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Virtually all biochemical reactions are catalyzed by proteins called enzymes. The catalytic power and specificity of enzymes is extraordinarily high. The reactions that they catalyze are generally enhanced in rate many orders of magnitude, often as much as $10^7$, over the nonenzymatic process. Consequently enzymatic reactions may occur under much milder conditions than comparable laboratory reactions. For example, the simple hydrolysis of an amide proceeds at a practical rate only on heating the amide in either strongly acidic or strongly basic aqueous solution, and even then reaction may not be complete for several hours. In contrast, hydrolysis of amide or peptide bonds catalyzed by typical proteolytic enzymes, such as trypsin, chymotrypsin, or carboxypeptidase A, occurs rapidly as physiological temperatures and physiological pH.$^7$ It is one of the remarkable attributes of many enzymes that they catalyze reactions that otherwise would require strongly acidic or basic conditions. Enzymes are strictly catalysts, however, and affect only the rate of reaction, not the position of equilibrium; they lower the energy of the transition state, not the energies of the reactants or products (see Figure 4-4).
Many enzymes appear to be tailor-made for one specific reaction involving only one reactant, which is called the substrate. Others can function more generally with different reactants (substrates). But there is no such thing as a universal enzyme that does all things for all substrates. However, nothing seems to be left to chance; even the equilibration of carbon dioxide with water is achieved with the aid of an enzyme known as carbonic anhydrase.$^8$ Clearly, the scope of enzyme chemistry is enormous, yet the structure and function of relatively few enzymes are understood in any detail. We can give here only a brief discussion of the mechanisms of enzyme action - first some general principles then some specific examples.
Aspects of the Mechanisms of Enzyme Reactions
An enzyme usually catalyzes a single chemical operation at a very specific position, which means that only a small part of the enzyme is intimately involved. The region of the enzyme structure where key reactions occur as the result of association of the substrate with the enzyme is called the active site. The initial association of the enzyme $\left( \ce{E} \right)$ and the substrate $\left( \ce{S} \right)$ is formation of an enzyme-substrate complex $\left( \ce{ES} \right)$:
$\ce{E} + \ce{S} \rightleftharpoons \ce{ES}$
Complexation could occur in many different ways, but for the intimate complexation required for catalysis, the enzyme must have, or must be able to assume, a shape complementary to that of the substrate. Originally, it was believed that the substrate fitted the enzyme somewhat like a key in a lock; this concept has been modified in recent years to the induced-fit theory, whereby the enzyme can adapt to fit the substrate by undergoing conformational changes (Figure 25-18). Alternatively, the substrate may be similarly induced to fit the enzyme. The complementarity is three-dimensional, an important factor in determining the specificity of enzymes to the structure and stereochemical configuration of the substrates.
Detailed structures for the active sites of enzymes are difficult to obtain and have been worked out only for a few enzymes that have been studied extensively by both chemical and x-ray methods. Very revealing information has been obtained by x-ray diffraction studies of complexes between the enzyme and nonsubstrates, which are molecules similar to actual substrates and complex with the enzyme at the active site, but do not react further. These substances often inhibit reaction of the normal substrate by associating strongly with the enzyme at the active site and not moving onward to products. The x-ray studies of enzymes complexed with nonsubstrates show that the active site generally is a cleft or cavity in the folded structure of the enzyme that is largely hydrophobic in character. The enzyme-substrate complex can be inferred to be held together largely by van der Waals attractive forces between like groups (Section 12-3C), hydrogen-bonding, and by electrostatic attraction between ionic or polar groups. To achieve a stereospecific catalyzed reaction, there must be at least three points of such interactions to align properly the substrate within the cavity of the enzyme.
The reaction of the $\ce{ES}$ complex may convert the substrate to product $\left( \ce{P} \right)$ directly, and simultaneously free the enzyme $\left( \ce{E} \right)$ to react with more of the substrate:
$\ce{ES} \rightarrow \ce{P} + \ce{E}$
However, the reaction between enzyme and substrate often is much more complex. In many cases, the substrate becomes covalently bound to the enzyme. Then, in a subsequent step, or steps, the enzyme-bound substrate $\left( \ce{ES'} \right)$ reacts to give products and regenerate the active enzyme $\left( \ce{E} \right)$:
$\ce{ES} \rightarrow \ce{ES'} \rightarrow \ce{P} + \ce{E}$
Carboxypeptidase A
The considerable detail to which we now can understand enzyme catalysis is well illustrated by what is known about the action of carboxypeptidase A. This enzyme (Section 25-7B and Table 25-3) is one of the digestive enzymes of the pancreas that specifically hydrolyze peptide bonds at the $\ce{C}$-terminal end. Both the amino-acid sequence and the three-dimensional structure of carboxypeptidase A are known. The enzyme is a single chain of 307 amino-acid residues. The chain has regions where it is associated as an $\alpha$ helix and others where it is associated as a $\beta$-pleated sheet. The prosthetic group is a zinc ion bound to three specific amino acids and one water molecule near the surface of the molecule. The amino acids bound ot zinc are His 69, His 196, and Glu 72; the numbering refers to the position of the amino acid along the chain, with the amino acid at the $\ce{N}$-terminus being number 1. The zinc ion is essential for the activity of the enzyme and is implicated, therefore, as part of the active site.
X-ray studies$^9$ of carboxypeptidase complexed with glycyltyrosine (with which it reacts only slowly) provide a detailed description of the active site, which is shown schematically in Figure 25-19a and is explained below.
1. The tyrosine carboxylate group of the substrate is associated by electrostatic attraction with the positively charged side chain of arginine 145 $\left( \ce{W} \right)$:
2. The tyrosine side chain of the substrate associates with a nonpolar pocket in the enzyme $\left( \ce{X} \right)$.
3. Hydrogen bonding possibly occurs between the substrate tyrosine amide unshared pair and the side-chain $\ce{HO}$ groups of the enzyme tyrosine 248 $\left( \ce{Y} \right)$.
4. The glycyl carbonyl oxygen in the substrate probably is coordinated with the zinc ion $\left( \ce{Z} \right)$, displacing the water molecule coordinated to the zinc in the uncomplexed enzyme.
5. A side-chain carboxylate anion of glutamic acid 270 is so situated with respect to the reaction center that it could well function as a nucleophile by attacking the glycine carbonyl carbon.
The arrangement of the enzyme-substrate complex suggests a plausible reaction mechanism analogous to nonenzymatic mechanisms of amide hydrolysis (Section 24-4). The carboxyl group of Glu 270 can add to the amide carbonyl to form a tetrahedral intermediate that then rapidly dissociates to release the terminal amino acid, leaving the rest of the substrate bound to the enzyme as a mixed anhydride which can be symbolized as $\ce{E(CO)OCOR_1}$. Reaction of the acyl-enzyme intermediate with water will release the peptide, minus the terminal amino acid, and regenerates the enzyme.
This postulated sequence of events may leave you wondering why the enzyme speeds up the hydrolysis, especially because the sequence proceeds through an energetically unfavorable reaction, the formation of a carboxylic anhydride from an amide and a carboxylic acid:
How, then, can we possibly expect that the enzyme could make the anhydride route be faster than the simple water route? The answer is the way in which interactions in the enzyme-substrate complex can stabilize the transition state for the anhydride-forming reaction. Thus for carboxypeptidase the zinc can act as a strong electrophile to facilitate attack on the amide carbonyl. Hydrogen bonding of the amide nitrogen to Tyr 248 both will facilitate attack on the carbonyl group and assist in the breaking of the $\ce{C-N}$ bond. Furthermore, the nonpolar environment of the alkyl side-chains of the enzyme will increase the nucleophilicity of the $\ce{-CO_2^-}$ group that forms the anhydride (see Section 8-7F). Inspection of Figure 25-20 shows qualitatively that if the energy of the transition state for formation of the anhydride is lowered greatly, the overall rate will be determined by the rate of hydrolysis of the anhydride! In this circumstance (assuming the anhydride hydrolysis is uncatalyzed), the efficiency of the enzyme in breaking the peptide bond is as great as it can be, at least by this particular pathway of peptide hydrolysis. Such efficiencies have been established for other enzymes.
Participation of Serine in Enzyme Action
Mechanisms similar to the one described for carboxypeptidase appear to operate in the hydrolysis of amide and ester bonds catalyzed by a number of proteinases and esterases. The substrate, here generalized as $\ce{RCOX}$, acts to acylate the enzyme, which subsequently reacts with water to give the observed products:
In the acylation step a nucleophilic group on one of the amino-acid side chains at the active site behaves as the nucleophile. As we have seen in Section 25-9B, the nucleophile of carboxypeptidase is the free carboxyl group of glutamic acid 270. In several other enzymes (chymotrypsin, subtilisin, trypsin, elastase, thrombin, acetylcholinesterase), it is the hydroxyl group of a serine residue:
This raises another question. Why is the serine hydroxyl an effective nucleophile when water and other hydroxylic compounds clearly are not similarly effective? Apparently, the nucleophilicity of the serine $\ce{-CH_2OH}$ is enhanced by acid-base catalysis involving proton transfers between acidic and basic side-chain functions in the vicinity of the active site. The serine is believed to transfer its $\ce{OH}$ proton to an amphoteric$^{10}$ site $\ce{B-A-H}$ on the enzyme at the same instant that the proton of $\ce{B-A-H}$ is transferred to another base $\ce{B'}^\ominus$ (Equation 25-8). These proton transfers are, of course, reversible:
$\tag{25-8}$
Loss of its proton makes the serine hydroxyl oxygen a much more powerful nucleophile, and even though the equilibrium of Equation 25-8 must lie far to the left at physiological pH, it can increase greatly the reactivity of the serine hydroxyl.
In chymotrypsin and subtilisin, this charge-relay network system, as it is called, is made up of a specific aspartic acid residue, acting as $\ce{B'}^\ominus$, and a specific histidine residue (acting as the amphoteric $\ce{B-A-H}$):
$^7$The slowness with which amide bonds are hydrolyzed in the presence of either strong acids or strong bases, and their susceptibility to hydrolysis under the influence of enzymes, clearly is a key advantage in the biological functioning of peptides. Amide hydrolysis in neutral solution has a favorable, but not large, equilibrium constant. Therefore it does not take a great deal of biochemical energy either to form or to hydrolyze peptide bonds. The resistance to ordinary hydrolysis provides needed stability for proteins, and yet when it is necessary to break down the peptide bonds of proteins, as in digestion, this can be done smoothly and efficiently with the aid of the proteolytic enzymes.
$^8$Many enzymes are named by adding the suffix -ase to a word, or words, descriptive of the type of enzymatic activity. Thus, esterases hydrolyze esters, proteinases hydrolyze proteins, reductases achieve reductions, and synthetases achieve synthesis of polypeptide chains, nucleic acid chains, and other molecules.
$^9$W. N. Lipscomb, Accounts of Chemical Research 3, 81 (1970); E. T. Kaiser and B. L. Kaiser, ibid. 5, 219 (1972). Lipscomb received the 1976 Nobel Prize in chemistry for structural work on boranes.
$^{10}$Amphoteric means that a substance can act either as an acid or as a base.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.09%3A_Enzymes.txt |
Many enzymes only operate in combination with organic molecules that are actually reagents for the reaction. These substances are called coenzymes or cofactors. Some coenzymes function with more than one enzyme and are involved in reactions with a number of different substrates.
Several of the B vitamins function as coenzymes or as precursors of coenzymes; some of these have been mentioned previously. Nicotinamide adenine dinucleotide $\left( \ce{NAD}^\oplus \right)$ which, in conjunction with the enzyme alcohol dehydrogenase, oxidizes ethanol to ethanal (Section 15-6C), also is the oxidant in the citric acid cycle (Section 20-10B). The precursor to $\ce{NAD}^\oplus$ is the B vitamin niacin or nicotinic acid (Section 23-2). Riboflavin (vitamin B$_2$) is a precursor of flavin adenine nucleotide $\left( \ce{FAD} \right)$, a coenzyme in redox processes rather like $\ce{NAD}^\oplus$ (Section 15-6C). Another example of a coenzyme is pyridoxal (vitamin B$_6$), mentioned in connection with the deamination and decarboxylation of amino acids (Section 25-5C). Yet another is coenzyme A $\left( \textbf{CoA} \ce{SH} \right)$, which is essential for metabolism and biosynthesis (Sections 18-8F, 20-10B, and 30-5A).
An especially interesting coenzyme is thiamine pyrophosphate (vitamin B$_1$) which, in conjunction with the appropriate enzyme, decarboxylates 2-oxopropanoic acid (pyruvic acid; Section 20-10B). We can write the overall reaction as follows:
Although we do not know just how thiamine binds to the enzyme, the essential features of the reaction are quite well understood. Thiamine has an acidic hydrogen at the 2-position of the azathiacyclopentadiene ring, and you should recognize that the conjugate base, $10$, is both a nitrogen ylide and a sulfur ylide (Section 16-4A):
The acidity of the ring proton of the thiamine ring is a consequence of the adjacent positive nitrogen and the known ability of sulfur to stabilize an adjacent carbanion. Nucleophilic attack of the anionic carbon of $10$ on $\ce{C_2}$ of 2-oxopropanoic acid is followed by decarboxylation:
The overall reaction introduces a two-carbon chain at the $\ce{C_2}$ position of the thiamine ring and the resulting modified coenzyme, $11$, functions in subsequent biological reactions as a carrier of a $\ce{CH_3-CHOH}-$ group and a potential source of a $\ce{CH_3CO}-$ group. The metabolism of glucose (Section 20-10) requires the conversion of pyruvate to ethanoyl CoA by way of $11$; and, in fermentation, the hydroxylethyl group of $11$ is released as ethanal, which is reduced to ethanol by $\ce{NADH}$ (see Section 15-6C for discussion of the reverse reaction).
Thiamine pyrophosphate also plays a key role in the biosynthetic reactions that build (or degrade) pentoses from hexoses. We have mentioned these reactions previously in connection with the Calvin cycle (Section 20-9) and the pentose-phosphate pathway (Section 20-10C).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
25.11: Enzyme Regulation
You may have wondered how the proteolytic enzymes such as trypsin, pepsin, chymotrypsin, carboxypeptidase, and others keep from self-destructing by catalyzing their own hydrolysis or by hydrolyzing each other. An interesting feature of the digestive enzymes is that they are produced in an inactive form in the stomach or the pancreas - presumably to protect the different kinds of proteolytic enzymes from attacking each other or other proteins.
The inactive precursors are called trypinogen, pepsinogen, chymotrypsinogen, and procarboxypeptidase. These precursors are converted to the active enzymes by hydrolytic cleavage of a few specific peptide bonds under the influence of other enzymes (trypsin, for example, converts chymotrypsinogen to chymotrypsin). The digestive enzymes do not appear to self-destruct, probably because they are so constructed that it is sterically impossible to fit a part of one enzyme molecule into the active site of another. In this connection, it is significant that chymotrypsin attacks denatured proteins more rapidly than natural proteins with their compact structures of precisely folded chains.
Presumably all enzymes must have some regulatory mechanism that turns them on and off as needed. Less is known about regulation mechanisms than about the enzymatic reactions themselves, but one type of control has been recognized. This occurs when a reaction product inhibits one of the reaction steps producing it by tying up the enzyme as a nonreactive complex (feedback inhibition). As the simplest example, suppose that the product $\left( \ce{P} \right)$ as well as the substrate $\left( \ce{S} \right)$ complexes with the enzyme $\left( \ce{E} \right)$; then we can write the following set of equilibria for the net reaction:
$\ce{E} + \ce{S} \rightleftharpoons \ce{ES} \rightarrow \ce{E} + \ce{P}$
$\ce{E} + \ce{P} \rightleftharpoons \ce{EP}$
Clearly, a reaction of this type will decrease in rate as the product accumulates. It may stop altogether if the active sites are saturated with the product, and it will start again only on removal of the product.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
25.12: Enzyme Technology
Because enzymes function nearly to perfection in living systems, there is great interest in how they might be harnessed to carry on desired reactions of practical value outside of living systems. The potential value in the use of enzymes (separate from the organisms that synthesize them) is undeniable, but how to realize this potential is another matter.
Practical use of separated enzymes is not new. Hydrolytic enzymes isolated from bacteria were widely used for a brief period to assist in removing food stains from clothing, but many people suffer allergic reactions to enzymes used in this way, and the practice was stopped. A major objective in enzyme technology is to develop an enzymatic process for the hydrolysis of cellulose to glucose (Section 20-7A). Some microorganisms do possess the requisite enzymes to catalyze the hydrolysis of the $\beta$-1,4 glucoside links in cellulose. If these enzymes could be harnessed for industrial production of glucose from cellulose, this could be an important supplementary food source. Technology already is available to convert glucose into ethanol and ethanoic acid, and from there to many chemicals now derived from petroleum.
A difficult problem in utilizing enzymes as catalysts for reactions in a noncellular environment is their instability. Most enzymes readily denature and become inactive on heating, exposure to air, or in organic solvents. An expensive catalyst that can be used only for one batch is not likely to be economical in an industrial process. Ideally, a catalyst, be it an enzyme or other, should be easily separable from the reaction mixtures and indefinitely reusable. A promising approach to the separation problem is to use the technique of enzyme immobilization. This means that the enzyme is modified by making it insoluble in the reaction medium. If the enzyme is insoluble and still able to manifest its catalytic activity, it can be separated from the reaction medium with minimum loss and reused. Immobilization can be achieved by linking the enzyme covalently to a polymer matrix in the same general manner as is used in solid-phase peptide synthesis (Section 25-7D).
Enzymes also have possible applications in organic synthesis. But there is another problem in addition to difficulties with enzyme stability. Enzymes that achieve carbon-carbon bond formation, the synthetases, normally require cofactors such as ATP. How to supply ATP in a commercial process and regenerate it continuously from ADP or AMP is a technical problem that has to be solved if the synthetases are to be economically useful. This is a challenging field of biological engineering.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.10%3A_Coenzymes.txt |
One of the most interesting and basic problems connected with the synthesis of proteins in living cells is how the component amino acids are induced to link together in the sequences that are specific for each type of protein. There also is the related problem of how the information as to the amino-acid sequences is perpetuated in each new generation of cells. We now know that the substances responsible for genetic control in plants and animals are present in and originate from the chromosomes of cell nuclei. Chemical analysis of the chromosomes has revealed them to be composed of giant molecules of deoxyribonucleoproteins, which are deoxyribonucleic acids (DNA) bonded to proteins. Since it is known that DNA rather than the protein component of a nucleoprotein contains the genetic information for the biosynthesis of enzymes and other proteins, we shall be interested mainly in DNA and will first discuss its structure. Part or perhaps all of a particular DNA is the chemical equivalent of the Mendelian gene - the unit of inheritance.
The Structure of DNA
The role of DNA in living cells is analogous to that of a punched tape used for controlling the operation of an automatic turret lathe - DNA supplies the information for the development of the cells, including synthesis of the necessary enzymes and such replicas of itself as are required for reproduction by cell division. Obviously, we would not expect the DNA of one kind of organism to be the same as DNA of another kind of organism. It is therefore impossible to be very specific about the structure of DNA without being specific about the organism from which it is derived. Nonetheless, the basic structural features of DNA are the same for many kinds of cells, and we mainly shall be concerned with these basic features in the following discussion.
In the first place, DNA molecules are quite large, sufficiently so to permit them to be seen individually in photographs taken with electron microscopes. The molecular weights vary considerably, but values of 1,000,000 to 4,000,000,000 are typical. X-ray diffraction indicates that DNA is made up of two long-chain molecules twisted around each other to form a double-stranded helix about $20 \: \text{Å}$ in diameter. The arrangement is shown schematically in $12$:
As we shall see, the components of the chains are such that the strands can be held together efficiently by hydrogen bonds. In agreement with this structure, it has been found that, when DNA is heated to about $80^\text{o}$ under proper conditions, the strands of the helix unwind and dissociate into two randomly coiled fragments. Furthermore, when the dissociated material is allowed to cool slowly under the proper conditions, the fragments recombine and regenerate the helical structure.
Chemical studies show that the strands of DNA have the structure of a long-chain polymer made of alternating phosphate and sugar residues carrying nitrogen bases, $13$:
The sugar is $D$-2-deoxyribofuranose, $14$, and each sugar residue is bonded to two phosphate groups by way of ester links involving the 3- and 5-hydroxyl groups:
The backbone of DNA is thus a polyphosphate ester of a 1,3-diol:
Each of the sugar residues of DNA is bonded at the 1-position of one of four bases: cytosine, $15$; thymine, $16$; adenine, $17$; and guanine, $18$. The four bases are derivatives of either pyrimidine or purine, both of which are heterocyclic nitrogen bases:
Unlike phenols (Section 26-1), structural analysis of many of the hydroxy-substituted aza-aromatic compounds is complicated by isomerism of the keto-enol type, sometimes called lactim-lactam isomerism. For 2-hydroxypyrimidine, $19$, these isomers are $19a$ and $19b$, and the lactam form is more stable, as also is true for cytosine, $15$, thymine, $16$, and the pyrimidine ring of guanine, $18$.
For the sake of simplicity in illustrating $\ce{N}$-glycoside formation in DNA, we shall show the type of bonding involved for the sugar and base components only (i.e., the deoxyribose nucleoside structure). Attachment of 2-deoxyribose is through a $\ce{NH}$ group to form the $\beta$-$\ce{N}$-deoxyribofuranoside (Section 20-5):
Esterification of the 5'-hydroxyl group of deoxyribose nucleosides, such as cytosine deoxyriboside, with phosphoric acid gives the corresponding nucleotides:$^{11}$
The number of nucleotide units in a DNA chain varies from about 3,000 to 10,000,000 Although the sequence of the purine and pyrimidine bases in the chains are not known, there is a striking equivalence between the numbers of certain of the bases regardless of the origin of DNA. Thus the number of adenine (A) groups equals the number of thymine (T) groups, and the number of guanine (G) groups equals the number of cytosine (C) groups: A = T and G = C. The bases of DNA therefore are half purines and half pyrimidines. Furthermore, although the ratios of A to G and T to C are constant for a given species, they vary widely from one species to another.
The equivalence between the purine and pyrimidine bases in DNA was accounted for by J. D. Watson and F. Crick (1953) through the suggestion that the two strands are constructed so that, when twisted together in the helical structure, hydrogen bonds are formed involving adenine in one chain and thymine in the other, or cytosine in one chain and guanine in the other. Thus each adenine occurs paired with a thymine and each cytosine with a guanine and the strands are said to have complementary structures. The postulated hydrogen bonds are shown in Figure 25-22, and the relationship of the bases to the strands in Figure 25-23.
Genetic Control and the Replication of DNA
It is now well established that DNA provides the genetic recipe that determines how cells reproduce. In the process of cell division, the DNA itself also is reproduced and thus perpetuates the information necessary to regulate the synthesis of specific enzymes and other proteins of the cell structure. In replicating itself prior to cell division, the DNA double helix evidently separates at least partly into two strands (see Figure 25-24). Each of the separated parts serves as a guide (template) for the assembly of a complementary sequence of nucleotides along its length. Ultimately, two new DNA double strands are formed, each of which contains one strand from the parent DNA.
The genetic information inherent in DNA depends on the arrangement of the bases (A, T, G, and C) along the phosphate-carbohydrate backbone - that is, on the arrangement of the four nucleotides specific to DNA. Thus the sequence A-G-C at a particular point conveys a different message than the sequence G-A-C.
It is quite certain that the code involves a particular sequence of three nucleotides for each amino acid. Thus the sequence A-A-A codes for lysine, and U-C-G codes for serine. The sequences or codons for all twenty amino acids are known.
Role of RNA in Synthesis of Proteins
It is clear that DNA does not play a direct role in the synthesis of proteins and enzymes because most of the protein synthesis takes place outside of the cell nucleus in the cellular cytoplasm, which does not contain DNA. Furthermore, it has been shown that protein synthesis can occur in the absence of a cell nucleus or, equally, in the absence of DNA. Therefore the genetic code in DNA must be passed on selectively to other substances that carry information from the nucleus to the sites of protein synthesis in the cytoplasm. These other substances are ribonucleic acids (RNA), which are polymeric molecules similar in structure to DNA, except that $D$-2-deoxyribofuranose is replaced by $D$-ribofuranose and the base thymine is replace by uracil, as shown in Figure 25-25.
RNA also differs from DNA in that there are not the same regularities in the overall composition of its bases and it usually consists of a single polynucleotide chain. There are different types of RNA, which fulfill different functions. About $80\%$ of the RNA in a cell is located in the cytoplasm in clusters closely associated with proteins. These ribonucleoprotein particles specifically are called ribosomes, and the ribosomes are the sites of most of the protein synthesis in the cell. In addition to the ribosomal RNA (rRNA), there are ribonucleic acids called messenger RNA (mRNA), which convey instructions as to what protein to make. In addition, there are ribonucleic acids called transfer RNA (tRNA), which actually guide the amino acids into place in protein synthesis. Much is now known about the structure and function of tRNA.
The principal structural features of tRNA molecules are shown schematically in Figure 25-26. Some of the important characteristics of tRNA molecules are summarized as follows.
1. There is at least one particular tRNA for each amino acid.
2. The tRNA molecules have single chains with 73-93 ribonucleotides. Most of the tRNA bases are adenine (A), cytosine (C), guanine (G), and uracil (U). There also are a number of unusual bases that are methylated derivatives of A, C, G, and U.
3. The clover-leaf pattern of Figure 25-26 shows the general structure of tRNA. There are regions of the chain where the bases are complementary to one another, which causes it to fold into two double-helical regions. The chain has three bends or loops separating the helical regions.
4. The 5'-terminal residue usually is a guanine nucleotide; it is phosphorylated at the 5'-$\ce{OH}$. The terminus at the 3' end has the same sequence of three nucleotides in all tRNA's, namely, CCA. The 3'-$\ce{OH}$ of the adenosine in this grouping is the point of attachment of the tRNA to its specific amino acid:
With this information on the structure of tRNA, we can proceed to a discussion of the essential features of biochemical protein synthesis.
The information that determines amino-acid sequence in a protein to be synthesized is contained in the DNA of a cell nucleus as a particular sequence of nucleotides derived from adenine, guanine, thymine, and cytosine. For each particular amino acid there is a sequence of three nucleotides called a codon.
The information on protein structure is transmitted from the DNA in the cell nucleus to the cytoplasm where the protein is assembled by messenger RNA. This messenger RNA, or at least part of it, is assembled in the nucleus with a base sequence that is complementary to the base sequence in the parent DNA. The assembly mechanism is similar to DNA replication except that thymine (T) is replaced by uracil (U). The uracil is complementary to adenine in the DNA chain (see Figure 25-27). After the mRNA is assembled, it is transported to the cytoplasm where it becomes attached to the ribosomes.
The amino acids in the cytoplasm will not form polypeptides unless activated by ester formation with appropriate tRNA molecules. The ester linkages are through the 3'-$\ce{OH}$ of the terminal adenosine nucleotide (Equation 25-9) and are formed only under the influence of a synthetase enzyme that is specific for the particular amino acid. The energy for ester formation comes from ATP hydrolysis (Sections 15-5F and 20-10). The product is called an aminoacyl-tRNA.
The aminoacyl-tRNA's form polypeptide chains in the order specified by codons of the mRNA bound to the ribosomes (see Figure 25-28). The order of incorporation of the amino acids depends on the recognition of a codon in mRNA by the corresponding anticodon in tRNA by a complementary base-pairing of the type A $\cdots$ U and C $\cdots$ G. The first two bases of the codon recognize only their complementary bases in the anticodon, but there is some flexibility in the identity of the third base. Thus phenylalanine tRNA has the anticodon A-A-G and responds to the codons U-U-C and U-U-U, but not U-U-A or U-U-G:
The codons of the mRNA on the ribosomes are read from the 5' to the 3' end. Thus the synthetic polynucleotide (5')A-A-A-(A-A-A)$_n$-A-A-C(3') contains the code for lysine (A-A-A) and asparagine (A-A-C); the actual polypeptide obtained using this mRNA in a cell-free system was Lys-(Lys)$_n$-Asn, and not Asn-(Lys)$_n$-Lys.
The start of protein synthesis is signaled by specific codon-anticodon interactions. Termination is also signaled by a codon in the mRNA, although the stop signal is not recognized by tRNA, but by proteins that then trigger the hydrolysis of the completed polypeptide chain from the tRNA. Just how the secondary and tertiary structures of the proteins are achieved is not yet clear, but certainly the mechanism of protein synthesis, which we have outlined here, requires little modification to account for preferential formation of particular conformations.
$^{11}$The positions on the sugar ring are primed to differentiate them from the positions of the nitrogen base.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.13%3A_Biosynthesis_of_Proteins.txt |
A problem of great interest to those curious about the evolution of life concerns the origins of biological molecules. When and how were the molecules of life, such as proteins, nucleic acids, and polysaccharides, first synthesized?
In the course of geological history, there must have been a prebiotic period when organic compounds were formed and converted to complex molecules similar to those we encounter in living systems. The composition of the earth's atmosphere in prebiotic times was almost certainly very different from what it is today. Probably it was a reducing atmosphere consisting primarily of methane, ammonia, water, and because there was little or no free oxygen, there was no stratosphere ozone layer and little, if any, screening from the sun's ultraviolet radiation. Starting with $\ce{CH_4}$, $\ce{NH_3}$, and $\ce{H_2O}$, it is plausible that photochemical processes would result in formation of hydrogen cyanide, $\ce{HCN}$, and methanal, $\ce{CH_2O}$, by reactions such as the following:
$\ce{CH_4} + \ce{NH_3} \overset{h \nu}{\longrightarrow} \ce{HCN} + 3 \ce{H_2}$
$\ce{CH_4} + \ce{H_2O} \overset{h \nu}{\longrightarrow} \ce{CH_2=O} + 2 \ce{H_2}$
Hydrogen cyanide and methanal are especially reasonable starting materials for the prebiotic synthesis of amino acids, purine and pyrimidine bases, ribose, and other sugars. Formation of glycine, for example, could have occurred by a Strecker synthesis (Section 25-6), whereby ammonia adds to methanal in the presence of $\ce{HCN}$. Subsequent hydrolysis of the intermediate aminoethanenitrile would produce glycine:
The plausibility of these reactions is strongly supported by the classic experiments of S. Miller (1953), who showed that a mixture of methane or ethane, ammonia, and water, on prolonged ultraviolet irradiation or exposure to an electric discharge, produced a wide range of compounds including racemic $\alpha$- and $\beta$-amino acids.
It is not difficult to visualize how sugars such as ribose may be formed. Methanal is known to be converted by bases through a series of aldol-type additions to a mixture of sugarlike molecules called "formose". Formation of racemic ribose along with its stereoisomers could occur as follows:
It is more difficult to conceive how the nucleic acid bases such as adenosine may be achieved in prebiotic syntheses. However, adenine, $\ce{C_5H_5N_5}$, corresponds in composition to a pentamer of hydrogen cyanide and could result by way of a trimer of $\ce{HCN}$, $20$, and the adduct of ammonia with $\ce{HCN}$, $21$:
Without worrying about the mechanistic details of how adenine could be formed from $20$ and $21$, you can see that the adenine ring system is equivalent to $20$ plus two $21$ minus two $\ce{NH_3}$:
How did these small prebiotic organic molecules grow into large polymeric substances such as peptides, RNA, and so on? It is important to recognize that by whatever reactions polymerization occurred, they had to be reactions that would occur in an essentially aqueous environment. This presents difficulties because condensation of amino acids to form peptides, or of nucleotides to form RNA or DNA, is not thermodynamically favorable in aqueous solution.
It is quite possible that primitive polypeptides were formed by the polymerization of aminoethanenitriles produced by the addition of ammonia and $\ce{HCN}$ to methanal or other aldehydes. The resulting imino polymers certainly would hydrolyze to polypeptides:
There are many other questions regarding the origin of the molecules of life for which we have only partial answers, or no answers at all.
It is difficult to imagine just how these molecules, once formed, somehow evolved further into the extraordinarily complex systems afforded by even the simplest bacterium able to utilize energy from the sun to support and reproduce itself. Nonetheless, synthetic peptides do coil and aggregate rather like natural proteins, and some also have shown catalytic activities characteristic of natural enzymes. One would hope that some kind of life would be found elsewhere in the solar system, the analysis of which would help us to better understand how life began on earth.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.14%3A_Chemical_Evolution.txt |
Generally, the reactivity of a substituent on an aromatic ring is greatly modified from that of its aliphatic counterpart. Likewise, the substituent can influence the reactivity of the ring. We have seen this interplay between ring and substituent in the chemistry of aryl halides (Section 14-6), of arenamines (Sections 23-7C and 23-9F), and in electrophilic substitution reactions of aromatic compounds (Section 22-5). It is particularly manifest in the chemistry of substances that have oxygen attached directly to arene rings. We shall discuss aryl oxygen compounds and some of their oxidation products called quinones in this chapter. We also shall discuss aromatic substances that have carbon substituents in the form of alkyl, haloalkyl (such as \(\ce{-CH_2Cl}\), \(\ce{-CHCl_2}\), \(\ce{-CCl_3}\)), aldehyde (\(\ce{-CHO}\)), and carboxylic acid (\(\ce{-CO_2H}\)) groups. We classify such substances as aromatic side-chain derivatives (for want of a better term)
• 26.1: Aryl Oxygen Compounds
Phenols are enols, and enols normally are unstable with respect to the corresponding carbonyl compounds. The situation is different for phenols because of the inclusion of the carbon-carbon double bond into the aromatic ring and the associated aromatic stabilization. Phenol (benzenol) exists exclusively in the enol form. In general, phenols are somewhat more polar than the corresponding saturated alcohols.
• 26.2: Quinones
Quinones are not aromatic compounds but are conjugated cyclic diketones. However, quinones and the related aromatic arenols are readily interconverted, and their chemistry is largely interdependent. A characteristic and important reaction of quinones is reduction to the corresponding arenediols. The reduction products of 1,4-quinones are called hydroquinones. Reduction can be achieved electrochemically and with differing reducing agents.
• 26.3: Tropolones and Related Compounds
The tropolones make up a very interesting class of nonbenzenoid aromatic compound that was discovered first in several quite different kinds of natural products. As one example, the substance called ββ -thujaplicin or hinokitiol has been isolated from the oil of the Formosan cedar and is 4-isopropyltropolone.
• 26.4: Some Aromatic Side-Chain Compounds
We have discussed in this chapter and in previous chapters how the reactivity of halogen, amino, and hydroxy substituents are modified when linked to aromatic carbons rather than to saturated carbons. Other substituents, particularly those linked to an aromatic ring through a carbon-carbon bond, also are influenced by the ring, although usually to a lesser degree. We shall refer to aromatic compounds containing substituents of this type as aromatic side-chain compounds.
• 26.5: Natural Occurrence and Uses of Some Aromatic Side-Chain Compounds
Derivatives of aromatic aldehydes occur naturally in the seeds of plants. For example, amygdalin is a substance occurring in the seeds of the bitter almond. It is a derivative of gentiobiose, which is a disaccharide made up of two glucose units; one of the glucose unites is bonded by a β-glucoside linkage to the OHOH group of the cyanohydrin of benzenecarbaldehyde
• 26.6: Correlations of Structure with Reactivity of Aromatic Compounds
This section is concerned with the quantitative correlation of reaction rates and equilibria of organic reactions with the structure of the reactants. We will restrict the discussion to benzene derivatives. The focus is on a remarkably simple treatment developed by L. P. Hammett in 1935, which has been tremendously influential. Hammett's correlation covers chemical reactivity, spectroscopy and other physical properties, and even the biological activity of drugs.
• 26.E: More on Aromatic Compounds (Exercises)
These are the homework exercises to accompany Chapter 26 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
26: More on Aromatic Compounds
Phenols (Arenols) - Physical Properties
Phenols are enols, and enols normally are unstable with respect to the corresponding carbonyl compounds (Section 17-1D). Thus
The situation is different for phenols because of the inclusion of the carbon-carbon double bond into the aromatic ring and the associated aromatic stabilization. Phenol (benzenol) exists exclusively in the enol form:
The physical properties of some representative phenols are summarized in Table 26-1. In general, phenols are somewhat more polar than the corresponding saturated alcohols. The magnitudes of the differences are well illustrated by comparison of the physical properties of benzenol and cyclohexanol, shown in Table 26-2. The determining factor appears to be the greater acidity of the phenolic hydroxyl group, which means that, in the undissociated form, the $\ce{O-H}$ bond is more strongly polarized as $\overset{\delta \ominus}{\ce{O}}-\overset{\delta \oplus}{\ce{H}}$ than for alcohols. Phenols therefore form stronger hydrogen bonds than alcohols, thereby resulting in higher boiling points, higher water solubility, and increased ability to act as solvents for reasonably polar organic molecules.
Table 26-1: Physical Properties of Some Representative Phenols
The wavelengths of the ultraviolet absorption maxima of the arenols shown in Table 26-1 indicate a considerable effect of substituents on these absorptions, which correspond to the $200$-$\text{nm}$ and $255$-$\text{nm}$ absorptions of benzene (Section 22-3B).
Table 26-2: Comparative Physical Properties of Benzenol and Cyclohexanol
Substances such as 2-hydroxybenzaldehyde, 2-hydroxybenzoic acid, and 2-nitrobenzenol form intra- rather than intermolecular hydrogen bonds. This effectively reduces intermolecular attraction, thereby reducing boiling points and increasing solubility in nonpolar solvents as compared to the meta and para isomers, which only form intermolecular hydrogen bonds:
Synthesis of Phenols
Benzenol and the 2-, 3-, and 4-methylbenzenols (cresols) can be isolated from coal tar (Section 22-11). Benzenol itself is used commercially in such large quantities that alternate methods of preparation are necessary and most of these start with benzene or alkylbenzenes. Direct oxidation of benzene is not satisfactory because benzenol is oxidized more readily than is benzene.
At one time, benzenol was made industrially by sulfonating or chlorinating benzene and then introducing the hydroxyl group by nucleophilic substitution with strong alkali:
Current commercial syntheses of benzenol involve oxidation of methylbenzene or isopropylbenzene (Section 16-9E). Oxidation of isopropylbenzene is economically feasible for the production of benzenol because 2-propanone (acetone) also is a valuable product:
A common laboratory procedure converts an aromatic amine to a phenol by way of the arenediazonium salt, $\ce{ArNH_2} \rightarrow \ce{ArN_2^+} \rightarrow \ce{ArOH}$ (Section 23-10B).
Reactions of Phenols Involving the $\ce{O-H}$ Bonds
The reactions of the hydroxyl groups of phenols, wherein the $\ce{O-H}$ bonds are broken, are similar to those of alcohols. Thus phenols are weak acids ($K_a = 10^{-10}$ to $10^{-8}$; Table 26-1), intermediate in strength between carboxylic acids and alcohols.
Enols are stronger acids than alcohols because of the increase in electron delocalization in enolate anions as compared to the neutral enols (see Section 15-8A). The stabilization energy of benzenol (Table 21-1) is $48 \: \text{kcal mol}^{-1}$, $5 \: \text{kcal}$ greater than that of benzene. We can ascribe this increase to delocalization of an unshared electron pair from oxygen:
When the $\ce{OH}$ proton is removed by a base the resulting anion has even greater stabilization, because the corresponding valence-bond structures do not involve charge separation:
We can be confident that substituent groups that stabilize the anion will increase the acidity. Thus 4-nitrobenzenol is about 500 times stronger as an acid than benzenol, because of the increased delocalization of charge to the nitro group:
It is possible to prepare esters of phenols with carboxylic acid anhydrides or acid halides, and phenyl ethers by reaction of benzenolate anion with halides, sulfate esters, sulfonates, or other alkyl derivatives that react well by the $S_\text{N}2$ mechanism:
Phenols (like carboxylic acids; Section 24-7C and Table 18-6) are converted to methoxy derivatives with diazomethane:
Almost all phenols and enols (such as those of 1,3-diketones) give colors with ferric chloride in dilute water or alcohol solutions. Benzenol itself produces a violet color with ferric chloride, and the methylbenzenols give a blue color. The products apparently are ferric arenolate salts, which absorb visible light to give an excited state having electrons delocalized over both the iron atom and the unsaturated system.
Reactions of Phenols Involving the $\ce{C-O}$ Bonds
In general, it is very difficult to break the aromatic $\ce{C-O}$ bond of arenols. Thus concentrated halogen acids do not convert simple arenols to aryl halides, and alkoxyarenes are cleaved with hydrogen bromide or hydrogen iodide in the manner $\ce{ArO-R}$ rather than $\ce{Ar-OR}$:
(Diaryl ethers, such as diphenyl ether, do not react with hydrogen iodide even at $200^\text{o}$.) There is no easy way to convert arenols to aryl halides, except where activation is provided by 2- or 4-nitro groups. Thus 2,4-dinitrobenzenol is converted to 1-chloro-2,4-dinitrobenzene with phosphorus pentachloride:
An exception to the generalization that $\ce{C-O}$ bonds to aromatic systems are difficult both to make and to break is provided by reversible conversion of benzenediols and 1- or 2-naphthalenols to the corresponding amines, usually at elevated temperatures with sodium hydrogen sulfite or an acidic catalyst. The sodium hydrogen sulfite-induced reaction is called the Bucherer reaction:
These reactions do not work well with simple benzenols because the key step is formation of the keto isomer of the arenol - a process that is unfavorable for simple benzenols.
$\tag{26-1}$
The role of the hydrogen sulfite is participation in a reversible 1,4-addition to the unsaturated ketone to hold it in the ketone form that then is converted to the imine by $\ce{NH_3}$ (Section 16-4C) and hence to the arenamine:
Substitution Reactions at the Ring Carbons of Arenols
The electron-rich $\pi$-orbital systems of arenols and especially of arenolate ions make these compounds very susceptible to electrophilic substitution. Arenols typically react rapidly with bromine in aqueous solution to substituted the positions ortho or para to the hydroxyl group. Benzenol itself gives 2,4,6-tribromobenzenol in high yield:
Several important reactions of arenols involve aromatic substitution of arenolate ions with carbon electrophiles. In a sense, these reactions are alkylation and acylation reactions as discussed for arenes (Sections 22-4E and 22-4F). In another sense, they are alkylation and acylation reactions of enolate anions and therefore could give rise to products by $\ce{C}$- and $\ce{O}$-alkylation, or $\ce{C}$- and $\ce{O}$-acylation (Section 17-4). Thus:
In most cases, $\ce{O}$-alkylation predominates. However, with 2-propenyl halides either reaction can be made essentially the exclusive reaction by proper choice of solvent. With sodium benzenolate the more polar solvents, such as 2-propanone, lead to 2-propenyloxybenzene, whereas in nonpolar solvents, such as benzene, 2-(2-propenyl)benzenol is the favored product:
It should be noted that formation of 2-(2-propenyl)benzenol in nonpolar solvents is not the result of $\ce{O}$-propenylation followed by rearrangement, even though the $\ce{C}$-propenylation product is thermodynamically more stable. Rearrangement in fact does occur, but at much higher temperatures (above $200^\text{o}$) than required to propenylate sodium benzenolate:
Such rearrangements are quite general for aryl allyl ethers and are called Claisen rearrangements. They are examples of the pericyclic reactions discussed in Section 21-10D.
The Kolbe-Schmitt reaction produces $\ce{O}$- and $\ce{C}$-carboxylation through the reaction of carbon dioxide with sodium benzenolate at $125^\text{o}$:
Sodium benzenolate absorbs carbon dioxide at room temperature to form sodium phenyl carbonate ($\ce{O}$-carboxylation) and, when this is heated to $125^\text{o}$ under a pressure of several atmospheres of carbon dioxide, it rearranges to sodium 2-hydroxybenzoate (sodium salicylate). However, there is no evidence that this reaction is other than a dissociation-recombination process, in which the important step involves electrophilic attack by carbon dioxide on the aromatic ring of the benzenolate ion ($\ce{C}$-carboxylation):
With the sodium benzenolate at temperatures of $125^\text{o}$ to $150^\text{o}$, ortho substitution occurs; at higher temperatures ($250^\text{o}$ to $300^\text{o}$), particularly with the potassium salt, the para isomer is favored.
The Kolbe-Schmitt reaction is related to enzymatic carboxylations as of $D$-ribulose-1,5-diphosphate with carbon dioxide, a key step in photosynthesis (Section 20-9). The overall result is $\ce{C-C}$ bond formation by addition of $\ce{CO_2}$ to an enolate salt or its enamine equivalent.
In the somewhat related Reimer-Tiemann reaction, sodium benzenolate with trichloromethane in alkaline solution forms the sodium salt of 2-hydroxybenzenecarbaldehyde (salicylaldehyde). The electrophile in this case probably is dichlorocarbene (Section 14-7B):
Many phenols undergo aldol-like addition reactions with carbonyl compounds in the presence of acids or bases. Thus benzenol reacts with methanal under mild alkaline conditions to form (4-hydroxyphenyl)methanol:
The use of this type of reaction in the formation of polymers will be discussed in Chapter 29.
Arenols usually will undergo diazo coupling reactions with aryldiazonium salts at pH values high enough to convert some of the arenol to the more powerfully nucleophilic arenolate anions:
However, if the pH is too high, coupling is inhibited because the diazonium salt is transformed into $\ce{ArN=N-O}^\ominus$, which is the nonelectrophilic conjugate base of a diazotic acid (Table 23-4).
Addition Reactions
Arenols can be reduced successfully with hydrogen over nickel catalysts to the corresponding cyclohexanols. A variety of alkyl-substituted cyclohexanols can be prepared in this way:
Oxidation of Arenols
Benzenol can be oxidized to 1,4-benzenedione (para-benzoquinone) by chromic acid. The reaction may proceed by way of phenyl hydrogen-chromate (Section 15-6B) as follows:
Oxidation reactions of arenols with other oxidants are complex. Oxidative attack seems to involve, as the first step, removal of the hydroxyl hydrogen to yield a phenoxy radical:
The subsequent course depends upon the substituents on the aromatic ring. With 2,4,6-tri-tert-butylbenzenol, the radical is reasonably stable in benzene solution and its presence is indicated by both its dark-blue color and the fact that it adds to 1,3-butadiene:
Apparently, dimerization of the above phenoxy radical through either oxygen or the ring is inhibited by the bulky tert-butyl groups. With fewer or smaller substituents, the phenoxy radicals may form dimerization or disproportionation products. Examples of these reactions follow.
Dimerization:
Disproportionation:
Arene Polyols
Several important aromatic compounds have more than one arene hydroxyl group. These most often are derivatives of the following dihydric and trihydric arenols, all of which have commonly used (but poorly descriptive) names:
All are exceptionally reactive towards electrophilic reagents, particularly in alkaline solution, and all are readily oxidized. The 1,2- and 1,4-benzenediols, but not 1,3-benzenediol, are oxidized to quinones:
The preparation of these substances can be achieved by standard methods for synthesizing arenols, but most of them actually are made on a commercial scale by rather special procedures, some of which are summarized as follows:
The gallic acid used in the preparation of 1,2,3-benzenetriol can be obtained by microbial degradation of tannins, which are complex combinations of glucose and gallic acid obtained from oak bark and gallnuts. A few other representatives of the many types of naturally occurring derivatives of polyhydric arenols are
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/26%3A_More_on_Aromatic_Compounds/26.01%3A_Aryl_Oxygen_Compounds.txt |
Quinones are not aromatic compounds but are conjugated cyclic diketones. Yet it is convenient to discuss their chemistry at this point because quinones and the related aromatic arenols are readily interconverted, and their chemistry is largely interdependent.
A variety of quinonelike structures have been prepared, the most common of which are the 1,2 - and 1,4-quinones as exemplified by 1,2- and 1,4-benzenediones. Usually the 1,2-quinones are more difficult to make and are more reactive than the 1,4-quinones. A few 1,6- and 1,8-quinones also are known.
No 1,3-quinones are known, possibly because they would have nonplanar, highly strained structures and therefore would be unstable:
A number of quinones are known in which the quinone arrangement extends over more than one ring. Examples are:
Reduction of Quinones
A characteristic and important reaction of quinones is reduction to the corresponding arenediols. The reduction products of 1,4-quinones are called hydroquinones:
$\tag{26-2}$
Reduction can be achieved electrochemically and with a variety of reducing agents (metals in aqueous acid, catalytic hydrogenation). Such reductions are unusual among organic reactions in being sufficiently rapid and reversible to give easily reproducible electrode potentials in an electrolytic cell. The position of the 1,4-benzenediol-1,4,-benzenedione equilibrium (Equation 26-2) is proportional to the square of the hydrogen-ion concentration. Therefore the electrode potential is sensitive to pH; a change of one unit of pH in water solution changes the potential of the electrode by $0.059 \: \text{V}$. Before the invention of the glass-electrode pH meter, the half-cell potential developed by this equilibrium was used widely to determine pH values of aqueous solutions. The method is not very useful above pH 9 because the quinone reacts irreversibly with alkali.
Numerous studies have been made of the relationship between half-cell reduction potentials and the structures of quinones. As might be expected, the potentials are greatest when the resonance stabilization associated with formation of the aromatic ring is greatest.
When alcohols solutions of hydroquinone and quinone are mixed, a brown-red color develops and a green-black 1:1 complex crystallizes that is known as quinhydrone. This substance is a charge-transfer complex (Section 24-6C), with the diol acting as the electron donor and the dione as the electron acceptor. Quinhydrone is not very soluble and dissociates considerably to its components in solution.
The reduction of a quinone requires two electrons, and it is possible that these electrons could be transferred either together or one at a time. The product of a single-electron transfer leads to what appropriately is called a semiquinone, $1$, with both a negative charge and an odd electron (a radical anion):
The formation of relatively stable semiquinone radicals by electrolytic reduction of quinones has been established by a variety of methods. Some semiquinone radicals undergo reversible dimerization reactions to form peroxides.
A particularly stable cation-radical of the semiquinone type is formed by mild oxidation of $\ce{N}$,$\ce{N}$,$\ce{N'}$,$\ce{N'}$-tetramethyl-1,4-benzenediamine. The cation, which is isolable as a brilliant-blue perchlorate salt, $2$, is called "Wurster's Blue":
Quinones of Biological Importance
Oxidation and reduction in biochemical systems involve many reactions that are similar to the arenediol-arenediol couple. We have mentioned several previously: $\ce{NADP}^\oplus \rightleftharpoons \ce{NADPH}$ (Section 20-9), and $\ce{FADH_2} \rightleftharpoons \ce{FAD}$ (Section 15-6C).
An important question in metabolic oxidation is just how reduction of oxygen $\left( \frac{1}{2} \ce{O_2} + 2 \ce{H}^\oplus + 2 \ce{e} \rightarrow \ce{H_2O} \right)$ is linked to the oxidation of $\ce{NADH}$ $\left( \ce{NADH} \rightarrow \ce{NAD}^\oplus + \ce{H}^\oplus + 2 \ce{e} \right)$. The route for transfer of electrons from $\ce{NADH}$ to oxygen (oxidation plus phosphorylation; Section 20-10) is indirect, complicated, and involves, in an early stage, oxidation of $\ce{NADH}$ by flavin mononucleotide $\left( \ce{FMN} \right)$ by the reaction $\ce{FMN} + \ce{NADH} + \ce{H}^\oplus \rightarrow \ce{FMNH_2} + \ce{NAD}^\oplus$. But the reduced form of $\ce{FMN}$, $\ce{FMNH_2}$, does not react directly with oxygen. Instead, it reduces a quinone called coenzyme Q (CoQ) to the corresponding arenediol:
The effect of this step is to form a slightly polar reductant (CoQ$\ce{H_2}$) from a strongly polar reductant $\left( \ce{FMNH_2} \right)$, and this permits the reduced material to penetrate into a less polar region of the oxidative apparatus. The reduced CoQ does not react directly with oxygen but is a participant in a chain of oxidation-reduction reactions involving electron transfer between a number of iron-containing proteins known as cytochromes. At the end of this chain of reaction, the reduced form of a copper-containing cytochrome actually reacts with oxygen. The sequence of electron-carriers may be summarized as
$\ce{NADH} \rightarrow \ce{FMNH_2} \rightarrow \textbf{CoQ}\ce{H_2} \rightarrow \text{cytochromes} \rightarrow \ce{O_2}$
A related process occurs in photosynthesis (Section 20-9). You will recall that a critical part of photosynthesis involves the transfer of electrons from the photosystem that oxidizes water $\left( \ce{H_2O} \rightarrow \frac{1}{2} \ce{O_2} + 2 \ce{H}^\oplus + 2 \ce{e} \right)$ to the photosystem that reduces $\ce{NADP}^\oplus$ $\left( \ce{NADP}^\oplus + 2 \ce{H}^\oplus + 2 \ce{e} \rightarrow \ce{NADPH} + \ce{H}^\oplus \right)$. As in oxidative phosphorylation, the electron-transfer route is complex. However, one of the electron carriers is a quinone called plastoquinone that closely resembles coenzyme Q found in animals. Plastoquinone, like coenzyme Q, is reduced to the hydroquinone form, which is part of an electron-transport chain involving iron- and copper-containing proteins:
Among other naturally occurring substances having quinone-type structures, one of the most important is the blood antihemorrhagic factor, vitamin K$_1$, which occurs in green plants and is a substituted 1,4-naphthalenedione:
The structure of vitamin K$_1$ has been established by degradation and by synthesis. Surprisingly, the long alkyl side chain of vitamin K$_1$ is not necessary for its action in aiding blood clotting because 2-methyl-1,4-naphthoquinone is almost equally active on a molar basis.
Besides playing a vital role in the oxidation-reduction processes of living organisms, quinones occur widely as natural pigments found mainly in plants, fungi, lichens, marine organisms, and insects (see alizarin, Section 28-4A, as representative of a natural anthraquinone-type dye).
Photographic Developers
Photography makes important practical use of the arenediol-arenediol oxidation-reduction system. Exposure of the minute grains of silver bromide in a photographic emulsion to blue light (or any visible light in the presence of suitably sensitizing dyes) produces a stable activated form of silver bromide, the activation involving generation of some sort of crystal defect. Subsequently, when the emulsion is brought into contact with a developer, which may be an alkaline aqueous solution of 1,4-benzenediol (hydroquinone) and sodium sulfite, the particles of activated silver bromide are reduced to silver metal much more rapidly than the ordinary silver bromide. Removal of the unreduced silver bromide with sodium thiosulfate ("fixing") leaves a suspension of finely divided silver in the emulsion in the form of the familiar photographic negative.
A variety of compounds are used as photographic developing agents. They are not all arenediols. In fact, most are aromatic aminoalcohols or diamines, but irrespective of their structural differences, they all possess the ability to undergo redox reactions of the type described for 1,4-benzenediol. Structural formulas and commercial names for several important developers are
1,4-Benzenediamine also is an effective developing agent, but it may cause dermatitis in sensitive individuals. $\ce{N}$,$\ce{N}$-Diethyl-1,4-benzenediamine is used as a developer in color photography. These substances react with silver bromide to produce benzenediamine derivatives:
Addition Reactions of Quinones
Being $\alpha$,$\beta$-unsaturated ketones, quinones have the potential of forming 1,4-addition products in the same way as their open-chain analogs (Section 17-5B). 1,4-Benzenedione undergoes such additions rather readily. The products are unstable and undergo enolization to give substituted 1,4-benzenediols. Two examples are the addition of hydrogen chloride and the acid-catalyzed addition of ethanoic anhydride. In the latter reaction, the hydroxyl groups of the adduct are acylated by the anhydride. Hydrolysis of the product yields 1,2,4-benzenetriol:
Quinones usually undergo Diels-Alder additions readily, provided that they have at least one double bond that is not part of an aromatic ring. With 1,4-benzenedione and 1,3-butadiene, either the mono- or diadduct can be obtained. The monoadduct enolizes under the influence of acid or base to a 1,4-benzenediol derivative:
Quinones of Cyclobutadiene
Benzoquinones owe their unusual properties as $\alpha$,$\beta$-unsaturated ketones to the ease by which they are transformed to stable aromatic systems. How would these properties change if the quinone were derived from nonaromatic structures, such as cyclobutadiene, cyclooctatetraene, or pentalene? There is no final answer to this question because few such substances have been prepared, the best known so far being the mono- and diphenylcyclobutenediones:
For example, $3$ can be prepared from sulfuric acid hydrolysis of the cycloaddition product of ethynylbenzene and trifluorochloroethene (Section 13-3D):
The dione, $3$, is a yellow crystalline solid that, despite its strained four-membered ring, is much less reactive than 1,2-benzenedione (ortho-benzoquinone). It cannot be reduced to a cyclobutenediol, does not undergo Diels-Alder reactions, and with bromine gives a substitution product rather than addition. The bromo compound so formed hydrolyzes rapidly to a hydroxy compound, $4$, which is an extraordinarily strong acid having an ionization constant about $10^9$ times that of benzenol:
A related compound, 3,4-dihydroxy-1,2-cyclobutenedione, $5$, also has been prepared and is a very strong dibasic acid. It is sometimes called "squaric acid":
From the data so far available, it appears that the quinones corresponding to cyclobutadiene have more aromatic character than do the cyclobutadienes themselves.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
26.03: Tropolones and Related Compounds
The tropolones make up a very interesting class of nonbenzenoid aromatic compound that was discovered first in several quite different kinds of natural products. As one example, the substance called $\beta$-thujaplicin or hinokitiol has been isolated from the oil of the Formosan cedar and is 4-isopropyltropolone:
Tropolone itself can be prepared in a number of ways, the most convenient of which involves oxidation of 1,3,5-cycloheptatriene with alkaline potassium permanganate. The yield is low but the product is isolated readily as the cupric salt:
The cycloheptatriene for this synthesis can be obtained best by thermal rearrangement of the Diels-Alder addition product of cyclopentadiene and ethyne:
Tropolone is an acid with an ionization constant of $10^{-7}$, which is intermediate between the $K_a$ of ethanoic acid and the $K_a$ of benzenol. Like most arenols, tropolones form colored complexes with ferric chloride solution. Tropolone has many properties that attest to its aromatic character - it resists hydrogenation, undergoes diazo coupling, and can be nitrated, sulfonated, and substituted with halogens. The aromaticity of tropolone can be attributed to resonance involving the two nonequivalent VB structures $6a$ and $6b$, and to several dipolar structures, such as $6c$ and $6d$, in which the ring has the stable tropylium cation structure with six $\pi$ electrons (Section 21-9B):
The tropylium cation is prepared easily by transfer of a hydride ion from cycloheptatriene to triphenylmethyl cation in sulfur dioxide solution. This reaction is related to the hydride ion transfer, $\ce{(CH_3)_3C}^\oplus + \ce{RH} \rightarrow \ce{(CH_3)_3CH} + \ce{R}^\oplus$, discussed in Section 10-9:
Seven equivalent VB structures can be written for the tropylium cation so only one seventh of the positive charge is expected to be on each carbon. Because the cation has six $\pi$ electrons, it is expected from Huckel's $\left( 4n + 2 \right)$ $\pi$-electron rule to be unusually stable for a carbocation.
The infrared and Raman spectra of tropylium bromide in hydrobromic acid solution have no common bands, which means that the cation exists in a highly symmetrical form in this solution (see Section 9-8). At higher pH, reversible formation of the hydroxy compound occurs:
$\ce{C_7H_7^+} + \ce{OH^-} \rightleftharpoons \ce{C_7H_7OH}$
The equilibrium constant for this reaction is such that the cation is half converted to the hydroxy compound at about pH 5.
Colchicine is an important naturally occurring tropolone derivative. It is isolated from the autumn crocus and is used in medicine for the treatment of gout. It also has an effect on cell division and is used in plant genetic studies to cause doubling of chromosomes. The structure has been confirmed by total synthesis.
Contributors
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/26%3A_More_on_Aromatic_Compounds/26.02%3A_Quinones.txt |
We have discussed in this chapter and in previous chapters how the reactivity of halogen, amino, and hydroxy substituents are modified when linked to aromatic carbons rather than to saturated carbons. Other substituents, particularly those linked to an aromatic ring through a carbon-carbon bond, also are influenced by the ring, although usually to a lesser degree. Examples include $\ce{-CH_2OH}$, $\ce{-CH_2OCH_3}$, $\ce{-CH_2Cl}$, $\ce{-CHO}$, $\ce{-COCH_3}$, $\ce{-CO_2H}$, and $\ce{-CN}$, and we shall refer to aromatic compounds containing substituents of this type as aromatic side-chain compounds. Our interest in such compounds will be directed mainly to reactions at the side chain, with particular reference to the effect of the aromatic ring on reactivity. In this connection, we shall discuss the relatively stable triarylmethyl cations, anions, and radicals, as well as quantitative correlation of rates of organic reactions by what is known as the Hammett equation.
Preparation of Aromatic Side-Chain Compounds
Carboxylic acids can be obtained from most alkylbenzenes by oxidation of the side chain with reagents such as potassium permanganate, potassium dichromate, or nitric acid:
Under the conditions of oxidation, higher alkyl or alkenyl groups are degraded and ring substituents, other than halogen and nitro groups, often fail to survive. As an example, oxidation of 5-nitro-2-indanone with dilute nitric acid leads to 4-nitro-1,2-benzenedicarboxylic acid:
To retain a side-chain substituent, selective methods of oxidation are required. For example, 4-methylbenzoic acid can be prepared from 1-(4-methylphenyl)ethanone by the haloform reaction (Section 17-2B):
Many side-chain halogen compounds can be synthesized by reactions that also are applicable to alkyl halides (see table 14-5), but there are other methods especially useful for the preparation of arylmethyl halides. The most important of these are the chloromethylation of aromatic compounds (to be discussed later in this section) and radical halogenation of alkylbenzenes.
Light-induced, radical chlorination or bromination of alkylbenzenes with molecular chlorine or bromine was discussed previously (Section 14-3C). Under these conditions, methylbenzene reacts with chlorine to give successively phenylchloromethane, phenyldichloromethane, and phenyltrichloromethane:
Related reactions occur with other reagents, notably sulfuryl chloride, $\ce{SO_2Cl_2}$; tert-butyl hypochlorite, $\ce{(CH_3)_3COCl}$; and $\ce{N}$-bromobutanimide, $\ce{(CH_2CO)_2NBr}$. The $\alpha$ substitution of alkylbenzenes is the result of radical-chain reactions (Section 14-3C).
Side-chain fluorine compounds with the groupings $\ce{-CHF_2}$, $\ce{-CF_2}-$, and $\ce{-CF_3}$ are available by the reaction of sulfur tetrafluoride or molybdenum hexafluoride with carbonyl compounds (see Section 16-4D):
Some specific examples follow:
Arylmethyl chlorides or bromides are quite reactive compounds that are readily available or easily prepared, and as a result they are useful intermediates for the synthesis of other side-chain derivatives. Thus phenylmethyl chloride can be hydrolyzed to phenylmethanol, converted to phenylethanenitrile with alkali-metal cyanides, and oxidized to benzenecarbaldehyde (benzaldehyde):
Phenyldichloromethane hydrolyzes readily to benzaldehyde, and phenyltrichloromethane to benzoic acid:
Carbon side chains also may be introduced by direct substitution and several such reactions have been discussed in detail previously. These include Friedel-Crafts alkylation and acylation (Section 22-4E and 22-4F), the Gattermann-Koch reaction for preparation of aldehydes from arenes and carbon monoxide (Section 22-4F), and the Kolbe-Schmitt, Reimer-Tiemann, and Gattermann reactions for synthesis of acids and aldehydes from arenols (Section 26-1E).
Chloromethylation is a useful method for substitution of $\ce{-CH_2Cl}$ for an aromatic hydrogen, provided one starts with a reasonably reactive arene. The reagents are methanol and hydrogen chloride in the presence of zinc chloride:
$\ce{C_6H_6} + \ce{CH_2O} + \ce{HCl} \overset{\ce{ZnCl_2}}{\longrightarrow} \ce{C_6H_5CH_2Cl} + \ce{H_2O}$
The mechanism of the chloromethylation reaction is related to that of Friedel-Crafts alkylation and acylation and probably involves an incipient chloromethyl cation, $^\oplus \ce{CH_2Cl}$:
Triarylmethyl Cations
Phenylmethyl halides are similar in $S_\text{N}1$ reactivity to 2-propenyl halides. The $S_\text{N}1$ reactivity of phenylmethyl derivatives can be ascribed mainly to stabilization of the cation by electron delocalization. Diphenylmethyl halides, $\ce{(C_6H_5)_2CHX}$, are still more reactive and this is reasonable because the diphenylmethyl cation has two phenyl groups over which the positive charge can be delocalized and therefore should be more stable relative to the starting halide than is the phenylmethyl cation:
Accordingly, we might expect triphenylmethyl (or trityl) halides, $\ce{(C_6H_5)_3C-X}$, to be even more reactive. In fact, the $\ce{C-X}$ bonds of such compounds are extremely labile. In liquid sulfur dioxide, triarylmethyl halides ionize reversibly, although the equilibria are complicated by ion-pair association:
Triarylmethyl cations are among the most stable carbocations known. They are intensely colored and are formed readily when the corresponding triarylmethanols are dissolved in strong acids:
Triarylmethyl Anions
In addition to stable cations, triarylmethyl compounds form stable carbanions. Because of this the corresponding hydrocarbons are relatively acidic compared to simple alkanes. They react readily with strong bases such as sodium amide, and the resulting carbanions, like the cations, are intensely colored:
The acid strengths of arylmethanes are listed in Table 26-3. All are quite weak acids relative to water but vary over many powers of ten relative to one another. The stronger acids form the more stable carbanions, and the carbanion stability generally is determined by the effectiveness with which the negative charge can be delocalized over the substituent aryl groups.
Table 26-3: Strengths of Some Hydrocarbon Acids
Triarylmethyl Radicals
Triarylmethyl compounds also form rather stable triarylmethyl radicals, and indeed the first stable carbon free radical to be reported was the triphenylmethyl radical, $\ce{(C_6H_5)_3C} \cdot$, prepared inadvertently by M. Gomberg in 1900. Gomberg's objective was to prepare hexaphenylethane by a Wurtz coupling reaction of triphenylmethyl chloride with metallic silver:
However, he found that unless air was carefully excluded from the system, the product was triphenylmethyl peroxide, $\ce{(C_6H_5)_3COOC(C_6H_5)_3}$, rather than the expected hexaphenylethane.
Gomberg believed that the yellow solution obtained from the reaction of triphenylmethyl chloride with silver in benzene in the absence of air contained the triphenylmethyl radical. However, subsequent investigations showed that the molecular weight of the dissolved substance was closer to that for $\ce{C_{38}H_{30}}$, hexaphenylethane. A bitter battle raged over the nature of the product and its reactions. The controversy finally was though to have been settled by the demonstration that the hydrocarbon $\ce{C_{38}H_{30}}$ dissociates rapidly, but only slightly, to triphenylmethyl radicals at room temperature in inert solvents ($K = 2.2 \times 10^{-4}$ at $24^\text{o}$ in benzene). For many years thereafter, the hydrocarbon $\ce{C_{38}H_{30}}$ was believed to be the hexaphenylethane. Now it is known that this conclusion was incorrect. The product is a dimer of triphenylmethyl, but it is formed by the addition of one radical to the 4-position of a phenyl ring of the other:
Formation of the peroxide in the presence of oxygen is explained as follows:
Although the foregoing reactions involving the triphenylmethyl radical seemed very unreasonable at the time they were discovered, the stability of the radical now has been established beyond question by a variety of methods such as esr spectroscopy (Section 27-9). This stability can be attributed to delocalization of the odd electron over the attached phenyl groups:
Aromatic Aldehydes. The Benzoin Condensation
Most of the reactions of aromatic aldehydes, $\ce{ArCHO}$, are those expected of aldehydes with no $\alpha$ hydrogens and most of these will not be reviewed here. One reaction that usually is regarded as being characteristic of aromatic aldehydes (although, in fact, it does occur with other aldehydes having no $\alpha$ hydrogens), is known as the benzoin condensation. This reaction essentially is a dimerization fo two aldehyde molecules through the catalytic action of sodium or potassium cyanide:
Unsymmetrical or mixed benzoins may be obtained in good yield from two different aldehydes:
As to the mechanism of benzoin formation, cyanide ion adds to the aldehyde to form $12$. This anion is in equilibrium with $13$, wherein the negative charge can be delocalized over the phenyl and nitrile groups. A subsequent aldol-type addition of $13$ to the carbonyl carbon of a second aldehyde molecule gives the addition product $14$, and loss of $\ce{HCN}$ from $14$ leads to the benzoin:
Benzoins are useful intermediates for the synthesis of other compounds because they can be oxidized to 1,2-diones and reduced in stages to various products, depending upon the reaction conditions. The 1,2-diketone known as benzil, which is obtained by nitric acid oxidation of benzoin, undergoes a base-catalyzed hydration rearrangement to form an $\alpha$-hydroxy acid, commonly called the benzilic acid rearrangement (see Section 17-7):
Benzils, like other 1,2-diones, react with 1,2-benzenediamines to form diazaarenes known as quinoxalines. This kind of reaction is an important general procedure for the synthesis of aromatic ring systems containing nitrogen:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
26.05: Natural Occurrence and Uses of Some Aromatic Side-Chain Compounds
Derivatives of aromatic aldehydes occur naturally in the seeds of plants. For example, amygdalin is a substance occurring in the seeds of the bitter almond. It is a derivative of gentiobiose, which is a disaccharide made up of two glucose units; one of the glucose unites is bonded by a $\beta$-glucoside linkage to the $\ce{OH}$ group of the cyanohydrin of benzenecarbaldehyde:
The flavoring vanillin occurs naturally as a glucovanillin (a glucoside) in the vanilla bean (Section 20-5). It is made commercially in several ways. One is from eugenol, itself a constituent of several essential oils:
Methyl 2-hydroxybenzoate (methyl salicylate, oil of wintergreen) occurs in many plants, but it also is readily prepared synthetically by esterification of 2-hydroxybenzoic acid, which in turn is made from benzenol (see Section 26-1E):
The ethanoyl derivative of 2-hydroxybenzoic acid is better known as aspirin and is prepared from the acid with ethanoic anhydride, using sulfuric acid as catalyst:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/26%3A_More_on_Aromatic_Compounds/26.04%3A_Some_Aromatic_Side-Chain_Compounds.txt |
This section is concerned with the quantitative correlation of reaction rates and equilibria of organic reactions with the structure of the reactants. We will restrict the discussion to benzene derivatives. The focus is on a remarkably simple treatment developed by L. P. Hammett in 1935, which has been tremendously influential. Hammett's correlation covers chemical reactivity, spectroscopy and other physical properties, and even the biological activity of drugs. Virtually all quantitative treatments of reactivity of organic compounds in solution start with the kinds of correlations that are discussed in this section.
The Hammett Equation
If we compare the acid strengths $\left( K_a \right)$ of a series of substituted benzoic acids with the strength of benzoic acid itself (Table 26-4), we see that there are considerable variations with the nature of the substituent and its ring position, ortho, meta, or para. Thus all three nitrobenzoic acids are appreciably stronger than benzoic acid in the order ortho $\gg$ para $>$ meta. A methoxy substituent in the ortho or meta position has a smaller acid-strengthening effect, and in the para position decreases the acid strength relative to benzoic acid. Rate effects also are produced by different substituents, as is evident from the data in Table 26-5 for basic hydrolysis of some substituted ethyl benzoates. A nitro substituent increases the rate, whereas methyl and methoxy substituents decrease the rate relative to that of the unsubstituted ester.
Table 26-4: Dissociation Constants $\left( 10^{-5} \times K_a \right)$ of Some Substituted Benzoic Acids in Water at $25^\text{o}$
Table 26-5: Specific Rate Constants$^a$ for Alkaline Hydrolysis of Some Substituted Ethyl Benzoates in $85\%$ Ethanol-Water Solution at $30^\text{o}$
The straight line in Figure 26-3 can be expressed conveniently by Equation 26-3, in which the two variables are log $k$ and log $K$, the slope of the line is $\rho$, and the intercept is $C$:
$\text{log} \: k = \rho \: \text{log} \: K + C \tag{26-3}$
For the particular case for which the ring substituent is hydrogen $\left( \ce{R=H} \right)$, Equation 26-3 becomes
$\text{log} \: k_0 = \rho \: \text{log} \: K_0 + C \tag{26-4}$
in which $K_0$ is the dissociation constant of benzoic acid and $k_0$ is the rate of hydrolysis of ethyl benzoate. Subtracting Equation 26-4 from Equation 26-3 we obtain
$\text{log} \: \frac{k}{k_0} = \rho \: \text{log} \: \frac{K}{K_0} \tag{26-5}$
This equation could be tested on the ratios of any rates or equilibrium constants, but it is convenient to reserve log $\left( K/K_0 \right)$ for the dissociation of benzoic acids in water at $25^\text{o}$ (Table 26-4) and correlate the rate or equilibrium constants for all other processes with log $\left( K/K_0 \right)$. The common procedure is to rewrite Equation 26-5 as Equation 26-6:
$\text{log} \: \frac{k}{k_0} = \rho \sigma \tag{26-6}$
in which $\sigma$ is defined as:
$\sigma = \text{log} \: \frac{K}{K_0} \tag{26-7}$
Equation 26-6 is known as the Hammett equation, but before we discuss its general applications, it will be helpful to say more about the $\sigma$ term in Equation 26-7.
The relative strength of a substituted benzoic acid and hence the value of $\sigma$ depends on the nature and position of the substituent in the ring. For this reason, $\sigma$ is called the substituent constant. Table 26-6 lists several substituent constants and these will be seen to correspond to the polar character of the respective substituents. Thus the more electron-attracting a substituent is, by either resonance or induction, the more acid-strengthening it is, and the more positive is its $\sigma$ value (relative to $\ce{H}$ as 0.000). Conversely, the more strongly a substituent donates electrons by resonance or induction, the more negative is its $\sigma$ value. We expect that among the more electron-attracting and electron-donating substituents will be those with electric charges, positive and negative respectively. Indeed, a diazonium group $\left( \ce{-N_2^+} \right)$ in the para position has a very large $\sigma$ value of +1.91, whereas a para $\ce{-O}^\ominus$ group has a $\sigma$ value of -1.00. In general, meta $\sigma$ constants correspond to the inductive effect of the substituent while the para $\sigma$ constants represent the net influence of inductive and resonance effects. If there is a substantial resonance effect, and it and the inductive effect operate in the same direction, $\sigma_\text{para}$ will have a considerably greater magnitude than $\sigma_\text{meta}$. The converse will be true if the resonance and inductive effects operate in opposite directions.
Table 26-6: Hammett Substituent Constants
Usually, $\rho$ for a given reaction is influenced by conditions such as the temperature and composition of the solvent. However, the changes normally are not large unless an actual change in mechanism occurs with a change in the reaction conditions.
Scope of the Hammett Equation
The Hammett treatment provides a correlation of much experimental data. Tables 26-6 and 26-7 contain 38 substituent constants and 16 reaction constants. This means that we can calculate relative $k$ or $K$ values for 608 individual reactions. To illustrate, let us suppose that we need to estimate the relative rates of Reaction 16 of Table 2-7 for the para-substituents $\ce{R} = \ce{OCH_3}$ and $\ce{R} = \ce{CF_3}$. According to the $\rho$ value of 4.92 for this reaction and the $\sigma$ values of p-$\ce{OCH_3}$ and p-$\ce{CF_3}$ in Table 26-6, we may write
$\text{log} \: \frac{k_{\text{p-OCH}_3}}{k_0} = 4.92 \times \left( -0.27 \right) \text{, and} \: \text{log} \: \frac{k_{\text{p-CF}_3}}{k_0} = 4.92 \times \left( 0.54 \right)$
Subtracting these two equations gives the result:
This then is a rate ratio of 1/10,000. If we have a further table of the sixteen $k_0$ or $K_0$ values for the reactions listed in Table 26-7, we can calculate actual $k$ or $K$ values for 608 different reactions. It should be recognized that neither Table 26-6 nor Table 26-7 is a complete list; at least 80 substituent constants$^1$ and several hundred $\rho$ constants are now available.
The Hammett relationship formalizes and puts into quantitative terms much of the qualitative reasoning we have used for reactions involving aliphatic, alicyclic, and aromatic compounds. Considerable effort has been made to extend the Hammett idea to cover reactions other than of meta- and para-substituted benzene derivatives, but these will not be discussed here.$^1$
Limitations of the Hammett Equation
The effects of substituents in ortho positions on the reactivity of benzene derivatives do not correlate well with the Hammett equation, as can be seen in Figure 26-3. The problem is that ortho substituents are close enough to the reaction site to exert significant "proximity" effects, which may be polar as well as steric in origin. Thus the enhanced acid strength of 2-nitrobenzoic acid over the 3- and 4-isomers (see Table 26-4) may be due to a polar stabilization of the acid anion by the neighboring positive nitrogen, which of course is not possible with the 3- and 4-isomers:
In contrast, the slower rate of alkaline hydrolysis of ethyl 2-nitrobenzoate than of its 3- and 4-isomers is more likely due to a steric hindrance effect of the 2-nitro group (see Table 26-5):
Because the effect of steric hindrance on different types of reactions is not expected to be the same, a given substituent is unlikely to exert the same relative steric effect in one reaction as in another. Consequently we cannot hope to find a very simple relationship such as the Hammett equation that will correlate structure and reactivity of ortho-substituted compounds.
The Hammett equation also fails for open-chain aliphatic derivatives. For example, there is no simple linear relationship between log $K$ for a series of substituted ethanoic acids $\left( \ce{RCH_2CO_2H} \right)$ and log $k$ for the hydrolysis rates of similarly substituted ethyl ethanoates $\left( \ce{RCH_2CO_2C_2H_5} \right)$. The freedom of motion available to a flexible open-chain compound permits a much wider range of variations in steric effects than for meta- and para-substituted aromatic compounds.
The Hammett equation sometimes fails for meta- and para-substituted aromatic compounds. This failure may be expected whenever the opportunity arises for strong electron delocalization between the substituent and the reaction site. Generally, reactions that are strongly assisted by donation of electrons to the reaction site, as in $S_\text{N}1$ reactions and electrophilic aromatic substitution, will be facilitated by electron-delocalization effects of substituents with unshared electron pairs adjacent to the aromatic group (e.g., $\ce{-OCH_3}$, $\ce{-OH}$, $\ce{-O}^\ominus$, $\ce{-NH_2}$, and $\ce{-Cl}$). Such reactions generally give a poor Hammett correlation. Thus a diphenylmethyl chloride with one 4-methoxy group solvolyzes in ethanol at $25^\text{o}$ at a rate much faster than predicted by the Hammett equation, because of the resonance stabilization provided by the substituent to the intermediate carbocation:
The same type of stabilization by a 4-methoxy group does not appear to be important in influencing the ionization of 4-methoxybenzoic acid.
Similarly, those reactions that are strongly assisted by withdrawal of electrons from the reaction site, such as nucleophilic aromatic substitution, give a poor fit to a Hammett plot for the substituents that are capable of withdrawing electrons by delocalization ($\ce{-NO_2}$, $\ce{-N_2^+}$, $\ce{-C \equiv N}$, and so on). An example is Reaction 16 in Table 26-7. To correlate reactivity data with structures where strong resonance effects operate, different sets of substituent constants are required.$^1$
$^1$J. Hine, Structural Effects on Equilibria in Organic Chemistry, Wiley-Interscience, New York, 1975, p. 65. This book offers a very broad coverage of quantitative correlations of substituent effects on processes as diverse as radical formation and rates of rotation around single $\ce{C-C}$ bonds.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/26%3A_More_on_Aromatic_Compounds/26.06%3A_Correlations_of_Structure_with_Reactivity_of_Aromatic_Compounds.txt |
In Chapter 9, we gave an exposition of the most generally useful and practical spectroscopic methods currently employed in modern organic laboratories. However, in our discussions of NMR spectra, we passed rather quickly over the basis of understanding why some lines are broad and others sharp, why rate effects can cause chemical shifts to be averaged, and how to correlate spin- spin splitting with the energies of NMR transitions. These topics will be discussed in this chapter along with a brief explanation of the remarkable effects on NMR spectra associated with some kinds of chemical reactions, namely, chemically induced dynamic nuclear polarization (CIDNP). In addition to the spectroscopic methods covered in Chapter 9, there are a number of other spectroscopic techniques that are less generally used, but can provide, and have provided, critical information with regard to specialized problems. Because some of these are relatively new and may become more widely used in the next few years, it is important that you be aware of them and their potentialities. However, because they may be peripheral to your present course of study, we have reserved consideration of them to this chapter.
• 27.1: Prelude to More Spectroscopy
In addition to the spectroscopic methods covered in Chapter 9, there are a number of other spectroscopic techniques that are less generally used, but can provide, and have provided, critical information with regard to specialized problems.
• 27.2: Line-Width Differences in NMR
Some resonances in NMR spectra are sharp and others are broad. We can understand these differences by consideration of the lifetimes of the magnetic states between which the NMR transitions occur. The lifetimes of the states can be related to the width of the lines by the Heisenberg uncertainty principle.
• 27.3: Use of the Uncertainty Principle to Measure the Rates of Chemical Transformations
We have seen how the uncertainty principle relates the attainable line widths in different kinds of spectroscopy to the lifetimes of the states - the shorter the lifetime, the greater the spread in energy of the states and the greater the spectroscopic line width. So far we have associated short lifetimes with excited states, but this need not necessarily be so. Short lifetimes also may be associated with chemical or conformational changes.
• 27.4: Why Spin-Spin Splitting?
The underlying basis for spin-spin splitting in NMR involves perturbation of the nuclear magnetic energy levels by magnetic interaction between the nuclei. This Module addresses the origin of spin-spin splittings (and is not needed for the qualitative use of spin-spin splitting in structural analysis) with an emphasis on understanding the origins of the line spacings and line multiplicities. We will confine our attention to protons, but the same considerations apply to other nuclei.
• 27.5: Chemically Induced Dynamic Nuclear Polarization (CIDNP)
One of the most startling developments in NMR spectroscopy since its inception has been the discovery of chemically induced dynamic nuclear polarization or CIDNP. An especially dramatic example is provided by irradiation of 3,3-dimethyl-2-butanone with ultraviolet light. The CIDNP effect is a complicated one and is observed exclusively for radical reactions. However, it is not expected for chain-propagation steps, but only for termination steps.
• 27.6: Photoelectron Spectroscopy
The excitation of electrons to higher energy states through absorption of visible and ultraviolet light (usually covering the range of wavelengths from 200 nm to 780 nm) is discussed elsewhere. We now will consider what happens on absorption of much shorter wavelength, more energetic, photons.
• 27.7: Mössbauer Spectroscopy
Mössbauer spectroscopy is a nuclear spectroscopy that is which is capable of giving chemical information. The technique would be used widely if there were more nuclei with the proper nuclear properties. For organic chemistry, probably the most important available nucleus is the iron nuclide Fe-57 ( 2.2% of the natural mixture of iron isotopes). Iron occurs in many biologically important substances, such as hemoglobin, myoglobin, cytochromes, the iron storage substance ferritin, etc.
• 27.8: Field- and Chemical-Ionization Mass Spectroscopy
There are two ways of achieving ion formation without imparting as much energy as by electron impact - in other words, "soft" rather than "hard" ionizations. Field ionization is one such method, in which ionization results from passing the molecules through a high electric field. An alternative method is chemical ionization that, as might be expected from its name, more chemically interesting and is closely allied to ion cyclotron resonance, which will be discussed in the next section.
• 27.9: Ion-Cyclotron Resonance
Ion-cyclotron resonance combines features of mass spectroscopy in that the charge/mass ratio is involved, and of NMR spectroscopy in that detection depends on absorption of energy from a RF oscillator. The applications depend on reactions between the ions during the time they remain in the cyclotron, which may be many seconds. It is possible to measure the concentrations of the ions as a function of time and determine the rates of reaction of ions with neutral molecules in the gas phase.
• 27.10: Electron-Spin Resonance (ESR) Spectroscopy of Organic Radicals
An important method of studying radicals is electron-spin resonance (ESR) spectroscopy. The principles of this form of spectroscopy are much the same as of NMR spectroscopy, but the language used by the practitioners of these two forms of magnetic resonance spectroscopy is different. The important point is that an unpaired electron, like a proton, has a spin and a magnetic moment such that it has two possible orientations in a magnetic field.
• 27.E: More about Spectroscopy (Exercises)
These are the homework exercises to accompany Chapter 27 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
27: More about Spectroscopy
In Chapter 9, we gave an exposition of the most generally useful and practical spectroscopic methods currently employed in modern organic laboratories. However, in our discussions of nmr spectra, we passed rather quickly over the basis of understanding why some lines are broad and others sharp, why rate effects can cause chemical shifts to be averaged, and how to correlate spin-spin splitting with the energies of nmr transitions. These topics will be discussed in this chapter along with a brief explanation of the remarkable effects on nmr spectra associated with some kinds of chemical reactions, namely, chemically induced dynamic nuclear polarization (CIDNP).
In addition to the spectroscopic methods covered in Chapter 9, there are a number of other spectroscopic techniques that are less generally used, but can provide, and have provided, critical information with regard to specialized problems. Because some of these are relatively new and may become more widely used in the next few years, it is important that you be aware of them and their potentialities. However, because, they may be peripheral to your present course of study, we have reserved consideration of them to this chapter.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.01%3A_Prelude_to_More_Spectroscopy.txt |
If you look at the NMR spectra of many different kinds of organic compounds, you will notice that some resonances are sharp and others are broad. In a few spectra, all of the peaks may be broad as the result of poor spectrometer performance, but this is not true for the spectra of Figures 9-29 and 24-2 where, within a given spectrum, some resonances will be seen to be sharp and others broad. We can understand these differences by consideration of the lifetimes of the magnetic states between which the NMR transitions occur.$^1$ The lifetimes of the states can be related to the width of the lines by the Heisenberg uncertainty principle.
You may have heard of the uncertainty principle, but if you have not studied chemical physics you may have little idea of its possible importance to organic chemistry. The usual statement of the principle is that there are limits to how precisely we can specify the momentum and the position of a particle at the same time. An alternative statement has more relevance to spectroscopy and chemistry, namely, that the precision with which we can define the energy of a state depends on the lifetime of the state. The shorter the lifetime, the less the certainty with which we can define the energy.$^2$
Let us consider an example. Suppose a magnetic nucleus in a ground state with a long lifetime and rather precisely defined energy goes to an excited state with a short lifetime, $\Delta t$.$^3$ The uncertainty principle tells us that the energy of the excited state cannot be defined precisely. It will have an inherent uncertainty in its energy so that an imprecise $\nu$, having uncertainty in frequency $\Delta \nu$, will take the nucleus from the ground state to the excited state. The imprecision of the energy $\Delta \Delta E$, or the imprecision $\Delta \nu$ in the transition frequency, $\nu$, depends on $\Delta t$, and is given approximately by the relationship
$\Delta \Delta E \sim \frac{h}{2 \pi} \times \frac{1}{\Delta t} \sim h \Delta \nu \tag{27-1}$
in which $h$ is Planck's constant. What this means is that the absorption line corresponding to the transition will have an uncertainty in line width that is inversely proportional to $\Delta t$ (see Figure 27-1).
