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Learning Objectives
• Describe the reactions of the citric acid cycle.
• Describe the function of the citric acid cycle and identify the products produced.
The acetyl group enters a cyclic sequence of reactions known collectively as the citric acid cycle (or Krebs cycle or tricarboxylic acid [TCA] cycle). The cyclical design of this complex series of reactions, which bring about the oxidation of the acetyl group of acetyl-CoA to carbon dioxide and water, was first proposed by Hans Krebs in 1937. (He was awarded the 1953 Nobel Prize in Physiology or Medicine.) Acetyl-CoA’s entrance into the citric acid cycle is the beginning of stage III of catabolism. The citric acid cycle produces adenosine triphosphate (ATP), reduced nicotinamide adenine dinucleotide (NADH), reduced flavin adenine dinucleotide (FADH2), and metabolic intermediates for the synthesis of needed compounds.
Steps of the Citric Acid Cycle
At first glance, the citric acid cycle appears rather complex (Figure \(1\)). All the reactions, however, are familiar types in organic chemistry: hydration, oxidation, decarboxylation, and hydrolysis. Each reaction of the citric acid cycle is numbered, and in Figure \(1\), the two acetyl carbon atoms are highlighted in red. Each intermediate in the cycle is a carboxylic acid, existing as an anion at physiological pH. All the reactions occur within the mitochondria, which are small organelles within the cells of plants and animals.
1. In the first step, acetyl-CoA enters the citric acid cycle, and the acetyl group is transferred onto oxaloacetate, yielding citrate. Note that this step releases coenzyme A. The reaction is catalyzed by citrate synthase.
2. In the next step, aconitase catalyzes the isomerization of citrate to isocitrate. In this reaction, a tertiary alcohol, which cannot be oxidized, is converted to a secondary alcohol, which can be oxidized in the next step.
3. Isocitrate then undergoes a reaction known as oxidative decarboxylation because the alcohol is oxidized and the molecule is shortened by one carbon atom with the release of carbon dioxide (decarboxylation). The reaction is catalyzed by isocitrate dehydrogenase, and the product of the reaction is α-ketoglutarate. An important reaction linked to this is the reduction of the coenzyme nicotinamide adenine dinucleotide (NAD+) to NADH. The NADH is ultimately reoxidized, and the energy released is used in the synthesis of ATP, as we shall see.
4. The fourth step is another oxidative decarboxylation. This time α-ketoglutarate is converted to succinyl-CoA, and another molecule of NAD+ is reduced to NADH. The α-ketoglutarate dehydrogenase complex catalyzes this reaction. This is the only irreversible reaction in the citric acid cycle. As such, it prevents the cycle from operating in the reverse direction, in which acetyl-CoA would be synthesized from carbon dioxide.
So far, in the first four steps, two carbon atoms have entered the cycle as an acetyl group, and two carbon atoms have been released as molecules of carbon dioxide. The remaining reactions of the citric acid cycle use the four carbon atoms of the succinyl group to resynthesize a molecule of oxaloacetate, which is the compound needed to combine with an incoming acetyl group and begin another round of the cycle.
In the fifth reaction, the energy released by the hydrolysis of the high-energy thioester bond of succinyl-CoA is used to form guanosine triphosphate (GTP) from guanosine diphosphate (GDP) and inorganic phosphate in a reaction catalyzed by succinyl-CoA synthetase. This step is the only reaction in the citric acid cycle that directly forms a high-energy phosphate compound. GTP can readily transfer its terminal phosphate group to adenosine diphosphate (ADP) to generate ATP in the presence of nucleoside diphosphokinase.
Succinate dehydrogenase then catalyzes the removal of two hydrogen atoms from succinate, forming fumarate. This oxidation-reduction reaction uses flavin adenine dinucleotide (FAD), rather than NAD+, as the oxidizing agent. Succinate dehydrogenase is the only enzyme of the citric acid cycle located within the inner mitochondrial membrane. We will see soon the importance of this.
In the following step, a molecule of water is added to the double bond of fumarate to form L-malate in a reaction catalyzed by fumarase.
One revolution of the cycle is completed with the oxidation of L-malate to oxaloacetate, brought about by malate dehydrogenase. This is the third oxidation-reduction reaction that uses NAD+ as the oxidizing agent. Oxaloacetate can accept an acetyl group from acetyl-CoA, allowing the cycle to begin again.
Video: "The Citric Acid Cycle: An Overview". In the matrix of the mitochondrion, the Citric Acid Cycle uses Acetyl CoA molecules to produce energy through eight chemical reactions. This animation provides an overview of the pathway and its products. NDSU VCell Production's animation; for more information please see http://vcell.ndsu.edu/animations. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/21%3A_The_Generation_of_Biochemical_Energy/21.07%3A_The_Citric_Acid_Cycle.txt |
Learning Outcomes
• Summarize the electron transport chain.
• Recognize that electron transport chain is the third and final stage of aerobic cellular respiration.
• Identify the products of the citric acid cycle.
What do trains, trucks, boats, and planes all have in common? They are ways to transport. And they all use a lot of energy. To make ATP, energy must be "transported" - first from glucose to NADH, and then somehow passed to ATP. How is this done? With an electron transport chain, the third stage of aerobic respiration. This third stage uses energy to make energy.
The Electron Transport Chain: ATP for Life in the Fast Lane
At the end of the Krebs Cycle, energy from the chemical bonds of glucose is stored in diverse energy carrier molecules: four ATPs, but also two FADH$_2$ and ten NADH molecules. The primary task of the last stage of cellular respiration, the electron transport chain, is to transfer energy from the electron carriers to even more ATP molecules, the "batteries" which power work within the cell.
Pathways for making ATP in stage 3 of aerobic respiration closely resemble the electron transport chains used in photosynthesis. In both electron transport chains, energy carrier molecules are arranged in sequence within a membrane so that energy-carrying electrons cascade from one to another, losing a little energy in each step. In both photosynthesis and aerobic respiration, the energy lost is harnessed to pump hydrogen ions into a compartment, creating an electrochemical gradient or chemiosmotic gradient across the enclosing membrane. And in both processes, the energy stored in the chemiosmotic gradient is used with ATP synthase to build ATP.
For aerobic respiration, the electron transport chain or "respiratory chain" is embedded in the inner membrane of the mitochondria (see figure below). The FADH$_2$ and NADH molecules produced in glycolysis and the Krebs Cycle, donate high-energy electrons to energy carrier molecules within the membrane. As they pass from one carrier to another, the energy they lose is used to pump hydrogen ions into the mitochondrial intermembrane space, creating an electrochemical gradient. Hydrogen ions flow "down" the gradient - from outer to inner compartment - through the ion channel/enzyme ATP synthase, which transfers their energy to ATP. Note the paradox that it requires energy to create and maintain a concentration gradient of hydrogen ions that are then used by ATP synthase to create stored energy (ATP). In broad terms, it takes energy to make energy. Coupling the electron transport chain to ATP synthesis with a hydrogen ion gradient is chemiosmosis, first described by Nobel laureate Peter D. Mitchell. This process, the use of energy to phosphorylate ADP and produce ATP is also known as oxidative phosphorylation.
After passing through the electron transport chain, low-energy electrons and low-energy hydrogen ions combine with oxygen to form water. Thus, oxygen's role is to drive the entire set of ATP-producing reactions within the mitochondrion by accepting "spent" hydrogens. Oxygen is the final electron acceptor, no part of the process - from the Krebs Cycle through the electron transport chain- can happen without oxygen.
The electron transport chain can convert the energy from one glucose molecule's worth of $FADH_2$ and $NADH$ + $\ce{H^+}$ into as many as 34 ATP. When the four ATP produced in glycolysis and the Krebs Cycle are added, the total of 38 ATP fits the overall equation for aerobic cellular respiration:
$\ce{6O2} + \underbrace{\ce{C6H12O6}}_{\text{stored chemical energy}} + \ce{38 ADP} + \text{39 P}_\text{i} \rightarrow \underbrace{\ce{38 ATP}}_{\text{stored chemical energy}} + \ce{6CO2} + \ce{6 H2O}$
Aerobic respiration is complete. If oxygen is available, cellular respiration transfers the energy from one molecule of glucose to 38 molecules of ATP, releasing carbon dioxide and water as waste. "Deliverable" food energy has become energy which can be used for work within the cell - transport within the cell, pumping ions and molecules across membranes, and building large organic molecules. Can you see how this could lead to "life in the fast lane" compared to anaerobic respiration (glycolysis alone)?
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/21%3A_The_Generation_of_Biochemical_Energy/21.08%3A_The_Electron-Transport_Chain_and_ATP_Production.txt |
Thumbnail: Pyruvic acid is the simplest of the alpha-keto acids, with a carboxylic acid and a ketone functional group. Pyruvic acid supplies energy to cells through the citric acid cycle (also known as the Krebs cycle) when oxygen is present (aerobic respiration), and alternatively ferments to produce lactate when oxygen is lacking (fermentation). Imag esed with permission (Public Domain; Benjah-bmm27).
22: Carbohydrate Metabolism
Learning Objectives
• Describe digestion of carbohydrates.
The first stage of catabolism is digestion, in which food molecules are broken down into smaller molecules. Carbohydrate digestion begins in the mouth (Figure $1$) where salivary $\alpha$-amylase attacks the $\alpha$-glycosidic linkages in starch, the main carbohydrate ingested by humans. Cleavage of the glycosidic linkages produces a mixture of dextrins, maltose, and glucose. The $\alpha$-amylase mixed into the food remains active as the food passes through the esophagus, but it is rapidly inactivated in the acidic environment of the stomach.
The primary site of carbohydrate digestion is the small intestine. The secretion of $\alpha$-amylase in the small intestine converts any remaining starch molecules, as well as the dextrins, to maltose. Maltose is then cleaved into two glucose molecules by maltase. Disaccharides such as sucrose and lactose are not digested until they reach the small intestine, where they are acted on by sucrase and lactase, respectively. The major products of the complete hydrolysis of disaccharides and polysaccharides are three monosaccharide units: glucose, fructose, and galactose. These are absorbed through the wall of the small intestine into the bloodstream.
22.03: Glycolysis
Learning Objectives
• Objective 1
• Objective 2
Glycolysis is the catabolic process in which glucose is converted into pyruvate via ten enzymatic steps. There are three regulatory steps, each of which is highly regulated.
Introduction
There are two phases of Glycolysis:
1. the "priming phase" because it requires an input of energy in the form of 2 ATPs per glucose molecule and
2. the "pay off phase" because energy is released in the form of 4 ATPs, 2 per glyceraldehyde molecule.
The end result of Glycolysis is two new pyruvate molecules which can then be fed into the Citric Acid cycle (also known as the Kreb's Cycle) if oxygen is present, or can be reduced to lactate or ethanol in the absence of of oxygen using a process known as Fermentation. Glycolysis occurs within almost all living cells and is the primary source of Acetyl-CoA, which is the molecule responsible for the majority of energy output under aerobic conditions. The structures of Glycolysis intermediates can be found in the following diagram:
Phase 1: The "Priming Step"
The first phase of Glycolysis requires an input of energy in the form of ATP (adenosine triphosphate).
1. alpha-D-Glucose is phosphorolated at the 6 carbon by ATP via the enzyme Hexokinase (Class: Transferase) to yield alpha-D-Glucose-6-phosphate (G-6-P). This is a regulatory step which is negatively regulated by the presence of glucose-6-phosphate.
2. alpha-D-Glucose-6-phosphate is then converted into D-Fructose-6-phosphate (F-6-P) by Phosphoglucoisomerase (Class: Isomerase)
3. D-Fructose-6-phosphate is once again phosphorolated this time at the 1 carbon position by ATP via the enzyme Phosphofructokinase (Class: Transferase) to yield D-Fructose-1,6-bisphosphate (FBP). This is the committed step of glycolysis because of its large \(\Delta G\) value.
4. D-Fructose-1,6-bisphosphate is then cleaved into two, three carbon molecules; Dihydroxyacetone phosphate (DHAP) and D-Glyceraldehyde-3-phosphate (G-3-P) by the enzyme Fructose bisphosphate aldolase (Class: Lyase)
5. Because the next portion of Glycolysis requires the molecule D-Glyceraldehyde-3-phosphate to continue Dihydroxyacetone phosphate is converted into D-Glyceraldehyde-3-phosphate by the enzyme Triose phosphate isomerase (Class: Isomerase)
Phase 2: The "Pay Off Step"
The second phase of Glycolysis where 4 molecules of ATP are produced per molecule of glucose. Enzymes appear in red:
1. D-Glyceraldehyde-3-phosphate is phosphorolated at the 1 carbon by the enzyme Glyceraldehyde-3-phosphate dehodrogenase to yield the high energy molecule 1,3-Bisphosphoglycerate (BPG)
2. ADP is then phosphorolated at the expense of 1,3-Bisphosphoglycerate by the enzyme Phosphoglycerate kinase (Class: Transferase) to yield ATP and 3-Phosphoglycerate (3-PG)
3. 3-Phosphoglycerate is then converted into 2-Phosphoglycerate by Phosphoglycerate mutase in preparation to yield another high energy molecule
4. 2-Phosphoglycerate is then converted to phosphoenolpyruvate (PEP) by Enolase. H2O, potassium, and magnesium are all released as a result.
5. ADP is once again phosphorolated, this time at the expense of PEP by the enzyme pyruvate kinase to yield another molecule of ATP and and pyruvate. This step is regulated by the energy in the cell. The higher the energy of the cell the more inhibited pyruvate kinase becomes. Indicators of high energy levels within the cell are high concentrations of ATP, Acetyl-CoA, Alanine, and cAMP.
Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the "Pay Off" phase occurs twice per molecule of glucose.
Problems
1. What is the net yield of Glycolysis as far as ATP?
2. Name the enzymes that are key regulatory sites in Glycolysis.
3. Why are the enzymes in the previous question targets for regulation?
4. Why is the priming phase necessary?
5. Draw the entire pathway for glycolysis including enzymes, reactants and products for each step.
Contributors and Attributions
• Darik Benson, (University California Davis) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/22%3A_Carbohydrate_Metabolism/22.01%3A_Digestion_of_Carbohydrates.txt |
Learning Objectives
• Define lipids and recognize the different classes.
Fats and oils, found in many of the foods we eat, belong to a class of biomolecules known as lipids. Gram for gram, they pack more than twice the caloric content of carbohydrates: the oxidation of fats and oils supplies about 9 kcal of energy for every gram oxidized, whereas the oxidation of carbohydrates supplies only 4 kcal/g. Although the high caloric content of fats may be bad news for the dieter, it says something about the efficiency of nature’s designs. Our bodies use carbohydrates, primarily in the form of glucose, for our immediate energy needs. Our capacity for storing carbohydrates for later use is limited to tucking away a bit of glycogen in the liver or in muscle tissue. We store our reserve energy in lipid form, which requires far less space than the same amount of energy stored in carbohydrate form. Lipids have other biological functions besides energy storage. They are a major component of the membranes of the 10 trillion cells in our bodies. They serve as protective padding and insulation for vital organs. Furthermore, without lipids in our diets, we would be deficient in the fat-soluble vitamins A, D, E, and K.
Lipids are not defined by the presence of specific functional groups, as carbohydrates are, but by a physical property—solubility. Compounds isolated from body tissues are classified as lipids if they are more soluble in organic solvents, such as dichloromethane, than in water. By this criterion, the lipid category includes not only fats and oils, which are esters of the trihydroxy alcohol glycerol and fatty acids, but also compounds that incorporate functional groups derived from phosphoric acid, carbohydrates, or amino alcohols, as well as steroid compounds such as cholesterol (Figure \(1\) presents one scheme for classifying the various kinds of lipids). We will discuss the various kinds of lipids by considering one subclass at a time and pointing out structural similarities and differences as we go.
Lipids Formed from Fatty Acids
Fatty acids are carboxylic acids with 12-22 carbon atoms connected in a long, unbranched chain. As shown in the diagram above, most lipids are classified as esters or amides of fatty acids.
• Waxes are esters formed from long-chain fatty acids and long-chain alcohols. Most natural waxes are mixtures of such esters.
• Triacylglycerol (triglycerides) are esters of glycerol, a trialcohol, and three fatty acids. Many organisms store energy in this form.
• Glycerophospholipids are esters of glycerol formed from two fatty acid chains and a charged phosphate.
• Sphingolipids are fatty acid amides formed from a fatty acid attached to an amino alcohol backbone, called sphingosine, along with either a phosphate (sphingomyelin) or a carbohydrate (glycolipid). These along with glycerophospholipids are important for the structure and function of cellular membranes.
Other Lipids
Not all lipids contain fatty acid groups:
• Sterols (also classified as steroids) all contain the steroid nucleus, which is four fused rings. Cholesterol is the most commonly known sterol and is also an important lipid in cell membranes.
• Eicosanoids are important chemical messengers that include prostaglandins, which have a five-member ring and a carboxylic acid chain.
In the next sections of this chapter you will learn more about the structure, properties, and functions of each of these types of lipid.
To Your Health: Prostaglandins
Prostaglandins are chemical messengers synthesized in the cells in which their physiological activity is expressed. They are unsaturated fatty acids containing 20 carbon atoms and are synthesized from arachidonic acid—a polyunsaturated fatty acid—when needed by a particular cell. They are called prostaglandins because they were originally isolated from semen found in the prostate gland. It is now known that they are synthesized in nearly all mammalian tissues and affect almost all organs in the body. The five major classes of prostaglandins are designated as PGA, PGB, PGE, PGF, and PGI. Subscripts are attached at the end of these abbreviations to denote the number of double bonds outside the five-carbon ring in a given prostaglandin.
The prostaglandins are among the most potent biological substances known. Slight structural differences give them highly distinct biological effects; however, all prostaglandins exhibit some ability to induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response. Aspirin and other nonsteroidal anti-inflammatory agents, such as ibuprofen, obstruct the synthesis of prostaglandins by inhibiting cyclooxygenase, the enzyme needed for the initial step in the conversion of arachidonic acid to prostaglandins.
Their wide range of physiological activity has led to the synthesis of hundreds of prostaglandins and their analogs. Derivatives of PGE2 are now used in the United States to induce labor. Other prostaglandins have been employed clinically to lower or increase blood pressure, inhibit stomach secretions, relieve nasal congestion, relieve asthma, and prevent the formation of blood clots, which are associated with heart attacks and strokes. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/23%3A_Lipids/23.01%3A_Structure_and_Classification_of_Lipids.txt |
Learning Objectives
• To recognize the structures of common fatty acids and classify them as saturated, monounsaturated, or polyunsaturated.
• Describe characteristics of fatty acid esters.
Fatty acids are carboxylic acids that are structural components of fats, oils, and all other categories of lipids, except steroids. More than 70 have been identified in nature. They usually contain an even number of carbon atoms (typically 12–20), are generally unbranched, and can be classified by the presence and number of carbon-to-carbon double bonds. Thus, Fatty acids may be saturated, containing all carbon-carbon single bonds, or unsaturated, containing at least one or more carbon-carbon double bonds. Saturated fatty acids are saturated with hydrogens; in other words, the number of hydrogen atoms attached to the carbon skeleton is maximized. Stearic acid is an example of a saturated fatty acid (Figure $1$).
When the hydrocarbon chain contains a double bond, there are less hydrogens attached to the carbon skeleton and the fatty acid is said to be unsaturated. Oleic acid is an example of a monounsaturated fatty acid (Figure $1$), because it contains one double bond. A fatty acid with two or more double bonds is a polyunsaturated fat. The atoms or groups around the double bonds in unsaturated fatty acids can be arranged in either the cis or trans isomeric form, compare cis vs. trans oleic acid in (Figure $1$). Naturally occurring fatty acids are generally in the cis configuration.
Table $1$ lists some common fatty acids. Chemists use a shorthand notation for fatty acids rather than the common name because they are all carboxylic acids with different numbers of carbons and degree of unsaturation (number of double bonds). This notation gives the number of carbons followed by the number of double bonds present. For example: palmitic acid, a 16 carbon saturated fatty acid, would be represented by C16:0 and palmitoleic acid, a 16 carbon monounsaturated fatty acid, would be C16:1.
Table $1$: Some Common Fatty Acids Found in Natural Fats
Name Number of Carbons Number of Double Bonds Condensed Structural Formula Melting Point (°C) Shorthand Notation Source
lauric acid 12 0 CH3(CH2)10COOH 44 C12:0 palm kernel oil
myristic acid 14 0 CH3(CH2)12COOH 58 C14:0 oil of nutmeg
palmitic acid 16 0 CH3(CH2)14COOH 63 C16:0 palm oil
palmitoleic acid 16 1 CH3(CH2)5CH=CH(CH2)7COOH 0.5 C16:1 macadamia oil
stearic acid 18 0 CH3(CH2)16COOH 70 C18:0 cocoa butter
oleic acid 18 1 CH3(CH2)7CH=CH(CH2)7COOH 16 C18:1 olive oil
linoleic acid 18 2 CH3(CH2)3(CH2CH=CH)2(CH2)7COOH −5 C18:2 canola oil
$\alpha$-linolenic acid 18 3 CH3(CH2CH=CH)3(CH2)7COOH −11 C18:3 flaxseed
arachidonic acid 20 4 CH3(CH2)4(CH2CH=CH)4(CH2)2COOH −50 C20:4 liver
Two polyunsaturated fatty acids—linoleic and $\alpha$-linolenic acids—are termed essential fatty acids because humans must obtain them from their diets. Both substances are required for normal growth and development, but the human body does not synthesize them. The body uses linoleic acid to synthesize many of the other unsaturated fatty acids, such as arachidonic acid, a precursor for the synthesis of prostaglandins. In addition, the essential fatty acids are necessary for the efficient transport and metabolism of cholesterol. The average daily diet should contain about 4–6 g of the essential fatty acids.
Waxes
Waxes are esters formed from long-chain fatty acids and long-chain alcohols. Most natural waxes are mixtures of such esters. Plant waxes on the surfaces of leaves, stems, flowers, and fruits protect the plant from dehydration and invasion by harmful microorganisms. Carnauba wax, used extensively in floor waxes, automobile waxes, and furniture polish, is largely myricyl cerotate, obtained from the leaves of certain Brazilian palm trees. Animals also produce waxes that serve as protective coatings, keeping the surfaces of feathers, skin, and hair pliable and water repellent. In fact, if the waxy coating on the feathers of a water bird is dissolved as a result of the bird swimming in an oil slick, the feathers become wet and heavy, and the bird, unable to maintain its buoyancy, drowns.
Triacylglycerols
Animal fats and vegetable oils are called triacylcylgerols (or triglycerides) because they are esters composed of three fatty acid units joined to glycerol, a trihydroxy alcohol:
If all three OH groups on the glycerol molecule are esterified with the same fatty acid, the resulting ester is called a simple triglyceride. Although simple triglycerides have been synthesized in the laboratory, they rarely occur in nature. Instead, a typical triacylglycerols obtained from naturally occurring fats and oils contains two or three different fatty acid components and is thus termed a mixed triglyceride.
Tristearin
a simple triglyceride
a mixed triglyceride
A triacylglycerol is called a fat if it is a solid at 25 °C; it is called an oil if it is a liquid at that temperature. These differences in melting points reflect differences in the degree of unsaturation and number of carbon atoms in the constituent fatty acids. Triglycerides obtained from animal sources are usually solids, while those of plant origin are generally oils. Therefore, we commonly speak of animal fats and vegetable oils.
No single formula can be written to represent the naturally occurring fats and oils because they are highly complex mixtures of triglycerides in which many different fatty acids are represented. Table $2$ shows the fatty acid compositions of some common fats and oils. The composition of any given fat or oil can vary depending on the plant or animal species it comes from as well as on dietetic and climatic factors. To cite just one example, lard from corn-fed hogs is more highly saturated than lard from peanut-fed hogs. Palmitic acid is the most abundant of the saturated fatty acids, while oleic acid is the most abundant unsaturated fatty acid.
Table $2$: Average Fatty Acid Composition of Some Common Fats and Oils (%)*
Source Lauric (C12:0) Myristic (C14:0) Palmitic (C16:0) Stearic (C18:0) Oleic (C18:1) Linoleic (C18:2) Linolenic (C18:3)
Fats
butter (cow) 3 11 27 12 29 2 1
tallow 3 24 19 43 3 1
lard 2 26 14 44 10
Oils
canola oil 4 2 62 22 10
coconut oil 47 18 9 3 6 2
corn oil 11 2 28 58 1
olive oil 13 3 71 10 1
peanut oil 11 2 48 32
soybean oil 11 4 24 54 7
*Totals less than 100% indicate the presence of fatty acids with fewer than 12 carbon atoms or more than 18 carbon atoms.
Coconut oil is highly saturated. It contains an unusually high percentage of the low-melting C8, C10, and C12 saturated fatty acids.
Terms such as saturated fat or unsaturated oil are often used to describe the fats or oils obtained from foods. Saturated fats contain a high proportion of saturated fatty acids, while unsaturated oils contain a high proportion of unsaturated fatty acids. The high consumption of saturated fats is a factor, along with the high consumption of cholesterol, in increased risks of heart disease. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/23%3A_Lipids/23.02%3A_Fatty_Acids_and_Their_Esters.txt |
Learning Objectives
• To identify the functions of steroids produced in mammals.
All the lipids discussed so far are saponifiable, reacting with aqueous alkali to yield simpler components, such as glycerol, fatty acids, amino alcohols, and sugars. Lipid samples extracted from cellular material, however, also contain a small but important fraction that does not react with alkali. The most important nonsaponifiable lipids are the steroids. These compounds include the bile salts, cholesterol and related compounds, and certain hormones (such as cortisone and the sex hormones).
Steroids occur in plants, animals, yeasts, and molds but not in bacteria. They may exist in free form or combined with fatty acids or carbohydrates. All steroids have a characteristic structural component consisting of four fused rings. Chemists identify the rings by capital letters and number the carbon atoms as shown in Figure \(\PageIndex{1a}\). Slight variations in this structure or in the atoms or groups attached to it produce profound differences in biological activity.
Cholesterol
Cholesterol (Figure \(\PageIndex{1b}\)) does not occur in plants, but it is the most abundant steroid in the human body (240 g is a typical amount). Excess cholesterol is believed to be a primary factor in the development of atherosclerosis and heart disease, which are major health problems in the United States today. About half of the body’s cholesterol is interspersed in the lipid bilayer of cell membranes. Much of the rest is converted to cholic acid, which is used in the formation of bile salts. Cholesterol is also a precursor in the synthesis of sex hormones, adrenal hormones, and vitamin D.
Excess cholesterol not metabolized by the body is released from the liver and transported by the blood to the gallbladder. Normally, it stays in solution there until being secreted into the intestine (as a component of bile) to be eliminated. Sometimes, however, cholesterol in the gallbladder precipitates in the form of gallstones (Figure \(2\)). Indeed, the name cholesterol is derived from the Greek chole, meaning “bile,” and stereos, meaning “solid.”
To Your Health: Cholesterol and Heart Disease
Heart disease is the leading cause of death in the United States for both men and women. The Centers for Disease Control and Prevention reported that heart disease claimed 631,636 lives in the United States (26% of all reported deaths) in 2006.
Scientists agree that elevated cholesterol levels in the blood, as well as high blood pressure, obesity, diabetes, and cigarette smoking, are associated with an increased risk of heart disease. A long-term investigation by the National Institutes of Health showed that among men ages 30 to 49, the incidence of heart disease was five times greater for those whose cholesterol levels were above 260 mg/100 mL of serum than for those with cholesterol levels of 200 mg/100 mL or less. The cholesterol content of blood varies considerably with age, diet, and sex. Young adults average about 170 mg of cholesterol per 100 mL of blood, whereas males at age 55 may have cholesterol levels at 250 mg/100 mL or higher because the rate of cholesterol breakdown decreases with age. Females tend to have lower blood cholesterol levels than males.
To understand the link between heart disease and cholesterol levels, it is important to understand how cholesterol and other lipids are transported in the body. Lipids, such as cholesterol, are not soluble in water and therefore cannot be transported in the blood (an aqueous medium) unless they are complexed with proteins that are soluble in water, forming assemblages called lipoproteins. Lipoproteins are classified according to their density, which is dependent on the relative amounts of protein and lipid they contain. Lipids are less dense than proteins, so lipoproteins containing a greater proportion of lipid are less dense than those containing a greater proportion of protein.
Research on cholesterol and its role in heart disease has focused on serum levels of low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs). One of the most fascinating discoveries is that high levels of HDLs reduce a person’s risk of developing heart disease, whereas high levels of LDLs increase that risk. Thus the serum LDL:HDL ratio is a better predictor of heart disease risk than the overall level of serum cholesterol. Persons who, because of hereditary or dietary factors, have high LDL:HDL ratios in their blood have a higher incidence of heart disease.
How do HDLs reduce the risk of developing heart disease? No one knows for sure, but one role of HDLs appears to be the transport of excess cholesterol to the liver, where it can be metabolized. Therefore, HDLs aid in removing cholesterol from blood and from the smooth muscle cells of the arterial wall.
Dietary modifications and increased physical activity can help lower total cholesterol and improve the LDL:HDL ratio. The average American consumes about 600 mg of cholesterol from animal products each day and also synthesizes approximately 1 g of cholesterol each day, mostly in the liver. The amount of cholesterol synthesized is controlled by the cholesterol level in the blood; when the blood cholesterol level exceeds 150 mg/100 mL, the rate of cholesterol biosynthesis is halved. Hence, if cholesterol is present in the diet, a feedback mechanism suppresses its synthesis in the liver. However, the ratio of suppression is not a 1:1 ratio; the reduction in biosynthesis does not equal the amount of cholesterol ingested. Thus, dietary substitutions of unsaturated fat for saturated fat, as well as a reduction in consumption of trans fatty acids, is recommended to help lower serum cholesterol and the risk of heart disease.
Steroid Hormones
Hormones are chemical messengers that are released in one tissue and transported through the circulatory system to one or more other tissues. One group of hormones is known as steroid hormones because these hormones are synthesized from cholesterol, which is also a steroid. There are two main groups of steroid hormones: adrenocortical hormones and sex hormones.
The adrenocortical hormones, such as aldosterone and cortisol (Table \(1\)), are produced by the adrenal gland, which is located adjacent to each kidney. Aldosterone acts on most cells in the body, but it is particularly effective at enhancing the rate of reabsorption of sodium ions in the kidney tubules and increasing the secretion of potassium ions and/or hydrogen ions by the tubules. Because the concentration of sodium ions is the major factor influencing water retention in tissues, aldosterone promotes water retention and reduces urine output. Cortisol regulates several key metabolic reactions (for example, increasing glucose production and mobilizing fatty acids and amino acids). It also inhibits the inflammatory response of tissue to injury or stress. Cortisol and its analogs are therefore used pharmacologically as immunosuppressants after transplant operations and in the treatment of severe skin allergies and autoimmune diseases, such as rheumatoid arthritis.
Table \(1\): Representative Steroid Hormones and Their Physiological Effects
Hormone Effect
regulates salt metabolism; stimulates kidneys to retain sodium and excrete potassium
stimulates the conversion of proteins to carbohydrates
regulates the menstrual cycle; maintains pregnancy
stimulates female sex characteristics; regulates changes during the menstrual cycle
stimulates and maintains male sex characteristics
The sex hormones are a class of steroid hormones secreted by the gonads (ovaries or testes), the placenta, and the adrenal glands. Testosterone and androstenedione are the primary male sex hormones, or androgens, controlling the primary sexual characteristics of males, or the development of the male genital organs and the continuous production of sperm. Androgens are also responsible for the development of secondary male characteristics, such as facial hair, deep voice, and muscle strength. Two kinds of sex hormones are of particular importance in females: progesterone, which prepares the uterus for pregnancy and prevents the further release of eggs from the ovaries during pregnancy, and the estrogens, which are mainly responsible for the development of female secondary sexual characteristics, such as breast development and increased deposition of fat tissue in the breasts, the buttocks, and the thighs. Both males and females produce androgens and estrogens, differing in the amounts of secreted hormones rather than in the presence or absence of one or the other.
Sex hormones, both natural and synthetic, are sometimes used therapeutically. For example, a woman who has had her ovaries removed may be given female hormones to compensate. Some of the earliest chemical compounds employed in cancer chemotherapy were sex hormones. For example, estrogens are one treatment option for prostate cancer because they block the release and activity of testosterone. Testosterone enhances prostate cancer growth. Sex hormones are also administered in preparation for sex-change operations, to promote the development of the proper secondary sexual characteristics. Oral contraceptives are synthetic derivatives of the female sex hormones; they work by preventing ovulation.
Bile Salts
Bile is a yellowish green liquid (pH 7.8–8.6) produced in the liver. The most important constituents of bile are bile salts, which are sodium salts of amidelike combinations of bile acids, such as cholic acid (part (a) of Figure \(3\)) and an amine such as the amino acid glycine (part (b) of Figure \(3\)). They are synthesized from cholesterol in the liver, stored in the gallbladder, and then secreted in bile into the small intestine. In the gallbladder, the composition of bile gradually changes as water is absorbed and the other components become more concentrated.
Because they contain both hydrophobic and hydrophilic groups, bile salts are highly effective detergents and emulsifying agents; they break down large fat globules into smaller ones and keep those smaller globules suspended in the aqueous digestive environment. Enzymes can then hydrolyze fat molecules more efficiently. Thus, the major function of bile salts is to aid in the digestion of dietary lipids.
Surgical removal is often advised for a gallbladder that becomes infected, inflamed, or perforated. This surgery does not seriously affect digestion because bile is still produced by the liver, but the liver’s bile is more dilute and its secretion into the small intestine is not as closely tied to the arrival of food.
Summary
Steroids have a four-fused-ring structure and have a variety of functions. Cholesterol is a steroid found in mammals that is needed for the formation of cell membranes, bile acids, and several hormones. Bile salts are secreted into the small intestine to aid in the digestion of fats. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/23%3A_Lipids/23.06%3A_Steroids.txt |
Learning Objectives
• Identify the distinguishing characteristics of membrane lipids.
• Describe membrane components and how they are arranged.
All living cells are surrounded by a cell membrane. Plant cells (Figure \(\PageIndex{1A}\)) and animal cells (Figure \(\PageIndex{1B}\)) contain a cell nucleus that is also surrounded by a membrane and holds the genetic information for the cell. Everything between the cell membrane and the nuclear membrane—including intracellular fluids and various subcellular components such as the mitochondria and ribosomes—is called the cytoplasm. The membranes of all cells have a fundamentally similar structure, but membrane function varies tremendously from one organism to another and even from one cell to another within a single organism. This diversity arises mainly from the presence of different proteins and lipids in the membrane.
The lipids in cell membranes are highly polar but have dual characteristics: part of the lipid is ionic and therefore dissolves in water, whereas the rest has a hydrocarbon structure and therefore dissolves in nonpolar substances. Often, the ionic part is referred to as hydrophilic, meaning “water loving,” and the nonpolar part as hydrophobic, meaning “water fearing” (repelled by water). Molecules that have both polar and nonpolar parts are called amphipathic. When allowed to float freely in water, lipids spontaneously cluster together in different arrangements: micelles, liposomes, monolayers, or bilayers (Figure \(2\)).
Micelles are aggregations in which the lipids’ hydrocarbon tails—being hydrophobic—are directed toward the center of the assemblage and away from the surrounding water while the hydrophilic heads are directed outward, in contact with the water. Each micelle may contain thousands of lipid molecules. Polar lipids may also form a monolayer, a layer one molecule thick on the surface of the water. The polar heads face into water, and the nonpolar tails stick up into the air. Bilayers are double layers of lipids arranged so that the hydrophobic tails are sandwiched between an inner surface and an outer surface consisting of hydrophilic heads. The hydrophilic heads are in contact with water on either side of the bilayer, whereas the tails, sequestered inside the bilayer, are prevented from having contact with the water. Bilayers like this make up every cell membrane (Figure \(3\)).
In the bilayer interior, the hydrophobic tails (that is, the fatty acid portions of lipid molecules) interact by means of dispersion forces. The interactions are weakened by the presence of unsaturated fatty acids. As a result, the membrane components are free to mill about to some extent, and the membrane is described as fluid.
Membrane Proteins
If membranes were composed only of lipids, very few ions or polar molecules could pass through their hydrophobic “sandwich filling” to enter or leave any cell. However, certain charged and polar species do cross the membrane, aided by proteins that move about in the lipid bilayer. The two major classes of proteins in the cell membrane are integral proteins, which span the hydrophobic interior of the bilayer, and peripheral proteins, which are more loosely associated with the surface of the lipid bilayer (Figure \(3\)). Peripheral proteins may be attached to integral proteins, to the polar head groups of phospholipids, or to both by hydrogen bonding and electrostatic forces.
Transport Across Cell Membranes
Plasma membranes must allow certain substances to enter and leave a cell, and prevent some harmful materials from entering and some essential materials from leaving. In other words, plasma membranes are selectively permeable—they allow some substances to pass through, but not others. If they were to lose this selectivity, the cell would no longer be able to sustain itself, and it would be destroyed. Some cells require larger amounts of specific substances than do other cells; they must have a way of obtaining these materials from extracellular fluids. This may happen passively, as certain materials move back and forth, or the cell may have special mechanisms that facilitate transport. Some materials are so important to a cell that it spends some of its energy, hydrolyzing adenosine triphosphate (ATP), to obtain these materials. Red blood cells use some of their energy doing just that. All cells spend the majority of their energy to maintain an imbalance of sodium and potassium ions between the interior and exterior of the cell.
There are two primary modes of transport across membranes. Passive transport is a naturally occurring phenomenon and does not require the cell to exert any of its energy to accomplish the movement. In passive transport, substances move from an area of higher concentration to an area of lower concentration. Active transport mechanisms require the use of the cell’s energy, usually in the form of adenosine triphosphate (ATP). If a substance must move into the cell against its concentration gradient—that is, if the concentration of the substance inside the cell is greater than its concentration in the extracellular fluid (and vice versa)—the cell must use energy to move the substance. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/23%3A_Lipids/23.07%3A_Cell_Membranes-_Structure_and_Transport.txt |
Learning Objectives
• Describe the process of digestion of lipids and their transport into the bloodstream.
Lipid digestion begins in the upper portion of the small intestine. A hormone secreted in this region stimulates the gallbladder to discharge bile into the duodenum. The principal constituents of bile are the bile salts, which emulsify large, water-insoluble lipid droplets, disrupting some of the hydrophobic interactions holding the lipid molecules together and suspending the resulting smaller globules (micelles) in the aqueous digestive medium. These changes greatly increase the surface area of the lipid particles, allowing for more intimate contact with the lipases and thus rapid digestion of the fats. Another hormone promotes the secretion of pancreatic juice, which contains these enzymes.
The lipases in pancreatic juice catalyze the digestion of triglycerides first to diglycerides and then to 2‑monoglycerides and fatty acids:
The large and hydrophobic long-chain fatty acids and monoacylglycerides are not so easily suspended in the watery intestinal chyme. However, bile salts and lecithin resolve this issue by enclosing them in a micelle, which is a tiny sphere with polar (hydrophilic) ends facing the watery environment and hydrophobic tails turned to the interior, creating a receptive environment for the long-chain fatty acids. The core also includes cholesterol and fat-soluble vitamins. Without micelles, lipids would sit on the surface of chyme and never come in contact with the absorptive surfaces of the epithelial cells. Micelles can easily squeeze between microvilli and get very near the luminal cell surface. At this point, lipid substances exit the micelle and are absorbed via simple diffusion.
The free fatty acids and monoacylglycerides that enter the epithelial cells are reincorporated into triglycerides. The triglycerides are mixed with phospholipids and cholesterol, and surrounded with a protein coat. This new complex, called a chylomicron, (see Figure \(3\)), is a water-soluble lipoprotein.
After being processed by the Golgi apparatus, chylomicrons are released from the cell. Too big to pass through the basement membranes of blood capillaries, chylomicrons instead enter the large pores of lacteals. The lacteals come together to form the lymphatic vessels. The chylomicrons are transported in the lymphatic vessels and empty through the thoracic duct into the circulatory system (Figure \(4\)).
24.02: Lipoproteins for Lipid Transport
Learning Objectives
• Describe the difference in composition and function between the five classes of lipoproteins.
During times of feasting, lipids enter the body's metabolic pathways through the process of digestion. Any excess energy will be stored in adipose tissues. If food has not been eaten for a while, fasting, lipids mobilized from stored adipose tissue and synthesized in the liver can be used for energy. Transportation of nonpolar lipids through the bloodstream requires binding to water-soluble proteins like serum albumin or packaging in lipoproteins. There are five categories of lipoprotein defined by their composition and density, or the ratio of lipid to protein. Chylomicrons, which have the highest lipid to protein ratio, have the lowest-density, are responsible for transporting lipids from digestion into the blood and eventually to the liver. Following is a list of the remaining four lipoproteins and their functions:
• Very low-density lipoproteins (VLDLs) (0.96-1.006 g/cm3) are made in the liver from remnants of chylomicrons and transport triglycerides from the liver to various tissues in the body for energy or storage. As the VLDLs travel through the circulatory system, the lipoprotein lipase strips the VLDL of triglycerides. As triglyceride removal persists, the VLDLs become intermediate-density lipoproteins.
• Intermediate-density lipoproteins (IDLs) (1.007-1.019 g/cm3) transport a variety of fats and cholesterol left from VLDLs from the tissues to the liver where they are transformed into low-density lipoprotein.
• Low-density lipoproteins (LDLs) (1.020-1.062 g/cm3) carry cholesterol and other lipids from the liver to tissue throughout the body. LDLs are comprised of very small amounts of triglycerides, and house over 50 percent cholesterol and cholesterol esters. Once inside the cell, the LDL is taken apart and its cholesterol is released and used for membrane formation, steroid synthesis, or if in excess forming arterial plaques.
• High-density lipoproteins (HDLs) (1.063-1.210 g/cm3) are responsible for carrying cholesterol out of the bloodstream and into the liver, where it is either reused or removed from the body with bile.
Note To Your Health: Good vs. Bad Cholesterol
You are probably familiar with HDL and LDL being referred to as "good cholesterol" and "bad cholesterol," respectively. This is an oversimplification to help the public interpret their blood lipid values, because cholesterol is cholesterol; it's not good or bad. LDL and HDL are lipoproteins, and as a result you can't consume good or bad cholesterol, you consume cholesterol. A more appropriate descriptor for these lipoproteins would be HDL "good cholesterol transporter" and LDL "bad cholesterol transporter."
For healthy total blood cholesterol, the desired range you would want to maintain is under 200 mg/dL. More specifically, when looking at individual lipid profiles, a low amount of LDL and a high amount of HDL prevents excess buildup of cholesterol in the arteries and wards off potential health hazards. An LDL level of less than 100 milligrams per deciliter is ideal while an LDL level above 160 mg/dL would be considered high. In contrast, a low value of HDL is a telltale sign that a person is living with major risks for disease. Values of less than 40 mg/dL for men and 50 mg/dL for women mark a risk factor for developing heart disease. In short, elevated LDL blood lipid profiles indicate an increased risk of heart attack, while elevated HDL blood lipid profiles indicate a reduced risk.The University of Maryland Medical Center reports that omega-3 fatty acids promote lower total cholesterol and lower triglycerides in people with high cholesterol.[1]
It is suggested that people consume omega-3 fatty acids such as alpha-linolenic acid in their diets regularly. Polyunsaturated fatty acids are especially beneficial to consume because they both lower LDL and elevate HDL, thus contributing to healthy blood cholesterol levels. The study also reveals that saturated and trans fatty acids serve as catalysts for the increase of LDL cholesterol. Additionally, trans fatty acids decrease HDL levels, which can impact negatively on total blood cholesterol.
1. Omega-3 fatty acids. University of Maryland Medical Center. http://www.umm.edu/altmed/articles/omega-3-000316.htm. Updated August 5, 2015. Accessed September 28, 2017. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/24%3A_Lipid_Metabolism/24.01%3A_Digestion_of_Triacylglycerols.txt |
Learning Objectives
• Identify the main metabolic pathways of triacylglycerols.
In the body, triacylglycerols are essential for long-term energy storage as well as for insulation and protection. Triacylglycerols can be ingested directly or synthesized from extra dietary protein or carbohydrates. Figure \(1\) is a summary of the metabolic pathways of triacylglycerols that will be discussed in more detail in the following sections.
Dietary Triacylglycerols
Dietary triacylglycerols are transported through the bloodstream in chylomicrons. When a chylomicron encounters the enzyme lipoprotein lipase, triacylglycerols are broken down by hydrolysis into fatty acids and glycerol. These breakdown products then pass through capillary walls to be used for energy by cells or stored in adipose tissue as fat. Liver cells combine the remaining chylomicron remnants with proteins, forming lipoproteins that transport cholesterol in the blood.
Stored Triacylglycerols
When energy is needed, lipases in adipose tissue (fat cells) are activated to hydrolyze stored triacylglycerols into fatty acids and glycerol and release them into the bloodstream. Once the fatty acids reach their destination (muscle or liver cells) they are used to generate acetyl-CoA and eventually ATP.
Glycerol from Triacylglycerols
The hydrolysis of triacylglycerols produce fatty acids, that are used for energy, and glycerol. The glycerol can be used to regenerate triacylglycerol or it can enter glycolysis or gluconeogenesis in the form dihydroxyacetone phosphate (DHAP).
24.05: Oxidation of Fatty Acids
Learning Objectives
• Describe the steps of fatty acid oxidation.
• Calculate the yield of ATP from fatty acid oxidation.
Fatty acids released in the digestion of triglycerides and other lipids are broken down in a series of sequential reactions accompanied by the gradual release of usable energy.
Activation of Fatty Acids
Fatty acid oxidation is initiated in the cytosol. There the fatty acids, which like carbohydrates are relatively inert, must first be activated by conversion to an energy-rich fatty acid derivative of coenzyme A called fatty acyl-coenzyme A (CoA). The activation is catalyzed by acyl-CoA synthetase. For each molecule of fatty acid activated, one molecule of coenzyme A and one molecule of adenosine triphosphate (ATP) are used, equaling a net utilization of the two high-energy bonds in one ATP molecule (which is therefore converted to adenosine monophosphate [AMP] rather than adenosine diphosphate [ADP]):
Transport into the Mitochondrial Matrix
The fatty acyl-CoA cannot cross the membrane by diffusion and therefore must be transported into the mitochondrial matrix using a carrier molecule known as carnitine. At the outer mitochondrial membrane, fatty acyl-CoA is attached to the hydroxyl group of carnitine in a transesterification reaction catalyzed by carnitine acyltransferase I (also called carnitine palmitoyl transferase I, CPTI). The acyl-carnitine derivative is transported into the mitochondrial matrix by facilitated diffusion through a translocase protein (carnitine-acylcarnitine translocase). Once in the matrix, the acyl-carnitine is converted back to the fatty acyl-CoA and carnitine in a reaction catalyzed by carnitine acyltransferase II (also called carnitine palmitoyl transferase II, CPTI).
Oxidation of Fatty Acyl-CoA
Once inside the mitochondria the fatty acyl-CoA can enter into beta-oxidation, which is a four-step cycle that removes carbons from the fatty acid two at a time to form acetyl-CoA (which eventually yields energy in the form of ATP).
STEP 1: First $\beta$-oxidation (dehydrogenation of an alkane)
A fatty acyl-CoA is oxidized by the enzyme, Acyl-CoA dehydrogenase, to yield a trans double bond between the $\alpha$- and $\beta$- carbons (the first and second carbons attached to the carbonyl group). The coenzyme FAD is reduced by accepting two hydrogen atoms and their electrons forming FADH2, which moves into the electron transport chain.
STEP 2: Hydration (addition of water across the double bond)
The trans alkene is then hydrated with the help of Enoyl-CoA hydratase. The hydroxyl group is placed on the $\beta$-carbon.
STEP 3: Second $\beta$-oxidation (oxidation of an alcohol)
The alcohol of the hydroxyacly-CoA is then oxidized to a carbonyl with the help of $\beta$-hydroxyacyl-CoA dehydrogenase and the coenzyme by NAD+.
STEP 4: Thiolytic Cleavage
The enzyme thiolase (acyl-CoA acetyltransferase) cleaves off the acetyl-CoA and attaches a new coenzyme A to the chain to yield an acyl-CoA that is two carbons shorter than before. The acyl-CoA will continue through the cycle and start back at step 1. The cleaved acetyl-CoA can then enter into the the citric acid cycle and electron transport chain because it is already within the mitochondria.
Additional steps are required for unsaturated fatty acids and those with an even number of carbons. The amount of energy released by a fatty acid is determined by the amount of ATP produced from the acetyl-CoA entering the citric acid cycle and from the reduced coenzymes NADH and FADH2 entering the electron transport chain.
ATP Yield from Fatty Acid Oxidation
The amount of ATP obtained from fatty acid oxidation depends on the size of the fatty acid being oxidized. For our purposes here. we’ll study palmitic acid, a saturated fatty acid with 16 carbon atoms, as a typical fatty acid in the human diet. Calculating its energy yield provides a model for determining the ATP yield of all other fatty acids.
The breakdown of 1 mol of palmitic acid requires 1 mol of ATP (for activation) and forms 8 mol of acetyl-CoA. Recall from that each mole of acetyl-CoA metabolized by the citric acid cycle yields 10 mol of ATP. The complete degradation of 1 mol of palmitic acid requires the $\beta$-oxidation reactions to be repeated seven times. Thus, 7 mol of NADH and 7 mol of FADH2 are produced. Reoxidation of these coenzymes through respiration yields 2.5–3 and 1.5–2 mol of ATP, respectively. The energy calculations can be summarized as follows:
1 mol of ATP is split to AMP and 2Pi −2 ATP
8 mol of acetyl-CoA formed (8 × 12) 96 ATP
7 mol of FADH2 formed (7 × 2) 14 ATP
7 mol of NADH formed (7 × 3) 21 ATP
Total 129 ATP
The number of times β-oxidation is repeated for a fatty acid containing n carbon atoms is n/2 – 1 because the final turn yields two acetyl-CoA molecules.
The combustion of 1 mol of palmitic acid releases a considerable amount of energy:
$C_{16}H_{32}O_2 + 23O_2 → 16CO_2 + 16H_2O + 2,340\; kcal \nonumber$
The percentage of this energy that is conserved by the cell in the form of ATP is as follows:
$\mathrm{\dfrac{energy\: conserved}{total\: energy\: available}\times100=\dfrac{(129\: ATP)(7.4\: kcal/ATP)}{2,340\: kcal}\times100=41\%} \nonumber$
The efficiency of fatty acid metabolism is comparable to that of carbohydrate metabolism, which we calculated to be 42%.
The oxidation of fatty acids produces large quantities of water. This water, which sustains migratory birds and animals (such as the camel) for long periods of time. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/24%3A_Lipid_Metabolism/24.03%3A_Triacylglycerol_Metabolism_-_An_Overview.txt |
Learning Objectives
• Describe the structure and function of ketone bodies.
In the liver, most of the acetyl-CoA obtained from fatty acid oxidation is oxidized by the citric acid cycle. However, some of the acetyl-CoA is used to synthesize a group of compounds known as ketone bodies: acetoacetate, β-hydroxybutyrate, and acetone. Two acetyl-CoA molecules combine, in a reversal of the final step of β-oxidation, to produce acetoacetyl-CoA. The acetoacetyl-CoA reacts with another molecule of acetyl-CoA and water to form β-hydroxy-β-methylglutaryl-CoA, which is then cleaved to acetoacetate and acetyl-CoA. Most of the acetoacetate is reduced to β-hydroxybutyrate, while a small amount is decarboxylated to carbon dioxide and acetone.
The acetoacetate and β-hydroxybutyrate synthesized by the liver are released into the blood for use as a metabolic fuel (to be converted back to acetyl-CoA) by other tissues, particularly the kidney and the heart. Thus, during prolonged starvation, ketone bodies provide about 70% of the energy requirements of the brain. Under normal conditions, the kidneys excrete about 20 mg of ketone bodies each day, and the blood levels are maintained at about 1 mg of ketone bodies per 100 mL of blood.
In starvation, diabetes mellitus, and certain other physiological conditions in which cells do not receive sufficient amounts of carbohydrate, the rate of fatty acid oxidation increases to provide energy. This leads to an increase in the concentration of acetyl-CoA. The increased acetyl-CoA cannot be oxidized by the citric acid cycle because of a decrease in the concentration of oxaloacetate, which is diverted to glucose synthesis. In response, the rate of ketone body formation in the liver increases further, to a level much higher than can be used by other tissues. The excess ketone bodies accumulate in the blood and the urine, a condition referred to as ketosis. When the acetone in the blood reaches the lungs, its volatility causes it to be expelled in the breath. The sweet smell of acetone, a characteristic of ketosis, is frequently noticed on the breath of severely diabetic patients.
Because two of the three kinds of ketone bodies are weak acids, their presence in the blood in excessive amounts overwhelms the blood buffers and causes a marked decrease in blood pH (to 6.9 from a normal value of 7.4). This decrease in pH leads to a serious condition known as acidosis. One of the effects of acidosis is a decrease in the ability of hemoglobin to transport oxygen in the blood. In moderate to severe acidosis, breathing becomes labored and very painful. The body also loses fluids and becomes dehydrated as the kidneys attempt to get rid of the acids by eliminating large quantities of water. The lowered oxygen supply and dehydration lead to depression; even mild acidosis leads to lethargy, loss of appetite, and a generally run-down feeling. Untreated patients may go into a coma. At that point, prompt treatment is necessary if the person’s life is to be saved.
24.07: Biosynthesis of Fatty Acids
Learning Objectives
• Describe how fatty acids are synthesized and compare to fatty acid break-down.
Lipogenesis or the synthesis of fatty acids occurs in the cytoplasm and endoplasmic reticulum of the cell and is chemically similar to the beta-oxidation process, but with a couple of key differences. The first of these occur in preparing substrates for the reactions that grow the fatty acid. Transport of acetyl-CoA from the mitochondria occurs when it begins to build up. Two molecules can play roles in moving it to the cytoplasm – citrate and acetylcarnitine. Joining of oxaloacetate with acetyl-CoA in the mitochondrion creates citrate which moves across the membrane, followed by action of citrate lyase in the cytoplasm of the cell to release acetyl-CoA and oxaloacetate. Additionally, when free acetyl-CoA accumulates in the mitochondrion, it may combine with carnitine and be transported out to the cytoplasm.
The fatty acid chain is built two carbons at a time (opposite of beta-oxidation), beginning with two acetyl-CoA molecules. One is converted to malonyl-CoA by carboxylation, catalyzed by the enzyme acetyl-CoA carboxylase (ACC), the only regulatory enzyme of fatty acid synthesis (Figure \(1\)). Next, both acetyl-CoA molecules have their CoA portions replaced by a carrier protein known as ACP (acyl-carrier protein) to form acetyl-ACP and malonyl-ACP. Joining of a fatty acyl-ACP (in this case, acetyl-ACP) with malonyl-ACP splits out the carboxyl that was added and creates the intermediate at the upper right in the figure at left.
From this point forward, the chemical reactions resemble those of beta oxidation reversed. First, the ketone is reduced to a hydroxyl using NADPH. In contrast to the hydroxylated intermediate of beta oxidation, the beta intermediate here is in the D-configuration. Next, water is removed from carbons 2 and 3 of the hydroxyl intermediate to produce a trans doubled bonded molecule. Last, the double bond is hydrogenated to yield a saturated intermediate. The process cycles with the addition of another malonyl-ACP to the growing chain until ultimately an intermediate with 16 carbons is produced (palmitoyl-CoA). At this point, the cytoplasmic synthesis ceases.
Enzymes of Fatty Acid Synthesis
Acetyl-CoA carboxylase, which catalyzes synthesis of malonyl-CoA, is the only regulated enzyme in fatty acid synthesis. Its regulation involves both allosteric control and covalent modification. The enzyme is known to be phosphorylated by both AMP Kinase and Protein Kinase A. Dephosphorylation is stimulated by phosphatases activated by insulin binding. Dephosphorylation activates the enzyme and favors its assembly into a long polymer, while phosphorylation reverses the process.Citrate acts as an allosteric activator and may also favor polymerization. Palmitoyl-CoA allosterically inactivates it.
In animals, six different catalytic activities necessary for the remaining catalytic actions to fully make palmitoyl-CoA are contained in a single complex called Fatty Acid Synthase (Figure \(2\)). These include transacylases for swapping CoA with ACP on acetyl-CoA and malonyl-CoA; a synthase to catalyze addition of the two carbon unit from the three carbon malonyl-ACP in the first step of the elongation process; a reductase to reduce the ketone; a dehydrase to catalyze removal of water, and a reductase to reduce the trans double bond. In bacteria, these activities are found on separate enzymes and are not part of a complex.
Elongation of Fatty Acids
Elongation to make fatty acids longer than 16 carbons occurs in the endoplasmic reticulum and is catalyzed by enzymes described as elongases. Mitochondria also can elongate fatty acids, but their starting materials are generally shorter than 16 carbons long. The mechanisms in both environments are similar to those in the cytoplasm (a malonyl group is used to add two carbons, for example), but CoA is attached to the intermediates, not ACP. Further, whereas cytoplasmic synthesis employs the fatty acid synthase complex (Figure \(2\)), the enzymes in these organelles are separable and not part of a complex.
Desaturation of Fatty Acids
Fatty acids are synthesized in the saturated form and desaturation occurs later. Enzymes called desaturases catalyze the formation of cis double bonds in mature fatty acids. These enzymes are found in the endoplasmic reticulum. Animals are limited in the desaturated fatty acids they can make, due to an inability to catalyze reactions beyond carbons 9 and 10. Thus, humans can make oleic acid, but cannot synthesis linoleic acid or linolenic acid. Consequently, these two must be provided in the diet and are referred to as essential fatty acids. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/24%3A_Lipid_Metabolism/24.06%3A_Ketone_Bodies_and_Ketoacidosis.txt |
Thumbnail: Tryptophan is an α-amino acid that is used in the biosynthesis of proteins. It is an essential amino acid, which means that the human body cannot synthesize it for itself, and need to be provided by the diet.
25: Protein and Amino Acid Metabolism
Learning Objectives
• List the steps of protein digestion.
Protein digestion begins in the stomach (Figure \(1\)), where the action of gastric juice hydrolyzes about 10% of the peptide bonds. Gastric juice is a mixture of water (more than 99%), inorganic ions, hydrochloric acid, and various enzymes and other proteins.
The hydrochloric acid (HCl) in gastric juice is secreted by glands in the stomach lining. The pH of freshly secreted gastric juice is about 1.0, but the contents of the stomach may raise the pH to between 1.5 and 2.5. HCl helps to denature food proteins; that is, it unfolds the protein molecules to expose their chains to more efficient enzyme action. The principal digestive component of gastric juice is pepsinogen, an inactive enzyme produced in cells located in the stomach wall. When food enters the stomach after a period of fasting, pepsinogen is converted to its active form—pepsin—in a series of steps initiated by the drop in pH. Pepsin catalyzes the hydrolysis of peptide linkages within protein molecules. It has a fairly broad specificity but acts preferentially on linkages involving the aromatic amino acids tryptophan, tyrosine, and phenylalanine, as well as methionine and leucine.
