present, show the mechanism of the reaction.
Problem 29-54
In step 6 of fatty-acid biosynthesis (Figure 29.6), acetoacetyl ACP is reduced stereospecifically by NADPH to yield an alcohol. Does hydride ion add to the Si face or the Re face of acetoacetyl ACP?
Problem 29-55
In step 7 of fatty-acid biosynthesis (Figure 29.6), dehydration of a β-hydroxy thioester occurs to give trans-crotonyl ACP. Is the dehydration a syn elimination or an anti elimination?
Problem 29-56
In step 8 of fatty-acid biosynthesis (Figure 29.6), reduction of trans-crotonyl ACP gives butyryl ACP. A hydride from NADPH adds to C3 of the crotonyl group from the Re face, and protonation on C2 occurs on the Si face. Is the reduction a syn addition or an anti addition? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.15%3A_Additional_Problems.txt |
• 30.1: Why This Chapter?
• 30.2: Molecular Orbitals of Conjugated Pi Systems
HOMO and LUMO are often referred to as frontier orbitals and their energy difference is termed the HOMO–LUMO gap. One common way of thinking about reactions in this way is through the concept of frontier orbitals. This idea says that if one species is going to donate electrons to another in order to form a new bond, then the donated electrons are most likely going to come from the highest occupied energy level.
• 30.3: Electrocyclic Reactions
An electrocyclic reaction is the concerted cyclization of a conjugated π-electron system by converting one π-bond to a ring forming σ-bond. The key sigma bond must be formed at the terminus of a pi system. These reactions classified by the number of pi electrons involved.
• 30.4: Stereochemistry of Thermal Electrocyclic Reactions
Frontier orbital theory can predict the stereochemistry of electrocyclic reactions. Electrons in the HOMO are the highest energy and therefore the most easily moved during a reaction. A molecular orbital diagram can be used to determine the orbital symmetry of a conjugated polyene's HOMO. Thermal reactions utilize the HOMO from the ground-state electron configuration of the molecular orbital diagram while photochemical reactions utilize the HOMO in the excited-state electron configuration.
• 30.5: Photochemical Electrocyclic Reactions
Electron excitation changes the symmetry of the new HOMO which has a corresponding effect on the reaction stereochemistry. Under photochemical reaction conditions conjugated dienes undergo disrotatory cyclization whereas under thermal conditions they underwent conrotatory cyclization. Likewise, conjugated triene undergo conrotatory photochemical cyclization while undergoing disrotatory thermal cyclization.
• 30.6: Cycloaddition Reactions
A concerted combination of two π-electron systems to form a ring of atoms having two new σ bonds and two fewer π bonds is called a cycloaddition reaction. The number of participating π-electrons in each component is given in brackets preceding the name of the reaction. The Diels-Alder reaction is the most useful cycloaddition reaction due to the ubiquity of 6-membered rings and its ability to reliably control stereochemistry in the product.
• 30.7: Stereochemistry of Cycloadditions
Frontier orbital theory can be used to predict if a given cycloaddition will occur with suprafacial or with antarafacial geometry. In a standard Diels-Alder reaction, bonding interactions are created when the electron containing HOMO of the diene donates electrons to the electron vacant LUMO of the other the dienophile. The dienophile has one pi bond, so it will use the pi MOs for a 2 atom system.
• 30.8: Sigmatropic Rearrangements
Molecular rearrangements in which a σ-bonded atom or group, flanked by one or more π-electron systems, shifts to a new location with a corresponding reorganization of the π-bonds are called sigmatropic reactions. The reactant and product have the same number and type of bonds, just different bond locations.
• 30.9: Some Examples of Sigmatropic Rearrangements
• 30.10: A Summary of Rules for Pericyclic Reactions
• 30.11: Chemistry Matters—Vitamin D, the Sunshine Vitamin
• 30.12: Key Terms
• 30.13: Summary
• 30.14: Additional Problems
30: Orbitals and Organic Chemistry - Pericyclic Reactions
Chapter Contents
30.1 Molecular Orbitals of Conjugated Pi Systems
30.2 Electrocyclic Reactions 30.3 Stereochemistry of Thermal Electrocyclic Reactions 30.4 Photochemical Electrocyclic Reactions 30.5 Cycloaddition Reactions 30.6 Stereochemistry of Cycloadditions 30.7 Sigmatropic Rearrangements 30.8 Some Examples of Sigmatropic Rearrangements 30.9 A Summary of Rules for Pericyclic Reactions
Broad outlines of both polar and radical reactions have been in place for more than a century, but our understanding of pericyclic reactions has emerged more recently. Prior to the mid-1960s, in fact, they were even occasionally referred to as “no-mechanism reactions.” They occur largely in the laboratory rather than in biological processes, but a knowledge of them is necessary, both for completeness in studying organic chemistry and in understanding biological pathways where they do occur.
Most organic reactions take place by polar mechanisms, in which a nucleophile donates two electrons to an electrophile in forming a new bond. Other reactions take place by radical mechanisms, in which each of two reactants donates one electron in forming a new bond. Both kinds of reactions occur frequently in the laboratory and in living organisms. Less common, however, is the third major class of organic reactions—pericyclic reactions.
A pericyclic reaction is one that occurs by a concerted process through a cyclic transition state. A concerted reaction is one in which all bonding changes occur simultaneously; no intermediates are involved. Rather than try to expand this definition now, we’ll begin by briefly reviewing some of the ideas of molecular orbital theory introduced in the chapters on Structure and Bonding and Conjugated Compounds and Ultraviolet Spectroscopy and then looking individually at the three main classes of pericyclic reactions: electrocyclic reactions, cycloadditions, and sigmatropic rearrangements. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.01%3A_Why_This_Chapter.txt |
A conjugated polyene, as we saw in Section 14.1, is one with alternating double and single bonds. According to molecular orbital (MO) theory, the p orbitals on the sp2-hybridized carbons of a conjugated polyene interact to form a set of π molecular orbitals whose energies depend on the number of nodes they have between nuclei. Molecular orbitals with fewer nodes are lower in energy than isolated p atomic orbitals and are bonding MOs; molecular orbitals with more nodes are higher in energy than isolated p orbitals and are antibonding MOs. Pi molecular orbitals of ethylene and 1,3-butadiene are shown in Figure 30.2.
A similar sort of molecular orbital description can be derived for any conjugated π electron system. 1,3,5-Hexatriene, for example, has three double bonds and six π MOs, as shown in Figure 30.3. In the ground state, only the three bonding orbitals, ψ1, ψ2, and ψ3, are filled. On irradiation with ultraviolet light, however, an electron is promoted from the highest-energy filled orbital (ψ3) to the lowest-energy unfilled orbital (ψ4*) to give an excited state (Section 14.7), in which ψ3 and ψ4* are each half-filled. (An asterisk denotes an antibonding orbital.)
What do molecular orbitals and their nodes have to do with pericyclic reactions? The answer is, everything. According to a series of rules formulated in the mid-1960s by R. B. Woodward and Roald Hoffmann at Harvard University, a pericyclic reaction can take place only if the symmetries of the reactant MOs are the same as the symmetries of the product MOs. In other words, the lobes of reactant MOs must be of the correct algebraic sign for bonding to occur in the transition state leading to product.
If the symmetries of reactant and product orbitals match up, or correlate, the reaction is said to be symmetry-allowed. If the symmetries of reactant and product orbitals don’t correlate, the reaction is symmetry-disallowed. Symmetry-allowed reactions often occur under relatively mild conditions, but symmetry-disallowed reactions can’t occur by concerted paths. They either take place by nonconcerted, higher-energy pathways, or they don’t take place at all.
The Woodward–Hoffmann rules for pericyclic reactions require an analysis of all reactant and product molecular orbitals, but Kenichi Fukui at Kyoto Imperial University in Japan introduced a simplified version. Hoffman and Fukui shared a Nobel Prize for their work, and Woodward would have joined them (for his second Nobel Prize) had he not passed away before the award was made. According to Fukui’s simplified version, we need to consider only two molecular orbitals, called the frontier orbitals. These frontier orbitals are the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). In ground-state 1,3,5-hexatriene, for example, ψ3 is the HOMO and ψ4* is the LUMO (Figure 30.3). In excited-state 1,3,5-hexatriene, however, ψ4* is the HOMO and ψ5* is the LUMO.
Problem 30-1
Look at Figure 30.2, and tell which molecular orbital is the HOMO and which is the LUMO for both ground and excited states of ethylene and 1,3-butadiene.
30.03: Electrocyclic Reactions
The best way to understand how orbital symmetry affects pericyclic reactions is to look at some examples. Let’s look first at a group of polyene rearrangements called electrocyclic reactions. An electrocyclic reaction is a pericyclic process that involves the cyclization of a conjugated acyclic polyene. One π bond is broken, the other π bonds change position, a new σ bond is formed, and a cyclic compound results. For example, a conjugated triene can be converted into a cyclohexadiene, and a conjugated diene can be converted into a cyclobutene.
Pericyclic reactions are reversible, and the position of the equilibrium depends on the specific case. In general, the triene $⇄⇄$ cyclohexadiene equilibrium favors the cyclic product, whereas the diene $⇄⇄$ cyclobutene equilibrium favors the less strained open-chain product.
The most striking feature of electrocyclic reactions is their stereochemistry. For example, (2E,4Z,6E)-2,4,6-octatriene yields only cis-5,6-dimethyl-1,3-cyclohexadiene when heated, and (2E,4Z,6Z)-2,4,6-octatriene yields only trans-5,6-dimethyl-1,3-cyclohexadiene. Remarkably, however, the stereochemical results change completely when the reactions are carried out under what are called photochemical, rather than thermal, conditions. Irradiation of (2E,4Z,6E)-2,4,6-octatriene with ultraviolet light (denoted hν) yields trans-5,6-dimethyl-1,3-cyclohexadiene (Figure 30.4).
A similar result is obtained for the thermal electrocyclic ring-opening of 3,4-dimethylcyclobutene. The trans isomer yields only (2E,4E)-2,4-hexadiene when heated, and the cis isomer yields only (2E,4Z)-2,4-hexadiene. On UV irradiation, however, the results are opposite. Cyclization of the 2E,4E isomer under photochemical conditions yields cis product (Figure 30.5).
To account for these results, we need to look at the two outermost lobes of the polyene MOs—the lobes that interact when cyclization occurs. There are two possibilities: lobes of like sign can be either on the same side of the molecule or on opposite sides.
For a bond to form, the outermost π lobes must rotate so that favorable bonding interaction is achieved—a positive lobe with a positive lobe or a negative lobe with a negative lobe. If two lobes of like sign are on the same side of the molecule, the two orbitals must rotate in opposite directions—one clockwise and one counterclockwise. This kind of motion is referred to as disrotatory.
Conversely, if lobes of like sign are on opposite sides of the molecule, both orbitals must rotate in the same direction, either both clockwise or both counterclockwise. This kind of motion is called conrotatory. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.02%3A_Molecular_Orbitals_of_Conjugated_Pi_Systems.txt |
How can we predict whether conrotatory or disrotatory motion will occur in a given case? According to frontier orbital theory, the stereochemistry of an electrocyclic reaction is determined by the symmetry of the polyene HOMO. The electrons in the HOMO are the highest-energy, most loosely held electrons and are therefore most easily moved during reaction. For thermal reactions, the ground-state electron configuration is used to identify the HOMO; for photochemical reactions, the excited-state electron configuration is used.
Let’s look again at the thermal ring-closure of conjugated trienes. According to Figure 30.3, the HOMO of a conjugated triene in its ground state has lobes of like sign on the same side of the molecule, a symmetry that predicts disrotatory ring-closure. This disrotatory cyclization is precisely what is observed in the thermal cyclization of 2,4,6-octatriene. The 2E,4Z,6E isomer yields cis product; the 2E,4Z,6Z isomer yields trans product (Figure 30.6).
In the same way, the ground-state HOMO of conjugated dienes (Figure 30.2) has a symmetry that predicts conrotatory ring-closure. In practice, however, the conjugated diene reaction can be observed only in the reverse direction (cyclobutene $→Figure 30.7).$
Note that a conjugated diene and a conjugated triene react with opposite stereochemistry. The diene opens and closes by a conrotatory path, whereas the triene opens and closes by a disrotatory path. This is due to the different symmetries of the diene and triene HOMOs.
It turns out that there is an alternating relationship between the number of electron pairs (double bonds) undergoing bond reorganization and the stereochemistry of ring-opening or -closure. Polyenes with an even number of electron pairs undergo thermal electrocyclic reactions in a conrotatory sense, whereas polyenes with an odd number of electron pairs undergo the same reactions in a disrotatory sense.
Problem 30-2
Draw the products you would expect from conrotatory and disrotatory cyclizations of (2Z,4Z,6Z)-2,4,6-octatriene. Which of the two paths would you expect the thermal reaction to follow?
Problem 30-3
trans-3,4-Dimethylcyclobutene can open by two conrotatory paths to give either (2E,4E)-2,4-hexadiene or (2Z,4Z)-2,4-hexadiene. Explain why both products are symmetry-allowed, and then account for the fact that only the 2E,4E isomer is obtained in practice.
30.05: Photochemical Electrocyclic Reactions
We noted previously that photochemical electrocyclic reactions take a different stereochemical course than their thermal counterparts, and we can now explain this difference. Ultraviolet irradiation of a polyene causes an excitation of one electron from the ground-state HOMO to the ground-state LUMO, thus changing their symmetries. But because electronic excitation changes the symmetries of HOMO and LUMO, it also changes the reaction stereochemistry. (2E,4E)-2,4-Hexadiene, for instance, undergoes photochemical cyclization by a disrotatory path, whereas the thermal reaction is conrotatory. Similarly, (2E,4Z,6E)-2,4,6-octatriene undergoes photochemical cyclization by a conrotatory path, whereas the thermal reaction is disrotatory (Figure 30.8).
Thermal and photochemical electrocyclic reactions always take place with opposite stereochemistry because the symmetries of the frontier orbitals are always different. Table 30.1 gives some simple rules that make it possible to predict the stereochemistry of electrocyclic reactions.
Table 30.1 Stereochemical Rules for Electrocyclic Reactions
Electron pairs (double bonds) Thermal reaction Photochemical reaction
Even number Conrotatory Disrotatory
Odd number Disrotatory Conrotatory
Problem 30-4
What product would you expect to obtain from the photochemical cyclization of (2E,4Z,6E)-2,4,6-octatriene? Of (2E,4Z,6Z)-2,4,6-octatriene?
30.06: Cycloaddition Reactions
A cycloaddition reaction is one in which two unsaturated molecules add to one another to yield a cyclic product. As with electrocyclic reactions, cycloadditions are governed by the orbital symmetry of the reactants. Symmetry-allowed processes often take place readily, but symmetry-disallowed processes take place with difficulty, if at all, and then only by nonconcerted pathways. Let’s look at two examples to see how they differ.
As we saw in Section 14.4, the Diels–Alder cycloaddition reaction is a pericyclic process that takes place between a diene (four π electrons) and a dienophile (two π electrons) to yield a cyclohexene product. Many thousands of Diels–Alder reactions are known. They often take place easily at room temperature or slightly above, and they are stereospecific with respect to substituents. For example, room-temperature reaction between 1,3-butadiene and diethyl maleate (cis) exclusively yields the cis-disubstituted cyclohexene product. A similar reaction between 1,3-butadiene and diethyl fumarate (trans) exclusively yields the trans-disubstituted product.
In contrast to the [4 + 2]-π-electron Diels–Alder reaction, the [2 + 2]-π-electron cycloaddition between two alkenes does not occur thermally. The [2 + 2] cycloaddition takes place only on irradiation, yielding cyclobutane products.
For a successful cycloaddition, the terminal π lobes of the two reactants must have the correct symmetry for bonding to occur. This can happen in either of two ways, called suprafacial and antarafacial. Suprafacial cycloadditions take place when a bonding interaction occurs between lobes on the same face of one reactant and lobes on the same face of the other reactant. Antarafacial cycloadditions take place when a bonding interaction occurs between lobes on the same face of one reactant and lobes on opposite faces of the other reactant (Figure 30.9).
Note that both suprafacial and antarafacial cycloadditions are symmetry-allowed. Geometric constraints often make antarafacial reactions difficult, however, because there must be a twisting of the π orbital system in one of the reactants. Thus, suprafacial cycloadditions are much more common for small π systems. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.04%3A_Stereochemistry_of_Thermal_Electrocyclic_Reactions.txt |
How can we predict whether a given cycloaddition reaction will occur with suprafacial or with antarafacial geometry? According to frontier orbital theory, a cycloaddition reaction takes place when a bonding interaction occurs between the HOMO of one reactant and the LUMO of the other. An intuitive explanation of this rule is to imagine that one reactant donates electrons to the other. As with electrocyclic reactions, it’s the electrons in the HOMO of the first reactant that are least tightly held and most likely to be donated. Of course when the second reactant accepts those electrons, they must go into a vacant, unoccupied orbital—the LUMO.
For a [4 + 2] cycloaddition (Diels–Alder reaction), let’s arbitrarily select the diene LUMO and the alkene HOMO. The symmetries of the two ground-state orbitals are such that bonding of the terminal lobes can occur with suprafacial geometry (Figure 30.10), so the Diels–Alder reaction takes place readily under thermal conditions. Note that, as with electrocyclic reactions, we need be concerned only with the terminal lobes. For purposes of prediction, interactions among the interior lobes need not be considered.
In contrast with the thermal [4 + 2] Diels–Alder reaction, the [2 + 2] cycloaddition of two alkenes to yield a cyclobutane can only occur photochemically. The explanation for this follows from orbital-symmetry arguments. Looking at the ground-state HOMO of one alkene and the LUMO of the second alkene, it’s apparent that a thermal [2 + 2] cycloaddition must take place by an antarafacial pathway (Figure 30.11a). Geometric constraints make the antarafacial transition state difficult, however, and so concerted thermal [2 + 2] cycloadditions are not observed.
In contrast with the thermal process, photochemical [2 + 2] cycloadditions are observed. Irradiation of an alkene with UV light excites an electron from ψ1, the ground-state HOMO, to ψ2*, which becomes the excited-state HOMO. Interaction between the excited-state HOMO of one alkene and the LUMO of the second alkene allows a photochemical [2 + 2] cycloaddition reaction to occur by a suprafacial pathway (Figure 30.11b).
The photochemical [2 + 2] cycloaddition reaction occurs smoothly, particularly with α,β-unsaturated carbonyl compounds, and represents one of the best methods known for synthesizing cyclobutane rings. For example:
Thermal and photochemical cycloaddition reactions always take place with opposite stereochemistry. As with electrocyclic reactions, we can categorize cycloadditions according to the total number of electron pairs (double bonds) involved in the rearrangement. Thus, a thermal [4 + 2] Diels–Alder reaction between a diene and a dienophile involves an odd number (three) of electron pairs and takes place by a suprafacial pathway. A thermal [2 + 2] reaction between two alkenes involves an even number (two) of electron pairs and must take place by an antarafacial pathway. For photochemical cyclizations, these selectivities are reversed. These general rules are given in Table 30.2.
Table 30.2 Stereochemical Rules for Cycloaddition Reactions
Electron pairs (double bonds) Thermal reaction Photochemical reaction
Even number Antarafacial Suprafacial
Odd number Suprafacial Antarafacial
Problem 30-5
What stereochemistry would you expect for the product of the Diels–Alder reaction between (2E,4E)-2,4-hexadiene and ethylene? What stereochemistry would you expect if (2E,4Z)-2,4-hexadiene were used instead?
Problem 30-6
1,3-Cyclopentadiene reacts with cycloheptatrienone to give the product shown. Tell what kind of reaction is involved, and explain the observed result. Is the reaction suprafacial or antarafacial? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.07%3A_Stereochemistry_of_Cycloadditions.txt |
A sigmatropic rearrangement, the third general kind of pericyclic reaction, is a process in which a σ-bonded substituent atom or group migrates across a π electron system from one position to another. A σ bond is broken in the reactant, the π bonds move, and a new σ bond is formed in the product. The σ-bonded group can be either at the end or in the middle of the π system, as the following [1,5] and [3,3] rearrangements illustrate:
The notations [1,5] and [3,3] describe the kind of rearrangement that is occurring. The numbers refer to the two groups connected by the σ bond in the reactant and designate the positions in those groups to which migration occurs. For example, in the [1,5] sigmatropic rearrangement of a 1,3-diene, the two groups connected by the σ bond are a hydrogen atom and a pentadienyl group. Migration occurs to position 1 of the H group (the only possibility) and to position 5 of the pentadienyl group. In the [3,3] rearrangement of an allylic vinylic ether, the two groups connected by the σ bond are an allylic group and the vinylic ether. Migration occurs to position 3 of the allylic group and also to position 3 of the vinylic ether.
Like electrocyclic reactions and cycloadditions, sigmatropic rearrangements are controlled by orbital symmetries. There are two possible modes of reaction: migration of a group across the same face of the π system is suprafacial, and migration of a group from one face of the π system to the other face is antarafacial (Figure 30.12).
Both suprafacial and antarafacial sigmatropic rearrangements are symmetry-allowed, but suprafacial rearrangements are often easier for geometric reasons. The rules for sigmatropic rearrangements are identical to those for cycloaddition reactions (Table 30.3).
Table 30.3 Stereochemical Rules for Sigmatropic Rearrangements
Electron pairs (double bonds) Thermal reaction Photochemical reaction
Even number Antarafacial Suprafacial
Odd number Suprafacial Antarafacial
Problem 30-7
Classify the following sigmatropic reaction by order [x,y], and tell whether it will proceed with suprafacial or antarafacial stereochemistry:
30.09: Some Examples of Sigmatropic Rearrangements
Because a [1,5] sigmatropic rearrangement involves three electron pairs (two π bonds and one σ bond), the orbital-symmetry rules in Table 30.3 predict a suprafacial reaction. In fact, the [1,5] suprafacial shift of a hydrogen atom across two double bonds of a π system is a commonly observed sigmatropic rearrangements. For example, 5-methyl-1,3-cyclopentadiene rapidly rearranges at room temperature to yield a mixture of 1-methyl-, 2-methyl-, and 5-methyl-isomers.
As another example, heating 5,5,5-trideuterio-(3Z)-1,3-pentadiene causes scrambling of deuterium between positions 1 and 5.
Both these [1,5] hydrogen shifts occur by a symmetry-allowed suprafacial pathway, as illustrated in Figure 30.13. In contrast with these thermal [1,5] sigmatropic hydrogen shifts, however, thermal [1,3] hydrogen shifts are unknown. If they were to occur, they would have to proceed by a strained antarafacial reaction pathway.
Two other important sigmatropic reactions are the Claisen rearrangement of either an allyl aryl ether ($H2C═CHCH2─O─ArH2C═CHCH2─O─Ar$) or an allyl vinyl ($H2C═CHCH2─O─CH═CH2H2C═CHCH2─O─CH═CH2$) ether, and the Cope rearrangement of a 1,5 hexadiene to an isomeric 1,5-diene. These two rearrangements, along with the Diels–Alder reaction, are among the most generally useful pericyclic reactions for organic synthesis. Thousands of examples of all three are known.
Like the Diels–Alder reaction discussed in Section 14.4 and Section 14.5, the Claisen rearrangement takes place in a single step through a pericyclic mechanism in which a reorganization of bonding electrons occurs in a six-membered, cyclic transition state. The 6-allyl-2,4-cyclohexadienone intermediate then isomerizes to o-allylphenol (Figure 30.14).
Evidence for this mechanism comes from the observation that the rearrangement takes place with transposition of the allyl group. That is, allyl phenyl ether containing a 14C label on the allyl ether carbon atom yields o-allylphenol in which the label is on the terminal vinylic carbon (green in Figure 30.14).
Problem 30-8
Draw a curved-arrow mechanism for the Cope arrangement just shown.
The Cope rearrangement that converts a 1,5-diene to an isomeric 1,5-diene is somewhat limited but a modification called the Oxy Cope Rearrangement is wider in scope. As shown in the following example, a 1-5 diene with an –OH next to the double bond can be converted into an oxy-anion by reaction with a strong base such as potassium hydride (KH). A Cope rearrangement then occurs, and reaction with aqueous acid gives an enol that tautomerizes to an aldehyde.
Although biological examples of pericyclic reactions are relatively rare, a much-studied example occurs in bacteria during biosynthesis of the essential amino acid phenylalanine. Phenylalanine arises from the precursor chorismate through a Claisen rearrangement to prephenate, followed by decarboxylation to phenylpyruvate and reductive amination (Figure 30.16). You might note that the reductive amination of phenylpyruvate is the exact reverse of the transamination process shown in Figure 29.17, by which amino acids are deaminated. In addition, the reductive amination of ketones is a standard method for preparing amines in the laboratory, as we saw in Section 24.6.
Problem 30-9
Propose a mechanism to account for the fact that heating 1-deuterioindene scrambles the isotope label to all three positions on the five-membered ring.
Problem 30-10
What product would you expect from Claisen rearrangement of 2-butenyl phenyl ether?
Problem 30-11
When a 2,6-disubstituted allyl phenyl ether is heated in an attempted Claisen rearrangement, migration occurs to give the p-allyl product as the result of two sequential pericyclic reactions. Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.08%3A_Sigmatropic_Rearrangements.txt |
How can you keep straight all the rules about pericyclic reactions? The summary in Table 30.1, Table 30.2, and Table 30.3 can be distilled into a mnemonic phrase that provides an easy way to predict the stereochemical outcome of any pericyclic reaction:
The Electrons Circle Around (TECA)
Thermal reactions with an Even number of electron pairs are Conrotatory or Antarafacial.
A change either from thermal to photochemical or from an even to an odd number of electron pairs changes the outcome from conrotatory/antarafacial to disrotatory/suprafacial. A change from both thermal and even to photochemical and odd causes no change because two negatives make a positive.
These selection rules are summarized in Table 30.4; knowing them gives you the ability to predict the stereochemistry of literally thousands of pericyclic reactions.
Table 30.4 Stereochemical Rules for Pericyclic Reactions
Electronic state Electron pairs Stereochemistry
Ground state (thermal) Even number
Odd number
Antara–con
Supra–dis
Excited state (photochemical) Even number
Odd number
Supra–dis
Antara–con
Problem 30-12
Predict the stereochemistry of the following pericyclic reactions:
1. The thermal cyclization of a conjugated tetraene
2. The photochemical cyclization of a conjugated tetraene
3. A photochemical [4 + 4] cycloaddition
4. A thermal [2 + 6] cycloaddition
5. A photochemical [3,5] sigmatropic rearrangement
30.11: Chemistry MattersVitamin D the Sunshine Vitamin
30 • Chemistry Matters 30 • Chemistry Matters
Vitamin D, discovered in 1918, is a general name for two related compounds, cholecalciferol (vitamin D3) and ergocalciferol (vitamin D2). Both are derived from steroids (Section 27.6) and differ only in the nature of the hydrocarbon side chain attached to the five-membered ring. Cholecalciferol comes primarily from dairy products and fish; ergocalciferol comes from some vegetables.
The function of vitamin D in the body is to control the calcification of bones by increasing intestinal absorption of calcium. When sufficient vitamin D is present, approximately 30% of ingested calcium is absorbed, but in the absence of vitamin D, calcium absorption falls to about 10%. A deficiency of vitamin D thus leads to poor bone growth and to the diseases rickets in children and osteoporosis in adults.
Actually, neither vitamin D2 nor D3 is present in foods. Rather, foods contain the precursor molecules 7-dehydrocholesterol and ergosterol. In the presence of sunlight, both precursors are converted in the outer, epidermal layer of skin to the active vitamins, hence the nickname for vitamin D, the “sunshine vitamin.”
Pericyclic reactions are unusual in living organisms, and the photochemical synthesis of vitamin D is one of only a few well-studied examples. The reaction takes place in two steps, an electrocyclic ring-opening of a cyclohexadiene to yield an open-chain hexatriene, followed by a sigmatropic [1,7] H shift to yield an isomeric hexatriene. Only the initial electrocyclic ring-opening requires irradiation by so-called UVB light of 295 to 300 nm wavelength. The subsequent sigmatropic [1,7] H shift occurs spontaneously by a thermal isomerization.
Following synthesis under the skin, further metabolic processing of cholecalciferol and ergocalciferol in the liver and kidney introduces two additional –OH groups to give the active forms of the vitamin, calcitriol and ergocalcitriol.
30.12: Key Terms
30 • Key Terms 30 • Key Terms
• antarafacial
• conrotatory
• cycloaddition reaction
• disrotatory
• electrocyclic reaction
• frontier orbitals
• highest occupied molecular orbital (HOMO)
• lowest unoccupied molecular orbital (LUMO)
• pericyclic reaction
• photochemical reaction
• sigmatropic rearrangement
• suprafacial
• symmetry-allowed
• symmetry-disallowed
30.13: Summary
30 • Summary 30 • Summary
A pericyclic reaction takes place in a single step through a cyclic transition state without intermediates. There are three major classes of pericyclic processes: electrocyclic reactions, cycloaddition reactions, and sigmatropic rearrangements. The stereochemistry of these reactions is controlled by the symmetry of the orbitals involved in bond reorganization.
Electrocyclic reactions involve the cyclization of conjugated acyclic polyenes. For example, 1,3,5-hexatriene cyclizes to 1,3-cyclohexadiene on heating. Electrocyclic reactions can occur by either conrotatory or disrotatory pathways, depending on the symmetry of the terminal lobes of the π system. Conrotatory cyclization requires that both lobes rotate in the same direction, whereas disrotatory cyclization requires that the lobes rotate in opposite directions. The reaction course in a specific case can be found by looking at the symmetry of the highest occupied molecular orbital (HOMO).
Cycloaddition reactions are those in which two unsaturated molecules add together to yield a cyclic product. For example, Diels–Alder reaction between a diene (four π electrons) and a dienophile (two π electrons) yields a cyclohexene. Cycloadditions can take place either by suprafacial or antarafacial pathways. Suprafacial cycloaddition involves interaction between lobes on the same face of one component and on the same face of the second component. Antarafacial cycloaddition involves interaction between lobes on the same face of one component and on opposite faces of the other component. The reaction course in a specific case can be found by looking at the symmetry of the HOMO of one component and the lowest unoccupied molecular orbital (LUMO) of the other.
Sigmatropic rearrangements involve the migration of a σ-bonded group across a π electron system. For example, Claisen rearrangement of an allylic vinylic ether yields an unsaturated carbonyl compound, and Cope rearrangement of a 1,5-hexadiene yields an isomeric 1,5-hexadiene. Sigmatropic rearrangements can occur with either suprafacial or antarafacial stereochemistry; the selection rules for a given case are the same as those for cycloaddition reactions.
The stereochemistry of any pericyclic reaction can be predicted by counting the total number of electron pairs (bonds) involved in bond reorganization and then applying the mnemonic “The Electrons Circle Around.” That is, thermal (ground-state) reactions involving an even number of electron pairs occur with either conrotatory or antarafacial stereochemistry. Exactly the opposite rules apply to photochemical (excited-state) reactions. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.10%3A_A_Summary_of_Rules_for_Pericyclic_Reactions.txt |
30 • Additional Problems 30 • Additional Problems
Visualizing Chemistry
Problem 30-13
Predict the product obtained when the following substance is heated:
Problem 30-14
The 13C NMR spectrum of homotropilidene taken at room temperature shows only three peaks. Explain.
Mechanism Problems
Problem 30-15
The following rearrangement of N-allyl-N,N-dimethylanilinium ion has been observed. Propose a mechanism.
Problem 30-16 Plastic photochromic sunglasses are based on the following reversible rearrangement of a dye inside the lenses that occurs when the lenses are exposed to sunlight. The original dye absorbs UV light but not visible light and is thus colorless, while the rearrangement product absorbs visible light and is thus darkened. (a)
Show the mechanism of the rearrangement.
(b) Why does the rearrangement product absorb at a longer wavelength (visible light) than the original dye (UV)?
Problem 30-17
The sex hormone estrone has been synthesized by a route that involves the following step. Identify the pericyclic reactions involved, and propose a mechanism.
Problem 30-18
Coronafacic acid, a bacterial toxin, was synthesized using a key step that involves three sequential pericyclic reactions. Identify them, and propose a mechanism for the overall transformation. How would you complete the synthesis?
Problem 30-19
The following thermal rearrangement involves two pericyclic reactions in sequence. Identify them, and propose a mechanism to account for the observed result.
Electrocyclic Reactions
Problem 30-20 Do the following electrocyclic reactions take place in a conrotatory or disrotatory manner? Under what conditions, thermal or photochemical, would you carry out each reaction? (a) (b) Problem 30-21 The following thermal isomerization occurs under relatively mild conditions. Identify the pericyclic reactions involved, and show how the rearrangement occurs. Problem 30-22 Would you expect the following reaction to proceed in a conrotatory or disrotatory manner? Show the stereochemistry of the cyclobutene product, and explain your answer. Problem 30-23 Heating (1Z,3Z,5Z)-1,3,5-cyclononatriene to 100 °C causes cyclization and formation of a bicyclic product. Is the reaction conrotatory or disrotatory? What is the stereochemical relationship of the two hydrogens at the ring junctions, cis or trans? Problem 30-24 (2E,4Z,6Z,8E)-2,4,6,8-Decatetraene has been cyclized to give 7,8-dimethyl-1,3,5-cyclooctatriene. Predict the manner of ring-closure—conrotatory or disrotatory—for both thermal and photochemical reactions, and predict the stereochemistry of the product in each case. Problem 30-25 Answer Problem 30-24 for the thermal and photochemical cyclizations of (2E,4Z,6Z,8Z)-2,4,6,8-decatetraene. Problem 30-26 The cyclohexadecaoctaene shown isomerizes to two different isomers, depending on reaction conditions. Explain the observed results, and indicate whether each reaction is conrotatory or disrotatory. Cycloaddition Reactions Problem 30-27 Which of the following reactions is more likely to occur? Explain. Problem 30-28 The following reaction takes place in two steps, one of which is a cycloaddition while the other is a reverse cycloaddition. Identify the two pericyclic reactions, and show how they occur. Problem 30-29 Two sequential pericyclic reactions are involved in the following furan synthesis. Identify them, and propose a mechanism for the transformation. Sigmatropic Rearrangements Problem 30-30 Predict the product of the following pericyclic reaction. Is this [5,5] shift a suprafacial or an antarafacial process? Problem 30-31 Propose a pericyclic mechanism to account for the following transformation: Problem 30-32 Vinyl-substituted cyclopropanes undergo thermal rearrangement to yield cyclopentenes. Propose a mechanism for the reaction, and identify the pericyclic process involved. Problem 30-33 The following synthesis of dienones occurs readily. Propose a mechanism to account for the results, and identify the kind of pericyclic reaction involved. Problem 30-34 Karahanaenone, a terpenoid isolated from oil of hops, has been synthesized by the thermal reaction shown. Identify the kind of pericyclic reaction, and explain how karahanaenone is formed. General Problems Problem 30-35 What stereochemistry—antarafacial or suprafacial—would you expect to observe in the following reactions? (a) A photochemical [1,5] sigmatropic rearrangement (b)
A thermal [4 + 6] cycloaddition
(c) A thermal [1,7] sigmatropic rearrangement (d) A photochemical [2 + 6] cycloaddition
Problem 30-36
Bicyclohexadiene, also known as Dewar benzene, is extremely stable despite the fact that its rearrangement to benzene is energetically favored. Explain why the rearrangement is so slow.
Problem 30-37
Ring-opening of the trans-cyclobutene isomer shown takes place at much lower temperature than a similar ring-opening of the cis-cyclobutene isomer. Explain the temperature effect, and identify the stereochemistry of each reaction as either conrotatory or disrotatory.
Problem 30-38
Photolysis of the cis-cyclobutene isomer in Problem 30-37 yields cis-cyclododecaen-7-yne, but photolysis of the trans isomer yields trans-cyclododecaen-7-yne. Explain these results, and identify the type and stereochemistry of the pericyclic reaction.
Problem 30-39
The 1H NMR spectrum of bullvalene at 100 °C consists only of a single peak at 4.22 δ. Explain.
Problem 30-40
The following rearrangement was devised and carried out to prove the stereochemistry of [1,5] sigmatropic hydrogen shifts. Explain how the observed result confirms the predictions of orbital symmetry.
Problem 30-41
The following reaction is an example of a [2,3] sigmatropic rearrangement. Would you expect the reaction to be suprafacial or antarafacial? Explain.
Problem 30-42
When the compound having a cyclobutene fused to a five-membered ring is heated, (1Z,3Z)-1,3-cycloheptadiene is formed. When the related compound having a cyclobutene fused to an eight-membered ring is heated, however, (1E,3Z)-1,3-cyclodecadiene is formed. Explain these results, and suggest a reason why opening of the eight-membered ring occurs at a lower temperature.
Problem 30-43
In light of your answer to Problem 30-42, explain why a mixture of products occurs in the following reaction:
Problem 30-44
In nature, the enzyme chorismate mutase catalyzes a Claisen rearrangement of chorismate that involves both the terminal double bond and the double bond with the highlighted carbon. What is the structure of prephenate, the biological precursor to the amino acids phenylalanine and tyrosine?
Problem 30-45 Predict the product(s) if the starting materials underwent a Claisen rearrangement. Draw arrows to illustrate the rearrangement of electrons. (a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.14%3A_Additional_Problems.txt |
Synthetic polymer are man-made polymer that is not a biopolymer (e..g, proteins or complex carbohydrates). Synthetic polymers are mostly non-biodegradable and often synthesized from petroleum. The eight most common types of synthetic organic polymers are: Low-density polyethylene (LDPE), High-density polyethylene (HDPE), Polypropylene (PP), Polyvinyl chloride (PVC), Polystyrene (PS), Nylon, nylon 6, nylon 6,6, Teflon (Polytetrafluoroethylene) and Thermoplastic polyurethanes (TPU).
• 31.1: Why This Chapter?
• 31.2: Chain-Growth Polymers
Polymers resulting from additions to alkenes monomers are chain-growth polymers. In these processes each addition step results in a longer chain which ends in a reactive site. The mechanism of each addition step is the same, and each addition step adds another monomer to extend the chain by one repeating unit. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions.
• 31.3: Stereochemistry of Polymerization- Ziegler-Natta Catalysts
An efficient and stereospecific catalytic polymerization procedure was developed by Karl Ziegler (Germany) and Giulio Natta (Italy) in the 1950's. Ziegler-Natta catalysts are prepared by reacting certain transition metal halides with organometallic reagents. The catalyst formed by reaction of triethylaluminum with titanium tetrachloride is commonly used. Ziegler-Natta catalysts allowed for the first time, the stereochemically controlled synthesis of polymers with virtually no branching.
• 31.4: Copolymers
Homopolymers are made with a single monomer and are made up of identical repeating units. Copolymers is made when two or more different monomers are polymerized together to create a polymer with variable repeating units. For example the monomers hexafluoropropene and vinylidene fluoride can be polymerized together to create the copolymer vitron which is used to create durable gaskets.
• 31.5: Step-Growth Polymers
Often, the reactions used to link these monomers include multiple nucleophilic acyl substitutions. Step-growth polymerizations usually use two different monomers, neither of which would undergo polymerization on its own. The two monomers are multifuntional and complementary to each other, such that each provides the other with a reactive partner. In this section, we will be focusing on monomers which are difunctional, meaning they contain two of the same reactive functional group.
• 31.6: Olefin Metathesis Polymerization
Alkene metathesis reactions are gaining wide popularity in synthesizing unsaturated olefinic compounds. Central to this catalysis is a metal carbene intermediate that reacts with olefins to give different olefinic compounds. When two different olefin substrates are used, the reaction is called the “cross metathesis” owing to the fact that the olefinic ends are exchanged. In a process called, olefin metathesis polymerization, unsaturated olefinic polymers can be created by a metathesis reaction.
• 31.7: Intramolecular Olefin Metathesis
• 31.8: Polymer Structure and Physical Properties
To account for the physical differences between the different types of polymers, the nature of the aggregate macromolecular structure, or morphology, of each substance must be considered. Because polymer molecules are so large, they generally pack together in a non-uniform fashion, with ordered or crystalline-like regions, called crystallites, mixed together with disordered or amorphous domains. In some cases the entire solid may be amorphous, composed entirely of tangled macromolecules.
• 31.9: Chemistry Matters—Degradable Polymers
• 31.10: Key Terms
• 31.11: Summary
• 31.12: Additional Problems
31: Synthetic Polymers
Chapter Contents
31.1 Chain-Growth Polymers
31.2 Stereochemistry of Polymerization: Ziegler–Natta Catalysts 31.3 Copolymers 31.4 Step-Growth Polymers 31.5 Olefin Metathesis Polymerization 31.6 Intramolecular Olefin Metathesis 31.7 Polymer Structure and Physical Properties
Our treatment of polymers has thus far been dispersed over several chapters, but it’s also important to take a comprehensive view. In the present chapter, we’ll look further at how polymers are made, and we’ll see how polymer structure correlates with physical properties. No course in organic chemistry would be complete without a look at polymers.
Polymers are a fundamental part of the modern world, used in everything from coffee cups to cars to clothing. In medicine, too, their importance is growing, with uses as diverse as cardiac pacemakers, artificial heart valves, and biodegradable sutures.
We’ve seen on several occasions in previous chapters that a polymer, whether synthetic or biological, is a large molecule built up by repetitive bonding of many smaller units, or monomers. Polyethylene, for instance, is a synthetic polymer made from ethylene (Section 8.10), nylon is a synthetic polyamide made from a diacid and a diamine (Section 21.9), and proteins are biological polyamides made from amino acids. Note that polymers are often drawn by indicating their repeating unit in parentheses. The repeating unit in polystyrene, for example, comes from the monomer styrene. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/31%3A_Synthetic_Polymers/31.01%3A_Why_This_Chapter.txt |
Synthetic polymers are classified by their method of synthesis as either chain-growth or step-growth. These categories are somewhat imprecise but nevertheless provide a useful distinction. Chain-growth polymers are produced by chain-reaction polymerization in which an initiator adds to the carbon–carbon double bond of an unsaturated substrate (a vinyl monomer) to yield a reactive intermediate. This intermediate reacts with a second molecule of monomer to yield a new intermediate, which reacts with a third monomer unit, and so on.
The initiator can be a radical, an acid, or a base. Historically, as we saw in Section 8.10, radical polymerization was the most common method because it can be carried out with practically any vinyl monomer.
Acid-catalyzed (cationic) polymerization, by contrast, is effective only with vinyl monomers that contain an electron-donating group (EDG) capable of stabilizing the chain-carrying carbocation intermediate.
Isobutylene (2-methylpropene) is a good example of a monomer that polymerizes rapidly under cationic conditions. The reaction is carried out commercially at –80 °C, using BF3 and a small amount of water to generate BF3OH H+ catalyst. The product is used in the manufacture of truck and bicycle inner tubes.
Vinyl monomers with electron-withdrawing groups (EWG) can be polymerized by basic (anionic) catalysts. The chain-carrying step is a conjugate nucleophilic addition of an anion to the unsaturated monomer (Section 19.13).
Acrylonitrile ($H2C═CHCNH2C═CHCN$), methyl methacrylate [$H2C═C(CH3)CO2CH3H2C═C(CH3)CO2CH3$], and styrene ($H2C═CHC6H5H2C═CHC6H5$) can all be polymerized anionically. The polystyrene used in foam coffee cups, for example, is prepared by anionic polymerization of styrene using butyllithium as catalyst.
An interesting example of anionic polymerization accounts for the remarkable properties of “super glue,” one drop of which can support up to 2000 lb. Super glue is simply a solution of pure methyl α-cyanoacrylate, which has two electron-withdrawing groups that make anionic addition particularly easy. Trace amounts of water or bases on the surface of an object are sufficient to initiate polymerization of the cyanoacrylate and bind articles together. Skin is a good source of the necessary basic initiators, and many people have found their fingers stuck together after inadvertently touching super glue. So good is super glue at binding tissues that related cyanoacrylate esters such as Dermabond are often used in place of sutures to close wounds.
Problem 31-1
Order the following monomers with respect to their expected reactivity toward cationic polymerization, and explain your answer:
H2C$\text{═}$CHCH3, H2C$\text{═}$CHCl, H2C$\text{═}$CH–C6H5, H2C$\text{═}$CHCO2CH3
Problem 31-2
Order the following monomers with respect to their expected reactivity toward anionic polymerization, and explain your answer:
H2C$\text{═}$CHCH3, H2C$\text{═}$CHC$\text{≡}$N, H2C$\text{═}$CHC6H5
Problem 31-3
Polystyrene is produced commercially by reaction of styrene with butyllithium as an anionic initiator. Using resonance structures, explain how the chain-carrying intermediate is stabilized.
31.03: Stereochemistry of Polymerization- Ziegler-Natta Catalysts
Although we didn’t point it out when discussing chain-growth polymers in Section 8.10, the polymerization of a substituted vinyl monomer can lead to a polymer with numerous chirality centers in its chain. Propylene, for example, might polymerize with any of the three stereochemical outcomes shown in Figure 31.2. A polymer with all methyl groups on the same side of the zigzag backbone is called isotactic, one in which the methyl groups alternate regularly on opposite sides of the backbone is called syndiotactic, and one with its methyl groups randomly oriented is called atactic.
The three different stereochemical forms of polypropylene all have somewhat different properties, and all can be made by using the right polymerization catalyst. Propylene polymerization using radical initiators does not work well, but polymerization using Ziegler–Natta catalysts allows preparation of isotactic, syndiotactic, and atactic polypropylene.
Ziegler–Natta catalysts—there are many different formulations—are organometallic transition-metal complexes prepared by treatment of an alkylaluminum with a titanium compound. Triethylaluminum with titanium tetrachloride is a typical preparation.
$(CH3CH2)3Al+TiCl4→A Ziegler–Natta catalyst(CH3CH2)3Al+TiCl4→A Ziegler–Natta catalyst$
Following their introduction in 1953, Ziegler–Natta catalysts revolutionized the field of polymer chemistry because of two advantages: first, the resultant polymers are linear, with practically no chain branching, and second, they are stereochemically controllable. Isotactic, syndiotactic, and atactic forms can all be produced, depending on the catalyst system used.
The active form of a Ziegler–Natta catalyst is an alkyltitanium intermediate with a vacant coordination site on the metal. Coordination of alkene monomer to the titanium occurs, and the coordinated alkene then inserts into the carbon–titanium bond to extend the alkyl chain. A new coordination site opens up during the insertion step, so the process repeats indefinitely.
The linear polyethylene produced by the Ziegler–Natta process, called high-density polyethylene, is a highly crystalline polymer with 4000 to 7000 ethylene units per chain and molecular weights in the range 100,000 to 200,000 amu. High-density polyethylene has greater strength and heat resistance than the branched product of radical-induced polymerization, called low-density polyethylene, and is used to produce plastic squeeze bottles and molded housewares.
Polyethylenes of even higher molecular weights are produced for specialty applications. So-called high-molecular-weight (HMW) polyethylene contains 10,000 to 18,000 monomer units per chain and molecular weights in the range of 300,000–500,000. It is used for underground pipes and large containers. Ultrahigh-molecular-weight (UHMW) polyethylene contains more than 100,000 monomer units per chain and has molecular weights ranging from 3,000,000 to 6,000,000 amu. It is used in bearings, conveyor belts, and bulletproof vests, among other applications requiring exceptional wear resistance.
Problem 31-4
Vinylidene chloride,
, does not polymerize in isotactic, syndiotactic, and atactic forms. Explain.
Problem 31-5
Polymers such as polypropylene contain a large number of chirality centers. Would you therefore expect samples of isotactic, syndiotactic, or atactic polypropylene to rotate plane-polarized light? Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/31%3A_Synthetic_Polymers/31.02%3A_Chain-Growth_Polymers.txt |
Up to this point we’ve discussed only homopolymers—polymers that are made up of identical repeating units. In practice, however, copolymers are more important commercially. Copolymers are obtained when two or more different monomers are allowed to polymerize together. For example, copolymerization of vinyl chloride with vinylidene chloride (1,1-dichloroethylene) in a 1 : 4 ratio leads to the polymer Saran.
Copolymerization of monomer mixtures often leads to materials with properties quite different from those of either corresponding homopolymer, giving the polymer chemist a vast flexibility for devising new materials. Table 31.1 lists some common copolymers and their commercial applications.
Table 31.1 Some Common Copolymers and Their Uses
Monomers Structures Trade name Uses
Vinyl chloride
Vinylidene chloride
Saran Fibers, food packaging
Styrene
1,3-Butadiene
SBR (styrene–butadiene rubber) Tires, rubber articles
Hexafluoropropene
Vinylidene fluoride
Viton Gaskets, seals
Acrylonitrile
1,3-Butadiene
Nitrile rubber Adhesives, hoses
Isobutylene
Isoprene
Butyl rubber Inner tubes
Acrylonitrile
1,3-Butadiene
Styrene
ABS (monomer initials) Pipes, high-impact applications
Several different types of copolymers can be defined, depending on the distribution of monomer units in the chain. If monomer A is copolymerized with monomer B, for instance, the resultant product might have a random distribution of the two units throughout the chain, or it might have an alternating distribution.
The exact distribution of monomer units depends on the initial proportions of the two reactant monomers and their relative reactivities. In practice, neither perfectly random nor perfectly alternating copolymers are usually found. Most copolymers have many random imperfections.
Two other forms of copolymers that can be prepared under certain conditions are called block copolymers and graft copolymers. Block copolymers are those in which different blocks of identical monomer units alternate with each other; graft copolymers are those in which homopolymer branches of one monomer unit are “grafted” onto a homopolymer chain of another monomer unit.
Block copolymers are prepared by initiating the polymerization of one monomer as if growing a homopolymer chain and then adding an excess of the second monomer to the still-active reaction mix. Graft copolymers are made by gamma irradiation of a completed homopolymer chain in the presence of the second monomer. The high-energy irradiation knocks hydrogen atoms off the homopolymer chain at random points, thus generating new radical sites that can initiate polymerization of the added monomer.
Problem 31-6
Draw the structure of an alternating segment of butyl rubber, a copolymer of isoprene (2-methyl-1,3-butadiene) and isobutylene (2-methylpropene) prepared using a cationic initiator.
Problem 31-7
Irradiation of poly(1,3-butadiene), followed by addition of styrene, yields a graft copolymer that is used to make rubber soles for shoes. Draw the structure of a representative segment of this styrene–butadiene graft copolymer.
31.05: Step-Growth Polymers
Step-growth polymers are produced by reactions in which each bond in the polymer is formed stepwise, independently of the other bonds. Like the polyamides (nylons) and polyesters that we saw in Section 21.9, most step-growth polymers are produced by reaction between two difunctional reactants. The polymer Nylon 66, for instance, is made by reaction between the six-carbon adipic acid and the six-carbon hexamethylenediamine (1,6-hexanediamine). Alternatively, a single reactant with two different functional groups can polymerize. Nylon 6 is made by polymerization of the six-carbon caprolactam. The reaction is initiated by adding a small amount of water, which hydrolyzes some caprolactam to 6-aminohexanoic acid. Nucleophilic addition of the amino group to caprolactam then propagates the polymerization.
Polycarbonates
Polycarbonates are like polyesters, but their carbonyl group is linked to two –OR groups, [$O═C(OR)2O═C(OR)2$]. Lexan, for instance, is a polycarbonate prepared from diphenyl carbonate and a diphenol called bisphenol A. Lexan has unusually high impact strength, making it valuable for use in machinery housings, telephones, bicycle safety helmets, and bulletproof glass.
Polyurethanes
A urethane is a carbonyl-containing functional group in which the carbonyl carbon is bonded to both an –OR group and an –NR2 group. As such, a urethane is halfway between a carbonate and a urea.
A urethane is typically prepared by nucleophilic addition reaction of an alcohol with an isocyanate ($R─N═C═OR─N═C═O$), so a polyurethane is prepared by reaction of a diol with a diisocyanate. The diol is usually a low-molecular-weight polymer (MW ≈ 1000 amu) with hydroxyl end-groups; the diisocyanate is often toluene-2,4-diisocyanate.