It is most convenient to think of line widths in frequency units because most of our spectra are plotted this way. If the scale is wavelength or energy, it can be converted to frequency by the procedures given previously (Section 9-3). Division of Equation 27-1 by $h$ leads to the relationship $\Delta \nu \sim 1/ \left(2 \pi \times \Delta t \right)$. In NMR spectroscopy, we may wish to consider spin-spin splittings or chemical shifts involving lines no farther than $1 \: \text{Hz}$ apart. However, two lines $1 \: \text{Hz}$ apart will not be clearly distinguishable unless $\Delta \nu$ of each is less than about $1 \: \text{Hz}$, which corresponds to a $\Delta t$, the lifetime of the excited state, of $1/ \left( 2 \pi \right) \cong 0.16 \: \text{sec}$. If $\Delta \nu$ is $\geq 2 \: \text{Hz}$, lines that are $1 \: \text{Hz}$ apart will be so poorly resolved as to appear as one line (cf. Figure 27-2). A $\Delta \nu$ of $2 \: \text{Hz}$ corresponds to a $\Delta t$ of $1/ \left( 2 \times 2 \pi \right) \cong 0.08 \: \text{sec}$. Clearly, line separations observed in NMR spectroscopy and, in fact, in all forms of spectroscopy, depend on the lifetimes of the states between which transitions take place. The lifetime of $0.16 \: \text{sec}$ required for $\Delta \nu$ to be $1 \: \text{Hz}$ is a long time for a molecule! During $0.16 \: \text{sec}$, a molecule such as ethanol in the liquid phase may undergo $10^{11}$ collisions with other molecules, $10^{10}$ rotations about the $\ce{C-C}$ bonds, and $10^{12}$ vibrations of each of the various bonds, and may even undergo a number of chemical changes. The properties of magnetic states that have lifetimes of this order clearly must be an average over all of these happenings.
It is possible to shorten the lifetime of an excited nuclear magnetic state (or increase its relaxation rate) in a number of ways. For a liquid, the simplest way is to dissolve in it paramagnetic metal ions, such as $\ce{Cu}$(II), $\ce{Fe}$(III), $\ce{Mn}$(II), and the like, or other substances ($\ce{O_2}$, $\ce{NO}$, and so on) that have unpaired electrons. Another way is to reduce the rate of motion of magnetic nuclei in different molecules with respect to one another, which is easily done by increasing the viscosity. Without going into details of the mechanisms by which substances with unpaired electrons or increased viscosity shorten the lifetime of excited nuclear magnetic states, it is important to know that dramatic line broadening thereby can be produced. Thus the proton resonance line of water is enormously broadened by adding paramagnetic $\ce{Mn}$(II) ions or by freezing (water molecules in ice move much more slowly relative to one another than in liquid water).
Notes
1. It may be helpful to you before proceeding to review the introductions to Section 9-10 and 9-10A in which the general characteristics of the nuclear magnetic states are described.
2. A brief exposition of the basis of the uncertainty principle is given by R. P. Feynman, Lectures in Physics, Addison-Wesley, Reading, Mass., 1963, Vol. 1, pp. 6-10.
3. The uncertainty principle will be applied in this section to NMR spectroscopy but, as we will see later, it is applicable to all other forms of spectroscopy.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.02%3A_Line-Width_Differences_in_NMR.txt |
We have seen how the uncertainty principle relates the attainable line widths in different kinds of spectroscopy to the lifetimes of the states - the shorter the lifetime, the greater the spread in energy of the states and the greater the spectroscopic line width. So far we have associated short lifetimes with excited states, but this need not necessarily be so. Short lifetimes also may be associated with chemical or conformational changes. As a specific example, suppose we have a magnetic nucleus in the $+\frac{1}{2}$ state located in a chemical environment whereby it experiences a magnetic field $H$ such that $H = H_0 \left( 1 - \sigma \right)$. This nucleus will have a particular magnetic energy, call it $E$. Now suppose the nucleus has a lifetime $\Delta t$ before it moves to a different chemical environment where it experiences a different field $H' = H_0 \left( 1 - \sigma' \right)$ and has a different energy $E'$. Clearly, there will be an uncertainty in the energy $E$ depending on the lifetime of the $+\frac{1}{2}$ nucleus in the particular chemical environment before it switches to the new environment with a different shielding and a different energy.
Consider a specific example, 2,2,3,3-tetrachlorobutane. This substance can exist in three different conformations, $1$, $2$, and $3$. By reference to the discussions in Section 5-2, you will recognize that $1$ is achiral, whereas $2$ and $3$ are enantiomers:
Clearly, if we could separate $1$ from $2$ and $3$, the protons of its methyl groups would have different chemical shifts from those of $2$ and $3$ (which, as enantiomers, would have their methyl proton resonances at the same frequency).
Now consider a mixture of the conformations $1$, $2$, and $3$ in which the lifetimes of the conformations before they convert one into the other are $\Delta t$. Assuming that the lifetimes of the $+\frac{1}{2}$ and $-\frac{1}{2}$ magnetic states are long compared to $\Delta t$, then uncertainty in the transition energies will depend on the lifetimes of the chemical states (conformations) with different chemical shifts for the protons. The chemical-shift difference between $1$ and $2$ or $3$ at $-44^\text{o}$, as shown by Figure 27-3, is about $5 \: \text{Hz}$. From Equation 27-1, we can see that $5 \: \text{Hz}$ also will be the degree of the uncertainty in the frequency when $\Delta t \sim 1/ \left( 2 \pi \Delta \nu \right) = 1/ \left( 2 \pi \times 5 \: \text{Hz} \right) = 0.03 \: \text{sec}$. Thus if $1$ has a lifetime much longer than $0.03 \: \text{sec}$, say $1 \: \text{sec}$, before going to $2$ or $3$, it will give a sharp resonance of its own and, of course, $2$ and $3$ will also. However, if $1$, $2$, and $3$ have lifetimes much shorter than $0.03 \: \text{sec}$, say $0.001 \: \text{sec}$, then we expect one average resonance for $1$, $2$, and $3$.
Either condition can be realized for 2,2,3,3-tetrachlorobutane by taking the proton nmr spectrum at different temperatures (Figure 27-3). At $-44^\text{o}$, at which $\Delta t$ is $1.0 \: \text{sec}$, we see the separate peaks for $1$ and for $2$ and $3$. At $-20^\text{o}$, at which $\Delta t$ is $0.045 \: \text{sec}$, the uncertainty is such that the lines have coalesced and we no longer can see the separate peaks. When the spectrum is taken at room temperature, at which $\Delta t$ is about $0.00005 \: \text{sec}$, a single very sharp line is observed. We get a sharp line at this temperature because, for practical purposes, there is no uncertainty about the average chemical shift of $1$, $2$, and $3$. The line width now is determined again by the lifetimes of the $+\frac{1}{2}$ and $-\frac{1}{2}$ magnetic states, not by the lifetimes of the conformations.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.03%3A_Use_of_the_Uncertainty_Principle_to_Measure_the_Rates_of_Chemical_Transformations.txt |
In Section 9-10G, we outlined the structural features that lead to observation of spin-spin splitting in the NMR spectra of organic compounds. Rules for predicting the multiplicities and intensities of spin-spin splitting patterns also were discussed. However, we did not discuss the underlying basis for spin-spin splitting, which involves perturbation of the nuclear magnetic energy levels shown in Figure 9-21 by magnetic interaction between the nuclei. You may wish to understand more about the origin of spin-spin splitting than is provided by the rules for correlating and predicting spin-spin splitting given previously, but having a command of what follows is not necessary to the qualitative use of spin-spin splitting in structural analysis. However, it will provide you with an understanding of the origins of the line spacings and line multiplicities. We will confine our attention to protons, but the same considerations apply to other nuclei ($\ce{^{13}C}$, $\ce{^{15}N}$, $\ce{^{19}F}$, and $\ce{^{31}P}$) that have the spin $I = \frac{1}{2}$. The main differences between proton-proton splittings and those of other nuclei are in the magnitudes of the splitting constants ($J$ values) and their variation with structure.
Why does splitting occur? Let us start by comparing the two-proton systems of $7$ and $8$:
The protons in each compound will have the shift differences typical of $\ce{Cl_2CH}-$ and $\ce{-CHO}$ and, at $60 \: \text{MHz}$, can be expected from the data in Table 9-4 and Equation 9-4 to be observed at about $350 \: \text{Hz}$ and $580 \: \text{Hz}$, respectively, from TMS. Now consider a frequency-sweep experiment$^4$ arranged so that the $\ce{-CH=O}$ proton will come into resonance first.
For $7$ the two protons are separated by seven bonds in all (five carbon-carbon and two carbon-hydrogen bonds), thus we expect spin-spin splitting to be negligible. We can construct an energy diagram (Figure 27-5) for the magnetic energies of the possible states of two protons at $60 \: \text{MHz}$ with the aid of Figure 9-24. (If this diagram is not clear to you, we suggest you review Section 9-10A before proceeding.)
Now consider $\ce{Cl_2CH-CH=O}$, in which the protons are in close proximity to one another, three bonds apart. Each of these protons has a magnetic field and two possible magnetic states that correspond to a compass needle pointing either north or south (see Figure 9-21). The interactions between a north-north set of orientations ($+\frac{1}{2}$, $+\frac{1}{2}$ or $\rightrightarrows$) of the two protons or a south-south set ($-\frac{1}{2}$, $-\frac{1}{2}$ or $\leftleftarrows$) will make these states less stable, whereas the interactions between either a north-south ($+\frac{1}{2}$, $-\frac{1}{2}$ or $\rightleftarrows$) or a south-north ($-\frac{1}{2}$, $+\frac{1}{2}$ or $\leftrightarrows$) orientation will make these states more stable. Why? Because north-south or south-north orientations of magnets attract each other, whereas north-north or south-south repel each other.$^5$ Let us suppose the $+\frac{1}{2}$, $+\frac{1}{2}$ or $-\frac{1}{2}$, $-\frac{1}{2}$ orientations are destabilized by $1.25 \: \text{Hz}$. The $+\frac{1}{2}$, $-\frac{1}{2}$ or $-\frac{1}{2}$, $+\frac{1}{2}$ states must then be stabilized $1.25 \: \text{Hz}$. Correction of the energy levels and the transition energies for these spin-spin magnetic interactions is shown in Figure 27-6.
There are four possible combinations of the magnetic quantum numbers of the two protons of $\ce{CHCl_2CHO}$, as shown in Figure 27-6. Because the differences in energy between the magnetic states corresponding to these four combinations is very small (see Section 9-10A), there will be almost equal numbers of $\ce{CHCl_2CHO}$ molecules with the $\left( +\frac{1}{2}, \: +\frac{1}{2} \right)$, $\left( -\frac{1}{2}, \: +\frac{1}{2} \right)$, $\left( +\frac{1}{2}, \: -\frac{1}{2} \right)$, and $\left( -\frac{1}{2}, \: -\frac{1}{2} \right)$ spin combinations. The transitions shown in Figure 27-6 will be observed for those molecules with the two protons in the $\left( +\frac{1}{2}, \: +\frac{1}{2} \right)$ state going to the $\left( -\frac{1}{2}, \: +\frac{1}{2} \right)$ state or for the molecules with $\left( +\frac{1}{2}, \: +\frac{1}{2} \right)$ to $\left( +\frac{1}{2}, \: -\frac{1}{2} \right)$ as well as from $\left( -\frac{1}{2}, \: +\frac{1}{2} \right) \rightarrow \left( -\frac{1}{2}, \: -\frac{1}{2} \right)$ and $\left( +\frac{1}{2}, \: -\frac{1}{2} \right) \rightarrow \left( -\frac{1}{2}, \: -\frac{1}{2} \right)$.$^6$ It is very important to remember that the transitions shown in Figure 27-6 involve molecules that have protons in different spin states, and by the uncertainty principle (Section 27-1) the lifetimes of these spin states must be long if sharp resonance lines are to be observed.
Now if we plot the energies of the transitions shown in Figures 27-5 and 27-6, we get the predicted line positions and intensities of Figure 27-7. Four lines in two equally spaced pairs appear for $\ce{Cl_CH-CH=O}$, as expected from the naive rules for spin-spin splitting.
A harder matter to explain, and what indeed is beyond the scope of this book to explain, is why, as the chemical shift is decreased at constant spin-spin interactions, the outside lines arising from a system of two nuclei of the type shown in Figure 27-7 become progressively weaker in intensity, as shown in Figure 9-44. Furthermore, the inside lines move closer together, become more intense, and finally coalesce into a single line as the chemical-shift difference, $\delta$, approaches zero. All we can give you here is the proposition that the outside lines become "forbidden" by spectroscopic selection rules as the chemical shift approaches zero. At the same time, the transitions leading to the inside lines become more favorable so that the integrated peak intensity of the overall system remains constant.$^7$
In a similar category of difficult explanations is the problem of why second-order splittings are observed, as in Figure 9-32. The roots of the explanation again lie in quantum mechanics which we cannot cover here, but which do permit very precise quantitative prediction and also qualitative understanding.$^7$ The important point to remember is that whenever the chemical shifts and couplings begin to be of similar magnitude, you can expect to encounter NMR spectra that will have more lines and lines in different positions than you would expect from the simple treatment we developed in this chapter and in Chapter 9.
In extreme cases, such as with the protons of 4-deuterio-1-buten-3-yne, shown in Figure 27-8, none of the line positions or spacings correspond directly to any one chemical shift or spin-spin coupling. However, it is important to recognize that such spectra by no means defy analysis and, as also is seen in Figure 27-8, excellent correspondence can be obtained between calculated and observed line positions and intensities by using appropriate chemical shift and coupling parameters. However, such calculations are numerically laborious and are best made with the aid of a high-speed digital computer.
$^4$We will use frequency sweep simply because it is easier to talk about energy changes in frequency units. However, the same arguments will hold for a field-sweep experiment.
$^5$Such interactions with simple magnets will average to zero if the magnets are free to move around each other at a fixed distance. However, when electrons are between the magnets, as they are for magnetic nuclei in molecules, a small residual stabilization (or destabilization) is possible. Because these magnetic interactions are "transmitted" through the bonding electrons, we can understand in principle why it is that the number of bonds between the nuclei, the bond angles, conjugation, and so on, is more important than just the average distance between the nuclei in determining the size of the splittings.
$^6$The transitions $\left( +\frac{1}{2}, \: +\frac{1}{2} \right) \rightarrow \left( -\frac{1}{2}, \: -\frac{1}{2} \right)$ and $\left( -\frac{1}{2}, \: +\frac{1}{2} \right) \rightarrow \left( +\frac{1}{2}, \: -\frac{1}{2} \right)$ are in quantum-mechanical terms known as "forbidden" transitions and are not normally observed. Notice that the net spin changes by $\pm$1 for "allowed" transitions.
$^7$A relatively elementary exposition of these matters is available in J. D. Roberts, An Introduction to Spin-Spin Splitting in High-Resolution Nuclear Magnetic Resonance Spectra, W. A. Benjamin, Inc., Menlo Park, Calif., 1961.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.04%3A_Why_Spin-Spin_Splitting.txt |
One of the most startling developments in NMR spectroscopy since its inception has been the discovery of chemically induced dynamic nuclear polarization or CIDNP. An especially dramatic example is provided by irradiation of 3,3-dimethyl-2-butanone with ultraviolet light.
The CIDNP effect is a complicated one and we will not attempt to explain it in detail. It is observed exclusively for radical reactions. However, it is not expected for chain-propagation steps, but only for termination steps. Furthermore, chemically dissimilar radicals have to be involved at some stage in the reaction sequence. Let us now consider how these considerations apply to the irradiation of 3,3-dimethyl-2-butanone.
Absorption of light by a ketone can give several reactions, but an especially important one, which will be discussed in more detail in Chapter 28, is cleavage of $\ce{C-C=O}$ bonds to give radical pairs. For 3,3-dimethyl-2-butanone there are two possible cleavage reactions of this type:
The heavy lines drawn over the radical pairs indicate that the radicals in the pairs are in close proximity to one another. Combination of the radicals in the pairs regenerates the ketone, whereas separation of the radicals can lead to formation of other products. The radicals in a pair can combine with each other only if the odd electron on one radical has its spin opposite to the spin of the odd electron on the other radical. This is necessary for formation of an electron-pair bond.
CIDNP arises because the radical combination products have nonequilibrium distributions of their proton magnetic states. How can nonequilibrium distributions arise? First, we must recognize that the radicals formed by irradiation of the ketone can have different proton magnetic states. For example, the methyl protons of any given $\ce{CH_3CO} \cdot$ radical will be in one of the proton states: $+\frac{1}{2}, \: +\frac{1}{2}, \: +\frac{1}{2}; \: -\frac{1}{2}, \: +\frac{1}{2}, \: +\frac{1}{2}; \cdots ; -\frac{1}{2}, \: -\frac{1}{2}, \: -\frac{1}{2}$ states (8 in all; see Section 27-3).
The effect of the different proton magnetic states is to cause the two unpaired electrons of the radical pairs to become unpaired at different rates. In other words, $\overline{\ce{R} \uparrow + \ce{R'} \downarrow}$ pairs produced by irradiation are converted to $\overline{\ce{R} \uparrow + \ce{R'} \uparrow}$ at different rates, depending on the proton magnetic states of $\ce{R} \uparrow$ and $\ce{R'} \downarrow$. Thus, a particular pair of proton magnetic states for $\ce{R} \uparrow$ and $\ce{R'} \downarrow$ can favor radical-pair recombination over radical-pair separation while another pair of proton magnetic states for $\ce{R} \uparrow$ and $\ce{R'} \downarrow$ can favor separation over combination. The result is a "sorting" of proton magnetic states, some appearing preferentially in particular products and others appearing in other products. Thus one product may have more than the normal equilibrium value of a higher-energy magnetic state and hence will emit radio-frequency energy to get back to equilibrium, while another product may have an abnormally low concentration of the higher-energy magnetic states and hence exhibit an enhanced absorption intensity. Figure 27-9 shows that recombination of the radical pairs produced in photolysis of 3,3-dimethyl-2-butanone forms ketone with a higher-than-normal magnetic energy in the protons of the methyl group (reduced absorption) and lower-than-normal magnetic energy in the protons of the tert-butyl group (enhanced absorption).
The other CIDNP peaks in Figure 27-9b arise from reactions of the separated radicals first formed, and show both enhanced absorption and enhanced emission. You should try to identify the origin of each of the CIDNP resonances with the expected reaction products:
Because thermodynamic equilibrium usually is established between magnetic states of protons in a few seconds, the enhanced-absorption and enhanced-emission resonances disappear quickly when irradiation is stopped.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.05%3A_Chemically_Induced_Dynamic_Nuclear_Polarization_%28CIDNP%29.txt |
The excitation of electrons to higher energy states through absorption of visible and ultraviolet light (usually covering the range of wavelengths from $200 \: \text{nm}$ to $780 \: \text{nm}$) is discussed in Sections 9-9 and 28-1. We now will consider what happens on absorption of much shorter wavelength, more energetic, photons.
When radiation of wavelengths on the order of $120 \: \text{nm}$ is absorbed by a molecule of ethene, the excited state has just sufficient energy (about $250 \: \text{kcal mol}^{-1}$) to cause the most loosely bound electron to be ejected. With radiation of still shorter wavelength, such as the $58.4 \: \text{nm}$ $\left( 490 \: \text{kcal mol}^{-1} \right)$ provided by a helium discharge tube, these electrons will have, by the Einstein law, a kinetic energy of $\left( 490 - 250 \right) = 240 \: \text{kcal mol}^{-1}$. More tightly bound electrons also can be ejected by $58.4 \: \text{nm}$ radiation, and they will have kinetic energies $E = h \nu - I$, in which $h \nu$ is the energy of the absorbed radiation $\left( 490 \: \text{kcal mol}^{-1} \right)$ and $I$ is the ionization energy. If we know $h \nu$ and measure the number of ejected electrons as a function of their kinetic energies, we can derive a spectrum that shows how the probability of excitation correlates with the ionization energy. Such spectra, called photoelectron spectra, are shown in Figure 27-10 for gaseous ethene, ethyne, and benzene and are quite individualistic. Considerable fine structure is observed as the consequence of a considerable spread in the vibrational levels of the excited state.
Substitution and conjugation have substantial effects on ionization energies. We have mentioned how methyl groups are able by their electron-donating power to stabilize carbon cations more than hydrogens do (Section 8-7B). The same effect is very prominent in the ionization of alkenes, the lowest energy required to eject an electron from 1-butene being about $11 \: \text{kcal mol}^{-1}$ greater than from cis- or trans-2-butene. The corresponding differences for 1-hexene and 2,3-dimethyl-2-butene are about $27 \: \text{kcal mol}^{-1}$. Conjugation produces similar effects. The lowest energy required to eject an electron from 1,4-pentadiene with isolated double bonds is $21 \: \text{kcal mol}^{-1}$ greater than for the isomeric 1,3-pentadiene with conjugated double bonds.
Photoelectron spectroscopy also can be carried on with x rays as the source of excitation and in this form often is called "ESCA" (Electron Spectroscopy for Chemical Analysis). The x rays used have wavelengths on the order of $0.9 \: \text{nm}$ $\left( 32,000 \: \text{kcal mol}^{-1} \right)$ and the energies involved are more than ample to cause ejection of electrons form inner shells as well as from valence shells. An x-ray photoelectron spectrum of the carbon $1s$ electrons of ethyl trifluoroethanoate is shown in Figure 27-11. This spectrum is extremely significant in that it shows four different peaks - one for each chemically different carbon present. What this means is that the energy required to eject a $1s$ electron of carbon depends on the chemical state of the carbon. The energy range for this compound is fully $185 \: \text{kcal mol}^{-1}$ of the approximately $6700 \: \text{kcal mol}^{-1}$ required to eject a $1s$ electron. This form of spectroscopy is especially well suited to the study of solid surfaces and is being used widely for the characterization of solid catalysts.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.06%3A_Photoelectron_Spectroscopy.txt |
A different form of molecular excitation is that of changes in the energies of the atomic nuclei. In general, enormous energies are involved, and such excitations will not be of interest to the study of organic chemistry unless the atomic energy levels are detectably influenced by the chemical surroundings of the nuclei. Usually this is not so, but there is one form of nuclear spectroscopy, known as Mössbauer spectroscopy, which is capable of giving chemical information. The technique would be used widely if there were more nuclei with the proper nuclear properties. For organic chemistry, probably the most important available nucleus is the iron nuclide $\ce{^{57}Fe}$ ($2.2\%$ of the natural mixture of iron isotopes). Iron occurs in many biologically important substances, such as hemoglobin, myoglobin, cytochromes, the iron storage substance ferritin, and so on, and there are a number of other types of stable organoiron compounds including ferrocene, cyclobutadiene iron tricarbonyl, and cyclooctatetraene iron tricarbonyl, which will be discussed in Chapter 31. These compounds present unusually difficult problems in how to formulate the bonding between carbon and iron. Important information has been obtained for such substances by Mössbauer spectroscopy.
The essence of the Mössbauer technique as applied to $\ce{^{57}Fe}$ follows. A radioactive $\ce{^{57}Co}$ nucleus captures an electron and is converted to an excited $\ce{^{57}Fe}$ nucleus, which then emits a $\gamma$ ray and becomes an ordinary $\ce{^{57}Fe}$ nucleus. If the excited $\ce{^{57}Fe}$ nucleus is in a ridig material so that there is no recoil motion associated with the emission of the $\gamma$ ray, then this ray is extraordinarily monochromatic (has a very small $\Delta \nu$, Section 27-1) even though of great energy $\left( 14.4 \: \text{KeV} = 3.3 \times 10^5 \: \text{kcal mol}^{-1} \right)$. When such a $\gamma$ ray passes through a sample containing $\ce{^{57}Fe}$ atoms (also held rigidly), the $\gamma$ ray can be absorbed to produce another excited $\ce{^{57}Fe}$ nucleus. The chemical environment of the iron atoms can change the wavelength at which this absorption occurs. The problem is how to vary the wavelength of the $\gamma$ rays to match the nuclear absorption frequency. The way this is done is almost unbelievably simple - move the sample back and forth a few $\text{mm sec}^{-1}$ in the path of the $\gamma$ rays and measure the velocities at which absorption takes place. The velocity of light is $3 \times 10^{11} \: \text{mm sec}^{-1}$. Therefore, a Doppler effect of $1 \: \text{mm sec}^{-1}$ corresponds to a difference of only one par in $3 \times 10^{11}$. However, the selectivity of the recoilless $\gamma$ rays emitted from excited $\ce{^{57}Fe}$ nuclei is on the order of one part in $5 \times 10^{+13}$ (equivalent to about a $7$-$\text{cm}$ variation in the distance from the earth to the sun!).
A Mössbauer spectrum that has helped to corroborate the structure of cyclooctatetraene iron tricarbonyl is shown in Figure 27-12. The separation of the two absorption peaks in Figure 27-12 corresponds to a sample Doppler velocity of $1.23 \: \text{mm sec}^{-1}$. This Doppler effect means that there is the very small energy difference of $1.4 \times 10^{-6} \: \text{kcal mol}^{-1}$ in the two transitions shown.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.07%3A_Mossbauer_Spectroscopy.txt |
As we mentioned in connection with our discussion of mass spectroscopy in Section 9-11, one problem with the practical application of mass spectra to structure analysis involving the production of ions by electron impact is that the $\ce{M^+}$ peak may be very weak. In many situations we would like to have mass spectra with less intensive fragmentation than that obtainable by electron impact. There are two ways of achieving ion formation without imparting as much energy as by electron impact - in other words, "soft" rather than "hard" ionizations. Field ionization is one such method, in which ionization results from passing the molecules through an electric field of $10^7$-$10^8 \: \text{V cm}^{-1}$. This may seem like a practically unattainable electric field. However, it can be achieved easily by impressing a potential of $10^4$ volts across a pair of electrodes, one of which has a very sharp radius of curvature $\left( \sim 10^{-5} \: \text{cm} \right)$, as can be achieved with a very fine metal point, very fine wire, or a sharp edge. Field ionization of a molecule differs from electron-impact ionization in that the electron normally is ejected from the molecule in its ground state. As a result, the parent ion $\ce{M^+}$ peak is very strong, even for molecules for which the $\ce{M^+}$ is virtually absent on electron impact.
Chemical ionization is, as might be expected from its name, more chemically interesting and is closely allied to ion cyclotron resonance, which will be discussed in the next section. The principle of chemical ionization is simple. The molecule to be studied is injected into the ionizing region of the mass spectrometer in the presence of $0.5$-$1.5 \: \text{mm} \: \ce{Hg}$ pressure of a gas, usually methane. Electron impact causes ionization of the methane, which is present in relatively large concentration. The ionization products of methane then react with the compound to be analyzed and convert it to ions. The gas mixture then exits into a low-pressure zone $\left( 10^{-4} \: \text{mm} \right)$ and the ions are analyzed according to m/e in the usual way.
What happens to $\ce{CH_4}$ when it is bombarded with electrons at, say, $1 \: \text{mm}$ pressure? The simplest reaction is formation of the $\ce{M^+}$ ion from $\ce{CH_4} + \ce{e^-} \rightarrow \ce{CH_4^+} + 2 \ce{e^-}$, but $\ce{CH_3^+}$ and $\ce{CH_2^+}$ also are produced by electron impact. If there is sufficient $\ce{CH_4}$, a variety of rapid transformations take place between each of the ions produced by electron impact and the neutral $\ce{CH_4}$. The principal ions formed are
Of these, the methonium cation, $\ce{CH_5^+}$, is formed in largest amounts, the ethyl cation, $\ce{C_2H_5^+}$, is next, and there is a smaller amount of the $\ce{C_3H_5^+}$ cation (2-propenyl cation; Section 8-7B). These ions then react with the substance to be analyzed, thereby converting it into ions. Different reactions are possible, but it we have an unsaturated compound, call it $\ce{RH}$, then
We then have a strong $\left( \ce{M} - 1 \right)^+$ peak and weaker $\left( \ce{M} + 29 \right)^+$ and $\left( \ce{M} + 41 \right)^+$ peaks. The larger cations probably are similar to those formed in cationic polymerization (Section 10-8B), whereas formation of the $\left( \ce{M} - 1 \right)^+$ cation corresponds to the hydrogen-transfer reaction discussed in Section 10-9.
With many compounds there is little fragmentation on chemical ionization. An example of a comparison of the spectra resulting from electron impact and chemical ionization is given in Figure 27-13. The simplicity of the spectra makes chemical-ionization mass spectroscopy especially useful for continuous analysis of the effluent from gas-liquid chromatographic columns (Section 9-2).
A problem with all mass spectroscopy of large molecules is how to get them into the vapor phase so that they can be ionized and their fragmentation patterns determined. Simple heating may cause excessive degradation and formation of ions not corresponding to the desired substance. Two useful methods that involve only intense short-term local heating of the sample appear to have promise in this connection. One method uses a burst from a powerful infrared laser to volatilize part of the sample, and the other uses bombardment by heavy and energetic particles from fission of californium-252 nuclei to raise the local temperature of the sample to about $10,000^\text{o}$. The latter technique both volatilizes and ionizes the sample molecules.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.08%3A_Field-_and_Chemical-Ionization_Mass_Spectroscopy.txt |
A gaseous ion in a magnetic field moves in a circular orbit with an angular frequency $\omega_c$ such that $\omega_c = \left( e/m \right) \left( H_0/c \right)$, in which $e/m$ is the ratio of charge to mass, $H_0$ is the applied magnetic field, and $c$ is the velocity of light. The frequency $\omega_c$ is called the "cyclotron frequency" and is the basis of the cyclotron particle accelerator used in nuclear physics. Now suppose a radio-frequency field is imposed on the ions from a variable oscillator, as shown in Figure 27-15. When the frequency of the oscillator $\omega$ equals $\omega_c$, the ions absorb energy and move faster through larger orbits, but at the same frequency $\omega_c$.
Ion-cyclotron resonance combines features of mass spectroscopy in that the ratio $e/m$ is involved, and of NMR spectroscopy in that detection depends on absorption of energy from a radio-frequency oscillator. The chemical applications depend on reactions between the ions during the time they remain in the cyclotron, which may be many seconds. Suppose then that we generate $\ce{OH}^\ominus$ by electron bombardment of a gaseous mixture of water and 2-methyl-2-propanol (tert-butyl alcohol). The $\ce{OH}^\ominus$ ion can be detected by its characteristic frequency $\omega = \left( e/m \right) \left( H_0/c \right)$, in which $e/m = 1/17$. Now, because the reaction $\ce{(CH_3)_3COH} + \ce{OH}^\ominus \rightarrow \ce{(CH_3)_3CO}^\ominus + \ce{H_2O}$ occurs, a new ion of $e/m = 1/73$ appears. The reverse reaction, $\ce{(CH_3)_3CO}^\ominus + \ce{H_2O} \rightarrow \ce{(CH_3)_3COH} + \ce{OH}^\ominus$, does not occur to a measurable extent. From this we can infer that $\ce{(CH_3)_3COH}$ is a stronger acid than $\ce{H_2O}$ in the gas phase. These experiments clearly are related to chemical-ionization mass spectroscopy (Section 27-7), and provide the basis for determining the gas-phase acidities of alkynes and water, discussed in Section 11-8. A detailed gas-phase acidity scale has been established by this means.
Many unusual reactions occur between ions and neutral molecules in the gas phase, which can be detected by ion-cyclotron resonance; a few examples are
$\ce{CH_3F^+} \: \text{(from electron impact)} + \ce{CH_3F} \rightarrow \ce{CH_3} \overset{+}{\ce{F}} \ce{H} + \cdot \ce{CH_2F} \: \text{(} \ce{H} \cdot \: \text{atom transfer)}$
$\ce{CH_3FH^+} + \ce{N_2} \rightarrow \ce{CH_3N_2^+} + \ce{HF} \: \: \: \: \: \text{(nucleophilic displacement)}$
$\ce{CH_3FH^+} + \ce{Xe} \rightarrow \ce{CH_3Xe^+} + \ce{HF} \: \: \: \: \: \text{(nucleophilic displacement)}$
Clearly, in gas-phase reactions $\ce{HF}$ is an extremely good leaving group in being rapidly displaced both by $\ce{Xe}$ and $\ce{N_2}$. From our discussions of leaving groups in Section 8-7C, we can infer that $\ce{H_2F}^\oplus$ must be a very strong acid in the gas phase and the available evidence indicates that this is so.
It is possible to measure the concentrations of the ions as a function of time and thus determine the rates of reaction of ions with neutral molecules in the gas phase. Figure 27-16 shows the results of a typical experiment wherein a sequence of reactions occurs that involves chloromethane as the neutral molecule and begins with the ion $\ce{CH_3Cl}^\oplus$ formed by a short burst $\left( 10 \: \text{msec} \right)$ of $16 \: \text{KeV}$ electrons. The originally formed $\ce{CH_3Cl}^\oplus$ ions react with $\ce{CH_3Cl}$ to yield $\ce{CH_3ClH}^\oplus + \cdot \ce{CH_2Cl}$. The buildup of $\ce{CH_3ClH}^\oplus$ and the disappearance of $\ce{CH_3Cl}^\oplus$ clearly are coupled. A slower reaction, $\ce{CH_3ClH}^\oplus + \ce{CH_3Cl} \rightarrow \ce{(CH_3)_2Cl}^\oplus + \ce{HCl}$, then takes over the action.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.09%3A_Ion-Cyclotron_Resonance.txt |
An important method of studying radicals is electron-spin resonance (ESR) spectroscopy. The principles of this form of spectroscopy are much the same as of NMR spectroscopy, but the language used by the practitioners of these two forms of magnetic resonance spectroscopy is different.
First, let us discuss the similarities. The important point is that an unpaired electron, like a proton, has a spin and a magnetic moment such that it has two possible orientations in a magnetic field. The two orientations correspond to magnetic quantum numbers $+\frac{1}{2}$ and $-\frac{1}{2}$ that define two energy states. These states differ in energy by $\Delta E = \left( h \gamma \right) H$, in which $\gamma$ is the gyromagnetic ratio of the electron. (See Section 9-10A for a discussion of the analogous situation for protons.) Transitions between these states occur with absorption of radiation of frequency $\nu = \gamma H$. Because $\gamma$ for free electrons is about 1000 times larger than $\gamma$ for protons, the frequency of absorption $nu$ of electrons is about 1000 times that of protons at the same magnetic field. At magnetic fields of 3600 gauss the absorption frequency of free electrons is about $10,000 \: \text{MHz}$, which falls in the microwave, rather than the radio-wave region.
The basic apparatus for ESR spectroscopy is similar to that shown in Figure 9-22 for NMR spectroscopy, except that the sample is irradiated with a microwave generator. The spectra produced by ESR absorptions of unpaired electrons are similar to those shown in Figure 9-25, except that ESR spectrometers normally are so arranged as to yield a plot of the first derivative of the curve of absorption against magnetic field rather than the absorption curve itself, as shown in Figure 27-17. This arrangement is used because it gives a better signal-to-noise ratio than a simple plot of absorption against magnetic field.
The sensitivity of ESR spectroscopy for detection of radicals is very high. Under favorable conditions, a concentration of radicals as low as $10^{-12} \: \text{M}$ can be detected readily. Identification of simple hydrocarbon radicals often is possible by analysis of the fine structure in their spectra, which arises from spin-spin splittings involving those protons that are reasonably close to the centers over which the unpaired electron is distributed. Methyl radicals, $\ce{CH_3} \cdot$, generated by x-ray bombardment of methyl iodide at $-196^\text{o}$ show four resonance lines of intensity 1:3:3:1, as expected for interaction of the electron with $n + 1$ protons (see Section 9-10G).
The chemical shift generally is much less important in ESR spectroscopy than in NMR. One reason is that the lifetimes of the electrons in the $+\frac{1}{2}$ and $-\frac{1}{2}$ states generally are very short ($10^{-6} \: \text{sec}$ or less) so ESR lines are quite broad by comparison with NMR lines (Section 27-1). ESR chemical shifts usually are measured in terms of "$g$ factors", which, like NMR $\delta$ values, are field-independent. The resonance frequency is given by $\nu = g \mu_0 H_0/h$, in which $\mu_0$ is the magnetic moment of the electron.
Spin-spin splittings arising from proton-electron interactions are very large in ESR spectra and usually are reported in gauss, under the heading hyperfine interactions. The proton-electron splitting in the methyl radical is 23 gauss $\left( 64.4 \: \text{MHz} \right)$, which is vastly larger than the $7$-$\text{Hz}$ proton-proton splitting in ethanol (Figure 9-23). The large splittings (and broad lines) typical of ESR make it possible to run ESR spectra on solids or highly viscous materials, for which the fine structure typical of high-resolution NMR spectra would be wholly washed out (Section 27-1).
ESR spectra are subject to exchange effects in the same way as NMR spectra. A specific example is provided by electron exchange between sodium naphthalenide and naphthalene. Naphthalene has a set of ten $\pi$-molecular orbitals, similar to the six $\pi$-molecular orbitals of benzene (Figure 21-5). The ten naphthalene $\pi$ electrons fill the lower five of these orbitals. In a solvent such as 1,2-dimethoxyethane, which solvates small metal ions well, naphthalene accepts an electron from a sodium atom and forms sodium naphthalenide, a radical anion:
The additional electron goes into the lowest unoccupied molecular orbital of the napthalene, which means the electron circulates over all of the carbons. The electron resonance is split into a total of 25 lines by electron-proton magnetic interactions. The reason for the complex splitting can be understood if we notice that there are eight protons in two sets of four. One set spits the electron signal into five lines $\left( n + 1 \right)$ of intensity 1:4:6:4:1 with a spacing of 5.0 gauss, while the second set splits each of the five lines into another 1:4:6:4:1 quintet with a spacing of 1.9 gauss. So, in all, there are twenty-five lines - five sets of five.
If excess naphthalene is added to a solution of sodium naphthalenide, intermolecular electron exchange occurs:
This means that the electron goes from naphthalene A with a particular set of $+\frac{1}{2}, -\frac{1}{2}$ proton nuclei to naphthalene B with a different set. The result is that the lines broaden and, if the exchange is very fast, the splitting vanishes. Because the splittings area bout 5 gauss $\left( 14 \: \text{MHz} \right)$, the mean lifetime before exchange has to be about $10^{-8} \: \text{sec}$ or less to obscure the splitting (see Sections 27-1 and 27-2).
The most exciting applications of ESR are in the study of radical intermediates in organic reactions. Considerable use has been made of the technique in biochemical reactions and it has been shown that radicals are generated and decay in oxidations brought about by enzymes. Radicals also have been detected by ESR measurements in algae that "fix" carbon dioxide in photosynthesis. The character of the radicals formed has been found to depend upon the wavelength of the light supplied for photosynthesis.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.10%3A_Electron-Spin_Resonance_%28ESR%29_Spectroscopy_of_Organic_Radicals.txt |
The role of light in effecting chemical change has been recognized for many years. Indeed, the connection between solar energy and the biosynthesis of plant carbohydrates from carbon dioxide and water was known by the early 1800's. Yet organic photochemistry was slow to develop as a well-understood and manageable science. Progress only became rapid following the development of spectroscopy and spectroscopic techniques for structure determination and the detection of transient species. For this reason photochemistry for many years was the domain of physical and theoretical chemists. Their work laid the foundation for modern organic photochemistry, which correlates the nature of excited electronic states of molecules with the reactions they undergo.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
28: Photochemistry
The role of light in effecting chemical change has been recognized for many years. Indeed, the connection between solar energy and the biosynthesis of plant carbohydrates from carbon dioxide and water was known by the early 1800's. Yet organic photochemistry was slow to develop as a well-understood and manageable science. Progress only became rapid following the development of spectroscopy and spectroscopic techniques for structure determination and the detection of transient species. For this reason photochemistry for many years was the domain of physical and theoretical chemists. Their work laid the foundation for modern organic photochemistry, which correlates the nature of excited electronic states of molecules with the reactions they undergo.
Quite apart from the unparalleled importance of photosynthesis, photochemical reactions have a great impact on biology and technology, both good and bad. Vision in all animals is triggered by photochemical reactions. The destructive effects of ultraviolet radiation on all forms of life can be traced to photochemical reactions that alter cellular DNA, and the harmful effects of overexposure to sunlight and the resulting incidence of skin cancer are well established. The technical applications of photochemistry are manifold. The dye industry is based on the fact that many organic compounds absorb particular wavelengths of visible light, and the search for better dyes and pigments around the turn of this century was largely responsible for the development of synthetic organic chemistry. Dye chemistry has helped establish the relationship between chemical structure and color, which also is important in color printing and color photography. We cover these important applications of photochemistry only briefly in this chapter, but we hope to convey some understanding of the fundamentals involved.
Most photochemical reactions can be considered to occur in three stages:
1. Absorption of electromagnetic radiation to produce electronically excited states.
2. Primary photochemical reactions involving excited electronic states.
3. Secondary or dark reactions whereby the products of the primary photochemical reactions are converted to stable products.
We shall begin with a closer look at electronic excitation, some aspects of which were discussed in Section 9-9. Because transfer of electronic energy from one molecule to another is a basic process in photochemistry, we will discuss energy transfer also before giving an overview of representative photochemical reactions. The closely related phenomena of chemiluminescence and bioluminescence then will be described. Finally, there will be a discussion of several important applications of photochemistry.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.01%3A_Prelude_to_Photochemistry.txt |
Electromagnetic radiation in the ultraviolet and visible region spans a wavelength range of about $800$-$100 \: \text{nm}$ corresponding to energies of $36$-$286 \: \text{kcal mol}^{-1}$. Absorption of such radiation by molecules is not to be regarded as equivalent to simple excitation by thermal energy of $36$-$286 \: \text{kcal mol}^{-1}$. Instead, all the energy of the light quantum is taken up in excitation of an electron to a high-energy, usually antibonding, orbital (Section 9-9). An important point about such processes is that they occur more rapidly than the atoms vibrate in the bonds (Franck-Condon principle). The short transition time of an electron between ground and excited states is in complete contrast to what happens during absorption of a quantum of radio-frequency energy in nmr spectroscopy, wherein the absorption process may be slow compared to chemical reactions (Section 27-1). Therefore an electronically excited molecule is, in the first instant that it is produced $\left( <10^{-13} \: \text{sec} \right)$, just like the ground-state molecule as far as positions and kinetic energies of the atoms go, but has a very different electronic configuration. What happens at this point depends on several factors, some of which can be best illustrated by energy diagrams of the type used previously (Section 21-1). We shall consider diatomic molecules, but the argument can be extended to more complicated systems.
Consider Figure 28-1, which shows schematic potential-energy curves for a molecule $\ce{A-B}$ in the ground state $\left( \ce{A-B} \right)$ and in excited electronic states $\left( \ce{A-B} \right)^*$. We have noted previously (Section 6-1) that in the ground states of most molecules all electrons are paired; excited states also can have all electrons paired. Such states with paired electrons are called singlet states. But, because the bonding is weaker in excited states, the average bond length $r_e$ between the nuclei is greater in the excited state than in the ground state. For this reason the upper curve $\left( S_1 \right)$ in Figure 28-1 is displaced toward a larger average bond length relative to the lower or ground-state curve $\left( S_0 \right)$.
Excited states also can have unpaired electrons. States with two unpaired electrons are called triplet states $\left( T \right)$ and normally are more stable than the corresponding singlet states because, by Hund's rule, less interelectronic repulsion is expected with unpaired than paired electrons (Sections 6-1 and 21-9A). (For clarity, the potential-energy curve for the excited triplet state $\left( T_1 \right)$ of $\ce{A-B}$ is given an unrealistically long equilibrium bond distance, which puts it to the right of the curve for the $S_1$ state in Figure 28-1.) The electronic configurations for ground singlet $\left( S_0 \right)$, excited singlet $\left( S_1 \right)$, and triplet $\left( T_1 \right)$ states of the $\sigma$ electrons of a diatomic molecule are shown in Figure 28-2. This diagram will be helpful in interpreting the transitions between $S_0$, $S_1$, and $T_1$ states shown in Figure 28-1, and which we will now discuss in more detail.
When a molecule absorbs sufficient radiant energy to cause electronic excitation, the spin of the excited electron remains unchanged in the transition. That is to say, ground-state molecules with paired electrons $\left( S_0 \right)$ give excited states with paired electrons $\left( S_1 \right)$, not triplet states $\left( T_1 \right)$. The transition marked $1$ in Figure 28-1 corresponds to a singlet-singlet $\left( S_0 \rightarrow S_1 \right)$ transition from a relatively high vibrational level of $\ce{A-B}$. The energy change occurs with no change in $r$ (Franck-Condon principle), and the electronic energy of the $\ce{A-B}^*$ molecule so produced is seen to be above the level required for dissociation of $\ce{A-B}^*$. The vibration of the excited molecule therefore has no restoring force and leads to dissociation to $\ce{A}$ and $\ce{B}$ atoms. In contrast, the transition marked $2$ leads to an excited vibrational state of $\ce{A-B}^*$, which is not expected to dissociate but can lose vibrational energy to the surroundings and come down to a lower vibrational state. This is called "vibrational relaxation" and usually requires about $10^{-12} \: \text{sec}$. The vibrationally "relaxed" excited state can return to the ground state with emission of radiation (transition $F$, $S_1 \rightarrow S_0$); this is known as fluorescence, the wavelength of fluorescence being different from that of the original light absorbed. Normally, fluorescence, if it occurs at all, occurs in $10^{-9}$ to $10^{-7} \: \text{sec}$ after absorption of the original radiation.
In many cases, the excited state $\left( S_1 \right)$ can return to the ground state $\left( S_0 \right)$ by nonradiative processes. The most important processes are:
1. By chemical reaction, often with surrounding molecules. This process forms the basis of much organic photochemistry, which will be described in a later section.
2. By transfer of its excess electronic energy to other molecules. This kind of energy transfer also is a very important aspect of photochemistry, and we shall return to it shortly.
3. By decay through a lower energy state. If, for example, the potential-energy curves for the upper and lower singlet states were closer together than shown in Figure 28-1, they may actually cross at some point, thus providing a pathway for $S_1$ to relax to $S_0$ without fluorescing. But what about decay of $S_1$ through the triplet state $\left( T_1 \right)$?
Conversion of a singlet excited state to a triplet state $\left( S_1 \rightarrow T_1 \right)$ is energetically favorable but usually occurs rather slowly, in accord with the spectroscopic selection rules, which predict that spontaneous changes of electron spin should have very low probabilities. Nonetheless, if the singlet state is sufficiently long-lived, the singlet-triplet change, $S_1 \rightarrow T_1$, (often called intersystem crossing) may occur for a very considerable proportion of the excited singlet molecules.
The triplet state, like the singlet state, can return to the ground state by nonradiative processes, but in many cases a radiative transition $\left( T_1 \rightarrow S_0 \right)$ occurs, even though it has low probability. Such transitions result in emission of light of considerably longer wavelength than either that absorbed originally or resulting from fluorescence. This type of radiative transition is called phosphorescence (transition $P$ in Figure 28-1). Because phosphorescence is a process with a low probability, the $T_1$ state may persist from fractions of a second to many seconds. For benzene at $-200^\text{o}$, the absorption of light at $254 \: \text{nm}$ leads to fluorescence centered on $290 \: \text{nm}$ and phosphorescence at $340 \: \text{nm}$. The half-life of the triplet state of benzene at $-200^\text{o}$ is $7 \: \text{sec}$.
The Carbonyl Group
In previous discussions of electronic absorption spectra (Section 9-9), we have identified two different kinds of transitions in the spectra of simple carbonyl compounds such as 2-propanone or methanal. One involves excitation of an electron in a nonbonding $n$ orbital on oxygen to an antibonding $\left( \pi^* \right)$ orbital of the carbon-oxygen double bond (an $n \rightarrow \pi^*$ transition), and the other involves excitation of an electron in the bonding $\left( \pi \right)$ orbital to the corresponding antibonding orbital (a $\pi \rightarrow \pi^*$ transition). These changes are shown for methanal in Figure 28-3. Besides the transitions already discussed, methanal shows strong absorption at $175 \: \text{nm}$, which possibly is $n \rightarrow \sigma^*$, or else $\sigma \rightarrow \sigma^*$.
Although the $n \rightarrow \pi^*$ and $\pi \rightarrow \pi^*$ transitions of Figure 28-3 are singlet-singlet transitions, each of the two singlet excited states produced has a corresponding triplet state. Accordingly, there are four easily accessible excited states of a carbonyl group - the $n \rightarrow \pi^*$ singlet $\left( S_1 \right)$, $n \rightarrow \pi^*$ triplet $\left( T_1 \right)$, $\pi \rightarrow \pi^*$ singlet $\left( S_2 \right)$, and $\pi \rightarrow \pi^*$ triplet $\left( T_2 \right)$. The energies of these electronic states for methanal decrease in the order $S_2 > T_2 > S_1 > T_1$, although this ordering may not hold for all carbonyl compounds.
As we shall see, $n \rightarrow \pi^*$ singlet and triplet states of carbonyl compounds play an important role in photochemistry. Aldehydes and ketones display all the characteristics of absorption, fluorescence, phosphorescence, and intersystem crossing $\left( S_1 \rightarrow T_1 \right)$ illustrated in Figure 28-1. Generally, they are more efficient at intersystem crossing than are unsaturated hydrocarbons, perhaps because the energies of the $S$ and $T$ states involved are not widely different.
Besides the bond lengths being longer in excited states of molecules, the molecular shapes differ from those of the ground states. Although the Franck-Condon principle requires that absorption produce excited states with the same geometry as the ground states, the excited molecules thereafter can relax to more stable shapes, which may be nonplanar and twisted about the erstwhile $\pi$ bonds. Methanal is planar with a $\ce{C-O}$ bond length of $1.21 \: \text{Å}$ in the ground state, but in the $n \rightarrow \pi^*$ singlet $\left( S_1 \right)$ state, methanal is pyramidal, with a $\ce{C-O}$ bond length of $1.32 \: \text{Å}$. Methanal is even more distorted in the $n \rightarrow \pi^*$ triplet state, although the bond length remains about the same at $1.32 \: \text{Å}$.
Indirect Electronic Excitation Energy Transfer
It is possible to produce electronic excited states of molecules indirectly by way of energy transfer from other excited molecules. An example is provided by excitation of naphthalene as the result of energy transfer from excited benzophenone. Benzophenone, $\ce{C_6H_5COC_6H_5}$, absorbs ultraviolet light with $\lambda_\text{max} = 330 \: \text{nm}$ in an $n \rightarrow \pi^*$ transition. Naphthalene does not absorb appreciably in this region. Yet irradiation of a mixture of benzophenone and naphthalene with $330$-$\text{nm}$ light produces phosphorescent emission from naphthalene. Thus benzophenone absorbs the light and transfers its excess energy to naphthalene, which returns to the ground state by emission. Because the emission is from the triplet state of naphthalene, benzophenone must be involved in exciting the naphthalene to the triplet state. We may write the process as follows:
Energy transfer does not involve a net change in electron spin. For this to hold for excitation of naphthalene from $S_0$ to $T_1$, the energy transfer must come from triplet (not singlet) benzophenone). The process of producing excited states in this way is called photosensitization. Singlet-singlet, as well as triplet-triplet, energy transfers are possible, but in all cases there is no net change in spin. Efficient energy transfer will only be possible if $\Delta G^0$ for the transfer is small or negative.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.02%3A_Light_Absorption_Flourescence_and_Phosphorescence.txt |
An extraordinary variety of reactions of organic compounds are known to occur under the influence of visible and ultraviolet light. Some of these, such as the photochemical halogenation of alkanes and photosynthesis in green plants, already have been discussed (see Sections 4-4D and 20-9). It is not our purpose here to review organic photochemistry in detail - rather, we shall mention a few types of important photochemical reactions and show how these can be explained by the principles discussed in the preceding section.
Compounds have very different chemical behavior in their excited states compared to their ground states. Not only is the energy much higher, but the molecular geometry and electronic configurations are different. Intuitively, we expect that excited states of molecules, in which two electrons occupy separate unfilled orbitals, would have substantial diradical character. This is the case, especially for triplet states, as we shall see.
Photodissociation Reactions
We have mentioned how chlorine molecules dissociate to chlorine atoms on absorption of near-ultraviolet light and thereby cause radical-chain chlorination of saturated hydrocarbons (Section 4-4D). Photochemical chlorination is an example of a photochemical reaction that can have a high quantum yield - that is, many molecules of chlorination product can be generated per quantum of light absorbed. The quantum yield of a reaction is said to be unity when $1 \: \text{mol}$ of reactant is converted to product(s) per einstein$^1$ of light absorbed. The symbol for quantum yield is usually $\Phi$.
2-Propanone (acetone) vapor undergoes a photodissociation reaction with $313$-$\text{nm}$ light with $\Phi$ somewhat less than unity. Absorption of light by 2-propanone results in the formation of an excited state that has sufficient energy to undergo cleavage of a $\ce{C-C}$ bond (the weakest bond in the molecule) and form a methyl radical and an ethanoyl radical. This is a primary photochemical reaction:
$\tag{28-1}$
The subsequent steps are dark reactions.
At temperatures much above room temperature, the ethanoyl radical breaks down to give another methyl radical and carbon monoxide:
$\tag{28-2}$
If this reaction goes to completion, the principal reaction products are ethane and carbon monoxide:
$2 \ce{CH_3} \cdot \rightarrow \ce{CH_3-CH_3} \tag{28-3}$
If the ethanoyl radical does not decompose completely, then some 2,3-butanedione also is formed. This reaction is quite important at room temperature or below:
$\tag{28-4}$
Lesser amounts of methane and ketene also are formed as the result of disproportionation reactions involving hydrogen-atom transfers of the types we have encountered previously in radical reactions (see Section 10-8C):
$\tag{28-5}$
The product-forming reactions, Equations 28-2 through 28-5, all depend on the primary photochemical event, Equation 28-1, which breaks the $\ce{C-C}$ bond to the carbonyl group. This cleavage has been termed a Norrish type I process after the eminent photochemist, R. G. W. Norrish:$^2$
Another photochemical reaction is important for ketones that have at least one $\gamma$ hydrogen on a chain connected to the carbonyl group, as in
In this pathway (Norrish type II process), cleavage occurs at the $\ce{C}_\alpha \ce{-C}_\beta$ bond to give, as the major product, a ketone of shorter chain length and an alkene. Thus for 2-pentanone:
This reaction occurs in an interesting way. Whatever the nature of then $n \rightarrow \pi^*$ excited state, $S_1$ or $T_1$, the primary photochemical reaction is the abstraction of a hydrogen atom from the $\gamma$ carbon by the carbonyl oxygen to give the diradical, $1$:
The subsequent dark reactions readily are understood as typical of diradicals. Cleavage of $1$ at $\ce{C}_\alpha \ce{-C}_\beta$ gives ethene and an enol, which rearranges to the ketone. Alternatively, $1$ can cyclize to a cyclobutanol:
A variety of photodissociation reactions have been found to take place with ketones, but the products almost always can be explained as the result of Norrish type I and/or II cleavage. Examples are:
Photoreduction of Diaryl Ketones
Diaryl ketones do not undergo photodissociation in the same way as alkyl ketones, probably because cleavage to phenyl and other aryl radicals is unfavorable (Table 4-6). Nevertheless, aromatic ketones are photochemically reactive in the presence of compounds that can donate a hydrogen atom, with the result that the carbonyl group is reduced. Indeed, one of the classic photochemical reactions of organic chemistry is the formation of 1,1,2,2-tetraphenyl-1,2-ethanediol ($3$, benzopinacol) by the action of light on a solution of diphenylmethanone ($2$, benzophenone) in isopropyl alcohol. The yield is quantitative.
The light is absorbed by $2$ and the resulting activated ketone, $2^*$, removes a hydrogen from isopropyl alcohol:
Benzopinacol results from dimerization of the radicals, $4$:
Since the quantum yields of 2-propanone and benzopinacol both are nearly unity when the light intensity is not high, it is clear that two of the radicals, $4$, must be formed for each molecule of $2$ that becomes activated by light. This is possible if the 2-hydroxy-2-propyl radical formed by Equation 28-7 reacts with $2$ to give a second diphenylhydroxymethyl radical:
This reaction is energetically favorable because of the greater possibility for delocalization of the odd electron in $4$ than in the 2-hydroxy-2-propyl radical.
Photochemical formation of $3$ also can be achieved from diphenylmethanone, $2$, and diphenylmethanol, $5$:
The mechanism is similar to that for isopropyl alcohol as the reducing agent:
This reduction is believed to involve the triplet state of $2$ by the following argument: Formation of $3$ is reasonably efficient even when the concentration of the alcohol, $5$ is low; therefore, whatever the excited state of the ketone, $2^*$, that accepts a hydrogen atom from $5$, it must be a fairly long-lived one. Because solutions of $2$ show no visible fluorescence, they must be converted rapidly to another state of longer life than the singlet $\left( S_1 \right)$. The long-lived state is then most reasonably a triplet state. In fact, if naphthalene is added to the reaction mixture, formation of benzopinacol, $3$ is drastically inhibited because the benzophenone triplet transfers energy to naphthalene more rapidly than it reacts with the alcohol, $5$ (see Section 28-1A).
Photochemical Isomerization of Cis and Trans Alkenes
An important problem in many syntheses is to produce the desired isomer of a cis-trans pair of alkenes. The problem would not arise if it were possible to isomerize the undesired isomer to the desired isomer. In many cases such isomerizations can be carried out photochemically. A typical example is afforded by cis- and trans-1,2-diphenylethene (stilbene):
Here the trans form is easily available by a variety of reactions and is more stable than the cis isomer because it is less sterically hindered. However, it is possible to produce a mixture containing mostly cis isomer by irradiating a solution of the trans isomer in the presence of a suitable photosensitizer. This process in no way contravenes the laws of thermodynamics because the input of radiant energy permits the equilibrium point to be shifted from what it would be normally.
Isomerization appears to occur by the following sequence: The sensitizer, usually a ketone such as benzophenone or 1-(2-naphthyl)ethanone, is raised by an $n \rightarrow \pi^*$ transition from the singlet ground state $\left( S_0 \right)$ to an excited state $\left( S_1 \right)$ by absorption of light. Intersystem crossing then occurs rapidly to give the triplet state $\left( T_1 \right)$ of the sensitizer:
The next step is excitation of the alkene by energy transfer from the triplet state of the sensitizer. Remember, the net electron spin is conserved during energy transfer, which means that the alkene will be excited to the triplet state:
$^3\text{Sens}^* \left( \uparrow \uparrow \right) + ^1\text{Alkene} \left( \uparrow \downarrow \right) \rightarrow ^1\text{Sens} \left( \uparrow \downarrow \right) + ^3\text{Alkene}^* \left( \uparrow \uparrow \right)$
The triplet state of the alkene is most stable when the $p$ orbitals, which make up the normal $\pi$ system of the double bond, are not parallel to one another (Figure 6-17). Therefore, if the energy-transfer process leads initially to a planar triplet, this is converted rapidly to the more stable nonplanar form. The excitation of either the cis or the trans isomer of the alkene appears to lead to a common triplet state, as shown in Figure 28-4.
The final step in the isomerization is decay of the alkene triplet to the ground state. This happens either by emission of light (phosphorescence) or by having the triplet energy converted to thermal energy without emission of light. Either way, the cis or trans isomer could be formed and, as can be seen from Figure 28-4, the ratio of isomers produced depends on the relative rates of decay of the alkene triplet to the ground-state isomers, $k_c/k_t$. This ratio turns out to favor formation of the less stable isomer. Therefore, provided both isomers can be photosensitized efficiently, sensitized irradiation of either one will lead ultimately to a mixture of both, in which the thermochemically less stable isomer predominates. The sensitizer must have a triplet energy in excess of the triplet energy of the alkene for energy transfer to occur, and the photostationary or equilibrium point is independent of the nature of the sensitizer when the latter transfers energy efficiently to both cis and trans isomers. In the practical use of the sensitized photochemical equilibrium of cis and trans isomers, it is normally necessary to carry out pilot experiments to determine what sensitizers are useful.
Another example of how photochemical isomerization can be used is provided by the equilibration of the $E$ and $Z$ form of 1-bromo-2-phenyl-1-propene:
The $E$ isomer is formed to the extent of $95\%$ in the dehydrohalogenation of 1,2-dibromo-2-phenylpropane:
Photoisomerization of the elimination product with 1-(2-naphthyl)ethanone as sensitizer produce a mixture containing $85\%$ of the $Z$ isomer.
Photochemical Cyclization Reactions
One may well ask why the isomerization of alkenes discussed in the preceding section requires a sensitizer. Why cannot the same result be achieved by direct irradiation? One reason is that a $\pi \rightarrow \pi^*$ singlet excited state $\left( S_1 \right)$ produced by direct irradiation of an alkene or arene crosses over to the triplet state $\left( T_1 \right)$ inefficiently (compared to $n \rightarrow \pi^*$ excitation of ketones). Also, the $S_1$ state leads to other reactions beside isomerization which, in the case of 1,2-diphenylethene and other conjugated hydrocarbons, produce cyclic products. For example, cis-1,2-diphenylethene irradiated in the presence of oxygen gives phenanthrene by the sequence of Equation 28-8. The primary photoreaction is cyclization to a dihydrophenanthrene intermediate, $6$, which, in the presence of oxygen, is converted to phenanthrene:
The cyclization step of Equation 28-8 is a photochemical counterpart of the electrocyclic reactions discussed in Section 21-10D. Many similar photochemical reactions of conjugated dienes and trienes are known, and they are of great interest because, like their thermal relatives, they often are stereospecific but tend to exhibit stereochemistry opposite to what is observed for formally similar thermal reactions. For example,
These reactions are $4n$-electron concerted processes controlled by the symmetry of the reacting orbitals. The thermal reaction is most favorable with a Mobius transition state (achieved by conrotation), whereas the photochemical reaction is most favorable with a Huckel transition state (achieved by disrotation).
Conjugated dienes also undergo photochemical cycloaddition reactions. Related thermal cycloadditions of alkadienes have been discussed in Sections 13-3A, 21-10A, and 21-10D, but the thermal and photochemical reactions frequently give different cyclic products. Butadiene provides an excellent example of the differences:
In the thermal reaction the [4 + 2] or Diels-Alder adduct is the major product, whereas in the photochemical reaction [2 + 2] cycloadditions dominate. Because the photochemical additions are sensitized by a ketone, $\ce{C_6H_5COCH_3}$, these cycloadditions occur through the triplet state of 1,3-butadiene and, as a result, it is not surprising that these cycloadditions are stepwise, nonstereospecific, and involve diradical intermediates.
Excitation:
Cycloaddition:
Direct irradiation of 1,3-butadiene with $254$-$\text{nm}$ light produces cyclobutene and small amounts of bicyclo[1.1.0]butane along with dimers.
In contrast to conjugated dienes, simple alkenes such as 2-butene do not react easily by photosensitized cycloaddition. But they will form [2 + 2] cycloadducts on direct irradiation. These additions occur by way of a singlet excited state and are stereospecific:
A related reaction, which has no precedent in thermal chemistry, is the cycloaddition of an alkene and an aldehyde or ketone to form an oxacyclobutane:
In this kind of addition the ground-state alkene $\left( S_0 \right)$ reacts with an excited state (usually $T_1$) of the carbonyl compound by way of a diradical intermediate:
Singlet Oxygen Reactions
The ground state of molecular oxygen is unusual because it is a triplet state. Two electrons of parallel spin occupy separate $\pi$ orbitals of equal energy (degenerate), as shown schematically in Figure 28-5,$^3$ The next two higher electronic states both are singlet states and lie respectively $24$ and $37 \: \text{kcal mol}^{-1}$ above the ground state. From this we can understand why ordinary oxygen has the properties of a diradical and reacts rapidly with many radicals, as in the radical-chain oxidation of hydrocarbons (autoxidation; Sections 15-10 and 16-9E):
Oxygen also efficiently quenches excited triplet states of other molecules $\left( ^3A^* \right)$ and, in accepting triplet energy, is itself promoted to an excited singlet state. Notice that the total spin orientation is conserved:
Singlet oxygen is highly reactive toward many organic molecules and will form oxygenated addition or substitution products. As one example, conjugated dienes react with singlet oxygen to give peroxides by [4 + 2] cycloaddition. Because only singlet states are involved, this addition is quite analogous to thermal Diels-Alder reactions (Section 21-10A):
If the alkene or alkadiene has at least one hydrogen on the carbon adjacent to the double bond, reaction with singlet oxygen may give hydroperoxides. The mechanism of this reaction is related to [4 + 2] cycloadditions and is presumed to occur through a Huckel pericyclic transition state (see Section 21-10D):
Many reactions of this type can be achieved by allowing the hydrocarbon to react with oxygen in the presence of a sensitizing dye that strongly absorbs visible light. The dyes most commonly used for this purpose include fluorescein (and its chlorinated analogs, eosin and rose bengal), methylene blue, and porphyrin pigments (such as chlorophyll).
The overall process of photosensitized oxygenation of a substrate $\left( \ce{A} \right)$ proceeds by the following steps:
$^1\text{Sens} \overset{h \nu}{\longrightarrow} \: ^1\text{Sens}^* \longrightarrow ^3\text{Sens}^*$
$^3\text{Sens}^* + ^3\ce{O_2} \longrightarrow ^1\text{Sens} + ^1\ce{O_2}*$
$^1\ce{O_2}* + \ce{A} \longrightarrow \ce{AO_2}$
Singlet oxygen can be produced chemically as well as by photochemical sensitization. There are several chemical methods available, one of the best known being the reaction of sodium hypochlorite with peroxide:
$\ce{NaOCl} + \ce{H_2O_2} \rightarrow \ce{NaCl} + \ce{H_2O} + ^1\ce{O_2} \tag{28-9}$
An alternative method of formation, which can be used in organic solvents at low temperatures, involves the thermal decomposition of triethyl phosphite ozonide (Equation 28-10):
Regardless of whether singlet oxygen is formed chemically or photochemically, it gives similar products in reactions with alkenes.
Photobiology
Photosensitized reactions of oxygen are largely damaging to living organisms. Indeed, singlet oxygen reacts destructively with amino acids, proteins, and nucleic acids. How does an organism protect itself against the damaging effects of oxygen? There are no simple answers, but green plants provide a clue. Chlorophyll is an excellent sensitizing dye for singlet oxygen; yet green plants evidently are not harmed because of it. A reason may be that singlet oxygen is quenched very efficiently by other plant pigments, especially the carotenoid pigments such as $\beta$-carotene (Section 2-1). That this is the case is indicated by the fact that mutant plants unable to synthesize carotene are killed rapidly by oxygen and light.
That direct irradiation with ultraviolet light is damaging to single-cell organisms is well known. It also is known that the nucleic acids, DNA and RNA, are the important targets of photochemical damage, and this knowledge has stimulated much research in the field of photobiology in the hope of unravelling the chemistry involved.
An interesting and significant outcome is the finding that the pyrimidine bases of nucleic acids (uracil, thymine, and cytosine) are photoreactive and undergo [2 + 2] cycloadditions on irradiation with ultraviolet light. Thymine, for example, gives a dimer of structure $7$:
Comparable experiments with the nucleic acids have confirmed that cycloaddition of their pyrimidine bases also occurs with ultraviolet light and effectively cross-links the chains, a process obviously quite inimical to the functioning of the DNA (Section 25-13B). A remarkable and not well understood aspect of photobiology is the repair and defense mechanism both plants and animals possess to minimize the damaging effects of radiation.
On the positive side, there are photochemical reactions that are essential for human health. One of these is the formation of vitamin D (the antirachitic vitamin) by irradiation of ergosterol. This photochemical reaction is an electrocyclic ring opening of the cyclohexadiene ring of ergosterol of the type described in Section 28-2D. The product, previtamin D2, subsequently rearranges thermally to vitamin D2:
$^1$The einstein unit is defined in Section 9-4.
$^2$Recipient with G. Porter of the Nobel Prize in chemistry in 1967 for work on photochemical reactions.
$^3$For a more detailed account of the electronic configuration of molecular oxygen, see M. Orchin and H. H. Jaffe, The importance of Antibonding Orbitals, Houghton Mifflin Co., Boston, 1967; or H. B. Gray, Chemical Bonds, W. A. Benjamin, Inc., Menlo Park, Calif., 1973.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.03%3A_Organic_Photochemistry.txt |
The most common means of generating electronically excited states of molecules is by the absorption of electromagnetic radiation. But excited states are accessible by other routes. Indeed, as shown in Section 28-2E, the excited singlet state of molecular oxygen can be produced by chemical reactions (Equations 28-9 and 28-10). Many other reactions are known that generate products in electronically excited states, and this is especially evident when the electronically excited products go to the ground state by the emission of visible light. This behavior is known as chemiluminescence and is transduction of chemical energy $\left( \Delta H \right)$ into radiant energy $\left( h \nu \right)$. Chemiluminescence is possible only when the $\Delta H$ of the reaction is sufficiently large to allow for production of at least one of the products in an electronically excited state $\left( ^* \right)$. Chemiluminescence amounts to $\Delta H \rightarrow ^* \rightarrow h \nu$, which is opposite to most photochemistry which involves $h \nu \rightarrow ^* \rightarrow \Delta H$.
A beautiful example of a chemiluminescent reaction is the thermal dissociation of the cyclic peroxide, $8$, into two molecules of 2-propanone:
We should not be surprised at the high exothermicity of Reaction 28-11. The peroxide is of high energy (thermochemically unstable) because it combines the strain-energy characteristics of small rings with the weakness of $\ce{O-O}$ bonds, whereas the product is a stable substance with a strong carbonyl bond.
Chemiluminescence in many reactions is hard to detect because the efficiency of light emission is low. Thus, even though the excited state may be formed in high yield, it may be quenched by other species more efficiently than it loses energy by emission. This fact can be used to advantage by adding a substance that quenches the excited state efficiently and, after energy transfer, gives bright fluorescence or phosphorescence:
$\Delta H \rightarrow ^* \underset{\text{energy transfer}}{\overset{\text{dye}}{\longrightarrow}} \text{dye}^* \overset{\text{luminescence}}{\longrightarrow} \text{dye} + h \nu$
Chemiluminescence can be greatly amplified by this process and it forms the basis of spectacular demonstrations of "cold light". An example is the perhydrolysis of ethanedioic (oxalic) esters with hydrogen peroxide in the presence of a fluorescent substance (Equation 28-12). The reaction is believed to pass through the highly unstable dioxacyclobutanedione, which dissociates into two moles of carbon dioxide with such exothermicity that electronic excitation occurs, as evident from the intense light produced in the presence of fluorescent dyes:
$\tag{28-12}$
This reaction has been developed into a commercial product, marketed under the trade name "Coolite", which can be used as an emergency light source by simply shaking a tube to bring the reactants in contact with one another.
Of major interest is the identity of the excited state (singlet or triplet) produced by chemiluminescent reactions. Little is known about excited states produced chemically except in a few cases, as in Reaction 28-11. Here the chemiluminescence dissociation gives a ratio of triplet 2-propanone to excited singlet 2-propanone of 100:1. This is a surprising result because it means that spectroscopic selection rules of electron-spin conservation are not followed in this chemiexcitation. The reaction has generated a triplet state from a singlet state. How can this be? Some idea of what is involved can be obtained from Figure 28-7, in which we see that breaking of the two sigma $\ce{C-C}$ and $\ce{O-O}$ bonds gives directly one molecule of ground-state ketone (all spins paired) and one molecule of triplet ketone. In this process, the electrons associated with the orbitals on one of the oxygen atoms appear to interact in such a way as to interchange electrons between orbitals on the same atoms with a spin inversion. This is called spin-orbit coupling.
Bioluminescence
The emission of visible light by living organisms is a mysterious and fascinating phenomenon. The magical glow of the firefly and of certain plants and marine animals is a familiar sight and one that has stimulated man's curiosity and imagination for centuries. Despite intense interest in bioluminescence, it is only recently that substantial progress has been made in our understanding of how it occurs.
One of the earliest studies of bioluminescence was made by the French scientist R. Dubois toward the end of the nineteenth century. He demonstrated that bioluminescent organisms emitted light as a consequence of chemical change. He succeeded in isolating the active chemical from fireflies (luciferin) and the activating enzyme (luciferase, named by Dubois from the Latin lucifer, meaning light bearer). Luciferin and the enzyme in the presence of oxygen were found to reproduce the natural bioluminescence:
$\text{luciferin} + \ce{O_2} \overset{\text{luciferase}}{\longrightarrow} \text{oxyluciferin} + h \nu$
Further progress required elucidation of the structures of luciferin and its oxidation product. It turned out that there are several luciferins, depending on the organism. Firefly luciferin has the benzothiazol structure, $9$; the luciferins from the marine crustacean Cypridina hilgendorfii and the sea pansy Renilla reformis have structures $10$ and $11$, respectively. Their oxidation products $12$, $13$, and $14$ also are shown:
Although the luciferins $9$-$11$ may not seem closely related, each appears to react with oxygen (at the direction of the appropriate enzyme) to give cyclic peroxylacetone intermediates $12$-$14$. Luminescence is the consequence of the energetically favorable dissociation of the dioxacyclobutanone ring system to carbon dioxide and a carbonyl component. This mechanism is suggested by experiments with the peroxy acid, $15$, which with $\ce{N}$,$\ce{N}$-dicyclohexylcarbodiimide gives a very reactive compound presumed to be the peroxylactone, $16$. This substance liberates $\ce{CO_2}$ rapidly at room temperature with luminescence:
$^4$Production of two molecules of excited 2-propanone per molecule of $8$ is not possible under the same conditions because this would correspond to a reaction with $\Delta H^0$ of at least $156 \: \text{kcal}$ above formation of two moles of ground-state ketone.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.04%3A_Chemiluminescence.txt |
Visible light is electromagnetic radiation having a rather narrow range of wavelengths ($400$-$800 \: \text{nm}$). A black substance absorbs all wavelengths of visible light. Selective absorption of visible light by a substance imparts color, but the color is not that of the light absorbed but instead of the residual light that the substance transmits or reflects. For example, a compound that absorbs in the region $435$-$480 \: \text{nm}$ removes blue light from the visible spectrum, and the residual light is recognized by the eye as being yellow. The relationship of the observed color to wavelength of light absorbed is shown in Table 28-1. It is customary to call the color observed the complementary color or the subtraction color to that absorbed.
Table 28-1: Color and Wavelength
Clearly, the color perceived, its brightness and its intensity, depends on the shape of the electronic spectral curve of the absorbing substance, which in turn depends on the chemical structure of the substance. A change in absorption from the blue to the red end of the spectrum corresponds to a decrease in the energy of the associated electronic transitions. We know also that this trend is associated with increasing conjugation of multiple bonds. For instance, 1,2-diphenylethene is colorless, whereas 1,10-diphenyl-1,3,5,7,9-decapentaene is yellow-orange:
The effect of substituents on colors associated with conjugated systems is of particular interest in the study of dyes, because most dyes have relatively short conjugated systems and would not be intensely colored in the absence of substituent groups. (The plant pigments $\beta$-carotene, Section 2-1, and lycopene, often used as food coloring, are exceptions.)
The conjugated $\pi$ system common to all four compounds is that of the benzenoid ring, which is described as the absorbing chromophore (Section 22-3B). The hydroxyl and nitro substituents can be seen individually to shift the $\lambda_\text{max}$ of the chromophore to longer wavelengths. However, the combined effect of the two substituents is much more dramatic, especially if the $\ce{OH}$ group is converted to the corresponding anion, 4-nitrobenzenolate. Now $\lambda_\text{max}$ is shifted into the visible region, giving a yellow color, and because $\epsilon$ is large, the color is intense. Thus, properly chosen substituents can shift the main benzenoid absorption band from the ultraviolet into the visible region of the spectrum. Such substituents are often called auxochromes. They act by extending the conjugation of the chromophore and are particularly effective in causing large shifts towards the visible when one substituent is a $\pi$-electron donor and the other a $\pi$-electron acceptor. Thus, with the 4-nitrobenzenolate ion, interaction between the strongly electron-donating $\ce{-O}^\ominus$ group and the strongly electron-accepting $\ce{-NO_2}$ group provides significant stabilization:
Hence, substitution of an electron-attracting group (such as $\ce{NO_2}$) at one end of such a system and an electron-donating group (such as $\ce{O}^\ominus$) at the other end should be particularly favorable for stabilization of the excited state (relative to the ground state, where $17a$, $17b$, etc., are of lesser importance). At the same time, we should not expect that two electron-attracting (or two electron-donating) groups at opposite ends would be nearly as effective.
We hope you will understand from the foregoing discussion why it is that many intensely colored substances of natural or synthetic origin have conjugated structures with substituents, often cationic or anionic substituents, that can donate or accept electrons from the conjugated system. Such compounds provide us with many useful dyes, pigments, indicators, and food-coloring agents, as well as conferring color on plants and animals. A few examples follow:
Dyes
Historically, the dye industry has been closely linked with the development of synthetic organic chemistry. Although dyes have been extracted from natural sources for centuries, it was not until 1856 that a synthetic dye was produced commercially. The previous year, William Henry Perkin - at age 17 - oxidized benzenamine (aniline) with potassium dichromate and isolated from the product (which was mostly aniline black; Section 23-11D) a purple compound that was excellent for dyeing silk. Perkin started commercial production of the dye under enormous difficulties. Because there was no organic chemical industry at the time, he had to design and build his own equipment as well as devise efficient syntheses for starting materials. His route to benzenamine stared with crude benzene from coal, which he nitrated and then reduced with iron and acid. He had to make the nitric acid (from nitrate salts and sulfuric acid) because concentrated nitric acid was not available. It was not until 1890 that the structure of Perkin's dye, called mauveine, was established by Otto Fischer. The dye was actually a mixture because the benzene used contained methylbenzene), but the product from the oxidation of benzenamine itself is structurally related to aniline black:
Although the mauveine dyes have been replaced with better dyes, they are representative of a group of useful dyes having the general structure
in which $\ce{X}$ and $\ce{Y}$ can be oxygen, nitrogen, sulfur, or carbon. The rings invariably carry substituents (hydroxyl or amino) that provide enhanced stabilization of the excited states. Examples of these ring systems follow:
A large number of useful dyes are substituted triphenylmethane derivatives. Crystal Violet (Section 28-4) and phenolphthalein are excellent examples of this kind of dye.
Other important dyes are derivatives of the following types of substances:
Examples are
There is more to a successful dye than just an attractive color.$^5$ If it is to be useful, say for coloring fabrics, some simple means must be available for introducing the color into the fiber and then, usually of greater difficulty and importance, the color must be reasonably permanent - that is, resistant to normal laundry or cleaning procedures (wash-fast) and stable to light (light-fast). Here again, fundamentally important problems are involved. The scientific approach to improving wash-fastness of fabric dyes has to be based on a knowledge of the structural factors bearing on the intermolecular forces that determine solubilities. Light-fastness is connected with the photochemistry of organic compounds.
Pigments
The distinction between a dye and a pigment is that a dye actually is absorbed by the material to be colored, whereas a pigment is applied with a binding material to the surface. Pigments usually are highly insoluble substances. Many insoluble inorganic substances that would be wholly unsatisfactory as dyes are useful pigments.
Copper phthalocyanine is an example of a very important class of organic pigments. These are tetraazatetrabenzo derivatives of the porphyrin compounds discussed in Sections 20-9 and 25-8B. Copper phthalocyanine arises from condensation of four molecules of 1,2-benzenedicarbonitrile in the presence of copper metal at $200^\text{o}$:
$^5$For a good account of dyes, see L. F. Fieser and M. Fieser, Organic Chemistry, D. C. Health & Co., Lexington, Mass., 1956, Chapter 36.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
28.06: The Sensation of Color
The sensation of color can be achieved in different ways. According to Table 28-1, which relates wavelength to color, we could recognize a given color, say yellow, by direct perception of light encompassing a narrow band of wavelengths around $580 \: \text{nm}$, or by subtraction of blue light ($435$-$480 \: \text{nm}$) from white light.
A third way of producing color is by an additive process. In fact, a wide range of colors can be achieved by the addition of three colors - red, green, and blue - as indicated in Figure 28-10. Mixing all three so-called primary additive colors, in the right intensities, gives white light; mixing only red and green gives yellow. It is important to recognize that addition of any two primary colors is equivalent to subtracting the third. This point is amplified in Figure 28-10. Subtraction of the three primary additive colors, red, green, and blue, from white light gives, respectively, the three primary subtraction colors, cyan, magenta, and yellow. Application of additive and subtractive processes in color perception is illustrated in the following sections. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.05%3A_Color_and_Constitution.txt |
Subtractive Color Process
Photography is a popular activity for many, but relatively few have an understanding of the chemistry involved, particularly in color photography. This is unfortunate because color photography represents an interesting combination of photochemistry (energy transfer), organic chemistry (dye formation), optics, psychology and physiology (color perception), and engineering (production and development of film).
The sensation of full color in color transparencies produced by photographic means is achieved by subtraction. As a simple example, let us suppose that the subject to be photographed is blue. To obtain a blue image by shining white light through a transparency, the transparency is made to subtract (absorb) yellow light - that is, to absorb strongly in the $580$-$\text{nm}$ region. How a color image of the subject is recorded chemically on the film and how the film is developed into a transparency will become clearer from the following discussion.
Color Film
The emulsion of a typical color film has three silver-bromide layers separately sensitized by suitable dyes to blue, green, and red light (Figure 28-11). When processed (Section 28-6C), the color formed in each layer is complementary to the color to which the layer is sensitive. Thus, if unexposed film is processed, intense yellow, magenta, and cyan colors are respectively formed in the blue-, green-, and red-sensitive layers. Then, when white light strikes this processed film, the yellow layer subtracts the blue, the magenta subtracts the green, and the cyan subtracts the red, with the result that the film appears black (or nearly so), as corresponds to no exposure to light. However, if the film is exposed to strong blue light before processing, the blue-sensitive layer responds, and when the film is processed, no yellow dye is formed in the blue-sensitive layer (see Figure 28-11). The transparency then contains only the subtraction colors, magenta and cyan. When white light enters a transparency of superimposed magenta and cyan dyes, only blue light is transmitted, as befits the color of the original sensitizing light. (From the right side of Figure 28-10, we see the overlap of 5 and 4 leads to blue.) Similarly, exposure of the film to strong yellow light (containing no blue), followed by processing, results in formation of yellow dye and no magenta nor cyan. This is because the green- and red-sensitive emulsions both are sensitive to yellow light, while the blue-sensitive emulsion does not respond to yellow light.
In summary, the overall process from color film to the projection of a color image involves two separate conversions of each color into its complement, the net result being an image that has the same colors as the original subject.
Chemistry of Color Developers
We have seen that full color perception can be achieved by subtraction methods using dyes in suitable combinations. We now have to consider how such dyes are formed on exposure and development of color film. First though, you should recognize that a photographic emulsion, whether for color or black-and-white film, is light-sensitive primarily because of the presence of sliver halide. You will recall from previous discussions (Section 26-2C) that the sequence from exposure to development involves the following:
The chemistry involved in the formation of the dyes in the usual color films is highly ingenious and is achieved through steps shown in Figure 28-12 which, for purposes of illustration, are carried through for an initial exposure to green light. The exposure activates only the green-sensitive emulsion, and development with an ordinary developer such as metol-hydroquinone (Section 26-2C) produces silver metal only in the green-sensitive layer. The film now has the visual appearance of a milky negative. The developer then is washed out and the film is fogged, a process that activates the silver bromide remaining unreduced in the first step. The activated silver bromide so formed then is reduced with a color developer, usually 4-amino-$\ce{N}$,$\ce{N}$-diethylbenzenamine, in the presence of a color coupler. Production of dye occurs in direct proportion to the amount of activated silver bromide present, in close conformity with the following equation:
$\tag{28-13}$
A different color coupler is used for each layer of the emulsion and, although the complete color picture is formed in this step, the film is coal black because of the metallic silver produced at the same time. The silver then is oxidized to silver bromide with dichromate solution containing bromide ion and removed with thiosulfate solution. The final image thus contains no silver.
The color-forming reactions obviously are critical to the success of the overall process and, of necessity, involve some degree of compromise between requirements for yield, reproducibility, suitability of color, and light-fastness. In reactions of the type shown in Equation 28-13, the color coupler is a methylene compound, $\ce{R_2CH_2}$, in which the $\ce{R}$ groups have sufficient electron-attracting character to undergo some degree of formation of $\ce{R_2CH}^\ominus$ in the alkaline medium used for color development. The first two steps in the overall sequence follow:
The product $24$ is a photographic developer and is oxidized either by activated silver bromide or by $23$ to a new quinonimmonium salt, $25$:
This substance has an acidic hydrogen at the $\ce{-CHR_2}$ and, on loss of this proton, a neutral conjugated imine results, $26$, which is the actual dye. Color is expected for these molecules because the $\ce{R}$ groups are electron-attracting and the diethylamino group at the other end of the conjugated system is electron-donating. The remaining important question is how to adjust the $\ce{R}$ groups to obtain the proper colors in each layer of the film. Compounds that give yellow, magenta, and cyan dyes with 4-amino-$\ce{N}$,$\ce{N}$-diethylbenzenamine and activated silver bromide are shown below.
In the case of $\ce{N}$-phenyl-3-phenyl-3-oxopropanamide, the yellow dye formed has the following structure:
Such dyes often are called azomethines. The nature of the exact color couplers used in color film is not well publicized. It is important that the couplers not diffuse out of the layers where they belong and, for this reason, insolubilizing substituent groups are attached to strategic positions on the color couplers.
The "instant" color process pioneered by Polaroid operates by a wonderfully ingenious, subtractive-color, three-layer scheme basically similar to the one shown in Figures 28-11 and 28-12. A very important difference is that the colors are transferred from the emulsion to a layer of white pigment $\left( \ce{TiO_2} \right)$, so the process actually is a printing process. The subtractive dyes (yellow, magenta, and cyan) are present in the emulsion as substituent groups on base-soluble photographic developers, schematically [dye$\ce{-(CH_2)}_n-$developer]. A copper phthalocyanine derivative (Section 28-4B) is used to provide the cyan color.
What occurs in a given layer of the film is roughly as follows: Activation of the silver bromide in the layer by the light to which it is sensitized, and then development with an alkaline developer solution converts the "dye developer" to a base-insoluble form (Section 26-2C) in proportion to the light absorbed:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.07%3A_Color_Photography.txt |
Vision is a process in which light is absorbed by a pigment in a photoreceptor cell (by a dye in the eye) and the photochemistry that ensues ultimately produces a transient electrical signal that is transmitted to the brain and interpreted as a visual image. There is much that is not fully understood about this process, but we shall discuss briefly the chemistry involved.