Protein digestion is completed in the small intestine. Pancreatic juice, carried from the pancreas via the pancreatic duct, contains inactive enzymes such as trypsinogen and chymotrypsinogen. They are activated in the small intestine as follows (Figure \(2\)): The intestinal mucosal cells secrete the proteolytic enzyme enteropeptidase, which converts trypsinogen to trypsin; trypsin then activates chymotrypsinogen to chymotrypsin (and also completes the activation of trypsinogen). Both of these active enzymes catalyze the hydrolysis of peptide bonds in protein chains. Chymotrypsin preferentially attacks peptide bonds involving the carboxyl groups of the aromatic amino acids (phenylalanine, tryptophan, and tyrosine). Trypsin attacks peptide bonds involving the carboxyl groups of the basic amino acids (lysine and arginine). Pancreatic juice also contains procarboxypeptidase, which is cleaved by trypsin to carboxypeptidase. The latter is an enzyme that catalyzes the hydrolysis of peptide linkages at the free carboxyl end of the peptide chain, resulting in the stepwise liberation of free amino acids from the carboxyl end of the polypeptide.
Aminopeptidases in the intestinal juice remove amino acids from the N-terminal end of peptides and proteins possessing a free amino group. Figure \(3\) illustrates the specificity of these protein-digesting enzymes. The amino acids that are released by protein digestion are absorbed across the intestinal wall into the circulatory system, where they can be used for protein synthesis.
This diagram illustrates where in a peptide the different peptidases we have discussed would catalyze hydrolysis the peptide bonds.
25.02: Amino Acid Metabolism - An Overview
Learning Objectives
• Objective 1
• Objective 2
Once the proteins in the diet have been hydrolyzed, the free amino acids join the non-essential amino acid synthesized in the liver and the amino acids recycled from the body's own proteins, constituting the amino acid pool now available for metabolic processes. Most of the amino acid pool is used for the synthesis of protein and other nitrogen-containing compounds such as DNA bases, neurotransmitters, hormones, etc. Under certain metabolic situations, amino acids can also be used as a source of energy by the body. It is worth mentioning that the human body cannot store amino acids. If the amino acids in the amino acid pool are not used for biological processes, they are degraded and the nitrogen excreted in the urine as urea.
Protein turnover
A balance between protein synthesis and protein degradation is required for good health and normal protein metabolism. Not all the amino acids needed for the biological function of the body need to be incorporated through the diet. When the proteins already present in the metabolism have complete their lifespan, they are also recycled. Protein turnover refers to the replacement of older proteins as they are broken down within the cell. Different types of proteins have very different turnover rates, depending on their particular function. Structural proteins such as college tend to have long half-life periods (in the range of years), while enzymatic protein have a shorter life span to adapt to the metabolic requirements of the body.
Example protein half-lives
Name Half-Life
Collagen 117 years
Eye lens crystallin >70 years
Replication factor C subunit 1 9 hours
40S ribosomal protein S8
3 hours
Ornithine decarboxylase 11 minutes
Once the protein have been hydrolyzed and amino acids recycled, these amino acids are added to the amino acid pool for further utilization.
Note To Your Health: Complete and Incomplete Proteins
Amino acids are classified into three groups namely: essential amino acids and nonessential amino acids
ESSENTIAL AMINO ACIDS
• Essential amino acids cannot be made by the body. As a result, they must come from food.
• The 9 essential amino acids are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
NONESSENTIAL AMINO ACIDS
Nonessential means that our bodies produce an amino acid, even if we do not get it from the food we eat. Nonessential amino acids include: alanine, arginine, asparagine, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine.
Based on this classification of amino acids, proteins can also be classified as either complete or incomplete. Complete proteins provide adequate amounts of all nine essential amino acids. Animal proteins such as meat, fish, milk, and eggs are good examples of complete proteins. Incomplete proteins do not contain adequate amounts of one or more of the essential amino acids. For example, if a protein doesn't provide enough of the essential amino acid leucine it would be considered incomplete. Leucine would be referred to as the limiting amino acid, because there is not enough of it for the protein to be complete. Most plant foods are incomplete proteins, with a few exceptions such as soy. Table \(1\) shows the limiting amino acids in some plant foods.
Table \(1\) Limiting Amino Acids in Some Common Plant Foods.
Even though most plant foods do not contain complete proteins, it does not mean that they should be sworn off as protein sources. It is possible to pair foods containing incomplete proteins with different limiting amino acids to provide adequate amounts of the essential amino acids. These two proteins are called complementary proteins, because they supply the amino acid(s) missing in the other protein. A simple analogy would be that of a 4 piece puzzle. If one person has 2 pieces of a puzzle, and another person has 2 remaining pieces, neither of them have a complete puzzle. But when they are combined, the two individuals create a complete puzzle.
Two examples of complementary proteins are shown below.
Figure \(1\) .Two complementary protein examples
It should be noted that complementary proteins do not need to be consumed at the same time or meal. It is currently recommended that essential amino acids be met on a daily basis, meaning that if a grain is consumed at one meal, a legume could be consumed at a later meal, and the proteins would still complement one another. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/25%3A_Protein_and_Amino_Acid_Metabolism/25.01%3A_Digestion_of_Proteins.txt |
Learning Objectives
• Describe how nitrogen is conserved from amino acids.
The liver is the principal site of amino acid metabolism, but other tissues, such as the kidney, the small intestine, muscles, and adipose tissue, take part. Generally, the first step in the breakdown of amino acids is the separation of the amino group from the carbon skeleton, usually by a transamination reaction. The carbon skeletons resulting from the deaminated amino acids are used to form either glucose or fats, or they are converted to a metabolic intermediate that can be oxidized by the citric acid cycle. The latter alternative, amino acid catabolism, is more likely to occur when glucose levels are low—for example, when a person is fasting or starving.
Transamination
Transamination is an exchange of functional groups between any amino acid (except lysine, proline, and threonine) and an α-keto acid. The amino group is usually transferred to the keto carbon atom of pyruvate, oxaloacetate, or α-ketoglutarate, converting the \(alpha\)-keto acid to alanine, aspartate, or glutamate, respectively. Transamination reactions are catalyzed by specific transaminases (also called aminotransferases), which require pyridoxal phosphate as a coenzyme.
In an \(alpha\)-keto acid, the carbonyl or keto group is located on the carbon atom adjacent to the carboxyl group of the acid.
Oxidative Deamination
In the breakdown of amino acids for energy, the final acceptor of the \(alpha\)-amino group is \(alpha\)-ketoglutarate, forming glutamate. Glutamate can then undergo oxidative deamination, in which it loses its amino group as an ammonium (NH4+) ion and is oxidized back to \(alpha\)-ketoglutarate (ready to accept another amino group):
This reaction occurs primarily in liver mitochondria. Most of the NH4+ ion formed by oxidative deamination of glutamate is converted to urea and excreted in the urine in a series of reactions known as the urea cycle.
The synthesis of glutamate occurs in animal cells by reversing the reaction catalyzed by glutamate dehydrogenase. For this reaction nicotinamide adenine dinucleotide phosphate (NADPH) acts as the reducing agent. The synthesis of glutamate is significant because it is one of the few reactions in animals that can incorporate inorganic nitrogen (NH4+) into an α-keto acid to form an amino acid. The amino group can then be passed on through transamination reactions, to produce other amino acids from the appropriate α-keto acids.
25.04: The Urea Cycle
Learning Objectives
• Describe the urea cycle reactions.
Deamination of amino acids results in the production of ammonia (as well as the ammonium ion, NH4+). Ammonia is an extremely toxic base and its accumulation in the body would quickly be fatal. Animals that live in aquatic environments tend to release ammonia into the water. Terrestrial organisms have evolved other mechanisms to excrete nitrogenous wastes. These animals must detoxify ammonia by converting it into a relatively nontoxic form such as urea or uric acid. Mammals, including humans, produce urea through the urea cycle, whereas reptiles and many terrestrial invertebrates produce uric acid. Animals that secrete urea as the primary nitrogenous waste material are called ureotelic animals.
The Urea Cycle
The urea cycle is the primary mechanism by which mammals convert ammonia to urea. Urea is made in the liver and excreted in urine. The overall chemical reaction by which ammonia is converted to urea is 2 NH3 (ammonia) + CO2 + 3 ATP + H2O → H2N-CO-NH2 (urea) + 2 ADP + 4 Pi + AMP.
The urea cycle utilizes five intermediate steps, catalyzed by five different enzymes, to convert ammonia to urea, as shown in Figure \(1\). The amino acid L-ornithine gets converted into different intermediates before being regenerated at the end of the urea cycle. Hence, the urea cycle is also referred to as the ornithine cycle. The enzyme ornithine transcarbamylase catalyzes a key step in the urea cycle and its deficiency can lead to accumulation of toxic levels of ammonia in the body. The first two reactions occur in the mitochondria and the last three reactions occur in the cytosol. Urea concentration in the blood, called blood urea nitrogen or BUN, is used as an indicator of kidney function.
Note Everyday Connection: Gout
Mammals use uric acid crystals as an antioxidant in their cells. However, too much uric acid tends to form kidney stones and may also cause a painful condition called gout, where uric acid crystals accumulate in the joints, as illustrated in Figure \(2\). Food choices that reduce the amount of nitrogenous bases in the diet help reduce the risk of gout. For example, tea, coffee, and chocolate have purine-like compounds, called xanthines, and should be avoided by people with gout and kidney stones.
Figure \(2\): Gout causes the inflammation visible in this person’s left big toe joint. (credit: "Gonzosft"/Wikimedia Commons) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/25%3A_Protein_and_Amino_Acid_Metabolism/25.03%3A_Amino_Acid_Catabolism_-_The_Amino_Group.txt |
Learning Objectives
• Objective 1
• Objective 2
Any amino acid can be converted into an intermediate of the citric acid cycle. Once the amino group is removed, usually by transamination, the α-keto acid that remains is catabolized by a pathway unique to that acid and consisting of one or more reactions. For example, phenylalanine undergoes a series of six reactions before it splits into fumarate and acetoacetate. Fumarate is an intermediate in the citric acid cycle, while acetoacetate must be converted to acetoacetyl-coenzymeA (CoA) and then to acetyl-CoA before it enters the citric acid cycle.
Those amino acids that can form any of the intermediates of carbohydrate metabolism can subsequently be converted to glucose via a metabolic pathway known as gluconeogenesis. These amino acids are called glucogenic amino acids. Amino acids that are converted to acetoacetyl-CoA or acetyl-CoA, which can be used for the synthesis of ketone bodies but not glucose, are called ketogenic amino acids. Some amino acids fall into both categories. Leucine and lysine are the only amino acids that are exclusively ketogenic. Figure \(1\) summarizes the ultimate fates of the carbon skeletons of the 20 amino acids.
25.06: Biosynthesis of Nonessential Amino Acids
Learning Objectives
• Objective 1
• Objective 2
In humans, only half of the standard amino acids (Glu, Gln, Pro, Asp, Asn, Ala, Gly, Ser, Tyr, Cys) can be synthesized (Figure \(12\) and 13), and are thus classified the nonessential amino acids. Within this group, the first three, glutamate, glutamine, and proline, have a shared anabolic pathway. It begins with glutamate dehydrogenase, which adds ammonia to α-ketoglutarate in the presence of NADPH to form glutamate. This is a key reaction for all amino acid synthesis: glutamate is a nitrogen (amino group) donor for the production of all the other amino acids.
Glutamine synthetase catalyzes the formation of glutamine from glutamate and ammonia. This is an important biochemical reaction for a completely different reason: it is the primary route for ammonia detoxification.
Proline is synthesized from glutamate in a two-step process that begins with the reduction of glutamate to a semialdehyde form that spontaneously cyclizes to D-pyrroline-5-carboxylate. This is reduced by pyrroline carboxylate reductase to proline.
Alanine and Aspartate are the products of glutamate-based transamination of pyruvate and oxaloacetate, respectively.
Asparagine is synthesized through one of two known pathways. In bacteria, an asparagine synthetase combines aspartate and ammonia. However, in mammals, the aspartate gets its amino group from glutamine.
The synthesis of serine begins with the metabolic intermediate 3-phosphoglycerate (glycolysis). Phosphoglycerate dehydrogenase oxidizes it to 3-phosphohydroxypyruvate. An amino group is donated by glutamate in a reaction catalyzed by phosphoserine transaminase, forming 3-phosphoserine, and finally the phosphate is removed by phosphoserine phosphatase to produce serine.
Serine is the immediate precursor to glycine, which is formed by serine hydroxymethyltransferase. This enzyme requires the coenzyme tetrahydrofolate (THF), which is a derivative of vitamin B9 (folic acid).
Serine is also a precursor for cysteine, although the synthesis of cysteine actually begins with the essential amino acid methionine. Methionine is converted to S-adenosylmethionine by methionine adenosyltransferase. This is then converted to S-adenosylhomocysteine by a member of the SAM-dependent methylase family. The sugar is removed by adenosylhomocysteinase, and the resultant homocysteine is connected by cystathionine synthase to the serine molecule to form cystathionine. Finally, cystathionine-g-lyase catalyzes the production of cysteine.
Tyrosine is another amino acid that depends on an essential amino acid as a precursor. In this case, phenylalanine oxygenase reduces phenylalanine to produce the tyrosine.
In general, the synthesis of essential amino acids, usually in microorganisms, is much more complex than for the nonessential amino acids and is best left to a full-fledged biochemistry course.
28: Chemical Messengers- Hormones Neurotransmitters and Drugs
Thumbnail: A ball-and-stick model of testosterone. Testosterone is the primary male sex hormone and an anabolic steroid. Image used with perspective (Public Domain; Ben Mills).
29.03: Blood
Thumbnail: Capillary blood from a bleeding finger. (CC BY 2.0; Crystal (Crystl) from Bloomington, USA).
29: Body Fluids
https://bio.libretexts.org/Bookshelv.../17.5%3A_Blood
29.05: Blood Clotting
https://bio.libretexts.org/TextMaps/...Blood_Clotting
29.08: Urine Composition and Function
Learning Objectives
• Describe how normal urine consists of water, urea, salts and pigment
Urine is a liquid byproduct of the body secreted by the kidneys through a process called urination and excreted through the urethra. The normal chemical composition of urine is mainly water content, but it also includes nitrogenous molecules, such as urea, as well as creatinine and other metabolic waste components. Other substances may be excreted in urine due to injury or infection of the glomeruli of the kidneys, which can alter the ability of the nephron to reabsorb or filter the different components of blood plasma.
Normal Chemical Composition of Urine
Urine is an aqueous solution of greater than 95% water, with a minimum of these remaining constituents, in order of decreasing concentration:
• Urea 9.3 g/L.
• Chloride 1.87 g/L.
• Sodium 1.17 g/L.
• Potassium 0.750 g/L.
• Creatinine 0.670 g/L .
• Other dissolved ions, inorganic and organic compounds (proteins, hormones, metabolites).
Urine is sterile until it reaches the urethra, where epithelial cells lining the urethra are colonized by facultatively anaerobic gram-negative rods and cocci. Urea is essentially a processed form of ammonia that is non-toxic to mammals, unlike ammonia, which can be highly toxic. It is processed from ammonia and carbon dioxide in the liver.
Abnormal Types of Urine
There are several conditions that can cause abnormal components to be excreted in urine or present as abnormal characteristics of urine. They are mostly referred to by the suffix -uria. Some of the more common types of abnormal urine include:
• Proteinuria—Protein content in urine, often due to leaky or damaged glomeruli.
• Oliguria—An abnormally small amount of urine, often due to shock or kidney damage.
• Polyuria—An abnormally large amount of urine, often caused by diabetes.
• Dysuria—Painful or uncomfortable urination, often from urinary tract infections.
• Hematuria—Red blood cells in urine, from infection or injury.
• Glycosuria—Glucose in urine, due to excess plasma glucose in diabetes, beyond the amount able to be reabsorbed in the proximal convoluted tubule.
Summary
Urine is a liquid by-product of the body secreted by the kidneys through a process called urination and excreted through the urethra. It is an aqueous solution of greater than 95% water. Other constituents include urea, chloride, sodium, potassium, creatinine and other dissolved ions, and inorganic and organic compounds. Urea is a non-toxic molecule made of toxic ammonia and carbon dioxide. Any abnormal constituents found in urine are an indication of disease.
Contributors and Attributions
• Boundless (www.boundless.com) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/25%3A_Protein_and_Amino_Acid_Metabolism/25.05%3A_Amino_Acid_Catabolism_-_The_Carbon_Atoms.txt |
If you are reading these words, you are likely starting a chemistry course. Get ready for a fantastic journey through a world of wonder, delight, and knowledge. One of the themes of this book is "chemistry is everywhere," and indeed it is; you would not be alive if it were not for chemistry, because your body is a big chemical machine.
If you do not believe it, do not worry. Every chapter in this book contains examples that will show you how chemistry is, in fact, everywhere. So enjoy the ride, and enjoy chemistry.
1.02: Modern Chemistry
Learning Objective
• Learn the basic terms used to describe matter
The definition of chemistry—the study of the interactions of matter with other matter and with energy—uses some terms that should also be defined. We start the study of chemistry by defining basic terms.
Matter
Matter is anything that has mass and takes up space. A book is matter, a computer is matter, food is matter, and dirt in the ground is matter. Sometimes matter may be difficult to identify. For example, air is matter, but because it is so thin compared to other matter (e.g., a book, a computer, food, and dirt), we sometimes forget that air has mass and takes up space. Things that are not matter include thoughts, ideas, emotions, and hopes.
Example \(1\)
Which of the following is matter and not matter?
1. a hot dog
2. love
3. a tree
Solution
1. A hot dog has mass and takes up space, so it is matter.
2. Love is an emotion, and emotions are not matter.
3. A tree has mass and takes up space, so it is matter.
Exercise \(1\)
Which of the following is matter and not matter?
1. the moon
2. an idea for a new invention
Answer a
The moon is matter.
Answer b
The invention itself may be matter, but the idea for it is not.
To understand matter and how it changes, we need to be able to describe matter. There are two basic ways to describe matter: physical properties and chemical properties.
Physical properties
Physical properties are characteristics that describe matter as it exists. Some physical characteristics of matter are shape, color, size, and temperature. An important physical property is the phase (or state) of matter. The three fundamental phases of matter are solid, liquid, and gas (Figure \(1\)).
Chemical Properties
Chemical properties are characteristics of matter that describe how matter changes form in the presence of other matter. Does a sample of matter burn? Burning is a chemical property. Does it behave violently when put in water? This reaction is a chemical property as well (Figure \(2\)). In the following chapters, we will see how descriptions of physical and chemical properties are important aspects of chemistry.
Physical Change
A physical change occurs when a sample of matter changes one or more of its physical properties. For example, a solid may melt (Figure \(3\)), or alcohol in a thermometer may change volume as the temperature changes. A physical change does not affect the chemical composition of matter.
Chemical Change
Chemical change is the process of demonstrating a chemical property, such as the burning match in Figure \(2\) "Chemical Properties". As the matter in the match burns, its chemical composition changes, and new forms of matter with new physical properties are created. Note that chemical changes are frequently accompanied by physical changes, as the new matter will likely have different physical properties from the original matter.
Example \(2\)
Describe each process as a physical change or a chemical change.
1. Water in the air turns into snow.
2. A person's hair is cut.
3. Bread dough becomes fresh bread in an oven.
Solution
1. Because the water is going from a gas phase to a solid phase, this is a physical change.
2. Your long hair is being shortened. This is a physical change.
3. Because of the oven's temperature, chemical changes are occurring in the bread dough to make fresh bread. These are chemical changes. (In fact, a lot of cooking involves chemical changes.)
Exercise \(2\)
Identify each process as a physical change or a chemical change.
1. A fire is raging in a fireplace.
2. Water is warmed to make a cup of coffee.
Answer a
chemical change
Answer b
physical change
Substance
A sample of matter that has the same physical and chemical properties throughout is called a substance. Sometimes the phrase pure substance is used, but the word pure isn't needed. The definition of the term substance is an example of how chemistry has a specific definition for a word that is used in everyday language with a different, vaguer definition. Here, we will use the term substance with its strict chemical definition.
Chemistry recognizes two different types of substances: elements and compounds.
Element
An element is the simplest type of chemical substance; it cannot be broken down into simpler chemical substances by ordinary chemical means. There are 118 elements known to science, of which 80 are stable. (The other elements are radioactive, a condition we will consider in Chapter 15.) Each element has its own unique set of physical and chemical properties. Examples of elements include iron, carbon, and gold.
Compound
A compound is a combination of more than one element. The physical and chemical properties of a compound are different from the physical and chemical properties of its constituent elements; that is, it behaves as a completely different substance. There are over 50 million compounds known, and more are being discovered daily. Examples of compounds include water, penicillin, and sodium chloride (the chemical name for common table salt).
Mixtures
Physical combinations of more than one substance are called mixtures. Elements and compounds are not the only ways in which matter can be present. We frequently encounter objects that are physical combinations of more than one element or compound—mixtures. There are two types of mixtures.
Heterogeneous Mixture
A heterogeneous mixture is a mixture composed of two or more substances. It is easy to tell, sometimes by the naked eye, that more than one substance is present.
Homogeneous Mixture/ Solution
A homogeneous mixture is a combination of two or more substances that is so intimately mixed, that the mixture behaves as a single substance. Another word for a homogeneous mixture is a solution. Thus, a combination of salt and steel wool is a heterogeneous mixture because it is easy to see which particles of the matter are salt crystals and which are steel wool. On the other hand, if you take salt crystals and dissolve them in water, it is very difficult to tell that you have more than one substance present just by looking—even if you use a powerful microscope. The salt dissolved in water is a homogeneous mixture, or a solution (Figure \(3\)).
Example \(3\)
Identify the following combinations as heterogeneous mixtures or homogenous mixtures.
1. soda water (carbon dioxide is dissolved in water)
2. a mixture of iron metal filings and sulfur powder (both iron and sulfur are elements)
Solution
1. Because carbon dioxide is dissolved in water, we can infer from the behavior of salt crystals dissolved in water that carbon dioxide dissolved in water is (also) a homogeneous mixture.
2. Assuming that the iron and sulfur are simply mixed together, it should be easy to see what is iron and what is sulfur, so this is a heterogeneous mixture.
Exercise \(3\)
1. the human body
2. an amalgam, a combination of some other metals dissolved in a small amount of mercury
Answer a
heterogeneous mixture
Answer b
homogeneous mixture
There are other descriptors that we can use to describe matter, especially elements. We can usually divide elements into metals and nonmetals, and each set shares certain (but not always all) properties.
Metal
A metal is an element that conducts electricity and heat well and is shiny, silvery, solid, ductile, and malleable. At room temperature, metals are solid (although mercury is a well-known exception). A metal is ductile because it can be drawn into thin wires (a property called ductility); and malleable because it can be pounded into thin sheets (a property called malleability).
Nonmetal
A non-metal is an element that is brittle when solid, and does not conduct electricity or heat very well. Non-metals cannot be made into thin sheets or wires (Figure \(4\)). Nonmetals also exist in a variety of phases and colors at room temperature.
Semi-metals
Some elements have properties of both metals and nonmetals and are called semi-metals (or metalloids). We will see later how these descriptions can be assigned rather easily to various elements.
Describing Matter Flowchart
"Describing Matter" is a flowchart of the relationships among the different ways of describing matter.
Example \(1\): Chemistry is Everywhere: In the Morning
Most people have a morning ritual, a process that they go through every morning to get ready for the day. Chemistry appears in many of these activities.
• If you take a shower or bath in the morning, you probably use soap, shampoo, or both. These items contain chemicals that interact with the oil and dirt on your body and hair to remove them and wash them away. Many of these products also contain chemicals that make you smell good; they are called fragrances.
• When you brush your teeth in the morning, you usually use toothpaste, a form of soap, to clean your teeth. Toothpastes typically contain tiny, hard particles called abrasives that physically scrub your teeth. Many toothpastes also contain fluoride, a substance that chemically interacts with the surface of the teeth to help prevent cavities.
• Perhaps you take vitamins, supplements, or medicines every morning. Vitamins and other supplements contain chemicals your body needs in small amounts to function properly. Medicines are chemicals that help combat diseases and promote health.
• Perhaps you make some fried eggs for breakfast. Frying eggs involves heating them enough so that a chemical reaction occurs to cook the eggs.
• After you eat, the food in your stomach is chemically reacted so that the body (mostly the intestines) can absorb food, water, and other nutrients.
• If you drive or take the bus to school or work, you are using a vehicle that probably burns gasoline, a material that burns fairly easily and provides energy to power the vehicle. Recall that burning is a chemical change.
These are just a few examples of how chemistry impacts your everyday life. And we haven't even made it to lunch yet!
Key Takeaways
• Chemistry is the study of matter and its interactions with other matter and energy.
• Matter is anything that has mass and takes up space.
• Matter can be described in terms of physical properties and chemical properties.
• Physical properties and chemical properties of matter can change.
• Matter is composed of elements and compounds.
• Combinations of different substances are called mixtures.
• Elements can be described as metals, nonmetals, and semi-metals. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/01%3A_Introduction_to_Chemistry/1.01%3A_Evolution_of_Chemistry.txt |
Learning Objective
• Learn what science is and how it works.
Chemistry is a branch of science. Although science itself is difficult to define exactly, the following definition can serve as a starting point. Science is the process of knowing about the natural universe through observation and experiment. Science is not the only process of knowing (e.g., the ancient Greeks simply sat and thought), but it has evolved over more than 350 years into the best process that humanity has devised, to date, to learn about the universe around us.
The process of science is usually stated as the scientific method, which is rather naively described as follows:
1. state a hypothesis,
2. test the hypothesis, and
3. refine the hypothesis
In actuality, the process is not that simple. (For example, a scientist does not go into their lab every day and exclaim, "I am going to state a hypothesis today and spend the day testing it!") The process is not that simple because science and scientists have a body of knowledge that has already been identified as coming from the highest level of understanding, and most scientists build from that body of knowledge.
An educated guess about how the natural universe works is called a hypothesis. A scientist who is familiar with how part of the natural universe works—say, a chemist—is interested in furthering that knowledge. That person makes a reasonable guess—a hypothesis—that is designed to see if the universe works in a new way as well. Here's an example of a hypothesis: "if I mix one part of hydrogen with one part of oxygen, I can make a substance that contains both elements."
For a hypothesis to be termed a scientific hypothesis, it has to be something that can be supported or refuted through carefully crafted experimentation or observation.
Most good hypotheses are grounded in previously understood knowledge and represent a testable extension of that knowledge. The scientist then devises ways to test if that hypothesis is correct or not. That is, the scientist plans experiments. Experiments are tests of the natural universe to see if a guess (hypothesis) is correct. An experiment to test our previous hypothesis would be to actually mix hydrogen and oxygen and see what happens. Most experiments include observations of small, well-defined parts of the natural universe designed to see results of the experiments.
A Scientific Hypothesis
A hypothesis is often written in the form of an if/then statement that gives a possibility (if) and explains what may happen because of the possibility (then). For example, if eating elemental sulfur repels ticks, then someone that is eating sulfur every day will not get ticks.
Why do we have to do experiments? Why do we have to test? Because the natural universe is not always so obvious, experiments are necessary. For example, it is fairly obvious that if you drop an object from a height, it will fall. Several hundred years ago (coincidentally, near the inception of modern science), the concept of gravity explained that test. However, is it obvious that the entire natural universe is composed of only about 115 fundamental chemical building blocks called elements? This wouldn't seem true if you looked at the world around you and saw all the different forms matter can take. In fact, the concept of the element is only about 200 years old, and the last naturally occurring element was identified about 80 years ago. It took decades of tests and millions of experiments to establish what the elements actually are. These are just two examples; a myriad of such examples exists in chemistry and science in general.
When enough evidence has been collected to establish a general principle of how the natural universe works, the evidence is summarized in a theory. A theory is a general statement that explains a large number of observations. "All matter is composed of atoms" is a general statement, a theory, that explains many observations in chemistry. A theory is a very powerful statement in science. There are many statements referred to as "the theory of _______" or the "______ theory" in science (where the blanks represent a word or concept). When written in this way, theories indicate that science has an overwhelming amount of evidence of its correctness. We will see several theories in the course of this text.
A specific statement that is thought to never be violated by the entire natural universe is called a law. A scientific law is the highest understanding of the natural universe that science has and is thought to be inviolate. The fact that all matter attracts all other matter—the law of gravitation—is one such law. Note that the terms theory and law used in science have slightly different meanings from those in common usage; where theory is often used to mean hypothesis ("I have a theory…"), and a law is an arbitrary limitation that can be broken but with potential consequences (such as speed limits). Here again, science uses these terms differently, and it is important to apply their proper definitions when you use these words in science. (Figure \(1\))
There is an additional phrase in our definition of science: "the natural universe." Science is concerned only with the natural universe. What is the natural universe? It's anything that occurs around us, well, naturally. Stars, planets, the appearance of life on earth; as well as how animals, plants, and other matter function are all part of the natural universe. Science is concerned with that—and only that.
Of course, there are other things that concern us. For example, is the English language part of science? Most of us can easily answer no; English is not science. English is certainly worth knowing (at least for people in predominantly English-speaking countries), but why isn't it science? English, or any human language, is not science because ultimately it is contrived; it is made up. Think of it: the word spelled b-l-u-e represents a certain color, and we all agree what color that is. But what if we used the word h-a-r-d to describe that color? (Figure \(2\)) That would be fine—as long as everyone agreed. Anyone who has learned a second language must initially wonder why a certain word is used to describe a certain concept; ultimately, the speakers of that language agreed that a particular word would represent a particular concept. It was contrived.
That doesn't mean language isn't worth knowing. It is very important in society. But it's not science. Science deals only with what occurs naturally.
Example \(1\): Identifying Science
Which of the following fields would be considered science?
1. geology, the study of the earth
2. ethics, the study of morality
3. political science, the study of governance
4. biology, the study of living organisms
Solution
1. Because the earth is a natural object, the study of it is indeed considered part of science.
2. Ethics is a branch of philosophy that deals with right and wrong. Although these are useful concepts, they are not science.
3. There are many forms of government, but all are created by humans. Despite the fact that the word science appears in its name, political science is not true science.
4. Living organisms are part of the natural universe, so the study of them is part of science.
Exercise \(1\)
Which is part of science, and which is not?
1. dynamics, the study of systems that change over time
2. aesthetics, the concept of beauty
Answer A
science
Answer B
not science
The field of science has gotten so big that it is common to separate it into more specific fields. First, there is mathematics, the language of science. All scientific fields use mathematics to express themselves—some more than others. Physics and astronomy are scientific fields concerned with the fundamental interactions between matter and energy. Chemistry, as defined previously, is the study of the interactions of matter with other matter and with energy. Biology is the study of living organisms, while geology is the study of the earth. Other sciences can be named as well. Understand that these fields are not always completely separate; the boundaries between scientific fields are not always readily apparent. A scientist may be labeled a biochemist if he or she studies the chemistry of biological organisms.
Finally, understand that science can be either qualitative or quantitative. Qualitative implies a description of the quality of an object. For example, physical properties are generally qualitative descriptions: sulfur is yellow, your math book is heavy, or that statue is pretty. A quantitative description represents the specific amount of something; it means knowing how much of something is present, usually by counting or measuring it. Some quantitative descriptions include: 25 students in a class, 650 pages in a book, or a velocity of 66 miles per hour. Quantitative expressions are very important in science; they are also very important in chemistry.
Example \(2\): qualitative vs. quantitative Descriptions
Identify each statement as either a qualitative description or a quantitative description.
1. Gold metal is yellow.
2. A ream of paper has 500 sheets in it.
3. The weather outside is snowy.
4. The temperature outside is 24 degrees Fahrenheit.
Solution
1. Because we are describing a physical property of gold, this statement is qualitative.
2. This statement mentions a specific amount, so it is quantitative.
3. The word snowy is a description of how the day is; therefore, it is a qualitative statement.
4. In this case, the weather is described with a specific quantity—the temperature. Therefore, it is quantitative.
Exercise \(2\)
Are these qualitative or quantitative statements?
1. Roses are red, and violets are blue.
2. Four score and seven years ago….
Answer A
qualitative
Answer B
quantitative
Food and Drink Application: Carbonated Beverages
Some of the simple chemical principles discussed in this chapter can be illustrated with carbonated beverages: sodas, beer, and sparkling wines. Each product is produced in a different way, but they all have one thing in common: they are solutions of carbon dioxide dissolved in water.
Carbon dioxide is a compound composed of carbon and oxygen. Under normal conditions, it is a gas. If you cool it down enough, it becomes a solid known as dry ice. Carbon dioxide is an important compound in the cycle of life on earth.
Even though it is a gas, carbon dioxide can dissolve in water, just like sugar or salt can dissolve in water. When that occurs, we have a homogeneous mixture, or a solution, of carbon dioxide in water. However, very little carbon dioxide can dissolve in water. If the atmosphere were pure carbon dioxide, the solution would be only about 0.07% carbon dioxide. In reality, the air is only about 0.03% carbon dioxide, so the amount of carbon dioxide in water is reduced proportionally.
However, when soda and beer are made, manufacturers do two important things: they use pure carbon dioxide gas, and they use it at very high pressures. With higher pressures, more carbon dioxide can dissolve in the water. When the soda or beer container is sealed, the high pressure of carbon dioxide gas remains inside the package. (Of course, there are more ingredients in soda and beer besides carbon dioxide and water.)
When you open a container of soda or beer, you hear a distinctive hiss as the excess carbon dioxide gas escapes. But something else happens as well. The carbon dioxide in the solution comes out of solution as a bunch of tiny bubbles. These bubbles impart a pleasing sensation in the mouth, so much so that the soda industry sold over 225 billion servings of soda in the United States alone in 2009.
Some sparkling wines are made in the same way—by forcing carbon dioxide into regular wine. Some sparkling wines (including champagne) are made by sealing a bottle of wine with some yeast in it. The yeast ferments, a process by which the yeast converts sugars into energy and excess carbon dioxide. The carbon dioxide produced by the yeast dissolves in the wine. Then, when the champagne bottle is opened, the increased pressure of carbon dioxide is released, and the drink bubbles just like an expensive glass of soda.
Key Takeaways
• Science is a process of knowing about the natural universe through observation and experiment.
• Scientists go through a rigorous process to determine new knowledge about the universe; this process is generally referred to as the scientific method.
• Science is broken down into various fields, of which chemistry is one.
• Science, including chemistry, is both qualitative and quantitative. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/01%3A_Introduction_to_Chemistry/1.03%3A_Learning_Chemistry.txt |
Learning Objective
• Learn the units that go with various quantities
• Express units using their abbreviations
• Make new units by combining numerical prefixes with units
A number indicates "how much," but the unit indicates "of what." The "of what" is important when communicating a quantity. For example, if you were to ask a friend how close you are to Lake Erie and your friend says "six," then your friend isn't giving you complete information. Six what? Six miles? Six inches? Six city blocks? The actual distance to the lake depends on what units you use.
Chemistry, like most sciences, uses the International System of Units, or SI for short. (The letters SI stand for the French "le Système International d'unités.") SI specifies certain units for various types of quantities, based on seven fundamental units. We will use most of the fundamental units in chemistry. Initially, we will deal with three fundamental units. The meter (m) is the SI unit of length. It is a little longer than a yard (Figure \(1\)). The SI unit of mass is the kilogram (kg), which is about 2.2 pounds (lb). The SI unit of time is the second (s).
To express a quantity, you need to combine a number with a unit. If you have a length that is 2.4 m, then you express that length as simply 2.4 m. A time of 15,000 s can be expressed as 1.5 × 104 s in scientific notation.
Sometimes, a given unit is not an appropriate size to easily express a quantity. For example, the width of a human hair is very small, and it doesn't make much sense to express it in meters. SI also defines a series of numerical prefixes, referring to multiples or fractions of a fundamental unit, to make a unit more conveniently sized for a specific quantity. Table \(1\) lists the prefixes, their abbreviations, and their multiplicative factors. Some of the prefixes, such as kilo-, mega-, and giga-, represent more than one of the fundamental unit, while other prefixes, such as centi-, milli-, and micro-, represent fractions of the original unit. Note, too, that once again we are using powers of 10. Each prefix is a multiple of or fraction of a power of 10.
Table \(1\): Multiplicative Prefixes for SI Units
Prefix Abbreviation Multiplicative Amount Power of Ten
giga- G 1,000,000,000 × 109 ×
mega- M 1,000,000 × 106 ×
kilo- k 1,000 × 103 ×
deci- d 1/10 × ×
centi- c 1/100 × ×
milli- m 1/1,000 × ×
micro- μ* 1/1,000,000 × ×
nano- n 1/1,000,000,000 × ×
pico- p 1/1,000,000,000,000 × x
* The letter μ is the Greek letter lowercase equivalent to an m and is called "mu" (pronounced "myoo").
To use the fractions to generate new units, simply combine the prefix with the unit itself; the abbreviation for the new unit is the combination of the abbreviation for the prefix and the abbreviation of the unit. For example, the kilometer (km) is 1,000 × meter, or 1,000 m. Thus, 5 kilometers (5 km) is equal to 5,000 m. Similarly, a millisecond (ms) is 1/1,000 × second, or one-thousandth of a second. Thus, 25 ms is 25 thousandths of a second. You will need to become proficient in combining prefixes and units. (You may recognize that one of our fundamental units, the kilogram, automatically has a prefix-unit combination. The word kilogram means 1,000 g.)
In addition to the fundamental units, SI also allows for derived units based on a fundamental unit or units. There are many derived units used in science. For example, the derived unit for area comes from the idea that area is defined as width times height. Because both width and height are lengths, they both have the fundamental unit of meter, so the unit of area is meter × meter, or meter2 (m2). This is sometimes spoken as "square meters." A unit with a prefix can also be used to derive a unit for area, so we can also have cm2, mm2, or km2 as acceptable units for area.
Volume is defined as length times width times height, so it has units of meter × meter × meter, or meter3 (m3)—sometimes spoken as "cubic meters." The cubic meter is a rather large unit, however, so another unit is defined that is somewhat more manageable: the liter (L). A liter is 1/1,000th of a cubic meter and is a little more than 1 quart in volume (Figure \(2\)). Prefixes can also be used with the liter unit, so we can speak of milliliters (1/1,000th of a liter; mL) and kiloliters (1,000 L; kL).
Another definition of a liter is one-tenth of a meter cubed. Because one-tenth of a meter is 10 cm, then a liter is equal to 1,000 cm3 (Figure \(3\)). Because 1 L equals 1,000 mL, we conclude that 1 mL equals 1 cm3; thus, these units are interchangeable.
Units are not only multiplied together—they can also be divided. For example, if you are traveling at one meter for every second of time elapsed, your velocity is 1 meter per second, or 1 m/s. The word per implies division, so velocity is determined by dividing a distance quantity by a time quantity. Other units for velocity include kilometers per hour (km/h) or even micrometers per nanosecond (μm/ns). Later, we will see other derived units that can be expressed as fractions.
Example \(1\)
1. A human hair has a diameter of about 6.0 × 10−5 m. Suggest an appropriate unit for this measurement and write the diameter of a human hair in terms of that unit.
2. What is the velocity of a car if it goes 25 m in 5.0 s?
Solution
1. The scientific notation 10−5 is close to 10−6, which defines the micro- prefix. Let us use micrometers as the unit for hair diameter. The number 6.0 × 10−5 can be written as 60 × 10−6, and a micrometer is 10−6 m, so the diameter of a human hair is about 60 μm.
2. If velocity is defined as a distance quantity divided by a time quantity, then velocity is 25 meters/5.0 seconds. Dividing the numbers gives us 25/5.0 = 5.0, and dividing the units gives us meters/second, or m/s. The velocity is 5.0 m/s.
Exercise \(1\)
1. Express the volume of an Olympic-sized swimming pool, 2,500,000 L, in more appropriate units.
2. A common garden snail moves about 6.1 m in 30 min. What is its velocity in meters per minute (m/min)?
Answer a
• 2.5 ML
Answer b
• 0.203 m/min
Key Takeaways
• Numbers tell "how much," and units tell "of what."
• Chemistry uses a set of fundamental units and derived units from SI units.
• Chemistry uses a set of prefixes that represent multiples or fractions of units.
• Units can be multiplied and divided to generate new units for quantities. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/02%3A_PSS-_Scientific_Measurements/2.01%3A_PSS.1-_Uncertainty_in_Measurements.txt |
Learning Objectives
• Identify the number of significant figures in a reported value.
The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain or estimated digit. In the ruler illustration below, the bottom ruler gave a length with 2 significant figures, while the top ruler gave a length with 3 significant figures. In a correctly reported measurement, the final digit is significant but not certain. Insignificant digits are not reported. With either ruler, it would not be possible to report the length at $2.553 \: \text{cm}$ as there is no possible way that the thousandths digit could be estimated. The 3 is not significant and would not be reported.
Measurement Uncertainty
Some error or uncertainty always exists in any measurement. The amount of uncertainty depends both upon the skill of the measurer and upon the quality of the measuring tool. While some balances are capable of measuring masses only to the nearest $0.1 \: \text{g}$, other highly sensitive balances are capable of measuring to the nearest $0.001 \: \text{g}$ or even better. Many measuring tools such as rulers and graduated cylinders have small lines which need to be carefully read in order to make a measurement. Figure $1$ shows two rulers making the same measurement of an object (indicated by the blue arrow).
With either ruler, it is clear that the length of the object is between $2$ and $3 \: \text{cm}$. The bottom ruler contains no millimeter markings. With that ruler, the tenths digit can be estimated and the length may be reported as $2.5 \: \text{cm}$. However, another person may judge that the measurement is $2.4 \: \text{cm}$ or perhaps $2.6 \: \text{cm}$. While the 2 is known for certain, the value of the tenths digit is uncertain.
The top ruler contains marks for tenths of a centimeter (millimeters). Now the same object may be measured as $2.55 \: \text{cm}$. The measurer is capable of estimating the hundredths digit because he can be certain that the tenths digit is a 5. Again, another measurer may report the length to be $2.54 \: \text{cm}$ or $2.56 \: \text{cm}$. In this case, there are two certain digits (the 2 and the 5), with the hundredths digit being uncertain. Clearly, the top ruler is a superior ruler for measuring lengths as precisely as possible.
Example $1$: Reporting Measurements to the Proper Number of Significant Figures
Use each diagram to report a measurement to the proper number of significant figures.
a.
b.
Solutions
Solutions to Example 2.3.1
Explanation Answer
a. The arrow is between 4.0 and 5.0, so the measurement is at least 4.0. The arrow is between the third and fourth small tick marks, so it’s at least 0.3. We will have to estimate the last place. It looks like about one-third of the way across the space, so let us estimate the hundredths place as 3. The symbol psi stands for “pounds per square inch” and is a unit of pressure, like air in a tire. The measurement is reported to three significant figures. 4.33 psi
b. The rectangle is at least 1.0 cm wide but certainly not 2.0 cm wide, so the first significant digit is 1. The rectangle’s width is past the second tick mark but not the third; if each tick mark represents 0.1, then the rectangle is at least 0.2 in the next significant digit. We have to estimate the next place because there are no markings to guide us. It appears to be about halfway between 0.2 and 0.3, so we will estimate the next place to be a 5. Thus, the measured width of the rectangle is 1.25 cm. The measurement is reported to three significant figures. 1.25 cm
Exercise $1$
What would be the reported width of this rectangle?
Answer
1.25 cm
When you look at a reported measurement, it is necessary to be able to count the number of significant figures. The table below details the rules for determining the number of significant figures in a reported measurement. For the examples in the table, assume that the quantities are correctly reported values of a measured quantity.
Table $1$: Significant Figure Rules
Rule Examples
1. All nonzero digits in a measurement are significant.
• 237 has three significant figures.
• 1.897 has four significant figures.
2. Zeros that appear between other nonzero digits (middle zeros) are always significant.
• 39,004 has five significant figures.
• 5.02 has three significant figures.
3. Zeros that appear in front of all of the nonzero digits are called leading zeros. Leading zeros are never significant.
• 0.008 has one significant figure.
• 0.000416 has three significant figures.
4. Zeros that appear after all nonzero digits are called trailing zeros. A number with trailing zeros that lacks a decimal point may or may not be significant. Use scientific notation to indicate the appropriate number of significant figures.
• 1400 is ambiguous.
• $1.4 \times 10^3$ has two significant figures.
• $1.40 \times 10^3$ three significant figures.
• $1.400 \times 10^3$ has four significant figures.
5. Trailing zeros in a number with a decimal point are significant. This is true whether the zeros occur before or after the decimal point.
• 620.0 has four significant figures.
• 19.000 has five significant figures.
Exact Numbers
Integers obtained either by counting objects or from definitions are exact numbers, which are considered to have infinitely many significant figures. If we have counted four objects, for example, then the number 4 has an infinite number of significant figures (i.e., it represents 4.000…). Similarly, 1 foot (ft) is defined to contain 12 inches (in), so the number 12 in the following equation has infinitely many significant figures:
Example $2$
Give the number of significant figures in each. Identify the rule for each.
1. 5.87
2. 0.031
3. 52.90
4. 00.2001
5. 500
6. 6 atoms
Solution
Solution to Example 2.3.2
Explanation Answer
a All three numbers are significant (rule 1). 5.87, three significant figures
b The leading zeros are not significant (rule 3). The 3 and the 1 are significant (rule 1). 0.031, two significant figures
c The 5, the 2 and the 9 are significant (rule 1). The trailing zero is also significant (rule 5). 52.90, four significant figures
d The leading zeros are not significant (rule 3). The 2 and the 1 are significant (rule 1) and the middle zeros are also significant (rule 2). 00.2001, four significant figures
e The number is ambiguous. It could have one, two or three significant figures. 500, ambiguous
f The 6 is a counting number. A counting number is an exact number. 6, infinite
Exercise $2$
Give the number of significant figures in each.
1. 36.7 m
2. 0.006606 s
3. 2,002 kg
4. 306,490,000 people
5. 3,800 g
Answer a
three significant figures
Answer b
four significant figures
Answer c
four significant figures
Answer d
infinite (exact number)
Answer e
Ambiguous, could be two, three or four significant figures.
Accuracy and Precision
Measurements may be accurate, meaning that the measured value is the same as the true value; they may be precise, meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise. The video below demonstrates the concepts of accuracy and precision.
Example $3$
The following archery targets show marks that represent the results of four sets of measurements.
Which target shows
1. a precise, but inaccurate set of measurements?
2. a set of measurements that is both precise and accurate?
3. a set of measurements that is neither precise nor accurate?
Solution
1. Set a is precise, but inaccurate.
2. Set c is both precise and accurate.
3. Set d is neither precise nor accurate.
Summary
Uncertainty exists in all measurements. The degree of uncertainty is affected in part by the quality of the measuring tool. Significant figures give an indication of the certainty of a measurement. Rules allow decisions to be made about how many digits to use in any given situation. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/02%3A_PSS-_Scientific_Measurements/2.02%3A_PSS.2-_Significant_Digits.txt |
Learning Objectives
• Give an example of a measurement whose number of significant digits is clearly too great, and explain why.
• State the purpose of rounding off, and describe the information that must be known to do it properly.
• Round off a number to a specified number of significant digits.
• Explain how to round off a number whose second-most-significant digit is 9.
• Carry out a simple calculation that involves two or more observed quantities, and express the result in the appropriate number of significant figures.
The numerical values we deal with in science (and in many other aspects of life) represent measurements whose values are never known exactly. Our pocket-calculators or computers don't know this; they treat the numbers we punch into them as "pure" mathematical entities, with the result that the operations of arithmetic frequently yield answers that are physically ridiculous even though mathematically correct. The purpose of this unit is to help you understand why this happens, and to show you what to do about it.
Digits: Significant and otherwise
Consider the two statements shown below:
• "The population of our city is 157,872."
• "The number of registered voters as of Jan 1 was 27,833.
Which of these would you be justified in dismissing immediately? Certainly not the second one, because it probably comes from a database which contains one record for each voter, so the number is found simply by counting the number of records. The first statement cannot possibly be correct. Even if a city’s population could be defined in a precise way (Permanent residents? Warm bodies?), how can we account for the minute-by minute changes that occur as people are born and die, or move in and move away?
What is the difference between the two population numbers stated above? The first one expresses a quantity that cannot be known exactly — that is, it carries with it a degree of uncertainty. It is quite possible that the last census yielded precisely 157,872 records, and that this might be the “population of the city” for legal purposes, but it is surely not the “true” population. To better reflect this fact, one might list the population (in an atlas, for example) as 157,900 or even 158,000. These two quantities have been rounded off to four and three significant figures, respectively, and the have the following meanings:
• 157900 (the significant digits are underlined here) implies that the population is believed to be within the range of about 157850 to about 157950. In other words, the population is 157900±50. The “plus-or-minus 50” appended to this number means that we consider the absolute uncertainty of the population measurement to be 50 – (–50) = 100. We can also say that the relative uncertainty is 100/157900, which we can also express as 1 part in 1579, or 1/1579 = 0.000633, or about 0.06 percent.
• The value 158000 implies that the population is likely between about 157500 and 158500, or 158000±500. The absolute uncertainty of 1000 translates into a relative uncertainty of 1000/158000 or 1 part in 158, or about 0.6 percent.
Which of these two values we would report as “the population” will depend on the degree of confidence we have in the original census figure; if the census was completed last week, we might round to four significant digits, but if it was a year or so ago, rounding to three places might be a more prudent choice. In a case such as this, there is no really objective way of choosing between the two alternatives.
This illustrates an important point: the concept of significant digits has less to do with mathematics than with our confidence in a measurement. This confidence can often be expressed numerically (for example, the height of a liquid in a measuring tube can be read to ±0.05 cm), but when it cannot, as in our population example, we must depend on our personal experience and judgment.
So, what is a significant digit? According to the usual definition, it is all the numerals in a measured quantity (counting from the left) whose values are considered as known exactly, plus one more whose value could be one more or one less:
• In “157900” (four significant digits), the left most three digits are known exactly, but the fourth digit, “9” could well be “8” if the “true value” is within the implied range of 157850 to 157950.
• In “158000” (three significant digits), the left most two digits are known exactly, while the third digit could be either “7” or “8” if the true value is within the implied range of 157500 to 158500.
Although rounding off always leads to the loss of numeric information, what we are getting rid of can be considered to be “numeric noise” that does not contribute to the quality of the measurement. The purpose in rounding off is to avoid expressing a value to a greater degree of precision than is consistent with the uncertainty in the measurement.
Implied Uncertainty
If you know that a balance is accurate to within 0.1 mg, say, then the uncertainty in any measurement of mass carried out on this balance will be ±0.1 mg. Suppose, however, that you are simply told that an object has a length of 0.42 cm, with no indication of its precision. In this case, all you have to go on is the number of digits contained in the data. Thus the quantity “0.42 cm” is specified to 0.01 unit in 0 42, or one part in 42 . The implied relative uncertainty in this figure is 1/42, or about 2%. The precision of any numeric answer calculated from this value is therefore limited to about the same amount.
Rounding Error
It is important to understand that the number of significant digits in a value provides only a rough indication of its precision, and that information is lost when rounding off occurs. Suppose, for example, that we measure the weight of an object as 3.28 g on a balance believed to be accurate to within ±0.05 gram. The resulting value of 3.28±.05 gram tells us that the true weight of the object could be anywhere between 3.23 g and 3.33 g. The absolute uncertainty here is 0.1 g (±0.05 g), and the relative uncertainty is 1 part in 32.8, or about 3 percent.
How many significant digits should there be in the reported measurement? Since only the left most “3” in “3.28” is certain, you would probably elect to round the value to 3.3 g. So far, so good. But what is someone else supposed to make of this figure when they see it in your report? The value “3.3 g” suggests an implied uncertainty of 3.3±0.05 g, meaning that the true value is likely between 3.25 g and 3.35 g. This range is 0.02 g below that associated with the original measurement, and so rounding off has introduced a bias of this amount into the result. Since this is less than half of the ±0.05 g uncertainty in the weighing, it is not a very serious matter in itself. However, if several values that were rounded in this way are combined in a calculation, the rounding-off errors could become significant.
Rules for Rounding
The standard rules for rounding off are well known. Before we set them out, let us agree on what to call the various components of a numeric value.
• The most significant digit is the left most digit (not counting any leading zeros which function only as placeholders and are never significant digits.)
• If you are rounding off to n significant digits, then the least significant digit is the nth digit from the most significant digit. The least significant digit can be a zero.
• The first non-significant digit is the n+1th digit.
Rounding-off rules
• If the first non-significant digit is less than 5, then the least significant digit remains unchanged.
• If the first non-significant digit is greater than 5, the least significant digit is incremented by 1.
• If the first non-significant digit is 5, the least significant digit can either be incremented or left unchanged (see below!)
• All non-significant digits are removed.
Fantasies about fives
Students are sometimes told to increment the least significant digit by 1 if it is odd, and to leave it unchanged if it is even. One wonders if this reflects some idea that even numbers are somehow “better” than odd ones! (The ancient superstition is just the opposite, that only the odd numbers are "lucky".)
In fact, you could do it equally the other way around, incrementing only the even numbers. If you are only rounding a single number, it doesn't really matter what you do. However, when you are rounding a series of numbers that will be used in a calculation, if you treated each first nonsignificant 5 in the same way, you would be over- or understating the value of the rounded number, thus accumulating round-off error. Since there are equal numbers of even and odd digits, incrementing only the one kind will keep this kind of error from building up. You could do just as well, of course, by flipping a coin!
Table \(1\): Examples of rounding-off
number to round
number of significant digits
result
comment
34.216 3 34.2 First non-significant digit (1) is less than 5, so number is simply truncated.
2.252 2 2.2 or 2.3 First non-significant digit is 5, so least sig. digit can either remain unchanged or be incremented.
39.99 3 40.0 Crossing "decimal boundary", so all numbers change.
85,381 3 85,400 The two zeros are just placeholders
0.04597 3 0.0460 The two leading zeros are not significant digits.
Rounding up the Nines
Suppose that an object is found to have a weight of 3.98 ± 0.05 g. This would place its true weight somewhere in the range of 3.93 g to 4.03 g. In judging how to round this number, you count the number of digits in “3.98” that are known exactly, and you find none! Since the “4” is the left most digit whose value is uncertain, this would imply that the result should be rounded to one significant figure and reported simply as 4 g. An alternative would be to bend the rule and round off to two significant digits, yielding 4.0 g. How can you decide what to do? In a case such as this, you should look at the implied uncertainties in the two values, and compare them with the uncertainty associated with the original measurement.
Table \(2\)
rounded value
implied max
implied min
absolute uncertainty
relative uncertainty
3.98 g 3.985 g 3.975 g ±.005 g or 0.01 g 1 in 400, or 0.25%
4 g 4.5 g 3.5 g ±.5 g or 1 g 1 in 4, 25%
4.0 g 4.05 g 3.95 g ±.05 g or 0.1 g 1 in 40, 2.5%
Clearly, rounding off to two digits is the only reasonable course in this example. Observed values should be rounded off to the number of digits that most accurately conveys the uncertainty in the measurement.
• Usually, this means rounding off to the number of significant digits in in the quantity; that is, the number of digits (counting from the left) that are known exactly, plus one more.
• When this cannot be applied (as in the example above when addition of subtraction of the absolute uncertainty bridges a power of ten), then we round in such a way that the relative implied uncertainty in the result is as close as possible to that of the observed value.
Rounding the Results of Calculations
When carrying out calculations that involve multiple steps, you should avoid doing any rounding until you obtain the final result. Suppose you use your calculator to work out the area of a rectangle:
rounded value
relative implied uncertainty
1.58 1 part in 158, or 0.6%
1.6 1 part in 16, or 6 %
Note
Your calculator is of course correct as far as the pure numbers go, but you would be wrong to write down "1.57676 cm2" as the answer. Two possible options for rounding off the calculator answer are shown at the right.
It is clear that neither option is entirely satisfactory; rounding to 3 significant digits overstates the precision of the answer, whereas following the rule and rounding to the two digits in ".42" has the effect of throwing away some precision. In this case, it could be argued that rounding to three digits is justified because the implied relative uncertainty in the answer, 0.6%, is more consistent with those of the two factors.
The "rules" for rounding off are generally useful, convenient guidelines, but they do not always yield the most desirable result. When in doubt, it is better to rely on relative implied uncertainties.
Addition and Subtraction
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise that the least precise number used to get the answer. When adding or subtracting, we go by the number of decimal places (i.e., the number of digits on the right side of the decimal point) rather than by the number of significant digits. Identify the quantity having the smallest number of decimal places, and use this number to set the number of decimal places in the answer.
Multiplication and Division
The result must contain the same number of significant figures as in the value having the least number of significant figures.
Logarithms and antilogarithms
If a number is expressed in the form a × 10b ("scientific notation") with the additional restriction that the coefficient a is no less than 1 and less than 10, the number is in its normalized form. Express the base-10 logarithm of a value using the same number of significant figures as is present in the normalized form of that value. Similarly, for antilogarithms (numbers expressed as powers of 10), use the same number of significant figures as are in that power.
Examples \(1\)
The following examples will illustrate the most common problems you are likely to encounter in rounding off the results of calculations. They deserve your careful study!
calculator result
rounded remarks
1.6 Rounding to two significant figures yields an implied uncertainty of 1/16 or 6%, three times greater than that in the least-precisely known factor. This is a good illustration of how rounding can lead to the loss of information.
1.9E6 The "3.1" factor is specified to 1 part in 31, or 3%. In the answer 1.9, the value is expressed to 1 part in 19, or 5%. These precisions are comparable, so the rounding-off rule has given us a reasonable result.
A certain book has a thickness of 117 mm; find the height of a stack of 24 identical books:
2810 mm The “24” and the “1” are exact, so the only uncertain value is the thickness of each book, given to 3 significant digits. The trailing zero in the answer is only a placeholder.
10.4 In addition or subtraction, look for the term having the smallest number of decimal places, and round off the answer to the same number of places.
23 cm see below
The last of the examples shown above represents the very common operation of converting one unit into another. There is a certain amount of ambiguity here; if we take "9 in" to mean a distance in the range 8.5 to 9.5 inches, then the implied uncertainty is ±0.5 in, which is 1 part in 18, or about ± 6%. The relative uncertainty in the answer must be the same, since all the values are multiplied by the same factor, 2.54 cm/in. In this case we are justified in writing the answer to two significant digits, yielding an uncertainty of about ±1 cm; if we had used the answer "20 cm" (one significant digit), its implied uncertainty would be ±5 cm, or ±25%.
When the appropriate number of significant digits is in question, calculating the relative uncertainty can help you decide. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/02%3A_PSS-_Scientific_Measurements/2.03%3A_PSS.3-_Rounding_Off_Nonsignificant_Digits.txt |
Learning Objectives
• Express a large number or a small number in scientific notation.
• Carry out arithmetical operations and express the final answer in scientific notation
Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form
$N \times 10^n \nonumber$
where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (100 = 1). The number 10 is called the base because it is this number that is raised to the power $n$. Although a base number may have values other than 10, the base number in scientific notation is always 10.