Several different kinds of polyurethanes can be produced, depending on the nature of the polymeric alcohol used. One major use of polyurethane is in the stretchable spandex fibers used for bathing suits and athletic gear. These polyurethanes have a fairly low degree of cross-linking so that the resultant polymer is soft and elastic. A second major use of polyurethanes is in the foams used for insulation. Foaming occurs when a small amount of water is added during polymerization, giving a carbamic acid intermediate that spontaneously loses bubbles of CO2.
Polyurethane foams are generally made using a polyalcohol rather than a diol as the monomer so that the polymer has a high amount of three-dimensional cross-linking. The result is a rigid but very light foam suitable for use as thermal insulation in building construction and portable ice chests.
Problem 31-8
Poly(ethylene terephthalate), or PET, is a polyester used to make soft-drink bottles. It is prepared by reaction of ethylene glycol with 1,4-benzenedicarboxylic acid (terephthalic acid). Draw the structure of PET.
Problem 31-9
Show the mechanism of the nucleophilic addition reaction of an alcohol with an isocyanate to yield a urethane.
31.06: Olefin Metathesis Polymerization
Perhaps the most important advance in polymer synthesis in recent years has been the development of olefin metathesis polymerization by Yves Chauvin of the Institut Français du Pétrole, Robert H. Grubbs of CalTech, and Richard R.Schrock of MIT who shared the 2005 Nobel Prize in Chemistry for their work. At its simplest, an olefin metathesis reaction is two olefins (alkenes) exchanging substituents on their double bonds.
Olefin metathesis catalysts, such as the Grubbs catalyst now in common use, contain a carbon–metal double bond (usually to ruthenium, Ru) and have the general structure $M═CH─RFigure 31.3.$
) reaction of the catalyst and olefin 1 to give a four-membered metallacycle intermediate, followed by (2) ring-opening to give a different form of catalyst that contains part of olefin 1. (3) Reaction of this new catalyst with olefin 2 gives another metallacycle intermediate, (4) which opens to give metathesis product and another form of catalyst. (5, 6) The repeating ring-forming and ring-opening steps then continue.
There are several methods for implementing the olefin metathesis reaction to prepare polymers. One method, called ring-opening metathesis polymerization, or ROMP, involves the use of a moderately strained cycloalkene, such as cyclopentene. The strain of the ring favors ring-opening, thereby driving formation of the open-chain product. The polymer that results has double bonds spaced regularly along the chain, allowing for either hydrogenation or further functionalization if desired.
A second method of using olefin metathesis to prepare polymers is by acyclic diene metathesis, or ADMET. As the name suggests, ADMET involves olefin metathesis of an open-chain substrate with two double bonds at the ends of a long chain, such as 1,8-nonadiene. As the reaction proceeds, the gaseous ethylene by-product escapes, thereby driving the equilibrium toward polymer product. So efficient is this reaction that polymers with molecular weights as high as 80,000 amu have been prepared.
The ROMP and ADMET procedures are particularly valuable because the metathesis reaction is compatible with the presence in the olefin monomer of many different functional groups. In addition, the double bonds in the polymers provide still more flexibility for further manipulations. Among the commercial polymers produced by olefin metathesis are Vestenamer, used in the manufacture of tires and other molded rubber objects, and Norsorex, used in the automobile industry as a sealing material.
Problem 31-10
Look at the structures of Vestenamer and Norsorex and show how they might be made by olefin metathesis polymerization.
31.07: Intramolecular Olefin Metathesis
The olefin metathesis reaction has been used primarily as an intermolecular process for polymer synthesis, but it is also finding increased usage in complex organic synthesis as an intramolecular process in what is called ring-closing metathesis (RCM). The optimum catalyst is an Ru(II) complex, and cyclic alkenes with up to 30 atoms in ring size are easily produced. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/31%3A_Synthetic_Polymers/31.04%3A_Copolymers.txt |
Polymers aren’t really that different from other organic molecules. They’re much larger, of course, but their chemistry is similar to that of analogous small molecules. Thus, the alkane chains of polyethylene undergo radical-initiated halogenation, the aromatic rings of polystyrene undergo typical electrophilic aromatic substitution reactions, and the amide linkages of nylon are hydrolyzed by aqueous base.
The major difference between small and large organic molecules is in their physical properties. For instance, their large size means that polymers experience substantially greater van der Waals forces than do small molecules (Section 2.12). But because van der Waals forces operate only at close distances, they are strongest in polymers like high-density polyethylene, in which chains can pack together closely in a regular way. Many polymers, in fact, have regions that are essentially crystalline. These regions, called crystallites, consist of highly ordered portions in which the zigzag polymer chains are held together by van der Waals forces (Figure 31.4).
As you might expect, polymer crystallinity is strongly affected by the steric requirements of substituent groups on the chains. Linear polyethylene is highly crystalline, but poly(methyl methacrylate) is noncrystalline because the chains can’t pack closely together in a regular way. Polymers with a high degree of crystallinity are generally hard and durable. When heated, the crystalline regions melt at the melt transition temperature, Tm, to give an amorphous material.
Noncrystalline, amorphous polymers like poly(methyl methacrylate), sold under the trade name Plexiglas, have little or no long-range ordering among chains but can nevertheless be very hard at room temperature. When heated, the hard amorphous polymer becomes soft and flexible at a point called the glass transition temperature, Tg. Much of the art in polymer synthesis lies in finding methods for controlling the degree of crystallinity and the glass transition temperature, thereby imparting useful properties to the polymer.
In general, polymers can be divided into four major categories, depending on their physical behavior: thermoplastics, fibers, elastomers, and thermosetting resins. Thermoplastics are the polymers most people think of when the word plastic is mentioned. These polymers have a high glass transition temperature and are therefore hard at room temperature but become soft and viscous when heated. As a result, they can be molded into toys, beads, telephone housings, or any of a thousand other items. Because thermoplastics have little or no cross-linking, the individual chains can slip past one another in the melt. Some thermoplastic polymers, such as poly(methyl methacrylate) and polystyrene, are amorphous and noncrystalline; others, such as polyethylene and nylon, are partially crystalline. Among the better-known thermoplastics is poly(ethylene terephthalate), or PET, used for making plastic soft-drink bottles.
Plasticizers—small organic molecules that act as lubricants between chains—are usually added to thermoplastics to keep them from becoming brittle at room temperature. An example is poly(vinyl chloride), which is brittle when pure but becomes supple and pliable when a plasticizer is added. In fact, most drip bags used in hospitals to deliver intravenous saline solutions are made of poly(vinyl chloride), although replacements such as polypropylene are appearing.
Dialkyl phthalates such as di(2-ethylhexyl) phthalate (generally called dioctyl phthalate) are commonly used as plasticizers, although questions about their safety have been raised. The U.S. Food and Drug Administration (FDA) has advised the use of alternative materials in compromised patients and infants but has found no evidence of toxicity for healthy individuals. In addition, children’s toys that contain phthalates have been banned in the United States.
Fibers are thin threads produced by extruding a molten polymer through small holes in a die, or spinneret. The fibers are then cooled and drawn out, which orients the crystallite regions along the axis of the fiber and adds considerable tensile strength Figure 31.5. Nylon, Dacron, and polyethylene all have the semicrystalline structure necessary for being drawn into oriented fibers.
Elastomers are amorphous polymers that have the ability to stretch out and spring back to their original shapes. These polymers must have low glass transition temperatures and a small amount of cross-linking to prevent the chains from slipping over one another. In addition, the chains must have an irregular shape to prevent crystallite formation. When stretched, the randomly coiled chains straighten out and orient along the direction of the pull. Van der Waals forces are too weak and too few to maintain this orientation, however, and the elastomer therefore reverts to its random coiled state when the stretching force is released (Figure 31.6).
Natural rubber (Section 14.6) is the most common example of an elastomer. Rubber has the long chains and occasional cross-links needed for elasticity, but its irregular geometry prevents close packing of the chains into crystallites. Gutta-percha, by contrast, is highly crystalline and is not an elastomer (Figure 31.7).
Thermosetting resins are polymers that become highly cross-linked and solidify into a hard, insoluble mass when heated. Bakelite, a thermosetting resin first produced in 1907, has been in commercial use longer than any other synthetic polymer. It is widely used for molded parts, adhesives, coatings, and even high-temperature applications such as missile nose cones.
Chemically, Bakelite is a phenolic resin, produced by reaction of phenol and formaldehyde. On heating, water is eliminated, many cross-links form, and the polymer sets into a rocklike mass. The cross-linking in Bakelite and other thermosetting resins is three-dimensional and is so extensive that we can’t really speak of polymer “chains.” A piece of Bakelite is essentially one large molecule.
Problem 31-11
What product would you expect to obtain from catalytic hydrogenation of natural rubber? Would the product be syndiotactic, atactic, or isotactic?
Problem 31-12
Propose a mechanism to account for the formation of Bakelite from acid-catalyzed polymerization of phenol and formaldehyde. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/31%3A_Synthetic_Polymers/31.08%3A_Polymer_Structure_and_Physical_Properties.txt |
31 • Chemistry Matters 31 • Chemistry Matters
The high chemical stability of many polymers is both a blessing and a curse. Heat resistance, wear resistance, and long life are valuable characteristics of clothing fibers, bicycle helmets, underground pipes, food wrappers, and many other items. Yet when these items outlive their usefulness, disposal becomes a problem.
Recycling of unwanted polymers is the best solution, and six types of plastics in common use are frequently stamped with identifying codes assigned by the Society of the Plastics Industry Table 31.2. After being sorted by type, the items to be recycled are shredded into small chips, washed, dried, and melted for reuse. Soft-drink bottles, for instance, are made from recycled poly(ethylene terephthalate), trash bags are made from recycled low-density polyethylene, and garden furniture is made from recycled polypropylene and mixed plastics.
Table 31.2 Recyclable Plastics
Polymer Recycling code Use
Poly(ethylene terephthalate) 1—PET Soft-drink bottles
High-density polyethylene 2—HDPE Bottles
Poly(vinyl chloride) 3—V Floor mats
Low-density polyethylene 4—LDPE Grocery bags
Polypropylene 5—PP Furniture
Polystyrene 6—PS Molded articles
Mixed plastics 7 Benches, plastic lumber
Frequently, however, plastics are simply thrown away rather than recycled, and much work has therefore been carried out on developing biodegradable polymers, which can be broken down by soil microorganisms. Among the most common biodegradable polymers are polyglycolic acid (PGA), polylactic acid (PLA), and polyhydroxybutyrate (PHB). All are polyesters and are therefore susceptible to hydrolysis of their ester links. Copolymers of PGA with PLA have found a particularly wide range of uses. A 90/10 copolymer of polyglycolic acid with polylactic acid is used to make absorbable sutures that are degraded and absorbed by the body within 90 days after surgery.
Biodegradation is important, but even better than the breakdown of a small group of specialized polymers would be the development of a generalized method of breakdown that could be used on any polymer, including even hydrocarbon polymers such as polyethylene and polypropylene. Such a method has recently been reported by chemists at the University of Delaware, using hydrogen at 200 °C and 2 atm pressure in the presence of an organozirconium catalyst to turn polyethylene into simple hydrocarbons. The method uses a large amount of hydrogen at present, but it is an impressive start.
31.10: Key Terms
31 • Key Terms 31 • Key Terms
• atactic polymer
• block copolymer
• chain-growth polymer
• copolymer
• crystallite
• elastomer
• fiber
• glass transition temperature (Tg)
• graft copolymer
• homopolymer
• isotactic polymer
• melt transition temperature (Tm)
• monomer
• olefin metathesis polymerization
• plasticizer
• polycarbonate
• polymer
• polyurethane
• syndiotactic polymer
• thermoplastic
• thermosetting resin
• urethane
• Ziegler–Natta catalyst
31.11: Summary
31 • Summary 31 • Summary
Synthetic polymers can be classified as either chain-growth or step-growth. Chain-growth polymers are prepared by chain-reaction polymerization of vinyl monomers in the presence of a radical, an anion, or a cation initiator. Radical polymerization is sometimes used, but alkenes such as 2-methylpropene that have electron-donating substituents on the double bond polymerize easily by a cationic route through carbocation intermediates. Similarly, monomers such as methyl α-cyanoacrylate that have electron-withdrawing substituents on the double bond polymerize by an anionic, conjugate addition pathway.
Copolymerization of two monomers gives a product with properties different from those of either homopolymer. Graft copolymers and block copolymers are two examples.
Alkene polymerization can be carried out in a controlled manner using a Ziegler–Natta catalyst. Ziegler–Natta polymerization minimizes the amount of chain branching in the polymer and leads to stereoregular chains—either isotactic (substituents on the same side of the chain) or syndiotactic (substituents on alternate sides of the chain), rather than atactic (substituents randomly disposed).
Step-growth polymers, the second major class of polymers, are prepared by reactions between difunctional molecules, with individual bonds in the polymer formed independently of one another. Polycarbonates are formed from a diester and a diol, and polyurethanes are formed from a diisocyanate and a diol.
The chemistry of synthetic polymers is similar to the chemistry of small molecules with the same functional groups, but the physical properties of polymers are greatly affected by size. Polymers can be classified by physical property into four groups: thermoplastics, fibers, elastomers, and thermosetting resins. The properties of each group can be accounted for by the structure, the degree of crystallinity, and the amount of cross-linking they contain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/31%3A_Synthetic_Polymers/31.09%3A_Chemistry_MattersDegradable_Polymers.txt |
31 • Additional Problems 31 • Additional Problems
Visualizing Chemistry
Problem 31-13
Identify the structural class to which the following polymer belongs, and show the structure of the monomer units used to make it:
Problem 31-14 Show the structures of the polymers that could be made from the following monomers (green = Cl): (a) (b) Mechanism Problems Problem 31-15
Poly(ethylene glycol), or Carbowax, is made by anionic polymerization of ethylene oxide using NaOH as catalyst. Propose a mechanism.
Problem 31-16
The polyurethane foam used for home insulation uses methanediphenyldiisocyanate (MDI) as monomer. The MDI is prepared by acid-catalyzed reaction of aniline with formaldehyde, followed by treatment with phosgene, COCl2. Propose mechanisms for both steps.
Problem 31-17
Write the structure of a representative segment of polyurethane prepared by reaction of ethylene glycol with MDI (Problem 31-16).
Problem 31-18
The polymeric resin used for Merrifield solid-phase peptide synthesis (Section 26.8) is prepared by treating polystyrene with N-(hydroxymethyl)phthalimide and trifluoromethanesulfonic acid, followed by reaction with hydrazine. Propose a mechanism for both steps.
Problem 31-19 Polydicyclopentadiene (PDCPD), marketed as Telene or Metton, is a highly cross-linked thermosetting resin used for molding such impact-resistant parts as cabs for large trucks and earth-moving equipment. PDCPD is prepared by ring-opening metathesis polymerization of dicyclopentadiene, which is itself prepared from 1,3-cyclopentadiene. The polymerization occurs by initial metathesis of the more highly strained double bond in the bicyclo[2.2.1]heptane part of the molecule (Section 4.9) to give a linear polymer, followed by cross-linking of different chains in a second metathesis of the remaining cyclopentene double bond. (a)
Show the mechanism of the formation of dicyclopentadiene from cyclopentadiene.
(b) Draw the structure of a representative sample of the initially formed linear polymer containing three monomer units. (c) Draw the structure of a representative sample of PDCPD that shows how cross-linking of the linear chains takes place.
General Problems
Problem 31-20 Identify the monomer units from which each of the following polymers is made, and tell whether each is a chain-growth or a step-growth polymer: (a) (b) (c) (d) (e) Problem 31-21 Draw a three-dimensional representation of segments of the following polymers: (a)
Syndiotactic polyacrylonitrile
(b) Atactic poly(methyl methacrylate) (c) Isotactic poly(vinyl chloride)
Problem 31-22
Draw the structure of Kodel, a polyester prepared by heating dimethyl 1,4-benzenedicarboxylate with 1,4-bis(hydroxymethyl)cyclohexane.
Problem 31-23
Show the structure of the polymer that results from heating the following diepoxide and diamine:
Problem 31-24
Nomex, a polyamide used in such applications as fire-retardant clothing, is prepared by reaction of 1,3-benzenediamine with 1,3-benzenedicarbonyl chloride. Show the structure of Nomex.
Problem 31-25
Nylon 10,10 is an extremely tough, strong polymer used to make reinforcing rods for concrete. Draw a segment of nylon 10,10, and show its monomer units.
Problem 31-26
1,3-Cyclopentadiene undergoes thermal polymerization to yield a polymer that has no double bonds in the chain. Upon strong heating, the polymer breaks down to regenerate cyclopentadiene. Propose a structure for the polymer.
Problem 31-27
When styrene, C6H5CH$\text{═}$CH2, is copolymerized in the presence of a few percent p-divinylbenzene, a hard, insoluble, cross-linked polymer is obtained. Show how this cross-linking of polystyrene chains occurs.
Problem 31-28
Nitroethylene, H2C$\text{═}$CHNO2, is a sensitive compound that must be prepared with great care. Attempted purification of nitroethylene by distillation often results in low recovery of product and a white coating on the inner walls of the distillation apparatus. Explain.
Problem 31-29
Poly(vinyl butyral) is used as the plastic laminate in the preparation of automobile windshield safety glass. How would you synthesize this polymer?
Problem 31-30
What is the structure of the polymer produced by anionic polymerization of β-propiolactone using NaOH as catalyst?
Problem 31-31
Glyptal is a highly cross-linked thermosetting resin produced by heating glycerol and phthalic anhydride (1,2-benzenedicarboxylic acid anhydride). Show the structure of a representative segment of glyptal.
Problem 31-32
Melmac, a thermosetting resin often used to make plastic dishes, is prepared by heating melamine with formaldehyde. Look at the structure of Bakelite shown in Section 31.7, and then propose a structure for Melmac.
Problem 31-33 Epoxy adhesives are cross-linked resins prepared in two steps. The first step involves SN2 reaction of the disodium salt of bisphenol A with epichlorohydrin to form a low-molecular-weight prepolymer. This prepolymer is then “cured” into a cross-linked resin by treatment with a triamine such as H2NCH2CH2NHCH2CH2NH2. (a)
What is the structure of the prepolymer?
(b) How does addition of the triamine to the prepolymer result in cross-linking?
Problem 31-34
The smoking salons of the Hindenburg and other hydrogen-filled dirigibles of the 1930s were insulated with urea–formaldehyde polymer foams. The structure of this polymer is highly cross-linked, like that of Bakelite (Section 31.7). Propose a structure.
Problem 31-35
2-Ethyl-1-hexanol, used in the synthesis of di(2-ethylhexyl) phthalate plasticizer, is made commercially from butanal. Show the likely synthesis route. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/31%3A_Synthetic_Polymers/31.12%3A_Additional_Problems.txt |
A • Nomenclature of Polyfunctional Organic Compounds A • Nomenclature of Polyfunctional Organic Compounds
With more than 40 million organic compounds now known and thousands more being created daily, naming them all is a real problem. Part of the problem is due to the sheer complexity of organic structures, but part is also due to the fact that chemical names have more than one purpose. For the Chemical Abstracts Service (CAS), which catalogs and indexes the worldwide chemical literature, each compound must have only one correct name. It would be chaos if half the entries for CH3Br were indexed under “M” for methyl bromide and half under “B” for bromomethane. Furthermore, a CAS name must be strictly systematic so that it can be assigned and interpreted by computers; common names are not allowed.
People, however, have different requirements than computers. For people—which is to say students and professional chemists in their spoken and written communications—it’s best that a chemical name be pronounceable and as easy as possible to assign and interpret. Furthermore, it’s convenient if names follow historical precedents, even if that means a particularly well-known compound might have more than one name. People can readily understand that bromomethane and methyl bromide both refer to CH3Br.
As noted in the text, chemists overwhelmingly use the nomenclature system devised and maintained by the International Union of Pure and Applied Chemistry, or IUPAC. Rules for naming monofunctional compounds were given throughout the text as each new functional group was introduced, and a list of where these rules can be found is given in Table A1.
Table A1 Nomenclature Rules for Functional Groups
Functional group Text section
Acid anhydrides 21-1
Acid halides 21-1
Acyl phosphates 21-1
Alcohols 17-1
Aldehydes 19-1
Alkanes 3-4
Alkenes 7-3
Alkyl halides 10-1
Alkynes 9-1
Amides 21-1
Amines 24-1
Aromatic compounds 15-1
Carboxylic acids 20-1
Cycloalkanes 4-1
Esters 21-1
Ethers 18-1
Ketones 19-1
Nitriles 20-1
Phenols 17-1
Sulfides 18-7
Thiols 18-7
Thioesters 21-1
Naming a monofunctional compound is reasonably straightforward, but even experienced chemists often encounter problems when faced with naming a complex polyfunctional compound. Take the following compound, for instance. It has three functional groups, ester, ketone, and $C═CC═C$, but how should it be named? As an ester with an -oate ending, a ketone with an -one ending, or an alkene with an -ene ending? It’s actually named methyl 3-(2-oxo-6-cyclohexenyl)propanoate.
The name of a polyfunctional organic molecule has four parts—suffix, parent, prefixes, and locants—which must be identified and expressed in the proper order and format. Let’s look at each of the four.
Name Part 1. The Suffix: Functional-Group Precedence
Although a polyfunctional organic molecule might contain several different functional groups, we must choose just one suffix for nomenclature purposes. It’s not correct to use two suffixes. Thus, keto ester 1 must be named either as a ketone with an -one suffix or as an ester with an -oate suffix, but it can’t be named as an -onoate. Similarly, amino alcohol 2 must be named either as an alcohol (-ol) or as an amine (-amine), but it can’t be named as an -olamine or -aminol.
The only exception to the rule requiring a single suffix is when naming compounds that have double or triple bonds. Thus, the unsaturated acid $H2C═CHCH2CO2HH2C═CHCH2CO2H$ is 3-butenoic acid, and the acetylenic alcohol $HC≡CCH2CH2CH2OHHC≡CCH2CH2CH2OH$ is 5-pentyn-1-ol.
How do we choose which suffix to use? Functional groups are divided into two classes, principal groups and subordinate groups, as shown in Table A2. Principal groups can be cited either as prefixes or as suffixes, while subordinate groups are cited only as prefixes. Within the principal groups, an order of priority has been established: the proper suffix for a given compound is determined by choosing the principal group of highest priority. For example, Table A2 indicates that keto ester 1 should be named as an ester rather than as a ketone because an ester functional group is higher in priority than a ketone. Similarly, amino alcohol 2 should be named as an alcohol rather than as an amine. Thus, the name for 1 is methyl 4-oxopentanoate and the name for 2 is 5-amino-2-pentanol. Further examples are shown:
Table A2 Classification of Functional Groups a
Functional group Name as suffix Name as prefix
Principal groups
Carboxylic acids -oic acid carboxy
-carboxylic acid
Acid anhydrides -oic anhydride
-carboxylic anhydride
Esters -oate alkoxycarbonyl
-carboxylate
Thioesters -thioate alkylthiocarbonyl
-carbothioate
Acid halides -oyl halide halocarbonyl
-carbonyl halide
Amides -amide carbamoyl
-carboxamide
Nitriles -nitrile cyano
-carbonitrile
Aldehydes -al oxo
-carbaldehyde
Ketones -one oxo
Alcohols -ol hydroxy
Phenols -ol hydroxy
Thiols -thiol mercapto
Amines -amine amino
Imines -imine imino
Ethers ether alkoxy
Sulfides sulfide alkylthio
Disulfides disulfide
Alkenes -ene
Alkynes -yne
Alkanes -ane
Subordinate groups
Azides azido
Halides halo
Nitro compounds nitro
aPrincipal groups are listed in order of decreasing priority; subordinate groups have no priority order.
Name Part 2. The Parent: Selecting the Main Chain or Ring
The parent, or base, name of a polyfunctional organic compound is usually easy to identify. If the principal group of highest priority is part of an open chain, the parent name is that of the longest chain containing the largest number of principal groups. For example, compounds 6 and 7 are isomeric aldehydo amides, which must be named as amides rather than as aldehydes according to Table A2. The longest chain in compound 6 has six carbons, and the substance is named 5-methyl-6-oxohexanamide. Compound 7 also has a chain of six carbons, but the longest chain that contains both principal functional groups has only four carbons. Thus, compound 7 is named 4-oxo-3-propylbutanamide.
If the highest-priority principal group is attached to a ring, the parent name is that of the ring system. Compounds 8 and 9, for instance, are isomeric keto nitriles and must both be named as nitriles according to Table A2. Substance 8 is named as a benzonitrile because the −CN functional group is a substituent on the aromatic ring, but substance 9 is named as an acetonitrile because the −CN functional group is on an open chain. Thus, their names are 2-acetyl-(4-bromomethyl)benzonitrile (8) and (2-acetyl-4-bromophenyl)acetonitrile (9). As further examples, compounds 10 and 11 are both keto acids and must be named as acids, but the parent name in 10 is that of a ring system (cyclohexanecarboxylic acid) and the parent name in 11 is that of an open chain (propanoic acid). Thus, their names are trans-2-(3-oxopropyl)cyclohexanecarboxylic acid (10) and 3-(2-oxocyclohexyl)propanoic acid (11).
Name Parts 3 and 4. The Prefixes and Locants
With the parent name and the suffix established, the next step is to identify and give numbers, or locants, to all substituents on the parent chain or ring. The substituents include all alkyl groups and all functional groups other than the one cited in the suffix. For example, compound 12 contains three different functional groups (carboxyl, keto, and double bond). Because the carboxyl group is highest in priority and the longest chain containing the functional groups has seven carbons, compound 12 is a heptenoic acid. In addition, the parent chain has a keto (oxo) substituent and three methyl groups. Numbering from the end nearer the highest-priority functional group gives the name (E)-2,5,5-trimethyl-4-oxo-2-heptenoic acid. Look back at some of the other compounds we’ve named to see other examples of how prefixes and locants are assigned.
Writing the Name
With the name parts established, the entire name can be written out. Several additional rules apply:
1. Order of prefixes. When the substituents have been identified, the parent chain has been numbered, and the proper multipliers such as di- and tri- have been assigned, the name is written with the substituents listed in alphabetical, rather than numerical, order. Multipliers such as di- and tri- are not used for alphabetization, but the italicized prefixes iso- and sec- are used.
2. Use of hyphens; single- and multiple-word names. The general rule is to determine whether the parent is itself an element or compound. If it is, then the name is written as a single word; if it isn’t, then the name is written as multiple words. Methylbenzene is written as one word, for instance, because the parent—benzene—is a compound. Diethyl ether, however, is written as two words because the parent—ether—is a class name rather than a compound name. Some further examples follow:
3. Parentheses. Parentheses are used to denote complex substituents when ambiguity would otherwise arise. For example, chloromethylbenzene has two substituents on a benzene ring, but (chloromethyl)benzene has only one complex substituent. Note that the expression in parentheses is not set off by hyphens from the rest of the name.
Additional Reading
Further explanations of the rules of organic nomenclature can be found online at ACD Labs (accessed May 2023) and in the following references:
1. “A Guide to IUPAC Nomenclature of Organic Compounds,” CRC Press, Boca Raton, FL, 1993.
2. “Nomenclature of Organic Chemistry, Sections A, B, C, D, E, F, and H,” International Union of Pure and Applied Chemistry, Pergamon Press, Oxford, 1979.
32.02: Appendix B - Acidity Constants for Some Organic Compounds
B • Acidity Constants for Some Organic Compounds B • Acidity Constants for Some Organic Compounds Table B1
Compound pKa
CH3SO3H −1.8
CH(NO2)3 0.1
0.3
CCl3CO2H 0.5
CF3CO2H 0.5
CBr3CO2H 0.7
$HO2CC≡CCO2HHO2CC≡CCO2H$ 1.2; 2.5
HO2CCO2H 1.2; 3.7
CHCl2CO2H 1.3
CH2(NO2)CO2H 1.3
$HC≡CCO2HHC≡CCO2H$ 1.9
(Z) $HO2CCH═CHCO2HHO2CCH═CHCO2H$ 1.9; 6.3
2.4
CH3COCO2H 2.4
NCCH2CO2H 2.5
$CH3C≡CCO2HCH3C≡CCO2H$ 2.6
CH2FCO2H 2.7
CH2ClCO2H 2.8
HO2CCH2CO2H 2.8; 5.6
CH2BrCO2H 2.9
3.0
3.0
CH2ICO2H 3.2
CHOCO2H 3.2
3.4
3.5
HSCH2CO2H 3.5; 10.2
CH2(NO2)2 3.6
CH3OCH2CO2H 3.6
CH3COCH2CO2H 3.6
HOCH2CO2H 3.7
HCO2H 3.7
3.8
4.0
CH2BrCH2CO2H 4.0
4.1
4.2
$H2C═CHCO2HH2C═CHCO2H$ 4.2
HO2CCH2CH2CO2H 4.2; 5.7
HO2CCH2CH2CH2CO2H 4.3; 5.4
4.5
$H2C═C(CH3)CO2HH2C═C(CH3)CO2H$ 4.7
CH3CO2H 4.8
CH3CH2CO2H 4.8
(CH3)3CCO2H 5.0
CH3COCH2NO2 5.1
5.3
O2NCH2CO2CH3 5.8
5.8
6.2
6.6
HCO3H 7.1
7.2
(CH3)2CHNO2 7.7
7.8
CH3CO3H 8.2
8.5
CH3CH2NO2 8.5
8.7
CH3COCH2COCH3 9.0
9.3; 11.1
9.3; 12.6
9.4
9.9; 11.5
9.9
CH3COCH2SOCH3 10.0
10.3
CH3NO2 10.3
CH3SH 10.3
CH3COCH2CO2CH3 10.6
CH3COCHO 11.0
CH2(CN)2 11.2
CCl3CH2OH 12.2
Glucose 12.3
$(CH3)2C═NOH(CH3)2C═NOH$ 12.4
CH2(CO2CH3)2 12.9
CHCl2CH2OH 12.9
CH2(OH)2 13.3
HOCH2CH(OH)CH2OH 14.1
CH2ClCH2OH 14.3
15.0
15.4
CH3OH 15.5
$H2C═CHCH2OHH2C═CHCH2OH$ 15.5
CH3CH2OH 16.0
CH3CH2CH2OH 16.1
CH3COCH2Br 16.1
16.7
CH3CHO 17
(CH3)2CHCHO 17
(CH3)2CHOH 17.1
(CH3)3COH 18.0
CH3COCH3 19.3
23
CH3CO2CH2CH3 25
$HC≡CHHC≡CH$ 25
CH3CN 25
CH3SO2CH3 28
(C6H5)3CH 32
(C6H5)2CH2 34
CH3SOCH3 35
NH3 36
CH3CH2NH2 36
(CH3CH2)2NH 40
41
43
$H2C═CH2H2C═CH2$ 44
CH4 ∼60
An acidity list covering more than 5000 organic compounds has been published: E.P. Serjeant and B. Dempsey (eds.), “Ionization Constants of Organic Acids in Aqueous Solution,” IUPAC Chemical Data Series No. 23, Pergamon Press, Oxford, 1979. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.01%3A_Appendix_A_-_Nomenclature_of_Polyfunctional_Organic_Compounds.txt |
• Absolute configuration (Section 5.5): The exact three-dimensional structure of a chiral molecule. Absolute configurations are specified verbally by the Cahn–Ingold–Prelog R,S convention.
• Absorbance (A) (Section 14.7): In optical spectroscopy, the logarithm of the intensity of the incident light divided by the intensity of the light transmitted through a sample; A = log I0/I.
• Absorption spectrum (Section 12.5): A plot of wavelength of incident light versus amount of light absorbed. Organic molecules show absorption spectra in both the infrared and the ultraviolet regions of the electromagnetic spectrum.
• Acetals, R2C(OR′)2 (Section 19.10): A type of functional group consisting of two −OR groups bonded to the same carbon, R2C(OR′)2. Acetals are often used as protecting groups for ketones and aldehydes.
• Acetoacetic ester synthesis (Section 22.7): The synthesis of a methyl ketone by alkylation of an alkyl halide with ethyl acetoacetate, followed by hydrolysis and decarboxylation.
• Acetyl group (Section 19.1): The CH3CO− group.
• Acetylide anion (Section 9.7): The anion formed by removal of a proton from a terminal alkyne, $R–C≡C:−R–C≡C:−$.
• Achiral (Section 5.2): Having a lack of handedness. A molecule is achiral if it has a plane of symmetry and is thus superimposable on its mirror image.
• Acid anhydrides (Chapter 21 Introduction): A type of functional group with two acyl groups bonded to a common oxygen atom, RCO2COR′.
• Acid halides (Chapter 21 Introduction): A type of functional group with an acyl group bonded to a halogen atom, RCOX.
• Acidity constant, Ka (Section 2.8): A measure of acid strength. For any acid HA, the acidity constant is given by the expression $Ka=[H3O+][A−][HA]Ka=[H3O+][A−][HA]$.
• Activating groups (Section 16.4): Electron-donating groups such as hydroxyl (−OH) or amino (−NH2) that increase the reactivity of an aromatic ring toward electrophilic aromatic substitution.
• Activation energy (Section 6.9): The difference in energy between ground state and transition state in a reaction. The amount of activation energy determines the rate at which the reaction proceeds. Most organic reactions have activation energies of 40–100 kJ/mol.
• Active site (Section 6.11, Section 26.11): The pocket in an enzyme where a substrate is bound and undergoes reaction.
• Acyclic diene metathesis (ADMET) (Section 31.5): A method of polymer synthesis that uses the olefin metathesis reaction of an open-chain diene.
• Acyl group (Section 16.3, Section 19.1): A −COR group.
• Acyl phosphates (Chapter 21 Introduction): A type of functional group with an acyl group bonded to a phosphate, RCO2PO32−.
• Acylation (Section 16.3, Section 24.7): The introduction of an acyl group, −COR, onto a molecule. For example, acylation of an alcohol yields an ester, acylation of an amine yields an amide, and acylation of an aromatic ring yields an alkyl aryl ketone.
• Acylium ion (Section 16.3): A resonance-stabilized carbocation in which the positive charge is located at a carbonyl-group carbon, $R–C+═O↔R–C≡O+R–C+═O↔R–C≡O+$. Acylium ions are intermediates in Friedel–Crafts acylation reactions.
• Adams’ catalyst (Section 6.11): The PtO2 catalyst used for alkene hydrogenations.
• 1,2-Addition (Section 14.2, Section 19.13): Addition of a reactant to the two ends of a double bond.
• 1,4-Addition (Section 14.2, Section 19.13): Addition of a reactant to the ends of a conjugated π system. Conjugated dienes yield 1,4 adducts when treated with electrophiles such as HCl. Conjugated enones yield 1,4 adducts when treated with nucleophiles such as amines.
• Addition reactions (Section 6.1): Occur when two reactants add together to form a single product with no atoms left over.
• Adrenocortical hormones (Section 27.6): Steroid hormones secreted by the adrenal glands. There are two types of these hormones: mineralocorticoids and glucocorticoids.
• Alcohols (Section 3.1, Chapter 17 Introduction): A class of compounds with an −OH group bonded to a saturated, sp3-hybridized carbon, ROH.
• Aldaric acid (Section 25.6): The dicarboxylic acid resulting from oxidation of an aldose.
• Aldehydes (RCHO) (Section 3.1, Chapter 19 Introduction): A class of compounds containing the −CHO functional group.
• Alditol (Section 25.6): The polyalcohol resulting from reduction of the carbonyl group of a sugar.
• Aldol reaction (Section 23.1): The carbonyl condensation reaction of an aldehyde or ketone to give a β-hydroxy carbonyl compound.
• Aldonic acids (Section 25.6): Monocarboxylic acids resulting from oxidation of the −CHO group of an aldose.
• Aldoses (Section 25.1): A type of carbohydrate with an aldehyde functional group.
• Alicyclic (Section 4.1): A nonaromatic cyclic hydrocarbon such as a cycloalkane or cycloalkene.
• Aliphatic (Section 3.2): A nonaromatic hydrocarbon such as a simple alkane, alkene, or alkyne.
• Alkaloids (Chapter 2 Chemistry Matters): Naturally occurring organic bases, such as morphine.
• Alkanes (Section 3.2): A class of compounds of carbon and hydrogen that contains only single bonds.
• Alkene (Section 3.1, Chapter 7 Introduction): A hydrocarbon that contains a carbon–carbon double bond, $R2C═CR2R2C═CR2$.
• Alkoxide ion, RO (Section 17.2): The anion formed by deprotonation of an alcohol.
• Alkoxymercuration (Section 18.2): A method for synthesizing ethers by mercuric-ion catalyzed addition of an alcohol to an alkene followed by demercuration on treatment with NaBH4.
• Alkyl group (Section 3.3): The partial structure that remains when a hydrogen atom is removed from an alkane.
• Alkyl halide (Section 3.1, Chapter 10 Introduction): A compound with a halogen atom bonded to a saturated, sp3-hybridized carbon atom.
• Alkylamines (Section 24.1): Amino-substituted alkanes RNH2, R2NH, or R3N.
• Alkylation (Section 9.8, Section 16.3, Section 18.2, Section 22.7): Introduction of an alkyl group onto a molecule. For example, aromatic rings can be alkylated to yield arenes, and enolate anions can be alkylated to yield α-substituted carbonyl compounds.
• Alkyne (Section 3.1, Chapter 9 Introduction): A hydrocarbon that contains a carbon–carbon triple bond, $RC≡CRRC≡CR$.
• Allyl group (Section 7.3): A $H2C═CHCH2−H2C═CHCH2−$ substituent.
• Allylic (Section 10.3): The position next to a double bond. For example, $H2C═CHCH2BrH2C═CHCH2Br$ is an allylic bromide.
• α-Amino acids (Section 26.1): A type of difunctional compound with an amino group on the carbon atom next to a carboxyl group, RCH(NH2)CO2H.
• α Anomer (Section 25.5): The cyclic hemiacetal form of a sugar that has the hemiacetal −OH group cis to the −OH at the lowest chirality center in a Fischer projection.
• α Helix (Section 26.9): The coiled secondary structure of a protein.
• α Position (Chapter 22 Introduction): The position next to a carbonyl group.
• α-Substitution reaction (Chapter 22 Introduction): The substitution of the α hydrogen atom of a carbonyl compound by reaction with an electrophile.
• Amides (Section 3.1, Chapter 21 Introduction): A class of compounds containing the −CONR2 functional group.
• Amidomalonate synthesis (Section 26.3): A method for preparing α-amino acids by alkylation of diethyl amidomalonate with an alkyl halide followed by deprotection and decarboxylation.
• Amines (Section 3.1, Chapter 24 Introduction): A class of compounds containing one or more organic substituents bonded to a nitrogen atom, RNH2, R2NH, or R3N.
• Amino acid (Section 26.1): See α-Amino acid.
• Amino sugar (Section 25.7): A sugar with one of its −OH groups replaced by −NH2.
• Amphiprotic (Section 26.1): Capable of acting either as an acid or as a base. Amino acids are amphiprotic.
• Amplitude (Section 12.5): The height of a wave measured from the midpoint to the maximum. The intensity of radiant energy is proportional to the square of the wave’s amplitude.
• Anabolic steroids (Section 27.6): Synthetic androgens that mimic the tissue-building effects of natural testosterone.
• Anabolism (Section 29.1): The group of metabolic pathways that build up larger molecules from smaller ones.
• Androgen (Section 27.6): A male steroid sex hormone.
• Angle strain (Section 4.3): The strain introduced into a molecule when a bond angle is deformed from its ideal value. Angle strain is particularly important in small-ring cycloalkanes, where it results from compression of bond angles to less than their ideal tetrahedral values.
• Annulation (Section 23.12): The building of a new ring onto an existing molecule.
• Anomeric center (Section 25.5): The hemiacetal carbon atom in the cyclic pyranose or furanose form of a sugar.
• Anomers (Section 25.5): Cyclic stereoisomers of sugars that differ only in their configuration at the hemiacetal (anomeric) carbon.
• Antarafacial (Section 30.5): A pericyclic reaction that takes place on opposite faces of the two ends of a π electron system.
• Anti conformation (Section 3.7): The geometric arrangement around a carbon–carbon single bond in which the two largest substituents are 180° apart as viewed in a Newman projection.
• Anti periplanar (Section 11.8): Describing the stereochemical relationship in which two bonds on adjacent carbons lie in the same plane at an angle of 180°.
• Anti stereochemistry (Section 8.2): The opposite of syn. An anti addition reaction is one in which the two ends of the double bond are attacked from different sides. An anti elimination reaction is one in which the two groups leave from opposite sides of the molecule.
• Antiaromatic (Section 15.3): Referring to a planar, conjugated molecule with 4n π electrons. Delocalization of the π electrons leads to an increase in energy.
• Antibonding MO (Section 1.11): A molecular orbital that is higher in energy than the atomic orbitals from which it is formed.
• Anticodon (Section 28.5): A sequence of three bases on tRNA that reads the codons on mRNA and brings the correct amino acids into position for protein synthesis.
• Antisense strand (Section 28.4): The template, noncoding strand of double-helical DNA that does not contain the gene.
• Arene (Section 15.1): An alkyl-substituted benzene.
• Arenediazonium salt (Section 24.8): An aromatic compound $Ar–N+≡N X−Ar–N+≡N X−$; used in the Sandmeyer reaction.
• Aromaticity (Chapter 15 Introduction): The special characteristics of cyclic conjugated molecules, including unusual stability and a tendency to undergo substitution reactions rather than addition reactions on treatment with electrophiles. Aromatic molecules are planar, cyclic, conjugated species with 4n + 2 π electrons.
• Arylamines (Section 24.1): Amino-substituted aromatic compounds, ArNH2.
• Atactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents is oriented randomly along the backbone.
• Atomic mass (Section 1.1): The weighted average mass of an element’s naturally occurring isotopes.
• Atomic number, Z (Section 1.1): The number of protons in the nucleus of an atom.
• ATZ Derivative (Section 26.6): An anilinothiazolinone, formed from an amino acid during Edman degradation of a peptide.
• Aufbau principle (Section 1.3): The rules for determining the electron configuration of an atom.
• Axial bonds (Section 4.6): Bonds or positions in chair cyclohexane that lie along the ring axis, perpendicular to the rough plane of the ring.
• Azide synthesis (Section 24.6): A method for preparing amines by SN2 reaction of an alkyl halide with azide ion, followed by reduction.
• Azo compounds (Section 24.8): A class of compounds with the general structure $R–N═N–R′R–N═N–R′$.
• Backbone (Section 26.4): The continuous chain of atoms running the length of a protein or other polymer.
• Base peak (Section 12.1): The most intense peak in a mass spectrum.
• Basicity constant, Kb (Section 24.3): A measure of base strength in water. For any base B, the basicity constant is given by the expression
• Bent bonds (Section 4.4): The bonds in small rings such as cyclopropane that bend away from the internuclear line and overlap at a slight angle, rather than head-on. Bent bonds are highly strained and highly reactive.
• Benzoyl (Section 19.1): The C6H5CO− group.
• Benzyl (Section 15.1): The C6H5CH2− group.
• Benzylic (Section 11.5): The position next to an aromatic ring.
• Benzyne (Section 16.7): An unstable compound having a triple bond in a benzene ring.
• β Anomer (Section 25.5): The cyclic hemiacetal form of a sugar that has the hemiacetal −OH group trans to the −OH at the lowest chirality center in a Fischer projection.
• β Diketone (Section 22.5): A 1,3-diketone.
• β-Keto ester (Section 22.5): A 3-oxoester.
• β Lactam (Chapter 21 Chemistry Matters): A four-membered lactam, or cyclic amide. Penicillin and cephalosporin antibiotics contain β-lactam rings.
• β-Oxidation pathway (Section 29.3): The metabolic pathway for degrading fatty acids.
• β-Pleated sheet (Section 26.9): A type of secondary structure of a protein.
• Betaine (Section 19.11): A neutral dipolar molecule with nonadjacent positive and negative charges. For example, the adduct of a Wittig reagent with a carbonyl compound is a betaine.
• Bicycloalkane (Section 4.9): A cycloalkane that contains two rings.
• Bimolecular reaction (Section 11.2): A reaction whose rate-limiting step occurs between two reactants.
• Block copolymers (Section 31.3): Polymers in which different blocks of identical monomer units alternate with one another.
• Boat cyclohexane (Section 4.5): A conformation of cyclohexane that bears a slight resemblance to a boat. Boat cyclohexane has no angle strain but has a large number of eclipsing interactions that make it less stable than chair cyclohexane.
• Boc derivative (Section 26.7): A butyloxycarbonyl N-protected amino acid.
• Bond angle (Section 1.6): The angle formed between two adjacent bonds.
• Bond dissociation energy, D (Section 6.8): The amount of energy needed to break a bond and produce two radical fragments.
• Bond length (Section 1.5): The equilibrium distance between the nuclei of two atoms that are bonded to each other.
• Bond strength (Section 1.5): An alternative name for bond dissociation energy.
• Bonding MO (Section 1.11): A molecular orbital that is lower in energy than the atomic orbitals from which it is formed.
• Branched-chain alkanes (Section 3.2): Alkanes that contain a branching connection of carbons as opposed to straight-chain alkanes.
• Bridgehead (Section 4.9): An atom that is shared by more than one ring in a polycyclic molecule.
• Bromohydrin (Section 8.3): A 1,2-bromoalcohol; obtained by addition of HOBr to an alkene.
• Bromonium ion (Section 8.2): A species with a divalent, positively charged bromine, R2Br+.
• Brønsted–Lowry acid (Section 2.7): A substance that donates a hydrogen ion (proton; H+) to a base.
• Brønsted–Lowry base (Section 2.7): A substance that accepts H+ from an acid.
• C-terminal amino acid (Section 26.4): The amino acid with a free −CO2H group at the end of a protein chain.
• Cahn–Ingold–Prelog sequence rules (Section 5.5, Section 7.5): A series of rules for assigning relative rankings to substituent groups on a chirality center or a double-bond carbon atom.
• Cannizzaro reaction (Section 19.12): The disproportionation reaction of an aldehyde on treatment with base to yield an alcohol and a carboxylic acid.
• Carbanion (Section 10.6, Section 19.7): A carbon anion, or substance that contains a trivalent, negatively charged carbon atom (R3C:). Alkyl carbanions are sp3-hybridized and have eight electrons in the outer shell of the negatively charged carbon.
• Carbene, R2C (Section 8.9): A neutral substance that contains a divalent carbon atom having only six electrons in its outer shell (R2C:).
• Carbinolamine (Section 19.8): A molecule that contains the R2C(OH)NH2 functional group. Carbinolamines are produced as intermediates during the nucleophilic addition of amines to carbonyl compounds.
• Carbocation (Section 6.4, Section 7.9): A carbon cation, or substance that contains a trivalent, positively charged carbon atom having six electrons in its outer shell (R3C+).
• Carbohydrates (Chapter 25 Introduction): Polyhydroxy aldehydes or ketones. Carbohydrates can be either simple sugars, such as glucose, or complex sugars, such as cellulose.
• Carbonyl condensation reactions (Section 23.1): A type of reaction that joins two carbonyl compounds together by a combination of α-substitution and nucleophilic addition reactions.
• Carbonyl group (Section 3.1, Preview of Carbonyl Chemistry): The $C═OC═O$ functional group.
• Carboxyl group (Section 20.1): The −CO2H functional group.
• Carboxylation (Section 20.5): The addition of CO2 to a molecule.
• Carboxylic acids, RCO2H (Section 3.1, Chapter 20 Introduction): Compounds containing the −CO2H functional group.
• Carboxylic acid derivative (Chapter 21 Introduction): A compound in which an acyl group is bonded to an electronegative atom or substituent that can act as a leaving group in a substitution reaction. Esters, amides, and acid halides are examples.
• Catabolism (Section 29.1): The group of metabolic pathways that break down larger molecules into smaller ones.
• Catalyst (Section 6.11): A substance that increases the rate of a chemical transformation by providing an alternative mechanism but is not itself changed in the reaction.
• Cation radical (Section 12.1): A reactive species, typically formed in a mass spectrometer by loss of an electron from a neutral molecule and having both a positive charge and an odd number of electrons.
• Chain-growth polymers (Section 8.10, Section 21.9, Section 31.1): Polymers whose bonds are produced by chain reaction mechanisms. Polyethylene and other alkene polymers are examples.
• Chain reaction (Section 6.6): A reaction that, once initiated, sustains itself in an endlessly repeating cycle of propagation steps. The radical chlorination of alkanes is an example of a chain reaction that is initiated by irradiation with light and then continues in a series of propagation steps.
• Chair conformation (Section 4.5): A three-dimensional conformation of cyclohexane that resembles the rough shape of a chair. The chair form of cyclohexane is the lowest-energy conformation of the molecule.
• Chemical shift (Section 13.3): The position on the NMR chart where a nucleus absorbs. By convention, the chemical shift of tetramethylsilane (TMS) is set at zero, and all other absorptions usually occur downfield (to the left on the chart). Chemical shifts are expressed in delta units (δ), where 1 δ equals 1 ppm of the spectrometer operating frequency.
• Chiral (Section 5.2): Having handedness. Chiral molecules are those that do not have a plane of symmetry and are therefore not superimposable on their mirror image. A chiral molecule thus exists in two forms, one right-handed and one left-handed. The most common cause of chirality in a molecule is the presence of a carbon atom that is bonded to four different substituents.
• Chiral environment (Section 5.12): The chiral surroundings or conditions in which a molecule resides.
• Chirality center (Section 5.2): An atom (usually carbon) that is bonded to four different groups.
• Chlorohydrin (Section 8.3): A 1,2-chloroalcohol; obtained by addition of HOCl to an alkene.
• Chromatography (Section 26.5): A technique for separating a mixture of compounds into pure components. Different compounds adsorb to a stationary support phase and are then carried along it at different rates by a mobile phase.
• Cis–trans isomers (Section 4.2, Section 7.4): Stereoisomers that differ in their stereochemistry about a ring or double bond.
• Citric acid cycle (Section 29.7): The metabolic pathway by which acetyl CoA is degraded to CO2.
• Claisen condensation reaction (Section 23.7): The carbonyl condensation reaction of two ester molecules to give a β-keto ester product.
• Claisen rearrangement (Section 30.8): The pericyclic conversion of an allyl phenyl ether to an o-allylphenol or an allyl vinyl ether to a γ,δ-unsaturated ketone by heating.
• Coding strand (Section 28.4): The sense strand of double-helical DNA that contains the gene.
• Codon (Section 28.5): A three-base sequence on a messenger RNA chain that encodes the genetic information necessary to cause a specific amino acid to be incorporated into a protein. Codons on mRNA are read by complementary anticodons on tRNA.
• Coenzyme (Section 26.10): A small organic molecule that acts as a cofactor in a biological reaction.
• Cofactor (Section 26.10): A small nonprotein part of an enzyme that is necessary for biological activity.
• Combinatorial chemistry (Chapter 16 Chemistry Matters): A procedure in which anywhere from a few dozen to several hundred thousand substances are prepared simultaneously.
• Complex carbohydrates (Section 25.1): Carbohydrates that are made of two or more simple sugars linked together by glycoside bonds.
• Concerted reaction (Chapter 30 Introduction): A reaction that takes place in a single step without intermediates. For example, the Diels–Alder cycloaddition reaction is a concerted process.
• Condensed structures (Section 1.12): A shorthand way of writing structures in which carbon–hydrogen and carbon–carbon bonds are understood rather than shown explicitly. Propane, for example, has the condensed structure CH3CH2CH3.
• Configuration (Section 5.5): The three-dimensional arrangement of atoms bonded to a chirality center.
• Conformations (Section 3.6): The three-dimensional shape of a molecule at any given instant, assuming that rotation around single bonds is frozen.
• Conformational analysis (Section 4.8): A means of assessing the energy of a substituted cycloalkane by totaling the steric interactions present in the molecule.
• Conformers (Section 3.6): Conformational isomers.
• Conjugate acid (Section 2.7): The product that results from protonation of a Brønsted–Lowry base.
• Conjugate addition (Section 19.13): Addition of a nucleophile to the β carbon atom of an α,β-unsaturated carbonyl compound.
• Conjugate base (Section 2.7): The product that results from deprotonation of a Brønsted–Lowry acid.