The eye is an extraordinarily sensitive instrument. To be sure, its wavelength response is restricted to $400$-$800 \: \text{nm}$, but its degree of sensitivity is such that a fully dark-adapted eye can clearly detect objects in light so dim as to correspond to a light input over the retina of only about 10,000 quanta per second - one light quantum per three minutes per receptor cell in the retina!
The retina is made up of two kinds of light-sensitive (photoreceptor) cells, known as rods and cones. The rods are the more sensitive and are responsible for vision in dim light. The cones are much fewer in number than the rods and provide detail and color vision in good light. The part of the retina that corresponds to the center of the visual field contains only cones. A red pigment called rhodopsin is the photosensitive substance in the rod cells of the retina. It absorbs most strongly in the blue-green region of the visible spectrum $\left( \lambda_\text{max} \: 500 \: \text{nm} \right)$ and is essentially unaffected by the far-red end of the spectrum. Cone vision appears to involve a different pigment called iodopsin, which absorbs farther toward the red than does rhodopsin.
Rhodopsin is a combination of a protein called opsin, and the highly conjugated aldehyde, 11-cis-retinal:
The structure of opsin is unknown, but its prosthetic group (11-cis-retinal) is bonded to it through an imine (Schiff base) function formed between the aldehyde group of the retinal and the side-chain amino function of a lysine unit of opsin:
Opsin itself is colorless, whereas 11-cis-retinal absorbs strongly at $370 \: \text{nm}$. The combination of opsin with 11-cis-retinal produces a remarkable shift of $\lambda_\text{max}$ to longer wavelengths ($430 \: \text{nm}$ to $620 \: \text{nm}$, depending on the species). Similar shifts in wavelength for 11-cis-retinal in combination with simple amines are observed only up to $\lambda_\text{max}$ of $440 \: \text{nm}$, and only then for the protonated Schiff base. From this evidence, the chromophore in rhodopsin is believed to be protonated and to be profoundly modified by the structure of the opsin.
Light striking the retina changes the color of rhodopsin from red to yellow. The primary photochemical event in this process was established by G. Wald (Nobel Laureate in Physiology and Medicine, 1967), who showed that light absorption led to a change of configuration about the $\ce{C_{11}}$-$\ce{C_{12}}$ double bond of the retinal chromophore from cis to trans:
The exact point at which the nerve impulse is transmitted is not established with certainty, but it has to occur before the hydrolysis step because hydrolysis is too slow to account for the nerve impulse. One theory suggests that an electrical signal is generated at the instant of light absorption by electron transfer to a $\pi \rightarrow \pi^*$ singlet excited state that has substantially charged carbon atoms.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.08%3A_Chemistry_of_Vision.txt |
Polymers are substances made up of recurring structural units, each of which can be regarded as derived from a specific compound called a monomer. The number of monomeric units usually is large and variable, each sample of a given polymer being characteristically a mixture of molecules with different molecular weights. The range of molecular weights is sometimes quite narrow, but is more often very broad. The concept of polymers being mixtures of molecules with long chains of atoms connected to one another seems simple and logical today, but was not accepted until the 1930's when the results of the extensive work of H. Staudinger, who received the Nobel Prize in Chemistry in 1953, finally became appreciated. Prior to Staudinger's work, polymers were believed to be colloidal aggregates of small molecules with quite nonspecific chemical structures.
• 29.1: Prelude
The adoption of definite chemical structures for polymers has had far-reaching practical applications, because it has led to an understanding of how and why the physical and chemical properties of polymers change with the nature of the monomers from which they are synthesized. To a very considerable degree the properties of a polymer can be tailored to specific applications. Much of the emphasis in this chapter will be on how the properties of polymers can be related to their structures.
• 29.2: A Simple Addition Polymerization. The Parts of a Polymer
The thermal polymerization of 1,3-cyclopentadiene by way of the Diels-Alder addition is not an important polymerization, but it does provide a simple concrete example of how a monomer and a polymer are related.
• 29.3: Types of Polymers
Polymers can be classified in several different ways - according to their structures, the types of reactions by which they are prepared, their physical properties, or their technological uses.
• 29.4: Forces Between Polymer Chains
Polymers are produced on an industrial scale primarily, although not exclusively, for use as structural materials. Their physical properties are particularly important in determining their usefulness, be it as rubber tires, sidings for buildings, or solid rocket fuels. Polymers that are not highly cross-linked have properties that depend greatly on the forces that act between the chains.
• 29.5: Correlation of Polymer Properties with Structure
The properties of many of the commercially important thermoplastic and elastic polymers can be understood in terms of their chemical structures by using the concepts developed in the preceding section.
• 29.6: Condensation Polymers
There is a wide variety of condensation reactions that, in principle, can be used to form high polymers. However, high polymers can be obtained only in high-yield reactions, and this limitation severely restricts the number of condensation reactions having any practical importance.
• 29.7: Addition Polymers
We have discussed the synthesis and properties of a considerable number of addition polymers in this and previous chapters. Our primary concern here will be with some aspects of the mechanism of addition polymerization that influence the character of the polymer formed. The most important type of addition polymerization is that of alkenes (usually called vinyl monomers) such as ethene, propene, ethenylbenzene, and so on
• 29.8: Block, Graft, and Ladder Polymers
A variation on the usual variety of copolymerization is the preparation of polymer chains made of rather long blocks of different kinds of monomers. A number of ingenious systems have been devised for making such polymers.
• 29.9: Naturally Occurring Polymers
There are a number of naturally occurring polymeric substances that have a high degree of technical importance. Some of these, such as natural rubber, cellulose, and starch, have regular structures and can be regarded as being made up of single monomer units. Others, such as wool, silk and deoxyribonucleic acid are copolymers. We shall confine our attention here to wool and collagen, which have properties related to topics discussed previously in this chapter.
• 29.10: Preparation of Synthetic Polymers
• 29.E: Polymers (Exercises)
These are the homework exercises to accompany Chapter 29 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Thumbnail: Polyethylene terephthalate polymer chain. (Public Domain; Jynto).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
29: Polymers
Polymers are substances made up of recurring structural units, each of which can be regarded as derived from a specific compound called a monomer. The number of monomeric units usually is large and variable, each sample of a given polymer being characteristically a mixture of molecules with different molecular weights. The range of molecular weights is sometimes quite narrow, but is more often very broad. The concept of polymers being mixtures of molecules with long chains of atoms connected to one another seems simple and logical today, but was not accepted until the 1930's when the results of the extensive work of H. Staudinger, who received the Nobel Prize in Chemistry in 1953, finally became appreciated. Prior to Staudinger's work, polymers were believed to be colloidal aggregates of small molecules with quite nonspecific chemical structures.
The adoption of definite chemical structures for polymers has had far-reaching practical applications, because it has led to an understanding of how and why the physical and chemical properties of polymers change with the nature of the monomers from which they are synthesized. This means that to a very considerable degree the properties of a polymer can be tailored to particular practical applications. Much of the emphasis in this chapter will be on how the properties of polymers can be related to their structures. This is appropriate because we already have given considerable attention in previous chapters to methods of synthesis of monomers and polymers, as well as to the mechanisms of polymerization reactions.
The special technical importance of polymers can be judged by the fact that half of the professional organic chemists employed by industry in the United States are engaged in research or development related to polymers.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.01%3A_Prelude.txt |
The thermal polymerization of 1,3-cyclopentadiene by way of the Diels-Alder addition is not an important polymerization, but it does provide a simple concrete example of how a monomer and a polymer are related:
The first step in this polymerization is formation of the dimer, which involves 1,3-cyclopentadiene acting as both diene and dienophile. This step occurs readily on heating, but slowly at room temperature. In subsequent steps, 1,3-cyclopentadiene adds to the relatively strained double bonds of the bicyclo[2.2.1]heptene part of the polymer. These additions to the growing chain require higher temperatures ($180$-$200^\text{o}$). If cyclopentadiene is heated to $200^\text{o}$ until substantially no further reaction occurs, the product is a waxy solid having a degree of polymerization $n$ ranging from two to greater than six.
Polycyclopentadiene molecules have two different kinds of double bonds for end groups and a complicated backbone of saturated fused rings. The polymerization is reversible and, on strong heating, the polymer reverts to cyclopentadiene.
There are two commonly used and numerically different ways of expressing the average molecular weight of a polymer such as polycyclopentadiene. One is the number-average molecular weight, $\overline{M_n}$, which is the total weight of a polymer sample, $m$, divided by the total number of moles of molecules it contains, $\Sigma N_i$. Thus
$\overline{M_n} = \frac{m}{\Sigma N_i} = \frac{\Sigma \left( N_i M_i \right)}{\Sigma N_i}$
in which $N_i$ is the number of moles of a single kind of molecular species, $i$, and $M_i$ is the molecular weight of that species.
An alternative way of expressing the molecular weight is by the weight average, $\overline{M_w}$, which can be computed by summing up the contribution (as measured by the weight fraction $w_i$) of each molecular species $i$ and its molecular weight $M_i$:
$\overline{M_w} = \Sigma \left( w_i M_i \right)$
The reason for using two different molecular weights is that some properties, such as freezing points, vapor pressure, and osmotic pressure of dilute solutions, are related directly to $\overline{M_w}$, whereas other properties, such as light-scattering, sedimentation, and diffusion constants, are related directly to $\overline{M_n}$.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
29.03: Types of Polymers
Polymers can be classified in several different ways - according to their structures, the types of reactions by which they are prepared, their physical properties, or their technological uses.
From the standpoint of general physical properties, we usually recognize three types of solid polymers: elastomers, thermoplastic polymers, and thermosetting polymers. Elastomers are rubbers or rubberlike elastic materials. Thermoplastic polymers are hard at room temperature, but on heating become soft and more or less fluid and can be molded. Thermosetting polymers can be molded at room temperature or above, but when heated more strongly become hard and infusible. These categories overlap considerably but are nonetheless helpful in defining general areas of utility and types of structures.
The structural characteristics that are most important to determining the properties of polymers are:
1. the degree of rigidity of the polymer molecules,
2. the electrostatic and van der Waals attractive forces between the chains,
3. the degree to which the chains tend to form crystalline domains, and
4. the degree of cross-linking between the chains.
Of these, cross-linking is perhaps the simplest and will be discussed next.
Consider a polymer made of a tangle of molecules with long linear chains of atoms. If the intermolecular forces between the chains are small and the material is subjected to pressure, the molecules will tend to move past one another in what is called plastic flow. Such a polymer usually is soluble in solvents that will dissolve short-chain molecules with chemical structures similar to those of the polymer. If the intermolecular forces between the chains are sufficiently strong to prevent motion of the molecules past one another the polymer will be solid at room temperature, but will usually lose strength and undergo plastic flow when heated. Such a polymer is thermoplastic. A cross-link is a chemical bond between polymer chains other than at the ends. Cross-links are extremely important in determining physical properties because they increase the molecular weight and limit the translational motions of the chains with respect to one another. Only two cross-links per polymer chain are required to connect all the polymer molecules in a given sample to produce one gigantic molecule. Only a few cross-links (Figure 29-1) reduce greatly the solubility of a polymer and tend to produce what is called a gel polymer, which, although insoluble, usually will absorb (be swelled by) solvents in which the uncross-linked polymer is soluble. The tendency to absorb solvents decreases as the degree of cross-linking is increased because the chains cannot move enough to allow the solvent molecules to penetrate between the chains.
Thermosetting polymers normally are made from relatively low-molecular-weight, usually semifluid substances, which when heated in a mold become highly cross-linked, thereby forming hard, infusible, and insoluble products having a three-dimensional network of bonds interconnecting the polymer chains (Figure 29-2).
Polymers usually are prepared by two different types of polymerization reactions - addition and condensation. In addition polymerization all of the atoms of the monomer molecules become part of the polymer; in condensation polymerization some of the atoms of the monomer are split off in the reaction as water, alcohol, ammonia, or carbon dioxide, and so on. Some polymers can be formed either by addition or condensation reactions. An example is polyethylene glycol, which, in principle, can form either by dehydration of 1,2-ethanediol (ethylene glycol), which is condensation, or by addition polymerization of oxacyclopropane (ethylene oxide):\(^1\)
Other addition polymerizations were discussed previously, including poly-1,3-cyclopentadiene, alkene polymers (Section 19-8), polyalkadienes (Section 13-4), polyfluoroalkenes (Section 14-7D), and polymethanal (Section 16-4B).
\(^1\)Regardless of whether the same polymer would be obtained by polymerization starting with different monomers, the products usually are named to correspond to the starting material. Thus polyethylene glycol and polyethylene oxide would not be used interchangeably for \(\ce{HO-(CH_2CH_2-O)}_n \ce{-H}\).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.02%3A_A_Simple_Addition_Polymerization._The_Parts_of_a_Polymer.txt |
Polymers are produced on an industrial scale primarily, although not exclusively, for use as structural materials. Their physical properties are particularly important in determining their usefulness, be it as rubber tires, sidings for buildings, or solid rocket fuels.
Polymers that are not highly cross-linked have properties that depend greatly on the forces that act between the chains. By way of example, consider a polymer such as polyethene which, in a normal commercial sample, will be made up of molecules having 1000 to 2000 $\ce{CH_2}$ groups in continuous chains. Because the material is a mixture of different molecules, it is not expected to crystallize in a conventional way.$^2$ Nonetheless, x-ray diffraction shows polyethene to have very considerable crystalline character, there being regions as large as several hundred angstrom units in length, which have ordered chains of $\ce{CH_2}$ groups oriented with respect to one another like the chains in crystalline low-molecular-weight hydrocarbons. These crystalline regions are called crystallites (Figure 29-3). Between the crystallites of polyethene are amorphous, noncrystalline regions in which the polymer chains are essentially randomly ordered with respect to one another (Figure 29-4). These regions constituted crystal defects.
The forces between the chains in the crystallites of polyethene are the so-called van der Waals or dispersion forces, which are the same forces acting between hydrocarbon molecules in the liquid and solid states, and, to a lesser extent, in the vapor state. These forces are relatively weak and arise through synchronization of the motions of the electrons in the separate atoms as they approach one another. The attractive force that results is rapidly overcome by repulsive forces when the atoms get very close to one another (see Figure 12-9, which shows how the potential energy between a pair of atoms varies with the internuclear distance). The attractive intermolecular forces between pairs of hydrogens in the crystallites of polyethene are only about $0.1$-$0.2 \: \text{kcal mol}^{-1}$ per pair, but for a crystalline segment of 1000 $\ce{CH_2}$ units, the sum of these interactions could well be greater than the $\ce{C-C}$ bond strengths. Thus when a sample of the crystalline polymer is stressed to the point at which it fractures, carbon-carbon bonds are broken and radicals that can be detected by esr spectroscopy (Section 27-9) are generated.
In other kinds of polymers, even stronger intermolecular forces can be produced by hydrogen bonding. This is especially important in the polyamides, such as the nylons, of which nylon 66 is most widely used (Figure 29-5).
The effect of temperature on the physical properties of polymers is very important to their practical uses. At low temperatures, polymers become hard and glasslike because the motions of the segments of the polymer chains with relation to each other are slow. The approximate temperature below which glasslike behavior is apparent is called the glass temperature and is symbolized by $T_g$. When a polymer containing crystallites is heated, the crystallites ultimately melt, and this temperature is usually called the melting temperature and is symbolized as $T_m$. Usually, the molding temperature will be above $T_m$ and the mechanical strength of the polymer will diminish rapidly as the temperature approaches $T_m$.
Another temperature of great importance in the practical use of polymers is the temperature at which thermal breakdown of the polymer chains occurs. Decomposition temperatures obviously will be sensitive to impurities, such as oxygen, and will be influenced strongly by the presence of inhibitors, antioxidants, and so on. Nonetheless, there will be a temperature (usually rather high, $200^\text{o}$ to $400^\text{o}$) at which uncatalyzed scission of the bonds in a chain will take place at an appreciable rate and, in general, one cannot expect to prevent this type of reaction from causing degradation of the polymer. Clearly, if this degradation temperature is comparable to $T_m$, as it is for polypropenenitrile (polyacrylonitrile), difficulties are to be expected in simple thermal molding of the plastic. This difficulty is overcome in making polypropenenitrile (Orlon) fibers by dissolving the polymer in $\ce{N}$,$\ce{N}$-dimethylmethanamide and forcing the solution through fine holes into a heated air space where the solvent evaporates.
Physical properties such as tensile strength, x-ray diffraction pattern, resistance to plastic flow, softening point, and elasticity of most polymers can be understood in a general way in terms of crystallites, amorphous regions, the degree of flexibility of the chains, cross-links, and the strength of the forces acting between the chains (dispersion forces, hydrogen bonding, etc.). A good way to appreciate the interaction between the physical properties and structure is to start with a rough classification of properties of solid polymers according to the way the chains are disposed in relation to each other.
1. An amorphous polymer is one with no crystallites. If the attractive forces between the chains are weak and if the motions of the chain are not in some way severely restricted as by cross-linking or large rotational barriers, such a polymer would be expected to have low tensile strength and when stressed undergo plastic flow in which the chains slip by one another.
2. An unoriented crystalline polymer is one which is considerably crystallized but has the crystallites essentially randomly oriented with respect to one another, as in Figure 29-4. When such polymers are heated they often show rather sharp $T_m$ points, which correspond to the melting of the crystallites. Above $T_m$, these polymers are amorphous and undergo plastic flow, which permits them to be molded. Other things being the same, we expect $T_m$ to be higher for polymers with stiff chains (high barriers to internal rotation).
3. An oriented crystalline polymer is one in which the crystallites are oriented with respect to one another, usually as the result of a cold-drawing process. Consider a polymer such as nylon, which has strong intermolecular forces and, when first prepared, is in an unoriented state like the one represented by Figure 29-4. When the material is subjected to strong stress in one direction, usually above $T_g$ so that some plastic flow can occur, the material elongates and the crystallites are drawn together and oriented along the direction of the applied stress (Figure 29-6).
An oriented crystalline polymer usually has a much higher tensile strength than the unoriented polymer. Cold drawing is an important step in the production of synthetic fibers.
1. Elastomers usually are amorphous polymers. The key to elastic behavior is to have highly flexible chains with either sufficiently weak forces between the chains or a sufficiently irregular structure to be unstable in the crystalline state. The tendency for the chains to crystallize often can be considerably reduced by random introduction of methyl groups, which by steric hindrance inhibit ordering of the chains. A useful elastomer needs to have some kind of cross-linked region to prevent plastic flow and flexible enough chains to have a low $T_g$. The structure of a polymer of this kind is shown schematically in Figure 29-7; the important difference between this elastomer and the crystalline polymer of Figure 29-4 is the size of the amorphous regions. When tension is applied and the material elongates, the chains in the amorphous regions straighten out and become more nearly parallel. At the elastic limit, a semicrystalline state is reached, which is different from the one produced by cold drawing of a crystalline polymer in that it is stable only while under tension. The forces between the chains are too weak to maintain the crystalline state in the absence of tension. Thus when tension is released, contraction occurs and the original, amorphous polymer is produced. The entropy (Section 4-4B) of the chains is more favorable in the relaxed state than in the stretched state.
A good elastomer should not undergo plastic flow in either the stretched or relaxed state, and when stretched should have a "memory" of its relaxed state. These conditions are best achieved with natural rubber (cis-poly-2-methyl-1,3-butadiene, cis-polyisoprene; Section 13-4) by curing (vulcanizing) with sulfur. Natural rubber is tacky and undergoes plastic flow rather readily, but when it is heated with $1$-$8\%$ by weight of elemental sulfur in the presence of an accelerator, sulfur cross-links are introduced between the chains. These cross-links reduce plastic flow and provide a reference framework for the stretched polymer to return to when it is allowed to relax. Too much sulfur completely destroys the elastic properties and produces hard rubber of the kind used in cases for storage batteries.
The chemistry of the vulcanization of rubber is complex. The reaction of rubber with sulfur is markedly expedited by substances called accelerators, of which those commonly known as mercaptobenzothiazole and tetramethylthiuram disulfide are examples:
Clearly, the double bonds in natural rubber are essential to vulcanization because hydrogenated rubber ("hydrorubber") is not vulcanized by sulfur. The degree of unsaturation decreases during vulcanization, although the decrease is much less than one double bond per atom of sulfur introduced. There is evidence that attack occurs both at the double bond and at the adjacent hydrogen (in a manner similar to some halogenations; Section 14-3A) giving cross-links possibly of the following types:
The accelerators probably function by acting as sulfur carriers from the elemental sulfur to the sites of the polymer where the cross-links are formed.
$^2$Quite good platelike crystals, about $100 \: \text{Å}$ thick, have been formed from dilute solutions of polyethene. In these crystals, $\ce{CH_2}$ chains in the anti conformation (Section 5-2) run between the large surfaces of the plates. However, the evidence is strong that when the $\ce{CH_2}$ chains reach the surface of the crystal they do not neatly fold over and run back down to the other surface. Instead, the parts of a given chain that are in the crystalline segments appear to be connected at the ends of the crystallites by random loops of disordered $\ce{CH_2}$ sequences, something like an old-fashioned telephone switchboard.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.04%3A_Forces_Between_Polymer_Chains.txt |
The properties of many of the commercially important thermoplastic and elastic polymers can be understood in terms of their chemical structures by using the concepts developed in the preceding section. Thus the simple linear polymers, polyethene $\ce{-(CH_2CH_2)}_n-$, polymethanal $\ce{-(CH_2-O)}_n-$, and polytetrafluoroethene $\ce{-(CF_2-CF_2)}_n-$, with regular chains and low barriers to rotation about the bonds in the chain tend to be largely crystalline with rather high melting points and low glass temperatures (see Table 29-1). The situation with polychloroethene (polyvinyl chloride), polyfluoroethene (polyvinyl fluoride), and polyethenylbenzene (polystyrene) as usually prepared is quite different. These polymers are much less crystalline and yet have rather high glass temperatures, which suggests that there is considerable attractive force between the chains. The low degree of crystallinity of these polymers is the result of their having a low degree of regularity of the stereochemical configuration of the chiral carbons in the chain. The discovery by G. Natta in 1954 that the stereochemical configurations of chiral centers in polymer chains could be crucial in determining their physical properties has had a profound impact on both the practical and theoretical aspects of polymer chemistry. Natta's work was done primarily with polypropene and this substance provides an excellent example of the importance of stereochemical configurations.
Table 29-1: Representative Synthetic Thermoplastic and Elastic Polymers and Their Uses$^a$
There are striking differences in physical properties between the atactic and isotactic forms. The atactic material is soft, elastic, somewhat sticky, and rather soluble in solvents such as 1,1,2,2-tetrachloroethane. Isotactic polypropene is a hard, clear, strong crystalline polymer that melts at $175^\text{o}$. It is practically insoluble in all organic solvents at room temperature, but will dissolve to the extent of a few percent in hot 1,1,2,2-tetrachloroethane. That the difference between the atactic and isotactic polymers irises from differences in the configurations of the methyl groups on the chains is shown in a striking way by $\ce{^{13}C}$ nmr spectra (Figure 29-9). The differences in these spectra result from differences in the interactions between the methyl groups for the different configurations, in the same way as we have shown you earlier for axial and equatorial methyl groups on cyclohexane rings (Section 12-3D).
Why should polypropene melt so much higher than polyethene ($175^\text{o}$ vs. $110^\text{o}$)? The answer lies in the differences between the way the polymers crystallize. Polyethene crystallites have extended zig-zag chains that have very low barriers to rotation about the $\ce{C-C}$ bonds. Because of interferences between the methyl groups, polypropene does not crystallize in extended zig-zag chains but instead forms a helix, something like the $\alpha$ helix (Section 25-8A), with the chain carbons on the inside and the methyl carbons on the outside. These coils are more rigid than the extended $\ce{CH_2}$ chains in polyethene and have stabilizing interchain $\ce{H} \cdots \ce{H}$ interactions so that a higher temperature is required for melting. Polypropene can be cold drawn to form fibers that resemble nylon fibers although, as might be expected, these fibers do not match the $270^\text{o}$ melting point of nylon and, because of their hydrocarbon character, are much more difficult to dye.
Although both linear polyethene and isotactic polypropene are crystalline polymers, ethene-propene copolymers prepared with the aid of Ziegler catalysts are excellent elastomers. Apparently, a more or less random introduction of methyl groups along a polyethene chain reduces the crystallinity sufficiently drastically to lead to an amorphous polymer. The ethene-propene copolymer is an inexpensive elastomer, but having no double bonds, is not capable of vulcanization. Polymerization of ethene and propene in the presence of a small amount of dicyclopentadiene or 1,4-hexadiene gives an unsaturated heteropolymer, which can be vulcanized with sulfur in the usual way.
The rationale in using these particular dienes is that only the strained double bond of dicyclopentadiene and the terminal double bond of 1,4-hexadiene undergo polymerization with Ziegler catalysts. Consequently the polymer chains contain one double bond for each molecule of dicyclopentadiene or 1,4-hexadiene that is incorporated. These double bonds later can be converted to cross-links by vulcanization with sulfur (Sections 13-4 and 29-3).
Polychloroethene (polyvinyl chloride), as usually prepared, is atactic and not very crystalline. It is relatively brittle and glassy. The properties of polyvinyl chloride can be improved by copolymerization, as with ethenyl ethanoate (vinyl acetate), which produces a softer polymer ("Vinylite") with better molding properties. Polyvinyl chloride also can be plasticized by blending it with substances of low volatility such as tris-(2-methylphenyl) phosphate (tricresyl phosphate) and dibutyl benzene-1,2-dicarboxylate (dibutyl phthalate) which, when dissolved in the polymer, tend to break down its glasslike structure. Plasticized polyvinyl chloride is reasonably flexible and is widely used as electrical insulation, plastic sheeting, and so on.
Table 29-1 contains information about a number of representative important polymers and their uses. Some similar data on other polymers already have been given (Section 13-4 and Table 10-4). The important use of modified polymers as ion-exchange resins is discussed in Section 25-4C.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.05%3A_Correlation_of_Polymer_Properties_with_Structure.txt |
There is a wide variety of condensation reactions that, in principle, can be used to form high polymers. However, high polymers can be obtained only in high-yield reactions, and this limitation severely restricts the number of condensation reactions having any practical importance. A specific example of an impractical reaction is the formation of poly-1,4-butanediol by reaction of 1,4-bromobutane with the disodium salt of the diol:
It is unlikely that this reaction would give useful yields of any very high polymer because $E2$ elimination, involving the dibromide, would give a double-bond end group and prevent the chain from growing.
Polyesters
A variety of polyester-condensation polymers are made commercially. Ester interchange (Section 18-7A) appears to be the most useful reaction for preparation of linear polymers:
Thermosetting space-network polymers can be prepared through the reaction of polybasic acid anhydrides with polyhydric alcohols. A linear polymer is obtained with a bifunctional anhydride and a bifunctional alcohol, but if either reactant has three or more reactive sites, then formation of a three-dimensional polymer is possible. For example, 2 moles of 1,2,3-propanetriol (glycerol) can react with 3 moles of 1,2-benzenedicarboxylic anhydride (phthalic anhydride) to give a highly cross-linked resin, which usually is called a glyptal:
The first stage of the reaction involves preferential esterification of the primary hydroxyl groups with the anhydride to give
In the next stage, in the formation of the resin, direct esterification occurs slowly, particularly at the secondary hydroxyls. Normally, when the resin is used for surface coatings, esterification is carried only to the point where the polymer is not so cross-linked as to be insoluble. It then is applied to the surface in a solvent and baked until esterification is complete. The product is hard, infusible, and insoluble, being cross-linked to the point of being essentially one large molecule.
A wide variety of thermosetting polyester (alkyd) resins can be made by similar procedures. The following polybasic acids and anhydrides and polyhydric alcohols are among the other popular ingredients in alkyd formulations:
Articles in which glass fibers are imbedded to improve impact strength often are made by mixing the fibers with an ethenylbenzene (styrene) solution of a linear glycol (usually 1,2-propanediol)-butendioic anhydride polyester and then producing a cross-linked polymer between the styrene and the double bonds in the polyester chains by a peroxide-induced radical polymerization (Section 29-6E).
Nylons
A variety of polyamides can be made by heating diamines with dicarboxylic acids. The most generally useful of these is nylon 66, the designation 66 arising from the fact that it is made from the six-carbon diamine, 1,6-hexanediamine, and a six-carbon diacid, hexanedioic acid:
The polymer can be converted into fibers by extruding it above its melting point through spinnerettes, then cooling and drawing the resulting filaments. It also is used to make molded articles. Nylon 66 is exceptionally strong and abrasion resistant.
The starting materials for nylon 66 can be made in many ways. Apparently, the best route to hexanedioic acid is by air oxidation of cyclohexane by way of cyclohexanone:
1,6-Hexanediamine can be prepared in many ways. One is from 1,3-butadiene by the following steps:
Nylon 6 can be prepared by polymerization of 1-aza-2-cycloheptanone ($\varepsilon$-caprolactam), obtained through the Beckmann rearrangement of cyclohexanone oxime (Section 24-3C):
Bakelite Resins
One of the oldest known thermosetting synthetic polymers is made by condensation of phenols with aldehydes using basic catalysts. The resins that are formed are known as Bakelites. The initial stage is the base-induced reaction of benzenol and methanal to give a (4-hydroxyphenyl)methanol, and this reaction closely resembles an aldol addition and can take place at either the 2- or the 4-position of the benzene ring:
The next step in the condensation is formation of a bis(hydroxyphenyl)methane derivative, which for convenience is here taken to be the 4,4'-isomer:
This reaction is probably a Michael type of addition to a base-induced dehydration product of the (4-hydroxyphenyl)methanol:
Continuation of these reactions at the 2-, 4-, and 6-positions of the benzenol leads to the cross-linked three-dimensional Bakelite resin:
As with the alkyd resins (Section 29-5A), the initial polymerization of a Bakelite resin usually is carried to only a relatively low stage of completion. The low-melting prepolymer (called a resole) then is heated in a mold to give the final insoluble, infusible polymer.
Urea-Methanal and Melamine Resins
Syntheses of a number of polymers are based on condensation of methanal with amino compounds by mechanisms at least formally analogous to those involved in the preparation of Bakelite resins. The key reactions are:
If the amino compound is urea, these types of reactions will lead to a three-dimensional polymer:
A similar polymer is made from 2,4,6-triamino-1,3,5-triazine (melamine) and methanal. It is used for plastic dishes under the name Melmac.
Epoxy Resins
A very useful group of adhesives and plastics is based on condensation polymers of bisphenol A and chloromethyloxacyclopropane (epichlorohydrin). The first step in the formation of epoxy resins is to form a prepolymer by condensation polymerization of the sodium salt of bisphenol A with the epoxide:
The formation of a prepolymer involves two different kinds of reactions. One is an $S_\text{N}2$-type displacement, and the other is oxide-ring opening of the product by attack of more bisphenol A. Usually, for practical purposes the degree of polymerization $n$ of the prepolymer is small (5 to 12 units).
The epoxy prepolymer can be cured, that is, converted to a three-dimensional network, in several different ways. A trifunctional amine, such as $\ce{NH_2CH_2CH_2NHCH_2CH_2NH_2}$, can be mixed in and will extend the chain of the polymer and form cross-links by reacting with the oxide rings:
Alternatively, a polybasic acid anhydride can be used to link the chains through combination with secondary alcohol functions and then the oxide rings.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.06%3A_Condensation_Polymers.txt |
We have discussed the synthesis and properties of a considerable number of addition polymers in this and previous chapters. Our primary concern here will be with some aspects of the mechanism of addition polymerization that influence the character of the polymer formed.
Alkene Polymerization
The most important type of addition polymerization is that of alkenes (usually called vinyl monomers) such as ethene, propene, ethenylbenzene, and so on. In general, we recognize four basic kinds of mechanisms for polymerization of vinyl monomers - radical, cationic, anionic, and coordination. The elements of the first three of these have been outlined (Section 10-8). The possibility, in fact the reality, of a fourth mechanism is essentially forced on us by the discovery of the Ziegler and other (mostly heterogeneous) catalysts, which apparently do not involve "free" radicals, cations, or anions, and which can and usually do lead to highly stereoregular polymers. With titanium-aluminum Ziegler catalysts, the growing chain has a $\ce{C-Ti}$ bond; further monomer units then are added to the growing chain by coordination with titanium, followed by an intramolecular rearrangement to give a new growing-chain end and a new vacant site on titanium where a new molecule of monomer can coordinate:
In the coordination of the monomer with the titanium, the metal is probably behaving as an electrophilic agent and the growing-chain end can be thought of as being transferred to the monomer as an anion. Because this mechanism gives no explicit role to the aluminum, it is surely oversimplified. Ziegler catalysts polymerize most monomers of the type $\ce{RCH=CH_2}$, provided the $\ce{R}$ group is one that does not react with the organometallic compounds present in the catalyst. More reactions of the type that occur in Ziegler polymerizations will be discussed in Chapter 31.
Radical Polymerization
In contrast to coordination polymerizations, formation of vinyl polymers by radical chain mechanisms is reasonably well understood - at least for the kinds of procedures used on the laboratory scale. The first step in the reaction is the production of radicals; this can be achieved in a number of different ways, the most common being the thermal decomposition of an initiator, usually a peroxide or an azo compound:
Many polymerizations are carried out on aqueous emulsions of monomers. For these, water-soluble inorganic peroxides, such as ammonium peroxysulfate, often are employed.
Other ways of obtaining initiator radicals include high-temperature decomposition of the monomer and photochemical processes, often involving a ketone as a photosensitizer. Addition of the initiator radicals to monomer produces a growing-chain radical that combines with successive molecules of monomer until, in some way, the chain is terminated. Addition to an unsymmetrical monomer can occur in two ways. Thus for ethenylbenzenes:
All evidence on addition of radicals to ethenylbenzene indicates that the process by which $\ce{X} \cdot$ adds to the $\ce{CH_2}$ end of the double bond is greatly favored over addition at the $\ce{CH}$ end. This direction of addition is in accord with the considerable stabilization of the phenylmethyl radicals relative to the alkyl radicals (see Sections 14-3C and 26-4D). Polymerization then will result in the addition of monomer units to give phenyl groups only on alternate carbons ("head-to-tail" addition):
In general, we predict that the direction of addition of an unsymmetrical monomer will be such as to give always the most stable growing-chain radical. Similar considerations were discussed previously (Section 21-11) in respect to how [2 + 2] cycloadditions occur.
The process of addition of monomer units to the growing chain can be interrupted in different ways. One is chain termination by combination or disproportionation of radicals. Explicitly, two growing-chain radicals can combine to form a carbon-carbon bond, or disproportionation can occur with a hydrogen atom being transferred from one chain to the other:
The disproportionation reaction is the radical equivalent of the $E2$ reaction:
Which mode of termination occurs can be determined by measuring the number of initiator fragments per polymer molecule. If there are two initiator fragments in each molecule, termination must have occurred by combination. One initiator fragment per molecule indicates disproportionation. Apparently, ethenylbenzene polymerizations terminate by combination, but with methyl 2-methylpropenoate, both reactions take place, disproportionation being favored.
Another very important way that a growing chain may be terminated is by chain transfer. This stops the chain but starts a new one. Thiols, such as phenylmethanethiol and dodecanethiol, are efficient chain-transferring agents. The reactions involved are as follows (where $\ce{M}$ represents monomer and $\ce{RSH}$ represents the chain-transfer reagent):
\begin{align} \ce{X-M-(M)}_n \ce{-M} \cdot + \ce{RSH} &\rightarrow \ce{X-M-(M)}_n \ce{-M-H} + \ce{RS} \cdot \ \ce{RS} \cdot + \ce{M} &\rightarrow \ce{RS-M} \cdot \ \ce{RS-M} \cdot + \left( n + 1 \right) \ce{M} &\rightarrow \ce{RS-M-(M)}_n \ce{-M} \cdot \: \text{(new growing chain)} \ \ce{RS-M-(M)}_n \ce{-M} \cdot + \ce{RSH} &\rightarrow \ce{RS-M-(M)}_n \ce{-M-H} + \ce{RS} \cdot \: \text{etc.} \end{align}
Chain transfer reduces the average molecular weight of the polymer without wasting initiator radicals. Dodecanethiol has considerable use in the manufacture of GRS rubber (Section 13-4) as a regulator to hold down the molecular weight in the emulsion polymerization of 1,3-butadiene and ethenylbenzene.
Polymerization inhibitors stop or slow down polymerization by reacting with the initiator or growing-chain radicals. A wide variety of substances can behave as inhibitors: quinones, hydroquinones, aromatic nitro compounds, aromatic amines, and so on. In cases where the inhibitor is a hydrogen donor (symbolized here by $\ce{InH}$), then for inhibition to occur, the radical resulting from hydrogen transfer $\left( \ce{In} \cdot \right)$ must be too stable to add to monomer. If it does add to monomer and starts a new chain, chain transfer occurs instead of inhibition. For perfect inhibition, the $\ce{In} \cdot$ radicals must combine with themselves (or initiator radicals) to give inert products:
\begin{align} \ce{X-(M)}_n \ce{-M} \cdot + \ce{InH} &\rightarrow \ce{X-(M)}_n \ce{-M-H} + \ce{In} \cdot \ 2 \ce{In} \cdot &\rightarrow \text{inert products (inhibition)} \ \ce{In} \cdot + \ce{M} &\rightarrow \ce{In-M} \cdot \: \text{(chain transfer)} \end{align}
Many compounds are known that fall in the intermediate zone between chain transfer and inhibition reagents.
Some inhibitors such as 2,3,5,6-tetrachloro-1,4-benzenedione (tetrachlorobenzoquinone) act as inhibitors by adding to the growing chain radicals to give radicals too stable to continue the chain:
Again, for inhibition to be effective there must be destruction of the stable radicals by dimerization or disproportionation.
The reactive vinyl monomers usually are stabilized against polymerization, while in storage, by addition of $0.1$ to $1\%$ of an inhibitor. 1,4-Benzenediol (hydroquinone), 2,6-di-tert-butyl-4-methylbenzenol, and 4-tert-butyl-1,2-benzenediol are used for this purpose. These substances are especially effective at scavenging $\ce{RO} \cdot$ radicals, which are formed by oxidation of the monomer with atmospheric oxygen.
Cationic Polymerization
Polymerization by the cationic mechanism is most important for 2-methylpropene (isobutylene), which does not polymerize well by other methods, and was discussed previously in considerable detail (Section 10-8B).
Anionic Polymerization
In general, we expect that anionic polymerization will be favorable when the monomer carries substituents that will stabilize the anion formed when a basic initiator such as amide ion adds to the double bond of the monomer:
Cyano and alkoxycarbonyl groups are favorable in this respect and propenenitrile and methyl 2-methylpropenoate can be polymerized with sodium amide in liquid ammonia. Ethenylbenzene and 2-methyl-1,3-butadiene undergo anionic polymerization under the influence of organolithium and organosodium compounds, such as butyllithium and phenylsodium.
An important development in anionic polymerization has been provided by M. Szwarc's "living polymers". The radical anion, sodium naphthalenide (Section 27-9), transfers an electron reversibly to ethenyl benzene to form a new radical anion, $1$, in solvents such as 1,2-dimethoxyethane or oxacyclopentane:
Dimerization of the sodium naphthalenide radical anion would result in a loss of aromatic stabilization, but this is not true for $1$, which can form a $\ce{C-C}$ bond and a resonance-stabilized bis-phenylmethyl dianion, $2$ (Section 26-4C).
The anionic ends of $2$ are equivalent and can add ethenylbenzene molecules to form a long-chain polymer with anionic end groups, $3$:
If moisture and oxygen are rigorously excluded, the anionic groups are stable indefinitely, and if more monomer is added polymerization will continue. Hence the name "living polymer", in contrast to a radical-induced polymerization, which only can be restarted with fresh monomer and fresh initiator, and even then not by growth on the ends of the existing chains.
The beauty of the Szwarc procedure is that the chains can be terminated by hydrolysis, oxidation, carboxylation with $\ce{CO_2}$, and so on, to give polymer with the same kind of groups on each end of the chain. Also, it is possible to form chains in which different monomers polymerize well by the anion mechanism and contain no groups or impurities that will destroy the active ends. Thus one can start with ethenylbenzene $\left( \ce{S} \right)$, and when the reaction is complete, add methyl 2-methylpropenoate $\left( \ce{M} \right)$ to obtain a block copolymer of the type
$\ce{M-M-M-M-M-S-S-S-S-S-S-M-M-M-M-M}$
Copolymers
When polymerization occurs in a mixture of monomers there will be competition between the different kinds of monomers to add to the growing chain and produce a copolymer. Such a polymer will be expected to have physical properties quite different from those of a mixture of the separate homopolymers. Many copolymers, such as GRS, ethene-propene, Viton rubbers, and Vinyon plastics are of considerable commercial importance.
The rates of incorporation of various monomers into growing radical chains have been studied in considerable detail. The rates depend markedly on the nature of the monomer being added and on the character of the radical at the end of the chain. Thus a 1-phenylethyl-type radical on the growing chain reacts about twice as readily with methyl 2-methylpropenoate as it does with ethenylbenzene; a methyl 2-methylpropenoate end shows the reverse behavior, being twice as reactive toward ethylbenzene as toward methyl 2-methylpropenoate. This kind of behavior favors alternation of the monomers in the chain and reaches an extreme in the case of 2-methylpropene and butenedioic anhydride. Neither of these monomers separately will polymerize well with radical initiators. Nonetheless, a mixture polymerizes very well with perfect alternation of the monomer units. It is possible that, in this case, a 1:1 complex of the two monomers is what polymerizes.
In general however, in a mixture of two monomers one is considerably more reactive than the other and the propagation reaction tends to favor incorporation of the more reactive monomer, although there usually is some bias toward alternation. Ethenylbenzene and 2-methyl-1,3-butadiene mixtures are almost unique in having a considerable bias toward forming the separate homopolymers.
One of the more amazing copolymerizations is that of ethenylbenzene and oxygen gas, which at one atmosphere oxygen pressure gives a peroxide with an average molecular weight of 3000 to 4000 and a composition approaching $\ce{C_8H_8O_2}$:
When heated rapidly in small portions the product undergoes a mild explosion and gives high yields ($80\%$ to $95\%$) of methanal and benzenecarbaldehyde. The mechanism may be a kind of unzipping process, starting from a break in the chain an spreading toward each end:
Another interesting copolymerization is of ethene and carbon monoxide by the radical mechanism. The polymer contains $\ce{-CH_2-CH_2-CO-CH_2-CH_2}-$ units, which are broken apart at a $\ce{CH_2} \vdots \ce{C=O}$ bond on absorption of ultraviolet light, thereby giving a polymer that has the possibility of degrading in the environment through the action of sunlight (see Section 28-2A).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.07%3A_Addition_Polymers.txt |
A variation on the usual variety of copolymerization is the preparation of polymer chains made of rather long blocks of different kinds of monomers. A number of ingenious systems have been devised for making such polymers, including the Szwarc method described in Section 29-6D. Another scheme, which will work with monomers that polymerize well by radical chains but not with anion chains, is to irradiate a stream of a particular monomer, flowing through a glass tube, with sufficient light to get polymerization well underway. The stream then is run into a dark flask containing a large excess of a second monomer. The growing chains started in the light-induced polymerization then add the second monomer to give a two-block polymer if termination is by disproportionation, or a three-block polymer if by combination. Thus, with $\ce{A}$ and $\ce{B}$ being the two different monomers,
Block polymers also can be made easily by condensation reactions:
The very widely used polyurethane foams can be considered to be either block polymers or copolymers. The essential ingredients are a diisocyanate and a diol. The diisocyanate most used is 2,4-diisocyano-1-methylbenzene, and the diol can be a polyether or a polyester with hydroxyl end groups. The isocyano groups react with the hydroxyl end groups to form initially an addition polymer, which has polycarbamate (polyurethane) links, and isocyano end groups:
A foam is formed by addition of the proper amount of water. The water reacts with the isocyanate end groups to form carbamic acids which decarboxylate to give amine groups:
The carbon dioxide evolved is the foaming agent, and the amino groups formed at the same time extend the polymer chains by reacting with the residual isocyano end groups to form urea linkages:
$\ce{R'N=C=O} + \ce{RNH_2} \rightarrow \ce{R'NHCONHR}$
Graft polymers can be made in great profusion by attaching chains of one kind of polymer to the middle of another. A particularly simple but uncontrollable way of doing this is to knock groups off a polymer chain with x-ray or $\gamma$ radiation in the presence of a monomer. The polymer radicals so produced then can grow side chains made of the new monomer.
A more elegant procedure is to use a photochemical reaction to dissociate groups from the polymer chains and form radicals capable of polymerization with an added monomer.
Modern technology has many uses for very strong and very heat-resistant polymers. The logical approach to preparing such polymers is to increase the rigidity of the chains, the strengths of the bonds in the chains, and the intermolecular forces. All of these should be possible if one were to make the polymer molecules in the form of a rigid ribbon rather than a more or less flexible chain. Many so-called ladder polymers with basic structures of the following type have been prepared for this purpose:
With the proper structures, such polymers can be very rigid and have strong intermolecular interactions. Appropriate syntheses of true ladder polymers in high yield usually employ difficultly obtainable starting materials. An example is
Although there seem to be no true ladder polymers in large-scale commercial production, several semi-ladder polymers that have rather rigid structures are employed where high-temperature strength is important. Among these are
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.08%3A_Block_Graft_and_Ladder_Polymers.txt |
There are a number of naturally occurring polymeric substances that have a high degree of technical importance. Some of these, such as natural rubber (Section 13-4), cellulose, and starch (Section 20-7), have regular structures and can be regarded as being made up of single monomer units. Others, such as wool, silk (Section 25-8A), and deoxyribonucleic acid (Section 25-13A) are copolymers. Because we already have considered the chemistry of most of these substances, we shall confine our attention here to wool and collagen, which have properties related to topics discussed previously in this chapter.
Wool
The structure of wool is more complicated than that of silk fibroin (Figure 25-13) because wool, like insulin (Figure 25-8) and lysozyme (Figure 25-15), contains a considerable quantity of cystine, which provides $\ce{-S-S}-$ (disulfide) cross-links between the peptide chains. These disulfide linkages play an important part in determining the mechanical properties of wool fibers because if the disulfide linkages are reduced, as with ammonium mercaptoethanoate solution, the fibers become much more pliable.
Advantage is taken of this reaction in the curling of hair, the reduction and curling being followed by restoration of the disulfide linkages through treatment with a mild oxidizing agent.
Collagen
The principal protein of skin and connective tissue is called collagen and is primarily constituted of glycine, proline, and hydroxyproline. Collagen is made up of tropocollagen, a substance with very long and thin molecules ($14 \times 2900 \: \text{Å}$, MW about 300,000). Each tropocollagen molecule consists of three twisted polypeptide strands. When collagen is boiled with water, the strands come apart and the product is ordinary cooking gelatin. Connective tissue and skin are made up of fibrils, $200 \: \text{Å}$ to $1000 \: \text{Å}$ wide, which are indicated by x-ray diffraction photographs to be composed of tropocollagen molecules running parallel to the long axis. Electron micrographs show regular bands, $640 \: \text{Å}$ apart, across the fibrils, and it is believed that these correspond to tropocollagen molecules, all heading in the same direction but regularly staggered by about a fourth of their length (Figure 29-10).
The conversion of collagen fibrils to leather presumably involves formation of cross-links between the tropocollagen molecules. Various substances can be used for the purpose, but chromium salts act particularly rapidly.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
29.10: Preparation of Synthetic Polymers
A prevalent but erroneous notion is that useful polymer, such as those given in Table 29-1, can be, and are, made by slap-dash procedures applied to impure starting materials. This is far from the truth; actually, the monomers used in most large-scale polymerizations are among the purest known organic substances. Furthermore, to obtain uniform commercially useful products, extraordinary care must be used in controlling the polymerization reactions. The reasons are simple - namely, formation of a high-molecular-weight polymer requires a reaction the proceeds in very high yields, and purification of the product by distillation, crystallization, and so on, is difficult, if not impossible. Even a minute contribution of any side reaction that stops polymer chains from growing further will seriously affect the yield of high polymer.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers/29.09%3A_Naturally_Occurring_Polymers.txt |
For an organic chemist, a natural product is one that is produced by a living organism. This definition encompasses many compounds already discussed, such as carbohydrates, proteins, lipids, and nucleic acids, all of which play an important and primary role in metabolic reactions. However, there are other organic compounds produced naturally, some of extraordinary complexity, which are not primary metabolites. Organic chemists always have been fascinated by the great diversity of these substances and particularly those that can be isolated from plants or are produced by microorganisms. Many of these compounds, such as the alkaloids and mold metabolites, do not seem to have any obvious metabolic or evolutionary function. In fact, some compounds may be formed as the result of a "metabolic accident" or are by-products of the synthesis machinery of the cellular enzymes. Regardless of their utility to the parent organism, their value to man as drugs, herbs, flavorings, poisons, dyes, and so on is undisputed.
• 30.1: Classification of Natural Products
There are several ways to categorize natural products. They may be grouped according to a recurring structural feature. Or they may be grouped according to the genus of their plant source.
• 30.2: Approaches to the Study of Natural Products
Chemists have a compelling curiosity to discover what compounds Nature provides, but to obtain this information it is necessary to isolate compounds from their natural source and to determine their structures.
• 30.3: Isoprenoid Compounds
The odor of a freshly crushed mint leaf is due to the presence in the plant of volatile compounds, which are called terpenes. Isolation of these substances from the various parts of plants, even from the wood in some cases, by steam distillation or ether extraction gives what are known as essential oils.
• 30.4: Steroids
The term steroid applies to compounds containing hydrogenated cyclopentanophenanthrene carbon skeleton. Most steroids are alcohols, and accordingly are named as sterols. Important examples include cholesterol, ergosterol, estradiol, and stigmasterol.
• 30.5: Biosynthesis
Biosynthesis is a multi-step, enzyme-catalyzed process where substrates are converted into more complex products in living organisms. In biosynthesis, simple compounds are modified, converted into other compounds, or joined together to form macromolecules.
• 30.6: Some Nitrogen-Containing Natural Products
Basic nitrogen compounds in plants are classified as alkaloids. Several examples were given previously of this large and remarkably heterogeneous class of compounds, many of which have very complex structures. It is difficult to give a coherent account of alkaloid chemistry in the limited space available to us here.
• 30.7: Prostaglandins
The prostaglandins are a group of physiologically active lipid compounds having diverse hormone-like effects in animals. Prostaglandins have been found in almost every tissue in humans and other animals. They are derived enzymatically from fatty acids. Every prostaglandin contains 20 carbon atoms, including a 5-carbon ring. They are a subclass of eicosanoids and of the prostanoid class of fatty acid derivatives.
• 30.E: Natural Products and Biosynthesis (Exercises)
These are the homework exercises to accompany Chapter 30 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
30: Natural Products and Biosynthesis
There are several ways to categorize natural products. They may be grouped according to a recurring structural feature. Flavonoid compounds, for example, are oxygenated derivatives of the aromatic ring structure \(1\); likewise, alkaloids having an indole ring, \(2\) are called indole alkaloids:
Or they may be grouped according to the genus of their plant source (morphine and codeine, Section 23-2, are examples of opium alkaloids), or by their physiological effects (antimicrobials, antibiotics, analgesics), or by similarities in the route by which they are synthesized by the organism (biosynthesis). The structural and biosynthetic classifications make the most sense to the chemist and is the organization chosen here.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
30.02: Approaches to the Study of Natural Products
Chemists have a compelling curiosity to discover what compounds Nature provides, but to obtain this information it is necessary to isolate compounds from their natural source and to determine their structures. This is seldom an easy task, especially when the compound of interest is present at low concentrations such that enormous quantities of source material are required to extract even a few micrograms of the desired product. In this circumstance a high degree of skill and technology is required in both the isolation procedures and the subsequent investigations to establish the chemical structure.
A second objective is the total synthesis of the compound from smaller molecules. Indeed, in the classical approach to structure determination, a structure was assigned to a natural product through chemical degradation studies to smaller, identifiable molecules. However, the assigned structure was not regarded as fully confirmed until the compound was synthesized and shown to be identical in all respects (composition, configuration, conformation) with the natural compound. This approach persists, although the enormous impact of modern methods of separation and spectroscopic analysis has made it possible to determine structure beyond a reasonable doubt in almost all cases without recourse to synthesis.
Nevertheless, the synthesis of natural products continues to be important. It provides new methodology, new reactions and techniques. It also provides alternative sources of natural compounds and offers routes to related but unnatural analogs. In the case of a useful drug, the synthetic objective is to find a related structure that is more potent at lower dosages with fewer side effects than the natural compound.
Yet another area of investigation in natural-product chemistry concerns the way in which the compound is synthesized biologically - that is, the biosynthesis of the compound. These are experimentally difficult studies and involve first identifying the starting materials (biological precursors). This can be done by feeding the organism isotopically labeled compounds suspected of being precursors and then determining where and how much of the labeled material is incorporated into the natural product. Ultimately, each step in the synthesis should be elucidated and each enzyme isolated and the entire sequence reconstructed in a cell-free system. From experiments of this type we now have a rather good understanding of the biosynthesis of fatty acids, terpenes, and steroids.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/30%3A_Natural_Products_and_Biosynthesis/30.01%3A_Classification_of_Natural_Products.txt |
The odor of a freshly crushed mint leaf, like many plant odors, is due to the presence in the plant of volatile $\ce{C_{10}}$ and $\ce{C_{15}}$ compounds, which are called terpenes. Isolation of these substances from the various parts of plants, even from the wood in some cases, by steam distillation or ether extraction gives what are known as essential oils. These are widely used in perfumery, as food flavorings and medicines, and as solvents. Among the typical essential oils are those obtained from cloves, roses, lavender, citronella, eucalyptus, peppermint, camphor, sandalwood, cedar, and turpentine. Such substances are of interest to use here because, as was pointed out by Wallach in 1887 and reemphasized by Ruzicka in 1935, the components of the essential oils can be regarded as derived from isoprene:
Not only are the carbon skeletons of these substances divisible into isoprene units, but the terpene hydrocarbons are usually exact multiples of $\ce{C_5H_8}$. An example is myrcene $\left( \ce{C_{10}H_{16}} \right)$, which occurs in the oils of bay and verbena and has a carbon skeleton divisible into two isoprene units.
The connection between the isoprene units in myrcene is between the 1- and 4-positions; this turns out to be more common than 1,1 and 4,4 linkages.
Terpene Hydrocarbons
A wide variety of cyclic terpene hydrocarbons are known and, as multiples of $\ce{C_5H_8}$, these have fewer double bonds than the open-chain terpenes. Because it is time consuming to show all the carbon and hydrogen atoms of such substances, the structures often are drawn in a convenient shorthand notation wherein the carbon-carbon bonds are represented by lines, carbon atoms being understood at the junctions or the ends of lines. By this notation, myrcene can be represented by formulas such as the following:
The left semicyclic structural formula is useful to show relationships with the open-chain (acyclic) and cyclic terpene hydrocarbons.
Table 30-1: Some Isoprenoid Hydrocarbons$^a$
A number of terpene hydrocarbons are shown in Table 30-1. The designation "terpene" is by custom specifically reserved for the $\ce{C_{10}}$ compounds, the $\ce{C_{15}}$ compounds being known as sesquiterpenes, the $\ce{C_{20}}$ as diterpenes, $\ce{C_{30}}$ as triterpenes, and so on. It should be apparent from Table 30-1 that the $\ce{C_{10}}$ and $\ce{C_{15}}$ compounds, which are the important components of essential oils, in reality are members of a much larger class of substances with carbon skeletons made up of isoprene units and occurring in both plants and animals. It is common to refer to all members of the group as isoprenoid compounds. The so-called isoprene rule, which correlates the structures of these substances, speaks for their synthesis in living systems from some common precursor with five carbon atoms. We can characterize the isoprenoid compounds as being biogenetically related. Isoprene itself does not occur naturally and appears to play no part in biosynthesis. The actual five-carbon intermediate appears to be isopentenyl pyrophosphate, and the role of this substance in biosynthesis will be discussed later:
Oxygenated Isoprenoid Compounds
A great profusion of oxygen-containing isoprenoid compounds are known. Of particular importance in the acyclic series are the alcohols geraniol, nerol, and linaloöl, and the aldehydes geranial (citral a), neral (citral b), and citronellal:
The alcohols occur in oil of rose and other flower essences. They have geranium or rose odors and are important perfume ingredients. The aldehydes have much stronger citruslike odors and occur as major or minor constituents in many essential oils, such as oil of citronella, oil of lemon, and so on.
Monocyclic and bicyclic oxygenated terpenes include some familiar and interesting substances such as menthone and menthol from peppermint oil, 1,8-cineole from eucalyptus, and ascaridole, which is a naturally occurring peroxide from chenopodium oil:
Camphor is a particularly well-known bicyclic terpene ketone, which has uses in medicine and as a plasticizer for nitrocellulose (Section 20-7):
For many years, the principal source of camphor was the Formosan camphor tree. It now can be synthesized on a large scale from $\alpha$-pinene. Some of the other types of naturally occurring bicyclic ketones follow:
Higher oxygenated terpenes include the sesquiterpene alcohol, farnesol, which has a lily-of-the-valley odor and occurs in ambrette-seed oil. On acid dehydration it gives $\alpha$-farnesene (Table 30-1) under some conditions, and bisabolene (a component of oil of bergamot) under others:
As we shall see, cyclization reactions of this general type seem to be important in terpene biosynthesis. The 6,7-trans-farnesol has been shown to have hormone action in some insects. It acts to regulate the changes from caterpillar to cocoon to moth.
Two important diterpene alcohols are vitamin A (Section 28-7) and phytol, which occurs as an ester of the propanoic acid side-chain of chlorophyll (Figure 20-6):
The phytyl group appears also as a side chain in vitamin K$_1$ (Section 26-2B) and $\beta$-Carotene (Section 2-1) has vitamin A activity and apparently is oxidized in the body at the central double bond to give one mole of vitamin A.
The diterpene acid, abietic acid, is a major constituent of rosin, which is obtained as a nonvolatile residue in the manufacture of turpentine by steam distillation of pine oleoresin or shredded pine stumps. Abietic acid is used extensively in varnishes and as its sodium salt in laundry soaps.
Isoprenoid Compounds of Animal Origin
A number of compounds important to animal physiology have been identified as isoprenoid compounds. Notable examples are vitamin A, retinal (Section 28-7), and squalene (Table 30-1). Also, terpene hydrocarbons and oxygenated terpenes have been isolated from insects and, like farnesol, show hormonal and pheromonal activity. As one example, the juvenile hormone isolated from Cecropia silk moths has the structure shown in $3$:
The structure was established by an impressive combination of chemical, spectroscopic, and synthetic methods with about $200 \: \mu \text{g}$ of pure compound isolated from the abdomens of a myriad of male moths.$^1$
Juvenile hormone plays a critical role in maintaining the juvenile or larval stage of insects, and if its secretion is not controlled, normal development to the adult stage is prevented. Use of hormones or substances with hormonelike activity to control insect populations is an area of intense research interest and activity.$^2$ The secretion of juvenile hormone is controlled by other hormones originating in the brain (brain hormone) and the phthoracic gland (molting hormone, ecdysone; see Table 30-2).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/30%3A_Natural_Products_and_Biosynthesis/30.03%3A_Isoprenoid_Compounds.txt |
The term steroid applies to compounds containing hydrogenated cyclopentanophenanthrene carbon skeleton:
Most steroids are alcohols, and accordingly are named as sterols. Important examples include cholesterol, ergosterol, estradiol, stigmasterol, and other representative sterols given in Table 30-2. As you can see from their structures, most possess the same ring skeleton but vary considerably in their peripheral structural features, stereochemistry, and in the degree of ring unsaturation.
Table 30-2: Representative Steroids
Sterols are widely distributed in both plants and animals. Many are of vital importance to animal physiology, such as cholesterol, the bile acids, vitamin D, sex hormones, and corticoid hormones. Many have value as medicinals, such as the cardiac glycosides, hormones, and steroidal antibiotics. The occurrence and physiological properties of representative steroids are included in Table 30-2.
Cholesterol
Cholesterol is an unsaturated alcohol of formula $\ce{C_{27}H_{45}OH}$ that has long been known to be the principal constituent of human gall stones and has received notoriety in recent years for its connection with circulatory ailments, particularly hardening of the arteries. Cholesterol, either free or in the form of esters, actually is widely distributed in the body, particularly in nerve and brain tissue, of which it makes up about one sixth of the dry weight. The function of cholesterol in the body is not understood; experiments with labeled cholesterol indicate that cholesterol in nerve and brain tissue is not rapidly equilibrated with cholesterol administered in the diet. Two things are clear: Cholesterol is synthesized in the body and its metabolism is regulated by a highly specific set of enzymes. The high specificity of these enzymes may be judged from the fact that the very closely related plant sterols, such as sitosterol, are not metabolized by the higher animals, even though they have the same stereochemical configuration of all the groups in the ring and differ in structure only near the end of the side chain:
The accepted numbering system for the steroid nucleus and attached side chains is illustrated for cholesterol in $4$. The methyl groups at the junction of rings $A$ and $B$ $\left( \ce{C_{10}} \right)$ and rings $C$ and $D$ $\left( \ce{C_{13}} \right)$ are called angular methyls. To avoid misinterpretation of structure and stereochemistry, methyl groups and hydrogens at ring junctions should be explicitly written as $\ce{CH_3}$ or $\ce{H}$. The stereochemistry is specified by a solid line if the atom or group is above the ring plane $\left( \beta \right)$, and by a dashed line if below the ring plane $\left( \alpha \right)$. Thus compound $5$, 5$\alpha$-cholestan-3$\beta$-ol, which is obtained by the reduction of cholesterol, implies in the name that the hydroxyl at $\ce{C_3}$ is above the ring plane and that the hydrogen at $\ce{C_5}$ is below the ring plane (i.e., the $A$/$B$ rings, have the trans-decalin stereochemistry; Section 12-9).
Structure of Cholesterol
Although cholesterol was recognized as an individual chemical substance in 1812, all aspects of its structure and stereochemical configuration were not settled until about 1955. The structural problem was a very difficult one, because most of cholesterol is saturated and not easily degraded. Fortunately, cholesterol is readily available, so that it was possible to use rather elaborate degradative sequences, which would have been quite out of the question with some of the rarer natural products.
The first step in the elucidation of the structure of cholesterol was the determination of the molecular formula, first incorrectly as $\ce{C_{26}H_{44}O}$ in 1859 and then correctly as $\ce{C_{27}H_{46}O}$ in 1888. The precision required to distinguish between these two formulas is quite high, because $\ce{C_{26}H_{44}O}$ has $83.80\%$ $\ce{C}$ and $11.90\%$ $\ce{H}$, whereas $\ce{C_{27}H_{46}O}$ has $83.87\%$ $\ce{C}$ and $11.99\%$ $\ce{H}$. Cholesterol was shown in 1859 to be an alcohol by formation of ester derivatives and in 1868 to possess a double bond by formation of a dibromide. By 1903 the alcohol function was indicated to be secondary by oxidation to a ketone rather than an aldehyde. The presence of the hydroxyl group and double bond when combined with the molecular formula showed the presence of four carbocyclic rings. Further progress was only possible by oxidative degradation.
The structure proof for cholesterol paralleled that for two other important steroids, the so-called bile acids, cholic and desoxycholic acid, which function to help solubilize fats in the intestinal tract. Proof that cholesterol and the bile acids have the same general ring system was achieved by dehydration and reduction of cholesterol to two different hydrocarbons, 5$\alpha$-cholestane and 5$\beta$-cholestane (coprostane), which differ only in the stereochemistry of the junction between rings $A$ and $B$:
Oxidation of 5$\beta$-cholestane, but not 5$\alpha$-cholestane, gave an acid that turned out to be identical with cholanic acid obtained by dehydration of cholic acid at $300^\text{o}$ followed by hydrogenation:
Once the connection between cholesterol and the bile acids was established, further work on the structure proof was directed towards degradation experiments on the bile acids which, with their hydroxyl groups on rings $B$ and $C$, offered more possible degradation reactions than cholesterol. Outstanding contributions toward the structure proof were made by the German chemists H. Wieland and A. Windaus, both of whom were honored by the award of the Nobel Prize in chemistry. Wieland received the award in 1927 and Windaus in 1928. Despite their many years of effort, the structure proposed by Windaus in 1928 for desoxycholic acid was only tentative and was unspecific as to the location of two carbons.
Serious doubt as to the correctness of the Windaus structure came as a result of an x-ray study of ergosterol by J. D. Bernal in 1932. He pointed out that the x-ray evidence indicated ergosterol to be a long, rather flat molecule. Steroids with the ring system corresponding to the Windaus structure would have a globular shape. This observation stimulated further work and a re-examination of the evidence that eventually led to the correct structure. The details of this research may be found elsewhere.$^3$
Synthesis of Steroids
One of the most notable achievements of the 1950's was the total synthesis of a number of important steroids, including estrone, cholesterol, cortisone, androsterone, and testosterone. The courses of some of these syntheses are extraordinarily complex and involve large numbers of steps. Although they have not surpassed Nature in providing practical quantities of synthetic steroids, they have led to the development of key reactions of general use in organic synthesis. An especially useful reaction for building fused ring systems is the so-called Robinson annelation reaction developed by Sir Robert Robinson (Nobel Prize, 1947) and J. W. Cornforth (Nobel Prize, 1975). The reaction involves a Michael addition to an $\alpha$,$\beta$-unsaturated ketone immediately followed by an aldol addition:
The pharmacological importance of many natural steroids has stimulated much synthetic work in an effort to obtain practical quantities of naturally occurring and unnatural steroids. Oftentimes a combination of biosynthesis and organic synthesis works best. For example, the need for large quantities of cortisone derivatives for therapeutic use in treatment of arthritis and similar metabolic diseases has led to intensive research on synthetic approaches for methods of producing steroids with oxygen functions at $\ce{C_{11}}$, which is not a particularly common point of substitution in steroids.
An efficient way of doing this is by microbiological oxidation. Cortisone can be manufactured on a relatively large scale from the saponin, diosgenin, which is isolated from tubers of a Mexican yam of the genus Dioscorea. Diosgenin is converted to progesterone, then by a high-yield ($80\%$-$90\%$) oxidation with the mold, Rhizopus nigricans, to 11-hydroxyprogesterone and finally to cortisone:
Especially important steroid derivatives in use today are the synthetic estrogens, 17-$\alpha$-ethynylestradiol and its 3-$\ce{OCH_3}$ derivative, mestranol:
Both of these compounds have potent estrogenic activity (inhibit ovulation) and are widely used as oral contraceptives. They are synthesize from the natural estrogen, estrone, by the following reaction:
A compound known as diethylstilbestrol (DES) also exhibits estrogenic activity even though it is unrelated structurally to steroidal estrogens. It has acquired notoriety as a possible cause of uterine cancer. Diethylstilbestrol has been used extensively as an additive in cattle and chicken feed because it gives a greater gain in weight for a given amount of feed.
$^3$See, for example, the excellent and authoritative account give by L. F. Fieser and M. Fieser, Steroids, Van Nostrand Reinhold Co., New York, 1959, Chapter 3.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/30%3A_Natural_Products_and_Biosynthesis/30.04%3A_Steroids.txt |
Fatty Acids
The idea that ethanoic acid (acetic acid) is a possible common starting material for the biosynthesis of many organic compounds was first proposed by Collie (1893) on purely structural grounds. He recognized a structural connection between a linear chain of recurring $\ce{CH_3CO}$ units (a polyketomethylene chain, $\ce{CH_3COCH_2COCH_2COCH_2CO}-$) and certain cyclic natural products. In the example given below, orsellinic acid is represented as if it were derived from a chain of four $\ce{CH_3CO}$ units by a condensation-cyclization reaction:
Experimental verification of Collie's hypothesis came many years later when isotopic hydrogen and carbon ($\ce{^2H}$, $\ce{^3H}$, $\ce{^{13}C}$, and $\ce{^{14}C}$) became available. Tracer studies showed that long-chain fatty acids are made by plants and animals from $\ce{CH_3CO}$ units by successively linking together the carbonyl group of one to the methyl group of another (K. Bloch and F. Lynen, Nobel Prize, 1964). If ethanoic acid supplied to the organism is labeled at the carboxyl group with $\ce{^{14}C} \: \left( \overset{*}{\ce{C}} \right)$, the fatty acid has the label at alternate carbons:
$\ce{CH_3} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{O_2H}$
However, if the carbon of the methyl group is labeled, the product comes out labeled at the other set of alternate carbons:
$\overset{*}{\ce{C}} \ce{H_3CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CO_2H}$
Ethanoic acid is activated for biosynthesis by combination with the thiol, coenzyme A (\textbf{CoA} \ce{SH}\), Figure 18-7) to give the thioester, ethanoyl (acetyl) coenzyme A $\left( \ce{CH_3COS} \textbf{CoA} \right)$. You may recall that the metabolic degradation of fats also involves this coenzyme (Section 18-8F) and it is tempting to assume that fatty acid biosynthesis is simply the reverse of fatty acid metabolism to $\ce{CH_3COS} \textbf{CoA}$. However, this is not quite the case. In fact, it is a general observation in biochemistry that primary metabolites are synthesized by different routes from those by which they are metabolized (for example, compare the pathways of carbon in photosynthesis and metabolism of carbohydrates in Sectionss 20-9 and 20-10).
A brief description of the main events in fatty-acid biosynthesis follows, and all of these steps must be understood to be under control of appropriate enzymes and their coenzymes even though they are omitted here.
The $\ce{CH_3CO}$ group of ethanoyl coenzyme A is first transferred to a protein having a free thiol $\left( \ce{SH} \right)$ group to make another thioester, represented here as $\ce{CH_3COS}-\textbf{ACP}$, where ACP stands for Acyl-Carrier-Protein. The growing carbon chain remains bound to this protein throughout the synthesis:
$\ce{CH_3COS} \textbf{CoA} + \ce{HS}-\textbf{ACP} \rightarrow \ce{CH_3COS}-\textbf{ACP} + \textbf{CoA} \ce{SH}$
Carboxylation of $\ce{CH_3COS}-\textbf{ACP}$ yields a propanedioyl thioester, $6$, which then undergoes a Claisen condensation with a second mole of $\ce{CH_3COS}-\textbf{ACP}$ accompanied by decarboxylation to yield a 3-oxobutanoyl thioester, $7$:
Reduction of the ketone group of the thioester (by $\ce{NADPH}$) leads to a thiol ester of a four-carbon carboxylic acid. Repetitive condensations with thioester $6$ followed by reduction eventually lead to fatty acids. Each repetition increases the chain length by two carbons:
The preceding scheme is representative of fatty acid biosynthesis in plants, animals, and bacteria. The major difference is that plant and bacterial fatty acids usually contain more double bonds (or even triple bonds) than do animal fatty acids.
Biosynthesis of Aromatic Rings
Collie's hypothesis that aromatic compounds are made biologically from ethanoic acid was greatly expanded by A. J. Birch to include an extraordinary number of diverse compounds. The generic name "acetogenin" has been suggested as a convenient classification for ethanoate (acetate)-derived natural products, but the name "polyketides" also is used. Naturally occurring aromatic compounds and quinones are largely made in this way. An example is 2-hydroxy-6-methylbenzoic acid formed as a metabolite of the mold Penicillium urticae; using $\ce{^{14}C}$-carboxyl-labeled ethanoic acid, the label has been shown to be at the positions indicated below:
Terpene Biosynthesis
The biosynthesis of terpenes clearly follows a somewhat different course from fatty acids in that branched-chain compounds are formed. One way that this can come about is for 2-oxobutanoyl coenzyme A to undergo an aldol addition at the keto carbonyl group with the ethanoyl coenzyme A to give the 3-methyl-3-hydroxypentanedioic acid derivative, $8$:
The next step is reduction of one of the carboxyl groups of $8$ to give mevalonic acid:
This substance has been shown by tracer studies to be an efficient precursor of terpenes and steroids. Mevalonic acid has six carbon atoms, whereas the isoprene unit has only five. Therefore, if mevalonic acid is the precursor of isoprene units, it must lose one carbon atom at some stage. Synthesis of mevalonic acid labeled at the carboxyl group with $\ce{^{14}C}$, and use of this material as a starting material for production of cholesterol, gives unlabeled cholesterol. Therefore, the carboxyl carbon is the one that is lost:
Formation of the "biological isoprene unit" from mevalonic acid has been shown to proceed by stepwise phosphorylation of both alcohol groups, then elimination and decarboxylation to yield 3-methyl-3-butenyl pyrophosphate, $9$ (often called $\Delta^3$-isopentenyl pyrophosphate):
The coupling of the five-carbon units, $9$, to give isoprenoid compounds has been suggested to proceed by the following steps. First, isomerization of the double bond is effected by an enzyme $\left( \ce{E} \right)$ carrying an $\ce{SH}$ group:
The ester, $10$, then becomes connected to the double bond of a molecule of $9$, probably in an enzyme-induced carbocation type of polymerization (Section 10-8B):
The product of the combination of two units of the pyrophosphate, $9$, through this sequence is geranyl pyrophosphate if, as shown, the proton is lost to give a trans double bond. Formation of a cis double bond would give neryl pyrophosphate (Section 30-3B).
Continuation of the head-to-tail addition of five-carbon units to geranyl (or neryl) pyrophosphate can proceed in the same way to farnesyl pyrophosphate and so to gutta-percha (or natural rubber). At some stage, a new process must be involved because, although many isoprenoid compounds are head-to-tail type polymers of isoprene, others, such as squalene, lycopene, and $\beta$- and $\gamma$-carotene (Table 30-1), are formed differently. Squalene, for example, has a structure formed from head-to-head reductive coupling of two farnesyl pyrophosphates:
Since squalene can be produced from farnesyl pyrophosphate with $\ce{NADPH}$ and a suitable enzyme system, the general features of the above scheme for terpene biosynthesis are well supported by experiment.
In summary, the sequence from ethanoate to squalene has been traced as
$\text{ethanoyl coenzyme A} \rightarrow \text{mevalonic acid} \rightarrow \text{isopentenyl pyrophosphate} \rightarrow \text{farnesyl pyrophosphate} \rightarrow \text{squalene}$
Cholesterol Biosynthesis
Isotopic labeling experiments show that cholesterol is derived from ethanoate by way of squalene and lanosterol. The evidence for this is that homogenized liver tissue is able to convert labeled squalene to labeled lanosterol and thence to labeled cholesterol. The conversion of squalene to lanosterol is particularly interesting because, although squalene is divisible into isoprene units, lanosterol is not - a methyl being required at $\ce{C_8}$ and not $\ce{C_{13}}$:
As a result, some kind of rearrangement must be required to get from squalene to lanosterol. The nature of this rearrangement becomes clearer if we write the squalene formula so as to take the shape of lanosterol:
When squalene is written in this form, we see that it is beautifully constructed for cyclization to lanosterol. The key intermediate that initiates the cyclization is the 2,3-epoxide of squalene. Enzymatic cleavage of the epoxide ring is followed by cyclization and then manifold hydride $\left( \ce{H} \colon \right)$ and methide $\left( \ce{CH_3} \colon \right)$ shifts to give lanosterol:
The evidence is strong that the biosynthesis of lanosterol actually proceeds by a route of this type. With squalene made from either methyl- or carboxyl-labeled ethanoate, all the carbons of lanosterol and cholesterol are labeled just as predicted from the mechanism. Furthermore, ingenious double-labeling experiments have shown that the methyl at $\ce{C_{13}}$ of lanosterol is the one that was originally located at $\ce{C_{14}}$, whereas the one at $\ce{C_{14}}$ is the one that came from $\ce{C_8}$.
The conversion of lanosterol to cholesterol involves removal of the three methyl groups at the 4,4- and 14-positions, shift of the double bond at the $B$/$C$ junction between $\ce{C_5}$ and $\ce{C_6}$, and reduction of the $\ce{C_{24}}$-$\ce{C_{25}}$ double bond. The methyl groups are indicated by tracer experiments to be eliminated by oxidation to carbon dioxide.
The biosynthetic connection between ethanoyl coenzyme A and the complex natural products briefly discussed is summarized in Figure 30-1.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/30%3A_Natural_Products_and_Biosynthesis/30.05%3A_Biosynthesis.txt |
Alkaloids
Basic nitrogen compounds in plants are classified as alkaloids. Several examples were given previously of this large and remarkably heterogeneous class of compounds, many of which have very complex structures (Section 23-2). It is difficult to give a coherent account of alkaloid chemistry in the limited space available to us here.
The biosynthesis of alkaloids has been extensively studied, and although for a time it was thought that alkaloids arose primarily from amino acid precursors, strong evidence now is available that ethanoate also is involved. The mode of alkaloid biosynthesis is not yet as well understood as that of the terpenes and steroids. One experimental problem is the difficulty of feeding suitably labeled precursors to plants.
Vitamin B$_{12}$
Vitamin B$_{12}$ is the nutritional factor required for the prevention of pernicious anemia. Its structure was determined in 1956 through the chemical studies of Alexander Todd (Nobel Prize, 1957) and the x-ray diffraction studies of Dorothy Hodgkin (Nobel Prize, 1964). It is one of the most complex natural products known, yet it has features that are not unfamiliar. It is related to the metalloporphyrins discussed previously (Section 25-8B), but the ring system surrounding the cobalt atom has one less carbon bridging two of the nitrogen-containing rings than the porphyrin ring of heme or chlorophyll. The B$_{12}$ ring system is called a corrin ring, and the vitamin is a cobalt-corrin complex.
The corrin ring includes methyl, ethanamide, and propanamide groups, and one of these is linked through a nucleotide residue to the cobalt atom. There are five nitrogen ligands around the cobalt, and a sixth ligand is attached through carbon - here a cyano group - so that an alternate name for vitamin B$_{12}$ is cyanocobalamin:
A total synthesis of vitamin B$_{12}$ was announced in 1972, as the result of a collaborative effort between R. B. Woodward (Harvard) and A. Eschenmoser (Zurich). The synthesis was completed after 11 years of effort involving 100 co-workers from 19 countries. A number of important techniques and reactions of synthetic value were developed during the course of this work, including the principle of conservation of orbital symmetry (the Woodward-Hoffman rules, Section 21-10). The biochemical action of vitamin B$_{12}$ is considered in Chapter 31.
Penicillins and Cephalosporins
The first antibiotics of medicinal value were discovered by Alexander Fleming in 1929 as metabolites of the microorganism Penicillium notatum. They became known as penicillins, but their development as useful drugs was slow in coming. However, the urgent need for nontoxic antibiotics was recognized during World War II, and resulted in a team effort by English and American scientists to develop efficient methods for preparing penicillin by fermentation and to undertake clinical and chemical studies. By 1943, penicillin was available in quantity for the treatment of war wounded. By 1945, the basic structure and stereochemistry was deduced through chemical degradation and x-ray diffraction studies:
The structure is unusual in that it has a four-membered cyclic amide ring ($\beta$-lactam). It was the first example to be discovered of a natural product with this ring structure.
Fermentation can produce penicillins that differ only in the nature of the side-chain group $\ce{R}$. The common natural penicillin is penicillin G, in which $\ce{R} \: =$ phenylmethyl (benzyl):
The cephalosporins are antibiotics produced by the bacterial strain cephalosporium. They are closely related to the penicillins. Thus cephalosporin C has a $\beta$-lactam ring but a six-membered sulfur-containing ring:
Both the cephalosporins and the penicillins owe their antibacterial action to their ability to block bacterial cell-wall biosynthesis. Cephalosporin C is less active than the penicillins, but is less susceptible to enzymatic destruction by $\beta$-lactamases, which are enzymes that cleave the lactam ring. In fact, the so-called resistance of staph bacteria to penicillins is attributed to the propagation of strains that produce $\beta$-lactamase. Numerous semisynthetic penicillins and cephalosporins have been made in the hope of finding new broad-spectrum antibiotics with high activity but with greater $\beta$-lactam stability. Several of these are in clinical use.
The total synthesis of penicillin V was achieved by J. C. Sheehan (1957) and of cephalosporin by R. B. Woodward (1966). Biosynthetic routes have been worked out in part, and the precursors to both ring systems are $L$-cysteine and $D$-valine:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
30.07: Prostaglandins
Some of the most recent and exciting developments in the field of natural products are related to the compounds known as prostaglandins. All are oxygenated unsaturated derivatives of prostanoic acid, which is a \(\ce{C_{20}}\) fatty acid in which there is a cyclopentane ring formed by connecting the \(\ce{C_8}\) and \(\ce{C_{12}}\) positions:
There are two main types of prostaglandins that differ in the oxygen function at \(\ce{C_9}\), which is a carbonyl in Series E (PGE) and hydroxyl in Series F (PGF). Examples follow:
Prostaglandins are found in low concentrations distributed in a large number of organs, tissues, and body fluids of mammals. They exhibit a broad spectrum of physiological activity and are remarkably potent. Their precise biological role is not entirely clear, but they are known to induce strong contractions of smooth muscle tissue (lungs, uterus) and to lower blood pressure and sodium levels. Prostaglandins also have been implicated in the control of pituitary hormones released from the hypothalamus, and in the incidence of "pain" as a response to fever and inflammation. In fact, the analgesic property of aspirin possibly may result from the inhibition of prostaglandin biosynthesis. Although prostaglandins are not yet in extensive clinical use, their wide-ranging physiological effects hold promise that they will become useful drugs for the treatment of high blood pressure, thrombosis, respiratory disease, hypertension, ulcers, and in the regulation of fertility in both men and women.
A number of brilliant total syntheses of natural prostaglandins have been developed and these also have provided a number of interesting prostaglandin analogs. The biosynthesis of prostaglandins proceeds by oxygenation at \(\ce{C_{11}}\) of unsaturated fatty acids. This is followed by cyclization (probably as the result of a radical addition mechanism) to a bicyclic peroxide. Cleavage of the peroxide ring leads to prostaglandins:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/30%3A_Natural_Products_and_Biosynthesis/30.06%3A_Some_Nitrogen-Containing_Natural_Products.txt |
In the years following Kekulé brilliant proposal for the structure of benzene, organic chemistry underwent a tremendous expansion, and in the process a wide variety of paradigms or working hypotheses were developed about what kinds of compounds could "exist" and what kinds of reactions could occur. In many cases, acceptance of these hypotheses appeared to stifle many possible lines of investigation and caused contrary evidence to be pigeonholed as "interesting but not conclusive." As one example, the paradigm of angle strain was believed to wholly preclude substances that we know now are either stable or important reaction intermediates, such as cubane (Section 12-10), cyclopropanone (Section 17-11), and benzyne (Sections 14-6C and 23-8). No paradigm did more to retard the development of organic chemistry than the notion that, with a "few" exceptions, compounds with bonds between carbon and transition metals (Fe, Co, Ni, Ti, and so on) are inherently unstable. This idea was swept away in 1951 with the discovery of ferrocene (\(C_5H_5)_2Fe\)) by P. L. Pauson. Ferrocene has unheard of properties for an organoiron compound, stable to more than 500° and able to be dissolved in, and recovered from, concentrated sulfuric acid! Pauson's work started an avalanche of research on transition metals in the general area between organic and inorganic chemistry, which has flourished ever since and has led to an improved understanding of important biochemical processes.
• 31.1: Metallocenes
A metallocene is a compound typically consisting of two cyclopentadienyl anions bound to a metal center (M) in the oxidation state II. Certain metallocenes and their derivatives exhibit catalytic properties, although metallocenes are rarely used industrially. The first metallocene to be classified was ferrocene.
• 31.2: Other Organometallic Compounds of Transition Metals
Not all organometallic compounds of transition metals date from the discovery of ferrocene. Many have been known for a long time but their structures were not understood.
• 31.3: Transition-Metal Compounds as Reagents for Organic Syntheses
Transition-Metal Compounds can act as reagents for organic syntheses. For example, some transition-metal hydrides show promise as synthetic reagents and sodium reacts with iron pentacarbonyl to produce a salt known as sodium tetracarbonylferrate(-II)2, which has been shown to have considerable potential as a reagent for organic synthesis.
• 31.4: Some Homogeneous Catalytic Reactions Involving Transition-Metal Complexes
Some transition-metal complexes can catalyze homogeneous reactions such as hydrogenation, Hydroformylation of Alkenes, Carbonylation of Methanol and alkene metathesis.
• 31.5: π-Propenyl Complexes of Nickel
A considerable body of highly useful chemistry based on nickel has been developed. Many of these reactions involve what are called π -propenyl complexes and their formation has a close analogy in the formation of ferrocene from cyclopentadienylmagnesium compounds and ferric chloride.
• 31.6: Vitamin B₁₂ as an Organometallic Compound
The structure of vitamin B12 with a cyanide ion coordinated with cobalt is not the active form of the vitamin but is a particularly stable form, convenient to isolate and handle. The active form is a coenzyme that is remarkable in having a carbon-cobalt bond to an essentially alkyl-type carbon.
• 31.E: Transition Metal Organic Compounds (Exercises)
These are the homework exercises to accompany Chapter 31 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Thumbnail: Ball-and-stick model of a metallocene molecule where the cyclopentadienyl anions are in a staggered conformation. The purple ball in the middle represents the metal cation. (Public Domain; Ben Mills).
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
31: Transition Metal Organic Compounds
The discovery of ferrocene was one of those fortuitous accidents that was wholly unforeseeable - the kind of discovery which, over and over again, has changed the course of science. Pauson was trying to synthesize fulvalene, $1$, by first coupling two molecules of cyclopentadienylmagnesium bromide with $\ce{FeCl_3}$ and then dehydrogenating the product:
The rationale for the coupling reaction was that phenylmagnesium bromide with $\ce{FeCl_3}$ gives high yields of biphenyl, presumably by way of an unstable phenyliron compound:
The reaction product was a beautifully crystalline, highly stable orange substance, $\ce{C_{10}H_{10}Fe}$, which Pauson formulated as a simple combination of two cyclopentadienide anions and ferrous ion with two $\ce{C-Fe}$ bonds, $2$. However, the product soon was shown by a variety of physical methods to have the "sandwich" structure, $3$:
The bonding between the metal and the cyclopentadiene rings involves the $\pi$ electrons of the two rings, all carbons being equally bonded to the central ferrous ion. The latter, in accepting a share of 12 $\pi$ electrons from two cyclopentadienyl anions, achieves the 18 outer-shell electron configuration$^1$ of the inert gas, krypton. Analysis of the structure of crystalline ferrocene shows that when you look down on the molecule along the ring-iron-ring axis the cyclopentadiene rings are seen to be staggered with respect to one another, as shown in $4$. Ferrocene has mp $173^\text{o}$ and, although stable to sulfuric acid, it is readily oxidized by nitric acid to the less stable ferricinium ion:
Like benzene, ferrocene does not react easily by addition but does undergo electrophilic substitution. For example, Friedel-Crafts acylation (Section 22-4F) with $\ce{CH_3COCl}$ gives both a monoethanoylferrocene and a diethanoylferrocene. The two acyl groups become attached to two different rings and, because only one diethanoylferrocene can be isolated, the cyclopentadienyl groups appear to be free to rotate about the axis of the carbon-iron bonds:
Ferrocene is only one of a large number of compounds of transition metals with the cyclopentadienyl anion. Other metals that form sandwich-type structures similar to ferrocene include nickel, titanium, cobalt, ruthenium, zirconium, and osmium. The stability of metallocenes varies greatly with the metal and its oxidation state; ferrocene, ruthenocene, and osmocene are particularly stable because in each the metal achieves the electronic configuration of an inert gas. Almost the ultimate in resistance to oxidative attack is reached in $\ce{(C_2H_5)_2Co}^\oplus$, cobalticinium ion, which can be recovered from boiling aqua regia (a mixture of concentrated nitric and hydrochloric acids named for its ability to dissolve platinum and gold). In cobalticinium ion, the metal has the 18 outer-shell electrons characteristic of krypton.