A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:
• If the decimal point is moved to the left n places, n is positive.
• If the decimal point is moved to the right n places, n is negative.
Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example $1$.
Example $1$: Expressing Numbers in Scientific Notation
Convert each number to scientific notation.
1. 637.8
2. 0.0479
3. 7.86
4. 12,378
5. 0.00032
6. 61.06700
7. 2002.080
8. 0.01020
Solution
Solutions to Example 2.2.1
Explanation Answer
a
To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8
Because the decimal point was moved two places to the left, n = 2.
$6.378 \times 10^2$
b
To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479
Because the decimal point was moved two places to the right, n = −2.
$4.79 \times 10^{−2}$
c This is usually expressed simply as 7.86. (Recall that 100 = 1.) $7.86 \times 10^0$
d Because the decimal point was moved four places to the left, n = 4. $1.2378 \times 10^4$
e Because the decimal point was moved four places to the right, n = −4. $3.2 \times 10^{−4}$
f Because the decimal point was moved one place to the left, n = 1. $6.106700 \times 10^1$
g Because the decimal point was moved three places to the left, n = 3. $2.002080 \times 10^3$
h Because the decimal point was moved two places to the right, n = -2. $1.020 \times 10^{−2}$
Addition and Subtraction
Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example $2$ illustrates how to do this.
Example $2$: Expressing Sums and Differences in Scientific Notation
Carry out the appropriate operation and then express the answer in scientific notation.
1. $(1.36 \times 10^2) + (4.73 \times 10^3) \nonumber$
2. $(6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber$
Solution
Solutions to Example 2.2.2.
Explanation Answer
a
Both exponents must have the same value, so these numbers are converted to either
$(1.36 \times 10^2) + (47.3 \times 10^2) = (1.36 + 47.3) \times 10^2 = 48.66 × 10^2$
or
$(0.136 \times 10^3) + (4.73 \times 10^3) = (0.136 + 4.73) \times 10^3) = 4.87 \times 10^3$.
Choosing either alternative gives the same answer, reported to two decimal places.
In converting 48.66 × 102 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$4.87 \times 10^3$
b
Converting the exponents to the same value gives either
$(6.923 \times 10^{-3}) − (0.8756 \times 10^{-3}) = (6.923 − 0.8756) \times 10^{−3}$
or
$(69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) = (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}$.
In converting 60.474 × 10-4 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$6.047 \times 10^{−3}$
Multiplication and Division
When multiplying numbers expressed in scientific notation, we multiply the values of $N$ and add together the values of $n$. Conversely, when dividing, we divide $N$ in the dividend (the number being divided) by $N$ in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example $3$.
Example $3$: Expressing Products and Quotients in Scientific Notation
Perform the appropriate operation and express your answer in scientific notation.
1. $(6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber$
2. $\dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber$
3. $\dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber$
Solution
Solution to Example 2.2.3
Explanation Answer
a
In multiplication, we add the exponents:
$(6.022 \times 10^{23})(6.42 \times 10^{−2})= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber$
In converting $38.7 \times 10^{21}$ to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$3.87 \times 10^{22}$ b
In division, we subtract the exponents:
${1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = {1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \nonumber$
In converting $0.183 \times 10^4$ to scientific notation, $n$ has become more negative by 1 because the value of $N$ has increased.
$1.83 \times 10^3$ c
This problem has both multiplication and division:
${(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = {39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber$
$4.7\times 10^{-31}$ | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/02%3A_PSS-_Scientific_Measurements/2.06%3A_PSS.6-_Exponential_Numbers.txt |
Learning Objectives
• Express a large number or a small number in scientific notation.
• Carry out arithmetical operations and express the final answer in scientific notation
Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form
$N \times 10^n \nonumber$
where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (100 = 1). The number 10 is called the base because it is this number that is raised to the power $n$. Although a base number may have values other than 10, the base number in scientific notation is always 10.
A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:
• If the decimal point is moved to the left n places, n is positive.
• If the decimal point is moved to the right n places, n is negative.
Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example $1$.
Example $1$: Expressing Numbers in Scientific Notation
Convert each number to scientific notation.
1. 637.8
2. 0.0479
3. 7.86
4. 12,378
5. 0.00032
6. 61.06700
7. 2002.080
8. 0.01020
Solution
Solutions to Example 2.2.1
Explanation Answer
a
To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8
Because the decimal point was moved two places to the left, n = 2.
$6.378 \times 10^2$
b
To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479
Because the decimal point was moved two places to the right, n = −2.
$4.79 \times 10^{−2}$
c This is usually expressed simply as 7.86. (Recall that 100 = 1.) $7.86 \times 10^0$
d Because the decimal point was moved four places to the left, n = 4. $1.2378 \times 10^4$
e Because the decimal point was moved four places to the right, n = −4. $3.2 \times 10^{−4}$
f Because the decimal point was moved one place to the left, n = 1. $6.106700 \times 10^1$
g Because the decimal point was moved three places to the left, n = 3. $2.002080 \times 10^3$
h Because the decimal point was moved two places to the right, n = -2. $1.020 \times 10^{−2}$
Addition and Subtraction
Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example $2$ illustrates how to do this.
Example $2$: Expressing Sums and Differences in Scientific Notation
Carry out the appropriate operation and then express the answer in scientific notation.
1. $(1.36 \times 10^2) + (4.73 \times 10^3) \nonumber$
2. $(6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber$
Solution
Solutions to Example 2.2.2.
Explanation Answer
a
Both exponents must have the same value, so these numbers are converted to either
$(1.36 \times 10^2) + (47.3 \times 10^2) = (1.36 + 47.3) \times 10^2 = 48.66 × 10^2$
or
$(0.136 \times 10^3) + (4.73 \times 10^3) = (0.136 + 4.73) \times 10^3) = 4.87 \times 10^3$.
Choosing either alternative gives the same answer, reported to two decimal places.
In converting 48.66 × 102 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$4.87 \times 10^3$
b
Converting the exponents to the same value gives either
$(6.923 \times 10^{-3}) − (0.8756 \times 10^{-3}) = (6.923 − 0.8756) \times 10^{−3}$
or
$(69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) = (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}$.
In converting 60.474 × 10-4 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$6.047 \times 10^{−3}$
Multiplication and Division
When multiplying numbers expressed in scientific notation, we multiply the values of $N$ and add together the values of $n$. Conversely, when dividing, we divide $N$ in the dividend (the number being divided) by $N$ in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example $3$.
Example $3$: Expressing Products and Quotients in Scientific Notation
Perform the appropriate operation and express your answer in scientific notation.
1. $(6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber$
2. $\dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber$
3. $\dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber$
Solution
Solution to Example 2.2.3
Explanation Answer
a
In multiplication, we add the exponents:
$(6.022 \times 10^{23})(6.42 \times 10^{−2})= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber$
In converting $38.7 \times 10^{21}$ to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$3.87 \times 10^{22}$ b
In division, we subtract the exponents:
${1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = {1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \nonumber$
In converting $0.183 \times 10^4$ to scientific notation, $n$ has become more negative by 1 because the value of $N$ has increased.
$1.83 \times 10^3$ c
This problem has both multiplication and division:
${(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = {39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber$
$4.7\times 10^{-31}$ | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/02%3A_PSS-_Scientific_Measurements/2.07%3A_PSS.7-Scientific_Notation.txt |
Learning Objective
• Convert from one unit to another unit of the same type.
In Section 2.2, we showed some examples of how to replace initial units with other units of the same type to get a numerical value that is easier to comprehend. In this section, we will formalize the process.
Consider a simple example: how many feet are there in 4 yards? Most people will almost automatically answer that there are 12 feet in 4 yards. How did you make this determination? Well, if there are 3 feet in 1 yard and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards.
This is correct, of course, but it is informal. Let us formalize it in a way that can be applied more generally. We know that 1 yard (yd) equals 3 feet (ft):
$1\, yd = 3\, ft\nonumber$
In math, this expression is called an equality. The rules of algebra say that you can change (i.e., multiply or divide or add or subtract) the equality (as long as you do not divide by zero) and the new expression will still be an equality. For example, if we divide both sides by 2, we get:
$\dfrac{1}{2}\,yd= \dfrac{3}{2}\, ft\nonumber$
We see that one-half of a yard equals 3/2, or one and a half, feet—something we also know to be true—so the above equation is still an equality. Going back to the original equality, suppose we divide both sides of the equation by 1 yard (number and unit):
$\dfrac{1\,yd}{1\,yd}= \dfrac{3\,ft}{1\,yd}\nonumber$
The expression is still an equality, by the rules of algebra. The left fraction equals 1. It has the same quantity in the numerator and the denominator, so it must equal 1. The quantities in the numerator and denominator cancel, both the number and the unit:
$\dfrac{1\,yd}{1\,yd}= \dfrac{3\,ft}{1\,yd}\nonumber$
When everything cancels in a fraction, the fraction reduces to 1:
$1= \dfrac{3\,ft}{1\,yd}\nonumber$
Conversion Factors
We have an expression that equals 1.
$\dfrac{3\,ft}{1\,yd}=1\nonumber$
This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units.
The expression
$\dfrac{3\,ft}{1\,yd}\nonumber$
is called a conversion factor and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.)
To see how this happens, let us start with the original quantity: 4 yd.
Now let us multiply this quantity by 1. When you multiply anything by 1, you do not change the value of the quantity. Rather than multiplying by just 1, let us write 1 as:
$\dfrac{3\,ft}{1\,yd}\nonumber$
$4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber$
The 4 yd term can be thought of as 4yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard:
$4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber$
That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:
$\dfrac{4\times 3\, ft}{1}= \dfrac{12\,ft}{1}= 12\,ft\nonumber$
Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.
How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter, which is:
$1\,mm= \dfrac{1}{1000\,m}\nonumber$
The 1/1000 is what the prefix milli- means. Most people are more comfortable working without fractions, so we will rewrite this equation by bringing the 1,000 into the numerator of the other side of the equation:
$1000\,mm=1\,m\nonumber$
Now we construct a conversion factor by dividing one quantity into both sides. But now a question arises: which quantity do we divide by? It turns out that we have two choices, and the two choices will give us different conversion factors, both of which equal 1:
$\dfrac{1000\,mm}{1000\,mm}= \dfrac{1\,m}{1000\,mm} \nonumber$
or
$\dfrac{1000\,mm}{1\,m}= \dfrac{1\,m}{1\,m}\nonumber$
$1=\dfrac{1\,m}{1000\,mm}\nonumber$
or
$\dfrac{1000\,mm}{1\,m}=1\nonumber$
Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Therefore, we will use the second conversion factor. Canceling units and performing the mathematics, we get:
$14.66m\times \dfrac{1000\,mm}{1\,m}= 14660\,mm\nonumber$
Note how $m$ cancels, leaving $mm$, which is the unit of interest.
The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses.
Example $1$
1. Convert 35.9 kL to liters.
2. Convert 555 nm to meters.
Solution
1. We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1000L/ 1kL. Applying this conversion factor, we get:
$35.9\, kL\times \dfrac{1000\,L}{1\,kL}= 35,900\, L \nonumber \nonumber$
1. We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator:
$\dfrac{1\,m}{10^{9}\,nm} \nonumber \nonumber$
Applying this conversion factor, we get:
$555\,nm\times \dfrac{1m}{10^{9}nm}= 0.000000555\,m= 5.55\times 10^{-7}\,m \nonumber \nonumber$
In the final step, we expressed the answer in scientific notation.
Exercise $1$
1. Convert 67.08 μL to liters.
2. Convert 56.8 m to kilometers.
Answer a
6.708 × 10−5 L
Answer b
5.68 × 10−2 km
What if we have a derived unit that is the product of more than one unit, such as m2? Suppose we want to convert square meters to square centimeters? The key is to remember that m2 means m × m, which means we have two meter units in our derived unit. That means we have to include two conversion factors, one for each unit. For example, to convert 17.6 m2 to square centimeters, we perform the conversion as follows:
\begin{align} 17.6m^{2} &= 17.6(m\times m)\times \dfrac{100cm}{1m}\times \dfrac{100cm}{1m} \nonumber \[4pt] &= 176000\,cm \times cm \nonumber \[4pt] &= 1.76\times 10^{5} \,cm^2\end{align}\nonumber
Example $2$
How many cubic centimeters are in 0.883 m3?
Solution
With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have:
$0.883m^{3}\times \dfrac{100\,cm}{1\,m}\times \dfrac{100\,cm}{1\,m} \times \dfrac{100\,cm}{1\,m}= 883000\,cm^{3} = 8.83\times 10^{5}\,cm^{3}\nonumber$
You should demonstrate to yourself that the three meter units do indeed cancel.
Exercise $2$
How many cubic millimeters are present in 0.0923 m3?
Answer
9.23 × 107 mm3
Suppose the unit you want to convert is in the denominator of a derived unit—what then? Then, in the conversion factor, the unit you want to remove must be in the numerator. This will cancel with the original unit in the denominator and introduce a new unit in the denominator. The following example illustrates this situation.
Example $3$
Convert 88.4 m/min to meters/second.
Solution
We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1min/60s. Apply and perform the math:
$\dfrac{88.4m}{min}\times \dfrac{1\,min}{60\,s}= 1.47\dfrac{m}{s}\nonumber$
Notice how the 88.4 automatically goes in the numerator. That's because any number can be thought of as being in the numerator of a fraction divided by 1.
Exercise $3$
Convert 0.203 m/min to meters/second.
Answer
0.00338 m/s
or
3.38 × 10−3 m/s
Sometimes there will be a need to convert from one unit with one numerical prefix to another unit with a different numerical prefix. How do we handle those conversions? Well, you could memorize the conversion factors that interrelate all numerical prefixes. Or you can go the easier route: first convert the quantity to the base unit—the unit with no numerical prefix—using the definition of the original prefix. Then, convert the quantity in the base unit to the desired unit using the definition of the second prefix. You can do the conversion in two separate steps or as one long algebraic step. For example, to convert 2.77 kg to milligrams:
$2.77\,kg\times \dfrac{1000\,g}{1\,kg}= 2770\,g\nonumber$ (convert to the base units of grams)
$2770\,g\times \dfrac{1000\,mg}{1\,g}= 2770000\,mg = 2.77\times 10^{6}\,mg\nonumber$ (convert to desired unit)
Alternatively, it can be done in a single multi-step process:
\begin{align} 2.77\, \cancel{kg}\times \dfrac{1000\,\cancel{g}}{1\,\cancel{kg}}\times \dfrac{1000\,mg}{1\,\cancel{g}} &= 2770000\, mg \nonumber \[4pt] &= 2.77\times 10^{6}\,mg \end{align}\nonumber
You get the same answer either way.
Example $4$
How many nanoseconds are in 368.09 μs?
Solution
You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-,
$368.0\,\mu s\times \dfrac{1\,s}{1000000\,\mu s}\times \dfrac{1000000000}{1\,s}= 3.6809\times 10^{5}\,ns\nonumber$
Exercise $4$
How many milliliters are in 607.8 kL?
Answer
6.078 × 108 mL
When considering the significant figures of a final numerical answer in a conversion, there is one important case where a number does not impact the number of significant figures in a final answer: the so-called exact number. An exact number is a number from a defined relationship, not a measured one. For example, the prefix kilo- means 1,000-exactly 1,000, no more or no less. Thus, in constructing the conversion factor:
$\dfrac{1000\,g}{1\,kg}\nonumber$
neither the 1,000 nor the 1 enter into our consideration of significant figures. The numbers in the numerator and denominator are defined exactly by what the prefix kilo- means. Another way of thinking about it is that these numbers can be thought of as having an infinite number of significant figures, such as:
$\dfrac{1000.0000000000 \dots \,g}{1.0000000000 \ldots \,kg}\nonumber$
The other numbers in the calculation will determine the number of significant figures in the final answer.
Example $5$
A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.
Solution
Area is defined as the product of the two dimensions, which we then have to convert to square meters, and express our final answer to the correct number of significant figures—which in this case will be three.
$36.7\,cm\times 128.8\,cm\times \dfrac{1\,m}{100\,cm}\times \dfrac{1\,m}{100\,cm}= 0.472696\,m^{2}= 0.473\,m^{2}\nonumber$
The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.
Exercise $5$
What is the volume of a block in cubic meters with the dimensions 2.1 cm × 34.0 cm × 118 cm?
Answer
0.0084 m3
Chemistry is Everywhere: The Gimli Glider
On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed "the Gimli Glider."
The Gimli Glider is the Boeing 767 that ran out of fuel and glided to safety at Gimli Airport. The aircraft ran out of fuel because of confusion over the units used to express the amount of fuel. Source: Photo courtesy of Will F., (CC BY-SA 2.5; Aero Icarus).
The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.
What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.
Key Takeaways
• Units can be converted to other units using the proper conversion factors.
• Conversion factors are constructed from equalities that relate two different units.
• Conversions can be a single step or multi-step.
• Unit conversion is a powerful mathematical technique in chemistry that must be mastered.
• Exact numbers do not affect the determination of significant figures. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/02%3A_PSS-_Scientific_Measurements/2.09%3A_PSS.9-_Unit_Analysis_Problem_Solving.txt |
Learning Objectives
• State the different measurement systems used in chemistry.
• Describe how prefixes are used in the metric system and identify how the prefixes milli-, centi-, and kilo- compare to the base unit.
How long is a yard? It depends on whom you ask and when you asked the question. Today we have a standard definition of the yard, which you can see marked on every football field. If you move the ball ten yards, you get a first down and it does not matter whether you are playing in Los Angeles, Dallas, or Green Bay. But at one time that yard was arbitrarily defined as the distance from the tip of the king's nose to the end of his outstretched hand. Of course, the problem there is simple: new king, new distance (and then you have to re-mark all of those football fields).
SI Base Units
All measurements depend on the use of units that are well known and understood. The English system of measurement units (inches, feet, ounces, etc.) are not used in science because of the difficulty in converting from one unit to another. The metric system is used because all metric units are based on multiples of 10, making conversions very simple. The metric system was originally established in France in 1795. The International System of Units is a system of measurement based on the metric system. The acronym SI is commonly used to refer to this system and stands for the French term, Le Système International d'Unités. The SI was adopted by international agreement in 1960 and is composed of seven base units in Table $1$.
Quantity SI Base Unit Symbol
Table $1$: SI Base Units of Measurement
Length meter $\text{m}$
Mass kilogram $\text{kg}$
Temperature kelvin $\text{K}$
Time second $\text{s}$
Amount of a Substance mole $\text{mol}$
Electric Current ampere $\text{A}$
Luminous Intensity candela $\text{cd}$
The first units are frequently encountered in chemistry. All other measurement quantities, such as volume, force, and energy, can be derived from these seven base units.
Unfortunately, the Metric System is Not Ubiquitous
The map below shows the adoption of the SI units in countries around the world. The United States has legally adopted the metric system for measurements, but does not use it in everyday practice. Great Britain and much of Canada use a combination of metric and imperial units.
Prefix Multipliers
Conversions between metric system units are straightforward because the system is based on powers of ten. For example, meters, centimeters, and millimeters are all metric units of length. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. Metric prefixes are used to distinguish between units of different size. These prefixes all derive from either Latin or Greek terms. For example, mega comes from the Greek word $\mu \varepsilon \gamma \alpha \varsigma$, meaning "great". Table $2$ lists the most common metric prefixes and their relationship to the central unit that has no prefix. Length is used as an example to demonstrate the relative size of each prefixed unit.
Prefix Unit Abbreviation Meaning Example
Table $2$: SI Prefixes
giga $\text{G}$ 1,000,000,000 1 gigameter $\left( \text{Gm} \right)=10^9 \: \text{m}$
mega $\text{M}$ 1,000,000 1 megameter $\left( \text{Mm} \right)=10^6 \: \text{m}$
kilo $\text{k}$ 1,000 1 kilometer $\left( \text{km} \right)=1,000 \: \text{m}$
hecto $\text{h}$ 100 1 hectometer $\left( \text{hm} \right)=100 \: \text{m}$
deka $\text{da}$ 10 1 dekameter $\left( \text{dam} \right)=10 \: \text{m}$
1 1 meter $\left( \text{m} \right)$
deci $\text{d}$ 1/10 1 decimeter $\left( \text{dm} \right)=0.1 \: \text{m}$
centi $\text{c}$ 1/100 1 centimeter $\left( \text{cm} \right)=0.01 \: \text{m}$
milli $\text{m}$ 1/1,000 1 millimeter $\left( \text{mm} \right)=0.001 \: \text{m}$
micro $\mu$ 1/1,000,000 1 micrometer $\left( \mu \text{m} \right)=10^{-6} \: \text{m}$
nano $\text{n}$ 1/1,000,000,000 1 nanometer $\left( \text{nm} \right)=10^{-9} \: \text{m}$
pico $\text{p}$ 1/1,000,000,000,000 1 picometer $\left( \text{pm} \right)=10^{-12} \: \text{m}$
There are a couple of odd little practices with the use of metric abbreviations. Most abbreviations are lowercase. We use "$\text{m}$" for meter and not "$\text{M}$". However, when it comes to volume, the base unit "liter" is abbreviated as "$\text{L}$" and not "$\text{l}$". So we would write 3.5 milliliters as $3.5 \: \text{mL}$.
As a practical matter, whenever possible you should express the units in a small and manageable number. If you are measuring the weight of a material that weighs $6.5 \: \text{kg}$, this is easier than saying it weighs $6500 \: \text{g}$ or $0.65 \: \text{dag}$. All three are correct, but the $\text{kg}$ units in this case make for a small and easily managed number. However, if a specific problem needs grams instead of kilograms, go with the grams for consistency.
Example $1$: Unit Abbreviations
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kiloliter
2. microsecond
3. decimeter
4. nanogram
Solutions
Solutions to Example 2.5.1
Explanation Answer
a The prefix kilo means “1,000 ×,” so 1 kL equals 1,000 L. kL
b The prefix micro implies 1/1,000,000th of a unit, so 1 µs equals 0.000001 s. µs
c The prefix deci means 1/10th, so 1 dm equals 0.1 m. dm
d The prefix nano means 1/1000000000, so a nanogram is equal to 0.000000001 g. ng
Exercise $1$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kilometer
2. milligram
3. nanosecond
4. centiliter
Answer a:
km
Answer b:
mg
Answer c:
ns
Answer d:
cL
Summary
• Metric prefixes derive from Latin or Greek terms. The prefixes are used to make the units manageable.
• The SI system is based on multiples of ten. There are seven basic units in the SI system. Five of these units are commonly used in chemistry. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/03%3A_The_Metric_System/3.01%3A_2.1-Basic_Units_and_Symbols.txt |
Learning Objectives
• Define density.
• Use density as a conversion factor.
Density ($\rho$) is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows:
$\text{Density}=\dfrac{\text{Mass}}{\text{Volume}} \nonumber$
or just
$\rho =\dfrac{m}{V} \label{eq2}$
Based on this equation, it's clear that density can, and does, vary from element to element and substance to substance due to differences in the relationship of mass and volume. Pure water, for example, has a density of 0.998 g/cm3 at 25° C. The average densities of some common substances are in Table $1$. Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Table $1$: Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Density can be measured for all substances—solids, liquids and gases. For solids and liquids, density is often reported using the units of g/cm3. Densities of gases, which are significantly lower than the densities of solids and liquids, are often given using units of g/L.
Example $1$: Ethyl Alcohol
Calculate the density of a 30.2 mL sample of ethyl alcohol with a mass of 23.71002 g
Solution
This is a direct application of Equation \ref{eq2}:
$\rho = \dfrac{23.71002\,g}{30.2\,mL} = 0.785\, g/mL \nonumber$
Exercise $1$
1. Find the density (in kg/L) of a sample that has a volume of 36.5 L and a mass of 10.0 kg.
2. If you have a 2.130 mL sample of acetic acid with mass 0.002234 kg, what is the density in kg/L?
Answer a
$0.274 \,kg/L$
Answer b
$1.049 \,kg/L$
Density as a Conversion Factor
Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
$\dfrac{13.6\:g}{1\:mL} = 1 \nonumber$
and
$\dfrac{1\:mL}{13.6\:g} = 1 \nonumber$
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 2.0 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top, so that our final answer has a unit of mass:
$\mathrm{2.0\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}=27.2\:g=27\:g} \nonumber$
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
Example $2$: Mercury Thermometer Steps for Problem Solving
A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury?
Solution
Solution to Example 2.9.2
Steps for Problem Solving Unit Conversion
Identify the "given" information and what the problem is asking you to "find."
Given: 0.750 g
Find: mL
List other known quantities. 13.6 g/mL (density of mercury)
Prepare a concept map.
Calculate.
$0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} = 0.055147 ... \; \rm{mL} \approx 0.0551\; \rm{mL} \nonumber$
We have limited the final answer to three significant figures.
Exercise $2$
What is the volume of 100.0 g of air if its density is 1.3 g/L?
Answer
$77 \, L$
Summary
• Density is defined as the mass of an object divided by its volume.
• Density can be used as a conversion factor between mass and volume. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/03%3A_The_Metric_System/3.07%3A_The_Density_Concept.txt |
Learning Objective
• Learn about the various temperature scales that are commonly used in chemistry.
• Define density and use it as a conversion factor.
There are other units in chemistry that are important, and we will cover others over the course of the entire book. One of the fundamental quantities in science is temperature. Temperature is a measure of the average amount of energy of motion, or kinetic energy, a system contains. Temperatures are expressed using scales that use units called degrees, and there are several temperature scales in use. In the United States, the commonly used temperature scale is the Fahrenheit scale (symbolized by °F and spoken as "degrees Fahrenheit"). On this scale, the freezing point of liquid water (the temperature at which liquid water turns to solid ice) is 32°F, and the boiling point of water (the temperature at which liquid water turns to steam) is 212°F.
Science also uses other scales to express temperature. The Celsius scale (symbolized by °C and spoken as "degrees Celsius") is a temperature scale where 0°C is the freezing point of water and 100°C is the boiling point of water; the scale is divided into 100 divisions between these two landmarks and extended higher and lower. By comparing the Fahrenheit and Celsius scales, a conversion between the two scales can be determined:
\begin{align} \ce{^{\circ}C} &= \ce{(^{\circ}F-32)\times 5/9} \label{eq1} \[4pt] \ce{ ^{\circ}F} &= \left(\ce{^{\circ}C \times 9/5 } \right)+32 \label{eq2} \end{align}
Example $1$: Conversions
1. What is 98.6 °F in degrees Celsius?
2. What is 25.0 °C in degrees Fahrenheit?
Solution
1. Using Equation \ref{eq1}, we have
\begin{align*} ^{\circ}C &=(98.6-32)\times \dfrac{5}{9} \[4pt] &= 66.6\times \dfrac{5}{9} \[4pt] &= 37.0^{\circ}C \end{align*}\nonumber
1. Using Equation \ref{eq2}, we have
\begin{align*} ^{\circ}F &= \left(25.0\times \dfrac{9}{5}\right)+32 \[4pt] &= 45.0+32 \[4pt] &= 77.0^{\circ}F \end{align*}\nonumber
Exercise $1$
1. Convert 0 °F to degrees Celsius.
2. Convert 212 °C to degrees Fahrenheit.
Answer a
−17.8 °C
Answer b
414 °F
The fundamental unit of temperature (another fundamental unit of science, bringing us to four) in SI is the kelvin (K). The Kelvin temperature scale (note that the name of the scale capitalizes the word Kelvin, but the unit itself is lowercase) uses degrees that are the same size as the Celsius degree, but the numerical scale is shifted up by 273.15 units. That is, the conversion between the Kelvin and Celsius scales is as follows:
$K = {^{\circ}C + 273.15}\nonumber$
For most purposes, it is acceptable to use 273 instead of 273.15. Note that the Kelvin scale does not use the word degrees; a temperature of 295 K is spoken of as "two hundred ninety-five kelvins" and not "two hundred ninety-five degrees Kelvin."
The reason that the Kelvin scale is defined this way is because there exists a minimum possible temperature called absolute zero (zero kelvins). The Kelvin temperature scale is set so that 0 K is absolute zero, and temperature is counted upward from there. Normal room temperature is about 295 K, as seen in the following example.
Example $2$: Room Temperature
If normal room temperature is 72.0°F, what is room temperature in degrees Celsius and kelvin?
Solution
First, we use the formula to determine the temperature in degrees Celsius:
\begin{align*} ^{\circ}C &= (72.0-32)\times \dfrac{5}{9} \nonumber \[4pt] &= 40.0\times \dfrac{5}{9} \nonumber \[4pt] &= 22.2^{\circ}C \end{align*}\nonumber
Then we use the appropriate formula above to determine the temperature in the Kelvin scale:
\begin{align*} K &= 22.2^{\circ}C+273.15 \nonumber \[4pt] &= 295.4K \end{align*}\nonumber
So, room temperature is about 295 K.
Exercise $2$
What is 98.6°F on the Kelvin scale?
Answer
310.2 K
Figure $1$ compares the three temperature scales. Note that science uses the Celsius and Kelvin scales almost exclusively; virtually no practicing chemist expresses laboratory-measured temperatures with the Fahrenheit scale. In fact, the United States is one of the few countries in the world that still uses the Fahrenheit scale on a daily basis. The other two countries are Liberia and Myanmar (formerly Burma). People driving near the borders of Canada or Mexico may pick up local radio stations on the other side of the border that express the daily weather in degrees Celsius, so do not get confused by their weather reports.
Density
Density is a physical property that is defined as a substance's mass divided by its volume:
$density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{v}\nonumber$
Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm3, g/L, kg/L, and even kg/m3. Densities for some common substances are listed in Table $1$.
Table $1$: Densities of Some Common Substances
Substance Density (g/mL or g/cm3)
water 1.0
gold 19.3
mercury 13.6
air 0.0012
cork 0.22–0.26
aluminum 2.7
iron 7.87
Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm3. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table $1$), the volume units will cancel and leave you with mass units, telling you the mass of the sample:
$7.88\,\cancel{cm^{3}}\times \dfrac{2.7\,g}{\cancel{cm^{3}}}= 21\, g \text{ of aluminium} \nonumber \nonumber$
where we have limited our answer to two significant figures.
Example $3$: Mercury
What is the mass of 44.6 mL of mercury?
Solution
Use the density from Table $1$ "Densities of Some Common Substances" as a conversion factor to go from volume to mass:
$44.6\,\cancel{mL}\times \dfrac{13.6\,g}{\cancel{mL}}= 607\,g \nonumber \nonumber$
The mass of the mercury is 607 g.
Exercise $3$
What is the mass of 25.0 cm3 of iron?
Answer
197 g
Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:
$d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \nonumber$
Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator:
$d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \nonumber$
That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units, but also the numbers:
$\dfrac{1}{d}= \dfrac{1\,mL}{19.3\,g} \nonumber \nonumber$
This reciprocal density is still a useful conversion factor, but now the mass unit will cancel and the volume unit will be introduced. Thus, if we want to know the volume of 45.9 g of gold, we would set up the conversion as follows:
$45.9\,\cancel{g}\times \dfrac{1\,mL}{19.3\cancel{g}}= 2.38\,mL \nonumber \nonumber$
Note how the mass units cancel, leaving the volume unit, which is what we are looking for.
Example $4$: Calculating Volume from Density
A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/cm3, what is the volume of the cork?
Solution
To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find:
$\dfrac{1}{d}= \dfrac{1\,cm^{3}}{0.22\,g} \nonumber \nonumber$
We can use this expression as the conversion factor. So
$3.78\,\cancel{g}\times \dfrac{1\,cm^{3}}{0.22\,\cancel{g}}= 17.2\,cm^{3} \nonumber \nonumber$
Exercise $4$
What is the volume of 3.78 g of gold?
Answer
0.196 cm3
Care must be used with density as a conversion factor. Make sure the mass units are the same, or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.
Food and Drink Application: Cooking Temperatures
Because degrees Fahrenheit is the common temperature scale in the United States, kitchen appliances, such as ovens, are calibrated in that scale. A cool oven may be only 150°F, while a cake may be baked at 350°F and a chicken roasted at 400°F. The broil setting on many ovens is 500°F, which is typically the highest temperature setting on a household oven.
People who live at high altitudes, typically 2,000 ft above sea level or higher, are sometimes urged to use slightly different cooking instructions on some products, such as cakes and bread, because water boils at a lower temperature the higher in altitude you go, meaning that foods cook slower. For example, in Cleveland water typically boils at 212°F (100°C), but in Denver, the Mile-High City, water boils at about 200°F (93.3°C), which can significantly lengthen cooking times. Good cooks need to be aware of this.
A meat thermometer with a dial. Notice the markings for Fahrenheit (outer scale) and Celsius (inner scale) temperatures. Recipes for cooking food in an oven can use very different numbers, depending on the country you're in. (CC BY2.0 Bev Sykes)
At the other end is pressure cooking. A pressure cooker is a closed vessel that allows steam to build up additional pressure, which increases the temperature at which water boils. A good pressure cooker can get to temperatures as high as 252°F (122°C); at these temperatures, food cooks much faster than it normally would. Great care must be used with pressure cookers because of the high pressure and high temperature. (When a pressure cooker is used to sterilize medical instruments, it is called an autoclave.)
Other countries use the Celsius scale for everyday purposes. Therefore, oven dials in their kitchens are marked in degrees Celsius. It can be confusing for US cooks to use ovens abroad—a 425°F oven in the United States is equivalent to a 220°C oven in other countries. These days, many oven thermometers are marked with both temperature scales.
Key Takeaways
• Chemistry uses the Celsius and Kelvin scales to express temperatures.
• A temperature on the Kelvin scale is the Celsius temperature plus 273.15.
• The minimum possible temperature is absolute zero and is assigned 0 K on the Kelvin scale.
• Density relates the mass and volume of a substance.
• Density can be used to calculate volume from a given mass or mass from a given volume.
3.09: 2.9-Heat and Specific Heat
If a swimming pool and wading pool, both full of water at the same temperature, were subjected to the same input of heat energy, the wading pool would certainly rise in temperature more quickly than the swimming pool. The heat capacity of an object depends on both its mass and its chemical composition. Because of its much larger mass, the swimming pool of water has a larger heat capacity than the wading pool.
Heat Capacity and Specific Heat
Different substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water in the same sun will not become nearly as hot. We would say that water has a high heat capacity (the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$). Water is very resistant to changes in temperature, while metals in general are not. The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. The symbol for specific heat is $c_p$, with the $p$ subscript referring to the fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram per degree $\left( \text{J/g}^\text{o} \text{C} \right)$ or calories per gram per degree $\left( \text{cal/g}^\text{o} \text{C} \right)$ (Table $1$). This text will use $\text{J/g}^\text{o} \text{C}$ for specific heat.
$\text{specific heat}= \dfrac{\text{heat}}{\text{mass} \times \text{cal/g}^\text{o} \text{C}} \nonumber$
Notice that water has a very high specific heat compared to most other substances.
Table $1$: Specific Heat Capacities
Substance Specific Heat Capacity
at 25oC in J/g oC
Substance Specific Heat Capacity
at 25oC in J/g oC
$\ce{H2}$ gas 14.267 steam @ 100oC 2.010
$\ce{He}$ gas 5.300 vegetable oil 2.000
$\ce{H2O(l)}$ 4.184 sodium 1.23
lithium
3.56
air 1.020
ethyl alcohol
2.460
magnesium 1.020
ethylene glycol
2.200
aluminum 0.900
ice @ 0oC
2.010
concrete 0.880
steam @ 100oC
2.010
glass 0.840
Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days.
Summary
• Heat capacity is the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$).
• The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/03%3A_The_Metric_System/3.08%3A_Temperature.txt |
Learning Objectives
• To describe the solid, liquid and gas phases.
Water can take many forms. At low temperatures (below $0^\text{o} \text{C}$), it is a solid. When at "normal" temperatures (between $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$), it is a liquid. While at temperatures above $100^\text{o} \text{C}$, water is a gas (steam). The state the water is in depends upon the temperature. Each state (solid, liquid, and gas) has its own unique set of physical properties. Matter typically exists in one of three states: solid, liquid, or gas.
The state a given substance exhibits is also a physical property. Some substances exist as gases at room temperature (oxygen and carbon dioxide), while others, like water and mercury metal, exist as liquids. Most metals exist as solids at room temperature. All substances can exist in any of these three states. Figure $2$ shows the differences among solids, liquids, and gases at the molecular level. A solid has definite volume and shape, a liquid has a definite volume but no definite shape, and a gas has neither a definite volume nor shape.
Plasma: A Fourth State of Matter
Technically speaking a fourth state of matter called plasma exists, but it does not naturally occur on earth, so we will omit it from our study here.
A plasma globe operating in a darkened room. (CC BY-SA 3.0; Chocolateoak).
Solids
In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume. Most solids are hard, but some (like waxes) are relatively soft. Many solids composed of ions can also be quite brittle.
Solids are defined by the following characteristics:
• Definite shape (rigid)
• Definite volume
• Particles vibrate around fixed axes
If we were to cool liquid mercury to its freezing point of $-39^\text{o} \text{C}$, and under the right pressure conditions, we would notice all of the liquid particles would go into the solid state. Mercury can be solidified when its temperature is brought to its freezing point. However, when returned to room temperature conditions, mercury does not exist in solid state for long, and returns back to its more common liquid form.
Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. The effect of this regular arrangement of particles is sometimes visible macroscopically, as shown in Figure $3$. Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, “without form”) solids. Glass is one example of an amorphous solid.
Liquids
If the particles of a substance have enough energy to partially overcome intermolecular interactions, then the particles can move about each other while remaining in contact. This describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.
Liquids have the following characteristics:
• No definite shape (takes the shape of its container)
• Has definite volume
• Particles are free to move over each other, but are still attracted to each other
A familiar liquid is mercury metal. Mercury is an anomaly. It is the only metal we know of that is liquid at room temperature. Mercury also has an ability to stick to itself (surface tension) - a property all liquids exhibit. Mercury has a relatively high surface tension, which makes it very unique. Here you see mercury in its common liquid form.
Video $1$: Mercury boiling to become a gas.
If we heat liquid mercury to its boiling point of $357^\text{o} \text{C}$, and under the right pressure conditions, we would notice all particles in the liquid state go into the gas state.
Gases
If the particles of a substance have enough energy to completely overcome intermolecular interactions, then the particles can separate from each other and move about randomly in space. This describes the gas state, which we will consider in more detail elsewhere. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either. The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Gases have the following characteristics:
• No definite shape (takes the shape of its container)
• No definite volume
• Particles move in random motion with little or no attraction to each other
• Highly compressible
Table $1$: Characteristics of the Three States of Matter
Characteristics Solids Liquids Gases
shape definite indefinite indefinite
volume definite definite indefinite
relative intermolecular interaction strength strong moderate weak
relative particle positions in contact and fixed in place in contact but not fixed not in contact, random positions
Example $1$
What state or states of matter does each statement, describe?
1. This state has a definite volume, but no definite shape.
2. This state has no definite volume.
3. This state allows the individual particles to move about while remaining in contact.
Solution
1. This statement describes the liquid state.
2. This statement describes the gas state.
3. This statement describes the liquid state.
Exercise $1$
What state or states of matter does each statement describe?
1. This state has individual particles in a fixed position with regard to each other.
2. This state has individual particles far apart from each other in space.
3. This state has a definite shape.
Answer a:
solid
Answer b:
gas
Answer c:
solid
Summary
• Three states of matter exist - solid, liquid, and gas.
• Solids have a definite shape and volume.
• Liquids have a definite volume, but take the shape of the container.
• Gases have no definite shape or volume
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/04%3A_Matter_and_Energy/4.03%3A_Physical_States_of_Matter.txt |
Learning Objectives
To separate physical from chemical properties.
All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property).
Physical Property
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Silver is a shiny metal that conducts electricity very well. It can be molded into thin sheets, a property called malleability. Salt is dull and brittle and conducts electricity when it has been dissolved into water, which it does quite easily. Physical properties of matter include color, hardness, malleability, solubility, electrical conductivity, density, melting point, and boiling point.
For the elements, color does not vary much from one element to the next. The vast majority of elements are colorless, silver, or gray. Some elements do have distinctive colors: sulfur and chlorine are yellow, copper is (of course) copper-colored, and elemental bromine is red. However, density can be a very useful parameter for identifying an element. Of the materials that exist as solids at room temperature, iodine has a very low density compared to zinc, chromium, and tin. Gold has a very high density, as does platinum. Pure water, for example, has a density of 0.998 g/cm3 at 25°C. The average densities of some common substances are in Table \(1\). Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Table \(1\): Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Hardness helps determine how an element (especially a metal) might be used. Many elements are fairly soft (silver and gold, for example) while others (such as titanium, tungsten, and chromium) are much harder. Carbon is an interesting example of hardness. In graphite, (the "lead" found in pencils) the carbon is very soft, while the carbon in a diamond is roughly seven times as hard.
Melting and boiling points are somewhat unique identifiers, especially of compounds. In addition to giving some idea as to the identity of the compound, important information can be obtained about the purity of the material.
Chemical Properties
Chemical properties of matter describe its potential to undergo some chemical change or reaction by virtue of its composition. The elements, electrons, and bonds that are present give the matter potential for chemical change. It is quite difficult to define a chemical property without using the word "change". Eventually, after studying chemistry for some time, you should be able to look at the formula of a compound and state some chemical property. For example, hydrogen has the potential to ignite and explode given the right conditions—this is a chemical property. Metals in general have the chemical property of reacting with an acid. Zinc reacts with hydrochloric acid to produce hydrogen gas—this is a chemical property.
A chemical property of iron is its capability of combining with oxygen to form iron oxide, the chemical name of rust (Figure \(2\)). The more general term for rusting and other similar processes is corrosion. Other terms that are commonly used in descriptions of chemical changes are burn, rot, explode, decompose, and ferment. Chemical properties are very useful in identifying substances. However, unlike physical properties, chemical properties can only be observed as the substance is in the process of being changed into a different substance.
Table \(2\): Contrasting Physical and Chemical Properties
Physical Properties Chemical Properties
Gallium metal melts at 30 oC. Iron metal rusts.
Mercury is a very dense liquid. A green banana turns yellow when it ripens.
Gold is shiny. A dry piece of paper burns.
Example \(1\)
Which of the following is a chemical property of iron?
1. Iron corrodes in moist air.
2. Density = 7.874 g/cm3
3. Iron is soft when pure.
4. Iron melts at 1808 K.
Solution
"Iron corrodes in moist air" is the only chemical property of iron from the list.
Exercise \(\PageIndex{1A}\)
Which of the following is a physical property of matter?
1. corrosiveness
2. pH (acidity)
3. density
4. flammability
Answer
c
Exercise \(\PageIndex{1B}\)
Which of the following is a chemical property?
1. flammability
2. melting point
3. boiling point
4. density
Answer
a
Summary
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change. To identify a chemical property, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. | textbooks/chem/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/04%3A_Matter_and_Energy/4.08%3A_Physical_and_Chemical_Properties.txt |
The learning about and learning how to use organic chemistry is not a straightforward process, wherein one step leads to another in a simple, logical way like Euclidean geometry. A more realistic analogy would be to consider yourself thrust into and required to deal successfully with a sizable group of strangers speaking a new and complex language. In such a situation, one has to make many decisions- how much of the language to learn at the outset? Which people are the best to interact with first? Which will be the most important to know in the long run? How well does one have to know each person? How much does one have to know about the history of the group to understand their interactions? These are difficult questions, and a period of confusion, if not anxiety, is expected in any attempt to complete a task of this kind in a set, brief period of time. Clearly, it would be difficult to learn all at once the language, the people, and the interactions between them. Nonetheless, this is pretty much what is expected of you in learning organic chemistry.
• 1.1: Prelude to Organic Chemistry
You now are starting the study of organic chemistry, which is the chemistry of compounds of carbon. In this introductory chapter, we will tell you something of the background and history of organic chemistry, something of the problems and the rewards involved, and something of our philosophy of what is important for you to learn so that you will have a reasonable working knowledge of the subject, whether you are just interested in chemistry or plan for a STEM career.
• 1.2: A Bit of History
Try to visualize the problems confronting the organic chemist of 100 years ago. You will have no more than reasonably pure samples of organic compounds, the common laboratory chemicals of today, glassware, balances, thermometers, means of measuring densities, and a few optical instruments. You also will have a relatively embryonic theory that there are molecules in those bottles and that one compound differs from another because its molecules have different members, kinds or arrangements of atom
• 1.3: What Preparation Should You Have?
We have tried to give you a taste of the beginnings of organic chemistry and a few of the important principles that brought order out of the confusion that existed as to the nature of organic compounds. Before moving on to other matters, it may be well to give you some ideas of what kind of preparation will be helpful to you in learning about organic chemistry from this textbook.
• 1.4: Why Is Organic Chemistry Special?
One very important factor that make so much of chemistry center on a single element is that carbon-carbon bonds are strong, so long chains or rings of bonded carbon atoms are possible. Carbon is not unique in forming bonds to itself because other elements such as boron, silicon, and phosphorus form strong bonds in the elementary state. The uniqueness of carbon stems more from the fact that it forms strong carbon-carbon bonds that also are strong when in combination with other elements.
• 1.5: The Breadth of Organic Chemistry
Organic chemistry originally was defined as the chemistry of those substances formed by living matter. However, after the discovery that a supposedly typical organic compound, urea, could be prepared by heating ammonium cyanate, this definition lost significance and organic chemistry now is broadly defined as the chemistry of carbon-containing compounds.
• 1.6: Some Philosophical Observations
As you proceed with your study of organic chemistry, you may well feel confused as to what it is you are actually dealing with. On the one hand, there will be exhortations to remember how organic chemistry pervades our everyday life. A useful method for developing this sort of feeling for the relationship between structures and actual compounds is to check your perception of particular substances with their properties as given in a chemical handbook.
• 1.E: Introduction to Organic Chemistry (Exercises)
These are the homework exercises to accompany Chapter 1 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Thumbnail: Close up of coffee beans, photo taken for espresso flickr group. (CC BY 2.0 Generic; Nate Steiner).
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."
01: Introduction to Organic Chemistry
You now are starting the study of organic chemistry, which is the chemistry of compounds of carbon. In this introductory chapter, we will tell you something of the background and history of organic chemistry, something of the problems and the rewards involved, and something of our philosophy of what is important for you to learn so that you will have a reasonable working knowledge of the subject, whether you are just interested in chemistry or plan for a career as a chemist, an engineer, a physician, a biologist, and so on. The subject is very large; more than two million organic compounds have been isolated or prepared and characterized, yet the number of guiding principles is relatively small. You certainly will not learn everything about organic chemistry from this book, but with a good knowledge of the guiding principles, you will be able later to find out what you need to know either from the chemical literature, or directly by experiment in the laboratory.
Unfortunately, learning about and learning how to use organic chemistry is not a straightforward process, wherein one step leads to another in a simple, logical way like Euclidean geometry. A more realistic analogy would be to consider yourself thrust into and required to deal successfully with a sizable group of strangers speaking a new and complex language. In such a situation, one has to make many decisions- how much of the language to learn at the outset? Which people are the best to interact with first? Which will be the most important to know in the long run? How well does one have to know each person? How much does one have to know about the history of the group to understand their interactions? These are difficult questions, and a period of confusion, if not anxiety, is expected in any attempt to complete a task of this kind in a set, brief period of time. Clearly, it would be difficult to learn all at once the language, the people, and the interactions between them. Nonetheless, this is pretty much what is expected of you in learning organic chemistry.
A number of approaches have been devised to help you become familiar with and use organic chemistry. In terms of our analogy, one way is to learn the language, then the relationships between the people, and finally, well prepared, to proceed to interact with the people singly and then in groups. Such an approach may be logical in concept, but is not to everyone's taste as a way to learn. Many of us do better with an interactive approach, where language, relationships, and people are worked out more or less in concert, with attendant misunderstandings and ambiguities.
What we will try to do is to introduce some of the important basic concepts and the elements of the language of organic chemistry, then show how these are used in connection with various classes of compounds. The initial round will be a fairly extensive one and you should not expect to be able to master everything at once. This will take practice and we will provide opportunity for practice.
One of the appealing yet bothersome features of modern organic chemistry is its extraordinary vitality. Unlike Euclidean geometry or classical mechanics, it is evolving rapidly and many of the concepts introduced in this book are either new or have been drastically modified in the past ten years. Every issue of the current chemical journals has material of such basic interest that one would like to include it in an introductory course. Truly, those who write organic textbooks write on water, with no hope of producing the definitive book. Things just change too fast. Despite this, one of the great ideas of modern civilization, namely that organic compounds can be described in terms of more or less simple three-dimensional molecular structures with atoms held together by chemical bonds, has persisted for more than one hundred years and seems unlikely to be superseded, no matter how much it is refined and modified.
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)/01%3A_Introduction_to_Organic_Chemistry/1.01%3A_Prelude_to_Organic_Chemistry.txt |
You may not be much interested in the way that organic chemistry developed, but if you skip to the next section without reading further, you will miss some of the flavor of a truly great achievement - of how a few highly creative chemists were able, with the aid of a few simple tools, to determine the structures of molecules, far too small and too elusive to be seen individually with the finest optical microscope, manifesting themselves only by the collective behavior of at least millions of millions at once.
Try to visualize the problems confronting the organic chemist of 100 years ago. You will have no more than reasonably pure samples of organic compounds, the common laboratory chemicals of today, glassware, balances, thermometers, means of measuring densities, and a few optical instruments. You also will have a relatively embryonic theory that there are molecules in those bottles and that one compound differs from another because its molecules have different members or kinds of atoms and different arrangements of bonds. Your task will be to determine what kinds and what numbers of atoms they contain, that is, to determine their molecular formulas. Obviously, a compound with formula $C_2H_6O$ and one with $C_2H_6O_2$ are not the same compound. But suppose two compounds from different sources both are $C_2H_6O$. To decide whether these are the same or different you could smell them (far better to sniff than to inhale), taste them (emphatically not recommended), see if they have the same appearance and viscosity (if liquids), or use more sophisticated criteria: boiling point, melting point, density, or refractive index. Other possibilities would be to see if they both have the same solubility in water or other solvents and whether they give the same reaction products with various reagents. Of course, all this gets a bit tough when the compounds are not pure and no good ways are available to purify them, but that is part of the job. Think about how you might proceed.
In retrospect it is surprising that in less than fifty years an enormous, even if incomplete, edifice of structural organic chemistry was constructed on the basis of the results of chemical reactions without determination of a single bond distance, and with no electronic theory as a guide. Interestingly, all of the subsequent developments of the quantum mechanical theory of chemical bonds has not altered this edifice in significant ways. Indeed, for a long time, a goal of molecular quantum mechanics was simply to be able to corroborate that when an organic chemist draws a single line between two carbon atoms to show that they are bonded, he in fact knows what he is doing. And that when he draws two (or three) bonds between the carbons to indicate a double (or triple) bond, quantum mechanics supports this also as a valid idea.
Furthermore, when modern tools for determining organic structures that involve actually measuring the distances between the atoms became available, these provided great convenience, but no great surprises. To be sure, a few structures turned out to be incorrect because they were based on faulty or inadequate experimental evidence. But, on the whole, the modern three-dimensional representations of molecules that accord with actual measurements of bond distances and angles are in no important respect different from the widely used three-dimensional ball-and-stick models of organic molecules, and these, in essentially their present form, date from at least as far back as E. Paterno, in 1869.
How was all of this achieved? Not by any very simple process. The essence of some of the important ideas follow, but it should be clear that what actually took place was far from straightforward. A diverse group of people was involved; many firmly committed to, if not having a vested interest in, earlier working hypotheses or paradigms that had served as useful bases for earlier experimentation, but were coming apart at the seams because they could not accommodate the new facts that kept emerging. As is usual in human endeavors, espousal of new and better ideas did not come equally quickly to all those used to thinking in particular ways. To illustrate, at least one famous chemist, Berthelot, still used $HO$ as the formula for water twenty-five years after it seemed clear that $H_2O$ was a better choice.
Determination of Molecular Formulas
Before structures of molecules could be established, there had to be a means of establishing molecular formulas and for this purpose the key concept was Avogadro's hypothesis, which can be stated in the form "equal volumes of gases at the same temperature and pressure contain the same number of molecules." Avogadro's hypothesis allowed assignment of relative molecular weights from measurements of gas densities. Then, with analytical techniques that permit determination of the weight percentages of the various elements in a compound, it became possible to set up a self-consistent set of relative atomic weights.$^1$ From these and the relative molecular weights, one can assign molecular formulas. For example, if one finds that a compound contains $22.0 \%$ carbon (atomic weight $= 12.00$), $4.6 \%$ hydrogen (atomic weight $= 1.008$), and $73.4 \%$ bromine (atomic weight $= 79.90$), then the ratios of the numbers of atoms are $\left( 22.0/12.00 \right) : \left( 4.6/1.008 \right) : \left( 73.4/79.90 \right) = 1.83:4.56:0.92$. Dividing each of the last set of numbers by the smallest ($0.92$) gives $1.99:4.96:1 \cong 2:5:1$, which suggests a molecular formula of $C_2H_5Br$ or a multiple thereof. If we know that hydrogen gas is $H_2$ and has a molecular weight of $2 \times 1.008 = 2.016$, we can compare the weight of a given volume of hydrogen with the weight of the same volume of our unknown in the gas phase at the same temperature and pressure. If the experimental ratio of these weights turns out to be $54$, then the molecular weight of the unknown would be $2.016 \times 54 = 109$ and the formula $C_2H_5Br$ would be correct.
Valence
If we assume that the molecule is held together by chemical bonds, without knowing more, we could write numerous structures such as $H-H-H-H-H-C-C-Br, H-C-Br-H-H-C-H-H$, and so on. However, if we also know of the existence of stable $H_2$, but not $H_3$; of stable $Br_2$, but not of $Br_3$; and of stable $CH_3Br$, $CH_2Br_2$, $CHBr_3$, and $CBr_4$, but not of $CH_4Br$, $CHBr$, $CBr$, and so on, a pattern of what is called valence emerges. It will be seen that the above formulas all are consistent if hydrogen atoms and bromine atoms form just one bond (are univalent) while carbon atoms form four bonds (are tetravalent). This may seem almost naively simple today, but a considerable period of doubt and uncertainty preceded the acceptance of the idea of definite valences for the elements that emerged about 1852.
Structural Formulas
If we accept hydrogen and bromine as being univalent and carbon as tetravalent, we can write
as a structural formula for $C_2H_5Br$.$^2$ However, we also might have written
There is a serious problem as to whether these formulas represent the same or different compounds. All that was known in the early days was that every purified sample of $C_2H_5Br$, no matter how prepared, had a boiling point of $38^\text{o}C$ and density of $1.460 \: \text{g} \: \text{ml}^{-1}$. Furthermore, all looked the same, all smelled the same, and all underwent the same chemical reactions. There was no evidence that $C_2H_5Br$ was a mixture or that more than one compound of this formula could be prepared. One might conclude, therefore, that all of the structural formulas above represent a single substance even though they superficially, at least, look different. Indeed, because $H-Br$ and $Br-H$ are two different ways of writing a formula for the same substance, we suspect that the same is true for
There are, though, two of these structures that could be different from one another, namely
In the first of these, $CH_3-$ is located opposite the $Br-$ and the $H-$'s on the carbon with the $Br$ also are opposite one another. In the second formula, $CH_3-$ and $Br-$ are located next to each other as are the $H-$'s on the same carbon. We therefore have a problem as to whether these two different formulas also represent different compounds.
Tetrahedral Carbon
A brilliant solution to the problem posed in the preceding section came in 1874 when J. H. van't Hoff proposed that all four valences of carbon are equivalent and directed to the corners of a regular tetrahedron.$^3$ If we redraw the structures for $C_2H_5Br$ as $1$, we see that there is only one possible arrangement and, contrary to the impression we got from our earlier structural formulas, the bromine is equivalently located with respect to each of the hydrogens on the same carbon.
A convenient way of representing organic molecules in three dimensions, which shows the tetrahedral relationships of the atoms very clearly, uses the so-called ball-and-stick models. The sticks that represent the bonds or valences form the tetrahedral angles of $109.47^\text{o}$.
The Question of Rotational Isomers
The tetrahedral carbon does not solve all problems without additional postulates. For example, there are two different compounds known with the same formula $C_2H_4Br_2$. These substances, which we call isomers, can be reasonably written as
However, ball-and-stick models suggest further possibilities for the second structure, for example $3$, $4$, and $5$:
This is a problem apparently first clearly recognized by Paterno, in 1869. We call these rotational (or conformational) isomers, because one is converted to another by rotation of the halves of the molecule with respect to one another, with the $C-C$ bond acting as an axle. If this is not clear, you should make a ball-and-stick model and see what rotation around the $C-C$ bond does to the relationships between the atoms on the carbons.
The difficulty presented by these possibilities finally was circumvented by a brilliant suggestion by van't Hoff of "free rotation," which holds that isomers corresponding to different rotational angles, such as $3$, $4$, and $5$, do not have separate stable existence, but are interconverted by rotation around the $C-C$ bond so rapidly that they are indistinguishable from one another. Thus there is only one isomer corresponding to the different possible rotational angles and a total of only two isomers of formula $C_2H_4Br_2$. As we shall see, the idea of free rotation required extensive modification some 50 years after it was first proposed, but it was an extremely important paradigm, which, as often happens, became so deeply rooted as to become essentially an article of faith for later organic chemists. Free rotation will be discussed in more detail in Chapters 5 and 27.
The Substitution Method for Proof of Structure
The problem of determining whether a particular isomer of $C_2H_4Br_2$ is
could be solved today in a few minutes by spectroscopic means, as will be explained in Chapter 9. However, at the time structure theory was being developed, the structure had to be deduced on the basis of chemical reactions, which could include either how the compound was formed or what it could be converted to. A virtually unassailable proof of structure, where it is applicable, is to determine how many different substitution products each of a given group of isomers can give. For the $C_2H_4Br_2$ pair of isomers, substitution of a bromine for a hydrogen will be seen to give only one possibility with one compound and two with the other:
Therefore, if we have two bottles, one containing one $C_2H_4Br_2$ isomer and one the other and run the substitution test, the compound that gives only one product is $6$ and the one that gives a mixture of two products is $7$. Further, it will be seen that the test, besides telling which isomer is $6$ and which is $7$, establishes the structures of the two possible $C_2H_3Br_3$ isomers, $8$ and $9$. Thus only $8$ can be formed from both of the different $C_2H_4Br_2$ isomers whereas $9$ is formed from only one of them.
The Benzene Problem
There were already many interconversion reactions of organic compounds known at the time that valence theory, structural formulas, and the concept of the tetrahedral carbon came into general use. As a result, it did not take long before much of organic chemistry could be fitted into a concordant whole. One difficult problem was posed by the structures of a group of substitution products of benzene, $C_6H_6$, called "aromatic compounds," which for a long time defied explanation. Benzene itself had been prepared first by Michael Faraday, in 1825. An ingenious solution for the benzene structure was provided by A. Kekule, in 1866, wherein he suggested (apparently as the result of a hallucinatory perception) that the six carbons were connected in a hexagonal ring with alternating single and double carbon-to-carbon bonds, and with each carbon connected to a single hydrogen, $10$:
This concept was controversial, to say the least, mainly on two counts. Benzene did not behave as expected, as judged by the behavior of other compounds with carbon-to-carbon double bonds and also because there should be two different dibromo substitution products of benzene with the bromine on adjacent carbons ($11$ and $12$) but only one such compound could be isolated.