• Conjugation (Chapter 14 Introduction): A series of overlapping p orbitals, usually in alternating single and multiple bonds. For example, 1,3-butadiene is a conjugated diene, 3-buten-2-one is a conjugated enone, and benzene is a cyclic conjugated triene.
• Conrotatory (Section 30.2): A term used to indicate that p orbitals must rotate in the same direction during electrocyclic ring-opening or ring-closure.
• Constitutional isomers (Section 3.2, Section 3.6, Section 5.9): Isomers that have their atoms connected in a different order. For example, butane and 2-methylpropane are constitutional isomers.
• Cope rearrangement (Section 30.8): The sigmatropic rearrangement of a 1,5-hexadiene.
• Copolymers (Section 31.3): Polymers obtained when two or more different monomers are allowed to polymerize together.
• Coupled reactions (Section 29.1): Two reactions that share a common intermediate so that the energy released in the favorable step allows the unfavorable step to occur.
• Coupling constant, J (Section 13.6): The magnitude (expressed in hertz) of the interaction between nuclei whose spins are coupled.
• Covalent bond (Section 1.4, Section 1.5): A bond formed by sharing electrons between atoms.
• Cracking (Chapter 3 Chemistry Matters): A process used in petroleum refining in which large alkanes are thermally cracked into smaller fragments.
• Crown ethers (Section 18.6): Large-ring polyethers; used as phase-transfer catalysts.
• Crystallites (Section 31.7): Highly ordered crystal-like regions within a long polymer chain.
• Curtius rearrangement (Section 24.6): The conversion of an acid chloride into an amine by reaction with azide ion, followed by heating with water.
• Cyanohydrins (Section 19.6): A class of compounds with an −OH group and a −CN group bonded to the same carbon atom; formed by addition of HCN to an aldehyde or ketone.
• Cycloaddition reaction (Section 14.4, Section 30.5): A pericyclic reaction in which two reactants add together in a single step to yield a cyclic product. The Diels–Alder reaction between a diene and a dienophile to give a cyclohexene is an example.
• Cycloalkane (Section 4.1): An alkane that contains a ring of carbons.
• D Sugars (Section 25.3): Sugars whose hydroxyl group at the chirality center farthest from the carbonyl group has the same configuration as D-glyceraldehyde and points to the right when drawn in Fischer projection.
• d,l form (Section 5.8): The racemic mixture of a chiral compound.
• Deactivating groups (Section 16.4): Electron-withdrawing substituents that decrease the reactivity of an aromatic ring toward electrophilic aromatic substitution.
• Deamination (Section 29.9): The removal of an amino group from a molecule, as occurs with amino acids during metabolic degradation.
• Debyes (D) (Section 2.2): Units for measuring dipole moments; 1 D = 3.336 × 10−30 coulomb meter (C ∙ m).
• Decarboxylation (Section 22.7): The loss of carbon dioxide from a molecule. β-Keto acids decarboxylate readily on heating.
• Degenerate orbitals (Section 15.2): Two or more orbitals that have the same energy level.
• Degree of unsaturation (Section 7.2): The number of rings and/or multiple bonds in a molecule.
• Dehydration (Section 8.1, Section 12.3, Section 17.6): The loss of water from an alcohol to yield an alkene.
• Dehydrohalogenation (Section 8.1, Section 9.2): The loss of HX from an alkyl halide. Alkyl halides undergo dehydrohalogenation to yield alkenes on treatment with strong base.
• Delocalization (Section 10.4, Section 15.2): A spreading out of electron density over a conjugated π electron system. For example, allylic cations and allylic anions are delocalized because their charges are spread out over the entire π electron system. Aromatic compounds have 4n + 2 π electrons delocalized over their ring.
• Delta (δ) scale (Section 13.3): An arbitrary scale used to calibrate NMR charts. One delta unit (δ) is equal to 1 part per million (ppm) of the spectrometer operating frequency.
• Denatured (Section 26.9): The physical changes that occur in a protein when secondary and tertiary structures are disrupted.
• Deoxy sugar (Section 25.7): A sugar with one of its −OH groups replaced by an −H.
• Deoxyribonucleic acid (DNA) (Chapter 28 Introduction): The biopolymer consisting of deoxyribonucleotide units linked together through phosphate–sugar bonds. Found in the nucleus of cells, DNA contains an organism’s genetic information.
• DEPT-NMR (Section 13.12): An NMR method for distinguishing among signals due to CH3, CH2, CH, and quaternary carbons. That is, the number of hydrogens attached to each carbon can be determined.
• Deshielding (Section 13.2): An effect observed in NMR that causes a nucleus to absorb toward the left (downfield) side of the chart. Deshielding is caused by a withdrawal of electron density from the nucleus.
• Dess–Martin periodinane (Section 17.7): An iodine-based reagent commonly used for the laboratory oxidation of a primary alcohol to an aldehyde or a secondary alcohol to a ketone.
• Deuterium isotope effect (Section 11.8): A tool used in mechanistic investigations to establish whether a C−H bond is broken in the rate-limiting step of a reaction.
• Dextrorotatory (Section 5.3): A word used to describe an optically active substance that rotates the plane of polarization of plane-polarized light in a right-handed (clockwise) direction.
• Diastereomers (Section 5.6): Non–mirror-image stereoisomers; diastereomers have the same configuration at one or more chirality centers but differ at other chirality centers.
• Diastereotopic (Section 13.7): Hydrogens in a molecule whose replacement by some other group leads to different diastereomers.
• 1,3-Diaxial interaction (Section 4.7): The strain energy caused by a steric interaction between axial groups three carbon atoms apart in chair cyclohexane.
• Diazonium salts (Section 24.8): A type of compound with the general structure RN2+ X.
• Diazotization (Section 24.8): The conversion of a primary amine, RNH2, into a diazonium ion, RN2+, by treatment with nitrous acid.
• Dieckmann cyclization reaction (Section 23.9): An intramolecular Claisen condensation reaction of a diester to give a cyclic β-keto ester.
• Diels–Alder reaction (Section 14.4, Section 30.5): The cycloaddition reaction of a diene with a dienophile to yield a cyclohexene.
• Dienophile (Section 14.5): A compound containing a double bond that can take part in the Diels–Alder cycloaddition reaction. The most reactive dienophiles are those that have electron-withdrawing groups on the double bond.
• Digestion (Section 29.1): The first stage of catabolism, in which food is broken down by hydrolysis of ester, glycoside (acetal), and peptide (amide) bonds to yield fatty acids, simple sugars, and amino acids.
• Dihedral angle (Section 3.6): The angle between two bonds on adjacent carbons as viewed along the C−C bond.
• Dipole moment, μ (Section 2.2): A measure of the net polarity of a molecule. A dipole moment arises when the centers of mass of positive and negative charges within a molecule do not coincide.
• Dipole–dipole forces (Section 2.12): Noncovalent electrostatic interactions between dipolar molecules.
• Disaccharide (Section 25.8): A carbohydrate formed by linking two simple sugars through an acetal bond.
• Dispersion forces (Section 2.12): Noncovalent interactions between molecules that arise because of constantly changing electron distributions within the molecules.
• Disrotatory (Section 30.2): A term used to indicate that p orbitals rotate in opposite directions during electrocyclic ring-opening or ring-closing reactions.
• Disulfides (RSSR′) (Section 18.7): A class of compounds of the general structure RSSR′.
• Deoxyribonucleic acid (DNA) (Chapter 28 Introduction): Chemical carriers of a cell’s genetic information.
• Double bond (Section 1.8): A covalent bond formed by sharing two electron pairs between atoms.
• Double helix (Section 28.2): The structure of DNA in which two polynucleotide strands coil around each other.
• Doublet (Section 13.6): A two-line NMR absorption caused by spin–spin splitting when the spin of the nucleus under observation couples with the spin of a neighboring magnetic nucleus.
• Downfield (Section 13.3): Referring to the left-hand portion of the NMR chart.
• E geometry (Section 7.5): A term used to describe the stereochemistry of a carbon–carbon double bond. The two groups on each carbon are ranked according to the Cahn–Ingold–Prelog sequence rules, and the two carbons are compared. If the higher-ranked groups on each carbon are on opposite sides of the double bond, the bond has E geometry.
• E1 reaction (Section 11.10): A unimolecular elimination reaction in which the substrate spontaneously dissociates to give a carbocation intermediate, which loses a proton in a separate step.
• E1cB reaction (Section 11.10): A unimolecular elimination reaction in which a proton is first removed to give a carbanion intermediate, which then expels the leaving group in a separate step.
• E2 reaction (Section 11.8): A bimolecular elimination reaction in which C−H and C−X bond cleavages are simultaneous.
• Eclipsed conformation (Section 3.6): The geometric arrangement around a carbon–carbon single bond in which the bonds to substituents on one carbon are parallel to the bonds to substituents on the neighboring carbon as viewed in a Newman projection.
• Eclipsing strain (Section 3.6): The strain energy in a molecule caused by electron repulsions between eclipsed bonds. Eclipsing strain is also called torsional strain.
• Edman degradation (Section 26.6): A method for N-terminal sequencing of peptide chains by treatment with N-phenylisothiocyanate.
• Eicosanoid (Section 27.4): A lipid derived biologically from 5,8,11,14-eicosatetraenoic acid, or arachidonic acid. Prostaglandins, thromboxanes, and leukotrienes are examples.
• Elastomer (Section 31.7): An amorphous polymer that has the ability to stretch out and spring back to its original shape.
• Electrocyclic reaction (Section 30.2): A unimolecular pericyclic reaction in which a ring is formed or broken by a concerted reorganization of electrons through a cyclic transition state. For example, the cyclization of 1,3,5-hexatriene to yield 1,3-cyclohexadiene is an electrocyclic reaction.
• Electromagnetic spectrum (Section 12.5): The range of electromagnetic energy, including infrared, ultraviolet, and visible radiation.
• Electron configuration (Section 1.3): A list of the orbitals occupied by electrons in an atom.
• Electron-dot structure (Section 1.4): A representation of a molecule showing valence electrons as dots.
• Electron shells (Section 1.2): A group of an atom’s electrons with the same principal quantum number.
• Electron-transport chain (Section 29.1): The final stage of catabolism in which ATP is produced.
• Electronegativity (EN) (Section 2.1): The ability of an atom to attract electrons in a covalent bond. Electronegativity increases across the periodic table from left to right and from bottom to top.
• Electrophile (Section 6.3, Section 6.4): An “electron-lover,” or substance that accepts an electron pair from a nucleophile in a polar bond-forming reaction.
• Electrophilic addition reactions (Section 7.7): Addition of an electrophile to a carbon–carbon double bond to yield a saturated product.
• Electrophilic aromatic substitution reaction (Chapter 16 Introduction): A reaction in which an electrophile (E+) reacts with an aromatic ring and substitutes for one of the ring hydrogens.
• Electrophoresis (Section 26.2, Section 28.6): A technique used for separating charged organic molecules, particularly proteins and DNA fragments. The mixture to be separated is placed on a buffered gel or paper, and an electric potential is applied across the ends of the apparatus. Negatively charged molecules migrate toward the positive electrode, and positively charged molecules migrate toward the negative electrode.
• Electrostatic potential maps (Section 2.1): Molecular representations that use color to indicate the charge distribution in molecules as derived from quantum-mechanical calculations.
• Elimination reactions (Section 6.1): What occurs when a single reactant splits into two products.
• Elution (Section 26.5): The passage of a substance from a chromatography column.
• Embden–Meyerhof pathway (Section 29.5): An alternative name for glycolysis.
• Enamines (Section 19.8): Compounds with the $R2N–CR═CR2R2N–CR═CR2$ functional group.
• Enantiomers (Section 5.1): Stereoisomers of a chiral substance that have a mirror-image relationship. Enantiomers have opposite configurations at all chirality centers.
• Enantioselective synthesis (Chapter 19 Chemistry Matters): A reaction method that yields only a single enantiomer of a chiral product starting from an achiral reactant.
• Enantiotopic (Section 13.7): Hydrogens in a molecule whose replacement by some other group leads to different enantiomers.
• 3′ End (Section 28.1): The end of a nucleic acid chain with a free hydroxyl group at C3′.
• 5′ End (Section 28.1): The end of a nucleic acid chain with a free hydroxyl group at C5′.
• Endergonic (Section 6.7): A reaction that has a positive free-energy change and is therefore nonspontaneous. In an energy diagram, the product of an endergonic reaction has a higher energy level than the reactants.
• Endo (Section 14.5): A term indicating the stereochemistry of a substituent in a bridged bicycloalkane. An endo substituent is syn to the larger of the two bridges.
• Endothermic (Section 6.7): A reaction that absorbs heat and therefore has a positive enthalpy change.
• Energy diagram (Section 6.9): A representation of the course of a reaction, in which free energy is plotted as a function of reaction progress. Reactants, transition states, intermediates, and products are represented, and their appropriate energy levels are indicated.
• Enol (Section 9.4, Section 22.1): A vinylic alcohol that is in equilibrium with a carbonyl compound, $C═C–OHC═C–OH$.
• Enolate ion (Section 22.1): The anion of an enol, $C═C–O−C═C–O−$.
• Enthalpy change (ΔH) (Section 6.7): The heat of reaction. The enthalpy change that occurs during a reaction is a measure of the difference in total bond energy between reactants and products.
• Entropy change (ΔS) (Section 6.7): The change in amount of molecular randomness. The entropy change that occurs during a reaction is a measure of the difference in randomness between reactants and products.
• Enzyme (Section 6.11, Section 26.10): A biological catalyst. Enzymes are large proteins that catalyze specific biochemical reactions.
• Epimers (Section 5.6): Diastereomers that differ in configuration at only one chirality center but are the same at all others.
• Epoxide (Section 8.7): A three-membered-ring ether functional group.
• Equatorial bonds (Section 4.6): Bonds or positions in chair cyclohexane that lie along the rough equator of the ring.
• ESI (Section 12.4): Electrospray ionization; a “soft” ionization method used for mass spectrometry of biological samples of very high molecular weight.
• Essential amino acid (Section 26.1): One of nine amino acids that are biosynthesized only in plants and microorganisms and must be obtained by humans in the diet.
• Essential monosaccharide (Section 25.7): One of eight simple sugars that is best obtained in the diet rather than by biosynthesis.
• Essential oil (Chapter 8 Chemistry Matters): The volatile oil obtained by steam distillation of a plant extract.
• Esters (Section 3.1, Chapter 21 Introduction): A class of compounds containing the −CO2R functional group.
• Estrogens (Section 27.6): Female steroid sex hormones.
• Ethers (Section 3.1, Chapter 18 Introduction): A class of compounds that has two organic substituents bonded to the same oxygen atom, ROR′.
• Exergonic (Section 6.7): A reaction that has a negative free-energy change and is therefore spontaneous. On an energy diagram, the product of an exergonic reaction has a lower energy level than that of the reactants.
• Exo (Section 14.5): A term indicating the stereochemistry of a substituent in a bridged bicycloalkane. An exo substituent is anti to the larger of the two bridges.
• Exon (Section 28.4): A section of DNA that contains genetic information.
• Exothermic (Section 6.7): A reaction that releases heat and therefore has a negative enthalpy change.
• Fats (Section 27.1): Solid triacylglycerols derived from an animal source.
• Fatty acids (Section 27.1): A long, straight-chain carboxylic acid found in fats and oils.
• Fiber (Section 31.7): A thin thread produced by extruding a molten polymer through small holes in a die.
• Fibrous proteins (Section 26.9): A type of protein that consists of polypeptide chains arranged side by side in long threads. Such proteins are tough, insoluble in water, and used in nature for structural materials such as hair, hooves, and fingernails.
• Fingerprint region (Section 12.7): The complex region of the infrared spectrum from 1500–400 cm−1.
• First-order reaction (Section 11.4): Designates a reaction whose rate-limiting step is unimolecular and whose kinetics therefore depend on the concentration of only one reactant.
• Fischer esterification reaction (Section 21.3): The acid-catalyzed nucleophilic acyl substitution reaction of a carboxylic acid with an alcohol to yield an ester.
• Fischer projections (Section 25.2): A means of depicting the absolute configuration of a chiral molecule on a flat page. A Fischer projection uses a cross to represent the chirality center. The horizontal arms of the cross represent bonds coming out of the plane of the page, and the vertical arms of the cross represent bonds going back into the plane of the page.
• Fmoc derivative (Section 26.7): A fluorenylmethyloxycarbonyl N-protected amino acid.
• Formal charges (Section 2.3): The difference in the number of electrons owned by an atom in a molecule and by the same atom in its elemental state.
• Formyl (Section 19.1): A −CHO group.
• Frequency, ν (Section 12.5): The number of electromagnetic wave cycles that travel past a fixed point in a given unit of time. Frequencies are expressed in units of cycles per second, or hertz.
• Friedel–Crafts reaction (Section 16.3): An electrophilic aromatic substitution reaction to alkylate or acylate an aromatic ring.
• Frontier orbitals (Section 30.1): The highest occupied (HOMO) and lowest unoccupied (LUMO) molecular orbitals.
• FT-NMR (Section 13.10): Fourier-transform NMR; a rapid technique for recording NMR spectra in which all magnetic nuclei absorb at the same time.
• Functional (Section 3.1): An atom or group of atoms that is part of a larger molecule and has a characteristic chemical reactivity.
• Functional RNAs (Section 28.4): An alternative name for small RNAs.
• Furanose (Section 25.5): The five-membered-ring form of a simple sugar.
• Gabriel amine synthesis (Section 24.6): A method for preparing an amine by SN2 reaction of an alkyl halide with potassium phthalimide, followed by hydrolysis.
• Gauche conformation (Section 3.7): The conformation of butane in which the two methyl groups lie 60° apart as viewed in a Newman projection. This conformation has 3.8 kJ/mol steric strain.
• Geminal (Section 19.5): Referring to two groups attached to the same carbon atom. For example, the hydrate formed by nucleophilic addition of water to an aldehyde or ketone is a geminal diol.
• Gibbs free-energy change (ΔG) (Section 6.7): The free-energy change that occurs during a reaction, given by the equation $ΔG=ΔH−TΔS.ΔG=ΔH−TΔS.$ A reaction with a negative free-energy change is spontaneous, and a reaction with a positive free-energy change is nonspontaneous.
• Gilman reagent (LiR2Cu) (Section 10.7): A diorganocopper reagent.
• Glass transition temperature, Tg (Section 31.7): The temperature at which a hard, amorphous polymer becomes soft and flexible.
• Globular proteins (Section 26.9): A type of protein that is coiled into a compact, nearly spherical shape. Globular proteins, which are generally water-soluble and mobile within the cell, are the structural class to which enzymes belong.
• Gluconeogenesis (Section 29.8): The anabolic pathway by which organisms make glucose from simple three-carbon precursors.
• Glycal (Section 25.9): An unsaturated sugar with a C1–C2 double bond.
• Glycal assembly method (Section 25.9): A method for linking monosaccharides together to synthesize polysaccharides.
• Glycerophospholipids (Section 27.3): Lipids that contain a glycerol backbone linked to two fatty acids and a phosphoric acid.
• Glycoconjugate (Section 25.6): A molecule in which a carbohydrate is linked through its anomeric center to another biological molecule such as a lipid or protein.
• Glycol (Section 8.7): A diol, such as ethylene glycol, HOCH2CH2OH.
• Glycolipid (Section 25.6): A biological molecule in which a carbohydrate is linked through a glycoside bond to a lipid.
• Glycolysis (Section 29.5): A series of ten enzyme-catalyzed reactions that break down glucose into 2 equivalents of pyruvate, CH3COCO2.
• Glycoprotein (Section 25.6): A biological molecule in which a carbohydrate is linked through a glycoside bond to a protein.
• Glycoside (Section 25.6): A cyclic acetal formed by reaction of a sugar with another alcohol.
• Graft copolymers (Section 31.3): Copolymers in which homopolymer branches of one monomer unit are “grafted” onto a homopolymer chain of another monomer unit.
• Green chemistry (Chapter 11 Chemistry Matters, Chapter 24 Chemistry Matters): The design and implementation of chemical products and processes that reduce waste and minimize or eliminate the generation of hazardous substances.
• Grignard reagent (RMgX) (Section 10.6): An organomagnesium halide.
• Ground-state electron configuration (Section 1.3): The most stable, lowest-energy electron configuration of a molecule or atom.
• Haloform reaction (Section 22.6): The reaction of a methyl ketone with halogen and base to yield a haloform (CHX3) and a carboxylic acid.
• Halogenation (Section 8.2, Section 16.1): The reaction of halogen with an alkene to yield a 1,2-dihalide addition product or with an aromatic compound to yield a substitution product.
• Halohydrin (Section 8.3): A 1,2-haloalcohol, such as that obtained on addition of HOBr to an alkene.
• Halonium ion (Section 8.2): A species containing a positively charged, divalent halogen. Three-membered-ring bromonium ions are intermediates in the electrophilic addition of Br2 to alkenes.
• Hammond postulate (Section 7.10): A postulate stating that we can get a picture of what a given transition state looks like by looking at the structure of the nearest stable species. Exergonic reactions have transition states that resemble reactant; endergonic reactions have transition states that resemble product.
• Heat of combustion (Section 4.3): The amount of heat released when a compound burns completely in oxygen.
• Heat of hydrogenation (Section 7.6): The amount of heat released when a carbon–carbon double bond is hydrogenated.
• Heat of reaction (Section 6.7): An alternative name for the enthalpy change in a reaction, ΔH.
• Hell–Volhard–Zelinskii (HVZ) reaction (Section 22.4): The reaction of a carboxylic acid with Br2 and phosphorus to give an α-bromo carboxylic acid.
• Hemiacetal (Section 19.10): A functional group having one −OR and one −OH group bonded to the same carbon.
• Henderson–Hasselbalch equation (Section 20.3, Section 24.5, Section 26.2): An equation for determining the extent of dissociation of a weak acid at various pH values.
• Hertz, Hz (Section 12.5): A unit of measure of electromagnetic frequency, the number of waves that pass by a fixed point per second.
• Heterocycle (Section 15.5, Section 24.9): A cyclic molecule whose ring contains more than one kind of atom. For example, pyridine is a heterocycle that contains five carbon atoms and one nitrogen atom in its ring.
• Heterolytic bond breakage (Section 6.2): The kind of bond-breaking that occurs in polar reactions when one fragment leaves with both of the bonding electrons: A : B → A+ + B:.
• Hofmann elimination reaction (Section 24.7): The elimination reaction of an amine to yield an alkene by reaction with iodomethane followed by heating with Ag2O.
• Hofmann rearrangement (Section 24.6): The conversion of an amide into an amine by reaction with Br2 and base.
• Highest occupied molecular orbital (HOMO) (Section 14.7, Section 30.1): The symmetries of the HOMO and LUMO are important in pericyclic reactions.
• Homolytic bond breakage (Section 6.2): The kind of bond-breaking that occurs in radical reactions when each fragment leaves with one bonding electron: A : B → A+ + B:.
• Homopolymers (Section 31.3): A polymer made up of identical repeating units.
• Homotopic (Section 13.7): Hydrogens in a molecule that give the identical structure on replacement by X and thus show identical NMR absorptions.
• Hormones (Section 27.6): Chemical messengers that are secreted by an endocrine gland and carried through the bloodstream to a target tissue.
• HPLC (Section 26.5): High-pressure liquid chromatography; a variant of column chromatography using high pressure to force solvent through very small absorbent particles.
• Hückel 4n + 2 rule (Section 15.3): A rule stating that monocyclic conjugated molecules having 4n + 2 π electrons (n = an integer) are aromatic.
• Hund’s rule (Section 1.3): If two or more empty orbitals of equal energy are available, one electron occupies each, with their spins parallel, until all are half-full.
• Hybrid orbital (Section 1.6): An orbital derived from a combination of atomic orbitals. Hybrid orbitals, such as the sp3, sp2, and sp hybrids of carbon, are strongly directed and form stronger bonds than atomic orbitals do.
• Hydration (Section 8.4): Addition of water to a molecule, such as occurs when alkenes are treated with aqueous sulfuric acid to give alcohols.
• Hydride shift (Section 7.11): The shift of a hydrogen atom and its electron pair to a nearby cationic center.
• Hydroboration (Section 8.5): Addition of borane (BH3) or an alkylborane to an alkene. The resultant trialkylborane products can be oxidized to yield alcohols.
• Hydrocarbons (Section 3.2): A class of compounds that contain only carbon and hydrogen.
• Hydrogen bond (Section 2.12): A weak attraction between a hydrogen atom bonded to an electronegative atom and an electron lone pair on another electronegative atom.
• Hydrogenated (Section 6.11): Addition of hydrogen to a double or triple bond to yield a saturated product.
• Hydrogenolysis (Section 26.7): Cleavage of a bond by reaction with hydrogen. Benzylic ethers and esters, for instance, are cleaved by hydrogenolysis.
• Hydrophilic (Section 2.12): Water-loving; attracted to water.
• Hydrophobic (Section 2.12): Water-fearing; repelled by water.
• Hydroquinones (Section 17.10): 1,4-dihydroxybenzene.
• Hydroxylation (Section 8.7): Addition of two −OH groups to a double bond.
• Hyperconjugation (Section 7.6, Section 7.9): An electronic interaction that results from overlap of a vacant p orbital on one atom with a neighboring C−H σ bond. Hyperconjugation is important in stabilizing carbocations and substituted alkenes.
• Imide (Section 24.6): A compound with the −CONHCO− functional group.
• Imines (Section 19.8): A class of compounds with the $R2C═NRR2C═NR$ functional group.
• Inductive effect (Section 2.1, Section 7.9, Section 16.4): The electron-attracting or electron-withdrawing effect transmitted through σ bonds. Electronegative elements have an electron-withdrawing inductive effect.
• Infrared (IR) spectroscopy (Section 12.6): A kind of optical spectroscopy that uses infrared energy. IR spectroscopy is particularly useful in organic chemistry for determining the kinds of functional groups present in molecules.
• Initiator (Section 8.10, Section 31.1): A substance that is used to initiate a radical chain reaction or polymerization. For example, radical chlorination of alkanes is initiated when light energy breaks the weak Cl−Cl bond to form Cl· radicals.
• Integrating (Section 13.5): A technique for measuring the area under an NMR peak to determine the relative number of each kind of proton in a molecule.
• Intermediate (Section 6.10): A species that is formed during the course of a multistep reaction but is not the final product. Intermediates are more stable than transition states but may or may not be stable enough to isolate.
• Intramolecular, intermolecular (Section 23.6): A reaction that occurs within the same molecule is intramolecular; a reaction that occurs between two molecules is intermolecular.
• Intron (Section 28.4): A section of DNA that does not contain genetic information.
• Ion pairs (Section 11.4): A loose association between two ions in solution. Ion pairs are implicated as intermediates in SN1 reactions to account for the partial retention of stereochemistry that is often observed.
• Ionic bond (Section 1.4): The electrostatic attraction between ions of unlike charge.
• Isoelectric point (pI) (Section 26.2): The pH at which the number of positive charges and the number of negative charges on a protein or an amino acid are equal.
• Isomers (Section 3.2, Section 5.9): Compounds that have the same molecular formula but different structures.
• Isoprene rule (Chapter 8 Chemistry Matters): An observation to the effect that terpenoids appear to be made up of isoprene (2-methyl-1,3-butadiene) units connected head-to-tail.
• Isotactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents is oriented regularly along the backbone.
• Isotopes (Section 1.1): Atoms of the same element that have different mass numbers.
• IUPAC system of nomenclature (Section 3.4): Rules for naming compounds, devised by the International Union of Pure and Applied Chemistry.
• Kekulé structure (Section 1.4): An alternative name for a line-bond structure, which represents a molecule by showing covalent bonds as lines between atoms.
• Ketals (Section 19.10): An alternative name for acetals derived from a ketone rather than an aldehyde and consisting of two −OR groups bonded to the same carbon, R2C(OR′)2. Ketals are often used as protecting groups for ketones.
• Keto–enol tautomerism (Section 9.4, Section 22.1): The equilibration between a carbonyl form and vinylic alcohol form of a molecule.
• Ketones (R2CO) (Section 3.1, Chapter 19 Introduction): A class of compounds with two organic substituents bonded to a carbonyl group, $R2C═OR2C═O$.
• Ketoses (Section 25.1): Carbohydrates with a ketone functional group.
• Kiliani–Fischer synthesis (Section 25.6): A method for lengthening the chain of an aldose sugar.
• Kinetic control (Section 14.3): A reaction that follows the lowest activation energy pathway is said to be kinetically controlled. The product is the most rapidly formed but is not necessarily the most stable.
• Kinetics (Section 11.2): Referring to reaction rates. Kinetic measurements are useful for helping to determine reaction mechanisms.
• Koenigs–Knorr reaction (Section 25.6): A method for the synthesis of glycosides by reaction of an alcohol with a pyranosyl bromide.
• Krebs cycle (Section 29.7): An alternative name for the citric acid cycle, by which acetyl CoA is degraded to CO2.
• L Sugar (Section 25.3): A sugar whose hydroxyl group at the chirality center farthest from the carbonyl group points to the left when drawn in Fischer projection.
• Lactams (Section 21.7): Cyclic amides.
• Lactones (Section 21.6): Cyclic esters.
• Lagging strand (Section 28.3): The complement of the original 3′ → 5′ DNA strand that is synthesized discontinuously in small pieces that are subsequently linked by DNA ligases.
• LD50 (Chapter 1 Chemistry Matters): The amount of a substance per kilogram body weight that is lethal to 50% of test animals.
• LDA (Section 22.5): Lithium diisopropylamide, LiN(i-C3H7)2, a strong base commonly used to convert carbonyl compounds into their enolate ions.
• Leading strand (Section 28.3): The complement of the original 5′ → 3′ DNA strand that is synthesized continuously in a single piece.
• Leaving group (Section 11.3): The group that is replaced in a substitution reaction.
• Levorotatory (Section 5.3): An optically active substance that rotates the plane of polarization of plane-polarized light in a left-handed (counterclockwise) direction.
• Lewis acid (Section 2.11): A substance with a vacant low-energy orbital that can accept an electron pair from a base. All electrophiles are Lewis acids.
• Lewis base (Section 2.11): A substance that donates an electron lone pair to an acid. All nucleophiles are Lewis bases.
• Lewis structures (Section 1.4): Representations of molecules showing valence electrons as dots.
• Lindlar catalyst (Section 9.5): A hydrogenation catalyst used to convert alkynes to cis alkenes.
• Line-bond structure (Section 1.4): An alternative name for a Kekulé structure, which represents a molecule by showing covalent bonds as lines between atoms.
• $1→4Section 25.8): A glycoside link between the C1 −OH group of one sugar and the C4 −OH group of another sugar.$
• Lipids (Chapter 27 Introduction): Naturally occurring substances isolated from cells and tissues by extraction with a nonpolar solvent. Lipids belong to many different structural classes, including fats, terpenoids, prostaglandins, and steroids.
• Lipid bilayer (Section 27.3): The ordered lipid structure that forms a cell membrane.
• Lipoprotein (Chapter 27 Chemistry Matters): A complex molecule with both lipid and protein parts that transports lipids through the body.
• Locant (Section 3.4): A number in a chemical name that locates the positions of the functional groups and substituents in the molecule.
• Lone-pair electrons (Section 1.4): Nonbonding valence-shell electron pairs. Lone-pair electrons are used by nucleophiles in their reactions with electrophiles.
• Lowest unoccupied molecular orbital (LUMO) (Section 14.7, Section 30.1): The symmetries of the LUMO and the HOMO are important in determining the stereochemistry of pericyclic reactions.
• Magnetic resonance imaging, MRI (Chapter 13 Chemistry Matters): A medical diagnostic technique based on nuclear magnetic resonance.
• Magnetogyric ratio (Section 13.1): A ratio of the isotope’s magnetic moment to its angular momentum.
• MALDI (Section 12.4): Matrix-assisted laser desorption ionization; a soft ionization method used for mass spectrometry of biological samples of very high molecular weight.
• Malonic ester synthesis (Section 22.7): The synthesis of a carboxylic acid by alkylation of an alkyl halide with diethyl malonate, followed by hydrolysis and decarboxylation.
• Markovnikov’s rule (Section 7.8): A guide for determining the regiochemistry (orientation) of electrophilic addition reactions. In the addition of HX to an alkene, the hydrogen atom bonds to the alkene carbon that has fewer alkyl substituents.
• Mass number (A) (Section 1.1): The total of protons plus neutrons in an atom.
• Mass spectrometry (MS) (Section 12.1): A technique for measuring the mass, and therefore the molecular weight (MW), of ions.
• McLafferty rearrangement (Section 12.3, Section 19.4): A mass-spectral fragmentation pathway for carbonyl compounds.
• Mechanism (Section 6.2): A complete description of how a reaction occurs. A mechanism accounts for all starting materials and all products and describes the details of each individual step in the overall reaction process.
• Meisenheimer complex (Section 16.6): An intermediate formed by addition of a nucleophile to a halo-substituted aromatic ring.
• Melt transition temperature, Tm (Section 31.7): The temperature at which crystalline regions of a polymer melt to give an amorphous material.
• Mercapto group (Section 18.7): An alternative name for the thiol group, −SH.
• Meso compounds (Section 5.7): Compounds that contain chirality centers but are nevertheless achiral because they contain a symmetry plane.
• Messenger RNA (mRNA) (Section 28.4): A kind of RNA formed by transcription of DNA and used to carry genetic messages from DNA to ribosomes.
• Meta (m) (Section 15.1): A naming prefix used for 1,3-disubstituted benzenes.
• Metabolism (Section 29.1): A collective name for the many reactions that go on in the cells of living organisms.
• Metallacycle (Section 31.5): A cyclic compound that contains a metal atom in its ring.
• Methylene group (Section 7.3): A −CH2− or $═CH2═CH2$ group.
• Micelles (Section 27.2): Spherical clusters of soaplike molecules that aggregate in aqueous solution. The ionic heads of the molecules lie on the outside, where they are solvated by water, and the organic tails bunch together on the inside of the micelle.
• Michael reaction (Section 23.10): The conjugate addition reaction of an enolate ion to an unsaturated carbonyl compound.
• Molar absorptivity (ɛ) (Section 14.7): A quantitative measure of the amount of UV light absorbed by a sample.
• Molecular ion (Section 12.1): The cation produced in a mass spectrometer by loss of an electron from the parent molecule. The mass of the molecular ion corresponds to the molecular weight of the sample.
• Molecular mechanics (Chapter 4 Chemistry Matters): A computer-based method for calculating the minimum-energy conformation of a molecule.
• Molecular orbital (MO) theory (Section 1.11, Section 14.1): A description of covalent bond formation as resulting from a mathematical combination of atomic orbitals (wave functions) to form molecular orbitals.
• Molecule (Section 1.4): A neutral collection of atoms held together by covalent bonds.
• Molozonide (Section 8.8): The initial addition product of ozone with an alkene.
• Monomers (Section 8.10, Section 21.9; Chapter 31 Introduction): The simple starting units from which polymers are made.
• Monosaccharides (Section 25.1): Simple sugars.
• Monoterpenoids (Chapter 8 Chemistry Matters, Section 27.5): Ten-carbon lipids.
• Multiplet (Section 13.6): A pattern of peaks in an NMR spectrum that arises by spin–spin splitting of a single absorption because of coupling between neighboring magnetic nuclei.
• Mutarotation (Section 25.5): The change in optical rotation observed when a pure anomer of a sugar is dissolved in water. Mutarotation is caused by the reversible opening and closing of the acetal linkage, which yields an equilibrium mixture of anomers.
• n + 1 rule (Section 13.6): A hydrogen with n other hydrogens on neighboring carbons shows n + 1 peaks in its 1H NMR spectrum.
• N-terminal amino acid (Section 26.4): The amino acid with a free −NH2 group at the end of a protein chain.
• Natural gas (Chapter 3 Chemistry Matters): A naturally occurring hydrocarbon mixture consisting chiefly of methane, along with smaller amounts of ethane, propane, and butane.
• Natural product (Chapter 7 Chemistry Matters): A catchall term generally taken to mean a secondary metabolite found in bacteria, plants, and other living organisms.
• New molecular entity, NME (Chapter 6 Chemistry Matters): A new biologically active chemical substance approved for sale as a drug by the U.S. Food and Drug Administration.
• Newman projection (Section 3.6): A means of indicating stereochemical relationships between substituent groups on neighboring carbons. The carbon–carbon bond is viewed end-on, and the carbons are indicated by a circle. Bonds radiating from the center of the circle are attached to the front carbon, and bonds radiating from the edge of the circle are attached to the rear carbon.
• Nitration (Section 16.2): The substitution of a nitro group onto an aromatic ring.
• Nitriles (Section 3.1, Section 20.1): A class of compounds containing the $C≡NC≡N$ functional group.
• Nitrogen rule (Section 24.10): A compound with an odd number of nitrogen atoms has an odd-numbered molecular weight.
• Node (Section 1.2): A surface of zero electron density within an orbital. For example, a p orbital has a nodal plane passing through the center of the nucleus, perpendicular to the axis of the orbital.
• Nonbonding electrons (Section 1.4): Valence electrons that are not used in forming covalent bonds.
• Noncoding strand (Section 28.4): An alternative name for the antisense strand of DNA.
• Noncovalent interactions (Section 2.12): One of a variety of nonbonding interactions between molecules, such as dipole–dipole forces, dispersion forces, and hydrogen bonds.
• Nonessential amino acid (Section 26.1): One of the eleven amino acids that are biosynthesized by humans.
• Normal alkanes (Section 3.2): Straight-chain alkanes, as opposed to branched alkanes. Normal alkanes are denoted by the suffix n, as in n-C4H10 (n-butane).
• NSAID (Chapter 15 Chemistry Matters): A nonsteroidal anti-inflammatory drug, such as aspirin or ibuprofen.
• Nuclear magnetic resonance (NMR) spectroscopy (Chapter 13 Introduction): A spectroscopic technique that provides information about the carbon–hydrogen framework of a molecule. NMR works by detecting the energy absorptions accompanying the transitions between nuclear spin states that occur when a molecule is placed in a strong magnetic field and irradiated with radiofrequency waves.
• Nucleic acid (Section 28.1): Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA); biological polymers made of nucleotides joined together to form long chains.
• Nucleophile (Section 6.3, Section 6.4): An electron-rich species that donates an electron pair to an electrophile in a polar bond-forming reaction. Nucleophiles are also Lewis bases.
• Nucleophilic acyl substitution reaction (Section 21.2): A reaction in which a nucleophile attacks a carbonyl compound and substitutes for a leaving group bonded to the carbonyl carbon.
• Nucleophilic addition reaction (Section 19.4): A reaction in which a nucleophile adds to the electrophilic carbonyl group of a ketone or aldehyde to give an alcohol.
• Nucleophilic aromatic substitution reactions (Section 16.6): The substitution reactions of an aryl halide by a nucleophile.
• Nucleophilic substitution reactions (Section 11.1): Reactions in which one nucleophile replaces another attached to a saturated carbon atom.
• Nucleophilicity (Section 11.3): The ability of a substance to act as a nucleophile in an SN2 reaction.
• Nucleoside (Section 28.1): A nucleic acid constituent consisting of a sugar residue bonded to a heterocyclic purine or pyrimidine base.
• Nucleotides (Section 28.1): Nucleic acid constituents consisting of a sugar residue bonded both to a heterocyclic purine or pyrimidine base and to a phosphoric acid. Nucleotides are the monomer units from which DNA and RNA are constructed.
• Nylons (Section 21.9): Synthetic polyamide step-growth polymers.
• Okazaki fragments (Section 28.3): Short segments of a DNA lagging strand that is biosynthesized discontinuously and then linked by DNA ligases.
• Olefin (Chapter 7 Introduction): An alternative name for an alkene.
• Olefin metathesis polymerization (Section 31.5): A method of polymer synthesis based on using an olefin metathesis reaction.
• Olefin metathesis reaction (Section 31.5): A reaction in which two olefins (alkenes) exchange substituents on their double bonds.
• Oligonucleotides (Section 28.7): Short segments of DNA.
• Optical isomers (Section 5.4): An alternative name for enantiomers. Optical isomers are isomers that have a mirror-image relationship.
• Optically active (Section 5.3): A property of some organic molecules wherein the plane of polarization is rotated through an angle when a beam of plane-polarized light is passed through a solution of the molecules.
• Orbital (Section 1.2): A wave function, which describes the volume of space around a nucleus in which an electron is most likely to be found.
• Organic chemistry (Chapter 1 Introduction): The study of carbon compounds.
• Organohalides (Chapter 10 Introduction): Compounds that contain one or more halogen atoms bonded to carbon.
• Organometallic compound (Section 10.6): A compound that contains a carbon–metal bond. Grignard reagents, RMgX, are examples.
• Organophosphate (Section 1.10): A compound that contains a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon.
• Ortho (o) (Section 15.1): A naming prefix used for 1,2-disubstituted benzenes.
• Oxidation (Section 8.7, Section 10.8): A reaction that causes a decrease in electron ownership by carbon, either by bond formation between carbon and a more electronegative atom (usually oxygen, nitrogen, or a halogen) or by bond-breaking between carbon and a less electronegative atom (usually hydrogen).
• Oximes (Section 19.8): Compounds with the $R2C═NOHR2C═NOH$ functional group.
• Oxirane (Section 8.7): An alternative name for an epoxide.
• Oxymercuration (Section 8.4): A method for double-bond hydration by reaction of an alkene with aqueous mercuric acetate followed by treatment with NaBH4.
• Ozonide (Section 8.8): The product initially formed by addition of ozone to a carbon–carbon double bond. Ozonides are usually treated with a reducing agent, such as zinc in acetic acid, to produce carbonyl compounds.
• Para (p) (Section 15.1): A naming prefix used for 1,4-disubstituted benzenes.
• Paraffins (Section 3.5): A common name for alkanes.
• Parent peak (Section 12.1): The peak in a mass spectrum corresponding to the molecular ion. The mass of the parent peak therefore represents the molecular weight of the compound.
• Pauli exclusion principle (Section 1.3): No more than two electrons can occupy the same orbital, and those two must have spins of opposite sign.
• Peptide bond (Section 26.4): An amide bond in a peptide chain.
• Peptides (Chapter 26 Introduction): A type of short amino acid polymer in which the individual amino acid residues are linked by amide bonds.
• Pericyclic reaction (Chapter 30 Introduction): A reaction that occurs in a single step by a reorganization of bonding electrons in a cyclic transition state.
• Periplanar (Section 11.8): A conformation in which bonds to neighboring atoms have a parallel arrangement. In an eclipsed conformation, the neighboring bonds are syn periplanar; in a staggered conformation, the bonds are anti periplanar.
• Peroxides (Section 18.1): Molecules containing an oxygen–oxygen bond functional group, ROOR′ or ROOH.
• Peroxyacid (Section 8.7): A compound with the −CO3H functional group. Peroxyacids react with alkenes to give epoxides.
• Phenols (Chapter 17 Introduction): A class of compounds with an −OH group directly bonded to an aromatic ring, ArOH.
• Phenoxide ion, ArO (Section 17.2): The anion of a phenol.
• Phenyl (Section 15.1): The name for the −C6H5 unit when the benzene ring is considered as a substituent. A phenyl group is abbreviated as −Ph.
• Phosphine (Section 5.10): A trivalent phosphorus compound, R3P.
• Phosphite (Section 28.7): A compound with the structure P(OR)3.
• Phospholipids (Section 27.3): Lipids that contain a phosphate residue. For example, glycerophospholipids contain a glycerol backbone linked to two fatty acids and a phosphoric acid.
• Phosphoramidite (Section 28.7): A compound with the structure R2NP(OR)2.
• Phosphoric acid anhydride (Section 29.1): A substance that contains PO2PO link, analogous to the CO2CO link in carboxylic acid anhydrides.
• Photochemical reactions (Section 30.3): A reaction carried out by irradiating the reactants with light.
• Physiological pH (Section 20.3): The pH of 7.3 that exists inside cells.
• Pi (π) bond (Section 1.8): The covalent bond formed by sideways overlap of atomic orbitals. For example, carbon–carbon double bonds contain a π bond formed by sideways overlap of two p orbitals.
• PITC (Section 26.6): Phenylisothiocyanate; used in the Edman degradation.
• pKa (Section 2.8): The negative common logarithm of the Ka; used to express acid strength.
• Plane of symmetry (Section 5.2): A plane that bisects a molecule such that one half of the molecule is the mirror image of the other half. Molecules containing a plane of symmetry are achiral.
• Plane-polarized light (Section 5.3): Light that has its electromagnetic waves oscillating in a single plane rather than in random planes. The plane of polarization is rotated when the light is passed through a solution of a chiral substance.
• Plasticizers (Section 31.7): Small organic molecules added to polymers to act as a lubricant between polymer chains.
• Polar aprotic solvents (Section 11.3): Polar solvents that can’t function as hydrogen ion donors. Polar aprotic solvents such as dimethyl sulfoxide (DMSO) and dimethylformamide (DMF) are particularly useful in SN2 reactions because of their ability to solvate cations.
• Polar covalent bond (Section 2.1): A covalent bond in which the electron distribution between atoms is unsymmetrical.
• Polar reactions (Section 6.2, Section 6.3): Reactions in which bonds are made when a nucleophile donates two electrons to an electrophile and in which bonds are broken when one fragment leaves with both electrons from the bond.
• Polarity (Section 2.1): The unsymmetrical distribution of electrons in a molecule that results when one atom attracts electrons more strongly than another.
• Polarizability (Section 6.3): The measure of the change in a molecule’s electron distribution in response to changing electrostatic interactions with solvents or ionic reagents.
• Polycarbonates (Section 31.4): Polyesters in which the carbonyl groups are linked to two −OR groups, [$O═C(OR)2O═C(OR)2$].
• Polycyclic (Section 4.9): Containing more than one ring.
• Polycyclic aromatic compound (Section 15.6): A compound with two or more benzene-like aromatic rings fused together.
• Polymer (Section 8.10, Section 21.9; Chapter 31 Introduction): A large molecule made up of repeating smaller units. For example, polyethylene is a synthetic polymer made from repeating ethylene units, and DNA is a biopolymer made of repeating deoxyribonucleotide units.
• Polymerase chain reaction (PCR) (Section 28.8): A method for amplifying small amounts of DNA to produce larger amounts.
• Polysaccharides (Section 25.9): A type of carbohydrate that is made of many simple sugars linked together by glycoside (acetal) bonds.
• Polyunsaturated fatty acids (Section 27.1): Fatty acids that contain more than one double bond.
• Polyurethane (Section 31.4): A step-growth polymer prepared by reaction between a diol and a diisocyanate.
• Posttranslational modification (Section 28.6): A chemical modification of a protein that occurs after translation from DNA.
• Primary, secondary, tertiary, and quaternary (Section 3.3): Terms used to describe the substitution pattern at a specific site. A primary site has one organic substituent attached to it, a secondary site has two organic substituents, a tertiary site has three, and a quaternary site has four.
Primary Secondary Tertiary Quaternary
Carbon RCH3 R2CH2 R3CH R4C
Carbocation RCH2+ R2CH+ R3C+
Hydrogen RCH3 R2CH2 R3CH
Alcohol RCH2OH R2CHOH R3COH
Amine RNH2 R2NH R3N
• Primary structure (Section 26.9): The amino acid sequence in a protein.
• pro-R (Section 5.11): One of two identical atoms or groups of atoms in a compound whose replacement leads to an R chirality center.
• pro-S (Section 5.11): One of two identical atoms or groups of atoms in a compound whose replacement leads to an S chirality center.
• Prochiral (Section 5.11): A molecule that can be converted from achiral to chiral in a single chemical step.
• Prochirality center (Section 5.11): An atom in a compound that can be converted into a chirality center by changing one of its attached substituents.
• Promoter sequence (Section 28.4): A short sequence on DNA located upstream of the transcription start site and recognized by RNA polymerase.
• Propagation step (Section 8.10): A step in a radical chain reaction that carries on the chain. The propagation steps must yield both product and a reactive intermediate.
• Prostaglandins (Section 27.4): Lipids derived from arachidonic acid. Prostaglandins are present in nearly all body tissues and fluids, where they serve many important hormonal functions.
• Protecting group (Section 17.8, Section 19.10, Section 26.7): A group that is introduced to protect a sensitive functional group toward reaction elsewhere in the molecule. After serving its protective function, the group is removed.
• Proteins (Chapter 26 Introduction): Large peptides containing 50 or more amino acid residues. Proteins serve both as structural materials and as enzymes that control an organism’s chemistry.
• Protein Data Bank (Chapter 26 Chemistry Matters): A worldwide online repository of X-ray and NMR structural data for biological macromolecules. To access the Protein Data Bank, go to https://www.rcsb.org.
• Protic solvents (Section 11.3): Solvents such as water or alcohol that can act as a proton donor.
• Pyramidal inversion (Section 24.2): The rapid stereochemical inversion of a trivalent nitrogen compound.
• Pyranose (Section 25.5): The six-membered, cyclic hemiacetal form of a simple sugar.
• Quadrupole mass analyzer (Section 12.1): A type of mass spectrometer that uses four cylindrical rods to create an oscillating electrostatic field. Ion trajectories are determined by their m/z ratios. At a given field, only one m/z value will make it through the quadrupole region—the others will crash into the quadrupole rods or the walls of the instrument and never reach the detector.
• Quartet (Section 13.6): A set of four peaks in an NMR spectrum, caused by spin–spin splitting of a signal by three adjacent nuclear spins.
• Quaternary: See Primary.
• Quaternary ammonium salt (Section 24.1): An ionic compound containing a positively charged nitrogen atom with four attached groups, R4N+ X.
• Quaternary structure (Section 26.9): The highest level of protein structure, involving an ordered aggregation of individual proteins into a larger cluster.
• Quinone (Section 17.10): A 2,5-cyclohexadiene-1,4-dione.
• R configuration (Section 5.5): The configuration at a chirality center as specified using the Cahn–Ingold–Prelog sequence rules.
• R (Section 3.3): A generalized abbreviation for an organic partial structure.
• Racemate (Section 5.8): A mixture consisting of equal parts (+) and (−) enantiomers of a chiral substance; also called a racemic mixture.
• Radical (Section 2.6, Section 6.2): A species that has an odd number of electrons, such as the chlorine radical, Cl·.
• Radical reactions (Section 6.2, Section 6.3): Reactions in which bonds are made by donation of one electron from each of two reactants and in which bonds are broken when each fragment leaves with one electron.
• Rate constant (Section 11.2): The constant k in a rate equation.
• Rate equation (Section 11.2): An equation that expresses the dependence of a reaction’s rate on the concentration of reactants.
• Rate-limiting step (Section 11.4): The slowest step in a multistep reaction sequence; also called the rate-determining step. The rate-limiting step acts as a kind of bottleneck in multistep reactions.
• Re face (Section 5.11): One of two faces of a planar, sp2-hybridized atom.
• Rearrangement reactions (Section 6.1): What occurs when a single reactant undergoes a reorganization of bonds and atoms to yield an isomeric product.
• Reducing sugars (Section 25.6): Sugars that reduce silver ion in the Tollens test or cupric ion in the Fehling or Benedict tests.
• Reduction (Section 8.6, Section 10.8): A reaction that causes an increase of electron ownership by carbon, either by bond-breaking between carbon and a more electronegative atom or by bond formation between carbon and a less electronegative atom.
• Reductive amination (Section 24.6, Section 26.3): A method for preparing an amine by reaction of an aldehyde or ketone with ammonia and a reducing agent.
• Refining (Chapter 3 Chemistry Matters): The process by which petroleum is converted into gasoline and other useful products.
• Regiochemistry (Section 7.8): A term describing the orientation of a reaction that occurs on an unsymmetrical substrate.
• Regiospecific (Section 7.8): A term describing a reaction that occurs with a specific regiochemistry to give a single product rather than a mixture of products.
• Replication (Section 28.3): The process by which double-stranded DNA uncoils and is replicated to produce two new copies.
• Replication forks (Section 28.3): The point of unraveling in a DNA chain where replication occurs.
• Residues (Section 26.4): Amino acids in a protein chain.