Many other unsaturated organic compounds can form $\pi$ complexes with transition metals. A substance that is in some ways analogous to ferrocene is the complex of two benzene molecules with chromium metal, called dibenzenechromium. The bonding involves zerovalent chromium and the $\pi$ electrons of the two benzene rings. In dibenzenechromium, the electronic configuration of the chromium atom is similar to that of krypton:
Although dibenzenechromium is thermally quite stable, it is less so than ferrocene and melts with decomposition at $285^\text{o}$ to give benzene and metallic chromium. Furthermore, it appears to lack the aromatic character of either benzene or ferrocene as judged by the fact that it is destroyed by reagents used for electrophilic substitution reactions.
Several transition-metal complexes of cyclobutadiene have been prepared, and this is all the more remarkable because of the instability of the parent hydrocarbon. Reactions that logically should lead to cyclobutadiene give dimeric products instead. Thus, 3,4-dichlorocyclobutene has been dechlorinated with lithium amalgam in ether, and the hydrocarbon product is a dimer of cyclobutadiene, $5$. However, 3,4-dichlorocyclobutene reacts with diiron nonacarbonyl, $\ce{Fe_2(CO)_9}$, to give a stable iron tricarbonyl complex of cyclobutadiene, $6$, whose structure has been established by x-ray analysis. The $\pi$-electron system of cyclobutadiene is considerably stabilized by complex formation with iron, which again attains the electronic configuration of krypton.
Oxidation of $6$ with ceric iron, $\ce{Ce}$(IV), releases cyclobutadiene which quickly dimerizes, but can be trapped by good dienophiles such as ethyl propynoate to give a cycloadduct.
Many metallocene derivatives are known of other conjugated cyclic polyenes. Examples are bis(cyclooctatetraene)uranium (uranocene, $7$) and bis(pentalenylnickel), $8$ (see Section 22-12B):
Many of the metallocene compounds display unusual reactivities and reactions, of which none is more startling than the discovery by the Russian chemist, M. E. Vol'pin, of the absorption of dinitrogen, $\ce{N_2}$, by titanocene, $\ce{(C_2H_5)_2Ti}$, to form a complex or complexes that can be reduced easily to form ammonia. The nature of these complexes is in doubt, but very clear evidence has been obtained by J. E. Bercaw for the structure of the complex $9$ formed from decamethylzirconocene and dinitrogen:
This complex treated with acids gives $\ce{NH_2-NH_2}$ and some $\ce{NH_3}$.
$^1$Figure 6-4 shows that iron(0) has 8 electrons in the $4s$ and $3d$ orbitals. Ferrous iron $\left( \ce{Fe^{2+}} \right)$ then will have 6 outer-shell electrons. This 6 plus the 12 $\pi$ electrons of the two cyclopentadienide rings makes the 18-electron total and the krypton electronic configuration.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/31%3A_Transition_Metal_Organic_Compounds/31.01%3A_Metallocenes.txt |
$\pi$-Type Alkene and Cycloalkene Complexes
Not all organometallic compounds of transition metals date from the discovery of ferrocene. Many have been known for a long time but their structures were not understood. A conspicuous example is the anion of a substance known as Zeise's salt, which is formed from the reaction of ethanol with chloroplatinic acid, $\ce{H_2PtCl_6}$. The anion has the formula $\ce{Pt(C_2H_4)Cl_3^-}$. Although known since 1830, it finally was shown to have a structure with both ethene carbons bonded to platinum. The carbon-to-metal bonding usually is formulated as a $\pi$ complex, $11a$, or charge-transfer complex (Section 22-4D). Alternatively, we can think of the bonding in $11a$ as between platinum as a Lewis acid (electron-pair acceptor) and ethene as a Lewis base (electron-pair donor):
The arrow in $11a$ symbolizes donation of $\pi$ electrons. However, because the stability of the ion is much greater than would be expected for either a simple acid-base or charge-transfer complex, it is postulated that unshared $d$ electrons from the metal participate in the bonding. This is symbolized by the dashed arrow in $11b$, which stands for donation of $d$ electrons into the $\pi^*$ orbital of the double bond or, as it is often called, "back bonding". Perhaps most simple is $11c$, where the $\ce{C-Pt}$ bonding is formulated as a three-membered ring with essentially $\ce{C-Pt}$ (\sigma\) bonds. In this formulation, full participation of a platinum electron pair is assumed.
Many complexes of alkenes, cycloalkenes, alkynes, and cycloalkynes with transition metals are now known. Some examples are:
Many substances of this type have potential synthetic usefulness as catalysts or as reagents.
Alkyl-Transition-Metal Bonds
Organometallic compounds of transition metals with alkyl-to-metal bonds for many years were regarded as highly unstable substances and prone to dissociation into radicals that would couple or disproportionate, as illustrated by the following sequence:
$\tag{31-1}$
The fact is that the stability depends on the character of the attached alkyl groups. Transition-metal compounds with $\ce{CH_3}-$, $\ce{(CH_3)_3CCH_2}-$, and $\ce{C_6H_5CH_2}-$ groups are in many cases very much more stable than those with $\ce{CH_3CH_2}-$, $\ce{(CH_3)_2CH}-$, and $\ce{(CH_3)_3C}-$ groups, even though the ease of formation of radicals by dissociation would be expected to be especially favorable with the $\ce{C_6H_5CH_2}-$ group. Furthermore, decompositions that give alkene and metal hydride or disproportionation products (alkene and alkane) may fail to give coupling products altogether. These facts and many others indicate that an important mode of decomposition of a variety of alkyl-substituted transition-metal compounds does not proceed by a radical mechanism. Instead, there is a transfer of a $\beta$ hydrogen from the alkyl group to the metal to form a $\pi$-type alkene complex and a metal-hydride bond. Decomposition of this complex produces the alkene and the metal hydride. These changes are reversible.
Formation of the alkane then can result from cleavage of an alkyl-metal bond by the metal hydride:
This mechanism but not the sequence of Equation 31-1 makes clear why $\ce{CH_3}-$, $\ce{(CH_3)_3CCH_2}-$, and $\ce{C_6H_5CH_2}-$ are stable attached to transition metals, because each of these substituents lacks a $\beta$ hydrogen that would permit formation of a $\pi$-bonded complex. The hydride-shift reaction is especially important in hydrogenation and carbonylation reactions, as will be shown in Section 31-3. In fact, the reversible rearrangement of $\pi$ to $\sigma$ complexes has wide generality and includes alkyl shifts as well as hydride shifts. An important example of the alkyl-shift mechanism is the polymerization of ethene by Ziegler-Natta catalysts ($\ce{R_3Al}$ and $\ce{TiCl_4}$); the chain-building sequence is given in Section 29-6A.
In understanding these reactions, it is helpful to view the metal-alkene $\pi$ complex as an incipient carbocation (just as $\pi$ complexes of halogens are incipient carbocations). Alkyl and hydride shifts then bear analogy to carbocation rearrangements. This may be an oversimplification but it makes the chemistry easier to follow.
In other reactions, we will see that the metal can act as a nucleophilic reagent.
Carbene-Metal Complexes
Bonding between a transition-metal atom and one $sp^2$-hybridized carbon can be represented by the following valence-bond structures:
Stable transition-metal complexes of this type are known and others have been recognized as likely intermediates in a number of reactions. Rightly or wrongly, they are called carbene-metal complexes, although they also can be regarded either as metal-stabilized carbocations or as metal-stabilized ylides (Section 16-4A).
Perhaps the most common examples of this type of carbon-metal bonding are the metal carbonyls, in which the carbon monoxide ligand functions as the "carbene":
When represented in this way the chemistry of carbonyl complexes of transition metals becomes easier to understand. Hydroformylation reactions and other carbonylations that are catalyzed by transition-metal complexes frequently involve hydride or alkyl transfers from the metal atom to the "positive" carbonyl carbon (Sections 16-9G, 31-3, and 31-4):
We now proceed to describe some selected reactions that can be understood within this framework of $\sigma$-, $\pi$-, and carbene-type bonding between the metal and carbon.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/31%3A_Transition_Metal_Organic_Compounds/31.02%3A_Other_Organometallic_Compounds_of_Transition_Metals.txt |
Reactions of Zirconocene Chlorohydride
Some transition-metal hydrides show promise as synthetic reagents of the same general applicability as the boron hydrides (Section 11-6). An excellent illustration is provided by the work of J. Schwarz with zirconocene chlorohydride, $13$, which is available by reduction of zirconocene dichloride:
(The cyclopentadienide rings in $13$ are shown as being nonparallel and this is in accord with x-ray diffraction studies of metallocenes that have extra substituents on the metal.) Henceforth we will abbreviate the structure $13$ by $\ce{(Cp)_2ZrClH}$.
Alkenes react with $\ce{(Cp)_2ZrClH}$ to form alkyl-$\ce{Zr}$ bonds with zirconium becoming attached to the least-hindered primary carbon:
The initial step in this kind of reaction is formation of the $\pi$-alkene complex followed by hydride transfer:
These reaction must be reversible for an alkene with an internal double bond to form an adduct with the metal atom at the end of the chain. The process is seen as a series of interconversions between $\pi$ and $\sigma$ complexes, which permits the metal atom to move to the least-hindered (primary) carbon:
One of the elegant features of these reactions is the formation of crystalline $\ce{[(Cp)_2Zr(Cl)]_2O}$ on treatment of the reaction products with water. This substance can be converted back to zirconocene dichloride with $\ce{HCl}$ and thence back to $\ce{(Cp)_2ZrClH}$:
This sequence of steps is an important part of the mechanism of the hydroformylation of alkenes (oxo reaction), to be discussed in Section 31-4B, and also is related to the carbonylation reactions of boranes discussed in Section 16-9G.
A Nucleophilic Transition-Metal Reagent. Sodium Tetracarbonylferrate(-II)
Sodium reacts with iron pentacarbonyl to produce a salt known as sodium tetracarbonylferrate(-II)$^2$, $\ce{Na_2Fe(CO)_4}$, which has been shown by J. P. Collman and co-workers to have considerable potential as a reagent for organic synthesis.
$2 \ce{Na} + \ce{Fe(CO)_5} \rightarrow \ce{Na_2Fe(CO)_4} + \ce{CO}$
The tetracarbonylferrate dianion is a good nucleophile and reacts with alkyl halides or alkyl sulfonate esters by the $S_\text{N}2$ mechanism (with inversion) to form $\ce{C-Fe}$ bonds:
The resulting anion undergoes insertion with carbon monoxide or ketone formation with acyl halides in a manner similar to alkylchlorozirconocenes (Section 31-3A):
The product of $\ce{CO}$ insertion has the potential of transferring $\ce{R}- \overset{\ominus}{\ce{C}} \ce{=O}$, and is converted to $\ce{RCHO}$ with acids, to $\ce{RCOX}$ with halogens, or to $\ce{RCO_2H}$ by oxidation:
$^2$The designation (-II) indicates that the iron in this substance can be regarded as being in the -2 oxidation state.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
31.04: Some Homogeneous Catalytic Reactions Involving Transition-Metal Complexes
Hydrogenation
The mechanisms of hydrogenation of alkenes over finely divided metals such as nickel, platinum, and so on (Section 11-2) now are understood in a general way. However, these reactions are extremely difficult to study because they occur on a metallic surface whose structure is hard to define. In contrast, the mechanisms of hydrogenation with homogeneous catalysts are known in considerable detail and provide insight into their heterogeneous counterparts.
Homogeneous hydrogenation catalyzed by the four-coordinated rhodium complex, $\ce{Rh[(C_6H_5)_3P]_3Cl}$, has been particularly well investigated. With this catalyst, the first step is formation of the six-coordinated rhodium hydride of known configuration, $16$, in which we abbreviate the ligand, triphenylphosphine, $\ce{(C_6H_5)_3P}$, as $\ce{L}$:
The next step is coordination of the alkene (here ethene) with $16$ with loss of $\ce{L}$ to give the $\pi$ complex, $17$, also of known configuration:
Stable ethene complexes of $\ce{Rh}$ similar to $17$ have been isolated and shown to have the $\pi$-complex structure. Formation of $17$ must be an equilibrium process because addition of extra $\ce{L}$ reduces the rate of hydrogenation by shifting the equilibrium to the left.
Hydrogenation proceeds by hydride rearrangement of $17$ to a five-coordinated ethyl-rhodium complex, $18$. This complex regains a ligand molecule to replace the one lost previously, thereby giving the six-coordinated complex, $19$:
Stable complexes with $\ce{Rh-CH_2CH_3}$ bonds similar to $19$ have been well characterized. The final step is formation of ethane from $19$ with regeneration of $\ce{RhL_3Cl}$:
$19 \rightleftharpoons \ce{RhL_3Cl} + \ce{CH_3CH_3}$
Although we abbreviate $\ce{(C_6H_5)_3P}$ as $\ce{L}$ and show little role for it or for the $\ce{Cl}$ attached to rhodium in the reaction, these ligands play a very important role in providing the electronic and steric environment around the rhodium, which makes efficient catalysis possible. A useful diagram of how the catalyst functions in the overall reaction is shown in Figure 31-2.
Hydroformylation of Alkenes (Oxo Reaction)
The conversion of alkenes to aldehydes with carbon monoxide and hydrogen in the presence of a cobalt catalyst is an important reaction (Section 16-9F):
Carbonylation of Methanol
A successful commercial synthesis of ethanoic acid starts with methanol and carbon monoxide in the presence of a rhodium catalyst and hydrogen iodide:
$\ce{CH_3OH} + \ce{CO} \underset{175^\text{o}, \: 400 \: \text{psi}}{\overset{\ce{RhCl_3} \cdot 3 \ce{H_2O}, \: \ce{HI}}{\longrightarrow}} \ce{CH_3CO_2H}$
The reaction has some similarity to the hydroformylation reaction described in Section 31-4B. The hydrogen iodide is required to transform methanol to methyl iodide. The rhodium catalyst then reacts with the methyl iodide as a nucleophilic reagent:
Complexation with carbon monoxide follows:
A shift of the methyl group from rhodium to carbon and then hydrolysis gives the acid and regenerates $\ce{HI}$ and the rhodium catalyst:
The Alkene Metathesis Reaction
One of the most curious catalytic reactions of alkenes ever discovered is alkene metathesis or alkene dismutation, in which two alkenes exchange alkylidene groups, usually over a tungsten catalyst. The essence of the reaction is illustrated by a commercial process for converting excess propene to a mixture of ethene and butenes:
The reaction products are those expected if cyclobutanes were intermediates, but formation and cleavage of cyclobutanes is not the correct mechanism because cyclobutanes generally are not converted to alkenes over alkene-metathesis catalysts.
After a great deal of research on the mechanism of this reaction, it now appears likely that the crucial step is the formation of carbene metal complexes and that the products are formed by recombination of the carbenes with alkene in the various possible ways:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/31%3A_Transition_Metal_Organic_Compounds/31.03%3A_Transition-Metal_Compounds_as_Reagents_for_Organic_Syntheses.txt |
A considerable body of highly useful chemistry based on nickel has been developed, largely by the German chemist, G. Wilke. Many of these reactions involve what are called $\pi$-propenyl ($\pi$-allyl) complexes and their formation has a close analogy in the formation of ferrocene from cyclopentadienylmagnesium compounds and ferric chloride (Section 31-1). Treatment of $\ce{NiBr_2}$ with two moles of 2-propenylmagnesium bromide gives a stable (albeit oxygen sensitive) substance of composition $\ce{(C_3H_5)_2Ni}$:
Unlike $\ce{C_3H_5MgBr}$, the metal compound has a very complex proton NMR spectrum. Analysis of the spectrum indicates it arises from a mixture (75:25) of two $\ce{(C_3H_5)_2Ni}$ isomers with each isomer having its $\ce{C_3H_5}$ groups in a rigid planar arrangement as follows:
These facts can be accommodated by the trans- and cis-di-$\pi$-propenylnickel structures, $20$. Di-$\pi$-propenylnickel has many interesting reactions, among which are the following examples:
The $\pi$-propenyl-type structures are more stable for nickel than for other metals such as iron. With 1,3-butadiene, $\ce{Fe(CO)_5}$ forms a double $\pi$ complex, whereas $\ce{Ni(CO)_4}$ produces a bis-$\pi$-propenyl-type structure, $21$:
With more 1,3-butadiene, $21$ is converted first to $22$, which after rearrangement reacts with 1,3-butadiene to give back $21$ with liberation of trans,trans,trans-1,5,9-cyclododecatriene:
The overall sequence thus provides a catalytic route for the cyclic trimerization of 1,3-butadiene.
Ethene and alkynes react with $21$ in the presence of excess carbon monoxide to give ten-membered ring compounds, whereas the reaction of $21$ with excess carbon monoxide results in formation of a mixture of six- and eight-membered rings:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
31.06: Vitamin B as an Organometallic Compound
The structure of vitamin B$_{12}$ shown in Section 30-6B with a cyanide ion coordinated with cobalt is not the active form of the vitamin but is a particularly stable form, convenient to isolate and handle. The active form is a coenzyme that is remarkable in having a carbon-cobalt bond to an essentially alkyl-type carbon. The carbon-cobalt bond is to a 5'-deoxyadenosyl group, and if we abbreviate vitamin B$_{12}$ coordinated to cyanide as $23$, the coenzyme can be written, in the same style, as $24$. (You will notice that $23$ is an abbreviation of the formula of Section 30-6B turned $180^\text{o}$.)
Both $23$ and the B$_{12}$ coenzyme, $24$, are compounds of $\ce{Co}$(III) and both substances have all electrons paired. B$_{12}$ can be reduced to a form with $\ce{Co}$(II) which has an unpaired electron and gives an ESR signal (Section 27-9). The cobalt-carbon bond of $24$ appears to be formed from $23$ by removal of the cyano group and a two-electron reduction to $\ce{Co}$(I). The reduced cobalt is powerfully nucleophilic and probably is alkylated with adenosine triphosphate (ATP, Section 15-5F) to form $24$:
Vitamin B$_{12}$ coenzyme participates in several biological reactions but none is more unusual, or as hard to rationalize, as its role in the interconversion of methylpropanedioyl CoA (methylmalonyl CoA, $25$) to butanedioyl CoA (succinyl CoA, $26$):$^3$
This rearrangement, which is important in the biochemical utilization of propanoic acid, has been shown to involve transfer of a hydrogen from the $\ce{CH_3}-$ group of $25$ to the $\ce{-CH_2R}$ group of $24$. Then rearrangement and formation of $26$ occurs along with reformation of $24$:
We formulate the intermediate oxidized forms of $25$ and $26$ with cobalt-to-carbon bonds, but there is no definitive evidence that this is correct. The overall reaction involves attack on the $\ce{CH_3}-$ of $25$, not an easy reaction to carry out in the laboratory, except with reagents such as $\ce{Cl} \cdot$, because this $\ce{CH_3}$ is not adjacent to a double bond or other activating group. Furthermore, there is no very good analogy for the rearrangement step. At present, although it is known that $24$ is reduced to give $\ce{CH_3-R}$, the details of this important biochemical mechanism remain to be elucidated by further research.
$^3$For the structure of CoA, see Section 18-8F.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/31%3A_Transition_Metal_Organic_Compounds/31.05%3A_-Propenyl_Complexes_of_Nickel.txt |
INTRODUCTION
Laboratory courses can be magical.They can be enlightening experiences that open your eye to a big picture. A laboratory experience should work in tandem with a lecture course, and fully realize concepts, techniques and reactions that you have heard of. Unfortunately, they can also be discouraging experiences. One of the key factors that dictates which experience you will have, is preparation.
A rewarding aspect of a well-prepared experiment is that it can firmly cement the information that you have obtained through studying in a way that is far superior to simply reading about it. Your knowledge evolves beyond routine memorizing to real understanding, because you have seen the reaction and principles with your own eyes. The synergy between a lecture course and a lab component should not be underestimated.
In this chapter I outline a systematic way of preparing for any organic chemistry experiment to ensure that you succeed in the laboratory and that you leave with an optimal experience. Seeing, after all, is believing. I discuss how to make and use the flow diagrams, how to obtain relevant safety information about the chemicals you are handling, and how to use your note-book to prepare efficiently.
01: HOW TO PREPARE FOR AN ORGANIC CHEMISTRY EXPERIMENT
There are several reasons why a systematic plan for preparation is so integral to a laboratory experience, but it boils down to the fact that an organic chemist must have information that far supersedes the information provided in a typical lab manual. When you are performing an experiment, there is little time to look up densities, perform calculations, look up safety and hazard information for chemicals and do any in-depth research on the theory behind the techniques you use. You must bring all this information with you. In many cases the lack of this information can have detrimental consequences for your experiment, your learning outcome and last but not least the safety of the laboratory.
So how can you prepare better?
Let us start by looking at the procedure. It is a recipe that describes the order of events, and the key operations that must be performed. However, just reading the procedure will never prepare you adequately.1 Why is that? The first reason, is that there is far too much material to remember, at any given time, and events that happen early in the experiment, will influence the outcome later. The second reason, is that any lab manual, even the ones designed for undergraduate use, never spells out everything that must be done. A third reason, is that any lab manual expects that the reader has some expertise and some level of experience.
This is something you most likely already know, from your everyday life.
Let’s say that you want to follow a recipe you are not familiar with to make a birthday cake. You start on step 1 and work your way through the recipe, but after a while, you run into a problem.
1 Every term we see many students that simply read the procedure (sometimes repeatedly), and end up confused and suffering through a sub-par experience.
The recipe calls for four cups of flour, but you only have two. You have a hot oven that is ready for the cake batter, a half-made frosting, and a batter that is not ready at all. You switch off the oven and rush to the store for more flour. By the time you come home, the oven is cold (and must be re-heated), and the frosting has split because you left it out in room temperature for half an hour.
This example might seem far removed from the lab, but it is perfectly transferrable. Because you had not read the recipe (or prepared adequately), you wasted time and part of your cake. Nobody wants to waste a cake.
If we take another cake-analogy and move closer to what an experiment in a lab might look like, we can say that you have a recipe for your great grandmother’s favorite chocolate cake that you wish to make. Here is the start of the recipe:
GREAT GRANDMOTHER ELSA’S FAVORITE CHOCOLATE CAKE:
1. Pre-heat oven.
2. Beat 1 stick of butter with 1 cup of sugar until mixture is light.
3. Add 3 eggs, one at the time, to the sugar and butter mixture.
4. In a bowl, add 2 cups of flour, 1 small spoon of baking powder and 1 cup of cacao-powder.
Let us think of this recipe as a lab procedure. What we absolutely do not want is to read the procedure once, and then go to the kitchen (lab) and start to make the cake (the experiment). Why not? Well, we will immediately run into many problems. In step one; Elsa says to pre-heat the oven. To what temperature, exactly? Moreover, in step 2, we beat together butter and sugar. With what? And in what? What size bowl do we need? What will we use to do the actual beating? In step 3, we add one egg at the time, but surely, we should carefully beat in each egg, before the next one is added? Elsa does not clarify that, but if we did not have to beat in each egg, there would be no point in adding one egg at the time.
As in Elsa’s recipe and in a lab manual (even one designed for undergraduate use) we absolutely need to visualize the recipe or procedure, before we move on. You might have noted that through the act of visualizing Elsa’s recipe, I identified several issues and questions. We want to be aware of these up front, so that we have the answers and solutions before we start cooking.
Visualization is the act of seeing the procedure before your eyes, as you read it. This is a very important part of preparation, as it allows you to identify any
problems, questions or issues before you go into the lab. It also allows you time to look up valuable information. Elsa did not give us the exact temperature for baking her cake, but we could always look up what baking times for similar cakes are.
At this stage, I want to introduce another powerful tool: the flow chart. The flow chart is a graphical over- view of a procedure that should contain the essentials needed to complete the procedure. The flow chart is a product created after visualization, and it should be something you bring with you to lab. It should outline the flow of the experiment, but also provide something additional to the procedure: namely, the answers to all those questions that came up during visualization.
In the figure shown I have shown a flow chart for Great Grandmother Elsa’s recipe. Note that I have included all the information from the questions above.
I have the temperature of the oven (based on a search I did for similar cakes), I have the sizes of the bowl, and the added clarification to beat in each egg, before adding the next. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/01%3A_HOW_TO_PREPARE_FOR_AN_ORGANIC_CHEMISTRY_EXPERIMENT/1.01%3A_READING_DRAWING_AND_UNDERSTANDING_THE_POWER_OF_VISUALIZATION.txt |
Safety is a number one concern for everyone working in a lab. Safety concerns have many different manifestations, and in general, we always work to make everyone as safe as possible by creating a standard procedure for whenever is in the lab. In any given situation in lab (and there are no exceptions), we use the following:
1. Lab goggles. Our eyes are very delicate, and almost all chemicals are detrimental for our eye health. Goggles are worn at all times, and they are never removed in the lab. Goggles must be splash proof, which means that they protect the eyes from multiple angles, should chemicals splash.
2. Lab coat. The coat is the first line of defense against chemicals that may get in contact with us, and the coat also provides a first line of defense for our clothes.
3. Fume-hoods. A fume-hood is a very important part of the lab environment. It is a hood equipped with an adjustable sash, where the air flow inside the hood is very high. This removes any vapors quickly and safely. Everything that you do in an organic chemistry lab, is done in a fume-hood to minimize your exposure to fumes.
In many situations, the use of gloves is also appropriate. This is covered in detail in chapter 3.
The scary reality of working in a lab is that chemicals are dangerous. All substances have some inherent dangers associated with them, and although we spend our times exposed to many chemicals outside of the lab, the lab is a place where we can expose ourselves to a greater variety of hazardous chemicals. As a base line, we will treat any chemical as dangerous, and the utmost of care should always be taken to ensure that exposure is at a minimum. Because chemicals do vary in terms of their properties and hazards, it is also necessary to know the exact hazards of each chemical.
Where should you go to look up, identify and familiarize yourself with this information?
The answer is Safety Data Sheets (SDS).
SDS are legally required information sheets that provide an overview of the safety and hazards associated with a given chemical or mixture. There may be thousands of data sheets available for any given chemical, so before I move on, I would recommend using Sigma Aldrich as the main supplier of relevant and reliable data sheets.2
The SDS are compiled by the Globally Harmonized System of classification and labeling of chemicals (GHS), an internationally recognized system that standardizes safety and hazard information for chemicals. That means that people from all over the world should be able to access any given SDS and find the same types of information, presented in the same way.
Knowing how to work with SDS is essential for anyone working in a laboratory environment, because they contain information on the risks, hazards and safety issues associated with the chemicals in use. SDS give an informed backdrop to the experiment and helps you make safe and good choices when handling chemicals.
To contextualize this, just imagine that you were planning to handle a chemical that is explosive in contact with water. Knowing this hazard in advance is obviously key to your safety. On the other hand, what if you were planning to handle a carcinogenic substance (a substance that causes cancer)? You would obviously want to know these risks beforehand so that you could work in the safest way possible. Furthermore, if you ever plan to work in a professional laboratory, consulting SDS will be a very important part of your preparation.
2 The site http://sigmaaldrich.com/ is usually a good starting point. You can search for chemicals by name or structure.
Let us have a closer look at how a SDS is formatted.
The SDS for any given compound is typically organized into several categories, such as hazards, toxicity, composition, and first aid measures, to name a few.
The information found in a SDS can be misleading if misinterpreted. As an example, let us examine the SDS of sodium chloride (NaCl), which is table salt. Under category 4, which is first-aid information, the SDS states that if the substance is in contact with skin to “Wash off with soap and plenty of water. Consult a physician.” That seems ridiculous. I doubt that any of us call a physician if we get salt on our fingers while we are eating fries, but the restaurant cook who is dipping their fingers into a bowl of salt eight hours a day might wonder if it will cause any harmful effects. The SDS also lists information about the “acute toxicity” of sodium chloride. The data stated is LD50 Oral Rat 3,550 mg/kg. How does one interpret this?
LD50 Oral Rat refers to the amount of salt that will kill 50% of the rats who swallow it. Because rats come in all sizes, the amount is expressed as ratio between mass of compound and mass of rate. In this case, 3550 mg (or 3.55 g) of sodium chloride will kill 50% of rats who weigh 1kg.
If we assume that human bodies work like rats, we can find a lethal dose for half a population of humans. If we assume that a human weigh 100 lb (which is a convenient number) or 45.4 kg, then 45.4 x 3.55 = 161 g NaCl might be lethal to 50% of the humans who swallow it. 161 g is over 1/3 lb of salt, so the fries would probably kill you before the salt did, but there is how you interpret this figure.
As you can see from this example, the reliable information in the SDS has to be worked out ina realistic context. In this particular case, we have learned that sodium chloride can be lethal if ingested in (relatively) large amounts.
Here are some of the hazards categories that SDS frequently identify for organic chemicals, and some notes about each category to help you get started. Please note that this list is not exhaustive.
Table 1. Some common safety categories
Category Type What does it mean? Additional comments.
Flammable Physical A liquid, gas, solid or solution is flammable. Usually a spark or Most organic liquids are flammable, so the risk of starting a fire is always present and necessary steps must always be taken to avoid such risks.
Oxidizing Physical The compound is a strong oxidant. If you mix an oxidant with a reductant, usually a strongly exothermic reaction happens. Care should be taken to avoid this. Some oxidizers must also be disposed of in a particular way.
Pyrophoric Physical The compound ignites spontaneously in contact with air. Pyrophoric substances are very dangerous, and usually require handling in inert gas.
Toxicity Health A substance is toxic.This category is vast, and requires specific Most organic molecules are toxic by numerous modes. We have dedicated a separate section to toxicity below this table.
Skin irritation Health A substance irritates the skin and/ or mucus membranes Most organic chemicals irritate the skin. If the compounds are solids, gloves often offer sufficient protection, assuming the substances do not permeate
Eye irritation /damage Health A substance irritates or permanently damages the eye Because the eye is so delicate, most chemicals and substances damage the eye. One of the most basic safety feature of any lab, is that the people working
Carcinogen Health A substance that is suspected or known to cause cancer in mammalian cells. This is a very important category, as substances that are either suspected or known carcinogens must be handled with extreme care. It is important to note the two subcategories of suspected and known. In practical lab work, you should deal with both these chemicals the same way: very carefully
Reproductive toxicity Health The substance damages the reproductive system This is a very important category, as chemicals that damages the reproductive system can have very detrimental effects.
How do you use the SDS information in your preparation phase?
1. Read all SDS for all chemicals encountered in the experiment. Pay particular attention to category 2, which contains a summary of the hazard statements for the chemical.
2. Analyze the information in the SDS. Think carefully about what measures can (and should!) be taken to avoid exposure, and to minimize the potential risks associated with the chemical.
3. Write it down. Include the major hazard statements for each chemical in your notebook. Also, include any important analysis you have done. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/01%3A_HOW_TO_PREPARE_FOR_AN_ORGANIC_CHEMISTRY_EXPERIMENT/1.02%3A__SAFETY_IS_KEY-_HOW_TO_FIND_RELEVANT_SAFETY_INFORMATION_AND_READ_SAFETY_DATA_SHEE.txt |
Let us use all the information we have accumulated in the last sub-chapter, and use a real case example to find the relevant information we need. We will analyze the SDS of toluene, as a specific example.
We will focus on category 2 in the SDS, which is a good starting point for most SDS. This category contains hazard information. Category 2.1 contains the most important summary of the hazards and safety information associated for that chemical. For toluene, we have the following seven items:
1. Flammable liquids (Category 2), H225
2. Skin irritation (Category 2), H315
3. Reproductive toxicity (Category 2), H361
4. Specific target organ toxicity -single exposure (Category 3), Central nervous system, H336
5. Specific target organ toxicity -repeated exposure (Category 2), H373
6. Aspiration hazard (Category 1), H304
7. Acute aquatic toxicity (Category 2), H401
Each of these items should go in the lab notebook, if toluene is used for that experiment. As a deeper analysis, we will also go through each of these seven items, and evaluate the relative hazard. As outlined in the previous sub-chapter, our aim is to analyze each item and think carefully about measures that we should take to avoid exposure and risk.
Items 1 and 2 (flammable liquid and skin irritation) are, as we have seen, typical for most organic chemicals and we follow good laboratory hygiene and work in a well ventilated fume-hood. We also wear goggles and a lab coat at all times. Item three, which deals with reproductive toxicity,3 is worth making a note of. That is an effect that we absolutely should be aware of when dealing with the chemical. Items 4 and 5 are also noteworthy. These two items say that the chemical has target toxicity both for single and repeated exposure. The first is more serious than the latter, because only one exposure of the chemical can have central nervous system toxicity. We therefore want to do everything we can to avoid contact with the chemicals. A further analysis shows that nitrile gloves offer fair protection, so we want to make sure to wear these gloves whenever handling toluene.4 Item 6 is common for most organic chemicals, as inhalation of most organic chemicals is damaging. Working in the fume-hood is one measure to limit inhalation. Item 7 is worth not- ing as well, and following proper chemical handling,5 we make sure to never pour chemicals down the drain.
The above analysis might be deeper than you are used to, but it is very important for your long-term health and well-being. One of the main goals of a chemistry lab course is to learn about safety in that environment.
3 Reproductive toxicity means a substance that in some way interfers with reproduction. It includes effects on sexual function, and fertility in both males and females, as well as developmental toxicity in the offspring.
4 Chapter 3 covers the use of gloves
5 Chapter 3 covers safe handling of chemicals
In conclusion, several safety-related issues are important for preparation.
1. Find and read SDS for all chemicals used
2. Find the main safety and hazard statements for each chemical used
3. Include the main safety and hazard statements in your note book
1.04: THE NOTEBOOK MAKING SURE THE NUMBERS ADD UP
At PSU, it is required to use and keep a notebook. This notebook will be used both in the preparation of the experiment, and during the experiment itself. We will focus on how to use the notebook in the prepara- tion phase of an experiment. In chapter 4.4 we will discuss the use of the notebook during the experiment.
When stepping into the lab, your notebook should contain the following:
1. A title and a balanced reaction scheme, if a synthesis is planned. If not, the relevant chemical structures of the compounds of interest should be shown.
2. A flow-diagram that outlines the experiment, where you have included the most relevant information.
3. A table or overview that shows all chemicals handled, and their risk and hazard statements.
4. A synthesis table that shows the quantities of the reagents/reactants that you plan to use, with room to fill in the quantities you actually used.
5. A table that shows the properties of the reagents/reactants and reaction solvents used in the procedure.
The synthesis table should contain the most important physical properties of the reactant(s), reagent(s) and product(s). This table should contain the quantities you should measure, with room to add your actual measurements. This format allows you to order all the necessary measurements in the same place. An example is shown in table 2. There are many exam- ples where this information is important in lab. If you for example measure 0.4 mL of benzaldehyde, instead of the 0.3 mL the procedure called for, you need to scale the procedure up. That means that the amount of NaBH4 also must be changed. For this process, you need the density of benzaldehyde, the molecular weight of benzaldehyde to find the moles, and the molecular weight of NaBH4.
Table 2. Example of a synthesis table for reagent(s), reactant(s) and product(s)
Compound Mw [g/mol] d [g/mL] V [mL] m [g] moles [mmol] Y ield [%]
Benzaldehyde
NaBH4
Benzyl alcohol
106.7
37.8
108.1
1.044
-
1.044
0.30 _____
-
-
0.31 _____
0.15 _____
0.21* _____
2.9 _____
3.9 _____
1.9*_____
-
-
65.6*_____
* The values for benzyl alcohol is based on the procedure, that states that a yield of 65.6% is expected.
The table for physical properties is used not only for the reagent(s), reactant(s) and product(s), but also reaction solvent, work-up chemicals and others. This table is very handy in the lab, as the physical properties of the chemicals play a large part in the work-up of any reaction.
Table 3. Example of a table for physical data
Compound State Boiling point [°C] d [g/mL] Solubility in water [g/100 mL] Solubility in diethyl ether
Benzaldehyde
NaBH4
Benzyl alcohol
Diethyl ether
Methanol
Liquid
Solid
Liquid
Liquid
Liquid
178
-
205
36
65
1.044
Not relevant
1.044
0.71
0.79
0.3
Decomposes
3.50
6.05
Soluble
Soluble
Non soluble
Soluble
-
Soluble | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/01%3A_HOW_TO_PREPARE_FOR_AN_ORGANIC_CHEMISTRY_EXPERIMENT/1.03%3A_A_SPECIFIC_SDS.txt |
Organic chemistry is an art-form and a craft. In research, we often talk about candidates that “have it”, the “magic touch” to get a compound analytically pure, or solve a very difficult synthetic challenge. In this chapter we will go through some common techniques that you will encounter in the organic chemistry labs. The aim of this chapter is not to give you a comprehensive introduction in the theory behind these techniques, but to highlight technical aspects to make sure you also develop this “magic touch.” We will discuss recrystallization, distillation, liquid-liquid extraction, TLC, chromatography and lastly sublimation.
02: COMMON ORGANIC CHEMISTRY LABORATORY TECHNIQUES
Recrystallization is a laboratory technique for purifying solids. The key features of this technique is causing a solid to go into solution, and then gradually allowing the dissolved solid to crystallize. Sounds easy, doesn’t it? It is actually a very challenging process to get completely right. The goal, is to obtain a compound in high purity as uniform crystals. Recrystallization is therefore a purification technique.
The key features necessary for a successful recrystallization process, are a very controlled temperature decrease and sufficient time. Because most solids have a better solubility at higher temperatures, we can sat- urate or almost saturate a solution at high temperature (usually the boiling temperature of the solvent), and then slowly allow the solution to reach room temperature. As the temperature starts to decrease, so does the solubility of the compound. As the solubility decreases, the solution at some point becomes supersaturated and crystals will start to form.
Before we move on, let me address the main problem associated with crystallization: the formation of precipitate, versus crystals. A precipitate is simply a mixture of compounds in the solution that crash out. This can happen for a variety of reasons, but a student may have taken the very hot solution and placed it directly on a cold surface to cool (a process called “shock cooling”) or even plunged the hot solution into an ice bath. A precipitate may not pure, because it can contain several compounds. Crystals, however, are often composed solely of one compound. It is very easy to get a precipitate, but very difficult to get crystals.
We already mentioned temperature and time being key factors for successful recrystallization. Precipitation typically occurs when the temperature has not been lowered gradually. Furthermore, we must not disturb the solution (or its container) as it cools (even though it is tempting). Disturbing the solution can break up any seed crystals6 that have started growing.7
6 A seed crystal is a small piece of crystal from which crystallization occurs.
7 A typical challenge in a research laboratory involves recrystallization to form a crystal suitable for x-ray analysis. Such crystals are often referred to as single crystals, and not only must they be completely pure, but also the crystal lattice and growth must be highly ordered. This task can be monumental, as very small variables can be detrimental to the growth of a single crystals. During grad-school, a post-doc told me to leave the crystals growing in a room where nobody ever went, as even the vibrations of footsteps or laboratory equipment could be detrimental.
Please also note that some compounds simpl crystallize more easily than others. More rigid molecules are, as a rule, easier to crystalize.8 Rigid, in this context, mean compounds that contain fewer bond capable of undergoing internal rotation, so that there are fewer possible conformers possible.
Let us go through a recrystallization process, focusing on technical aspects and trouble shooting.
How to perform a recrystallization:
1. The crude impure solid is dissolved in hot solvent. If some solid remains undissolved after adding solvent, it is likely to be an impurity and should be removed by filtrering the (hot!) solution.
Typical problems: Adding too much solvent so that the product does not crystallize later. Filtering the hot solution too slowly so that the solvent cools and the solid starts crystallizing in the funnel and/or on the sides of glassware.
2. The solution is allowed to stand without being disturbed. The temperature is allowed to gradually drop, leading to growth of large crystals. The flask should not be placed on a surface (it will shock-cool the solution), but either placed in an insulated jar, or clamped.
Typical problems: Crystals do not form at all (too much solvent), precipitate forms instead of crystals (temperature has dropped too quickly, or an oil forms).
3. The solution is allowed to stand until crystallization is complete.
Typical problems: Crystallization can be a slow process, and impatience can lead to low recovery.
4. The solution is placed in an ice-water bath to lower the temperature even further, and allow more crystals to form. At this point, most crystals should already have formed.
5. The crystals are filtered and air-dried.
How can we tell if a recrystallization has been a success? Sim- ple visual inspection is a good start: The crystals should have shiny surfaces and catch the light. They should appear uniform, and you should have crystals of similar structure and size. A melting point analysis should also show a narrower and elevated melting point range compared to the crude material.
Like any purification technique, recrystallization has some limitations. First of all the compound you crystallize should be a solid at standard conditions. Greases, waxes and oils cannot be crystallized at standard conditions. Secondly, the crude material should be mostly pure. There is not any minimum purity standard for any crude material, because the success of any recrystallization depends on the identities of the other constituents and their respective solubilities, but in general the crude material should contain about 80% of the desired compound.
The crude material is transferred to a suitable crystallization vessel. The crude material is dissolved in a solvent, and gently heated.
The solution is allowed to gently and slowly cool down. Notice the crystals growing in the solution. The solution is cooled to room temperature, leading to the formation of large crystals.
8 I have prepared new compounds that I have tried on and off to recrystallize for the better part of a whole year, without success. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/02%3A_COMMON_ORGANIC_CHEMISTRY_LABORATORY_TECHNIQUES/2.01%3A_RECRYSTALLIZATION.txt |
Distillation is a purification technique for a liquid or a mixture of liquids. We utilize the difference in boiling points of liquids as a basis of separation. The core of a distillation process, is selective evaporation and condensation of particular components. Our overall goal is to evaporate and condense only one component from a mixture, but to attain this goal, we must allow many, many cycles of evaporation and condensation to take place. This process gradually enriches the vapor phase in favor of the most volatile component. After a sufficient number of evaporation and condensation cycles have taken place, the final condensate contains a liquid that is en- riched in the more volatile component.
Distillation is easier to understand if we envision a spesific mixture of two liquids, say diethyl ether and ethanol. The boiling points of the two liquids are 36°C and 78°C, respectively. When we boil this mixture, we observe the following: the entire mixture (both compounds) boils, but the vapor phase is enriched in the more volatile component (diethyl ether). As this vapor mixture rises, cools, and condenses, the resulting liquid is enriched in diethyl ether too. If we attach a column to the flask so that the vapor enters this column, the condensing liquid will be heated by rising vapors, and it will boil again producing a vapor that is even more enriched in diethyl ether. The higher the column, the more times this cycle of evaporation-condensation can be repeated, and the higher up we sample the vapor, the more enriched the vapor phase will be in the more volatile component (diethyl ether). Ideally, with a long enough column, one could obtain a vapor that is nearly pure diethyl ether, and leave behind a liquid that is nearly pure ethanol, the less volatile component.
Let us discuss a typical distillation apparatus (shown above) We start with a flask containing the solution (often called a still pot), which is heated. You can see from our discussion above, that a key feature of a distillation apparatus must be a column, where many cycles of condensation and evaporation can take place. The column can be just a short tube (an “unpacked” col- umn), which is the basis of a simple distillation. This is typically less efficient. The column can be packed with an inert material, which is the basis of a fractional distillation, and typically constitutes a more efficient distillation. The inert material, such as copper sponge or glass beads, provides a large surface area, thus allowing many more evaporation-condensation cycles.
The column is attached to an adapter. In the top of this connector is the thermometer, which is used to read the temperature of the vapor, just as it condenses. The temperature reading is important, because, at normal conditions, the temperature of the vapor passing through is the same as the boiling point of the substance being collected. Vapor that passes the thermometer condenses in the condenser, a double-walled tube that is cooled by water flowing through the outer layer, and drips into the receiver.
Before moving on, let us look at benefits and drawbacks to a simple distillation versus a fractional distil- lation. We have already said that the simple distillation is less efficient at separating liquids, because there is a smaller surface area inside the column, but it is usually much faster. For mixtures that contain only one volatile component, a simple distillation can be more than sufficient. The fractional distillation is more efficient, and is suited for mixtures of volatile liquids. The closer the difference in the boiling points, the more demanding the distillation. The drawback is that fractional distillations typically take longer, because we want to achieve pseudo-equilibrium between vapor and liquid throughout this system. Slow boiling and ample time is important to achieve this goal.
Sounds easy, right? What can go wrong? Poor separation is the most likely problem you will face. This means that the fractions obtained are impure, and often contain traces of the other liquids found in the original mixture. We will assume that we are performing a fractional distillation, and that we have several liquids that must be separated. Let us look at some of the technical aspects that are important for acceptable separation, and some key features for the assembly of the distillation apparatus.
Troubleshooting Distillation
1. The distillation result is poor: the fractions obtained are not of acceptable purity.
Typical problems: Distillation too fast. The components require time to separate. We need many evaporation-condensation cycles for good separation, and pseudo-equilibrium between vapor and liquid throughout the system. If we supply too much energy to the system (i.e., too much heat), we allow even the less volatile components enough energy to keep evaporating. The vapor phase is therefore not enriched with the more volatile component.
A certain amount of time is also required. Allowing the mixture to gently reflux for a while (30 minutes) before gradually increasing the energy supplied to the system through heating is typically a good strategy.
2. You collect distillate, but the temperature reading does not correspond to the boiling point of the component. Typically, the temperature reading is much lower.
Typical problems: The thermometers found in a standard organic chemistry teaching laboratory is not of the highest quality, and they are often mistreated through many terms. They should therefore be calibrated by reading boiling distilled water. Furthermore, the location of the thermometer bulb is essential. If the bulb is too high, the vapor condenses before the thermometer can read the temperature. Typically, the bulb should rest right above the lowest part of the adapter (see the set-up above).
3. Even though the liquid in the still pot is boiling, no distillate is being collected.
Typical problems: Insulation is a key feature to consider whenever distillation does not occur. Have a look at the entire apparatus and see where the vapor has reached, by looking for drops of condensation. Insulating the top of the still pot, as well as the column and top of the three-way adapter can be beneficial. Remember that the vapor must heat the glassware to the boiling point of the condensing liquid before it can evaporate again and rise through the apparatus. Because the distillation is performed in a fume-hood, the constant air-flow inside the hood also cools the apparatus.
4. Nothing distills but the amount of liquid in the still pot mysteriously goes down.
Typical problems: There is a leak in the system, or several leaks. The vapor is escaping though openings between joints. Make sure that all joints are properly sealed. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/02%3A_COMMON_ORGANIC_CHEMISTRY_LABORATORY_TECHNIQUES/2.02%3A_Distillation.txt |
Liquid-liquid extraction involves the exchange of certain com- pounds between two solvents that are immiscible or only partially miscible. Liquid-liquid extraction is also very commonly used for washing an organic phase, for example to remove inorganic compounds, or to protonate or deprotonate bases or acids, respectively, so they become soluble in the aqueous phase. A very typical extraction flow diagram is shown below, where a reaction mixture is quenched with water, extracted (several times), washed with brine, dried, filtered and finally evaporated to yield a crude product or a pure product.
Liquid-liquid extraction (we will refer to it simply as extraction from now on) is typically conducted with one aqueous phase (either pure water, or an aqueous solution) and one organic phase. It is important to note that the desired compound (usually an organic molecule) can in theory be in either phase. It depends on both the nature of the compound, but also the nature of the aqueous phase. We will first examine some typical scenarios involving extraction, then move on to discuss technical details to keep in mind.
1. Extraction of neutral compounds. If the desired organic compound is neutral (i.e. is neither acidic nor basic), the extraction sequence usually involves simply extracting with an organic solvent several times.
2. Extraction of acidic compounds. If the desired compound is acidic, we can selectively deprotonate that compound by using an aqueous base. This will pull the deprotonated compound into the aqueous phase. Because very few organic compounds are soluble in water, we can discard the organic phase which now contains any byprod- ucts and/or unreacted starting materials. Acidification of the aque- ous phase can precipitate the desired product.
3. Extraction of basic compounds. If the product is basic, we can perform a sequence very similar to the acidic compound above. We can protonate the basic compound by using an aqueous acid, pulling the protonated compound into the aqueous phase and discarding the organic phase. Neutralizing the acidic phase will depro- tonate the basic compound, which may precipitate, or may require a second extraction with an organic solvent.
The typical apparatus used in an extraction, is the separatory funnel. These come in many different sizes, but the typical size is 100 mL, which you will encounter in several labs at PSU. The funnel is equipped with a top and a nozzle, and the nozzle can be closed or opened with a stopcock. The funnel is narrow at the nozzle, which is a structural feature that allows us to accurately separate the two phases.
How to perform an extraction. For simplicity’s sake, we will assume that the product is a neutral compound.
1. Make sure the stopcock is closed. Add a small amount of water or the aqueous solution you are using. Some- times the stopcock is improperly sealed, and you will find that it leaks. You would rather know that now, rather than after the organic material has been added.
2. Add the reaction mixture, as well as water if necessary, and the required organic solvent. Make sure that the solutions added do not touch the inside of the top of the funnel, as solids can crystallize here and prevent the stopper to seal properly, leading to leaks.
3. The funnel is closed with a stopper, turned on its head, and shaken very carefully. The stopper is always supported by your hand.The pressure is released after each round of shake by opening the stopcock (point it upwards into the hood, but never towards yourself or someone else), and then closed again. Now the two layers are shaken vigorously. It is imperative that the two phases are mixed, because no compounds will exchange without a good mixing.
4. The funnel is rested in a ring stand, and the stopper removed. The two phases are allowed to separate. The phases are removed by draining them out, one at a time.
5. The aqueous phase can be extracted again, up to three times. The combined organic fractions are typically dried with a drying agent,9 then filtered, and the solvent evaporated.
9 A classic problem is that a student either has very little product after extraction ( because of improper mixing of phases) or that the student requires a vast amount of drying agent (not allowing the two phases to separate properly).
One common problem with extractions is one that does not arise from improper technique, but from the annoying behavior of certain compounds: they form emulsions. These usually appear between the two lay- ers, and are chemically composed of the two solvents, and other compounds found in the original solution. The emulsion can resolve itself over a few minutes, or be very persistent. In the latter case, it can sometimes be a good strategy to drain the emulsion out in the phase that does not contain the product, and then re-extract it several times
Let us say that we are extracting an aqueous solution with the organic solvent diethyl ether. Let us also say that the desired compound is organic, and in this example will be dissolved in the organic phase. A very persistent emulsion is formed. If we drain the aqueous phase and as little of the organic phase as we can (while still draining the emulsion out), we can obtain some organic phase without emulsion that is set aside. We then add the phase back to the separatory funnel and re-extract the aqueous phase that still contains some organic phase, as well as emulsion again. This usually disperse the emulsion. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/02%3A_COMMON_ORGANIC_CHEMISTRY_LABORATORY_TECHNIQUES/2.03%3A_LIQUID-LIQUID_EXTRACTION.txt |
Thin-layer chromatography (or TLC) is a quick, cheap, and reliable way to evaluate the composition of a sample, and the identity of a given compound. The TLC plate works by chromatographic principles: a mobile phase (solvent or solvent mixtures) will climb up the plate material (stationary phase). Compounds that are less polar will wander farther on the plate, because they experience les1s attraction to the stationary material. Compounds that are more polar will wander lower on the plate, because they experience more at- traction to the stationary material. This process is separating the different compounds present in the crude sample
Most compounds that we handle in the o-chem labs at PSU should give a single spot on the TLC plate. Ideally, each compound in a mixture will produce a distinct spot so a sample with two compounds will give two different spots, and so on.
An important property of any compound, is its Rf-value (retention factor). In simple terms, this value is an indication of how far up a TLC-plate a compound has wandered. A high Rf -value indicates that the compound has travelled far up the plate and is less polar, while a lower Rf -value indicates that the compound has not travelled far, and is more polar. This value is easily calculated by measuring the distance the spot has wandered, and dividing this by the distance the solvent has traveled. The Rf-value is dependent on both the compound and the solvent used for development. If we analyze two compounds and they give the same Rf -value with the same solvent system, the two compounds are most likely identical.
The apparatus for doing a TLC-experiment is very simple. We need a developing chamber where the mobile phase is kept. The chamber should be sealed, and although there are special TLC-chambers one can use, a beaker covered by a watch glass to create a seal often suffices.
How to prepare a TLC plate:
1. A horizontal line is drawn at the bottom of the plate (5-8 mm from the bottom edge) and another horizontal line is drawn at the top of the plate (8 mm from the top edge). A gentle soft and a very soft pencil must be used because too much pressure or harder pencils will crack the very thin coating on the plate, leading to unacceptable TLC results after development (ink pens should never be used because the ink will wander up the plate).
2. Vertical ticks are drawn on the bottom line, depending on how many samples are analyzed. These are usually labeled (A, B, C, for example)
3. Solutions are prepared of all the samples that must be analyzed. These must neither be too concentrated, nor too dilute and a concentration of about 5% is a good starting point. The samples are applied to the ticks with capillary tubes without damaging the coating on the plate.
4. The TLC plate is placed into the developing chamber, and care is taken to make sure that the bottom edge of the plate is perpendicular to the solvent. If not, parts of the plate will overdevelop, and we will get poor resolution and unclear results.
5. The chamber is sealed with the watch glass and the solvent front is allowed to climb up to the topmost line. We then remove the plate and allow it to dry before we observe the plate under UV-light.
We will discuss two common ways to use a TLC- analysis. The first is to verify the identify of a com- pound. A quick TLC analysis can be used to identify whether or not an unknown compound is the same as another known compound. We typically make three applications: one of the unknown sample, one of a stock (known) compound, and a third with one spot of both (called a co-spot). If we find that the two spots have the same Rf-values, and the third spot only shows one spot, the two compounds are identical.
The second common way to use a TLC- plate, is to monitor a reaction. We can use TLC-analyses to see whether or not a reaction has gone to completion. We typically see the disappearance of the starting material(s) and the emergence of a new spot that corresponds to the product. We usually have to make controls as described above, with spots of the starting material(s) to learn their Rf-values.
Like any of the techniques we have discussed in this chapter, TLC-analyses require a fair amount of care to get right. The most likely problems can be traced to the application and preparation of the TLC-plate. Let us have a closer look at some TLC plates that are of inadequate quality and their causes.
2.05: SUBLIMATION
Sublimation is a purification technique for solids and in the context of this book, for organic compounds with lower melting points. Sublimation describes the process of a solid becoming a gas, without passing through the liquid state. The gas phase is then typically crystallized on a cold surface. However, many sublimations require reduced pressure, and we will primarily focus on that variation of sublimations. The reason for many sublimations happening under reduced pressure, is because the sublimation point is decreased with decreasing pressure. To allow organic compounds to sublimate, and not simply melt, a reduced pressure is often necessary.
A crude, but efficient sublimation apparatus can be made from of a filter flask, where the side neck is connected to an aspirator or vacuum trap, and the neck is equipped with a cold finger (a tube with ice water, or other cooling material). The impure solids are placed in the bottom of the filter flask, and the cold finger is inserted. While an aspirator or pump is reducing the pressure inside the filter flask, the material is carefully heated on a hot plate (take care to avoid melting or boiling the material). The organic compound will start to sublime forming a gas. Once this gas reaches the cold finger, it will immediately crystallize on the cold finger, where it can be collected.
Troubleshooting sublimation
1. The crude material does not sublime.
The most common reasons are insufficient heat, and/or the pressure is not low enough.
2. The sample sublimes, but the material crystallizes on the sides of the filter flask, and not on the cold finger.
The bottom of the flask might be hot enough, but the sides are not. The material is allowed to sublimate, but cannot reach the cold trap and is able to crystallize earlier. Insulation of the lower part of the filter flask can be appropriate.
3. I don’t know where the sample went!
The crude material was added, but during the process of heating the flask, the material is gone, and there are no crystals. There can be several reasons, but the most likely contender is that the cold finger did not efficiently work as a trap. Lowering the position of the cold finger is a good solution.
4. The crude material is boiling!
The system can be too hot, and less energy should be supplied to heat the system. Potentially the pressure should be reduced.
5. The product crystallizes on the cold finger, but looks pasty and wet and not crystalline.
The ice cold water in the cold finger has been in the finger for too long. The cold surface has started condensing water from the atmosphere, and once the product starts to reach the cold surface it is met by water. In general, the ice cold water should be added to the cold finger right before the sublimation starts to eliminate this problem. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/02%3A_COMMON_ORGANIC_CHEMISTRY_LABORATORY_TECHNIQUES/2.04%3A_TLC_-ANALYSIS.txt |
In chapter 1 we discussed some of the possible dangers associated with exposure to the chemicals used in organic chemistry labs. In this chapter, we will examine techniques to reduce such exposures. We will also discuss waste and waste management, safety in the lab, and lastly the mental health of the student during the lab experience.
03: GETTING YOUR HANDS DIRTY - CHEMICAL HANDLING WASHING WASTE AND SAFETY
When you start to think about ways to handle chemicals to ensure your safety, you likely think about wearing gloves to minimize contact between your skin and the chemical. Let us get one thing straight im- mediately: the gloves10 may offer only limited protection. This is an important fact because a common misconception is that the gloves you find in lab effectively protect the skin from solvents and other substances handled. This is simply not true.
To give you one example, the nitrile gloves that are typically made available are easily penetrated, and eventually broken down, by two common (and hazardous) solvents, acetone and dichloromethane.
There are several excellent online resources to find information about the permeability of various solvents and reactants.11
10 The most commonly employed gloves in many teaching labs, are nitrile gloves. There are many different gloves, in general the more cumbersome they are to wear, the better protection they offer. Unfortunately, there are no gloves on the market that are both practical and offer protection against most organic chemicals encountered in undergraduate organic chemistry labs.
11 Ansell has a very good overview. www.ansellpro.com/download/A...esistanceGuide. pdf
Another reason we wish to alert you to the limitations of lab gloves, is that wearing them can provide a false sense of security, and this can be very dangerous. Imagine you are wearing gloves, First, you now think that they are protecting you. You are perfectly safe. As a result, you might be less careful, or you may pay less attention to the risks involved in tyour experiment. Not only that, once you put on gloves, you probably pay little attention to how you handle the gloves themselves, what we call “glove hygiene”.
Common examples of poor glove hygiene (things you must always avoid) include: touching (and contaminating) your face, skin, hair, or clothes with your gloves, touching (and contaminating) objects like phones, pens, and notebooks, that you will use outside the lab, failing to check gloves periodically for leaks and cracks, and failing to remove gloves each time you leave the lab
In general, gloves should only be worn for short periods of time, and exchanged routinely, especially whenever a chemical has been in contact with them. The contaminated exterior of the gloves should never come in contact with your skin. Whenever you are done with your gloves, you should remove them in a safe manner and wash your hands (never use bare hands to remove gloves).
3.02: GETTING THE CHEMICALS TO YOUR REACTION FLASK- POURING, SYRIN
Laboratory chemicals are typically stored, and dispensed, in a variety of containers and locations. Therefore, early hurdles in every experiment are 1) figuring out how to obtain the desired amount of the chemical, 2) figuring out how to transfer this material safely from the location where it is obtained to the location where it will be used, and 3) figuring out how to put the materials in the apparatus.
Solids: Solids are usually stored in wide-mouthed bottles and weighed before use. The solid is weighed us- ing a weigh boat, or a piece of weighing paper, and it is usually possible to transfer the solid from the bottle to the boat (or paper) using a clean spatula or lab scoop. A good laboratory practice is to remove the weigh boat (or paper) from the balance each time you add solid to it, and then return the boat (or paper) to the scale. This practice reduces the likelihood that any chemicals will get spilled on the balance.
Liquids: Liquids are typically more difficult to handle, and the appropriate method will depend on how much liquid is needed and the specific properties of that compound. In most cases, it will be safer and more convenient to obtain a specific volume instead of weighing the liquid. For reagent quantities, a pump sys- tem is often employed to deliver calibrated volumes, or a graduated cylinder can be used. Solvents, whether needed for reactions or extractions, tend to be used in larger quantities and a graduated cylinder will nearly always be more conventient.
3.03: WASHING YOUR GLASSWARE
Cooking a dinner creates dirty dishes and doing an organic chemistry lab experiment creates dirty glass- ware. Cleaning your glassware properly is an important aspect of the lab experience. This should be a priority, as impurities or trace amounts of chemicals from previous labs can ruin or complicate your experiment. Starting with visibly dirty or contaminated glassware is also considered poor lab practive, as the outcome of the reaction cannot be trusted: was the outcome due to the reagents you combined, or was it in some way influenced by the contaminants?
We will cover two good cleaning protocols, one for glassware that has been used for organic samples, and one for inorganic solutions.
A good cleaning protocol for glassware that has contained organic solutions or samples: the glassware is rinsed with acetone in the hood, and the contaminated acetone solution is then poured into the organic waste. The glassware is then allowed to air dry. If the glassware is still dirty, try this: wash with soap and water, rinse with distilled water, then rinse with acetone again, and air-dry.
This might seem like a very involved process, but in most cases the glassware is perfectly clean after the initial acetone rinses and requires no further cleanings. Also, by performing an acetone rinse first, it is usual- ly safe to remove still dirty glassware from the hood for further cleaning because the acetone rinse will most likely have washed away potentially harmful organic materials. Then, when you wash with soap and water, you do not need to worry about working with potentially harmful organic materials outside of the hood, or pouring them into the drain.
Glassware that has contained only non-hazardous inorganic solutions (such as dilute solutions of NaHCO3, HCl, NaOH, NaCl, and so on) can be rinsed with water in the sink, and then soap, if necessary, washed with soap and water. The glassware is then lastly rinsed with distilled water and allowed to air dry. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/03%3A_GETTING_YOUR_HANDS_DIRTY_-_CHEMICAL_HANDLING_WASHING_WASTE_AND_SAFETY/3.01%3A_GLOVES_-_THE_GOOD%2C_THE_BAD_AND_THE_UGLY.txt |
Waste is a philosophical concept, because we determine, somewhat arbitrarily, when a solution or a sub- stance has become waste. Once you no longer have a use for a chemical, it becomes a ‘waste’ that must be properly disposed of.
You might argue that this is a flawed concept (pouring 20 mL of NaCl-solution into a beaker, for example, does not change it from said solution to waste), but we employ this waste policy to ensure that we never contaminate any of the stock solutions, or samples.
In the PSU organic chemistry labs, we have three distinct types of waste containers:
Organic waste: Anything organic goes in this waste container, including organic solvents, solids that are left over from reactions or analyses, products obtained, organic extraction solvents and so on.
Aqueous waste: All aqueous solution used or obtained are poured in this waste container. You might be tempted to pour it straight in the drain, but the solutions could have been in contact with organic material. Let us for example say that you wash an organic extract with a saturated NaCl solution (brine), and you drain it out. This has been in contact with organic solvent and material, and even if we consider a relatively low solubility, the aqueous solution will still contain some organic molecules.
Solid waste: Solids, such as drying agents, are disposed of in these jars, but so are TLC plates, and any paper tissues that have been in contact with your fume hood.
Obviously we provide these waste containers as part of our safe laboratory hygiene practices, and there- fore we wish to train our students in the proper disposal of chemicals, but you might ask, “What are the implications of not following these procedures?”
Let’s look at the alternative.
As we have covered earlier, most organic chemicals have a number of safety hazards associated with them, either immediate, long-term, or both. Failing to dispose of these chemicals properly might create health hazards for the workers who unknowingly come into contact with these chemicals. Also, many chemicals that might seem relatively safe in a lab environment, are bio-accumulated by other organisms. These chem- icals can produce disastrous effects for when they enter the environment beyond the lab. Aquatic organisms are especially vulnerable, and many organic chemicals have harmful toxicity with long lasting effects for these organisms.
The handling and disposal of chemicals is therefore not only an important aspect of the day-to-day work in the lab, but also for the environment, ecology, and the workers and other organisms that surround us.
3.05: WHAT CAN GO WRONG? SOME SAFETY SCENARIOS TO CONSIDER.
Even with full awareness of safety, and the utmost of care, accidents can happen. It is important to consider some safety issues that can arise, and discuss how best to approach them. One of the most important things to note, is that your health is the most important consideration for everyone in the lab: instructors (TAs and professor in charge), support staff and lab mates. If you have an accident, you must alert your TA immediately. Do not try to remedy the problem on your own, and do not try to cover it up. Every second may make the difference between a good and a bad outcome when an accident has happened, and it is imperative that you get the help and assistance you need, immediately. Your safety is always more important than your pride.
3.06: STRESS AND ANXIETY
It is not unusual to feel worried and anxious about your performance in the organic chemistry labs. Your concerns might range from anticipated difficulties in the lab (handling of chemicals, dangerous substances) or just your desire to do a good job.
Here are a few strategies that can help mitigate any feelings of anxiety and stress:
1. Plan ahead. It sounds simple, but as we have seen from the first chapter, planning is the most important step before undertaking an experiment. Remember to visualize the experiment prior to the lab so you are mentally prepared. This will prepare ou for the things you need to do and the decisions you will need to make.
2. Calm yourself down. If you feel overwhelmed or anxious, a very effective trick is to make sure your appara- tus and chemicals are taken care of, and then simply taking a step back and taking five slow, deep breaths. This calming exercise can mitigate any imminent feelings of discomfort, and help you reassert yourself before moving on.
3. Communicate with your TA. If you find yourself feeling overwhelmed or deeply uncomfortable while in the lab, communicate this to your TA. Your TA will help secure your apparatus and material, and potentially have a quick chat about your current state of mind. If necessary, the TA, student and professor will work together to find a plan to assist the student.
4. Take a break. If the experience simply gets overwhelming, tell your TA, and take a short break. Leave your lab coat and goggles behind, and go out in the hall, or take a bathroom break.
Remember, lab should be a positive, engaging and interactive learning experience. Although it is natural (and sometimes even helpful) to compare yourself to peers, experiments by their very nature have a variety of outcomes, and you can learn something from all of them. The primary objective in the lab is to focus on what you are doing.12
12 It is also worth mentioning that we have seen incorrect advice spread in the lab through this interaction. This happens because of discussion, when one individual presents an idea or concept to his or her peers, and the other students accepts it. Although someone is free to give their opinion, it is always worth taking it with a grain of salt, and ask your TA if you are unsure. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/03%3A_GETTING_YOUR_HANDS_DIRTY_-_CHEMICAL_HANDLING_WASHING_WASTE_AND_SAFETY/3.04%3A_WASTE.txt |
We have talked about preparation, safety and all the processes that should happen before an experiment. Now we will move on to the actual experiment itself. The aim of this chapter is to contextualize the organic chemistry experiment, and discuss the different phases of the experiment. You will find that most experi- ments follow the same general outline, which should make it easier to prepare.
04: HOW TO SURVIVE AN ORGANIC CHEMISTRY EXPERIMENT
There are typically three phases associated with an organic chemistry experiment:
phase 1: Set-up and reaction
The reaction apparatus is set up, and the reactants and solvents are measured and mixed together. The reac- tion mixture is stirred, and heating the reaction mixture is often necessary.
phase 2: Work-up and purification
The reaction mixture is often quenched with water, and then work-up is performed. Work-up can be a combination of the purification techniques discussed in chapter 2, such as extraction, filtration and/or distillation. The aim of this phase is to separate the main product from the other components, and isolate it in high purity.
phase 3: Characterization
The obtained product is analyzed by means of different techniques, such as melting point analyses, and IR spectroscopy. The aim of this phase, is to obtain information about the identity and the purity of the product.
Keeping this organization in mind can be valuable, as it firmly cements the purpose of the experiment. Each phase has distinct purposes, which all funnel into the overall goal of the experiment.
The set-up and reaction phase is all about setting up the conditions of the experiment. We want a certain chemical reaction to take place, and must add reactants in exact proportions, with solvents and potential catalysts. The reaction conditions, i.e. the temperature, the time and other parameters, are usually defined at the outset. This phase is also crucial in determining the success of the experiment. Without weighing out correct quantities of reactants, for example, or heating the mixture to the correct temperature, the desired reaction may not find place, or competing side-reactions might happen instead
The work-up and purification phase usually involves the most skill and technique. After the reaction stage, we have a mixture of potentially unreacted starting materials, products, solvents and byproducts. At this stage, a series of purification techniques are employed to isolate the product, from this mixture.
The analysis and spectroscopy phase has one aim: to establish the identity and the purity of the compound prepared. Several techniques are involved, usually melting point analyses (if the product is a solid), and IR spectroscopy. In some cases NMR spectroscopy is also used in the teaching lab.
Please notice that while information about the final purity and identity of the compound is obtained in phase 3, the actual outcome of the experiment is determined by what took place during the reaction and the purification stage, phases 1 and 2. We will explore this matterfurther in the next subchapter.
4.02: WHY DID MY EXPERIMENT FAIL TO WORk THE WAY I EXPECTED?
Experiments fail all the time to work in the ways we expect, especially in a teaching lab. Countless students each week in numerous colleges and universities do not get their product.13 Why not?
13 There are no overall statistics as far as I am aware, and this is simply based on my experience, but usually in a class of 20 students, anywhere from 1-5 of them have experiments fail in any given week, this often happens for unknown reasons. Happily, only once in a while is a student so unlucky to have two failed experiments in a given quarter.
The answer is complex, but we will try to go through some common reasons experiments fail in the teaching labs.
Previously, we said that the outcome of the experiment is set in phase 1, and that very often the purity of the product is set in phase 2. Let us examine these statements closer. What can be the reason for a failed experiment, an experiment in which no product is obtained?
Is it because the chemical theories underlying the experiment are fundamentally wrong? Or is it because the procedure given to the students contains a fatal error? If either of these possibilities were true, we would expect the experiment to fail for the entire class because these are fundamental flaws in the design of the experiment and should apply to everyone equally. If, however, we see that the experiment works for some students and not others, we must seek an explanation at the level of the individual, i.e., we must assume that the different outcomes reflect different choices that individual students made.
If absolutely no product is obtained, the mistake or error can usually be traced back to stage 1. Very typical mistakes (that the student might not have been aware of ) are calculation errors and/or errors in measuring out reactants. A misplaced decimal point, e.g. 0.1 g versus 0.01 g is easy to overlook. Other common errors are improper heating during the reaction stage, and letting the reaction continue for a wrong period of time. Another error that we see quite often, is use of the wrong reagents. Chemical names are strange, and it is easy to overlook subtle differences such as “conc.” versus “6M” sulfuric acid, or acetic anhydride versus acetic acid. The reagents in each pair sound very similar, but may react completely differently. Of course, with so many possibilities for error, it can be very difficult to backtrack and find an error after the experiment is complete.
Simple misunderstandings often occur, even with a very carefully written procedure, and a very careful TA. To give you an idea of the level of involvement required in designing lab manuals, all of the PSU organic chemistry labs have been piloted at least three times, by three different people. The manual is constantly updated and refined, and even after several terms, minor changes are still implemented. This means that people read and interpret differently. Something that might seem very clearly worded to you might be very confusing to someone else. In this case your TA will aid you in figuring out what has gone wrong.
Sadly, even when a reaction has been performed properly, it is possible for things to go wrong in stage 2, the work-up and purification. In fact, this is often a greater source of trouble because purification, by its very nature, produces two things: the desired (pure) material and the undesired contaminants. Unless one pays careful attention, one can easily toss out the purified product and hold on to the contaminants. Also, as you probably noticed in chapter 2, purification procedures can be complicated and there are many ways for them to go astray.
Getting two phases confused during a liquid-liquid extraction, for example, or failing to get crystal formation during recrystallization, are examples of mistakes that can prevent you from obtaining product. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/04%3A_HOW_TO_SURVIVE_AN_ORGANIC_CHEMISTRY_EXPERIMENT/4.01%3A_TYPICAL_FLOW_OF_AN_ORGANIC_CHEMISTRY_EXPERIENCE.txt |
I believe that the following will not be a surprise to anyone, but it is perhaps the primary reason why stu- dents suffer through sub-par lab experiences: lack of understanding. One piece of advice I usually give is to make sure that you understand each step in the procedure.
A chemical procedure is in many ways like engineering: every element serves a purpose, and has a specific function. A very fulfilling and rewarding exercise is to make sure that you have this correlation firmly in your mind before you enter the lab. This will give you a deep understanding, and make the outcome of your experience more positive, both scientifically but also personally.
Do you really understand why we added the base in step 3? Do you really understand why we shake the separatory funnel, or why we add the anhydrous sodium sulfate? Do you really understand why we distilled the product, and why we did not recrystallize it?
A key feature of surviving (and enjoying!) an organic chemistry lab experience is therefore to attain a deeper understanding. That way, you will see how the elements of the procedure come together, and how each aspect has a vital and important purpose. This can be a valuable way to prepare, and to make sure that you are stacking the odds in your favor, to make sure that the experiment goes well.
4.04: TAKING NOTES AND USING THE NOTEBOOK
One of your key tools in the lab, is your notebook. This is equally true of real research situations, where a scientist takes careful notes during an experiment. The point of the notebook is to collect data on your procedure, as well as observations you make. In theory, you can fill the entirety of a notebook in only one experiment, if you uncritically write down anything that you can think of. The key to using the notebook is therefore to collect not only data, but relevant data. This data will be imperative for you when you compile your laboratory report later.
The distinction between irrelevant- and relevant data can be challenging, but we will cover some situa- tions here. The most important aspect to remember, is that relevant data provides proof and indication of the success and outcome of an experiment.
1. Numbers and units. Masses and/or volumes of starting materials, catalyst solutions and reaction solvents are always important. They will deviate slightly from the quantities given in the procedure, and you should record what you actually use because you will also use these numbers later. Many other numbers are unimportant. The amount of drying agent (you add enough), the amount of water in your water bath (you add enough), the setting on your hot plate, the size of the Erlenmeyer flask you collected your organic extract in, and the size of the filter paper are not important.
2. Logic. When you are asked to write a procedure as part of your lab report, you must provide a text describing what you did. The procedure in your report must match the information in your notebook; you should never draw on the lab instructions to fill in missing gaps. That said, many essebtuak steps are unnecessary to note. For example, if you filter a solution using a Buchner apparatus, it is not necessary to make note of the fact that you clamped the neck of the filter flask, and wet the filter paper before you turned the water aspirator on. These are standard steps that are automatically included in “the solution was filtered using Buchner filtration” is sufficient.
3. Observations. This is in many ways a very difficult category, because you can make an infinite amount of observations over the span of an experiment, but many of them are secondary or irrelevant. For example, a reac- tion mixture might change color from clear to yellow, and finally to brown. That may or may not be important, depending on the specific experiment. Some color changes indicate the formation of key intermediates and provide useful guides to an expert reader. Other color changes may not be so readily interpreted, but may still be useful because they serve as landmarks of the chemical changes that are occurring. Always pay attention to your experiment, as small subtle changes can be important. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/04%3A_HOW_TO_SURVIVE_AN_ORGANIC_CHEMISTRY_EXPERIMENT/4.03%3A_THE_DEEP_UNDERSTANDING.txt |
INTRODUCTION
Upon completion of the lab, you will move into a new phase of your lab experience. You now must now carefully evaluate your results with care and interpret them in a logical way. In this chapter, we will focus on the interpretation of the three most important types of results.
05: HOW TO INTERPRET YOUR RESULTS
We have already said that any experiment has some inherent goal.
Many organic chemistry experiments are performed for the express purpose of obtaining a target com- pound. The obvious goal of these experiments is the target compound, but how can you satisfy yourself that the goal has been achieved? Of equal importance to the target are data and measurements that prove the identity and purity of the target. Some results that indicate the identity and purity of a compound are IR spectroscopy, melting point analysis, and TLC-analysis. As you will see, these results provide different kinds of information so you will normally collect more than one type of data. Let’s go through some typical scenarios, that illustrate the ways that data can be interpreted.
5.02: IR -SPECTROSCOPY- THE WORKHORSE
IR-spectroscopy is perhaps the most frequently used technique in the organic chemistry labs at PSU. It is routinely used to identify products and to verify that an experiment has succeeded. One limitation of IR -spectroscopy that is worth remembering is that while the data can be used to distinguish clearly between many functional groups, the data is nearly useless for distinguishing between structural isomers, or characterizing carbon skeletons that contain the same functional groups
An IR -spectrum routinely shows peaks from the range of 3600 to 500 cm-1.
IR -frequencies correspond to the frequencies of molecular vibrations. Molecules vibrate at many frequencies, because each vibration involves a particular group of atoms, and a particular motion such as stretches, bending, wagging etc. The same group of atoms will often vibrate at the same frequency, regardless of the surrounding molecular structure. This fact enables us to assign specific peaks to specific functional groups.
There are several useful sources of IR-spectroscopy correlation tables, and a simplified version is supplied here. Although not very extensive, this table gives you information about the most important functional groups encountered in the organic chemistry labs at PSU.
Something that is worth noting is that many functional groups undergo two types of vibrations and produce peaks in two or more areas of the IR- spectrum. In these cases we should look for, and find, all of the peaks expected for the functional group. For example, if we use the presence of an O-H peak to conclude “alcohol”, we expect to find (and should look for) the presence of a C-O peak to confirm the conclusion. Likewise, the presence of a C-H sp2 peak, must mean that the spectrum contains a C=C peak. Whenever possible, base our conclusion on all of the relevant data in the spectrum. Do not “cherry pick” your results.
Table 1. Simplified IR spectroscopy correlation table
Bond Absorption [cm-1] Frequency Comments
O-H 3600-3400 The peak is usually broader than most, and its intensity and frequency vary with concentration. If present, also look for C-O peak.
C-H (sp2) 3080-3010 Alkene or benzene. If present, also look for C=C peak.
C-H (sp3) 2960-2820 If the sample is a DCM-solution, you will find a very strong solvent peak here that obscures any C-H (sp3) peaks that might be generated by the target compound.
C-H (aldehyde) 2720 and/or 2820 Very characteristic. If present, also look for C=O peak.
C=O (carbonyl) 1750-1680 Very characteristic. A large number of functional groups contain a carbonyl group, so, if present, also look for other peaks that might identify the functional group
C=C 1680-1600 Alkene or benzene.
C-O (alcohol, ester, ether) 1300-1100 Complementary bond in alcohol, esters and ethers (but not only those bonds). If an O-H peak is present, this might indicate an alcohol or carboxylic acid group, while the presence of a carbonyl peak and absence of O-H peak can indicate an ester group. The absence of both O-H and carbonyl peaks would indicate an ether group.
In general, it is a good idea to obtain an IR -spectrum of the starting material as well as an IR- spectrum of the product. This allows for a direct comparison of the two spectra and makes it easier to establish that the starting material is no longer present, and to find new peaks that might correspond to the product.
The spectra shown below come from an experiment in which a student tried to oxidize a secondary alcohol to a ketone. The top spectrum belongs to the starting material (alcohol), while the middle spectrum came from an authentic sample of the product. The bottom spectrum was obtained from a sample of the student’s product.
The student’s spectrum (bottom) contains a C=O peak at 1747 cm-1, but a comparison of this spectrum with the starting material (top) reveals an alternative interpretation. The starting material’s spectrum has a clear and distinct OH-peak (and the corresponding C-O peak). The spectrum from the stock sample of product, lacks the OH-peak, but a very characteristic C=O peak is readily observed. The IR-spectrum of the student’s product also indicates the presence of the desired ketone (strong C=O peak), but it also reveals the same O-H peak found in the starting material’s spectrum.
One logical conclusion from this analysis is that the student’s oxidation has not gone to completion or that the purification has not been satisfactory.
This is a microcosm of the entirety of IR -analysis: looking for the appearance of disappearance of specific peaks to answer the question of whether a reaction has worked or not. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/05%3A_HOW_TO_INTERPRET_YOUR_RESULTS/5.01%3A__WHAT_DOES_IT_ALL_MEAN%3F.txt |
The melting point of a compound is useful in two ways: it says something about the identity of a compound, and something about the purity of a compound. The presence of impurities will influence the melting point of a compound, leading to wider and depressed melting point ranges. Because even small deviations in melting point ranges is meaningful, it is worth mentioning that a control should be made when running the melting point of an isolated compound, against a stock sample of the same compound. In general, a deviation of more than 1 oC in obtained melting point usually means the material is not of acceptable purity. Because the margin is relatively small, it is always worth running your sample with an authentic sample of the compound in the same melting point apparatus. If there are even small deviations in the apparatus, you will find that the authentic sample’s melting point is also deviating from the literature value.
Melting points are usually not the only analytical data obtained for a sample; routinely IR-spectroscopy is also performed. Any abnormality in either data set should reflect in the other data source. For example, an IR spectrum corresponding to a product with a depressed melting point will often show peaks that are not expected for a pure sample. It is worth noting that this is not necessarily the case, as the impurities must also produce strong peaks in non-crowded areas of the IR-spectrum to be detectable.
Let us look at a specific example. A student recrystallized a sample of 2-naphthol from ethyl acetate. The published literature on 2-naphthol lists its melting point as 121-123 °C, but the student’s sample melted from 115- 121 °C, a range that was much broader and much lower than the literature one.
These data are a clear indication that the product is not pure, but it always is preferable to be able to identify the source of the impurity as opposed to simply stating that there were impurities present. The key to this puzzle is found in the student’s IR -spectrum.
You will notice a clear indication of an OH-bond (around 3200 cm-1) that is likely due to 2-naphthol, but you will also notice something else: a clear C=O peak. It is very likely that this is trace ethyl acetate left over from the recrystallization, which indicates that the sample was not allowed to air-dry sufficiently.
A powerful observation is found when the two analyses are combined and viewed together.
5.04: TLC- IDENTITY AND PURITY
TLC can be a very useful resource to obtain quick and usefyul data about purity and identity. Unlike melting point analysis, where measurements can be compared to literature values, control TLC – experiments are essential to support the presence of com- pounds of interest. TLC has an advantage over melting point analysis, however, in that it can be applied to both liquids and solids.
Let us say that a student brominated acetophenone using bromine (scheme 1). Because the product contains a C-Br bond, IR -spectroscopy is not very useful (the peak assosicated with a C-Br lies in a region of the spectrum that is hard to observe). Instead, the student performed a TLC analysis.
In order to get conclusive data from the TLC analysis, the student made up a TLC – plate with four samples. Spot A is the starting material acetophenone, spot B is a stock solution of the product, spot C is the product obtained by the student, and spot D is a co-spot of two solutions, the stock solution of product (spot B) and the student’s product (spot C).
The TLC -analysis gives us important information about the experiment. The first result is the student’s sample, C, which appears to be two compounds. One of those (the darker spot) agrees with authentic product (B),but the other (the lighter spot) agrees with the starting material (A). This suggests that the student’s product is contaminated with unreacted starting material. Spot D supports this interpretation: samples B and C produce a single dark spot in D. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/05%3A_HOW_TO_INTERPRET_YOUR_RESULTS/5.03%3A_MELTING_POINT_ANALYSIS-_IDENTITY_AND_PURITY.txt |
INTRODUCTION
You prepared for the lab, finished the experiment, and now you have to compile a report. This writing assignment can in many ways be equally as challenging as the actual experiment itself. In this chapter, we will cover the best way to write a sound, well-written and scientifically coherent report.
06: HOW TO WRITE A REPORT
Different lab courses14 will have different report requirements, but the organic chemistry reports at PSU are based on the framework followed in most scientific research articles. These are typically divided into three parts:
1. The why: gives the information necessary to contextualize the experiment, states the goal of the experiment.
2. The how: gives information about how the experiment was performed
3. The what: provides relevant data and interpretation as it relates the goal of the experiment
Below is a table showing the different pieces that need to be included in your lab reports.
Table 1. Key goals of a report
# Section Name Key goals of section
1 Why Title A short title of the experiment.
2 Why Purpose of The main goal and purpose of the experiment.
3 How Reaction scheme The reaction scheme for the transformation attempted.
4 How Procedure A description of the procedure is provided so that a professional can reproduce your experiment.
5 What Data The key data obtained are presented, usually in tabulated form.
6 What Discussion The results and findings from #5 are discussed, and the outcome of the experiment as it pertains to #2 is explained
7 What Conclusion A conclusion is stated, based on the results.
8 What References Relevant references are provided.
As the table shows, each section serves a distinct purpose and function. We will now look at each section in detail, and focus on the general goals of each, followed by things to watch out for, and common mistakes. The section numbers #1-8 follow the sections used in the report forms at PSU.
14 Different scientific disciplines, and even sub-disciplines, have different reporting requirements, different ways of pre- senting information, and different ways of using technical language.
6.02: THE WHY- THE REASONS FOR DOING THE EXPERIMENT.
Section 1: Title
A single sentence that fully captures the main goal of the experiment.
Section 2: The purpose of the Experiment
(1) General: This section should be a brief statement about the ultimate goal of the experiment. Thisaection should never be longer than two sentences, and in many instances kept to only one sentence. Common goals are the identification of a compound, the synthesis of a compound, or the purification of a mixture. Tech- niques (TLC analysis, IR spectroscopy, extractions) should be mentioned only when these were the means to achieve the goal, and not a goal, and not the goal in themselves.
(2) Things to watch out for: The purpose must be result oriented, which means that general tasks like collecting data, performing calculations, getting results, and learning techniques, are inappropriate. Although learning to do new things is a top priority, it is never the purpose.
(3) Some examples:
Good: The purpose of the experiment was to synthesize benzoic acid through an oxidation reaction using bleach
Good: The purpose was to identify an unknown compound using TLC-analysis and melting point analyses.
Bad: The purpose of the experiment was to collect data and learn about recrystallization.
Bad: The purpose was to perform a four step synthesis that involved acetylation, hydrolysis, condensation and reduction, and use several new lab techniques such as extraction and sublimation. The student should learn about IR spectroscopy too, and analyze the data obtained. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/06%3A_HOW_TO_WRITE_A_REPORT/6.01%3A_THE_BASICS.txt |
Section 3: Relevant Structures or Reaction scheme
(1) General: Depending on the experiment, you should either supply structural formulas of the compounds you investigated, or provide a reaction scheme for a synthesis. Structural formulas should incorporate standard bond angles and can be hand-drawn. Usually the required structures are the compounds you investigated and listed in the purpose, or the reactants and the products you obtained through a chemical synthesis. Solvents are typically not required, nor are drying agents or other work-up compounds.