Kekule explained the second objection away by maintaining that $11$ and $12$ were in rapid equilibrium through concerted bond shifts, in something like the same manner as the free-rotation hypothesis mentioned previously:
However, the first objection could not be dismissed so easily and quite a number of alternative structures were proposed over the ensuing years. The controversy was not really resolved until it was established that benzene is a regular planar hexagon, which means that all of its $C-C$ bonds have the same length, in best accord with a structure written not with double, not with single, but with 1.5 bonds between the carbons, as in $13$:
This. in turn, generated a massive further theoretical controversy over just how $13$ should be interpreted, which, for a time, even became a part of "Cold-War" politics!$^4$ We shall examine experimental and theoretical aspects of the benzene structure in some detail later. It is interesting that more than 100 years after Kekule's proposal the final story on the benzene structure is yet to be told.$^5$
Proof of Structure through Reactions
The combination of valence theory and the substitution method as described in Section 1-1F gives, for many compounds, quite unequivocal proofs of structure. Use of chemical transformations for proofs of structure depends on the applicability of a simple guiding principle, often called the "principle of least structural change." As we shall see later, many exceptions are known and care is required to keep from making serious errors. With this caution, let us see how the principle may be applied. The compound $C_2H_5Br$ discussed in Section 1-1A reacts slowly with water to give a product of formula $C_2H_6O$. The normal valence of oxygen is two, and we can write two, and only two, different structures, $19$ and $20$, for $C_2H_6O$:
The principle of least structural change favors $19$ as the product, because the reaction to form it is a simple replacement of bromine bonded to carbon by $-OH$, whereas formation of $20$ would entail a much more drastic rearrangement of bonds. The argument is really a subtle one, involving an assessment of the reasonableness of various possible reactions. On the whole, however, it works rather well and, in the specific case of the $C_2H_6O$ isomers, is strongly supported by the fact that treatment of $19$ with strong hydrobromic acid ($HBr$) converts it back to $C_2H_5Br$. In contrast, the isomer of structure $20$ reacts with $HBr$ to form two molecules of $CH_3Br$:
In each case, $C-O$ bonds are broken and $C-Br$ bonds are formed.
We could conceive of many other possible reactions of $C_2H_6O$ with $HBr$, for example
which, as indicated by $\nrightarrow$, does not occur, but hardly can be ruled out by the principle of least structural change itself. Showing how the probability of such alternative reactions can be evaluated will be a very large part of our later discussions.
Reactivity, Saturation, Unsaturation, and Reaction Mechanisms
The substitution method and the interconversion reactions discussed for proof of structure possibly may give you erroneous ideas about the reactions and reactivity of organic compounds. We certainly do not wish to imply that it is a simple, straightforward process to make all of the possible substitution products of a compound such as
In fact, as will be shown later, direct substitution of bromine for hydrogen with compounds such as this does not occur readily, and when it does occur, the four possible substitution products indeed are formed, but in far from equal amounts because there are differences in reactivity for substitution at the different positions. Actually, some of the substitution products are formed only in very small quantities. Fortunately, this does not destroy the validity of the substitution method but does make it more difficult to apply. If direct substitution fails, some (or all) of the possible substitution products may have to be produced by indirect means. Nonetheless, you must understand that the success of the substitution method depends on determination of the total number of possible isomers - it does not depend on how the isomers are prepared.
Later, you will hear a lot about compounds or reagents being "reactive" and "unreactive." You may be exasperated by the loose way that these terms are used by organic chemists to characterize how fast various chemical changes occur. Many familiar inorganic reactions, such as the neutralization of hydrochloric acid with sodium hydroxide solution, are extremely fast at ordinary temperatures. But the same is not often true of reactions of organic compounds. For example, $C_2H_5Br$ treated in two different ways is converted to gaseous compounds, one having the formula $C_2H_6$ and the other $C_2H_4$. The $C_2H_4$ compound, ethene, reacts very quickly with bromine to give $C_2H_4Br_2$, but the $C_2H_6$ compound, ethane, does not react with bromine except at high temperatures or when exposed to sunlight (or similar intense light). The reaction products then are $HBr$ and $C_2H_5Br$, and later, $HBr$ and $C_2H_4Br_2$, $C_2H_3Br_3$, and so on.
We clearly can characterize $C_2H_4$ as "reactive" and $C_2H_6$ as "unreactive" toward bromine. The early organic chemists also used the terms "unsaturated" and "saturated" for this behavior, and these terms are still in wide use today. But we need to distinguish between "unsaturated" and "reactive," and between "saturated" and "unreactive," because these pairs of terms are not synonymous. The equations for the reactions of ethene and ethane with bromine are different in that ethene adds bromine, $C_2H_4 + Br_2 \rightarrow C_2H_4Br_2$, whereas ethane substitutes bromine, $C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr$.
You should reserve the term "unsaturated" for compounds that can, at least potentially, react by addition, and "saturated' for compounds that can only be expected to react by substitution. The difference between addition and substitution became much clearer with the development of the structure theory that called for carbon to be tetravalent and hydrogen univalent. Ethene then was assigned a structure with a carbon-to-carbon double bond, and ethane a structure with a carbon-to-carbon single bond:
Addition of bromine to ethene subsequently was formulated as breaking one of the carbon-carbon bonds of the double bond and attaching bromine to these valences. Substitution was written similarly but here bromine and a $C-H$ bond are involved:
We will see later that the way in which these reactions actually occur is much more complicated than these simple equations indicate. In fact, such equations are regarded best as chemical accounting operations. The number of bonds is shown correctly for both the reactants and the products, and there is an indication of which bonds break and which bonds are formed in the overall process. However, do not make the mistake of assuming that no other bonds are broken or made in intermediate stages of the reaction.
Much of what comes later in this book will be concerned with what we know, or can find out, about the mechanisms of such reactions - a reaction mechanism being the actual sequence of events by which the reactants become converted to the products. Such information is of extraordinary value in defining and understanding the range of applicability of given reactions for practical preparations of desired compounds.
The distinction we have made between "unsaturated" and "reactive" is best illustrated by a definite example. Ethene is "unsaturated" (and "reactive") toward bromine, but tetrachloroethene, $C_2Cl_4$, will not add bromine at all under the same conditions and is clearly "unreactive." But is it also "saturated"?
The answer is definitely no, because if we add a small amount of aluminum bromide, $AlBr_3$, to a mixture of tetrachloroethene and bromine, addition does occur, although sluggishly:
Obviously, tetrachloroethene is "unsaturated" in the sense it can undergo addition, even if it is unreactive to bromine in the absence of aluminum bromide.
The aluminum bromide functions in the addition of bromine to tetrachloroethene as a catalyst, which is something that facilitates the conversion of reactants to products. The study of the nature and uses of catalysts will concern us throughout this book. Catalysis is our principal means of controlling organic reactions to help form the product we want in the shortest possible time.
$^1$We will finesse here the long and important struggle of getting a truly self-consistent table of atomic weights. If you are interested in the complex history of this problem and the clear solution to it proposed by S. Cannizzaro in 1860, there are many accounts available in books on the history of chemistry. One example is J. R. Partington, A History of Chemistry, Vol. IV, Macmillan, London, 1964. Relative atomic weights now are based on $^{12}C = 12$ (exactly).
$^2$Formulas such as this appear to have been used first by Crum Brown, in 1864, after the originators of structural formulas, A. Kekule and A. Couper (1858), came up with rather awkward, impractical representations. It seems incredible today that even the drawing of these formulas was severely criticized for many years. The pot was kept boiling mainly by H. Kolbe, a productive German chemist with a gift for colorful invective and the advantage of a podium provided by being editor of an influential chemical journal.
$^3$The name of J. A. Le Bel also is associated with this particular idea, but the record shows that Le Bel actually opposed the tetrahedral formulations, although, simultaneously with van't Hoff, he made a related very important contribution, as will be discussed in Chapter 5.
$^4$The "resonance theory," to be discussed in detail in Chapters 6 and 21, was characterized in 1949 as a physically and ideologically inadmissible theory formulated by "decadent bourgeois scientists." See L. R. Graham, Science and Philosophy in the Soviet Union, Vintage Books, New York, 1973, Chapter VIII, for an interesting account of this controversy.
$^5$Modern organic chemistry should not be regarded at all as a settled science, free of controversy. To be sure, personal attacks of the kind indulged in by Kolbe and others often are not published, but profound and indeed acrimonious differences of scientific interpretation exist and can persist for many years.
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)/01%3A_Introduction_to_Organic_Chemistry/1.02%3A_A_Bit_of_History.txt |
We have tried to give you a taste of the beginnings of organic chemistry and a few of the important principles that brought order out of the confusion that existed as to the nature of organic compounds. Before moving on to other matters, it may be well to give you some ideas of what kind of preparation will be helpful to you in learning about organic chemistry from this textbook.
The most important thing you can bring is a strong desire to master the subject. We hope you already have some knowledge of general chemistry and that you already will have had experience with simple inorganic compounds. That you will know, for example, that elemental bromine is $Br_2$ and a noxious, dark red-brown, corrosive liquid; that sulfuric acid is $H_2SO_4$, a syrupy colorless liquid that reacts with water with the evolution of considerable heat and is a strong acid; that sodium hydroxide is $NaOH$, a colorless solid that dissolves in water to give a strongly alkaline solution. It is important to know the characteristics of acids and bases, how to write simple, balanced chemical reactions, such as $2H_2 + O_2 \rightarrow 2H_2O$, and $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$, what the concept of a mole of a chemical substance is, and to be somewhat familiar with the periodic table of the elements as well as with the metric system, at least insofar as grams, liters, and degrees centigrade are concerned. Among other things, you also should understand the basic ideas of the differences between salts and covalent compounds, as well as between gases, liquids, and solids; what a solution is; the laws of conservation of mass and energy; the elements of how to derive the Lewis electron structures of simple molecules such as $H : \underset{\cdot \cdot}{\ddot{O}} : H =$ water; that $PV = nRT$; and how to calculate molecular formulas from percentage compositions and molecular weights. We shall use no mathematics more advanced than simple algebra but we do expect that you can use logarithms and are able to carry through the following conversions forward and backward:
$\text{log}_{10} \: 510,000 = \: \text{log}_{10} \left( 5.1 \times 10^5 \right) = 5.708$
The above is an incomplete list, given to illustrate the level of preparation we are presuming in this text. If you find very much of this list partly or wholly unfamiliar, you don't have to give up, but have a good general chemistry textbook available for study and reference - and use it! Some useful general chemistry books are listed at the end of the chapter. A four-place table of logarithms will be necessary; a set of ball-and-stick models and a chemical handbook will be very helpful, as would be a small electronic calculator or slide rule to carry out the simple arithmetic required for many of the exercises.
In the next section, we review some general chemistry regarding saltlike and covalent compounds that will be of special relevance to our later discussions.
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)/01%3A_Introduction_to_Organic_Chemistry/1.03%3A_What_Preparation_Should_You_Have.txt |
Let us consider some of the factors that make so much of chemistry center on a single element, carbon. One very important feature is that carbon-carbon bonds are strong, so long chains or rings of carbon atoms bonded to one another are possible. Diamond and graphite are two familiar examples, the diamond lattice being a three-dimensional network of carbon atoms, whereas graphite actually more closely resembles a planar network. The lubricating properties of graphite actually are related to its structure, which permits the planes to slide one past the other.
But carbon is not unique in forming bonds to itself because other elements such as boron, silicon, and phosphorus form strong bonds in the elementary state. The uniqueness of carbon stems more from the fact that it forms strong carbon-carbon bonds that also are strong when in combination with other elements. For example, the combination of hydrogen with carbon affords a remarkable variety of carbon hydrides, or hydrocarbons as they usually are called. In contrast, none of the other second-row elements except boron gives a very extensive system of stable hydrides, and most of the boron hydrides are much more reactive than hydrocarbons, especially to water and air.
Carbon forms bonds not only with itself and with hydrogen but also with many other elements, including strongly electron-attracting elements such as fluorine and strongly electropositive metals such as lithium:
Why is carbon so versatile in its ability to bond to very different kinds of elements? The special properties of carbon can be attributed to its being a relatively small atom with four valence electrons. To form simple saltlike compounds such as sodium chloride, $Na^\oplus Cl^\ominus$, carbon would have to either lose the four valence electrons to an element such as fluorine and be converted to a quadripositive ion, $C^{4 \oplus}$, or acquire four electrons from an element such as lithium and form a quadrinegative ion, $C^{4 \ominus}$. Gain of four electrons would be energetically very unfavorable because of mutual repulsion between the electrons.
Customarily, carbon completes its valence-shell octet by sharing electrons with other atoms. In compounds with shared electron bonds (or covalent bonds) such as methane, ethane, or tetrafluoromethane, each of the bonded atoms including carbon has its valence shell filled, as shown in the following electron-pair or Lewis$^6$ structures:
In this way, repulsions between electrons associated with completion of the valence shell of carbon are compensated by the electron-attracting powers of the positively charged nuclei of the atoms to which the carbon is bonded.
However, the electrons of a covalent bond are not necessarily shared equally by the bonded atoms, especially when the affinities of the atoms for electrons are very different. Thus, carbon-fluorine and carbon-lithium bonds, although they are not ionic, are polarized such that the electrons are associated more with the atom of higher electron affinity. This is usually the atom with the higher effective nuclear charge.
$\overset{\delta \oplus}{C} \: \: \: \: \: : \overset{\delta \ominus}{F} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \overset{\delta \ominus}{C:} \: \: \: \: \: \overset{\delta \oplus}{Li} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \scriptsize{\left( \delta \oplus, \: \delta \ominus \: \text{denote partial ionic bonds} \right)}$
We see then a gradation from purely ionic to purely covalent bonding in different molecules, and this is manifest in their chemical and physical properties. Consider, for instance, the hydrides of the elements in the second horizontal row of the periodic table. Their melting and boiling points,$^7$ where known, are given below.
Lithium hydride can be regarded as a saltlike ionic compound, $\overset{\oplus}{Li} \: \: \: : \overset{\ominus}{H}$. Electrostatic attractions between oppositely charged ions in the crystal lattice are strong, thereby causing lithium hydride to be a high-melting, nonvolatile solid like sodium chloride, lithium fluoride, and so on.
Methane, $CH_4$, is at the other extreme. It boils at $-161^\text{o}$, which is about $800^\text{o}$ lower even than the melting point of lithium hydride. Because carbon and hydrogen have about the same electron-attracting power, $C-H$ bonds have little ionic character, and methane may be characterized as a nonpolar substance. As a result, there is relatively little electrostatic attraction between methane molecules and this allows them to "escape" more easily from each other as gaseous molecules - hence the low boiling point.
Hydrogen fluoride has a boiling point some $200^\text{o}$ higher than that of methane. The bonding electron pair of $HF$ is drawn more toward fluorine than to hydrogen so the bond may be formulated as $\overset{\delta \oplus}{H}$ ---- $\overset{\delta \ominus}{F}$. In liquid hydrogen fluoride, the molecules tend to aggregate through what is called hydrogen bonding in chains and rings arranged so the positive hydrogen on one molecule attracts a negative fluorine on the next:
When liquid hydrogen fluoride is vaporized, the temperature must be raised sufficiently to overcome these intermolecular electrostatic attractions; hence the boiling point is high compared to liquid methane. Hydrogen fluoride is best characterized as a polar, but not ionic, substance. Although the $O-H$ and $N-H$ bonds of water and ammonia have somewhat less ionic character than the $H-F$ bonds of hydrogen fluoride, these substances also are relatively polar in nature and also associate through hydrogen bonding in the same way as does hydrogen fluoride.
The chemical properties of lithium hydride, methane, and hydrogen fluoride are in accord with the above formulations. Thus, when the bond to the hydrogen is broken, we might expect it to break in the sense $\begin{array}{c:c} Li^\oplus & :H^\ominus \end{array}$ for lithium hydride, and $\begin{array}{c:c} \overset{\delta \oplus}{H} & : \underset{\cdot \cdot}{\ddot{F}}:^{\delta \ominus} \end{array}$ for hydrogen fluoride so that the electron pair goes with the atom of highest electron affinity. This is indeed the case as the following reaction indicates:
$\begin{array}{c:c} Li^\oplus & :H^\ominus \end{array} + \begin{array}{c:c} H & : \underset{\cdot \cdot}{\ddot{F}} : \end{array} \longrightarrow Li^\oplus : \underset{\cdot \cdot}{\ddot{F}}:^\oplus + \: H : H$
Methane, with its relatively nonpolar bonds, is inert to almost all reagents that could remove hydrogen as $H^\oplus$ or $H:^\ominus$ except under anything but extreme conditions. As would be expected, methyl cations $CH_3^\oplus$ and methyl anions $CH_3:^\ominus$ are very difficult to generate and are extremely reactive. For this reason, the following reactions are not observed:
From the foregoing you may anticipate that the chemistry of carbon compounds will be largely the chemistry of covalent compounds and will not at all resemble the chemistry of inorganic salts such as sodium chloride. You also may anticipate that the major differences in chemical and physical properties of organic compounds will arise from the nature of the other elements bonded to carbon. Thus methane is not expected to, nor does it have, the same chemistry as other one-carbon compounds such as methyllithium, $CH_3Li$, or methyl fluoride, $CH_3F$.
$^6$G. N. Lewis (1876-1946), the renowned U.S. chemist, was the first to grasp the significance of the electron-pair in molecular structure. He laid the foundation for modern theory of structure and bonding in his treatise on Valence and the Structure of Atoms and Molecules (1923).
$^7$Throughout this text all temperatures not otherwise designated should be understood to be in $^\text{o}C$; absolute temperatures will be shown as $^\text{o}K$.
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)/01%3A_Introduction_to_Organic_Chemistry/1.04%3A_Why_Is_Organic_Chemistry_Special.txt |
Organic chemistry originally was defined as the chemistry of those substances formed by living matter and, for quite a while, there was a firm belief that it would never be possible to prepare organic compounds in the laboratory outside of a living system. However, after the discovery by Wohler, in 1828, that a supposedly typical organic compound, urea, could be prepared by heating an inorganic salt, ammonium cyanate, this definition gradually lost significance and organic chemistry now is broadly defined as the chemistry of carbon-containing compounds. Nonetheless, the designation "organic" is still very pertinent because the chemistry of organic compounds is also the chemistry of living organisms.
Each of us and every other living organism is comprised of, and endlessly manufactures, organic compounds. Further, all organisms consume organic compounds as raw materials, except for those plants that use photosynthesis or related processes to synthesize their own from carbon dioxide. To understand every important aspect of this chemistry, be it the details of photosynthesis, digestion, reproduction, muscle action, memory or even the thought process itself, is a primary goal of science and it should be recognized that only through application of organic chemistry will this goal be achieved.
Modern civilization consumes vast quantities of organic compounds. Coal, petroleum, and natural gas are primary sources of carbon compounds for use in production of energy and as starting materials for the preparation of plastics, synthetic fibers, dyes, agricultural chemicals, pesticides, fertilizers, detergents, rubbers and other elastomers, paints and other surface coatings, medicines and drugs, perfumes and flavors, antioxidants and other preservatives, as well as asphalts, lubricants, and solvents that are derived from petroleum.
Much has been done and you soon may infer from the breadth of the material that we will cover that most everything worth doing already has been done. However, many unsolved scientific problems remain and others have not even been thought of but, in addition, there are many technical and social problems to which answers are badly needed. Some of these include problems of pollution of the environment, energy sources, overpopulation and food production, insect control, medicine, drug action, and improved utilization of natural resources.
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."
1.06: Some Philosophical Observations
As you proceed with your study of organic chemistry, you may well feel confused as to what it is you are actually dealing with. On the one hand, there will be exhortations to remember how organic chemistry pervades our everyday life. And yet, on the other hand, you also will be exhorted to think about organic compounds in terms of abstract structural formulas representing molecules when there is absolutely no way at all to deal with molecules as single entities. Especially if you are not studying organic compounds in the laboratory concurrently, you may come to confuse the abstraction of formulas and ball-and-stick models of the molecules with the reality of organic compounds, and this would be most undesirable. At each stage of the way, you should try to make, or at least visualize, a juncture between a structural formula and an actual substance in a bottle. This will not be easy - it takes time to reach the level of experience that a practicing organic chemist has so that he can tell you with some certainty that the structural formula $21$ represents in actuality, a limpid, colorless liquid with a pleasant odor, slightly soluble in water, boiling somewhere about $100^\text{o}$.
A useful method for developing this sort of feeling for the relationship between structures and actual compounds is to check your perception of particular substances with their properties as given in a chemical handbook.
One, perhaps comforting, thought for you at this time is that differences between the chemical behaviors of relatively similar organic compounds usually are ascribed to just three important and different kinds of effects - two of which have root in common experience. One, called steric hindrance, is a manifestation of experience that two solid objects cannot occupy the same space at once. Another is the electrical effect, which boils down to a familiar catechism that like electrical charges repel each other and unlike charges attract each other. The remaining important effect, the one that has no basis in common experience, derives from quantum mechanics. The quantum mechanical effect explains why benzene is unusually stable, how and why many reactions occur in special ways and, probably most important of all, the ways that organic compounds interact with electromagnetic radiation of all kinds - from radio waves to x rays.
We shall try to give as clear explanations as possible of the quantum mechanical effects, but some of it will just have to be accepted as fact that we cannot ourselves experience directly nor understand intuitively. For example, when a grindstone rotates, so far as our experience goes, it can have an infinitely variable rate of rotation and, consequently, infinitely variable rotation (angular) momentum. However, molecules in the gas phase have only specific rotation rates and corresponding specific rotational momentum values. No measurement technique can detect in-between values of these quantities. Molecules are "quantized rotators." About all you can do is try to accept this fact, and if you try long enough, you may be able to substitute familiarity for understanding and be happy with that.
All of us have some concepts we use continually (even perhaps unconsciously) about energy and work. Thermodynamics makes these concepts quantitative and provides very useful information about what might be called the potential for any process to occur, be it production of electricity from a battery, water running uphill, photosynthesis, or formation of nitrogen oxides in combustion of gasoline. In the past, most organic chemists seldom tried to apply thermodynamics to the reactions in which they were interested. Much of this was due to the paucity of thermodynamic data for more than a few organic compounds, but some was because organic chemists often liked to think of themselves as artistic types with little use for quantitative data on their reactions (which may have meant that they didn't really know about thermodynamics and were afraid to ask).
Times have changed. Extensive thermochemical data are now available, the procedures are well understood, and the results both useful and interesting. We shall make considerable use of thermodynamics in our exposition of organic chemistry. We believe it will greatly improve your understanding of why some reactions go and others do not.
Finally, you should recognize that you almost surely will have some problems with the following chapters in making decisions as to how much time and emphasis you should put on the various concepts, principles, facts, and so on, that we will present for you. As best we can, we try to help you by pointing out that this idea, fact, and so on, is "especially important," or words to that effect. Also, we have tried to underscore important information by indicating the breadth of its application to other scientific disciplines as well as to technology. In addition, we have caused considerable material to be set in smaller type and indented. Such material includes extensions of basic ideas and departments of fuller explanation. In many places, the exposition is more complete than it needs to be for you at the particular location in the book. However, you will have need for the extra material later and it will be easier to locate and easier to refresh your memory on what came before, if it is one place. We will try to indicate clearly what you should learn immediately and what you will want to come back for later.
The problem is, no matter what we think is important, you or your professor will have your own judgments about relevance. And because it is quite impossible to write an individual text for your particular interests and needs, we have tried to accommodate a range of interests and needs through providing a rather rich buffet of knowledge about modern organic chemistry. Hopefully, all you will need is here, but there is surely much more, too. So, to avoid intellectual indigestion, we suggest you not try to learn everything as it comes, but rather try hardest to understand the basic ideas and concepts to which we give the greatest emphasis. As you proceed further, the really important facts, nomenclature, and so on (the kind of material that basically requires memorization), will emerge as that which, in your own course of study, you will find you use over and over again. In hope that you may wish either to learn more about particular topics or perhaps gain better understanding through exposure to a different perspective on how they can be presented, we have provided supplementary reading lists at the end of each chapter.
Our text contains many exercises. You will encounter some in the middle of the chapters arranged to be closely allied to the subject at hand. Others will be in the form of supplementary exercises at the end of the chapters. Many of the exercises will be drill; many others will extend and enlarge upon the text. The more difficult problems are marked with a star ($^\mathbf{*}$).
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)/01%3A_Introduction_to_Organic_Chemistry/1.05%3A_The_Breadth_of_Organic_Chemistry.txt |
In this chapter, we first briefly review the most important types of covalent bonds encountered in organic substances and the ways in which these bonds are represented in structural formulas. Next we consider the sizes and shapes of organic molecules and how structural formulas written in two dimensions can be translated into three-dimensional models that show the relative positions of the atoms in space. We also discuss models that reflect the relative sizes of the atoms and the way in which the atoms may interfere with each other when in close quarters (steric hindrance). Then we go on to further important aspects of structure-the functional group concept and position isomerism. Our aim is to have you become more familiar with the various kinds of organic compounds and begin to see how the practicing organic chemist visualizes molecules and correlates the diverse kinds of structures that he has to deal with in his work.
• 2.1: Structural Formulas
With few exceptions, carbon compounds can be formulated with four covalent bonds to each carbon, regardless of whether the combination is with carbon or some other element. The two-electron bond is called a single bond. Compounds in which two electrons from each of the carbon atoms are mutually shared, produce two two-electron bonds, which are called double bonds. Similarly, when three electrons from each carbon atom are mutually shared, three two-electron bonds are made, called triple bonds.
• 2.2: The Sizes and Shapes of Organic Molecules
The size and shape of molecules are as much a part of molecular structure as is the order in which the component atoms are bonded. Contrary to the impression you may get from structural formulas, complex molecules are not flat and formless, but have well-defined spatial arrangements that are determined by the lengths and directional character of their chemical bonds. Mechanical model are useful that address molecular geometry, including approximations of relative bond lengths and orientation.
• 2.3: Classification by Functional Groups
There are a number of recurring types of structural features in organic compounds that commonly are known as functional groups. In fact, a traditional approach to the subject of organic chemistry involves the classification of compounds according to their functional groups. It will be helpful to look at the structural features of some of the major types of organic compounds even though the details of their chemistry will not be discussed until later chapters.
• 2.4: Isomerization in Organic Compounds
More than one stable substance can correspond to a given molecular formula. Compounds having the same number and kinds of atoms are called isomers. Compounds having the same number and kind of atoms but having different bonding arrangements between the atoms are called position isomers.
• 2.E: Structural Organic Chemistry (Exercises)
These are the homework exercises to accompany Chapter 2 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."
02: Structural Organic Chemistry
The building block of structural organic chemistry is the tetravalent carbon atom. With few exceptions, carbon compounds can be formulated with four covalent bonds to each carbon, regardless of whether the combination is with carbon or some other element. The two-electron bond, which is illustrated by the carbon-hydrogen bonds in methane or ethane and the carbon-carbon bond in ethane, is called a single bond. In these and many related substances, each carbon is attached to four other atoms:
There exist, however, compounds such as ethene (ethylene), $C_2H_4$, in which two electrons from each of the carbon atoms are mutually shared, thereby producing two two-electron bonds, an arrangement which is called a double bond. Each carbon in ethene is attached to only three other atoms:
Similarly, in ethyne (acetylene), $C_2H_2$, three electrons from each carbon atom are mutually shared, producing three two-electron bonds, called a triple bond, in which each carbon is attached to only two other atoms:
Of course, in all cases each carbon has a full octet of electrons. Carbon also forms double and triple bonds with several other elements that can exhibit a covalence of two or three. The carbon-oxygen (or carbonyl) double bond appears in carbon dioxide and many important organic compounds such as methanal (formaldehyde) and ethanoic acid (acetic acid). Similarly, a carbon-nitrogen triple bond appears in methanenitrile (hydrogen cyanide) and ethanenitrile (acetonitrile).
By convention, a single straight line connecting the atomic symbols is used to represent a single (two-electron) bond, two such lines to represent a double (four-electron) bond, and three lines a triple (six-electron) bond. Representations of compounds by these symbols are called structural formulas; some examples are
A point worth noting is that structural formulas usually do not indicate the nonbonding electron pairs. This is perhaps unfortunate because they play as much a part in the chemistry of organic molecules as do the bonding electrons and their omission may lead the unwary reader to overlook them. However, when it is important to represent them, this can be done best with pairs of dots, although a few authors use lines:
To save space and time in the representation of organic structures, it is common practice to use "condensed formulas" in which the bonds are not shown explicitly. In using condensed formulas, normal atomic valences are understood throughout. Examples of condensed formulas are
Another type of abbreviation that often is used, particularly for ring compounds, dispenses with the symbols for carbon and hydrogen atoms and leaves only the lines in a structural formula. For instance, cyclopentane, $C_5H_{10}$, often is represented as a regular pentagon in which it is understood that each apex represents a carbon atom with the requisite number of hydrogens to satisfy the tetravalence of carbon:
Likewise, cyclopropane, $C_3H_6$; cyclobutane, $C_4H_8$; and cyclohexane, $C_6H_{12}$, are drawn as regular polygons:
Although this type of line drawing is employed most commonly for cyclic structures, its use for open chain (acyclic) structures is becoming increasingly widespread. There is no special merit to this abbreviation for simple structures such as butane, $C_4H_{10}$; 1-butene, $C_4H_8$; or 1,3-butadiene, $C_4H_6$, but it is of value in representing more complex molecules such as $\beta$-carotene, $C_{40}H_{56}$:
Line structures also can be modified to represent the three-dimensional shapes of molecules, and the way that this is done will be discussed in detail in Chapter 5. At the onset of you study of organic chemistry, you should write out the formulas rather completely until you are thoroughly familiar with what these abbreviations stand for.
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)/02%3A_Structural_Organic_Chemistry/2.01%3A_Structural_Formulas.txt |
The size and shape of molecules are as much a part of molecular structure as is the order in which the component atoms are bonded. Contrary to the impression you may get from structural formulas, complex molecules are not flat and formless, but have well-defined spatial arrangements that are determined by the lengths and directional character of their chemical bonds. It is not easy to visualize the possible arrangements of the bonds in space and it is very helpful to have some kind of mechanical model that reflects the molecular geometry, including at least an approximation to the relative lengths of the bonds. "Ball-and-stick" models such as the ones used by Paterno (Section 1-1D) fill this purpose admirably.
Bond Angles and Ball-and-Stick Models
It is well established that the normal carbon atom forms its four single bonds in compounds of the type $CX_4$ so that the four attached atoms lie at the corners of a regular tetrahedron. The bond angles $X-C-X$ are $109.5^\text{o}$ and this value is the "normal" valence angle of carbon. For many purposes, ball-and-stick models of organic compounds give useful information about the spatial relationships of the atoms, and for $CX_4$ the angles between sticks are set at $109.5^\text{o}$ (Figure 2-1). Organic molecules strongly resist deformation forces that alter their valence angles from normal values. Therefore ball-and-stick models correspond better to the behavior of actual molecules if the connectors representing single bonds are made to be rather stiff.
Whereas methane, $CH_4$, is tetrahedral, ethene, $C_2H_4$, is not. According to the best available physical measurements, all six atoms of ethene lie in a single plane and the $H-C-H$ bond angles are $117.3^\text{o}$. Methanal (formaldehyde) also is a planar molecule with an $H-C-H$ bond angle of $118^\text{o}$.
Ethyne, $C_2H_2$, has been established experimentally to be a linear molecule; that is, the $H-C-C$ bond angle is $180^\text{o}$:
Structural units that have $C-C-C$ valence angles substantially less than the tetrahedral value include double and triple bonds, and small rings such as cyclopropane. Several bent bonds are required to construct models of compounds containing these units. Interestingly, such compounds are much less stable and more reactive than otherwise similar molecules for which models can be constructed with straight sticks at tetrahedral angles.
Bond Lengths and Space-Filling Models
The length of a chemical bond is the average distance between the nuclei of two bonded atoms, regardless of where the bonding electrons happen to be. The customary unit of length is the angstrom$^1$ ($\text{A} = 10^{-10} \: \text{m}$), and measurements often can be made with an accuracy of $0.001 \: \text{A}$ by using the techniques of molecular spectroscopy, x-ray diffraction (for crystalline solids), and electron diffraction (for volatile compounds). Bond lengths vary considerably with structure and depend on the identity of both atoms, the type of bonding (single, double, or triple), and the nature of other atoms or groups bonded to the two atoms in question. These effects are apparent in the data of Table 2-1, which lists the bond lengths in several simple organic compounds. Multiple bonds, double or triple, clearly are shorter than single bonds, and it can be stated as a general observation that the more bonding electrons in a given bond, the shorter (and stronger) the bond. The lengths of single $C-C$ bonds also vary significantly depending on what other atoms or groups are attached to the carbons. Thus Table 2-1 shows that single $C-C$ bonds become progressively shorter as the number of multiple bonds or electronegative atoms attached to the carbons increases.
Although molecular models cannot represent the subtle variations in bond lengths and bond angles that actual molecules exhibit, most kinds of commercially available molecular models do attempt to reproduce relative bond lengths with some degree of reality. In the ball-and-stick type, the sticks usually come in various lengths to simulate different kinds of bonds; $C-H$ bonds typically are scaled to represent $1.1 \: \text{A}$, $C-C$ bonds to be $1.54 \: \text{A}$, and $C=C$ and $C \equiv C$ to be correspondingly shorter. In some model sets the bonds can be cut to any desired length.
While the ball-and-stick models of molecules are very useful for visualizing the relative positions of the atoms in space, they are unsatisfactory whenever we also want to show how large the atoms are. Actually, atomic radii are so large relative to the lengths of chemical bonds that when a model of a molecule such as methyl chloride is constructed with atomic radii and bond lengths, both to scale, the bonds connecting the atoms are not clearly evident. None-
As we shall see, such crowding has many chemical consequences.
Ideally, a model should reflect not only the size and shape of the molecule it represents but also the flexibility of the molecule. By this we mean that
it should simulate the type of motions available to the molecule, particularly bond rotation. For example, it is known that rotation normally occurs about single bonds in open-chain compounds but is restricted about double bonds. Motions of this kind are demonstrated easily with ball-and-stick models, but are not at all obvious with the space-filling type. For this reason, ball-and-stick models or their equivalent are more generally useful than the space-filling models for visualizing structures and the positions of the atoms relative to one another.
$^1$The angstrom unit likely will be replaced eventually by the nanometer ($1 \: \text{nm} = 10^{-9} \: \text{m} = 10 \: \text{A}$).
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)/02%3A_Structural_Organic_Chemistry/2.02%3A_The_Sizes_and_Shapes_of_Organic_Molecules.txt |
There are a number of recurring types of structural features in organic compounds that commonly are known as functional groups. In fact, a traditional approach to the subject of organic chemistry involves the classification of compounds according to their functional groups. Thus the structural features $C = C$, $C \equiv C$, $C = O$, $OH$, $NH_2$, and $C \equiv N$ are the functional groups of alkenes, alkynes, carbonyl compounds, alcohols, amines, and nitriles, respectively. It will be helpful to look at the structural features of some of the major types of organic compounds even though the details of their chemistry will not be discussed until later chapters. Examples of the structures arranged in accord with their functional groups are given in Table 2-2. The examples chosen are representative of compounds containing carbon and hydrogen (hydrocarbons) as well as compounds containing halogens, oxygen, nitrogen, and sulfur. We do not expect you to memorize this table. In time you will become familiar with all of the types of structures in it.
In Table 2-2 we generally have used systematic names as first-choice names because these names emphasize the relationships between the compounds and ease the burden fo the beginning student in having to remember many special names. We have little hope that systematic names such as methanal, 2-propanone, and ethanoic acid soon will replace the commonly used nonsystematic names formaldehyde, acetone, and acetic acid. But there is no question that every organic chemist knows what compounds the names methanal, 2-propanone, and ethanoic acid represent, so the beginner can communicate with these names and later become familiar with and use the special names. We will have more to say on this subject in Chapter 3.
How Are Functional-Group Classifications Useful?
One of the main reasons for classifying compounds by their functional groups is that it also classifies their chemical behavior. By this we mean that the reactions of compounds and, to some extent, their physical properties are influenced profoundly by the nature of the functional groups present. Indeed, many organic reactions involve transformations of the functional group that do not affect the rest of the molecule. For instance, alcohols, $R-OH$, can be transformed into a number of other compounds, such as organic halides, $R-Cl$ or $R-Br$; ethers, $R-O-R$; and amines, $R-NH_2$ without changing the structure of the hydrocarbon group $R$. Furthermore, any compound possessing a particular functional group may be expected to exhibit reactions characteristic of that group and, to some extent at least, of inorganic compounds with similar functional groups.
A good example of the use of the functional-group concept is for acid-base properties. Alcohols, $ROH$, are structurally related to water, $HOH$, in that both possess a hydroxyl function. We may then expect the chemistry of alcohols to be similar to that of water. In fact, both are weak acids because the $OH$ group has a reactive proton that it can donate to a sufficiently strongly basic substance, written as $:B$ here:
Water and alcohols both are weak bases because the oxygens of their $OH$ groups have unshared electron pairs to use in bonding with a proton donated by an acid, $HA$:
We can carry the analogy further to include carboxylic acids, $RCO_2H$, which also have a hydroxyl function. They also should possess acidic and basic properties. They do have these properties and they are, in fact, stronger acids than either water or alcohols and form salts with bases:
Amines, $RNH_2$, are structurally related to ammonia, $NH_3$, and we therefore may predict that they will have similar properties. A property of ammonia that you probably will have encountered earlier is that it acts as a base and forms salts with acids. Amines behave likewise:
It is with logic of this kind - inferring chemical behavior from structural analogies - that much of organic chemistry can be understood. There are other logical classification schemes, however, and one of these depends more on types of reactions than on functional groups.
Classifications by Reaction Types
The rationale of classification by reaction types is that different functional groups may show the same kinds of reactions. Thus, as we have just seen, alcohols, carboxylic acids, and amines all can accept a proton from a suitably strong acid. Fortunately, there are very few different types of organic reactions - at least as far as the overall result that they produce. The most important are acid-base, substitution, addition, elimination, and rearrangement reactions. Some examples of these are given below, and you should understand that these are descriptive of the overall chemical change and nothing is implied as to how or why the reaction occurs (also see Section 1-1I).
Substitution of one atom or group of atoms for another:
Addition, usually to a double or triple bond:
Elimination, which is the reverse of addition:
Rearrangement where one structure is converted to an isomeric structure:
Certain reactions commonly are described as either oxidation or reduction reactions and most simply can be thought of as reactions that result in changes in the oxygen or hydrogen content of a molecule by direct or indirect reactions with oxygen or hydrogen, respectively. They frequently fall into one of the categories already mentioned. Reduction of ethene to ethane is clearly addition, as is oxidation of ethene to oxacyclopropane:
Reactions that lead to substantial degradation of molecules into smaller fragments are more difficult to classify. An example is the combustion of ethane to carbon dioxide and water. All of the chemical bonds in the reactants are broken in this reaction. It seems pointless to try to classify this as anything but a complete combustion or oxidation reaction:
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."
2.04: Isomerization in Organic Compounds
More than one stable substance can correspond to a given molecular formula. Examples are butane and 2-methylpropane (isobutane), each of which has the molecular formula \(C_4H_{10}\). Similarly, methoxymethane (dimethyl ether) and ethanol have the same formula, \(C_2H_6O\):
Compounds having the same number and kinds of atoms are called isomers.\(^2\) Whereas only one stable substance is known corresponding to the formula \(CH_4\), thirty-five stable isomers have been prepared of the formula \(C_9H_{20}\). From this one may begin to sense the rich variety of organic chemistry, which leads to many problems - in telling one compound from another, in determining structures, and also in finding suitable names for compounds. In the rest of this chapter we will describe one type of isomer - the position isomer - and in later chapters we will discuss another type of isomer - the stereoisomer - and the experimental approaches that are used to establish the purity, identity, structure, and stereochemistry of organic compounds.
Position Isomers
Compounds having the same number and kind of atoms but having different bonding arrangements between the atoms are called position isomers. Butane and 2-methylpropane are examples of position isomers. The atoms are connected differently in the two structures because the carbon chain in butane is a straight or continuous chain, whereas in 2-methylpropane it is branched:
Therefore these two molecules are structurally different and, accordingly, do not have the same chemical and physical properties. They cannot be converted one into the other without breaking and remaking \(C-C\) and \(C-H\) bonds. Methoxymethane and ethanol are also position isomers because the oxygen clearly is connected differently in the two molecules:
The term position isomer means the same as constitutional isomer. The designation structural isomer also is used, but this term is taken by some to include both position isomers and stereoisomers; that is, "structure" can mean both the way in which atoms are connected and their different arrangements in space.
The number of position isomers possible for a given formula rapidly increases with the increasing number of carbon atoms, as can be seen from the number of theoretically possible structures of formula \(C_nH_{2n+2}\) up to \(n = 10\) given in Table 2-3. In 1946, it was reported that all of the 75 compounds with values of \(n = 1\) to \(n = 9\) had been prepared in the laboratory. Before we can begin to discuss the chemistry of these compounds it is necessary to know how
to name them; without convenient and systematic rules for nomenclature that are adopted universally, catastrophic confusion would result. We shall tackle this problem in the next chapter.
\(^2\)The prefix iso is from the Greek word meaning the same or alike.
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)/02%3A_Structural_Organic_Chemistry/2.03%3A_Classification_by_Functional_Groups.txt |
A chemical nomenclature is a set of rules to generate systematic names for chemical compounds. The nomenclature used most frequently worldwide is the one created and developed by the International Union of Pure and Applied Chemistry (IUPAC). The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning which chemical compound the name refers to: each chemical name should refer to a single substance.
• 3.1: Prelude to Organic Nomenclature
Organic chemists, regardless of what languages they speak, can communicate with one another about their chemical work simply by writing equations and structural formulas. For more rapid and efficient communication we need to have names for compounds that would contain enough information so we could generate the proper structures from them, and conversely, if we know the structures then the system would have simple enough rules that we could construct universally recognized and accepted names.
• 3.2: Alkanes
The most definitive set of organic nomenclature rules currently in use were evolved through several international conferences and are known as the International Union of Pure and Applied Chemistry Rules (IUPAC rules). We first shall describe this system for naming the hydrocarbons known as alkanes - the so-called saturated paraffin hydrocarbons that have no double or triple bonds, or rings.
• 3.3: Cycloalkanes
The cycloalkanes with one ring and are named by adding the prefix cyclo- to the name of the corresponding continuous-chain alkane having the same number of carbon atoms as the ring. Substituents are assigned numbers consistent with their position in such a way as to give the lowest numbers possible for the substituent positions. The substituent groups derived from cycloalkanes by removing one hydrogen are named by replacing the ending -ane of the hydrocarbon with -yl to give cycloalkyl.
• 3.4: Alkenes, Cycloalkenes, and Alkadienes
The open-chain hydrocarbons with one double bond are called alkenes. The carbon-carbon double bond often is called an "olefinic linkage" and the alkenes designated as olefins (oil-formers). The IUPAC system for naming alkenes, the longest continuous chain containing the double bond is given the name of the corresponding alkane with the ending -ane changed to -ene. This chain then is numbered so that the position of the first carbon of the double bond is indicated by the lowest possible number.
• 3.5: Alkynes
A number of hydrocarbons, called alkynes or acetylenes, have triple bonds between carbon atoms. The IUPAC system for naming alkynes employs the ending -yne instead of the -ane used for naming of the corresponding saturated hydrocarbon. Hydrocarbons with more than one triple bond are called alkadiynes, alkatriynes, and so on, according to the number of triple bonds Hydrocarbons with both double and triple bonds are called alkenynes (not alkynenes).
• 3.6: Arenes
The so-called aromatic hydrocarbons, or arenes, are cyclic unsaturated compounds that have such strikingly different chemical properties from conjugated alkenes (polyenes) that it is convenient to consider them as a separate class of hydrocarbon. The simplest member is benzene, which frequently is represented as a cyclic conjugated molecule of three single and three double carbon-carbon bonds. The hydrocarbon group from benzene itself is called a phenyl group.
• 3.E: Organic Nomenclature (Exercises)
These are the homework exercises to accompany Chapter 3 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Thumbnail: A chemical nomenclature is a set of rules to generate systematic names for chemical compounds. The nomenclature used most frequently worldwide is the one created and developed by the International Union of Pure and Applied Chemistry (IUPAC).
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."
03: Organic Nomenclature
Organic chemists, regardless of what languages they speak, can communicate with one another about their chemical work simply by writing equations and structural formulas. But this is a slow process if the molecules are complicated, and is not well suited for conversation (try describing a structural formula of a complex molecule to someone). For more rapid and efficient communication we need to have names for compounds, and we should have every reason to hope that, after 100 years, the names now in use would be clear, unambiguous, easy to pronounce, easy to spell and to remember, as well as being amenable to arrangement in alphabetical order. But, even more, we should hope that the names of organic compounds would contain enough information so we could generate the proper structures from them, and conversely, if we know the structures then the system would have simple enough rules that we could construct universally recognized and accepted names.
Unfortunately, these splendid ideals have not yet been realized. A good part of the problem is that people are resistant to change and especially resistant to changes in names. To give an example, the carboxylic acid, $\ce{CH_3-CO_2H}$, commonly is known as "acetic acid". The name arises from the Latin word acetum, for sour wine or vinegar, and acetic acid is the principal constituent, besides, water, of vinegar. A similarly common compound is called "acetone" and, in the ideal world, acetone should be structurally related to acetic acid. But acetone is $\ce{CH_3-CO-CH_3}$, and the name arises only because acetone is formed by strong heating of the calcium salt of acetic acid, $\ce{(CH_3-CO_2)_2Ca} \rightarrow \ce{CH_3-CO-CH_3} + \ce{CaCO_3}$, a reaction that is of no current importance whatsoever. Better nomenclature systems use names based on the name of the hydrocarbon with the same number of carbons in the longest continuous chain in the molecule. On this basis, $\ce{CH_3CO_2H}$ is related to ethane and is called ethanoic acid, whereas $\ce{CH_3-CO-CH_3}$ with three carbons is related to propane and called 2-propanone.
As far as possible, we shall use these names as our first choices, because organic chemistry is growing too fast to sustain the present chaos of nonsystematic nomenclatures in current use. You might well ask why nonsystematic names persist for so long. The reasons are complex and variable. Alchemists intentionally used abstruse names and symbolism to disguise what they really were working with. Chemical industry, especially in the drug area, has practiced much the same thing in using unintelligible trade names and codes for proprietary products. Obviously, everyone who handles or sells chemicals is not a chemist, and to the nonchemist, a short nonsystematic name will make more sense than a longer systematic name. A salesman who markets tons of "acrolein", $\ce{CH_2=CH-CH=O}$, has little reason to adopt the systematic name, 2-propenal.
People probably persist in using nicknames for chemicals for much the same reason that they use nicknames for people. Nicknames are less formal, usually shorter, and imply familiarity with the subject. Another cogent reason to resist dramatic changes in chemical nomenclature is that it would make the current and earlier literature archaic or even unintelligible. Universal adoption tomorrow of a nomenclature system different from the one we use here would render this book instantly obsolete. As a result, changes usually are made in small steps and may not be really effective until a generation or more passes. (Consider in this context the efforts to convert monetary systems and weights and measures to the decimal system.)
Ideally, every organic substance should have a completely descriptive, systematic name to permit only one structural formula to be written for it. This ideal has been approached closely in some of the current nomenclature systems but, unfortunately, truly systematic nomenclature for very complicated compounds is often hopeless for conversational or routine scriptorial purposes. As a result, we will at times resort to using (common) trivial names, especially if it is impractical to do otherwise. Clearly, the description 9-(2,6,6-trimethyl-1-cyclohexenyl)-3,7-dimethyl-2,4,6,8-nonatetraen-1-ol has phonetic disadvantages as a handy name for vitamin A:
A very important consideration for becoming more familiar with the systematic names is their increasing use in indexing systems. When organic chemists dealt with relatively few compounds, it was possible to accommodate a wide variety of special nomenclature customs. However, the rapid growth of knowledge in the past twenty years, which probably has doubled the number of organic compounds, has also enormously increased the burden on those who dedicate themselves to making this knowledge easily available to others by indexing the current literature. A natural reaction is to discard common names in favor of more systematic ones and to develop numerical designations suitable for computer processing. The difference in sizes of the Chemical Abstracts$^1$ indexes for the years 1907-1916 and for the current year should be convincing as to the need for having systematic names become more widely used and important. But the fact remains that the naming systems used in indexing are not always the same as those used in practice, and we are left with the necessity of having to know both.
Learning the nomenclature of organic compounds has many of the elements of learning a language, be it Latin or Fortran. Fortunately, like a language it does not have to be learned all at once. One can become familiar with naming of simple hydrocarbons, then study their chemistry (avoiding that part which involves compounds with as yet unlearned names), proceed to the naming of alkenes, study their chemistry, and so on. This is a very simple and natural way but can be inconvenient in a textbook if one wants to review the nomenclature of more than one class of compounds at a time.
In this chapter, we consolidate the nomenclature of a number of classes of compounds - an undertaking that may not seem very logical to someone who will soon be troubled enough with the chemistry of these compounds let alone their names. We recommend, however, a thorough study now of alkane and haloalkane nomenclature (Section 3-1) followed by a more cursory examination of the rest of the chapter. Then, as unfamiliar names arise, you can quickly review the basic rules for alkanes and proceed to the new class you have encountered. The idea is to have many of the important rules in one place. Nomenclature rules for other types of compounds are given in Chapter 7.
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)/03%3A_Organic_Nomenclature/3.01%3A_Prelude_to_Organic_Nomenclature.txt |
The most definitive set of organic nomenclature rules currently in use were evolved through several international conferences and are known as the International Union of Pure and Applied Chemistry Rules (IUPAC rules). We first shall describe this system for naming the hydrocarbons known as alkanes - the so-called saturated paraffin hydrocarbons that have no double or triple bonds, or rings, and conform to the general formula \(C_nH_{2n+2}\).
The alkanes are classified as "continuous chain" (that is, "unbranched") if all the carbon atoms in the chain are linked to no more than two other carbons; or "branched chain" if one or more carbon atoms are linked to more than two other carbons:
The first four continuous-chain hydrocarbons have nonsystematic names:
The higher members, beginning with pentane, are named systematically with a numerical prefix (pent-, hex-, hept-, etc., to denote the number of carbon atoms) and with the ending -ane to classify the compound as a paraffin hydrocarbon, as in Table 3-1. To specify a continuous-chain hydrocarbon, the prefix n- (for normal) sometimes is used. However, in the absence of any qualifying prefix, the hydrocarbon is considered to be "normal" or unbranched and we shall not use this prefix henceforth. You should memorize the names up to \(C_{10}H_{22}\).
The possibility of having branched-chain hydrocarbons that are structural isomers of the continuous-chain hydrocarbons begins with butane (\(n = 4\)). The IUPAC rules for the systematic naming of these hydrocarbons follow.
1. The longest continuous chain of carbon atoms is taken as the parent hydrocarbon and is the framework on which the various substituent groups are attached. Thus the hydrocarbon \(1\) is a substituted pentane rather than a substituted butane because the longest chain has five carbons:
2. The substituent groups attached to the main chain are named by replacing the ending -ane of the alkane by -yl. We then have the alkyl groups
(or alkyl radicals, the simplest examples being the methyl (\(CH_3-\)) and ethyl (\(CH_3CH_3-\)) groups.
3. The parent hydrocarbon then is numbered starting from the end of the chain, and the substituent groups are assigned numbers corresponding to their positions on the chain. The direction of numbering is chosen to give the lowest numbers to the side-chain substituents.\(^2\) Thus hydrocarbon \(1\) is 2,3-dimethylpentane rather than 3,4-dimethylpentane. The prefix di- signifies that there are two identical substituents:
The prefixes used to designate the number of substituents follow up to ten.
Mono- is not used to designate a single substituent in systematic nomenclature, but may be used in conversation for emphasis.
4. Where there are two identical substituents at one position as in \(2\), numbers are supplied for each, and the prefix, di-, tri-, and so on, is included to signify the number of groups of the same kind:
5. Branched-chain substituent groups are given appropriate names by a simple extension of the system used for branched-chain hydrocarbons. The longest chain of the substituent is numbered starting with the carbon attached directly to the parent hydrocarbon chain. Parentheses are used to separate the numbering of the substituent and of the main hydrocarbon chain:
Additional examples of alkyl substituents and their names are listed in Table 3-2. These are further classified according to whether they are primary, secondary, or tertiary. An alkyl group is described as primary if the carbon at the point of attachment is bonded to only one other carbon, as secondary if bonded to two other carbons, and tertiary if bonded to three other carbons. Thus if \(R\) is any hydrocarbon radical, the different kinds of alkyl groups are
Confusion can arise here because we often refer to specific carbons rather than whole alkyl groups as primary, secondary, and so on. In this context, a carbon is primary if it is bonded to one other carbon, secondary if bonded to two, tertiary if bonded to three, and quaternary if bonded to four. Thus, either carbon of ethane is primary, the \(C2\) carbon in propane, \(CH_3CH_2CH_3\), is secondary, and the \(C2\) carbon of 2,2-dimethylpropane, \((CH_3)_4C\), is quaternary.
The situation with regard to naming alkyl substituents has been muddied considerably by the fact that the IUPAC rules allow use of trivial names for a few alkyl groups. Thus sec-butyl sometimes is used in place of 1-methylpropyl, and tert-butyl in place of 1,1-dimethylethyl These and other examples are included in parentheses in Table 3-2. Further odd-ball but less official customs are the prefix iso, which is reserved for substituents with two methyl groups at the end of an otherwise straight chain (e.g., isopropyl), and the prefix neo,
which is used to denote three methyl groups at the end of a chain (e.g., neopentyl, which is more properly called 2,2-dimethylpropyl). Also in common use are the names isobutane and neopentane for the hydrocarbons 2-methyl-propane and 2,2-dimethylpropane, respectively There is no ambiguity involved in the use of iso and neo prefixes here, but the practice of using the name "isooctane" for 2,2,4-trimethylpentane is erroneous. Fortunately, use of these special names is declining.
6. When there are two or more different substituents present, the question arises as to what order they should be cited in naming the compound. The system adopted by IUPAC and long practiced by Chemical Abstracts cites them in alphabetical order without regard for whether there is a multiplying prefix such as di- or tri-. Examples are given below.
When a hydrocarbon is substituted with other than alkyl groups a new problem arises, which can be illustrated by \(CH_3CH_2Cl\) This substance can be called either chloroethane or ethyl chloride, and both names are used in conversation and in print almost interchangeably. In the IUPAC system, halogens, nitro groups, and a few other monovalent groups are considered to be substituent groups on hydrocarbons and are named as haloalkanes, nitroalkanes, and so on.
The alphabetical order of precedence is preferred for substituents of different types when two or more are attached to a hydrocarbon chain because this makes indexing and using indexes more straightforward:
The rest of this chapter is devoted to names of compounds we will not discuss for several chapters ahead and you may wish to stop at this point and return later as necessary. However, you should test your knowledge of alkane nomenclature by doing Exercises 3-1, 3-2, 3-3, 3-9, 3-10, and 3-11.
\(^2\)Confusion is possible when the numbering from each end is similar. The rule is that when the series of substituent locants are compared term by term, the "lowest" series has the lowest number at the first point of difference. The compound
is 2,3,5-trimethylhexane, not 2,4,5-trimethylhexane.
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)/03%3A_Organic_Nomenclature/3.02%3A_Alkanes.txt |
The cycloalkanes with one ring have the general formula \(C_nH_{2n}\), and are named by adding the prefix cyclo- to the name of the corresponding continuous-chain alkane having the same number of carbon atoms as the ring. Substituents are assigned numbers consistent with their position in such a way as to give the lowest numbers possible for the substituent positions:
The substituent groups derived from cycloalkanes by removing one hydrogen are named by replacing the ending -ane of the hydrocarbon with -yl to give cycloalkyl. Thus cyclohexane becomes cyclohexyl, cyclopentane becomes cyclopentyl, and so on. Remember, the numbering of the cycloalkyl substituent starts at the position of attachment, and larger rings take precedence over smaller rings:
When a cycloalkane has an alkyl substituent, the compound could be called either an alkylcycloalkane or a cycloalkylalkane. The alkylcycloalkane name is the proper one in naming alkyl substituted cycloalkanes.
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."
3.04: Alkenes Cycloalkenes and Alkadienes
The open-chain hydrocarbons with one double bond have the general formula \(C_nH_{2n}\) and are called alkenes. The carbon-carbon double bond often is called an "olefinic linkage" and the alkenes designated as olefins (oil-formers). These terms arose because the gaseous lower-molecular-weight alkenes yield "oily" products on treatment with chlorine or bromine. The term "unsaturated" hydrocarbon also is used - again because these substances normally react with bromine and chlorine and are hence "unsaturated" with reference to reagents of this type (also see Section 1-1I).
According to the IUPAC system for naming alkenes, the longest continuous chain containing the double bond is given the name of the corresponding alkane with the ending -ane changed to -ene. This chain then is numbered so that the position of the first carbon of the double bond is indicated by the lowest possible number:
A few very common alkenes also are called "alkylenes" by appending the suffix -ene to the name of the hydrocarbon radical with the same carbon skeleton. Examples are shown below with their alkylene names in parentheses. We shall continue to use the IUPAC names whenever possible.
The hydrocarbon groups derived from alkenes have the suffix -enyl, as in alkenyl, and numbering of the group starts with the carbon atom attached to the main chain:
A few alkenyl groups have trivial names that commonly are used in place of systematic names. These are vinyl, allyl, and isopropenyl. And again we shall avoid using these names, except parenthetically:
Also, hydrogen atoms that are bonded directly to the unsaturated carbon atoms of a double bond often are called vinyl hydrogens, although the term alkenic hydrogens is more accurate and therefore preferable.
Cycloalkenes are named by the system used for the open-chain alkenes, except that numbering always is started at one of the carbons of the double bond and continued around the ring through the double bond so as to keep the index numbers as small as possible:
When a hydrocarbon group is double-bonded to a single carbon of a cycloalkane ring, the suffix -ylidene, as in alkylidene, is used:
Many compounds contain two or more double bonds and are known as alkadienes, alkatrienes, alkatetraenes, and so on, the suffix denoting the number of double bonds. The location of each double bond is specified by appropriate numbers, as illustrated below:
A further classification is used for the relationships of the double bonds to each other. Thus 1,2-alkadienes and similar substances are said to have cumulated double bonds:
1,3-Alkadienes and other compounds with alternating double and single bonds are said to have conjugated double bonds:
Compounds with double bonds that are neither cumulated nor conjugated are classified as having isolated double-bond systems:
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)/03%3A_Organic_Nomenclature/3.03%3A_Cycloalkanes.txt |
A number of hydrocarbons, called alkynes or acetylenes, have triple bonds between carbon atoms.\(^3\) They conform to the general formula \(C_nH_{2n-2}\) for one triple bond. The IUPAC system for naming alkynes employs the ending -yne instead of the -ane used for naming of the corresponding saturated hydrocarbon:
The numbering system for locating the triple bond and substituent groups is analogous to that used for the corresponding alkenes:
Hydrocarbons with more than one triple bond are called alkadiynes, alkatriynes, and so on, according to the number of triple bonds Hydrocarbons with both double and triple bonds are called alkenynes (not alkynenes). The chain always should be numbered to give the multiple bonds the lowest possible numbers, and when there is a choice, double bonds are given lower numbers than triple bonds. For example,
The hydrocarbon substituents derived from alkynes are called alkynyl groups:
\(^3\)Alkyne rhymes with "mine" and "thine."