• Resolution (Section 5.8): The process by which a racemate is separated into its two pure enantiomers.
• Resonance effect (Section 16.4): The donation or withdrawal of electrons through orbital overlap with neighboring π bonds. For example, an oxygen or nitrogen substituent donates electrons to an aromatic ring by overlap of the O or N orbital with the aromatic ring p orbitals.
• Resonance forms (Section 2.4): Individual structural forms of a resonance hybrid.
• Resonance hybrid (Section 2.4): A molecule, such as benzene, that can’t be represented adequately by a single Kekulé structure but must instead be considered as an average of two or more resonance forms. The resonance forms themselves differ only in the positions of their electrons, not their nuclei.
• Restriction endonucleases (Section 28.6): Enzymes that are able to cleave a DNA molecule at points in the chain where a specific base sequence occurs.
• Retrosynthetic (Section 9.9, Section 16.10): Planning an organic synthesis by working backward from the final product to the starting material.
• Ribonucleic acid (RNA) (Chapter 28 Introduction): The biopolymer found in cells that serves to transcribe the genetic information found in DNA and uses that information to direct the synthesis of proteins.
• Ribosomal RNA (rRNA) (Section 28.4): A kind of RNA used in the physical makeup of ribosomes.
• Ring-current (Section 15.7): The circulation of π electrons induced in aromatic rings by an external magnetic field. This effect accounts for the downfield shift of aromatic ring protons in the 1H NMR spectrum.
• Ring-flip (Section 4.6): A molecular motion that interconverts two chair conformations of cyclohexane. The effect of a ring-flip is to convert an axial substituent into an equatorial substituent.
• Ring-opening metathesis polymerization (ROMP) (Section 31.5): A method of polymer synthesis that uses an olefin metathesis reaction of a cycloalkene.
• RNA (Section 28.1): Ribonucleic acid.
• Robinson annulation reaction (Section 23.12): A method for synthesis of cyclohexenones by sequential Michael reaction and intramolecular aldol reaction.
• S configuration (Section 5.5): The configuration at a chirality center as specified using the Cahn–Ingold–Prelog sequence rules.
• s-Cis conformation (Section 14.5): The conformation of a conjugated diene that is cis-like around the single bond.
• Saccharide (Section 25.1): A sugar.
• Salt bridge (Section 26.9): An ionic attraction between two oppositely charged groups in a protein chain.
• Sandmeyer reaction (Section 24.8): The nucleophilic substitution reaction of an arenediazonium salt with a cuprous halide to yield an aryl halide.
• Sanger dideoxy method (Section 28.6): A commonly used method of DNA sequencing.
• Saponification (Section 21.6, Section 27.2): An old term for the base-induced hydrolysis of an ester to yield a carboxylic acid salt.
• Saturated (Section 3.2): A molecule that has only single bonds and thus can’t undergo addition reactions. Alkanes are saturated, but alkenes are unsaturated.
• Sawhorse representations (Section 3.6): A manner of representing stereochemistry that uses a stick drawing and gives a perspective view of the conformation around a single bond.
• Schiff bases (Section 19.8, Section 29.5): An alternative name for an imine, $R2C═NR′R2C═NR′$, used primarily in biochemistry.
• Second-order reaction (Section 11.2): A reaction whose rate-limiting step is bimolecular and whose kinetics are therefore dependent on the concentration of two reactants.
• Secondary: See Primary.
• Secondary metabolite (Chapter 7 Chemistry Matters): A small naturally occurring molecule that is not essential to the growth and development of the producing organism and is not classified by structure.
• Secondary structure (Section 26.9): The level of protein substructure that involves organization of chain sections into ordered arrangements such as β-pleated sheets or α helices.
• Semiconservative replication (Section 28.3): The process by which DNA molecules are made containing one strand of old DNA and one strand of new DNA.
• Sense strand (Section 28.4): The coding strand of double-helical DNA that contains the gene.
• Sequence rules (Section 5.5, Section 7.5): A series of rules for assigning relative rankings to substituent groups on a double-bond carbon atom or on a chirality center.
• Sesquiterpenoids (Section 27.5): 15-carbon lipids.
• Sharpless epoxidation (Chapter 19 Chemistry Matters): A method for enantioselective synthesis of a chiral epoxide by treatment of an allylic alcohol with tert-butyl hydroperoxide, (CH3)3C–OOH, in the presence of titanium tetraisopropoxide and diethyl tartrate.
• Shielding (Section 13.2): An effect observed in NMR that causes a nucleus to absorb toward the right (upfield) side of the chart. Shielding is caused by donation of electron density to the nucleus.
• Si face (Section 5.11): One of two faces of a planar, sp2-hybridized atom.
• Sialic acid (Section 25.7): One of a group of more than 300 carbohydrates based on acetylneuramic acid.
• Side chain (Section 26.1): The substituent attached to the α carbon of an amino acid.
• Sigma (σ) bond (Section 1.5): A covalent bond formed by head-on overlap of atomic orbitals.
• Sigmatropic reaction (Section 30.8): A pericyclic reaction that involves the migration of a group from one end of a π electron system to the other.
• Silyl ether (Section 17.8): A substance with the structure R3Si–O–R. The silyl ether acts as a protecting group for alcohols.
• Simmons–Smith reaction (Section 8.9): The reaction of an alkene with CH2I2 and Zn−Cu to yield a cyclopropane.
• Simple sugars (Section 25.1): Carbohydrates that cannot be broken down into smaller sugars by hydrolysis.
• Single bond (Section 1.8): A covalent bond formed by sharing one electron pair between atoms.
• Skeletal structures (Section 1.12): A shorthand way of writing structures in which carbon atoms are assumed to be at each intersection of two lines (bonds) and at the end of each line.
• Small RNAs (Section 28.4): A type of RNA that has a variety of functions within the cell, including silencing transcription and catalyzing chemical modifications of other RNA molecules.
• SN1 reaction (Section 11.4): A unimolecular nucleophilic substitution reaction.
• SN2 reaction (Section 11.2): A bimolecular nucleophilic substitution reaction.
• Solid-phase synthesis (Section 26.8): A technique of synthesis whereby the starting material is covalently bound to a solid polymer bead and reactions are carried out on the bound substrate. After the desired transformations have been effected, the product is cleaved from the polymer.
• Solvation (Section 11.3): The clustering of solvent molecules around a solute particle to stabilize it.
• sp hybrid orbitals (Section 1.9): Hybrid orbitals derived from the combination of an s and a p atomic orbital. The two sp orbitals that result from hybridization are oriented at an angle of 180° to each other.
• sp2 hybrid orbitals (Section 1.8): Hybrid orbitals derived by combination of an s atomic orbital with two p atomic orbitals. The three sp2 hybrid orbitals that result lie in a plane at angles of 120° to each other.
• sp3 hybrid orbitals (Section 1.6): Hybrid orbitals derived by combination of an s atomic orbital with three p atomic orbitals. The four sp3 hybrid orbitals that result are directed toward the corners of a regular tetrahedron at angles of 109° to each other.
• Specific rotation, [α]D (Section 5.3): The optical rotation of a chiral compound under standard conditions.
• Sphingomyelins (Section 27.3): Phospholipids that have sphingosine as the backbone rather than glycerol.
• Spin–spin splitting (Section 13.6): The splitting of an NMR signal into a multiplet because of an interaction between nearby magnetic nuclei whose spins are coupled. The magnitude of spin–spin splitting is given by the coupling constant, J.
• Staggered conformation (Section 3.6): The three-dimensional arrangement of atoms around a carbon–carbon single bond in which the bonds on one carbon bisect the bond angles on the second carbon as viewed end-on.
• Statin (Chapter 29 Chemistry Matters): A drug that controls cholesterol biosynthesis in the body by blocking the HMG-CoA reductase enzyme.
• Step-growth polymers (Section 21.9, Section 31.4): Polymers in which each bond is formed independently of the others. Polyesters and polyamides (nylons) are examples.
• Stereocenter (Section 5.2): An alternative name for a chirality center.
• Stereochemistry (Section 3.5; Chapters 3, 4, 5): The branch of chemistry concerned with the three-dimensional arrangement of atoms in molecules.
• Stereogenic center (Section 5.2): An alternative name for a chirality center.
• Stereoisomers (Section 4.2): Isomers that have their atoms connected in the same order but have different three-dimensional arrangements. The term stereoisomer includes both enantiomers and diastereomers.
• Stereospecific (Section 8.9, Section 14.5): A term indicating that only a single stereoisomer is produced in a given reaction rather than a mixture.
• Steric strain (Section 3.7, Section 4.3, Section 4.7): The strain imposed on a molecule when two groups are too close together and try to occupy the same space. Steric strain is responsible both for the greater stability of trans versus cis alkenes and for the greater stability of equatorially substituted versus axially substituted cyclohexanes.
• Steroids (Section 27.6): Lipids whose structure is based on a tetracyclic carbon skeleton with three 6-membered and one 5-membered ring. Steroids occur in both plants and animals and have a variety of important hormonal functions.
• Stork enamine reaction (Section 23.11): The conjugate addition of an enamine to an α,β-unsaturated carbonyl compound, followed by hydrolysis to yield a 1,5-dicarbonyl product.
• STR loci (Chapter 28 Chemistry Matters): Short tandem repeat sequences of noncoding DNA that are unique to every individual and allow DNA fingerprinting.
• Straight-chain alkanes (Section 3.2): Alkanes whose carbon atoms are connected without branching.
• Substitution reactions (Section 6.1): What occurs when two reactants exchange parts to give two new products. SN1 and SN2 reactions are examples.
• Sulfides (Section 3.1, Section 18.7): A class of compounds that has two organic substituents bonded to the same sulfur atom, RSR′.
• Sulfonation (Section 16.2): The substitution of a sulfonic acid group (−SO3H) onto an aromatic ring.
• Sulfone (Section 18.7): A compound of the general structure RSO2R′.
• Sulfonium ions (Section 18.7): A species containing a positively charged, trivalent sulfur atom, R3S+.
• Sulfoxide (Section 18.7): A compound of the general structure RSOR′.
• Suprafacial (Section 30.5): A word used to describe the geometry of pericyclic reactions. Suprafacial reactions take place on the same side of the two ends of a π electron system.
• Suzuki–Miyaura reaction (Section 10.7): The palladium-catalyzed coupling reaction of an aromatic or vinylic halide with an aromatic or vinylic boronic acid.
• Symmetry-allowed, symmetry-disallowed (Section 30.1): A symmetry-allowed reaction is a pericyclic process that has a favorable orbital symmetry for reaction through a concerted pathway. A symmetry-disallowed reaction is one that does not have favorable orbital symmetry for reaction through a concerted pathway.
• Symmetry plane (Section 5.2): A plane that bisects a molecule such that one half of the molecule is the mirror image of the other half. Molecules containing a plane of symmetry are achiral.
• Syn periplanar (Section 11.8): Describing a stereochemical relationship in which two bonds on adjacent carbons lie in the same plane and are eclipsed.
• Syn stereochemistry (Section 8.5): The opposite of anti. A syn addition reaction is one in which the two ends of the double bond react from the same side. A syn elimination is one in which the two groups leave from the same side of the molecule.
• Syndiotactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents alternates regularly on opposite sides of the backbone.
• Tautomers (Section 9.4, Section 22.1): Isomers that interconvert spontaneously, usually with the change in position of a hydrogen.
• Terpenoids (Chapter 8 Chemistry Matters, Section 27.5): Lipids that are formally derived by head-to-tail polymerization of isoprene units.
• Tertiary: See Primary.
• Tertiary structure (Section 26.9): The level of protein structure that involves the manner in which the entire protein chain is folded into a specific three-dimensional arrangement.
• Thermodynamic control (Section 14.3): An equilibrium reaction that yields the lowest-energy, most stable product is said to be thermodynamically controlled.
• Thermoplastics (Section 31.7): Polymers that have a high Tg and are hard at room temperature but become soft and viscous when heated.
• Thermosetting resins (Section 31.7): Polymers that become highly cross-linked and solidify into a hard, insoluble mass when heated.
• Thioesters (Chapter 21 Introduction): A class of compounds with the RCOSR′ functional group.
• Thiols (Section 3.1, Chapter 18 Introduction): A class of compounds containing the −SH functional group.
• Thiolate ion (Section 18.7): The anion of a thiol, RS.
• TMS (Section 13.3): Tetramethylsilane; used as an NMR calibration standard.
• TOF (Section 12.4): Time-of-flight mass spectrometry; a sensitive method of mass detection accurate to about 3 ppm.
• Tollens’ reagent (Section 25.6): A solution of Ag2O in aqueous ammonia; used to oxidize aldehydes to carboxylic acids.
• Torsional strain (Section 3.6, Section 4.3): The strain in a molecule caused by electron repulsion between eclipsed bonds. Torsional strain is also called eclipsing strain.
• Tosylate (Section 11.1): A p-toluenesulfonate ester; useful as a leaving group in nucleophilic substitution reactions.
• Transamination (Section 29.9): The exchange of an amino group and a keto group between reactants.
• Transcription (Section 28.4): The process by which the genetic information encoded in DNA is read and used to synthesize RNA in the nucleus of the cell. A small portion of double-stranded DNA uncoils, and complementary ribonucleotides line up in the correct sequence for RNA synthesis.
• Transfer RNA (tRNA) (Section 28.4): A kind of RNA that transports amino acids to the ribosomes, where they are joined together to make proteins.
• Transimination (Section 29.9): The exchange of an amino group and an imine group between reactants.
• Transition state (Section 6.9): An activated complex between reactants, representing the highest energy point on a reaction curve. Transition states are unstable complexes that can’t be isolated.
• Translation (Section 28.5): The process by which the genetic information transcribed from DNA onto mRNA is read by tRNA and used to direct protein synthesis.
• Tree diagram (Section 13.8): A diagram used in NMR to sort out the complicated splitting patterns that can arise from multiple couplings.
• Triacylglycerols (Section 27.1): Lipids, such as those found in animal fat and vegetable oil, that are a triester of glycerol with long-chain fatty acids.
• Tricarboxylic acid cycle (Section 29.7): An alternative name for the citric acid cycle by which acetyl CoA is degraded to CO2.
• Triple bonds (Section 1.8): A type of covalent bond formed by sharing three electron pairs between atoms.
• Triplet (Section 13.6): A symmetrical three-line splitting pattern observed in the 1H NMR spectrum when a proton has two equivalent neighbor protons.
• Turnover number (Section 26.10): The number of substrate molecules acted on by an enzyme molecule per unit time.
• Twist-boat conformation (Section 4.5): A conformation of cyclohexane that is somewhat more stable than a pure boat conformation.
• Ultraviolet (UV) spectroscopy (Section 14.7): An optical spectroscopy employing ultraviolet irradiation. UV spectroscopy provides structural information about the extent of π electron conjugation in organic molecules.
• Unimolecular reaction (Section 11.4): A reaction that occurs by spontaneous transformation of the starting material without the intervention of other reactants. For example, the dissociation of a tertiary alkyl halide in the SN1 reaction is a unimolecular process.
• Unsaturated (Section 7.2): A molecule that has one or more multiple bonds.
• Upfield (Section 13.3): The right-hand portion of the NMR chart.
• Urethane (Section 31.4): A functional group in which a carbonyl group is bonded to both an −OR and an −NR2.
• Uronic acid (Section 25.6): A monocarboxylic acid formed by oxidizing the −CH2OH end of an aldose without affecting the −CHO end.
• Valence bond theory (Section 1.5): A bonding theory that describes a covalent bond as resulting from the overlap of two atomic orbitals.
• Valence shell (Section 1.4): The outermost electron shell of an atom.
• van der Waals forces (Section 2.12): Intermolecular forces that are responsible for holding molecules together in the liquid and solid states.
• Vegetable oils (Section 27.1): Liquid triacylglycerols derived from a plant source.
• Vicinal (Section 9.2): A term used to refer to a 1,2-disubstitution pattern. For example, 1,2-dibromoethane is a vicinal dibromide.
• Vinyl group (Section 7.3): A $H2C═CH–H2C═CH–$ substituent.
• Vinyl monomer (Section 8.10, Section 31.1): A substituted alkene monomer used to make a chain-growth polymer.
• Vinylic (Section 9.3): A term that refers to a substituent at a double-bond carbon atom. For example, chloroethylene is a vinylic chloride, and enols are vinylic alcohols.
• Vitamin (Section 26.10): A small organic molecule that must be obtained in the diet and is required in trace amounts for proper growth and function.
• Vulcanization (Section 14.6): A technique for cross-linking and hardening a diene polymer by heating with a few percent by weight of sulfur.
• Walden inversion (Section 11.1): The inversion of configuration at a chirality center that accompanies an SN2 reaction.
• Wave equation (Section 1.2): A mathematical expression that defines the behavior of an electron in an atom.
• Wave function (Section 1.2): A solution to the wave equation for defining the behavior of an electron in an atom. The square of the wave function defines the shape of an orbital.
• Wavelength, λ (Section 12.5): The length of a wave from peak to peak. The wavelength of electromagnetic radiation is inversely proportional to frequency and inversely proportional to energy.
• Wavenumber, $ν˜Section 12.6): The reciprocal of the wavelength in centimeters.$
• Waxes (Section 27.1): A mixture of esters of long-chain carboxylic acids with long-chain alcohols.
• Williamson ether synthesis (Section 18.2): A method for synthesizing ethers by SN2 reaction of an alkyl halide with an alkoxide ion.
• Wittig reaction (Section 19.11): The reaction of a phosphorus ylide with a ketone or aldehyde to yield an alkene.
• Wohl degradation (Section 25.6): A method for shortening the chain of an aldose sugar by one carbon.
• Wolff–Kishner reaction (Section 19.9): The conversion of an aldehyde or ketone into an alkane by reaction with hydrazine and base.
• X-ray crystallography (Chapter 12 Chemistry Matters): A technique that uses X rays to determine the structure of molecules.
• Ylide (Section 19.11): A neutral species with adjacent + and − charges, such as the phosphoranes used in Wittig reactions.
• Z geometry (Section 7.5): A term used to describe the stereochemistry of a carbon–carbon double bond. The two groups on each carbon are ranked according to the Cahn–Ingold–Prelog sequence rules, and the two carbons are compared. If the higher ranked groups on each carbon are on the same side of the double bond, the bond has Z geometry.
• Zaitsev’s rule (Section 11.7): A rule stating that E2 elimination reactions normally yield the more highly substituted alkene as major product.
• Ziegler–Natta catalysts (Section 31.2): Catalysts of an alkylaluminum and a titanium compound used for preparing alkene polymers.
• Zwitterion (Section 26.1): A neutral dipolar molecule in which the positive and negative charges are not adjacent. For example, amino acids exist as zwitterions, $H3CN+–CHR–CO2−H3CN+–CHR–CO2−$ | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.03%3A_Appendix_C_-_Glossary.txt |
D • Periodic Table D • Periodic Table
32.5.01: Chapter 1
Problem 1-1
(a) 1s2 2s2 2p4
(b) 1s2 2s2 2p3
(c) 1s2 2s2 2p6 3s6 3p4
Problem 1-2
(a) 2
(b) 2
(c) 6
Problem 1-5
(a) CCl4
(b) AlH3
(c) CH2Cl2
(d) SiF4
(e) CH3NH2
Problem 1-6
(a)
(b)
(c)
(d)
Problem 1-7 C2H7 has too many hydrogens for a compound with two carbons.
Problem 1-10
The CH3 carbon is sp3; the double-bond carbons are sp2; the C$\text{=}$C−C and C$\text{=}$C−H bond angles are approximately 120°; other bond angles are near 109°.
Problem 1-11
All carbons are sp2, and all bond angles are near 120°.
Problem 1-12
All carbons except CH3 are sp2.
Problem 1-13
The CH3 carbon is sp3; the triple-bond carbons are sp; the C$\text{≡}$C−C and H−C$\text{≡}$C bond angles are approximately 180°.
Problem 1-14
(a) O has 2 lone pairs and is sp3-hybridized.
(b) N has 1 lone pair and is sp3-hybridized.
(c) P has 1 lone pair and is sp3-hybridized.
(d) S has 2 lone pairs and is sp3-hybridized.
Problem 1-15
(a)
(b)
Problem 1-16
(a)
There are numerous possibilities, such as:
(b)
There are numerous possibilities, such as:
(c)
There are numerous possibilities, such as:
(d)
There are numerous possibilities, such as:
32.5.02: Chapter 2
Problem 2-1
(a) H
(b) Br
(c) Cl
(d) C
Problem 2-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 2-3 H3C−OH < H3C−MgBr < H3C−Li = H3C−F < H3C−K
Problem 2-4
The nitrogen is electron-rich, and the carbon is electron-poor.
Problem 2-5
The two C–O dipoles cancel because of the symmetry of the molecule:
Problem 2-6
(a)
(b)
(c)
(d)
Problem 2-7
(a) For carbon: FC = 4 − 8/2 − 0 = 0 For the middle nitrogen: FC = 5 − 8/2 − 0 = +1 For the end nitrogen: FC = 5 − 4/2 − 4 = −1
(b) For nitrogen: FC = 5 − 8/2 − 0 = +1 For oxygen: FC = 6 − 2/2 − 6 = −1
(c) For nitrogen: FC = 5 − 8/2 − 0 = +1 For the triply bonded carbon: FC = 4 − 6/2 − 2 = −1
Problem 2-9 The structures in (a) are resonance forms.
Problem 2-10
(a)
(b)
(c)
(d)
Problem 2-12 Phenylalanine is stronger.
Problem 2-13 Water is a stronger acid.
Problem 2-14 Neither reaction will take place.
Problem 2-15 Reaction will take place.
Problem 2-16 Ka = 4.9 × 10–10
Problem 2-17
(a)
(b)
Problem 2-18
(a)
(b)
Problem 2-19 Vitamin C is water-soluble (hydrophilic); vitamin A is fat-soluble (hydrophilic).
32.5.03: Chapter 3
Problem 3-1
(a) Sulfide, carboxylic acid, amine
(b) Aromatic ring, carboxylic acid
(c) Ether, alcohol, aromatic ring, amide, C$\text{=}$C bond
Problem 3-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 3-5
(a)
(b)
(c)
Problem 3-6
(a) Two
(b) Four
(c) Four
Problem 3-8
(a)
(b)
(c)
Problem 3-9 Primary carbons have primary hydrogens, secondary carbons have secondary hydrogens, and tertiary carbons have tertiary hydrogens.
Problem 3-10
(a)
(b)
(c)
Problem 3-11
(a) Pentane, 2-methylbutane, 2,2-dimethylpropane
(b) 2,3-Dimethylpentane
(c) 2,4-Dimethylpentane
(d) 2,2,5-Trimethylhexane
Problem 3-12
(a)
(b)
(c)
(d)
Problem 3-13 Pentyl, 1-methylbutyl, 1-ethylpropyl, 2-methylbutyl, 3-methylbutyl, 1,1-dimethylpropyl, 1,2-dimethylpropyl, 2,2-dimethylpropyl
Problem 3-16
(a)
(b)
(c)
(d) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.04%3A_Appendix_D_-_Periodic_Table.txt |
Problem 4-1
(a) 1,4-Dimethylcyclohexane
(b) 1-Methyl-3-propylcyclopentane
(c) 3-Cyclobutylpentane
(d) 1-Bromo-4-ethylcyclodecane
(e) 1-Isopropyl-2-methylcyclohexane
(f) 4-Bromo-1-tert-butyl-2-methylcycloheptane
Problem 4-2
(a)
(b)
(c)
(d)
Problem 4-3 3-Ethyl-1,1-dimethylcyclopentane
Problem 4-4
(a) trans-1-Chloro-4-methylcyclohexane
(b) cis-1-Ethyl-3-methylcycloheptane
Problem 4-5
(a)
(b)
(c)
Problem 4-6 The two hydroxyl groups are cis. The two side chains are trans.
Problem 4-7
(a) cis-1,2-Dimethylcyclopentane
(b) cis-1-Bromo-3-methylcyclobutane
Problem 4-8 Six interactions; 21% of strain
Problem 4-9 The cis isomer is less stable because the methyl groups nearly eclipse each other.
Problem 4-10 Ten eclipsing interactions; 40 kJ/mol; 35% is relieved.
Problem 4-11 Conformation (a) is more stable because the methyl groups are farther apart.
Problem 4-14 Before the ring-flip, red and blue are equatorial and green is axial. After the ring-flip, red and blue are axial and green is equatorial.
Problem 4-15 4.2 kJ/mol
Problem 4-16 Cyano group points straight up.
Problem 4-17 Equatorial = 70%; axial = 30%
Problem 4-18
(a) 2.0 kJ/mol (axial Cl)
(b) 11.4 kJ/mol (axial CH3)
(c) 2.0 kJ/mol (axial Br)
(d) 8.0 kJ/mol (axial CH2CH3)
Problem 4-20 trans-Decalin is more stable because it has no 1,3-diaxial interactions.
Problem 4-21 Both ring-fusions are trans.
32.5.05: Chapter 5
Problem 5-1 Chiral: screw, shoe
Problem 5-2
(a)
(b)
(c)
Problem 5-4
(a)
(b)
Problem 5-5 Levorotatory
Problem 5-6 +16.1°
Problem 5-7
(a) −Br
(b) −Br
(c) −CH2CH3
(d) −OH
(e) −CH2OH
(f) −CH$\text{=}$O
Problem 5-8
(a) −OH, −CH2CH2OH, −CH2CH3, −H
(b) −OH, −CO2CH3, −CO2H, −CH2OH
(c) −NH2, −CN, −CH2NHCH3, −CH2NH2
(d) −SSCH3, −SH, −CH2SCH3, −CH3
Problem 5-9
(a) S
(b) R
(c) S
Problem 5-10
(a) S
(b) S
(c) R
Problem 5-13 Compound (a) is D-erythrose 4-phosphate, (d) is its enantiomer, and (b) and (c) are diastereomers.
Problem 5-14 Five chirality centers and 25 = 32 stereoisomers
Problem 5-15 S,S
Problem 5-16 Compounds (a) and (d) are meso.
Problem 5-17 Compounds (a) and (c) have meso forms.
Problem 5-20 Two diastereomeric salts: (R)-lactic acid plus (S)-1-phenylethylamine and (S)-lactic acid plus (S)-1-phenylethylamine
Problem 5-21
(a) Constitutional isomers
(b) Diastereomers
Problem 5-22
(a)
(b)
Problem 5-23
(a)
(b)
Problem 5-24 (S)-Lactate
Problem 5-25 The −OH adds to the Re face of C2, and −H adds to the Re face of C3. The overall addition has anti stereochemistry.
32.5.06: Chapter 6
Problem 6-1
(a) Substitution
(b) Elimination
(c) Addition
Problem 6-2
(a) Carbon is electrophilic.
(b) Sulfur is nucleophilic.
(c) Nitrogens are nucleophilic.
(d) Oxygen is nucleophilic; carbon is electrophilic.
Problem 6-4 Bromocyclohexane; chlorocyclohexane
Problem 6-5
The mechanism is shown in Figure 6.4
Problem 6-6
(a)
(b)
(c)
Problem 6-8 1-Chloro-2-methylpentane, 2-chloro-2-methylpentane, 3-chloro-2-methylpentane, 2-chloro-4-methylpentane, 1-chloro- 4-methylpentane
Problem 6-10 Negative ΔG° is favored.
Problem 6-11 Larger Keq is more exergonic.
Problem 6-12 Lower ΔG is faster.
32.5.07: Chapter 7
Problem 7-1
(a) 1
(b) 2
(c) 2
Problem 7-2
(a) 5
(b) 5
(c) 3
(d) 1
(e) 6
(f) 5
Problem 7-3 C16H13ClN2O
Problem 7-4
(a) 3,4,4-Trimethyl-1-pentene
(b) 3-Methyl-3-hexene
(c) 4,7-Dimethyl-2,5-octadiene
(d) 6-Ethyl-7-methyl-4-nonene
Problem 7-5
(a)
(b)
(c)
(d)
Problem 7-6
(a) 1,2-Dimethylcyclohexene
(b) 4,4-Dimethylcycloheptene
(c) 3-Isopropylcyclopentene
Problem 7-7
(a) 2,5,5-Trimethylhex-2-ene
(b) 2,3-Dimethylcyclohexa-1,3-diene
Problem 7-9 Compounds (c), (e), and (f) have cis–trans isomers.
Problem 7-10
(a) cis-4,5-Dimethyl-2-hexene
(b) trans-6-Methyl-3-heptene
Problem 7-11
(a) −CH3
(b) −Cl
(c) −CH$\text{=}$CH2
(d) −OCH3
(e) −CH$\text{=}$O
(f) −CH$\text{=}$O
Problem 7-12
(a) −Cl, −OH, −CH3, −H
(b) −CH2OH, −CH$\text{=}$CH2, −CH2CH3, −CH3
(c) −CO2H, −CH2OH, −C$\text{≡}$N, −CH2NH2
(d) −CH2OCH3, −C$\text{≡}$N, −C$\text{≡}$CH, −CH2CH3
Problem 7-13
(a) Z
(b) E
(c) Z
(d) E
Problem 7-15
(a) 2-Methylpropene is more stable than 1-butene.
(b) trans-2-Hexene is more stable than cis-2-hexene.
(c) 1-Methylcyclohexene is more stable than 3-methylcyclohexene.
Problem 7-16
(a) Chlorocyclohexane
(b) 2-Bromo-2-methylpentane
(c) 4-Methyl-2-pentanol
(d) 1-Bromo-1-methylcyclohexane
Problem 7-17
(a) Cyclopentene
(b) 1-Ethylcyclohexene or ethylidenecyclohexane
(c) 3-Hexene
(d) Vinylcyclohexane (cyclohexylethylene)
Problem 7-18
(a)
(b)
Problem 7-19 In the conformation shown, only the methyl- group C−H that is parallel to the carbocation p orbital can show hyperconjugation.
Problem 7-20 The second step is exergonic; the transition state resembles the carbocation. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.04%3A_Chapter_4.txt |
Problem 8-1 2-Methyl-2-butene and 2-methyl-1-butene
Problem 8-2 Five
Problem 8-3 trans-1,2-Dichloro-1,2-dimethylcyclohexane
Problem 8-5 trans-2-Bromocyclopentanol
Problem 8-6 Markovnikov
Problem 8-7
(a) 2-Pentanol
(b) 2-Methyl-2-pentanol
Problem 8-8
(a) Oxymercuration of 2-methyl-1-hexene or 2-methyl-2-hexene
(b) Oxymercuration of cyclohexylethylene or hydroboration of ethylidenecyclohexane
Problem 8-9
(a)
(b)
Problem 8-10
(a) 3-Methyl-1-butene
(b) 2-Methyl-2-butene
(c) Methylenecyclohexane
Problem 8-12
(a) 2-Methylpentane
(b) 1,1-Dimethylcyclopentane
(c) tert-Butylcyclohexane
Problem 8-14
(a) 1-Methylcyclohexene
(b) 2-Methyl-2-pentene
(c) 1,3-Butadiene
Problem 8-15
(a) CH3COCH2CH2CH2CH2CO2H
(b) CH3COCH2CH2CH2CH2CHO
Problem 8-16
(a) 2-Methylpropene
(b) 3-Hexene
Problem 8-17
(a)
(b)
Problem 8-18
(a) H2C ═ CHOCH3
(b) ClCH ═ CHCl
Problem 8-20
An optically inactive, non-50 : 50 mixture of two racemic pairs: (2R,4R) + (2S,4S) and (2R,4S) + (2S,4R)
Problem 8-21 Non-50 : 50 mixture of two racemic pairs: (1S,3R) + (1R,3S) and (1S,3S) + (1R,3R)
32.5.09: Chapter 9
Problem 9-1
(a) 2,5-Dimethyl-3-hexyne
(b) 3,3-Dimethyl-1-butyne
(c) 3,3-Dimethyl-4-octyne
(d) 2,5,5-Trimethyl-3-heptyne
(e) 2,4-Octadiene-6-yne
Problem 9-2 1-Hexyne, 2-hexyne, 3-hexyne, 3-methyl-1-pentyne, 4-methyl-1-pentyne, 4-methyl-2-pentyne, 3,3-dimethyl-1-butyne
Problem 9-3
(a) 1,1,2,2-Tetrachloropentane
(b) 1-Bromo-1-cyclopentylethylene
(c) 2-Bromo-2-heptene and 3-bromo-2-heptene
Problem 9-4
(a) 4-Octanone
(b) 2-Methyl-4-octanone and 7-methyl-4-octanone
Problem 9-5
(a) 1-Pentyne
(b) 2-Pentyne
Problem 9-6
(a) C6H5C $\text{≡}$ CH
(b) 2,5-Dimethyl-3-hexyne
Problem 9-7
(a) Mercuric sulfate–catalyzed hydration of phenylacetylene
(b) Hydroboration/oxidation of cyclopentylacetylene
Problem 9-8
(a) Reduce 2-octyne with Li/NH3.
(b) Reduce 3-heptyne with H2/Lindlar catalyst.
(c) Reduce 3-methyl-1-pentyne.
Problem 9-9 No: (a), (c), (d); yes: (b)
Problem 9-10
(a) 1-Pentyne + CH3I, or propyne + CH3CH2CH2I
(b) 3-Methyl-1-butyne + CH3CH2I
(c) Cyclohexylacetylene + CH3I
Problem 9-12
(a) KMnO4, H3O+
(b) H2/Lindlar
(c) 1. H2/Lindlar; 2. HBr
(d) 1. H2/Lindlar; 2. BH3; 3. NaOH, H2O2
(e) 1. H2/Lindlar; 2. Cl2
(f) O3
Problem 9-13
(a) 1. HC$\text{≡}$CH + NaNH2 ; 2. CH3(CH2)6CH2Br; 3. 2 H2/Pd
(b) HC$\text{≡}$CH + NaNH2 ; 2. (CH3)3CCH2CH2I; 3. 2 H2/Pd
(c) 1. HC$\text{≡}$CH + NaNH2 ; 2. CH3CH2CH2CH2I; 3. BH3; 4. H2O2
(d) 1. HC$\text{≡}$CH + NaNH2 ; 2. CH3CH2CH2CH2CH2I; 3. HgSO4, H3O+
32.5.10: Chapter 10
Problem 10-1
(a) 1-Iodobutane
(b) 1-Chloro-3-methylbutane
(c) 1,5-Dibromo-2,2-dimethylpentane
(d) 1,3-Dichloro-3-methylbutane
(e) 1-Chloro-3-ethyl-4-iodopentane
(f) 2-Bromo-5-chlorohexane
Problem 10-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 10-3 Chiral: 1-chloro-2-methylpentane, 3-chloro-2-methylpentane, 2-chloro-4-methylpentane Achiral: 2-chloro-2-methylpentane, 1-chloro-4-methylpentane
Problem 10-4 1-Chloro-2-methylbutane (29%), 1-chloro-3-methylbutane (14%), 2-chloro-2-methylbutane (24%), 2-chloro-3-methylbutane (33%)
Problem 10-6 The intermediate allylic radical reacts at the more accessible site and gives the more highly substituted double bond.
Problem 10-7
(a) 3-Bromo-5-methylcycloheptene and 3-bromo-6-methylcycloheptene
(b) Four products
Problem 10-8
(a) 2-Methyl-2-propanol + HCl
(b) 4-Methyl-2-pentanol + PBr3
(c) 5-Methyl-1-pentanol + PBr3
(d) 3,3-Dimethyl-cyclopentanol + HF, pyridine
Problem 10-9 Both reactions occur.
Problem 10-10 React Grignard reagent with D2O.
Problem 10-11
(a) 1. NBS; 2. (CH3)2CuLi
(b) 1. Li; 2. CuI; 3. CH3CH2CH2CH2Br
(c) 1. BH3; 2. H2O2, NaOH; 3. PBr3; 4. Li, then CuI; 5. CH3(CH2)4Br
Problem 10-12
(a)
(b)
Problem 10-13
(a) Reduction
(b) Neither
32.5.11: Chapter 11
Problem 11-1 (R)-1-Methylpentyl acetate, CH3CO2CH(CH3)CH2CH2CH2CH3
Problem 11-2 (S)-2-Butanol
Problem 11-4
(a) 1-Iodobutane
(b) 1-Butanol
(c) 1-Hexyne
(d) Butylammonium bromide
Problem 11-5
(a) (CH3)2N
(b) (CH3)3N
(c) H2S
Problem 11-6 CH3OTos > CH3Br > (CH3)2CHCl > (CH3)3CCl
Problem 11-7 Similar to protic solvents
Problem 11-8 Racemic 1-ethyl-1-methylhexyl acetate
Problem 11-9 90.1% racemization, 9.9% inversion
Problem 11-11 H2C ═ CHCH(Br)CH3 > CH3CH(Br)CH3 > CH3CH2Br > H2C ═ CHBr
Problem 11-12 The same allylic carbocation intermediate is formed.
Problem 11-13
(a) SN1
(b) SN2
Problem 11-15
(a) Major: 2-methyl-2-pentene; minor: 4-methyl-2-pentene
(b) Major: 2,3,5-trimethyl-2-hexene; minor: 2,3,5-trimethyl-3-hexene and 2-isopropyl-4-methyl-1-pentene
(c) Major: ethylidenecyclohexane; minor: cyclohexylethylene
Problem 11-16
(a) 1-Bromo-3,6-dimethylheptane
(b) 4-Bromo-1,2-dimethylcyclopentane
Problem 11-17 (Z)-1-Bromo-1,2-diphenylethylene
Problem 11-18 (Z)-3-Methyl-2-pentene
Problem 11-19 Cis isomer reacts faster because the bromine is axial.
Problem 11-20
(a) SN2
(b) E2
(c) SN1
(d) E1cB
32.5.12: Chapter 12
Problem 12-1 C19H28O2
Problem 12-2
(a) 2-Methyl-2-pentene
(b) 2-Hexene
Problem 12-3
(a) 43, 71
(b) 82
(c) 58
(d) 86
Problem 12-4 102 (M+), 84 (dehydration), 87 (alpha cleavage), 59 (alpha cleavage)
Problem 12-5 X-ray energy is higher; λ = 9.0 × 10−6 m is higher in energy.
Problem 12-6
(a) 2.4 × 106 kJ/mol]
(b) 4.0 × 104 kJ/mol
(c) 2.4 × 103 kJ/mol
(d) 2.8 × 102 kJ/mol
(e) 6.0 kJ/mol
(f) 4.0 × 10−2 kJ/mol
Problem 12-7
(a) Ketone or aldehyde
(b) Nitro compound
(c) Carboxylic acid
Problem 12-8
(a) CH3CH2OH has an −OH absorption.
(b) 1-Hexene has a double-bond absorption.
(c) CH3CH2CO2H has a very broad −OH absorption.
Problem 12-9 1450–1600 cm−1: aromatic ring; 2100 cm−1: C$\text{≡}$C; 3300 cm−1: C$\text{≡}$C−H
Problem 12-10
(a) 1715, 1640, 1250 cm−1
(b) 1730, 2100, 3300 cm−1
(c) 1720, 2500–3100, 3400–3650 cm−1
Problem 12-11 1690, 1650, 2230 cm−1 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.08%3A_Chapter_8.txt |
Problem 13-1 7.5 × 10−5 kJ/mol for 19F; 8.0 × 10−5 kJ/mol for 1H
Problem 13-2 1.2 × 10−4 kJ/mol
Problem 13-3
The vinylic C − H protons are nonequivalent.
Problem 13-4
(a) 7.27 δ
(b) 3.05 δ
(c) 3.46 δ
(d) 5.30 δ
Problem 13-5
(a) 420 Hz
(b) 2.1 δ
(c) 1050 Hz
Problem 13-6
(a) 1.43 δ
(b) 2.17 δ
(c) 7.37 δ
(d) 5.30 δ
(e) 9.70 δ
(f) 2.12 δ
Problem 13-7
There are seven kinds of protons labeled. The types and expected range of absorption of each follow. a: ether, 3.5–4.5 δ; b: aryl, 6.5–8.0 δ; c: aryl, 6.5–8.0; d: vinylic, 4.5–6.5 δ; e: vinylic, 4.5–6.5 δ; f: alkyl (secondary), 1.2–1.6 δ; g: alkyl (primary), 0.7–1.3 δ.
Problem 13-8 Two peaks; 3 : 2 ratio
Problem 13-9
(a) −CHBr2, quartet; −CH3, doublet
(b) CH3O−, singlet; −OCH2 −, triplet; −CH2Br, triplet
(c) ClCH2− , triplet; −CH2−, quintet
(d) CH3− , triplet; −CH2− , quartet; −CH− , septet; (CH3)2, doublet
(e) CH3−, triplet; −CH2−, quartet; −CH−, septet; (CH3)2, doublet
(f) $\text{=}$CH, triplet, −CH2−, doublet, aromatic C−H, two multiplets
Problem 13-10
(a) CH3OCH3
(b) CH3CH(Cl)CH3
(c) ClCH2CH2OCH2CH2Cl
(d) CH3CH2CO2CH3 or CH3CO2CH2CH3
Problem 13-11 CH3CH2OCH2CH3
Problem 13-12
(a) Enantiotopic
(b) Diastereotopic
(c) Diastereotopic
(d) Diastereotopic
(e) Diastereotopic
(f) Homotopic
Problem 13-13
(a) 2
(b) 4
(c) 3
(d) 4
(e) 5
(f) 3
Problem 13-15
J1–2 = 16 Hz; J2–3 = 8 Hz
Problem 13-16 1-Chloro-1-methylcyclohexane has a singlet methyl absorption.
Problem 13-17
(a) 4
(b) 7
(c) 4
(d) 5
(e) 5
(f) 7
Problem 13-18
(a) 1,3-Dimethylcyclopentene
(b) 2-Methylpentane
(c) 1-Chloro-2-methylpropane
Problem 13-19 −CH3, 9.3 δ; −CH2− , 27.6 δ; C=O, 174.6 δ;− OCH3, 51.4 δ
Problem 13-23 A DEPT-90 spectrum would show two absorptions for the non-Markovnikov product (RCH$\text{=}$CHBr) but no absorptions for the Markovnikov product (RBrC$\text{=}$CH2).
32.5.14: Chapter 14
Problem 14-1 Expected ΔH°hydrog for allene is −252 kJ/mol. Allene is less stable than a nonconjugated diene, which is less stable than a conjugated diene.
Problem 14-2 1-Chloro-2-pentene, 3-chloro-1-pentene, 4-chloro-2-pentene
Problem 14-3 4-Chloro-2-pentene predominates in both.
Problem 14-4 1,2 Addition: 6-bromo-1,6-dimethylcyclohexene 1,4 Addition: 3-bromo-1,2-dimethylcyclohexene
Problem 14-5 Interconversion occurs by SN1 dissociation to a common intermediate cation.
Problem 14-6 The double bond is more highly substituted.
Problem 14-8 Good dienophiles: (a), (d)
Problem 14-9 Compound (a) is s-cis. Compound (c) can rotate to s-cis.
Problem 14-13 300–600 kJ/mol; UV energy is greater than IR or NMR energy.
Problem 14-14 1.46 × 10−5 M
Problem 14-15 All except (a) have UV absorptions.
32.5.15: Chapter 15
Problem 15-1
(a) Meta
(b) Para
(c) Ortho
Problem 15-2
(a) m-Bromochlorobenzene
(b) (3-Methylbutyl)benzene
(c) p-Bromoaniline
(d) 2,5-Dichlorotoluene
(e) 1-Ethyl-2,4-dinitrobenzene
(f) 1,2,3,5-Tetramethylbenzene
Problem 15-3
(a)
(b)
(c)
(d)
Problem 15-4
Pyridine has an aromatic sextet of electrons.
Problem 15-5 Cyclodecapentaene is not flat because of steric interactions.
Problem 15-6 All C – C bonds are equivalent; one resonance line in both 1H and 13C NMR spectra.
Problem 15-7 The cyclooctatetraenyl dianion is aromatic (ten π electrons) and flat.
Problem 15-10
The thiazolium ring has six π electrons.
Problem 15-11
Yes, it's aromatic.
Problem 15-12 The three nitrogens in double bonds each contribute one; the remaining nitrogen contributes two. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.13%3A_Chapter_13.txt |
Problem 16-1 o-, m-, and p-Bromotoluene
Problem 16-3 o-xylene: 2; p-xylene: 1; m-xylene: 3
Problem 16-4 D+ does electrophilic substitutions on the ring.
Problem 16-5 No rearrangement: (a), (b), (e)
Problem 16-6 tert-Butylbenzene
Problem 16-7
(a) (CH3)2CHCOCl
(b) PhCOCl
Problem 16-8
(a) Phenol > Toluene > Benzene > Nitrobenzene
(b) Phenol > Benzene > Chlorobenzene > Benzoic acid
(c) Aniline > Benzene > Bromobenzene > Benzaldehyde
Problem 16-9
(a) o- and p-Bromonitrobenzene
(b) m-Bromonitrobenzene
(c) o- and p-Chlorophenol
(d) o- and p-Bromoaniline
Problem 16-10 Alkylbenzenes are more reactive than benzene itself, but acylbenzenes are less reactive.
Problem 16-11 Toluene is more reactive; the trifluoromethyl group is electron-withdrawing.
Problem 16-12 The nitrogen electrons are donated to the nearby carbonyl group by resonance and are less available to the ring.
Problem 16-13 The meta intermediate is most favored.
Problem 16-14
(a) Ortho and para to − OCH3
(b) Ortho and para to − NH2
(c) Ortho and para to − Cl
Problem 16-15
(a) Reaction occurs ortho and para to the − CH3 group.
(b) Reaction occurs ortho and para to the − OCH3 group.
Problem 16-16 The phenol is deprotonated by KOH to give an anion that carries out a nucleophilic acyl substitution reaction on the fluoronitrobenzene.
Problem 16-17 Only one benzyne intermediate can form from p-bromotoluene; two different benzyne intermediates can form from m-bromotoluene.
Problem 16-18
(a) m-Nitrobenzoic acid
(b) p-tert-Butylbenzoic acid
Problem 16-19 A benzyl radical is more stable than a primary alkyl radical by 52 kJ/mol and is similar in stability to an allyl radical.
Problem 16-20 1. CH3CH2Cl, AlCl3; 2. NBS; 3. KOH, ethanol
Problem 16-21 1. PhCOCl, AlCl3; 2. H2/Pd
Problem 16-22
(a) 1. HNO3, H2SO4; 2. Cl2, FeCl3
(b) 1. CH3COCl, AlCl3; 2. Cl2, FeCl3; 3. H2/Pd
(c) 1. CH3CH2COCl, AlCl3; 2. Cl2, FeCl3; 3. H2/Pd; 4. HNO3, H2SO4
(d) 1. CH3Cl, AlCl3; 2. Br2, FeBr3; 3. SO3, H2SO4
Problem 16-23
(a) Friedel–Crafts acylation does not occur on a deactivated ring.
(b) Rearrangement occurs during Friedel–Crafts alkylation with primary halides; chlorination occurs ortho to the alkyl group.
32.5.17: Chapter 17
Problem 17-1
(a) 5-Methyl-2,4-hexanediol
(b) 2-Methyl-4-phenyl-2-butanol
(c) 4,4-Dimethylcyclohexanol
(d) trans-2-Bromocyclopentanol
(e) 4-Bromo-3-methylphenol
(f) 2-Cyclopenten-1-ol
Problem 17-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 17-3 Hydrogen-bonding is more difficult in hindered alcohols.
Problem 17-4
(a) HC$\text{≡}$CH < (CH3)2CHOH < CH3OH < (CF3)2CHOH
(b) p-Methylphenol < Phenol < p-(Trifluoromethyl)phenol
(c) Benzyl alcohol < Phenol < p-Hydroxybenzoic acid
Problem 17-5 The electron-withdrawing nitro group stabilizes an alkoxide ion, but the electron-donating methoxyl group destabilizes the anion.
Problem 17-6
(a) 2-Methyl-3-pentanol
(b) 2-Methyl-4-phenyl-2-butanol
(c) meso-5,6-Decanediol
Problem 17-7
(a) NaBH4
(b) LiAlH4
(c) LiAlH4
Problem 17-8
(a) Benzaldehyde or benzoic acid (or ester)
(b) Acetophenone
(c) Cyclohexanone
(d) 2-Methylpropanal or 2-methylpropanoic acid (or ester)
Problem 17-9
(a) 1-Methylcyclopentanol
(b) 1,1-Diphenylethanol
(c) 3-Methyl-3-hexanol
Problem 17-10
(a)
(b)
(c)
(d)
(e)
(f)
Problem 17-11 Cyclohexanone + CH3CH2MgBr
Problem 17-12 1. p-TosCl, pyridine; 2. NaCN
Problem 17-13
(a) 2-Methyl-2-pentene
(b) 3-Methylcyclohexene
(c) 1-Methylcyclohexene
(d) 2,3-Dimethyl-2-pentene
(e) 2-Methyl-2-pentene
Problem 17-14
(a) 1-Phenylethanol
(b) 2-Methyl-1-propanol
(c) Cyclopentanol
Problem 17-15
(a) Hexanoic acid, hexanal
(b) 2-Hexanone
(c) Hexanoic acid, no reaction
Problem 17-16 SN2 reaction of F on silicon with displacement of alkoxide ion.
Problem 17-17 Protonation of 2-methylpropene gives the tert-butyl cation, which carries out an electrophilic aromatic substitution reaction.
Problem 17-18 Disappearance of –OH absorption; appearance of C$\text{=}$O
Problem 17-19
(a) Singlet
(b) Doublet
(c) Triplet
(d) Doublet
(e) Doublet
(f) Singlet
32.5.18: Chapter 18
Problem 18-1
(a) Diisopropyl ether
(b) Cyclopentyl propyl ether
(c) p-Bromoanisole or 4-bromo-1-methoxybenzene
(d) 1-Methoxycyclohexene
(e) Ethyl isobutyl ether
(f) Allyl vinyl ether
Problem 18-2 A mixture of diethyl ether, dipropyl ether, and ethyl propyl ether is formed in a 1 : 1 : 2 ratio.
Problem 18-3
(a) CH3CH2CH2O + CH3Br
(b) PhO + CH3Br
(c) (CH3)2CHO + PhCH2Br
(d) (CH3)3CCH2O + CH3CH2Br
Problem 18-5
(a) Either method
(b) Williamson
(c) Alkoxymercuration
(d) Williamson
Problem 18-6
(a) Bromoethane > 2-Bromopropane > Bromobenzene
(b) Bromoethane > Chloroethane > 1-Iodopropene
Problem 18-7
(a)
(b)
Problem 18-8 Protonation of the oxygen atom, followed by E1 reaction
Problem 18-9 Br and I are better nucleophiles than Cl.
Problem 18-11 Epoxidation of cis-2-butene yields cis-2,3-epoxybutane, while epoxidation of trans-2-butene yields trans-2,3-epoxybutane.
Problem 18-12
(a)
(b)
Problem 18-13
(a) 1-Methylcyclohexene + OsO4; then NaHSO3
(b) 1-Methylcyclohexene + m-chloroperoxybenzoic acid, then H3O+
Problem 18-14
(a)
(b)
(c)
Problem 18-16
(a) 2-Butanethiol
(b) 2,2,6-Trimethyl-4-heptanethiol
(c) 2-Cyclopentene-1-thiol
(d) Ethyl isopropyl sulfide
(e) o-Di(methylthio)benzene
(f) 3-(Ethylthio)cyclohexanone
Problem 18-17
• (a) 1. LiAlH4; 2. PBr3; 3. ${({\rm H}_{2}{\rm N})_{2}{\rm C}\text{═}{\rm S}}$; 4. H2O, NaOH
• (b) 1. HBr; 2. ${({\rm H}_{2}{\rm N})_{2}{\rm C}\text{═}{\rm S}}$; 3. H2O, NaOH
Problem 18-18 1,2-Epoxybutane
Problem 18-69 Acetyl chloride is more electrophilic than acetone.