(2) Things to watch out for: Make sure that you provide structural, and not molecular formulae An example of the latter would be C6H12O6 for glucose. Molecular formulas have their place in organic chemistry, but are relatively useless for your report, as they do not provide any clarity in which bonds have been formed or broken. Take care to draw your formulae thatobey standard drawing conventions, i.e. with correct bond-angles, and/or correct atom hybridization. If you are drawing a reaction, draw molecules in the correct order: reactants on the left side of the reaction arrow, products on the right. Reagents, catalysts, solvents, and other important reaction conditions are drawn above and below the arrow.
Section 4: procedure
(1) General: The procedure should encompass the core of the actions performed in the experimentation. It should contain the essential information as well as your exact measurements. The procedure must be written in past tense, and other verb forms such as present tense, or imperative form, must be avoided. You should assume that the reader has an extensive chemistry background, and is reading your procedure in order to learn how to repeat the experiment (note: the procedural format used in the lab manual is far too detailed for this reader). A challenging part of the procedure is to omit unnecessary details and focus on the necessities, see below for specific examples.
(2) Things to watch out for: Do not explain the reason for why anything is done in the procedure. That means that you might have added an acid to precipitate a product, but the reason for this action, is not appropriate. Furthermore, many actions that a chemist performs as part of an experiment should not be mentioned in the procedure. An example: before I can swirl a round bottom flask I must loosen the clamp, but the clamp adjustment should not be in the procedure because it follows logically. Likewise, I close the stop-cock on my separatory funnel before I pour materials into the funnel, but there is no need to mention this, or any of the other details on how I handled the funnel. A trained chemist will know how to handle a separatory funnel during a liquid-liquid extraction.
(3) Some examples:
Bad (explanations): Na2SO4 was added because it is a drying agent and therefore removes water from the crude reaction mixture.
Bad (wordy, imprecise): The distillation apparatus was assembled the way we have learned before and the reaction mixture was heated on a sand bath on a jack and the sand bath was set to approximately 350°C. When I heated the mixture, it turned milky white, like a beautiful pearl that had been ground up and it smelled gross. I got a fraction boiling at 95-98°C that I think was the product but I’m not sure.
Good: Acetic acid (1 mL) and ethanol (1 mL) were mixed in a round bottom flask and heated to reflux.
Good: The reaction mixture was gravity filtered and the solvent removed using rotary evaporation. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/06%3A_HOW_TO_WRITE_A_REPORT/6.03%3A_THE_HOW-_HOW_WAS_THE_EXPERIMENT_PERFORMED%3F.txt |
Section 5. Data
(1) General: Data that you collect must be presented clearly. In almost all cases, a tabulated form is required. The data that you should present is intimately tied to the purpose of the experiment. If the purpose is to synthesize a compound, the data are related to that purpose, i.e. they indicate whether or not the synthesis was successful.
Tables should be numbered and have captions. Units should be given in the title cells, in parentheses or brackets.
Most likely, the data will be analytical and spectroscopic data. Examples are:
• TLC-analysis (spots, distance travelled, Rf-values),
• Melting point data (melting point ranges, literature values)
• IR-spectroscopy data (functional groups, wavenumber, literature values)
• Synthesis data (only appropriate when a chemical synthesis has been performed): masses, volumes (if applicable), densities (if applicable), moles, yields
• NMR data (assignment, shift values, integrals)
In each table, the most important data should be presented. The task is to filter out the important data points. For example, when providing a table with IR data, you can readily find many peaks, but most of them are irrelevant to the purpose of an experiment or do not provide any information about the success of the experiment
A key task is therefore to select and present meaningful data, and not just dump all your data into the report.
If you have acquired an IR (or NMR) spectrum, a copy of the spectrum should be included in this section.
(2) Things to watch out for: Two common examples of irrelevant data include mass of glassware and mass of filter paper. Appearance, consistency of a product and other physical properties of a similar nature, are also inappropriate. Make sure that the tables are formatted correctly.
(3) Some examples:
Good Table:
Table 1. Synthesis table
Compound Mw, [g/mol], m, [g], n, [mmol], Yield, [%],
Benzaldehyde
NaBH4
1,1-diphenylmethanol
182
38
184
0.21
0.10
0.19
1.2
2.6
1.0
-
-
83
Good Table:
Table 2. IR data
Bond Observed value [cm-1] Literature value [cm-1]
O-H
C-H (sp2)
C-H (sp3)
3550
3010
2985
3200-3600
3090-3005
2985-2800
SECTION 6. INTERPRET THE OBTAINED RESULTS (OR WHAT DID YOUR DATA MEAN?)
(1) General: In many cases, interpretation of data is the most challenging component of any report. The key feature of an interpretation is that it says something about the data that relates to the purpose of that experiment.
There are two important scenarios: If the purpose of the experiment is to identify an unknown sample, then the interpretation extract information from the data that says something about the identity of the unknown. If the purpose of the experiment is to synthesize a compound, the interpretation must extract information that describes the identity and purity of the target compound.
The natural tendency is to write interpretations that are far too long. A good interpretation is exact, precise, and addresses the purpose of the experiment.
Such information can be:
1. TLC (says something about purity and identity)
2. Melting point (says something about purity and identity, especially when compared to a known melting point)
3. IR data (says something about which functional groups contained in the sample, which might indicate the presence or lack of the suspected product)
4. Yield (says something about the success of the reaction)
5. Boiling point (says something about the identity of the compound)
Your data may not always support the expected outcome of an experiment. For example, your data may indicate that no reaction took place, or a product was very impure, and so on. These types of interpretations are appropriate as long as they are based on actual observations.
An example of a data interpretation that does not support the expected outcome is when you call attention to an OH peak in the IR spectrum that is not consistent with the structure of the desired product. That should be addressed, and a likely scenario to explain the presence of that peak should be provided.
(2) Things to watch out for: Unfocused or off-topic interpretations. Examples: interpretations that discuss what you learned, experiences you had or comments about your success as a lab student. None of these examples relates to the purpose of the experiment, and none of them deal with the objective data you collected. The interpretation should not deal with unimportant or irrelevant observations.
For example, if you say that the IR spectrum did not show the correct peaks because you collected the spectrum in a wrong way, that becomes a meaningless statement (how is it possible to collect a spectrum wrong, how do you know it is wrong, and why did you not address that issue if you knew about it?).
Here are some other examples with speculative conclusions that should be avoided.
• I weighed something wrong, or the scales were wrong
• The yield was low because I transferred the sample between vials
• The glassware used has a high degree of uncertainty
Furthermore, any error must have a direction and observations cannot contradict themselves.
If we come back to the example above with the OH peak that was found in your IR, that should not be there, surely that must mean that your yield also cannot be correct. And what about your melting point? That must also be off, or broader than it should be.
(3) Some examples:
Bad (imprecise, does not refer to specific data, no clear connection to the purpose of the experiment): The yield was higher than I wanted because of impurities and I weighed the sample wrong. The IR does not show what I should have because of something happening during the collection of the spectrum that should not have happened. The cylinder I used to measure was wet so that led to not getting correct data.
Bad (does not refer to any specific data, does not connect data to the purpose of the experiment, irrelevant):
IR spectroscopy is used to saying something about which functional groups are found in an organic mol- ecule. It can be used to find bonds that correspond to different functional groups. In my molecule, I have many bonds, and that can be seen in the spectrum. Furthermore, I have some strong peaks, and some weak ones, which is to be expected.
Good (Discusses specific data, interprets these data to support specific conclusion that relate to the pur- pose of the experiment): The yield was higher than expected (104%) due to alcohol contaminants in the isolated product. This is evident from the IR-spectrum of the sample (O-H stretch at 3551 cm-1), which is a bond not present in the product. The melting point mirrors this (74.1°C - 84.9°C), as it is broader than expected (81.5-85.0°C)1.
1. Correct citation to a literature source for the melting point.
SECTION 7. CONCLUSIONS (OR WHAT WAS THE PRIMARy RESULT OF YOUR EXPERIMENT)
(1) General: The conclusion will be the final statement based on the success (or lack thereof ) in the experiment. It will pertain to the purpose of the experiment. It will draw on the interpretation under #6, and lead you to say something logical about the success of the experiment as a whole.
Some points that often are relevant are:
• Did you synthesize the suspected product (why, why not?)
• What was the purity of the compound isolated, what was the identity of the compound?
(2) Things to watch out for: Conclusions that are not anchored in the purpose of the experiment should be avoided. If you have recrystallized a product, concluding about how much you liked or did not like the lab, is not appropriate. Subjective opinions and learning outcomes, although important to us, are not things that should be placed in the conclusion. Also, avoid using statements about what you would have done differently.
(3) Some examples:
Bad (irrelevant): I really liked learning about melting point analysis and I can see that it will be valuable for me in the future, because I am becoming a melting point expert.
Good (says something about the success of the experiment and its purpose): Benzoic acid was prepared in a moderate yield. The product isolated is likely benzoic acid, as it shows a similar IR spectrum that contains the key functional groups. | textbooks/chem/Organic_Chemistry/Book%3A_How_to_be_a_Successful_Organic_Chemist_(Sandtorv)/06%3A_HOW_TO_WRITE_A_REPORT/6.04%3A__THE_WHAT-_WHAT_WERE_THE_MAJOR_FINDINGS_OF_THE_EXPERIMENT%3F.txt |
• 1.1: Carbonyl Group - Notation, Structure, and Bonding
You will be learning and applying the principles which govern the structure of organic compound and relating your understanding of structure to the reactions--the changes in structure--which happen when specific portions of organic compounds interact with other chemical substances. We will spend the first several weeks of the semester looking at a group of organic compounds which share a common structural element--the carbonyl group.
• 1.2: Functional Groups, Hybridization, Naming
First, let's look at the various structural representations you were asked to develop for the molecules given at the end of the last lecture. Here they are:
• 1.3: Additions- Electrophilic and Nucleophilic
Last time we listed three reasons to expect that the carbonyl (C=O) group would be a functional group, a reactive part of the molecule. All of these reasons were connected with the way electrons are distributed in the group. Functional groups are the places where changing the location of electrons can happen fairly easily, which means that the distribution of electrons in a functional group is a key to its reactivity. We need a specific example to make these ideas useful.
• 1.4: Acetal Formation, Mechanism, Resonance
Last time I left you with a problem, "what is the mechanism for the base catalyzed addition of water to a carbonyl group?" Let's go through that and see how it goes.
• 1.5: Nitrogen Nucleophiles - Imine Formation
In many of the biological reactions of carbonyl groups the nucleophile is a nitrogen atom. The eventual outcome is different, so let's take a look at the details. The specific molecule as an example of a nitrogen nucleophile is methylamine. What happens if Nu = CH3NH2? Here's the complete mechanism.
• 1.6: Addition of Organometallics - Grignard
Last time we looked at a reaction in which a new carbon-carbon bond was made. Today, we'll look at another such reaction, one which is generally quite useful for synthesis, the assembly of larger carbon structures from smaller molecules.
• 1.7: Oxidation and Reduction, alpha-C-H acidity
Last time we saw how a nucleophilic addition of a carbon atom to the carbonyl carbon could be carried out through the use of a Grignard reagent. This time we'll look at oxidations and reductions of carbonyl groups and at the acidity of the alpha hydrogen atom
• 1.8: Enolates, Aldol Condensation, Synthesis
Last time we worked through the reagents which oxidize aldehydes to carboxylic acids and the reagents which reduce aldehydes to primary alcohols and ketones to secondary alcohols. We also learned how enolate ions can be formed by the removal of an alpha hydrogen atom and how these enolate ions can act as nucleophiles toward bromine.
• 1.9: Carboxylic Acid Derivatives- Interconversion
Today we'll look at carboxylic acid derivatives. This group of compounds also contains a carbonyl group, but now there is an electronegative atom (oxygen, nitrogen, or a halogen) attached to the carbonyl carbon. This difference in structure leads to a major change in reactivity. Here we find that the reactions of this group of compounds typically involve substitution of the electronegative atom by a nucleophile.
• 1.10: Carboxylic Acid Derivatives - Alpha Carbon Reactions
Last time we looked at several reactions of carboxylic acid derivatives and learned how to make esters and amides. Today we'll look at some further reactions of esters, including some which make new carbon-carbon bonds.
• 1.11: Fats, Fatty Acids, Detergents
Last time we looked at several reactions of carboxylic acid derivatives and learned how to make esters and amides. Today we'll look at some further reactions of esters. We'll also examine the chemistry of the ester functional group in fats.
• 1.12: Carboxylic Acids
Last time we looked the reactions of esters with lithium aluminum hydride and with Grignard reagents. We followed that with the chemistry of enolates formed from esters and then looked at fats and soaps. Today we'll look at the parent functional group of carboxylic acid derivatives, the carboxylic acid group itself. We'll see how we can make carboxylic acids and ask why they are acids.
• 1.13: Alcohols
Today, we'll examine the chemistry of alcohols. First, we'll review the reactions we've already seen which make alcohols. Then we'll look at the reactions of the alcohol functional group.
• 1.14: Ethers, Epoxides, Thiols
Today, we'll go on to look at the acidity of alcohols and the uses of their conjugate bases as nucleophiles. That will take us to the chemistry of ethers. We'll finish with a look at thiols, close relatives of alcohols in which a sulfur atom has replaced the oxygen.
• 1.15: Chirality, Three Dimensional Structure
Today we'll take a look at structure. We'll find that we have to begin to think in three dimensions to understand some structural differences between very similar compounds.
• 1.16: R/S Naming, Two or More Stereogenic Centers
Today, we'll look at naming compounds with stereocenters, and then we'll examine the complications which arise when a molecule has more than one stereocenter in it.
• 1.17: Carbohydrates- Monosaccharides
Last time we learned how a chiral compound's absolute configuration can be described by the R/S naming system. We also considered the situations which can arise when a compound has two (or more) stereogenic carbons. Our examples for that were in fact sugars; monosaccharide aldotetroses. We'll begin by making some structural sense of those terms.
• 1.18: Glycosides, Disaccharides, Polysaccharides
Today we'll look in more detail at the chemistry of that hemiacetal linkage. In particular, we'll recall how hemiacetals are converted to acetals. We'll find that these acetal linkages are what holds di- and polysaccharides together.
• 1.19: Amines- Structure and Synthesis
Last time we completed our study of carbohydrates. Now we are turning our attention to another important class of organic compounds, amines. Many important drugs are amines, the bases present in RNA and DNA are amines, and the fundamental building blocks of proteins are amino acids.
• 1.20: Amines- Reactions
Last time we looked at the behavior of amines as bases, at their involvement in hydrogen bonds, and at the ways they can be synthesized. This time, we'll continue our study of amines by examining some of their reactions.
• 1.21: Amino Acids and Peptides
Now we'll look at what happens when a carboxylic acid functional group and an amine functional group are in the same molecule. Our focus will be on the alpha amino acids, those in which the amino group is bonded to the alpha carbon -- the one next to the carbonyl group -- of the carboxylic acid. These are the basic building blocks of proteins and are the most important type of amino acid.
• 1.22: Proteins
Last time we looked at the structural characteristics of amino acids and the peptide bond which joins individual amino acids together to make proteins and peptides. We also learned about the sequence (order) in which amino acid units are joined in peptides. Today we'll study the ways in which the specific sequence of a peptide may be discovered and the methods which are used to synthesize such a peptide.
• 1.23: Nucleic Acids
Today we'll study the chemistry of the molecule which carries the information necessary for directing the biosynthesis of proteins and peptides. This is DNA, and we'll learn that the structure of DNA provides a very strong rationale for its function.
• 1.24: Nucleophilic Substitution, SN2, SN1
Today's topic takes us back to an important organic reaction mechanism. We've studied a few reactions which proceed by this mechanism. Now it's time to examine it in detail.
• 1.25: Elimination - E2 and E1
Last time we saw an overview of the nucleophilic substitution mechanisms of alkyl halides. We examined one of these, the SN2 mechanism in detail. Today we'll examine the other, the SN1 mechanism, and then go on to look at elimination reactions, the major competition for substitutions.
• 1.26: Alkenes and Alkyne Structure
Today we'll begin by looking at the structural characteristics of alkenes, the products of elimination reactions. Then we'll return to the topic of elimination reactions and examine their reaction mechanisms in more detail.
• 1.27: Electrophilic Additions
Last time we learned how the carbon-carbon double bond introduces stereochemical distinctions into alkenes, and we looked at the mechanisms of the elimination reactions which are important in making alkenes. Today we'll examine the characteristic reactions of alkenes -- additions
• 1.28: Polymers
Last time we examined the characteristic reactions of alkenes -- additions. Today, we'll see how reactions like these and some familiar reactions of carboxylic acid derivatives can be used to make very long chains -- polymers. We have already looked at some important polymers from biological systems; starch, cellulose, proteins, and nucleic acids.
• 1.29: Metabolic Organic Reactions
Today we're going to examine a selection of processes which occur in metabolism. We will focus on comparing these reactions to reactions we have already studied. In particular we will see that the reactions which break carbon-carbon bonds are just reverse versions of the aldol and Claisen condensations which we have studied earlier. Keep in mind that while we are looking for connections between these reactions and familiar organic reactions, all steps in these schemes are catalyzed by enzymes.
• 1.30: Aromatic Compounds
Today we'll find that resonance is very important in understanding both the structure and the reactions of aromatic compounds. First, let's take a look at the structural representations which distinguish aromatic compounds from those that aren't aromatic.
• 1.31: Electrophilic Substitution
Today we'll look at examples of electrophilic aromatic substitution reactions, learn what we can make with them, and see how the prior presence of a substituent on the aromatic ring influences where the attacking electrophile becomes attached.
• 1.32: Side Chain Oxidations, Phenols, Arylamines
Misc.
• 1.33: Radical Reactions
Last time we looked at how the benzene ring changes the reactivity of an atom or group to which it is directly attached. Today we'll finish presenting new material in the course by taking a brief look at reactions in which bond making and bond breaking events involve electrons moving singly rather than as pairs.
01: Chapters
Introduction
Organic chemistry is the chemistry of the element carbon. The compounds formed from carbon and a few other elements (O, N, P, S and H) form the chemical basis for living systems. Most therapeutic drugs are organic compounds. Organic polymers, whether obtained from nature or by synthetic means, are extremely important economic materials. These include plastics, rubber, glues, starch, cotton, and wood, as well as the proteins we must have in our diet.
Why is it called "organic" chemistry? Historically, when chemists discovered the Law of Definite Proportions at the beginning of the 19th century, it appeared that this law did not apply to the various compounds they had isolated from plant and animal sources. Carbon compounds can be so complex that the ratios of elements in them did not appear to be simple numbers. For example, ordinary table sugar has the molecular formula C12H22O11, not the kind of simple ratio seen with the oxides of copper, Cu2O or CuO, for example. Chemists imagined that organic compounds were held together by a mysterious "vital force".
The beginning of the end of the "vital force" hypothesis is generally considered to be Friedrich Wöhler's synthesis of urea in 1828. He started with lead cyanate, which is about as "dead" as any chemical can be, and ammonium hydroxide or chloride, also "dead", which generated ammonium cyanate, NH4+OCN- (empirical formula CH4N2O) (still "dead"). When he heated the ammonium cyanate, he got urea, H2NCONH2 (molecular formula also CH4N2O, but atoms arranged differently). Urea is just what the name sounds like, a major ingredient in urine, and was thought at the time to be a purely "organic" chemical. Other syntheses of "organic" compounds from "inorganic" materials soon convinced chemists that organic compounds obeyed the same laws of chemistry as other chemicals.
Although chemists gave up the "vital force" hypothesis at least 150 years ago, a shadow of it lingers on in the popular notion that "natural" organic materials are somehow safer or more healthful than synthetic chemicals. This popular notion ignores the fact that we would not know, for example, what vitamin C is if we could not find out its molecular structure, synthesize it, and show that the synthetic material is in every way identical with the vitamin C that the famous Hungarian chemist Szent-Gyorgy (pronounced "Saint-George") first isolated from Hungarian peppers. This bit of popular culture also ignores the toxicity of nicotine, strychnine, pufferfish toxin, and botulism toxin, the last of which is the most poisonous chemical known. You might even call the AIDS virus an "organic chemical", since it has a known chemical structure, though its ability to reproduce itself in the human body (and to undergo rapid molecular evolution to defeat immune response or chemical inhibitors) makes it far more sinister than a mere poison.
You will be learning and applying the principles which govern the structure of organic compound and relating your understanding of structure to the reactions--the changes in structure--which happen when specific portions of organic compounds interact with other chemical substances. We will spend the first several weeks of the semester looking at a group of organic compounds which share a common structural element--the carbonyl group.
Structural Principles
First, though, we need to review a few structural characteristics of the carbon atom. These are ideas which were part of your general chemistry courses, but it will help if we briefly restate them.
1. Carbon is tetracovalent. That means that a carbon atom typically makes four bonds to other atoms and that these bonds are covalent--formed by sharing an electron pair between the two atoms joined by the bond. Such arrangements provide eight valence electrons for a carbon atom, so that it's electronic configuration is like that of the very stable noble gas neon. Similarly, hydrogen forms one covalent bond, oxygen two, and nitrogen three.
2. Carbon can form multiple covalent bonds. That is, a single carbon atom can form a double (to C, O or N) or triple (to C or N) bond to another atom. A double bond would involve two electron pairs between the bonded atoms and a triple bond would involve three electron pairs.
3. Bonds between carbon and atoms other than carbon or hydrogen are polar. That is, in a bond between carbon and oxygen or nitrogen the electrons are closer to the more electronegative element (oxygen or nitrogen) than to the carbon, so the carbon has a slightly positive charge. (Fluorine is the most electronegative element, and the elements close to fluorine in the periodic table are also quite electronegative.)
4. Bonds between one carbon atom and another and between a carbon and a hydrogen are non polar. That is, the electron pair forming the bond is quite evenly shared by the atoms.
5. We can predict the geometry of the bonds around an atom by using the idea that electron pairs and groups of electron pairs (such as in double or triple bonds) repel each other (Valence Shell Electron Pair Repulsion--VSEPR--Theory).
Representing Structures
Now, let's apply some of these ideas to a small organic compound, formaldehyde. The molecular formula (composition by element) of formaldehyde is CH2O. If we interpret this literally, reading from left to right, we get something like this (bonds are indicated by lines):
If we check this against our understanding of how many bonds each atom should form, we find that the carbon has one bond where we expect four, the hydrogens have two bonds instead of one, and the oxygen has only one instead of the expected two bonds.
A better approach is to draw four bonds to carbon and then think how the hydrogens and the oxygen might be linked to the carbon. If we connect the carbon and oxygen by a single bond, we get:
Here the lines which don't connect to more than one atom represent unused valences. We have four such unused valences and only two hydrogens to use them, so we'd be stuck with something like (there are other possibilities, but none based on this skeleton that work well):
If we remember the possibility that carbon and oxygen can make a double bond, we can check out a skeleton like:
Now we have two unused valences and two hydrogens to connect to them, so by doing so we arrive at this structure for formaldehyde.
The structural unit made up of a carbon joined by a double bond to oxygen is known as a carbonyl group (often represented as "C=O"). We will spend the next several weeks on the chemistry of this group as it is found in somewhat different structural situations
We can learn from this process that converting a molecular formula to a structure is best done by working the atoms which can form more than one bond first, then checking each trial skeleton against the typical numbers of bonds for a particular atom. Those that pass that test can be further tested by balancing the number of unused valences against the remaining atoms so as to come out even.
More Carbons
Let's look at a more complex case, one with several carbon atoms. As you will learn from studying the section on isomerism in Brown (Section 3.2), a molecular formula is not enough to specify the structure of an organic compound which includes more than three carbons. We need information about which atoms, particularly carbons, nitrogens and oxygens, are connected to each other. This is often represented in a condensed formula like:
CH3CH2CH2CHO
To take a more detailed look at the structure of this compound, we need to expand its representation. (See Brown, Section 3.4) As we have learned to do, we begin by ignoring the hydrogens and focusing on the carbons and the oxygen to arrive at a skeleton. A trial skeleton might be (where the unused valences are again shown as lines that connect to only one atom):
If we compare this skeleton to the condensed formula above, the left end carbon seems to be associated with three hydrogens in the condensed formula and there are three unused valences on the left end carbon as shown in the skeleton, so that matches well. Similarly, the two middle carbons have two hydrogens each, and there are two unused valences on each of those carbons in the skeleton. If we match all these up we arrive at:
The right end carbon and the oxygen are troublesome, though. We have only one hydrogen left and there are three unused valences to deal with. We've seen this situation before when we were working on formaldehyde and we can use the same idea here, so that we try a double bond between carbon and oxygen. That gives us this skeleton:
Adding the final hydrogen, we arrive at this expanded structure:
Bond-Line Structures
It is often inconvenient to show all the carbons and hydrogens in detail, especially since we will learn soon that the parts of a molecule which are made up of only carbon and hydrogens joined by single bonds do not play a significant role in the reactions of that molecule. (These portions of the molecule are known as "R-groups.") We can represent these atoms and how they are connected by using an abbreviated structural type known as a bond-line structure. Such a structure for the compound we just worked with is shown below:
In a bond-line structure, carbons are shown by the "empty" ends of lines and by junctions (corners) between lines. Letters at the end of a line represent atoms of the designated heteroatom (Heteroatoms are atoms other than carbon or hydrogen. Hydrogen is often shown where necessary for clarity.) Double or triple bonds are represented by two or three parallel lines joining the same two atoms. Hydrogens are added as needed to fill up the remaining unused valences.
We can "flesh out" the skeleton above by first filling in the carbons at the corners and "empty" ends:
Then we count how many bond each carbon has showing, and add enough hydrogens to each carbon to bring its number of bonds up to four. (For other atoms like oxygen or nitrogen which can make more than one bond, hydrogens are added as needed to arrive at an appropriate number of bonds.) For example, the left end carbon has one bond showing, so we need to add three hydrogens:
Continuing in the same way with the middle carbons, each of which has two bonds showing, we arrive at the final expanded structure:
Notice that we didn't need to do anything to the right end carbon (the "carbonyl" carbon) or the oxygen, since these atoms already have the appropriate number of bonds showing.
Sometimes we'll represent a molecule by showing its R-groups in a condensed fashion and its reactive parts (functional groups, see next lecture) in an expanded fashion:
All these representations are useful, but we will commonly use the stick or line representation because it is economical to draw. You should practice converting between stick or line structures, condensed structures, expanded structure and the structures which show R-groups in condensed way and functional groups expanded. Try it with these examples: | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.01%3A_Carbonyl_Group-_Notation_Structure_Bonding.txt |
Representation Answers
First, let's look at the various structural representations you were asked to develop for the molecules given at the end of the last lecture. Here they are:
Compare your answers with these representations. Please see me if there are puzzles.
Naming
Naming organic compounds is a necessity, and the names of large molecules can be fairly complex. The rules are simple - but picky. You can learn the basics of these rules by studying on your own using the appropriate sections in Brown. The principles are outlined in Section 3.5. The application of these rules to aldehydes and ketones are discussed in Section 11.2. Study these, practice applying them by doing problems, and bring up puzzles in class for discussion. There will be naming questions on the exams, so you will profit by working on naming. These questions will not tackle complex examples.
Shape
Now let's look at the carbonyl group so as to understand why it is a site for chemical reactivity. We'll start by examining its geometry, specifically the bond angles around the carbon atom. We'll use formaldehyde as our example:
In the context of VSEPR theory we notice that there are three groups of electrons associated with that carbon, two single bonds to hydrogen and one double bond to oxygen. The best way for these clusters of negative charge to be as far apart from each other so as to minimize their mutual repulsion is to adopt bond angles of approximately 120o:
Such an arrangement is given the name "trigonal."
The problem with this is that if we think of the orbitals used by the valence electrons of a carbon atom, they are the 2s and 2p orbitals. You will remember that the 2p orbitals are arranged at 90o angles from each other.
This doesn't fit the 120o bond angles we need for our carbonyl group.
Hybridization
Fortunately, the theory of quantum mechanics tells us that we can mix the 2s and 2p orbitals in suitable proportions without violating the mathematical rules of differential equations. If we mix the 2s orbital and two of the 2p orbitals in this way, the resulting orbitals are pointed at from each other. Orbitals formed in this way are called hybrid orbitals and the process is called hybridization. These particular hybrid orbitals are called sp2 orbitals since they are made by hybridizing one "s" orbital and two "p" orbitals and they have the appropriate geometry for a trigonal carbon atom such as is found in the carbonyl group. (See Section 1.14 in Atkins and Carey for the same ideas applied to the doubly bonded carbons in ethylene.)
To summarize these connections, when we see a carbon involved in a double bond, its geometry will be trigonal, with 120o bond angles, and it will have sp2 hybridization. In a shorthand way trigonal and sp2 are synonyms.
Bonding
The sp2 hybrid orbitals formed in this fashion form single bonds. In the case of formaldehyde, each bond to hydrogen is formed by overlap between the sp2 hybrid orbitals on the carbonyl carbon and 1s orbitals on the hydrogen. This forms what is called a molecular orbital, one which involves atomic orbitals from two or (less commonly) more atoms. When two electrons, usually one from each atom, occupy this molecular orbital, we have a covalent bond. This bond is referred to as a sigma bond since it is shaped like a cylinder (hot dog) and the letter sigma is the first letter of the greek word for cylinder.
In a similar way, one of the bonds beween the carbon and the oxygen is formed by overlap of an sp2 hybrid orbital from carbon and a similar orbital from oxygen. (The oxygen orbital is also an sp2 hybrid orbital, but we will not pursue that.) This is also a sigma bond. What about the second carbon-oxygen bond?
When we discussed the hybridization process earlier, you may have wondered what happened to the carbon 2p orbital which wasn't used to form the sp2 hybrid orbital. It is still there, and it is used to make the second bond to oxygen. It overlaps with a similar 2p orbital from the oxygen atom to form what is called a pi bond. The term pi is used because half of this bond extends above the plane of all three sigma bonds and half extends below that plane. The first letter of the greek word for plane is pi. The pi molecular orbital can be visualized as a hot dog bun, which would lead us to visualize the double bond as a complete hot dog. One bond is a sigma bond, represented by the sausage, and the other is a pi bond, represented by the bun.
This description of the bonding in formaldehyde can be represented this way (see also Figure 11.1 in Atkins and Carey):
Similar representations apply to the double bonded carbons in ethylene (Atkins and Carey, Section 1.14) and to carbons involved in one double bond wherever they are found.
Reaction Sites
Why does this bonding scheme make the carbonyl group a reactive group, a functional group? There are three things involved here, at least one of which is seen in every functional group.
First, notice that the pi bond is above and below the plane of the sigma bonds. That means that the electrons in this bond (the pi electrons) are not located between the positively charged carbon and oxygen nuclei, but are instead above and below a line joining them. The pi electrons are farther away from the nuclei and are less strongly held to them than are sigma electrons. These pi electrons are consequently easier to move into a different location than are sigma electrons. If they are moved so as to make a new bond involving another atom, the pi bond has been broken, the structure has changed, and a reaction has occured. The short statement: pi bonds are easier to break than sigma bonds. We'll look at several examples in the next few lectures.
Second, both carbon-oxygen bonds are polar (see Atkins and Carey, Sections 1.5 and 11.2). That means that the electrons in both bonds are more strongly attracted to the more electronegative oxygen atom than the less electronegative carbon. This results in an oxygen which is slightly negatively charged and a carbon which is slightly positively charged. An atom which has an electron pair to donate (a Lewis base) will be attracted the electron-poor carbon and will show a tendency to make a new bond there. An atom which is deficient in electrons (a Lewis acid) will correspondingly be attracted to the electron-rich oxygen and show a tendency to bond there. We can indicate this polarity using the symbol +-->, where the arrow points towards the more negative end. We can also use the greek letter delta to describe a small amount (much less than an electron's worth) of charge:
The short statement: polar bonds are places where reactions occur.
Third, there are unshared pairs of electrons (Lewis base structural sites) on the oxygen. This can be seen if we take our expanded structure for formaldehyde and convert it to a Lewis dot structure by replacing each line with two dots to represent electrons:
This picture is incomplete. The oxygen does not have a full octet and should not be neutral in a formal charge sense. We can correct this by showing the two unshared pairs on the oxygen atom. Doing so represents the full octet and the formally neutral oxygen, and most importantly, tells us that the oxygen is a Lewis base.
We will then expect that the oxygen will be the site of reaction with Lewis acids. This reinforces the conclusion we arrived at on the basis of polarity.
For contrast, consider that the "R-group" parts of molecules are made of carbons and hydrogens which are connected by sigma bonds. The carbons are sp3 hybridized (Atkins and Carey, Section 1.12) and cannot make pi bonds. Carbon and hydrogen have very similar electronegativity, so the sigma bond between them is not at all polar. All the electrons associated with the carbon and hydrogen atoms are involved in bonding, and each has a completely occupied valence shell. There are no unshared pairs or vacancies, so neither atom will behave as a Lewis acid or base.
To put these ideas into action, go through the functional groups listed inside the front cover of Atkins and Carey (ignore the alkanes). Each group is reactive for one or more of the reasons listed above. List the reasons for each group, and be ready to discuss any puzzling issues in class. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.02%3A_Functional_Groups_Hybridization_Naming.txt |
Addition of Water
Last time we listed three reasons to expect that the carbonyl (C=O) group would be a functional group, a reactive part of the molecule. All of these reasons were connected with the way electrons are distributed in the group. This makes sense because reactions involve making and breaking bonds. Bonds are electrons, so making and breaking bonds will change the location of electrons. Functional groups are the places where changing the location of electrons can happen fairly easily, which means that the distribution of electrons in a functional group is a key to its reactivity.
We need a specific example to make these ideas useful. We'll begin with the addition of water to a carbonyl group, specifically the aldehyde carbonyl group in acetaldehyde. The overall reaction (from reactants to products) is:
This type of reaction is known as an addition reaction. The name fits, because the product is the sum (or adduct) of the reactants. Addition reactions occur typically with functional groups which include pi bonds. Functional groups which include pi bonds are called unsaturated functional groups because some of the atoms in them have fewer than the maximum number of sigma bonds. For contrast, those which have no pi bonds do have the maximum number (four for carbon) of sigma bonds and are called saturated. (Take another look at the table of functional groups inside the front cover of Brown and note which ones are unsaturated.)
As we study more addition reactions of unsaturated functional groups during the semester, we'll notice that the pi bond in the reactant is typically broken. In the product we find that there are now two groups or atoms attached where the pi bond had been. In our instance, the pi bond between carbon and oxygen has disappeared and the hydrogen from water has added to the oxygen while the OH group from water has added to the carbon.
We can also notice that nothing happened to the CH3 group. It is saturated so it lacks a pi bond and cannot undergo an addition reaction. It is unreactive and is simply carried along from the reactant to the product.
Polarity Matching
Let's probe a little deeper into this reaction. Does it make sense that the OH group from water attaches to the carbon of the carbonyl? Does is make sense that the H from water attaches to the oxygen of the carbonyl? Let's look at the polarity of these materials.
We notice that the positive (carbon) end of the carbonyl dipole becomes attached to the negative (oxygen) end of the OH dipole. Put another way, the electron-rich oxygen of water attacks the electron-poor carbon of the carbonyl group. Much of what we will learn in organic chemistry can be related to this idea, and we will develop it in more detail later in this lecture.
We noticed earlier that the CH3 group was not involved in any bond breaking or bond making. In this respect, it is very much like the "spectator ion" which did nothing in an inorganic reaction. These groups are caled "R-groups." Since these groups do not change in a reaction, we should focus our attention on the groups which do, the functional groups. For example, let's predict what the product of the following reaction would be:
We might remember that the carbonyl group of acetaldehyde (above) adds water, the similar carbonyl group of butanal (below) should do so as well. We would look again at the acetaldehyde reaction, notice where the OH and H go, and put the OH and H in the same positions on the carbonyl group of butanal. We would get the following answer:
General Reaction
Notice again that the CH3CH2CH2 group didn't change and is merely copied from the reactant to the product. It helps immensely to think of this reaction as a reaction of the carbonyl group, not of the whole compound. In that way we can focus our attention on the part of the molecule which reacts and regard the rest as carried along for the ride. To express this in symbols, we can write the following equation which says that addition of water is a general reaction of the carbonyl group which occurs to any aldehyde.
"R" is a "stand-in" for any group which might be attached to an aldehyde carbonyl. If a specific case tells us that R = CH3, then the complete equation is the first one we looked at (acetaldehyde). If a question tells us that R = CH3CH2CH2, as the butanal question did above, then we can use our general"R" reaction (directly above) and just replace "R" with CH3CH2CH2 whereever we see "R." This means that instead of learning every reaction of every compound, we only need to learn the reactions of the functional groups and how to apply them to specific cases as needed.
Mechanism
Now that we know something about how to use general ("R") reactions to tackle specific cases, let's turn to another question. How does this reaction take place? What sequence of events results in the breaking of the C-O pi bond and the O-H bond in water and the making of a new C-O sigma bond and a new O-H sigma bond?
Some experimental observations will help. We find that the addition of water to an aldehyde is rather slow if the solution is neutral (pH = 7, neither acidic nor basic), and that it is much more rapid if acid or base is added. In fact, the more acid or base is added, the faster the reaction goes. The acid or base is not used up, so what we are seeing is that the acid or base is acting as a catalyst.
Let's look at the acid catalyized case first and ask how an H+ might play a role in this reaction. Where would an H+ attack the carbonyl group? From our analysis of the C=O structure, we'd expect the electron-poor H+ to attack electrons on the electron rich oxygen of the C=O group. Since we need to break the pi bond, let's have the electrons of the pi bond move to make a new bond between O and H.
We symbolize that shift of electrons from between the carbonyl carbon and the carbonyl oxygen by drawing a curved arrow leading from the pi bond location to the new bond location between the carbonyl oxygen and the H+. That arrow can be interpreted to mean that the carbon loses one electron (one half of one pair) and the hydrogen gains one electron. The oxygen neither gains nor loses electrons since the electron pair stays connected to it. The outcome of this transaction is:
Keeping score, we have broken the carbon-oxygen pi bond, and we have made the new carbon-hydrogen sigma bond. We still have to make another bond, and the positively charged carbon atom with only three bonds looks like a reactive place. We need to make a bond between that carbon and the oxygen of water, but where do we get the electron pair needed to make that bond? The carbon is a poor candidate, so let's look at the oxygen. When we're looking for electrons to make a bond, we should consider unshared pairs. These are typically found on atoms like oxygen and nitrogen, although we don't usually draw them in unless we need them. Does such a pair exist on the oxygen of water? If we do our electron calculations carefully, the answer is yes, and that pair attacks the positively charged carbon atom. (Water is acting as a Lewis base, an electron pair donor, and the carbon is acting as a Lewis acid, and electron pair acceptor. The carbon has only six electrons in its sigma bonds and has a vacancy for an electron pair. Check the formal charge to verify this.)
This makes the final carbon-oxygen bond. What remains (if we compare this structure with our product) is to break the final carbon hydrogen bond. Doing so, with that bond's electrons becoming an unshared pair on oxygen, gives us an H+ to replace the one which started the reaction so that the catalytic H+ is not used up. This sequence of steps by which the breaking and making of the requisite bonds is accomplished is called a mechanism.
If we summarize, another way to describe this is to say that the Lewis acid H+ attacks the electron pair of the carbon-oxygen pi bond. The positively charged carbon (carbocation) which results is also a Lewis acid and is attacked by the unshared pair of the oxygen of water, acting as a Lewis base. Finally the Lewis acid H+ is regenerated by cleavage of the oxygen-hydrogen bond in much the same way as H3O+ serves as a source of H+
Electrophile-Nucleophile
The terms Lewis acid and Lewis base are useful, but when we are talking about making and breaking bonds to carbons, we find that two other terms are more general. We use the term electrophile to designate atoms or groups which form bonds by using electron pairs from another atom. The positively charged carbon above is an example. It is attacked by the oxygen of water, using the oxygen's unshared pair. We use the term nucleophile to designate the atom or group which donates the electrons to make such a new bond to carbon. In our example, the oxygen atom is serving as a nucleophile. Another way to say this is that nucleophiles make bonds using their own electron pairs. Electrophiles make bonds using the electron pairs of nucleophiles. We can identify those roles in our mechanism as follows:
A summary of the reactivity of the carbonyl group is that electrophiles attack the oxygen; nucleophiles attack the carbon. We will find this to be a very useful way to organize what we learn about many other reactions of carbonyl groups.
Exercise \(1\)
Now, use these principles to work out a mechanism for the base catalyzed addition of water to a carbonyl group. Some suggestions: categorize the OH-- as a nucleophile or electrophile. Remember to break the pi bond to avoid having five bonds to carbon. Keep track of changes in the charge on particular atoms. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.03%3A_Additions-_Electrophilic_and_Nucleophilic.txt |
Base-Catalyzed Hydration
Last time I left you with a problem, "what is the mechanism for the base catalyzed addition of water to a carbonyl group?" Let's go through that and see how it goes.
First let's check out the electronic structure of the OH-. If we find unshared pairs we can say that the OH- is a Lewis base and a nucleophile. That would tell us that an electron pair is ready to be used to make a new bond. Then we have the question of where that bond will go. In the carbonyl group, we know that the carbon is the more positive end of the C=O dipole, so let's try to make our new bond there.
This seems to be the right place to make the bond, but if we do that, the carbonyl carbon has five bonds (*). That means ten electrons in the valence shell, so it doesn't happen. We must find a way to reduce the number of bonds to four. We can think back to the mechanism we worked out for the acid catalyzed addition of water. In that mechanism we broke the pi bond by using its electrons to make a new bond to H+.
Let's look into the possibility that the C=O pi bond breaks as the C-O sigma bond forms. After all, the electrons in the pi bond are farther from the nuclei than those in a sigma bond, so they should be easier to push around.
Another way to say this is that the pi bond is weaker than the sigma bond, so it takes less energy to break it. The energy required to break the pi bond comes from making the new sigma bond to the OH-. (In the acid catalyzed case, the new bond which is being formed is the bond beween the carbonyl oxygen and the H+. Breaking old bonds is usually assisted by the formation of new bonds.
We finish this mechanism by making the only bond which is left to do, the O-H bond.
Here again, the breaking of one bond is assisted by the formation of another bond. Also, this mechanism makes a new OH- to replace the one which was used in the first step, consistent with the observation that the reaction is base catalyzed, which means that the OH- is not used up.
Hemiacetal Formation
Now let's use what we know about the acid catalyzed addition of water to make a prediction of what will happen when we mix an aldehyde with an alcohol and add a drop or two of an acid catalyst.
We want to use our mechanism to predict the structure of the product. Recall the mechanism of acid-catalyzed addition of water
If we follow closely the bonding changes that happen around the water oxygen, we notice that one of the hydrogens always stays attached. That bond doesn't break. That bond could just as well be a C-O bond as is the situation in an alcohol. What would happen if we just replaced water by methanol in the mechanism for acid-catalyzed hydration?
Indeed, the same mechanism seems to work just fine. The OH group in methanol is a nucleophile just like the OH group in water. Another way to say this is that the functional group of water is the same as the functional group of any alcohol, an OH group. If we learn a reaction for one alcohol, it will work very much the same way for any other alcohol - often including water as the smallest possible alcohol.
Acetal Formation
When we do the experiment to test our prediction, we find that yes, the product we have predicted is formed. But, we also find another product, one which hadn't been predicted.
(The new product is called an acetal. The one we predicted is called a hemiacetal.) Since the hemiacetal seems to be about halfway to the acetal, we'll explore converting the hemiacetal into the acetal.
If we look closely at the differences between these two products, we see that to do this, we need to replace the OH group of the hemiacetal with the OCH3 group of the acetal. (We can rule out the seemingly simpler alternative of replacing the H with the CH3 because isotopic labelling experiments show that the O-CH3 bond is not broken and the bond between the central carbon and the OH oxygen is broken.) This reaction is also acid catalyzed, so we may begin by making a bond between H+ and the OH oxygen, using the Lewis base electrons of the oxygen (electron pairs are often symbolized by bars).
In the next step the bond between the central carbon and the oxygen we have just broken is broken. This produces water and leaves the central carbon atom with only three bonds and a vacancy in its valence shell. A formal charge calculation tells us that this atom is also positively charged, but it is the electron pair vacancy which is more important.
The central atom is consequently an electrophile, open to making a bond with a nucleophile. The nucleophile is the oxygen of another molecule of methanol, whose unshared electron pair becomes the new carbon oxygen bond.
It remains to break the bond between the hydrogen and the positively charged oxygen which produces the acetal and replaces the H+ to regenerate the catalyst.
Just to have the whole process in one place, here's the full mechanism from the aldehyde through the hemiacetal to the acetal:
Some Principles
There are a couple of general ideas we can extract from what we did in working out this mechanism.
Charge and Reactivity: Let's contrast two positively charged molecules we found in this mechanism, one with three bonds to oxygen and one with three bonds to carbon.
Both molecules are positively charged, and formal charge calculations tell us where that charge is. But there is an important difference in their structures and that difference makes an important difference in their reactions.
The positively charged cation (we call it a carbocation) with three bonds has only six electrons in its valence shell. It is an electrophile. It makes a fourth bond using electrons from a nucleophile (the methanol oxygen atom here).
The positively charged oxygen (we call it an oxonium ion) has three bonds and an unshared pair. It has a complete octet in its valence shell. It needs no more electrons so it makes no more bonds. It is not an electrophile. Its reaction is to break a bond, keeping the electrons, by dropping off an H+.
This distinction is important. The charge alone does not tell us what to expect. It does provide a means of determining whether a positively charged atom has a vancancy or not. From that we can decide whether a such an atom is an electrophile -- ready to accept electrons to make a bond -- or not.
ROH as Nucleophile: Notice that whenever we used an alcohol (general formula, ROH) as a nucleophile, the order of the steps was:
1. Reaction of the unshared pair on oxygen to form a new covalent bond. The oxygen gets a positive charge.
2. Then an H+ is lost from the positively charged oxygen (oxonium ion).
Students often wonder why we don't reverse the order so that the H+ comes off first to make RO-, which would then act as the nucleophile. This does not happen because the reaction is occuring in an acidic solution. A srong base like RO- (about as basic as OH-) would be immediately consumed by reaction with the acid and would not survive. The alcohol (weaker base, about as basic as water) is the solvent, so there are many alcohol molecules surrounding the carbocation. This makes its reaction with an alcohol rather than RO- (called an alkoxide ion) very likely.
Intermediates: The aldehyde, the methanol, the hemiacetal, and the acetal are all stable molecules. They can be isolated and studied over a period of time. All the other molecules in this mechanism are much less stable. They have relatively high energies and thus short lifetimes. When they are formed, they react quickly. The main clue to this in their structures is that they have unusual bonding patterns such as three bonds to carbon (with a positive charge) instead of four, or three bonds to oxygen instead of two. Such molecules are called reactive intermediates; reactive because their unusual bonding pattern suggest that a change in bonding will happen, and intermediates because they appear between the reactants and the products.
Resonance
To finish up, let's return to one of our carbocation intermediates. Recall that we began thinking about acid catalyzed reactions of aldehydes by using a C-O pi bond to supply the electrons to make a new bond with H+.
There is an alternative. We could also think about using one of the unshared electron pairs on the oxygen atom.
Let's compare the two molecules we obtain in this way.
Notice first that the "connectivity" of these two structures is identical. That is, each atom is connected to exactly the same atoms in one structure as it is in the other. Then, notice that the only difference is in the location of an electron pair. On the left, an electron pair is shown as unshared on the oxygen. On the right, that pair is shown as making a pi bond.
When two (or more) structures differ only in the location of electrons, without changing which atoms are bonded to which other atoms, those structures are different ways of describing a single molecule. The symbols differ, but there is only one molecule being described. The notation for this is to connect the two structures by a double headed arrow, which does not imply equilibrium. (We can use the curved arrow symbol to keep track of the formal electron motions.)
This situation is called resonance. When resonance is involved the real structure (called a resonance hybrid) is more stable than any of the formal (called contributing) structures we might draw. This is called resonance stabilization.
Often, I will only draw one structure and expect that you will recognize that resonance is involved and know what other structures might be included in describing the resonance hybrid. The structure presented will be the one which most directly connects to the reaction being described.
We will use resonance many more times during the semester, so there will be many opportunities to clarify your understanding. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.04%3A_Acetal_Formation_Mechanism_Resonance.txt |
Imine Formation
Last time We looked at the formation of a hemiacetal and an acetal from an aldehyde and an alcohol. In both this case and the previous case in which we added water to an aldehyde, the first steps involved acid-catalyzed addition of a nucleophile to the carbonyl carbon of an aldehyde.
In many of the biological reactions of carbonyl groups the nucleophile is a nitrogen atom. The eventual outcome is different, so let's take a look at the details. The specific molecule as an example of a nitrogen nucleophile is methylamine. What happens if Nu = CH3NH2? Here's the complete mechanism.
Compare this mechanism to that for the formation of an acetal.
If we compare the individual steps in these mechanisms, we notice that they are very similar up until the last steps. Let's describe them in general terms:
1. Formation of a bond between the carbonyl oxygen and H+.
2. Attack of the nucleophile (N: or O:) on the electrophilic carbonyl carbon.
3. Loss of an H+ from the now positively charged N or O atom.
4. Formation of a bond between the OH oxygen and H+.
5. Cleavage of the bond between the now positively charged H2O and the central carbon atom. This forms a resonance stabilized intermediate.
6. Here the patterns diverge. With methylamine as the nucleophile, the nitrogen is still bonded to another hydrogen, which can be lost as H+. This forms a carbon-nitrogen double bond and the nitrogen forms its normal three covalent bonds. With oxygen from the alcohol as the nucleophile, there isn't an O-H bond to break at this point, so the reaction continues with a second molecule of alcohol reacting. Another way to say this is that forming three covalent bonds to oxygen results in a charged reactive molecule which doesn't persist.
We can learn from this that when a nucleophile adds to a carbonyl group in an aldehyde or ketone, it always adds to the electrophilic carbonyl carbon. What happens next depends on the structure of the nucleophile. If there are two hydrogens on its nucleophilic atom, it will eventually lose them with the formation of a double bond between the nucleophilic atom and the carbonyl carbon.
[A question -- Water has two hydrogen atoms. What happens if you apply the pattern just discussed to water and an aldehyde? Does this provide a pathway for exchange of oxygen between water and the aldehyde? What would happen if this were done with water labeled with Oxygen-18?]
Biological Example
There is a biological example of this reaction. In the metabolic synthesis of amino acids (we'll study these later, they're the components of proteins) the key molecule is glutamic acid. This is formed from a ketone (alpha keto-glutaric acid) by a two step process, the first of which is the reaction of the ketone carbonyl with ammonia to generate a carbon-nitrogen double bond. This double bond is then "reduced" (more about this later) by the addition of two hydrogen atoms to form glutamic acid. [In this description we have ignored the details of the carboxylic acid groups.]
You may wish to work through the mechanism of the first step of this example using the pattern above. Notice also that even though the carboxylic acid groups at the end of the carbon chains in these molecules are functional groups by any structural definition, they don't react here. We can learn from this that a reaction needs both the right functional group and the right circumstances to occur.
Another biological example is part of the process by which nitrogen atoms are transferred from glutamic acid to other carbonyl-containing molecules in the formation of other important amino acids. Here is an example of the reactants:
What do you anticipate that the product of this reaction would be? Base your answer on the fact that a carbon-nitrogen bond needs to be formed and that we have a pattern for what happens when a nitrogen nucleophile reacts with a carbonyl group. The completion of this process, which is called transamination, involves several other steps and will be covered when you take Biochemistry.
Cyanohydrin
If we look back over the reactions we've studied, we can see some consistent patterns emerging in the addition reactions of carbonyl groups.
1. The carbonyl carbon is electrophilic. Nucleophiles add there.
2. If acid is present, the first step is attack of H+ at the oxygen atom. The nucleophile adds to the carbonyl carbon, which is now quite electrophilic, in the second step of the reaction.
3. If acid is not present, the first step is the reaction of the nucleophile with the carbonyl carbon, a process in which the carbonyl oxygen becomes negatively charged. This step is followed by attachment of an H+ to that oxygen.
4. Later steps, which depend upon the structure of the nucleophile, determine whether the overall reaction is addition or replacement of the oxygen by the nucleophilic atom.
Equilibrium
Now that we have learned several reactions, we can take a look at two general questions:
1. What controls whether a reaction proceeds to the left or to the right? In other words, what controls the equilibrium constant for a reaction?
2. What controls whether a reaction goes rapidly or slowly? In other words, what controls the kinetics of a reaction?
The answer to the first question is simple in principle, but difficult to predict in practice. Rather, we commonly look at the outcomes of similar reactions and decide what factors seem to be important in explaining those outcomes. In doing so, it is important to examine the situation for contrasts -- what's different between the cases we are examinining -- and for similarities -- what's the same between the cases we are examining.
For example, if we look at the addition of water to formaldehyde, we find that almost all of the formaldehyde has been reacted when equilibrium is reached. The equilibrium constant is quite large, much greater than one. In contrast, if we look at the same reaction with acetone (2-propanone), we find that only a very small fraction of the acetone has reacted to add water. (This is so regardless of whether the reaction is catalyzed by acid or base. Remember that catalysts do not change the position of equilibrium.) We can symbolize these statements by making the arrow pointing towards the predominant product larger.
What might be responsible for the change in equilibrium behavior? The important idea here is that at equilibrium, more stable (lower energy) molecules predominate. We have to look at the changes in structure between reactants and products and the changes in structure between our two cases to understand what changes the relative energies of the molecules involved. Changes which raise a molecule's energy will reduce its concentration at equilibrium. Changes which raise a molecule's energy will increase its concentration at equilibrium.
Steric Effects
In working out such explanations we ignore things that are the same in the cases we are examining. For instance, the bonds made and the bonds broken are the same whether the reaction involves formaldehyde or acetone. The change which does occur in this reaction can be described as a change in the geometry of the carbonyl carbon as it adds water. In the reactant, it is trigonal, with 120o bond angles and sp2 hybridization. After water is added, the same carbon is tetrahedral with 109.5o bond angles and sp3 hybridization. This change in bond angle means that the groups attached to the carbon atom are closer together after water is added than before, so the electrons in those groups repel each other more in the product than in the reactant. The stronger this repulsion the higher the energy of such a molecule.
The most obvious structural difference between the formaldehyde and acetone cases is that in formaldehyde, the atoms attached to the carbonyl carbon are hydrogens, which are about as small as atoms get. In the same location in acetone, we have methyl (CH3 groups, which are considerably larger. Pushing the larger methyl groups closer together in the product of adding water to acetone requires more energy that pushing hydrogens close together in the formaldehyde case. Consequently, the equilibrium is much less favorable for the addition of water to acetone.
This kind of explanation uses steric effects, effects on a reaction which arise from the size of atoms or groups.
Polar Effects
Another important explanation of changes in equilibrium constant lies in polar effects. This can be seen in the following example:
Why is hexafluoroacetone more completely hydrated than acetone? Since a trifluoromethyl group (CF3) is about the same size as a methyl (CH3 group, steric effects are not involved. The structural difference between these two cases is the electronegativity difference between H and F. Consider that the carbonyl carbon is already somewhat electron deficient since it is associated with the more electronegative oxygen and two of its valence electrons are relatively far away in a pi orbital. When the electronegativity of six fluorine atoms is included in hexafluoroacetone, it becomes even more electron deficient. This raises its energy in comparison to that of acetone, so there is less of it at equilibrium. On the product side of the reaction, no pi bonds are involved, so the polar effect on energy is less pronounced.
We will use steric and polar effects to explain how energies change in a reaction and between reactions. Keep in mind that steric effects involve changes in size between comparable situations, and polar effects involve changes in electronegativity or similar electronic characteristics.
Rates - Activation Energy
Next, let's take up the question of what makes a reaction faster or slower. Several variables are involved here. If two molecules must collide (such as hydroxide ion and formaldehyde for example) then higher concentrations of either make collisions more frequent and the reaction more rapid. If we raise the temperature, most reactions go faster. These are variables can be studied by experiments in which concentrations and temperatures are changed and the number of moles of product formed in a given time is measured. From such experiments we can obtain a measure of the reaction's activation energy. The activation energy is the energy which must be put into a collision of two reacting molecules in order for the reaction step to occur. If the activation energy is high, few molecules collide with enough energy to react and the reaction is slow since few molecules of reactant become product in a given time interval. If the activation energy is low, many molecules collide with enough energy and become product in a given time interval. The following diagram illustrates these points.
A molecular collision which packs enough energy for the molecules to reach the top of the hill -- the transition state -- makes product. If the transition state has a lower energy, then the activation energy is lower and the more molecules will collide with the necessary energy; more molecules will become product. Since higher temperatures produce more energetic conditions, we have an explanation for the fact that reactions go faster at higher temperatures.
It follows that if we are to understand why one reaction is faster than another, we have to think about how structural differences between one reaction and another influence the relative energy of the transition states and thus the activation energies. The following two reactions are the first steps of possible mechanisms for the hydration of an aldehyde.
In the "coupled" mechanism to the left, the bond between carbonyl oxygen and the carbonyl oxygen is broken at the same time as the bond between the OH- group and the carbonyl carbon is formed. Bond breaking requires an energy input -- it is moving electrons away from a stable position -- while bond making produces an energy output. If the energy required for breaking the C=O pi bond can come largely from the energy released by making the new C-O sigma bond, the overall energy of the transition state will be lowered, the activation energy is lower and the reaction is faster..
Contrast this with the "uncoupled" mechanism at the right. Here the C=O pi bond is being broken without the energy input from a forming C-O sigma bond. The transition state is much higher in energy than that for the coupled mechanism. This mechanism has a prohibitively high activation energy, so the reaction without base is very slow. Adding base allows the coupled reaction mechanism to proceed over its much lower energy transition state. Look back over the mechanisms we have examined. Are bond breaking processes commonly coupled with bond breaking processes? One of the reasons for the dramatic success of enzymes in making reactions go very fast is that they use the energy of forming bonds very efficiently to drive bond breaking.
We'll find that explaining faster and slower reactions by looking at what happens to transition state energies to be very useful. I hope you'll use opportunities for questions to clarify these ideas in your own mind. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.05%3A_Nitrogen_Nucleophiles_-_Imine_Formation.txt |
Review of Reactivity
Last time we looked at a reaction in which a new carbon-carbon bond was made. Today, we'll look at another such reaction, one which is generally quite useful for synthesis, the assembly of larger carbon structures from smaller molecules.
First, let's look back over the reactions we've studied. We see some consistent patterns emerging in the addition reactions of carbonyl groups.
1. The carbonyl carbon is electrophilic. Nucleophiles add there.
2. If acid is present, the first step is attack of H+ at the oxygen atom. The nucleophile adds to the carbonyl carbon, which is now quite electrophilic, in the second step of the reaction.
3. If acid is not present, the first step is the reaction of the nucleophile with the carbonyl carbon, a process in which the carbonyl oxygen becomes negatively charged. This step is followed by attachment of an H+ to that oxygen.
4. Later steps, which depend upon the structure of the nucleophile, determine whether the overall reaction is addition or replacement of the oxygen by the nucleophilic atom.
Let's think a bit about the relationship between the presence of acid and the sequence of events. Remember that nucleophiles and Lewis bases react in the same way, by using an unshared electron pair to make a new bond. It isn't a surprise that molecules which are good (strong) Lewis bases are also good (strong) nucleophiles. Hydroxide ion (OH-) is a strong base, as are most compounds which share the -O- functional group. It is also a strong nucleophile, which we see in its unassisted reaction with a carbonyl carbon in the base-catalyzed addition of water.
It is tempting to think that we could set up a really fast reaction if we used both acid to attack the carbonyl oxygen with H+, which would make the carbonyl carbon really electrophilic, and base to attack the carbon with the strong nucleophile OH-. This sounds really attractive, but it doesn't work. What we've forgotten with this idea is that significant concentrations of acid and base can't exist in the same solution because they neutralize each other. We have to conclude that in acidic solutions, only weak bases like water can exist (the conjugate strong bases like those which include -O- functional group would be neutralized to give -OH groups). Such weak bases are also weak nucleophiles and need the increased electrophilic character which comes when the carbonyl oxygen is attached to an H+. This is the pattern we saw when weak nucleophiles like water and alcohols reacted in acid-catalyzed addition of water and in acetal/hemiacetal formation.
Another way to look at this is to say that if a strong nucleophile is to be used we must stay away from acidic solutions. We've seen this pattern both in the base-catalyzed addition of water and in the formation of a cyanohydrin. Addition of acid would have destroyed the nucleophiles (converted OH- to H2O and CN- to HCN) in these cases.
Grignard Reagent
We saw that the cyanide ion is a useful nucleophile and that its addition to a carbonyl group makes a carbon-carbon bond. Making carbon-carbon bonds is the central concern in organic synthesis, so it is important to find other compounds in which a carbon atom serves as a nucleophile. Let's think a bit about what that might mean.
A nucleophile needs to have a pair of electrons to donate in order to make a new covalent bond. A carbon nucleophile would need to have an unshared pair or a bonding pair in which the polarity of the bond was such that the carbon was a strongly negative end of the dipole. That would imply that the carbon should be bonded to an atom which is less electronegative than carbon itself. A quick glance at the periodic table suggests that the bond will have to be between carbon and a metal. While there are many metals, we will look at only one, magnesium.
In the early part of the 20th century, Victor Grignard, a French organic chemist (the French pronunciation of his name can be approximated as "greenyard") studied the reactions of bromoalkanes with magnesium metal. When he carried out these reactions in solutions with ether, he found that the magnesium dissolved, heat was released and the solution turned dark gray. If he added a ketone or aldehyde to this mixture, heat was again evolved and a light gray precipitate was formed. When he finished the reaction by adding aqueous acid to the mixture, he found that he had made an alcohol in which the carbon to which the bromine atom had been attached had now become bonded to the carbonyl carbon. The process is outlined as follows:
Since we have learned that the carbonyl carbon of aldehydes reacts as an electrophile, we must conclude that the carbon which started out attached to the bromine is behaving as a nucleophile. This can be understood if the magnesium is inserted between the bromine and the carbon, making a carbon magnesium bond which would be polarized so that the carbon is the negative end of the dipole. A compound with a carbon-magnesium bond is called a Grignard reagent. The details of the insertion of the magnesium atom into the carbon-bromine bond are not well understood, nor is the exact structure of the Grignard reagent itself. However, the reactivity of this reagent is symbolized effectively by the formula given.
Addition to Carbonyl Group
Let's look at the reaction of the Grignard reagent with the carbonyl carbon in a little more detail.
In the top reaction, the nucleophilic electron pair is shown as coming from the carbon-magnesium sigma bond, which is strongly polarized so that the electrons are much closer to the carbon. This makes the carbon nucleophilic. In the bottom reaction, the depiction of this bond is taken to an ionic extreme in which the electron pair is shown as entirely belonging to the carbon, which emphasizes the carbon's nucleophilic character. In this picture, the magnesium ion is "in the neighborhood" rather than being covalently bonded to the carbon. Such an extreme picture is probably and exaggeration, but it does emphasize the consideration of the attacking carbon as a nucleophile.
The picture of a carbon bearing an unshared electron pair also tells us that such a carbon would be a very strong base, much stronger than needed to take an H+ from water to generate the weaker base OH-. A practical consequence of this is that Grignard reagents must be kept dry, away from even the slightest traces of moisture, lest they be destroyed by reaction with water.
Now let's examine how we can use this overall reaction in synthesis. First, for economy in notation, let's develop a shorthand for the preparation of a Grignard reagent, its addition to a carbonyl group, and the reaction of that product with acid to make an alcohol. Here's the longhand version (repeated):
Now here's the shorthand version:
The shorthand version is interpreted to mean that
1. First we react bromomethane with magnesium metal using ether as a solvent. When this reaction is complete --
2. Second we add propanal (CH3CH2CHO) to this solution. After the second reaction is complete --
3. Third we add HCl to neutralize the O- and make the alcohol.
This represent a sequence of events in carried out in the laboratory. In each numbered step all the reactants present are allowed to react before the next reaction is started. At the completion of each numbered step, the product could (in principle) be isolated and stored to be used later. In practice the high reactivity of water with Grignard reagents makes this very difficult, so it is not done.
This way of describing a sequence of laboratory events must not be confused with the sequence of steps which we used to describe a mechanism. When describing a mechanism we are tracing the path of single molecule at a time. At a given time, very few molecules are actually reacting; most are "resting" as either reactants which haven't gotten enough energy to proceed or products which have finished passing through the series of mechanistic steps. Usually, context will tell you which type of interpretation is meant. If you are in doubt, please ask.
Thinking from Products to Reactants
Now, lets see how the addition of a Grignard reagent can be used in synthesis. If we look at the product of our shorthand description and remember that the addition of a Grignard reagent makes a new carbon-carbon bond in which one of the carbons is attached to an OH group, we can see that there are two such bonds (thicker and longer in the drawing) in our product molecule. Either is a candidate for being formed in the addition of the Grignard reagent. We can imagine the reagents needed for this to happen by simply erasing either of those bonds and examining what we get.
We remember that the bond came from an unshared electron pair on the Grignard reagent, and that the carbonyl carbon is electrophilic, so we show the appropriate charges. The structures we show here are not those of real molecules, but they serve to tell us which parts play what roles in the reactions. (The double-line arrows represent the direction of our thinking, not the direction of the actual reaction.)
In this way we arrive at a sketch of the reactants needed, which we need to turn into real reagents. We do this by remembering that the electrophilic carbon is provided by the carbonyl carbon and that the nucleophilic carbon comes from the Grignard reagent and then writing down what that tells us.
We finish up by placing the formation of the necessary Grignard reagent, its addition to the appropriate carbonyl compound, and the hydrolysis of the addition product in the appropriate sequence. This whole thought process is called "retrosynthetic analysis," but it is more simply regarded as thinking backward from a goal to a process which will get you there. It's a useful skill in organic chemistry and in many other areas.
If we do a few of these problems, we notice a pattern. If the carbonyl compound is formaldehyde, in the product there will be only one bond between the carbon attached to the OH group and another carbon (from the Grignard reagent). There will be two bonds between the OH-bearing carbon and hydrogens (from formaldehyde). Such an alcohol is called a primary alcohol, because the OH-bearing carbon is bonded to only one other carbon atom.
Similarly, if an aldehyde other than formaldehyde is used, the OH-bearing carbon in the product is bonded to two other carbons, and the alcohol formed is called secondary. Carrying this process one step further, the reaction of a ketone with a Grignard reagent gives a tertiary alcohol. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.06%3A_Addition_of_Organometallics_-_Grignard.txt |
Last time we saw how a nucleophilic addition of a carbon atom to the carbonyl carbon could be carried out through the use of a Grignard reagent. This time we'll look at oxidations and reductions of carbonyl groups and at the acidity of the alpha hydrogen atom
Symbolizing Oxidations & Reductions
We'll start by recalling what the terms "oxidation" and "reduction" meant in inorganic chemistry. Oxidation is usually used to describe a process in which electrons are removed from a molecule or atom. Here's an example:
$Fe^{2+} \rightarrow Fe^{3+} + e^-$
This is interpreted to mean that a ferrous (+2 ) ion has been oxidized to a ferric (+3) ion by the removal of one electron (e-). Similarly, reduction is used to describe a process in which electrons are added to a molecule or atom. An example might be:
$Cr^{6+} + 3e^- \rightarrow Cr^{3+}$
We would interpret this to mean that a chromium atom in oxidation level six is being reduced to chromium in oxidation level three by the addition of three electrons. Using these symbols we can keep track of the oxidation levels of atoms as electrons are added or removed. Keep in mind that these equations are "half reactions" and they don't indicate where the electrons come from or where they go.
There are important areas of biochemistry (photosynthesis, oxidative phosphorylation) where this symbolism for oxidation and reduction is very useful, but for most of organic chemistry where molecules have many atoms, keeping track of oxidation levels this way is cumbersome and not very useful. Instead, we will use the following definitions:
• Oxidation means the addition of oxygen to a molecule or the removal of hydrogen from a molecule.
• Reduction means the addition of hydrogen to a molecule or the removal of oxygen from a molecule.
Let's look at some examples:
What about a reaction in which both oxygen and hydrogen are added or subtracted. If the ratio is one oxygen to two hydrogens (in other words, water), neither oxidation or reduction is happening. Addition or removal of water does not involve, by itself, an oxidation or a reduction reaction. The addition of water to an aldehyde to form a hydrate does not involve oxidation or reduction. You may wish to look at the formation of an acetal or hemiacetal in this way. If the net change in the number of hydrogens and oxygens comes out to be a ratio of two hydrogens to one oxygen, neither oxidation nor reduction is involved.
Oxidizing Agents
Now, let's turn to what we have to do to make an oxidation reaction go. If we return to the example we had earlier, we are trying to figure out what reagent to add to an aldehyde in order to oxidize it to a carboxylic acid. Such a reagent is called an oxidizing agent. Experiments have shown that there are many such reagents, but one which is commonly and effectively used for this purpose is chromic acid (H2CrO4). This reagent is prepared in situ (in the reaction mixture) by mixing a strong acid such as H2SO4and a sodium or potassium salt of the chromate (CrO42-) or dichromate (Cr2O72-) ion. The reaction and the necessary reagents are shown as follows:
If a question asks for a reagent which will carry out a reaction, it is necessary to answer with a specific reagent. That means that you will need to develop a personal list of reagents and what they do. Both chromate and dichromate ions contain CrVI, so we recognize them as potential oxidizing agents based on the ability of CrVI to absorb electrons (and be reduced). Similarly, KMnO4- (permanganate) ions contain MnVII and are good oxidizing agents for many organic reactions.
A final point: most of the reactions we have illustrated for aldehydes also work with ketones. We have made note of the generally less favorable equilibrium constant for additions to ketones, but the additions of water, alcohols, amines, cyanide and Grignard reagents all proceed to measurable extents with ketones. This is not so with oxidation. To oxidize a ketone would require breaking a carbon-carbon bond between the carbonyl carbon and a carbon bonded to it. Mechanisms to do this involve such high activation energies that they do not occur to any practical extent.
Reducing Agents
What about reduction? Similarly, there are specific reagents for the reduction of aldehydes and ketones to alcohols. (See Brown, Sec. 11.10B). Two important ones are sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4). Their use is illustrated in the following two examples.
Since there is no carbon-carbon bond breaking which occurs in these reactions, we might expect that these reagents would also reduce ketones to alcohols. That is indeed the case, as seen in these examples.
If we compare the outcome of reduction of aldehydes to that of reduction of ketones, we notice that aldehydes produce primary alcohols while ketones produce secondary alcohols. This give us an alternate method to the Grignard addition for making these types of compounds. (Now would be a good time to make a personal list of reactions which produce alcohols, together with the necessary reagents and the specific structural types involved.)
Mechanism of Reduction
The comparison of these reduction reactions with the Grignard addition suggests another important point. In these reductions, hydrogen atoms are added to the carbonyl carbon and to the carbonyl oxygen. Since there is no hint of acid present, these reactions resemble base-catalyzed addition of water, addition of HCN, and the addition of a Grignard reagent rather than acid catalyzed reactions. We may expect them to begin with the attack of a nucleophile at the carbonyl carbon. In order to arrive at the observed product, the effective nucleophile has to be :H-, just as in the addition of a Grignard reagent, the nucleophile was effectively :R-. We justified that statement for the Grignard reagent by examining the carbon-magnesium bond and arguing that it was very strongly polarized so that the carbon effectively controlled the bonding pair of electrons, making it behave like :C-. Can we do something similar for the hydrogen in sodium borohydride or lithium aluminum hydride?
First, we need to realize that sodium borohydride is a salt which is made up of a sodium cation (Na+) and a borohydride anion (BH4-) The sodium ion plays no important role in the reaction, so we will ignore it (ions like sodium and potassium are seldom directly involved in reactions. They are present merely to maintain charge balance so that stable compounds can be added to reaction mixtures. They are often called spectator ions.) The borohydride ion is the important player in this process, and it is the B-H bond that we want to examine.
A quick glance at the periodic table to review electronegativities tells us that both boron and aluminum are metals with relatively low electronegativities. Each is less electronegative than carbon, and since hydrogen has about the same electronegativity as carbon we can conclude that the B-H bond is polarized with the boron positive and the hydrogen negative. This is very similar to the way we understood the electronic situation in a Grignard reagent, so we conclude that the B-H bond effectively serves as a source of hydride ion (:H-). This is summarized below.
With this information and the addition of a Grignard reagent as a pattern, we can arrive at the following mechanism for the reduction of an aldehyde by sodium borohydride.
Notice the familiar pattern: Attack by a nucleophile at the carbonyl carbon, followed by protonation of the carbonyl oxygen.
The same mechanism applies to the reduction of ketones by sodium borohydride and to reactions of carbonyl groups with lithium aluminum hydride. In the latter case, lithium aluminum hydride is itself highly reactive with water, so the water is added after the lithium aluminum hydride has reacted.
Enolate Ions
We have one major topic in carbonyl chemistry to introduce -- the reactions which occur at the carbon attached to the carbonyl carbon. This carbon is called the alpha-carbon. Here's an example of such a reaction.
We can pick out several important features of this reaction from this example. First, there are two alpha carbons, but only one undergoes a reaction. That one (on the left) is attached to a hydrogen in the reactant. Second, we notice that there is no reaction at the beta-C-H bonds. Reactivity is restricted to the C-H bond on a carbon directly to the carbonyl carbon. This type of reaction requires an alpha-C-H bond. In a sense, the presence of the carbonyl group next to this C-H bond makes it a functional group.
Now to consider a mechanism. Our mechanism must account for the formation of a carbon-bromine bond and an oxygen-hydrogen bond (in water) and the cleavage of a carbon-hydrogen bond (alpha) and a bromine-bromine bond. It must also suggest a role for the OH-. so let's begin by speculating that the hydroxide ion acting as a base makes a bond to the alpha-hydrogen. This process is coupled to the breaking of the alpha-C-H bond, whose electrons are then transferred to the alpha-carbon.
This is striking, because we are proposing that the alpha-hydrogen is acidic enough to be removed by a base no stronger than OH-. This is certainly not true of other C-H bonds, and it is only true here to the extent that the equilibrium constant for this step is very small. There is very little of the product of this step at equilibrium (we'll call this product an enolate, for reasons which will make more sense a bit later). What is it about an enolate ion which allows it to be made at all. Put another way, why is the enolate ion stable enough to be formed, while a similar ion made from another C-H bond isn't. When we're looking for reasons of stability, resonance is one of the ideas we should explore. Since this process requires the carbonyl group next door, that resonance should involve the carbonyl group.
Sure enough, if we move electrons as indicated by the curved arrows, we find an alternative resonance structure. This structure has the energetic advantage that the unshared electron pair is on the oxygen (the more electronegative atom) rather than the carbon. And, the resonance itself contributes to lowering the energy of the enolate, which makes it stable enough for a little of it to be formed at equilibrium. It is important to re-emphasize that none of this can happen if the carbonyl group is not next door to reactive C-H bond.
The alpha carbon in the enolate is all set up to behave as a nucleophile, providing a pair of electrons to make a new bond to a molecule of bromine. This requires that the bromine-bromine bond be broken at the same time. Notice that either resonance structure can be used to deliver the needed pair of electrons to make the carbon-bromine bond. This is necessarily so, since the two resonance structures are alternate descriptions of the same resonance hybrid.
Bromine is behaving as an electrophile, not because it has a vacancy in a valence orbital, but because it can simulate such a vacancy by breaking a bond. This reminds us of the behavior of the pi bond in a carbonyl group when it reacts with a nucleophile at the carbon atom.
Example $1$:
There are some questions that are routinely asked, so let's take them up here.
1. Why does the OH- attack the alpha C-H bond rather than the carbonyl carbon?
1. It attacks both, but the attack at the carbonyl carbon leads to the hydrate - the product of addition of water. This certainly occurs, but it leads nowhere since the hydrate is present in very small quantities at equilibrium for ketones and most aldehydes.
2. Why does the reaction with molecular bromine take place at the carbon only? After all, there is a potentially nucleophilic unshared electron pair on oxygen as well.
1. Reaction can take place at the oxygen, but the compound formed with a covalent bond between two rather electronegative elements is less stable than the one which results from reaction at the alpha-carbon.
Contributors
• Kirk McMichael (Washington State University) | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.07%3A_Oxidation_and_Reduction_alpha-C-H_acidity.txt |
Last time we worked through the reagents which oxidize aldehydes to carboxylic acids and the reagents which reduce aldehydes to primary alcohols and ketones to secondary alcohols. We also learned how enolate ions can be formed by the removal of an alpha hydrogen atom and how these enolate ions can act as nucleophiles toward bromine.
Enols
What happens if we leave out bromine? The alternate electrophile is effectively a H+ from water. It can react either at the alpha carbon (which bears some electron density) or at the oxygen (which also is a basic site). The first reaction just reverses the formation of the enolate, but the second reaction makes a new structure, an enol. Since either reaction produces an OH- ion, either reaction is catalyzed by base.
Notice that the product of this mechanism, an enol, is an isomer of the original ketone. It is not a resonance structure, because there is no OH bond in the ketone, so the two structures are differ in how they are connected, and resonance structures may not be that different. Notice also that this is an equilibrium between isomers. Isomers that are in equilibrium with each other are called tautomers. This situation is called keto-enol tautomerism (the same term is used when the carbonyl component is an aldehyde.) Generally, there is very little enol in equilibrium with the keto tautomer, and we will generally write an enol structure when we need it.
Since addition of water was catalyzed by both acid and base, we can ask whether keto-enol tautomerism can also be catalyzed by acid. We know what the first step must be, because an H+ must react with the carbonyl oxygen. This moves electrons away from the carbonyl carbon, which in turn attracts electrons from it's immediate neighborhood, particularly the alpha C-H bond. This bond is thus susceptible to attack by the weakly basic water molecule (there's very little hydroxide ion in an acidic solution) to form the enol.
Of course, since a catalyst cannot change the position of equilibrium, the amount of the enol tautomer formed is small whether the reaction is acid catalyzed or base catalyzed.
The term "enol" is a combination of the "ene" ending typical for alkenes (compounds with C=C double bonds) and the "ol" ending typical for alcohols. The anion derived from an enol has the ending "ate" appended as is the case for many polyatomic anions (consider sulfate as an example).
Aldol Addition
One way to describe what we have just done is to say that we have worked out a way to make the alpha-carbon of a carbonyl compound into a carbon nucleophile. Earlier we found that the carbon nucleophile of a Grignard reagent was very important as a synthetic tool. Can we use this alpha-carbon nucleophile similarly? To some extent, the answer is yes. There are a few restrictions:<
1. To use the strongest nucleophile we can form at the alpha carbon, we will use base catalysis so as to make enolate ions.
2. Since there is little enolate ion at equilibrium, we will use the more reactive carbonyl compound -- an aldehyde, rather than a ketone.
The mechanistic pattern is as follows:
The first step forms the enolate ion. The second step makes a carbon-carbon bond by nucleophilic attack of the enolate alpha-carbon on the carbonyl carbon of a second molecule of the aldehyde. The third step places a proton on the now negatively charged oxygen and regenerates the basic catalyst.
Let's look at the overall reaction and relate the structure of the product to that of the reactant.
Two molecules of an aldehyde combine so that the alpha-carbon of one becomes attached to the carbonyl carbon of the other. The carbonyl oxygen becomes an alcohol group, located on the beta carbon (relative to the remaining carbonyl group) of the product. The overall reaction is an equilibrium, as we would expect since all steps in our mechanism are equilibria. The equilibrium favors aldol product when the carbonyl group is an aldehyde, but does not for ketones.
If we are seeking to use the aldol addition for synthetic purposes, we look for the characteristic beta-hydroxy-aldehyde arrangement of functional groups. Then we check to see that the R-groups are in fact identical. If these criteria are met, then we can work backwards to the needed aldehyde by mentally breaking the "thick" bond -- the one which was formed in the actual addition step. That gives us our reactant. Incidentally, the term aldol stems from the "ald" of aldehyde and the "ol" of alcohol.
Aldol Condensation
The particular example we have discussed is capable of one more reaction. If we heat the aldol addition product, it usually loses a molecule of water to make a double bond between the alpha and beta carbons of the former aldol. This is a dehydration reaction, and it is also an example of an elimination reaction. An elimination reaction is the structural opposite of an addition reaction, and we'll look at typical mechanisms for these reactions later
Notice that this elimination reaction requires that the alpha carbon of the aldol product have attached a hydrogen. Since that carbon already lost one hydrogen in forming the enolate intermediate essential to the aldol addition, it must have had two hydrogens in the original aldehyde. The product of this second reaction is called an alpha-beta-unsaturated aldehyde (unsaturated because of the carbon-carbon double bond), and the overall reaction is called an aldol condensation. The term condensation is used to describe a reaction in which two molecules combine to form a larger product with the loss of a molecule of water.
The key structural feature of a molecule which might be made by way of an aldol condensation is the carbon-carbon double bond between the alpha and beta carbons of an aldehyde. If a synthetic target has that feature, we need to check to see if the R-groups are identical. If so, we can make that molecule by the aldol condensation.
In your later biochemical studies you will find that carbon-carbon bonds are often made (in photosynthesis for example) by enzymatically catalyzed aldol additions. The reverse reaction is also common (called a retro-aldol reaction), particularly in glycolysis, the metabolism of glucose. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.08%3A_Enolates_Aldol_Condensation_Synthesis.txt |
Structures of Carboxylic Acid Derivatives
Last time we completed our study of the reactions of aldehydes and ketones, compounds in which a carbonyl group is bonded either to carbons or hydrogens. The typical reaction pattern for these compounds was addition, with a nucleophile adding to the carbonyl carbon and an electrophile adding to the carbonyl oxygen. Today we'll look at carboxylic acid derivatives. This group of compounds also contains a carbonyl group, but now there is an electronegative atom (oxygen, nitrogen, or a halogen) attached to the carbonyl carbon. This difference in structure leads to a major change in reactivity. Here we find that the reactions of this group of compounds typically involve substitution of the electronegative atom by a nucleophile. Before looking at that reaction in detail, though, let's see what kind of compounds we're talking about.
Notice that each of these functional groups has either an oxygen, a nitrogen, or a halogen attached to the carbonyl carbon.
Substitution by Addition-Elimination
The typical reactions of these compounds are substitutions -- replacing one of these heteroatoms by a another atom. Here's an example:
The chlorine of the acyl chloride has been replaced by the -OCH2CH3, more specifically by the oxygen atom of that group. This type of reaction, in which an atom or group is replaced by another atom or group, is called a substitution reaction.
We can begin to connect this reaction type with what we have seen earlier by thinking about the mechanism. We notice that the O- end of the group (called an alkoxide) which is doing the substituting is very much like the oxygen in an OH-. Since we've seen the OH- act as a nucleophile when it attacked the electrophilic carbon of a carbonyl group, let's begin by seeing what happens if we use the same approach here.
The first step is familiar from aldehyde and ketone chemistry. The nucleophilic oxygen uses its electrons to make a new bond to the electrophilic carbonyl carbon while the pi bond's electrons move to the carbonyl oxygen. We've made the necessary oxygen-carbon bond. In the next step, the pi bond is reformed, and the carbon-chlorine bond is broken. This is a new type of step, and it happens when breaking this bond is eased by the electron pair being attracted to an electronegative atom such as oxygen, nitrogen or a halogen. This step is called an elimination. The overall substitution process occurs by an addition-elimination mechanism which begins with a nucleophilic addition to the carbonyl group and finishes with the departure of an atom with the bonding pair of electrons. This atom or group is called a "leaving group."
Here's a general statement of the mechanism:
As we look at some specific examples, keep this pattern in mind. There will be some elaborations on it, but we will always find an addition step in which the nucleophile attacks and an elimination step in which the leaving group leaves.
Esters - Preparation and Mechanisms
The conversion of acyl chlorides to esters is more commonly carried out by using an alcohol rather than an alkoxide (RO-).
The mechanism of this reaction starts just the same way the earlier one did; the first step is attack of the nucleophile at the carbonyl carbon. In this instance, the nucleophile is an unshared electron pair on a neutral oxygen atom. The intermediate formed in this step rapidly shifts a proton (H+) to the O-. Such transfers of protons between oxygen atoms or nitrogen atoms are fast. (These intermediates are called "tetrahedral intermediates" since carbonyl carbon has been changed to a tetrahedral geometry and an sp3 hybridization.)
The tetrahedral intermediate loses HCl in a single step, one in which the H+ is transferred to a second molecule of alcohol and the Cl comes off as Cl-. It is important to notice that the neutral alcohol oxygen serves as the nucleophile. The O-H bond is not broken until after the C-O bond is formed. There is never any alkoxide in this reaction. Indeed, an alkoxide ion could not survive in the strongly acidic (HCl) solution. This pattern, neutral nucleophile attacks first, then the proton is removed, is very common for neutral nucleophiles and must be followed.
This is a practical and useful method for making esters, but it does make the strong acid HCl, which is often troublesome. A more practical variation is to add a weak base such as pyridine to react with the HCl and neutralize it. This gives us a procedure for making esters from acyl chlorides which uses the following reaction statement:
A similar procedure is used to make amides from acyl chlorides and amines (the amine must have at least one hydrogen attached to the nitrogen).
Acyl chlorides are the most reactive carboxylic acid derivatives. The electronegative chlorine atom pulls electrons toward it in the C-Cl bond, which makes the carbonyl carbon more electrophilic. This makes nucleophilic attack easier. Also, the Cl- is an excellent leaving group, so that step is also fast. Because of their reactivity, acyl chlorides are easily converted into esters and amides and are thus valuable synthetic intermediates. They are made from carboxylic acids by this reaction (Atkins & Carey, Sec 12:10):
In the mechanism we just looked at, the key steps (attack of a nucleophile and departure of a leaving group) were accompanied by steps in which protons moved from one location to another. Such proton transfers are very common in acid catalyzed reactions. Here's the mechanism for the acid catalyzed formation of an ester from a carboxylic acid and an alcohol:
Notice that the nucleophilic attack is preceded by protonation of the carbonyl oxygen. We've seen this step before in the acid-catalyzed additions of nucleophiles to carbonyl groups. Its purpose is to increase the reactivity of the carbonyl carbon as an electrophile, so that it can be easily attacked by the alcohol oxygen. After nucleophilic attack, there is a proton transfer. Its purpose is to make one of the OH groups (either will do) into a good leaving group, water.
Think back to the addition of alcohols to aldehydes to produce hemiacetals. The two mechanisms start off identically. Compare them in detail, and work out the reasons for the different outcome.
Notice that each step in this mechanism is presented as an equilibrium. That means that the whole reaction represents an equilibrium in which significant amounts of carboxylic acid and alcohol co-exist with ester and water.