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."
3.06: Arenes
The so-called aromatic hydrocarbons, or arenes, are cyclic unsaturated compounds that have such strikingly different chemical properties from conjugated alkenes (polyenes) that it is convenient to consider them as a separate class of hydrocarbon. The simplest member is benzene, $C_6H_6$, which frequently is represented as a cyclic conjugated molecule of three single and three double carbon-carbon bonds. Actually, all the carbon-carbon bonds are equivalent (see Chapter 1) but it is convenient to represent the structure in the manner shown:
A variety of substituted benzenes are known that have one or more of the hydrogen atoms of the ring replaced with other atoms or groups. In almost all of these compounds the special properties associated with the benzene nucleus are retained. A few examples of "benzenoid" hydrocarbons follow, and it will be noticed that the hydrocarbon substituents include alkyl, alkenyl, and alkynyl groups. Many have trivial names indicated in parentheses:
The hydrocarbon group from benzene itself ($C_6H_5-$) is called a phenyl group and often is abbreviated as $Ph$ or less preferably by the symbol $\phi$. Generally, aryl groups are abbreviated as $Ar$, in contrast to alkyl groups for which we use $R$. Thus $CH_3Ar$ is a methyl-substituted arene, whereas $RC_6H_5$ is an alkyl-substituted benzene.
When there are two or more substituents on a benzene ring, position isomerism arises. Thus there are three possible isomeric disubstituted benzene derivatives according to whether the substituents have the 1,2, 1,3, or 1,4 relationship. The isomers commonly are designated as ortho, meta, and para (or o, m, and p) for the 1,2-, 1,3-, and 1,4-isomers, respectively. The actual symmetry of the benzene ring is such that only one 1,2-disubstitution product is found, despite the fact that two would be predicted if benzene had the 1,3,5-cyclohexatriene structure:
When the benzene ring carries different substituents we shall cite them in alphabetical order (disregarding multiplying prefixes) and assign their positions on the ring with the lowest possible numbers. Examples are
The IUPAC nomenclature system for other types of compounds is given in Chapter 7 and is based on the fundamental rules described in 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)/03%3A_Organic_Nomenclature/3.05%3A_Alkynes.txt |
Alkanes are the simplest organic molecules, consisting of only carbon and hydrogen and with only single bonds between carbon atoms. Alkanes are used as the basis for naming the majority of organic compounds (their nomenclature). Alkanes have the general formula \(C_nH_{2n+2}\).
• 4.1: Prelude to Alkanes
Although this chapter is concerned with the chemistry of only one class of compounds: alkanes, several fundamental principles are developed that we shall use extensively in later chapters. The study of some of these principles has been associated traditionally more with physical chemistry than with organic chemistry. We include them here, at the beginning of out discussion of organic reactions, because they provide a sound basis for understanding the practical use of organic reactions.
• 4.2: Physical Properties of Alkanes and The Concept of Homology
The series of straight-chain alkanes, in which n is the number of carbons in the chain, shows a remarkably smooth gradation of physical properties. As n increases, each additional \(CH_2\) group contributes a fairly constant increment to the boiling point and density, and to a lesser extent to the melting point. This makes it possible to estimate the properties of an unknown member of the series from those of its neighbors.
• 4.3: Chemical Reactions of Alkanes. Combustion of Alkanes
As a class, alkanes generally are unreactive because their chemical "affinity" for most common reagents may be regarded as "saturated" or satisfied. Thus none of the C−H or C−C bonds in a typical saturated hydrocarbon, for example ethane, are attacked at ordinary temperatures by a strong acid, such as sulfuric acid, or by an oxidizing agent, such as bromine (in the dark), oxygen, or potassium permanganate. Under ordinary conditions, ethane is similarly stable to reducing agents.
• 4.4: Combustion. Heats of Reaction. Bond Energies
All hydrocarbons are attacked by oxygen at elevated temperatures and, if oxygen is in excess, complete combustion occurs to carbon dioxide and water. The heat evolved in this process - the heat of the combustion reaction, is a measure of the amount of energy stored in the C-C and C-H bonds of the hydrocarbon COMPARED to the energy stored in the products, carbon dioxide and water.
• 4.5: Halogenation of Alkanes. Energies and Rates of Reactions
The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by substitution reactions of alkanes. In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place
• 4.6: Practical Halogenations and Problems of Selectivity
Given the knowledge that a particular reaction will proceed at a suitable rate, a host of practical considerations are necessary for satisfactory operation. These considerations include interference by possible side reactions that give products other than those desired, the ease of separation of the desired product from the reaction mixture, and costs of materials, apparatus, and labor. We shall consider these problems in connection with the important synthetic reactions discussed in this book.
• 4.7: Nitration of Alkanes
Another reaction of commercial importance is the nitration of alkanes to give nitroparaffins. Such reactions usually are carried out in the vapor phase at elevated temperatures using nitric acid or nitrogen tetroxide as the nitrating agent.
• 4.E: Alkanes (Exercises)
These are the homework exercises to accompany Chapter 4 of the Textmap for Basic Principles of Organic Chemistry (Roberts and Caserio).
Thumbnail: Structure of propane.
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."
04: Alkanes
Although this chapter is concerned with the chemistry of only one class of compounds, saturated hydrocarbons or alkanes, several fundamental principles are developed that we shall use extensively in later chapters. The study of some of these principles has been associated traditionally more with physical chemistry than with organic chemistry. We include them here, at the beginning of out discussion of organic reactions, because they provide a sound basis for understanding the key questions concerning the practical use of organic reactions. Is the equilibrium point of a given reaction far enough toward the desired products to be useful? Can conditions be found in which the reaction will take place at a practical rate? How can unwanted side reactions be suppressed?
Initially, we will be concerned with the physical properties of alkanes and how these properties can be correlated by the important concept of homology. This will be followed by a brief survey of the occurrence and uses of hydrocarbons, with special reference to the petroleum industry. Chemical reactions of alkanes then will be discussed, with special emphasis on combustion and substitution reactions, These reactions are employed to illustrate how we can predict and use energy changes - particularly \(\Delta H\), the heat evolved or absorbed by a reacting system, which often can be estimated from bond energies. Then we consider some of the problems involved in predicting reaction rates in the context of a specific reaction, the chlorination of methane. The example is complex, but it has the virtue that we are able to break the overall reaction into quite simple steps.
Before proceeding further, it will be well to reiterate what an alkane is, lest you be confused as to the difference between alkanes and alkenes. Alkanes are compounds of carbon and hydrogen only, without double bonds, triple bonds, or rings. They all conform to the general formula \(\ce{C_nH_{2n+2}}\) and sometimes are called paraffin hydrocarbons, open-chain saturated hydrocarbons, or acyclic hydrocarbons. The nomenclature of alkanes has been discussed in Chapter 3, and you may find it well to review Section 3-1 before proceeding.
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)/04%3A_Alkanes/4.01%3A_Prelude_to_Alkanes.txt |
The series of straight-chain alkanes, in which $n$ is the number of carbons in the chain, shows a remarkably smooth gradation of physical properties (see Table 4-1 and Figure 4-1). As $n$ increases, each additional $CH_2$ group contributes a fairly constant increment to the boiling point and density, and to a lesser extent to the melting point. This makes it possible to estimate the properties of an unknown member of the series from those of its neighbors. For example, the boiling points of hexane and heptane are $69^\text{o}$ and $98^\text{o}$, respectively. Thus a difference in structure of one $CH_2$ group for these compounds makes a difference in boiling point of $29^\text{o}$; we would predict the boiling point of the next higher member, octane, to be $98^\text{o} + 29^\text{o} = 127^\text{o}$, which is close to the actual boiling point of $126^\text{o}$.
Members of a group of compounds, such as the alkanes, that have similar chemical structures and graded physical properties, and which differ from one another by the number of atoms in the structural backbone, are said to constitute a homologous series. When used to forecast the properties of unknown members of the series, the concept of homology works most satisfactorily for the higher-molecular-weight members because the introduction of additional $CH_2$ groups makes a smaller relative change in the overall composition of such molecules. This is better seen from Figure 4-2, which shows
how $\Delta T$, the differences in boiling points and melting points between consecutive members of the homologous series of continuous-chain alkanes, changes with the number of carbons, $n$.
Branched-chain alkanes do not exhibit the same smooth gradation of physical properties as do the continuous-chain alkanes. Usually there is too great a variation in molecular structure for regularities to be apparent. Nevertheless, in any one set of isomeric hydrocarbons, volatility increases with increased branching. This can be seen from the data in Table 4-2, which lists the physical properties of the five hexane isomers. The most striking feature of the data is the $19^\text{o}$ difference between the boiling points of hexane and 2,2-dimethylbutane.
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."
4.03: Chemical Reactions of Alkanes. Combustion of Alkanes
As a class, alkanes generally are unreactive. The names saturated hydrocarbon, or "paraffin," which literally means "not enough affinity" [L. par(um), not enough, $+$ affins, affinity], arise because their chemical "affinity" for most common reagents may be regarded as "saturated" or satisfied. Thus none of the $C-H$ or $C-C$ bonds in a typical saturated hydrocarbon, for example ethane, are attacked at ordinary temperatures by a strong acid, such as sulfuric acid ($H_2SO_4$), or by an oxidizing agent, such as bromine (in the dark), oxygen, or potassium permanganate ($KMnO_4$). Under ordinary conditions, ethane is similarly stable to reducing agents such as hydrogen, even in the presence of catalysts such as platinum, palladium, or nickel.
However, all saturated hydrocarbons are attacked by oxygen at elevated temperatures and, if oxygen is in excess, complete combustion to carbon dioxide and water occurs. Vast quantities of hydrocarbons from petroleum are utilized as fuels for the production of heat and power by combustion, although it is becoming quite clear that few of the nations of the world are going to continue to satisfy their needs (or desires) for energy through the use of petroleum the way it has been possible in the past.
Petroleums differ considerably in composition depending on their source. However, a representative petroleum$^1$ on distillation yields the following fractions:
1. Gas fraction, boiling up to $40^\text{o}$, contains normal and branched alkanes from $C_1$ to $C_5$. Natural gas is mainly methane and ethane. "Bottled" gas (liquefied petroleum gas) is mainly propane and butane.
2. Gasoline, boiling point from $40^\text{o}$ to $180^\text{o}$, contains mostly hydrocarbons from $C_6$ to $C_{10}$. Over 100 compounds have been identified in gasoline, and these include continuous-chain and branched alkanes, cycloalkanes, and alkylbenzenes (arenes). The branched alkanes make better gasoline than their continuous-chain isomers because they give less "knock" in high-compression gasoline engines.
3. Kerosine, boiling point $180^\text{o}$ to $230^\text{o}$, contains hydrocarbons from $C_{11}$ to $C_{12}$. Much of this fraction is utilized as jet engine fuels or is "cracked" to simpler alkanes (and alkenes).
4. Light gas oil, boiling point $230^\text{o}$ to $305^\text{o}$, $C_{13}$ to $C_{17}$, is utilized as diesel and furnace fuels.
5. Heavy gas oil and light lubricating distillate, boiling point $305^\text{o}$ to $405^\text{o}$, $C_{18}$ to $C_{25}$.
6. Lubricants, boiling point $405^\text{o}$ to $515^\text{o}$, $C_{26}$ to $C_{38}$, familiarly encountered as paraffin was and petroleum jelly (Vaseline).
7. The distillation residues known as asphalts.
The way in which petroleum is refined and the uses for it depend very much on supply and demand, which always are changing. However, the situation for the United States in 1974 is summarized in Figure 4-3, which shows roughly how much of one barrel of oil (160 liters) is used for specific purposes.
In the past three decades, petroleum technology has outpaced coal technology, and we now are reliant on petroleum as the major source of fuels and chemicals. Faced with dwindling oil reserves, however, it is inevitable that coal again will become a major source of raw materials. When coal is heated at high temperatures in the absence of air, it carbonizes to coke and gives off a gaseous mixture of compounds. Some of these gases condense to a black viscous oil (coal tar), others produce an aqueous condensate called ammoniacal liquors, and some remain gaseous (coal gas). The residue is coke, which is used both as a fuel and as a source of carbon for the production of steel. The major component in coal gas is methane. Coal tar is an incredible mixture of compounds, mostly hydrocarbons, a substantial number of which are arenes. Coal and coal tar can be utilized to produce alkanes, but the technology involved is more complex and costly than petroleum refining. It seems inevitable that the cost of hydrocarbon fuel will continue to rise as supply problems become more difficult. And there is yet no answer to what will happen when the world's limited quantities of petroleum and coal are exhausted.
$^1$See F. D. Rossini, "Hydrocarbons in Petroleum," J. Chem. Educ. 37, 554 (1960).
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)/04%3A_Alkanes/4.02%3A_Physical_Properties_of_Alkanes_and_The_Concept_of_Homology.txt |
All hydrocarbons are attacked by oxygen at elevated temperatures and, if oxygen is in excess, complete combustion occurs to carbon dioxide and water:
$CH_4 + 2O_2 \rightarrow CO_Edit2 + 2H_2O$
The heat evolved in this process - the heat of the combustion reaction, $\Delta H$ - is a measure of the amount of energy stored in the $C-C$ and $C-H$ bonds of the hydrocarbon compared to the energy stored in the products, carbon dioxide and water. It can be measured experimentally with considerable accuracy and generally is reported as $\Delta H^\text{0}$ the amount of heat (in kilocalories)$^2$ liberated on complete combustion of one mole of hydrocarbon when the reactants and the products are in standard states, and at the same temperature, usually $25^\text{o}$.$^3$ Not all chemical reactions that occur spontaneously liberate heat - some actually absorb heat. By convention, $\Delta H^\text{0}$ is given a negative sign when heat is evolved (exothermic reaction) and a positive sign when heat is absorbed (endothermic reaction). The heat evolved or absorbed also is called the enthalpy change.
For the combustion of $1 \: \text{mol}$ of methane at $15^\text{o}$, we find by experiment (corrected from constant volume to constant pressure, if necessary) that the reaction is exothermic by $212.8 \: \text{kcal}$. This statement can be expressed as follows:
$CH_4(g) + 2O_2(g) \rightarrow CO_2 (g) + 2H_2O (l)$
with $\Delta H^o = -212.8 \, kcal$.
The symbol $\left( g \right)$ denotes that the reactants and products are in the gaseous state except for the water, which is liquid $\left( l \right)$. If we wish to have $\Delta H^\text{0}$ with gaseous water $H_2O \: \left( g \right)$ as the product we have to make a correction for the heat of vaporization of water ($10.5 \: \text{kcal} \: \text{mol}^{-1}$ at $25^\text{o}$):
The task of measuring the heats of all chemical reactions is a formidable one and about as practical as counting grains of sand on the beach. However, it is of practical interest to be able to estimate heats of reaction, and this can be done quite simply with the aid of bond energies. The necessary bond energies are given in Table 4-3, and it is important to notice that they apply only to complete dissociation of gaseous substances to gaseous atoms at $25^\text{o}C$. Also, they do not apply, without suitable corrections, to many compounds, such as benzene, that have more than one double bond. This limitation will be discussed in Chapters 6 and 21.
To calculate $\Delta H^\text{0}$ for the combustion of one mole of methane, first we break bonds as follows, using $98.7 \: \text{kcal} \: \text{mol}^{-1}$ for the energy of each of the
$C-H$ bonds,
and then $118.9 \: \text{kcal} \: \text{mol}^{-1}$ for the energy of the double bond in oxygen:
Then we make bonds, using $192 \: \text{kcal} \: \text{mol}^{-1}$ for each $O=C$ bond in carbon dioxide,
and $110.6 \: \text{kcal} \: \text{mol}^{-1}$ for each of the $H-O$ bonds in water:
The net sum of these $\Delta H^\text{0}$ values is $394.8 + 237.8 - 384.0 - 442.4 = -193.8 \: \text{kcal}$, which is reasonably close to the value of $-191.8 \: \text{kcal}$ for the heat of combustion of one mole of methane determined experimentally.
The same type of procedure can be used to estimate $\Delta H^\text{0}$ values for many other kinds of reactions of organic compounds in the vapor phase at $25^\text{o}$. Moreover, if appropriate heats of vaporization are available, it is straightforward to compute $\Delta H^\text{0}$ for vapor-phase reactions of substances which are normally liquids or solids at $25^\text{o}$. The special problems that arise when solutions and ionic substances are involved are considered in Chapters 8 and 11.
It is important to recognize that the bond energies listed in Table 4-3 for all molecules other than diatomic molecules are average values. That the $C-H$ bond energy is stated to be $98.7 \: \text{kcal}$ does not mean that, if the hydrogens of methane were detached one by one, $98.7 \: \text{kcal}$ would have to be put in at each step. Actually, the experimental evidence is in accord with quite different energies for the separate dissociation steps:
The moral is that we should try to avoid using the bond energies in Table 4-3 as a measure of $\Delta H^\text{0}$ for the dissociation of just one bond in a polyatomic molecule. For this we need what are called bond-dissociation energies, some of which are given in Table 4-6. The values given have been selected to emphasize how structure influences bond energy. Thus, $C-H$ bond energies in alkanes decrease in the order primary $>$ secondary $>$ tertiary; likewise, $C-H$ bonds decrease in strength along the series $C \equiv C-H \: > \: C=C-H \: > \: C-C-H$.
How accurate are $\Delta H^\text{0}$ values calculated from bond energies? Generally quite good provided nonbonded interactions between the atoms are small and the bond angles and distances are close to the normal values (Section 2-2B). A few examples of calculated and experimental heats of combustion of some hydrocarbons are given in Table 4-4. Negative discrepancies represent heats of combustion smaller than expected from the average bond energies and positive values correspond to larger than expected heats of combustion.
Comparing isomers in Table 4-4, we see that 2-methylpropane and 2,2,3,3-tetramethylbutane give off less heat when burned than do butane and octane, and this is a rather general characteristic result of chain branching.
Cyclopropane has a $\Delta H^\text{0}$ of combustion of $27.7 \: \text{kcal} \: \text{mol}^{-1}$ greater than expected from bond energies, and this clearly is associated with the abnormal $C-C-C$ bond angles in the ring. These matters will be discussed in detail in Chapter 12. For cyclohexane, which has normal bond angles, the heat of combustion is close to the calculated value.
$^2$In this book we use kilocalories in place of the presently recommended (SI) joules for units of energy. As of the date of writing, it is not clear just how general the use of the joule will become among chemists. To convert calories to joules (or $\text{kcal}$ to $\text{kJ}$), multiply by 4.184.
$^3$You may wonder how a reaction, such as combustion of methane, can occur at $25^\text{o}$. The fact is that the reaction can be carried out at any desired temperature. The important thing is that the $\Delta H^\text{0}$ value we are talking about here is the heat liberated or absorbed when you start with the reactants at $25^\text{o}$ and finish with the products at $25^\text{o}$. As long as $\Delta H^\text{0}$ is defined this way, it does not matter at what temperature the reaction actually occurs. Standard states for gases are $1 \: \text{atm}$ partial pressure. Standard states for liquids or solids usually are the pure liquid or solid at $1 \: \text{atm}$ external pressure.
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)/04%3A_Alkanes/4.04%3A_Combustion._Heats_of_Reaction._Bond_Energies.txt |
The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by substitution reactions of alkanes. In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place.
A simple example of a substitution reaction is the formation of chloromethane and chlorine:
$\ce{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}$
The equation for the reaction is simple, the ingredients are cheap, and the product is useful. However, if we want to decide in advance whether such a reaction is actually feasible, we have to know more. Particularly, we have to know whether the reaction proceeds in the direction it is written and, if so, whether conditions can be found under which it proceeds at a convenient rate. Obviously, if one were to mix methane and chlorine and find that, at most, only $1 \%$ conversion to the desired product occurred and that the $1 \%$ conversion could be achieved only after a day or so of strong heating, this reaction would be both too unfavorable and too slow for an industrial process.
One way of visualizing the problems involved is with energy diagrams, which show the energy in terms of some arbitrary reaction coordinate that is a measure of progress between the initial and final states (Figure 4-4). Diagrams such as Figure 4-4 may not be familiar to you, and a mechanical analogy may be helpful to provide better understanding of the very important ideas involved. Consider a two-level box containing a number of tennis balls. An analog to an energetically favorable reaction would be to have all of the balls
on the upper level where any disturbance would cause them to roll down to the lower level under the influence of gravity, thereby losing energy.
If the upper level is modified and a low fence added to hold the balls in place, it will be just as energetically favorable as when the fence is not there for the balls to be at the lower level. The difference is that the process will not occur spontaneously without some major disturbance. We can say there is an energy barrier to occurrence of the favorable process.
Now, if we shake the box hard enough, the balls on the upper level can acquire enough energy to bounce over the barrier and drop to the lower level. The balls then can be said to acquire enough activation energy to surmount the barrier. At the molecular level, the activation energy must be acquired either by collisions between molecules as the result of their thermal motions, or from some external agency, to permit the reactants to get over the barrier and be transformed into products. We shortly will discuss this more, but first we wish to illustrate another important concept with our mechanical analogy, that of equilibrium and equilibration.
With gentle shaking of our two-level box, all of the balls on the upper level are expected to wind up on the lower level. There will not be enough activation to have them go from the lower to the upper level. In this circumstance, we can say that the balls are not equilibrated between the lower and upper levels. However, if we shake the box vigorously and continuously, no matter whether we start with all of the balls on the lower or upper level, an equilibrium will be set up with, on the average, most of the balls in the energetically more favorable lower level, but some in the upper level as well.
To maintain a constant average fraction of the balls at each level with vigorous and continued shaking, the rate at which balls go from the upper to the lower level must be equal to the rate that they go in the opposite direction. The balls now will be equilibrated between the two levels. At equilibrium, the fraction of the balls on each of the two levels is wholly independent of the height of the barrier, just as long as the activation (shaking) is sufficient to permit the balls to go both ways.
The diagrams of Figure 4-4 are to be interpreted in the same general way. If thermal agitation of the molecules is sufficient, then equilibrium can be expected to be established between the reactants and the products, whether the overall reaction is energetically favorable (left side of Figure 4-4) or energetically unfavorable (right side of Figure 4-4). But as with our analogy, when equilibrium is established we expect the major portion of the molecules to be in the more favorable energy state.
What happens when methane is mixed with chlorine? No measurable reaction occurs when the gases are mixed and kept in the dark at room temperature. Clearly, either the reaction is energetically unfavorable or the energy barrier is high. The answer as to which becomes clear when the mixture is heated to temperatures in excess of $300^\text{o}$ or when exposed to strong violet or ultraviolet light, whereby a rapid or even explosive reaction takes place. Therefore the reaction is energetically favorable, but the activation energy is greater than can be attained by thermal agitation alone at room temperature. Heat or light therefore must initiate a pathway for the reactants to be converted to products that has a low barrier or activation energy.
Could we have predicted the results of this experiment ahead of time? First, we must recognize that there really are several questions here. Could we have decided whether the reaction was energetically favorable? That the dark reaction would be slow at room temperature? That light would cause the reaction to be fast? We consider these and some related questions in detail because they are important questions and the answers to them are relevant in one way or another to the study of all reactions in organic chemistry.
The Question of the Equilibrium Constant
Presumably, methane could react with chlorine to give chloromethane and hydrogen chloride, or chloromethane could react with hydrogen chloride to give methane and chlorine. If conditions were found for which both reactions proceeded at a finite rate, equilibrium finally would be established when the rates of the reactions in each direction became equal:
$\ce{CH_4 + Cl_2 \rightleftharpoons CH_3Cl + HCl}$
At equilibrium, the relationship among the amounts of reactants and products is given by the equilibrium constant expression
$K_{eq} = \dfrac{[CH_3Cl][HCl]}{[CH_4][Cl_2]} \label{4-1}$
in which $K_\text{eq}$ is the equilibrium constant.
The quantities within the brackets of Equation $\ref{4-1}$ denote either concentrations for liquid reactants or partial pressures for gaseous substances. If the equilibrium constant $K_\text{eq}$ is greater than $1$, then on mixing equal volumes of each of the participant substances (all are gases above $-24^\text{o}$), reaction to the right will be initially faster than reaction to the left, until equilibrium is established; at this point there will be more chloromethane and hydrogen chloride present than methane and chlorine. However, if the equilibrium constant were less than $1$, the reaction initially would proceed faster to the left and, at equilibrium, there would be more methane and chlorine present than chloromethane and hydrogen chloride.$^4$ For methane chlorination, we know from experiment that the reaction goes to the right and that $K_\text{eq}$ is much greater than unity. Naturally, it would be helpful in planning other organic preparations to be able to estimate $K_\text{eq}$ in advance.
It is a common experience to associate chemical reactions with equilibrium constants greater than one with the evolution of heat, in other words, with negative $\Delta H^\text{0}$ values. There are, in fact, many striking examples. Formation of chloromethane and hydrogen chloride from methane and chlorine has a $K_\text{eq}$ of $10^{18}$ and $\Delta H^\text{0}$ of $-24 \: \text{kcal}$ per mole of $CH_3Cl$ formed at $25^\text{o}$. Combustion of hydrogen with oxygen to give water has a $K_\text{eq}$ of $10^{40}$ and $\Delta H^\text{0} = -57 \: \text{kcal}$ per mole of water formed at $25^\text{o}$. However, this correlation between $K_\text{eq}$ and $\Delta H^\text{0}$ is neither universal nor rigorous. Reactions are known that absorb heat (are endothermic) and yet have $K_\text{eq} > 1$. Other reactions have large $\Delta H^\text{0}$ values and equilibrium constants much less than $1$.
The problem is that the energy change that correlates with $K_\text{eq}$ is not $\Delta H^\text{0}$ but $\Delta G^\text{0}$ (the so-called change of "standard Gibbs energy")$^5$, and if we know $\Delta G^\text{0}$, we can calculate $K_\text{eq}$ by the equation
$\Delta G^o =-2.303 RT \log_{10} K_{eq} \label{4-2}$
in which $R$ is the gas constant and $T$ is the absolute temperature in degrees Kelvin. For our calculations, we shall use $R$ as $1.987 \: \text{cal} \: \text{deg}^{-1} \: \text{mol}^{-1}$ and you should not forget to convert $\Delta G^\text{0}$ to $\text{cal}$.
Tables of $\Delta G^\text{0}$ values for formation of particular compounds (at various temperatures and states) from the elements are available in handbooks and the literature. With these, we can calculate equilibrium constants quite accurately. For example, handbooks give the following data, which are useful for methane chlorination:
Combining these with proper regard for sign gives
and $\text{log} \: K_\text{eq} = -\left( -24.7 \times 1000 \right)/ \left(2.303 \times 1.987 \times 298.2 \right)$, so $K_\text{eq} = 1.3 \times 10^{18}$. Unfortunately, insufficient $\Delta G^\text{0}$ values for formation reactions are available to make this a widely applicable method for calculating $K_\text{eq}$ values.
The situation is not wholly hopeless, because there is a relationship between $\Delta G^\text{0}$ and $\Delta H^\text{0}$ that also involves $T$ and another quantity, $\Delta S^\text{0}$, the standard entropy change of the process:
$\Delta G^o = \Delta H^o -T \Delta S^o \label{4-3}$
This equation shows that $\Delta G^\text{0}$ and $\Delta H^\text{0}$ are equal when $\Delta S^\text{0}$ is zero. Therefore the sign and magnitude of $T \Delta S^\text{0}$ determine how well $K_\text{eq}$ correlates with $\Delta H^\text{0}$. Now, we have to give attention to whether we can estimate $T \Delta S^\text{0}$ values well enough to decide whether the $\Delta H^\text{0}$ of a given reaction (calculated from bond energies or other information) will give a good or poor measure of $\Delta G^\text{0}$.
Entropy and Molecular Disorder
To decide whether we need to worry about $\Delta S^\text{0}$ with regard to any particular reaction, we have to have some idea what physical meaning entropy has. To be very detailed about this subject is beyond the scope of this book, but you should try to understand the physical basis of entropy, because if you do, then you will be able to predict at least qualitatively whether $\Delta H^\text{0}$ will be about the same or very different from $\Delta G^\text{0}$. Essentially, the entropy of a chemical system is a measure of its molecular disorder or randomness. Other things being the same, the more random the system is, the more favorable the system is.
Different kinds of molecules have different degrees of translational, vibrational, and rotational freedom and, hence, different average degrees of molecular disorder or randomness. Now, if for a chemical reaction the degree of molecular disorder is different for the products than for the reactants, there will be a change in entropy and $\Delta S^\text{0} \neq 0$.
A spectacular example of the effect of molecular disorder in contributing to the difference between $\Delta H^\text{0}$ and $\Delta G^\text{0}$ is afforded by the formation of liquid nonane, $C_9H_{20}$, from solid carbon and hydrogen gas at $25^\text{o}$:
$\ce{9C(s) + 10H_2(g) \rightarrow C_910_{20}(l)}$
with $\Delta H^o = -54.7 \, kcal$ and $\Delta S^o = 5.0 \, kcal$.
Equations $\ref{4-2}$ and $\ref{4-3}$ can be rearranged to calculate $\Delta S^\text{0}$ and $K_\text{eq}$ from $\Delta H^\text{0}$ and $\Delta G^\text{0}$:
and
$K_{eq} = 10^{-\Delta G^o/2.303 \,RT} = 10^{-5.900/(2.303 \times 1.987 \times 298.2)} = 4.7 \times 10^{-5}$
These $\Delta H^\text{0}$, $\Delta S^\text{0}$, and $K_\text{eq}$ values can be compared to those for $H_2 + \frac{1}{2} O_2 \longrightarrow H_2O$, for which $\Delta H^\text{0}$ is $-57 \: \text{kcal}$, $\Delta S^\text{0}$ is $8.6 \: \text{e.u.}$, and $K_\text{eq}$ is $10^{40}$. Obviously, there is something unfavorable about the entropy change from carbon and hydrogen to nonane. The important thing is that there is a great difference in the constraints on the atoms on each side of the equation. In particular, hydrogen molecules in the gaseous state have great translational freedom and a high degree of disorder, the greater part of which is lost when the hydrogen atoms become attached to a chain of carbons. This makes for a large negative $\Delta S^\text{0}$, which corresponds to a decrease in $K_\text{eq}$. The differences in constraints of the carbons are less important. Solid carbon has an ordered, rigid structure with little freedom of motion of the individual carbon atoms. These carbons are less constrained in nonane, and this would tend to make $\Delta S^\text{0}$ more positive and $\Delta G^\text{0}$ more negative, corresponding to an increase in $K_\text{eq}$ (Equations $\ref{4-2}$ and $\ref{4-3}$). However, this is a small effect on $\Delta S^\text{0}$ compared to the enormous difference in the degree of disorder of hydrogen between hydrogen gas and hydrogen bound to carbon in nonane.
Negative entropy effects usually are observed in ring-closure reactions such as the formation of cyclohexane from 1-hexene, which occur with substantial loss of rotational freedom (disorder) about the $C-C$ bonds:
There is an even greater loss in entropy on forming cyclohexane from ethene because substantially more freedom is lost in orienting three ethene molecules to form a ring:
For simple reactions, with the same number of molecules on each side of the equation, with no ring formation or other unusual changes in the constraints between the products and reactants, $\Delta S^\text{0}$ usually is relatively small. In general, for such processes, we know from experience that $K_\text{eq}$ usually is greater than 1 if $\Delta H^\text{0}$ is more negative than $-15 \: \text{kcal}$ and usually is less than 1 for $\Delta H^\text{0}$ more positive than $+15 \: \text{kcal}$. We can use this as a "rule of thumb" to predict whether $K_\text{eq}$ should be greater or less than unity for vapor-phase reactions involving simple molecules. Some idea of the degree of success to be expected from this rule may be inferred from the examples in Table 4-5, which also contains a further comparison of some experimental $\Delta H^\text{0}$ values with those calculated from bond energies.
Suppose $\Delta G^\text{0}$ is positive, what hope do we have of obtaining a useful conversion to a desired product? There is no simple straightforward and general answer to this question. When the reaction is reversible the classic procedure of removing one or more of the products to prevent equilibrium from being established has many applications in organic chemistry, as will be seen later. When this approach is inapplicable, a change in reagents is necessary. Thus, iodine does not give a useful conversion with 2,2-dimethylpropane, $1$, to give 1-iodo-2,2-dimethylpropane, $2$, because the position of equilibrium is too far to the left ($K_\text{eq} \cong 10^{-5}$):
Alternative routes with favorable $\Delta G^\text{0}$ values are required. Development of ways to make indirectly, by efficient processes, what cannot be made directly is one of the most interesting and challenging activities of organic chemists.
Why Do Methane and Chlorine Fail to React in the Dark at $25^\text{o}$?
To reach an understanding of why methane and chlorine do not react in the dark, we must consider the details of how the reaction occurs - that is, the reaction mechanism. The simplest mechanism would be for a chlorine molecule to collide with a methane molecule in such a way as to have chloromethane and hydrogen chloride formed directly as a result of a concerted breaking of the $Cl-Cl$ and $C-H$ bonds and making of the $C-Cl$ and $H-Cl$ bonds (see Figure 4-5). The failure to react indicates that there must be an energy barrier too high for this mechanism to operate. Why should this be so?
First, this mechanism involves a very precisely oriented "four-center" collision between chlorine and methane that would have a low probability of occurrence (i.e., a large decrease in entropy because a precise orientation means high molecular ordering). Second, it requires pushing a chlorine molecule sufficiently deeply into a methane molecule so one of the chlorine atoms comes close enough to the carbon to form a bond and yield chloromethane.
Generally, to bring nonbonded atoms to near-bonding distances ($1.2 \: \text{A}$ to $1.8 \: \text{A}$) requires a large expenditure of energy, as can be seen in Figure 4-6. Interatomic repulsive forces increase rapidly at short distances, and pushing a chlorine molecule into a methane molecule to attain distances similar to the $1.77$-$\text{A}$ carbon-chlorine bond distance in chloromethane would require a considerable amount of compression (see Figure 4-7). Valuable information
about interatomic repulsions can be obtained with space-filling models of the CPK type (Section 2-2), which have radii scaled to correspond to actual atomic interference radii, that is, the interatomic distance at the point where curves of the type of Figure 4-6 start to rise steeply. With such models, the degree of atomic compression required to bring the nonbonded atoms to within near-bonding distance is more evident than with the ball-and-stick models. It may be noted that four-center reactions of the type postulated in Figure 4-5 are encountered only rarely.
If the concerted four-center mechanism for formation of chloromethane and hydrogen chloride from chlorine and methane is discarded, all the remaining possibilities are stepwise reaction mechanisms. A slow stepwise reaction is dynamically analogous to the flow of sand through a succession of funnels with different stem diameters. The funnel with the smallest stem will be the most important bottleneck and, if its stem diameter is much smaller than the others, it alone will determine the flow rate. Generally, a multistep chemical reaction will have a slow rate-determining step (analogous to the funnel with the small stem) and other relatively fast steps, which may occur either before or after the slow step.
A possible set of steps for the chlorination of methane follows:
Reactions (1) and (2) involve dissociation of chlorine into chlorine atoms and the breaking of a $C-H$ bond of methane to give a methyl radical and a hydrogen atom. The methyl radical, like chlorine and hydrogen atoms, has one election not involved in bonding. Atoms and radicals usually are highly reactive, so formation of chloromethane and hydrogen chloride should proceed readily by Reactions (3) and (4). The crux then will be whether Steps (1) and (2) are reasonable under the reaction conditions.
In the absence of some external stimulus, only collisions due to the usual thermal motions of the molecules can provide the energy needed to break the bonds. At temperatures below $100^\text{o}$, it is very rare indeed that the thermal agitation alone can supply sufficient energy to break any significant number of bonds stronger than $30$ to $35 \: \text{kcal mol}^{-1}$.
The $Cl-Cl$ bond energy from Table 4-3 is $58.1 \: \text{kcal}$, which is much too great to allow bond breaking from thermal agitation at $25^\text{o}$ in accord with Reaction (1). For Reaction (2) it is not advisable to use the $98.7 \: \text{kcal} \: C-H$ bond energy from Table 4-3 because this is one fourth of the energy required to break all four $C-H$ bonds (Section 4-3). More specific bond-dissociation energies are given in Table 4-5, and it will be seen that to break one $C-H$ bond of methane requires $104 \: \text{kcal}$ at $25^\text{o}$, which again is too much to be gained by thermal agitation. Therefore we can conclude that Reactions (1)-(4) can not be an important mechanism for chlorination of methane at room temperature.
One might ask whether dissociation into ions would provide viable mechanisms for methane chlorination. Part of the answer certainly is: Not in the vapor phase, as the following thermochemical data show:
Ionic dissociation simply does not occur at ordinarily accessible temperatures by collisions between molecules in the vapor state. What is needed for formation of ions is either a highly energetic external stimulus, such as bombardment with fast-moving electrons, or an ionizing solvent that will assist ionization. Both of these processes will be discussed later. The point here is that ionic dissociation is not a viable step for the vapor-phase chlorination of methane.
Why Does Light Induce the Chlorination of Methane?
First, we should make clear that the light does more than provide energy merely to lift the molecules of methane and chlorine over the barrier of Figure 4-4. This is evident from the fact that very little light is needed, far less than one light photon per molecule of chloromethane produced. The light could activate either methane or chlorine, or both. However, methane is colorless and chlorine is yellow-green. This indicates that chlorine, not methane, interacts with visible light. A photon of near-ultraviolet light, such as is absorbed by chlorine gas, provides more than enough energy to split the molecule into two chlorine atoms:
Once produced, a chlorine atom can remove a hydrogen atom from a methane molecule and form a methyl radical and a hydrogen chloride molecule. The bond-dissociation energies of $CH_4$ ($104 \: \text{kcal}$) and $HCl$ ($103.1 \: \text{kcal}$) suggest that this reaction is endothermic by about $1 \: \text{kcal}$:
Use of bond-dissociation energies gives a calculated $\Delta H^\text{0}$ of $-26 \: \text{kcal}$ for this reaction, which is certainly large enough, by our rule of thumb, to predict that $K_\text{eq}$ will be greater than 1. Attack of a methyl radical on molecular chlorine is expected to require somewhat more oriented collision than for a chlorine atom reacting with methane (the chlorine molecule probably should be endwise, not sidewise, to the radical) but the interatomic repulsion probably should not be much different.
The net result of $CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl$ and $CH_3 \cdot + Cl_2 \longrightarrow CH_3Cl + Cl \cdot$ is formation of chloromethane and hydrogen chloride from methane and chlorine. Notice that the chlorine atom consumed in the first step is replaced by another one in the second step. This kind of sequence of reactions is called a chain reaction because, in principle, one atom can induce the reaction of an infinite number of molecules through operation of a "chain" or cycle of reactions. In our example, chlorine atoms formed by the action of light on
$Cl_2$ can induce the chlorination of methane by the chain-propagating steps:
In practice, chain reactions are limited by so-called termination processes. In our example, chlorine atoms or methyl radicals are destroyed by reacting with one another, as shown in the following equations:
Chain reactions may be considered to involve three phases: First, chain initiation must occur, which for methane chlorination is activation and conversion of chlorine molecules to chlorine atoms by light. Second, chain-propagation steps convert reactants to products with no net consumption of atoms or radicals. The propagation reactions occur in competition with chain-terminating steps, which result in destruction of atoms or radicals. Putting everything together, we can write:
The chain-termination reactions are expected to be exceedingly fast because atoms and radicals have electrons in unfilled shells that normally are bonding. As a result, bond formation can begin as soon as the atoms or radicals approach one another closely, without need for other bonds to begin to break. The evidence is strong that bond-forming reactions between atoms and radicals usually are diffusion-controlled, that there is almost no barrier or activation energy required, and the rates of combination are simply the rates at which encounters between radicals or atoms occur.
If the rates of combination of radicals or atoms are so fast, you might well wonder how chain propagation ever could compete. Of course, competition will be possible if the propagation reactions themselves are fast, but another important consideration is the fact that the atom or radical concentrations are very low. Suppose that the concentration of $Cl \cdot$ is $10^{-11} \: \text{M}$ and the $CH_4$ concentration $1 \: \text{M}$. The probability of encounters between two $Cl \cdot$ atoms will be proportional to $10^{-11} \times 10^{-11}$, and between $CH_4$ and $Cl \cdot$ atoms it will be $10^{-11} \times 1$. Thus, other things being the same, $CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl$ (propagation) would be favored over $2Cl \cdot \longrightarrow Cl_2$ (termination) by a factor of $10^{11}$. Under favorable conditions, the methane-chlorination chain may go through 100 to 10,000 cycles before termination occurs by radical or atom combination. Consequently the efficiency (or quantum yield) of the reaction is very high in terms of the amount of chlorination that occurs relative to the amount of the light absorbed.
The overall rates of chain reactions usually are slowed very much by substances that can combine with atoms or radicals and convert them into species incapable of participating in the chain-propagation steps. Such substances are called radical traps, or inhibitors. Oxygen acts as an inhibitor in the chlorination of methane by rapidly combining with a methyl radical to form the comparatively stable (less reactive) peroxymethyl radical, $CH_3OO \cdot$. This effectively terminates the chain:
Can We Predict Whether Reactions Will Be Fast or Slow?
To a considerable degree, we can predict relative reactivities, provided we use common sense to limit our efforts to reasonable situations. In the preceding section, we argued that reactions in which atoms or radicals combine can well be expected to be extremely fast because each entity has a potentially bonding electron in an outer unfilled shell, and bringing these together to form a bond does not require that other bonds be broken:
The difference between the average energy of the reactants and the energy of the transition state is called the activation energy (Figure 4-4). We expect this energy to be smaller (lower barrier) if a weak bond is being broken and a strong bond is being made. The perceptive reader will notice that we are suggesting a parallel between reaction rate and $\Delta H^\text{0}$ because $\Delta H^\text{0}$ depends on the difference in strengths of the bonds being broken and formed. Yet previously (Section 4-4A), we pointed out that the energy barrier for a reaction need bear no relationship to how energetically feasible the reaction is, and this is indeed true for complex reactions involving many steps. But our intuitive parallel between rate and $\Delta H^\text{0}$ usually works quite well for the rates of individual steps. This is borne out by experimental data on rates of removal of a hydrogen atom from methane by atoms or radicals ($X \cdot$), such as $F \cdot$, $Cl \cdot$, $Br \cdot$, $HO \cdot$, $H_2N \cdot$, which generally parallel the strength of the new bond formed:
Similarly, if we look at the $H-C$ bond-dissociation energies of the hydrocarbons shown in Table 4-6, we would infer that $Cl \cdot$ would remove a hydrogen most rapidly from the carbon forming the weakest $C-H$ bond and, again, this is very much in accord with experience. For example, the chlorination of methylbenzene (toluene) in sunlight leads to the substitution of a methyl hydrogen rather than a ring hydrogen for the reason that the methyl $C-H$ bonds are weaker and are attacked more rapidly than the ring $C-H$ bonds. This can be seen explicitly in the $\Delta H^\text{0}$ values for the chain-propagation steps calculated from the bond-dissociation energies of Table 4-6.
Methyl substitution (observed):
Ring substitution (not observed):
The $\Delta H^\text{0}$ of ring-hydrogen abstraction is unfavorable by $+7 \: \text{kcal}$ because of the high $C-H$ bond energy ($110 \: \text{kcal}$). Thus this step is not observed. It is too slow in comparison with the more favorable reaction at the methyl group even though the second propagation step is energetically favorable by $-37 \: \text{kcal}$ and presumably would occur very rapidly. Use of bond-dissociation energies to predict relative reaction rates becomes much less valid when we try to compare different kinds of reactions. To illustrate, ethane might react with $F \cdot$ to give fluoromethane or hydrogen fluoride:
It is not a good idea to try to predict the relative rates of these two reactions on the basis of their overall $\Delta H^\text{0}$ values because the nature of the bonds made and broken is too different.
How Should We Go about Formulating a Reaction Mechanism?
Faced with proposing a mechanism for a reaction that involves overall making or breaking of more than two bonds, the beginner almost invariably tries to concoct a process wherein, with a single step, all of the right bonds break and all of the right bonds form. Such mechanisms, called concerted mechanisms, have three disadvantages. First, they are almost impossible to prove correct. Second, prediction of the relative rates of reactions involving concerted mechanisms is especially difficult. Third, concerted mechanisms have a certain sterility in that one has no control over what happens while they are taking place, except an overall control of rate by regulating concentrations, temperature, pressure, choice of solvents, and so on.
To illustrate, suppose that methane chlorination appeared to proceed by way of a one-step concerted mechanism:
At the instant of reaction, the reactant molecules in effect would disappear into a dark closet and later emerge as product molecules. There is no way to prove experimentally that all of the bonds were made and formed simultaneously. All one could do would be to use the most searching possible tests to probe for the existence of discrete steps. If these tests fail, the reaction still would not be proved concerted because other, still more searching tests might be developed later that would give a different answer. The fact is, once you accept that a particular reaction is concerted, you, in effect, accept the proposition that further work on its mechanism is futile, no matter how important you might feel that other studies would be regarding the factors affecting the reaction rate.
The experienced practitioner in reaction mechanisms accepts a concerted mechanism for a reaction involving the breaking and making of more than two bonds as a last resort. He first will try to analyze the overall transformation in terms of discrete steps that are individually simple enough surely to be concerted and that also involves energetically reasonable intermediates.
Such an analysis of a reaction in terms of discrete mechanistic steps offers many possibilities for experimental studies, especially in development of procedures for detecting the existence, even if highly transitory, of the proposed intermediates. We shall give many examples of the fruitfulness of this kind of approach in subsequent discussions.
$^4$If calculations based on chemical equilibrium constants are unfamiliar to you, we suggest you study one of the general chemistry texts listed for supplemental reading at the end of Chapter 1.
$^5$Many books and references use $\Delta F^\text{0}$ instead of $\Delta G^\text{0}$. The difference between standard Gibbs energy $\Delta G^\text{0}$ and the Gibbs energy $\Delta G$ is that $\Delta G^\text{0}$ is defined as the value of the free energy when all of the participants are in standard states. The free energy for $\Delta G$ for a reaction $\text{A} + \text{B} + \cdots \longrightarrow \text{X} + \text{Y} + \cdots$ is equal to $\Delta G^\text{0} - 2.303 RT \: \text{log} \: \frac{\left[ \text{X} \right] \left[ \text{Y} \right] \cdots}{\left[ \text{A} \right] \left[ \text{B} \right] \cdots}$ where the products, $\left[ \text{X} \right], \left[ \text{Y} \right] \cdots$, and the reactants, $\left[ \text{A} \right], \left[ \text{B} \right] \cdots$, do not have to be in standard states. We shall use only $\Delta G^\text{0}$ in this book.
$^6$The entropy unit $\text{e.u.}$ has the dimensions calorie per degree or $\text{cal deg}^{-1}$.
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)/04%3A_Alkanes/4.05%3A_Halogenation_of_Alkanes._Energies_and_Rates_of_Reactions.txt |
Given the knowledge that a particular reaction will proceed at a suitable rate, a host of practical considerations are necessary for satisfactory operation. These considerations include interference by possible side reactions that give products other than those desired, the ease of separation of the desired product from the reaction mixture, and costs of materials, apparatus, and labor. We shall consider these problems in connection with the important synthetic reactions discussed in this book.
The chlorination of saturated hydrocarbons can be induced by light, but also can be carried out at temperatures of about $300^\text{o}$ in the dark. Under such circumstances the mechanism is similar to that of light-induced chlorination, except that the chlorine atoms are formed by thermal dissociation of chlorine molecules. Solid carbon surfaces catalyze thermal chlorination, possibly by aiding in the cleavage of the chlorine molecules.
Direct monohalogenation of saturated hydrocarbons works satisfactorily only with chlorine and bromine. For the general reaction
the calculated $\Delta H^\text{0}$ value is negative and very large for fluorine, negative and moderate for chlorine and bromine, and positive for iodine (see Table 4-7). With fluorine, the reaction evolves so much heat that it may be difficult to control, and products from cleavage of carbon-carbon as well as of carbon-hydrogen bonds may be obtained. The only successful, direct fluorination procedure for hydrocarbons involves diffusion of minute amounts of fluorine mixed with helium into liquid or solid hydrocarbons at low temperatures, typically $-78^\text{o}$ (Dry Ice temperature). As fluorination proceeds, the concentration of fluorine can be increased. The process is best suited for preparation of completely fluorinated compounds, and it has been possible to obtain in this way amounts of $\left( CF_3 \right)_4C$ and $\left( CF_3 \right)_3 C-C \left( CF_3 \right)_3$ from 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane corresponding to $10$-$15\%$ yields based on the fluorine used.
Bromine generally is much less reactive toward hydrocarbons than chlorine is, both at high temperatures and with activation by light. Nonetheless, it usually is possible to brominate saturated hydrocarbons successfully. Iodine is unreactive.
Table 4-7: Calculated Heat of Reaction for Halogenation fo Hydrocarbons
Halogen (X) $\Delta H^o$ (kcal/mole)a
F -116
Cl -27
Br -10
I 13
aCalculated from the bond energies of Table 4-3.
The chlorination of methane does not have to stop with the formation of chloromethane (methyl chloride). It is usual when chlorinating methane to obtain some of the higher chlorination products: dichloromethane (methylene chloride), trichloromethane (chloroform), and tetrachloromethane (carbon tetrachloride):
In practice, one can control the degree of substitution to a considerable extent by controlling the methane-chlorine ratio. For example, for monochlorination to predominate, a high methane-chlorine ratio is necessary such that the chlorine atoms react with $CH_4$ and not with $CH_3Cl$.
Selectivity in Alkane Halogenation
For propane and higher hydrocarbons for which more than one monosubstitution product is generally possible, difficult separation problems bay arise when a particular product is desired. For example, the chlorination of 2-methylbutane $3$ at $300^\text{o}$ gives all four possible monosubstitution products, $4$, $5$, $6$, and $7$:
On a purely statistical basis, we may expect the ratio of products from $3$ to correlate with the number of available hydrogens at the various positions of substitution. That is, $4$, $5$, $6$, and $7$ would be formed in the ratio 6:3:2:1 ($50\%$:$25\%$:$17\%$:$8\%$). However, as can be seen from Table 4-6, the strengths of hydrogen bonds to primary, secondary, and tertiary carbons are not the same and, from the argument given in Section 4-4E we would expect the weaker $C-H$ bonds to be preferentially attacked by $Cl \cdot$. The proportion of $7$ formed is about three times that expected on a statistical basis which is in accord with our expectation that the tertiary $C-H$ bond of 2-methylbutane should be the weakest of the $C-H$ bonds. (See Table 4-6.)
The factors governing selectivity in halogenation of alkanes follow:
1. The rates at which the various $C-H$ bonds of 2-methylbutane are broken by attack of chlorine atoms approach 1:1:1 as the temperature is raised above $300^\text{o}$. At higher temperatures both chlorine atoms and hydrocarbons become more reactive because of increases in their thermal energies. Ultimately, temperatures are attained where a chlorine atom essentially removes the first hydrogen with which it collides regardless of position on the hydrocarbon chain. In such circumstances, the composition of monochlorination products will correspond to that expected from simple statistics.
2. Bromine atoms are far more selective than chlorine atoms. This is not unexpected because is endothermic, whereas corresponding reactions with a chlorine atoms usually are exothermic (data from Table 4-6). Bromine removes only those hydrogens that are relatively weakly bonded to a carbon atom. As predicted, attack of $Br \cdot$ on 2-methylbutane leads mostly to 2-bromo-2-methylbutane, some secondary bromide, and essentially no primary bromides:
3. The selectivity of chlorination reactions carried on in solution is increased markedly in the presence of benzene or alkyl-substituted benzenes because benzene and other arenes form loose complexes with chlorine atoms. This substantially cuts down chlorine-atom reactivity, thereby making the chlorine atoms behave more like bromine atoms.
Chemical Initiation of Radical-Chain Substitution
It is possible to achieve chlorination of alkanes using sulfuryl chloride ($SO_2Cl_2$, bp $69^\text{o}$) in place of chlorine:
The reaction has a radical-chain mechanism and the chains can be initiated by light or by chemicals, usually peroxides, $ROOR$. Chemical initiation requires an initiator with a weak bond that dissociates at temperatures between $40$-$80^\text{o}$. Peroxides are good examples. The $O-O$ bond is very weak ($30$-$50 \: \text{kcal}$) and on heating dissociates to alkoxyl radicals, $RO \cdot$, which are reactive enough to generate the chain-propagating radicals from the reactants. The exact sequence of chemical initiation is not always known, but a plausible route in the present case would have $RO \cdot$ abstract hydrogen from the alkane:
The propagation steps that would follow are
Chlorination with sulfuryl chloride of alkanes with more than one kind of hydrogen gives a mixture of alkyl chlorides resembling that obtained with chlorine itself. However, in some circumstances the mixture of chlorides is not the same mixture obtained with chlorine itself and when this is true, the hydrogen-abstraction step probably involves $\cdot SO_2Cl$ rather than $Cl \cdot$. The alternative propagation steps then are
Different product ratios are expected from $Cl \cdot$ and $ClSO_2 \cdot$ for the same reason that $Cl \cdot$ and $Br \cdot$ lead to different product ratios (Section 4-5A). Other reagents that sometimes are useful halogenating agents in radical-chain reactions include
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."
4.07: Nitration of Alkanes
Another reaction of commercial importance is the nitration of alkanes to give nitroparaffins. Such reactions usually are carried out in the vapor phase at elevated temperatures using nitric acid (\(HNO_3\)) or nitrogen tetroxide (\(N_2O_4\)) as the nitrating agent:
All available evidence points to a radical mechanism for nitration, but many aspects of the reaction are not fully understood. Mixtures are obtained; nitration of propane gives not only 1- and 2-nitropropanes but nitroethane and nitromethane:
In commercial practice, the yield and product distribution in nitration of alkanes is controlled as far as possible by the judicious addition of catalysts (e.g., oxygen and halogens), which are believed to raise the concentration of alkyl radicals. The products are separated from the mixtures by fractional distillation.
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)/04%3A_Alkanes/4.06%3A_Practical_Halogenations_and_Problems_of_Selectivity.txt |
By now you should be familiar with position isomers wherein compounds of the same molecular formula differ because substituents, chain branches, and so on, are not at the same positions in the molecules. 1-Chloropropane and 2-chloropropane are straightforward examples of position isomers. A much more subtle form of isomerism is present when two different compounds have the same molecular formulas, the same substituent and chain-branching positions, and, indeed, even have the same names by all of the nomenclature rules we have given you so far. Such isomers are different because their molecules have different arrangements of the atoms in space. These are stereoisomers and this type of isomerism, called stereoisomerism, is of enormous importance to all areas of organic chemistry and biochemistry.
• 5.1: Prelude to Stereoisomerism
• 5.2: Configurational Isomers
We have defined isomers in a very general way as nonidentical molecules that possess the same number and kind of atoms. However, there are several ways in which isomers can be nonidentical. Most, but not all alkenes, have stereoisomers that are not identical because of different spatial arrangements of the component atoms. Stereoisomers that do not interconvert rapidly under normal conditions, and therefore are stable enough to be separated, specifically are called configurational isomers.
• 5.3: Conformational Isomers
An infinite number of different atomic orientations are possible in many organic molecules, depending on the angular relationship between the hydrogens on each carbon. Two extreme orientations or conformations are the eclipsed conformation and the staggered conformation. It has not been possible to obtain separate samples of ethane that correspond to these or intermediate orientations because actual ethane molecules appear to have essentially "free rotation" about the single bond between carbons
• 5.4: Representation of Organic Structure
Many problems in organic chemistry require consideration of structures in three dimensions, and it is very helpful to use molecular models to visualize the relative positions of the atoms in space.
• 5.5: The D, L Convention for Designating Stereochemical Configurations
We pointed out in Chapter 3 the importance of using systematic names for compounds such that the name uniquely describes the structure. It is equally important to be able to unambiguously describe the configuration of a compound. The convention that is used to designate the configurations of chiral carbons of naturally occurring compounds is called the \(D, L\) system. To use it, we view the molecule of interest according to the rules presented here.
• 5.6: Molecules with More Than One Chiral Center. Diastereomers
We have seen examples of molecules with one chiral center that exist in two mirror-image configurations, which we call enantiomers. What happens when there is more than one chiral center? How many stereoisomers should we expect?
• 5.7: Some Examples of the Importance of Stereoisomerism to Biology. Biological Stereospeciflcity
Asymmetric or chiral reagents can differentiate between enantiomers, especially by having at least some difference in reactivity toward them. The property of being able to discriminate between diastereomers is called stereospecificity, and this is an especially important characteristic of biological systems.
• 5.E: Stereoisomerism of Organic Molecules (Exercises)
These are the homework exercises to accompany Chapter 5 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."
05: Stereoisomerism of Organic Molecules
By now you should be familiar with position isomers wherein compounds of the same molecular formula differ because substituents, chain branches, and so on, are not at the same positions in the molecules. 1-Chloropropane and 2-chloropropane are straightforward examples of position isomers. A much more subtle form of isomerism is present when two different compounds have the same molecular formulas, the same substituent and chain-branching positions, and, indeed, even have the same names by all of the nomenclature rules we have given you so far. Such isomers are different because their molecules have different arrangements of the atoms in space. These are stereoisomers and this type of isomerism, called stereoisomerism, is of enormous importance to all areas of organic chemistry and biochemistry.
To understand stereoisomerism of carbon compounds, we must understand the ways in which the bonds to carbon atoms are arranged in space. As shown in Section 2-2A, this depends on whether the carbon atoms form single, double, or triple bonds to another atom. Thus, four single bonds to a carbon form a tetrahedral arrangement; two single bonds and one double bond to a carbon give a planar array with bond angles near $120^\text{o}$, while one single bond and one triple bond (or two double bonds) to a carbon are arranged linearly:
Finally, if you have not studied the material already, you may wish to return to the last part of Chapter 3 and become acquainted with the nomenclature of cycloalkanes, alkenes, cycloalkenes, and alkynes (Sections 3-2 to 3-4).