Problem 18-71
(a) Nucleophilic acyl substitution
(b) Nucleophilic addition
(c) Carbonyl condensation | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.16%3A_Chapter_16.txt |
Problem 19-1
(a) 2-Methyl-3-pentanone
(b) 3-Phenylpropanal
(c) 2,6-Octanedione
(d) trans-2-Methylcyclohexanecarbaldehyde
(e) 4-Hexenal
(f) cis-2,5-Dimethylcyclohexanone
Problem 19-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 19-3
(a) Dess–Martin periodinane
(b) 1. O3; 2. Zn
(c) DIBAH
(d) 1. BH3, then H2O2, NaOH; 2. Dess–Martin periodinane
Problem 19-4
(a) HgSO4, H3O+
(b) 1. CH3COCl, AlCl3; 2. Br2, FeBr3
(c) 1. Mg; 2. CH3CHO; 3. H3O+; 4. CrO3
(d) 1. BH3; 2. H2O2, NaOH; 3. CrO3
Problem 19-6 The electron-withdrawing nitro group in p-nitrobenzaldehyde polarizes the carbonyl group.
Problem 19-7 CCl3CH(OH)2
Problem 19-8 Labeled water adds reversibly to the carbonyl group.
Problem 19-9 The equilibrium is unfavorable for sterically hindered ketones.
Problem 19-11 The steps are the exact reverse of the forward reaction shown in Figure 19.7.
Problem 19-13
(a) H2/Pd
(b) N2H4, KOH
(c) 1. H2/Pd; 2. N2H4, KOH
Problem 19-14 The mechanism is identical to that between a ketone and 2 equivalents of a monoalcohol, shown in Figure 19.11.
Problem 19-16
(a) Cyclohexanone + (Ph)3P$\text{=}$CHCH3
(b) Cyclohexanecarbaldehyde + (Ph)3P$\text{=}$CH2
(c) Acetone + (Ph)3P$\text{=}$CHCH2CH2CH3
(d) Acetone + (Ph)3P$\text{=}$CHPh
(e) PhCOCH3 + (Ph)3P$\text{=}$CHPh
(f) 2-Cyclohexenone + (Ph)3P$\text{=}$CH2
Problem 19-18 Intramolecular Cannizzaro reaction
Problem 19-19 Addition of the pro-R hydrogen of NADH takes place on the Re face of pyruvate.
Problem 19-20 The −OH group adds to the Re face at C2, and −H adds to the Re face at C3, to yield (2R,3S)-isocitrate.
Problem 19-22
(a) 3-Buten-2-one + (CH3CH2CH2)2CuLi
(b) 3-Methyl-2-cyclohexenone + (CH3)2CuLi
(c) 4-tert-Butyl-2-cyclohexenone + (CH3CH2)2CuLi
(d) Unsaturated ketone + (H2C$\text{=}$CH)2CuLi
Problem 19-23 Look for appearance of either an alcohol or a saturated ketone in the product.
Problem 19-24
(a) 1715 cm−1
(b) 1685 cm−1
(c) 1750 cm−1
(d) 1705 cm−1
(e) 1715 cm−1
(f) 1705 cm−1
Problem 19-25
(a) Different peaks due to McLafferty rearrangement
(b) Different peaks due to α cleavage and McLafferty rearrangement
(c) Different peaks due to McLafferty rearrangement
Problem 19-26 IR: 1750 cm−1; MS: 140, 84
32.5.20: Chapter 20
Problem 20-1
(a) 3-Methylbutanoic acid
(b) 4-Bromopentanoic acid
(c) 2-Ethylpentanoic acid
(d) cis-4-Hexenoic acid
(e) 2,4-Dimethylpentanenitrile
(f) cis-1,3-Cyclopentanedicarboxylic acid
Problem 20-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 20-3 Dissolve the mixture in ether, extract with aqueous NaOH, separate and acidify the aqueous layer, and extract with ether.
Problem 20-4 43%
Problem 20-5
(a) 82% dissociation
(b) 73% dissociation
Problem 20-6 Lactic acid is stronger because of the inductive effect of the −OH group.
Problem 20-7 The dianion is destabilized by repulsion between charges.
Problem 20-8 More reactive
Problem 20-9
(a) p-Methylbenzoic acid < Benzoic acid < p-Chlorobenzoic acid
(b) Acetic acid < Benzoic acid < p-Nitrobenzoic acid
Problem 20-10
(a) 1. Mg; 2. CO2; 3. H3O+
(b) 1. Mg; 2. CO2; 3. H3O+ or 1. NaCN; 2. H3O+
Problem 20-11 1. NaCN; 2. H3O+; 3. LiAlH4
Problem 20-12 1. PBr3; 2. NaCN; 3. H3O+; 4. LiAlH4
Problem 20-13
(a) Propanenitrile + CH3CH2MgBr, then H3O+
(b) p-Chlorobenzonitrile + CH3MgBr, then H3O+
Problem 20-14 1. NaCN; 2. CH3CH2MgBr, then H3O+
Problem 20-15 A carboxylic acid has a very broad − OH absorption at 2500–3300 cm−1.
Problem 20-16 4-Hydroxycyclohexanone: H–C–O absorption near 4 δ in the 1H spectrum and C═O absorption near 210 δ in the 13C spectrum. Cyclopentanecarboxylic acid: –CO2H absorption near 12 δ in the 1H spectrum and –CO2H absorption near 170 δ in the 13C spectrum.
32.5.21: Chapter 21
Problem 21-1
(a) 4-Methylpentanoyl chloride
(b) Cyclohexylacetamide
(c) Isopropyl 2-methylpropanoate
(d) Benzoic anhydride
(e) Isopropyl cyclopentanecarboxylate
(f) Cyclopentyl 2-methylpropanoate
(g) N-Methyl-4-pentenamide
(h) (R)-2-Hydroxypropanoyl phosphate
(i) Ethyl 2,3-dimethyl-2-butenethioate
Problem 21-2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Problem 21-4
(a) Acetyl chloride > Methyl acetate > Acetamide
(b) Hexafluoroisopropyl acetate > 2,2,2-Trichloroethyl acetate > Ethyl acetate
Problem 21-5
(a) CH3CO2 Na+
(b) CH3CONH2
(c) CH3CO2CH3 + CH3CO2 Na+
(d) CH3CONHCH3
Problem 21-7
(a) Acetic acid + 1-butanol
(b) Butanoic acid + methanol
(c) Cyclopentanecarboxylic acid + isopropyl alcohol
Problem 21-9
(a) Propanoyl chloride + methanol
(b) Acetyl chloride + ethanol
(c) Benzoyl chloride + ethanol
Problem 21-10 Benzoyl chloride + cyclohexanol
Problem 21-11 This is a typical nucleophilic acyl substitution reaction, with morpholine as the nucleophile and chloride as the leaving group.
Problem 21-12
(a) Propanoyl chloride + methylamine
(b) Benzoyl chloride + diethylamine
(c) Propanoyl chloride + ammonia
Problem 21-13
(a) Benzoyl chloride + [(CH3)2CH]2CuLi, or 2-methylpropanoyl chloride + Ph2CuLi
(b) 2-Propenoyl chloride + (CH3CH2CH2)2CuLi, or butanoyl chloride + (H2C$\text{=}$CH)2CuLi
Problem 21-14 This is a typical nucleophilic acyl substitution reaction, with p-hydroxyaniline as the nucleophile and acetate ion as the leaving group.
Problem 21-15 Monomethyl ester of benzene-1,2-dicarboxylic acid
Problem 21-16 Reaction of a carboxylic acid with an alkoxide ion gives the carboxylate ion.
Problem 21-17 LiAlH4 gives HOCH2CH2CH2CH2OH; DIBAH gives HOCH2CH2CH2CHO.
Problem 21-18
(a) CH3CH2CH2CH(CH3)CH2OH + CH3OH
(b) PhOH + PhCH2OH
Problem 21-19
(a) Ethyl benzoate + 2 CH3MgBr
(b) Ethyl acetate + 2 PhMgBr
(c) Ethyl pentanoate + 2 CH3CH2MgBr
Problem 21-20
(a) H2O, NaOH
(b) Benzoic acid + LiAlH4
(c) LiAlH4
Problem 21-21 1. Mg; 2. CO2, then H3O+; 3. SOCl2; 4. (CH3)2NH; 5. LiAlH4
Problem 21-23
(a)
(b)
(c)
Problem 21-25
(a) Ester
(b) Acid chloride
(c) Carboxylic acid
(d) Aliphatic ketone or cyclohexanone
Problem 21-26
(a) CH3CH2CH2CO2CH2CH3 and other possibilities
(b) CH3CON(CH3)2
(c) CH3CH$\text{=}$CHCOCl or H2C$\text{=}$C(CH3)COCl | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.19%3A_Chapter_19.txt |
Problem 22-1
(a)
(b)
(c)
(d)
(e)
(f)
Problem 22-2 (a) 4 (b) 3 (c) 3 (d) 2 (e) 4 (f) 5
Problem 22-4 Acid-catalyzed formation of an enol is followed by deuteronation of the enol double bond and dedeuteronation of oxygen.
Problem 22-5 1. Br2; 2. Pyridine, heat
Problem 22-6 The intermediate α-bromo acid bromide undergoes a nucleophilic acyl substitution reaction with methanol to give an α-bromo ester.
Problem 22-7
(a) CH3CH2CHO
(b) (CH3)3CCOCH3
(c) CH3CO2H
(d) PhCONH2
(e) CH3CH2CH2CN
(f) CH3CON(CH3)2
Problem 22-9 Acid is regenerated, but base is used stoichiometrically.
Problem 22-10
(a) 1. Na+ −OEt; 2. PhCH2Br; 3. H3O+
(b) 1. Na+ −OEt; 2. CH3CH2CH2Br; 3. Na+ −OEt; 4. CH3Br; 5. H3O+
(c) 1. Na+ −OEt; 2. (CH3)2CHCH2Br; 3. H3O+
Problem 22-11 Malonic ester has only two acidic hydrogens to be replaced.
Problem 22-12 1. Na+ −OEt; 2. (CH3)2CHCH2Br; 3. Na+ −OEt; 4. CH3Br; 5. H3O+
Problem 22-13
(a) (CH3)2CHCH2Br
(b) PhCH2CH2Br
Problem 22-14 None can be prepared.
Problem 22-15 1. 2 Na+ −OEt; 2. BrCH2CH2CH2CH2Br; 3. H3O+
Problem 22-16
(a) Alkylate phenylacetone with CH3I
(b) Alkylate pentanenitrile with CH3CH2I
(c) Alkylate cyclohexanone with H2C$\text{═}$CHCH2Br
(d) Alkylate cyclohexanone with excess CH3I
(e) Alkylate C6H5COCH2CH3 with CH3I
(f) Alkylate methyl 3-methylbutanoate with CH3CH2I
32.5.23: Chapter 23
Problem 23-1
(a)
(b)
(c)
Problem 23-2 The reverse reaction is the exact opposite of the forward reaction shown in Figure 23.2.
Problem 23-3
(a)
(b)
(c)
Problem 23-5
(a) Not an aldol product
(b) 3-Pentanone
Problem 23-6 1. NaOH; 2. LiAlH4; 3. H2/Pd
Problem 23-8
(a) C6H5CHO + CH3COCH3
(b) Not easily prepared
(c) Not easily prepared
Problem 23-9 The CH2 position between the two carbonyl groups is so acidic that it is completely deprotonated to give a stable enolate ion.
Problem 23-11
(a)
(b)
(c)
Problem 23-12 The cleavage reaction is the exact reverse of the forward reaction.
Problem 23-16
(a)
(b)
(c)
Problem 23-17
(a)
(b)
Problem 23-18 CH3CH2COCH$\text{═}$CH2 + CH3CH2NO2
Problem 23-19
(a)
(b)
(c)
Problem 23-20
(a) Cyclopentanone enamine + propenenitrile
(b) Cyclohexanone enamine + methyl propenoate
Problem 23-22 2,5,5-Trimethyl-1,3-cyclohexanedione + 1-penten-3-one
32.5.24: Chapter 24
Problem 24-1
(a) N-Methylethylamine
(b) Tricyclohexylamine
(c) N-Ethyl-N-methylcyclohexylamine
(d) N-Methylpyrrolidine
(e) Diisopropylamine
(f) 1,3-Butanediamine
Problem 24-2
(a)
(b)
(c)
(d)
(e)
(f)
Problem 24-3
(a)
(b)
(c)
(d)
Problem 24-4
(a) CH3CH2NH2
(b) NaOH
(c) CH3NHCH3
Problem 24-5 Propylamine is stronger; benzylamine pKb = 4.67; propylamine pKb = 3.29
Problem 24-6
(a) p-Nitroaniline < p-Aminobenzaldehyde < p-Bromoaniline
(b) p-Aminoacetophenone < p-Chloroaniline < p-Methylaniline
(c) p-(Trifluoromethyl)aniline < p-(Fluoromethyl)aniline < p-Methylaniline
Problem 24-7 Pyrimidine is essentially 100% neutral (unprotonated).
Problem 24-8
(a) Propanenitrile or propanamide
(b) N-Propylpropanamide
(c) Benzonitrile or benzamide
(d) N-Phenylacetamide
Problem 24-9 The reaction takes place by two nucleophilic acyl substitution reactions.
Problem 24-11
(a) Ethylamine + acetone, or isopropylamine + acetaldehyde
(b) Aniline + acetaldehyde
(c) Cyclopentylamine + formaldehyde, or methylamine + cyclopentanone
Problem 24-13
(a) 4,4-Dimethylpentanamide or 4,4-dimethylpentanoyl azide
(b) p-Methylbenzamide or p-methylbenzoyl azide
Problem 24-14
(a) 3-Octene and 4-octene
(b) Cyclohexene
(c) 3-Heptene
(d) Ethylene and cyclohexene
Problem 24-15 H2C$\text{═}$CHCH2CH2CH2N(CH3)2
Problem 24-16 1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. HOSO2Cl; 5. aminothiazole; 6. H2O, NaOH
Problem 24-17
(a) 1. HNO3, H2SO4; 2. H2/PtO2; 3. 2 CH3Br
(b) 1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. Cl2; 5. H2O, NaOH
(c) 1. HNO3, H2SO4; 2. Cl2, FeCl3; 3. SnCl2
(d) 1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. 2 CH3Cl, AlCl3; 5. H2O, NaOH
Problem 24-18
(a) 1. CH3Cl, AlCl3; 2. HNO3, H2SO4; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuBr; 6. KMnO4, H2O
(b) 1. HNO3, H2SO4; 2. Br2, FeBr3; 3. SnCl2, H3O+; 4. NaNO2, H2SO4; 5. CuCN; 6. H3O+
(c) 1. HNO3, H2SO4; 2. Cl2, FeCl3; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuBr
(d) 1. CH3Cl, AlCl3; 2. HNO3, H2SO4; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuCN; 6. H3O+
(e) 1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. 2 Br2; 5. H2O, NaOH; 6. NaNO2, H2SO4; 7. CuBr
Problem 24-19 1. HNO3, H2SO4; 2. SnCl2; 3a. 2 equiv. CH3I; 3b. NaNO2, H2SO4; 4. product of 3a + product of 3b
Problem 24-21 4.1% protonated
Problem 24-23 The side-chain nitrogen is more basic than the ring nitrogen.
Problem 24-24
Reaction at C2 is disfavored because the aromaticity of the benzene ring is lost.
Problem 24-25 (CH3)3CCOCH3 → (CH3)3CCH(NH2)CH3 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.22%3A_Chapter_22.txt |
Problem 25-1
(a) Aldotetrose
(b) Ketopentose
(c) Ketohexose
(d) Aldopentose
Problem 25-2
(a) S
(b) R
(c) S
Problem 25-3 A, B, and C are the same.
Problem 25-6
(a) L-Erythrose; 2S,3S
(b) D-Xylose; 2R,3S,4R
(c) D-Xylulose; 3S,4R
Problem 25-8
(a)
(b)
(c)
Problem 25-9 16 D and 16 L aldoheptoses
Problem 25-15 α-D-Allopyranose
Problem 25-17 D-Galactitol has a plane of symmetry and is a meso compound, whereas D-glucitol is chiral.
Problem 25-18 The −CHO end of L-gulose corresponds to the −CH2OH end of D-glucose after reduction.
Problem 25-19 D-Allaric acid has a symmetry plane and is a meso compound, but D-glucaric acid is chiral.
Problem 25-20 D-Allose and D-galactose yield meso aldaric acids; the other six D-hexoses yield optically active aldaric acids.
Problem 25-21 D-Allose + D-altrose
Problem 25-22 L-Xylose
Problem 25-23 D-Xylose and D-lyxose
Problem 25-25
(a) The hemiacetal ring is reduced.
(b) The hemiacetal ring is oxidized.
(c) All hydroxyl groups are acetylated.
32.5.26: Chapter 26
Problem 26-1 Aromatic: Phe, Tyr, Trp, His; sulfur-containing: Cys, Met; alcohols: Ser, Thr; hydrocarbon side chains: Ala, Ile, Leu, Val, Phe
Problem 26-2 The sulfur atom in the −CH2SH group of cysteine makes the side chain higher in ranking than the −CO2H group.
Problem 26-4 Net positive at pH = 5.3; net negative at pH = 7.3
Problem 26-5
(a) Start with 3-phenylpropanoic acid: 1. Br2, PBr3; 2. NH3
(b) Start with 3-methylbutanoic acid: 1. Br2, PBr3; 2. NH3
Problem 26-6
(a)
(b)
(c)
(d)
Problem 26-8 Val-Tyr-Gly (VYG), Tyr-Gly-Val (YGV), Gly-Val-Tyr (GVY), Val-Gly-Tyr (VGY), Tyr-Val-Gly (YVG), Gly-Tyr-Val (GYV)
Problem 26-12
Trypsin: Asp-Arg + Val-Tyr-Ile-His-Pro-Phe
Chymotrypsin: Asp-Arg-Val-Tyr + Ile-His-Pro-Phe
Problem 26-13 Methionine
Problem 26-15
(a) Arg-Pro-Leu-Gly-Ile-Val
(b) Val-Met-Trp-Asp-Val-Leu (VMWNVL)
Problem 26-16 This is a typical nucleophilic acyl substitution reaction, with the amine of the amino acid as the nucleophile and tert-butyl carbonate as the leaving group. The tert-butyl carbonate then loses CO2 and gives tert-butoxide, which is protonated.
Problem 26-17
(1) Protect the amino group of leucine.
(2) Protect the carboxylic acid group of alanine.
(3) Couple the protected amino acids with DCC.
(4) Remove the leucine protecting group.
(5) Remove the alanine protecting group.
Problem 26-18
(a) Lyase
(b) Hydrolase
(c) Oxidoreductase
32.5.27: Chapter 27
Problem 27-1 CH3(CH2)18CO2CH2(CH2)30CH3
Problem 27-2 Glyceryl tripalmitate is higher melting.
Problem 27-3 $[{\rm CH}_{3}({\rm CH}_{2})_{7}{\rm CH}\text{═}{\rm CH}({\rm CH}_{2})_{7}{\rm CO}_{2}\!^{-}]{2}\ {\rm Mg}^{2+}$
Problem 27-4 Glyceryl dioleate monopalmitate → glycerol + 2 sodium oleate + sodium palmitate
Problem 27-6 The pro-S hydrogen is cis to the −CH3 group; the pro-R hydrogen is trans.
Problem 27-7
(a)
(b)
Problem 27-8
(a)
(b)
Problem 27-10 Three methyl groups are removed, the side-chain double bond is reduced, and the double bond in the B ring is migrated.
32.5.28: Chapter 28
Problem 28-3 (5′) ACGGATTAGCC (3′)
Problem 28-5 (3′) CUAAUGGCAU (5′)
Problem 28-6 (5′) ACTCTGCGAA (3′)
Problem 28-7
(a) GCU, GCC, GCA, GCG
(b) UUU, UUC
(c) UUA, UUG, CUU, CUC, CUA, CUG
(d) UAU, UAC
Problem 28-8
(a) AGC, GGC, UGC, CGC
(b) AAA, GAA
(c) UAA, CAA, GAA, GAG, UAG, CAG
(d) AUA, GUA
Problem 28-9 Leu-Met-Ala-Trp-Pro-Stop
Problem 28-10 (5′) TTA-GGG-CCA-AGC-CAT-AAG (3′)
Problem 28-11 The cleavage is an SN1 reaction that occurs by protonation of the oxygen atom followed by loss of the stable triarylmethyl carbocation.
32.5.29: Chapter 29
Problem 29-1 HOCH2CH(OH)CH2OH + ATP → HOCH2CH(OH)CH2OPO32 + ADP
Problem 29-2 Caprylyl CoA → Hexanoyl CoA → Butyryl CoA → 2 Acetyl CoA
Problem 29-3
(a) 8 acetyl CoA; 7 passages
(b) 10 acetyl CoA; 9 passages
Problem 29-4 The dehydration is an E1cB reaction.
Problem 29-5 At C2, C4, C6, C8, and so forth
Problem 29-6 The Si face
Problem 29-7 Steps 7 and 10
Problem 29-8 Steps 1, 3: Phosphate transfers; steps 2, 5, 8: isomerizations; step 4: retro-aldol reaction; step 5: oxidation and nucleophilic acyl substitution; steps 7, 10: phosphate transfers; step 9: E1cB dehydration
Problem 29-9 C1 and C6 of glucose become –CH3 groups; C3 and C4 become CO2.
Problem 29-10 Citrate and isocitrate
Problem 29-11 E1cB elimination of water, followed by conjugate addition
Problem 29-12 pro-R; anti geometry
Problem 29-13 The reaction occurs by two sequential nucleophilic acyl substitutions, the first by a cysteine residue in the enzyme, with phosphate as leaving group, and the second by hydride donation from NADH, with the cysteine residue as leaving group.
Problem 29-14 Initial imine formation between PMP and α-ketoglutarate is followed by double-bond rearrangement to an isomeric imine and hydrolysis.
Problem 29-15 (CH3)2CHCH2COCO2
Problem 29-16 Asparagine
32.5.30: Chapter 30
Problem 30-1 Ethylene: ψ1 is the HOMO and ψ2* is the LUMO in the ground state; ψ2* is the HOMO and there is no LUMO in the excited state. 1,3-Butadiene: ψ2 is the HOMO and ψ3* is the LUMO in the ground state; ψ3* is the HOMO and ψ4* is the LUMO in the excited state.
Problem 30-2 Disrotatory: cis-5,6-dimethyl-1,3-cyclohexadiene;
conrotatory: trans-5,6-dimethyl-1,3-cyclohexadiene. Disrotatory closure occurs.
Problem 30-3 The more stable of two allowed products is formed.
Problem 30-4 trans-5,6-Dimethyl-1,3-cyclohexadiene; cis-5,6-dimethyl-1,3-cyclohexadiene
Problem 30-5 cis-3,6-Dimethylcyclohexene; trans-3,6-dimethylcyclohexene
Problem 30-6 A [6 + 4] suprafacial cycloaddition
Problem 30-7 An antarafacial [1,7] sigmatropic rearrangement
Problem 30-9 A series of [1,5] hydrogen shifts occur.
Problem 30-10 o-(1-Methylallyl)phenol
Problem 30-11 Claisen rearrangement is followed by a Cope rearrangement.
Problem 30-12
(a) Conrotatory
(b) Disrotatory
(c) Suprafacial
(d) Antarafacial
(e) Suprafacial
32.5.31: Chapter 31
Problem 31-1 H2C$\text{═}$CHCO2CH3 < H2C$\text{═}$CHCl < H2C$\text{═}$CHCH3 < H2C$\text{═}$CH–C6H5
Problem 31-2 H2C$\text{═}$CHCH3 < H2C$\text{═}$CHC6H5 < H2C$\text{═}$CHC$\text{≡}$N
Problem 31-3
The intermediate is a resonance-stabilized benzylic carbanion,
Problem 31-4 The polymer has no chirality centers.
Problem 31-5 The polymers are racemic and have no optical rotation.
Problem 31-10
Vestenamer: ADMET polymerization of 1,9-decadiene or ROMP of cyclooctene; Norsorex: ROMP of norbornene. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/32%3A_Appendix/32.05%3A_Answer_Key/32.5.25%3A_Chapter_25.txt |
The organic chemistry course is organized as three modules. The chapters referred to below are from L.G. Wade, Jr. Organic Chemistry. 5th ed. Prentice Hall, 2003.
I. MOLECULAR ARCHITECTURE
CHAPTER TOPIC
1: Atomic structure review
Covalent bonding and chemical formulas
Polarity and physical properties of molecules
Lewis formulas, structural isomers, and resonance structures
2: Orbital view of bonding, orbital combinations, hybridization theory
Electron mobility and delocalization in resonance structures
Introduction to functional groups: Alkanes, alkenes, and alkynes
3: Introduction to organic nomenclature: Alkanes and alkyl halides
Structure of alkanes and cycloalkanes,
Conformational analysis
5: Stereochemistry: Molecular architecture in three dimensions
II. MOLECULAR TRANSFORMATIONS: INTRODUCTION TO ORGANIC SYNTHESIS AND REACTION MECHANISMS
CHAPTER TOPIC
4: Energetics of bond breaking and bond formation, reaction mechanisms
Free radical mechanisms
Alkane halogenation as a major tool for alkane functionalization
1 Introduction to ionic mechanisms: Bronsted-Lowry acid-base chemistry
Electron movement in ionic mechanisms
6 / 7 Introduction to Lewis acid-base chemistry and reaction kinetics
Nucleophilic substitution reactions of alkyl halides as a synthetic tool
Elimination reactions and alkene synthesis
III. CHEMISTRY OF COMMON FUNCTIONAL GROUPS
CHAPTER TOPIC
8 / 9 Chemistry of pi-bonds: Alkenes and alkynes
Organometallics, carbon nucleophiles, and their use in organic synthesis
10 Structure and synthesis of alcohols, chemistry of Grignard reagents
2.01: Learning Objectives and Basic Concepts
LEARNING OBJECTIVES
To review the basics concepts of atomic structure that have direct relevance to the fundamental concepts of organic chemistry. This material is essential to the understanding of organic molecular structure and, later on, reaction mechanisms.
BASIC CONCEPTS
Most of this material is a review of general chemistry. You might find it helpful to keep a general chemistry textbook available for reference purposes throughout the organic chemistry course.
The following diagram summarizes the basic facts of the structure of the atom.
A simplified view of the hydrogen atom, which consists of only one electron outside the nucleus. The nucleus contains only one proton and no neutrons. All other elements contain neutrons in their nuclei.
Elements in the periodic table are indicated by SYMBOLS. To the left of the symbol we find the atomic mass (A) at the upper corner, and the atomic number (Z) at the lower corner.
Examples:
Electron trade constitutes the currency of chemical reactions. The number of electrons in a neutral atom (that is, the atomic number) gives the element its unique identity. No two different elements can have the same atomic number. The periodic table is arranged by order of increasing atomic number, which is always an integer.
In contrast to the atomic number, different forms of the same element can have different masses. They are called isotopes. The following are representations for some of the isotopes of hydrogen and carbon.
The atomic mass reported in the periodic table for any given element is actually a weighted average of the masses of its isotopes as found in nature. Thus the mass of carbon is reported as 12.01115 rather than 12.00000 because it contains the relative contributions of both isotopes. The natural abundance of carbon12 is nearly 100%, whereas that of carbon-13 is only about 1%. The reported mass is slightly greater than 12.00000 because of the small contribution of carbon-13. Therefore the mass number, as found in periodic tables, does not have to be an integer like the atomic number. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/01%3A_Introduction_and_Course_Organization/1.01%3A_Introduction.txt |
The quantum numbers are parameters that describe the distribution of electrons in the atom, and therefore its fundamental nature. They are:
1. PRINCIPAL QUANTUM NUMBER (n) - Represents the main energy level, or shell, occupied by an electron. It is always a positive integer, that is n = 1, 2, 3 ...
2. SECONDARY QUANTUM NUMBER (l ) - Represents the energy sublevel, or type of orbital, occupied by the electron. The value of l depends on the value of n such that l = 0, 1, ... n-1. This number is sometimes also called azimuthal, or subsidiary.
3. MAGNETIC QUANTUM NUMBER (ml ) - Represents the number of possible orientations in 3-D space for each type of orbital. Since the type of orbital is determined by l, the value of ml ranges between -l and +l such that ml = -l, ...0, ...+l.
4. SPIN QUANTUM NUMBER (mS ) - Represents the two possible orientations that an electron can have in the presence of a magnetic field, or in relation to another electron occupying the same orbital. Only two electrons can occupy the same orbital, and they must have opposite spins. When this happens, the electrons are said to be paired. The allowed values for the spin quantum number ms are +1/2 and -1/2.
According to Heisenberg’s uncertainty principle, it is impossible to know the electron’s velocity and its position simultaneously. The exact position of the electron at any given time cannot be known. Therefore, it is impossible to obtain a photographic picture of the atom like we could of a busy street. Electrons are more like fast-moving mosquitoes in a swarm that cannot be photographed without appearing blurred. The uncertainty about their position persists even in the photograph. An alternative picture of the swarm can be obtained by describing the area where the mosquitoes tend to be concentrated and the factors that determine their preference for certain locations, and that’s the best we can do.
The quantum numbers provide us with a picture of the electronic arrangement in the atom relative to the nucleus. This arrangement is not given in terms of exact positions, like the photograph of a street, but rather in terms of probability distributions and potential energy levels, much like the mosquito swarm. The potential energy levels are described by the main quantum number n and by the secondary quantum number l. The probability distributions are given by the secondary quantum number l and by the magnetic quantum number ml .
The now outdated solar system model of the atom allows us to visualize the meaning of the potential energy levels. The main energy levels, also called shells, are given by the main quantum number n.
THE RELATIONSHIP BETWEEN POTENTIAL ENERGY AND STABILITY IS INVERSE
As the potential energy of a system increases, the system’s stability is more easily disrupted. As an example consider the objects on the earth. Objects that are positioned at ground level have lower potential energy than objects placed at high altitudes. The object that’s placed at high altitude, be it a plane or a rock at the top of a mountain, has a higher “potential” to fall (lower stability) than the object that’s placed at ground level. Systems tend towards lower levels of potential energy, thus the tendency of the plane or the rock to fall. Conversely, an object placed in a hole on the ground does not have a tendency to “climb out” because its potential energy is even lower than the object placed at ground level. Systems do not naturally tend towards states of higher potential energy. Another way of saying the same thing is to say that systems tend towards states of higher stability.
In the case of the electrons in the atom, those at lower levels of potential energy (lower shells, or lower n) are more stable and less easily disrupted than those at higher levels of potential energy. Chemical reactions are fundamentally electron transfers between atoms. In a chemical reaction, it is the electrons in the outermost shell that react, that is to say, get transferred from one atom to another. That’s because they are the most easily disrupted, or the most available for reactions. The outermost shell is the marketplace where all electron trade takes place. Accordingly, it has a special name. It is called the valence shell.
Now, the solar system model of the atom is outmoded because it does not accurately depict the electronic distribution in the atom. Electrons do not revolve around the nucleus following elliptical, planar paths. They reside in 3-D regions of space of various shapes called orbitals.
An orbital is a region in 3-D space where there is a high probability of finding the electron.
An orbital is, so to speak, a house where the electron resides. Only two electrons can occupy an orbital, and they must do so with opposite spin quantum numbers ms . In other words, they must be paired.
The type and shape of orbital is given by the secondary quantum number l. As we know, this number has values that depend on n such that l = 0, 1, ... n-1. Furthermore, orbitals are not referred to by their numerical l values, but rather by small case letters associated with those values. Thus, when l = 0 we talk about s orbitals. When l = 1 we talk about p orbitals. When l = 2 we talk about d orbitals, and so on. In organic chemistry, we are mostly concerned with the elements of the second row and therefore will seldom refer to l values greater than 1. We’ll be talking mostly about s and p orbitals, and occasionally about d orbitals in reference to third row elements.
Since the value of l depends on the value of n, only certain types of orbitals are possible for each n, as follows (only the highest energy level is shown for each row of elements):
FIRST ROW ELEMENTS: = 1 = 0 only s orbitals are possible, denoted as 1s orbitals.
SECOND ROW ELEMENTS: =2 = 0 orbitals are possible, denoted as 2s orbitals.
= 1 orbitals are possible, denoted as 2p orbitals.
THIRD ROW ELEMENTS: = 3 =0 s orbitals are possible, denoted as 3s orbitals
=1 orbitals are possible, denoted as 3p orbitals,
=2 and d orbitals are possible, denoted as 3d orbitals.
The shapes associated with s and p orbitals are shown below. For d orbitals refer to your general chemistry textbook. The red dot represents the nucleus.
Finally, the orientations of each orbital in 3-D space are given by the magnetic quantum number ml . This number depends on the value of l such that ml = -l, ...0, ...+l. Thus, when l = 0, ml = 0. There is only one value, or only one possible orientation in 3-D space for s-orbitals. That stands to reason, since they are spherical. In the case of p-orbitals l = 1, so ml = -1, 0, and +1. Therefore, there are three possible orientations in 3-D space for p-orbitals, namely along the x, y, and z axes of the Cartesian coordinate system. More specifically, those orbitals are designated as px, py, and pz respectively | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/02%3A_Atomic_Structure/2.02%3A_The_Four_Quantum_Numbers.txt |
To indicate the electronic configuration of the atom, that is to say, where the electrons reside, we use the following notation.
Given a periodic table, all we need to know to write the electronic configuration for a given atom is the atomic number Z, which tells us the number of electrons in the neutral atom. We start by writing the first potential energy level (n=1), then the possible types of orbitals in this level (s, p, etc.), and then the number of electrons occupying that orbital, which is always either 1or 2. It will always be 2 unless Z is an odd number and we’re down to the last electron in the valence shell. In this course we will seldom be concerned with elements beyond the second row, so we’ll keep it simple. The electronic configurations for the nonmetals of the second row are shown below.
BORON B Z=5 1s2, 2s2, 2pX1 Total of 5 electrons, 3 in the valence shell
CARBON C Z=6 1s2, 2s2, 2pX1, 2pY1 Total of 6 electrons, 4 in the valence shell
NITROGEN N Z=7 1s2, 2s2, 2pX1, 2pY1, 2pZ1 Total of 7 electrons, 5 in the valence shell
OXYGEN O Z=8 1s2, 2s2, 2pX2, 2pY1, 2pZ1 Total of 8 electrons, 6 in the valence shell
FLUORINE F Z=9 1s2, 2s2, 2pX2, 2pY2, 2pZ1 Total of 9 electrons, 7 in the valence shell
NEON Ne Z=10 1s2, 2s2, 2pX2, 2pY2, 2pZ2 Total of 10 electrons, 8 in the valence shell
We can also write electronic configurations where electrons are shown as half-arrows and potential energy levels are shown as horizontal dashes positioned at different heights to indicate those levels. The following diagram shows the electronic configuration for carbon.
The half-arrows shown together in opposite directions indicate that the electrons are paired. Single arrows indicate unpaired electrons. Notice that an empty orbital does not mean that such orbital does not exist. It only means there are no electrons in it. Given the right circumstances, it could hold electrons. This is in opposition to an orbital whose existence is not possible. For example d orbitals are not possible for second row elements and therefore are nonexistent in those elements.
Orbitals which are of exactly the same energy, such as the 2pX, 2pY, and 2pZ orbitals, are said to be degenerate.
In writing electronic configurations, we follow the Aufbau principle, Hund’s rule, and the Pauli exclusion principle.
The Aufbau principle (German for building up) makes reference to the process of building an atom from the ground up. That is to say, the manner in which electrons are placed in the atom one by one. We start by placing the first electron at the lowest potential energy level, which is the 1s orbital, and then following with the rest of the electrons always placing them at the lowest available level of potential energy. In other words, electrons always go into orbitals with the lowest possible energy.
Hund’s rule says that when electrons go into degenerate orbitals, they occupy them singly before pairing begins. That’s the reason why in the carbon atom shown in the previous page, the electrons in the 2pX and 2pY orbitals are placed one in each orbital, unpaired, rather than two in one orbital and paired.
Finally, the Pauli exclusion principle states that only electrons with opposite spins can occupy the same orbital. In other words, if two electrons must go into the same orbital, they must be paired. In the example shown above we have
2.04: Chemical Bonding
As stated earlier, all chemistry except for nuclear reactions involve electron transfers from one atom to another. More specifically, this electron trade takes place at the valence shells of the atoms. In general chemistry, we learn about two major types of chemical bonds, namely ionic and covalent.
Ionic bonding is more likely to take place between elements of highly different electronegativities, especially between metals and nonmetals. We can use Lewis formulas to indicate the manner of the electron transfer in a highly simplified fashion. The dots around the atomic symbol represent the electrons in the valence shell of each element. In the classic example of the reaction between sodium and chlorine, the only electron in the valence shell of sodium is completely transferred to the valence shell of chlorine. Sodium then forms a positive ion and chlorine a negative ion. The electrostatic attraction between oppositely charged ions forms the basis of the ionic bond and is the force that keeps the atoms together.
Covalent bonding is more likely to take place between elements of similar electronegativities, especially between nonmetals. In the reaction between two hydrogen atoms, for instance, the electron transfer is never complete. Instead, the two atoms share the electrons, which in this case spend equal amounts of time around both nuclei. There is no formation of full positive or negative charges and therefore there is no electrostatic attraction. The force that keeps the atoms together is the fulfillment of the octet rule, which will be discussed shortly. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/02%3A_Atomic_Structure/2.03%3A_Electronic_Configurations.txt |
LEARNING OBJECTIVES
To introduce the basic principles of covalent bonding, different types of molecular representations, bond polarity and its role in electronic density distributions, and physical properties of molecules.
VALENCE ELECTRONS
They are those found in the highest energy level of the atom, or outer shell. In the periodic table, the number of valence electrons is given by the group number. For example, in the second row, the nonmetals are:
BORON Group III 3 valence electrons 2s2, 2p1
CARBON Group IV 4 valence electrons 2s2, 2p2
NITROGEN Group V 5 valence electrons 2s2, 2p3
OXYGEN Group VI 6 valence electrons 2s2, 2p4
FLUORINE Group VII 7 valence electrons 2s2, 2p5
OCTET RULE
The atoms that participate in covalent bonding share electrons in a way that enables them to acquire a stable electronic configuration, or full valence shell. This means that they want to acquire the electronic configuration of the noble gas of their row. Obviously the name of this rule is a misnomer. Helium, the noble gas of the first row, has only two electrons. Hydrogen, the only element in the first row besides Helium, fulfills the “octet rule” by sharing two electrons only.
ELECTRON SHARING IN THE HYDROGEN MOLECULE
Two hydrogen atoms form a covalent bond to make a hydrogen molecule. Each contributes one electron and forms a system that is much more stable than the isolated atoms. Although the orbital representation is more visually telling, the Lewis formula representation is easier to write, and therefore will be used from now on, unless there is reason to do otherwise.
The elements of the second row fulfill the octet rule by sharing eight electrons, thus acquiring the electronic configuration of neon, the noble gas of this row. Besides hydrogen, most of the elements of interest in this course are the second row nonmetals: C, N, O, and the halogens. As the building block of all organic molecules, carbon is of particular interest to us. Carbon (4 electrons in the valence shell) combines with four hydrogen atoms to form a stable covalent compound where it shares 8 electrons, while each hydrogen shares 2. Thus every atom in this stable molecule fulfills the octet rule.
ELECTRON SHARING IN THE METHANE (CH4) MOLECULE
3.02: Building Simple Molecules
Among the simplest covalent compounds that the second row nonmetals can form are those that result from combination with hydrogen. Based on the number of electrons in their valence shells and the octet rule, we can predict how many hydrogen atoms will be needed to combine with each of those elements. Carbon, with 4 electrons in its valence shell, will need another four electrons to fulfill the octet rule. Thus it needs to combine with 4 hydrogen atoms to form a stable compound called methane (CH4) as shown above.
Nitrogen, the next nonmetal, has 5 electrons in the valence shell, so it needs to combine with 3 hydrogen atoms to fulfill the octet rule and form a stable compound called ammonia (NH3). This leaves two electrons that cannot be used for bonding (otherwise nitrogen would have to share more than 8 electrons, which is impossible). In the ammonia molecule, these electrons are paired and unshared, meaning that they are not engaged in bonding. Such electron pairs are referred to as lone pairs, unshared electrons, or nonbonding electrons.
A similar process leads to the formation of stable hydrogen compounds for the next two nonmetals, oxygen and fluorine. We see that the water molecule contains two pairs of nonbonding electrons, and hydrogen fluoride contains three pairs. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/03%3A_Covalent_Bonding/3.01%3A_Learning_Objectives_Valence_Electrons_Octet_Rule.txt |
The maximum number of electrons possible in the valence shell of the second row elements is eight. However, the elements of the third row, such as phosphorus and sulfur, can form stable systems by sharing eight or more electrons. The presence of d-orbitals, which can accommodate up to ten electrons, makes this possible.
Now, back to the second row, what happens when the first nonmetal, boron (Z=3), combines with hydrogen? By repeating the process outlined before for carbon, nitrogen, oxygen, and fluorine, we conclude that boron needs to bond to 5 hydrogen atoms to fulfill the octet rule. The problem is that with only three electrons in the valence shell this is impossible:
The only possibility for boron is to bond to three hydrogen atoms, in which case it forms a compound (borane, BH3) that does not fulfill the octet rule. The compound actually exists, but it is highly reactive, that is to say, unstable. Substances such as BH3 are referred to as electron-deficient molecules, and are very reactive towards electron-rich substances.
Aluminum, which is also in group III, exhibits similar behaviour.
3.04: Formal Charge
Sometimes atoms engage in covalent bonding by contributing more or less electrons than they have in their valence shell (we’ll examine the processes that lead to the loss or gain of electrons later). For example nitrogen can actually combine with four hydrogen atoms to form a stable species called ammonium ion (NH4). In this species, nitrogen still shares eight electrons, but contributes only four of its own. Since electrons are negative charges and this nitrogen is missing one, it acquires a net charge of +1 (in other words, there is a proton in the nucleus that is not matched by an electron outside the nucleus). This net charge is referred to as formal charge, and it must be indicated as part of the notation for the NH4 formula, as shown.
In another species known as a carbanion, carbon forms only three bonds and carries a pair of unshared electrons. In this species, carbon shares eight electrons, but it is contributing five of its own. Since it has a surplus of one electron (a negative charge), it carries a net charge of -1.
Obviously the concept of formal charge refers to a specific atom. Formulas should show these charges on the atoms where they belong. Other examples of covalent species with charged atoms are the hydronium ion and the amide ion.
3.05: Connectivity or Bonding Sequence Types of Formulas
CONNECTIVITY OR BONDING SEQUENCE
The term connectivity, or bonding sequence, describes the way atoms are connected together, or their bonding relationships to one another, in covalent compounds. For example, in the methane molecule one carbon is connected to four hydrogen atoms simultaneously, while each hydrogen atom is connected to only one carbon. No hydrogen atoms are connected together. In complex molecules the complete connectivity map is given by structural formulas (see below).
TYPES OF FORMULAS
The simplest type of formula for a compound indicates the types of atoms that make it up and their numbers. This is called a molecular formula. Examples of molecular formulas are BH3, C6H6, or C3H5ClO. Chemical catalogs such as the Aldrich catalog, scientific manuals, and databases such as Chemical Abstracts typically contain molecular formula indices to help locate substances whose elemental makeup is known.
Condensed structural formulas give some idea of the connectivity, but are still largely abbreviated. For example the ethane molecule, which has molecular formula C2H6 can be represented by the condensed formula CH3CH3, This at least tell us that each carbon is connected to three hydrogen atoms, and that two carbon atoms are connected together.
Lewis formulas are a second type of structural formulas. They give the most complete representation of the connectivity that is possible in two dimensions. The three types of formulas mentioned so far are shown below for the ethane molecule.
3.06: Types of Covalent Bonds
In the ethane Lewis formula shown above all bonds are represented as single lines called single bonds. Each single bond is made up of two electrons, called bonding electrons. It is also possible for two atoms bonded together to share 4 electrons. This bonding pattern is represented by two lines, each representing two electrons, and is called a double bond. The ethylene molecule shown below is an example. Finally, sharing of 6 electrons between two atoms is also possible. In such case, the representation uses three single lines, an arrangement called a triple bond. The acetylene molecule provides an example of a triple bond.
This terminology (single, double, or triple bond) is very loose and informal. The formulas shown above do not do justice to the actual nature of the bonds. All they do is show how many electrons are being shared between the two atoms (2, 4, or 6) but they say nothing about the electronic distribution, or the relative energies of the bonds, or the types of orbitals involved. They are, however, very useful in many situations. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/03%3A_Covalent_Bonding/3.03%3A_Exceptions_to_the_Octet_Rule.txt |
ELECTRONIC DISTRIBUTION AND BOND POLARITY
As we already learned, the atoms engaged in covalent bonding share electrons in order to fulfill the octet rule. However, this electron sharing can take place on an equal or unequal basis. If the atoms involved in covalent bonding are of equal electronegativities (which occurs only if they are the same atoms), then sharing takes place on an equal basis and there is no bias in the amount of time the bonding electrons spend around each atom. The hydrogen molecule (H2) shown below is an example of this. The electronic cloud surrounding the two atoms is highly symmetrical, and the H-H bond is said to be nonpolar.
Now consider the case of hydrogen chloride, H-Cl. Hydrogen and chlorine are engaged in covalent bonding, but the electronegativity of chlorine is higher than that of hydrogen. The greater tendency of chlorine to attract electrons results in unequal sharing between the two atoms. The bonding electrons spend more time around chlorine than around hydrogen. They are still being shared, but chlorine behaves as if it carried a negative charge, and hydrogen behaves as if it carried a positive charge. These charges are not full charges as is the case in ionic molecules. In covalent molecules they are referred to as partial charges, or poles, because they are analogous of the poles of a magnet. The positive pole is indicated by ∂ +, and the negative pole by ∂ -. The two together constitute a dipole, and the bond in question is said to be polar.
A polar bond is sometimes represented as a vector, with an arrow pointing in the direction of the more electronegative atom. The following are valid representations for polar bonds.
Electronegativity is the tendency of an atom to attract bonding electrons. Since the difference in electronegativity between two bonding atoms can be zero or very large, there is a polarity continuum, ranging from nonpolar to highly polar bonds. In an extreme case where the difference in electronegativity is vary large, the bond ceases to be covalent and becomes ionic.
Bond polarity is measured by the dipole moment. This parameter is reported in Debye units (D). General Chemistry textbooks typically contain tables of dipole moments for different types of bonds. For example, the dipole moment for the C-H bond is 0.3 D, whereas that for the H-Cl bond is 1.09 D.
POLARITY IN ORGANIC MOLECULES
Every covalent bond is either polar or nonpolar. When all the dipoles for all the covalent bonds that make up a molecule are added together as vectors, the result is the net dipole moment of the entire molecule. When its value is zero, the molecule is said to be nonpolar, otherwise it’s said to be polar. Obviously, it is possible to have nonpolar molecules made up of polar bonds, as long as the corresponding dipoles add up to zero. Some examples are shown below. Refer to chapter 2 in your textbook for a more comprehensive discussion of polarity and dipoles.
One must be careful in deciding whether a molecule is polar or nonpolar based purely on a two-dimensional representation. Molecules are three-dimensional, and direction is as important as magnitude when it comes to adding vectors. For example, a two-dimensional representation of the methylene chloride molecule (CH2Cl2) shown below might lead to the erroneous conclusion that it is nonpolar when in fact it is polar.
Many organic molecules are made up of long hydrocarbon chains with many C-H bonds. Since the difference in electronegativity between carbon and hydrogen is very small, the C-H bond has a very small dipole moment, and hydrocarbons are for the most part considered nonpolar molecules. However, the introduction of a relatively polar bond in such structures dominates the entire molecule, rendering it polar.
POLARITY AND PHYSICAL PROPERTIES
The polarity of molecules affects their physical properties. As a rule of thumb and other factors being similar, the higher the polarity of the molecule, the higher the value of properties such as melting and boiling point. The solubility of molecules in solvents is also largely determined by polarity. The rule “like dissolves like” makes reference to the fact that polar molecules dissolve better in polar solvents, and nonpolar molecules dissolve better in nonpolar solvents. Water and oil don’t mix because water is highly polar and oil is largely made up of hydrocarbon chains, which are nonpolar. Conversely, water and alcohol do mix because they are both of very similar polarities. For a more comprehensive discussion refer to chapter 2 of your textbook. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/03%3A_Covalent_Bonding/3.07%3A_Polarity.txt |
Learning Objectives
• To understand the uses and limitations of Lewis formulas, to introduce structural isomerIsm, and to learn the basic concept of resonance structures.
04: Lewis Formulas Structural Isomerism and Resonance Structures
Lewis formulas are structures that show the connectivity, or bonding sequence of the atoms, indicating single, double, or triple bonds. They should also show any formal charges and unshared electrons that might be present in the molecule. Additional examples of Lewis formulas follow.\
These examples were deliberately chosen because all three molecules shown have the same molecular formula, but different connectivities, or bonding sequences. Such substances are called structural isomers, or sometimes constitutional isomers.
Notice that only the first structure shows the unshared electrons of chlorine. In Lewis formulas of organic compounds, it is customary to omit the lone electron pairs on the halogens unless there is a reason to show them explicitly.
Lewis formulas are mostly used for covalent substances, but occasionally they also show ionic bonds that might be present in certain compounds.
4.02: Common Bonding Patterns for 1st and 2nd Row Elements
Once we write enough Lewis formulas containing the elements of interest in organic chemistry, which are mostly the second row elements, we find that certain bonding patterns occur over and over. Learning these patterns is useful when trying to write Lewis formulas because they provide a convenient starting point. For example, in several of the structures given in the previous section, we find that the carbon bonded to three hydrogens is a unit that occurs quite frequently. It is called the methyl group, represented by CH3. It is so common that it is valid to write it as such in Lewis formulas, even though it is in fact an abbreviated form, because everybody knows what it stands for.
Other common bonding patterns are shown below.
HYDROGEN:
Usually forms only one bond.
CARBON:
Forms four bonds when neutral, but it can also have only three bonds by bearing a positive or a negative charge. When it bears a negative charge it should also carry a pair of unshared electrons.
Neutral carbon
A carbocation has a central carbon with an incomplete octet and a formal +1 charge.
A carbanion has a central carbon with an unshared electron pair and a formal -1 charge.
NITROGEN:
Forms three bonds and carries a lone pair of electrons when neutral. It can also form four bonds by bearing a positive charge, in which case it carries no unshared electrons. Finally, it can also form two bonds as it carries two unshared electron pairs and a negative charge.
Neutral nitrogen
Positively charged nitrogen
Negatively charged nitrogen
OXYGEN:
Forms two bonds and carries two lone pairs when neutral. It can form three bonds with a positive charge, or one bond with a negative charge. In each case it must carry the appropriate number of unshared electron pairs to complete the octet.
HALOGENS:
Form one bond and carry three electron pairs when neutral. Can carry a negative charge with no bonds. They are rarely seen with positive charges.
THIRD ROW ELEMENTS:
They behave like their second-row counterparts, except that they can expand their valence shells if needed.
Electron pairs on oxygen are not shown for clarity.
ELECTRON DEFICIENCY IN SECOND ROW ELEMENTS:
One thing worth noting is that, in the second row, only boron and carbon can form relatively stable species in which they bond with an incomplete octet. Examples have already been discussed. Boron has no choice but to be electron deficient. Carbon can bond with a complete octet or with an incomplete octet. Obviously bonding with a complete octet provides higher stability.
Boron has no choice but to have an incomplete octet
An electron deficient carbon in a carbocation
It is however very rare to observe species where nitrogen or oxygen bond with incomplete octets. Their high electronegativity renders such situation high energy, and therefore very unstable. For all intents and purposes, avoid writing formulas where oxygen or nitrogen are shown with incomplete octets, even if they carry a positive charge.
To write this structure without the lone pair of electrons on oxygen is unacceptable.
This structure is unacceptable and indeed it looks quite awkward
This species might exist in the high energy environment of a mass spectrometer, but it is not frequently observed in common organic reactions. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/04%3A_Lewis_Formulas_Structural_Isomerism_and_Resonance_Structures/4.01%3A_Characteristics_of_Lewis_Formulas.txt |
RESONANCE STRUCTURES AND SOME LIMITATIONS OF LEWIS FORMULAS:
Lewis formulas are misleading in the sense that atoms and electrons are shown as being static. By being essentially two-dimensional representations they also fail to give an accurate idea of the three-dimensional features of the molecule, such as actual bond angles and topography of the molecular frame.