This allows us to push the reaction one way or the other by controlling concentrations, particularly of water. If we remove water from the reaction mixture, more ester is formed because carboxylic acid and alcohol react to replace the water we have removed. The resulting formation of ester is called Fischer esterification.
If we add water to the reaction mixture, equilibrium is restored by the production of more carboxylic acid and alcohol. This is called acid catalyzed ester hydrolysis.
Amides - Preparations
Esters can also react with amines or ammonia to form amides. This reaction doesn't involve acid catalysis, so the first step is nucleophilic attack at the carbonyl carbon. Proton transfer follows and loss of the alcohol portion of the ester.
This gives us two ways to make amides, this one from esters and the earlier one from acyl chlorides. Here's a third and very direct way to make amides, by heating carboxylic acids and amines together. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.09%3A_Carboxylic_Acid_Derivatives-_Interconversion.txt |
Reduction of Esters - Lithium Aluminum Hydride
We've seen that carboxylic acid derivatives react with nucleophiles to give substitution products in which the leaving group is replaced by the attacking nucleophile.
This same pattern describes the first steps in the reaction of esters with lithium aluminum hydride and Grignard reagents, but in both cases the reaction proceeds further because the first product formed also reacts with the reagent. For an example, lets look at the reduction of an ester with lithium aluminum hydride.
When the "hydride ion" (H:-) from lithium aluminum hydride replaces the OR' group of the ester, an aldehyde is formed. We've already seen that and aldehyde is reduced by lithium aluminum hydride, so it comes as no surprise that the aldehyde is immediately reduced to a primary alcohol. In fact, the aldehyde is more electrophilic than the ester, so as soon as a few molecules of aldehyde are formed, they are attacked by the hydride in preference to the ester. The reaction is completed by the later addition of aqueous acid to protonate the O- atoms.
The result is that esters are reduced by lithium aluminum hydride to primary alcohols in which the ester carbon has become the alcohol carbon. Sodium borohydride is not reactive enough to carry out this reduction. This is a useful way to make primary alcohols.
Reaction of Esters with Grignard Reagents
A very similar process occurs when an ester reacts with a Grignard reagent. Remember that the Grignard reagent serves as a source of carbon nucleophiles. We can use the same pattern as worked for the lithium aluminum hydride reduction by replacing the "H:-" with the "C:-" from the Grignard reagent. Here's an example in which the Grignard reagent is methylmagnesium bromide:
Here the first product is a ketone in which the OR' group of the ester has been replaced by the alkyl group of the Grignard reagent. This ketone is immediately attacked by another molecule of the Grignard reagent. After neutralization with aqueous acid the product is an alcohol in which the two identical groups attached to the alcohol carbon are from the Grignard reagent. If the R group of the ester is a carbon group this will be a tertiary alcohol.
In both these cases, we've combined a the general pattern for carboxylic acid derivatives, substitution of the nucleophile in place of the leaving group, with a pattern which applies to aldehydes and ketones, to arrive at "double" reaction. The first part is substitution and the second part is addition. We can't stop the reaction halfway, because aldehydes and ketones are more reactive than esters, so the aldehydes and ketones gobble up the reagent faster than the esters.
Claisen and Dieckmann Condensations
We spent some time on the aldol condensation when we were studying aldehydes, and we learned that the alpha C-H bonds of aldehydes and ketones can be attacked by base to form enolates. The same thing can occur with esters, although we need to be very specific about the identity of the base we use -- it needs to be the same as the OR' (alcohol derived) portion of the ester. (Consider what would happen if the base and the OR' group were different. The base would attack the carbonyl carbon and would replace the OR' group. A new and unwanted product would appear. If the base is the same as the OR' group, attack by the base on the carbonyl carbon results in no net change and keeps the reaction simpler.)
The first step in such a reaction is strictly analogous to making an enolate from an aldehyde. This is followed by attack of the nucleophilic alpha carbon of the enolate on the carbonyl carbon of a second molecule of the ester. Again, this is strictly analogous to the situation in the aldol reaction, and it has resulted in a nucleophilic addition to the carbonyl carbon. As is typical for carboxylic acid derivatives, the next step is loss of the leaving group, so that the carbonyl group which has been added to shows up as a ketone. (Since the newly-formed ketone is in a beta position relative to the ester functional group these compounds are called beta-keto-esters.) This overall reaction is called the Claisen condensation after Ludwig Claisen, a prominent German organic chemist of the late 19th and early 20th centuries.
When there are two ester groups on the same carbon chain (a diester), the enolate formed at the alpha carbon atom of one ester group can react with the carbonyl carbon of the other ester group.
The resulting Claisen condesation makes a ring of carbon atoms. If this ring has five or six carbons in it, it forms readily. This (intramolecular variation of the Claisen condensation is called the Dieckmann condensation. Since many organic compounds which occur in nature have five or six membered rings, the Dieckmann condensation has been widely used in synthesis.
Decarboxylation and Ketone Synthesis
Notice that both the Claisen condensation and its Dieckmann condensation variant make a compound we can call a beta-keto-ester, that is, an ester with a keto group on the carbon two atoms away from the carbonyl carbon. Such beta-keto-esters can be hydrolyzed to beta-keto-acids by base-promoted hydrolysis (saponification) and acidification. The beta-keto-acids readily lose a molecule of carbon dioxide when heated to form a ketone. This means that a Dieckmann or Claisen condensation can be followed by a hydrolysis-decarboxylation step to form a ketone. This gives us a way to make ketones. Notice that the new carbon-carbon formed in this sequence is one which joins a carbonyl carbon to one of its alpha carbons. Here's an example of the use of this synthetic sequence of reactions to make cyclohexanone:
We'll look at some of the details of the saponification reaction next time. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.10%3A_Carboxylic_Acid_Derivatives_-_Alpha_Carbon_Reactions.txt |
Biosynthesis of Fats
We've seen that carboxylic acid derivatives react with nucleophiles to give substitution products in which the leaving group is replaced by the attacking nucleophile.
This same pattern describes the first steps in the reaction of esters with lithium aluminum hydride and Grignard reagents, but in both cases the reaction proceeds further because the first product formed also reacts with the reagent. For an example, lets look at the reduction of an ester with lithium aluminum hydride.
When the "hydride ion" (H:-) from lithium aluminum hydride replaces the OR' group of the ester, an aldehyde is formed. We've already seen that and aldehyde is reduced by lithium aluminum hydride, so it comes as no surprise that the aldehyde is immediately reduced to a primary alcohol. In fact, the aldehyde is more electrophilic than the ester, so as soon as a few molecules of aldehyde are formed, they are attacked by the hydride in preference to the ester. The reaction is completed by the later addition of aqueous acid to protonate the O- atoms.
The result is that esters are reduced by lithium aluminum hydride to primary alcohols in which the ester carbon has become the alcohol carbon. Sodium borohydride is not reactive enough to carry out this reduction. This is a useful way to make primary alcohols.
A very similar process occurs when an ester reacts with a Grignard reagent. Remember that the Grignard reagent serves as a source of carbon nucleophiles. We can use the same pattern as worked for the lithium aluminum hydride reduction by replacing the "H:-" with the "C:-" from the Grignard reagent. Here's an example in which the Grignard reagent is methylmagnesium bromide:
Here the first product is a ketone in which the OR' group of the ester has been replaced by the alkyl group of the Grignard reagent. This ketone is immediately attacked by another molecule of the Grignard reagent. After neutralization with aqueous acid the product is an alcohol in which the two identical groups attached to the alcohol carbon are from the Grignard reagent. If the R group of the ester is a carbon group this will be a tertiary alcohol.
In both these cases, we've combined a the general pattern for carboxylic acid derivatives, substitution of the nucleophile in place of the leaving group, with a pattern which applies to aldehydes and ketones, to arrive at "double" reaction. The first part is substitution and the second part is addition. We can't stop the reaction halfway, because aldehydes and ketones are more reactive than esters, so the aldehydes and ketones gobble up the reagent faster than the esters.
We spent some time on the aldol condensation when we were studying aldehydes, and we learned that the alpha C-H bonds of aldehydes and ketones can be attacked by base to form enolates. The same thing can occur with esters, although we need to be very specific about the identity of the base we use -- it needs to be the same as the OR' (alcohol derived) portion of the ester. (That will make a bit more sense later).
The first step in such a reaction is strictly analogous to making an enolate from an aldehyde. This is followed by attack of the nucleophilic alpha carbon of the enolate on the carbonyl carbon of a second molecule of the ester. Again, this is strictly analogous to the situation in the aldol reaction, and it has resulted in a nucleophilic addition to the carbonyl carbon. As is typical for carboxylic acid derivatives, the next step is loss of the leaving group, so that the carbonyl group which has been added to shows up as a ketone. (Since the newly-formed ketone is in a beta position relative to the ester functional group these compounds are called beta-keto-esters.) This overall reaction is called the Claisen condensation after Ludwig Claisen, a prominent German organic chemist of the late 19th and early 20th centuries.
While this reaction is of considerable use for synthesis in organic chemistry, we are going to turn our attention to the involvement of this pattern in the biological synthesis of fatty acids.
First we need to know something about the structures of fats. Fats are esters in which the OR' (alcohol) portion of the molecule is the trihydroxy compound, glycerol (HOCH2CHOHCH2OH). There are three carboxylic acid portion in these molecules since each glycerol molecule can be esterified with a different carboxylic acid. They are consequently called triacylglycerols or triglycerides. These carboxylic acids typically have long hydrocarbon chains of fourteen to twenty carbons (including the carbonyl carbon). These chains are typically unbranched and, most strikingly, they contain even numbers of carbon atoms. The carboxylic acids obtained by hydrolysis of fats are called fatty acids.
Since most carbons in fats are completely reduced (maximum number of hydrogens, no oxygens), the oxidation of fats produces substantial quantities of energy, about double the energy per gram as is obtained from the more oxidized carbohydrates and proteins. Fats are thus used for long-term energy storage in organisms.
The observation that fatty acids have even numbers of carbons in them provides a clue as to the mechanism of their biological synthesis. The most obvious idea is that fatty acids are built up two carbons at a time, which would mean that each fatty acid would have to have an even number of carbons. A common two carbon molecule in biochemistry is acetic acid. (At the near acid/base neutrality which prevails in biological systems, acetic acid is present as it conjugate base, acetate ion).
This speculation was verified by feeding organisms with radioactively labeled (14C) acetate. The radioactive label appeared in the fats. More specifically, if the 14C was placed in the carbonyl carbon of the acetate, it appeared in the odd numbered locations in the fatty acid. If it was placed in the methyl carbon of the acetate, it appeared in the even numbered carbons in the fatty acid.
This is interpreted to mean that the fatty acid is built up two carbons at a time. Each two carbon unit is added by reacting the alpha carbon of one acetate unit with the carbonyl carbon of the next -- exactly the pattern which we see in the Claisen condensation.
Of course, there is much more to the process than this, but the key bond forming step has the same pattern as the Claisen condensation. That is, attack by nucleophilic carbon atom at the carbonyl carbon of an ester, which is followed by loss of the alcohol portion of the ester as a leaving group. In the the biological reaction, the esters have sulfur in them rather than oxygen and are attached to enzymes or other proteins, and the nucleophilic carbon is flanked by not one but two carbonyl groups. Nevertheless, the underlying pattern is the same.
The biological Claisen condensation we have looked at is followed by further biochemical steps which replace the carbonyl oxygen with two hydrogens. The four carbon ester which is thus produced is attacked at its carbonyl carbon by another molecule of the carbon nucleophile and the cycle continues.
It is important to realize that biological chemistry uses the same mechanisms that we use in the laboratory. Organic reaction mechanisms underlie all of biochemistry, particularly metabolism.
Saponification and Soaps
Next, we'll look at a reaction of fats which is of great economic importance. Nobody knows when the first soap was made, but it was very important in the Middle Ages and early industrial period because it was used to wash out wool fat from wool in preparation for dyeing and spinning. It was too expensive in those times to use for general cleanliness. The general availability of soap has led to a rise in standards of cleanliness and a corresponding rise in general health.
Soap is made by treating fats with a strong solution of lye (NaOH). The ester functional groups are hydrolyzed releasing its alcohol portion as glycerol and the acid portions as the a mixture of the sodium salts of the fatty acid. The glycerol is water soluble and is separated from the fatty acid salts. These are nearly insoluble in water, and are washed and compressed into a cake -- soap.
The mechanism of this reaction, which is called saponification) can be illustrated more easily on a simple ester than a fat. It follows the normal pattern for a carboxylic acid derivative and the first step is analogous to base-catalyzed hydration of aldehydes -- attack of the nucleophile hydroxide ion at the carbonyl carbon atom. This is followed by the usual departure of the leaving group (in this case, the OR'- of the ester). Since the OR'- is more basic than RCOO-, the last step is a neutralization reaction. (The last step will make more sense after next time.)
Detergent Structure
To finish up, we need to look at how soaps work and how that relates to their structures. Soaps are "bridge" molecules. Their long hydrocarbon chains are very much like oils, so they mix well with oils. In fact, we can say that the chains of soap dissolve in oils. At the same time, their ionic heads are polar enough to dissolve in water. If we imagine a small globule of oily material which has dissolved a bunch of soap molecules, we can see that the polar ionic heads will stick out (insoluble in oil) at the surface. These polar bumps on the surface attract water molecules, so that the oily glob is coated with water molecules that are held there by the ions. The glob is now ready to be washed away in water. Since much of the dirt we hope to get rid of is held in place by oily films, removing the oils removes the dirt.
The structural requirements for a soap (or more generally, for a detergent) are a long hydrocarbon tail (12 or more carbons) and a polar (often ionic) head. In synthetic detergents, the hydrocarbon tail is usually formed by linking several ethylene molecules together and attaching this to a benzene ring. The polar part is derived by covalently bonding the sulfur of a sulfate ion to the benzene ring. We'll learn more about the chemistry of making these later. If there are few branches on the hydrocarbon tail, the synthetic detergent will be biodegradable, an important characteristic. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.11%3A_Fats_Fatty_Acids_Detergents.txt |
Oxidation of Primary Alcohols
There are two types of reactions which can make carboxylic acids. We will look first at those that depend upon oxidation of groups from a lower oxidation level (fewer oxygens, more hydrogens). Then we'll see that there are reactions which make a carbon carbon bond on the way to making a carboxylic acid.
As you might suspect, carboxylic acids can be made by oxidizing groups which are less oxidized. The most important of these is the primary alcohol group. A typical reaction is:
We recognize the oxidizing agent, potassium dichromate, as the same reagent that was used in oxidizing aldehydes to carboxylic acids. Chromium is in oxidation level six in dichromates and it is reduced to oxidation level three in this process. The details of the reduction of chromate are complex, and we will not be concerned with them. The oxidation of aldehydes also produces carboxylic acids, but since aldehydes are less readily made than carboxylic acids, this process is not used much.
Oxidation of Side Chains on Aromatic Rings
Another oxidative process seems at first not to involve a functional group. Alkyl groups (usually methyl groups) attached directly to an aromatic ring are also oxidized to carboxylic acids. Since the methyl group contains only sigma C-H bonds, it doesn't look like a likely place for reaction. It is influenced by the neigboring aromatic ring, though, so reaction does occur. This is somewhat like the special reactivity of the alpha C-H bonds of a ketone or aldehyde. Here's an example:
Remember, this reaction requires the presence of an aromatic ring next to the alkyl group which is to be oxidized. Any carbons beyond the first one are lost in this process, which is one of the few reactions which breaks a carbon-carbon bond.
Making a Carbon-Carbon Bond
There are two reaction sequences which make carbon-carbon bonds on the way to carboxylic acids. The first is another use of the Grignard reagent. Remember that Grignard reagents react with carbonyl compounds to make alcohols. While we don't usually think of it that way, carbon dioxide is a a carbonyl compound (O=C=O). If a Grignard reagent is used to deliver a nucleophilic carbon atom to the carbonyl carbon of the carbon dioxide, we get a carboxlic acid (after quenching with aqueous acid).
The attack of the Grignard reagent on the carbon dioxide is directly analogous to the same step in reactions of other carbonyl compounds:
Here's the entire sequence presented in the more compact format we used for Grignard additions earlier:
Notice that the product has one more carbon than the bromoalkane we started with, and that this carbon is in a carboxylic acid group.
There is also another way to make a new carbon-bond and wind up with a carboxylic acid. If we have a primary alkyl halide (primary means that the carbon which is bonded to the halogen is bonded to only one other carbon atom, it's other two bonds are to hydrogen), we can react it with sodium cyanide. The cyanide ion will replace the halogen, and this makes a new carbon-carbon bond. The product is called a nitrile. Its carbon-nitrogen triple bond can be hydrolyzed with aqueous acid to produce a carboxylic acid. The sequence is as follows (normally, the nitrile is isolated and then hydrolyzed in a separate reaction):
We won't look at the details of the mechanisms of either of these reactions now. You might usefully speculate about the hydrolysis of the nitrile. You could begin by thinking of the C-N triple bond as being like a carbonyl group. After all, it does include a pi bond between carbon and an electronegative atom, much like the pi bond in a carbonyl group.
Acid and Base Strengths
The remaining issue with carboxylic acids is to understand why they are acidic. We can experience their acidity by tasting vinegar -- which is a dilute solution of acetic acid in water. The Lowry-Bronsted model of acids is useful for this. Let's refresh our memories on this model and on the interpretation of the term pKa.
When we think about acids using the Lowry-Bronsted model, we think of a molecule which can donate a proton (H+). A base is a molecule which can accept a proton (using the pair of electrons which are the defining feature of a Lewis base). Stronger acids donate protons readily to stronger bases. The products of this transaction are weaker acids and weaker bases. Here's the pattern.
Notice particularly that a strong acid is strong because it readily donates a proton. That means that its conjugate base (the base which remains after the proton is gone) is weak. After all, if the base were strong, it would grab the proton back and the acid couldn't give it away. When we describe an acid as strong, by saying that it has a small pKa, we are also saying that its conjugate base is weak. HCl is a strong acid (pKa -7). When we say that, we are also saying that Cl- is a weak base. The conjugate base of an acid whose pKa is small or negative is a weak base. This means that we can use a pKa table like Table 2.1 on p 43 of Brown to keep track of both acidity (strong acids have small or negative pKa's) and basicity (weak bases come from strong acids with small or negative pKa's).
Carboxylic Acid Acidity - Carboxylate Ion Basicity
This also means that we can restate the question "Why are carboxylic acids acidic?" to say "Why are the conjugate bases of carboxylic acids such weak bases?" To put this into context, notice that carboxylic acids have pKa's of about 5, while water and alcohols have pKa's of about 16. What makes the conjugate base (we'll call it a carboxylate ion) of a carboxylic acid so weak compared to a hydroxide (OH-) or alkoxide (RO-) ion?
As usual, we'll try to find our explanations in structure. There is one obvious difference between a carboxylate ion and an alkoxide ion. The carboxylate ion has two electronegative oxygens to only one for the alkoxide ion. These electronegative atoms would hold the electron pairs more tightly, which means that the electron pairs would be less available to make a bond to a proton. Less available electron pairs means a weaker base.
That accounts for some of the weakness of carboxylate ions as bases, but there's also a more subtle feature. Notice that we can move electrons between pi-bonding and unshared pair situations without changing the structure of the carboxylate ion. We recognize this as resonance and we recognize that it will lower the energy of a carboxylate ion compared to that of an alkoxide ion in which such resonance is not possible. Lower energy means more stable, more easily formed and less reactive, all of which adds up to a weaker base. Consequently the conjugate carboxylic acid is stronger than one whose conjugate base doesn't have the resonance possibility.
The somewhat paradoxical outcome of this is that carboxylic acids are stronger acids than alcohols because carboxylate ions, their conjugate bases, are weaker bases than alkoxides. This is due in large part to resonance stabilization of the carboxylate ions, which cannot happen in alkoxides.
This understanding of the structure of carboxylate ions also helps us understand how it is that when a Grignard reagent reacts with carbon dioxide, only one of the two carbonyl groups reacts. The product of this reaction is a carboxylate ion. It is resonance stabilized so the "real" structure is about halfway between the two resonance structures. That means that each C-O bond is halfway between a single and a double bond -- a bond and a half. Such a bond would be much less reactive than the double bond of a carbonyl group, so it isn't surprising that the Grignard reagent reacts with carbon dioxide in preference to the carboxylate ion. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.12%3A_Carboxylic_Acids.txt |
Making Alcohols
Here's a list of reactions we've already seen which make alcohols. The links will take you back to a previous lecture where the specific reactions are discussed.
Reaction Reactants Product
Grignard Addition Ketone, Grignard Reagent Tertiary Alcohol
Grignard Addition Aldehyde, Grignard Reagent Secondary Alcohol
Grignard Addition Formaldehyde, Grignard Reagent Primary Alcohol
Hydride Reduction Ketone, NaBH4 or LiAlH4 Secondary Alcohol
Hydride Reduction Aldehyde, NaBH4 or LiAlH4 Primary Alcohol
Hydride Reduction Ester, LiAlH4 Primary Alcohol
Hydride Reduction Carboxylic Acid, LiAlH4 Primary Alcohol
The last reaction needs some further comment. Here's how we would do it:
The mechanism is similar to the mechanism for the reduction of esters by LiAlH4, so we will not be concerned with the details. We can conclude that LiAlH4 can be used to make a primary alcohol from either an ester or the corresponding carboxylic acid.
We can extend our understanding of the use of this reaction in synthesis by asking "How could we make the carboxylic acid we need for this reaction?" One particularly important way to do that, important because it makes a carbon-carbon bond, is to make a carboxylic acid by use of the addition of a Grignard reagent to carbon dioxide.
This combination gives us a two step way to convert an alkyl halide (RBr) to a primary alcohol with the addition of one more carbon to the chain.
Alcohols to Alkyl Halides and Grignard Reagents
Of course, this raises the question: "Where do we get the RBr?" That takes us into the reactions of alcohols, because the most effective way to make alkyl halides is from alcohols. If we wish to make alkyl bromides, there are two reactions to consider.
If the alcohol is primary or secondary, the reagent of choice is phosphorous tribromide (PBr3). If the alcohol is tertiary, we use hydrogen bromide (HBr) to ake the alkyl halide. The situation is similar if we wish to make an alkyl chloride. HCl is used for tertiary alcohols and SOCl2 is used for primary and secondary alcohols. We'll take up the mechanisms of these reactions later. (If you wish to look ahead, see Chapter 7 in Brown.)
We now have an answer to the question of how to make alkyl bromides, so we can add another reaction to the beginning of the sequence we started with:
The alcohol product of this sequence can be used to make an alkyl halide to start a new sequence, which could be the starting point for a further sequence, etc. This makes alcohols extremely valuable synthetic reagents.
Including Aldehydes or Ketones
Suppose we used an aldehyde or ketone instead of carbon dioxide in such a sequence:
Here the product alcohol is produced directly rather than through a carboxylic acid. Notice that it is a secondary (from an aldehyde) or tertiary (from a ketone) alcohol, as we had seen when we looked at Grignard additions to aldehydes and ketones.
Oxidation of Alcohols to Aldehydes or Ketones
Since this has shown that one of the important components in a Grignard addition can be made from an alcohol, it seems natural to wonder whether the other major component, the aldehyde or ketone can be made from an alcohol. (We won't worry about how to make carbon dioxide -- there's plenty of it around.) The answer is yes, we can make an aldehyde or a ketone from an alcohol. The desired transformation is:
The products of these reactions have fewer hydrogens than the reactants, so these are oxidations. In the case of the ketone, further oxidation would require breaking a carbon-carbon bond to one of the R groups which is quite difficult. We can use chromic acid (K2Cr2O7 - H2SO4). The situation with the aldehyde is more difficult, since we already know that the use of chromic acid will oxidize it to a carboxylic acid. It took considerable research to work out, but the reagent which works here is called pyridinium chlorochromate (PCC for short). Our synthetic reactions for the oxidation of alcohols are then:
We can add the above reactions to our growing synthetic scheme.
Since the product here is a secondary alcohol, the scheme could be carried onwards through its oxidation to a ketone and/or its conversion to an alkyl bromide, etc.
Let's apply this to a specific example. Suppose we wished to make this compound:
We can compare its structure to the general product of the scheme above and figure out what R and R' are. We then plug the specific identities of R and R' into the scheme to arrive at:
To summarize, since alcohols can be made into alkyl halides and can also be made into aldehydes and ketones, they are important starting points for carrying out a Grignard synthesis. As products of Grignard syntheses, they are also useful materials to begin a new cycle.
The reactions we have looked at today have been those that affected the C-O bond of an alcohol. In making alkyl halides, we broke that bond and made a new bond between the carbon and a halogen. In oxidation, we added a pi bond to the C-O sigma bond of an alcohol.
Alcohols as Nucleophiles
We'll finish today by reminding ourselves of a couple of reactions of alcohols which use their unshared electron pairs, acting as nucleophiles, to make new bonds to carbonyl carbons.
Reaction Alcohol Reacts With Product
Hemiacetal Formation Aldehyde Hemiacetal
Ester Formation Acyl Chloride Ester
One thing to notice about these mechanisms is that the ushared electron pair is used to make the new bond before the O-H bond is broken. That's generally true unless there is a strong base present, something we'll take up next time. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.13%3A_Alcohols.txt |
Alcohol Acidity - Making Alkoxide Ions
The acidity of alcohols is very comparable to that of water. That is, both water and most alcohols have pKa's in the range of 15-16. As acids go, this makes them quite weak. Another way to say that the acidities of water and of alcohols are about the same is to say that the equilibrium constant for the following reaction is about 1.
\[ H_2O + CH_3O^- \rightleftharpoons OH^- + CH_3OH\]
Of course, saying that water and alcohols have about the same acidity also means that hydroxide ion and alkoxide ions have about the same base strength. One practical consequence of this is that we can't completely convert an alcohol (ROH) to its conjugate base, the alkoxide (RO-), by using hydroxide ion (HO-) as the base because hydroxide ion isn't any stronger than a typical alkoxide ion. Instead there are two other reactions which work and which are commonly used
In the first of these, it is hard to see what is serving as the base which removes the H+ from the alcohol. In a formal sense, two sodium atoms provide an electron each to make the new sigma bond in H2, so perhaps it makes sense to call those electrons the strong base.
In the second reaction, the identity of the strong base is easier to see. In sodium hydride (NaH), the NaH bond is very strongly polarized, much more than the B-H bond in sodium borohydride, so that we can realistically regard it as an ionic compound (Na+:H-) so that the :H_ serves as the base. Since the hydrogen molecule (H2) is such a weak acid that we never think of it as an acid, its conjugate base :H- is a very strong base.
Making Ethers
Now that we know about the acidity of alcohols and how to make alkoxide ions, their conjugate bases, we can ask "What are these alkoxide ions good for?" Since they are strong bases, we would also expect them to be good nucleophiles, and they are. We've seen their basic properties used in carrying out the Claisen condensation and occasional uses in other reactions with carboxylic acid derivatives, but their most direct use as synthetic reagents comes in their reactions with primary alkyl halides.
It is important to understand that this reaction only succeeds if the alkyl halide is primary. We'll take up the reasons for this later. Keep in mind that the most useful way to make an alkyl halide is from an alcohol, so that we could extend our reaction sequence by adding steps at the beginning to show how the alkoxide and the alkyl halide are made from alcohols:
The product of this reaction sequence is an ether, ROCH2R', so we have a synthetic method for making an ether from two alcohols, with the important restriction that one of the alcohols is primary.
There is also an acid catalyzed reaction which makes a symmetrical ether, one in which the two alkyl groups attached to the oxygen are identical. In this reaction a primary alcohol is heated with an acid catalyst (usually sulfuric acid). A molecule of water is lost and the ether is formed from two molecules of the alcohol.
Again, it is important to realize that this reaction can only be used on primary alcohols and that the ether produced this way is symmetrical.
Ethers as Solvents
Ethers are generally unreactive compounds. The C-O bond is polar, but breaking it is difficult. We will look at the reasons for this later.
The lack of reactivity of the ether functional group is one reason for the common use of ethers as solvents. A solvent has to dissolve the reactants to be useful, and it also must not react with the reactants present. In many cases, ethers meet these requirements well. They are fairly non-polar, since only the ether functional group contributes any polarity, so they dissolve most organic compounds easily. The unshared electron pairs on oxygen make them weak bases which allows them to dissolve some fairly polar reagents like Grignard reagents and lithium aluminum hydride. The lack of any acidic hydrogens, even those which are fairly weakly acidic like the OH hydrogens of alcohols, means that ethers are compatible with materials like Grignard reagents and lithium aluminum hydride which react with and are destroyed by alcohols. You will have seen ethers as solvents in many reactions.Two ethers which are commonly used as reaction solvents are:
In lab, you also have used diethyl ether as a solvent for extractions. This takes advantage of ether's ability to dissolve most non-ionic organic compounds, its immiscibility (doesn't mix) with water, and the fact that ionic compounds are generally insoluble in diethyl ether. Since THF is miscible with water, it is not useful as an extraction solvent
Epoxides - Ring Strain
There is one type of ether which is quite reactive. The key structural feature of these ethers is that the oxygen atom is contained in a three-membered ring. Such ethers are called epoxides. The internal angles of such a ring will be near 60o, and this pushes the bonding electrons much closer together than the 109.5o we would expect from the a VSEPR estimate or from sp3 hybridization. Thus epoxides have high energies compared to "ordinary" ethers. One result is that epoxides are, in contrast to "ordinary" ethers, rather reactive compounds. Such rings are regarded as "strained" because the deviation of their bond angles from normal values leads to extra reactivity.
We'll look at two examples:
Acid-Catalyzed Hydrolysis: The name of this reaction recalls the acid-catalyzed hydration of ketones and aldehydes -- the first reaction we studied this semester. In fact, if we look at the mechanism of the acid catalyzed hydrolysis of the simplest epoxide:
and compare it to the acid catalyzed hydration of an aldehyde:
we see that the steps in the two mechanisms are identical. In each case, the acidic H+ adds to the oxygen atom. This pulls electron density away from the attached carbon atom, making it susceptible to attack by the weakly nucleophilic oxygen of water. A carbon-oxygen bond is broken, and the reaction is completed by regeneration of the H+ catalyst. (We'll be in a position to look at the stereochemistry of this reaction in about three weeks).
The resemblance between these two mechanisms suggests that the reactions of epoxides are similar to those of carbonyl groups. More precisely, epoxide reactivity is intermediate between the reactivity of ethers (pretty unreactive) and carbonyl groups (reactive with selected reagents). Another example of this lies in the addition of Grignard reagents to epoxides. Again, we'll look at the simplest case (ethylene oxide), where there are no alkyl groups on the carbons of the epoxide.
Here again, we have a direct analogy with the reaction of an aldehyde or ketone with a Grignard reagent. As a synthetic reaction, this is best used with ethylene oxide, since complications are prevalent with more complex epoxide. The synthetic outcome of this reaction is to make a new primary alcohol which is two carbons longer than the Grignard reagent.
We'll defer discussion of the synthesis of epoxides until we look at the addition reactions of alkenes.
Thiols and Crosslinks
Our last topic for today is thiols -- the group of compounds in which the oxygen of an alcohol has been replaced by a sulfur. As one might expect given the close relationship between oxygen and sulfur in the periodic table, there are some similarities between alcohols and thiols. We will not examine the synthesis of thiols; rather, we'll look at some of their properties.
Perhaps the most obvious property of thiols, at least those which are small enough to be volatile, is their extremely disagreeable odor. This is understandable if we regard them as derived from hydrogen sulfide (rotten egg odor) in the same way that alcohols can be regarded as derived from water. Butanethiol (CH3CH2CH2CH2SH) is actually added to natural gas before it is distributed by pipeline so that the odor will signal the presence of a leak before it leads to dangerous concentrations of the gas. Chemists who work with thiols are careful to use them in the hood!
The SH group in a thiol is more acidic than the OH group in an alcohol. That means that the RS:- (thiolate ion) can be conveniently made by reacting a thiol (RSH) with hydroxide ion (OH-). The thiolate ion is a good nucleophile, so it can react readily with a primary alkyl bromide to produce a thioether:
The SH group has important roles to play in biochemistry. We've seen how the sulfur serves as a leaving group in the Claisen-like reaction which makes the carbon-carbon bonds in fatty acid biosynthesis. This is quite general, since thioesters (the name for an ester containing a sulfur in the leaving group position) are very common in biochemical processes.
The thiol group is also important in the structure of proteins. Keep in mind that proteins are polyamides -- long chains of amino acids linked together by amide linkages. These chains can coil around and adopt a variety of shapes, some fairly straight and regular and others snarled like a tangled fishing line. For many purposes such as those in which proteins serve as enzymes (biochemical catalysts), it is essential that the protein have only one shape. Thiols contribute to maintaining that shape
The chemistry of this depends on the fact that the S-H bond can be oxidized by very mild oxidizing agents such as oxygen (O2). This reaction produces a sulfur atom which has an unpaired electron. Two such sulfur atoms can join to produce a sulfur-sulfur bond. The reaction is outlined as follows:
The new functional group is called a disulfide and the bond between the two sulfurs is called a disulfide bond.
If two polymer chains which include SH groups are positioned so those groups are close together, the oxidation reaction makes a link between the two chains, much like rungs link the side rails of a ladder. The new links are called crosslinks and the new structure is much more rigid than was the case with the flexible polymers prior to crosslinking.
The original process for the vulcanization of natural rubber (which involved heating the gummy natural material with sulfur) probably produced such crosslinks, although the details of the chemistry are different. In proteins the amino acid which provides the SH groups is cysteine (HSCH2CHNH3+COO-). The protein keratin, found in hair, skin and feathers, is rich in cysteine. Crosslinks through disulfide bonds formed by the mild oxidation of cysteine are important in maintaining the shape of these systems.
Home permanents work by controlling crosslinking. The first step is to apply a mild reducing agent which converts the disulfide bonds to two SH groups. The hair is "clamped" into the desired shape, then a mild oxidizing agent is added to form disulfide bonds between SH groups which have been moved close together in the new shape. These bonds "lock" neighboring keratin protein chains into the new shape, rendering it "permanent," at least until it grows out. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.14%3A_Ethers_Epoxides_Thiols.txt |
Isomerism
Let's begin by reminding ourselves of some of the ideas involved in the topic of isomerism. Structural understanding of organic chemistry begins with the statements that carbon makes four bonds and each carbon can bond to another carbon (the Kekule-Couper-Butlerov theory of organic structure). In even very small molecules (try C2H6O) this means that there is more than one way to connect the atoms. Each of these connection patterns represents a different compound. Here are two for C2H6O:
There are two things to notice about this. First, the structural difference between the two molecules can be described in terms of "what's connected to what." For example, the oxygen in ethanol is connected to a CH3CH2 group and to a hydrogen while the oxygen in dimethyl ether is connected to two methyl groups. Connectivity is the key difference here. Second, and this is new, we can describe this connectivity quite nicely using just the usual two dimensions we can conveniently represent on paper, a chalkboard, or a computer monitor.
There are many types of organic compounds for which this description is inadequate. This is particularly true for compounds involved in biological processes such as sugars and amino acids (carbohydrates and proteins). For these compounds and many others we will need to use three dimensions in order to have an adequate description of the structure.
Chiral Objects
Let's look at some familiar objects so as to get a sense of when we will need to think about three dimensions. Consider the way in which a sock differs from a glove. It doesn't matter which foot you put a sock on, it will fit just as well. That certainly can't be said about a glove. A glove fits one hand much better than the other.
One quick way to differentiate between sock-like objects and glove-like objects is to ask whether an object is identical with its mirror image or not. Consider a pair of socks. One of the socks is a mirror image of the other (Think of holding one sock up to a mirror. What you see is identical to the other sock, and to the original sock.) Now consider a pair of gloves. The right glove is the mirror image of the left glove, but they are not identical. They can't be superimposed -- merged so they completely match. Objects like gloves which cannot be superimposed upon their mirror images are called chiral, from the Greek word for hand. Objects like socks which can be superimposed upon their mirror images are called achiral.
Now notice that we need three dimensions to describe the difference between a left glove and a right glove. If you hold up two gloves so that you are looking at the edge nearest the thumb, you notice that the difference lies in which way the thumb points. The rest of the glove can be pretty well described by a plane (two dimensions), just as a sock can be described by a plane, but the describing which way the thumb points requires a third dimension, out of the plane. Make up a list of your own of familiar objects and decide whether they are chiral or achiral by testing each for superimposability on its mirror image.
Stereogenic Centers
Now, how does this apply to molecules? You can satisfy yourself by playing with models that the following statement is true: A molecule which contains one carbon which is directly bonded to four different groups or atoms is not superimposable on its mirror image. Such a molecule is chiral just the same way that a glove is chiraland a carbon which is bonded to four different groups is called a stereogenic center. The lab for next week has a good example of a carbon bonded to four different atoms. You can move one of the two molecules to verify the mirror image relationship and you can check for superimposability.
So, if we wish to find out whether a candidate molecule is chiral or achiral, we check for stereogenic centers -- carbons which are bonded to four different groups or atoms. One question which always comes up at this point is "How different do the groups or atoms have to be?" If the groups have the same composition they are different if they are isomeric -- if they are connected differently. For example, propyl and iso-propyl are different.
Structural Representations
Once we've convinced ourselves that a stereogenic carbon atom makes a compound chiral, we need to think about how we will represent the three dimensional structures of these compounds when we are restricted to two dimensions. For example, the following sketch represents a 2-bromobutane. It is not superimposable upon its mirror image (see the three dimensional representations earlier).
How do we represent this structure? There are two conventional representations. The first is called a "perspective view." In this view, bonds represented by solid wedges are taken to point toward the viewer (in front of the plane of the paper), bonds represented by dashed line wedges are taken to point away from the viewer (behind the plane of the paper, and bonds represented by ordinary straight lines are taken to lie in the plane of the paper. In a perspective view the 2-bromobutane above would be represented as:
The Br is in front of the plane, the H is behind it, and the methyl (to the left) and ethyl (to the right) groups are taken as being in the plane (at least the carbons of those groups).
The other representation is known as the Fischer projection. This is a convention in which the stereogenic carbon atom is represented by the junction of two crossed lines. The horizontal line is taken to represent bonds which come from the carbon to an atom toward the viewer (in front of the plane) and the vertical lines are taken to represent a bond which goes from the carbon to an atom away from the viewer (behind the plane). This can be converted to a perspective view by drawing in the appropriate solid and dashed wedges:
The last "equals" sign says that the Fischer projection and the perspective view are representations of the same molecule. You may wish to verify this by imagining yourself as looking down from above the perspective view with the methyl group toward the top of your head. Can you see the wedge interpretation of the Fischer projection?
Because of the special meaning of horizontal and vertical lines in the Fischer projection, they can only maintain their meaning if we restrict manipulations to 180o rotations in the plane of the paper. You can demonstrate this for yourself by converting a Fischer projection to a perspective view, then rotating the Fischer projection by 90o and converting the new Fischer projection to another perspective view. The two perspective views are mirror images.
The term which describes the relationship between an object (molecule) and its non-superimposable mirror image is enantiomeric. An object and its non-superimposable mirror image are entantiomers of each other. If the object is superimposable on its mirror image, then there is really only one object and the term used is identical. These relationships are illustrated below: | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.15%3A_Chirality_Three_Dimensional_Structure.txt |
Optical Activity
First, though, let's look at a property in which one enantiomer differs from another. Enantiomers are alike in all respects but one. They have the same melting point, the same boiling point and the same solubility in common solvents. The difference between the two enantiomers only shows up when we put them in a chiral environment. Our analogy with gloves can help here. In a pair of gloves, the left glove weighs the same as the right glove, it is the same size and typically the same color. It is made of the same material. The difference between them only shows up when we try one on. One fits the right hand better than the other. In this instance, the right hand is a chiral environment and the properties of the left and right gloves differ in that environment.
A specific "chiral environment" for molecules is provided by polarized light. The outcome of this is that if we pass a beam of polarized light through a solution of one enantiomer, the plane of that polarization will be rotated either to the left or the right. This phenomenon is called optical activity. If we do the same experiment, but use the other enantiomer (the mirror image of the first one) the plane of polarization will be rotated in the opposite direction. In either case, the amount of the rotation (the number of degrees in the angle between the plane of polarization before passage through the sample and the plane of polarization after passage through the sample) is the same but the directions are opposite.
If the sample causes a rotation of the plane of polarization of the light in the clockwise direction (from the viewer's point of view) we say that the compound is dextrorotatory and we designate this by a plus sign in parentheses (+). Correspondingly, the terms for counter-clockwise rotation are levorotatory and (-).
R/S Naming
It has been possible to determine the absolute configuration of a chiral molecule since 1954. That is, we can know for a specific molecule which of the two mirror image structures is the one which represents the actual arrangement in three dimensions. We need a way to designate that information in the compound's name.
The system we use has two components. First we need to be able to list the four atoms or groups connected to the stereogenic carbon in a specific rank order. Then we need to have a way to distinguish the orientation of these groups or atoms in one enantiomer from the orientation in the other.
Ranking Groups or Atoms: We rank groups or atoms by the atomic number of the atom directly attached to the stereogenic carbon. The group or atom with the highest atomic number gets the highest rank number (4). In 2-bromobutane (below), this is bromine whose atomic number is 35. The lowest rank goes to hydrogen (atomic number one), so it gets rank number (1). That is fairly straightforward, but what do we do about the two carbons? They both have atomic numbers of six and are tied for ranking. The tie-breaker is to look next at the three atoms attached to each of these two carbons. We compare their atomic numbers until we find a difference. In 2-bromobutane one of these carbons is in a methyl group, so it is bonded to three hydrogens. The other "tied" carbon is carbon one of an ethyl group, so it is bonded to two hydrogens and a carbon. If we compare these two situations we find that there is no breaking the tie by comparing hydrogens, but the second carbon of the ethyl group has a higher atomic number than the hydrogen which is its competition on the methyl group. The tie is broken, the ethyl group has a higher (3) rank than the methyl group (2), so we have all of the atoms or groups attached to the stereogenic carbon ranked.
The "tie breaking" process can be extended to more complex situations. Study how it works for doubly bonded carbons in Section 4.3 (Rule 3) of Brown.
Orientation: Now that we have ranking for the four groups or atoms attached to the stereogenic carbon, we need to describe how they are oriented in space. We do this by turning our molecule so that the lowest (1) ranked atom or group is pointed behind the paper, away from us. Then we imagine a curved arrow which starts at the highest (4) ranked group, passes by the (3) group and ends at the (2) group. If this arrow points in a clockwise direction, we use the letter R (from Latin rectus) to describe the compound. If the arrow points counter-clockwise, we use the letter L (Latin sinistrus). These letters are prefixed to the IUPAC name of the compound, so our compound is R-2-bromobutane.
Practice this process using the problems in the text and bring up any questions in class. You can also practice with the on-line lab problems. Like other naming situations, it sometimes seems arbitrary and picky, but it isn't fundamentally hard. Practice is very necessary so that you can use the R/S system quickly and effectively.
One last point about the R/S system and optical activity: There is no direct relationship between a compound's absolute configuration (designated by R or S) and its optical activity. Some R compounds have clockwise (+) rotations, some have counter-clockwise (-) rotations, and the same is necessarily true of S compounds.
Diastereoisomerism
What happens if we have two (or more) stereogenic centers in the same compound?
Next time, we'll look at how these stereogenic atoms are included in naming, and then we'll go on to consider what complications arise if there are more than one stereogenic atom in a molecule. We'll use Fischer projections for this, since the next topic is sugars and Fischer projections are used routinely to describe sugar structures.
Let's take a compound with two stereogenic centers (designated with asterisks *):
Does it have a mirror image? Yes. Is it superimposable on its mirror image? No. (Remember that we can rotate Fischer projections 180o in the plane of the paper.)
Notice that neither the mirror image nor the rotated mirror image is superimposable on the original structure. The original structure and its mirror image are enantiomers. Each will be optically active.
Now let's look at what the relationship would be if we did changed the configuration at one of the stereogenic carbon atoms, but left the other one alone.
There isn't a mirror image relationship here, and the structures are not superimposable. Neither of the terms "enantiomer" or "identical" applies. The compounds are isomers though, and since they are connected identically, they are stereoisomers. The word we use for this relationship is diastereoisomeric. Diastereoisomers may be recognized because they are connected identically, they have two (or more) stereogenic atoms and comparison of those atoms reveals that the relationship at one (or more) atom is identical and the relationship at the other (or more) atom is mirror-image.
Since the diastereoisomer we made by changing the top stereogenic carbon of our original compound is not superimposable upon its mirror image, it too is optically active and has a mirror image enantiomer. Here is the complete set of enantiomeric and diastereoisomeric relationships for this case:
The horizontal relationships here are between enantiomers, compounds which are non-superimposable mirror images. The vertical and diagonal relationships are between diastereoisomers, stereoisomers which are neither identical nor mirror images.
Diastereoisomers show differences in properties other than optical activity. They typically have different melting and boiling points and solubilities and they may show differences in how fast they react with other reagents. These differences are not usually as large as those shown by compounds which are not isomers. Here are some on-line problems on this topic
Meso Compounds
There is one more case to consider. What if the four atoms or groups bonded to one of the stereogenic atoms are the same as those bonded to the other. We can explore this in our example by considering what happens if we reduce the aldehyde (CHO) group with NaBH4. The product is a primary alcohol (-CH2OH) group, the same group as is at the bottom of this compound. The four atoms or groups attached to the top stereogenic carbon are the same as those attached to the bottom stereogenic carbon.
Now, let's test this molecule to see if it is superimposable upon its mirror image. As before, we make the mirror image and then we rotate it 180o in the plane of the paper. The original structure is identical to its mirror image, so we do not have enantiomers here. A compound like this is called a meso compound. Since it is superimposable upon its mirror image, it is not optically active, even though it has two (or more) stereogenic carbon atoms.
Another way to detect a meso compound is to look for a plane of symmetry within the molecule. This is particularly easy with Fischer projections, since what we have to do is to imagine a plane cutting the molecule precisely in two so that there is a bottom half and a top half. If the bottom half is the mirror image of the top half, we have a meso compound.
A meso compound will have diastereoisomers as well. We can generate these by changing one of the stereogenic carbons and then making its mirror image. The latter two compounds are enantiomers. Each is a diastereoisomer of the original meso compound. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.16%3A_R_S_Naming_Two_or_More_Stereogenic_Centers.txt |
Mono-, Di-, and Polysaccharides
Sugars are small molecules which belong to the class of carbohydrates. As the name implies, a carbohydrate is a molecule whose molecular formula can be expressed in terms of just carbon and water. For example, glucose has the formula C6(H2O)6 and sucrose (table sugar) has the formula C6(H2O)11. More complex carbohydrates such as starch and cellulose are polymers of glucose. Their formulas can be be expressed as Cn(H2O)n-1. We'll look at them in more detail next time.
The difference between a monosaccharide and a disaccharide can be seen in the following example:
A quick glance tells us that a monosaccharide has just one ring, a disaccharide has two, and a polysaccharide has many. Beyond that, though, there's another important structural feature. Look at the disaccharide and focus on the oxygen which links the two rings together. The atom above it is connected to two oxygens, both of which are in ether-type situations. The carbon and these oxygens are in an acetal linkage. (The bonds are heavier and in blue.)
If we look at the corresponding location in the monosaccharide and ask what the functional group might be, we see that it is a hemiacetal. (Here the bonds are heavier and in red.) So, another way to describe the situation is that a monosaccharide has a single ring with a hemiacetal in it, a disaccharide has two rings linked by an acetal functional group, and a polysaccharide has many rings linked by many acetal functional groups. (Check this last statement against the polysaccharide structure above).
How about the "sugars" we saw last time with just 4 carbons. Why are they monosaccharides when there is no ring? If we consider that the OH group on the bottom carbon could form a hemiacetal with the aldehyde function, then we get a ring, and that structure fits our description of a monosaccharide.
We'll take a more detailed look at the cyclic and non-cyclic structures of sugars shortly.
"Oses" and D-Sugars
Now let's see what aldotetrose means. Taking the name apart from right to left, the ending "ose" means a sugar, which may be a monosaccharide, a disaccharide or an oligosaccharide (a "short" polysaccharide). The middle part "tetr" means that our sugar has four carbons linked in a straight unbranched chain. Terms like "pent" for five carbons and "hex" for six carbons are also in common use. The beginning "aldo" means that there is an aldehyde in the compound. The alternative would be a ketone group, for which we would use the prefix "keto."
Glucose, the most common monosaccharide, is an aldohexose. We understand that to mean that it is a sugar having six carbons in a straight unbranched chain which ends in an aldehyde group. We can represent that structure in this fashion:
This structure includes four stereogenic carbon atoms (marked with an asterisk *). There are a total of sixteen stereoisomers possible. Eight of these are enantiomers of the other eight. The rest of the relationships are diastereoisomeric. Since the groups at the top and the bottom of the chain are not the same, there are no meso isomers. Eight of the isomers are shown here. The other eight are mirror images of these and may be readily drawn.
The question is "which of the sixteen stereochemical representations (Fischer projections, remember that each stereoisomer shown also has an enantiomer which is not shown) describes the absolute configuration of glucose? When Emil Fischer took up this problem about 100 years ago, he realized that there was no way to determine if glucose was one of the eight structures above or one of the unshown enantiomers. He made the assumption that it was one of the ones above so that he could work on the diastereoisomeric part of the problem, hoping that later work would resolve the question of which enantiomer best represented glucose.
Fischer also developed the D/L system for specifying the structures of sugars. If the OH group on the stereogenic carbon farthest from the aldehyde group is to the right in the Fischer projection, then the compound is a D-sugar. All of the sugars in the figure above are D-sugars. If the OH group on the stereogenic carbon farthest from the aldehyde group is to the left in the Fischer projection, then the compound is an L sugar. The enantiomers of all the sugars in the figure above are L sugars. Fischer's assumption amounts to saying that glucose is a D-sugar. Later work resolved this issue, and Fischer was right.
Which D-aldohexose is Glucose?
How did Fischer determine which of the eight structures above was glucose? He had available samples of glucose and mannose, both aldohexoses, and arabinose, an aldopentose. He also learned how to reduce the aldehyde functional group to a primary alcohol. (We'll illustrate this with NaBH4 to avoid learning a new reaction, but he used another reagent.) He developed a method for extending the carbon chain of an aldose (called the Kiliani-Fischer chain extension). He also had a polarimeter so he could determine whether a sample was optically active or not. Perhaps most importantly, he had a group of talented and dedicated students.
Now, some data.
Experimental result: When the aldehyde group of arabinose was reduced to a primary alcohol group, the product was optically active.
Conclusion: Arabinose has either structure 2 or 4 in the scheme below. This is because if arabinose were either 1 or 3, the product would have a plane of symmetry (mirror plane) and would be optically inactive.
Experimental result: When the aldehyde group of glucose was reduced to a primary alcohol group, the product was optically active. The same result was obtained for mannose.
Conclusion: The structures "X'd" out below do not represent either glucose or mannose since the products from these structures would be meso compounds.
Experimental result: Kiliani-Fischer chain extension applied to arabinose produces glucose and mannose.
Conclusion: The bottom three stereogenic carbon atoms of glucose and mannose are have identical configurations to the three stereogenic carbon atoms of arabinose. This means that glucose and mannose differ only in the configuration of the stereogenic carbon atom nearest the aldehyde functional group. We can further conclude that if one member of a pair of aldohexoses (paired because their bottom three stereogenic carbons are identical) is ruled out, so is the other
Now let's see what we have left. There are four structures remaining as candidates. They are on the right below. If we go back to the possibilities for arabinose, we find that the two on the top come from structure 2 for arabinose, which was a possibility, while the two on the bottom come from structure 3, which was ruled out earlier. The conclusion is that arabinose is represented by structure 2, and glucose and mannose are the two structures to its right.
But which is glucose and which is mannose? Fischer noticed that if reactions could be developed which changed the aldehyde group into a primary alcohol and the primary alcohol into an aldehyde (switch ends) one of these structures would give itself, and the other would give back a new L sugar. The reactions are complex and we will not look at them, but when the chemistry was applied to the sample called mannose, the product was mannose. When the chemistry was applied to the sample called glucose, a new sugar was formed.
There was a great deal more to be done to confirm this conclusion and to synthesize the other six aldohexoses, but Fischer's exercise in logic and dedicated experimentation led to the conclusion that the eight D-aldohexoses are:
Notice that the new sugar which was produced from glucose by the "exchange ends" experiment is L-gulose. The names of the hexoses tell us which diastereoisomer we have; the D or L designation gives us which enantiomer we have.
The corresponding names for the aldopentoses are:
Cyclic Structures - Anomers
To finish today, we'll see what happens when a hemiacetal is formed between the aldehyde carbon and one of the OH groups on the chain. We'll look at two examples, ribose, which is a key component of RNA, and glucose because of its abundance. (You may wish to review the mechanism for hemiacetal formation.)
Since there are four OH groups in ribose, we could anticipate four different ring sizes. In three atom rings and four atom rings the bond angles are far from 109.5o, so these rings are strained, have higher energies and are hard to form. Remember that there is an equilibrium between a hemiacetal and the aldehyde/alcohol it comes from, and that high energy materials don't persist at equilibrium. We are left with rings which have either five or six atoms in them. In the case of ribose the important ring (found in RNA) is the five membered ring.
Notice that the carbon in the newly formed hemiacetal group is stereogenic. This means that there are two possible diastereoisomers for the cyclic structure. Usually both are formed, and they have a special name -- they are anomers of each other. The carbon between the two oxygens in the hemiacetal group is called the anomeric carbon. If the OH group is down (in a drawing with the ring oxygen to the rear center or right), the designation for that anomer is alpha. If the OH group is up, the designation is beta. Since the alpha and beta anomers are diastereoisomers, they have different properties; in particular, different optical activities. The term for a five atom sugar ring is "furanose."
Glucose usually makes hemiacetal cyclic structures with six atom rings, although five membered rings can also be formed when the six membered rings are precluded. Such six membered rings are named by the term "pyranose." The ring forms look like this, keeping in mind that alpha and beta anomers are also involved here:
Again, we have an equilibrium between the open chain form and the two diastereoisomeric anomers.
There is one further point to be made about these glucopyranoses. The structure we have drawn for the ring is flat. The bond angles would be 120o, quite far from the normal tetrahedral value of 109.5o. The atoms in the ring can have bond angles of about 109.5o if the ring puckers as shown here:
Of course, molecules adopt these puckered shapes (called chair conformations from their resemblance to a rather wide lounge chair) automatically. You can review the material on cyclohexane for a more detailed analysis of this material. It has been established that in these chair conformations, the molecules have a lower energy if the larger substituents on the carbons are roughly in the plane of the ring itself. These positions are called "equatorial" to distinguish them from the other positions (roughly perpendicular to the ring, called "axial"). A compound which can have all of its larger substituents (everything is larger than hydrogen) in an equatorial position is more stable than one which cannot. beta-D-glucose has all of its substituents in equatorial positions, and is thus the most stable hexopyranose. It is also the most abundant. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.17%3A_Carbohydrates-_Monosaccharides.txt |
Glycosides
Last time we explored the structural characteristics of monosaccharides. We saw that the major stereochemical features of aldohexoses and aldopentoses are usefully described by Fischer projection formulas, but we learned that the structures of these compounds must also be understood as cyclic hemiacetals. Today we'll look in more detail at the chemistry of that hemiacetal linkage. In particular, we'll recall how hemiacetals are converted to acetals. We'll find that these acetal linkages are what holds di- and polysaccharides together.
Let's begin by remembering the reaction sequence which links aldehydes and alcohols, hemiacetals, and acetals.
For our purposes, the key feature is the conversion of a hemiacetal and an alcohol to an acetal, with the concurrent release of a molecule of water. If we apply this feature of the scheme to a solution of glucose in methanol (with a trace of acid catalyst included), we get:
The acetal products are called "glycosides." If the sugar used is glucose, they are "glucosides." There are several reasonable mechanisms for these conversions and we will not look at them in detail. Keep in mind that the conversion between a hemiacetal and an acetal requires an acid catalyst. The conversion between an aldehyde and a hemiacetal is catalyzed either by base or by acid. Conditions can be arranged to produce either the alpha or beta stereochemistry in the glycoside.
Glycosides are very common in nature. Besides the di- and polysaccharides we will look at later, it is very common for glucose (or other sugars) and an alcohol to form an acetal linkage. Often this improves the water solubility of the alcohol and makes it easier to excrete. This is the case with cholesterol:
Reducing and Non-Reducing Sugars
There is another important difference between the hemiacetal and acetal linkages in sugars and saccharides, and that is their reaction with mild oxidizing agents. Aldehydes are fairly easy to oxidize to carboxylic acids, while acetals (which have no carbonyl group) are quite difficult to oxidize. The oxidizing agents used in carbohydrate chemistry are typically copper(II) compounds which are reduced to copper(I) oxide. Sugars which are oxidized by these reagents are called reducing sugars because they reduce the copper(II) to copper(I).
These reagents are used in basic solution, so that hemiacetals and aldehydes are in equilibrium. This means that the cyclic hemiacetal form of a sugar will produce an equilibrium amount of the open-chain aldehyde form, which will then reduce the copper(II) to copper (I) and give a positive test. A hemiacetal form is thus a reducing sugar.
In contrast, acetal forms (glycosides) are not reducing sugars, since with base present, the acetal linkage is stable and is not converted to the aldehyde or hemiacetal. The outcome is that in a reducing sugar the anomeric carbon is in an aldehyde or hemiacetal. In a non-reducing sugar, the anomeric carbon is in an acetal.
The characterization of sugars as reducing or non-reducing is gives useful clues as to their structures. Consider the disaccharides maltose and fructose. Maltose contains a hemiacetal functional group and is a reducing sugar. In fructose, both anomeric carbons are in acetal functional groups, so fructose is a non-reducing sugar.
Disaccharides
This brings us to the topic of disaccharides. The linkages between the monosaccharide ring units in disaccharides are acetal linkages. We can envision them as being made by the formation of an acetal from a hemiacetal and an alcohol. For this purpose, the hemiacetal includes the anomeric carbon of a monosaccharide and the alcohol role is played by a specific OH group of a second monosaccharide. The formation of maltose from two molecules of glucose is an example of this:
There are several intriguing features of this conversion. First, it is catalyzed by the enzyme maltase. The term "catalyzed" implies that enzyme speeds up the reaction in both directions, so that both formation and hydrolysis (conversion from acetal to hemiacetal using a molecule of water) are faster with the enzyme. Enzymatic catalysis is usually also very specific. In this case, that specificity shows up in the fact that the new acetal linkage has the alpha configuration, not the beta (and correspondingly, maltase catalyzes the hydrolysis of an alpha linkage but does nothing to the beta linkage). Also, only the OH group on the number four carbon atom is used as the alcohol when others, such as the ones on carbons 1, 2, 3 and 6 might have been used. This suggests that the enzyme holds the two molecules of glucose in specific positions so that only the OH on carbon 4 of one glucose can reach the anomeric carbon of the other glucose.
Fructose provides an example of a disaccharide in which the acetal linkage joins the anomeric carbons of a glucose molecule to the anomeric carbon of a fructose molecule. In this case there is no hemiacetal functional group, so fructose is a non-reducing sugar. Also, here one of the rings has five members rather than six, showing that the cyclization of fructose from the open-chain form to the hemiacetal cyclic form uses the OH at carbon 5 and the carbonyl carbon 2.
We can also look more carefully at fructose. In its cyclic form the anomeric (hemiacetal) carbon is involved in two carbon-carbon bonds. This means that when we open the molecule up to its open chain form the anomeric carbon becomes a keto carbonyl group. Fructose is thus an example of a ketose, a sugar in which the carbonyl group is a ketone rather than an aldehyde.
Polysaccharides
If we now return to our first look at polysaccharides, we can see that amylose starch is composed of many glucose monosaccharide units which are linked together by acetal functional groups involving the anomeric carbon of one glucose and the number four carbon of the next glucose. Repetition of this pattern many times gives the polymer.
Amylose is a linear polymer with few branches. In amylopectin, another type of starch, there are branches which involve acetal linkages through the oxygen on carbon 6. Glycogen, sometimes called animal starch, is a similar polymer found in animals as a storage medium for glucose. Glycogen is even more highly branched than amylopectin.
Hydrolysis of starch involves the cleavage of the acetal functional groups with the addition of a molecule of water for each acetal linkage and the production of many molecules of glucose. This is done by the enzymes called glycosidases which are found in saliva. These enzymes work only on alpha acetal linkages and do not attack beta linkages. Such beta linkages are found in cellulose. Since our glycosidases are unable to hydrolyze the beta linkages in cellulose, we cannot digest cellulose, even though it is also a polymer of glucose.
Of course, there are enzymes which hydrolyze the beta linkages in cellulose. Such enzymes are found in the bacteria which inhabit the stomachs of ruminants such as cattle and sheep, which makes cellulose digestible by ruminants. They are also found in fungi which rot wood.
The specificity of enzymes allows one monosaccharide, glucose, to be the building block for both starch, which we think of as a major source of energy in our foods, and cellulose, which we regard as a structural material in trees and a major component of paper. If we look at this in the context of the use of these materials in a plant, starch is found as a storage medium for glucose in seeds and tubers. It is used as a source of glucose both for energy and as a raw material for cellulose as the plant sprouts and enters its initial growth period. Enzymes specific for alpha linkages present in the sprouting plant hydrolyze the starch to glucose, as they do in the malting process used in beer and whisky production.
The cellulose produced as the plant grows is a major structural component of the plant. It must be quite stable if it is to serve that purpose, so enzymes specific for the alpha linkage do not attack its beta acetal functional groups and it is not readily hydrolyzed. The small stereochemical distinction between the alpha and beta linkages leads to very large consequences in the chemistry and function of starch and cellulose. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.18%3A_Glycosides_Disaccharides_Polysaccharides.txt |
Structural Types
Last time we completed our study of carbohydrates. Now we are turning our attention to another important class of organic compounds, amines. Many important drugs are amines, the bases present in RNA and DNA are amines, and the fundamental building blocks of proteins are amino acids.
The functional group of an amine is the nitrogen atom connected by three sigma bonds to alkyl groups or hydrogen atoms. (Aryl groups - benzene rings - can also be connected to a nitrogen in amines, but we will not study these until later in the course.) The chemistry of amides is different enough from that of amines that they are normally not included among the amines. The nitrogen atom of an amine can also be included in a ring. Such amines are called "heterocycles." (Recall that nitrogen is a heteroatom -- not a carbon or a hydrogen.)
Amines are classified as primary, secondary, or tertiary according to the number of carbons bonded directly to the nitrogen atom. Primary amines have one carbon bonded to the nitrogen. Secondary amines have two carbons bonded to the nitrogen, and tertiary amines have three carbons bonded to the nitrogen. The system is superficially similar to the way we have classified alcohols, but the important difference is that in alcohols we were counting bonds to the carbon which carried the OH group. For amines, we are counting the carbons bonded to the nitrogen.
Since nitrogen has a normal valence of three, we can also conclude that there are two N-H bonds in primary amines and one N-H bond in secondary amines. In tertiary amines there are no N-H bonds.
Bonding, Shape and Hybridization
Now, let's look at the bonding around the nitrogen atom of an amine. First, we need to remember that the nitrogen, in addition to forming three sigma bonds, also carries an unshared electron pair. This means that there are four groups of electrons associated with the nitrogen. Mutual repulsion of these groups leads to a tetrahedral arrangment, much like that of a typical sp3 carbon atom. This predicts a bond angle between the N-H bonds of 109.5o which agrees pretty well with the observed value of 107o found in ammonia (which is the "smallest" amine, like water is the smallest alcohol).
Since the bonds connected to the nitrogen are shaped like a flattened pyramid, the arrangement is often called pyramidal. This ignores the unshared electron pair, whose inclusion leads to the tetrahedral description and the corresponding understanding of the nitrogen's hybridization as sp3. The example used to illustrate this is ammonia, but the nitrogen is also well described as having sp3 hybridization primary, secondary and tertiary amines as well.
Hydrogen Bonds
The N-H bonds in amines are somewhat polar. As we might guess from considering electronegativities (estimated from positions in the periodic table), the N-H bond is more polar than the C-H bond and less polar than the O-H bond. This polarity shows up in a comparision of physical properties of amines and alcohols.
The simplest examples are water and ammonia. Water boils at 100oC, while ammonia boils at -33oC. This is interpreted to mean that it takes a good deal more energy to boil water than ammonia. Correspondingly, the forces between molecules of water (which resist separating them in the boiling process) are much stronger than those between molecules of ammonia. The most important of these forces is called the hydrogen bond.
The hydrogen bond is much weaker than a covalent bond. Breaking a hydrogen bond requires about 10% of the energy required to break a typical covalent bond. This is consistent with the fact that we can boil water without breaking any covalent bonds (the water molecules remain intact). Separation of one molecule from another only requires breaking the weaker hydrogen bonds.
One useful picture of a hydrogen bond is electrostatic -- an attraction between the positive end of a dipole on one molecule and the negative end of a dipole on another. The more polar the molecules, the greater the degree of positive and negative charge associated with the dipoles, and the stronger the hydrogen bonds. In ammonia and water, the most concentrated and available negatively charged region is where the unshared electron pair is. The positively charged regions are the hydrogen ends of the N-H bonds. We can envision the hydrogen bond as a weak (compared to covalent bonds) attraction between the unshared electron pairs on of a nitrogen in one molecule and a hydrogen (covalently bonded to nitrogen) on another.
Hydrogen bonds are extremely important in the chemistry of the genetic code. As we will study later, the double strands of DNA are held together by hydrogen bonds. The replication of DNA depends on hydrogen bonds which selectively connect specific base pairs, as do the several steps by which the genetic message determines the specific order of amino acids in a protein.
Amines as Bases
The distinguishing chemical property of amines is that they are bases. This is a direct consequence of the presence of the unshared electron pair on the nitrogen, which makes them Lewis bases. The basic unshared electron pair is less tightly held by the nitrogen of an amine than the corresponding oxygen of an alcohol, which makes it more available to act as a base. Consequently amines (and ammonia) are more basic than alcohols (and water), and less basic than alkoxide (RO-) and hydroxide (OH-) ions. It is convenient to think about the base strength of amines and ammonia in terms of the pKa of their conjugate acids, the ammonium ions. These ideas can be summarized in the following scheme, where we use water and ammonia as stand-ins for alcohols and amines:
As we have come to expect, these reactions go in the direction which consumes the stronger bases and acids and generates the weaker ones. Ammonium ions (pKa ~10) are stronger acids than water (pKa ~16, so water is produced when an ammonium ion is treated with hydroxide ion. Ammonium ions (pKa ~10) are weaker acids than H3O+ (pKa ~-2), so they are produced when amines are treated with aqueous solutions of strong mineral acids like sulfuric or hydrochloric acids. If the water is removed, there remains an ammonium salt which incorporates the negative counterion of the mineral acid (sulfate or chloride).
Since the basic properties of amines arise from the presence of the unshared electron pair on nitrogen, the strengths of primary, secondary and tertiary amines are quite similar. Aryl amines (those where the nitrogen is connected directly to an aromatic amine) weaker bases, but we will take that topic up later in the course.
Synthesis of Amines
Our last topic for today is the synthesis of amines. While it is possible to make alkyl amines (an example which is a primary amine with a primary alkyl group) would be RCH2NH2) by reaction of a primary halide with ammonia, these reactions are seldom very practical.
The more practical approches to making alkyl amines involve reactions which are reductions. The first of these is a reaction of an amide with lithium aluminum hydride. The overall effect of this reaction is to replace the carbonyl oxygen of the amide with two hydrogens from lithium aluminum hydride. We will not discuss the mechanism, but it is likely that it begins with the familiar nucleophilic attack of the hydride (H:- of lithium aluminum hydride on the carbonyl carbon of the amide. (This reaction is discussed in Section 13.8B of Brown.)
Notice that any other atoms or alkyl groups attached to the amide nitrogen are not changed. Since amides are commonly made by the reaction of acid chlorides with an amine, the two step sequence which begins with reacting an amine (or ammonia) with an acid chloride and follows with reduction of the amide using lithium aluminum hydride results in the substitution of an alkyl group on the nitrogen. Here's an example:
Another reductive synthesis of amines reacts a nitrile (RCN) with hydrogen in the presence of a catalyst which is very finely divided nickel. This reaction adds two molecules of H2 to the triple bond of the nitrile, and produces a primary alkyl primary amine (RCH2NH2. Lithium aluminum hydride will also carry out this reduction.
Recall that nitriles are commonly made by the reaction of primary alkyl halides with sodium or potassium cyanide, and we again have a two step sequence as in this example:
Finally, there is also a reductive method which involves a ketone or aldehyde carbonyl reaction. Early in the course we studied a reaction between primary amines and ketones or aldehydes which resulted in the replacement of the carbonyl oxygen by the nitrogen of the amine. Water was eliminated and the product was an imine.
If this reaction is carried out in a solution containing sodium cyanoborohydride (NaBH3CN, a less reactive analog of sodium borohydride) the imine is reduced to an amine even though there is also a reducible carbonyl group in the reaction mixture. The formation of glutamic acid (from which the amino groups of all other amino acids arise by transamination) is a biological analog of this "reductive amination." Here the reducing agent is NADPH+, one of several reducing agents which are common in biochemical reactions.
Again, we see that biochemical reactions use organic reaction mechanisms.
Contributors
• Kirk McMichael (Washington State University) | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.19%3A_Amines-_Structure_and_Synthesis.txt |
Amines and Carbonyls - Imine Formation
Last time we looked at the behavior of amines as bases, at their involvement in hydrogen bonds, and at the ways they can be synthesized. This time, we'll continue our study of amines by examining some of their reactions.
Let's begin reviewing reactions of amines with carbonyl compounds. When we first looked at aldehydes and ketones, we learned that the characteristic pattern of many reactions of the carbonyl group begins with the formation of a bond between the carbonyl carbon and an attacking nucleophile. The nucleophile provides the electrons to form the new bond and the pi bond of the carbonyl group is broken as it "gets out of the way." The electrons move from this pi bond onto what was the carbonyl oxygen. Here's an early example in which the nucleophile is an OH- group.
This reaction step works because the OH- group is a strong nucleophile (and a strong base) very capable of using one of its unshared pairs of electrons to make a new covalent bond. If a weak nucleophile is involved, like water, the reaction needs help in the form of acid catalysis. In this pattern, the H+ begins the mechanism by making a bond with the carbonyl oxygen. The electrons which make this bond can be envisioned as coming from the carbonyl pi bond, which leaves a positive charge on the what was the carbonyl carbon.
Once the pi electrons have been "gotten out of the way" by forming a new bond to the hydrogen, even a fairly weak nucleophile like water can us one of its unshared electron pairs to make a new bond to the former carbonyl carbon. The hydration reaction is completed by losing an H+, which also keeps thing tidy by replacing the H+ which was used to start the reaction.
The idea which emerges from this is that a strong nucleophile can attack directly, without help from an acid catalyst. For a weak nucleophile, an acid catalyst is needed so that the carbonyl carbon is prepared to share a pair of electrons as a new covalent bond. If we look at the mechanism of reaction between an aldehyde and an amine, we see how these factors balance. Here's the mechanism:
It is an experimental fact that this reaction -- imine formation -- is acid catalyzed. That suggests that the unshared pair of electrons on an amine nitrogen is not sufficiently nucleophilic to push the carbonyl pi electrons "out of the way" without help from an H+ which breaks that pi bond in an earlier step. Since we know that an amine (pKa of the conjugate acid ~ 10) is a weaker base than hydroxide or alkoxide ion (pKa of the conjugate acid ~ 16), it makes sense that an amine would also be a weaker nucleophile than hydroxide ion. The weaker nucleophile would be more likely to need a little help from acid catalysis.
It's also an experimental fact that if we put in too much acid, the reaction stops. How do we make sense of this? The key is to remember that an amine is a base. (Yes, sometimes it's easy to forget that something is a base if we've gotten fixated on its nucleophilic behavior, but that's our problem, not the amine's problem.) Being a base means that an amine will react with an acid to form an ammonium ion.
For each molecule of amine which does this, the unshared electron pair has been used to make the N-H bond and is not available to act as a nucleophile. That molecule of amine has been "benched" and is not available to react with the carbonyl compound. If this happens to all of the amine molecules (we've added too much acid) the reaction has to stop since one of its reactants is gone.
What's the best compromise? We need some amine to make the reaction go, so we want to add fewer acid molecules than there are amine molecules. We need some acid, though because it is important both to "jump start" the reaction and to catalyze the removal of the water molecule later in the mechanism. It turns out that the fastest rate happens if we control the pH so that half of the amine molecules are available to act as nucleophiles and the other half are present as the conjugate acid (ammonium salt). The ammonium ion (RNH3+) actually serves as the acid catalyst since it is the strongest acid which can co-exist with the amine. (Any stronger acid would just react with the amine to make more ammonium ion.)
Amines and Carboxylic Acid Derivatives - Amide Formation
Now let's turn our attention to the reactions of amines with carboxylic acids and their derivatives. Again, the nitrogen serves as a nucleophile in making a new bond to the carbonyl carbon. The pi bond is broken to "make room for" the nitrogen's pair of electrons. This step is just like the attack of a nitrogen nucleophile on a carbonyl carbon in an aldehyde or ketone, but what happens next is different.
The structural difference between aldehydes and ketones on one hand and carboxylic acid derivatives on the other is that a carboxylic acid derivative has a "leaving group." Leaving groups are distinguished from alkyl groups or hydrogen atoms by having an electronegative atom bonded to the carbonyl carbon. Pertinent examples include the chlorine in an acyl chloride and the -OR' group in an ester. Since the bond between one of these groups and the carbonyl carbon is polarized so that the electrons are closer to the leaving group atom than to the carbonyl carbon, it is already somewhat ionic and can cleave more easily than a carbon-carbon or carbon-hydrogen bond. This pathway is not available to aldehydes and ketones, but it dominates the reaction of carboxylic acid derivatives. The overall result is that when an amine (or any nucleophile) reacts with a carboxylic acid derivative the outcome is that the amine replaces the leaving group (a hydrogen is lost from the amine nitrogen too). The overall reaction is a substitution.
Now let's recall some examples of the reaction of amines with carboxylic acid derivatives. The details here are usually designed to overcome the fact that carboxylic acids and esters (and amides too) are less reactive than aldehydes or ketones. This is due to the fact that the "leaving group" atom in these derivatives also is electron rich (one or more unshared electron pairs) which tends to make the carbonyl carbon less "accepting" of a nucleophile's attempt to add an electron pair to it. Thus successful reactions between amines and carboxylic acid derivatives need to overcome the rather low reactivity of the carbonyl carbon in these compounds.
One very good way to do this is to put a very good leaving group in the carboxylic acid derivative. This is what is done with acyl chlorides.
We can get a sense of how good a leaving group might be by considering how strong a base is formed when the leaving group leaves. Remember that strong bases are difficult to form and weaker bases are easier to form. Chloride ion is the conjugate base of HCl, a very strong acid, so it is a very weak base and a very good leaving group.
Once we have the acyl chloride with its very good leaving group, we can use a moderately effective nucleophile like an amine to get a satisfactory method for making amides.
How about reactions between amines and esters. Here, the reaction is accelerated by heating it moderately. Notice that a stronger base (amine) is used up and a weaker base (alcohol) is produced. Notice also that before the alcohol (leaving group) portion of the ester departs, it picks up an H+ so that it can leave as the weak base alcohol (R'OH) rather than as the strong base alkoxide ion (R'O-). Again, weaker bases make better leaving groups.
In the case of making amides from carboxylic acids, the difficulty comes because the carboxylic acid is a stronger acid (pKa ~5) than the ammonium salt (pKa ~10). The result is that there is very little amine and carboxylic acid at equilibrium. so there is very little nucleophile present. Also, the O- in the carboxylic acid is a very poor leaving group. This reaction doesn't look promising at all, but it can be made to work by heating the ammonium salt strongly.
You may have noticed that we haven't tried acid catalysis of any of these reactions between amines and carboxylic acid derivatives. That's because any acid we add will react with the amine, so the strongest acid we can have in the reaction is the conjugate acid of the amine. That isn't a strong enough acid to "jump start" the lower reactivity carbonyl group of a carboxylic acid derivative.
Nitrosation - Nitroso Compounds and Diazonium Salts
Now to a reaction we haven't seen before. We will look at nitrosation because it follows on fairly naturally after the reactions of amines with carbonyl groups. Nitrosation reaction mechanisms begin with addition of a strong acid to sodium nitrite (NaNO2). Nitrous acid is formed, but it reacts further with acid to make water and the nitrosyl cation.
The nitrosyl cation is electron deficient. Its nitrogen has only three pairs of electrons in the valence shell, so it is a very good electrophile, very susceptible to attack by a nucleophile. When the nucleophile is a secondary amine, the product (after loss of an H+ from the amine nitrogen) is called an N-nitrosoamine.
Some of these N-nitrosoamines have been shown to be carcinogenic in animals, so there is concern regarding the possibility that they may be formed when sodium nitrite (added to some meats to prevent botulism) reacts with acid in the stomach and amines present in the body. The beneficial effect of sodium nitrite in preventing botulism poisoning must be weighed against the potential hazard of N-nitrosoamine carcinogenesis. As with many dietary hazards, the longer term effects are difficult to determine.
When the amine is primary, its reaction takes a different course. We will look at an example where the R group is the phenyl group (a benzene ring), since that is the most important application of this reaction.
The aromatic diazonium ions produced by this reaction are stable enough to persist in a cold acidic aqueous solution. They are important as synthetic intermediates in the preparation of a variety of aromatic compounds, including dyes and photographic chemicals. We'll take a longer look at what we can do with these compounds when we study aromatic chemistry in a few weeks. In the meantime, keep this reaction in mind. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.20%3A_Amines-_Reactions.txt |
Amino Acids and Side Chain Structure
Last time we finished our examination of amines. Now we'll look at what happens when a carboxylic acid functional group and an amine functional group are in the same molecule. Our focus will be on the alpha amino acids, those in which the amino group is bonded to the alpha carbon -- the one next to the carbonyl group -- of the carboxylic acid. These are the basic building blocks of proteins and are the most important type of amino acid. While there are many other ways to link an amino group and a carboxylic acid group in a single molecule, we will concern ourselves only with the alpha amino acids.
There are 20 alpha amino acids commonly found in proteins. They are listed in Table 18.1 (p 503) in Brown. When the structures of these molecules are examined, it becomes clear that they share the common structural unit RCH(NH3+)CO2- in which R can be either hydrogen (the amino acid is glycine) or one of 19 other possibilities. The one exception to this pattern is proline, in which the R group makes up part of a ring which also includes the amino group and the alpha carbon atom. Since the amino group in proline is involved in two carbon-nitrogen bonds, it is a secondary amino group.
The table is further divided into groups according to the structure of the R group. (The R group is often called the "side chain.") If the R group is made up of only carbon and hydrogen (no heteroatoms), the side chain is regarded as non-polar since there is very little polarity associated with carbon-carbon and carbon-hydrogen bonds. These side chains are hydrophobic (water avoiding) in much the same way that the long hydrocarbon tail of a soap or detergent is hydrophobic. This will be important when we consider how the characteristics of proteins depend upon their folding patterns in an aqueous environment. There are heteroatoms in methionine (sulfur) and tryptophan (nitrogen) but the overall behavior of these amino acids suggests that these heteroatoms contribute very little polarity to the side chain. Side chains which contain more polar functional groups such as amide, alcohol and thiol provide locations for a polar water molecule to hydrogen bond. They are thus somewhat hydrophilic, like the OH groups in a sugar. These side chains are important in making a protein sufficiently water soluble to operate effectively inside a cell.
In two cases (aspartic acid and glutamic acid) the side chain includes a carboxylic acid group in addition to the one next to the amino group. These groups are ionized (present as the carboxylate anion) when the pH is near neutral (pH ~ 7). (We'll take up the acid-base behavior of amino acids shortly.)
Similarly, there are three amino acids whose side chains include an amino group. These amino groups are also ionized (present as the ammonium ion) at neutral pH. The ionized groups are quite polar, and like the ionized ends of soaps or detergents, they make the side chain quite hydrophilic.
Acid Base Chemistry
At neutral pH (around 7, the typical pH of most body fluids and the pH at which biochemical reactions usually happen) the amino groups in amino acids are protonated to make ammonium ions and the carboxylic acids are ionized to their conjugate bases (carboxylate ions). One way to look at this is to look at a water solution at pH = 7 as a large reservoir of acid whose pKa is maintained at 7. If an acid with a pKa lower than 7 (like a carboxylic acid, pKa ~ 5) is dissolved in such a solution, it is the stronger acid and will transfer a proton to the solution and become the carboxylate ion. Thus when the pH is maintained at 7, carboxylic acids are ionized.
In the same way, when an amine (typical ammonium ion pKa ~ 10) is dissolved in water which is held at pH = 7, the water is the stronger acid so the amine is protonated to make the weaker acid. Amines when held at pH = 7 are protonated to make ammonium ions. Practically, holding the pH at 7 means that the solution is buffered by the inclusion of weak acids and weak bases in sufficient concentration so that the transfer of a few protons does not materially change the H+ concentration.
We can use this idea at any pH. For example, if an amino acid is dissolved in water which is held at pH = 2, the solution is a stronger acid than the carboxylic acid which would be formed by transferring a proton to the carboxylate ion. The carboxylic acid (pKa ~ 5) is formed as the weaker acid. Such a molecule would have only a positive charge from the ammonium ion. Similarly, in basic solution (pH > 11) the solution is a weaker acid than the ammonium ion, so the ammonium ion transfers a proton to the solution and becomes the amino group.
Since there are small variations in the specific pKa values of amino and carboxylic acid groups in amino acids, the exact pH at which the predominant species is the zwitterion (the molecule with one positive ammonium ion and one negative carboxylate ion) varies somewhat. This pH is called the isoelectric point (pI) because it is the pH at which the amino acid is as likely to be attracted to a positive electrode as to a negative one. The pI values for the common amino acids are given in Table 18.2 (p 506 in Brown).
Notice that the acidic amino acids have low pI numbers. This makes sense because it will take a fairly strongly acidic solution to ensure that one of the carboxylate ions is protonated. Similarly, for basic amino acids the pI values are higher since it will take a fairly basic solution to ensure that one of the ammonium ions has lost a proton and the positive charge.
Stereochemistry
For all of the amino acids except glycine, the alpha carbon atom is a stereogenic carbon atom (four different groups attached). In two cases there is also another stereogenic carbon atom in the molecule. Only one of the two possible enantiomers is found in nature in the cases of the amino acids which include stereogenic carbon atoms. In all these cases the absolute configuration of the alpha stereogenic carbon is S.
It became possible to determine absolute configurations well after the stereochemical relationships between amino acids and sugars had been worked out. That work showed that if we orient a Fischer projection of an amino acid with the carboxylate ion group at the top and the R group at the bottom, we find that the ammonium ion is pointed to the left. For this reason the amino acids are considered to have the L configuration (opposite to the D configurations assigned to common sugars). You may wish to verify that that an L-amino acid is also an S amino acid.
Amino acids are produced in living systems by biochemical pathways which involve multiple enzymes. The enzymes are proteins, themselves made up of L-amino acids so they provide a chiral environment in which only one of the two enantiomers is formed. Laboratory synthesis of amino acids typically does not involve a chiral environment, so equal amounts of the L- and D-amino acids are formed in typical laboratory syntheses. A mixture of equal amounts of enantiomers is called a racemic mixture.
Synthesis
Laboratory syntheses of amino acids are usually related to syntheses of amines and/or carboxylic acids. We'll take a look at one such synthesis, the Strecker synthesis. We won't look at it's mechanism in detail, but we will look for similarities with reactions we've seen before.
The reaction begins with imine formation from an aldehyde and ammonia. The acid catalysis required for this comes from ammonium chloride, a weak acid. An addition of hydrogen cyanide to the imine follows. This is analogous to the additions of nucleophiles to an aldehyde or ketone which we studied earlier. In this instance, the cyanide ion serves as the nucleophile.
The amino nitrile which results from these steps is purified and treated with aqueous HCl, followed by OH-. This converts the nitrile to a carboxylate salt. We can put this reaction in context by thinking of the C-N triple bond as being much like a carbonyl group. That suggests that the electrophilic H+ attacks the nitrogen, which is followed by a nucleophilic attack of water on the nitrile carbon. A C=N double bond remains, and it's reaction with water is the reverse of imine formation. The outcome is that the C=N double bond is hydrolyzed to a C=O double bond. Finally, neutralization with just enough base gives us the amino acid zwitterion.
Peptides and the Peptide Bond
Now let's turn our attention to the way in which amino acids are linked together to form proteins. The key structural element here is the peptide bond. This is an amide linkage which joins the ammonium group of one amino acid to the carboxylate group of another by a new covalent bond. The O- of the carboxylate is lost along with two H+ ions from the ammonium group to form water. This is quite analogous to the formation of an amide by heating a carboxylic acid and an amine. The specific reaction conditions and processes required to do this may be (as we will see) quite sophisticated, but it helps to remember that what is being done is the joining of a carboxylate carbon and an ammonium nitrogen by a new C-N (peptide) bond.
The new compound formed in this way is called a peptide. Our example is a dipeptide, formed from two amino acids. If a third amino acid is connected to the dipeptide by forming a new peptide bond at either the ammonium group or the carboxylate group of the dipeptide, we obtain a tripeptide, and so on. Polypeptides may have many amino acids. Polypeptides with more than 100 amino acids are considered to be proteins.
Since the amino acid whose carboxylic acid group participated in the formation of the peptide bond still has an ammonium group which contains a nitrogen atom, it is called the N terminus of the peptide. The N terminus is conventionally written to the left. Correspondingly, the amino acid which still has a free carboxylate group is called the C terminus and is written to the right. When the order of amino acids in a peptide is written out, it is conventional to write it left to right from the N terminus to the C terminus. The complete order of amino acids in a protein is called its sequence and is conveniently expressed by using the abbreviated names of the amino acids read from N to C terminus.
The sequence is held together by peptide bonds. As a part of an amide functional group, these bonds are difficult to break, so the sequence of a protein is quite stable. While there are many possible ways that a protein chain could be folded, the particular folding pattern adopted by the protein is completely determined by the its sequence.
In many cases the folding pattern is "locked in" by disulfide links. As we discussed when we were studying thiols, the presence of SH groups along a protein chain provides an opportunity for crosslinking between chains or the formation of loops within a chain. These disulfide bridges are important in holding the protein chain in a specific folding pattern.
A glance at Table 18.1 tells us that the SH groups necessary to make the disulfide links are found in the amino acid cysteine. Proteins which are stiff and used primarily for structural purposes (keratin in hair, skin and feathers, for example) usually have many disulfide links and thus have high contents of cysteine. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.21%3A_Amino_Acids_and_Peptides.txt |
Sequencing a Peptide
Last time we looked at the structural characteristics of amino acids and the peptide bond which joins individual amino acids together to make proteins and peptides. We also learned about the sequence (order) in which amino acid units are joined in peptides. Today we'll study the ways in which the specific sequence of a peptide may be discovered and the methods which are used to synthesize such a peptide.