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.01%3A_Prelude_to_Stereoisomerism.txt |
Geometric Isomerism
We have defined isomers in a very general way as nonidentical molecules that possess the same number and kind of atoms. However, there are several ways in which isomers can be nonidentical. Among the alkenes, 1- and 2-butene are position isomers, because in these compounds the double bond has a different position in the carbon chain:
Most, but not all alkenes, have stereoisomers that are not identical because of different spatial arrangements of the component atoms. Thus there are two stereoisomers of 2-butene that differ in the geometric arrangement of the groups attached to the double bond. In one isomer, both methyl groups are on the same side of the double bond (cis-2-butene) and in the other, the methyl groups are on opposite sides of the double bond (trans-2-butene):
It should be clear to you that there will be no cis-trans isomers of alkenes in which one end of the double bond carries identical groups. Thus we don not expect there to be cis-trans isomers of 1-butene or 2-methylpropene, and
indeed none are known:
You may wish to verify this by making ball-and-stick models of these substances.
Ring formation also confers rigidity on molecular structure such that rotation about the ring bonds is prevented. As a result, stereoisomerism of the cis-trans type is possible. For example, 1,2-dimethylcyclopropane exists in two forms that differ in the arrangement of the two methyl groups with
respect to the ring. In the cis isomer, the methyl groups both are situated above (or below) the plane of the ring and in the trans isomer they are situated one above and one below, as shown in Figure 5-2. Interconversion of these isomers does not occur without breaking one or more chemical bonds.
Stereoisomers that do not interconvert rapidly under normal conditions, and therefore are stable enough to be separated, specifically are called configurational isomers. Thus cis- and trans-2-butene are configurational isomers, as are cis- and trans-1,2-dimethylcyclopropane. The terms cis-trans isomerism or geometric isomerism commonly are used to describe configurational isomerism in compounds with double bonds and rings. When referring to the configuration of a particular isomer, we mean to specify its geometry. For instance, the isomer of 1,2-dichloroethene shown below has the trans configuration; the isomer of 1,3-dichlorocyclobutane has the cis configuration:
Cis-trans isomerism is encountered very frequently. By one convention, the configuration of a complex alkene is taken to correspond to the configuration of the longest continuous chain as it passes through the double bond. Thus the following compound is trans-4-ethyl-3-methyl-3-heptene, despite the fact that two identical groups are cis with respect to each other, because the longest continuous chain is trans as it passes through the double bond:
Notice that cis-trans isomerism is not possible at a carbon-carbon triple bond, as for 2-butyne, because the bonding arrangement at the triply bonded carbons is linear:
Many compounds have more than one double bond and each may have the potential for the cis or trans arrangement. For example, 2,4-hexadiene has three different configurations, which are designated as trans-trans, cis-cis, and trans-cis. Because the two ends of this molecule are identically substituted, the trans-cis becomes identical with cis-trans:
Chirality
The most important type of stereoisomerism is that which arises when molecules possess two structures that are not identical and also are mirror images of one another. This is not a difficult or unfamiliar concept. Many things around us, such as our hands, and pairs of shoes, are not identical and also are mirror images of one another. In the same way, nonidentical molecules exist in which the only distinction between them is that one is the mirror image of the other. A common statement is that such isomers are mirror images of one another, but these images are not "superimposable." A simple example of this type of stereoisomerism is 2-chlorobutane, , which can exist in two spatial configurations, $1$ and $2$, that correspond to reflections of each other. These isomers are specifically called enantiomers.
Compounds that lack reflection symmetry - meaning that they are not identical with their mirror images - are said to be chiral (pronounced "ki-rall", rhymes with spiral). This term is derived from the Greek word $\chi \epsilon \iota \rho =$ hand; and "handedness" or chirality is a property of dissymmetric molecules such that two configurational isomers are possible that are nonidentical mirror images. Compounds that possess reflection symmetry - meaning that they are identical with their mirror images - are said to be achiral. Enantiomers are not possible for achiral compounds. An enantiomeric pair is a pair of substances whose molecules are nonidentical mirror images.
The pressing question at this point is how can we tell whether a substances will be chiral or achiral. The most common origin of chirality in molecules, and the one originally recognized by van't Hoff and Le Bel, is the presence of one or more atoms, usually carbon atoms, each of which forms coplanar bonds to four different atoms or groups. This is the case for 2-chlorobutane, because the second tetrahedral carbon along the chain is bonded to four different groups: hydrogen, chlorine, methyl, and ethyl. Therefore there is a pair of enantiomers, $1$ and $2$. Another example is 2-bromo-2-chloro-1,1,1-trifluoroethane, which is a widely used inhalation anaesthetic. The four different groups in this case are hydrogen, chlorine, bromine, and trifluoromethyl; the pair of enantiomers is shown in Structures $3$ and $4$:
The atom that carries the four different substituents in $1$ and $2$, or $3$ and $4$, is called a chiral atoms or chiral center. The latter is the more general term because, as we shall see later (Section 13-5A), dissymmetry in molecules need not be centered at an atom.$^1$
In evaluating a chemical structure for chirality, you should look for carbons carrying four different attached groups. There may be more than one chiral carbon, and you should be alert to the fact that structural differences in the attached groups do not necessarily show up at the first, or even the second, position along a chain. As an example, consider the chirality of 1,1,3-trimethylcyclohexane,
Carbons $C2$, $C4$, $C5$, and $C6$ are clearly achiral because each is connected to two identical groups, which for these atoms are hydrogens. The same is true for $C1$ because it is connected to two $CH_3$ groups. You might conclude that $C3$ also is an achiral position because it is connected to two $CH_2$ groups. But this would be wrong. If you look farther, you will see that the groups attached to $C3$ actually are different and are $H$, $CH_3$, $-CH_2CH_2CH_2-$, and $-CH_2C \left( CH_3 \right)_2$. Therefore 1,1,3-trimethylcyclohexane has a chiral center at $C3$. In contrast, the 1,1,4-isomer has no chiral centers because the groups attached to the ring at $C4$ are identical:
Several other terms that we shall use frequently in addition to chirality are racemic mixture, resolution, and racemization. A mixture of equal amounts of both enantiomers is a racemic mixture; separation of a racemic mixture into its component enantiomers is a resolution, and the conversion of either enantiomer into equal parts of both is called racemization.
Optical Activity
Until recently, the phenomenon of chirality has been better known as optical isomerism, and configurational isomers that are enantiomers were referred to as optical antipodes. The reasons for this are mainly historical. It was discovered early in the nineteenth century that many compounds, whether solid, liquid, or gas, have the property of rotating the plane of polarization of polarized light and can be said to be "optically active." A satisfactory explanation of the origin of optical activity came much later and developed in its modern form from the classic researches of Louis Pasteur, and from the concept of the three-dimensional carbon atoms expressed independently by J. H. van't Hoff and J. A. Le Bel.$^2$
Pasteur's contribution to stereochemistry came as a result of his studies of the shapes of crystals of tartaric acid, $\ce{HO_2C-CHOH-CHOH-CO_2H}$, and its salts. Tartaric acid, a by-product of wine production, was known to be optically active, and Pasteur showed that it, and nineteen different salts of it, all formed crystals that were not identical with their mirror images. A different substance known as "racemic acid," for which we can write the same condensed formula, $\ce{HO_2C-CHOH-CHOH-CO_2H}$, was known to be optically inactive, and Pasteur expected that when he crystallized this acid or its salts he would obtain crystals that would be identical with their mirror images. However, crystallization of the sodium ammonium salt of racemic acid from water at temperatures below $28^\text{o}$ gave crystals of two different shapes and these shapes were mirror images of one another. Pasteur carefully picked apart the two kinds of crystals and showed that one of them was identical with the corresponding salt of tartaric acid, except that it rotated the plane of polarization of polarized light in the opposite direction. This separation of racemic acid into two optically active forms now is called a "resolution of racemic acid."
On the basis of his discoveries, Pasteur postulated that "optical isomerism" had to be related to the molecular dissymmetry of substances such that nonidentical mirror-image forms could exist. However, it remained for van' t Hoff and Le Bel to provide, almost simultaneously, a satisfactory explanation at the molecular level. In his first published work on tetrahedral carbon van't Hoff said "...it appears more and more that the present constitutional formulae are incapable of explaining certain cases of isomerism; the reason for this is perhaps the fact that we need a more definite statement about the actual positions of the atoms."$^3$ He goes on to discuss the consequences of the tetrahedral arrangements of atoms about carbon, explicitly in connection with optical isomerism and geometric, or cis-trans, isomerism.
It is not easy for the chemist of today to appreciate fully the contributions of these early chemists because we have long accepted the tetrahedral carbon as an experimentally established fact. At the time the concept was enunciated, however, even the existence of atoms and molecules was questioned openly by many scientists, and to ascribe "shapes" to what in the first place seemed like metaphysical conceptions was too much for many to accept.
Properties of Enantiomers
Optical activity is an experimentally useful property and usually is measured as the angle of rotation ($\alpha$) of the plane of polarization of polarized light passing through solutions of the substances under investigation (Figure 5-4). Where measurable optical activity is present, it is found that one enantiomer rotates the plane of polarization in one direction, whereas the other causes the plane to rotate equally but in the opposite direction. With reference to the plane of incident light, the enantiomer that rotates the plane to the right is called dextrorotatory and is symbolized by either d or ($+$); the enantiomer that rotates the plane to the left is levorotatory, symbolized by l or ($-$). A racemic mixture then can be designated as dl or ($\pm$), and will have no net optical rotation. It is very important to know that d, l, ($+$), or ($-$) do not designate configurations. Thus, although ($+$)-2-butanol actually has configuration $5$ and ($-$)-2-butanol has configuration $6$, there is no simple way to predict that a particular sign of rotation will be associated with a particular configuration. Methods used in assigning the true configurations to enantiomers will be discussed later.
A very important point to keep in mind about any pair of enantiomers is that they will have identical chemical and physical properties, except for the signs of their optical rotations, with one important proviso: All of the properties to be compared must be determined using achiral reagents in a solvent made up of achiral molecules or, in short, in an achiral environment. Thus the melting and boiling points (but not the optical rotations) of $5$ and $6$ will be identical in an achiral environment. How a chiral environment or chiral reagents influence the properties of substances such as $5$ and $6$ will be considered in Chapter 19.
$^1$In the older literature, chiral centers often are called asymmetric centers and you may be confused by the difference between asymmetric and dissymmetric. Both asymmetric and dissymmetric molecules (or objects) are chiral. An asymmetric object has no symmetry at all and looks different from all angles of view. Formulas $3$ and $4$ represent asymmetric molecules. A dissymmetric molecule is chiral, but looks the same from more than one angle of view. A helical spring is dissymmetric - it looks the same from each end. We will encounter dissymmetric molecules later.
$^2$The tetrahedral carbon was first proposed by E. Paterno in 1869 (see Section1-1E), but he apparently did not recognize its implications for chirality. These implications were recognized first by van't Hoff and Le Bel, with van't Hoff proceeding on the basis of bonds to carbon being directed to the corners of a regular tetrahedron. Le Bel was opposed to such a rigid formulation of the bonds to carbon.
$^3$An interesting account and references to van't Hoff's early work can be found in "The Reception of J. H. van't Hoff's Theory of the Asymmetric Carbon" by H. A. M. Snelders, J. Chem. Educ. 51, 2 (1974). A century has passed since van't Hoff first published his theory, which he did before he obtained his doctoral degree from the University of Utrecht. van't Hoff was the first recipient of the Nobel Prize in chemistry (1901) for his later work in thermodynamics and chemical kinetics.
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.02%3A_Configurational_Isomers.txt |
When using ball-and-stick models, if one allows the sticks to rotate in the holes, it will be found that for ethane, $CH_3-CH_3$, an infinite number of different atomic orientations are possible, depending on the angular relationship (the so-called torsional angle) between the hydrogens on each carbon. Two extreme orientations or conformations are shown in Figure 5-5. In end-on views of the models, the eclipsed conformation is seen to have the hydrogens on the forward carbon directly in front of those on the back carbon. The staggered conformation has each of the hydrogens on the forward carbon set between each of the hydrogens on the back carbon. It has not been possible to obtain separate samples of ethane that correspond to these or intermediate orientations because actual ethane molecules appear to have essentially "free rotation" about the single bond joining the carbons. Free, or at least rapid, rotation is possible around all $C-C$ single bonds, except when the carbons are part of a ring as in cyclopropane or cyclohexane.
For ethane and its derivatives, the staggered conformations are more stable than the eclipsed conformations. The reason for this in ethane is not wholly clear, but doubtless depends on the fact that, in the staggered conformation, the $C-H$ bonding electrons are as far away from one another as possible and give the least interelectronic repulsion. With groups larger than hydrogen atoms substituted on ethane carbons, space-filling models usually show less interference (steric hindrance) for staggered conformations than for eclipsed conformations.
The energy difference between eclipsed and staggered ethane is approximately $3 \: \text{kcal mol}^{-1}$.$^4$ This is shown in Figure 5-6 as the height of the peaks (eclipsed forms) separating the valleys (staggered forms) on a curve showing the potential energy of ethane as the methyl groups rotate with respect to each other through $360^\text{o}$. Rotation then is not strictly "free" because there is a $3$-$\text{kcal mol}^{-1}$ energy barrier to overcome on eclipsing the hydrogens. Even so, the barrier is low enough that rotation is very rapid at room temperature, occurring on the order of $10^{10}$ times per second.
In butane, $CH_3CH_2CH_2CH_3$, a $360^\text{o}$ rotation about the central $C-C$ bond allows the molecule to pass through three different eclipsed arrangements ($8$, $10$, $12$), and three different staggered arrangements ($7$, $9$, $11$), as shown in Figure 5-7. Experiment shows that butane favors the staggered form $7$ in which the methyl groups are farthest apart. This form is called the anti (or trans) conformation (sometimes conformer), and $63\%$ of the molecules of butane exist in this form at room temperature. The other two staggered forms $9$ and $11$ are called gauche (syn or skew) conformations and have a torsional angle of $60^\text{o}$ between the two methyl groups. Forms $9$ and $11$ actually are nonidentical mirror images, but bond rotation is so rapid that the separate enantiomeric conformations cannot be isolated. The populations of the two gauche forms are equal at room temperature ($18.5\%$ of each) so any optical rotation caused by one form is exactly canceled by an opposite rotation caused by the other.
The populations of the eclipsed forms of butane, like the eclipsed forms of ethane, are small and represent energy maxima for the molecule as rotation occurs about the central $C-C$ bond. The energy differences between the butane conformations are represented diagrammatically in Figure 5-8. The valleys correspond to staggered forms and the energy difference between the anti and gauche forms is $0.8$-$0.9 \: \text{kcal mol}^{-1}$.
Pioneering work in the field of conformational analysis was contributed by O. Hassel (Norway) and D. R. H. Barton (Britain), for which they shared the Nobel Prize in chemistry in 1969. Hassel's work involved the physical determination of preferred conformations of small molecules, whereas Barton was the first to show the general importance of conformation to chemical reactivity. Study of conformations and conformational equilibria has direct application to explaining the extraordinary specificity exhibited by compounds of biological importance. The compounds of living systems are tailor-made to perform highly specific or even unique functions by virtue of their particular configurations and conformations.
$^4$This is by no means a trivial amount of energy - the difference in energy between the staggered and eclipsed forms of $1 \: \text{mol}$ ($30 \: \text{g}$) of ethane being enough to heat $30 \: \text{g}$ of water from $0^\text{o}$ to $100^\text{o}$.
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.03%3A_Conformational_Isomers.txt |
Many problems in organic chemistry require consideration of structures in three dimensions, and it is very helpful to use molecular models to visualize the relative positions of the atoms in space. Unfortunately, we are forced to communicate three-dimensional concepts by means of drawings in two dimensions, and not all of us are equally gifted in making or visualizing such drawings. Obviously, communication by means of drawings, such as those in Figure 5-5 and 5-7, would be impractically difficult and time consuming, thus some form of abbreviation is necessary.
Conformational Drawings
Two styles of abbreviating the eclipsed and staggered conformations of ethane are shown in Figure 5-9; in each, the junction of lines representing bonds is assumed to be a carbon atom. Using the "sawhorse" convention, we always consider that we are viewing the molecule slightly from above and from the right, and it is understood that the central $C-C$ bond is perpendicular to the plane of the paper. With the "Newman" convention, we view the molecule directly down the $C-C$ bond axis so the carbon in front hides the carbon behind. The circle is only a visual aid to help distinguish the bonds of the back carbon from those of the front carbon. The rear atoms in the eclipsed conformation are drawn slightly offset from a truly eclipsed view so the bonds to them can be seen.
The staggered conformations of butane are shown in Figure 5-10 in both the sawhorse and Newman conventions. There is little to choose between the two conventions for simple ethane derivatives, but the sawhorse convention is strongly favored for representing the conformations of ring compounds such as cyclohexane. The resemblance between the gauche forms of butane and the most stable conformation of cyclohexane is strikingly apparent in the sawhorse representations of both, as shown in Figure 5-10. Notice that the ring carbons of cyclohexane do not lie in one plane and that all the bond angles are tetrahedral. The conformations of this interesting and important molecule are discussed in detail in Chapter 12.
Despite the usefulness of the sawhorse-type drawing, cyclic molecules often are drawn with planar rings and distorted bond angles even though the rings actually may not be planar. The reason for this is partly that planar rings are easier to draw and partly to emphasize the configuration of attached groups, irrespective of the conformation. Typical examples follow:
Generally we shall avoid such drawings and suggest that it is much better to learn to draw molecules in as nearly correct perspective as possible. Once the sawhorse representation of cyclohexane is mastered, it is almost as easy to drawn $14$ as $13$, and $14$ is much more informative about the shape of the molecule:
We have indicated how the enantiomers of 2-butanol differ by drawing their strutures $5$ and $6$ (Section 5-1D) in perspective to show the tetrahedral configuration of substituents at the chiral carbon. This configuration also can be represented by the sawhorse or Newman formulas using any one of the several possible staggered conformations such as $5a$ and $6a$ or $5b$ and $6b$:
These drawings are clear but can be cumbersome, particularly for more complex molecules, and we shortly shall describe other means of representing the configurations of chiral molecules.
Planar Structures
Planar molecules such as benzene, ethene, and methanal are best drawn in the plane of the paper with bond angles of about $120^\text{o}$. When it is desired to draw them as viewed on edge (out of plane) care must be taken to provide proper perspective. The forward bonds can be drawn with slightly heavier lines; a tapered bond indicates direction, the wide end pointing toward the viewer and the narrow end away from the viewer (Figure 5-11). Barred lines are used here to indicate a rear or receding bond (many writers use dashed lines, but these may be confused with other uses of dashed lines, as for partial bonds).
However, you will find other representations of planar carbons with rather grossly distorted bond angles. For example, methanoic acid is planar with nearly $120^\text{o}$ bond angles, but often is drawn with $H-C-O$ angles of $90^\text{o}$ and $180^\text{o}$:
The distorted structures commonly are used to save space and, regretfully, we have to use them very frequently for this reason.
Projection Formulas
The sawhorse or Newman representations of 2-butanol, $5a$ and $5b$ and $6a$ and $6b$, are excellent for showing the arrangements of the atoms in conformations, but are needlessly complex for representing the stereochemical configuration. Fischer projection formulas are widely used to show configurations and are quite straightforward, once one gets the idea of what they represent..
The projection formulas of 2-butanol are $5c$ and $6c$:
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.04%3A_Representation_of_Organic_Structure.txt |
We pointed out in Chapter 3 the importance of using systematic names for compounds such that the name uniquely describes the structure. It is equally important to be able to unambiguously describe the configuration of a compound. The convention that is used to designate the configurations of chiral carbons of naturally occurring compounds is called the \(D, L\) system. To use it, we view the molecule of interest according to the following rules:
1. The main carbon chain is oriented vertically with the lowest numbered carbon at the top. The numbering used for this purpose must follow the IUPAC rules:
2. Next, the structure must be arranged at the particular chiral carbon whose configuration is to be assigned so the horizontal bonds to that carbon extend toward you and the vertical bonds extend away from you. This arrangement will be seen to be precisely the same as the convention of projection formulas such as \(5c\) and \(6c\) (Section 5-3C).
3. Now the relative positions of the substituents on the horizontal bonds at the chiral centers are examined. If the main substituent is the left of the main chain, the \(L\) configuration is assigned; if this substituent is on the right, the \(D\) configuration is assigned.
For example, the two configurations of the amino acid, alanine, would be represented in perspective or projection as \(15\) and \(16\). The carboxyl carbon is \(C1\) and is placed at the top. The substituents at the chiral carbon connected to the horizontal bonds are amino (\(-NH_2\)) and hydrogen. The amino substituent is taken to be the main substituent; when this is on the left the acid has the \(L\) configuration, and when it is on the right, the \(D\) configuration. All of the amino acids that occur in natural proteins have been shown to have the \(L\) configuration.
Glyceraldehyde, \(CH_2OHCHOHCHO\), which has one chiral carbon bonded to an aldehyde function, hydrogen, hydroxyl, and hydroxymethyl (\(CH_2OH\)), is of special interest as the simplest chiral prototype of sugars (carbohydrates). Perspective views and Fischer projections of the \(D\) and \(L\) forms correspond to \(17\) and \(18\), respectively, where the carbon of the aldehyde function (\(-CH=O\)) is \(C1\):
The \(D,L\) system of designating configuration only can be applied when there is a main chain, and when we can make an unambiguous choice of the main substituent groups. Try, for instance, to assign \(D\) and \(L\) configurations to enantiomers of bromochlorofluoromethane. An excellent set of rules has been worked out for such cases that leads to unambiguous configurational assignments by what is called the \(R,S\) convention. We discuss the \(R,S\) system in detail in Chapter 19 and, if you wish, you can turn to it now. However, for the next several chapters, assigning configurations is much less important to us than knowing what kinds of stereoisomers are possible.
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.05%3A_The_D_L_Convention_for_Designating_Stereochemical_Configurations.txt |
We have seen examples of molecules with one chiral center that exist in two mirror-image configurations, which we call enantiomers. What happens when there is more than one chiral center? How many stereoisomers should we expect? Consider the stereoisomers of the important amino acid, threonine, (2-amino-3-hydroxybutanoic acid). For this substance, if we write all of the possible configurations of its two chiral carbons, we have four different projection formulas, $19$-$22$, corresponding to four different stereoisomers:
Because each chiral center added to a chain doubles the number of possible configurations, we expect eight different stereoisomers with three chiral carbons, sixteen with four, and so on, the simple rule the is $2^n$ possible different stereoisomers for $n$ chiral centers. As we shall see later, this rule has to be modified in some special cases.
What is the relationship between stereoisomers $19$-$22$? This will be clearer if we translate each of the projection formulas into a three-dimensional representation, as shown in Figure 5-13. You will be helped greatly if you work through the sequence yourself with a ball-and-stick model. Drawn as Newman projections, $19$-$22$ come out as shown in $19a$-$22a$:
It should be clear (and, if it isn't, ball-and-stick models will be invaluable) that $19a$ and (20a\) are mirror images of one another and that $21a$ and $22a$ are similarly mirror images.$^5$ What about other combinations such as $19a$ and $21a$ or $20a$ and $22a$? If you look at the pairs closely you will find that they are not mirror images and are not identical. Such substances, related to each other in this way and which can be converted one into the other only by changing the configurations at one or more chiral centers, are called diastereomers.
The difference between enantiomers and diastereomers is more than just geometry. Diastereomers have substantially different chemical and physical properties, whereas enantiomers have identical physical properties (apart from their optical rotations). This is illustrated in Table 5-1 for the threonine stereoisomers. The reason for the difference in physical properties between diastereomers can be seen very simply for a substance with two chiral centers by noting that a right shoe on a right foot ($D,D$) is a mirror image, or has the same physical properties, as a left shoe on a left foot ($L,L$), but is not a mirror image, nor does it have the same physical properties, as a left show on a right foot ($L,D$), or a right shoe on a left foot ($D,L$).
Meso Compounds (Achiral Diastereomers)
All of the threonine stereoisomers $19$-$22$ are chiral substances; that is, they are not identical with their mirror images. However, it is important to recognize that not all diastereomers are chiral. To illustrate this point, we return to the tartaric acids mentioned previously in connection is Pasteur's discoveries (Section 5-1C).
Proceeding as we did for threonine, we can write four projection formulas for tartaric acid, 2,3-dihydroxybutanedioic acid, as shown by $23$-$26$:
There are two pairs of mirror images $23a$ and $24a$, as well as $25a$ and $26a$. However, what will not be so immediately clear, but what you must verify for yourself is that $25a$ and $26a$ are, in fact, identical. This means that $25a$ and $26a$ are representations of a single achiral substance, identical with its mirror image. Substances that have chiral centers but are themselves achiral are called meso compounds.
The condition that makes possible the existence of meso compounds is an appropriate degree of molecular symmetry. There are several kinds of such molecular symmetry. In the case of projection formulas $25$ (or $26$) there is a plane of symmetry, which means that a plane can be placed through the molecule such that one half of the molecule is a mirror image of the other half. The mirror plane for meso-tartaric acid can be seen easily from its projection formulas $25b$ and $26b$. These two formulas are superimposable if one is rotated $180^\text{o}$ in the plane of the paper.
The Newman representations $25a$ and $26a$ of meso-tartaric acid does not have a mirror plane. Why is it different from the Fischer projections in this respect? The reason is that the projection formulas represent a particular eclipsed conformation $27$ of meso-tartaric acid that does have a mirror plane:
Therefore, if you are confronted with a particular sawhorse or Newman formula and you have to decide whether it represents a meso compound, the best procedure is to make a ball-and-stick model of the conformation and then rotate around the bonds to see if it can be brought into a conformation (staggered or eclipsed) that has a plane of symmetry (such as $27$) or is identical with its mirror image.
As expected from our previous discussions diastereomers of tartaric acid have different physical properties (Table 5-2).
If you find yourself confused about the $D,L$ and meso forms of tartaric acid, a simple analogy may help keep matters straight. Consider three sets of shoes. A right shoe beside a left shoe is a meso combination with a plane of symmetry. A left shoe next to a left shoe is not identical with, but is
the mirror image of, a right shoe next to a right shoe. None of the three combinations are identical. Each right or left shoe corresponds to a right or left configuration of a tartaric acid carbon so the three sets correspond to meso-, $L$-, and $D$-tartaric acid, respectively.
There is another symmetry test for meso configurations that is applicable to staggered conformations and can be illustrated with the tartaric acids. If you make models of $25a$ and $26a$ you will find that they are mirror images and identical but, as we have said, they have no plane of symmetry. In this conformation, the molecules do have a center of symmetry. Thus a line drawn at any angle through the midpoint of the central $C-C$ bond of $25a$ (or $26a$) has an identical environment on each side of the midpoint. Another way of putting it is that each half of the molecule is the photographic image (i.e., reverse) of the other half. For a molecule with chiral centers, if its projection formula has a plane of symmetry or if we can find a rotational conformation with either a plane or center of symmetry, then it will be meso and achiral.
The idea that for every $n$ chiral centers there can be $2^n$ different configurations will be true only if none of the configurations has sufficient symmetry to be identical with its mirror image. For every meso form there will be one less pair of enantiomers and one less total number of possible configurations than is theoretically possible according to the number of chiral centers. At most, one meso compounds is possible for structures with four chiral centers. An example is offered by the meso forms of tetrahydroxyhexanedioic acid which, with four chiral atoms, have configurations $28$ and $29$:
$^5$The same information can be obtained from projection formulas. You can see that projections $19$ and $20$ are mirror images and that $20$, $21$, or $22$ can not be superimposed on $19$. However, in some situations confusion can result in making such comparisons and it is important to be able to translate the projection formulas into ball-and-stick models or perspective drawings.
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.06%3A_Molecules_with_More_Than_One_Chiral_Center._Diastereomers.txt |
Symmetrical reagents do not differentiate between the members of a pair of enantiomers for the same reason that an ordinary sock fits equally well on a right foot as on a left foot. However, asymmetric or chiral reagents can differentiate between enantiomers, especially by having at least some difference in reactivity toward them. A good analogy is the comparison between the ease of putting a left shoe on a left foot and a left shoe on a right foot. The difference may not be very pronounced for simple compounds with only one or two chiral centers, but generally the larger and more complex that chiral reagent becomes, the greater is its selectivity or power to discriminate between enantiomers and diastereomers as well. The property of being able to discriminate between diastereomers is called stereospecificity, and this is an especially important characteristic of biological systems.
For example, our ability to taste and smell is regulated by chiral molecules in our mouths and noses that act as receptors to "sense" foreign substances. We can anticipate, then, that enantiomers may interact differently with the receptor molecules and induce different sensations. This appears to be the case. The two enantiomers of the amino acid, leucine, for example, have different tastes - one is bitter, whereas the other is sweet. Enantiomers also can smell different, as is known from the odors of the two carvones. One has the odor of caraway and the other of spearmint.
Some animals, and especially insects, rely on what amounts to a "sense-of-smell" for communication with others of their species. Substances synthesized by a particular species, and used to send messages in this way, are called pheromones. Many of these substances have rather simple molecular structures because they must be reasonably volatile and yet they are remarkably specific in the response they induce. When stereoisomerism is possible, usually only one isomer is effective. The sex attractant of the silkworm moth Bombyx mori has been identified as trans-10-cis-12-hexadecaden-1-ol, \(30\), familiarly known as "bombykol," and that of the gypsy moth is 2-methyl-cis-7-epoxy-octadecane, \(31\), or "disparlure":
There is hope that insect sex lures can be used to disrupt the mating pattern of insects and thereby control insect population. This approach to pest control has important advantages over conventional insecticides in that the chemical lures are specific for a particular species; also they are effective in remarkably low concentrations and are relatively nontoxic. There are problems, however, not the least of which is the isolation and identification of the sex attractant that is produced by the insects only in minute quantities. Also, synergistic effects are known to operate in several insect species such that not one but several pheromones act in concert to attract the opposite sex. Two notable pests, the European corn borer and the red-banded leaf roller, both use cis-11-tetradecenyl ethanoate, \(32\), as the primary sex attractant, but the pure cis isomer is ineffective unless a small amount of trans isomer also is present. The optimum amount appears to be between \(4\%\) and \(7\%\) of the trans isomer.
We shall discuss many other examples of biological stereospecificity in later chapters.
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)/05%3A_Stereoisomerism_of_Organic_Molecules/5.07%3A_Some_Examples_of_the_Importance_of_Stereoisomerism_to_Biology._Biological_Stereospeciflcity.txt |
In previous chapters, we have shown how you can use ball-and-stick models to predict the general arrangements in space of organic molecules. The sticks correspond to chemical bonds, which we represent in structural formulas as lines, or in Lewis structures as pairs of dots denoting shared pairs of electrons. Remembering that electrons and nuclei are charged particles, and that it is electrical forces of attraction and repulsion between the electrons and nuclei that determine the bonding, perhaps we should be surprised that such simple mechanical models provide so much useful information. What we will try to do in this chapter is to show you how the modern electronic theory of chemical bonding provides strong support for the use of ball-and-stick models for many organic molecules, and also where it indicates that the models need to be modified or cannot properly represent the structural arrangements.
• 6.1: Prelude to Bonding
here are several qualitative approaches to bonding in polyatomic molecules, but we shall discuss here the most widely used and currently popular approach. This approach involves setting up appropriate atomic orbitals for the atoms and considering that each bond arises from the attractive electrical forces of two or more nuclei for a pair of electrons in overlapping atomic orbitals, with each orbital on a different atom.
• 6.2: Hydrogenlike Atomic Orbitals
With the modern concept of a hydrogen atom we do not visualize the orbital electron traversing a simple planetary orbit. Rather, we speak of an atomic orbital, in which there is only a probability of finding the electron in a particular volume a given distance and direction from the nucleus. The boundaries of such an orbital are not distinct because there always remains a finite, even if small, probability of finding the electron relatively far from the nucleus.
• 6.3: Bond Formation Using Atomic Orbitals
In writing the conventional Lewis structures for molecules, we assume that a covalent chemical bond between two atoms involves sharing a pair of elections, one from each atom. Figure 6-5 shows how atomic orbitals can be considered to be used in bond formation. Here, we postulate that a single bond is formed by the pulling together of two atomic nuclei by attractive forces exerted by the nuclei for the two paired electrons in overlapping atomic orbitals.
• 6.4: Electron Repulsion and Bond Angles. Orbital Hybridization
Page notifications Off Save as PDF Share In predicting bond angles in small molecules, we find we can do a great deal with the simple idea that unlike charges produce attractive forces while like charges produce repulsive forces. We will have electron-nuclear attractions, electron-electron repulsions, and nucleus-nucleus repulsions.
• 6.5: Atomic-Orbital Models
The construction of several atomic orbital models for differing classes of organic molecules is presented.
• 6.6: Resonance
Bonding electrons can be associated with more than two nuclei, and there is a measure of stability to be gained by this because the degree of bonding increases when the electrons can distribute themselves over a greater volume. This effect often is called electron delocalization or resonance. It is important only if the component atomic orbitals overlap significantly, and this will depend in large part on the molecular geometry.
• 6.7: Advanced Quantum Theory of Organic Molecules
In recent years, great progress has been made in quantum-mechanical calculations of the properties of small organic molecules by so-called ab initio methods, which means calculations from basic physical theory using only fundamental constants, without calibration from known molecular constants. Calculations that are calibrated by one or more known properties and then used to compute other properties are called "semiempirical" calculations.
• 6.E: Bonding in Organic Molecules (Exercises)
These are the homework exercises to accompany Chapter 6 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."
06: Bonding in Organic Molecules
In previous chapters, we have shown how you can use ball-and-stick models to predict the general arrangements in space of organic molecules. The sticks correspond to chemical bonds, which we represent in structural formulas as lines, or in Lewis structures as pairs of dots denoting shared pairs of electrons. Remembering that electrons and nuclei are charged particles, and that it is electrical forces of attraction and repulsion between the electrons and nuclei that determine the bonding, perhaps we should be surprised that such simple mechanical models provide so much useful information. What we will try to do in this chapter is to show you how the modern electronic theory of chemical bonding provides strong support for the use of ball-and-stick models for many organic molecules, and also where it indicates that the models need to be modified or cannot properly represent the structural arrangements.
There are several qualitative approaches to bonding in polyatomic molecules, but we shall discuss here the most widely used and currently popular approach. This approach involves setting up appropriate atomic orbitals for the atoms and considering that each bond arises from the attractive electrical forces of two or more nuclei for a pair of electrons in overlapping atomic orbitals, with each orbital on a different atom. The geometry of the bonds is assumed to be determined by the geometry of the orbitals and by the repulsive forces between the electrons. In the course of showing how this approach can be applied, we shall discuss ways of formulating bonding and geometries for several important kinds of organic compounds. Finally, we will show you some of the results currently being obtained by sophisticated quantum-mechanical calculations, which provide strong support for our qualitative formulations.
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)/06%3A_Bonding_in_Organic_Molecules/6.01%3A_Prelude_to_Bonding.txt |
With the modern concept of a hydrogen atom we do not visualize the orbital electron traversing a simple planetary orbit. Rather, we speak of an atomic orbital, in which there is only a probability of finding the electron in a particular volume a given distance and direction from the nucleus. The boundaries of such an orbital are not distinct because there always remains a finite, even if small, probability of finding the electron relatively far from the nucleus.
There are several discrete atomic orbitals available to the electron of a hydrogen atom. These orbitals differ in energy, size, and shape, and exact mathematical descriptions for each are possible. Following is a qualitative description of the nature of some of the hydrogen atomic orbitals.
The most stable or ground state of a hydrogen atom is designated $1s$.$^1$ In the $1s$ state the electron is, on the average, closest to the nucleus (i.e., it is the state with the smallest atomic orbital). The $1s$ orbital is spherically symmetrical. This means that the probability of finding the electron at a given distance $r$ from the nucleus is independent of the direction from the nucleus. We shall represent the $1s$ orbital as a sphere centered on the nucleus with a radius such that the probability of finding the electron within the boundary surface is high (0.80 to 0.95); see Figure 6-1. This may seem arbitrary, but an orbital representation that would have a probability of 1 for finding the electron within the boundary surface would have an infinite radius. The reason is that there is a finite, even if small, probability of finding the electron at any given distance from the nucleus. The boundary surfaces we choose turn out to have sizes consistent with the distances between the nuclei of bonded atoms.
The $2s$ orbital is very much like the $1s$ orbital except that it is larger and therefore more diffuse, and it has a higher energy. For principal quantum number 2, there are also three orbitals of equal energies called $2p$ orbitals, which have different geometry than the $s$ orbitals. These are shown in Figure
6-2, in which we see that the respective axes passing through the tangent spheres of the three $p$ orbitals lie at right angles to one another. The $p$ orbitals are not spherically symmetrical.
The $3s$ and $3p$ states are similar to the $2s$ and $2p$ states but are of higher energy. The $3d$, $4d$, $4f$, $\cdots$, orbitals have still higher energies and quite different geometries; they are not important for bonding in most organic substances, at least for carbon compounds with hydrogen and elements in the first main row ($Li$-$Ne$) of the periodic table. The sequence of orbital energies is shown in Figure 6-3.
The famous Pauli exclusion principle states that no more than two elections can occupy a given orbital and then only if they differ with respect to a property of electrons called electron spin. An electron can have only one of two possible orientations of electron spin, as may be symbolized by $\uparrow$ and $\downarrow$. Two electrons with "paired" spins often are represented as $\uparrow \downarrow$. Such a pair of electrons can occupy a single orbital. The symbols $\uparrow \uparrow$ (or $\downarrow \downarrow$) represent two unpaired electrons, which may not go into a single orbital.
If we assume that all atomic nuclei have orbitals like those of the hydrogen atom, we can see how atoms more complex than hydrogen can be built up by adding electrons to the orbitals in accord with the Pauli exclusion principle. The lowest-energy states will be those in which the electrons are added to the lowest-energy orbitals. For example, the electronic configuration of the lowest-energy state of a carbon atom is shown in Figure 6-4, which also shows the relative energies of the $1s$ through $4p$ atomic orbitals. The orbitals of lowest energy are populated with the proper number of electrons to balance the nuclear charge of $+6$ for carbon and to preserve the Pauli condition of no more than two paired electrons per orbital. However, the two highest-energy electrons are put into different $2p$ orbitals with unpaired spins in accordance with Hund's rule. The rationale of Hund's rule depends on the fact that electrons come closer together. Now, suppose there are two electrons that can go into two different orbitals of the same energy (so-called degenerate orbitals). Hund's rule tells us that the repulsion energy between these electrons will be less if they have unpaired spins ($\uparrow \uparrow$). Why is this so? Because if they have unpaired
spins they cannot be in the same orbital at the same time. Therefore they will not be able to approach each other as closely as they would if they could be in the same orbital at the same time. For this reason the electronic configuration
is expected to be more stable than the configuration
if both orbitals have the same energy.
States such as the one shown in Figure 6-4 for carbon are built up through the following steps. Helium has two paired electrons in the $1s$ orbital; its configuration can be written as $\left( 1s \right)^2$, the superscript outside the parentheses denoting two paired electrons in the $1s$ orbital. For lithium, we expect $Li \: \left( 1s \right)^2 \left( 2s \right)^1$ to be the ground state, in which the $1s$ electrons must be paired according to the exclusion principle. Continuing in this way, we can derive the electronic configurations for the elements in the first three rows of the periodic table, as shown in Table 6-1. These configurations conform to the
$^1$The index number refers to the principal quantum number and corresponds to the "$K$ shell" designation often used for the electron of the normal hydrogen atom. The principal quantum number 2 corresponds to the $L$ shell, 2 to the $M$ shell, and so on. The notation $s$ (also $p$, $d$, $f$ to come later) has been carried over from the early days of atomic spectroscopy and was derived from descriptions of spectroscopic lines as "sharp", "principal", "diffuse", and "fundamental," which once were used to identify transitions from particular atomic states.
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)/06%3A_Bonding_in_Organic_Molecules/6.02%3A_Hydrogenlike_Atomic_Orbitals.txt |
In writing the conventional Lewis structures for molecules, we assume that a covalent chemical bond between two atoms involves sharing a pair of elections, one from each atom. Figure 6-5 shows how atomic orbitals can be considered to be used in bond formation. Here, we postulate that a single bond is formed by the pulling together of two atomic nuclei by attractive forces exerted by the nuclei for the two paired electrons in overlapping atomic orbitals.
Because two atomic orbitals can hold a maximum of four electrons, it is reasonable to ask why it is that two rather than one, three, or four electrons normally are involved in a bond. The answer is that two overlapping atomic
orbitals can be considered to combine to give one low-energy bonding molecular orbital and one high-energy antibonding molecular orbital (see the top part of Figure 6-6(a)).$^2$ Orbitals that overlap as shown in Figure 6-6(a) are said to overlap in the sigma manner,$^3$ and the bonding orbital is called a sigma orbital ($\sigma$); the antibonding orbital is called a $\sigma^*$ orbital (read "sigma star"). Two paired electrons suffice to fill the $\sigma$ orbital. Any additional electrons must go into the high-energy $\sigma^*$ orbital and contribute not to bonding but to repulsion between the atoms.
The hydrogen molecule-ion, $H_2^\oplus$, can be regarded as having one electron in a $\sigma$ orbital. It has been studied in the vapor state by spectroscopic means and found to have a dissociation energy to $H^\oplus$ and $H \cdot$ of $61 \: \text{kcal mol}^{-1}$ compared to the $104.2 \: \text{kcal mol}^{-1}$ bond energy for $H_2$. Several possible combinations of two hydrogen orbitals and from one to four electrons are shown in Figure 6-6(b).
$^2$More about the difference between bonding and antibonding orbitals is given in Section 21-2. For now we will say that the property of orbitals that leads to bonding or antibonding is a property analogous to phase. An in-phase combination of two orbitals is bonding, and an out-of-phase combination is antibonding.
$^3$The designation sigma ($\sigma$) denotes that orbital overlap and electron density are greatest along the internuclear axis.
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."
6.04: Electron Repulsion and Bond Angles. Orbital Hybridization
In predicting bond angles in small molecules, we find we can do a great deal with the simple idea that unlike charges produce attractive forces while like charges produce repulsive forces. We will have electron-nuclear attractions, electron-electron repulsions, and nucleus-nucleus repulsions. Let us first consider the case of a molecule with just two electron-pair bonds, as might be expected to be formed by combination of beryllium and hydrogen to give beryllium hydride, $H:Be:H$. The problem will be how to formulate the bonds and how to predict what the $H-Be-H$ angle, $\theta$, will be:
We might formulate a second $\sigma$ bond involving the $2p$ orbital, but a new problem arises as to where the hydrogen should be located relative to the beryllium orbital. Is it as in $2$, $3$, or some other way?
The $Be$ and $H$ nuclei will be farther apart in $2$ than they will be in $3$ or any other similar arrangement, so there will be less internuclear repulsion with $2$. We therefore expect the hydrogen to locate along a line going through the greatest extension of the $2p$ orbital.
According to this simple picture, beryllium hydride should have two different types of $H-Be$ bonds - one as in $1$ and the other as in $2$. This is intuitively unreasonable for such a simple compound. Furthermore, the $H-Be-H$ bond angle is unspecified by this picture because the $2s$ $Be$ orbital is spherically symmetrical and could form bonds equally well in any direction.
However, if we forget about the orbitals and only consider the possible repulsions between the electron pairs, and between the hydrogen nuclei, we can see that these repulsions will be minimized when the $H-Be-H$ bond angle is $180^\text{o}$. Thus arrangement $5$ should be more favorable than $4$, with a $H-Be-H$ angle less than $180^\text{o}$:
Unfortunately, we cannot check this particular bond angle by experiment because $BeH_2$ is unstable and reacts with itself to give a high-molecular-weight solid. However, a number of other compounds, such as $\left( CH_3 \right)_2 Be$, $BeCl_2$, $\left( CH_3 \right)_2 Hg$, $HgF_2$, and $\left( CH_3 \right)_2 Zn$, are known to have $\sigma$ bonds involving $\left( s \right)^1 \left( p \right)^1$ valence states. Measurements of the bond angles at the metal of these substances in the vapor state has shown them to be uniformly $180^\text{o}$.
How are the $s$ and $p$ orbitals deployed in this kind of bonding? It turns out that stronger bonds are formed when the degree of overlap of the orbitals is high. The degree of overlap will depend on the sizes of the orbital and, particularly, on how far out they extend from the nucleus. Figure 6-7 shows how far $2s$ and $2p$ orbitals extend relative to one another. Bonding with these orbitals as in $1$ and $2$ does not utilize the overlapping power of the orbitals to the fullest extent. With $1$ we have overlap that uses only part of the $2s$ orbital, and with $2$, only a part of the $2p$ orbital. Molecules such as $BeH_2$ can be formulated with better overlap and equivalent bonds with the aid of the concept of orbital hybridization. This concept, published independently by L. Pauling and J. C. Slater in 1931, involves determining which (if any) combinations of $s$ and $p$ orbitals may overlap better and make more effective bonds than do the individual $s$ and $p$ orbitals. The mathematical procedure for orbital hybridization predicts that an $s$ and a $p$ orbital of one atom can form two stronger covalent bonds if they combine to form two new orbitals called $sp$-hybridized orbitals (Figure 6-8). Each $sp$-hybrid orbital has an overlapping power of 1.93, compared to the pure $s$ orbital taken as unity and a pure $p$ orbital as 1.73. Bond angles of $180^\text{o}$ are expected for bonds to an atom using $sp$-hybrid orbitals and, of course, this also is the angle we expect on the basis of our consideration of minimum electron-pair and internuclear repulsions. Henceforth, we will proceed on the basis that molecules of the type $X:M:X$ may form $sp$-hybrid bonds.
On the basis of repulsion between electron pairs and between nuclei, molecules such as $BH_3$, $B \left( CH_3 \right)_3$, $BF_3$, and $AlCl_3$, in which the central atom forms three covalent bonds using the valence-state electronic configuration
$\left( s \right)^1 \left( p_x \right)^1 \left( p_y \right)^1$, are expected to be planar with bond angles of $120^\text{o}$. For example,
With atoms such as carbon and silicon, the valence-state electronic configuration to form four covalent bonds has to be $\left( s \right)^1 \left( p_x \right)^1 \left( p_y \right)^1 \left( p_z \right)^1$. Repulsion between the electron pairs and between the attached nuclei will be minimized by formation of a tetrahedral arrangement of the bonds. the same geometry is predicted from hybridization one one $s$ and three $p$ orbitals, which gives four $sp^3$-hybrid orbitals directed at angles of $109.5^\text{o}$ to each other. The predicted relative overlapping power of $sp^3$-hybrid orbitals is 2.00 (Figure 6-10).
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)/06%3A_Bonding_in_Organic_Molecules/6.03%3A_Bond_Formation_Using_Atomic_Orbitals.txt |
Alkanes
Saturated compounds such as the alkanes and their derivatives, which have normal tetrahedral angles for the bonds to carbon, can be formulated readily in terms of atomic orbitals with $sp^3 \: \sigma$ bonds to carbon. An example is shown in Figure 6-11, which also shows how an atomic-orbital model can be drawn in abbreviated style. The lines in this drawing correspond to bonds and are labeled as $sp^3$ with $sp^3$ (the overlapping orbitals of the $C-C$ bond) or as $sp^3$ with $s$ (the overlapping orbitals of the $C-H$ bonds).
Atoms with Unshared Electron Pairs
Many important molecules such as ammonia, water, and hydrogen fluoride have atoms with unshared pairs of electrons:
If we formulate each of these molecules in such a way to minimize repulsions between like charges, a basically tetrahedral arrangement will be expected because this will place the nuclei (and electron pairs) as widely separated as possible.
The water molecule could be formulated this way, as in $6$, with the oxygen at the center of the tetrahedron:
The simple picture predicts that the $H-O-H$ bond angle should be tetrahedral, $109.5^\text{o}$. But actually it is $104.5^\text{o}$.
There are two schools of thought as to why the angle is $104.5^\text{o}$. One idea is that the repulsion model is too simple and has to be modified to take into account that the repulsion is more severe between pairs of unshared electrons than between electrons in bonding orbitals on the same atom. This is because when a bond is formed between two nuclei, the attraction of the nuclei for the electrons shrinks the orbitals available to the bonding electrons, thereby reducing their electrostatic repulsion with other pairs. The degree of repulsion between electron pairs diminishes in the sequence: unshared pairs vs. unshared pairs $>$ unshared pairs vs. bonding pairs $>$ bonding pairs vs. bonding pairs. From this, we expect that in water the $H-O-H$ angle will be less than tetrahedral, because the larger repulsion between the two unshared pairs will tend to push the bonding pairs closer together.
A similar, but smaller, effect is expected for ammonia because now the repulsion is only between the one unshared pair and bonding pairs. The ammonia $H-N-H$ angle is $107.3^\text{o}$, which is only slightly smaller than the tetrahedral value of $109.5^\text{o}$.
The alternative point of view of why the bond angle of water is $104.5^\text{o}$ starts with the premise that, in the simplest approximation, the angle should be $90^\text{o}$! To see how this comes about let us compare $H:Be:H$ with $H:\underset{\cdot \cdot}{\ddot{O}}:H$. You will recall that to form two bonds to $Be$, we had to promote an electron and change the electronic configuration to the valence configuration, $\left( 2s \right)^1 \left( 2p \right)^1$. The situation with $H_2O$ is different in that the oxygen ground state and valence state are the same, $\left( 2s \right)^2 \left( 2p_x \right)^1 \left( 2p_y \right)^1 \left( 2p_z \right)^1$. This means we could form two equivalent bonds to oxygen using the $2p_y$ and $2p_z$ orbitals at an angle of $90^\text{o}$ (Figure 6-12).
Now, to explain why the $H-O-H$ bond angles are $104.5^\text{o}$ instead of $90^\text{o}$, we can say that the repulsion between the hydrogen nuclei is expected to widen the bond angle. An argument in favor of this formulation is provided by the bond angle in $H_2S$, which is $92.2^\text{o}$. This is much closer to the $90^\text{o}$ expected for $p$-bond orbitals and the hydrogens in $H_2S$ would not be expected to repel each other as much as in $H_2O$ because sulfur is a larger atom than oxygen.
Both ways of formulating the orbitals used in the bonding of water molecules are in current use. Arguments can be advanced in favor of both. Highly sophisticated quantum-mechanical calculations, which we will say more about later, suggest that oxygen in water molecules uses orbitals that are $18\% \: s$ and $82\% \: p$ in its bonds ($sp^{4.5}$), and furthermore, that the unshared pairs are in equivalent hybrid orbitals [not one pair as $\left( 2s \right)^2$ and the other as $\left( 2p \right)^2$]. Each of the unshared electron-pair orbitals of oxygen in water is calculated to be about $40\% \: s$ and $60\% \: p$ ($sp^{1.5}$).
The results are hardly clearcut, but the bonding orbitals are considerably closer to $sp^3$ ($25\% \: s$ and $75\% \: p$) than they are to $100\% \: p$. We recommend that the bonding orbitals of nitrogen and oxygen be considered to be $sp^3$ and the unshared pairs designated simply as $\left( n \right)^2$. An abbreviated atomic orbital model of methanol, $CH_3OH$, made on this basis is shown in Figure 6-13.
Compounds with Double Bonds
Recall from Chapter 2 that bond angles in compounds with carbon-carbon double bonds such as ethene are closer to $120^\text{o}$ than to the normal tetrahedral value of $109.5^\text{o}$. There are several way sin which a carbon-carbon double bond can be formulated in terms of atomic-orbital models. One very popular approach is to consider that ethene has two $sp^2$-hybridized carbons that form one carbon-carbon $\sigma$ bond and four carbon-hydrogen $\sigma$ bonds by overlap of the six $sp^2$ orbitals, as shown in Figure 6-14. The remaining carbon-carbon bond is formulated as arising from sidewise overlap of the two $p$ orbitals, one on each carbon, that are not utilized in making the $sp^2$ hybrids. Sidewise overlap of $p$ orbitals is called $\pi$ overlap to distinguish it from the endwise $\sigma$ overlap of the type we have discussed previously (Figure 6-15). The resulting $\pi$ bond differs from the $\sigma$ bond in that electron density is concentrated in the regions above and below the bond axis rather than along the bond axis.
Formulations of ethene in this way suggests that it should be a planar molecule with $H-C-H$ angles of $120^\text{o}$. Ethene is indeed planar, but its $H-C-H$ angles are found to be $117^\text{o}$, rather than the $120^\text{o}$ predicted for $sp^2$ bonds. An explanation of this discrepancy using further electron-repulsion arguments will be discussed later in the chapter.
The simple elegance of the $\sigma$-$\pi$ model of ethene should not be taken as proving that there actually are two different kinds of bonds between the carbons. The $\sigma$-$\pi$ representation of double bonds is not really unique. Given $sp^2$ hybridization of the carbons so there are $sp^2$-$\sigma$ bonds to the hydrogens, it is possible to take the $sp^2$ and $p$ orbitals used for the $\sigma$ and $\pi$ bonds, rehybridize them, and so derive a new set of overlapping orbitals for the double bond. These orbitals are called $\tau$ (tau) bonding orbitals and can be represented by two banana-shaped orbitals between the carbons (Figure 6-16). The result is two completely equivalent $C-C$ bonds. The $\tau$ model has the advantage of offering a striking parallel to ball-and-stick models, whereas the $\sigma$-$\pi$ model is of particular value as a basis for quantitative calculations, as will be discussed in Chapter 21.
Using the $\sigma$-$\pi$ model of double bonds, we conclude that the twisted configuration shown in Figure 6-17 should not be very stable. Here the $p$ orbitals are not in position to overlap effectively in the $\pi$ manner. The favored configuration is expected to have the axes of the $p$-$\pi$ orbitals parallel. Because considerable energy would have to be expended to break the $p$-$\pi$ double bond and to permit rotation about the remaining $sp^2$-$\sigma$ bond, restricted rotation and stable cis-trans isomers are expected. Similar conclusions can be reached on the basis of the $\tau$ model of the double bond.
Compounds with Triple Bonds
Ethyne, $C_2H_2$, is an organic compound that usually is formulated with $sp$ hybrid bonds. The carbon-hydrogen framework is built up through $\sigma$ overlap of two $sp$-hybrid orbitals, one from each carbon atom, to form a $C-C$ bond, and $\sigma$ overlap of the remaining $sp$ orbitals with the $s$ orbital of two hydrogens to form $C-H$ bonds. The remaining two carbon-carbon bonds result through sidewise $\pi$ overlap of the pure $p$ orbitals, as shown in Figure 6-18. This model fits well with the properties of the ethyne molecule being linear (bond angles of $180^\text{o}$. Also, the $C-H$ bonds in ethyne are different from those in ethene or ethane, as judged by their $C-H$ stretching and bending frequencies in the
infrared (Chapter 9), their bond energies, (Table 4-6), and their acidities (Section 11-8). These differences in properties are in keeping with the different states of hybridization of the carbon orbitals that we have postulated for ethane, ethene, and ethyne.
More on Hybrid Bond Orbitals and Molecular Geometry
A summary of the directional character of the $s$-$p$ hybrid atomic orbitals discussed so far is given in Table 6-2. By referring to this table, it usually is possible to deduce the nature of the bonding orbitals for most organic compounds from the molecular geometry, if this is known, Thus a tetrahedral molecule $AX_4$ with four attached ligands uses $sp^3$ hybrid orbitals localized on atom $A$; a planar triangular molecule $AX_3$ with three attached ligands at angles of $120^\text{o}$ is $sp^2$ hybridized at atom $A$; a linear molecule $AX_2$ with two ligands is $sp$ hybridized at $A$.
Applying the converse of these rules, one should be able to predict molecular geometry by making reasonable assumptions as to the state of hybridization for each atom in the molecule. Obviously in doing this we have to take account of unshared electron pairs. Prediction is easy if unshared pairs are absent. Thus, four attached ligands, as in $CH_4$, $CCl_4$, or $BF_4^\ominus$, imply $sp^3$ hybridization at the central atom and therefore a tetrahedral arrangement of ligands. Three ligands, as bonded to carbon in $CH_3^\oplus$ or to boron in $BF_3$, imply $sp^2$ hybridization for the central atom and a planar triangular arrangement if ligands. Two ligands, as in $CO_2$, imply $sp$ hybridization and linear geometry.
In many of our later discussions of organic reactions, we will be concerned with cationic, radical, and anionic carbon species that are substitution products of $CH_3^\oplus$, $CH_3 \cdot$, and $CH_3:^\ominus$. Because of the importance of these entities, you should know how to formulate them and related substances, such as $^\ominus: \ddot{N}H_2$, with atomic orbitals. Perhaps the most straightforward way is to start from $CH_4$ and see what changes in the $C-H$ bonds we would expect as the result of the hypothetical processes: $CH_4 \rightarrow CH_3:^\ominus + H^\oplus$, $CH_4 \rightarrow CH_3^\oplus + H:^\ominus$, and $CH_4 \rightarrow CH_3 \cdot + H \cdot$.
Methane is tetrahedral with $sp^3$ carbon bonding orbitals. Removal of $H^\oplus$ gives $CH_3:^\ominus$, which corresponds in electronic structure to $H_3N:$ and, for the same reasons, should have a pyramidal shape with nearly tetrahedral $H-C-H$ angles. Removal of $H:^\ominus$ from $CH_4$ to give $CH_3^\oplus$ with six bonding electrons, suggests a change to $sp^2$ bonding orbitals for the carbon and planar geometry with $H-C-H$ angles of $120^\text{o}$.
The radical, $CH_3 \cdot$ presents a special problem. We can think of it as being formed by the loss of $H \cdot$ from $CH_4$, by adding an electron to planar $CH_3^\oplus$, or by removing an electron from pyramidal $CH_3:^\ominus$. We can formulate $CH_3 \cdot$ with $Sp^2$ orbitals for the $C-H$ bonds and the extra electron in a $p$ orbital, or with $sp^3$ orbitals for the $C-H$ bonds and the extra electron in an $sp^3$ orbital:
The actual structure of $CH_3 \cdot$ has the hydrogens and carbons in a plane (left). Therefore it appears that the repulsions between the bonding electron pairs is greater than the repulsions between the extra electron and the bonding pairs. The actual structure corresponds to the one in which the bonding pairs are as far apart as possible.
More on Interelectronic Repulsion and Bond Angles
Molecules of the type $AX_4$, which have four identical ligands on the central atom and no unshared electrons on $A$ (e.g., $CH_4$ and $CCl_4$), are expected to be, and are, tetrahedral. By the same reasoning, three electron pairs around one atom should seek a planar arrangement with $120^\text{o}$ angles to minimize electron repulsion; accordingly, species of the type $AX_3$, which have no unshared pairs on $A$ (e.g., $BF_3$ and $CH_3^\oplus$), have this geometry. With only two electron pairs, the preferred arrangement is linear.
The bond angles of compounds with multiple bonds can be explained similarly. For example, in ethene the four electrons of the double bond occupy the region in space between the two carbon nuclei. The situation at either carbon is rather like the $AX_3$ case, except that one of the ligands now has a double complement of bonding electrons:
Therefore the carbon orbitals are expected to be directed in one plane to give bond angles that deviate somewhat from $120^\text{o}$ because of the high density of electrons in the multiple bond. Thus the $H-C-H$ angle shrinks to $117^\text{o}$, whereas the $H-C=C$ angles open up to $122^\text{o}$, because repulsion between electrons in the $H-C=C$ bonds is greater than between electrons in the $H-C-H$ bonds.