Furthermore, a given compound can have several valid Lewis formulas. For example CH3CNO can be represented by at least three different but valid Lewis structures called resonance forms, or resonance structures, shown below.
However, a stable compound such as the above does not exist in multiple states represented by structures I, or II, or III. The compound exists in a single state called a hybrid of all three structures. That is, it contains contributions of all three resonance forms, much like a person might have physical features inherited from each parent to varying degrees.
In the resonance forms shown above the atoms remain in one place. The basic bonding pattern, or connectivity, is the same in all structures, but some electrons have changed locations. This means that there are certain rules for electron mobility that enable us to “push” electrons around to arrive from one resonance structure to another. These rules will be examined in detail in a later paper.
ALL RESONANCE STRUCTURES MUST BE VALID LEWIS FORMULAS:
By convention, we use double-headed arrows to indicate that several resonance structures contribute to the same hybrid. Continuing with the example we’ve been using, the resonance structures for CH3CNO should be written in this way if we want to emphasize that they represent the same hybrid.
Do not confuse double-headed arrows with double arrows. A double arrow indicates that two or more species are in equilibrium with each other and therefore have a separate existence. Double-headed arrows indicate resonance structures that do not exist by themselves. They simply represent features that the actual molecule, the hybrid, possesses to one extent or another.
When writing resonance structures keep in mind that THEY ALL MUST BE VALID LEWIS FORMULAS. The factors that make up valid Lewis formulas are as follows.
1. Observe the rules of covalent bonding, including common patterns as discussed previously. Make sure to show all single, double, and triple bonds.
2. Account for the total number of valence electrons being shared (from all the elements), including bonding and nonbonding electrons. Make sure to show these nonbonding electrons.
3. Account for the net charge of the molecule or species, showing formal charges where they belong.
4. Observe the octet rule as much as possible, but also understand that there are instances where some atoms may not fulfill this rule.
5. Avoid having unpaired electrons (single electrons with no partners) unless the total number of valence electrons for all elements is an odd number. This is not a very frequent occurrence, but the following example shows a species that could exist as a reaction intermediate in some high energy environments.
The total number of valence electrons being shared for all atoms is 4 from carbon and 3 from the three hydrogens, for a total of 7. Because it is an odd number, it is impossible to have all these electrons paired. Therefore the presence of a single electron cannot be avoided. Notice that there is no formal charge on carbon, since it has no surplus or deficit of valence electrons.
RELATIVE ENERGIES OF RESONANCE STRUCTURES.
From the examples given so far it can be seen that some resonance forms are structurally equivalent and others are not. The potential energy associated with equivalent Lewis structures is the same. If the Lewis structures are not equivalent, then the potential energy associated with them is most likely different. This means that equivalent resonance structures are also equivalent in stability and nonequivalent structures have different stabilities. This in turn means that equivalent structures contribute equally to the hybrid and nonequivalent structures do not contribute equally to the hybrid.
In the example below, the two structures are equivalent. Therefore they make equal contributions to the hybrid.
It is very difficult to accurately represent the hybrid with drawings because it is a composite of all the resonance contributors. Some representations such as the first one shown below are sometimes given. In this case the broken line represents the electrons and the negative charge which are spread over three atoms (O-C-O). Another commonly used representation of the hybrid is given on the right, showing that each oxygen atom shares a -1/2 charge. However, none of them accurately conveys the true picture. For convenience, the unshared electrons are sometimes omitted from some representations, just like it’s done with hydrogen atoms. This must be kept in mind when examining the different structures.
There is a third resonance form that can be drawn for the acetate ion hybrid. The structure shown below is structurally different from the ones shown above. This means that it is of different energy and therefore does not contribute to the hybrid to the same extent as the others. In this case, it happens to be less stable than the other two and therefore does not make a significant contribution to the hybrid.
What factors dictate the relative stabilities of different resonance structures? Charge separation is an important one. In the example just given there are too many charges present in the structure (even though the net charge is still the same). This results in a structure with higher potential energy than others with fewer or no formal charges. This structure is less stable and its contribution to the hybrid is probably minor.
Another important factor that increases potential energy (lowers stability) is the presence of atoms with an incomplete octet. In the example below structure I has a carbon atom with a positive charge and therefore an incomplete octet. Based on this criterion, this structure is less stable than structure II and makes a less significant contribution to the hybrid.
Finally, another factor that comes into play when determining the relative energies of resonance structures is the relative electronegativities of atoms that bear charges. More electronegative atoms are more comfortable with negative charges. Less electronegative atoms are more comfortable with positive charges. In the example below, structure I is less stable than II. The negative charge is on carbon, which is less electronegative than oxygen. Therefore structure II makes a larger contribution to the hybrid.
Some texts refer to the different contributors as “important” or “unimportant.” This can be misleading, because “importance” is context-dependent. In the example above, structure I is less important in terms of its contribution to the hybrid, but in terms of reactivity, it is very important. It will be seen later that this structure provides a better indication of how this species reacts with electrophiles in certain types of reactions.
Based on the charge separation criterion, structure II below might be labeled “unimportant,” or “less important” because it is less stable than I. But if we are trying to assess the polarity of this molecule, structure II becomes very important because it reveals that the carbon atom has positive character and the oxygen has negative character. The hybrid representation on the right then portrays the molecule as a polar molecule, which structure I alone does not give much indication of.
The following additional examples further illustrate the relative importance of the factors that contribute to structural energy and stability. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/04%3A_Lewis_Formulas_Structural_Isomerism_and_Resonance_Structures/4.03%3A_Resonance_Structures.txt |
Learning Objectives
• To introduce the basic principles of molecular orbital theory and electronic geometry of molecules.
05: Orbital Picture of Bonding- Orbital Combinations Hybridization Theory and Molecular Orbitals
Atomic orbitals can be combined and reshaped –much like dough– to make other orbitals of different shapes and properties. There are two basic types of orbitals that can result from such processes. They are:
1. HYBRID ORBITALS. They result from combinations of orbitals within a given atom, either prior to or as bonding with another atom takes place.
2. MOLECULAR ORBITALS. They result from combinations of orbitals between atoms as bonding takes place to form molecules.
5.02: Orbital Hybridization Theory
If we look at the valence shell configuration of carbon, we find two paired electrons in the 2s orbital, and two unpaired electrons in the 2pX and 2pY orbitals, one in each:
In order to fulfill the octet rule, carbon must use its 4 valence electrons when bonding to other atoms. However, only unpaired electrons can bond. That means that the two paired electrons occupying the 2s orbital must become unpaired before they can bond. Since the energy gap between the 2s and 2p orbitals is very small, one of the 2s electrons can be promoted to the empty 2p orbital, leading to the following situation
Now the four electrons appear to be ready for bonding, but there is a problem. The 2p orbitals are known to be at right angles to each other. If bonding occurs in this state, the 3 equivalent p electrons would form 3 equivalent bonds oriented at 90o to each other, and the s electron would form a bond of a different type and orientation from the other three. No such compound exists. The simplest hydrocarbon –methane (CH4)– is known to have tetrahedral geometry, where the four C–H bonds are all equivalent and positioned at 109.5o angles to each other. In addition, there are some carbon compounds where the bond angles are 120o or even 180o. The shapes and relative positions of the valence orbitals in atomic carbon do not explain the shapes and relative positions of the bonds in carbon compounds.
HYBRIDIZATION THEORY ATTEMPTS TO EXPLAIN THE ACTUAL SHAPES OF MOLECULES BY INVOKING THE FORMATION OF HYBRID ORBITALS DURING, OR PRIOR TO, THE BONDING PROCESS.
Going back to the carbon model with four unpaired electrons in the valence shell, we can take it as a point of departure for formation of hybrid orbitals. The first step is to take either 2, 3, or all four of those orbitals and equalize their energies. Let’s say that we take all four of them and form 4 equivalent new orbitals. These orbitals are now of the same energy, which is intermediate between those of the original 2s and 2p orbitals. At the same time, we cannot name the new orbitals s or p, for they’re neither. We have to find a new name that reflects the fact that they were created from one s orbital and three p orbitals. We will call them sp3 orbitals. The process that leads to their formation is called sp3 hybridization.
All four sp3 orbitals that result from this process are equivalent. That means that they have the same size, shape, and energy. According to VSEPR (valence shell electron pair repulsion) theory, such orbitals will orient themselves in 3-D space to be as far apart from each other as possible. The resulting shape is then a tetrahedron, where the carbon nucleus is at the center and the orbitals point to the corners of the tetrahedron. The ideal angle between orbitals is then 109.5 degrees.
When an sp3 hybridized carbon bonds to hydrogen, it forms methane, whose geometry is known to be tetrahedral.
A 3-D representation of methane. The single lines represent bonds that are positioned on the plane of the paper. The solid wedge represents a bond coming out of the plane of the paper towards the front. The broken wedge represents a bond going behind the plane of the paper towards the back.
HYDROCARBONS
Hydrocarbons are substances containing only carbon and hydrogen. Hydrocarbons are classified into the following major categories: alkanes, alkenes, alkynes, and aromatic hydrocarbons. In the following pages we will do an overview of the basic characteristics of the first three, but will postpone the study of aromatics until later.
ALKANES AND sp3 HYBRIDIZATION OF CARBON
Alkanes are hydrocarbons where all the carbon atoms are sp3-hybridized, all bonds are single bonds, and all carbons are tetrahedral. Methane is the simplest alkane, followed by ethane, propane, butane, etc. The carbon chain constitutes the basic skeleton of alkanes. Carbon chains with four or more atoms can be linear or branched. Some examples are shown below including Lewis, molecular, and condensed formulas. Refer to chapter 2 of the Wade textbook for additional examples.
Linear alkanes
Some examples of branched alkanes are shown below. Notice that sometimes Lewis formulas become cumbersome and difficult to write without cluttering. Condensed formulas are more convenient to use in such situations.
Branched alkanes
As can be seen from the above examples, all alkanes have the general formula CnH2n+2 where n is the total number of carbon atoms. This holds regardless of whether the alkane is linear or branched. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/05%3A_Orbital_Picture_of_Bonding-_Orbital_Combinations_Hybridization_Theory_and_Molecular_Orbitals/5.01%3A_Orbital_Combinations.txt |
When atomic orbitals (pure or hybrid) of different atoms overlap to form covalent bonds, they may approach each other in two major ways: head to head, or sideways. Only head to head overlap is possible with s orbitals because they are spherical. Hybrid orbitals also undergo mostly head to head overlap when forming covalent bonds. p-orbitals, on the other hand, can approach each other either sideways or head to head. For now, however, we are concerned only with head to head overlap because that’s the only type that occurs in alkanes. We’ll discuss sideways overlap later in connection with alkenes and alkynes, that is, hydrocarbons that have double and triple bonds respectively.
When orbitals approach each other in a head to head fashion, the resulting covalent bonds are called sigma bonds.
As illustrations, consider the bonds that have already been studied. The bond between two hydrogen atoms is an example of sigma bonding. The bonds between the sp3 orbitals of hybridized carbon and the s orbitals of hydrogen in methane are also example of sigma bonds.
sigma bond between s orbitals
sigma bonds between s and sp3 orbitals
Two sp3 carbons can also overlap to form a C–C sigma bond where two sp3 orbitals overlap head to head, such as in the formation of the ethane molecule:
It can be easily seen that the only type of covalent bonds present in alkanes are sigma bonds, also loosely known as single bonds.
LINE-ANGLE FORMULAS
In alkanes of 3 carbon atoms or more, the main carbon chain acquires a zig-zag structure due to the 109.5o angle between C–C bonds, such as in propane:
Two representations of propane, where the zig-zag structure of the carbon chain becomes apparent
Writing Lewis formulas, or even condensed formulas, for alkanes of many carbon atoms can quickly become cumbersome. A short hand notation that uses zig-zag lines has been developed. The resulting representations are known as line-angle formulas. The beginning and the end of the zig-zag line, as well as any breaks in direction represent carbon atoms.
Line angle representation for propane equivalent to or CH3CH2CH3 The arrows point to the positions of the carbon atoms.
Every carbon atom has to form 4 bonds. The bonds that are not shown are assumed to be bonds to hydrogen. Other examples are:
5.04: Hybridization of Carbon
ALKENES AND sp2 HYBRIDIZATION OF CARBON
We will now reproduce the sp3 hybridization process for carbon, but instead of taking one s and three p orbitals to make four equivalent sp3 orbitals, this time we’ll take only one s and two p orbitals to make three equivalent sp2 orbitals, leaving one p orbital untouched. The process is shown below.
As shown, the three resulting sp2 orbitals are equivalent in energy, but the remaining p orbital has not been affected. It still retains its original energy and shape. Again, according to VSEPR theory, equivalent orbitals will arrange themselves in 3-D space to be as far apart from each other as possible. Therefore, the three equivalent sp2 orbitals will arrange themselves in a trigonal planar configuration. That is to say, the carbon nucleus will be at the center of an equilateral triangle, and the three sp2 orbitals will point to the corners of that triangle. The ideal angle between sp2 orbitals is therefore 120o. A top view of this arrangement is shown below.
In this top view, the unhybridized p orbital cannot be seen because it also arranges itself to be as far apart from the sp2 orbitals as possible. That is to say, it is positioned at right angles to those orbitals, with one lobe coming out of the plane of the page and the other going behind the page. To see this arrangement clearly, we must switch to a side view of the orbital system.
When two sp2 hybridized carbon atoms approach each other to bond, two sp2 orbitals approach each other head to head, and two p orbitals approach each other sideways. The bond formed by the sp2 orbitals is a sigma bond, and the bond formed by the p orbitals is called a pi bond. The process is shown below.
The illustration above tries to convey a basic feature of the pi bond as compared to the sigma bond. The sigma bond is short and strong. As a rule, head to head overlap is the most efficient way to bond and results in relatively strong and stable bonds. The pi bond, on the other hand, is relatively long and diffuse. Sideways overlap is less efficient than head to head overlap and results in formation of weaker bonds.
This has some implications in the properties and chemical reactivity of sigma and pi bonds. The electrons in the sigma bond (or sigma electrons) are more tightly bound to the nucleus and don’t move too much. In other words, they are more LOCALIZED. The electrons in the pi bond (or pi electrons) are less tightly bound by the nucleus, and therefore they are relatively mobile. Under certain conditions, they have the capability to become DELOCALIZED, that is to say, they can move in the molecular skeleton from one atom to another, or even become spread over several atoms, according to principles we’ll study later.
At the same time, in chemical reactions where electrons are to be traded, the pi electrons are more readily available because they are more exposed and less tightly bound by the nucleus. It is relatively easy to break a pi bond compared to the sigma bond. The principles of all this chemistry will be discussed later in the course.
ALKENES ARE HYDROCARBONS THAT CONTAIN AT LEAST ONE PI BOND AS PART OF THEIR MOLECULAR STRUCTURE. By this definition, the simplest possible alkene must contain two carbon atoms. It is called ethene. Below is a Lewis and a line-angle representation of ethene, which is sometimes informally called ethylene. Notice that although C–H bonds are not usually shown in line-angle formulas, sometimes they are included for enhanced clarity. In this case a pure line-angle formula for ethene would look awkward because it would resemble an equal sign (=).
Notice that a Lewis representation does not differentiate between the sigma and the pi bonds in the so-called “double bond.” It simply shows the two together as two equal dashes. The orbital picture better represents the actual nature of the two types of bonds. Additional examples are shown below. For a full discussion of the structure of alkenes refer to chapter 7 of the Wade textbook.
Observe that the general formula for open chain monoalkenes –that is, alkenes that do not form cyclic structures and which contain only one pi bond– is CnH2n where n is the total number of carbon atoms.
ALKYNES AND sp HYBRIDIZATION OF CARBON
The process for understanding the sp hybridization process for carbon is basically an extension of the other two types (sp3 and sp2). You should try to work out this scheme on your own and see if your predictions agree with those presented in the textbook. sp hybridization gives rise to the formation of hydrocarbons known as alkynes. Alkynes contain at least one triple bond, and have linear geometry around the carbons comprising the triple bond. Therefore, the ideal angle between the sp hybrid orbitals is 180o. Some examples of alkynes are shown below. For additional information refer to chapter 9 of the Wade textbook.
Observe that the general formula for open chain monoalkynes is CnH2n-2 where n is the total number of carbon atoms. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/05%3A_Orbital_Picture_of_Bonding-_Orbital_Combinations_Hybridization_Theory_and_Molecular_Orbitals/5.03%3A_Sigma_Bonding.txt |
The hybridization schemes for nitrogen and oxygen follow the same guidelines as for carbon. For example, sp3 hybridization for nitrogen results in formation of four equivalent sp3 orbitals, except that this time only three of them contain unpaired electrons, and one of them contains paired electrons. A similar situation holds true for oxygen, which ends up with two of the sp3 orbitals occupied with unpaired electrons, and the other two with paired electrons.
The four sp3 orbitals again orient themselves in 3-D space to be as far apart from each other as possible, but the ideal 109.5o angle becomes distorted because the orbital with two electrons repels the others more strongly than they repel themselves. However the geometry of this arrangement is still fundamentally tetrahedral. When this sp3 hybridized nitrogen bonds to hydrogen, the three unpaired electrons are used for bonding, and the remaining pair remains as nonbonding electrons.
A similar analysis for oxygen should lead to formation of two sp3 orbitals with unpaired electrons, and two with paired electrons. Thus, when sp3 oxygen bonds with hydrogen it forms water, which has a distorted tetrahedral angle, and two pairs of nonbonding electrons in the structure.
Now, what happens when nitrogen or oxygen become sp2 hybridized? The exact same thing that happened with carbon, with some minor changes. Let’s go through the process with oxygen to illustrate how it bonds to carbon to make a class of substances known as carbonyl compounds.
Three equivalent sp2 orbitals have formed, two of them containing paired electrons, and one containing a single unpaired electron. The unhybridized p orbital remains, with an unpaired electron in it. Again, VSEPR theory dictates that the three equivalent sp2 orbitals will acquire a trigonal planar arrangement, while the unhybridized p orbital will remain at right angles to the sp2 orbitals.
When an sp2 oxygen approaches an sp2 carbon for bonding, the process is analogous to that followed by two sp2 carbons that bond to form a “double bond.” Only those orbitals containing unpaired electrons can bond. In this case, the sp2 orbitals from oxygen and carbon that contain unpaired electrons will overlap head to head to form a sigma bond. At the same time, the p orbitals will overlap sideways to form a pi bond. Thus, we have the effective formation of a C=O “double bond.”
Notice that the pi bond in the C=O “double bond” appears distorted, indicating higher electron density around the oxygen than around the carbon. This is because the C=O bond is polar. The more electronegative oxygen atom attracts bonding electrons towards itself more strongly than carbon.
A similar analysis for nitrogen leads to the picture and geometry of a C=N double bond:
Some examples of compounds containing sp2 carbon, nitrogen, and oxygen are shown below. Notice that Lewis and line-angle formulas frequently neglect showing the lone pairs of electrons. Unless there is a specific purpose for showing nonbonding electrons, they are usually omitted and assumed to be present.
By going through an analogous process for sp hybridization of nitrogen and oxygen we can arrive at the molecular structure of species containing carbon-nitrogen and carbon-oxygen triple bonds. Notice that an oxygen containing a triple bond must also carry a positive charge in observance of the rules of covalent bonding.
Also notice that when two sp or sp2 atoms are bonded together to form a double or triple bond, they both must be of the same hybridization. There are some exceptions, but this is true in most common situations.
5.06: Summary
The preceding discussion covers the most common cases of atom hybridization encountered in this course. The following summarizes the concepts associated with each type of hybridization.
sp3 hybridization 4 orbitals tetrahedral geometry ideal angle: 109.5° single bonds
sp2 hybridization 3 orbitals trigonal planar geometry ideal angle: 120° double bonds
sp hybridization 2 orbitals linear geometry ideal angle 180° triple bonds | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/05%3A_Orbital_Picture_of_Bonding-_Orbital_Combinations_Hybridization_Theory_and_Molecular_Orbitals/5.05%3A_Orbital_Hybridization_in_Nitrogen_and_Oxygen.txt |
Learning Objectives
• To introduce the concept of electron delocalization from the perspective of molecular orbitals, to understand the relationship between electron delocalization and resonance, and to learn the principles of electron movement used in writing resonance structures in Lewis notation, known as the curved arrow formalism.
06: Electron Delocalization and Resonance
Now that we understand the difference between sigma and pi electrons, we remember that the pi bond is made up of loosely held electrons that form a diffuse cloud which can be easily distorted. This can be illustrated by comparing two types of double bonds, one polar and one nonpolar. The C=C double bond on the left below is nonpolar. Therefore the pi electrons occupy a relatively symmetric molecular orbital that’s evenly distributed (shared) over the two carbon atoms. The C=O double bond, on the other hand, is polar due to the higher electronegativity of oxygen. The pi cloud is distorted in a way that results in higher electron density around oxygen compared to carbon. Both atoms still share electrons, but the electrons spend more time around oxygen. The drawing on the right tries to illustrate that concept.
Using simple Lewis formulas, or even line-angle formulas, we can also draw some representations of the two cases above, as follows.
The dynamic nature of pi electrons can be further illustrated with the use of arrows, as indicated below for the polar C=O bond:
The CURVED ARROW FORMALISM is a convention used to represent the movement of electrons in molecules and reactions according to certain rules. We’ll study those rules in some detail. For now, we keep a few things in mind:
a) Curved arrows always represent the movement of electrons, not atoms.
b) Electrons always move towards more electronegative atoms or towards positive charges.
We notice that the two structures shown above as a result of “pushing electrons” towards the oxygen are RESONANCE STRUCTURES. That is to say, they are both valid Lewis representations of the same species. The actual species is therefore a hybrid of the two structures. We conclude that:
Curved arrows can be used to arrive from one resonance structure to another by following certain rules.
Just like pi electrons have a certain degree of mobility due to the diffuse nature of pi molecular orbitals, unshared electron pairs can also be moved with relative ease because they are not engaged in bonding. No bonds have to be broken to move those electrons. As a result, we keep in mind the following principle:
Curved arrows usually originate with pi electrons or unshared electron pairs, and point towards more electronegative atoms, or towards partial or full positive charges.
Going back to the two resonance structures shown before, we can use the curved arrow formalism either to arrive from structure I to structure II, or vice versa.
In case A, the arrow originates with pi electrons, which move towards the more electronegative oxygen. In case B, the arrow originates with one of the unshared electron pairs, which moves towards the positive charge on carbon. We further notice that pi electrons from one structure can become unshared electrons in another, and vice versa. We’ll look at additional guidelines for how to use mobile electrons later.
Finally, in addition to the above, we notice that the oxygen atom, for example, is sp2 hybridized (trigonal planar) in structure I, but sp3 hybridized (tetrahedral) in structure II. So, which one is it? Again, what we are talking about is the real species. The real species is a hybrid that contains contributions from both resonance structures. In this particular case, the best we can do for now is issue a qualitative statement: since structure I is the major contributor to the hybrid, we can say that the oxygen atom in the actual species is mostly trigonal planar because it has greater sp2 character, but it still has some tetrahedral character due to the minor contribution from structure II. We’ll explore and expand on this concept in a variety of contexts throughout the course.
What about sigma electrons, that is to say those forming part of single bonds? These bonds represent the “glue” that holds the atoms together and are a lot more difficult to disrupt. As a result, they are not as mobile as pi electrons or unshared electrons, and are therefore rarely moved. There are however some exceptions, notably with highly polar bonds, such as in the case of HCl illustrated below. We will not encounter such situations very frequently.
This representation better conveys the idea that the H–Cl bond is highly polar. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/06%3A_Electron_Delocalization_and_Resonance/6.01%3A_Mobility_of_Pi_Electrons_and_Unshared_Electron_Pairs.txt |
We now go back to an old friend of ours, CH3CNO, which we introduced when we first talked about resonance structures. We use this compound to further illustrate how mobile electrons are “pushed” to arrive from one resonance structure to another.
The movement of electrons that takes place to arrive at structure II from structure I starts with the triple bond between carbon and nitrogen. We’ll move one of the two pi bonds that form part of the triple bond towards the positive charge on nitrogen, as shown:
When we do this, we pay close attention to the new status of the affected atoms and make any necessary adjustments to the charges, bonds, and unshared electrons to preserve the validity of the resulting formulas. In this case, for example, the carbon that forms part of the triple bond in structure I has to acquire a positive charge in structure II because it’s lost one electron. The nitrogen, on the other hand, is now neutral because it gained one electron and it’s forming three bonds instead of four.
We can also arrive from structure I to structure III by pushing electrons in the following manner. The arrows have been numbered in this example to indicate which movement starts first, but that’s not part of the conventions used in the curved arrow formalism.
As we move a pair of unshared electrons from oxygen towards the nitrogen atom as shown in step 1, we are forced to displace electrons from nitrogen towards carbon as shown in step 2. Otherwise we would end up with a nitrogen with 5 bonds, which is impossible, even if only momentarily. Again, notice that in step 1 the arrow originates with an unshared electron pair from oxygen and moves towards the positive charge on nitrogen. A new pi bond forms between nitrogen and oxygen. At the same time, the pi electrons being displaced towards carbon in step 2 become a pair of unshared electrons in structure III. Finally, the hybridization state of some atoms also changes. For example the carbon atom in structure I is sp hybridized, but in structure III it is sp3 hybridized.
You may want to play around some more and see if you can arrive from structure II to structure III, etc. However, be warned that sometimes it is trickier than it may seem at first sight.
Here are some additional rules for moving electrons to write resonance structures:
1. Electron pairs can only move to adjacent positions. Adjacent positions means neighboring atoms and/or bonds.
2. The Lewis structures that result from moving electrons must be valid and must contain the same net charge as all the other resonance structures.
The following example illustrates how a lone pair of electrons from carbon can be moved to make a new pi bond to an adjacent carbon, and how the pi electrons between carbon and oxygen can be moved to become a pair of unshared electrons on oxygen. None of the previous rules has been violated in any of these examples.
Now let’s look at some examples of HOW NOT TO MOVE ELECTRONS. Using the same example, but moving electrons in a different way, illustrates how such movement would result in invalid Lewis formulas, and therefore is unacceptable. Not only are we moving electrons in the wrong direction (away from a more electronegative atom), but the resulting structure violates several conventions. First, the central carbon has five bonds and therefore violates the octet rule. Second, the overall charge of the second structure is different from the first. To avoid having a carbon with five bonds we would have to destroy one of the C–C single bonds, destroying the molecular skeleton in the process.
In the example below electrons are being moved towards an area of high electron density (a negative charge), rather than towards a positive charge. In addition, the octet rule is violated for carbon in the resulting structure, where it shares more than eight electrons.
Additional examples further illustrate the rules we’ve been talking about.
(a) Unshared electron pairs (lone pairs) located on a given atom can only move to an adjacent position to make a new pi bond to the next atom.
(b) Unless there is a positive charge on the next atom (carbon above), other electrons will have to be displaced to preserve the octet rule. In resonance structures these are almost always pi electrons, and almost never sigma electrons.
As the electrons from the nitrogen lone pair move towards the neighboring carbon to make a new pi bond, the pi electrons making up the C=O bond must be displaced towards the oxygen to avoid ending up with five bonds to the central carbon.
(c) As can be seen above, pi electrons can move towards one of the two atoms they share to form a new lone pair. In the example above, the pi electrons from the C=O bond moved towards the oxygen to form a new lone pair. Another example is:t
(d) pi electrons can also move to an adjacent position to make new pi bond. Once again, the octet rule must be observed:
One of the most common examples of this feature is observed when writing resonance forms for benzene and similar rings. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/06%3A_Electron_Delocalization_and_Resonance/6.02%3A_Using_Curved_Arrows--Additional_Concepts.txt |
The presence of alternating pi and sigma bonds in a molecule such as benzene is known as a conjugated system, or conjugated pi bonds. Conjugated systems can extend across the entire molecule, as in benzene, or they can comprise only part of a molecule. A conjugated system always starts and ends with a pi bond (i.e. an sp2 or an sp-hybridized atom), or sometimes with a charge. The atoms that form part of a conjugated system in the examples below are shown in blue, and the ones that do not are shown in red. Most of the times it is sp3 hybridized atoms that break a conjugated system.
Practically every time there are pi bonds in a molecule, especially if they form part of a conjugated system, there is a possibility for having resonance structures, that is, several valid Lewis formulas for the same compound. What resonance forms show is that there is electron delocalization, and sometimes charge delocalization. All the examples we have seen so far show that electrons move around and are not static, that is, they are delocalized. Charge delocalization is a stabilizing force because it spreads energy over a larger area rather than keeping it confined to a small area. Since electrons are charges, the presence of delocalized electrons brings extra stability to a system compared to a similar system where electrons are localized. The stabilizing effect of charge and electron delocalization is known as resonance energy.
Since conjugation brings up electron delocalization, it follows that the more extensive the conjugated system, the more stable the molecule (i.e. the lower its potential energy). If there are positive or negative charges, they also spread out as a result of resonance.The corollary is that
the more resonance forms one can write for a given system, the more stable it is. That is, the greater its resonance energy.
Examine the following examples and write as many resonance structures as you can for each to further explore these points:
Let’s look for a moment at the three structures in the last row above. In the first structure, delocalization of the positive charge and the pi bonds occurs over the entire ring. This becomes apparent when we look at all the possible resonance structures as shown below.
In the second structure, delocalization is only possible over three carbon atoms. This is demonstrated by writing all the possible resonance forms below, which now number only two.
Finally, the third structure has no delocalization of charge or electrons because no resonance forms are possible. Therefore, it is the least stable of the three. This brings us to the last topic. How do we recognize when delocalization is possible? Let’s look at some delocalization setups, that is to say, structural features that result in delocalization of electrons.
6.04: Delocalization Setups
There are specific structural features that bring up electron or charge delocalization. The presence of a conjugated system is one of them. Other common arrangements are:
(a) The presence of a positive charge next to a pi bond. The positive charge can be on one of the atoms that make up the pi bond, or on an adjacent atom.
(b) The presence of a positive charge next to an atom bearing lone pairs of electrons.
(c) The presence of a pi bond next to an atom bearing lone pairs of electrons.
6.05: Orbital Picture of Delocalization
The orbital view of delocalization can get somewhat complicated. For now we’re going to keep it at a basic level. We start by noting that sp2 carbons actually come in several varieties. Two of the most important and common are neutral sp2 carbons and positively charged sp2 carbons. Substances containing neutral sp2 carbons are regular alkenes. Species containing positively charged sp2 carbons are called carbocations. The central carbon in a carbocation has trigonal planar geometry, and the unhybridized p orbital is empty. The following representations convey these concepts.
Top and side view of a carbocation in Lewis and 3-D notation
Orbital view of a carbocation. The unhybridized p orbital is empty
A combination of orbital and Lewis or 3-D formulas is a popular means of representing certain features that we may want to highlight. For example, if we’re not interested in the sp2 orbitals and we just want to focus on what the p orbitals are doing we can use the following notation.
Let’s now focus on two simple systems where we know delocalization of pi electrons exists. One is a system containing two pi bonds in conjugation, and the other has a pi bond next to a positively charged carbon. We can represent these systems as follows.
Line angle and orbital picture of a simple conjugated system
Line angle and orbital picture of a system containing a carbocation next to a pi bond
If we focus on the orbital pictures, we can immediately see the potential for electron delocalization. The two pi molecular orbitals shown in red on the left below are close enough to overlap. Overlapping is a good thing because it delocalizes the electrons and spreads them over a larger area, bringing added stability to the system.
It is however time-consuming to draw orbitals all the time. The following representations are used to represent the delocalized system.
A similar process applied to the carbocation leads to a similar picture.
The resonance representation conveys the idea of delocalization of charge and electrons rather well.
Finally, the following representations are sometimes used, but again, the simpler they are, the less accurately they represent the delocalization picture.
There will be plenty of opportunity to observe more complex situations as the course progresses. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/06%3A_Electron_Delocalization_and_Resonance/6.03%3A_Delocalization_Conjugated_Systems_and_Resonance_Energy.txt |
All organic compounds are made up of at least carbon and hydrogen. The most basic type of organic compound is one made up exclusively of sp3 carbons covalently bonded to other carbons and hydrogens through sigma bonds only. The generic name for this family of compounds is alkanes. Alkanes are part of a more general category of compounds known as hydrocarbons. Some hydrocarbons such as alkenes and alkynes contain sp2 or sp-hybridized carbon atoms.
Alkanes are of great importance to the different classification systems and the naming of organic compounds because they consist of a carbon chain that forms the main structural unit of all organic substances. When an alkane carbon chain is modified in any way, even by the mere introduction of an sp2 carbon or a heteroatom (atoms other than carbon and hydrogen), is said to be functionalized. In other words, a functional group has been introduced and a new class of organic substances has been created.
A functional group is a specific arrangement of certain atoms in an organic molecule that becomes the center of reactivity. That is, it is the portion of the structure that controls the reactivity of the entire molecule and much of its physical properties. An entire classificaton system of functional groups is based on atom hybridization. Some of these functional groups are presented on page 2. They are the most commonly studied in introductory organic chemistry courses.
7.02: Alkane Nomenclature
Since alkanes are the most fundamental types of organic compounds, their structural features (a basic carbon chain, or skeleton) provide the basis for the nomenclature of all organic compounds. The earliest nomenclature systems followed almost no systematic rules. Substances were named based on their smell, or their natural source, etc. Many of those names are still in use today and are collectively known as common names. As organic chemistry developed and structures became more complex, a systematic method for naming organic compounds became necessary. The International Union of Pure and Applied Chemistry (IUPAC) is the organism that sets the rules for nomenclature of organic compounds today. Names that follow IUPAC rules are known as systematic names, or IUPAC names.
According to IUPAC rules, the first four alkanes are called methane, ethane, propane, and butane. They contain one, two, three, and four carbon atoms respectively, in a linear arrangement. Beginning with the fifth member of the series, pentane, the number of carbons is indicated by a greek prefix (penta, hexa, hepta, etc.). Most organic chemistry textbooks contain tables of at least the first ten members of the series, along with their structures and physical properties. Thus, the first two and most elementary rules for naming alkanes are to identify the length of the carbon chain, start the name with the appropriate greek prefix, and end the name with the suffix -ane. All this is obvious from the examples just given. Other rules are discussed next.
7.03: Overview of Functional Groups Based on Atom Hybridization
I. HYDROCARBONS
- Substances containing only carbon and hydrogen.
Alkanes C-C Only sp3 carbon present (technically not a functional group)
Alkenes C=C At least one $\pi$ -bond between two spcarbons present
Alkynes C $\equiv$ C At least one triple bond between two sp carbons present
II. ALKYL HALIDES, OR HALOALKANES
- Substances containing at least one C-X bond, where X=F, Cl, Br, or I.
ALKYL FLUORIDES C-F
ALKYL BROMIDES C-Br
ALKYL CHLORIDES C-Cl
ALKYL IODIDES C-I
III. GROUPS CONTAINING OXYGEN
- Both carbon and oxygen can be sp3 or sp2 hybridized, or a combination of both.
ALCOHOLS and ETHERS sp3 Oxygen
ALDEHYDES and KETONES sp2 Oxygen
CARBOXYLIC ACIDS , ESTERS , and ANHYDRIDES
These functional groups contain both sp3 and sp2 oxygen.
ACID, OR ACYL HALIDES The most common halogens used are chlorine and bromine
IV. GROUPS CONTAINING NITROGEN
- They may also contain other elements. For example amides also contain oxygen
AMINES (sp3 Nitrogen), AMIDES and NITRILES (sp Nitrogen)
Before proceeding, it is important to emphasize that beginning organic chemistry students must get used to seeing alkane chains from different angles, perspectives, and positions, as shown below using line-angle formulas for linear alkanes.
Hexane
Octane
BRANCHED ALKANES and ALKYL GROUPS.
When naming branched alkanes by IUPAC rules, identify and name the longest continuous carbon chain first. Then identify the branch, or branches. The branches are called alkyl groups. For example, a one carbon branch is called a methyl group. The names of alkyl groups are the same as those of analogous alkanes, except that their names end in -yl, instead of -ane. The following are examples of the most common alkyl groups encountered in introductory organic chemistry courses. Alky groups never exist by themselves. In alkanes they are always attached to a higher priority chain and are therefore sometime called substituents. The point of attachement is indicated by a dash.
In the following example, the longest continuous carbon chain has five carbons. Therefore the parent alkane is pentane. There is a methyl group attached to this chain. The molecule is then named methylpentane. Finally, the exact position of the methyl group is specified by numbering the main chain from the end closest to the methyl group. The complete IUPAC name for this alkane is 2-methylpentane.
In the following examples the alkyl groups are shown in red. Students must get used to line-angle formulas as early as possible.
Make sure you’ve identified the longest continuous carbon chain correctly. Otherwise you might end up with the wrong name. The longest continuous carbon chain doesn’t always have to be a horizontal row of carbons. It can take twists and turns. That’s why one must get used to visualize molecules from different angles and perspectives. The following example illustrates the correct name and an incorrect name based on a horizontal row of carbons.
The correct name above illustrates what’s done when there are several substituents of the same kind. First identify their positions, then use the prefixes di-, tri-, tetra-, etc. to indicate how many are present.
If there are substituents of different kinds present, name them in alphabetical order (e.g. ethyl before methyl). Prefixes such as di-, tri-, tetra-, etc. are ignored when alphabetizing.
If there are several options for choosing the longest continuous carbon chain, choose the one that yields the simplest name. That means:
(a) The longest continuous carbon chain has the greatest number of substituents, and
(b) The name has the lowest set of numbers indicating the substituent positions.
Complex alky groups (those which are branched themselves) are named as if they were alkanes, but the name ends in -yl and is enclosed in parenthesis. The carbon from which the substituent atttaches to the main chain is automatically number 1.
Although common alkyl group names such as isopropyl and t-butyl are commonly used as part of IUPAC names, strict IUPAC rules call for naming them as complex substituents. Common names are used because they are easier to say, shorter, and save paper and ink in scientific publications when they have to be used repeatedly in manuscripts.
ALKANES: COMMON NAMES
Like alkyl groups, alkanes can also have common names. Common, or “trivial” names are rarely used for straight chain alkanes, but are frequently used for branched alkanes. The following pages contain some terminology that organic chemistry students must memorize or become familiar with, due to the fact that it is widely used in textbooks and in the chemical literature.
ISOALKANES and ISOALKYL GROUPS
The iso structural unit consists of two methyl groups attached to a common carbon. When this unit is present in an alkane or alky group, the common name starts with the prefix iso.
When this unit is present in alkanes, the molecule is named according to the total number of carbon atoms present, but the name starts with the prefix iso. The smallest possible isoalkane is then isobutane.
The same rules apply to isoalkyl groups, but their names end in -yl. The smallest possible isoalky group is the isopropyl group, because alkyl groups are always attached to another carbon chain.
PRIMARY, SECONDARY, AND TERTIARY CARBONS.
Another bit of terminology associated with common names refers to the connectivity of sp3 carbons in alkanes and alkyl groups. A primary carbon is one that is covalently attached to only one other carbon. A secondary carbon is one attached to two other carbons. A tertiary carbon is attached to three other carbons. This definition implies that methane cannot have any such carbons, since it consists of only one carbon atom. Likewise, this terminology applies only to sp3 carbons. Other types are not defined in this way. In the following examples the carbons in question are indicated as dots.
The carbon atoms directly engaged π-bonding in alkenes are referred to as vinylic. An sp3 carbon directly attached to a vinylic carbon is referred to as allylic. Finally, an sp3 carbon directly attached to a benzene ring is referred to as benzylic.
Although it is emphasized that this is terminology used in common names, it is widely used and therefore important. It constitutes the basis for characterizing not only alkanes, but also other types of compounds where the main functionality is attached to a carbon that belongs to one of the types described. Thus we can have primary alcohols, secondary chlorides, allylic hydrogens, etc. based on the type of carbon to which they are attached.
This terminology is also used to identify the point of attachment of certain alkyl groups to the main chain of a molecule, based on the type of carbon from which the alkyl group connects to the main chain. For example four different butyl groups are possible, depending on their structure and their point of attachemnt to the main carbon chain. Two of them are named sec-butyl and tert-butyl (or t-butyl) because they connect to the main chain from a secondary and a tertiary carbon respectively.
The square represents the main carbon chain of a molecule
While there are four possible butyl groups, only two possible butanes exist. They are n-butane and isobutane (once again, these are common names). Likewise, there are three possible pentanes called n-pentane, isopentane, and neopentane.
The nomenclature concepts presented in the previous sections also apply to cycloalkanes. Please consult your organic chemistry textbook for examples and additional practice exercises.
7.04: Haloalkanes or Alkyl Halides
Like alkanes, alkyl halides, called haloalkanes by IUPAC rules, can have common names as well as systematic names. Common names are widely used because they are short and easy to say and write. Once you know what chloroform is, it’s easier to refer to it as such, rather than as trichloromethane.
COMMON NAMES -
Most common names refer to the main carbon frame as an alkyl group, and then add the halogen name, ending in -ide. Names such as isopropyl, or sec-butyl are extensively used. Likewise, primary halides have the halogen attached to a primary carbon, secondary halides have the halogen attached to a secondary carbon, etc. Refer to chapter 6 in your textbook. Examples are:
IUPAC NAMES -
The systematic naming of haloalkanes follows the same basic principles as that of alkanes. Find the longest carbon chain and number it trying to obtain the lowest set of numbers. Follow the alphabetical rule in naming the substituents, whether they are alkyl groups or halogens. Use prefixes such as di-, tri, as needed. The following examples illustrate these points.
If two substituents are equally far from either end, number the chain according to alphabetical order.
Name complex substituents using parenthesis and the guidelines discussed before for alkanes.
All these rules apply only if the substituents on the main carbon chain are alkyl groups and/or halogens. Other functional groups have different priorities, depending on their nature.
To name cyclohaloalkanes that contain halogen and alky substituents, number from the carbon bearing the halogen. Remember to indicate cis or trans when applicable. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/07%3A_Introduction_to_Organic_Chemistry/7.01%3A_Alkanes_Hydrocarbons_and_Functional_Groups.txt |
1. STRUCTURAL ISOMERS
2. CONFORMERS
3. NEWMAN PROJECTIONS & DIHEDRAL ANGLE
4. RELATIONSHIP BETWEEN STABILITY AND POTENTIAL ENERGY LEVEL IN MOLECULAR SYSTEMS
5. FACTORS THAT INCREASE POTENTIAL ENERGY (decrease stability) OF CONFORMERS
a) Steric factors (crowding of alkyl groups or other substituents)
b) Torsional strain (resistance of σ-bonds to freely rotate)
c) Angle strain in cycloalkanes and other ring systems (bond angles forced to depart from ideal values)
6. CONFORMATIONS OF n-BUTANE AND THEIR RELATIVE ENERGIES
a) Totally eclipsed
b) Gauche
c) Eclipsed
d) Anti
7. CYCLOALKANES
a) Factors that increase their potential energy
b) Relationship between ring strain, stability, and heat of combustion
8. CONFORMATIONS OF CYCLOHEXANE AND THEIR RELATIVE ENERGIES
a) Chair
b) Boat
c) Half-chair
d) Twist
9. AXIAL & EQUATORIAL BONDS IN CYCLOHEXANE
10. ENERGY DIFFERENCES BETWEEN AXIAL AND EQUATORIAL CONFORMATIONS IN MONOSUBTITUTED CYCLOHEXANES
11. STERIC MEANING OF 1,3-DIAXIAL INTERACTIONS
12. GEOMETRIC (cis and trans) ISOMERISM IN DISUBSTITUTED CYCLOHEXANE RINGS
Most of these topics are quite adequately covered in standard organic chemistry textbooks. A few of these topics will be further elaborated on here to illustrate certain concepts.
8.02: Conformational Analysis
Conformational analysis is the study of the different energy levels associated with the different conformations of a molecule. Conformations are the different 3-dimensional arrangements that the molecule can acquire by freely rotating around σ-bonds. One must keep in mind that conformations are not isomers. Isomers are different molecules. Conformations are simply different structural arrangements of the same molecule. The example below illustrates four out of an infinite number of conformations that the n-butane molecule can acquire by freely rotating around the C2-C3 bond, indicated in yellow. The red arrow indicates the direction of the rotation. The methyl carbons (C1 and C4) are indicated in red and all the hydrogen atoms in white. In this example C1, C2 and C3 remain stationary while C4 gradually moves down from top right to bottom right.
NEWMAN PROJECTIONS AND DIHEDRAL (TORSIONAL) ANGLE
The type of drawing used at the bottom of the figure in the last page is called a Newman projection, or representation. It is widely used to represent the conformations of open-chain alkanes that result as rotation around a sigma bond occurs. Another way to visualize Newman projections is to imagine them as the shadow a molecule would cast against a wall if a light beam was directed along the carbon-carbon axis under consideration.
A concept strongly associated with Newman projections is dihedral angle. It is defined as the angle between any two atoms attached to neighboring carbons in the Newman projection. For example, the two methyl groups above (in red) are attached to neighboring carbons. The angle between the two as seen in the Newman projection is zero (that is, they overlap, or” eclipse” each other). This is then the dihedral angle between those methyl groups.
CONFORMATIONAL ENERGY
The relationship between molecular structure and potential energy is a major area of study in organic chemistry. It has applications in conformational analysis because we are interested in relating the structure of conformers to their energy, and therefore to their relative stabilities. Remember that the relationship between potential energy and stability is inverse. The higher the potential energy of a system the lower its stability. Also remember that these are always relative concepts. There is no absolute energy or stability. They are always measured relative to a previously agreed upon standard.
There are three factors that increase the potential energy of conformers and therefore decrease their stability. They are:
a) Steric interactions - Crowding of alkyl groups or other substituents as they come too close together.
b) Torsional strain - Tendency of s-bonds to rotate in order to acquire a more stable conformation.
c) Angle strain - Increase in potential energy due to bond angles being forced to depart from ideal values in cycloalkanes and other rings.
STERIC INTERACTIONS
The term steric interactions refers to interference that occurs between atoms or groups of atoms by virtue of their size, or volume. When bulky groups or large molecules get too close to each other, steric interactions can become severe, raising the potential energy of the system. When these groups are allowed to drift far apart, steric interactions are relieved and the system gains stability.
Three representations of steric interactions, or “crowding,” between the two methyl groups in the eclipsed conformation of n-butane.
TORSIONAL STRAIN
This term is closely related to dihedral angle (also called torsional angle). Free rotation around sigma bonds changes the dihedral angle and therefore can bring bulky groups together or apart. When the dihedral angle is small and bulky groups interact (such as in the illustration above), there is a driving force for the molecule to rotate around the C2-C3 axis to relieve the steric interactions between the methyl groups. If this drive is impeded and the molecule is forced to acquire the eclipsed conformation, or a small dihedral angles between bulky groups, then torsional strain results. This might be the case for example in cyclic systems, where free rotation around sigma bonds is limited or impeded. Obviously, torsional strain increases the energy of a system and therefore decreases its stability.
As the molecule rotates around the C2-C3 bond, steric crowding between the methyl groups gets relieved. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/08%3A_Conformational_Analysis_of_Alkanes/8.01%3A_Important_Concepts.txt |
1. RELATIONSHIP BETWEEN SYMMETRY AND CHIRALITY
Asymmetric objects are chiral
Symmetric objects are achiral
2. RELATIONSHIP BETWEEN OBJECTS AND THEIR MIRROR IMAGES
Symmetric objects are superposable with their mirror images. They are one and the same. Asymmetric objects are nonsuperposable with their mirror images. They are different objects. In the case of molecules, chiral molecules and their mirror images are different molecules. Chiral molecules and their mirror images are a kind of stereoisomers called enantiomers.
3. DEFINITIONS
Stereoisomers - Compounds that have the same molecular formula and the same connectivity, but different arrangement of the atoms in 3-dimensional space. Stereoisomers cannot be converted into each other without breaking bonds.
Enantiomers - Nonsuperposable mirror images, or chiral molecules which are mirror images.
Chiral, or asymmetric carbon - A tetrahedral carbon atom bearing four different substituents.
Chirality centers, or stereocenters - Asymmetrically substituted atoms in a molecular structure. The most common type encountered in this course will be the chiral carbon described above.
Diastereomers - Stereoisomers which are not enantiomers (or mirror images).
Meso compounds, or meso forms - Symmetric, or achiral molecules that contain stereocenters. Meso compounds and their mirror images are not stereoisomers, since they are identical.
Optical activity - The ability of chiral substances to rotate the plane of polarized light by a specific angle.
Dextrorotatory - Ability of chiral substances to rotate the plane of polarized light to the right.
Levorotatory - Ability of chiral substances to rotate the plane of polarized light to the left.
Specific rotation - The measured angle of rotation of polarized light by a pure chiral sample under specified standard conditions (refer to textbook for a description of these).
Racemic mixture, racemic modification, or racemate - A mixture consisting of equal amounts of enantiomers. A racemic mixture exhibits no optical activity because the activities of the individual enantiomers are equal and opposite in value, therby canceling each other out.
Optical purity - The difference in percent between two enantiomers present in a mixture in unequal amounts. For example, if a mixture contains 75% of one enantiomer and 25% of the other, the optical purity is 75-25 = 50%.
Absolute configuration - A description of the precise 3-dimensional topography of the molecule.
Relative configuration - A description of the 3-dimensional topography of the molecule relative to an arbitrary standard. Absolute and relative configurations may or may not coincide
4. RELATIONSHIPS BETWEEN CHIRAL CENTERS AND CHIRAL MOLECULES - The term chiral center refers to an atom in the molecular structure. The term chiral molecule refers to the entire molecule.
The presence of one chiral center renders the entire molecule chiral. The presence of two or more chiral centers may or may not result in the molecule being chiral. In the examples given below the chiral centers are indicated with an asterisk. The vertical broken line represents a plane of symmetry.
5. RELATIONSHIPS BETWEEN CONFORMATIONS AND CHIRALITY - The primary criterion to determine molecular chirality is the absence of any symmetry elements. However, some achiral molecules have chiral conformations. For example the chair conformations of 1,2-disubstituted cyclohexanes are chiral, yet the molecule as a whole is considered achiral. On the whole, we can apply the following criteria.
a) If the contributing conformations average out to an achiral conformation, then the molecule is considered achiral. Such molecules do not show optical activity. In the case of 1,2-disubstituted cyclohexanes the two most stable conformations are chiral. If we could freeze and isolate one of them, it would exhibit optical activity. But because they are mirror images in equilibrium, their optical activities cancel out and the sample is optically inactive. A similar example is illustrated by the conformations of (2R,3S)-1,2-dichlorobutane, which again is achiral, even though some of its conformations are chiral.
(2R,3S)-2,3-dichlorobutane
b) If a chiral conformation prevails over the others, then the molecule is considered chiral and it will show optical activity. The most common situations of this type involve molecules which are locked up into a chiral conformation due to steric interactions that impede free rotation around sigma bonds. In the example shown below, the two benzene rings cannot be coplanar because the steric interactions between the methyl and chlorine groups are too severe. The molecule is locked up in a conformation that has no symmetry, therefore it is chiral. Also notice that the molecule does not have any chiral centers. Its chirality is strictly due to a conformational effect. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/09%3A_Supplementary_Notes_for_Stereochemistry/9.01%3A_Some_Important_Concepts_in_Stereochemistry.txt |
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9.02: R
The complete set of rules is given in the textbook, but here are some things to keep in mind when assigning configuration to chirality centers.
1. Make sure you have chiral centers in the molecule. The fact that a 3-dimensional formula is given does not imply that there are chiral centers.
2. Assign priorities to the atoms directly attached to the chirality center. The highest priority goes to the atom with the highest atomic number. In case there are isotopes, use the mass number instead, since they have the same atomic number.
Notice that since the atomic number of hydrogen is 1, it will always be the lowest priority group, as long as it is present.
3. If two or more of the atoms directly attached to the chiral center are of the same type, look at the next atom to break the tie. Do not do this unless there is a tie. Repeat this process until the tie is broken.