The first thing that is done in determining the sequence of a peptide is to find out which amino acids are present and in what ratios. This is much like beginning the process of determining the structure of an organic compound by determining the ratios of atoms such as carbon, hydrogen and oxygen. This is done by hydrolyzing the peptide bonds which hold the peptide together using HCl as an acid catalyst. (The mechanism is very much like acid catalyzed ester hydrolysis.)
This "amino acid analysis" tells us what the building blocks are in the peptide, but it tells us nothing about their sequence, the order in which they are joined. This information is lost when the peptide bonds which preserve that sequence are hydrolyzed. Even with as few as two amino acids, there are two possible sequences. Consider a dipeptide which amino acid analysis gives us gly and ala. Either of these could be the N terminus, so the dipeptide could be either gly-ala or ala-gly. Problem 18.6 in Brown gives you some experience with a pentapeptide, and things rapidly get more complex as the number of amino acid units in the peptide increases.
The next step is to determine which amino acids occupy the N terminus and C terminus positions in the peptide. N terminus determination is commonly done by a process called the Edman degradation. The chemistry is outlined as follows:
This reaction can be understood if we look for some analogies that will help us apply the patterns we used in the past. The -N=C=S group resembles a CO2 (O=C=O) molecule in that the carbon atom is connected to two electronegative atoms by a double (sigma and pi) bond. We know from reacting Grignard reagents with CO2 that the nucleophile attacks the carbon in CO2, so we can expect the same type of pattern in the Edman degradation. The nucleophile is the free NH2 group at the N terminus of the peptide, formed by loss of a proton from the NH3+to some unspecified base. As we have seen with other reactions of NH2 groups, this step is followed by a proton shift.
The product of the addition of N and H to the C=N double bond has a nucleophilic sulfur atom located just in reach of the carbonyl carbon at the other end of the N terminal amino acid. Attack of this sulfur at that carbonyl group is followed by departure of the NH group of the next amino acid. This cleaves the peptide bond between the N terminal amino acid and the next amino acid. Further reshuffling of protons yields an isomer of the phenylthiohydantoin. This isomer is converted to the phenylthiohydantoin during the treatment with HCl and the phenylthiohydantoin is identified. Since the phenylthiohydantoin includes the R group of the N terminal amino acid, identification of the phenylthiohydantoin also identifies the N terminal amino acid.
The other product of the Edman degradation is also a peptide -- it is the original peptide minus the original N terminal amino acid. It now has a new N terminal amino acid, which was adjacent to the N terminal amino acid in the original peptide. The new peptide can also be subjected to Edman degradation. When this is done, we learn the identity of the second amino acid (from the N terminal end) of the original peptide and obtain again a peptide which is now two amino acids shorter than the original. In principle, repetition of this sequence would allow us to run successive Edman degradations, clipping off an N terminal amino acid with each degradation, and thus learn the entire sequence of of a peptide or protein. In practice, such a process is practical only for about 20 to 40 amino acids.
Since the laboratory steps in an Edman degradation are very repetitive -- the same thing is done at each cycle, it has been possible to automate this process. Computer controlled "protein sequenators" are common in biochemistry laboratories.
Overlapping Sequences
When peptides and proteins larger than 20 to 40 amino acid units are to be sequenced, they are first broken into smaller fragments either by chemical or enzymatic partial hydrolysis. Here's an example of how partial chemical hydrolysis might be used.
Of course, angiotensin II (a peptide involved in blood pressure regulation) is small enough that a protein sequenator could give its sequence directly, but the example illustrates the way fragments may be overlapped to give a complete sequence for a larger protein. The complete sequence of a protein is called its primary structure
Peptide Synthesis
When a sequence has been obtained for a peptide, attention can be turned to its synthesis. There are two issues to resolve in synthesizing a peptide. One is to develop a method for making the peptide bond which does not damage anything else in the peptide. This is called "coupling" the two amino acids. The other is to be sure that the amino acids are added to the peptide in the proper sequence.
The key to the first issue is to convert the O- an amino acid's carboxylate group into a better leaving group. We've seen something similar when we've converted an OH into a Cl as we made the reactive acyl chlorides. This is done by the use of a reagent called dicyclohexyl carbodiimide, or DCC for short. DCC works by bonding to the O- and converting it to a good leaving group -- good because it has many electronegative atoms which can help stabilize the negative charge as it leaves. An amino acid which has a good leaving group is said to be "activated."
The actual coupling reaction occurs when the amino group of the amino acid "to the right" in the sequence attacks the carbonyl carbon of the "activated" amino acid and the DCC leaves with the oxygen it is bonded to. As usual, there are some proton shifts needed to tidy things up.
The second issue, adding amino acids in desired sequence, can be illustrated by considering the synthesis of a dipeptide such as Ala-Gly. If we simply mix equal quantities of glycine and alanine and run a DDC coupling reaction, we will get glycines reacting with glycines to give Gly-Gly, alanines reacting with alanines to give Ala-Ala, and glycines reacting with alanines in two ways to give Ala-Gly and Gly-Ala. This is a mess and it would be better to develop a more specific approach.
To do this we need to arrange things so that only one of the two carboxyl groups and only one of the two amino groups are free to engage in the coupling reaction. This is done by the use of protecting groups. If we make an amino group into an amide, it is much less reactive as a nucleophile. This can be done by reaction with an acid chloride.
Carboxyl groups are normally protected by conversion to a benzyl ester. This reaction is a Fischer esterification
After coupling, the protecting groups can be removed by hydrogenation.
This is a specific reaction in which only the bonds between the benzyl groups and the oxygens are broken, so the amide bond we have just made by coupling is not affected. (The carbonyl group on the formerly protected amino group is lost as CO2.) More sophisticated protection schemes in which either the N protection or the O protection can be selectively removed have been developed.
In practice, proteins are now synthesized by molecular biological techniques in which the gene which encodes the sequence of amino acids is isolated and used to direct the synthesis of the protein by a bacterium or a yeast.
Secondary, Tertiary and Quaternary Structure
When we are thinking of a peptide's sequence it is convenient to think of it as a chain which is stretched out, peptides are more commonly coiled (alpha helix, Brown p 518) or folded (beta sheet, Brown p 519). These shapes are called a peptide or protein'ssecondary structure and they are held in place primarily by hydrogen bonds. Recall that hydrogen bonds are much weaker than covalent bonds, but strong enough to resist rupture by mild temperatures. Hydrogen bonds are attractive interactions between the positive end of dipoles like the N-H and O-H bonds and negatively charged locations such as the unshared electron pairs on atoms like oxygen or nitrogen. In peptides it is commonly the N-H bonds of an amine and the oxygens of the carbonyl groups which participate in hydrogen bonds.
Regions of alpha helix or beta sheet are often combined by further folding patterns which make up a protein's tertiary structure. Structural proteins such as keratin or fibroin often have large regions of alpha-helix and make fibers. Enzymes are more often globular with hydrophilic amino acids on the outside and hydrophobic amino acids folded in towards the middle.
Individual proteins are often combined into clusters, which may include non-protein molecules such as heme (Fig 18.14, p 522 in Brown). Often such combinations are necessary so that the protein can carry out its biological function. Such clusters constitute the protein's quaternary structure.
Experiments have established that a protein's primary structure is enough, by itself, to determine how it will fold and combine with other proteins to make the appropriate secondary, tertiary and quaternary structures. It is not clear how this happens, and this is an area of active study. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.22%3A_Proteins.txt |
Functions of DNA
Last time we examined how a the amino acid sequence of a peptide or protein might be discovered. We also learned how a chemical synthesis of a small peptide can be carried through.
Today we'll study the chemistry of the molecule which carries the information necessary for directing the biosynthesis of proteins and peptides. This is DNA, and we'll learn that the structure of DNA provides a very strong rationale for its function.
First, let's think a little about what a "molecule of heredity" needs to do. It must store an immense amount of information -- the directions for synthesizing all of the proteins necessary to for the successful functioning of a living organism. It must be able to transmit that information faithfully, with an extremely low error rate, to the protein synthesis system, to both daughter cells upon cell division, and to a future generation upon reproduction of the organism.
Prior to the discovery of the structure of DNA, there was much speculation regarding possible molecules which might meet these requirements. Proteins themselves were seriously considered as candidates, since they are stable, they can be large enough to hold a large amount of information, and they are certainly intimately involved in biology. The information might be encoded in the sequence of amino acids in a protein, much like information in a word is encoded in a sequence of letters. We might think of the amino acids as a 20 character alphabet, with which we could make a very large number of words, sentences, paragraphs, books, etc.
When the structure of DNA was established by James Watson and Francis Crick in 1953 it was immediately clear that it had the necessary characteristics to meet the specifications of a molecule of heredity. It is a polymer which can be extremely long. (The DNA in the single molecule which makes up a human chromosome is about 12 cm. long.) This provides the capacity to store large amounts of information. Like a protein this polymer backbone is carries an alphabet. In this case, the alphabet consists of only four letters, A, C, G, and T. Let's look at the details.
DNA Polymer Backbone
The polymer backbone is made up of two types of structure. One is a modification of the sugar ribose. The modification is that the OH group on the carbon next to the anomeric carbon has been replaced by a hydrogen. This is called 2-deoxy-D-ribose.
These 2-deoxyribose units are linked together by phosphate esters which link the 3' oxygen of one sugar with the 5' oxygen of the next. (The "primes" ' are there to differentiate atoms in the sugar ring from those in the bases, which we'll take up next.) This gives us a "backbone" for DNA which looks like this:
The "alphabet" molecules, A, C, G, and T, are attached to this backbone at the anomeric (1') carbons. Recall that these carbons are the ones where other groups can be attached by reactions such as those that convert hemiacetals to acetals. We'll next look at the structures of these molecules, which are called bases since they contain nitrogen atoms which make them mildly basic.
The Bases A, C, G, and T
Two ring structures are found in the bases. C (cytosine) and T (thymine) have a single six membered ring, called a pyrimidine ring. A (adenine) and G (guanidine) have two rings joined together. This unit is called a purine ring. C and T are called pyrimidine bases; A and G are called purine bases. Here are the structures:
In each of these bases there is a secondary amine whose nitrogen forms a bond to the anomeric carbon of a deoxyribose in the DNA backbone. We can relate the chemistry of the formation of this linkage to the formation of a glycoside (acetal) from glucose (hemiacetal) and an alcohol. The difference is that in the current case the nucleophile is the secondary amine nitrogen of a base rather than the oxygen of an alcohol. An example of four bases attached in this way is:
The "word" here is CACT. Recall that the DNA backbone is very long, and it is clear that even with only a four letter alphabet, a great deal of information can be carried by DNA
Replication - Two DNA's from One
The next issue is transmission of the information to a daughter cell, to a succeeding generation, or to the protein synthesis machinery of a cell. The other key feature of the structure of DNA as discovered by Watson and Crick is that DNA molecules come in pairs, twisted together in the "double helix" (Figs. 19.8, 19.9; pp. 541,2 in Brown). Each of these molecules is a single long strand, held together by the covalent bonds along its backbone. The connections between the DNA strands are made by hydrogen bonds between the bases. Hydrogen bonds (as we learned when we studied amines) are much weaker than covalent bonds, but since there are many of them connecting the two DNA strands in the double helix, they serve very well to maintain that structure until there is a need for separation of the two chains.
Not only do the hydrogen bonds hold the chains together, they also are very specific in which bases are connected by the hydrogen bonds. Adenine (A) forms two hydrogen bonds only with thymine (T). Guanidine (G) forms three hydrogen bonds only with cytosine (C).
In each case, the hydrogen bond is formed between the positive hydrogen end of a polar N-H bond and a pair of electrons on either a nitrogen or a carbonyl oxygen. These "complementary" base pairs also have another important feature: a purine base (adenine or guanidine) always bonds to a pyrimidine base (cytosine or thymine). This means that the distance between the two strands is always the same (three rings and the hydrogen bonds). Hydrogen bonding between two purine bases, for example, would put four rings into the base pair, and the fit would be poor. You can try to put together other hydrogen bonding patterns, but these two are the ones which fit best.
Watson and Crick realized that the specificity of this base pairing scheme was the key to replication of DNA and the transmission of information from one generation to the next. This is done in three steps. First the double helix is separated into the individual DNA strands by successively breaking hydrogen bonds between the base pairs.
Second, as a segment of "unwound" DNA is exposed, bases from the solution encounter it, align with the complementary bases on the exposed DNA strands and form the proper base pairs, A with T and C with G. These bases are already joined to the necessary ribose and phosphate groups in molecules called nucleotides, so that as they line up in the proper arrangement, the materials for the formation of the backbone of a new polymer are in the proper locations.
Third, as separation and hydrogen bonding with new bases proceed, the individual nucleotides are joined together by the formation of new bonds between a phosphate of one nucleotide and the 3' OH group of the next nucleotide.
The outcome of these process is that each strand of the original DNA double helix has been used as a template upon which a copy of its former partner has been constructed. There are now two identical double helices which are the same as the original.
This is known as replication. In cell division each of these DNA copies would become part of one of the daughter cells. Each step in this process is assisted and controlled by enzymes, and there is also a "proofreading" function involved so that mismatched base pairs (such as an A-G pair) are excised and repaired.
Transcription and Translation - DNA to mRNA to Protein
There are two successive processes by which the information contained on a DNA strand is used to determine the amino acid sequence of a protein. In the first of these, called transcription, a copy of the DNA strand is made, but in this case the copy is RNA. There are two structural differences between DNA and RNA. In RNA the sugar is ribose (with the 2' OH group) while in DNA it is 2-deoxyribose (without the 2' OH group). Also, where T (thymine) would occur in DNA, U (uracil) occurs in RNA.
The transcription process by which a RNA copy is made is very similar to the process by which DNA replicates. In this case, only a partial unwinding of the DNA helix occurs, and the appropriate bases hydrogen bond to the separated DNA, U with A, C with G. The RNA nucleotides which are now lined up on the DNA template are then joined together to form the RNA strand, which is a copy of the DNA strand which was not the template, and is complementary to the strand which was used as a template.
In this way an RNA strand which carries the genetic message (called "messenger" or mRNA) to the protein synthesis machinery (called a ribosome) is made. Its base sequence specifies the amino acid sequence in the protein to be made. The codes for each amino acid use three bases in a row and are given in table 19.3 (p 547 of Brown). Since there are 64 ways to make three letter "words" (called codons) with a four letter alphabet, many amino acids are coded for by more than one "word."
In the ribosome, the codons on mRNA are matched with anticodons (A with U, G with C) on transfer RNA (tRNA) molecules. Each tRNA molecule carries the appropriate amino acid to the enzyme which links them together to make the protein. This process is called translation. We won't look at the linkage process in detail, but it does include protection and activation steps much like the chemical synthesis we studied earlier.
The information flow in the overall process is: Codon sequence in DNA determines codon sequence in mRNA. Codon sequence in mRNA determines the order in which tRNA molecules line up. The order of tRNA line-up determines the sequence in which amino acids are linked to make the protein. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.23%3A_Nucleic_Acids.txt |
Recall Nucleophilic Substitution Examples
Today's topic takes us back to an important organic reaction mechanism. We've studied a few reactions which proceed by this mechanism. Now it's time to examine it in detail.
Let's begin by recalling a couple of reactions which occur with alkyl halides, but only work well when the alkyl halide is primary (the halogen is bonded to a carbon which is directly bonded to only one other carbon.) One such reaction involved cyanide ion and resulted in a nitrile which was then converted to a carboxylic acid:
Another similar reaction used an alkoxide (the conjugate base of an alcohol) and resulted in an ether. You may recall this as the Williamson ether synthesis:
In both of these examples the bond between the carbon and the halogen (usually bromine or chlorine) has broken and its pair of electrons has remained with the halide. If we compare what happens here with what happened with the chlorine in an acyl chloride, we recognize that in both situations the halide has behaved as a leaving group.
Similarly, we can focus our attention on the new bond that is being made. The electrons which form this bond in the product have come from the attacking reagent -- the cyanide in making a nitrile and the alkoxide in the Williamson ether synthesis. We recognize this behavior as that of a nucleophile, an atom or group which supplies a pair of electrons to form a new covalent bond. We've seen nucleophiles add to carbonyl carbons in both ketones and aldehydes and in carboxylic acid derivatives.
These reactions are known as Nucleophilic Substitution Reactions, substitution reactions because one atom or group has been substituted for another, and nucleophilic because the substituting atom or group has supplied the electrons for the new bond.
Possible Mechanisms
The typical reactions of carboxylic acid derivatives are also nucleophilic substitution reactions, but these are different. Remember that in the reactions of carboxylic acid derivatives there was first an addition to a the carbonyl group in which the carbon-oxygen pi bond was broken. This step was followed by one in which the leaving group departed and the carbon-oxygen pi bond was re-formed:
In alkyl halides this mechanism is not available since there is no carbon-oxygen pi bond to break and reform. This leaves us with two possibilities:
1. The bond between the carbon atom and the leaving group breaks first. The carbon is left with six bonding electrons, an empty orbital, and a positive charge.
2. In a second step, the nucleophile reacts to form a bond, much like the pattern which we saw in the acid catalyzed reactions of carbonyl groups. We'll study these reactions next time.
3. The bond between the carbon atom and the leaving group breaks at the same time as the bond between the nucleophile and the carbon atom is formed.
4. There is only one step and it requires the nucleophile and the alkyl halide (also called the substrate) to collide for it to take place.
(A third possibility, that the nucleophile attacks first to form an intermediate which later loses the leaving group, is not possible because the carbon in the intermediate would have ten bonding electrons and five bonds -- a very high energy situation.)
Today we'll consider the second of these possibilities. This mechanism is called an SN2 mechanism; S for substitution, N for nucleophilic and 2 because two molecules collide at the critical point in the reaction.
SN2 Mechanism
First, let's look at what happens in a little more detail. This is a one step reaction. In order for it to take place, the nucleophile must come close enough for bonding to begin to happen, but this also means that the carbon-halogen bond must begin to break. Both of these actions increase the energy of the combination - bond breaking requires energy as does overcoming the repulsion which results from moving the nucleophile's electron pair into close contact with the electrons in the carbon's bonding shell.
As the reaction process goes forward, the energy increases until a significant bonding begins to occur between the nucleophile and the carbon. This releases enough energy to balance the energy required to break the carbon-halogen bond. At this point. called a transition state, the energy is at a maximum. There is roughly a half bond between the nucleophile and the carbon and a half bond between the carbon and the halogen. Since the energy is at a maximum, any slight movement will cause it to decrease, either to go back to reactants or to go on to products. This picture of the transition state is the key to understanding the characteristics of the SN2 mechanism.
The energy required to boost the nucleophile and the alkyl halide to the transition state energy level is called the activation energy. It comes from the energy with which the molecules collide. If the activation energy is low, many collisions will provide enough energy and the reaction will be fast. If the activation energy is high, few collisions will provide enough energy and the reaction will be slow. Another way to say this is that slow reactions have high activation energies (and high energy transition states) and fast reactions have low activation energies (and low energy transition states). In order to understand what makes a reaction go slow or go fast, we examine the transition state to see what changes will increase or decrease its energy.
Effect of Alkyl Halide Structure
With this background, we can look back at the restriction that our examples of SN2 reactions (nitrile and ether synthesis) only work well on primary alkyl halides. In the structure of the SN2 transition state, there are 90o bond angles between the breaking bond to the leaving group and the three bonds which remain connected to the carbon as well as between the bond being made to the nucleophile and those same three bonds.
As long as the two of the groups attached to the carbon being attacked are small hydrogens, the repulsions which happen do not require much energy. If the groups attached to the carbon are larger, though, like methyl groups, the transition state energy increases, the activation energy increases, and the reaction becomes much slower.
This means that the reactivity order for alkyl halides in SN2 reactions is:
methyl > primary > secondary > tertiary
The practical outcome of this is that SN2 reactions are generally reliable only when the alkyl halide is primary.
Stereochemical Inversion
The structural picture of the SN2 transition state also accounts for another characteristic of SN2 reactions. Notice that the nucleophile is bringing its bond-forming electrons into the transition state on the side of the carbon opposite to the position of the leaving group. We can understand this if we remember that the nucleophile and the leaving group are both electron rich atoms. We would expect them to repel each other and stay as far apart as possible while remaining connected by their half bonds to the central carbon atom.
The result of this is that the carbon atom is inverted. If we examine a three-dimensional picture of this, we can see that the three groups which remain connected to the carbon throughout the reaction move away from the entering nucleophile and towards the position occupied by the departing leaving group.
If the nucleophile and the leaving group are both high in the R/S priority order, this means that an R alkyl halide gives an S product, and vice-versa. The term for this is inversion of configuration and it is an inherent and consistent characteristic of the SN2 mechanism. If we know the configuration of the alkyl halide before reaction, we know that the configuration of the product will be the opposite. Conversely, if we determine that a nucleophilic substitution reaction proceeds with inversion of configuration, we conclude that its mechanism is SN2.
Our picture of this reaction starts with a tetrahedral sp3 carbon in the alkyl halide and ends with a tetrahedral sp3 in the product. In the transition state the three bonds to carbon which don't react are approximately flat, it makes sense to regard the carbon atom as sp2 hybridized at this point. One consequence of this is that the SN2 mechanism is restricted to halides which are sp3 hybridized at the reactive carbon. Carbons which are sp2 hybridized at the halogen-bearing carbon do not react by this mechanism.
Kinetics, Alkyl Halide and Nucleophile Effects
Earlier we saw that the energy required to reach the transition state comes from the energy with which the nucleophile and the alkyl halide collide. The requirement for a collision also means that the frequency with which the nucleophile and the alkyl halide collide is important. This frequency is primarily controlled by concentration.
If the concentration of the alkyl halide is high, then there will be many opportunities for a nucleophile to collide with an alkyl halide molecule. The rate of the reaction will increase proportionately as the alkyl halide concentration is made higher. When this is the case the reaction is said to be first order in alkyl halide. Similarly an increase in the nucleophile concentration will result in a proportionate increase in the rate, so the reaction is also first order in nucleophile. Overall, the reaction is said to be second order. This can be summarized in the rate equation.
Rate = k[RBr][Nu-]
It is the second order behavior (requirement for two molecules to collide in the critical transition state) which is designated by the "2" in SN2
Since the bond between the carbon and the leaving group is being broken in the transition state, the weaker this bond is the lower the activation energy and the faster the reaction. This leads to the following reactivity order for alkyl halides
RI > RBr > RCl > RF
Practically, alkyl fluorides are not used for SN2 reactions because the C-F bond is too strong. Often alkyl iodides are reactive enough to be difficult to store, so the the common choices for reactions are alkyl chlorides and alkyl bromides.
If we remember that the function of the nucleophile is to provide an electron pair to make a new bond, we can see a similarity between a nucleophile and a base. A Lewis base makes a new bond, typically to hydrogen, using its own electron pair. A nucleophile makes a new bond to carbon, using its own electron pair. As this suggests, good nucleophiles are typically strong bases. There are are other factors, but this is a good starting place and it reminds us to review base strengths, perhaps by reviewing Table 2.1 on p p 43 of Brown. Keep in mind that high pKa numbers mean weak acids which have strong conjugate bases.
Next time we'll take a look at the other mechanism for nucleophilic substitution, the SN1 mechanism. We'll also see how elimination reactions fit into this picture. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.24%3A_Nucleophilic_Substitution_SN2_SN1.txt |
SN1 Energy and Kinetics
Last time we saw an overview of the nucleophilic substitution mechanisms of alkyl halides. We examined one of these, the SN2 mechanism in detail. Today we'll examine the other, the SN1 mechanism, and then go on to look at elimination reactions, the major competition for substitutions. Here's the outline of the SN1 mechanism:
Recalling what the "2" in SN2 meant -- that the reaction was second order, two molecules had to collide to provide the activation energy needed to reach and pass through the transition state -- we can guess that the "1" in SN1 means that only one molecule needs to be "activated" in order to reach the transition state. That molecule is the alkyl halide. The critical step in this mechanism is the first step, in which the bond between the carbon atom and the halogen leaving group is broken. The transition state for this step has the bond stretched far enough that the halide ion is balanced between leaving as a stable chloride or bromide ion or slipping back into a covalent bond.
(The energy required to break this bond comes from random collisions with the solvent without the solvent reacting.) The activation energy for the first step is higher than for the second step, so the rate of the reaction is controlled by how many molecules get through the first step. Once a molecule is through the first step, it can react rapidly in the second step. We call the step which is slowest the "rate determining step." Notice that the rate determining step for this reaction doesn't involve the nucleophile. That means that changing the concentration of the nucleophile doesn't affect the rate. The only concentration which affects the rate is the concentration of the alkyl halide. We thus have a first order reaction:
Rate = k[RX]
This also means that the strength of the nucleophile -- its ability to use its electron pair to make a bond -- isn't important in determining how fast the reaction goes. It's not involved in the rate determining step, so it has no effect on the energy of that transition state.
Effect of Alkyl Group Structure
What does affect the energy of the rate determining transition state? If we examine its structure in more detail, we notice that there is considerable positive charge on the carbon atom and the carbon-halogen bond is nearly broken.
The nearly broken bond tells us that the effect of changing the halogen is the same as it was for the SN2 reaction:
RI > RBr > RCl > RF
The substantial degree of positive charge on the carbon is important in explaining how the structure of the alkyl group affects the rate. A large number of experiments have established that the reactivity order for alkyl halides in the SN1 mechanism is:
tertiary > secondary > primary > methyl
This is just the opposite of the order for the SN2 reaction. The outcome of this contrast is that tertiary alkyl halides consistently use the SN1 pathway. Primary and methyl alkyl halides use th SN2 pathway. For secondary alkyl halides either pathway is possible, and we have to look at other information to make a decision.
The reactivity order tells us that the transition state for the SN1 reaction of a tertiary alkyl halide has a lower energy than the transition state for a correspinding secondary alkyl halide. Perhaps the easiest way to understand this is to recognize that the transition state, with its substantial positive charge on carbon, is very much like the carbocation intermediate which is formed in the first step. Changes which make the carbocation intermediate, with its positively charged carbon, more stable will also make the transition state, with its nearly positively charged carbon, more stable.
The major thing which determines the energy of the carbocation intermediate is that it has only six electrons in its valence shell. It is an electrophile, a Lewis acid, and a seriously electron deficient molecule. The more electron density which can be shifted towards the positively charged carbon, the lower the energy. In a tertiary carbocation, there are three carbon atoms, each bonded to three other atoms, connected to the electron deficient carbocation carbon. This adds up to 18 electrons in the bonds adjacent to the carbocation, all of which can shift slightly to help neutralize the positive charge. Contrast this to the situation in a methy carbocation. Here there are no valence electrons other than the ones holding the hydrogens to the carbon, so there is a very poor supply of electrons to assist with lowering the energy of the carbocation.
While this discussion has focused on the carbocation intermediate with its full positive charge, the same principles apply in the case of the partial positive charge in the transition state. The more electrons there are in the near neighborhood, the more stable the transition state. There are more electrons available on the carbon atoms attached to a tertiarly carbocation center than there are on the hydrogens attached to a methyl carbocation center.
If there are pi bonds involved with a carbon attached to the carbocation carbon, the energy is reduced even more. Recall that pi electrons are less tightly bound than sigma electrons, so they are easier to move towards the electron deficiency. This can be symbolized in resonance terms:
The effect of this is that alkyl halides which have carbon-carbon pi bonds located one atom away from the carbon bearing the halogen are quite reactive in SN1 reactions.
Stereochemical Outcome
The product of an SN1 reaction is formed in the second step. This step starts with the carbocation intermediate, so let's look at its structure to see what we can learn from it. First, notice the distinction between an intermediate and a transition state. Remember that a transition state is at a maximum energy. Any slight change will cause it to fall from it's precarious perch and become either the product or the reactant of its step. In contrast, an intermediate is at an energy minimum so that most small changes only push it a little way up a hill from its lowest energy situation. It takes a substantial collision to provide it with the energy needed to reach a transition state so as to pass through and become a product. If the activation energy is fairly small, as it is with the second step of the SN1 mechanism, the reaction may well be fast, but it doesn't occur with every small wiggle as it does with a transition state.
Another way to say this is that a transition state has a very, very short lifetime, while an intermediate may exist for micro- or milliseconds, which is a long time and provides opportunities for many collisions with other molecules.
One outcome of this is that the carbocation intermediate "lives" long enough for a nucleophile to approach it on either face of the molecule. The carbocation is flat with 120o angles between its three bonds. That means that it is trigonal like the carbon of a carbonyl group and is sp2 hybridized. The "empty" orbital which we can associate with the electrophilic characteristics of the carbocation is a p orbital.
If the carbocation intermediate has the opportunity to engage in many collisions with potential nucleophiles, there is an equal chance that a nucleophile will attack at one lobe or the other of the p orbital. If there are three different groups attached to the carbocation center, the nucleophile will provide a fourth and generate a stereogenic carbon. Equal quantities of the two enantiomers will be expected so that the product mixture will not be optically active.
This is a somewhat idealized situation, since in practice the halide leaving group is "loitering" near one lobe of the p orbital. This makes reaction with the nucleophile easier on the other lobe, so there is usually some net inversion. The stereochemical oucome of the SN1 is not as clearcut as that of the SN2 mechanism's inversion, but it is still a distinguishing feature of the mechanism, one which can be used to decide which mechanism is operating in a particular reaction.
Solvolysis
The fact that the nucleophile is not involved in the rate determining step of an SN1 reaction also means that it proceeds well with a relatively weak nucleophiles. When the nucleophile is also the solvent, the reaction is called "solvolysis." Solvents like water and alcohols are particularly useful here, because they provide both a nucleophilic pair of electrons on the oxygen atom of the OH group and a fairly polar solvent which helps to stabilize the strongly polar transition state. The latter effect is much like the way in which a polar solvent dissolves a very polar substance like salt by surrounding its charged ions with polar solvent molecules.
This process resembles the way in which an alcohol or water might attack the electrophilic carbon of a carbonyl group after an acid had added an H+ to the oxygen. Remember that in these reactions the neutral nucleophilic atom attacks first; then the H+ is lost.
A summary of the important differences between the SN1 and SN2 mechanisms is in Table 7.5 (Brown, p 189).
Competition with Elimination
One of the things we dealt with last time was the fact that nucleophiles are also Lewis bases. One outcome of this is that the same atom or group can attack a carbon in an SN1 or SN2 reaction -- behaving as a nucleophile -- or attack a hydrogen atom -- behaving as a Lewis base. The latter attack can lead to an elimination reaction. We will look at elimination reactions in more detail in a week or so, but we can usefully examine them as competitors for SN1 and SN2 reactions just now.
When we first learned about the Williamson ether synthesis, we learned that it works best when the alkyl halide is primary. We now understand that as a characteristic of the SN2 mechanism -- primary alkyl halides react faster than secondary or tertiary alkyl halides. However, a patient person might suggest that even a slow SN2 reaction might succeed if we were willing to wait a while. What defeats this strategy is that the alkoxide ion can also react as a base. This elimination reaction (an E2 reaction for future reference) is fast enough that it uses up the secondary or tertiary alkyl halide long before the much slower SN2 reaction produces any useful amount of product.
Elimination reactions are always potential competitors for substitution reactions. The key factors which alert us to situations favorable to eliminations are:
1. An alkyl halide which is slow in SN2 reactions, i.e., tertiary and secondary alkyl halides.
2. The presence of a strong base like an alkoxide or hydroxide ion.
These conditions will typically produce much more elimination product (an alkene) than substitution product.
Can we use an SN1 pathway to avoid this difficulty? Yes -- and the key here is that since the rate of an SN1 reaction is not sensitive to the concentration or strenght of the nucleophile we can avoid the strong bases which promote elimination. Solvolysis of a tertiary alkyl halide using an alcohol as both nucleophile and solvent can make an ether very effectively.
Of course, this is only possible when the desired nucleophile can be used as a solvent as is the case for alcohols and water.
SN1 and SN2 Reactions of Alcohols
To finish up today, let's revisit some reactions of alcohols and see if we can use the SN1 or SN2 pattern to understand them a little better. Recall that a useful method for making an alcohol into an alkyl halide was to treat the alcohol with the hydrogen halide, particularly when the alcohol was tertiary:
This looks like a nucleophilic substitution and since the alkyl group is tertiary, the SN1 pathway through a carbocation intermediate looks like a good guess. We can also identify the nucleophile as the bromide ion (Br-), but what about the leaving group? The obvious answer is that the OH- serves as the leaving group, but this is worrisome since we would be making a strong base in the presence of an acid. The solution appears when we remember that the unshared electron pairs on the alcohol oxygen are also weak bases. They can accept a proton (H+) from the strong acid HBr. When this is done, the leaving group is water, a weak base.
When the OH group of an alcohol is replaced by another nucleophile, we can be sure that the OH group is first converted to a good leaving group before the C-O bond is broken. That is the function of the H+ in an acid, the SO2 part of thionyl chloride and the phosphorus in PBr3. We won't concern ourselves with the details of these processes, but we can notice that the need to make the OH group into a good leaving group is the same whether the reaction is SN1 as we would expect for a tertiary alcohol or SN2 as we would expect for a primary alcohol. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.25%3A_Elimination_-_E2_and_E1.txt |
Alkene Double Bond Structure
Today we'll begin by looking at the structural characteristics of alkenes, the products of elimination reactions. Then we'll return to the topic of elimination reactions and examine their reaction mechanisms in more detail.
The functional group of an alkene is the carbon-carbon double bond. In common with the double bond we studied at the beginining of the course (the carbon-oxygen double bond in the carbonyl group) this double bond consists of a sigma bond, viewed as "end to end" overlap between sp2 hybridized orbitals on the carbon atoms, and a pi bond, viewed as "side to side" overlap between the p orbitals on the same carbon atoms.
Stereoisomerism
The presence of the pi bond in the alkene functional group has two important consequences. First, the reactions of alkenes are essentially the reactions of this pi bond. We'll look at these reactions in some detail next time. Second, the pi bond means that rotation of one end of the double bond relative to the other requires so much energy that it does not happen at ordinary temperatures. This is because such rotation would destroy the "side to side" overlap of the p orbitals which make up the pi bond and would effectively break the pi bond. Breaking bonds requires energy input.
The result of this can be seen in the fact that there are two different substances which have the same "connectivity" structure and are both called 2-butene. (Alkene naming is treated in Section 5.2 of Brown). A sample of one of these compounds does not become the other, since to do so would require breaking the pi bond and there isn't enough energy available to do that. Since the difference is one of spatial arrangement, this is a type of stereoisomerism.
The absence of rotation about the double bond explains why there are two different 2-butene molecules. These differences must be reflected in the names given to the compounds. If there are two identical groups, one on each carbon involved in the double bond, then we can use the terms cis (which means that the two identical groups -- hydrogens or methyls in this case -- are on the same side of the double bond) or trans (means that the two identical groups are on opposite sides of the double bond).
This system breaks down when there are not identical groups on the two carbon atoms. In this type of situation we use a system which is derived from the same ranking rules we used in assigning R and S configurations to stereogenic carbon atoms. To use this system we look at each of the two doubly bonded carbons independently. For one carbon we examine the two groups or atoms which are connected to it by single bonds. We use the ranking rules to decide which of these groups or atoms has the higher ranking. Then weapply the same process to the second carbon atom. If the higher ranked group on one carbon is on the same side of the double bond as the higher ranked group on the other carbon, then the designation is Z (from the German word "Zusammen"). If the higher ranked group on one carbon is on the other side from the higher ranked group on the other carbon, the designation is E (from "Entgegen"). Here's an example:
Like all naming situations, you learn by doing problems, so do the naming problems listed in the internet syllabus and bring up points that need clarification. See Section 5.2C in Brown for further explanation.
Elimination Reactions
Now let's turn our attention to making this double bond. The reactions which do this are called elimination reactions, because two atoms or groups are "eliminated" from an alkyl halide or alcohol so that the double bond can be formed. A formal example is:
Notice that the two atoms eliminated were attached to adjacent carbon atoms. This must be so if we are to make an new pi bond between those atoms. If the atom bearing the bromine is designated the alpha carbon atom, then the one next to it is a beta carbon atom. These eliminations are often called beta eliminations.
In many cases there are more than one beta carbon atom. This can lead to situations where more than one beta-elimination product is possible. Here's an example:
Notice that the major product is the one which has the most substituents (non-hydrogen atoms, in this case, methyl groups) attached to the doubly bonded carbons. This is generally the case, and it is called Zaitsev's rule after Alexander Zaitsev the 19th century Russian chemist who first proposed the general statement. In elimination reactions the major product is the one in which the maximum number of substituents is attached to the doubly bonded carbons. (Notice that the carbon skeleton is not changed in this reaction.) Zaitsev's rule enables us to predict which of two or more possible beta-elimination products can be expected to predominate.
Often we can choose an alkyl halide such that only one beta-elimination product is possible. For example, if we wanted to make the "minor" product from the example above, we might choose the following approach:
Here we've chosen an alkyl halide which has only one beta hydrogen, so it can give only one alkene in a beta-elimination reaction. We've also chosen a base which is bulky to slow down the competing substitution reaction.
The E2 Mechanism - Dehydrohalogenation
The most important elimination reaction mechanisms are closely related to substitution mechanisms. The first one we'll study is called the E2 mechanism, E for elimination and 2 for second order. In studying the SN2 mechanism we learned that second order meant that the concentrations of both the alkyl halide and the nucleophile were important in determining the rate of the reaction.
rate = k[RX]{Nu]
We interpreted this to mean that the alkyl halide and the nucleophile must collide in order to form the transition state, which smoothly completes the reaction and becomes the product. The same interpretation occurs here. The important difference is that the reactant we considered to be a nucleophile attacking carbon in the SN2 reaction is now acting as a base and attacking a beta hydrogen.
If we follow the curved arrow notation we see that formation of the O-H bond releases the C-H bonding electrons to begin to form the new C-C pi bond. The C-Br bond breaks at the same time to provide room for the new C-C pi bond to develop. All of these bonding changes occur in one step. The transition state (top of the energy hill) has three partial bonds in it. Since it includes both the alkyl halide and the base, it is consistent with the need for them to collide in order for the reaction to occur and with the second order kinetic behavior.
Since the transition state structure has a partial broken C-Br bond just like the partially broken C-Br bond in the SN2 transition state, we expect that changing from one halide to another will produce the same change in rate that we saw for the SN2 mechanism:
RI > RBr > RCl > RF
This is what is observed.
Since the base plays such an essential role in this mechanism, it is very likely that the elimination reactions we considered to be unwanted complications when we were studying SN2 reactions were in fact E2 reactions. Recall that we learned that these eliminations are most prevalent when we have strong base (essential for an E2 mechanism) and a tertiary alkyl halide (which would make the competing SN2 mechanism very slow). The combination would be good for a fast E2 elimination and a slow SN2 substitution and would predict lots of elimination product.
The E2 mechanism is very closely related to the SN2 mechanism. In much the same way, there is an E1 mechanism which involves the same first step and the same carbocation intermediate as the SN1 mechanism. Here's the pattern:
Notice that the first step is identical to the first step in the SN1: dissociation of a halide ion to form a carbocation. The E1 mechanism continues by loss of a proton and the formation of a new pi bond. The first step determines how fast the reaction goes. The relative proportions of substistution and elimination product are determined by the relative rates of nucleophilic attack on the carbocation carbon (substitution by SN1) or loss of the proton (elimination by E1).
Since the rate of reactions proceeding by this mechanism is determined by the rate at which the carbocation is formed, the effect of changing alkyl group structure is the same as it was for the SN1 mechanism. (Notice that a methyl halide only has one carbon and cannot make a double bond.)
tertiary > secondary > primary
Notice that the rate of this reaction does not depend upon the base concentration. The base is not involved until after the carbocation is formed in the rate determining step. This means that the E1 mechanism is likely to be used where no strong base is present.
The E1 Mechanism - Dehydration
Alcohols can also be used to make alkenes by elimination reactions. We know that the OH- group is a poor leaving group (it's a strong base and strong bases make poor leaving groups). To permit an alcohol to react by breaking the C-O bond, the OH group must first be changed into a better leaving group. This can be done using an acid to place a proton on the OH oxygen. This makes an H2O of the OH, and since water is a much weaker base than OH-, it is a much better leaving group.
Once the alcohol has been protonated, what happens next depends upon the alkyl group structure and what else is in the reaction mixture. If there are bromide or chloride ions present (perhaps because the acid used was HBr or HCl) then those bromide or chloride ions can serve as good nucleophiles and complete a substitution reaction. If we have used only a small, catalytic, amount of an acid such as sulfuric acid, then there is only a small amount of bisulfate ion present to act as a nucleophile. The reaction is more likely to result in elimination. One important factor in this outcome is heating the reaction. Alkenes have lower boiling points than alcohols, so once an alkene is produced, it boils out of the reaction mixture and is collected by distillation. Removing the alkene as it is formed protects it from other possible reactions.
If the alcohol is primary an elimination uses the E2 pathway, primarily since forming primary carbocations is so slow. Ethylene can be made this way. Notice the close resemblance to the E2 mechanism for alky halides, with the major difference that no strong base is involved here since a strong base cannot exist in an acid solution.
Secondary and tertiary alcohols are more likely to use the E1 pathway. Here's a typical mechanism:
Like alkyl halide dehydrohalogenation (elimination), dehydration of alcohols follows Zaitsev's rule -- the more highly substituted alkene is the major product. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.26%3A_Alkenes_and_Alkyne_Structure.txt |
In the functional group of an alkene - the carbon-carbon double bond -- the most readily available electrons are those in the pi bond. They are farther from the nuclei than the electrons in a sigma bond, so they are more readily attracted to an electrophile if one approaches. Another way to say this is that the pi electrons are weakly nucleophilic. We'll begin by looking at a few reactions which begin with the attack of an electrophile on the pi part of the double bond.
First, let's briefly review a reaction of the carbonyl group with an electrophile. The acid catalyzed addition of water to an aldehyde is one such reaction discussed earlier. The mechanism is:
The first step is electrophilic attack on the carbonyl pi bond by the electrophilic, acid H+. This step makes a carbocation, which is then attacked by the weak but very abundant nucleophile water. (Do not use OH- as the nucleophile. There is very little OH- in an acidic solution!) The final "mop-up" step gives the product and returns an H+ to regenerate the catalyst.
Now let's apply this same mechanism to the addition of water to ethylene, the smallest alkene.
Notice that the steps are the same and that the last two steps (what happens to the carbocation) are the same steps which occured in the SN1 hydrolysis (solvolysis in water) of an alkyl halide.
These steps -- first an electrophile attacks the pi bond to form a carbocation, second a nucleophile attacks the carbocation -- are the key steps in the most important reactions of alkenes, electrophilic addition reactions. The first step is the slow one, so it is the one which determines the rate of the reaction. This has very important consequences when we introduce a slight complication into the structure of the alkene.
Orientation - Markovnikov's Rule
What happens if we apply this reaction to an alkene like propene? (Such an alkene is called "unsymmetrical" since the substitution patterns on the two alkene carbons are different.) There are two possible products since the OH could wind up on either the CH2 carbon or the CH3CH carbon.
We may reasonably expect that the product which is formed fastest will be the one which predominates in the product mixture. In fact, if one product is formed more than 100 times faster than the other, it is the only product we will observe in a practical sense. This means that predicting which product is formed comes down to predicting which product is formed faster. In turn, this question becomes "which product is formed by a pathway with a lower activation energy?" We estimate relative activation energies by looking at transition state structures. The pathway with the lower energy transition state will be the faster pathway. It will produce more product in a given time than the slower pathway, so its product will be found in the greater amount.
How do we make such a prediction? We know the mechanism, so we can apply it to the pathways which would lead to our potential products. Fortunately, the rates associated with these two pathways are determined by their first steps, so we need only concern ourselves with the activation energies for those steps.
We know that a secondary carbocation (on the left) is more stable (has a lower energy) than a primary carbocation (on the right). We have explained this by saying that the the electrons in a methyl group are better at partially relieving the carbocation's electron deficiency than the relative lack of electrons around a hydrogen. Since the transition states for these two steps are also electron deficient partially formed carbocations, we would expect the same effects to prevail in the transition states. This tells us that the reaction to the left will have the lower activation energy and will occur faster. We then expect to see the product in which the attacking H+ is attached to the less substituted carbon and the OH is attached to the more substituted carbon.
More generally, the electrophile attacks the less substituted carbon atom in the first step and the nucleophile attacks the more substituted product in the second step.
We can use this to predict which product will be formed in an electrophilic addition to an unsymmetrical alkene. All we have to do is identify the electrophile and the nucleophile in the compound which is adding. For example consider the addition of HBr. The HBr bond is polarized so that the H is positive and the Br is negative. The H+ is thus the electrophile and the Br- is the nucleophile. Application of the pattern above gives us:
Historically, this pattern was observed by Vladimir Markovnikov in 1870, long before the mechanism was understood. The generalization that hydrogen adds to the carbon with the most hydrogens (another way to say what is in bold above) is known as Markovnikov's rule. Common applications include the additions of HBr and water (discussed above) and HCl.
Hydroboration-Oxidation
Notice that we cannot make a primary alcohol by adding water to any other alkene than ethylene by an electrophilic addition reaction as outlined above. An alternative reaction sequence which does achieve this result was developed about 40 years ago by H. C. Brown. It is outlined as follows.
We will focus our attention on the first step. We know the product of this step, so we can identify the atoms which are acting as the electrophile and the nucleophile. The boron adds to the least substituted carbon, so it is acting as the electrophile. The hydrogen adds to the most substituted carbon, so it is acting as the nucleophile. Does this make sense? We have seen the B-H bond act this way before in the reduction of ketones and aldehydes by BH4- (in sodium borohydride). There the hydrogen acted as a nucleophile and we explained that by saying that hydrogen is more electronegative than boron. We can understand the reaction of boron as an electrophile if we consider that B2H6 dissociates into two molecules of BH3. A quick check of the electronic configuration of BH3 shows that the boron has an empty 2p orbital, so BH3 has the structural characteristics needed to qualify it as an electrophile.
Practically, the overall result of this reaction sequence is to add water to an alkene double bond in an orientation which is opposite to that predicted by Markovnikov's rule. For that reason it is referred to as an "anti-Markovnikov" addition. It is a powerful method for making primary alcohols and other alcohols where it is desired to locate the OH group on the less substituted carbon atom.
Halogen Addition
Halogen molecules such as Cl2 and Br2 also add to alkene double bonds. Here we need not be concerned with orientation since the two ends of the adding molecule are identical, but the electrophilic addition mechanism helps us understand another characteristic of this reaction, its stereochemistry.
If bromine is added to cyclopentene, we might anticipate two products which differ in how the bromine atoms are geometrically related to each other. If the two bromines are on the same face of the ring, the compound is called cis. If they are on opposite faces, the compound is called trans. The experimental result is that only the trans product is formed.
One of the bromine atoms is acting as an electrophile. If we apply the usual mechanism, it's first step it would go like this:
As we learned in our study of SN1 reactions, carbocations are attacked by nucleophiles on both faces. If a carbocation is present in this system, we'd expect to find both the cis and trans products.
This is not what happens when the experiment is done, so we conclude that the carbocation is not present. Something else is happening, something which prevents the nucleophilic bromide ion from attacking the face of the carbocation which already is attached to the first bromine. We understand that by envisioning that the electrophilic bromine atom attaches itself to both alkene carbon atoms. One bond is made using the electrons from the pi bond, and the other is made using an unshared electron pair from the bromine. This results in a new ring formed from the bromine and the two alkene carbon atoms. It is called a "bromonium" ion.
Attack by the nucleophilic bromide ion on the bromine in this ring would only result in cyclopentene and bromine, so no reaction would occur. Attack by the bromide ion on the either of the alkene carbons would be like an SN2 reaction. The attacked carbon would invert, and the product would have the trans configuration. (Notice that there are two such products, which are enantiomers, so we get a racemic mixture.)
Hydrogenation
There is another reaction of alkenes, hydrogenation, which deserves mention but which is not related to the electrophilic addition mechanism. Hydrogenation is the addition of molecular hydrogen (H22) to the alkene double bond. This converts a simple alkene into an alkane.
Hydrogenation reactions are carried out in the presence of a solid catalyst such as finely divided platinum (Pt) metal. The reaction occurs on the surface of the metal and involves temporary bonding of both the alkene and the hydrogen molecule to the atoms on the metal surface.
Hydrogenation is important in the conversion of highly unsaturated (many double bonds) fats to fats with fewer double bonds. This is done so that the product will have a higher melting point than the reactant and be more convenient as a dietary spread for bread. Also, the products are less susceptible to reaction with oxygen and do not turn rancid as rapidly. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.27%3A_Electrophilic_Additions.txt |
Polymers - Structure and Response to Heat
Let's begin by noticing some important real-world characteristics of polymers. While they all contain molecules with very long chains, there are some important differences between the properties of different types of polymers. Most polymers are formed into the desired shapes after softening or melting by heating. Some, like the familiar polyethylene and polystyrene, may be melted and reshaped again and again. These are called thermoplastic polymers.
Others char or burn when reheated. These are called thermosetting polymers. Examples include Bakelite and vulcanized rubber. The structural difference between these polymers is that the thermosetting polymers have crosslinks between the chains and the thermoplastic polymers do not. When a thermoplastic polymer is heated the chains are free to move past each other making the sample less rigid and eventually melting it. This cannot happen with a thermosetting polymer, since its chains are locked together by the cross links. The energy from the heat must eventually go into breaking bonds which leads to decomposition of the polymer.
This schematic view suggests the difference:
We noticed crosslinking earlier when we saw how the disulfide crosslink formed by oxidation of the SH group in cysteine was important in maintaining protein structure.
Repeating Units and Monomers
Since polymers are made by linking together many identical small molecules, there are repeating units in polymers. Here's an example, polyvinyl chloride, in which the repeating unit is -CH2-CHCl-.
In poly(vinyl chloride) the repeating unit comes directly from the end-to-end linking of many vinyl chloride molecules. A molecule from which a polymer is made is called a monomer. Each vinyl chloride monomer molecule contributes a CH2 group joined to a CHCl unit by a single bond. This single bond is a remnant of the double bond which joined those groups in the vinyl chloride molecule. This is just what happens in an addition reaction of an alkene. We'll see how an addition reaction leads to such polymers in a few paragraphs.
Repeating units can also be made from two monomers. Here's an example:
We notice that the repeating unit is linked to the rest of the chain by amide functional groups, and that the repeating unit contains an amide group. We can deduce the structrure of the monomers by imagining the compounds which might be used to make the amide group. In the industrial process these are a dicarboxylic acid (adipic acid) and a diamine (1,6-hexanediamine). These form a salt when dissoved in alcohol. The salt is heated at 250o under pressure to form the amide bonds. Nylon 66 is the result.
Step-Growth Polymers
Polymers formed in this way, where both ends of the growing chain have functional groups which can react with a monomer or with the appropriate functional group on another chain, are called step-growth polymers. Let's look at another step-growth polymer, but this time we'll look at the monomers and propose a structure for the polymer that would result.
The monomers are a dicarboxylic acid (terephthalic acid) and a dialcohol, also called a diol (ethylene glycol). We immediately recognize that these monomers can make an ester, so that we expect our polymer to be linked by ester functional groups. The resulting polymer is called polyethylene terephthalate and is the common polyester of plastic pop bottles and polyester fabrics.
Notice that in choosing how to represent the repeating unit in step-growth polymers we have picked the particular repeating unit (out of several possibilities) which is linked to the rest of the polymer through functional group bonds. In the case of Nylon, the functional group bond is an amide bond, so Nylon is a polyamide (in that way like a protein). The term polyester also puts our focus on the functional group which is made in the polymerization reaction. Step-growth polymerization involves normal functional group reactions. Polymers result because the monomers have two functional groups per molecule.
Chain-Growth Polymers
Polymers resulting from additions to alkenes 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.
Each step in this polymer formation is an addition to an alkene. The mechanism is in most cases a free radical addition. In free radical reactions the pi pair of electrons separates. One of these electrons pairs with an electron from the attacking reagent to form a sigma bond with one of the alkene carbons. and the other electron remains attached to the other alkene carbon. (Curved arrows with only one "barb" on a point are used to follow the path of a single electron in the same way that "double-headed" arrows follow the path of an electron pair.) Intermedates with an unpaired electron are called free radicals, so this step can be described as adding a free radical to an alkene to lengthen the chain by two carbons and generate a new free radical. In its turn this new free radical can add to another molecule of monomer and continue the process.
The most important monomers for this process are ethylene (which makes the polymer polyethylene) and substituted ethylenes like vinyl choride (polyvinyl chloride), styrene (phenylethylene, polystyrene), methyl methacrylate (Plexiglas), and acrylonitrile (cyanoethylene, acrylic fibers). Table 15.1, p 427 in Brown lists the structures of these monomers, from which you can deduce the structures of the polylmers.
We might ask about the orientation of attack of a radical on a substituted ethylene (vinyl) monomer. It has been shown that the orientation is the same as it is for an electrophilic addition, that is, the free radical attacks the less substituted of the two alkene carbons so as to produce the new free radical at the more substited carbon. Here's an example for styrene:
Other methods of polymerization are also known. The most important is the Ziegler-Natta polymerization which uses a combination of TiCl4 and an alkyl aluminum compound such as Al(CH2CH3)3. This process is discussed in more detail in Sec. 15.5B of Brown. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.28%3A_Polymers.txt |
Beta-Oxidation of Fatty Acids
Today we're going to examine a selection of processes which occur in metabolism. We will focus on comparing these reactions to reactions we have already studied. In particular we will see that the reactions which break carbon-carbon bonds are just reverse versions of the aldol and Claisen condensations which we have studied earlier. Keep in mind that while we are looking for connections between these reactions and familiar organic reactions, all steps in these schemes are catalyzed by enzymes.
The two processes we are going to study are both catabolic processes, that is, they are processes that break down and oxidize larger molecules to produce smaller molecules and energy. The first, beta-oxidation, is a key part of the process by which fatty acids are broken down to acetate. [Acetate is the conjugate base of acetic acid. Since a neutral pH is more basic than the pKa of acetic acid (~5), in neutral solution acetic acid is predominantly ionized and acetate is the major form present.] The overall scheme of beta-oxidation looks like this:
From this we can see that the outcome of a beta-oxidation "event" is that two carbon atoms are cleaved from a fatty acid. The bond broken is between the alpha and beta carbons. (This accounts for the term "beta-oxidation.") The gamma carbon shows up in the product as a carboxylic acid. This carboxylic acid, two carbons shorter than its "parent," can be shortened by another trip through the beta-oxidation process, with the production of another molecule of acetate and a new fatty acid, again two carbons shorter. That the reactions all occur at the carboxylic acid end of the molecule rather than at the CH3 end is not surprising, since a fundamental idea of organic chemistry is that reactions occur at functional groups rather than elsewhere.
Prior to the commencement of the actual beta-oxidation cycle, the carboxylic acid end of the fatty acid is esterified with the SH group of coenzyme-A. This reaction is discussed in more detail in Sec 20.3A in Brown. Here is the scheme as it more realistically looks with the coenzyme-A included.
Now let's look at the individual reactions of this process. (Notice the slightly different notation scheme. The principal reactant and product are connected by a straight arrow. The necessary reagents and their byproducts are connected by a curved arrow.) The first reaction results in the removal of hydrogen atoms from the alpha and beta carbo atoms. Its effects are opposite to those of hydrogenation of a double bond. The removal of hydrogen atoms makes this an oxidation reaction. The oxidizing agent is a molecule called "FAD" (for flavine adenine dinucleotide). We'll look at its structure later.
Water is then added to the alkene pi bond which results from the first reaction. This is analogous to the addition of water to alkenes we studied recently. Since we know that a carbon alpha to a carbonyl group is a rather nuclophilic place (remember enolates?), it makes sense that the electrophilic hydrogen from water would add there and the nucleophilic OH would add at the beta carbon.
As soon as we see the third step,
we recognize the oxidation of a a secondary alcohol to a ketone. This is a reaction for which we used chromium-6 reagents earlier, but now the oxidizing agent is NAD+ (nicotine adenine dinucleotide cation). The hydrogen attached to the OH-bearing carbon is transferred to the NAD+, and the OH hydrogen comes away as an H+.
The final step is the actual cleavage of the bond between the beta-carbon and the gamma carbonyl group. To put this in context, we have to think about it in reverse. When we do that, we see a pattern which is identical to the Claisen condensation.
The conclusion is that the final step in the beta-oxidation cycle is a reverse Claisen condensation. Like the Claisen condensation itself, this step is possible because the enolate ion obtained as the acetate fragment breaks away is stabilized by resonance. This enolate ion is neutralized by a proton source (acid). The shortened fatty acid is released from the enzyme as CoA?SH replaces the sulfur of the enzyme. This last step goes through a tetrahedral intermediate (not shown) as we would expect for a reaction which converts one carboxylic acid derivative to another.
The acetyl-coenzyme-A formed in this cycle enters the tricarboxylic acid cycle where it is oxidized to two molecules of CO2. The NADH and FADH2 produced in beta-oxidation and the tricarboxylic acid cycle enter a process called oxidative phosphorylation which results in the formation of ATP (adenosine triphosphate) for use in providing energy within the cell. We will not take up these latter processes, but they are an important part of a biochemistry course.
Glycolysis
Our next topic is glycolysis. This is the conversion of glucose to pyruvate. In the larger scheme of things the pyruvate produced is then converted to acetate, which like the acetate from beta-oxidation of fatty acids, enters the tricarboxylic acid cycle. Again, we will be looking at the reactions involved trying to find analogies in organic reactions we have studied. The chart in Fig 20.2 (p 571) in Brown lays out the whole scheme of glycolysis. Please refer to that chart to identify the reactions we are considering.
Reaction 1 is an esterification. It resembles the conversion of an alcohol to an ester by an acid chloride. In Reaction 1, ATP is used rather than an acid chloride, and the result is an ester of phosphoric acid rather than an ester of a carboxylic acid.
Reaction 2 is formally an internal oxidation-reduction reaction. The carbonyl group of glucose is reduced to a primary alcohol while the OH group at carbon-2 of glucose is oxidized to a ketone. The reaction goes through an enol as shown on p 572 of Brown.
Reaction 3 is similar to reaction 1, the formation of a phosphate ester (this is called a phosphorylation reaction).
Reaction 4 involves cleavage of a carbon-carbon bond and is the reaction in which the carbon skeleton changes from a six carbon chain to two three carbon chains. The first thing to notice is that the carbon-carbon bond broken joins a carbon which is alpha to a carbonyl group to one which is beta. This suggests that enol and enolate reactivity is going to be important. Like we did with the beta-oxidation of fatty acids, we can see a familiar reaction if we think about this one in the reverse direction, that is, look at it as a bond-making reaction rather than a bond-breaking reaction. If we ignore the groups which don't change, we can see that our reaction is a reverse aldol addition.
In more detail, the reverse aldol steps in this reaction are preceded by conversion of the carbonyl group of fructose 1,6-bisphosphate to an imine, and followed by the hydrolysis of that imine back to a carbonyl group and the amino group. The amino group is attached to the enzyme which catalyzes this reaction, and the formation of the imine helps anchor the fructose 1,6-bisphosphate molecule to the enzyme in the proper position to bring the bases and acids on the enzyme to the right location for the mechanism to go forward. The sequence of mechanistic steps is:
Reaction 5 converts dihydroxy acetone phosphate to glyceraldehyde 3-phosphate. This is like the reverse of reaction 2 in that it goes through an enol intermediate to oxidize an OH group to a carbonyl on one atom and reduce a carbonyl to an OH on the adjacent carbon. The sequence is on p 574 in Brown. At this point in glycolysis a glucose molecule has been converted to two molecules of glyceraldehyde 3-phosphate.
Reaction 6 is an oxidation of the aldehyde in glyceraldehyde 3-phosphate to a carboxylic acid. The odxidizing agent is NAD+, which is reduced to NADH. The process also includes formation of an anhydride between the newly formed carboxylic acid and a molecule of monohydrogen phosphate ion. This anhydride is quite reactive (since phosphate is a stable anion, it's a good leaving group much like chloride) along the lines of an acid chloride.
Reaction 7 exploits the reactivity of this mixed anhydride by using it to transfer the phosphate to ADP. The resulting ATP is used as an energy source for many cell processes, so this step in glycolysis directly produces energy for the cell.
Reaction 8 shifts the phosphate group from the OH on carbon 3 to the OH on carbon 2. It is like an ester hydrolysis at one carbon and an esterification at the other.
Reaction 9 is a dehydration reaction. It resembles the dehydration of cyclohexanol to cyclohexene we did in lab, although the conditions in a cell are much milder since enzyme efficiencies make it unnecessary to employ strong acids and heat. The product is an enol phosphate (the phosphoric acid of a ketone).
Reaction 10 does two things. Like reaction 7 transfers a phosphate to ADP, making ATP, and it produces the enol form of pyruvate which rapidly equilibrates to the keto form.
If we add up all the balanced reactions, we find that one glucose molecule, two ADP molecules, two NAD+ molecules, and two hydrogen phosphate molecules have been converted to two pyruvates, two ATP's, two NADH's and two hydronium ions.
Fates of Pyruvate
What happens to the pyruvate? This depends on the local conditions. In muscle tissue there may be a limited supply of oxygen (due to the demands of exercise outpacing the body's capacity to supply oxygen to the tissue). This means that the NADH produced in glycolysis not oxidized back to NAD+. Glycolysis requires NAD+, so it would stop and along with it production of energy in the form of ATP. This can be circumvented by lactate fermentation, which reduces pyrvate to lactate and oxidizes NADH to NAD+. Energy production can continue until the build-up of lactate and acid as a result of this reaction (Brown, p 577) exhausts the muscle.
In yeasts in the absence of oxygen, fermentation to alcohol occurs. This is the basis of the fermentation processes which produce beverage alcohol in beer and wine. The byproduct is CO2 which is responsible for the carbonation of beer and sparkling wines. If the yeast is used in baking, the CO2 expands the dough to produce the "rise" of bread dough. In this case the alcohol is the byproduct and it is largely driven off by the heat of baking. It is also responsible for much of the pleasant odor of baking bread.
If there is a good oxygen supply, pyruvate is oxidized to acetyl CoA and CO2. As we have seen now several times, the oxidizing agent is NAD+ which is reduced to NADH. This reaction is more complex than it looks at first glance, but we will leave those complexities for a biochemistry course. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.29%3A_Metabolic_Organic_Reactions.txt |
Symbols for the Benzene Ring
Today we'll find that resonance is very important in understanding both the structure and the reactions of aromatic compounds. First, let's take a look at the structural representations which distinguish aromatic compounds from those that aren't aromatic.
The most commonly encountered aromatic compound is benzene. The usual structural representation for benzene is a six carbon ring (represented by a hexagon) which includes three double bonds. Each of the carbons represented by a corner is also bonded to one other atom. In benzene itself, these atoms are hydrogens. The double bonds are separated by single bonds so we recognize the arrangement as involving conjugated double bonds. An alternative symbol uses a circle inside the hexagon to represent the six pi electrons. Each of these symbols has good and bad features. We'll use the three double bond symbol simply because it is also routinely used in the text.
Keep in mind that if the hexagon contains neither the three double bonds nor the circle, the compound is not aromatic. It is simply cyclohexane and there are two hydrogens on each carbon atom. This is easy to mistake when hurrying, so be careful when you are intepreting any structural formulas which include hexagons.
Kekule Structures
The structure with three double bonds was proposed by Kekule as an attempt to explain how a molecule whose molecular formula was C6H6 could be built out of carbons which make four bonds. The ring and the three double bonds fit the molecular formula, but the structure doesn't explain the chemical behavior of benzene at all well. Each of the double bonds would be expected to show the characteristic behavior of an alkene and undergo addition reactions, but this is not how benzene reacts.
In particular, we would expect a carbon-carbon double bond to react quickly with bromine to make a dibromo compound. This is what alkenes do very readily, and in fact it is a useful test for alkenes in the laboratory. Benzene does not react with bromine unless a very bright light or a strong catalyst is used, and then the reaction is not an addition reaction. We conclude that there is something quite unusual about the double bonds in benzene.
Kekule (thinking about this problem before bonds were understood as pairs of electrons) suggested that there are two forms of benzene which differ in the locations of the double bonds. His idea was that these were in rapid equilibrium, so rapid that there was never a fixed location for the double bond. One could say that an approaching bromine molecule could not "find" a double bond to react with.
Resonance
There were several other structures proposed for benzene, but a much more satisfactory approach became possible when we began to understand that covalent bonds consist of pairs of electrons shared between atoms. The difference between the two structures Kekule envisioned (called Kekule structures) is only the difference between the locations of three pairs of electrons. This is exactly the type of situation where resonance must be involved. The hybrid or "average" of the two Kekule structures has one sigma bond and one-half of a pi bond between each two carbon atoms. Thus each carbon is joined to each of its neighbors by a one-and-half bond. Each bond in the benzene ring has the same number of electrons and is the same length. This picture is in complete accord with experiments which show that all carbon-carbon bonds in benzene are the same length, with no hint of shorter (double) or longer (single) bonds. It also helps explain why benzene does not undergo addition reactions: there are no simple pi bonds.
Recall that resonance has another important feature: when resonance is involved, the real structure is more stable than we would expect from any of the structures we write using the one line = two electrons symbolism. This extra lowering of energy, which for benzene is about one-third as much as making a typical covalent bond, is quite important in the reactions of benzene and other aromatic compounds. As we will see, reactions of the benzene ring almost always result in products which in which the benzene ring persists -- an outcome of its stability.
Huckel Rule
When resonance theory was first applied to understanding the structure of benzene, the key feature seemed to be a resonance hybrid of ring structures containing alternating single and double bonds. This immediately led to attempts to make and study compounds like cyclooctatetraene and cyclobutane. These compounds also have ring structures with alternating single and double bonds.
Cyclooctatetraene has been made, but it does not posess the properties of extra stability and resistance to addition reactions which distinquish aromatic compounds. It readily adds bromine, for example. Cyclobutadiene is extremely unstable -- one cyclobutadiene molecule reacts with another cyclobutadiene molecule instantaneously even at very low temperatures -- so it certainly does not act like an aromatic molecule and it has been called "antiaromatic" as a result.
It seems that there is more to being aromatic than simply a ring with alternating single and double bonds. After considerable development of the underlying theory, the pattern which has emerged is that aromatic characteristics are only expected when there is a ring of pi electrons in which the number of pi electrons is equal to 4n + 2 (where n is an integer, 0, 1, 2, etc.). (This is known as the Huckel rule after its discoverer.) We can check this against the compounds we have considered so far: Benzene has 6 pi electrons (two for each pi bond) which is the number we get from 4n + 2 if n = 1. Cyclooctatetraene has 8 pi electrons, and there is no integer "n" which will make 4n + 2 = 8. Cyclobutadiene has 4 pi electrons and also doesn't fit 4n + 2. There are many other examples which support Huckel's rule.
It is important to be sure that the ring of alternating single and double bonds is complete. If there is an sp3 hybridized carbon in the ring, the conditions for aromatic character are not present, and we do not worry about checking for 4n + 2. Here's an example:
Another way to see this is to look at the p orbitals which combine to make the pi bonds. If these p orbitals combine to form an uninterrupted ring as is the case in benzene, then we can go ahead to use Huckel's rule to check for the proper number of pi electrons for aromatic character. If the ring of p orbitals is broken by a CH2 (group or another tetrahedral carbon) with no p orbital, then the compound cannot be aromatic and we need not try to apply Huckel's rule.
The p orbitals which make up the unbroken p orbital ring can be associated with other atoms than carbon. Two examples are furan and pyrrole, in which two of the six electrons needed come formally from unshared electron pairs on oxygen.
Such an unshared pair can also come from a carbon atom, which will have to have a negative charge. An example of this is the cyclopentadienide ion which can be made by treating cyclopentadiene with a moderately strong base. Cyclopentadienide ion is sufficiently stabilized by its aromatic character that cyclopentadiene (its conjugate acid) has a pKa of 16, close to that of water. Cyclopentadiene is a remarkably strong acid for a hydrocarbon because its conjugate base has the extra stability of an aromatic compound.
Extraordinarily stable cations can also be made if their structures are aromatic. Here are two:
Notice that here the formally positively charged carbon atoms are sp2 hybridized and have an empty p orbital which completes the cyclic arrangement of p orbitals.
Electrophilic Aromatic Substitution -- Mechanism
Let's finish up today by looking at the general mechanism for the characteristic reactions of aromatic compounds -- electrophilic aromatic substitution. The most important characteristics of these reactions follow directly from the stability of the aromatic ring. First, these reactions are typically catalyzed by strong electrophilic (Lewis acidic) catalysts like H2SO4, AlCl3, and FeCl3 which are required to overcome the stability of the aromatic ring. Second, these are substitution reactions since addition reactions would interrupt the p orbital ring and destroy the aromatic stability.
Even though the outcome of the attack of electrophiles on benzene is substitution rather than addition, the first step is the same as in electrophilic addition to alkenes -- attack of the electrophile on a pi bond and the formation of a new sigma bond between a carbon atom and the electrophile. The carbocation which is formed undergoes loss of the H+ from the carbon which was attacked. The electrons from the C-H bond are returned to the aromatic pi electron ring and aromatic stability is restored.
Notice that the intermediate here is a carbocation, but it is not aromatic. The carbon bearing the hydrogen and the electrophile is sp3 hybridized and has no p orbital to contribute to a cyclic p orbital system. The carbocation intermediate is somewhat resonance stabilized, though, by a resonance arrangement which is very similar to the one we saw in the addition of electrophiles to conjugated dienes.
This intermediate is a carbocation, and as we will see next time, its stability is important in determining how fast the reaction goes and (in benzene rings which bear substituents at one of the carbons) where the electrophile attacks. The key thing to recognize now is that the positive charge and the corresponding carbocation characteristics only appear at positions ortho and para relative to the point at which the electrophile attack. (Nomenclature is treated in Sec 6.3 of Atkins & Carey.) This will turn out to be quite important, so verify this for yourself. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.30%3A_Aromatic_Compounds.txt |
Electrophiles and Products
Let's begin by recalling the key steps in an electrophilic aromatic substitution mechanism.
An important feature of this mechanism is that we can identify the electrophile if we know the product because it is the atom or group which replaces the H+. Conversely, if we know the electrophile, we can predict the structure of the product. The following table outlines these relationships in more detail for several reactions which follow the electrophilic aromatic substitution pathway.
The connection between the electrophile and the product is clear from these examples. What is not clear is how the catalyst transforms the reagent into the electrophile. There are two patterns here. One applies when the electrophile is made by removing a halide ion from the reagent (halogenation and the two Friedel-Crafts reactions). The formation of a carbocation from an alkyl halide and aluminum trichloride is a typical example of the first process:
Notice that the role of the catalyst is to bond with the leaving group and make it into a better leaving group. This makes the formation of the electrophile (carbocation in this case) much easier so that there is more electrophile around to attack the benzene ring. Similar mechanisms are used in the halogenation and Friedel-Crafts acylation reactions.
For sulfonation and nitration, the catalyst is sulfuric acid. Just as was the case in the SN1 and SN2 reactions of alcohols, the function of the acid is to protonate an unshared electron pair on oxygen. When the OH group of nitric acid is involved, the OH2+group which is formed is a good leaving group. Its departure makes the active electrophile:
To summarize, the function of the catalyst is to convert the reagent into a strong electrophile, which then attacks the pi electrons on the aromatic ring to make a new covalent bond. The reaction is completed by the loss of an H+.
Ortho-Para Directive Effects
The reagent - catalyst - electrophile - product pattern works well when the aromatic compound is benzene itself, but things become more complicated when a substituted benzene (a molecule in which one of the hydrogens of benzene has been replaced by another atom or group) is used. There are three products which can arise in such a system. As an example let's look at toluene -- which is methyl benzene. We'll use nitration as our example of electrophilic aromatic substitution here.
Although there are five hydrogens available to be replaced on the benzene ring of toluene , two of those are directly adjacent (ortho) to the methyl group. Attack at either of these positions gives the same product. Similar considerations apply to the meta position. The result is that only three products are obtained. The actual distribution of these products shows that very little of the meta product is obtained. Instead almost all of the product consists of the ortho and para isomers. For this reason the methyl group is called an "ortho-para directing group." It "directs" the incoming electrophile to attack at the ortho and para positions relative to itself.
How does the methyl group exert this "directive" effect? We can begin to make sense of this by remembering that we explained Markovnikov's rule by considering that the product which is most abundant is formed fastest through a pathway which has the lowest activation energy. That lead us to a consideration of the relative energies of competing transition states and the use of intermediates as useful "stand-ins" for those transition states. We concluded that Markovnikov's rule is explained by the idea that the more abundant products are formed by pathways which go through lower energy intermediates.
If we apply the same reasoning to these directive effects, the question becomes "What makes the intermediates for the formation of ortho and para products more stable than those for para products? As usual, we will look for the answers by comparing the structures of the intermediates.
Each of the intermediates is described by resonance hybrid which includes three carbocations. However, for the intermediates leading to the ortho and para products, one of the contributing carbocationic structures is tertiary. This structure is more stable than the others because the electrons on the methyl group can directly stabilize the electron deficient carbocationic carbon. This stability is passed on to the resonance hybrid, which makes the intermediates for attack at the more stable than that for attack at the meta position. More stable intermediates mean lower energy transition states and faster reactions. A faster reaction means more product is formed through that pathway.
The result of this analysis is that groups which can donate electrons stabilize carbocations. Electrophilic attack will be faster at positions such that the carbocations produced have positive charges on carbons which are bonded to electron donating groups like methy groups. This occurs when attack happens at positions ortho and para to the electron donating group. Such electron donation happens with alkyl groups in general and when atoms like oxygen and amino-nitrogen are directly attacked to the ring. In the latter cases it is the unshared pair on the oxygen or nitrogen which helps reduce the energy of the electron deficient carbon atom. (See the figure at the bottom of p 165 in Atkins & Carey for an example.) Examine the ortho-para directing groups in Table 6.3 (p 162 of Atkins & Carey) and decide how each one acts as an electron donating group.
Meta Directive Effects
Can we extend this analysis to groups which are meta directing? Let's try an example. In the nitration of benzaldehyde the product mixture is 19% ortho, 72% meta, and 9% para. The aldehyde carbonyl group (and carbonyl groups in general) is a metadirecting substituent. We'll look at the resonance structures of the intermediate for an explanation:
We see that attack at positions ortho and para to the carbonyl group places the positive charge directly adjacent to the positive end of the carbonyl group's dipole. The energy of such an intermediate will be higher than one in which the positive charge never approaches the carbonyl carbon, which is the case in the intermediate for meta attack. The consequence is that attack at the meta position is faster and more meta product is formed. Groups in which the atom directly attached to the benzene ring have a partial or complete positive charge tend to pull electrons toward themselves. Such groups are called electron withdrawing groups and are meta directing groups in electrophilic aromatic substitution. Look at the meta directing groups in Table 6.3 (p 162 in Atkins & Carey) and see if the preceding statement is borne out.
Activation and Deactivation
The underlying idea in the forgoing analysis of directive effects in electrophilic was understanding the relative rates of reaction when the competition was between different sites in the same molecule. This is termed intramolecular competition and it was also used in our understanding of Markovnikov's rule. We can also look at competition between molecules -- intermolecular competition -- using the same ideas, comparing the energies of intermediates in competing pathways. For example, let's imagine a reaction in which we mix toluene and benzene and then do a nitration. If we get more nitrotoluene (all three isomers) than nitrobenzene, the toluene has reacted faster than the benzene. We say that the methyl group "activated" the benzene ring. We understand this to mean that the methyl group lowered the energy of intermediates and transition states along the pathway to product as compared to the effect of a hydrogen (in benzene).
A comparision of the intermediates along the pathways from benzene and toluene to their respective products shows us the reason. (Only the para pathway is shown; the ortho pathway is similar.) The intermediate for attack para to the methyl group includes one resonance structure which is tertiary and thus is stabilized by the electron donating character of the methyl group. There is no such structure in the intermediate for nitration of benzene, so the intermediate for toluene nitration is more stable and the reaction which goes through it is faster. We call the methy group (and alkyl groups in general) an "activating" group for electrophilic aromatic substitution.
Since the electron donating characteristics of the methyl group are responsible for both its activating properties and its ortho-para directing properties, we might suspect that ortho-para directing groups will also be activating groups. An examination of Table 6.3 (p 162, Atkins & Carey) shows that this is generally true. (The exceptions, the halogens, are treated in Section 6.10 of Atkins & Carey. We will not discuss them further)
This conclusion also suggests that meta directing groups will also be "deactivating" groups; that is, that benzene rings which carry a meta directing group will be less reactive than those which do not. Table 6.3 also bears this out. Both the meta directing characteristics and the deactivating characteristics are outcomes of the electron withdrawing nature of these groups and the consequent increase in the energy of electrophilic aromatic substitution intermediates when they are present. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.31%3A_Electrophilic_Substitution.txt |
Carbocations Stabilized by Resonance
Let's begin by looking at the addition of HCl to an alkene in which one of the alkene carbons is directly attached to the benzene ring.
In this example each of the alkene carbons has the same number of hydrogens so the direct application of Markovnikov's rule is stymied. We find that we must go beyond Markovnikov's rule to its explanantion in terms of the relative stability of the two carbocations which might be formed. Here are the two carbocations which would be formed by the attack of H+ on the alkene:
Which of these carbocations has the lower energy? Will the phenyl group donate electrons more effectively than the methyl group? The key thing to notice in answering these questions is that the pi electrons of the benzene ring are directly adjacent to the carbocation carbon in the structure on the left. This is reminiscent of the situation we saw in the addition of HCl to 1,3-butadiene where we encountered resonance stabilization of the carbocation by the pi electrons of the neighboring double bond. The same situation occurs here, and we can symbolize the resonance stabilization with the following resonance structures.
The effect of this resonance is to make the carbocation more stable when the charge and the electron deficiency are located on a carbon which is directly bonded to the phenyl group. Such a carbon is called a "benzylic" carbon. Since there is no such resonance stabilization available when a simple alkyl group is bonded to the carbocationic carbon, the benzylic carbocation is more stable, is formed faster, and the product formed from it is the only one we see.
With this background, we can quickly decide which of the two alkyl halides below will react faster by way of an SN1 mechanism.
The more reactive alkyl halide is the one which give the more stable carbocation. The more stable carbocation (since both will be secondary carbocations) is the one where benzylic resonance is possible. (The resonance structures are given above.) Only the alkyl halide on the right will make a carbocation in which the carbocation carbon is directly attached to a phenyl group. Consequently we understand why the alkyl halide on the right is more reactive.
Bases Stabilized by Resonance
We've seen that benzylic carbocations are stabilized by resonance. What happens if the atom directly attached to the benzene ring has an unshared pair of electrons -- in other words, it is electron rich rather than electron poor. We can see the effects of this arrangement in the acid-base properties of phenols and aromatic amines.
A phenol is an alcohol whose R group is a phenyl group. It is essential that the oxygen and the benzene ring be directly attached to each other. If there is a tetrahedral carbon between them, the compound is not a phenol but an alcohol.
An important difference between phenols and alcohols lies in their acidities. Recall that alcohols, like water, have pKa's of about 16. Phenols in contrast typically have pKa's of about 10, making them considerably stronger acids than alcohols. Since the stronger acid has the weaker conjugate base, we need to ask why a phenoxide (the conjugate base of a phenol) is weaker than an alkoxide (the conjugate base of an alcohol).
Since it is the unshared electron pair which makes either of the bases a base, we can ask why the unshared electron pair on the phenoxide ion is less available to bond with a proton than the unshared electron pair on the alkoxide ion. Again, the reason is found in resonance.
The resonance structures which improve the description of the phenoxide ion show that the electron pair on oxygen is partly moved into a pi bond position between the oxygen and the phenyl carbon it is attached to. To that extent the electron pair on a phenoxide ion is less available for attachment to a proton and the phenoxide ion is a weaker base than an alkoxide ion in which no such resonance is possible.
The same situation occurs in when an amino nitrogen is directly connected to an aromatic ring.
We see that the ammonium ion in which the nitrogen is directly attached to the phenyl group is a stronger acid than the one in which the nitrogen is attached to an sp3 hybridized carbon atom. This means that the conjugate base, the aryl amine, in which the nitrogen is attached to the phenyl group is also weaker than its alkyl amine counterpart. This is understood by resonance between the unshared electron pair on nitrogen and the aromatic ring, only possible if the connection is direct, which makes the unshared electron pair less available to serve as a base.
The resonance possibilities for an aromatic group allow it to serve either as an electron donating substituent when the attached atom needs electrons (carbocation) or as an electron withdrawing substituent when the attached atom has an unshared pair to share (oxygen or nitrogen).
Side Chain Oxidation
There is another reaction which occurs at the atom directly attached to an aromatic ring. This is known as "side chain oxidation." When a compound which has an alkyl group directly attached to an aryl group is treated with a strong oxidizing agent like chromic acid, the benzylic carbon is oxidized to a carboxylic acid group which remains attached to the aryl group. Any other carbon-carbon bonds are broken. There is one further restriction -- the benzylic carbon must have a hydrogen attached.
The mechanism of this reaction is obscure, but the fact that it specficially requires that there be a benzylic C-H bond suggests that breaking this bond is essential. Any intermediate that might be formed by breaking this bond will be stabilized by resonance with the aryl group, which provides an explanation for the specificity of attack at the benzylic position. Such reactions also occur in a biological context. Enzymes oxidize alkyl side chains on aromatic rings as part of making such compounds soluble enough to be eliminated.
Diazonium Ions
When primary aryl amines are reacted with nitrous acid (HNO2, generated from sodium nitrite (NaNO2 and HCl) a reaction occurs which makes a new nitrogen-nitrogen triple bond. We will not be concerned with the mechanism, but the product (called a diazonium ion) is a valuable synthetic intermediate.
The diazonium ion is seldom isolated (diazonium ions can be explosive) so instead it is used in the solution it is made in. The N2 group can be replaced by a variety of other groups and can also be an electrophile itself toward a sufficiently activated aromatic ring. The possibilities are outlined in the following chart:
The first four reactions on this chart are examples of the replacement of the N2 group by a nucleophile. These reactions provide a way to do nucleophilic substitutions on a carbon of an aromatic ring which are not possible by normal SN1 and SN2 mechanisms. The great stability of molecular nitrogen (N2) makes it a very good leaving group, but the mechanisms of these reactions are not simple.
The last reaction in the table is one in which the diazonium group acts as an electrophile. It is not a particularly good electrophile, so it requires a reactive ring as a target. Such a ring needs to have a strong electron donating group like OH or NH2 to increase its reactivity in an electrophilic aromatic substitution mechanism.
The product of the electrophilic coupling of a diazonium ion with a reactive aromatic ring includes a nitrogen-nitrogen double bond. These compounds are often intensely colored and are useful as dyes. A variety of other functional groups, often carrying ionic charges, may be included to make them stick tighter to cloth. Terms such as aniline dyes or azo dyes are often used to refer to these compounds.
Diazonium salts are versatile synthetic intermediate materials and we know that they are made from aniline or similar molecules. Where do aniline and its analogs come from? The most common route is by reduction of nitrobenzene or substituted nitrobenzenes. The usual reducing agent is tin (in the laboratory) or iron (in industrial practice) used in aqueous HCl solution. Since nitration is a commonly used route to nitrobenzene, the usual sequence for the production of aniline is:
The use of NaOH in the final step is to neutralize the HCl so that the basic amine (which would otherwise be in the form of its conjugate acid ammonium ion) is non-polar enough to be extracted from the water. | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.32%3A_Side_Chain_Oxidations_Phenols_Arylamines.txt |
Homolytic Bond Cleavage -- Fishhook Symbols
Last time we looked at how the benzene ring changes the reactivity of an atom or group to which it is directly attached. Today we'll finish presenting new material in the course by taking a brief look at reactions in which bond making and bond breaking events involve electrons moving singly rather than as pairs. To put this in perspective, lets recall how we envisioned the breaking of a covalent bond in the reactions we've encountered so far. The dissociation of HCl into H+ and Cl- is an example.
We interpreted this to mean that the shared electron pair which formed the covalent bond between the hydrogen and the chlorine atoms moved together to become an unshared electron pair on chlorine. This is often called a "heterolytic" bond cleavage since in the products the electron pair is distributed quite unevenly. We have used the curved arrow symbol to show the origin and the destination of electron pairs in these steps. The reaction mechanisms we have worked with so far have all involved making and breaking bonds by processes like this in which pairs of electrons move together.
Now let's look at a bond cleavage in which each partner in the bond takes one electron of the bonding electron pair. The dissociation of molecular chlorine (Cl2) is a good example. This occurs when Cl2 is heated strongly or when it is illuminated by bright light.
This is called "homolytic" bond cleavage since in the products the distribution of the electron pair is quite even. We use the "fishhook" (curved arrow with only one "barb" on the arrowhead) to show the motion of one electron. Such homolytic bond cleavages occur when the bond involved is not polar and there is no electrophile or nucleophile at hand to promote heterolytic patterns.
Bond Strength and Radical Energy
When a bond is made, the product has a lower energy than the reactants. It follows that breaking bonds requires energy. When the bond breaking process is homolytic there is no residual ionic attraction because there are no charges on the products, so the energy required to dissociate the bond is a good measure of how strong that bond is. Table 9.1 on page 237 of Atkins & Carey lists such "Bond Dissociation Energies."
A brief glance at this table shows us some interesting trends. First, all the halogen-halogen bonds (the Cl-Cl bond for example) are relatively weak. Carbon-carbon and carbon hydrogen bonds are stronger as are carbon-oxygen and hydrogen-oxygen bonds. Second, if we examine the trend in C-H bond dissociation energies as the structure of the alkyl group is changed, we notice that the strongest (435 kJ/mol) C-H bond is between the carbon of a methyl group and a hydrogen atom. The weakest (343 kJ/mol) is between the central carbon of a tertiary butyl group and a hydrogen atom.
We can interpret this to mean that the tertiary butyl free radical is more stable than the methyl free radical. (The term "free radical" or "radical" is used to mean an atom or group in which one of the bonding orbitals is occupied by a single electron. Radicals are normally uncharged.) That would explain why it is easier to break the tertiary C-H bond than the methyl C-H bond; the products from breaking the tertiary C-H bond are more stable.
Looking at the structures which would be formed by breaking the C-H bonds which lie between the methyl-H bond and the tert-butyl-H bond we can see that we have a familiar pattern. Just like carbocations, the order of stability of carbon free radicals is:
tertiary > secondary > primary > methyl
We have seen that intermediates which are more stable are also those which are formed most rapidly and that this fact often dominates the composition of product mixtures. Recall that this idea was the basis for our explanation of Markovnikov's rule and ofdirective and activation/deactivation effects in electrophilic aromatic substitution. The more stable intermediate (carbocation in those cases) was formed faster. More reaction occured through the faster pathway so the product mixture was dominated by the product formed through the more stable intermediate. In the next section we will see how this idea can be used to understand some free radical reactions.