Electron-attracting power (or electronegativity) of the ligands also is important in determining bond angles. Thus for compounds of the type $CH_3X$, in which $X$ is a more electron-attracting group than carbon, the $C-X$ bond is polarized in the sense $H_3 \overset{\delta \oplus}{C} - - - \overset{\delta \ominus}{X}$, and the carbon then should have some of the character of $CH_3^\oplus$. Thus the $H-C-H$ angles are expected to be greater than $109.5^\text{o}$, as in fact they are. In chloromethane, for example, the $H-C-H$ angle is $111^\text{o}$.
Also, we can explain on the basis of electron repulsions why the bond angle in phosphine, $:PH_3$ ($93^\text{o}$), is less than that in ammonia, $:NH_3$ ($107.3^\text{o}$), and the bond angle in $H: \underset{\cdot \cdot}{\ddot{S}} :H$ ($92.2^\text{o}$) is less than that in $H: \underset{\cdot \cdot}{\ddot{O}}$ ($104.5^\text{o}$). The important point is that phosphorus and sulfur are larger atoms than nitrogen and oxygen. This means than the $H-S-H$ and $H-P-H$ bond angles can be about $90^\text{o}$ without bringing the hydrogens and the bonding pairs as close together as they are in $H_2O$ and $NH_3$ where the bond angles are near to the tetrahedral value.
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)/06%3A_Bonding_in_Organic_Molecules/6.05%3A_Atomic-Orbital_Models.txt |
An Atomic-Orbital Model of Benzene
Until now, we have discussed bonding only in terms of electron pair associated with two nuclei. These we may call localized electrons. In fact, bonding electrons can be associated with more than two nuclei, and there is a measure of stability to be gained by this because the degree of bonding increases when the electrons can distribute themselves over a greater volume. This effect often is called electron delocalization or resonance. It is important only if the component atomic orbitals overlap significantly, and this will depend in large part on the molecular geometry.
The classic example of resonance is provided by the $\pi$ bonding of benzene. This compound was shown in Chapter 1 to have the molecular formula $C_6H_6$, to be planar, and hexagonal with bond angles of $120^\text{o}$, and to possess six equivalent $C-C$ bonds and six equivalent $C-H$ bonds. Benzene usually is written with a structural formula proposed by Kekulé:
That benzene is more stable than a single Kekulé, or 1,3,5-cyclohexatriene, structure can be gauged by comparing the experimental heat of combustion
of benzene with the calculated value based on the average bond energies of Table 4-3:
$\ce{C6H6(g) + 15/2 O2 -> 6CO2(g) + 3H2O(g)}$
with
\begin{aligned} &\Delta H_{\exp }^{0}=-789 \mathrm{kcal} \ &\Delta H_{\text {calc }}^{0}=-827 \mathrm{kcal} \end{aligned}
About $38 \: \text{kcal}$ less energy is released on combustion than calculated. Benzene, therefore, is $38 \: \text{kcal mol}^{-1}$ more stable than the cyclohexatriene structure predicts.
Representation of Resonance
Atomic-orbital models, like that shown for benzene, are useful descriptions of bonding from which to evaluate the potential for electron delocalization. But they are cumbersome to draw routinely. We need a simpler representation of electron delocalization.
The method that commonly is used is to draw a set of structures, each of which represents a reasonable way in which the electrons (usually in $p$ orbitals) could be paired. If more than one such structure can be written, the actual molecule, ion, or radical will have properties corresponding to some hybrid of these structures. A double-headed arrow $\leftrightarrow$ is written between the structures that we consider to contribute to the hybrid. For example, the two Kekulé forms are two possible electron-pairing schemes or valence-bond structures that could contribute to the resonance hybrid of benzene:
It is very important to know what attributes a reasonable set of valence-bond structures has to have to contribute to a hybrid structure. It is equally important to understand what is and what is not implied in writing a set of structures. Therefore we shall emphasize the main points to remember in the rest of this section.
1. The members of a set of structures, as the two Kekulé structures for benzene, have no individual reality. They are hypothetical structures representing different electron-pairing schemes. We are not to think of benzene as a 50:50 mixture of equilibrating Kekulé forms.
2. To be reasonable, all structures in a set representing a resonance hybrid must have exactly the same locations of the atoms in space. For example, formula $7$ does not represent a valid member of the set of valence-bond structures of benzene, because the atoms of $7$ have different positions from those of benzene (e.g., $7$ is not planar):
Structure $7$ actually represents a known $C_6H_6$ isomer that has a very different chemistry from that of benzene.
3. All members of the set must have the same number of paired or unpaired electrons. For the normal state of benzene, the six $\pi$ electrons have three of one spin and three of the other. Structures such as $8$, with four electrons of one spin and two of the other, are not valid contributors to the ground state of benzene:
4. The importance of resonance in any given case will depend on the energies of the contributing structures. The lower and more nearly equivalent the members of the set are in energy, the more important resonance becomes. That is to say, electron stabilization is greatest when there are two or more structures of lowest energy (as for the two Kekulé structures of benzene). As a corollary, the structure of a molecule is least likely to be satisfactorily represented by a conventional structural formula when two (or more) energetically equivalent, low-energy structures may be written.
5. If there is only one low-energy structure in the set then, to a first approximation, the resonance hybrid may be assigned properties like those expected for that structure. As an example, we show three possible pairing schemes for ethene, $9$, $10$, and $11$:
Although $10$ and $11$ are equivalent, they are much higher in energy than $9$ (see discussion in Section 4-4C). Therefore they do not contribute substantially to the structure of ethene that is best represented by $9$.
Resonance is by no means restricted to organic molecules. The following sets of valence-bond structures represent the hybrid structures of nitrate ion, $NO_3^\ominus$, carbonate ion $CO_3^{2 \ominus}$, and nitrous oxide, $N_2O$. These are only representative examples. We suggest that you check these structures carefully to verify that each member of a set conforms to the general rules for resonance summarized above.
A shorthand notation of hybrid structures frequently is used in which the delocalized $\pi$-bonding is shown as a broken line. For benzene, an inscribed circle also is used to indicate continuous $\pi$ bonding:
Resonance and Reactivity
Electron delocalization is an important factor in the reactivity (or lack of it) of organic molecules. As an example, recall from Chapter 4 that the bond energies of various types of $C-H$ bonds differ considerably (see Table 4-6). In particular, the methyl $C-H$ bond in propene is about $9 \: \text{kcal}$ weaker than the methyl $C-H$ bond of ethane or propane, and this difference can be explained by the use of the resonance concept. The following bond dissociations are involved:
delocalization is possible for the propyl radical, propane, or propene. Accordingly, the methyl $C-H$ bond strength in propene is less than in propane because of stabilization of the 2-propenyl radical.
The foregoing discussion adds further to our understanding of the selectivity observed in the halogenation reactions discussed in Chapter 4. When propene is chlorinated in sunlight, the product is 3-chloropropene, and we may explain this on the basis that the radical-chain reaction involves propagation steps in which a chlorine atom attacks the hydrogen corresponding to the weakest $C-H$ bond:
The resonance theory is very useful in accounting for, and in many cases predicting, the behavior of substances with $\pi$ bonds. However, it is not omnipotent. One example where it fails is cyclobutadiene, for which we can write two equivalent valence-bond structures corresponding to the Kekulé structures for benzene:
Despite this, cyclobutadiene is an extremely unstable substance, reacting with itself almost instantly at temperatures above $-250^\text{o}$. For better understanding of this and some related problems, we provide a more detailed discussion of electron delocalization in Chapter 21.
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)/06%3A_Bonding_in_Organic_Molecules/6.06%3A_Resonance.txt |
In recent years, great progress has been made in quantum-mechanical calculations of the properties of small organic molecules by so-called ab initio methods, which means calculations from basic physical theory using only fundamental constants, without calibration from known molecular constants. Calculations that are calibrated by one or more known properties and then used to compute other properties are called "semiempirical" calculations.
It should be made clear that there is no single, unique ab initio method. Rather, there is a multitude of approaches, all directed toward obtaining useful approximations to mathematical problems for which no solution in closed form is known or foreseeable. The calculations are formidable, because account must be taken of several factors: the attractive forces between the electrons and the nuclei, the interelectronic repulsions between individual electrons, the internuclear repulsions, and the electron spins.
The success of any given ab initio method usually is judged by how well it reproduces known molecular properties with considerable premium for use of tolerable amounts of computer time. Unfortunately, many ab initio calculations do not start from a readily visualized physical model and hence give numbers that, although agreeing well with experiment, cannot be used to enhance one's qualitative understanding of chemical bonding. To be sure, this should not be regarded as a necessary condition for making calculations. But it also must be recognized that the whole qualitative orbital and hybridization approach to chemical bonding presented in this chapter was evolved from mathematical models used as starting points for early ab initio and semiempirical calculations.
The success of any given ab initio method is judged by how well it reproduces known molecular properties.
The efforts of many chemical theorists now are being directed to making calculations that could lead to useful new qualitative concepts of bonding capable of increasing our ability to predict the properties of complex molecules. One very successful ab initio procedure, called the "generalized valence-bond" (GVB) method, avoids specific hybridization assignments for the orbitals and calculates an optimum set of orbitals to give the most stable possible electronic configuration for the specified positions of the atomic nuclei. Each chemical bond in the GVB method involves two electrons with paired spins in two more or less localized atomic orbitals, one on each atom. Thus the bonds correspond rather closely to the qualitative formulations used previously in this chapter, for example Figure 6-14.
The "generalized valence-bond" (GVB) method, avoids specific hybridization assignments for the orbitals and calculates an optimum set of orbitals to give the most stable possible electronic configuration for the specified positions of the atomic nuclei.
Electron-amplitude contour diagrams of the GVB orbitals for ethene are shown in Figure 6-22. Let us be clear about what these contour lines represent. They are lines of equal electron amplitude analogous to topological maps for which contour lines are equal-altitude lines. The electron amplitudes shown are those calculated in the planes containing the nuclei whose positions are shown with crosses. In general, the amplitudes decrease with distance from the nucleus. The regions of equal-electron amplitude for $s$-like orbitals (middle-right of Figure 6-22) surround the nuclei as a set of concentric shells corresponding to the surfaces of the layers of an onion (Figure 6-23). With the $sp^2$-like orbitals, the amplitude is zero at the nucleus of the atom to which the orbital belongs.
The physical significance of electron amplitude is that its square corresponds to the electron density, a matter that we will discuss further in Chapter 21. The amplitude can be either positive or negative, but its square (the electron density) is positive, and this is the physical property that can be measured by appropriate experiments.
Looking down on ethene, we see at the top of Figure 6-22 two identical $C-C$ $\sigma$-bonding orbitals, one on each carbon, directed toward each other. The long dashed lines divide the space around the atom into regions of opposite orbital phase (solid is positive and dotted is negative). The contours for one of the $C-H$ bonding orbitals are in the middle of the figure, and you will see that the orbital centered on the hydrogen is very much like an $s$ orbital, while the one on the carbon is a hybrid orbital with considerable $p$ character. There are three other similar sets of orbitals for the other ethene $C-H$ bonds.
When we look at the molecule edgewise, perpendicular to the $C-C$ $\sigma$ bond, we see the contours of the individual, essentially $p$-type, orbitals for $\pi$ bonding. Ethyne shows two sets of these orbitals, as expected.
What is the difference between the GVB orbitals and the ordinary hybrid orbitals we have discussed previously in this chapter? Consider the $sp^2$-like orbitals (upper part of Figure 6-22) and the $sp^2$ hybrids shown in Figure 6-9. The important point is that the $sp^2$ hybrid in Figure 6-9 is an atomic orbital calculated for a single electron on a single atom alone in space. The GVB orbital is much more physically realistic, because it is an orbital derived for a molecule with all of the nuclei and other electrons present. Nonetheless, the general shape of the GVB $sp^2$-like orbitals will be seen to correspond rather closely to the simple $sp^2$ orbital in Figure 6-9. This should give us confidence in the qualitative use of our simple atomic-orbital models.
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)/06%3A_Bonding_in_Organic_Molecules/6.07%3A_Advanced_Quantum_Theory_of_Organic_Molecules.txt |
We begin our study of compounds with a range of functional groups other than hydrocarbons and consider the nomenclature of organic compounds of oxygen, nitrogen, and halogens. Many of the principles you have learned in connection with naming hydrocarbons will have direct application to these compounds. We will use systematic nomenclature to obtain first-choice names, but we also will indicate common usage, at least parenthetically.
• 7.1: General Approaches to Naming Organic Compounds
There are two aspects to consider: how to derive the name from the structure, and how to derive the structure from the name.
• 7.2: Alcohols and Phenols- ROH, ArOH
The IUPAC nomenclature system for alcohols and phenols, involves adding the suffix -ol for \(OH\) to the name of the parent hydrocarbon. Notice that alkane \(+\) -ol becomes alkanol, with the e omitted.
• 7.3: Ethers, ROR'
The substituent name for the \(RO-\) function is alkoxy, and it is correct to name \(R-O-R'\) compounds as alkoxy derivatives of hydrocarbons. In the common nomenclature for ethers, each of the \(R\) groups in \(R-O-R'\) is named as a separate word, except when the groups are identical, in which case the prefix di or bis may be used (di is used for simple groups, bis for substituted groups).
• 7.4: Aldehydes
The suffix -al is appended to the name of the hydrocarbon corresponding to the longest carbon chain that includes the aldehyde carbon. Remember that alkane + -al becomes alkanal with the e omitted, and because the al function is necessarily at C1, the -1- is redundant and is omitted.
• 7.5: Ketones, RCOR'
The IUPAC system employs the suffix -one added to the prefix identifying the longest carbon chain of \(RCOR'\) that includes the carbonyl group. The chain is numbered to give the carbonyl group. The chain is numbered to give the carbonyl group the lowest possible number.
• 7.6: Carboxylic Acids
By the IUPAC system, the suffix -oic is added to the prefix identifying the hydrocarbon chain that includes the carboxyl carbon. Situations arise when it is necessary to consider the parent as a one-carbon chain. In such circumstances, \(RCO_2H\) becomes a substituted carboxylic acid. This variation is met most frequently when \(R\) is a cycloalkyl or aryl group.
• 7.7: Acyl Groups, RCO-
An acyl group and in specific cases in named by adding the suffix -oyl to the appropriate hydrocarbon prefix. That is, alkane- ++ -oyl becomes alkanoyl. Acyl groups also may be called alkanecarbonyl or cycloalkanecarbonyl groups.
• 7.8: Amines
Primary amines, having only one substituent on nitrogen, are named with the substituent as a prefix. More systematic nomenclature appends -amine to the longest chain, as for alcohols. Secondary and tertiary amines, which have two and three substituents on nitrogen, commonly are named as N-substituted amines. As for substituted amides, N is included to indicate that the substituent is on the nitrogen atom unless there is no ambiguity as to where the substituent is located.
• 7.9: Nitriles, RCN
Compounds with the ≡N function are named by adding the suffix nitrile to the main-chain hydrocarbon that includes the carbon linked to the nitrile (≡N) function. The chain is numbered so the CN carbon is C1.
• 7.10: The Use of Greek Letters to Denote Substituent Positions
In the older literature, considerable use is made of the Greek letters α, β, γ, and so on, to designate successive positions along a hydrocarbon chain. The carbon directly attached to the principal function group is denoted as α, the second carbon is β, and so on down the chain. Because the usage is widespread, cognizance of the system is important, but systematic naming and numbering systems should be used whenever possible.
• 7.11: Single- or Multiple-Word Names
A troublesome point in naming chemical compounds concerns the rules governing when a compound is to be written as a single word (as methylamine) or as two or more words (as methyl chloride). To solve this problem, you must determine whether the principle or parent function is an element or a compounds in its own right; if it is either one, then the name is written as a single word.
• 7.E: Other Compounds than Hydrocarbons (Exercises)
These are the homework exercises to accompany Chapter 7 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."
07: Other Compounds than Hydrocarbons
There are two aspects to consider: how to derive the name from the structure, and how to derive the structure from the name. We will discuss each by example.
Naming a Compound of Known Structure
You first should decide what type of compound it is. The decision usually is straightforward for hydrocarbons, which will fall in one or the other of the categories alkanes, alkenes, alkynes, arenes, cycloalkanes, and so on. But when the compound has more than one functional group it is not always obvious which is the parent function. For example, Compound \(1\) could be named as an alkene (because of the double-bond function) or as an alcohol (because of the \(OH\) function):
There are no simple rules to follow that dictate which is the parent function, and we suggest that the order of precedence of functional groups set by Chemical Abstracts be used whenever possible (see Table 7-1). By this system, the \(OH\) group takes precedence over hydrocarbons, and Compound \(1\) therefore is named as an alcohol, not as an alkene.
Having decided on the main classification, our next step is to identify the longest chain that includes the main functional group. Then this chain is numbered, starting at the end that gives the main function the lowest possible number. The remaining groups, functional or nonfunctional, are taken as substituents and are assigned numbers according to their position along the chain. Thus for Compound \(1\):
1. The longest continuous carbon chain carrying the \(OH\) group is a six-carbon unit. The prefix for a six-carbon hydrocarbon is hex-.
2. The chain is numbered so the \(OH\) group is at \(C2\), the lowest possible number. Therefore the IUPAC suffix is -2-ol, in which ol signifies alcohol (see Section 7-2).
3. The remaining functions are methyl (at \(C5\)) and -en(e) (at \(C4\)). The complete name is
(Notice that the final e is dropped from the suffix -ene when followed by another suffix beginning with a vowel.)
One further point of possible confusion is where to locate the numerical symbol for the main functional group in the name. For instance, if the double bond in \(1\) were absent, we could name the compound either 5-methylhexan-2-ol or 5-methyl-2-hexanol. The rule is to not divide the name unnecessarily. Thus 5-methyl-2-hexanol would be correct and 5-methylhexan-2-ol would be incorrect:
Translating a Name into its Chemical Structure
1. The first step is to identify the parent function, which usually is determined from the suffix or word at the end of the name. Suppose, for example, that a structure is to be written for a compound having the name 3-methoxybutanal. The suffix -al is the IUPAC suffix for aldehyde; therefore the compound is an aldehyde and the function is \(-CHO\).
3. The rest of the name, which generally precedes the parent name, describes the substituent and its position on the parent chain. In our example, 3-methoxy means a \(CH_3O-\) group at \(C3\). Thus the complete structure of 3-methoxybutanal is
The foregoing examples illustrate that naming compounds from structures or deducing structures from names requires knowledge of both the parent names and the substituent names of the important types of functional and nonfunctional groups. This information is summarized in the following sections and in Table 7-1.
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)/07%3A_Other_Compounds_than_Hydrocarbons/7.01%3A_General_Approaches_to_Naming_Organic_Compounds.txt |
1. By the IUPAC system, the suffix -ol for \(OH\) is added to the name of the parent hydrocarbon. Notice that alkane \(+\) -ol becomes alkanol, with the e omitted:
2. The substituent name for the \(OH\) group is hydroxy and should be used whenever the \(OH\) group is not the parent function (see Table 7-1). Notice how the precedence rules apply - hydroxy below carboxylic acid and hydroxy below ketone:
3. Many trivial names persist, particularly for aromatic, or arene alcohols (phenols):
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."
7.03: Ethers ROR'
The substituent name for the \(RO-\) function is alkoxy, and it is correct to name \(R-O-R'\) compounds as alkoxy derivatives of hydrocarbons:
In the common nomenclature for ethers, each of the \(R\) groups in \(R-O-R'\) is named as a separate word, except when the groups are identical, in which case the prefix di or bis may be used (di is used for simple groups, bis for substituted groups):
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."
7.04: Aldehydes
1. The suffix -al is appended to the name of the hydrocarbon corresponding to the longest carbon chain that includes the aldehyde carbon. Remember that alkane \(+\) -al becomes alkanal with the e omitted, and because the al function is necessarily at \(C1\), the -1- is redundant and is omitted:
Dialdehydes are named as -dials. Thus \(OHCCH_2CH_2CH_2CH_2CHO\) is hexanedial.
2. The simplest aldehyde is methanal, \(HCHO\), which is familiarly known as formaldehyde. However, when aldehydes are named as derivatives of methanal, they usually are called carbaldehydes, and the suffix "carbaldehyde" refers to the \(-CHO\) group. This system is used where the hydrocarbon group is not a chain, but a ring, and the \(CHO\) group can be thought of as a one-carbon chain:
3. When the \(-CHO\) group is a substituent on the parent chain or ring and it ranks below another functional group, it properly is designated by the prefix methanoyl. However, the prefix formyl also is used:
(The naming of acids will be discussed more in detail in Section 7-6.)
4. Trivial names are used for many simple aldehydes, some of which are shown below in parentheses:
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."
7.05: Ketones RCOR'
The IUPAC system employs the suffix -one added to the prefix identifying the longest carbon chain of \(RCOR'\) that includes the carbonyl group. The chain is numbered to give the carbonyl group. The chain is numbered to give the carbonyl group the lowest possible number. In the examples given, the names in parentheses correspond to a less systematic nomenclature of ketones by which the \(R\) groups each are named separately:
When the doubly bonded oxygen is regarded as a substituent along the parent chain or ring, it is called an oxo group, \(=O\),
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)/07%3A_Other_Compounds_than_Hydrocarbons/7.02%3A_Alcohols_and_Phenols-_ROH_ArOH.txt |
By the IUPAC system, the suffix -oic is added to the prefix identifying the hydrocarbon chain that includes the carboxyl carbon:
Situations arise when it is necessary to consider the parent as a one-carbon chain. In such circumstances, \(RCO_2H\) becomes a substituted carboxylic acid. This variation is met most frequently when \(R\) is a cycloalkyl or aryl group:
The substituent name for \(-CO_2H\) is carboxy:
For salts of carboxylic acids, the -oic suffix of the acid becomes -oate with the counter ion named as a separate word:
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."
7.07: Acyl Groups RCO-
1. The function is called an acyl group and in specific cases in named by adding the suffix -oyl to the appropriate hydrocarbon prefix. That is, alkane- \(+\) -oyl becomes alkanoyl:
Acyl groups also may be called alkanecarbonyl or cycloalkanecarbonyl groups:
2. When an acyl group replaces the hydrogen of alcohols, carboxylic acids, hydrogen halides, ammonia or amines, we have the acyl compounds known as esters, anhydrides, halides, and amides, respectively.
Each of these types of compounds are named as follows.Image
Carboxylic Esters, \(RCO_2R'\)
1. The name of the parent carboxylic acid (alkanoic) is changed to alkanoate and is preceded, as in a separate word, by the name of the ester alkyl group \(R'\):
2. When appropriate, esters also are named as carboxylates:
3. When it is necessary to name the \(-CO_2R'\) function as a substituent, it becomes alkoxycarbonyl, . (Notice that this is structurally different from .)
Notice the use of parentheses to separate the numbering of \(C3\) of the cyclohexane ring from the numbering of the chain.
4. It also may be necessary at times to name the \(R'CO_2-\) group as a substituent, in which case it becomes acyloxy- or \(R'\)-carbonyloxy-. For example,
Carboxylic Anhydrides, \(RCOOCOR'\)
Symmetrical anhydrides (\(R=R'\)) are named after the parent acid; unsymmetrical or "mixed" anhydrides (\(R \neq R'\)) cite each of the parent acids:
Acyl Halides, \(RCOX\)
The acyl group, \(RCO-\), and the halogen (as halide) are cited separately:
Amides, \(RCONH_2\)
1. The suffix amide is appended to the name of the hydrocarbon corresponding to the carbon chain that includes the carbonyl group. That is, alkan(e) \(+\) amide \(=\) alkanamide. A one-carbon chain is a carboxamide:
2. When the amide nitrogen is substituted with lower-ranking groups than the acyl group, the substituents are designated as prefixes. The letter N is used to show that the substitution is on nitrogen:
3. Names for amides as substituents include the following:
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."
7.08: Amines
The word "amine" is derived from ammonia, and the class of compounds known as amines therefore are commonly named as substituted ammonias. In this system, primary amines (\(RNH_2\)), having only one substituent on nitrogen, are named with the substituent as a prefix. More systematic nomenclature appends -amine to the longest chain, as for alcohols:
Secondary (\(R_2NH\)) and tertiary amines (\(R_3N\)), which have two and three substituents on nitrogen, commonly are named as N-substituted amines. As for substituted amides, N is included to indicate that the substituent is on the nitrogen atom unless there is no ambiguity as to where the substituent is located. Systematic nomenclature of secondary and tertiary amines is related to the systematic ether nomenclature discussed in Section 7-3:
As a substituent, the \(-NH_2\) group is called amino. N-Substituted amino groups are named accordingly:
\(^1\)Alphabetical order puts chloro- ahead of chloroethyl-.
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)/07%3A_Other_Compounds_than_Hydrocarbons/7.06%3A_Carboxylic_Acids.txt |
1. Compounds with the $\equiv N$ function are named by adding the suffix nitrile to the main-chain hydrocarbon that includes the carbon linked to the nitrile ($\equiv N$) function. The chain is numbered so the $CN$ carbon is $C1$:
2. Compounds of the type $RCN$ have to be called carbonitriles when $R$ is a cycloalkane or similar group:
3. Nitriles can be regarded as derivatives of carboxylic acids because the acid, $RCO_2H$, usually can be obtained from the nitrile, $RCN$:
A common system of naming nitriles takes the name of the corresponding carboxylic acid and changes the suffix -oic to -onitrile:
4. The substituent name for $-CN$ is cyano. For example,
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."
7.10: The Use of Greek Letters to Denote Substituent Positions
In the older literature, considerable use is made of the Greek letters $\alpha$, $\beta$, $\gamma$, and so on, to designate successive positions along a hydrocarbon chain. The carbon directly attached to the principal function group is denoted as $\alpha$, the second carbon is $\beta$, and so on down the chain:
The omega ($\omega$) position is sometimes used to designate the last position along the chain regardless of its length. Thus $\omega$-bromohexanoic acid is 6-bromohexanoic acid. In general, the use of Greek letters in the naming of compounds is to be avoided. Because the usage is widespread, cognizance of the system is important, but systematic naming and numbering systems should be used whenever possible.
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."
7.11: Single- or Multiple-Word Names
A troublesome point in naming chemical compounds concerns the rules governing when a compound is to be written as a single word (as methylamine) or as two or more words (as methyl chloride). To solve this problem, you must determine whether the principle or parent function is an element or a compounds in its own right; if it is either one, then the name is written as a single word.
The following examples should help to clarify the system. In each name, the part of the name that denotes the parent compound\(^2\) is italicized:
However, if the parent function cannot be constructed as being a real compound, the name is correctly written as two or more words. For example, \(CH_3Cl\) could be named as a chloride, in which case we use two words, methyl chloride, to describe it. A chloride, or any halide, is a class of compound, not a specific compound. To identify a specific halide, the adjective that describes the halide is written as a separate word preceding the class name. Examples follow in which the class name is italicized.\(^3\)
\(^2\)The parent compounds designated here as amine, carbaldehyde, and sulfonic acid are properly ammonia, methanal, and sulfurous acid (\(HSO_3H\)) when no substituent groups are attached.
\(^3\)These word-separated names sometimes are called radicofunctional names.
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)/07%3A_Other_Compounds_than_Hydrocarbons/7.09%3A_Nitriles_RCN.txt |
Nucleophilic substitution is a fundamental class of reactions in which an electron rich nucleophile selectively bonds with or attacks the positive or partially positive charge of an atom or a group of atoms to replace a so-called leaving group.
• 8.1: Prelude to Nucleophilic Substitution and Elimination Reactions
Substitution reactions involve the replacement of one atom or group by another. The halogenation of alkanes is one important type of substitution reaction, which proceeds by radical-chain mechanisms in which the bonds are broken and formed by atoms or radicals as reactive intermediates. This mode of bond-breaking is called homolytic bond cleavage. There are many reactions, often in solution, that do not involve atoms or radicals, but rather involve ions. They occur by heterolytic cleavage.
• 8.2: Classification of Reagents as Electrophiles and Nucleophiles. Acids and Bases
To understand ionic reactions, we need to be able to recognize whether a particular reagent will act to acquire an electron pair or to donate an electron pair. Reagents that acquire an electron pair in chemical reactions are said to be electrophilic ("electron-loving"). We can picture this in a general way as a heterolytic bond breaking of compound X:Y by an electrophile E such that E becomes bonded to Y by the electron pair of the XY bo
• 8.3: Thermochemistry of Substitution Reactions
Ionic or polar reactions of alkyl halides rarely are observed in the vapor phase because the energy required to dissociate a carbon-halogen bond heterolytically is almost prohibitively high. The reason is that ions are much more stable in water than in the gas phase; for example, the transfer of a chloride ion from the gas to water is exothermic with large ionic solvation energies.
• 8.4: General Considerations of Substitution Reactions
We now wish to discuss displacements by nucleophilic reagents (Y:) on alkyl derivatives (RX). These are ionic or polar reactions involving attack by a nucleophile at a carbon. Reactions of this type are very useful and can lead to compounds in which the new bond to carbon in the alkyl group. Nucleophilic substitutions are especially important for alkyl halides, but they should not be considered to be confined to alkyl halides.
• 8.5: Mechanisms of Nucleophilic Substitution Reactions
Two simple mechanisms can be written for the reaction of chloromethane with hydroxide ion in aqueous solution that differ in the timing of bond breaking relative to bond making. In the first mechanism, A , the overall reaction is the result of two steps, the first of which involves a slow dissociation of chloromethane to solvated methyl carbocation and solvated chloride ion. The second step involves a fast reaction between the carbocation and hydroxide ion (or water) to yield methanol.
• 8.6: Stereochemistry of \(S_N2\) Reactions
There are two simple ways in which the SN2 reaction of methyl chloride could occur with hydroxide ion; they differ in the direction of approach of the reagents. The hydroxide ion could attack chloromethane at the front side of the carbon where the chlorine is attached or, alternatively, the hydroxide ion could approach the carbon on the side opposite from the chlorine in what is called the back-side approach. The stereochemical consequences of front- and back-side displacements are different.
• 8.7: Stereochemistry of \(S_N1\) Reactions
Theoretically, if an SN1 reaction is carried out with a single pure enantiomer since a carbocation is most stable in the planar configuration and hence should lead to exactly equal amounts of the two enantiomers, regardless of the chiral configuration of the starting material. However, the extent of configuration change that actually results in an SN1 reaction depends upon the degree of "shielding" of the front side of the reacting carbon by the leaving group and its associated solvent molecule
• 8.8: Structural and Solvent Effects in \(S_N\) Reactions
We shall consider first the relationship between the structures of alkyl derivatives and their reaction rates toward a given nucleophile. This will be followed by a discussion of the relative reactivities of various nucleophiles toward a given alkyl derivative. Finally, we shall comment in more detail on the role of the solvent in nucleophilic substitution reactions.
• 8.9: The E2 Reaction
The conditions used for substitution reactions by the SN2 mechanism very often lead to elimination.
• 8.10: The E1 Reaction
Many secondary and tertiary halides undergo E1 elimination in competition with the SN1 reaction in neutral or acidic solutions. The SN1 and E1 reactions have a common rate-determining step, namely, slow ionization of the halide.
• 8.11: Elimination Reactions
• 8.E: Nucleophilic Substitution and Elimination Reactions (Exercises)
These are the homework exercises to accompany Chapter 8 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."
08: Nucleophilic Substitution and Elimination Reactions
Substitution reactions involve the replacement of one atom or group $\left( \ce{X} \right)$ by another $\left( \ce{Y} \right)$:
$\ce{RX} + \ce{Y} \rightarrow \ce{RY} + \ce{X}$
We already have described one very important type of substitution reaction, the halogenation of alkanes (Section 4-4), in which a hydrogen atom is replaced by a halogen atom ($\ce{X} = \ce{H}$, $\ce{Y} =$ halogen). The chlorination of 2,2-dimethylpropane is an example:
Reactions of this type proceed by radical-chain mechanisms in which the bonds are broken and formed by atoms or radicals as reactive intermediates. This mode of bond-breaking, in which one electron goes with $\ce{R}$ and the other with $\ce{X}$, is called homolytic bond cleavage:
There are a large number of reactions, usually occurring in solution, that do not involve atoms or radicals but rather involve ions. They occur by heterolytic cleavage as opposed to homolytic cleavage of electron-pair bonds. In heterolytic bond cleavage, the electron pair can be considered to go with one or the other of the groups $\ce{R}$ and $\ce{X}$ when the bond is broken. As one example, $\ce{Y}$ is a group such that it has an unshared electron pair and also is a negative ion. A heterolytic substitution reaction in which the $\ce{R} \colon \ce{X}$ bonding pair goes with $\ce{X}$ would lead to $\ce{RY}$ and $\colon \ce{X}^\ominus$,
A specific substitution reaction of this type is that of chloromethane with hydroxide ion to form methanol:
In this chapter, we shall discuss substitution reactions that proceed by ionic or polar mechanisms in which the bonds cleave heterolytically. We also will discuss the mechanistically related elimination reactions that result in the formation of carbon-carbon multiple bonds:
These reactions often are influenced profoundly by seemingly minor variations in the structure of the reactants, in the solvent, or in the temperature. It is our purpose to show how these reactions can be understood and how they can be used to prepare other useful organic compounds. But first it will be helpful to introduce the concepts of nucleophilic and electrophilic reagents, and to consider the $\Delta H$ values for heterolytic bond breaking.
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.01%3A_Prelude_to_Nucleophilic_Substitution_and_Elimination_Reactions.txt |
To understand ionic reactions, we need to be able to recognize whether a particular reagent will act to acquire an electron pair or to donate an electron pair. Reagents that acquire an electron pair in chemical reactions are said to be electrophilic ("electron-loving"). We can picture this in a general way as a heterolytic bond breaking of compound $X:Y$ by an electrophile $E$ such that $E$ becomes bonded to $Y$ by the electron pair of the $XY$ bond. Thus
Reagents that donate an electron pair in chemical reactions are said to be nucleophilic ("nucleus loving"). Thus the $X:Y$ bond also can be considered to be broken by the nucleophile $Nu:$, which donates its electron pair to $X$ while $Y$ leaves as $Y:^\ominus$ with the electrons of the $X:Y$ bond:
Thus, by definition, electrophiles are electron-pair acceptors and nucleophiles are electron-pair donors. These definitions correspond closely to definitions used in the generalized theory of acids and bases proposed by G. N. Lewis (1923). According to Lewis, an acid is any substance that can accept an electron pair, and a base is any substance that can donated an electron pair to form a covalent bond. Therefore acids must be electrophiles and bases must be nucleophiles. For example, the methyl cation may be regarded as a Lewis acid, or an electrophile, because it accepts electrons from reagents such as chloride ion or methanol. In turn, because chloride ion and methanol donate electrons to the methyl cation they are classified as Lewis bases, or nucleophiles:
The generalized Lewis concept of acids and bases also includes common proton-transfer reactions.$^1$ Thus water acts as a base because one of the electron pairs on oxygen can abstract a proton from a reagent such as hydrogen fluoride:
Alternatively, the hydronium ion ($H_3O^\oplus$) is an acid because it can accept electrons from another reagent (e.g., fluoride ion) by donating a proton.
A proton donor can be classified as an electrophile and a proton acceptor as a nucleophile. For example, hydrogen chloride can transfer a proton to ethene to form the ethyl cation. Therefore hydrogen chloride functions as the electrophile, or acid, and ethene functions as the nucleophile, or base:
What then is the difference between an acid and an electrophile, or between a base and nucleophile? No great difference until we try to use the terms in a quantitative sense. For example, if we refer to acid strength, or acidity, this means the position of equilibrium in an acid-base reaction. The equilibrium constant $K_a$ for the dissociation of an acid $HA$, or the $pK_a$, is a quantitative measure of acid strength. The larger the value of $K_a$ or the smaller the $pK_a$, the stronger the acid.
A summary of the relationships between $K_a$ and $pK_a$ follow, where the quantities in brackets are concentrations:
or
By definition, $-\text{log} \: K_a = pK_a$ and $-\text{log} \: \left[ H_3O^\oplus \right] = pH$; hence
or
However, in referring to the strength of reagents as electrophiles or nucleophiles we usually are not referring to chemical equilibria but to reaction rates. A good nucleophile is a reagent that reacts rapidly with a particular electrophile. In contrast, a poor nucleophile reacts only slowly with the same electrophile. Consequently, it should not then be taken for granted that there is a parallel between the acidity or basicity of a reagent and its reactivity as an electrophile or nucleophile. For instance, it is incorrect to assume that the strengths of a series of bases, $B:$, in aqueous solution will necessarily parallel their nucleophilicities toward a carbon electrophile, such as methyl chloride:
In what follows we will be concerned with the rates of ionic reactions under nonequilibrium conditions. We shall use the term nucleophilic repeatedly and we want you to understand that a nucleophile is any neutral or charged reagent that supplies a pair of electrons, either bonding or nonbonding, to form a new covalent bond. In substitution reactions the nucleophile usually is an anion, $Y:^\ominus$, or a neutral molecule, $Y:$ or $HY:$. The operation of each of these is illustrated in the following equations for reactions of the general compound $RX$ and some specific examples:
An electrophile is any neutral or charged reagent that accepts an electron pair (from a nucleophile) to form a new bond. In the preceding substitution reactions, the electrophile is $RX$. The electrophile in other reactions may be a carbon cation or a proton donor, as in the following examples:
$^1$The concept of an acid as a proton donor and a base as a proton acceptor is due to Bronsted and Lowry (1923). Previous to this time, acids and bases generally were defined as substances that functioned by forming $H^\oplus$ or $OH^\ominus$ in water solutions. The Bronsted-Lowry concept was important because it liberated acid-base phenomena from the confines of water-containing solvents by focusing attention on proton transfers rather than the formation of $H^\oplus$ or $OH^\ominus$. The Lewis concept of generalized acids and bases further broadened the picture by showing the relationship between proton transfers and reactions where an electron-pair acceptor is transferred from one electron-pair donor to another.
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.02%3A_Classification_of_Reagents_as_Electrophiles_and_Nucleophiles._Acids_and_Bases.txt |
Ionic or polar reactions of alkyl halides rarely are observed in the vapor phase because the energy required to dissociate a carbon-halogen bond heterolytically is almost prohibitively high. For example, while the heat of dissociation of chloromethane to a methyl radical and a chlorine atom is $84 \: \text{kcal mol}^{-1}$ (Table 4-6), dissociation to a methyl cation and a chloride ion requires about $227 \: \text{kcal mol}^{-1}$:
However, the heat of ionic dissociation of methyl chloride in aqueous solution is estimated to be $63 \: \text{kcal}$, and while this reaction is still substantially endothermic, it requires about $227 - 63 = 164 \: \text{kcal}$ less energy than in the gas phase:
The reason is that ions are much more stable in water than in the gas phase; for example, the transfer of a chloride ion from the gas to water is exothermic by $-85 \: \text{kcal}$. The $\Delta H^\text{0}$ value for the corresponding transfer of a methyl cation, $CH_3^\oplus$, is not known with certainty, but is about $-80 \: \text{kcal}$. These ionic solvation energies are clearly large. In contrast, the $\Delta H^\text{0}$ for solution of methyl chloride in water is small (about $1 \: \text{kcal}$). We can use these data to calculate the heat of ionic dissociation of chloromethane in water:
Thermochemical data for the solvation of ions as used in the preceding calculations are difficult to measure and even to estimate. Therefore this kind of calculation of $\Delta H^\text{0}$ for ionic reactions involving organic molecules in solution usually cannot be made. As a result, we have considerably fewer possibilities to assess the thermodynamic feasibility of the individual steps of polar reactions in solution than we do of vapor-phase radical processes. Bond energies are not of much use in predicting or explaining reactivity in ionic reactions unless we have some information that can be used to translate gas-phase $\Delta H^\text{0}$ values to solution $\Delta H^\text{0}$ values.
$^2$Calculated from the following data:
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."
8.04: General Considerations of Substitution Reactions
We now wish to discuss displacements by nucleophilic reagents ($Y:$) on alkyl derivatives ($RX$). These are ionic or polar reactions involving attack by a nucleophile at carbon. A typical example is the reaction of hydroxide ion with bromomethane to displace bromide ion:
The electron pair of the $C-O$ bond can be regarded as having been donated by the hydroxide ion, while the electron pair of the $C-Br$ bond departs with the leaving bromide ion. The name for this type of reaction is abbreviated $\text{S}_\text{N}$, $\text{S}$ for substitution and $\text{N}$ for nucleophilic.
Reactions of this type are very useful. They can lead to compounds in which the new bond to carbon in the alkyl group, $R$ is to chlorine, bromine, iodine, oxygen, sulfur, carbon, nitrogen, or phosphorus, depending on the nature of the nucleophile used.
Nucleophilic substitutions are especially important for alkyl halides, but they should not be considered to be confined to alkyl halides. Many other alkyl derivatives such as alcohols, ethers, esters, and "onium ions"$^3$ also can undergo $\text{S}_\text{N}$ reactions if conditions are appropriate. The scope of $\text{S}_\text{N}$ reactions is so broad that it is impossible to include all the various alkyl compounds and nucleophiles that react in this manner. Rather we shall approach the subject here through consideration of the mechanisms of $\text{S}_\text{N}$ reactions, and then develop the scope of the reactions in later chapters.
The mechanism of an $\text{S}_\text{N}$ reaction and the reactivity of a given alkyl compound $RX$ toward a nucleophile $Y$ depend upon the nature of $R$, $X$, and $Y$, and upon the nature of the solvent. For an $\text{S}_\text{N}$ reaction to occur at a reasonable rate, it is very important to select a solvent that will dissolve both the alkyl compound and the nucleophilic reagent; considerable assistance may be required from both the solvent and the nucleophile to break what usually is a slightly polar $C-X$ bond. However, the solvents that best dissolve slightly polar organic compounds seldom will dissolve the common, rather highly polar, nucleophilic agents such as $NaBr$, $NaCN$, and $H_2O$. In practice, relatively polar solvents, or solvent mixtures, such as 2-propanone (acetone), aqueous 2-propanone, ethanol, aqueous 1,4-dioxacyclohexane (dioxane), and so on, provide the best compromise for reactions between alkyl compounds and saltlike nucleophilic reagents. The importance of the solvent in stabilizing ions can be evaluated from the estimated thermochemistry of ionic reactions discussed in Section 8-2.
$^3$Examples of -onium cations 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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.03%3A_Thermochemistry_of_Substitution_Reactions.txt |
Two simple mechanisms can be written for the reaction of chloromethane with hydroxide ion in aqueous solution that differ in the timing of bond breaking relative to bond making. In the first mechanism, $A$, the overall reaction is the result of two steps, the first of which involves a slow dissociation of chloromethane to solvated methyl carbocation$^4$ and solvated chloride ion. The second step involves a fast reaction between the carbocation and hydroxide ion (or water) to yield methanol.
Mechanism $A$:
or
In the second mechanism, $B$, the reaction proceeds in a single step. Attack of hydroxide ion at carbon occurs simultaneously with the loss of chloride ion; that is, the carbon-oxygen bond is formed as the carbon-chlorine bond is broken:
Mechanism $B$:
Both of these mechanisms are important in the displacement reactions of alkyl compounds, although chloromethane appears to react only by Mechanism $B$. Now we will discuss the criteria for distinguishing between the concerted and stepwise mechanisms.
Kinetics of Substitution Mechanisms
Of the two mechanisms, $A$ requires that the reaction rate be determined solely by the rate of the first step (cf. earlier discussion in Section 4-4C). This means that the rate at which methanol is formed (measured in moles per unit volume per unit time) will depend on the chloromethane concentration, but not on the hydroxide ion concentration, because hydroxide ion is not utilized except in a fast secondary reaction. In contrast, Mechanism $B$ requires the rate to depend on the concentrations of both reagents because the slow step involves collisions between hydroxide ions and chloromethane molecules.
$v = k_A [CH_3Cl] \label{8-1}$
$v = k_B [CH_3Cl][OH^-] \label{8-2}$
More precisely, the reaction rate ($\nu$) may be expressed in terms of Equation $\ref{8-1}$ for Mechanism $A$ and Equation $\ref{8-2}$ for Mechanism $B$:
Customarily, $\nu$ is expressed in moles of product formed per liter of solution per unit of time (most frequently in seconds). The concentration terms $\left[ CH_3Cl \right]$ and $\left[ OH^\ominus \right]$ are then in units of moles per liter, and the proportionality constant $k$ (called the specific rate constant) has the units of $\text{sec}^{-1}$ for Mechanism $A$ and $\text{mol}^{-1} \times \text{L} \times \text{sec}^{-1}$ for Mechanism $B$.
It is important to recognize the difference between the order of a reaction with respect to a specific reactant and the overall order of a reaction. The order of a reaction with respect to a particular reactant is the power to which the concentration of that reactant must be raised to have direct proportionality between concentration and reaction rate. According to Equation $\ref{8-2}$ the rate of the chloromethane-hydroxide ion reaction is first order with respect to chloromethane and first order with respect to hydroxide ion. In Equation $\ref{8-1}$ the rate is first order with respect to chloromethane and zero order with respect to hydroxide ion because $\left[ OH^\ominus \right]^0 = 1$. The overall order of reaction is the sum of the orders of the respective reactants. Thus Equations $\ref{8-1}$ and $\ref{8-2}$ express the rates of overall first-order and second-order reactions, respectively.
We can use the overall reaction order to distinguish between the two possible mechanisms, $A$ and $B$. Experimentally, the rate of formation of methanol is found to be proportional to the concentrations both of chloromethane and of hydroxide ion. Therefore the reaction rate is second order overall and is expressed correctly by Equation $\ref{8-2}$. This means that the mechanism of the reaction is the single-step process $B$. Such reactions generally are classified as bimolecular nucleophilic substitutions, often designated $\text{S}_\text{N}2$, $\text{S}$ for substitution, $\text{N}$ for nucleophilic, and $2$ for bimolecular, because there are two reactant molecules in the transition state. To summarize: For an $\text{S}_\text{N}2$ reaction,
• rate:
$v = k[RX][Y] \label{8-3}$
• mechanism:
The stepwise Mechanism $A$ is a unimolecular nucleophilic substitution and accordingly is designated $\text{S}_\text{N}1$. The numeral $1$ (or $2$) used in these designations does not refer to the kinetic order of the reaction, but refers to the number of molecules (not including solvent molecules) that make up the transition state. Thus for $\text{S}_\text{N}1$,
• rate:
$v = k[RX] \label{8-4}$
• mechanism:
or
Solvolysis
Many $\text{S}_\text{N}$ reactions are carried out using the solvent as the nucleophilic agent. They are called solvolysis reactions and involve solvents such as water, ethanol, ethanoic acid, and methanoic acid. Two examples are
In these examples, solvolysis is necessarily a first-order reaction, because normally the solvent is in such great excess that its concentration does not change appreciably during reaction, and hence its contribution to the rate does not change. However, that the overall rate is first order does not mean the reaction necessarily proceeds by an $\text{S}_\text{N}1$ mechanism, particularly in solvents such as water, alcohols, or amines, which are reasonably good nucleophilic agents. The solvent can act as the displacing agent in an $\text{S}_\text{N}2$ reaction.
To distinguish between $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ mechanisms of solvolysis requires other criteria, notably stereochemistry (Sections 8-5 and 8-6), and the effect of added nucleophiles on the rate and nature of the reaction products. For example, it often is possible to distinguish between $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ solvolysis by adding to the reaction mixture a relatively small concentration of a substance that is expected to be a more powerful nucleophile than the solvent. If the reaction is strictly $\text{S}_\text{N}1$, the rate at which $RX$ disappears should remain essentially unchanged because it reacts only as fast as $R^\oplus$ forms, and the rate of this step is not changed by addition of the nucleophile, even if the nucleophile reacts with $R^\oplus$. However, if the reaction is $\text{S}_\text{N}2$, the rate of disappearance of $RX$ should increase because $RX$ reacts with the nucleophile in an $\text{S}_\text{N}2$ reaction and now the rate depends on both the nature and the concentration of the nucleophile.
$^4$Many organic chemists, and indeed the previous versions of this book, use the term "carbonium ion" for species of this kind. However, there is well-established usage of the -onium suffix, for ammonium, oxonium, chloronium, and so on, to denote positively charged atoms with filled valence shells. In the interest of greater uniformity of nomenclature we shall use "carbocation" for carbon positive ions that have unfilled valence shells (6 electrons).
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.05%3A_Mechanisms_of_Nucleophilic_Substitution_Reactions.txt |
There are two simple ways in which the $\text{S}_\text{N}2$ reaction of methyl chloride could occur with hydroxide ion. These differ in the direction of approach of the reagents (Figure 8-1). The hydroxide ion could attack chloromethane at the front side of the carbon where the chlorine is attached or, alternatively, the hydroxide ion could approach the carbon on the side opposite from the chlorine in what is called the back-side approach. In either case, the making of the $C-O$ bond is essentially simultaneous with the breaking of the $C-Cl$ bond. The difference is that for the back-side mechanism the carbon and the attached hydrogens become planar in the transition state.
The stereochemical consequences of front- and back-side displacements are different. With cyclic compounds, the two types of displacement lead to different products. For example, an $\text{S}_\text{N}2$ reaction between cis-3-methylcyclopentyl chloride and hydroxide ion would give the cis alcohol by front-side approach but the trans alcohol by back-side approach. The actual product is the trans alcohol, from which we know that reaction occurs by back-side displacement:
displacement to give 2-butanol with the inverted configuration. Similar studies of a wide variety of displacements have established that $\text{S}_\text{N}2$ reactions invariably proceed with inversion of configuration via back-side attack. This stereochemical course commonly is known as the Walden inversion.$^5$ An orbital picture of the transition state of an $\text{S}_\text{N}2$ reaction that leads to inversion of configuration follows:
$^5$The first documented observation that optically active compounds could react to give products having the opposite configuration was made by P. Walden, in 1895. The implications were not understood, however, until the mechanisms of nucleophilic substitution were elucidated in the 1930's, largely through the work of E. D. Hughes and C. K. Ingold, who established that $\text{S}_\text{N}2$ substitutions give products of inverted 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."
8.07: Stereochemistry of (S N1) Reactions
When an $\text{S}_\text{N}1$ reaction is carried out starting with a single pure enantiomer, such as $D$-2-chlorobutane, the product usually is a mixture of the enantiomeric substitution products with a slight predominance of that isomer which corresponds to inversion. Theoretically, a carbocation is expected to be most stable in the planar configuration (Section 6-4E) and hence should lead to exactly equal amounts of the two enantiomers, regardless of the chiral configuration of the starting material (Figure 8-3). However, the extent of configuration change that actually results in an $\text{S}_\text{N}1$ reaction depends upon the degree of "shielding" of the front side of the reacting carbon by the leaving group and its associated solvent molecules. If the leaving group does not get away from the carbocation before the product-determining step takes place, there will be some preference for nucleophilic attack at the back side of the carbon, which results in a predominance of the product of inverted configuration.
Other things being equal, the amount of inversion decreases as the stability of the carbocation intermediate increases, because the more stable the ion the longer is its lifetime, and the more chance it has of getting away from the leaving anion and becoming a relatively "free" ion. The solvent usually has a large influence on the stereochemical results of $\text{S}_\text{N}1$ reactions because the stability and lifetime of the carbocations depend upon the nature of the solvent (Section 8-7F).
An orbital picture of $\text{S}_\text{N}1$ ionization leading to a racemic product may be drawn as follows:
It should be clear that complete racemization is unlikely to be observed if $X^\ominus$ stays in close proximity to the side of the positive carbon that it originally departed from. We can say that $X^\ominus$ "shields" the front side, thereby favoring a predominance of inversion. If $X^\ominus$ gets far away before $:YH$ comes in, then there should be no favoritism for one or the other of the possible substitutions.
If $X^\ominus$ and the carbocation, $R^\oplus$, stay in close proximity, as is likely to be the case in a solvent that does not promote ionic dissociation, then a more or less "tight" ion pair is formed, $R^\oplus \: \cdot \: \cdot \: \cdot \: X^\ominus$. Such ion pairs often play an important role in ionic reactions in solvents of low dielectric constant (Section 8-7F).
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.06%3A_Stereochemistry_of_%28S_N2%29_Reactions.txt |
We shall consider first the relationship between the structures of alkyl derivatives and their reaction rates toward a given nucleophile. This will be followed by a discussion of the relative reactivities of various nucleophiles toward a given alkyl derivative. Finally, we shall comment in more detail on the role of the solvent in $\text{S}_\text{N}$ reactions.
Structure of the Alkyl Group, $R$, in $\text{S}_\text{N}2$ Reactions
The rates of $\text{S}_\text{N}2$-displacement reactions of simple alkyl derivatives, $RX$, follow the order primary $R$ $>$ secondary $R$ $\gg$ tertiary $R$. In practical syntheses involving $\text{S}_\text{N}2$ reactions, the primary compounds generally work very well, secondary isomers are fair, and the tertiary isomers are almost completely impractical. Steric hindrance appears to be particularly important in determining $\text{S}_\text{N}2$ reaction rates, and the slowness of tertiary halides seems best accounted for by steric hindrance to the back-side approach of an attacking nucleophile by the alkyl groups on the reacting carbon. Pertinent data, which show how alkyl groups affect $\text{S}_\text{N}2$ reactivity toward iodide ion, are given in Table 8-1. Not only do alkyl groups suppress reactivity when on the same carbon as the leaving group $X$, as in tert-butyl bromide, but they also have retarding effects when located one carbon away from the leaving group. This is evident in the data of Table 8-1 for 1-bromo-2,2-dimethylpropane (neopentyl bromide), which is very unreactive in $\text{S}_\text{N}2$ reactions. Scale models indicate the retardation to be the result of steric hindrance by the methyl groups located on the adjacent $\beta$ carbon to the approaching nucleophile:
In addition to steric effects, other structural effects of $R$ influence the $\text{S}_\text{N}2$ reactivity of $RX$. A double bond $\beta$ to the halogen,$^6$ as in 2-propenyl, phenylmethyl (benzyl), and 2-oxopropyl chlorides enhances the reactivity of the compounds toward nucleophiles. Thus the relative reactivities toward $I^\ominus$ in 2-propanone are
Possible reasons for these high reactivities will be discussed later (Section 14-3B).
Structure of the Alkyl Group, $R$, in $\text{S}_\text{N}1$ Reactions
The rates of $\text{S}_\text{N}1$ reactions of simple alkyl derivatives follow the order tertiary $R$ $\gg$ secondary $R$ $>$ primary $R$, which is exactly opposite that of $\text{S}_\text{N}2$ reactions. This is evident from the data in Table 8-2, which lists the relative rates of hydrolysis of some alkyl bromides; only the secondary and tertiary bromides react at measurable rates, and the tertiary bromide reacts some $10^5$ times faster than the secondary bromide.
Why do tertiary alkyl compounds ionize so much more rapidly than either secondary or primary compounds? The reason is that tertiary alkyl cations are more stable than either secondary or primary cations and therefore are formed more easily. You will appreciate this better by looking at the energy diagram of Figure 8-4, which shows the profile of energy changes for hydrolysis of an alkyl compound, $RX$, by the $\text{S}_\text{N}1$ mechanism. The rate of
reaction is determined by the ionization step, or by the energy of the transition state relative to that of the reactants. Actually, the energy of the transition state is only slightly higher than the energy of the ionic intermediates $R^\oplus X^\ominus$. Thus to a first approximation, we can say that the rate of ionization of $RX$ will depend on the energies of the ions formed. Now if we compare the rates for a series of compounds, $RX$, all having the same leaving group, $X$, but differing only in the structure of $R$, their relative rates of ionization will correspond to the relative stabilities of $R^\oplus$. The lower energy of $R^\oplus$, the faster will be the rate of ionization. Therefore the experimental results suggest that the sequence of carbocation stabilities is tertiary $R^\oplus$ $\gg$ secondary $R^\oplus$ $\gg$ primary $R^\oplus$.
Just why this sequence is observed is a more difficult question to answer. Notice in the following stability sequence that alkyl cations are more stable the more alkyl groups there are on the positive carbon:
The simplest explanation for why this is so is that alkyl groups are more polarizable than hydrogens. In this case, more polarizable means the electrons of the alkyl groups tend to move more readily toward the positive carbon than do those of the hydrogens. Such movements of electrons transfer part of the charge on the cationic carbon to the alkyl groups, thereby spreading the charge over a greater volume. This constitutes electron delocalization, which results in enhanced stability (see Section 6-5A).
An alternative way of explaining how the cationic charge is spread over the alkyl groups of a tertiary cation, such as the tert-butyl cation, is to write the cation as a hybrid of the following structures:
Other organohalogen compounds besides secondary and tertiary alkyl compounds can react by $\text{S}_\text{N}1$ mechanisms provided they have the ability to form reasonably stabilized carbon cations. Examples include 2-propenyl (allylic) and phenylmethyl (benzylic) compounds, which on ionization give cations that have delocalized electrons (see Section 6-6):
In general, the more stabilized the carbon cation from an alkyl halide, the more reactive the compound will be in $\text{S}_\text{N}1$-type reactions. This is especially apparent in the reactivities of compounds with phenyl groups on the reacting carbon. As the number of phenyl groups increases from zero to three, the $\text{S}_\text{N}1$ reactivity of the chlorides increases by more than $10^7$ because of increasing stabilization of the carbon cation by the phenyl groups:
In contrast, compounds such as chlorobenzene and chloroethene, in which the halogen is attached directly to a multiply bonded carbon atom, do not exhibit $\text{S}_\text{N}1$-type reactions. Evidently then, unsaturated carbon cations such as phenyl or ethenyl are appreciably less stable (more difficult to form) than tert-alkyl cations:
Reasons for this will be considered in Section 14-4B.
Steric hindrance is relatively unimportant in $\text{S}_\text{N}1$ reactions because the rate is independent of the nucleophile. In fact, steric acceleration is possible in the solvolysis of highly branched alkyl halides through relief of steric compression between the alkyl groups in the halide by formation of a planar cation:
Along with the effect $R$ has on the rate at which an alkyl compound $RX$ reacts by an $\text{S}_\text{N}1$ mechanism, the group $R$ also affects the nature of the products obtained. The intermediate alkyl cations $R^\oplus$ may react in various ways to give products of substitution, elimination, and rearrangement. Elimination pathways are discussed more fully starting in Section 8-8, and rearrangement of carbon cations in Section 8-9B.
The Leaving Group
The reactivity of a given alkyl derivative, $RX$, in either $\text{S}_\text{N}1$ or $\text{S}_\text{N}2$ reactions, is influenced strongly by the leaving group, $X$. The choice of leaving group is therefore an important consideration in any synthesis involving $\text{S}_\text{N}$ reactions.
From the foregoing discussion of structural effects in the $R$ group on $\text{S}_\text{N}$ reactivity, particularly in $\text{S}_\text{N}1$ reactions, we might expect the stability of $:X$ as an ion or neutral molecule to play a major role in determining how good or poor $X$ is as a leaving group. The stability of $:X$ is indeed important - the problem is that there are several factors that contribute to the stavility and hence the lability of the leaving group.