It is important to emphasize that in trying to break ties, one looks at the atoms directly attached to the element under observation before looking at any others. Study the examples on the following page very carefully to make sure this point is clear.
4. If there are atoms containing double or triple bonds, count them twice or thrice respectively. This holds for each of the atoms involved in the double or triple bonding.
5. Although not obvious from the above examples, when duplicating the atoms involved in double or triple bonding they are also being crossed over at the same time. This only becomes apparent when the atoms involved in multiple bonding are not of the same kind, as in the examples shown on the following page.
9.03: Shortcuts for Assigning Absolute Configuration on Paper
According to the Cahn-Ingold-Prelog convention, when assigning absolute configuration to a chiral carbon the lowest priority group that’s attached to that carbon must be pointing away from an observer who is looking at the carbon in question. On paper, that usually means that if the observer is the person looking at the page, then the lowest priority group is pointing away from the observer, going behind the plane of the paper. In a 3-D formula this is indicated thus:
When the formula is given to us in this way, it’s easy to assign configuration. All we have to do is assign priorities to the other three substituents and see if they are arranged clockwise or counterclockwise when the observer follows them in order of decreasing priorities. We don’t’ have to mentally reposition either ourselves or the molecule in any way. In the example given above we can see that the central carbon has the (S) configuration.
If the lowest priority group is not presented to us already positioned towards the back of the chiral carbon, then it is useful to remember the following basic principle:
Every time any two substituents are exchanged, the opposite configuration results
With this in mind, we can encounter two possible scenarios: Either the lowest priority group is positioned in front of the chiral carbon, or on the plane of the paper. If the lowest priority group is positioned in front of the chiral carbon (that is, opposite where it should be, according to the rules) we can still assign configuration by following the arrangement of the other three groups as given to us, but the configuration we obtain will be the opposite of the actual one. Following the same example given above we have:
With the lowest priority group positioned in the front rather than towards the back, the central carbon appears to have the (R) configuration. The actual configuration is therefore (S). This is best seen when we rotate the molecule until the H atom is in the correct orientation.
If the lowest priority group is positioned on the plane of the paper, we can momentarily exchange it with whatever group happens to be positioned in the back, then assign configuration, then reverse it.
ASSIGNING ABSOLUTE CONFIGURATIONS IN CYCLIC MOLECULES
Cyclic molecules are frequently represented on paper in such a way that the ring atoms are all lying on the plane of the paper, and substituents are either coming out of the paper towards the front or towards the back. It is therefore easy to assign configuration to any chiral centers forming part of the ring, since the lowest priority substituent will be either pointing to the front or to the back. However, always make sure there is in fact a chiral center present. The fact that a 3-D representation is given does not necessarily mean there is a chiral center in the molecule.
ASSIGNING ABSOLUTE CONFIGURATIONS IN FISCHER FORMULAS
The key points to keep in mind regarding Fischer projection formulas are:
1. Horizontal lines represent bonds to the chiral carbon that are coming out of the plane of the paper towards the front, whereas vertical lines represent bonds going behind the plane of the paper towards the back. Thus, Fischer formulas are easily translated into “bow tie” formulas, which are 3-D formulas.
2. The lowest priority group bonded to the chiral carbon must always be shown as a horizontal bond.
The process of assigning (R) or (S) configuration to the chiral carbon is the same as outlined before, but since the lowest priority group is pointing towards the front, the configuration obtained directly from a Fischer formula is the opposite of the actual one.
The order of priorities follows a clockwise direction in the Fischer formula. Therefore the actual configuration of this molecule is (S).
Once we know the actual configuration, we can represent the molecule in any of several possible ways using 3-D formulas. Thus the formulas shown below all represent the same molecule as given above in Fischer projection form. That is to say, all have the (S) configuration at the central carbon. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/09%3A_Supplementary_Notes_for_Stereochemistry/9.02%3A_R/S_Nomenclature_System_%28CahnIngoldPrelog_convention%29.txt |
Once we become skilled at assigning configuration to chiral centers represented in 3-D notation, we can easily translate those into Fischer formulas. All we have to do is draw the cross with the four substituents attached to the chiral carbon, making sure the lowest priority group is lying on a horizontal line. At first we don’t need to worry about where the groups are, for if it’s wrong, all we have to do is exchange any two groups (make sure the lowest priority group remains on a horizontal line) to change configuration.
Just as was shown before using 3-D formulas, the same molecule can be represented in Fischer notation in several different ways, all showing the same molecule. In the example below all the compounds represent the same (S) isomer.
Notice that we did not forget to always put the lowest priority group on a horizontal bond.
FISCHER NOTATION OF MOLECULES CONTAINING TWO OR MORE CHIRAL CENTERS
Open chain molecules containing two or more chiral centers are traditionally represented in Fischer notation showing the main carbon chain in a vertical arrangement. The D-(+)-glucose molecule below illustrates this convention, shown in both Fischer projection and the bow tie equivalent. Notice that in the latter only the end atoms have bonds shown with broken wedge lines. Also notice that all the hydrogen atoms bonded to chiral carbons are shown lying along horizontal lines.
Identifying planes of symmetry in Fischer formulas is relatively easy, since they are planar representations. The following illustrations show examples of both chiral and achiral molecules.
The process of assigning absolute configurations to the chiral centers in molecules containing two or more of them is basically an extension of the process followed for molecules containing only one. However, it helps to isolate the chiral centers and deal with one at a time to avoid confusion.
The following example illustrates this point. In this example we have numbered the carbon atoms in the main chain according to IUPAC rules that will be studied later. We have also marked the chiral carbons with asterisks.
Once the chiral centers have been identified, we focus on one at a time (shown as a red dot). First, we isolate carbon-2, then we assign priorities to the groups bonded to it, and assign configuration. In this case the configuration of carbon-2 turns out to be (R). A similar process for carbon-3 also leads to (R) configuration.
Carbon-2 (red dot) has the (R) configuration
Carbon-3 (red dot) also has the (R) configuration
The IUPAC notation used to indicate the configurations at carbons 2 and 3 is therefore (2R, 3R)
9.05: Assigning Configuration to Conformationally Mobile Systems
This is probably one of the trickiest situations to deal with, especially when the molecule is shown to us in conformations such as a cyclohexane chair, or represented by Newman projections. It is a good idea in these cases to work with models, because one cannot help but to turn the molecule around until the of the lowest priority groups are positioned where they should be, and the priorities of the groups attached to the chiral centers can be clearly seen.
In the case of cyclohexane chairs and other rings, it’s a good idea to flatten the ring and position it on the plane of the paper, with the lowest priority groups pointing towards the back when possible.
In the case of Newman projections, it helps to rotate one of the carbons around the C–C bond under consideration until as many similar groups as possible are aligned (eclipsing each other), then rotate the structure sideways to obtain a side view, rather than a projection, then assign priorities and configuration.
9.06: Molecules With 2 or More Chiral Centers- Diastereomers and Meso Forms
As was stated before, molecules containing 2 or more chiral centers may or may not be chiral themselves. Let’s consider the case of achiral molecules first. Molecules that contain two or more chiral centers and at least one plane of symmetry are called meso forms. Cis-disubstituted cyclohexanes are examples of meso forms. Since these molecules have symmetry, they cannot have enantiomers. They are one and the same with their mirror images.
Meso forms can also be open chain, as illustrated below.
Now let’s consider the case of chiral molecules that contain two or more stereocenters. Such molecules can have enantiomers because they are not the same as their mirror images.
Trans-1,2-dichlorocyclohexane. The mirror images are different compounds.
Chiral molecules with two or more chiral centers can also have stereoisomers which are not their mirror images. Such sets of stereoisomers are called diastereomers. In the following example the two molecules shown are stereoisomers (same connectivity but different spacial arrangement of the atoms) but they are not mirror images. Their relationship is one of diastereomers.
Cis- and trans-1,2-dichlorocyclohexane. are examples of diasteromers
As a corollary, we can state that cis/trans pairs of disubstituted cyclohexanes (or any other rings for this matter) are always diastereomers. Notice that we have referred to such sets before as geometric isomers. Geometric isomers are in fact a subcategory of diastereomers.
The following example is an illustration of open chain molecules with a diastereomeric relationship.
Also notice that in the above examples one of the members in each pair is chiral and the other is not. Diastereomeric sets are frequently made up of molecules where one of the molecules is chiral and the other is not. Also notice that although they are not mirror images, part of their structures do mirror each other. It is frequently the case that one half of one molecule mirrors one half of the other one, but the other halves are identical.
In this pair of cis/trans isomers, the top half of one molecule mirrors the top half of the other one (the chiral centers have opposite configurations), while the bottom halves are the same (the chiral centers have the same configuration).
SUMMARY OF RELATIONSHIPS BETWEEN MOLECULES WITH TWO OR MORE CHIRAL CENTERS
If n = number of chiral centers, the maximum possible number of stereoisomers is 2n
EXAMPLE 1: Possible combinations for 2,3-dibromobutane.
EXAMPLE 2: Possible combinations for 2-bromo-3-chlorobutane.
The diagram below shows the possible combinations of configurations for molecules with 2 chiral centers. Since each chiral center can be (R) or (S), the possible combinations are (R,R), (S,S), (R,S), (S,R).
If two molecules are mirror images, then their configurations are exactly opposite and they are enantiomers (E). If they are not mirror images but still they are stereoisomers then they are diastereomers (D). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/09%3A_Supplementary_Notes_for_Stereochemistry/9.04%3A_Translating_3-D_Formulas_Into_Fischer_Formulas.txt |
The introductory part of the organic chemistry course has three major modules: Molecular architecture (structure), molecular dynamics (conformational analysis), and molecular transformations (chemical reactions). An understanding of the first two is crucial to an understanding of the third one. The rest of the organic chemistry course will be largely spent on studying different kinds of reactions and mechanisms. A summary of key concepts follows.
1. MOLECULAR ARCHITECTURE - Basic principles of molecular structure.
a) Atomic structure
b) Orbitals and hybridization
c) Covalent bonding
d) Lewis structures and resonance forms
e) Isomerism, structural and geometric isomers
f) Polarity, functional groups, nomenclature systems
g) Three-dimensional structures, stereochemistry, stereoisomers.
2. MOLECULAR DYNAMICS - Basic principles of molecular motion involving rotation around single bonds and no bond breakage. The focus is on conformations and their energy relationships, especially in reference to alkanes and cycloalkanes (conformational analysis).
a) Steric interactions
b) Torsional strain and Newman projections
c) Angle strain in cycloalkanes
d) Conformations of cyclohexane and terminology associated with it
3. MOLECULAR TRANSFORMATIONS - This is the part that comprises the bulk of organic chemistry courses. It is the study of chemical reactions and the principles that rule transformations. There are three major aspects of this module. In organic chemistry I we will focus largely on the first two, and leave the study of synthetic strategy for later.
a) Reaction mechanisms - Step by step accounts of how electron movement takes place when bonds are broken and formed, and the conditions that favor these processes (driving forces). An understanding of the basic concepts of thermodynamics and kinetics is important here.
b) Energetics - An account of energy changes and/or requirements that take place during the course of a transformation.
c) Synthetic strategy - An understanding of the specific chemistry of functional groups and how it can be used to design molecules with specific structural features.
10.02: Energy Profiles and Reaction Profiles
A reaction profile is a plot of free energy (y-axis) vs. reaction progress (x-axis, or time axis). It is basically a movie played frame by frame. The x-axis represents sequential time events, or stages of the reaction. The y-axis represents the free energy associated with the structural changes taking place during the transformation.
As an example, consider the potential energy diagram of n-butane for rotation around the C2-C3 axis. For each stage during the rotation sequence (x-axis) there is a potential energy level associated with the particular structure that results from the rotation.
As an analogy (and an excuse to have a little fun), play a car chase sequence from an action movie frame by frame. The “low energy” images (e.g. when the chase is just beginning) have more definition than the “high energy” images (at the height of the chase), which appear more blurred.
10.03: Factors That Affect Potential Energy
The potential energy level of a structure or a system reflects its stability under ordinary conditions (e.g. standard temperature and pressure).The relationship is inverse. The higher the potential energy, the lower the stability of the system, and viceversa. There are many factors that affect the potential energy of a chemical structure or system. They can roughly be divided into those due to mass interactions and those due to electronic (charge) interactions.
Examples of mass interactions are steric effects. Whenever bulky groups get too close to each other a source of strain is introduced in the system and the potential energy goes up. Torsional strain is an instance of this type of interaction. Examples of electronic interactions are the repulsions between covalent bonds and lone electron pairs that compel the molecule to acquire a specific geometry in which all these entities are as far appart from each other as possible. Whenever we force these bonds to be closer to each other than in the optimal geometry, a source of strain is introduced.
Examples of charge (electrostatic) interactions are resonance structures where like charges are too close to each other. The repulsions between like charges raises the potential energy of the structure, making it into a minor contributor to the hybrid. On the other hand, the solvation process, where the partial dipoles of the solvent interact with those of the solute, but of opposite sign, results in a favorable interaction that stabilizes the system (lowers its potential energy). Hydrogen bonding is a similar interaction that brings added stability to structural arrangements such as the DNA helix and many proteins.
Resonance (electronic and charge delocalization) stabilizes a system by spreading charges and electrons over a larger area as compared to a system where electrons and charges are localized. Delocalization lowers the potential energy of the system. Finally, many chemical bonds form because the process of bond formation lowers the energy of the system compared to the isolated atoms. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/10%3A_Intro_to_Theory_of_Chemical_Reactions/10.01%3A_Background.txt |
The simplest and first process that takes place during a chemical transformation is the breaking of a bond. The following discussion is confined to covalent bonds because they are the most prevalent in organic molecules. Breaking bonds always requires an energy input. A free energy diagram for the breaking of a covalent bond between atoms A and B would have the following appearance.
If we were to visualize the breaking of this bond “frame by frame,” we would see the bond stretching gradually until it snapped. At the same time we would see a gradual increase of the free energy of the system.
The process of bond formation, being the opposite of bond breaking, always releases energy. That is, lowers the energy of the system.
NOTE: This discussion assumes that the reader is familiar with basic thermodynamic concepts such as free energy, enthalpy, entropy, etc. Refer to the appropriate sections in your textbook for detailed explanations (Sections 4-4 and 4-5 in the Wade textbook, 4th and 5th editions)
10.05: Transition States
One of the simplest types of organic reactions is the Sn2 reaction (Ch. 6 in the Wade textbook). In this reaction a bond is being broken as another bond is formed almost simultaneously. Atom A moves towards molecule B-C. As A bonds to B, C is breaking away.
$A + B-C \rightarrow A-B + C$
If we were to view this process “frame by frame,” we would see an energy diagram showing each stage of the process and the corresponding energy levels associated with each stage of the process. This is one of the simplest types of reaction profiles.
A point of particular interest in this reaction profile is the energy maximum. The structural species that exists at this point is called a transition state (shown in red). It is customary to enclose this structure in brackets, with a double dagger at the top right. The transition state represents a point of such high energy that it cannot possibly exist for any length of time. As an analogy, think of a rubber band that gets stretched to the point where it’s about to snap. The state of maximum energy occurs a fraction of a millisecond before it snaps. The rubber band cannot stay in that state for any length of time. Either it relaxes and goes back to its original state, or it snaps. Either way it must go down to a more stable state. Therefore, transition states cannot be physically or experimentally observed, much less studied. Their existence and their structure must be inferred from other information, especially that which pertains the surrounding states.
10.06: Covalent Bond Cleavage- Outcomes and Reaction Intermediates
What we refer to as the reactants and the products of a chemical reaction represent stable substances that can exist for a finite period of time under standard conditions. They can typically be kept in containers without significant decomposition as long as appropriate measures are observed. In other words, the beginning and the end of a chemical reaction (reactants and products), represent the most stable states that can exist in reference to the reaction profile. Any other intermediate structures or states must be of higher energy, or lower stability. Two types of points in particular merit special attention: energy maxima and energy minima. We’ve already discussed energy maxima above in relation to transition states.
An energy minimum that occurs between reactants and products represents a reaction intermediate. Reaction intermediates represent structures that do have some degree of stability, but not as much as reactants or products. They can live long enough that they can in principle be detected, observed, and even studied under the right conditions. However they are not stable enough to keep for indefinite periods of time.
In organic chemistry we are mainly concerned with three types of reaction intermediates that result from breaking a covalent bond in two different ways, referred to as homolytic cleavage and heterolytic cleavage. In the former, the electrons that make up the covalent bond get equally distributed between the atoms that make up the bond. In heterolytic cleavage, one of the two atoms gets the two electrons and usually develops a negative charge. The other atom gets no electrons and typically develops a positive charge. Homolytic cleavage gives rise to the formation of free radicals, that is, neutral species that carry an unpaired electron. Heterolytic cleavage gives rise to the formation of ions, that is, species that carry a positive or a negative charge.
In the curved arrow formalism, the movement of single electrons is indicated by half-headed arrows. The movement of electron pairs is indicated by full-headed arrows. The following examples illustrate the outcomes of homolytic and heterolytic bond cleavage.
When the atom in question is a positively charged carbon, the resulting species is called a carbocation. If it is negatively charged, it is called a carbanion. The following represent all three species with carbon as the central atom.
These three species are the most common types of reaction intermediates that occur in organic reactions. Again, they occur at energy minima located between reactants and products in the reaction profile. The hybridization state of the central carbon in each of these intermediates is important. This and other structural features are further discussed in your organic chemistry textbook (Section 4-16 in the Wade textbook). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/10%3A_Intro_to_Theory_of_Chemical_Reactions/10.04%3A_Bond_Breaking_and_Bond_Formation.txt |
The amount of energy required to break a covalent bond homolytically is referred to as bond dissociation energy (BDE). By definition, BDEs are always positive, since they represent the change in free energy resulting from the breaking of a bond, and breaking bonds always requires an enegy input.
10.08: Classification of Organic Reaction Mechanisms
All chemistry, except for nuclear reactions, involves electron transfers. Reaction mechanisms are step by step accounts of how electrons are transferred when bonds are broken and formed. Therefore they embody the rules of electron movement during chemical reactions. The following chart shows how mechanisms are classified depending on whether electrons are transferred singly or in pairs, and also according to other older criteria which nonetheless are still in wide use.
10.09: Activation Energies
The majority of chemical transformations do not amount to only breaking or forming bonds. The two processes go together, either simultaneously or sequentially. Since breaking bonds requires energy and forming bonds returns energy, most chemical reactions proceed with either a net output or a net input of energy. A few reactions can occur with no net energy imbalance either way.
Lilkewise, bond breaking typically leads (runs ahead of) bond formation. An old bond must be at least partially broken before a new one can begin to form. The energy cost of breaking the old bond must be supplied before we can get a return from bond formation. This is the reason for the activation energy requirement of many chemical reactions. The more the bond formation process lags behind bond breaking, the higher the activation energy requirement. However, one must be careful not to try to relate bond dissociation energies with activation energies. Unless an extremely simple process involving small atoms or molecules is under consideration, the two parameters are very different. One such process might be the breaking of the H-H bond in the hydrogen molecule to produce hydrogen atoms, or the breaking of the Cl-Cl bond in the chlorine molecule to produce chlorine atoms. Provided these processes are taking place in the gas phase and the bonds are being broken homolytically, the activation energy is the same as the bond dissociation energy.
But a large number of chemical reactions take place in solution, or with bond formation taking place as bond breaking is also occurring. A transition state is reached as two (or more) reactants approach each other and their potential energy rises to a maximum. At the potential energy maximum they have become so closely associated in a configuration that is so highly disorted that any further change will cause them to either revert to reactants, or to form products. Any factors (other than the bond dissociation energies) that stabilize or destabilize the transition state, such as involvement of the solvent (solvation) or the development (or relief) of steric strain can affect the energy level associated with this transition state. In a single step reaction the energy associated with the transition state is also the activation energy. In a multistep reaction, where several transition states are possible for each step, the energy associated with the highest transition state is the activation energy of the overall reaction. The activation energy can also be viewed as a kinetic parameter if it is defined as the minimum kinetic energy with which the reactants should approach each other for reaction to occur.
We will come back to further elaborate on some of these points as we expand on the study of different types of reactions and examine the meaning of the Hammond postulate. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/10%3A_Intro_to_Theory_of_Chemical_Reactions/10.07%3A_Bond_Dissociation_Energies.txt |
Having identified basic sites as areas of high electron density and acidic protons as hydrogen atoms of low electron density, we can now establish how curved arrows are used to indicate the movement of electrons in acid-base reactions. The rules are basically the same as for resonance structures.
(a) Full headed arrows indicate electron pair movement.
(b) The arrow always originates at the basic site, i.e. the source of electrons (typically nonbonding electrons, but can also be a p-bond).
(c) The arrow indicates a newly formed σ-bond between the basic site of one molecule and the acidic proton of another molecule. As this new bond is being formed, the atom (or group of atoms) to which the acidic proton is attached leaves as the conjugate base. Sometimes this atom or group of atoms is referred to as a leaving group.
SINCE STRONG ACIDS HAVE WEAK CONJUGATE BASES, THE BEST LEAVING GROUPS ARE WEAK BASES. IN OTHER WORDS, EQUILIBRIUM FAVORS DISPLACEMENT OF THE WEAKER BASE BY THE STRONGER BASE.
The red arrow originates at the electron source and moves towards the acidic proton. It indicates a new bond that forms between the oxygen and the acidic proton in HCl.
As oxygen bonds to the acidic proton, chlorine leaves with the electrons as chloride ion, the conjugate base of HCl. It can therefore be referred to as the leaving group.
In this case, water acts as the leaving group.
11.02: Using Resonance Structures to Predict Relative Reactivities of Basic Sites
Sometimes a molecule can have several atoms which could be potentially basic. The question then is, is one more basic than the other, or are they of equal basicity? Sometimes resonance structures can reveal relative reactivities that are not apparent from the main, or most stable contributor. An example is acetic acid, which contains two oxygen atoms that could in principle become protonated by acid. The problem can be approached in two ways. One is to look at the relative stabilities of the conjugate bases. It is shown below that the conjugate base that results from protonation of oxygen (a) is resonance stabilized, whereas the one that results from protonation of oxygen (b) is not. Based on this criterion, one predicts that oxygen (a) is more basic than oxygen (b), and that protonation will occur predominantly at oxygen (a).
One can arrive at the same conclusion by examining the resonance structures of acetic acid, which reveal that oxygen (a) carries a partial negative charge and that oxygen (b) carries a partial positive charge. The more negative character of (a) makes it the more basic of the two.
Resonance structures can be used in conjunction with other information, such as periodic trends, to make a prediction. In the example below, the two resonance structures of the cyclohexanone enolate ion show carbon and oxygen sharing the negative charge. However, basicity trends in the periodic table predict that in comparable situations, carbon is more basic than oxygen. Remember that we’re not looking at electronegativity trends here, but basicity, which runs opposite to electronegativity across the same row. The molecule will react with acids to favor protonation of the carbon over the oxygen atom.
Writing resonance structures involving σ-bonds is not as common as doing it with π-bonds. However it’s perfectly acceptable to do it in the case of highly polarized σ-bonds, such as might be the case in H-Cl.
Although fairly obvious to a chemistry student, the structure on the right makes the acidic nature of the hydrogen atom apparent. The same can be done with bases. One way to have a highly polarized σ-bond involving carbon is by bonding it to a metal. The higher electronegativity of carbon biases the electron density distribution in its favor. In this case, resonance structures reveal the negative character of such carbon, and therefore its basic character.
11.03: Alkenes as Weak Bases Outcomes of Protonation at Pi Bonds
As stated before, the double bond in alkenes is a source of electrons. Alkenes are weak bases because the π-electrons are only available after breaking the π-bond first. Nonetheless, alkenes are capable of becoming protonated by strong acids. The most important principle to keep in mind is that with unsymmetrical alkenes, formation of the most stable carbocation is the preferred outcome.
Protonation of the unsymmetrical alkene favors formation of the most stable carbocation. In this example the tertiary cation is favored over the secondary one.
Can you show the electron movement taking place here and the exact location of the new C-H bond in the carbocation that forms? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/11%3A_Bronsted_Acid-Base_Chemistry/11.01%3A_Electron_Movement_in_Ionic_Mechanisms.txt |
pKa VALUES OF COMMON ORGANIC AND INORGANIC ACIDS
R = alkyl group
Acid pKa Conjugate Base
-10
HBr -9
HCl -7
HO-SO3H (H2SO4) -10
-2
-1.7
0
HF 3
5
4-5
5
H2CO3 6.4
H2S 7
HCN 9.2
9
9.2 H3N
10
RSH 11
H2O 15.7
ROH (alcohols) 16-17
20
25
25
NH3 38
ArCH3 40
44
~ 48–53
11.05: Hydrogen Atoms and Protons in Organic Molecules
A hydrogen atom that has lost its only electron is sometimes referred to as a proton. That is because once the electron is lost, all that remains is the nucleus, which in the case of hydrogen consists of only one proton.
The large majority of organic reactions, or transformations, involve breaking old bonds and forming new ones. If a covalent bond is broken heterolytically, the products are ions. In the following example, the bond between carbon and oxygen in the t-butyl alcohol molecule breaks to yield a carbocation and hydroxide ion.
The full-headed curved arrow is being used to indicate the movement of an electron pair. In this case, the two electrons that make up the carbon-oxygen bond move towards the oxygen. The bond breaks, leaving the carbon with a positive charge, and the oxygen with a negative charge. In the absence of other factors, it is the difference in electronegativity between the two atoms that drives the direction of electron movement. When pushing arrows, remember that electrons move towards electronegative atoms, or towards areas of electron deficiency (positive, or partial positive charges). The electron pair moves towards the oxygen because it is the more electronegative of the two atoms.
If we examine the outcome of heterolytic bond cleavage between oxygen and hydrogen, we see that, once again, oxygen takes the two electrons because it is the more electronegative atom. Hydrogen is left with only a positive charge. In other words, it becomes a proton.
11.06: Bronsted-Lowry Acids and Acidic Protons
A hydrogen bonded to a very electronegative atom makes for a highly polar bond. The dipole moment favors electron density around the more electronegative atom, leaving the hydrogen with a partial positive charge. This bond is different from other bonds in the molecule because of its propensity to break into a negative ion and a positive hydrogen ion. This propensity is driven by the tendency of the more electronegative atom to take up the electrons that make up the covalent bond. In fact, one can write resonance structures for such molecule that show the bond in question already broken. The t-butyl alcohol molecule can be used again to illustrate this point.
The greatest contributor to the hybrid is obviously structure I because it is neutral. Structure II has charge separation and therefore is a minor contributor. However, the significance of structure II is that it shows the negative character of the oxygen and the positive character of the hydrogen, and therefore the polarity of this bond. A better representation of the hybrid could be structure III, which shows the oxygen with partial negative character, and the hydrogen with partial positive character.
In the Bronsted-Lowry theory of acids and bases, an acid is a hydrogen ion donor, or proton donor, and a base is a hydrogen ion acceptor, or proton acceptor. Hydrogen atoms that have a substantial degree of partial positive charge (i.e. low electron density around them) are commonly referred to as acidic protons. In the example above, the hydrogen bonded to oxygen is considered to be acidic, and the molecule as a whole is considered a Bronsted acid because it has a propensity to release a hydrogen ion, or proton. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/11%3A_Bronsted_Acid-Base_Chemistry/11.04%3A_pKa_Table.txt |
When a Bronsted acid (or simply acid) reacts with a Bronsted base (or simply base) a proton is transferred from the acid to the base. This results in formation of another acid, called the conjugate acid, and another base, called the conjugate base. For example, when hydroxide ion (a base) reacts with hydrogen chloride (an acid), a new acid (water) is formed. Water is then the conjugate acid of hydroxide ion. Likewise, a new base (chloride ion) is formed. Chloride ion is then the conjugate base of hydrogen chloride. The reaction is an equilibrium process because the new acid and the new base can react together to revert to the original reactants. Therefore we can also say that hydroxide ion is the conjugate base of water, and that hydrogen chloride is the conjugate acid of chloride ion. This relativity of concepts is characteristic of the Bronsted-Lowry acid-base theory.
Conjugate acid-base pairs differ only by a proton. Other examples of conjugate acid-base pairs are H2O / H3O+ and NH3 / NH4+. The above reaction also shows the direction of electron movement. In acid-base reactions, electron movement always originates at the base and moves towards the acidic proton in the acid. The base is the electron-rich species. We can identify bases because they usually have an atom with unshared electron pairs (lone pairs). Sometimes this atom also carries a negative charge, but this is not a requirement. Likewise, we can identify the acid because it is the molecule that has acidic protons (hydrogens that carry a strong partial positive charge). The following are examples of other acids and bases. The acidic protons are shown in red, and the basic atoms in blue. Keep in mind that the concept of acid or base is always relative in the Bronsted theory. Some molecules such as water can act as acids or as bases, depending on who they are reacting with. We will expand on this later.
11.08: The Scale of Acidity- pKa Values
Many acid-base reactions take place in water, one of the most universal solvents. Water also has the dual capability of acting as a proton donor or as a proton acceptor. It makes sense, then, to develop a scale of acidity based on the behavior of the substance of interest towards water. Since most acid-base reactions are equilibrium processes, the equilibrium constant of the reaction between an acid (or base) and water forms the basis for the pKa scale. Most general and organic chemistry textbooks contain adequate discussions of this parameter. What is of interest to us here is the relationship between pKa and acidity. The equation
$pK_{a} = -log K_{a} = log(\frac{1}{K_{a}})$
shows that such relationship is inverse. The stronger the acid (i.e. the higher its acidity constant Ka), the lower its pKa value, and viceversa. Tables of pKa values usually show the acids and their conjugate bases arranged by order of decreasing (or increasing) acidity. Chemistry students should become proficient at reading and using data presented in such tables.
11.09: Predicting Equilibrium in Acid-Base Reactions
Equilibrium in acid-base reactions always favors the weaker side. In the following example the pKa values for the substances acting as acids are shown under their structures. Equilibrium favors the left side because the substances on the left are the weaker acid and the weaker base.
We can arrive at the same conclusion looking at the bases. The strength of bases is measured by the pKa of their conjugate acids. To understand how this works, we must remember that the relative strengths of the acid and the base in a conjugate pair hold an inverse relationship. The stronger the acid, the weaker its conjugate base, and viceversa. Therefore, the relationship between the pKa of an acid and the strength of its conjugate base is direct: the stronger the base, the higher the pKa value of its conjugate acid. We can look at the same reaction again, except that now we’re focusing on the bases, and arrive at the same conclusion that equilibrium favors the left side.
The weaker acid and the weaker base are always on the same side. If you arrive at a different conclusion, something is not right. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/11%3A_Bronsted_Acid-Base_Chemistry/11.07%3A_Acid-Base_Reactions_as_Proton_Transfers.txt |
The most general principle ruling acid strength can be stated thus: strong acids have relatively stable conjugate bases. In general, the more stable the conjugate base, the stronger the acid. An important thing to remember is that stability and reactivity are inverse. The more stable a substance is, the less reactive it is, and viceversa. Therefore, another way of stating the rule above is by saying that strong acids have weak conjugate bases. HCl and H3O+ are strong acids. Accordingly, the corresponding conjugate bases, Cl- and H2O, are weak (very stable). Chloride ion is stable because the negative charge resides on a very electronegative atom. Water molecule is one of the most stable substances known.
How do we know which proton is the most acidic in a molecule (such as acetic acid) that contains more than one type of proton? Remember that the higher the degree of positive character on the proton, the more acidic it is. Examination of a pKa table reveals some trends for acidic protons. The following guidelines can be used to predict acidity.
1. Hydrogens directly attached to very electronegative atoms such as oxygen, sulphur, and the halogens carry a substantial degree of acidity.
2. Hydrogens attached to a positively charged nitrogen, oxygen, or sulfur are acidic. The high electronegativity of these atoms makes them uncomfortable with the positive charge. They seek to diffuse the charge among the neighboring atoms by withdrawing electron density from them. This can be shown by drawing resonance structures as shown.
In all cases structure B reveals the positive character of hydrogen, and therefore its acidic nature.
3. As evidenced by the pKa values of alkanes and alkenes, hydrogens attached to carbon are of very low acidity. Such substances are not normally considered acids at all. However, some hydrocarbons can be weakly acidic if their conjugate bases are stable ions. This can happen in the following cases.
a) There is one or more electronegative atoms near the proton under consideration. The inductive effect of these electronegative atoms leaves the hydrogens in the vicinity deprived of electron density, and therefore with partial positive character.
b) A hydrogen atom bonded to a carbon which is in turn bonded to another carbon that carries a partial or a full positive charge is acidic.
The acidity of the protons shown becomes apparent in elimination reactions (chapter 6) and in the chemistry of enols (chapter 22), when the presence of a base leads to formation of alkenes or enolate ions through a step involving a proton transfer.
c) The conjugate base is resonance-stabilized. This effect is most important when there is another factor enhancing the acidity, such as the presence of a dipole or electronegative atom (as in the nitrile functional group, –CN). Otherwise resonance stabilization alone is not enough to dramatically increase the acidity of a hydrogen attached to carbon (as in toluene, where the pKa is only 40).
d) The hydrogen is attached to an sp-hybridized carbon. Hybridization effects on acidity are discussed in chapter 9. The trends in hybridization can be extended to oxygen and nitrogen besides carbon, as in the example on the right. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/11%3A_Bronsted_Acid-Base_Chemistry/11.10%3A_Identifying_Acidic_Protons.txt |
The relative nature of Bronsted acid-base terminology becomes apparent when we consider substances that can act as either proton donors or acceptors. When two such substances react, how can we predict which will be the proton donor (acid) and which will be the proton acceptor (base)? The answer is that the stronger acid will force the other substance to act as a base. In other words, the substance with the lower pKa will act as the acid, and the other as the base.
In the first example below, methanol is forced to act as a proton acceptor by the strong sulfuric acid. However, in the second example methanol is the proton donor because it is a stronger acid than the amide ion.
11.12: Functional Groups and Reactivity Sites in Organic Molecules
As explained before in relation to functional groups, the very definition of this term is as a reactive center, or site, in the molecule. We can view an organic molecule as consisting of two major structural categories: the basic carbon skeleton, and functional groups.
The basic carbon skeleton is an alkane-like structure, made up of only sp3 carbons and hydrogen atoms, and therefore only sigma (single) bonds. It constitutes the framework that supports the reactive sites, or functional groups. The carbon skeleton is for most practical purposes considered a nonpolar part of the molecule. It is also considered to be nonreactive, except under very harsh, extreme, or special conditions.
Alkanes contain no functional groups, but as soon as their structure is chemically modified in any way that leads to relatively stable products, functional groups are introduced. Some of the simplest modifications that introduce functional groups are the creation of a π-bond and the introduction of a heteroatom. Technically speaking, heteroatoms are atoms other than carbon and hydrogen. The term is most commonly used in reference to nonmetals of the second and third rows. That is, N, O, S, P, and the halogens.
The introduction of π-bonds, as in alkenes and alkynes, does not affect the polarity of the molecule very much, but it affects the flexibility of it by restricting free rotation around carbon-carbon bonds. The presence of a π-bond introduces a center of reactivity by increasing the electron density at that site. At the same time, “mobile,” or “available” electrons are introduced. Remember that π-bonds are weaker than σ-bonds, and therefore π-electrons are easier to move for use in chemical reactions.
The π-bond is a high electron density region. π-electrons can be used for chemical reactions under the right conditions
The introduction of heteroatoms such as oxygen almost invariably introduces a dipole, since most of these elements are more electronegative than carbon and hydrogen. By introducing a dipole, an electron imbalance is created that results in formation of a low electron density area and a high electron density area. The low electron density area will then be reactive towards electron-rich molecules. The high electron density area will be reactive towards electron-poor (deficient) molecules.
Identifying reactive sites in organic molecules amounts to identifying areas of low or high electron density. In the case of carboncarbon double or triple bonds (alkenes and alkynes), the region is an area of high electron density. Such molecules typically act as electron donors, even though the π-bond must be broken before those electrons become available. If we’re not dealing with π-bonds, then we must identify the strongest dipoles in the molecule and the areas of low or high electron density they create, such as in the examples below.
Obviously, when it is a hydrogen atom that is positioned at the δ+ end of the dipole, we have an acidic proton that can engage in reactions with Bronsted bases. We have already learned to identify acidic protons, but how do we identify basic sites in molecules?
11.13: Identifying Bronsted Bases
Basic sites in organic molecules are areas of high electron density. We’ve already talked about π-bonds as examples of such areas. However, the large majority of Bronsted bases comprises molecules that have atoms with nonbonding electrons (lone pairs). The following trends can be observed.
1. Nonbonding electrons are the most readily available for reaction with acids because no energy has to be invested in breaking a bond before they can be used. The strongest Bronsted bases contain atoms with unshared electrons which are localized. Examples are:
2. In a conjugate acid-base pair, the more negatively charged (or less positively charged) partner is the most basic. Examples of relative basicities in such pairs are:
3. The periodic trend in basicity of atoms is the reverse of the acidity trend. This is particularly useful when comparing conjugate bases. Other factors being comparable, the basicity of the atom increases from bottom to top and from right to left. Examples are:
4. The above can also be applied to resonance structures. This concept can be used to predict the outcome of an acid-base reaction when there are several possible reactive sites.
5. Electron delocalization decreases the basic character of the atom. Delocalization stabilizes the molecule and makes it less reactive. Delocalized electrons are not as available as localized electrons because they are diffused over a larger area.
6. Resonance structures involving π-bonds are the most commonly seen type. However, resonance structures involving σ-bonds can be written when the bond is highly polarized, as in the case of a C-Li bond. Such structures often reveal reactivity sites not obvious from the main contributor. In the example below, structure II reveals the basic character of the carbon atom.
7. Alkenes are considered weak bases, since the π-bond must be broken before its π-electrons can be used for reaction with an acid.
8. Finally, a pKa table can provide the best measure of basicity when properly used. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/11%3A_Bronsted_Acid-Base_Chemistry/11.11%3A_Some_Acids_Can_Act_as_Bases_and_Vice_Versa.txt |
Lewis acids and bases are defined in terms of electron pair transfers. A Lewis base is an electron pair donor, and a Lewis acid is an electron pair acceptor. An organic transformation (the creation of products from reactants) essentially results from a process of breaking bonds and forming new ones. This process basically amounts to electron pair transfers. Ionic mechanisms involve electron pair transfers and are therefore described by the Lewis acid-base theory.
The Lewis definition implies the presence of high electron density centers in Lewis bases, and low electron density centers in Lewis acids. In a reaction between a Lewis acid and a Lewis base the electron pair donated by the base is used to form a new sigma bond to the electron deficient center in the acid. The identification of Lewis bases follows basically the same guidelines as the identification of Bronsted bases. They frequently contain atoms that have nonbonding electrons, or lone pairs. On the other hand, Lewis acids frequently contain atoms with an incomplete octet, a full positive charge, or a partial positive charge.
Water is an example of a Lewis base. Carbocations are examples of Lewis acids. When water reacts with a carbocation as shown below, one of the electron pairs from oxygen is used to form a new sigma bond to the central carbon in the carbocation. As with Bronsted acid-base reactions, curved arrow are used to indicate the movement of electron pairs during the reaction process. The arrow always originates with the Lewis base and moves towards the area of electron deficiency in the Lewis acid.
To avoid confusion between the Lewis and the Bronsted definitions of acids and bases, Lewis bases are sometimes called nucleophiles, and Lewis acids are called electrophiles. In the example above, water acts as a nucleophile (donates electrons), and the carbocation acts as an electrophile (receives electrons).
Since Bronsted acids and bases are a subcategory of the more encompassing Lewis definition, it can be said that most Bronsted bases are also nucleophiles, and that the proton is a Lewis acid, or an electrophile. There are however some subtle differences to keep in mind.
12.02: Recognizing Nucleophiles
For the most part, the same criteria used for identifying Bronsted bases can also be used to identify nucleophiles. Here is a summary.
1. The most common type of nucleophiles are those containing atoms with unshared electron pairs, such as the following.
Notice that nonbonding electrons are frequently omitted from formulas. Always remember their presence on oxygen, nitrogen, and negatively charged carbon atoms.
2. A negatively charged species is usually a stronger nucleophile or base than its neutral analog. Thus, hydroxide ion is stronger, both as a base and as a nucleophile, than water.
3. Carbon bonded to a metal has strong negative character, revealed when writing resonance structures. The carbon atom in such molecules is considered a strong nucleophile. Recall the structures of n-butyllithium and Grignard (organomagnesium) reagents from previous notes.
4. The pi bond is a region of high electron density. Pi bonds are not as strong as sigma bonds, which means that pi electrons are more available for reactions because pi bonds ore more easily broken. Molecules containing pi bonds are considered weak nucleophiles or bases, because they can react with strong acids or electrophiles. See examples under Bronsted acid-base chemistry.
5. Periodic trends in nucleophilicity are such that, other factors being similar, nucleophilicity increases from right to left across the same row, and from top to bottom across the same period or group. Make a note that this trend is different from the basicity trend, which increases in the same way across a row, but from bottom to top within a period.
6. Basic vs. nucleophilic behavior. There are some differences between Bronsted bases and nucleophiles. First, the term base is reserved for substances seeking acidic protons, whereas the term nucleophile is used for substances seeking electron-deficient centers, be they protons or other atoms (most commonly carbon).
Second, in acid-base reactions the size (or bulk) of the base seldom matters. In nucleophilic reactions, the size of the nucleophile can be an important (steric) factor. Smaller is more effective. Thus, although the nucleophilic atom in the two species below is the same (oxygen) and they have similar structures, methoxide ion is a more effective nucleophile than t-butoxide ion, even though they are about equally effective as bases. The larger t-butoxide ion has more difficulty reaching an electrophilic center, which is typically a carbon buried at the core of the molecular structure. That is not a problem when it acts as a base because acidic protons usually lie in the periphery of the molecule and are easily accessible.
methoxide ion smaller ion, better nucleophile
t-butoxide ion larger ion, poorer nucleophile | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/12%3A_Introduction_to_Lewis_Acid-Base_Chemistry/12.01%3A_Definitions.txt |
There are two requirements for a molecule to be considered a good electrophile. First, it must contain an electrophilic center or atom. Second, the electrophilic atom must be able to accommodate a new sigma bond. Please keep in mind the difference between electrophile and electrophilic center. The term electrophile refers to the molecule. The term electrophilic center refers to the particular part of the molecule susceptible to nucleophilic attack.
To avoid confusion, the term substrate is frequently used in reference to electrophiles. This term denotes a molecule being acted upon by another agent. For example, an enzyme substrate is a molecule being modified by an enzyme. Likewise, an electrophile can be thought of as the substrate of a nucleophile when the latter “attacks” its electrophilic center.
Electrophilic centers are areas of low electron density. Most often they are atoms which (a) contain an incomplete octet, and/or (b) carry a full or a partial positive charge. A partial positive charge can be revealed by writing resonance structures, or by identifying a polar bond.
1. The following are examples of electrophiles containing atoms with incomplete octets:
2.These are examples of electrophiles containing atoms with partial positive charges:
3. Atoms of the second row like oxygen and nitrogen, which are more electronegative than carbon, seldom act as electrophilic centers, even if they carry a positive charge. In that situation they seek to lessen their positive character by sharing the charge with adjacent atoms, causing them go become acidic (protons) or electrophilic (carbon for instance). Resonance structures can reveal this shift of positive charge.
The ammonium and hydronium ions are of moderate to high acidity because the highly electronegative oxygen and nitrogen seek to transfer their positive charge to the adjacent proton, making it acidic. In addition, those atoms cannot accomodate another bond without violating the octet rule.
In addition to making the protons acidic, a positive charge on oxygen can also make the adjacent carbon electrophilic by a similar transfer of positive character. Moreover, the oxygen atom becomes part of a good leaving group, in this case water. The stage is set for either a nucleophilic attack on carbon, or a reaction with a Bronsted base.
The reaction with a base (B - ) is an equilibrium process that normally has a low activation energy and is therefore relatively fast.
The reaction with a nucleophile (Nu - ) is a kinetic process that normally has a higher activation energy than a proton transfer and is therefore slower. If the nucleophile being used is also a good base, it will prefer to take the proton.
An example of the above is the reaction of protonated n-butanol with either bromide ion or ammonia. Bromide ion is one of the best nucleophiles, but a weak base. It prefers to act as a nucleophile. Ammonia is a moderately good nucleophile but also a good base. Given the choice, it prefers to act as a base.
acid-base equilibrium favors the left side because bromide ion is a weak base
Bromide ion is an effective nucleophile, preferring to attack the electrophilic carbon displacing the water
Ammonia is a weak nucleophile and a moderate base. As long as it can act as a base, it will prefer to do so rather than engaging in a higher energy nucleophilic displacement.
4. Unlike their second row counterparts, some electronegative elements of the third row such as sulfur and phosphorus can sometimes act as electrophilic centers due to their larger size and the ability to accommodate a new sigma bond using their d-orbitals:
Unlike oxygen, sulfur can be electrophilic because it can accommodate a new sigma bond using its d-orbitals without violating the octet rule.
12.04: Reactions Between Nucleophiles and Electrophiles
As mentioned earlier, a good electrophile must be able to accommodate a new sigma bond between its electrophilic center and the nucleophile. When the electrophilic center is an atom with an incomplete octet, this is no problem.
For electrophiles containing polarized pi bonds such as carbonyl groups, at least one resonance form shows an atom with an incomplete octet.
resonance forms of acetone
You can use either resonance structure to write the reaction between the nucleophile and the electrophile. The following are acceptable representations of a nucleophilic attack of hydroxide ion on acetone, but the second one makes it more apparent that the central carbon can take the extra bond.
Two acceptable representations of the nucleophilic attack of hydroxide ion on acetone using different resonance structures.
Sometimes the substrate has an electrophilic atom which is sp3-hybridized and already has a complete octet. In this case there are no pi electrons to displace as the new σ-bond forms. The nucleophile must displace another group as it bonds to the electrophile. The displaced group is called a leaving group. The leaving group can be displaced only if it leaves as a weak base, because weak bases are stable molecules that can take the electrons with them. In the following example, hydroxide ion is the attacking nucleophile. As it bonds to the sp3 electrophilic carbon, it must displace another group. The leaving group in this case is the bromine atom. It is a good leaving group because it leaves as bromide ion, which is a weak base and can take the electrons with it.
The reverse reaction, however, could not happen. Although bromide is a good nucleophile and methyl alcohol contains an electron deficient center (the carbon bonded to oxygen), the molecule does not contain a good leaving group. Hydroxide ion is a strong base, therefore it cannot be displaced by bromide.
Although the hydroxide ion is not a good leaving group, it is possible to do nucleophilic displacements on alcohols by protonating them with acid first. The protonated hydroxyl group is a potential water molecule, which is a weak base and therefore a good leaving group.
This approach has limitations. The most important is that the nucleophile must be a weak base, or it will prefer to react with the acidic protons. For all practical purposes, the only nucleophiles that can be used in this way are chloride and bromide ions. But this provides a good way to convert alcohols into primary, secondary, or tertiary chlorides and bromides.
Notice that in these case the reverse reaction can happen. Water is a good nucleophile and chloride and bromide are good leaving groups. One must isolate the product as it forms to keep it from reacting with water and go back to alcohol.
13.01: Study Guide for Chapters 6 and 7
Certain topics like the structure, synthesis, and nomenclature of alkyl halides have already been covered previously. This part of the course is primarily concerned with Sn1 and Sn2 reactions, E1 and E2 reactions, and alkene synthesis, structure, and nomenclature.
The following is the suggested order of readings and practice problems. It’s important to attend lectures because the material will be presented in a certain order. Most of the material is in the book, but a few additional guidelines, summaries, and suggestions may be given in class.
Overall, if you follow this study guide, you should be well prepared for the quizzes and the test. Notice that the amount of suggested problems is somewhat large. The best strategy is probably to do the problems in the suggested sections first, then tackle as many from the end of the chapters as time allows you. Needless to say, the more you practice, and the better prepared you’ll be for the test and quizzes.
I. Sn1 and Sn2 REACTIONS
a) Read the class notes titled Introduction to Lewis Acid-Base Chemistry.
b) Read chapter 4, pages 137-139 and do the problems in those sections.
c) Read sections 8 through16 from chapter 6 and do the problems in those sections. For section 16 do only problem 27.
d) Read class notes titled Highlights of Nucleophilic Substitution Reactions Involving sp3 Carbon
II. E1 and E2 REACTIONS
a) Read sections 17 – 21 from chapter 6 and do the problems in those sections. b) Read the following sections from chapter 7 and do the problems in those sections: 7-7(A-D), 7-9(A-E), and 7-10.
III. ALKENE STRUCTURE and NOMENCLATURE
a) From chapter 7, read sections 1, 4, 5, and 8, and do the problems in those sections
IV. ADDITIONAL SUGGESTED PROBLEMS FROM CHAPTERS 6 and 7
a) From the end of chapter 6: 41, 44-56, 60, 62, 66, 68.
b) From the end of chapter 7: 30-33, 36, 37(a, b), 38, 41, 42, 43a, 44, 45, 46, 53(a-c). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/12%3A_Introduction_to_Lewis_Acid-Base_Chemistry/12.03%3A_Recognizing_Electrophiles.txt |
From a synthetic point of view, this is the most useful reaction. It provides a means to prepare many functional groups from alkyl halides, and therefore from alkanes through the free radical halogenation reaction.
The Sn2 mechanism: a) is a single step process b) involves no intermediates c) involves only one transition state, which is of low polarity d) follows second order (bimolecular) kinetics. That is, rate=k[substrate][nucleophile]
In nucleophilic substitutions at sp3 carbon, Sn2 mechanisms are favored by using: a) sterically accessible substrates b) strong (negatively charged), small nucleophiles c) low to moderate polarity solvents.
Stereochemically, if the electrophilic center in the substrate is chiral, the Sn2 reaction produces a product with inverted configuration.
14.02: Sn1 Reactions
From a synthetic point of view, the Sn1 reaction is less useful. It is prone to side reactions such as eliminations and carbocation rearrangements.
The Sn1 mechanism:
a) is a multistep process
b) occurs with formation of carbocation intermediates in the rate determining step
c) involves one transition state per step. The rate-determining step involves a high polarity transition state
d) follows first order (unimolecular) kinetics. That is, rate=k[substrate]
In nucleophilic substitutions at sp3 carbon, Sn1 mechanisms are favored by using:
a) sterically hindered substrates
b) weak (neutral), small nucleophiles
c) moderate to high polarity solvents that can stabilize the transition state and the carbocation intermediate
Stereochemically, if the electrophilic center in the substrate is chiral, the Sn1 reaction produces a racemic product. The relative proportions of the enantiomers depend on the specific reaction, but will typically be close to 50/50.
14.03: Factors That Affect The Course of Nucleophilic Substitutions at sp3 Carbon
1. STERIC NATURE OF THE SUBSTRATE. Steric accessibility of the electrophilic center in the substrate is probably the most important factor that determines if a nucleophilic substitution will follow an Sn1 or an Sn2 mechanism.
EXAMPLES OF Sn2 (sterically accessible) SUBSTRATES
EXAMPLES OF Sn1 (sterically hindered) SUBSTRATES
Some substrates, whether they are sterically hindered or not, may prefer to undergo Sn1 reactions if they can dissociate into very stable carbocations in the presence of the solvent. In most cases this involves resonance-stabilized cations.
EXAMPLES OF Sn1 SUBSTRATES THAT FORM STABLE CARBOCATIONS
2. NATURE OF THE NUCLEOPHILE. Both Sn1 and Sn2 reactions prefer small nucleophiles. Large nucleophiles have more difficulty accessing the electrophilic center in the substrate. They also have increased tendency to act as Bronsted bases, seeking acidic protons rather than electrophilic centers due to the lower activation energy of acid-base reactions compared to nucleophilic substitutions.
Weak, small nucleophiles that favor Sn1 reactions are shown below. Notice that several are the conjugate acids of strong nucleophiles. They are also typically neutral, but some have a delocalized negative charge.