Radical Addition to Alkenes
We have made much of Markovnikov's rule. When we apply this rule to the addition of HBr to propene, we confidently predict that the product will be 2-bromopropane. It happens that occasionally considerable amounts of 1-bromopropane are formed. In fact it can happen that 1-bromopropane is the only product under some circumstances. This is puzzling and it tells us that something unusual is going on.
It took many experiments to pin down the details which control whether the reaction follows the Markovnikov pattern or the unusual course. It was eventually determined that the Markovnikov pattern is followed if the reagents are carefully purified just before they are used. The unusual pattern is followed if the reagents are "aged" before use. It was later learned that "aging" produces peroxides, compounds in which two alkyl groups or similar groups are joined through two oxygen atoms which share an O-O bond.
The O-O bond is weak (bond dissociation energy = 154 kJ/mol, much like the I-I bond) and it is non-polar so that we are not surprised that it breaks homolytically to give two oxygen free radicals.
Next one of these oxygen free radicals reacts with HBr to give an alcohol (strong O-H bond) and a bromine atom radical.
These steps (called chain initiation steps for reasons we'll come to later) set the stage for the reaction with propene. The bromine atom radical can either attack propene's primary carbon to give a secondary free radical or propene's secondary carbon to give a primary free radical. We know that the secondary free radical is the lower energy intermediate and we know that reactions proceeding through lower energy intermediates are faster than those going through higher energy intermediates. We predict that attack of the bromine on the primary carbon to give the secondary free radical will be faster and the major product will be formed through this pathway.
The reaction is completed when the secondary free radical reacts with another HBr molecule to produce the product -- 1-bromopropane -- and a new bromine atom.
Notice that the second step regenerates a bromine atom which can begin a new cycle by attacking a fresh molecule of propene. The two steps together are called a chain reaction mechanism because a product of each one is a reactant of the other, creating a chain of events which stops when the reagents are used up. The sum of these two steps is an anti-Markovnikov addition of HBr to propene. While the outcome is different from the Markovnikov pattern we studied earlier, the underlying principle is the same: We see the products of faster reactions, reactions which proceed by way of more stable intermediates. In this instance the intermediates are free radicals and the secondary free radical is formed faster than the primary free radical. In the reactions which proceeded by way of carbocations (Markovnikov's rule) the more stable carbocation was formed.
Radical Substitution on Alkanes
There are also free radical mechanisms for substitution reactions of alkanes. These are of use for synthesis only in rather restricted cases, but the products we see are also controlled by competition in which the more stable intermediates are formed faster. An example is the chlorination of butane.
We see that a secondary hydrogen has been replaced by a chlorine more often than a primary hydrogen, even though there are six primary hydrogens and only four secondary hydrogens. This would make sense if the reaction mechanism went through a free radical, since we would expect a more stable secondary free radical to be formed faster (more often) than a less stable primary free radical. Such a mechanism would be:
Again the more abundant product is produced through the more stable (secondary) free radical rather than through the less stable (primary) free radical. Notice that this reaction is also a chain reaction, since the second step produces a chlorine atom which can react with another butane molecule to continue the chain. The reaction is initiated in this case by homolytic cleavage of the weak Cl-Cl bond by light to produce chlorine atoms.
Radical Polymerization
Finally, let's return to another application of free radical additions to alkenes. We saw earlier that the HBr adds to alkenes through a peroxide initiated free radical chain mechanism to produce an anti-Markovnikov product. The key step in this reaction was the addition of a bromine atom radical to the alkene to give the more stable of the two possible free radicals. Suppose that there was no HBr present. Then a carbon free radical, once formed, would have to react with an alkene to make a new carbon-carbon bond. Here's an example where ethylene is the alkene.
The new molecule would also be a free radical and could add to another molecule of ethylene. The product of that reaction would also be a free radical and could add to another molecule of ethylene, etc. The outcome is that the each reaction extends the growing chain by two carbons and produces a free radical intermediate at the end of the chain which can continue the reaction. In this way very long molecules are produced. These polymers such as polyethylene and polystyrene are produced in billion pound per year amounts and are very important commercial products.
Other polymerization reactions also are important, and Table 9.2 on p 251 of Atkins & Carey lists many of them. We can figure out the structure of the alkene from which the polymer was made by looking for the "repeating unit" in the polymer and thinking backwards to the alkene which was "added to" in making the polymer. Here's how it works for poly(vinyl chloride):
Since the repeating unit is formed by adding to each end of the double bond of the monomer, the structure of the monomer can be discovered by mentally cutting loose a repeating unit from the polymer and placing a double bond between its carbon atoms.
Contributors
• Kirk McMichael (Washington State University) | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_-_A_Carbonyl_Early_Approach_(McMichael)/1.33%3A_Radical_Reactions.txt |
In this chapter, you will be introduced to some of the most fundamental principles of organic chemistry. With the concepts we learn about, we can begin to understand how carbon and a very small number of other elements in the periodic table can combine in predictable ways to produce a virtually limitless chemical repertoire. As you read through, you will recognize that the chapter contains a lot of review of topics you have probably learned already in an introductory chemistry course, but there will likely also be a few concepts that are new to you, as well as some topics which are already familiar to you but covered at a greater depth and with more of an emphasis on biologically relevant organic compounds.
• 1.1: Prelude to Atoms, Electron Configurations, and Lewis Structures
We will begin with a reminder of how chemists depict bonding in organic molecules with the 'Lewis structure' drawing convention, focusing on the concept of 'formal charge'. We will review the common bonding patterns of the six elements necessary for all forms of life on earth - carbon, hydrogen, nitrogen, oxygen, sulfur, and phosphorus - plus the halogens (fluorine, chlorine, bromine, and iodine).
• 1.2: Drawing Organic Structures
Now that you have had a chance to go back to your introductory chemistry textbook to review some basic information about atoms, orbitals, bonds, and molecules, let's direct our attention a little more closely to the idea of charged species. You know that an ion is a molecule or atom that has an associated positive or negative charge.
• 1.3: Functional groups and organic nomenclature
Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms. The structure of capsaicin, the compound discussed in the beginning of this chapter, incorporates several functional groups, labeled in the figure below and explained throughout this section.
• 1.4: Structures of some important biomolecules
Because we are focusing in this textbook on biologically relevant organic chemistry, we will frequently be alluding to important classes of biological molecules such as lipids, carbohydrates, proteins, and nucleic acids (DNA and RNA). Now is a good time to go through a quick overview of what these molecules look like.
• 1.5: Solutions to Chapter 1 Exercises
• 1.E: Problems for Chapter 1
01: Introduction to Organic Structure and Bonding I
It's a hot August evening at a park in the middle of North Hudson, Wisconsin, a village of just under 4000 people on the St. Croix river in the western edge of the state. A line of people are seated at tables set up inside a canvas tent. In front of a cheering crowd of friends, family, and neighbors, these brave souls are about to do battle . . .with a fruit plate.
Unfortunately for the contestants, the fruit in question is the habanero, one of the hotter varieties of chili pepper commonly found in markets in North America. In this particular event, teams of five people will race to be the first to eat a full pound of peppers. As the eating begins, all seems well at first. Within thirty seconds, though, what begins to happen is completely predictable and understandable to anyone who has ever mistakenly poured a little to much hot sauce on the dinner plate. Faces turn red, sweat and tears begin to flow, and a copious amount of cold water is gulped down.
Although technically the contestants are competing against each other, the real opponent in this contest - the cause of all the pain and suffering - is the chemical compound 'capsaicin', the source of the heat in hot chili peppers.
Composed of the four elements carbon, hydrogen, oxygen and nitrogen, capsaicin is produced by the pepper plant for the purpose of warding off hungry mammals. The molecule binds to and activates a mammalian receptor protein called TrpV1, which in normal circumstances has the job of detecting high temperatures and sending a signal to the brain - 'it's hot, stay away!' This strategy works quite well on all mammalian species except one: we humans (some of us, at least) appear to be alone in our tendency to actually seek out the burn of the hot pepper in our food.
Interestingly, birds also have a heat receptor protein which is very similar to the TrpV1 receptor in mammals, but birds are not at all sensitive to capsaicin. There is an evolutionary logic to this: it is to the pepper's advantage to be eaten by a bird rather than a mammal, because a bird can spread the pepper seeds over a much wider area. The region of the receptor which is responsible for capsaicin sensitivity appears to be quite specific - in 2002, scientists were able to insert a small segment of the (capsaicin-sensitive) rat TrpV1 receptor gene into the non-sensitive chicken version of the gene, and the resulting chimeric (mixed species) receptor was sensitive to capsaicin (Cell 2002, 108, 421).
Back at the North Hudson Pepperfest, those with a little more common sense are foregoing the painful effects of capsaicin overload and are instead indulging in more pleasant chemical phenomena. A little girl enjoying an ice cream cone is responding in part to the chemical action of another organic compound called vanillin.
What is it about capsaicin and vanillin that causes these two compounds to have such dramatically different effects on our sensory perceptions? Both are produced by plants, and both are composed of the elements carbon, hydrogen, oxygen, and (in the case of capsaicin) nitrogen. Since the birth of chemistry as a science, chemists have been fascinated - and for much of that history, mystified - by the myriad properties of compounds that come from living things. The term 'organic', from the Greek organikos, was applied to these compounds, and it was thought that they contained some kind of 'vital force' which set them apart from 'inorganic' compounds such as minerals, salts, and metals, and which allowed them to operate by a completely different set of chemical principles. How else but through the action of a 'vital force' could such a small subgroup of the elements combine to form compounds with so many different properties?
Today, as you are probably already aware, the term 'organic,' - when applied to chemistry - refers not just to molecules from living things, but to all compounds containing the element carbon, regardless of origin. Beginning early in the 19th century, as chemists learned through careful experimentation about the composition and properties of 'organic' compounds such as fatty acids, acetic acid and urea, and even figured out how to synthesize some of them starting with exclusively 'inorganic' components, they began to realize that the 'vital force' concept was not valid, and that the properties of both organic and inorganic molecules could in fact be understood using the same fundamental chemical principles.
They also began to more fully appreciate the unique features of the element carbon which makes it so central to the chemistry of living things, to the extent that it warrants its own subfield of chemistry. Carbon forms four stable bonds, either to other carbon atoms or to hydrogen, oxygen, nitrogen, sulfur, phosphorus, or a halogen. The characteristic bonding modes of carbon allow it to serve as a skeleton, or framework, for building large, complex molecules that incorporate chains, branches and ring structures.
Although 'organic chemistry' no longer means exclusively the study of compounds from living things, it is nonetheless the desire to understand and influence the chemistry of life that drives much of the work of organic chemists, whether the goal is to learn something fundamentally new about the reactivity of a carbon-oxygen bond, to discover a new laboratory method that could be used to synthesize a life-saving drug, or to better understand the intricate chemical dance that goes on in the active site of an enzyme or receptor protein. Although humans have been eating hot peppers and vanilla-flavored foods for centuries, we are just now, in the past few decades, beginning to understand how and why one causes searing pain, and the other pure gustatory pleasure. We understand that the precise geometric arrangement of the four elements in capsaicin allows it to fit inside the binding pocket of the TrpV1 heat receptor - but, as of today, we do not yet have a detailed three dimensional picture of the TrpVI protein bound to capsaicin. We also know that the different arrangement of carbon, hydrogen and oxygen atoms in vanillin allows it to bind to specific olfactory receptors, but again, there is much yet to be discovered about exactly how this happens.
In this chapter, you will be introduced to some of the most fundamental principles of organic chemistry. With the concepts we learn about, we can begin to understand how carbon and a very small number of other elements in the periodic table can combine in predictable ways to produce a virtually limitless chemical repertoire.
As you read through, you will recognize that the chapter contains a lot of review of topics you have probably learned already in an introductory chemistry course, but there will likely also be a few concepts that are new to you, as well as some topics which are already familiar to you but covered at a greater depth and with more of an emphasis on biologically relevant organic compounds.
We will begin with a reminder of how chemists depict bonding in organic molecules with the 'Lewis structure' drawing convention, focusing on the concept of 'formal charge'. We will review the common bonding patterns of the six elements necessary for all forms of life on earth - carbon, hydrogen, nitrogen, oxygen, sulfur, and phosphorus - plus the halogens (fluorine, chlorine, bromine, and iodine). We'll then continue on with some of the basic skills involved in drawing and talking about organic molecules: understanding the 'line structure' drawing convention and other useful ways to abbreviate and simplify structural drawings, learning about functional groups and isomers, and looking at how to systematically name simple organic molecules. Finally, we'll bring it all together with a review of the structures of the most important classes of biological molecules - lipids, carbohydrates, proteins, and nucleic acids - which we will be referring to constantly throughout the rest of the book.
Before you begin your study of organic chemistry, you may need to do some review of General Chemistry because it will be assumed that you already understand some basic chemistry concepts. A great way to review is to watch the following series of tutorials from Khan Academy:
Review tutorials
Here are some practice exercises to try before moving on:
Exercise 1.1
How many protons and neutrons do the following isotopes have?
1. 31P, the most common isotope of phosphorus
2. 32P, a radioactive isotope of phosphorus used often in the study of DNA and RNA.
3. 37Cl, one of the two common isotopes of chlorine.
4. tritium (3H), a radioactive isotope of hydrogen, used often by biochemists as a 'tracer' atom.
5. 14C, a radioactive isotope of carbon, also used as a tracer in biochemistry.
Exercise 1.2
The electron configuration of a carbon atom is 1s22s22p2, and that of a sodium cation (Na+) is 1s22s22p6. Show the electron configuration for:
1. a nitrogen atom
2. an oxygen atom
3. a fluorine atom
4. a magnesium atom
5. a magnesium cation (Mg2+)
6. a potassium atom
7. a potassium ion (K+)
8. a chloride anion (Cl-)
9. a sulfur atom
10. a lithium cation (Li+)
11. a calcium cation (Ca2+)
Exercise 1.3
Draw Lewis structures for the following species (use lines to denote bonds, dots for lone-pair electrons). All atoms should have a complete valence shell of electrons.
1. ammonia, NH3
2. ammonium ion, NH4+
3. amide ion, NH2-
4. formaldehyde, HCOH
5. acetate ion, CH3COO-
6. methyl amine, CH3NH2
7. ethanol, CH3CH2OH
8. diethylether, CH3CH2OCH2CH3
9. cyclohexanol (molecular formula C6H12O, with six carbons bonded in a ring and an OH group)
10. propene, CH2CHCH3
11. pyruvate, CH3COCO2H
Exercise solutions | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.01%3A_Prelude_to_Atoms_Electron_Configurations_and_Lewis_Structures.txt |
Formal Charges
Now that you have had a chance to go back to your introductory chemistry textbook to review some basic information about atoms, orbitals, bonds, and molecules, let's direct our attention a little more closely to the idea of charged species. You know that an ion is a molecule or atom that has an associated positive or negative charge. Copper, for example, can be found in both its neutral state (Cu0, which is the metal), or in its Cu+2 state, as a component of an ionic compound like copper carbonate (CuCO3), the green substance called 'patina' that forms on the surface of copper objects.
Organic molecules can also have positive or negative charges associated with them. Consider the Lewis structure of methanol, CH3OH (methanol is the so-called ‘wood alcohol’ that unscrupulous bootleggers sometimes sold during the prohibition days in the 1920's, often causing the people who drank it to go blind). Methanol itself is a neutral molecule, but can lose a proton to become a molecular anion (CH3O-), or gain a proton to become a molecular cation (CH3OH2+).
The molecular anion and cation have overall charges of -1 and +1, respectively. But we can be more specific than that - we can also state for each molecular ion that a formal charge is located specifically on the oxygen atom, rather than on the carbon or any of the hydrogen atoms.
Figuring out the formal charge on different atoms of a molecule is a straightforward process - it’s simply a matter of adding up valence electrons.
A unbound oxygen atom has 6 valence electrons. When it is bound as part of a methanol molecule, however, an oxygen atom is surrounded by 8 valence electrons: 4 nonbonding electrons (two 'lone pairs') and 2 electrons in each of its two covalent bonds (one to carbon, one to hydrogen). In the formal charge convention, we say that the oxygen 'owns' all 4 nonbonding electrons. However, it only 'owns' one electron from each of the two covalent bonds, because covalent bonds involve the sharing of electrons between atoms. Therefore, the oxygen atom in methanol owns 2 + 2 + (½ x 4) = 6 valence electrons.
The formal charge on an atom is calculated as the number of valence electrons owned by the isolated atom minus the number of valence electrons owned by the bound atom in the molecule:
Determining formal charge on an atom
formal charge =
(number of valence electrons owned by the isolated atom)
- (number of valence electrons owned by the bound atom)
or . . .
formal charge =
(number of valence electrons owned by the isolated atom)
- (number of non-bonding electrons on the bound atom)
- ( ½ the number of bonding electrons on the bound atom)
Using this formula for the oxygen atom of methanol, we have:
formal charge on oxygen =
(6 valence electrons on isolated atom)
- (4 non-bonding electrons)
- (½ x 4 bonding electrons)
= 6 - 4 - 2 = 0. Thus, oxygen in methanol has a formal charge of zero (in other words, it has no formal charge).
How about the carbon atom in methanol? An isolated carbon owns 4 valence electrons. The bound carbon in methanol owns (½ x 8) = 4 valence electrons:
formal charge on carbon =
(4 valence electron on isolated atom)
- (0 nonbonding electrons)
- (½ x 8 bonding electrons)
= 4 - 0 - 4 = 0. So the formal charge on carbon is zero.
For each of the hydrogens in methanol, we also get a formal charge of zero:
formal charge on hydrogen =
(1 valence electron on isolated atom)
- (0 nonbonding electrons)
- (½ x 2 bonding electrons)
= 1 - 0 - 1 = 0
Now, let's look at the cationic form of methanol, CH3OH2+. The bonding picture has not changed for carbon or for any of the hydrogen atoms, so we will focus on the oxygen atom.
The oxygen owns 2 non-bonding electrons and 3 bonding elections, so the formal charge calculations becomes:
formal charge on oxygen =
(6 valence electrons in isolated atom)
- (2 non-bonding electrons)
- (½ x 6 bonding electrons)
= 6 - 2 - 3 = 1. A formal charge of +1 is located on the oxygen atom.
For methoxide, the anionic form of methanol, the calculation for the oxygen atom is:
formal charge on oxygen =
(6 valence electrons in isolated atom)
- (6 non-bonding electrons)
- (½ x 2 bonding electrons)
= 6 - 6 - 1 = -1. A formal charge of -1 is located on the oxygen atom.
A very important rule to keep in mind is that the sum of the formal charges on all atoms of a molecule must equal the net charge on the whole molecule.
When drawing the structures of organic molecules, it is very important to show all non-zero formal charges, being clear about where the charges are located. A structure that is missing non-zero formal charges is not correctly drawn, and will probably be marked as such on an exam!
At this point, thinking back to what you learned in general chemistry, you are probably asking “What about dipoles? Doesn’t an oxygen atom in an O-H bond ‘own’ more of the electron density than the hydrogen, because of its greater electronegativity?” This is absolutely correct, and we will be reviewing the concept of bond dipoles later on. For the purpose of calculating formal charges, however, bond dipoles don’t matter - we always consider the two electrons in a bond to be shared equally, even if that is not an accurate reflection of chemical reality. Formal charges are just that - a formality, a method of electron book-keeping that is tied into the Lewis system for drawing the structures of organic compounds and ions. Later, we will see how the concept of formal charge can help us to visualize how organic molecules react.
Finally, don't be lured into thinking that just because the net charge on a structure is zero there are no atoms with formal charges: one atom could have a positive formal charge and another a negative formal charge, and the net charge would still be zero. Zwitterions, such as amino acids, have both positive and negative formal charges on different atoms:
Even though the net charge on glycine is zero, it is still neccessary to show the location of the positive and negative formal charges.
Exercise 1.4
Fill in all missing lone pair electrons and formal charges in the structures below. Assume that all atoms have a complete valence shell of electrons. Net charges are shown outside the brackets.
Solutions to exercises
Common bonding patterns in organic structures
The methods reviewed above for drawing Lewis structures and determining formal charges on atoms are an essential starting point for a novice organic chemist, and work quite will when dealing with small, simple structures. But as you can imagine, these methods become unreasonably tedious and time-consuming when you start dealing with larger structures. It would be unrealistic, for example, to ask you to draw the Lewis structure below (of one of the four nucleoside building blocks that make up DNA) and determine all formal charges by adding up, on an atom-by-atom basis, the valence electrons.
And yet, as organic chemists, and especially as organic chemists dealing with biological molecules, you will be expected soon to draw the structure of large molecules such as this on a regular basis. Clearly, you need to develop the ability to quickly and efficiently draw large structures and determine formal charges. Fortunately, this ability is not terribly hard to come by - all it takes is a few shortcuts and some practice at recognizing common bonding patterns.
Let’s start with carbon, the most important element for organic chemists. Carbon is said to be tetravalent, meaning that it tends to form four bonds. If you look at the simple structures of methane, methanol, ethane, ethene, and ethyne in the figures from the previous section, you should quickly recognize that in each molecule, the carbon atom has four bonds, and a formal charge of zero.
This is a pattern that holds throughout most of the organic molecules we will see, but there are also exceptions.
In carbon dioxide, the carbon atom has double bonds to oxygen on both sides (O=C=O). Later on in this chapter and throughout this book we will see examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom bears a positive or negative formal charge, respectively. If a carbon has only three bonds and an unfilled valence shell (in other words, if it does not fulfill the octet rule), it will have a positive formal charge.
If, on the other hand, it has three bonds plus a lone pair of electrons, it will have a formal charge of -1. Another possibility is a carbon with three bonds and a single, unpaired (free radical) electron: in this case, the carbon has a formal charge of zero. (One last possibility is a highly reactive species called a ‘carbene’, in which a carbon has two bonds and one lone pair of electrons, giving it a formal charge of zero. You may encounter carbenes in more advanced chemistry courses, but they will not be discussed any further in this book).
You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify by quick inspection carbons that bear positive and negative formal charges.
The pattern for hydrogens is easy: hydrogen atoms have only one bond, and no formal charge. The exceptions to this rule are the proton, H+, and the hydride ion, H-, which is a proton plus two electrons. Because we are concentrating in this book on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge.
Let us next turn to oxygen atoms. Typically, you will see an oxygen bonding in three ways, all of which fulfill the octet rule.
If an oxygen atom t has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of-1. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of +1.
When we get to our discussion of free radical chemistry in chapter 17, we will see other possibilities, such as where an oxygen atom has one bond, one lone pair, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for virtually everything we see until chapter 17.
Nitrogen has two major bonding patterns, both of which fulfill the octet rule:
If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of +1. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs.
Two third row elements are commonly found in biological organic molecules: sulfur and phosphorus. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO43-), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that atoms of elements in the third row and below in the periodic table have 'expanded valence shells' with d orbitals available for bonding, and the the octet rule does not apply.
Finally, the halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of +1.
These rules, if learned and internalized so that you don’t even need to think about them, will allow you to draw large organic structures, complete with formal charges, quite quickly.
Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear.
Exercise 1.5
Draw one structure that corresponds to each of the following molecular formulas, using the common bonding patters covered above. Be sure to include all lone pairs and formal charges where applicable, and assume that all atoms have a full valence shell of electrons. More than one correct answer is possible for each, so you will want to check your answers with your instructor or tutor.
a) C5H10O b) C5H8O c) C6H8NO+ d) C4H3O2-
Solutions to exercises
Using the 'line structure' convention
If you look ahead in this and other books at the way organic compounds are drawn, you will see that the figures are somewhat different from the Lewis structures you are used to seeing in your general chemistry book. In some sources, you will see condensed structures for smaller molecules instead of full Lewis structures:
More commonly, organic and biological chemists use an abbreviated drawing convention called line structures. The convention is quite simple and makes it easier to draw molecules, but line structures do take a little bit of getting used to. Carbon atoms are depicted not by a capital C, but by a ‘corner’ between two bonds, or a free end of a bond. Open-chain molecules are usually drawn out in a 'zig-zig' shape. Hydrogens attached to carbons are generally not shown: rather, like lone pairs, they are simply implied (unless a positive formal charge is shown, all carbons are assumed to have a full octet of valence electrons). Hydrogens bonded to nitrogen, oxygen, sulfur, or anything other than carbon are shown, but are usually drawn without showing the bond. The following examples illustrate the convention.
As you can see, the 'pared down' line structure makes it much easier to see the basic structure of the molecule and the locations where there is something other than C-C and C-H single bonds. For larger, more complex biological molecules, it becomes impractical to use full Lewis structures. Conversely, very small molecules such as ethane should be drawn with their full Lewis or condensed structures.
Sometimes, one or more carbon atoms in a line structure will be depicted with a capital C, if doing so makes an explanation easier to follow. If you label a carbon with a C, you also must draw in the hydrogens for that carbon.
Exercise 1.6
A good way to test your understanding of the line structure convention is to determine the number of hydrogen atoms in a molecule from its line structure. Do this for the structures below.
Exercise 1.7
a) Draw a line structure for the DNA base 2-deoxycytidine (the full structure was shown earlier)
b) Draw line structures for histidine (an amino acid) and pyridoxine (Vitamin B6).
Exercise 1.8
Add non-zero formal charges to the structural drawing below. The net charge is -2.
Exercise 1.9
Find, anywhere in chapters 2-17 of this textbook, one example of each of the common bonding patterns specified below. Check your answers with your instructor or tutor.
a) carbon with one double bond, two single bonds, no lone pairs, and zero formal charge
b) oxygen with two single bonds, two lone pairs, and zero formal charge
c) oxygen with one double bond, two lone pairs, and zero formal charge
d) nitrogen with one double bond, two single bonds, and a +1 formal charge
e) oxygen with one single bond, three lone pairs, and a negative formal charge
Solutions to exercises
Constitutional isomers
Now that we have reviewed how to draw Lewis structures and learned the line structure shortcut, it is a good time to learn about the concept of constitutional isomers. Imagine if you were asked to draw a structure (Lewis or line) for a compound with the molecular formula C4H10. This would not be difficult - you could simply draw:
But when you compared your answer with that of a classmate, she may have drawn this structure:
Who is correct? The answer, of course, is that both of you are. A molecular formula only tells you how many atoms of each element are present in the compound, not what the actual atom-to-atom connectivity is. There are often many different possible structures for one molecular formula. Compounds that have the same molecular formula but different connectivity are called constitutional isomers (sometimes the term ‘structural isomer’ is also used). The Greek term ‘iso’ means ‘same’.
Fructose and glucose, two kinds of sugar molecules, are constitutional isomers with the molecular formula C6H12O6.
Later, we will see other types of isomers that have the same molecular formula and the same connectivity, but are different in other respects.
Exercise 1.10
Draw a constitutional isomer of ethanol, CH3CH2OH.
Exercise 1.11
Draw all of the possible constitutional isomers with the given molecular formula.
a) C5H12
b) C4H10O
c) C3H9N
Solutions to exercises
Khan Academy video tutorials:
Representing structures of organic molecules
Drawing Lewis Dot structures and determining formal charges | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.02%3A_Drawing_organic_structures.txt |
Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms. The structure of capsaicin, the compound discussed in the beginning of this chapter, incorporates several functional groups, labeled in the figure below and explained throughout this section.
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups behave in organic reactions.
The 'default' in organic chemistry (essentially, the lack of any functional groups) is given the term alkane, characterized by single bonds between carbon and carbon, or between carbon and hydrogen. Methane, CH4, is the natural gas you may burn in your furnace. Octane, C8H18, is a component of gasoline.
Alkanes
Alkenes (sometimes called olefins) have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds. Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening. (If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas). Ethyne, commonly called acetylene, is used as a fuel in welding blow torches.
Alkenes and alkynes
In chapter 2, we will study the nature of the bonding on alkenes and alkynes, and learn that that the bonding in alkenes is trigonal planar in in alkynes is linear. Furthermore, many alkenes can take two geometric forms: cis or trans. The cis and trans forms of a given alkene are different molecules with different physical properties because, as we will learn in chapter 2, there is a very high energy barrier to rotation about a double bond. In the example below, the difference between cis and trans alkenes is readily apparent.
We will have more to say about the subject of cis and trans alkenes in chapter 3, and we will learn much more about the reactivity of alkenes in chapter 14.
Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens - in other words, they are saturated with hydrogen atoms. The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. As we will see in chapter 15, hydrogen can be added to double and triple bonds, in a type of reaction called 'hydrogenation'.
The aromatic group is exemplified by benzene (which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic), and naphthalene, a compound with a distinctive 'mothball' smell. Aromatic groups are planar (flat) ring structures, and are widespread in nature. We will learn more about the structure and reactions of aromatic groups in chapters 2 and 14.
Aromatics
When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or haloalkane. Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules.
Haloalkanes
In the alcohol functional group, a carbon is single-bonded to an OH group (the OH group, by itself, is referred to as a hydroxyl). Except for methanol, all alcohols can be classified as primary, secondary, or tertiary. In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon. In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol. The sulfur analog of an alcohol is called a thiol (from the Greek thio, for sulfur).
Alcohols, phenols, and thiols
Note that the definition of a phenol states that the hydroxyl oxygen must be directly attached to one of the carbons of the aromatic ring. The compound below, therefore, is not a phenol - it is a primary alcohol.
The distinction is important, because as we will see later, there is a significant difference in the reactivity of alcohols and phenols.
The deprotonated forms of alcohols, phenols, and thiols are called alkoxides, phenolates, and thiolates, respectively. A protonated alcohol is an oxonium ion.
In an ether functional group, a central oxygen is bonded to two carbons. Below is the structure of diethyl ether, a common laboratory solvent and also one of the first compounds to be used as an anesthetic during operations. The sulfur analog of an ether is called a thioether or sulfide.
Ethers and sulfides
Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations. In the case where a nitrogen has four bonds to carbon (which is somewhat unusual in biomolecules), it is called a quaternary ammonium ion.
Note
Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols versus amines - the definitions are different. In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.
Phosphate and its derivative functional groups are ubiquitous in biomolecules. Phosphate linked to a single organic group is called a phosphate ester; when it has two links to organic groups it is called a phosphate diester. A linkage between two phosphates creates a phosphate anhydride.
Organic phosphates
Chapter 9 of this book is devoted to the structure and reactivity of the phosphate group.
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
A group with a carbon-nitrogen double bond is called an imine, or sometimes a Schiff base (in this book we will use the term 'imine'). The chemistry of aldehydes, ketones, and imines will be covered in chapter 10.
Aldehydes, ketones, and imines
When a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to an oxygen, nitrogen, or sulfur, the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a set of related functional groups. The eponymous member of this family is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl group. The conjugate base of a carboxylic acid is a carboxylate. Other derivatives are carboxylic esters (usually just called 'esters'), thioesters, amides, acyl phosphates, acid chlorides, and acid anhydrides. With the exception of acid chlorides and acid anhydrides, the carboxylic acid derivatives are very common in biological molecules and/or metabolic pathways, and their structure and reactivity will be discussed in detail in chapter 11.
Carboxylic acid derivatives
Finally, a nitrile group is characterized by a carbon triple-bonded to a nitrogen.
Nitriles
A single compound often contains several functional groups, particularly in biological organic chemistry. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
The hormone testosterone, the amino acid phenylalanine, and the glycolysis metabolite dihydroxyacetone phosphate all contain multiple functional groups, as labeled below.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological organic chemistry. Table 9 in the tables section at the back of this book provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
Exercise 1.12
Identify the functional groups (other than alkanes) in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Solutions to exercises
Exercise 1.13
Draw one example each of compounds fitting the descriptions below, using line structures. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule). There are many possible correct answers for these, so be sure to check your structures with your instructor or tutor.
a) a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
b) an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
c) A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group.
Naming organic compounds
A system has been devised by the International Union of Pure and Applied Chemistry (IUPAC, usually pronounced eye-you-pack) for naming organic compounds. While the IUPAC system is convenient for naming relatively small, simple organic compounds, it is not generally used in the naming of biomolecules, which tend to be quite large and complex. It is, however, a good idea (even for biologists) to become familiar with the basic structure of the IUPAC system, and be able to draw simple structures based on IUPAC names.
Naming an organic compound usually begins with identify what is referred to as the 'parent chain', which is the longest straight chain of carbon atoms. We’ll start with the simplest straight chain alkane structures. CH4 is called methane, and C2H6 ethane. The table below continues with the names of longer straight-chain alkanes: be sure to commit these to memory, as they are the basis for the rest of the IUPAC nomenclature system (and are widely used in naming biomolecules as well).
Names for straight-chain alkanes:
1 carbon: methane
2 carbons: ethane
3 carbons: propane
4 carbons: butane
5 carbons: pentane
6 carbons: hexane
7 carbons: heptane
8 carbons: octane
9 carbons: nonane
10 carbons: decane
Substituents branching from the main parent chain are located by a carbon number, with the lowest possible numbers being used (for example, notice in the example below that the compound on the left is named 1-chlorobutane, not 4-chlorobutane). When the substituents are small alkyl groups, the terms methyl, ethyl, and propyl are used.
Other common names for hydrocarbon substituent groups are isopropyl, tert-butyl and phenyl.
Notice in the example below, an ‘ethyl group’ (in blue) is not treated as a substituent, rather it is included as part of the parent chain, and the methyl group is treated as a substituent. The IUPAC name for straight-chain hydrocarbons is always based on the longest possible parent chain, which in this case is four carbons, not three.
Cyclic alkanes are called cyclopropane, cyclobutane, cyclopentane, cyclohexane, and so on:
In the case of multiple substituents, the prefixes di, tri, and tetra are used.
Functional groups have characteristic suffixes. Alcohols, for example, have ‘ol’ appended to the parent chain name, along with a number designating the location of the hydroxyl group. Ketones are designated by ‘one’.
Alkenes are designated with an 'ene' ending, and when necessary the location and geometry of the double bond are indicated. Compounds with multiple double bonds are called dienes, trienes, etc.
Some groups can only be present on a terminal carbon, and thus a locating number is not necessary: aldehydes end in ‘al’, carboxylic acids in ‘oic acid’, and carboxylates in ‘oate’.
Ethers and sulfides are designated by naming the two groups on either side of the oxygen or sulfur.
If an amide has an unsubstituted –NH2 group, the suffix is simply ‘amide’. In the case of a substituted amide, the group attached to the amide nitrogen is named first, along with the letter ‘N’ to clarify where this group is located. Note that the structures below are both based on a three-carbon (propan) parent chain.
For esters, the suffix is 'oate'. The group attached to the oxygen is named first.
All of the examples we have seen so far have been simple in the sense that only one functional group was present on each molecule. There are of course many more rules in the IUPAC system, and as you can imagine, the IUPAC naming of larger molecules with multiple functional groups, ring structures, and substituents can get very unwieldy very quickly. The illicit drug cocaine, for example, has the IUPAC name 'methyl (1R,2R,3S,5S)-3-(benzoyloxy)-8-methyl-8-azabicyclo[3.2.1] octane-2-carboxylate' (this name includes designations for stereochemistry, which is a structural issue that we will not tackle until chapter 3).
You can see why the IUPAC system is not used very much in biological organic chemistry - the molecules are just too big and complex. A further complication is that, even outside of a biological context, many simple organic molecules are known almost universally by their ‘common’, rather than IUPAC names. The compounds acetic acid, chloroform, and acetone are only a few examples.
In biochemistry, nonsystematic names (like 'cocaine', 'capsaicin', 'pyruvate' or 'ascorbic acid') are usually used, and when systematic nomenclature is employed it is often specific to the class of molecule in question: different systems have evolved, for example, for fats and for carbohydrates. We will not focus very intensively in this text on IUPAC nomenclature or any other nomenclature system, but if you undertake a more advanced study in organic or biological chemistry you may be expected to learn one or more naming systems in some detail.
Exercise 1.14
Give IUPAC names for acetic acid, chloroform, and acetone.
Exercise 1.15
Draw line structures of the following compounds, based on what you have learned about the IUPAC nomenclature system:
1. methylcyclohexane
2. 5-methyl-1-hexanol
3. 2-methyl-2-butene
4. 5-chloropentanal
5. 2,2-dimethylcyclohexanone
6. 4-penteneoic acid
7. N-ethyl-N-cyclopentylbutanamide
Solutions to exercises
Drawing abbreviated organic structures
Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest:
The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule. For example, in chapter 15 when we look at biochemical oxidation-reduction reactions involving the flavin molecule, we will abbreviate a large part of the flavin structure which does not change at all in the reactions of interest:
As an alternative, we can use a 'break' symbol to indicate that we are looking at a small piece or section of a larger molecule. This is used commonly in the context of drawing groups on large polymers such as proteins or DNA.
Finally, 'R' groups can be used to concisely illustrate a series of related compounds, such as the family of penicillin-based antibiotics.
Using abbreviations appropriately is a very important skill to develop when studying organic chemistry in a biological context, because although many biomolecules are very large and complex (and take forever to draw!), usually we are focusing on just one small part of the molecule where a change is taking place.
As a rule, you should never abbreviate any atom involved in a bond-breaking or bond-forming event that is being illustrated: only abbreviate that part of the molecule which is not involved in the reaction of interest.
For example, carbon #2 in the reactant/product below most definitely is involved in bonding changes, and therefore should not be included in the 'R' group.
If you are unsure whether to draw out part of a structure or abbreviate it, the safest thing to do is to draw it out.
:
a) If you intend to draw out the chemical details of a reaction in which the methyl ester functional group of cocaine (see earlier figure) was converted to a carboxylate plus methanol, what would be an appropriate abbreviation to use for the cocaine structure (assuming that you only wanted to discuss the chemistry specifically occurring at the ester group)?
b) Below is the (somewhat complicated) reaction catalyzed by an enzyme known as 'Rubisco', by which plants 'fix' carbon dioxide. Carbon dioxide and the oxygen of water are colored red and blue respectively to help you see where those atoms are incorporated into the products. Propose an appropriate abbreviation for the starting compound (ribulose 1,5-bisphosphate), using two different 'R' groups, R1 and R2.
Solutions to exercises
Khan Academy video tutorial on functional groups | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.03%3A_Functional_groups_and_organic_nomenclature.txt |
Because we are focusing in this textbook on biologically relevant organic chemistry, we will frequently be alluding to important classes of biological molecules such as lipids, carbohydrates, proteins, and nucleic acids (DNA and RNA). Now is a good time to go through a quick overview of what these molecules look like. These are large, complex molecules and there is a lot of information here: you are not expected to memorize these structures or even, at this point, to fully understand everything presented in this section. For now, just read through the section and get what you can out of it, and work on recognizing the fundamental things you have just learned: common bonding patterns, formal charges, functional groups, and so forth. Later, you can come back to this section for review when these biomolecules are referred to in different contexts throughout the remainder of the book.
Polymers
Many of the biomolecules that we will be talking in this section are polymers. To understand what a polymer is, simply picture a long chain made by connecting lots of individual beads, each of which is equipped with two hooks. In chemical terminology, each bead is a monomer compound, the hooks are linking groups, and the whole chain is a polymer.
Although lipids can be described as biopolymers, we will use the monomer-linking group terminology in particular when we talk about carbohydrates, protein, and nucleic acids.
Click on the links below to learn about each of these important types of biological molecules:
• Lipids
• Carbohydrates
• Amino acids and proteins
• Nucleic acids
1.04: Structures of some important biomolecules
Proteins are polymers of amino acids, linked by amide groups known as peptide bonds. An amino acid can be thought of as having two components: a 'backbone', or 'main chain', composed of an ammonium group, an 'alpha-carbon', and a carboxylate, and a variable 'side chain' (in green below) bonded to the alpha-carbon.
There are twenty different side chains in naturally occurring amino acids, and it is the identity of the side chain that determines the identity of the amino acid: for example, if the side chain is a -CH3 group, the amino acid is alanine, and if the side chain is a -CH2OH group, the amino acid is serine. Many amino acid side chains contain a functional group (the side chain of serine, for example, contains a primary alcohol), while others, like alanine, lack a functional group, and contain only a simple alkane.
The two 'hooks' on an amino acid monomer are the amine and carboxylate groups. Proteins (polymers of ~50 amino acids or more) and peptides (shorter polymers) are formed when the amino group of one amino acid monomer reacts with the carboxylate carbon of another amino acid to form an amide linkage, which in protein terminology is a peptide bond. Which amino acids are linked, and in what order - the protein sequence - is what distinguishes one protein from another, and is coded for by an organism's DNA. Protein sequences are written in the amino terminal (N-terminal) to carboxylate terminal (C-terminal) direction, with either three-letter or single-letter abbreviations for the amino acids (see amino acid table). Below is a four amino acid peptide with the sequence "cysteine - histidine - glutamate - methionine". Using the single-letter code, the sequence is abbreviated CHEM.
When an amino acid is incorporated into a protein it loses a molecule of water and what remains is called a residue of the original amino acid. Thus we might refer to the 'glutamate residue' at position 3 of the CHEM peptide above.
Once a protein polymer is constructed, it in many cases folds up very specifically into a three-dimensional structure, which often includes one or more 'binding pockets' in which other molecules can be bound. It is this shape of this folded structure, and the precise arrangement of the functional groups within the structure (especially in the area of the binding pocket) that determines the function of the protein.
Enzymes are proteins which catalyze biochemical reactions. One or more reacting molecules - often called substrates - become bound in the active site pocket of an enzyme, where the actual reaction takes place. Receptors are proteins that bind specifically to one or more molecules - referred to as ligands - to initiate a biochemical process. For example, we saw in the introduction to this chapter that the TrpVI receptor in mammalian tissues binds capsaicin (from hot chili peppers) in its binding pocket and initiates a heat/pain signal which is sent to the brain.
Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase (in grey), with the substrate molecule bound inside the active site pocket.
(x-ray crystallographic data are from Protein Science 1999, 8, 291; pdb code 4ALD. Image produced with JMol First Glance)
Intro to nucleic acids ⇒ | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.04%3A_Structures_of_some_important_biomolecules/1.4.01%3A_Introduction_to_amin.txt |
The term 'carbohydrate', which literally means 'hydrated carbons', broadly refers to monosaccharides, disaccharides, oligosaccharides (shorter polymers) and polysaccharides (longer polymers). We will cover the chemistry of carbohydrates more completely in chapter 10, but the following is a quick overview.
Monosaccharides (commonly called 'sugars') are four- to six-carbon molecules with multiple alcohol groups and a single aldehyde or ketone group. Many monosaccharides exist in aqueous solution as a rapid equilibrium between an open chain and one or more cyclic forms. Two forms of a six-carbon monosaccharide are shown below.
Disaccharides are two monosaccharides linked together: for example, sucrose, or table sugar, is a disaccharide of glucose and fructose.
Oligosaccharides and polysaccharides are longer polymers of monosaccharides. Cellulose is a polysaccharide of repeating glucose monomers. As a major component of the cell walls of plants, cellulose is the most abundant organic molecule on the planet! A two-glucose stretch of a cellulose polymer is shown below.
The linking group in carbohydrates is not one that we have covered in this chapter - in organic chemistry this group is called an acetal, while biochemists usually use the term glycosidic bond when talking about carbohydrates (again, the chemistry of these groups in the context of carbohydrate structures will be covered in detail in chapter 10).
The possibilities for carbohydrate structures are vast, depending on which monomers are used (there are many monosaccharides in addition to glucose and fructose), which carbons are linked, and other geometric factors which we will learn about later. Multiple linking (branching) is also common, so many carbohydrates are not simply linear chains. In addition, carbohydrate chains are often attached to proteins and/or lipids, especially on the surface of cells. All in all, carbohydrates are an immensely rich and diverse subfield of biological chemistry.
Intro to amino acids and proteins
1.4.03: Introduction to lipi
Lipids are a class of biomolecules which includes fats, oils, waxes, and compounds such as cholesterol that are referred to as 'isoprenoids'.. Fats, oils, and waxes all incorporate fatty acids, which are composed of hydrocarbon chains terminating in a carboxylic acid/carboxylate group (we will learn in Chapter 7 that carboxylic acids are predominantly in their anionic, carboxylate form in biological environments). Saturated fatty acids contain only alkane carbons (single bonds only), mononsaturated fatty acids contain a single double bond, and polyunsaturated fatty acids contain two or more double bonds. The double bonds in naturally occurring fatty acids are predominantly in the cis configuration.
Fatty acids are synthesized in the body by a process in which the hydrocarbon chain is elongated two carbons at a time. Each two-carbon unit is derived from a metabolic intermediate called acetyl-coA, which is essentially an acetic acid (vinegar) molecule linked to a large 'carrier' molecule, called coenzyme A, by a thioester functional group. We will see much more of coenzyme A when we study the chemistry of thioesters in chapter 11.
The breakdown of fatty acids in the body also occurs two carbons at a time, and the endpoint is again acetyl-coenzyme A. We will learn about the details of all of the reactions in these metabolic pathways at various places in this book. If you go on to take a biochemistry course, you will learn more about the big picture of fatty acid metabolism - how it is regulated, and how is fits together with other pathways of central metabolism.
Fats and oils are forms of triacylglycerol, a molecule composed of a glycerol backbone with three fatty acids linked by ester functional groups.
Solid fats (predominant in animals) are triacylglycerols with long (16-18 carbon) saturated fatty acids. Liquid oils (predominant in plants) have unsaturated fatty acids, sometimes with shorter hydrocarbon chains. In chapter 2 we will learn about how chain length and degree of unsaturation influences the physical properties of fats and oils.
Cell membranes are composed of membrane lipids, which are diacylglycerols linked to a hydrophilic 'head group' on the third carbon of the glycerol backbone. The fatty acid chains can be of various lengths and degrees of saturation, and the two chains combined make up the hydrophobic 'tail' of each membrane lipid molecule.
In chapter 2 we will see how these molecules come together to form a cell membrane.
Exercise 1.17
What functional group links the phosphatidylcholine 'head' group to glycerol in the membrane lipid structure shown above?
Solutions to exercises
Waxes are composed of fatty acids linked to long chain alcohols through an ester group. Tricontanyl palmitate is a major component of beeswax, and is constituted of a 16-carbon fatty acid linked to a 30-carbon alcohol.
Isoprenoids, a broad class of lipids present in all forms of life, are based on a five-carbon, branched-chain building block called isoprene. In humans, cholesterol and hormones such as testosterone are examples of isoprenoid biomolecules. In plants, isoprenoids include the deeply colored compounds such as lycopene (the red in tomatoes) and carotenoids (the yellows and oranges in autumn leaves).
In almost all eukaryotes, isopentenyl diphosphate (the building block molecule for all isoprenoid compounds) is synthesized from three acetyl-Coenzyme A molecules. Bacteria and the plastid organelles in plants have a different biosynthetic pathway to isopentenyl diphosphate, starting with pyruvate and glyceraldehyde phosphate.
Intro to carbohydrates⇒
1.4.04: Introduction to nucl
Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are polymers composed of monomers called nucleotides. An RNA nucleotide consists of a five-carbon sugar phosphate linked to one of four nucleic acid bases: guanine (G), cytosine (C), adenine (A) and uracil (U).
In a DNA nucleototide, the sugar is missing the hydroxyl group at the 2' position, and the thymine base (T) is used instead of uracil. The conventional numbering system used for DNA and RNA is shown here for reference - the prime (') symbol is used to distinguish the sugar carbon numbers from the base carbon numbers.
The two 'hooks' on the RNA or DNA monomer are the 5' phosphate and the 3' hydroxyl on the sugar, which in DNA polymer synthesis are linked by a 'phosphate diester' group. By convention, DNA and RNA sequences are written in 5' to 3' direction.
Back to chapter 1 main page ⇒ | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.04%3A_Structures_of_some_important_biomolecules/1.4.02%3A_Introduction_to_carb.txt |
E1.1:
a) The atomic number of P (phosphorus) is 15, meaning there are 15 protons. The mass number for the 31P isotope is 31, so:
15 protons + 16 neutrons = mass number 31
(recall that mass number is number of protons and neutrons).
(for parts b-d, use the same reasoning as above)
b) 15 protons + 17 neutrons = mass number 32
c) 17 protons + 20 neutrons = mass number 37
d) 1 proton + 2 neutrons = mass number 3
e) 6 protons + 8 neutrons = mass number 14
E1.2:
a) 1s22s22p3
b) 1s22s22p4
c) 1s22s22p5
d) 1s22s22p63s2
e) 1s22s22p6 (same as Neon atom)
f) 1s22s22p63s23p64s1
g) 1s22s22p63s23p6 (same as Argon atom)
h) 1s22s22p63s23p6 (same as Argon atom)
i) 1s22s22p63s23p4
j) 1s2 (same as Helium atom)
k) 1s22s22p63s23p6 (same as Argon atom)
E1.3:
E1.4:
E1.6: Below are full structural drawings, showing all carbons and hydrogens:
E1.8:
E1.10: There is only one constitutional isomer of ethanol: dimethyl ether CH3OCH3
E1.11:
E1.12:
a) carboxylate, sulfide, aromatic, two amide groups (one of which is cyclic)
b) tertiary alcohol, thioester
c) carboxylate, ketone
d) ether, primary amine, alkene
E1.14:
acetic acid: ethanoic acid
chloroform: trichloromethane
acetone: propanone (not 2-propanone, because the '2' in this case would be redundant: if the carbonyl carbon were not in the #2 position, the compound would be an aldehyde not a ketone)
E1.17: The linking group is a phosphate diester
1.01: Solutions to selected Chapter 1 pr
Solutions to selected Chapter 1 problems
P1.1: The figure below illustrates a section of an intermediate compound that forms during the protein synthesis process in the cell. Lone pairs are not shown, as is typical in drawings of organic compounds.
a) The structure as drawn is incomplete, because it is missing formal charges - fill them in.
b) How many hydrogen atoms are on this structure?
c) Identify the two important biomolecule classes (covered in section 1.3) in the structure.
P1.2: Find, in Table 6 ('Structures of common coenzymes', in the tables section at the back of this book), examples of the following:
a) a thiol b) an amide c) a secondary alcohol d) an aldehyde
e) a methyl substituent on a ring f) a primary ammonium ion
g) a phosphate anhydride h) a phosphate ester
P1.3: Draw line structures corresponding to the following compounds. Show all lone pair electrons (and don't forget that non-zero formal charges are part of a correctly drawn structure!)
a) 2,2,4-trimethylpentane b) 3-phenyl-2-propenal
c) 6-methyl-2,5-cyclohexadienone d) 3-methylbutanenitrile
e) 2,6-dimethyldecane f) 2,2,5,5-tetramethyl-3-hexanol
g) methyl butanoate h) N-ethylhexanamide
i) 7-fluoroheptanoate j) 1-ethyl-3,3-dimethylcyclohexene
P1.4: Reaction A below is part of the biosynthetic pathway for the amino acid methionine, and reaction B is part of the pentose phosphate pathway of sugar metabolism.
a) What is the functional group transformation that is taking place in each reaction?
b) Keeping in mind that the 'R' abbreviation is often used to denote parts of a larger molecule which are not the focus of a particular process, which of the following abbreviated structures could be appropriate to use for aspartate semialdehyde when drawing out details of reaction A?
c) Again using the 'R' convention, suggest an appropriate abbreviation for the reactant in reaction B.
P1.5: Find, in the table of amino acid structures (Table 5), examples of the following:
a) a secondary alcohol b) an amide c) a thiol
d) a sulfide e) a phenol f) a side chain primary ammonium
g) a side chain carboxylate h) a secondary amine
P1.6: Draw correct Lewis structures for ozone (O3), azide ion, (N3-), and bicarbonate ion, HCO3-. Include lone pair electrons and formal charges, and use your General Chemistry textbook to review VSEPR theory, which will enable you to draw correct bond geometries.
P1.7: Draw one example each of compounds fitting the descriptions below, using line structures. Be sure to include all non-zero formal charges. All atoms should fit one of the common bonding patters discussed in this chapter. There are many possible correct answers - be sure to check your drawings with your instructor or tutor.
a) an 8-carbon molecule with secondary alcohol, primary amine, amide, and cis-alkene groups
b) a 12-carbon molecule with carboxylate, diphosphate, and lactone (cyclic ester) groups.
c) a 9-carbon molecule with cyclopentane, alkene, ether, and aldehyde groups
P1.8: Three of the four structures below are missing formal charges.
a) Fill in all missing formal charges (assume all atoms have a complete octet of valence electrons).
b) Identify the following functional groups or structural elements (there may be more than one of each): carboxylate, carboxylic acid, cyclopropyl, amide, ketone, secondary ammonium ion, tertiary alcohol.
c) Determine the number of hydrogen atoms in each compound.
P1.9:
a) Draw four constitutional isomers with the molecular formula C4H8. (
b) Draw two open-chain (non-cyclic) constitutional isomers of cyclohexanol (there are more than two possible answers).
P1.10: Draw structures of four different amides with molecular formula C3H7NO.
Solutions to selected Chapter 1 problems
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
1.0P: 1.P: Problems for Chapter 1
P1.1:
a) Formal charges are located as shown.
b) There are 16 hydrogen atoms:
c) The structure contains a nucleotide segment and an amino acid segment:
P1.4:
a)
Reaction A: aldehyde to primary alcohol
Reaction B: Secondary alcohol to ketone; aldehyde to primary alcohol
b) The second structure from the right is an appropriate abbreviation. The part of the molecule in the box does not change in the reaction, and this can be abbreviated with 'R'.
c) The part of the molecule in the box does not change in the reaction, and this can be abbreviated with 'R'.
P1.5:
a) Threonine contains a secondary alcohol.
b) Glutamine and asparagine contain amides.
c) Cysteine contains a thiol.
d) Methionine contains a sulfide.
e) Tyrosine contains a phenol.
f) The lysine side chain contains a primary ammonium.
g) The glutamate and aspartate side chains contain carboxylates.
h) Proline contains a secondary amine.
P1.6: Note that according to VSEPR theory, ozone has bent geometry, azide ion is linear, and the geometry around the oxygen and carbon atoms of bicarbonate is bent.
P1.8:
P1.10: | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.05%3A_Solutions_to_Chapter_1_Exercises.txt |
While organic and biological chemistry is a very diverse field of study, one fundamental question that interests all organic chemists is how the structure of an organic molecule determines its physical properties. We will look more closely at the nature of single and double covalent bonds, using the concepts of 'hybrid orbitals' and 'resonance' to attempt to explain how orbital overlap results in characteristic geometries and rotational behavior for single and double bonds, as well as bonds that have characteristics of somewhere in between single and double. Then we will move on to a review of the noncovalent interactions between molecules - Van der Waals, ion-ion, dipole-dipole and ion-dipole interactions, and hydrogen bonds - and how they are manifested in the observable physical properties of all organic substances.
• 2.1: Prelude to Organic Structure and Bonding II
To understand why sperm oil has properties that made it both a useful industrial lubricant for humans and an effective buoyancy control and/or sonic lens for a hunting sperm whale, we first have to understand the nature of both the forces holding each wax molecule together and also the forces governing the noncovalent interactions between one wax molecule and all the others around it which determine physical properties such as viscosity, melting point, and density.
• 2.2: Valence Bond Theory
Valence bond theory is most often used to describe bonding in organic molecules. In this model, bonds are considered to form from the overlap of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as molecular hydrogen (H2) or hydrogen fluoride (HF), our present understanding of s and p atomic orbitals will suffice. In order to explain the bonding in organic molecules, however, we will need to introduce hybrid orbitals
• 2.3: Molecular orbital theory- conjugation and aromaticity
Valence bond theory does a remarkably good job at explaining the bonding geometry of many of the functional groups in organic compounds. There are some areas, however, where the valence bond theory falls short. It fails to adequately account, for example, for some interesting properties of compounds that contain alternating double and single bonds. In order to understand these properties, we need to think about chemical bonding in a new way, using the ideas of molecular orbital (MO) theory.
• 2.4: Resonance
These two drawings are an example of what is referred to in organic chemistry as resonance contributors: two or more different Lewis structures depicting the same molecule or ion that, when considered together, do a better job of approximating delocalized pi-bonding than any single structure. By convention, resonance contributors are linked by a double-headed arrow, and are sometimes enclosed by brackets:
• 2.5: Non-covalent interactions
To understand the nature of noncovalent interactions, we first must return to covalent bonds and delve into the subject of dipoles. Many of the covalent bonds that we have seen – between two carbons, for example, or between a carbon and a hydrogen –involve the approximately equal sharing of electrons between the two atoms in the bond. In these examples, the two atoms have approximately the same electronegativity.
• 2.6: Physical properties of organic compounds
Virtually all of the organic chemistry that you will see in this course takes place in the solution phase. In the organic laboratory, reactions are often run in nonpolar or slightly polar solvents such as toluene (methylbenzene), dichloromethane, or diethylether.
• 2.7: Solutions to Chapter 2 exercises
• 2.P: Problems for Chapter 2
02: Introduction to Organic Structure and Bonding II
What does Moby Dick have in common with train engines and skin cream?
“Towards thee I roll, thou all-destroying but unconquering whale; to the last I grapple with thee; from hell's heart I stab at thee; for hate's sake I spit my last breath at thee. Sink all coffins and all hearses to one common pool! and since neither can be mine, let me then tow to pieces, while still chasing thee, though tied to thee, thou damned whale! THUS, I give up the spear!" The harpoon was darted; the stricken whale flew forward; with igniting velocity the line ran through the grooves;--ran foul. Ahab stooped to clear it; he did clear it; but the flying turn caught him round the neck, and voicelessly as Turkish mutes bowstring their victim, he was shot out of the boat, ere the crew knew he was gone. Next instant, the heavy eye-splice in the rope's final end flew out of the stark-empty tub, knocked down an oarsman, and smiting the sea, disappeared in its depths.
Herman Melville, Moby Dick
In the classic 19th century novel 'Moby Dick', Herman Melville's Captain Ahab obsessively hunts down the enormous albino sperm whale which years before had taken one of his legs, the monomaniacal quest ending with Ahab being dragged by the neck to the bottom of the sea by his enormous white nemesis. It is fitting, somehow, that one of the most memorable fictional characters in modern literature should be a real-life 50 ton monster – sperm whales are such fantastic creatures that if they didn't in fact exist, it would stretch the imagination to make them up. They are the largest predator on the planet, diving to depths of up to three kilometers and staying down as long as 90 minutes to hunt the giant squid and other deep-dwelling species that make up the bulk of their diet.
It would be hard for anyone to mistake a sperm whale for any other creature in the ocean, due to their enormous, squared-off foreheads. It is what is inside this distinctive physical feature, though, that brought them to the edge of extinction in the middle of the 20th century. For over 200 years, sperm whales had been prized by whalers for the oil that fills the 'spermaceti' and 'melon' compartments which make up the bulk of the front part of their bodies. Whalers in the 17th and 18th centuries would lower one of their crew into a hole cut into a captured whale, and he would literally ladle out the 'sperm oil ' by the bucketful, often filling eight barrels from the head of one animal. The different processed components obtained from raw sperm oil had properties that were ideal for a multitude of applications: as a lubricant for everything from sewing machines to train engines, as a fuel for lamps, and as a prized ingredient in cosmetics and skin products.
'Sperm oil' is not really an oil – it is mostly liquid wax. The composition of waxes in sperm oil is complex and variable throughout the life of the animal, but in general contains waxes with saturated and unsaturated hydrocarbon chains ranging from 16 to 24 carbons.
Remarkably, scientists are still not sure about the function of the enormous wax-filled reservoirs in the sperm whale's forehead. The most prevalent hypothesis holds that they play a role in echolocation. In the pitch-black void of the deep ocean a whale's eyes are useless, but it is able to navigate and locate prey in the same way that a bat does, using the reflection of sound waves. In fact, the sonic clicks generated by the sperm whale are the loudest sounds generated by any animal on earth. The wax reservoirs may be used somehow for the directional focusing of these sound waves.
Another intriguing but unproven hypothesis is that the reservoir serves as a buoyancy control device. The wax is normally liquid and buoyant at the whale's normal body temperature, but solidifies and becomes denser than water at lower temperatures. If the diving whale could cool the wax by directing cold seawater around the reservoir and restricting blood flow to the region, it could achieve negative buoyancy and thus conserve energy that otherwise would be expended in swimming down. When it needs to return to the surface, blood could be redirected to the wax, which would melt and become positively buoyant again, thus conserving energy on the upward trip.
Whatever its natural function, it is inarguable that the physical and chemical properties of sperm oil make it valuable, both to the whale and to humans. Fortunately for the world's population of whales, both economic forces and conservation efforts have made virtually all trade in sperm oil a thing of the past. Beginning in the late 19th century, the discovery of new oil fields and advances in petroleum processing led to the use of cheaper mineral oil alternatives for many of the major applications of sperm oil, one of the most notable substitutions being the use of kerosene for lamps. More recently, the 'oil' from the Jojoba plant, a native of the American southwest, has been found to be an excellent substitute for sperm oil in cosmetics and skin products, exhibiting many of the same desirable characteristics. Jojoba oil, like sperm oil, is composed primarily of liquid waxes rather than actual oils, and a major selling point of both is that the oily substance produced by human skin, called sebum, is also composed of about 25% wax.
While organic and biological chemistry is a very diverse field of study, one fundamental question that interests all organic chemists is how the structure of an organic molecule determines its physical properties. To understand why sperm oil has properties that made it both a useful industrial lubricant for humans and an effective buoyancy control and/or sonic lens for a hunting sperm whale, we first have to understand the nature of both the forces holding each wax molecule together – the covalent single and double bonds between atoms – and also the forces governing the noncovalent interactions between one wax molecule and all the others around it – the so-called 'intermolecular forces' which determine physical properties such as viscosity, melting point, and density.
That is what we will learn about in this chapter. First, we will look more closely at the nature of single and double covalent bonds, using the concepts of 'hybrid orbitals' and 'resonance' to attempt to explain how orbital overlap results in characteristic geometries and rotational behavior for single and double bonds, as well as bonds that have characteristics of somewhere in between single and double. Then we will move on to a review of the noncovalent interactions between molecules - Van der Waals, ion-ion, dipole-dipole and ion-dipole interactions, and hydrogen bonds - and how they are manifested in the observable physical properties of all organic substances.
Before reading any further on this chapter, you will probably need to go back and review some topics from your Introductory Chemistry course. Be sure that you understand the concepts of atomic orbitals, atomic electron configuration, and that you are able to describe s and p orbitals and orbital lobes and nodes. Now would also be a very good time to review VSEPR theory. You might want to watch review tutorials from Kahn academy on atomic orbitals and electron configuration and dot structures and VSEPR theory.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/02%3A_Introduction_to_Organic_Structure_and_Bonding_II/2.01%3A_Prelude_to_Organic_Structure_and_Bonding_II.txt |
As we have been discussing how to use Lewis structures to depict the bonding in organic compounds, we have been very vague so far in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together?
Valence bond theory is most often used to describe bonding in organic molecules. In this model, bonds are considered to form from the overlap of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as molecular hydrogen (H2) or hydrogen fluoride (HF), our present understanding of s and p atomic orbitals will suffice. In order to explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals.
The sigma bond in the H2 molecule
The simplest case to consider is the hydrogen molecule, H2. When we say that the two hydrogen nuclei share their electrons to form a covalent bond, what we mean in valence bond theory terms is that the two spherical 1s orbitals (the grey spheres in the figure below) overlap, and contain two electrons with opposite spin.
These two electrons are now attracted to the positive charge of both of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together.
How far apart are the two nuclei? If they are too far apart, their respective 1s orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, attractive positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second.
But something else is happening at the same time: as the atoms get closer, the repulsive positive-positive interaction between the two nuclei also begins to increase.
At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have an unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the bond length. For the H2 molecule, the distance is 74 pm (picometers, 10-12 meters). Likewise, the difference in potential energy between the lowest energy state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the bond dissociation energy, or, more simply, bond strength. For the hydrogen molecule, the H-H bond strength is equal to about 435 kJ/mol.
Every covalent bond in a given molecule has a characteristic length and strength. In general, the length of a typical carbon-carbon single bond in an organic molecule is about 150 pm, while carbon-carbon double bonds are about 130 pm, carbon-oxygen double bonds are about 120 pm, and carbon-hydrogen bonds are in the range of 100 to 110 pm. The strength of covalent bonds in organic molecules ranges from about 234 kJ/mol for a carbon-iodine bond (in thyroid hormone, for example), about 410 kJ/mole for a typical carbon-hydrogen bond, and up to over 800 kJ/mole for a carbon-carbon triple bond.
Table of bond lengths and bond energies
It is not accurate, however, to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as springs which have a defined length when relaxed, but which can be compressed, extended, and bent. This ‘springy’ picture of covalent bonds will become very important in chapter 4, when we study the analytical technique known as infrared (IR) spectroscopy.
One more characteristic of the covalent bond in H2 is important to consider at this point. The two overlapping 1s orbitals can be visualized as two spherical balloons being pressed together. This means that the bond has cylindrical symmetry: if we were to take a cross-sectional plane of the bond at any point, it would form a circle. This type of bond is referred to as a sigma (σ) bond.
A sigma bond can be formed by overlap of an s atomic orbital with a p atomic orbital. Hydrogen fluoride (HF) is an example:
A sigma bond can also be formed by the overlap of two p orbitals. The covalent bond in molecular fluorine, F2, is a sigma bond formed by the overlap of two half-filled 2p orbitals, one from each fluorine atom.
sp3 hybrid orbitals and tetrahedral bonding
Now let’s look more carefully at bonding in organic molecules, starting with methane, CH4. Recall the valence electron configuration of a carbon atom:
This picture is problematic when it comes to describing the bonding in methane. How does the carbon form four bonds if it has only two half-filled p orbitals available for bonding? A hint comes from the experimental observation that the four C-H bonds in methane are arranged with tetrahedral geometry about the central carbon, and that each bond has the same length and strength. In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by wave equations) to form four equivalent hybrid orbitals, which are called sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital.
interactive 3D model
(select 'load sp3' and 'load H 1s' to see orbitals)
This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible. This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.
Each C-H bond in methane, then, can be described as a sigma bond formed by overlap between a half-filled 1s orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3 hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 109 pm.
While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry. To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid / dashed wedge system. In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page. A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page. Normal lines imply bonds that lie in the plane of the page. This system takes a little bit of getting used to, but with practice your eye will learn to immediately ‘see’ the third dimension being depicted.
Exercise 2.1
Imagine that you could distinguish between the four hydrogen atoms in a methane molecule, and labeled them Ha through Hd. In the images below, the exact same methane molecule is rotated and flipped in various positions. Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)
Exercise 2.2
What kind of orbitals overlap to form the C-Cl bonds in chloroform, CHCl3?
Solutions to exercises
How does this bonding picture extend to compounds containing carbon-carbon bonds? In ethane (CH3CH3), both carbons are sp3-hybridized, meaning that both have four bonds with tetrahedral geometry. The carbon-carbon bond, with a bond length of 154 pm, is formed by overlap of one sp3 orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining sp3 orbitals on the two carbons and the 1s orbitals of hydrogen atoms. All of these are sigma bonds.
Interactive model of ethane
Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate. This means, in the case of ethane molecule, that the two methyl (CH3) groups can be pictured as two wheels on an axle, each one able to rotate with respect to the other.
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Video
In chapter 3 we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the ‘conformation’ of organic molecules.
The sp3 bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine. Just like the carbon atom in methane, the central nitrogen in ammonia is sp3-hybridized. With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a nonbonding pair (lone pair) of electrons.
space-filling image
The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid. Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, ‘pushes’ the three N-H s bonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3˚ rather than 109.5˚.
VSEPR theory also predicts, accurately, that a water molecule is ‘bent’ at an angle of approximately 104.5˚. The bonding in water results from overlap of two of the four sp3 hybrid orbitals on oxygen with 1s orbitals on the two hydrogen atoms. The two nonbonding electron pairs on oxygen are located in the two remaining sp3 orbitals.
Exercise 2.3
Draw, in the same style as the figures above, orbital pictures for the bonding in a) methylamine, and b) ethanol.
Solutions to exercises
video tutorial on sp3 orbitals and sigma bonds
sp2 and sp hybrid orbitals and pi bonds
The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:
1. Ethene is a planar (flat) molecule.
2. Bond angles in ethene are approximately 120o, and the carbon-carbon bond length is 134 pm, significantly shorter than the 154 pm single carbon-carbon bond in ethane.
3. There is a significant barrier to rotation about the carbon-carbon double bond.
Clearly, these characteristics are not consistent with an sp3 hybrid bonding picture for the two carbon atoms. Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py orbitals – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized.
The three sp2 hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120° between them. The unhybridized 2pz orbital is perpendicular to this plane (in the next several figures, sp2 orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2pz orbitals are shown in the 'space-filling' mode).
The carbon-carbon double bond in ethene consists of one sigma bond, formed by the overlap of two sp2 orbitals, and a second bond, called a pi (π) bond, which is formed by the side-by-side overlap of the two unhybridized 2pz orbitals from each carbon.
animation
Interactive 3D model (select 'show resulting pi orbital')
Unlike a sigma bond, a pi bond does not have cylindrical symmetry. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, six atoms lie in the same plane.
Exercise 2.4
Redraw the structures below, indicating the six atoms that lie in the same plane due to the carbon-carbon double bond.
Exercise 2.5
What is wrong with the way the following structure is drawn?
Solutions to exercises
A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde. In this molecule, the carbon is sp2-hybridized, and we will assume that the oxygen atom is also sp2 hybridized. The carbon has three sigma bonds: two are formed by overlap between sp2 orbitals with 1s orbitals from hydrogen atoms, and the third sigma bond is formed by overlap between the remaining carbon sp2 orbital and an sp2 orbital on the oxygen. The two lone pairs on oxygen occupy its other two sp2 orbitals.
spacefilling image
interactive 3D model
The pi bond is formed by side-by-side overlap of the unhybridized 2pz orbitals on the carbon and the oxygen. Just like in alkenes, the 2pz orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.
Exercise 2.6
a: Draw a diagram of hybrid orbitals in an sp2-hybridized nitrogen.
b: Draw a figure showing the bonding picture for the imine below.
c: In your drawing for part b, what kind of orbital holds the nitrogen lone pair?
Solutions to exercises
Recall that carbocations are transient, high-energy species in which a carbon only has three bonds (rather than the usual four) and a positive formal charge. We will have much more to say about carbocations in this and later chapters. For now, though, the important thing to understand is that a carbocation can be described as an sp2-hybridized carbon with an empty 2p orbital perpendicular to the plane of the sigma bonds.
Finally, the hybrid orbital concept applies as well to triple-bonded groups, such as alkynes and nitriles. Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne.
Both the VSEPR theory and experimental evidence tells us that the molecule is linear: all four atoms lie in a straight line. The carbon-carbon triple bond is only 120 pm long, shorter than the double bond in ethene, and is very strong, about 837 kJ/mol. In the hybrid orbital picture of acetylene, both carbons are sp-hybridized. In an sp-hybridized carbon, the 2s orbital combines with the 2px orbital to form two sp hybrid orbitals that are oriented at an angle of 180° with respect to each other (eg. along the x axis). The 2py and 2pz orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.
The carbon-carbon sigma bond, then, is formed by the overlap of one sp orbital from each of the carbons, while the two carbon-hydrogen sigma bonds are formed by the overlap of the second sp orbital on each carbon with a 1s orbital on a hydrogen. Each carbon atom still has two half-filled 2py and 2pz orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds. These two perpendicular pairs of 2p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).
interactive 3D model
Exercise 2.7
Look at the structure of thiamine diphosphate in the 'structures of common coenzymes' table. Identify the hybridization of all carbon atoms in the molecule.
Solutions to exercises
The hybrid orbital concept nicely explains another experimental observation: single bonds adjacent to double and triple bonds are progressively shorter and stronger than single bonds adjacent to other single bonds. Consider for example, the carbon-carbon single bonds in propane, propene, and propyne.
All three are single (sigma) bonds; the bond in propyne is shortest and strongest, while the bond in propane is longest and weakest. The explanation is relatively straightforward. An sp orbital is composed of one s orbital and one p orbital, and thus it has 50% s character and 50% p character. sp2 orbitals, by comparison, have 33% s character and 67% p character, while sp3 orbitals have 25% s character and 75% p character. Because of their spherical shape, 2s orbitals are smaller, and hold electrons closer and ‘tighter’ to the nucleus, compared to 2p orbitals. It follows that electrons in an sp orbital, with its greater s character, are closer to the nucleus than electrons in an sp2 or sp3 orbital. Consequently, bonds involving sp-sp3 overlap (as in propyne) are shorter and stronger than bonds involving sp2-sp3 overlap (as in propene). Bonds involving sp3-sp3 overlap (as in propane) are the longest and weakest of the three.
Exercise 2.8
a) What kinds of orbitals are overlapping in bonds b-i indicated below? Be sure to distinguish between sigma and pi bonds. An example is provided for bond 'a'.
b) In what kind of orbital is the lone pair of electrons located on the nitrogen atom of bond a? Of bond e?
Solutions to exercises
Khan Academy video tutorial on valence bond theory / hybrid orbitals | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/02%3A_Introduction_to_Organic_Structure_and_Bonding_II/2.02%3A_Valence_Bond_Theory.txt |
Valence bond theory does a remarkably good job at explaining the bonding geometry of many of the functional groups in organic compounds. There are some areas, however, where the valence bond theory falls short. It fails to adequately account, for example, for some interesting properties of compounds that contain alternating double and single bonds. In order to understand these properties, we need to think about chemical bonding in a new way, using the ideas of molecular orbital (MO) theory.
Let’s go back and consider again the simplest possible covalent bond: the one in molecular hydrogen (H2). When we described the hydrogen molecule using valence bond theory, we said that the two 1s orbitals from each atom overlap, allowing the two electrons to be shared and thus forming a covalent bond. In molecular orbital theory, we make a further statement: we say that the two atomic 1s orbitals mathematically combine to form two new orbitals. Recall that an atomic orbital (such as the 1s orbital of a hydrogen atom) describes a region of space around a single atom inside which electrons are likely to be found. A molecular orbital describes a region of space around two or more atoms inside which electrons are likely to be found.
Mathematical principles tell us that when orbitals combine, the number of orbitals before the combination takes place must equal the number of new orbitals that result from the combination – orbitals don’t just disappear! We saw this previously when we discussed hybrid orbitals: one s and three p orbitals make four sp3 hybrids. When two atomic 1s orbitals combine in the formation of H2, the result is two sigma (σ) orbitals.
Molecular orbitals for H2
According to MO theory, one sigma orbital is lower in energy than either of the two isolated atomic 1s orbitals –this lower sigma orbital is referred to as a bonding molecular orbital. The second, 'sigma star' (sigma*, σ*) orbital is higher in energy than the two atomic 1s orbitals, and is referred to as an antibonding molecular orbital.
The bonding sigma orbital, which holds both electrons in the ground state of the molecule, is egg-shaped, encompassing the two nuclei, and with the highest likelihood of electrons being in the area between the two nuclei. The high-energy, antibonding sigma* orbital can be visualized as a pair of droplets, with areas of higher electron density near each nucleus and a ‘node’, (area of zero electron density) midway between the two nuclei.
Remember that we are thinking here about electron behavior as wave behavior. When two separate waves combine, they can do so with constructive interference, where the two amplitudes build up and reinforce one another, or destructive interference, where the two amplitudes cancel one another out. Bonding MOs are the consequence of constructive interference between two atomic orbitals, which results in an attractive interaction and an increase in electron density between the nuclei. Antibonding MO’s are the consequence of destructive interference which results in a repulsive interaction and a region of zero electron density between the nuclei (in other words, a node).
Following the same aufbau ('building up') principle you learned in General Chemistry for writing out electron configurations, we place the two electrons in the H2 molecule in the lowest energy molecular orbital, which is the (bonding) sigma orbital. The bonding (attracting) MO is full, and the antibonding (repulsing) MO is empty.
MO theory and conjugated pi bonds
The advantage of using MO theory to understand bonding in organic molecules becomes more apparent when we think about pi bonds. Let’s first consider the pi bond in ethene from an MO theory standpoint (in this example we will be disregarding the sigma bonds in the molecule, and thinking only about the pi bond). We start with two atomic orbitals: one unhybridized 2p orbital from each carbon. Each contains a single electron. In MO theory, the two atomic combine mathematically to form two pi (π) molecular orbitals, one a low-energy pi bonding orbital and one a high-energy pi* (π*) antibonding orbital.
Molecular orbitals for ethene (ethylene)
In the bonding pi orbital, the two shaded lobes of the p orbitals interact constructively with each other, as do the two unshaded lobes (remember, the arbitrary shading choice represents mathematical (+) and (-) signs for the mathematical wavefunction describing the orbital). There is increased electron density between the two carbon nuclei in the molecular orbital - it is a bonding interaction.
In the higher-energy antibonding pi* orbital, the shaded lobe of one p orbital interacts destructively with the unshaded lobe of the second p orbital, leading to a node between the two nuclei and overall repulsion between the carbon nuclei.
Again using the 'building up' principle, we place the two electrons in the lower-energy, bonding pi molecular orbital. The antibonding pi* orbital remains empty.
Next, we'll consider the 1,3-butadiene molecule. From valence orbital theory alone we might expect that the C2-C3 bond in this molecule, because it is a sigma bond, would be able to rotate freely.
Experimentally, however, it is observed that there is a significant barrier to rotation about the C2-C3 bond, and that the entire molecule is planar. In addition, the C2-C3 bond is 148 pm long, shorter than a typical carbon-carbon single bond (about 154 pm), though longer than a typical double bond (about 134 pm).
Molecular orbital theory accounts for these observations with the concept of delocalized pi bonds. In this picture, the four 2p atomic orbitals combine mathematically to form four pi molecular orbitals of increasing energy. Two of these - the bonding pi orbitals - are lower in energy than the p atomic orbitals from which they are formed, while two - the antibonding pi* orbitals - are higher in energy.
The lowest energy molecular orbital, pi1, has only constructive interaction and zero nodes. Higher in energy, but still lower than the isolated p orbitals, the pi2 orbital has one node but two constructive interactions - thus it is still a bonding orbital overall. Looking at the two antibonding orbitals, pi3* has two nodes and one constructive interaction, while pi4* has three nodes and zero constructive interactions.
By the aufbau principle, the four electrons from the isolated 2pz atomic orbitals are placed in the bonding pi1 and pi2 MO’s. Because pi1 includes constructive interaction between C2 and C3, there is a degree, in the 1,3-butadiene molecule, of pi-bonding interaction between these two carbons, which accounts for its shorter length and the barrier to rotation. The valence bond picture of 1,3-butadiene shows the two pi bonds as being isolated from one another, with each pair of pi electrons ‘stuck’ in its own pi bond. However, molecular orbital theory predicts (accurately) that the four pi electrons are to some extent delocalized, or ‘spread out’, over the whole pi system.
space-filling view
1,3-butadiene is the simplest example of a system of conjugated pi bonds. To be considered conjugated, two or more pi bonds must be separated by only one single bond – in other words, there cannot be an intervening sp3-hybridized carbon, because this would break up the overlapping system of parallel p orbitals. In the compound below, for example, the C1-C2 and C3-C4 double bonds are conjugated, while the C6-C7 double bond is isolated from the other two pi bonds by sp3-hybridized C5.
A very important concept to keep in mind is that there is an inherent thermodynamic stability associated with conjugation. This stability can be measured experimentally by comparing the heat of hydrogenation of two different dienes. (Hydrogenation is a reaction type that we will learn much more about in chapter 15: essentially, it is the process of adding a hydrogen molecule - two protons and two electrons - to a pi bond). When the two conjugated double bonds of 1,3-pentadiene are 'hydrogenated' to produce pentane, about 225 kJ is released per mole of pentane formed. Compare that to the approximately 250 kJ/mol released when the two isolated double bonds in 1,4-pentadiene are hydrogenated, also forming pentane.
The conjugated diene is lower in energy: in other words, it is more stable. In general, conjugated pi bonds are more stable than isolated pi bonds.
Conjugated pi systems can involve oxygen and nitrogen atoms as well as carbon. In the metabolism of fat molecules, some of the key reactions involve alkenes that are conjugated to carbonyl groups.
In chapter 4, we will see that MO theory is very useful in explaining why organic molecules that contain extended systems of conjugated pi bonds often have distinctive colors. beta-carotene, the compound responsible for the orange color of carrots, has an extended system of 11 conjugated pi bonds.
Exercise 2.9
Identify all conjugated and isolated double bonds in the structures below. For each conjugated pi system, specify the number of overlapping p orbitals, and how many pi electrons are shared among them.
Exercise 2.10
Identify all isolated and conjugated double bonds in lycopene, the red-colored compound in tomatoes. How many pi electrons are contained in the conjugated pi system?
Solutions to exercises
Aromaticity
Molecular orbital theory is especially helpful in explaining the unique properties of aromatic compounds such as benzene:
Although benzene is most often drawn with three double bonds and three single bonds, in fact all of the carbon-carbon bonds are exactly the same length (138 pm). In addition, the pi bonds in benzene are significantly less reactive than 'normal' pi bonds, either isolated or conjugated. Something about the structure of benzene makes its pi bonding arrangement especially stable. This ‘something’ has a name: it is called ‘aromaticity’.
What exactly is this ‘aromatic’ property that makes the pi bonds in benzene so stable? In large part, the answer to this question lies in the fact that benzene is a cyclic molecule in which all of the ring atoms are sp2-hybridized. This allows the pi electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane. For this to happen, of course, the ring must be planar – otherwise the p orbitals couldn’t overlap properly. Benzene is indeed known to be a flat molecule.
Do all cyclic molecules with alternating single and double bonds have this same aromatic stability? The answer, in fact, is ‘no’. The eight-membered cyclooctatetraene ring shown below is not flat, and its pi bonds react like 'normal' alkenes. π
Clearly it takes something more to be aromatic, and this can best be explained with molecular orbital theory. Let’s look at an energy diagram of the pi molecular orbitals in benzene.
Quantum mechanical calculations tell us that the six pi molecular orbitals in benzene, formed from six atomic p orbitals, occupy four separate energy levels. pi1 and pi6* have unique energy levels, while the pi2 - pi3 and pi4*- pi5* pairs are degenerate, meaning they are at the same energy level. When we use the aufbau principle to fill up these orbitals with the six pi electrons in benzene, we see that the bonding orbitals are completely filled, and the antibonding orbitals are empty. This gives us a good clue to the source of the special stability of benzene: a full set of bonding MO’s is similar in many ways to the ‘full shell’ of electrons in the atomic orbitals of the stable noble gases helium, neon, and argon.
Now, let’s do the same thing for cyclooctatetraene, which we have already learned is not aromatic.
The result of molecular orbital calculations tells us that the lowest and highest energy MOs (pi1 and pi8*) have unique energy levels, while the other six form degenerate pairs. Notice that pi4 and pi5 are at the same energy level as the isolated 2pz atomic orbitals: these are therefore neither bonding nor antibonding, rather they are referred to as nonbonding MOs. Filling up the MOs with the eight pi electrons in the molecule, we find that the last two electrons are unpaired and fall into the two degenerate nonbonding orbitals. Because we don't have a perfect filled shell of bonding MOs, our molecule is not aromatic. As a consequence, each of the double bonds in cyclooctatetraene acts more like an isolated double bond.
Here, then, are the conditions that must be satisfied for a molecule or group to be considered aromatic:
Criteria for aromaticity:
The molecule or group must be cyclic.
The ring must be planar.
Each atom in the ring must be sp2-hybridized.
The number of pi electrons in the ring must equal 4n+2, where n is any positive integer including zero.
Rule #4 is known as the Hückel rule, named after Erich Hückel, a German scientist who studied aromatic compounds in the 1930’s. If n = 0, the Hückel number is 2. If n = 1, the Hückel number is 6 (the Hückel number for benzene). The series continues with 10, 14, 18, 22, and so on. Cyclooctatetraene has eight pi electrons, which is not a Hückel number. Because six is such a common Hückel number, chemists often use the term 'aromatic sextet'.
Benzene rings are ubiquitous in biomolecules and drugs - below are just a few examples.
Recall that a benzene ring with a hydoxyl substituent -such as seen in the tyrosine structure above - is called a phenol.
Heterocycles - cyclic structures in which the ring atoms may include oxygen or nitrogen - can also be aromatic. Pyridine, for example, is an aromatic heterocycle. In the bonding picture for pyridine, the nitrogen is sp2-hybridized, with two of the three sp2 orbitals forming sigma overlaps with the sp2 orbitals of neighboring carbon atoms, and the third nitrogen sp2 orbital containing the lone pair. The unhybridized p orbital contains a single electron, which is part of the 6 pi-electron system delocalized around the ring.
another image of orbitals in pyridine
Pyridoxine, commonly known as vitamin B6, and nicotine are both substituted pyridines.
Pyrrole is a five-membered aromatic heterocycle. In pyrrole, the lone pair electrons on the sp2-hybridized nitrogen are part of the aromatic sextet (contrast this to pyridine, where the lone pair occupies one of the sp2 hybrid orbitals).
Why don't we assume that the nitrogen in pyrrole is sp3-hybridized, like a normal secondary amine? The answer is simple: if it were, then pyrrole could not be aromatic, and thus it would not have the stability associated with aromaticity. In general, if a molecule or group can be aromatic, it will be, just as water will always flow downhill if there is a downhill pathway available.
Imidazole is another important example of an aromatic heterocycle found in biomolecules - the side chain of the amino acid histidine contains an imidazole ring.
In imidazole, one nitrogen is 'pyrrole-like' (the lone pair contributes to the aromatic sextet) and one is 'pyridine-like' (the lone pair is located in an sp2 orbital, and is not part of the aromatic sextet).
Fused-ring structures can also fulfill the Hückel criteria, and often have many of the same properties as monocyclic aromatic compounds, including a planar structure. Indole (a functional group on the side chain of the amino acid tryptophan) and purine (found in guanine and adenine nucleotide bases) both have a total of ten pi electrons delocalized around two rings.
The nucleic acid bases of DNA and RNA - guanine, adenine, cytosine, thymine, and uracil - are all aromatic systems, with the characteristic aromatic properties of planarity and delocalized pi electron density. When you study the structure and function of DNA and RNA in a biochemistry or molecular biology course, you will see that the planar shape of the bases plays a critically important role.
Exercise 2.11
Classify the nitrogen atoms of indole and purine as either 'pyrrole-like' or 'pyridine-like', in terms of where the lone pair electrons are located.
Exercise 2.12
Are the following molecules/ions aromatic? Explain, using criteria you learned in this section.
Solutions to exercises | textbooks/chem/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/02%3A_Introduction_to_Organic_Structure_and_Bonding_II/2.03%3A_Molecular_orbital_theory-_conjugation_and_aromaticity.txt |
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