Large nucleophiles, especially if they are strong, have a tendency to act as Bronsted bases rather than as nucleophiles. They should be avoided if a nucleophilic reaction is desired. Examples are:
3. SOLVENT USED. It has already been mentioned that Sn2 mechanisms are favored by low to moderate polarity solvents such as acetone and N,N-dimethylformamide (DMF). Sn1 mechanisms are favored by moderate to high polarity solvents such as water and alcohols. It is frequently the case that in Sn1 reactions the solvent also doubles as the nucleophile. Water and alcohols are prime examples of this practice.
Sn2 Solvents
Sn1 Solvents
4. LEAVING GROUP. The nature of the leaving group has more of an effect on the reaction rate (faster or slower) than it does on whether the reaction will follow an Sn1 or an Sn2 mechanism. The most important thing to remember in this regard is that good leaving groups are weak bases.
a) All halogens, except for fluorine, are good leaving groups
b) Groups that leave as resonance stabilized ions are also weak bases and therefore good leaving groups.
c) Water is a good leaving group frequently used to prepare alkyl chlorides and bromides from alcohols.
The OH group in alcohols is not a good leaving group because it leaves as hydroxide ion, which is a strong base. However, if the hydroxyl group is protonated first with strong acid, it can leave as a water molecule, which is a good leaving group. Refer to the manuscript titled Introduction to Lewis Acid-Base Chemistry for a discussion and examples of this approach.
14.04: Competing (Side) Reactions in Nucleophilic Substitutions
There are two major reactions that compete with nucleophilic substitutions. They are:
1. CARBOCATION REARRANGEMENTS (Sn1 only)
2. ELIMINATION REACTIONS (Sn1 and Sn2)
Carbocation rearrangements are examined first. Eliminations are examined in a separate paper.
14.05: Carbocation Rearrangements
Carbocations only form in Sn1 reactions. Carbocations are prone to skeletal rearrangements if this produces a more stable cation. Carbocation rearrangements occur mainly by two processes:
a) Hydride shift - migration of a hydrogen atom with electrons to an adjacent carbon
b) Alky shift - migration of a carbon (usually as part of an alky group) with electrons to an adjacent carbon.
A quick way to tell whether a substrate will produce a carbocation prone to rearrangement is to look at the carbon that bears the leaving group. If this carbon is next to a higher order carbon (meaning secondary, tertiary, allylic, etc.) then the carbocation that results can rearrange to a more stable one, and will do so, probably yielding a product with different carbon connectivity. Examples follow.
The leaving group (Br) is on a secondary carbon, but this carbon is next to a tertiary carbon. The nucleophile is water, therefore the expected product is an alcohol. The product will consist of a mixture of the expected secondary alcohol (minor) and a tertiary alcohol (major) due to the rearrangement to a more stable cation shown below.
The above example also shows the reason why, when the nucleophile is water or an alcohol, the group that replaces the leaving group in the product is the conjugate base of water (OH) or the alcohol (RO respectively).
Another example illustrates a similar point. Can you provide a step by step mechanism (it might be in the test, you never know)?
notice that the conjugate base of the nucleophile (in red) has replaced the leaving group
14.06: Sn1 Reactions and Rearrangements Involving Primary Substrates
Primary substrates normally do not follow Sn1 mechanisms because they do not form stable cations. However, a hindered primary substrate can be forced into an Sn1 mechanism if sufficient energy and time are allowed, for example boiling the substrate in a nucleophilic solvent such as ethanol.
The nucleophile cannot do a backside attack, and the substrate cannot form stable cations. In this case the substrate will begin to rearrange as the leaving group departs. This avoids formation of a primary cation. As the leaving group departs, a positive charge begins to develop on the carbon bearing the leaving group, and the rearrangement process starts, all in unison. The reaction of neopentyl bromide with ethanol illustrates this point.
As the positive charge develops on the primary carbon while bromine leaves, the methyl group is migrating to an adjacent position to forma more stable cation. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/14%3A_Highlights_of_Nucleophilic_Substitution_Reactions_Involving_sp3_Carbon/14.01%3A_Sn2_Reactions.txt |
In terms of synthetic value, any reactions whose mechanism involves carbocation formation suffer from some drawbacks. Once formed, carbocations can undergo several process that may result in formation of undesired side products. In the context of Sn1 reactions, some of the things that carbocations can do are:
1) They can go on to form the expected Sn1 products.
2) They can rearrange to form products whose connectivities are changed relative to that of the original substrate.
3) They can undergo elimination (E!) reactions to form alkenes
Examples:
This limits the synthetic usefulness of such reactions, for one has to deal with mixtures of products and the separation of the desired ones.
In fact, Sn1 and E1 reactions typically go hand in hand and are difficult to disassociate, because they share similar characteristics, and the conditions that favor one also favor the other. We’ve already learned the characteristics of Sn1 reactions and the factors that favor them. We can extend that to E1 reactions as well:
Characteristics of the Sn1 and E1 mechanisms:
a) They are multistep processes
b) They occur with formation of carbocation intermediates in the rate determining step
c) They follow first order (unimolecular) kinetics. That is, rate=k[substrate]
Sn1 and E1 mechanisms are favored by using:
a) Sterically hindered substrates
b) Weak (neutral), small nucleophiles and heating
c) Moderate to high polarity solvents
The Sn1 mechanism leads to substitution products, and the E1 mechanism leads to formation of alkenes.
The same substrates that are prone to undergo Sn1 reactions also undergo E1 reactions. They are of two major types:
a) Secondary and tertiary alky halides
b) Secondary and tertiary alcohols.
15.02: Alkyl Halides as Sn1 and E1 Substrates
As mentioned before, conditions that favor Sn1 also favor E1 reactions. The first and rate-determining step in the process is departure of the leaving group to form a carbocation. Let’s look at one of the examples from the previous page, the reaction between 2-bromo-3-methylbutane and methanol. As has been mentioned before, commonly used solvents in Sn1 reactions are water and alcohols. They frequently also double as nucleophiles. In E1 reactions these same substances would act as bases to capture a proton in the elimination step.
Once formed, the carbocation can:
a) Go on to form the Sn1 product. Remember that when the nucleophile is a neutral protic solvent, its conjugate base replaces the leaving group. This implies that a proton (H+) gets released in the process.
The bromide ion released in the first step and the proton released in the second step can then get together to form HBr, which is an inorganic product of the reaction. However, we will not focus on inorganic products. As a matter of fact, inorganic products are frequently left out when writing organic reactions and mechanisms to avoid clutter and keep the focus on the organic products.
b) Rearrange to a more stable cation. We already mentioned that this can happen by an alkyl or a hydride shift, depending on which process yields a more stable cation. This cation can in turn form another Sn1 product. This Sn1 product is a structural isomer of the one formed in (a) because its connectivity has changed.
c) Eliminate a proton to form an alkene. Remember that when there is a positive charge on carbon, the neighboring protons become highly acidic. The solvent –acting as a weak base– can capture one of these protons and cause an electron shift towards the positive charge that results in formation of a new pi bond, or alkene product. When the protons surrounding the positive charge are nonequivalent, several alkenes are possible, as illustrated below.
Secondary cation before rearrangement. The protons shown in red are now acidic due to the presence of the positive charge on carbon.
The rearranged cation can undergo similar elimination processes to yield two possible alkenes, but one of them is the same as one of the alkenes formed above, with the pi bond located between carbons 2 and 3. As an exercise, see if you can identify the protons which are acidic, how they are eliminated, and how electron movement takes place to form the alkene structures shown below.
Thus we have the net reaction we introduced on page 1, showing the formation of Sn1 and E1 products:
Can you now explain (mechanistically) how the various products form in the second example given on p.1? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/15%3A_Relationship_Between_Sn1_and_E1_Reactions/15.01%3A_Synthetic_Drawbacks_of_Sn1_Reactions.txt |
The classic textbook example of E1 elimination reactions is the acid-catalyzed alcohol dehydration. Strong acid catalysis is needed to protonate the hydroxyl group of alcohols and turn it into a good leaving group. For alcohols which are soluble in water, an aqueous solution of strong acid is usually used. Such solution contains a high concentration of hydronium ion (the conjugate acid of water), which acts as the proton donor. This first protonation step is an acid-base reaction, and as such it takes place very rapidly. The protonation of 4-methylcyclohexanol illustrates this step.
Once protonated, the hydroxyl group can leave as water, leaving behind a positively charged carbon. This is the rate determining step in the sequence.
The elimination step takes place after formation of the carbocation, with water acting as a base.
Notice that each step in this mechanism is reversible. This means that the acid-catalyzed alcohol dehydration is an equilibrium process. As such, equilibrium must be manipulated to shift the outcome towards the desired product. This reaction enables us to make alkenes from alcohols, or alcohols from alkenes. Typically, formation of alkenes is favored by use of concentrated acid, whereas formation of alcohols is favored by use of dilute acid. The following sequence shows the steps in reverse, starting with an alcohol, and arriving at an alkene.
15.04: Eliminations Involving Asymmetrical Substrates
In some eliminations the products include several possible alkenes. The next question is then, which ones form preferentially? Is there a rule for predicting which alkene will predominate? Let’s look at the following examples given before, but this time let’s focus only on the alkene products.
Saytzeff’s rule enables us to make a prediction. According to this rule, in elimination reactions, the most highly substituted alkene usually predominates. For details, refer to the Wade textbook, section 6-19 (5th ed.) or 6-18 (6th ed.).
The most highly substituted alkene is the one with the most alkyl groups directly attached to the carbon-carbon double bond. This of course does not include hydrogens attached to the carbon-carbon double bond. In the above example, we have mono, di, and trisubstituted products. According to Saytzeff’s rule, the trisubstituted product will predominate because it is the most stable.
Why is the trisubstituted alkene more stable than the other two? The relative stabilities of alkenes is measured by their heat of hydrogenation (much like the relative stabilities of cycloalkanes is measured by their heats of combustion, see ch. 3). The relationship is inverse: the higher the heat of hydrogenation, the lower the stability, because the higher the heat of hydrogenation, the higher the potential energy of the alkene.
For a full discussion of this trend with tables, refer to the Wade textbook, section 7-7 (A-D) in both the 5th and the 6th eds. Please note that in the case of cis and trans isomers, the higher stability of the trans isomer is due to a steric effect. The alkyl substituents in the cis isomer are closer to each other than in the trans isomer, leading to increased steric crowding.
Let’s look at the last example. In this case we have di and trisubstituted products. Once again, the trisubstituted product will be predominant because it is the most stable. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/15%3A_Relationship_Between_Sn1_and_E1_Reactions/15.03%3A_Alcohols_As_Sn1_and_E1_Substrates.txt |
Chapter 8 is mostly about alkene reactions. That is, how one can transform alkenes into other functional groups. Most of these reactons are electrophilic additions, or the addition of electrophiles across the double bond. Several of these reactions amount to addition of water to the π-bond. The result is the transformation of alkenes into alcohols. We now look at the generalities of this type of reaction (electrophilic addition) and then we look at each reaction individually.
The C=C π-bond of alkenes is a source of electrons. It is considered a weak base or nucleophile. As such, it can react with strong electrophiles. We have already learned how to identify electrophiles, but the π-bond requires strong electrophiles to react with. These can be strong proton acids, or species containing atoms with incomplete octets (Lewis acids). Examples of strong proton acids are HBr and H2SO4. Examples of Lewis acids are BH3, transition metal salts such as HgSO4, and carbocations.
The addition of strong electrophiles to the C=C π-bond can be viewed as the opposite of the elimination reaction, as illustrated below.
Elimination results in net loss of HBr to form a new C=C bond.
Addition of HBr (a strong electrophile) across the π-bond forms a new functional group, in this case an alkyl halide.
According to Saytzeff’s rule, in elimination reactions where formation of several alkenes is possible, the most highly substituted alkene predominates as a product.
Conversely, Markovnikov’s rule says that in addition reactions of proton acids to alkenes, the proton of the strong acid preferentially bonds to the carbon in the π-bond that already holds the greater number of hydrogens on it. In the alkene shown below, C-2 has more protons attached to it (one) than C-1 (none). Accordingly, the hydrogen from HBr bonds to C-2 and the bromine bonds to C-1
Markovnikov's product, a tertiary bromide
This preferred orientation is referred to as Markovnikov orientation. This suggests the possibility that another product with opposite orientation, called the anti-Markovnikov product, might form. Given that this product does not normally form under ordinary conditions, the question then is, are there special conditions under which it could form? The answer is yes, but with a very limited scope. We’ll address that point later.
Anti-Markovnikov's product, a secondary bromide
Another way to state Markovnikov’s rule, in this case, is to say the the tertiary bromide will form preferentially over the secondary bromide. If the electrophilic part of the reactant (H+) adds to the least substituted carbon (C-2), then the nucleophilic part (Br-) adds to the most substituted carbon (C-1). The degree of substitution in this case refers to the number of alkyl groups originally attached to the π-bond. This is another way by which Markovnikov’s rule becomes the counterpart of Saytzeff’s rule. The two statements are complementary.
Modern understanding of ionic mechanisms provides the key to Markovnikov’s rule. In the first step of the reaction, which is protonation of the π-bond, the most stable carbocation forms. This leads directly to formation of the tertiary bromide, i.e. Markovnikov’s product.
Here is another example involving a cyclic alkene:
16.02: Regioselectivity and Formation of Anti-Markovnikov Products Via Free Radical Mechanisms
REGIOSELECTIVITY -
A reaction such as the above that leads to preferential formation of one of several possible structural isomers is said to be regioselective. The addition of proton acids to alkenes, which follows an ionic mechanism, is an example of a regioselective reaction.
FORMATION OF ANTI-MARKOVNIKOV PRODUCTS VIA FREE RADICAL MECHANISMS.
Ionic mechanisms favor formation of the Markovnikov product through formation of the most stable carbocation. However, one can manipulate conditions to favor formation of the anti-Markovnikov product through a different mechanism, or a different series of steps. This is the case in the addition of HBr to alkenes in the presence of peroxides. Peroxides are well known free radical initiators. This promotes a reaction where the reactants add in a different order and by a different mechanism, in this case with formation of the most stable free radical.
2o bromide, anti-Markovnikov product.
Another example:
Refer to section 8-3B of the Wade textbook (p. 319 of the 5th ed.) and remember the following points.
a) The anti-Markovnikov addition of proton acids to alkenes works only with HBr.
b) It requires the presence of peroxides, which are free radical initiators.
c) The reaction follows a free radical mechanism, where the bromine is first to add to the alkene with formation of the most stable free radical, which eventually leads to the anti-Markovnikov product.
Other proton acids that follow Markovnikov’s rule include HCl. HI, and H2O (acid-catalyzed) to form alcohols. We now summarize all the important addition reactions of alkenes, including the markovnikov addition of proton acids and water. Then we address some concepts and mechanistic considerations relevant to these reactions. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/16%3A_Electrophilic_Additions_of_Alkenes_as_the_Counterpart_of_Eliminations/16.01%3A_Introduction.txt |
Alkenes are primarily prepared by elimination reactions of molecules that contain good leaving groups attached to sp3 carbons. Examples of such reactions are dehydrohalogenations with strong base, and acid-catalyzed dehydrations of alcohols. The opposite of an elimination is an addition reaction. In an addition reaction an alkene adds elements to each of the carbons involved in the π-bond, resulting in formation of sp3 carbons from sp2 carbons. This is one of the most important types of reactions that alkenes undergo.
Another important type of reaction involving alkenes is oxidative cleavage. In such reactions the carbon-carbon π-bond is completely broken by the action of an oxidizing agent, resulting in more oxidized forms of carbon, such as aldehydes, ketones, and carboxylic acids. All these reactions are covered in chapter 8 of the Wade text, 4th ed.
The following is a summary of the most important representative types of these reactions. For full details, please refer to the textbook.
1. ADDITION OF HBr, HCl, and HI.
The addition of these substances to an alkene proceeds by an ionic mechanism, with formation of the most stable carbocation. Therefore, it follows Markovnikov’s Rule.
An Anti-Markovnikov variation requires the presence of peroxides as free radical initiators, and can only be performed with HBr. The mechanism proceeds with formation of the most stable free radical, which results in formation of the Anti-Markovnikov product.
THE NEXT THREE REACTIONS ADD THE ELEMENTS OF WATER (H/OH) ACROSS THE DOUBLE BOND
2. ACID-CATALYZED ADDITION OF WATER.
Produces Markovnikov alcohols, it is an equilibrium reaction, and proceeds with formation of carbocations in the rate-determining step.
3. OXYMERCURATION-DEMERCURATION SEQUENCE.
This sequence consists of two steps. It is the most effective way to prepare Markovnikov alcohols from alkenes.
4. HYDROBORATION-OXIDATION SEQUENCE.
This sequence also consists of two steps. It is the most effective way to prepare anti-Markovnikov alcohols from alkenes. The components of water (H/OH) add to the π-bond with syn-stereochemistry. If chiral centers result from this reaction, the product is obtained as an enantiomeric mixture.
5. HALOGEN ADDITION.
This reaction is performed most frequently with bromine and chlorine. It adds a halogen atom to each of the two carbons comprising the π-bond with anti-stereochemistry. If chiral centers result from this reaction, the product is obtained as an enantiomeric mixture.
6. HALOGEN ADDITION IN THE PRESENCE OF WATER
(halohydrin formation). Produces Markovnikov alcohols with a neighboring halogen. The OH and the halogen add with anti -stereochemistry. If chiral centers result from this reaction, the product is obtained as an enantiomeric mixture.
THE NEXT TWO REACTIONS ADD TWO HYDROXYL GROUPS (OH) ACROSS THE DOUBLE BOND
7. SYN HYDROXYLATION.
Adds two hydroxyl groups accross the π-bond with syn-stereochemistry. It can be performed with two reagent mixtures: pot assium permanganate in basic medium, or osmium tetroxide in the presence of hydrogen peroxide. If chiral centers result from this reaction, the product is obtained as an enantiomeric mixture.
8. ANTI-HYDROXYLATION SEQUENCE.
Adds two hydroxyl groups accross the π-bond with anti-stereochemistry. It proceeds in two steps via an epoxide (three-membered ring ether). If chiral centers result from this reaction, the product is obtained as an enantiomeric mixture.
9. CATALYTIC HYDROGENATION.
Adds hydrogen accross the π-bond with syn-stereochemistry. It basically transforms an akene into an alkane.
MAIN OXIDATIVE CLEAVAGE REACTIONS
1. CLEAVAGE WITH HOT, CONCENTRATED POTASSIUM PERMANGANATE
yields ketones and/or carboxylic acids, depending on whether the carbon in question is in the middle of a chain or at the end of a chain in the product.
2. OZONOLYSIS.
Similar to the above, but it yields only carbonyl compounds (aldehydes or ketones). Ozone is required for the reaction, but the other reagent may vary from one context to another. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/16%3A_Electrophilic_Additions_of_Alkenes_as_the_Counterpart_of_Eliminations/16.03%3A_Summary_of_Alkene_Reactions.txt |
The C=C bond is considered to be a weak base/nucleophile. The high concentration of electron density makes the pi bond a Lewis base, but in order to donate electrons the pi bond must be broken first. In a reaction which is basically the reverse of elimination, an electrophile of general formula E-Y can add to the C=C double bond. E represents the electrophilic center, and Y is the rest of the molecule, or nucleophilic part. Proton acids such as HCl fit this description, with E = H and Y = Cl. Water can also be viewed as an electrophile with E = H and Y = OH. In a generalized representation, the electrophile E-Y adds to the pi bond, converting the sp2 carbons into sp3 carbons.
Examples: Addition of HBr and water to the π-bond
17.02: Markovnikov Orientation vs. Syn or Anti Addition
The term Markovnikov orientation refers to the bonding preference of E and Y for carbon atoms a or b. The following example shows how a proton acid HY can add to the π-bond of an unsymmetrical alkene with either Markovnikov or anti-Markovnikov orientation, depending on the reaction conditions used.
The terms syn addition and anti addition refer to the preference of H and Y to add to the π-bond from the same side of the C=C plane (syn addition), or from opposite sides (anti addition):
The following example showing the addition of water to 1-methylcyclohexene (deuterium-labeled) illustrates the four possible combinations.
Two reactions of interest to us, oxymercuration-demercuration and hydroboration-oxidation, follow paths (2) and (3) respectively.
17.03: Stepwise Addition of Electrophiles to Bonds with Formation of 3-Membered Ring Intermediates
Sometimes the electrophilic part E of an electrophile E-Y approaches the C=C π-bond to simultaneously bond to both carbons and form a three membered ring intermediate with departure of Y as a leaving group. In a second step, Y can act as a nucleophile to perform backside attack on one of the carbons, prompting ring opening that results in net anti addition of E-Y to the C=C bond.
The oxymercuration-demercuration reaction (addition of water in the presence of mercury salts), halogen addition, and the anti-hydroxylation (via epoxides) reaction are examples of this outcome.
17.04: Concerted Addition of Electrophiles to Bonds with Formation of 4-Membered Ring Transition States
The electrophile E-Y can also approach the C=C π-bond in such a way that both E and Y bond to each of the two carbons simultanesouly. This requires the formation of a four-membered ring transition state and necessarily results in net syn addition of E-Y to the π-bond.
The hydroboration-oxidation (addition of water using via borane) is an example of this outcome.
17.05: More on the Chemistry of 3-Membered Rings
Three-membered rings such as epoxides (cyclic ethers) are strained and susceptible to ring-opening. This can happen for example when a nucleophile attacks one of the ring carbons, breaking the C-O bond. The alkoxide group (oxygen with a negative charge) is normally not a good leaving group, but the driving force for the reaction is release of ring strain.
Weak nucleophiles, which may not be strong enough to open the ring, can be assisted by acid catalysis. Protonation of the oxygen with strong acid enhances its leaving group ability, and it also makes the ring carbons more electrophilic (see below).
The positive charge in the protonated epoxide is actually shared by the oxygen and the two carbons. The resonance structures show partial ring opening with positive charges on each of the two carbons.
In an unsymmetrical epoxide, the greatest resonance contributor would be the one forming the most stable cation. That is, the one where the positive charge is on the carbon bearing the most alkyl groups.
The attack of the nucleophile on the protonated epoxide will therefore take place on the carbon that carries the greater amount of positive charge, even though it is seemingly more sterically hindered. To visualize this, remember that this is not a normal sp3 carbon with an ideal bond angle is 109.5 degrees. Forcing the internal ring angle to a theoretical value of 60 degrees in the epoxide forces the external angle to open up. Also, the resonance structures of the protonated epoxide show that such carbons have substantial sp2 character and therefore are more planar than a regular sp3 carbon.
The product then loses a proton (in an acid-base reaction with water, for example) and forms a vicinal dialcohol, or vic diol. The following example shows formation of the trans diol that results from backside attack of water on the epoxide.
Several of the reactions from chapter 8 will further illustrate different aspects of the chemistry of three membered rings to explain the observed products
17.06: Reaction Examples
SOLVED PROBLEM 8-1, p. 329
Analysis: Markovnikov addition of HBr to the C=C bond. Use HBr:
Complement:
Analysis: Anti-Markovnikov addition of HBr to the C=C bond. Use HBr in the presence of peroxides
Retrosynthetic analysis, or retrosynthesis:
Recommended problem: 8-4, p. 330
OXYMERCURATION - DEMERCURATION, ALSO CALLED OXYMERCURATION - REDUCTION
Markovnikov alcohol
Markovnikov alcohol without rearrangement
Contrast with addition of water in the presence of strong acid:
HYDROBORATION - OXIDATION SEQUENCE.
An effective way to make Anti-Markovnikov alcohols. Water adds to the double bond with syn-stereochemistry.
Anti-Markovnikov alcohol
SOLVED PROBLEM 8-3, p. 339
Problem 8-15 (b)
Can I make B from A by hydroboration - oxidation?
Notice that the methyl and the alcohol group are cis to each other.
Analysis:
Answer: NO
Recommended: 8-15 (c), p. 342
ADDITION OF CHLORINE OR BROMINE TO THE C=C BOND
Anti addition yields the trans product, which is chiral. Therefore the enantiomer also forms.
Recommended: 8-17, p. 345
Variation with water:
Recommended: 8-5 and 8-6, p. 347
CATALYTIC HYDROGENATION - Transformation of alkenes into alkanes (syn addition of hydrogen).
Recommended: 8-23, p. 350
ANTI HYDROXYLATION SEQUENCE- Anti addition of OH / OH to the C=C bond.
Recommended for syn and anti hydroxilation: 8-34 (all), p. 359
OXIDATIVE CLEAVAGE: Strong oxidation with potassium permanganate.
In this reaction each of the sp2 carbons involved in the pi bond gets oxidized to its maximum possible oxidation state. Refer to notes set # 20 (Oxidation and Reduction in Organic Chemistry) to find out what these states are.
OXIDATIVE CLEAVAGE: Ozonolysis
In this reaction each of the sp2 carbons involved in the pi bond gets oxidized either to aldehyde or ketone, depending on whether it ends up at the end of a carbon chain or in the middle after the pi bond cleaves. If the oxidized carbon ends up at the end of a carbon chain it becomes an aldehyde, otherwise it becomes a ketone.
Recommended strong oxidation (oxidative cleavage) with KMnO4 and with ozone: 8-7, 8-36, and 8-37, p. 362-363.
Recommended problems from the end of the chapter: 47 (all), 49 ( a-f ), 58 (all), 63 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/17%3A_Alkene_Reactions_Part_2/17.01%3A_Some_Mechanistic_Concepts_in_Electrophilic_Addition_Reactions_to_CC_Bonds.txt |
1. NOMENCLATURE - Refer to section 9-2 of the textbook for IUPAC and common names, and to the chart of functional group order of precedence on page 2 of this manuscript.
2. ACID-BASE REACTIONS OF TERMINAL ALKYNES - Refer to section 9-6 of the textbook, including 9-6A and 9-6B.
The acetylide ion can be used as a strong base or nucelophile.
3. USING ACETYLIDE IONS AS CARBON NUCLEOPHILES - Carbon nucleophiles are used for carbon chain expansion by creating new carbon-carbon bonds. In this chapter they are used in Sn2 reactions to make internal alkynes, and in reactions with carbonyl compounds to make alcohols.
(a) SYNTHESIS OF INTERNAL ALKYNES BY Sn2 REACTIONS (Sect. 9-6A).
(b) SYNTHESIS OF ALCOHOLS BY NUCLEOPHILIC ADDITION TO CARBONYL COMPOUNDS (Sect. 9-7B).
4. ADDITION OF ELECTROPHILES TO THE CARBON-CARBON TRIPLE BOND - Similar to addition of electrophiles to the carbon-carbon double bond, with some differences (Sect. 9-9A to 9-9F).
5. OXIDATION REACTIONS OF THE CARBON-CARBON TRIPLE BOND - Similar to oxidation reactions of the carbon-carbon double bond, with some differences (Sect. 9-10).
18.02: Functional Group Order of Precedence For Organic Nomenclature
In the IUPAC nomenclature system, organic molecules are grouped into specific classes of compounds determined by the main functional group present in the structure. A system of priorities is used to determine the main functional group, which determines the identity of the compound. All other functional groups are treated as substituents. The following order of precedence refers to functional groups containing carbon as the central atom. As a rule of thumb, the higher the oxidation state of the central carbon, the higher the priority of the functional group. Thus, carboxylic acids have higher priority than alcohols, and so on (See also table 21-1 in your textbook).
1. CARBOXYLIC ACIDS (highest priority among carbon-containing functional groups).
2. CARBOXYLIC ACID DERIVATIVES
3. OTHER GROUPS CONTAINING OXYGEN OR NITROGEN
4. ALKENES AND ALKYNES
Note: substances containing double and triple bonds are called alkenynes. (notice that the name ends in yne). Chain numbering starts from the end closest to either group, unless they’re both equidistant from the chain ends, in which case the double bond takes priority and is given the lower number. See examples in the textbook.
5. LOWEST PRIORITY. These groups are usually considered substituents in the main chain.
18.03: Use of Carbon Nucleophiles in Organic Synthesis
Carbon nucleophiles are widely used in organic synthesis to create new carbon-carbon bonds when they react with electrophiles, and therefore exapand a carbon chain. To be nucleophilic, the carbon atom must be bonded to a less electronegative atom to create a dipole favoring higher electron density on carbon. In most cases this atom is a metal, typically an alkali (group I) or alkaline earth (group II) metal. The difference in electronegativity between the carbon and the metal dictates the degree of nucleophilicity (or basicity) of the carbon atom. Some of the most commonly used metals are Li, Na, K, and Mg. Such species that contain a carbon-metal bond are known as organometallic reagents, or just organometallics, in organic chemistry. Examples are:
Organolithium reagents
Sodium acetylides
Organomagnesium halides (Grignard reagents)
Two important types of organometallics used as carbon nuclephiles are the acetylides (ch. 9) and the Grignard reagents (ch. 10). More specifically, we focus on the reactions between these substances and carbonyl compounds (aldehydes and ketones) to produce primary, secondary, and tertiary alcohols. Before we proceed, we must revisit the characteristics of electophiles, since it is a reaction between a nucleophile and an electrophile that we are considering now.
18.04: Classification of Electrophiles By Their Carbon Hybridization
There are two major types of carbon electrophiles, those containing sp3 carbon and those containing sp2 carbon. We now look at the characteristics of each.
1. ELECTROPHILES CONTAINING sp3 CARBON
To be of use in synthesis, the electrophilic center of substrates containing sp3 carbon must be sterically accessible and contain a good leaving group. This is the type encountered in ch. 6 in Sn2 reactions. Examples are many primary and secondary halides. The outcome of the reaction between a nucleophile and this type of electrophile is a substitution product.
If the nucleophile used is a carbon nucleophile, the product has an expanded carbon chain because a new carbon-carbon bond has been created. For example, the acetylide ion is used as a carbon nucleophile to make internal alkynes from terminal alkynes. The complete sequence is shown below. This type of reaction is discussed in sections 9-6 and 9-7A of the textbook.
2. ELECTROPHILES CONTAINING sp2 CARBON
To be of use in organic synthesis, electrophiles containing sp2 carbon must contain a polarized C=X bond, where X can be O, N, or S. In this discussion we consider only carbonyl compounds, where X=O. The polarization of the C=O bond leaves the carbon with low electron density. At the same time it enables the displacement of the π-electrons towards the oxygen to make room for the new carbon-carbon bond that forms when the nucleophile attacks. Examples of electrophiles containing sp2 carbon are most aldehydes and ketones.
The outcome of the nucleophilic attack is an addition product. The sp2 carbon changes hybridization to sp3 (tetrahedral), and π-electrons are displaced towards the oxygen atom, resulting in formation of an alkoxide ion, which is the conjugate base of the alcohol. The last step in the sequence then calls for treatment of the alkoxide ion with water or dilute acid to convert it into the alcohol. As long as the alcohol is desired, this last protonation step is taken for granted and sometimes left out in reaction sequences shown in textbooks. This doesn’t mean that the step was actually omitted, but simply that the author assumes awareness on the reader’s part of such obvious step.
Abbreviated synthetic sequence, as shown in most textbooks
An interesting trend can be observed in regards to the type of alcohol obtained as a product. Carbon nuclephiles, when reacted with formaldehyde (as in the first example), yield primary alcohols. When reacted with any other aldehyde (acetaldehyde in the second example) they yield secondary alcohols. Finally, when reacted with ketones (acetone in the third example), they yield tertiary alcohols.
CARBON NUCLEOPHILE + FORMALDEHYDE -------------> PRIMARY ALCOHOL
CARBON NUCLEOPHILE + OTHER ALDEHYDES ---------> SECONDARY ALCOHOL
CARBON NUCLEOPHILE + KETONE --------------------------> TERTIARY ALCOHOL
Refer to pages 379 - 382 of the Wade textbook (5th ed.) for additional examples using alkynide ions as carbon nucleophiles, and to pages 422 - 429 for examples using Grignard reagents as carbon nucleophiles (including the summary on p. 429). Mention should be made that Grignard chemistry is an important topic that gets heavily emphasized in organic chemistry II. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/18%3A_Important_Concepts_in_Alkyne_Chemistry/18.01%3A_Summary_of_Important_Topics_for_Alkynes_and_Alkyne_Chemistry.txt |
From the pattern outlined above, one can devise a synthesis for a primary, secondary, or tertiary alcohol by reversing the steps shown in the examples in the previous page. In other words, knowing the structure of the target molecule, one can reason backwards and arrive at the starting materials needed for the synthesis. This approach is called retrosynthetic analysis, and it is also highly emphasized in organic chemistry courses. An example will serve to illustrate this point.
PROBLEM: How can the following alcohol be synthesized from simpler starting materials?
SOLUTION: First, note that the alcohol is tertiary. Tertiary alcohols can be synthesized from carbon nucleophiles and ketones. Second, it is also an internal alkyne. This suggests that an alkynide ion can be used as the nucleophile that reacts with the ketone.
To arrive at the starting materials, first identify the carbon bearing the hydroxyl group. This is the electrophilic carbon in the ketone. Next, identify the sp-hybridized carbon bonded to the hydroxyl-bearing carbon. This is the nucleophilic carbon in the alkynide ion. This leads directly to the fragments that represent the starting materials.
The complete synthesis starting from a terminal alkyne could be written thus:
See problem 10-2 (p. 437 in your textbook) for a similar example using Grignard reagents as the carbon nucleophiles.
18.06: Keto-Enol Tautomerism
An organic structure that consists of a hydroxyl group (OH) attached to a carbon which is also part of a double bond is called an enol. This is because it consists of an alkene and an alcohol at the same time. Such arrangement is susceptible to a rearrangement called tautomerization.
Tautomerization is a shift of the hydroxyl hydrogen to carbon 2 of the alkene, and viceversa. This means that it is an equilibrium process which is either acid or base catalyzed. In the acid-catalyzed tautomerization, the alkene acts as a base and a proton from the acid is added to the p-bond with formation of the most stable carbocation.
EXAMPLE: Acid-catalyzed tautomerization
The second resonance structure shows what is essentially a protonated ketone. The positive charge on oxygen makes the proton highly acidic. It is readily lost to a base (water for example) to yield the free ketone.
The net equilibrium can be represented as in the example below. In the absence of special stabilizing factors, equilibrium usually favors the “keto” form (so called, even though it could be an aldehyde) over the enol form.
The tautomerizaton process can also be base-catalyzed. In this case the base takes up the acidic proton from the hydroxyl group, leading to formation of an enolate ion, which becomes reprotonated at the carbon atom.
EXAMPLE: Base-catalyzed tautomerization
The second resonance structure shows what is essentially a ketone.
The highly basic negatively charged carbon captures a proton to form the free ketone.
The following examples show how an enol whose hydroxyl group is at the end of a carbon chain tautomerizes into an aldehyde, whereas an enol whose hydroxyl group is in the middle of a carbon chain tautomerizes into a ketone. In both cases the “keto form” is favored by equilibrium.
Of course, when the hydroxyl group is attached to a cyclic structure, the “keto form” is always a ketone. The two forms in equilibrium constitute a tautomeric pair, and they are referred to as tautomers.
Needless to say, tautomers are not resonance structures. They are independent species in equilibrim with each other. To be tautomers, the two species must feature a central carbon which in the enol form contains both the hydroxyl group and the alkene, and in the keto form contains the carbonyl group. In the above example that carbon is number 1. The example below shows species which are NOT tautomers. The carbon that contains the hydroxyl group (or the carbonyl group) is not the same as the carbon that contains the double bond. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/18%3A_Important_Concepts_in_Alkyne_Chemistry/18.05%3A_Strategy_For_Alcohol_Synthesis_%28Retrosynthetic_Analysis%29.txt |
In ionic and free radical reactions, oxidation and reduction are defined as processes by which an element undergoes a net loss or gain of electrons, respectively. The concept as applied to organic covalent compounds, where elements share electrons rather than losing or gaining them is the same, but it’s frequently simplified and narrowed down to make it easier to recognize these processes. Therefore it must be kept in mind that, while the following definition is grossly simplified, it serves the goal of quickly identifying oxidation and reduction processes in many organic reactions.
In reference to organic molecules, oxidation is a process by which a carbon atom gains bonds to more electronegative elements, most commonly oxygen. Reduction is a process by which a carbon atom gains bonds to less electronegative elements, most commonly hydrogen.
The following chart summarizes these concepts when applied to organic transformations. [ox] stands for oxidation, and [red] stands for reduction.
Some points must be noted. First, a gain or loss of bonds means simply more or less bonds. The double bond counts as two bonds and the triple bond counts as three bonds. Thus in the carbonyl group (C=O) carbon is considered to have two bonds to oxygen. Therefore, this carbon has a higher oxidation state than the alcohol carbon, which has only one bond to oxygen. In routine terminology, it is said that the aldehyde is a more highly oxidized functional group than the alcohol. Oxidation reactions are therefore those in which the central carbon of a functional group is transformed into a more highly oxidized form, and reduction reactions are those in which the central carbon is transformed into a more highly reduced form.
Second, there can be several functional groups where the central carbon has the same oxidation state. For example, the carbons bonded to oxygen in alcohols and ethers have the same oxidation state. Likewise for aldehydes and their hydrated forms, and for carboxylic acids and their derivatives. However, most references to oxidation and reduction reactions in organic chemistry textbooks involve the functional groups presented in the chart above.
The chart presented before shows the oxidation and reduction states for a molecule that contains only one carbon. But most organic compounds contain more than one carbon. The maximum oxidation state that a particular carbon can attain depends on how many other carbons it must remain attached to. For example, a molecule with two carbon atoms could not be oxidized all the way to carbon dioxide because the carbon atom in CO2 must have four bonds to oxygen, leaving no room for bonds to other carbons. The maximum oxidation state that a carbon can attain decreases gradually as the number of bonds to other carbons increases. Thus, the maximum oxidation state possible for a carbon that’s bonded to one other carbon is the carboxylic acid stage, and so on. The following chart illustrates this idea.
ONE SINGLE CARBON
CARBON BONDED TO ONE OTHER CARBON
CARBON BONDED TO TWO OTHER CARBONS
CARBON BONDED TO THREE OTHER CARBONS
20.01: Common Synthetic Sequences For OChem I
Students frequently get overwhelmed by the increasingly large number of organic reactions and mechanisms they must learn. This task is made particularly difficult by the perception that it is just “busy work,” and that it serves no purpose other than to meet an academic requirement. It is true that a lot of organic reactions seem to lack meaning in and of themselves, but obviously they are not useless. Unfortunately students do not get exposed to the contexts in which this information comes to life until later, or sometimes never. The context of most immediate relevance from the point of view of the organic chemistry student is in the application of these reactions in synthesis.
A synthesis is a series of two or more reactions designed to obtain a specific final product. A synthetic step (not to be confused with a mechanistic step, which is something entirely different) is a single reaction that must be conducted separately from the others in a synthesis. Therefore the number of steps in a synthetic sequence is the same as the number of reactions that must be conducted separately, that is to say, the number of reactions that make up the sequence. The concept of synthetic strategy refers to the design of the most efficient combination of reactions that will yield the desired final product. This concept is widely used in R&D departments in the pharmaceutical industry to obtain synthetic drugs of many kinds. This type of research brings together widely different fields of science such as biochemistry, organic chemistry, biology, and even computer science into a single integrated task: the task of drug discovery.
Another context in which organic reactions acquire meaning is in biochemistry. Many of the reaction types discussed in introductory organic chemistry, such as nucleophilic substitutions, eliminations, and oxidations and reductions actually take place in biological systems. There are some differences in the way these reactions happen in biological systems as opposed to the organic chemistry lab. For example most biological reactions take place in water as the medium, not in organic solvents like methylene chloride. Another difference is in the catalysis. In the vast majority of biological systems reactions are catalyzed by enzymes, which are organic macromolecules that have a high degree of specificity and precision. For example, at the stereochemical level they are capable of distinguishing one enantiomer from another. They can catalyze the formation of a specific stereoisomer with 100% efficiency. By way of contrast, one can use many catalysts in the organic chemistry lab that could not possibly be used in biological systems. However, it is surprising that the same mechanistic principles studied in organic chemistry courses, such as the rules of proton transfers, nucleophilic attacks, and steric interactions actually apply in almost the same form in bioorganic mechanisms. Thus, an understanding of the physiology of an organism at the molecular level requires a solid understanding of organic mechanisms and reaction types.
With that in mind, we now turn to the task of learning a few basic synthetic sequences that a beginning organic chemistry student can use to get started in the art of synthetic design. A typical synthetic problem requires the design of a specific molecule or functional group from simple starting materials that can be obtained commercially, or which are readily available in the lab. It is then up to the synthetic chemist to choose from a variety of organic reactions those that will accomplish the task most effectively. Needless to say, this is an art as much as it is a science, and only experience can bring about improved synthetic skills. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/19%3A_Oxidation_States_of_Carbon/19.01%3A_Oxidation_and_Reduction_in_Organic_Chemistry.txt |
First thing to remember is that in nucleophilic reactions, the nucleophile replaces the leaving group unchanged only if it originally carried a negative charge. A neutral nucleophile, typically water or an alcohol, will lose a proton in the process of replacing the leaving group. Therefore, it is the conjugate base of such nucleophile that actually replaces the leaving group.
Sn2 reaction with a negatively charged nucleophile
Sn1 reaction with a neutral nucleophile (most frequently H2O or ROH)
At this level we are acquainted with only one major type of leaving groups, which are the halogens (Cl, Br, and I). So if we’re supposed to start the synthesis from an alkane, we must first fit it with a good leaving group, namely a halogen. Once we have a leaving group in place we can bring in a number of different nucleophiles to replace it, depending on which functional group we’re trying to obtain at the end.
EXAMPLE 1. From propane, make 2-propanol (isopropanol). In other words, how can the following transformation be accomplished?
Obviously we don’t have the tools to do this in one step. So we’ll need at least two steps to do it. A retrosynthetic analysis starts from the target product and works backwards step by step until it arrives at the desired starting material. In a retrosynthesis we start from the end and worry only about the step we’re working on, momentarily forgetting the rest of the requirements. In this case, we notice that the target product is an alcohol, and alcohols contain the OH group, which can be a nucleophile. To use a nucleophile we must first have a molecule with a good leaving group (halogen). Therefore the last step in the synthesis can look something like this:
Having solved the last step, now we worry about how we’re going to obtain the bromide (isopropyl bromide). In a flash of insight, we recall that in a previous part of the course we learned how to make bromides from alkanes (free radical halogenation reaction – bingo!). Feverishly, we try to recall the conditions that led to it, and suddenly we recall that all we need is an alkane and bromine in the presence of light. We desperately scramble for a pencil to jot that down before we forget it and come up with the following:
Not bad, we pat ourselves on the back. Now we’re through with the second to the last step of the synthesis. With a rush of adrenaline now pumping through our veins, we now ask what the next step would be. How can we obtain propane from...? Wait a minute! We’re supposed to start the synthesis with propane. Does that mean we’re done? You betcha! The complete synthesis would then look something like this:
or
Notice that a synthetic sequence does not show all the substances that are actually present in the reaction medium, especially not the inorganic ones such as HBr. For the sake of clarity, the sequence shows only those materials of direct relevance to the task at hand, namely those organic substances whose fate we’re interested in following due to their relationship to the end product.
Also notice that a synthetic problem typically has many solutions. For example, in the synthesis above we could’ve used chlorine as the leaving group instead of bromine. We can use water as the nucleophile or OH ion. These are all equally effective solutions. The final choice is therefore frequently based on factors other then those directly related to the chemistry, for example price and availability of the substances required, their toxicity, etc.
EXAMPLE 2. From cyclohexane prepare cyclohexane nitrile:
Nothing can stop us now. We immediately realize that CN is one of the nucleophiles we learned about in the chapter on nucleophilic substitutions. After a quick retrosynthetic analysis we whip out the following scheme (shown in reverse because we’re working backwards, remember?): | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/20%3A_Common_Synthetic_Sequences/20.02%3A_Starting_with_an_Alkane_Provide_a_Synthesis_for_a_Molecule_That_Has_a_Functional_Group_Z_Where_Z_is_a_Nucleophile.txt |
First question to ask in a lot of synthetic problems is, given the functional group I’m supposed to prepare, how many ways do I know right now that I can use to form that particular functional group? In this case I’m supposed to prepare an alkene. At this point the only way I know to prepare alkenes is by means of elimination reactions. According to the lecture notes, the most efficient approaches are the E2 reaction (dehydrohalogenation of bulky halides in the presence of strong bases), and the acid-catalyzed E1 dehydration of secondary and tertiary alcohols.
The E2 reaction requires a halogen as the leaving group, so I can prepare an alkyl halide from the required alkane by the free radical halogenation reaction, just as I did before. The E1 reaction on the other hand requires the preparation of an alcohol as the starting material. However, I find that at this point the only preparations of alcohols that I know start with alkenes (acid catalyzed addition of water, oxymercuration reaction, and hydroboration sequence). It would be redundant to start with an alkene to prepare an alcohol, only to dehydrate it back to alkene. Therefore, at this point, I will stick with the E2 reaction because it is more compatible with the requirements of the problem.
EXAMPLE 3. Starting with cyclohexane, provide a synthesis for cyclohexene.
No problemo. By now I’m a whiz of organic synthesis. so here is the quick solution.
EXAMPLE 4. From n-butane, prepare 2-butene
Again, no problem. A retrosynthetic analysis yields the following solution.
The only thing to note in this approach is that the last step above would yield a mixture of cis and trans isomers, since in this particular case, that step would not involve a stereospecific reaction.
20.04: Prepare a Substituted Acetylene
A monosubstituted acetylene can be prepared from acetylene by a combination of acid-base reaction followed by an Sn2 displacement. A disubstituted acetylene can be prepared from a monosubstituted acetylene by a similar reaction sequence. Before proceeding, make sure you understand this nomenclature by consulting the relevant section in the chapter on alkynes.
EXAMPLE 5. From acetylene prepare methylacetylene.
My first task is to remove the acidic proton from acetylene to convert it into its conjugate base. I can then use this conjugate base as a nucleophile in an Sn2 reaction with the relevant alkyl halide to yield the desired product.
Something to note here: NaNH2 is an ionic salt. The actual base is NH2 (amide ion). Sodium remains a spectator ion and as such does not directly participate in the process. Second thing to note is that to perform the first step efficiently, a base whose pKa is higher than the pKa of acetylene must be used. Otherwise equilibrium would not be favorable to formation of the conjugate base. In this case, acetylene has a pKa of 25, and NH2 has a pKa of about 38 (as measured by the pKa of its conjugate acid, ammonia). Because 38 is greater than 25, NH2 can be effectively used to deprotonate acetylene.
Disubstituted acetylenes follow a similar reaction sequence, but the deprotonation steps must be conducted separately. Trying to remove both protons from acetylene simultaneously would not work very well because the potential energy of the resulting species is too high. That is to say, it’s not very stable.
Possible but impractical. This species is high energy, therefore very unstable.
EXAMPLE 6. From acetylene prepare methylethylacetylene.
The synthetic sequence has four steps total.
20.05: Specifically Prepare Primary Secondary or Tertiary Alcohols Using Carbon Nucleophiles Such as Alkynide Ions and Grignard Reagents
These sequences are extensively discussed in the notes and in the Wade textbook (5th ed.). Please refer to the following sections and exercises from the textbook to illustrate this type of synthetic problem.
Chapter 9: Section 9–7B. See solved problem 9–2, and problem 9–8.
Chapter 10: Section 10–9. See solved problem 10–2, and problems 10–13 through 10–15.
In addition, work as many related problems as possible from the end of the chapters, such as problem 10–39. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/20%3A_Common_Synthetic_Sequences/20.03%3A_Starting_With_an_Alkane_Provide_a_Synthesis_For_an_Alkane_With_the_Same_Number_of_Carbons.txt |
Like carbon, hydrogen can be used as a nucleophile if it is bonded to a metal in such a way that the electron density balance favors the hydrogen side. A hydrogen atom that carries a net negative charge and bears a pair of unshared electrons is called a hydride ion. How much negative charge density resides on hydrogen depends on the difference in electronegativity between hydrogen and the metal it’s bonded to.
The following table shows the difference in electronegativity between hydrogen and some common metals. Obviously, the greater the difference in electronegativity, the greater the density of negative charge on hydrogen, and the greater the reactivity of the compound as a hydride delivering agent.
For many routine synthetic purposes, sodium and lithium hydrides are simply too reactive, requiring special handling such as inert atmosphere, and careful control of reaction conditions. Calcium hydride is more manageable because it is less reactive, and it is preferred in many reactions. However, many reductions of organic compounds such as carbonyl and carboxyl compounds use aluminum and boron hydride reagents.They are manageable in the laboratory, they are commercially available, and they can be modified to fine-tune their reactivity to various degrees for specific uses. Two of the most widely used hydride reagents in organic synthesis are lithium aluminum hydride, and sodium borohydride, shown below.
As can be seen from their structure, lithium and sodium are not bonded to hydrogen. They are merely counterions for the negative portion, which is the actual hydride–delivering agent. Second, they are each capable of delivering up to 4 hydride equivalents. Last, we expect sodium borohydride to be less reactive, and therefore more selective, than lithium aluminum hydride. This is in fact the case.
Another way to control the reactivity of these compounds is to replace two hydrogens with bulky alkyl groups, as in the following structures.
Diisobutylaluminum hydride (DIBAL, or DIBAL-H)
Lithium–tri–t–butoxyaluminum hydride LiAlH(OtBu)3
These modifications do two things. The bulky groups prevent fast access of the hydride reagent to the electrophile by a steric effect, and each of them is capable of delivering only one hydride ion instead of four. The mechanism of the hydride attack on a carbonyl carbon shown below demonstrates how these reagents in general work. The hydride-delivering agent must approach the carbonyl carbon until it’s close enough to deliver the hydride ion. At the same time, the pi electrons from the C=O bond move to create a new bond to the metal, forming an alkoxide ion as the product, which is an alcohol functional group equivalent.
The last step towards formation of the alcohol is then protonation of the alkoxide using water or dilute acid. As in the case of Grignard reactions, this step is always assumed, and as such it may or may not be shown explicitly.
Balanced equations reflect the number of hydride ions delivered for complete reduction:
21.02: Synthetic Outcomes
The attack of hydride ions as nucleophiles on electrophilic carbon creates a new C–H bond, therefore the carbon atom undergoes a reduction (gain of bonds to hydrogen). Because of this, this type of reactions is commonly referred to as reductions, even though the mechanism is a nucleophilic addition. Likewise, the hydride delivering agent is more commonly referred to as a hydride reducing agent, or just reducing agent.
Unlike the attack of carbon nucleophiles on a carbonyl carbon, the attack of a hydride ion produces no new C–C bonds and therefore there is no expansion of the carbon chain. Consequently, only primary and secondary alcohols can be made by this approach.
21.03: Reactions with Acid Chlorides and Esters
The mechanism of action of hydride reductions on acid chlorides and esters (carboxyl groups) is similar to that taking place with carbonyl compounds, except that acid chlorides and esters have a leaving group (–Cl and –OR). So the reaction does not stop at formation of the alkoxide ion as a tetrahedral intermediate, but keeps going with an internal nucleophilic displacement of the leaving group. The direct outcome of this process is formation of the corresponding carbonyl compound (aldehyde or ketone), which may or may not undergo further reduction to alcohol, depending on the nature of the reagents used and reaction conditions. The following mechanism illustrates this concept. For simplicity, only the hydride ion is shown.
If a full reactivity reducing agent such as LiAlH4 is used, the reaction does not stop at the aldehyde stage, since the carbonyl carbon of the aldehyde can be attacked by another hydride equivalent. This results in formation of the primary alcohol (after hydrolysis of the alkoxide ion) as the final product.
The net reaction then is:
The reaction with an ester is similar, but the leaving group is different (R’O ). Can you draw the mechanism that leads to formation of the products shown?
Notice that with both (and all) carboxyl groups, hydride reductions lead to formation of primary alcohols only. There is no possibility of forming secondary alcohols by this method because the carboxyl group is at the end of the carbon chain, or else the chain gets broken so that the carboxyl carbon ends up at the end of a chain in the final product. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/21%3A_Hydride_Reactions/21.01%3A_Using_Hydrogen_as_a_Nucleophile_in_Hydride_Reductions.txt |
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