For the purpose of initially identifying good and poor leaving groups, consider development of a practical synthesis of diethyl ether. One route is by way of $\text{S}_\text{N}2$ displacement using an ethyl compound, $CH_3CH_2X$, and ethoxide ion:
Many $CH_3CH_2X$ compounds have $X$ groups that are quite unsatisfactory in this reaction. They include compounds such as ethane, propane, ethanol, ethyl methyl ether, ethylamine, and ethyl ethanoate; the respective groups, $H^\ominus$, $CH_3^\ominus$, $HO^\ominus$, $CH_3O^\ominus$, $NH_2^\ominus$, and $CH_3CO_2^\ominus$ all can be classified as very poor leaving groups. The more reactive ethyl derivatives (see Table 8-3) include the halides, particularly ethyl iodide, and sulfonic acid derivatives; the corresponding anions $Cl^\ominus$, $Br^\ominus$, $I^\ominus$, and $RS \left( O_2 \right) O^\ominus$ therefore are moderate to good leaving groups. Table 8-3 includes pertinent data for the rates of ether formation from various alkyl compounds and illustrates that the relative abilities of groups to leave are about the same in $\text{S}_\text{N}1$ reactions as they are in $\text{S}_\text{N}2$ reactions.
Why are groups such as $I^\ominus$ and $RSO_3^\ominus$ good leaving groups, whereas others such as $H^\ominus$, $HO^\ominus$ and $NH_2^\ominus$ are poor? The simplest correlation is with the strength of $HX$ as an acid. This is very reasonable because the ease of loss of $X^\ominus$, as from $\left( CH_3 \right) C-X$ in an $\text{S}_\text{N}1$ reaction, would be expected to be related, to some degree at least, to the ease of ionization of $H-X$ to $H^\oplus$ and $X^\ominus$. Therefore the stronger $HX$ is as an acid, the better $X$ will be as a leaving group. Thus $HF$ is a relatively weak acid and $F^\ominus$ is not a very good leaving group; $H-I$ is a very strong acid and $I^\ominus$ is a good leaving group. The usual order of reactivity of alkyl halides, $R-I$ $>$ $R-Br$ $>$ $R-Cl$ $>$ $R-F$ (when $R$ is the same group throughout), is in accord with the acid strengths of the halogen acids. SImilarly, $CF_3CO_2-$ is a much better leaving group than $CH_3CO_2-$, and we find that trifluoroethanoic acid, $CF_3CO_2H$ is a several thousand times stronger acid than ethanoic acid, $CH_3CO_2H$. For the same reason, $CF_3SO_3^\ominus$ is a better leaving group than $CH_3XO_3^\ominus$.
This correlation can be extended easily to groups that leave as neutral $X:$. For example, $ROH_2^\oplus \rightarrow R^\oplus + H_2O$ occurs far more readily than $ROH \rightarrow R^\oplus + OH^\ominus$ and we know that $H_3O^\oplus$ is a stronger acid (or better proton donor) than $H_2O$.
The relationship between $X^\ominus$ as a leaving group and $HX$ as an acid is very useful because much information is available on acid strengths. However, it is not a very fundamental explanation unless we can explain why some acids are strong acids and others are weak acids. One factor is the strength of the $H-X$ bond, but here we need to remember that the usual bond strengths are for dissociation to radicals or atoms, not ions, and for the gas, not for solutions. If we write the steps relating the bond-dissociation energy to the energy of ionic dissociation in solution, we see that for variations in $X$, in addition to the bond energy, the electron affinity of $X \cdot$, the solvation energy of $X^\ominus$, and the solvation energy of $HX$, also will be contributing factors.
Enhancement of Leaving Group Abilities by Electrophilic Catalysis
In general, a leaving group that leaves as a neutral molecule is a much better leaving group than one that leaves as an anion. Alcohols, $ROH$, are particularly unreactive in $\text{S}_\text{N}$ reactions because $OH^\ominus$ is a very poor leaving group. However, if a strong acid is present, the reactivity of the alcohol is enhanced greatly. The acid functions by donating a proton to the oxygen of the alcohol, thereby transforming the hydroxyl function into $ROH_2^\oplus$, which has a much better leaving group, $H_2O$, in place of $OH^\ominus$. The $\text{S}_\text{N}$ reactions of ethers and esters are acid-catalyzed for the same reason.
Heavy-metal salts, particularly those of silver, mercury, and copper, catalyze $\text{S}_\text{N}1$ reactions of alkyl halides in much the same way that acids catalyze the $\text{S}_\text{N}$ reactions of alcohols. A heavy-metal ion functions by complexing with the unshared electrons of the halide, thereby making the leaving group a metal halide rather than a halide ion. This acceleration of the rates of halide reactions is the basis for a qualitative test for alkyl halides with silver nitrate in ethanol solution:
Silver halide precipitates at a rate that depends upon the structure of the alkyl group, tertiary $>$ secondary $>$ primary. Tertiary halides usually react immediately at room temperature, whereas primary halides require heating. That complexes actually are formed between organic halides and silver ion is indicated by an increase in water solubility in the presence of silver ion for those halides that are slow in forming carboncations.
The Nucleophilic Reagent
The nucleophilicty of a particular reagent ($:Y$, $:Y^\ominus$, or $H \ddot{Y}$) can be defined as its ability to donate an electron pair to another atom (see Section 8-1). In fact, the $\text{S}_\text{N}2$ reactivity of a reagent toward a methyl derivative can be taken to measure its nucleophilicty toward carbon. The relative reaction rates of some nucleophiles toward methyl bromide are listed in order of increasing nucleophilicity in Table 8-4, together with their basicities as measured by $K_b$. Important generalizations can be made from these data provided that one recognizes that they may apply only to hydroxylic solvents.
1. For the atoms representing any one group (column) of the periodic table, nucleophilicty increases with increasing atomic number: $I^\ominus$ $>$ $Br^\ominus$ $>$
$Cl^\ominus$ $>$ $F^\ominus$; $HS^\ominus$ $>$ $HO^\ominus$; $PH_3$ $>$ $NH_3$. Thus, other things being equal, larger atoms are better nucleophiles.
2. For nucleophiles having the same atomic number of the entering atom (e.g., oxygen nucleophiles), there is usually a good correlation between the basicity of the reagent and its nucleophilicity. Thus a weak base such as $CH_3CO_2^\ominus$ is a poorer nucleophile than a strong base such as $^\ominus OH$. The poorer $X^\ominus$ is as a leaving group, the better it is as an entering group.
3. For nucleophiles of different atomic numbers, nucleophilicity usually does not parallel basicity. For example, for the halogens the reactivity sequence $I^\ominus$ $>$ $Br^\ominus$ $>$ $Cl^\ominus$ is opposite to the sequence for basicity $Cl^\ominus$ $>$ $Br^\ominus$ $>$ $I^\ominus$. Similarly, sulfur anions such as $HS^\ominus$ are better nucleophiles but weaker bases than corresponding oxyanions such as $HO^\ominus$.
4. A number of nucleophilic agents, which are very reactive in $\text{S}_\text{N}2$ reactions, are of the type $X-Y$, where both atoms have unshared electron pairs. Examples include $HOO^\ominus$, $H_2NO^\ominus$, $ClO^\ominus$, and $H_2NNH_2$, all of which are more reactive than the closely related nucleophiles $HO^\ominus$ and $NH_3$.
Why is the correlation between basicity and nucleophilcity so poor for atoms of different atomic number? it is now clear from much research that the dominant effect is associated with differences in the solvation energies of the ions, as defined for halide ions by the following equations:
The solvation energies of small ions with concentrated charge always are greater than those of large ions with diffuse charge.
When an ion participates in a nucleophilic attack on carbon, it must slough off some of the solvent molecules that stabilize it in solution. Otherwise, the ion cannot get close enough to the carbon, to which it will become attached, to begin forming a bond. Sloughing off solvent molecules will be less favorable for a small ion than a large ion. Consequently, we expect $Cl^\ominus$ to be less reactive than $I^\ominus$.
Strong evidence for solvation effects on reactivity is provided by the fact that chloride ion is more reactive than iodide in solvents that have low solvation energies for anions (see Section 8-7F). Furthermore, in the gas phase where solvation effects are absent, $F^\ominus$ is more reactive than any of the other halide ions toward chloromethane:
It should be recognized that $\text{S}_\text{N}$ reactions may be reversible when both the leaving group $X$ and the entering group $Y$ are good entering and leaving groups, respectively. In such circumstances, the position of the equilibrium often can be changed by suitably adjusting the reaction conditions. Thus $48\%$ aqueous hydrogen bromide can convert alcohols to alkyl bromides (Equation 8-6, forward direction), whereas the reverse reaction (hydrolysis) is achieved by high water concentration:
The Nature of the Solvent
The rates of $\text{S}_\text{N}$ reactions are sensitive to the nature and composition of the solvent. This is easy to understand for $\text{S}_\text{N}1$ reactions because the ionizing power of a solvent is crucial to the ease of formation of ions $\overset{\oplus}{R}$ and $\overset{\ominus}{X}$ from $RX$.
Actually, two factors are relevant in regard to the ionizing ability of solvents. First, a high dielectric constant increases ionizing power by making it easier to separate ions. This is because the force between charged particles varies inversely with the dielectric constant of the medium.$^7$ Thus water, with a dielectric constant of 80, is 40 times more effective than a hydrocarbon with a dielectric constant of 2. Second, and usually more important, is the ability of the solvent to solvate the separated ions. Cations are solvated most effectively by compounds of elements in the first row of the periodic table that have unshared electron pairs. Examples are ammonia, water, alcohols, carboxylic acids, sulfur dioxide, and methylsulfinylmethane [dimethyl sulfoxide, $\left( CH_3 \right)_2 SO$]. Anoins are solvated most efficiently by solvents having hydrogen attached to a strongly elecronegative element $Y$ so the $H-Y$ bond is strongly polarized as $\overset{\delta \oplus}{H} \: \cdot \: \cdot \: \cdot \: \overset{\delta \ominus}{Y}$. Such solvents usually are called protic solvents. Protic solvents form hydrogen bonds to the leaving group, which assist ionization in much the same way that silver ion catalyzes ionization of alkyl halides (Section 8-7D). We can represent solvation by the following structural formulas, but it must be recognized that the number of solvent molecules involved in close interacitons can be as large as four or six, or as small as one:
The most effective ionizing solvents are those that effectively solvate both anions and cations. Water strikes an excellent compromise with regard to the structural features that make up ionizing power, that is, dielectric constant and solvating ability. From this, we expect tert-butyl chloride to ionize much more readily in water than in ether, because ethers can solvate only cations effectively, whereas water can solvate both anions and cations. The fact is that $\text{S}_\text{N}1$ ionizations usually are so difficult that $\text{S}_\text{N}1$ reactions seldom occur in solvents that cannot effectively solvate both anions and cations, even if the dielectric constant of the solvent is high. Solvation by hydrogen bonding is especially helpful in assisting ionization. Solvents that cannot provide such hydrogen bonding [e.g., $CH_3OCH_3$, $\left( CH_3 \right)_3 N$, $CH_3NO_2$, $CH_3CN$, $\left( CH_3 \right)_2 SO$] generally are poor for $\text{S}_\text{N}1$ reactions. These solvents are called aprotic solvents. An important exception is liquid sulfur dioxide, $SO_2$, which promotes $\text{S}_\text{N}1$ ionization by having a high dielectric constant and being able to solvate both anions and cations.
A list of protic and aprotic solvents, their dielectric constants, boiling points, and melting points is given in Table 8-5. This table will be useful in selecting solvents for nucleophilic substitution reactions.
With regard to $\text{S}_\text{N}2$ reactions, the solvent can affect profoundly the reactivity of a given nucleophile. Thus anions such as $Cl^\ominus$ and $CN^\ominus$, which are weakly nucleophilic in hydroxylic solvents and in poor ionizing solvents such as 2-propanone (acetone), become very significantly nucleophilic in polar aprotic sovlents such as $\left( CH_3 \right)_2 SO$. The reason is that for salts such as $NaCl$ and $NaCN$ the aprotic solvent preferentially solvates the cation, leaving the anion relatively bare. This dissociation of the anion from the cation together with its poor solvation makes the anion abnormally reactive as a nucleophile.
$^6$The Greek letters $\alpha$, $\beta$, $\gamma$ are used here not as nomenclature, but to designated the positions along a carbon chain from a functional group, $X$: $C_\omega \cdots C_\delta-C_\gamma-C_\beta-C_\alpha-X$ (also see Section 7-10).
$^7$Specifically, electrostatic force $= q_1 q_2/r_{12}^2 \epsilon$ in which $q_1$ and $q_2$ are the charges, $r_{12}$ is the distance between the charges, and $\epsilon$ is the dielectric constant of the medium; $\epsilon = 1$ for a vacuum.
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.08%3A_Structural_and_Solvent_Effects_in_%28S_N%29_Reactions.txt |
Kinetics and Mechanism
The conditions used for substitution reactions by the $\text{S}_\text{N}2$ mechanism very often lead to elimination. The reaction of 2-bromopropane with sodium ethoxide in ethanol provides a good example:
Elimination to give propene competes with substitution to give ethyl isopropyl ether. Furthermore, the rate of elimination, like the rate of substitution, is proportional to the concentrations of 2-bromopropane and ethoxide ion. Thus elimination here is a second-order reaction (it may be helpful to review Section 8-4 at this point):
rate of substitution $= k_S \left[ R Br \right] \left[ \overset{\ominus}{O} C_2H_5 \right]$
rate of elimination $= k_E \left[ R Br \right] \left[ \overset{\ominus}{O} C_2H_5 \right]$
As to the mechanism of this kind of elimination, the attacking base, $^\ominus OC_2H_5$, removes a proton from the $\beta$ carbon more or less simultaneously with the formation of the double bond and the loss of bromide ion from the neighboring carbon:
The abbreviation for this mechanism is $\text{E}2$, $\text{E}$ for elimination and $2$ for bimolecular, there being two reactants involved in the transition state.
Structural Effects
Structural influences on $\text{E}2$ reactions have been studied in some detail. Like the competing $\text{S}_\text{N}2$ process, a good leaving group is necessary and of these, the most commonly used are the halides, $Cl$, $Br$, and $I$; sulfonate esters, $RS \left( O_2 \right) O-$; and -onium ions such as ammonium, $\overset{\oplus}{N} R_4$, and sulfonium, $\overset{\oplus}{S} R_3$:
Rather strong bases generally are required to bring about the $\text{E}2$ reaction. The effectiveness of a series of bases generally parallels their base strengths, and the order $\overset{\ominus}{N} H_2$ $>$ $\overset{\ominus}{O} C_2H_5$ $>$ $\overset{\ominus}{O} H$ $>$ $\overset{\ominus}{O} CCH_3$ is observed for $\text{E}2$ reactions. This fact is important in planning practical syntheses, because the $\text{E}2$ reaction tends to predominate with strongly basic, slightly polarizable reagents such as amide ion, $\overset{\ominus}{N} H_2$, or ethoxide ion, $\overset{\ominus}{O} C_2H_5$. In contrast, $\text{S}_\text{N}2$ reactions tend to be favored with weakly basic nucleophiles such as iodide ion or ethanoate ion (unless dipolar aprotic solvents are used, which may markedly change the reactivity of anionic nucleophiles).
As for the alkyl group, there are two important structural effects to notice. First, at least one $C-H$ bond adjacent ($\beta$) to the leaving group is required. Second, the ease of $\text{E}2$ elimination follows the order tertiary $R$ $>$ secondary $R$ $>$ primary $R$. Unlike $\text{S}_\text{N}2$ reactions, which are not observed for tertiary alkyl compounds because of steric hindrance to the approach of the nucleophile to carbon, the related $\text{E}2$ reaction usually occurs readily with tertiary $RX$ compounds. The reason is that little or no steric hindrance is likely for the approach of a base to a hydrogen unless the base is exceptionally bulky:
The reactivity order also appears to correlate with the $C-X$ bond energy, inasmuch as the tertiary alkyl halides both are more reactive and have weaker carbon-halogen bonds than either primary or secondary halides (see Table 4-6). In fact, elimination of $HX$ from haloalkenes or haloarenes with relatively strong $C-X$ bonds, such as chloroethene or chlorobenzene, is much less facile than for haloalkanes. Nonetheless, elimination does occur under the right conditions and constitutes one of the most useful general methods for the synthesis of alkynes. For example,
The conditions and reagents used for $\text{E}2$ and $\text{S}_\text{N}2$ reactions are similar enough that it is difficult to have one occur without the other. However, $\text{E}2$ elimination is favored over $\text{S}_\text{N}2$ substitution by (a) strongly basic nucleophiles, (b) bulky nucleophiles, and (c) increasing alkyl substitution at the $\alpha$ carbon. It also is observed that increasing the reaction temperature generally leads to an increase in elimination at the expense of substitution. In fact, surprisingly good yields of alkene or alkyne can be obtained by adding a halogen compound directly to molten or very hot $KOH$ with no solvent present, whereupon the product is formed rapidly and distills immediately from the hot reaction mixture:
Orientation Effects in Elimination Reactions
With halides having unsymmetrical $R$ groups, such as 2-chloro-2-methylbutane, it is possible to form two or more different alkenes, the proportion depending on the relative rates at which the different $\beta$ hydrogens are removed. Most $\text{E}2$ eliminations of alkyl halides with common bases, such as $HO^\ominus$, $C_2H_5O^\ominus$, and $NH_2^\ominus$, tend to give mixtures of alkenes with a preference for the most stable alkene, which usually is the one with the fewest hydrogens or most alkyl groups attached to the carbons of the double bond. Thus
However, the precise distribution of alkenes formed is found to vary enough with the nature of the leaving group, or the base used, so either product will predominated with some combination of reagents or conditions. For example, a change in the base alone can be decisive:
Stereochemistry of $\text{E}2$ Reactions
The $\text{E}2$ reaction occurs most easily if the molecule undergoing reaction can assume a conformation, $2$, in which the leaving groups, $H$ and $X$, are trans to each other and the atoms $H-C_\beta-C_\alpha-X$ lie in one plane. Elimination then proceeds from opposite sides of the incipient double bond to give an alkene of structure $3$. We shall call this mode of elimination antarafacial to distinguish
it from another possible mode of elimination that is called suprafacial. (See Figure 8-6).$^8$
The transition state for conversion of $2$ to $3$ is particularly reasonable because it combines some of the geometry of both the reactants and the products and therefore gives the best overlap of the reacting orbitals necessary for the formation of the $\pi$ bond. This is shown more explicitly below.$^9$
As an illustration of the stereospecificity of eliminations, the meso compound $4$ gives the cis-alkene $5$, whereas the $D,L$ isomers $6$ give the trans-alkene $7$ with ethoxide. Both reactions clearly proceed by antarafacial elimination:
When antarafacial elimination is rendered difficult by the inability of the reacting groups to acquire the desired trans arrangement, then suprafacial elimination can occur, although less readily. An example is chlorocyclopentane, in which $H$ and $X$ cannot assume a trans configuration without very considerable strain but which does undergo suprafacial elimination at a reasonable rate:
$^9$Persuasive arguments have been made that many $\text{E}2$ reactions proceed by the sequence
If this is so, antarafacial elimination still is predicted to be favored.
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.09%3A_The_E2_Reaction.txt |
Scope and Mechanism
Many secondary and tertiary halides undergo $\text{E}1$ elimination in competition with the $\text{S}_\text{N}1$ reaction in neutral or acidic solutions. For example, when tert-butyl chloride solvolyzes in $80\%$ aqueous ethanol at $25^\text{o}$, it gives $83\%$ tert-butyl alcohol by substitution and $17\%$ 2-methylpropene by elimination:
The ratio of substitution and elimination remains constant throughout the reaction, which means that each process has the kinetic order with respect to the concentration of tert-butyl halide. The $\text{S}_\text{N}1$ and $\text{E}1$ reactions have a common rate-determining step, namely, slow ionization of the halide. The solvent then has the choice of attacking the intermediate carbocation at the positive carbon to effect substitution, or at a $\beta$ hydrogen to effect elimination:
Factors influencing the $\text{E}1$ reactions are expected to be similar to those for the $\text{S}_\text{N}1$ reactions. An ionizing solvent is necessary, and for easy reaction the $RX$ compound must have a good leaving group and form a relatively stable $R^\oplus$ cation. Therefore the $\text{E}1$ orders of reaction rates are $X = I$ $>$ $Br$ $>$ $Cl$ $>$ $F$ and tertiary $R$ $>$ secondary $R$ $>$ primary $R$.
With halides such as 2-chloro-2-methylbutane, which can give different alkenes depending on the direction of elimination, the $\text{E}1$ reaction is like the $\text{E}2$ reaction in tending to favor the most stable or highly substituted alkene:
Rearrangement of Carbon Cations
Another feature of $\text{E}1$ reactions (and also of $\text{S}_\text{N}1$ reactions) is the tendency of the initially formed carbocation to rearrange, especially if a more stable carbocation is formed thereby. For example, the very slow $\text{S}_\text{N}1$ solvolysis of neopentyl iodide in methanoic acid leads predominantly to 2-methyl-2-butene:
In this reaction, ionization results in migration of a methyl group with its bonding pair of electrons from the $\beta$ to the $\alpha$ carbon, thereby transforming an unstable primary carbocation to a relatively stable tertiary carbocation. Elimination of a proton completes the reaction.
Rearrangements involving shifts of hydrogen (as $H:^\ominus$) occur with comparable ease if a more stable carbocation can be formed thereby:
Rearrangements of carbocations are among the fastest organic reactions known and must be reckoned with as a possibility whenever carbocation intermediates are involved.
Acid-Catalyzed Elimination Reactions
Alcohols and ethers rarely undergo substitution or elimination unless strong acid is present. As we noted in Section 8-7D the acid is necessary to convert a relatively poor leaving group ($HO^\ominus$, $CH_3O^\ominus$) into a relatively good one ($H_2O$, $CH_3OH$). Thus the dehydration of alcohols to alkenes is an acid-catalyzed reaction requiring strong acids such as sulfuric or phosphoric acid:
These are synthetically useful reactions for the preparation of alkenes when the alkene is less available than the alcohol. They can occur by either the $\text{E}1$ or $\text{E}2$ mechanism depending on the alcohol, the acid catalyst, the solvent, and the temperature.
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."
8.11: Elimination Reactions
Generally, an alkyl derivative, under appropriate conditions, will eliminate \(HX\), where \(X\) is commonly a halide, ester, or -onium function, provided that there is a hydrogen located on the carbon adjacent to that bearing the \(X\) function:
An important feature of many elimination reactions is that they occur with the same combination of reagents that cause nucleophilic substitution. In fact, elimination and substitution often are competitive reactions. Therefore it should be no surprise that substitution and elimination have closely related mechanisms.
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.10%3A_The_E1_Reaction.txt |
Exercise 8-1 Write Lewis structures for each of the following reagents and classify them as either electrophilic, nucleophilic, both, or neither by evaluating whether they will react appreciably with hydroxide ion, $\ce{HO}^\ominus$, or hydronium ion, $\ce{H_3O}^\oplus$. Write equations for each of the reactions involved.
a. $\ce{NH_3}$
b. $\ce{NH_2^-}$
c. $\ce{Na}^\oplus$
d. $\ce{Cl}^\ominus$
e. $\ce{Cl_2}$
f. $\ce{CH_4}$
g. $\ce{CN}^\ominus$
h. $\ce{CH_3OH}$
i. $\ce{CH_3} \overset{\oplus}{\ce{O}} \ce{H_2}$
j. $\ce{BF_4^-}$
k. $\ce{HBr}$
l. $\ce{HC \equiv C} \colon^\ominus$
m. $\colon \ce{CH_2}$
n. $\ce{FSO_3H}$
o. $\ce{SO_3}$
Exercise 8-2 Identify the electrophile and the nucleophile in each of the following reactions:
a. $\ce{CH_3I} + \overset{\ominus}{\ce{O}} \ce{CH_3} \rightarrow \ce{CH_3-O-CH_3} + \ce{I}^\ominus$
b. $\ce{CH_2=CH_2} + \ce{Br_2} \rightarrow \overset{\oplus}{\ce{C}} \ce{H_2-CH_2-Br} + \ce{Br}^\ominus$
c. $\ce{CH_3NH_2} + \ce{CH_3I} \rightarrow \ce{(CH_3)_2NH} + \ce{HI}$
Exercise 8-3 Calculate $\Delta H^0$ for the polar reaction of one mole of bromomethane with water in accord with the equation $\ce{CH_3Br} + \ce{H_2O} \rightarrow \ce{CH_3OH} + \ce{HBr}$ (a) in the gas phase and (b) with all of the participants in dilute aqueous solution. For Part (a) you will need the bond energies of Table 4-3 and for Part (b) you will require the bond energies and the following $\Delta H^0$ values:
Exercise 8-4 Ethyl chloride $\left( 0.1 \: \text{M} \right)$ reacts with potassium iodide $\left( 0.1 \: \text{M} \right)$ in 2-propanone (acetone) solution at $60^\text{o}$ to form ethyl iodide and potassium chloride at a rate $\left( v \right)$ of $5.44 \times 10^{-7} \: \text{mol L}^{-1} \: \text{s}^{-1}$.
a. If the reaction proceeded by an $S_\text{N}2$ mechanism, what is the value of $k$ (in proper units) and what would be the rate of the reaction in moles per liter per sec at $0.01 \: \text{M}$ concentrations of both reactants? Show your method of calculation.
b. Suppose the rate were proportional to the square of the potassium iodide concentration and the first power of the ethyl chloride concentration. What would be the rate with $0.01 \: \text{M}$ reactants?
c. It one starts with solutions initially $0.1 \: \text{M}$ in both reactants, the rate of formation of ethyl iodide initially is $5.44 \times 10^{-7} \: \text{mol L}^{-1} \: \text{s}^{-1}$, but falls as the reaction proceeds and the reactants are used up. Plot the rate of formation of ethyl iodide against the concentration of ethyl chloride as the reaction proceeds (remembering that one molecule of ethyl chloride consumes one molecule of potassium iodide). Assume that the rate of reaction is proportional to the first power of the ethyl chloride concentration; and to (1) the zeroth power, (2) the first power, and (3) the second power of the potassium iodide concentration.
d. What kind of experimental data would one need to determine whether the rate of the reaction of ethyl chloride with potassium iodide is first order in each reactant or second order in ethyl chloride and zero order in potassium iodide?
Exercise 8-5 The rate of solvolysis of tert-butyl chloride in aqueous solution is unaffected by having sodium azide, $\overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{N}} \ce{=} \overset{\oplus}{\ce{N}} \ce{=} \overset{\oplus}{\ce{N}}$, in the solution, yet the products include both 2-azido-2-methylpropane, $1$, and tert-butyl alcohol:
Show how this information can be used to determine whether an $S_\text{N}1$ or an $S_\text{N}2$ mechanism occurs in the solvolysis of tert-butyl chloride in aqueous solution.
Exercise 8-6 What inference as to reaction mechanism might you make from the observation that the rate of hydrolysis of a certain alkyl chloride in aqueous 2-propanone is retarded by having a moderate concentration of lithium chloride in the solution?
Exercise 8-7 Equations 8-3 through 8-5 show how Kenyon and Phillips established that inversion of configuration accompanies what we now recognize to be $S_\text{N}2$ substitutions. For each reaction, we indicate whether $\ce{R-O}$ or $\ce{O-H}$ is broken by an appropriately placed vertical line. Explain how the sequence of steps shows that inversion occurs in the $S_\text{N}2$ reaction of Equation 8-5. The symbols $\left( + \right)$ or $\left( - \right)$ designate for each compounds the sign of the rotation $\alpha$ of the plane of polarized light that it produces.
$\tag{8-3}$
$\tag{8-4}$
$\tag{8-5}$
Exercise 8-8 Explain how, in the presence of bromide ion, either enantiomer of 2-bromobutane racemizes (Section 5-1B) in 2-propanone solution at a rate that is first order in $\ce{Br}^\ominus$ and first order in 2-bromobutane.
Exercise 8-9* When either of the enantiomers of 1-deuterio-1-bromobutane is heated with bromide ion in 2-propanone, it undergoes an $S_\text{N}2$ reaction that results in a slow loss of its optical activity. If radioactive bromide ion $\left( \ce{Br}^* \: ^\ominus \right)$ is present in the solution, radioactive 1-deuterio-1-bromobutane is formed by the same $S_\text{N}2$ mechanism in accord with the following equation:
$\ce{CH_3CH_2CH_2CHDBr} + \ce{Br}^* \: ^\ominus \overset{\text{2-propanone}}{\longrightarrow} \ce{CH_3CH_2CH_2CHDBr}^* + \ce{Br}^\ominus$
Within experimental error, the time required to lose $10\%$ of the optical activity is just equal to the time required to have $5\%$ of the $\ce{CH_3CH_2CH_2CHDBr}$ molecules converted to $\ce{CH_3CH_2CH_2CHDBr}^*$ with radioactive bromide ion. Explain what we can conclude from these results as to the degree to which the $S_\text{N}2$ reaction produces inversion of configuration of the primary carbon of 1-deuterio-1-bromobutane.
Exercise 8-10 What can be concluded about the mechanism of the solvolysis of 1-butyl derivatives in ethanoic acid from the projection formulas of the starting material and product of the following reaction?
Would you call this an $S_\text{N}1$ or $S_\text{N}2$ reaction?
Exercise 8-11 In the reaction of 1-phenylethanol with concentrated $\ce{HCl}$, 1-phenylethyl chloride is formed:
If the alcohol originally has the $D$ configuration, what configuration would the resulting chloride have if formed (a) by the $S_\text{N}2$ mechanism and (b) by the $S_\text{N}1$ mechanism?
Exercise 8-12 Predict which compound in each of the following groups reacts most rapidly with potassium iodide in 2-propanone as solvent by the $S_\text{N}2$ mechanism. Give your reasoning and name the substitution product by the IUPAC system.
a. $\ce{(CH_3)_3CCH_2Cl}$ $\ce{(CH_3)_3CCl}$ $\ce{CH_3CH_2CH_2CH_2Cl}$
b.
c.
Exercise 8-13 Which of the monobromine-substituted methylcyclohexanes would you judge to be the most reactive in (a) $S_\text{N}2$-type displacement and (b) $S_\text{N}1$-type displacement?
Exercise 8-14 Answer the question in the preceding exercise, but with the monobromine-substituted 1-methylcyclohexenes.
Exercise 8-15 Select the compounds from the following list that would be expected to hydrolyze more rapidly than phenylmethyl (benzyl) chloride by the $S_\text{N}1$ mechanism:
a. 2-phenylmethyl chloride
b. diphenylmethyl chloride
c. 1-phenylmethyl chloride
d. (chloromethyl)cyclohexane
e. 1-chloro-4-methylbenzene
Exercise 8-16 Explain the following observations:
a. The tertiary chloride, apocamphyl chloride, is unreactive in either $S_\text{N}1$ or $S_\text{N}2$ reactions. For example, no reaction occurs when its solution in aqueous ethanol containing $30\%$ potassium hydroxide is refluxed for 20 hours.
b. Chloromethyl alkyl (or aryl) ethers, $\ce{ROCH_2Cl}$, are very reactive in $S_\text{N}1$ solvolysis reactions. Compared to chloromethane, the rate of hydrolysis of chloromethyl phenyl ether is about $10^{14}$. Also, the rate of hydrolysis is retarded significantly by lithium chloride.
Exercise 8-17 Methyl sulfides are prepared conveniently by the $S_\text{N}2$ reaction of a $\ce{CH_3X}$ derivative with a sulfur nucleophile: $\ce{RS}^\ominus + \ce{CH_3X} \rightarrow \ce{RSCH_3} + \ce{X}^\ominus$. The rate of the reaction with a given $\ce{RS}^\ominus$. If you are uncertain of the p$K_a$ of the acids $\ce{HX}$, look them up in an appropriate reference, such as the CRC Handbook of Physics and Chemistry.
a. dimethyl sulfate, $\ce{(CH_3O)_2SO_2}$
b. methyl nitrate, $\ce{CH_3ONO_2}$
c. methyl cyanide (ethanenitrile), $\ce{CH_3CN}$
d. methyl fluoride, $\ce{CH_3F}$
e. methyl iodide, $\ce{CH_3I}$
f. methyl fluorosulfonate,
g. methyl acetate (ethanoate),
h. methanol, $\ce{CH_3OH}$
Exercise 8-18 Account for the following observations:
a. tert-Alkyl fluorides are unreactive in $S_\text{N}1$ solvolysis reactions unless a strong acid is present.
b. $D$-1-Phenylethyl chloride dissolved in aqueous 2-propanone containing mercuric chloride loses much of its optical activity before undergoing hydrolysis to give racemic 1-phenylethanol.
c. 1-Bromobutane can be prepared by heating 1-butanol with a mixture of sodium bromide and sulfuric acid. The reaction fails, however, if the sulfuric acid is omitted.
d. Benzenoxide (phenoxide) ion, $\ce{C_6H_5O}^\ominus$, is a better leaving group than ethoxide, $\ce{C_2H_5O}^\ominus$.
Exercise 8-19 Using the discussion in Section 8-7 of how the structure of $\ce{R}$ and $\ce{X}$ influence the $S_\text{N}$ reactivity of $\ce{RX}$, predict the favored course of each of the following reactions. Give your reasoning.
a.
b. $\ce{(CH_3)_3C-O-CH_3} + \ce{HI} \rightarrow$
c. $\ce{CF_3-O-CH_3} + \ce{HBr} \rightarrow$
d.
Exercise 8-20 Methyl ethers of the type $\ce{R-O-CH_3}$ cannot be prepared by the reaction of the alcohol $\ce{ROH}$ with $\ce{CH_3I}$, but if $\ce{Ag_2O}$ is present the following reaction occurs under mild conditions:
$2 \ce{R-OH} + \ce{Ag_2O} + 2 \ce{CH_3I} \rightarrow 2 \ce{ROCH_3} + 2 \ce{AgI} + \ce{H_2O}$
Explain how $\ce{Ag_2O}$ promotes this reaction.
Exercise 8-21 Explain each of the following observations:
a. Methyl sulfide $\ce{(CH_3)_2S}$ reacts with $\ce{C_6H_5COCH_2Cl}$ in benzene to give the sulfonium salt, $\ce{C_6H_5COCH_2} \overset{\oplus}{\ce{S}} \ce{(CH_3)_2Cl}^\ominus$, which precipitates as it is formed. Attempts to recrystallize the product from ethanol result in formation of methyl sulfide and $\ce{C_6H_5COCH_2Cl}$.
b. $S_\text{N}2$ displacements of alkyl chlorides by $\overset{\ominus}{\ce{O}} \ce{H}$ often are catalyzed by iodide ion, $\ce{RCl} + \colon \overset{\ominus}{\ce{O}} \ce{H} \overset{\ce{I}^\ominus}{\longrightarrow} \ce{ROH} + \ce{Cl}^\ominus$ and may result in a product with less than $100\%$ of inverted configuration at the carbon carrying the chlorine.
c.* Tris(trifluoromethyl)amine, $\ce{(CF_3)_3N}$, is completely nonnucleophilic, whereas trimethylamine is a good nucleophile.
Exercise 8-22 Classify the following solvents according to effectiveness for solvation of (i) cations and (ii) anions:
a. 2-propanone, $\ce{CH_3COCH_3}$
b. tetrachloromethane, $\ce{CCl_4}$
c. anhydrous hydrogen fluoride, $\ce{HF}$
d. trichloromethane, $\ce{CHCl_3}$
e. trimethylamine, $\ce{(CH_3)_3N}$
f. trimethylamine oxide, $\ce{(CH_3)_3} \overset{\oplus}{\ce{N}} - \overset{\ominus}{\ce{O}}$
Exercise 8-23* Would you expect the $S_\text{N}2$ reaction of sodium cyanide with methyl bromide to be faster, slower, or about the same with $\ce{(CH_3)_2S=O}$ or ethanol as solvent? Explain.
Exercise 8-24 An alternative mechanism for $E2$ elimination is the following:
$\ce{CH_3CH_2Cl} + \ce{OH}^\ominus \overset{\text{fast}}{\rightleftharpoons} \: ^\ominus \colon \ce{CH_2CH_2Cl} + \ce{H_2O} \overset{\text{slow}}{\longrightarrow} \ce{CH_2=CH_2} + \ce{Cl}^\ominus$
a. Would this mechanism lead to overall second-order kinetics with respect to the concentrations of $\ce{OH}^\ominus$ and ethyl chloride? Explain.
b. This mechanism as written has been excluded for several halides by carrying out the reaction in deuterated solvents such as $\ce{D_2O}$ and $\ce{C_2H_5OD}$. Explain how such experiments could be relevant to the reaction mechanism.
c. Does the test in Part b also rule out $\ce{CH_3CH_2Cl} + \overset{\ominus}{\ce{O}} \ce{H} \overset{\text{slow}}{\longrightarrow} \colon \overset{\ominus}{\ce{C}} \ce{H_2CH_2Cl} + \ce{H_2O} \overset{\text{fast}}{\longrightarrow} \ce{CH_2=CH_2} + \ce{Cl}^\ominus$? Explain.
Exercise 8-25 Write equations and mechanisms for all the products that might reasonably be expected from the reaction of 2-chlorobutane with a solution of potassium hydroxide in ethanol.
Exercise 8-26
a. Why is potassium tert-butoxide, $\overset{\oplus}{\ce{K}} \overset{\ominus}{\ce{O}} \ce{C(CH_3)_3}$, an excellent base for promoting elimination reactions of alkyl halides, whereas ethylamine, $\ce{CH_3CH_2NH_2}$, is relatively poor for the same purpose?
b. Potassium tert-butoxide is many powers of ten more effective a reagent for achieving $E2$ eliminations in methylsulfinylmethane (dimethyl sulfoxide) than in tert-butyl alcohol. Explain.
Exercise 8-27 Which one of the following groups of compounds would eliminate $\ce{HCl}$ most readily on reaction with potassium hydroxide? Draw the structure of the product and name it.
a. $\ce{(CH_3)_3CCl}$ $\ce{CH_3CH_2CH_2CH_2Cl}$ $\ce{CH_3CH(Cl)CH_2CH_3}$
b. $\ce{(CH_3)_3CCH_2Cl}$ $\ce{(CH_3)_2CHCH_2Cl}$
c.
Exercise 8-28 Which one of each of the following pairs of compounds would react most rapidly with potassium hydroxide in an $E2$-type elimination? Draw the structure of the product and name it.
a. $\ce{(CH_3)_3COH}$ $\ce{(CH_3)_3CCl}$
b.
c.
Exercise 8-29 Write all the possible staggered conformations for each of the isomers of 2,3-dibromobutane, $8$ and $9$:
Show the structures of the alkenes that could be formed from each by antarafacial $E2$ elimination of one mole of hydrogen bromide with hydroxide ion. Which alkene should more readily eliminate further to form 2-butyne? Explain.
Exercise 8-30 For the reaction of Equation 8-7, would you expect the ratio of tert-butyl alcohol to 2-methylpropene to change significantly with changes in the nature of the leaving group [i.e., $\ce{Cl}$, $\ce{Br}$, $\ce{I}$, or $\overset{\oplus}{\ce{S}} \ce{(CH_3)_2}$]? Give your reasoning.
Would you expect the same or different behavior as $\ce{X}$ is changed, if elimination were occurring by an $E2$ mechanism with the solvent acting as the base? Explain.
Exercise 8-31 The reaction of tert-butyl chloride with water is accelerated strongly by sodium hydroxide. How would the ratio of elimination to substitution products be affected thereby? Explain.
Exercise 8-32 Explain how $\ce{(CH_3)_2CDCHBrCH_3}$ (where $\ce{D}$ is the hydrogen isotope of mass 2) might be used to determine whether 2-methyl-2-butene is formed directly from the bromide in an $E1$ reaction, or by rearrangement and elimination as shown in the preceding equations.
Exercise 8-33 Predict the products of the following reactions:
a. $\ce{CH_3CH_2CBr(CH_3)CH_2CH_3} \underset{S_\text{N}1, \: E1}{\overset{\ce{H_2O}}{\longrightarrow}}$
b. $\ce{(CH_3)_3CCH(CH_3)Cl} \underset{S_\text{N}1, \: E1}{\overset{\ce{H_2O}}{\longrightarrow}}$
c.
Exercise 8-34 Write reaction sequences, using specific and appropriate compounds, that illustrate the following conversions:
a. alcohol $\rightarrow$ ether
b. alcohol $\rightarrow$ alkene
c. alcohol $\rightarrow$ alkyl chloride
d. alcohol $\rightarrow$ nitrile $\left( \ce{ROH} \rightarrow \ce{RCN} \right)$
e. alkyl chloride $\rightarrow$ sulfonium salt $\rightarrow$ alkene
Exercise 8-35 $\ce{S}$-Adenosylmethionine is a biologically important compound that reacts in the $S_\text{N}2$ manner with the amino group of phosphorylated 2-aminoethanol, $\ce{NH_2CH_2CH_2OPO_3H_2}$. Which carbon of $\ce{S}$-adenosylmethionine would be most likely to undergo an $S_\text{N}2$ reaction with an $\ce{RNH_2}$ compound? Give your reasoning and write the structures of the expected products.
Exercise 8-36 Nitriles, $\ce{RCN}$, can be prepared by $S_\text{N}2$ displacement of alkyl derivatives, $\ce{RX}$, by using sodium or potassium cyanide:
$\ce{RX} + \ce{NaCN} \rightarrow \ce{RCN} + \ce{NaX}$
a. Which of the following solvents would be most suitable for this reaction: water, 2-propanone, ethanol, benzene, $\ce{(CH_3)_2S=O}$, or pentane? Give reasons for your choice.
b. Which of the six isomeric monobromoderivatives of 1-methylcyclohexane would you expect to react most rapidly with sodium cyanide? Why?
c. If you wished to make 2-phenylethanenitrile, $\ce{C_6H_5CH_2CN}$, which of the following phenylmethyl compounds, $\ce{RCH_2X}$, would you select to convert to the nitrile? $\ce{X} = \ce{-F}$, $\ce{-OH}$, $\ce{-OCOCH_3}$, $\ce{-H}$, $\ce{-NH_2}$, $\ce{-O_3SCH_3}$, $\ce{-SO_3-CH_3}$. Why?
Exercise 8-37 Give a plausible explanation for each of the following observations:
a. Aqueous sodium chloride will not convert tert-butyl alcohol to tert-butyl chloride but concentrated hydrochloric acid will.
b. Better yields are obtained in the synthesis of isopropyl methyl ether starting with methyl iodide rather than sodium methoxide:
$\ce{CH_3I} + \ce{(CH_3)_2CHO}^\ominus \ce{Na}^\oplus \rightarrow \ce{(CH_3)_2CHOCH_3} + \ce{Na}^\oplus \ce{I}^\ominus$
$\ce{(CH_3)_2CHI} + \ce{CH_3O}^\ominus \ce{Na}^\oplus \rightarrow \ce{(CH_3)_2CHOCH_3} + \ce{Na}^\oplus \ce{I}^\ominus$
c. The following reaction proceeds only if an equivalent amount of silver fluoroborate, $\ce{Ag}^\oplus \ce{BF_4} \: ^\ominus$, is added to the reaction mixture:
d. 1-Bromo-2-butene reacts with water to give a mixture of 2-buten-1-ol, 3-buten-2-ol, and some 1,3-butadiene.
Exercise 8-38 Which compound in the following pairs would react faster under the reaction conditions? Draw the structures of the major products expected.
a. $\ce{C_6H_5CH_2CH_2Br}$ or $\ce{C_6H_5C(CH_3)(Br)CH_3}$ in ethanol-water solution.
b. Same as in Part a, but with potassium iodide in acetone.
c. Same as in Part a, but with potassium hydroxide in ethanol.
d. $\ce{CH_3CH_2} \overset{\oplus}{\ce{N}} \ce{(CH_3)_3} \: \overset{\ominus}{\ce{B}} \ce{F_4}$ or $\ce{CH_3CH_2} \overset{\oplus}{\ce{N}} \ce{(CH_3)_3} \: \overset{\ominus}{\ce{O}} \ce{CH_3}$ on heating in methanol solution.
Exercise 8-39 Show the products of the following reactions and indicate the stereochemistry where important.
a. trans-1-bromo-3-methylcyclopentane $\underset{\text{acetone}}{\overset{\ce{KI}}{\longrightarrow}}$
b. trans-1-chloro-2-methylcyclopentane $\underset{\textit{tert} \text{-butyl alcohol}}{\overset{^\oplus \ce{K} ^\ominus \ce{O} \ce{C(CH_3)_3}}{\longrightarrow}}$
c.
d. $D$-2-butanol $\overset{\text{potassium metal}}{\longrightarrow}$ potassium salt $\overset{L \text{-2-chlorobutane}}{\longrightarrow}$
e.
f. meso-1,2-dibromo-1,2-diphenylethane $\overset{^\oplus \ce{K} ^\ominus \ce{O} \ce{C(CH_3)_3}}{\longrightarrow}$
Exercise 8-40 The $S_\text{N}1$ reactions of many $\ce{RX}$ derivatives that form moderately stable carbocations are substantially retarded by adding $\ce{X}^\ominus$ ions. However, such retardation is diminished, at given $\ce{X}^\ominus$ concentrations, by adding another nucleophile such as $\ce{N_3} \: ^\ominus$. Explain.
Exercise 8-41 The reaction of 1-chlorobutane with sodium hydroxide to give 1-butanol is catalyzed by sodium iodide.
a. Work out the stereochemistry to be expected for both the catalyzed and the uncatalyzed reactions if (optically active) were used as the starting material. Show your reasoning.
b. Does retention of configuration, as the overall result of an $S_\text{N}2$, automatically preclude operation of the usual mechanism? Explain.
Exercise 8-42* Suppose a water solution was made up initially to be $0.01 \: \text{M}$ in methyl bromide and $1.0 \: \text{M}$ in sodium ethanoate at $50^\text{o}$. In water, the $S_\text{N}2$ rate constant for reaction of hydroxide ion with methyl bromide at $50^\text{o}$ is $30 \times 10^{-4} \: \text{L mol}^{-1} \: \text{sec}^{-1}$, whereas that of ethanoate ion at $50^\text{o}$ is $1.0 \times 10^{-4} \: \text{L mol}^{-1} \: \text{sec}^{-1}$. The ionization constant of ethanoic acid at $50^\text{o}$ is $1.8 \times 10^{-5}$. In the following, neglect the rates of the reactions of methyl bromide with water or ethanoic acid and any further reactions of ethanoate:
a. Calculate the hydroxide-ion concentration in the initial solution.
b. Calculate the initial rates of formation of methyl ethanoate and methanol.
c. Compute the concentrations of the organic products when the reaction is complete. Show your reasoning and justify any assumptions.
d. What kind of information would be needed to predict what products would be expected from a solution of methyl bromide and sodium hydroxide in methanol? Explain.
Exercise 8-43 Indicate how you would synthesize each of the following substances from the given organic starting materials and any other necessary organic or inorganic reagents. Specify reagents and conditions. (You may have to use reactions discussed in Chapter 4.)
a. $\ce{CH_2=CH-CH_2OCOCH_3}$ from propene
b. $\ce{CH_3-O-CH_2CH_3}$ from ethanol
c. $\ce{CH_3-O-C(CH_3)_3}$ from tert-butyl alcohol
d. cyclohexene from cyclohexane
Exercise 8-44 Which compound in each of the following pairs would you expect to react more readily with (A) potassium iodide in 2-propanone, (B) concentrated sodium hydroxide in ethanol, and (C) silver nitrate in aqueous ethanol? Write equations for all the reactions involved and give your reasoning with respect to the predicted orders of reactivity.
a. methyl chloride and isobutyl chloride with A, B, and C
b. methyl chloride and tert-butyl chloride with A, B, and C
c. tert-butyl chloride and 1-fluoro-2-chloro-2-methylpropane with B and C
d. 1-chloro-2-butene and 4-chloro-1-butene with A, B, and C
Exercise 8-45 Classify each of the following reactions from the standpoint of yield, side reactions, and reaction rate as good, fair, or bad synthetic procedures for preparation of the indicated products under the given conditions. Show your reasoning and designate any important side reactions.
a.
b.
c.
d.
e.
f.
g.
h.
Exercise 8-46 Consider each of the following compounds to be in unlabeled bottles in pairs as indicated. For each pair give a chemical test (preferably a test-tube reaction) that will distinguish between the two substances. Write equations for the reactions involved.
$\begin{array}{lll} & \textit{Bottle A} & \textit{Bottle B} \ \textbf{a.} & \ce{(CH_3)_3CCH_2Cl} & \ce{CH_3CH_2CH_2CH_2Cl} \ \textbf{b.} & \ce{BrCH=CHCH_2Cl} & \ce{ClCH=CHCH_2Br} \ \textbf{c.} & \ce{(CH_3)_3CCl} & \ce{(CH_3)_2CHCH_2Cl} \ \textbf{d.} & \ce{CH_3CH=CHCl} & \ce{CH_2=CHCH_2Cl} \ \textbf{e.} & \ce{(CH_3)_2C=CHCl} & \ce{CH_3CH_2CH=CHCl} \ \textbf{f.} & \ce{CH_3CH_2CH=CHCl} & \ce{CH_2=CHCH_2CH_2Cl} \end{array}$
Exercise 8-47 Why does the following $E1$ reaction give more of the least substituted alkene? (Use models.)
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)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.E%3A_Nucleophilic_Substitution_and_Elimination_Reactions_%28Exercises%29.txt |
Separations can be achieved by differences in physical properties, such as differences in boiling point, or by chemical means, wherein differences in physical properties are enhanced by chemical reactions. In this chapter we will consider some separations of compounds based on differences in physical properties. Chemical procedures will be discussed elsewhere in connection with the appropriate classes of compounds.
• 9.1: Prelude to Separation, Purification, and Identification
The separation of mixtures of compounds to give the pure components is of great practical importance in chemistry. Many synthetic reactions give mixtures of products and it is necessary for you to have a reasonably clear idea of how mixtures of compounds can be separated. Almost all compounds of biochemical interest occur naturally as components of very complex mixtures from which they can be separated only with considerable difficulty.
• 9.2: How do we know when an Organic Compounds is Pure?
The classical criteria for determining the purity of organic compounds are correct elemental compositions and sharpness of melting point or constancy of boiling point. Important though these analytical and physical criteria are, they can be misleading or even useless. For instance, the analytical criterion is of no help with possible mixtures of isomers because these mixtures have the same elemental composition.
• 9.3: Chromatographic Separation Procedures
Many separation methods are based on chromatography, that is, separation of the components of a mixture by differences in the way they become distributed (or partitioned) between two different phases. Liquid-solid chromatography originally was developed for the separation of colored substances, hence the name chromatography, which stems from the Greek word chroma meaning color.
• 9.4: Why Can't We See Molecules?
The most straightforward way to determine the structures of molecules would be to "see" how the nuclei are arranged and how the electrons are distributed. This is not possible with visible light, because the wavelengths of visible light are very much longer than the usual molecular dimensions. A beam of electrons can have the requisite short wavelengths, but small organic molecules are destroyed rapidly by irradiation with electrons of the proper wavelengths.
• 9.5: Atomic Energy States and Line Spectra
A spectroscopic change related to a change in energy associated with the absorption of a quantum of energy. Spectra are the result of searches for such absorptions over a range of wavelengths. If one determines and plots the degree of absorption by a monoatomic gas, a series of very sharp absorption bands or lines are observed. The lines are sharp because they correspond to specific changes in electronic configuration without complication from other possible energy changes.
• 9.6: Energy States of Molecules
The energy states and spectra of molecules are much more complex than those of isolated atoms. In addition to the energies associated with molecular electronic states, there is kinetic energy associated with vibrational and rotational motions.
• 9.7: Microwave (Rotational) Spectra
ecause electronic and vibrational energy levels are spaced much more widely, and because changes between them, are induced only by higher-energy radiation, microwave absorptions by gaseous substances can be characterized as essentially pure “rotational spectra.” It is possible to obtain rotational moments of inertia from microwave spectra, and from these moments to obtain bond angles and bond distances for simple molecules.
• 9.8: Infrared (Rovibrational) Spectroscopy
Infrared spectroscopy was the province of physicists and physical chemists until about 1940. At that time, the potential of infrared spectroscopy as an analytical tool began to be recognized by organic chemists. The change was due largely to the production of small, quite rugged infrared spectrophotometers and instruments of this kind now are virtually indispensable for chemical analysis. A brief description of the principles and practice of this spectroscopic method is the topic of this section
• 9.9: Raman Spectroscopy
Raman spectroscopy often is a highly useful adjunct to infrared spectroscopy. The experimental arrangement for Raman spectra is quite simple in principle. Monochromatic light, such as from an argon-gas laser, is passed through a sample, and the light scattered at right angles to the incident beam is analyzed by an optical spectrometer.
• 9.10: Electronic Spectra of Organic Molecules
Absorption of light in the ultraviolet and visible regions produces changes in the electronic energies of molecules associated with excitation of an electron from a stable to an unstable orbital. Because the energy required to excite the valence-shell electrons of molecules is comparable to the strengths of chemical bonds, absorption may lead to chemical reactions. We discussed this briefly in connection with photochemical halogenation of alkanes.
• 9.11: Nuclear Magnetic Resonance Spectroscopy
uclear magnetic resonance (NMR) spectroscopy is extremely useful for identification and analysis of organic compounds. The principle on which this form of spectroscopy is based is simple. The nuclei of many kinds of atoms act like tiny magnets and tend to become aligned in a magnetic field. In NMR spectroscopy, we measure the energy required to change the alignment of magnetic nuclei in a magnetic field.
• 9.12: Mass Spectroscopy
The usual application of mass spectroscopy to organic molecules involves bombardment with a beam of medium-energy electrons in high vacuum, and analysis of the charged particles and fragments so produced. Most mass spectrometers are set up to analyze positively charged fragments, although negative-ion mass spectrometry also is possible.
• 9.E: Separation, Purification, and Identification of Organic Compounds (Exercises)
These are the homework exercises to accompany Chapter 9 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."
09: Separation Purification and Identification of Organic Compounds
The separation of mixtures of compounds to give the pure components is of great practical importance in chemistry. Many synthetic reactions give mixtures of products and it is necessary for you to have a reasonably clear idea of how mixtures of compounds can be separated. Almost all compounds of biochemical interest occur naturally as components of very complex mixtures from which they can be separated only with considerable difficulty.
Separations can be achieved by differences in physical properties, such as differences in boiling point, or by chemical means, wherein differences in physical properties are enhanced by chemical reactions. In this chapter we will consider some separations of compounds based on differences in physical properties. Chemical procedures will be discussed elsewhere in connection with the appropriate classes of compounds.
Identification and structure determination are often closely allied to the problem of separation. Once a compound is separated, how do we determine whether it is identical to some previously known compound (identification) or, if that can't be done, how do we determine its chemical structure? The spectroscopic properties of molecules have proven to be extremely informative for both identification and structure determination and this chapter is mainly concerned with the application of spectroscopy for such purposes. We will give you now an overview of the spectroscopic properties of the major classes of organic compounds. In subsequent chapter, spectroscopic properties will be discussed in the context of the class of compounds under consideration.
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)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.01%3A_Prelude_to_Separation_Purification_and_Identification.txt |
The classical criteria for determining the purity of organic compounds are correct elemental compositions (Section 1-1A) and sharpness of melting point or constancy of boiling point. Important though these analytical and physical criteria are, they can be misleading or even useless. For instance, the analytical criterion is of no help with possible mixtures of isomers because these mixtures have the same elemental composition. The simple physical criteria are not applicable to substances that decompose when one attempts to determine the melting point or boiling point. Furthermore, boiling points are not very helpful for liquids that are mixtures of substances with nearly the same boiling point or are azeotropes.$^1$ Similar difficulties may be encountered with mixtures of solid substances that form mixed crystals or are eutectics.$^2$ Much sharper criteria for the purity of organic compounds now are provided through use of "super-separation" methods to see if any contaminants can be separated, or by spectroscopic techniques, as will be discussed later in this chapter. We begin here with a brief description of chromatographic methods of separation.
$^1$An azeotrope is a mixture of two or more substances that boils at a constant temperature, either higher or lower than any of its constituents. Thus an 8.5:1 mole mixture of ethanol and water boils like a pure substance, distilling at $78.2^\text{o}$, which is lower than the boiling point of ethanol ($78.5^\text{o}$) or of water ($100^\text{o}$). In contrast, a 1.35:1 mole mixture of methanoic (formic) acid and water boils at $107.1^\text{o}$, which is higher than the boiling points of either methanoic acid ($100.7^\text{o}$) or water ($100^\text{o}$).
$^2$When solid substances are mixed, the melting point of each normally is depressed. The eutectic mixture is the mixture of the solids with the lowest melting point.
9.03: Chromatographic Separation Procedures
Gas-Liquid Chromatography
Many separation methods are based on chromatography, that is, separation of the components of a mixture by differences in the way they become distributed (or partitioned) between two different phases. To illustrate with an extreme example, suppose we have a mixture of gaseous methane and ammonia and contact this mixture with water. Ammonia, being very soluble in water (~$90 \: \text{g}$ per $100 \: \text{g}$ water at $1 \: \text{atm}$ pressure), will mostly go into the water phase, whereas the methane, being almost insoluble (~$0.003 \: \text{g}$ per $100 \: \text{g}$ of water) will essentially remain entirely in the gas phase. Such a separation of methane and ammonia would be a one-stage partitioning between gas and liquid phases and, clearly, could be made much more efficient by contacting the gas layer repeatedly with fresh water. Carried through many separate operations, this partitioning procedure is, at best, a tedious process, especially if the compounds to be separated are similar in their distributions between the phases. However, partitioning can be achieved nearly automatically by using chromatographic columns, which permit a stationary phase to be contacted by a moving phase. To illustrate, suppose a sample of a gaseous mixture of ammonia and methane is injected into a long tube (column) filled with glass beads moistened with water (the stationary phase), and a slow stream of an inert carrier gas, such as nitrogen or helium, is passed in to push the other gases through. A multistage partitioning would occur as the ammonia dissolves in the water and the resulting gas stream encounters fresh water as it moves along the column. Carrier gas enriched with methane would emerge first and effluent gas containing ammonia would come out later. This is a crude description of the method of gas-liquid chromatography (abbreviated often as glc, GC, or called vapor-phase chromatography, vpc). This technique has become so efficient as to revolutionize the analysis and separation of almost any organic substance that has even a slight degree of volatility at some reasonably attainable temperature. The most modern glc equipment runs wholly under computer control, with preprogrammed temperatures and digital integration of the detector output. A wide variety of schemes is available for measuring the concentration of materials in the effluent carrier gas, and some of these are of such extraordinary sensitivity that only very small samples are necessary ($10^{-9} \: \text{g}$, or less).
In the usual glc procedure, a few microliters of an organic liquid to be analyzed are injected into a vaporizer and carried with a stream of gas (usually helium) into a long heated column that is packed with a porous solid (such as crushed firebrick) impregnated with a nonvolatile liquid. Gas-liquid partitioning occurs, and small differences between partitioning of the components can be magnified by the large number of repetitive partitions possible in a long column. Detection often is achieved simply by measuring changes in thermal conductivity of the effluent gases. A schematic diagram of the apparatus and a typical separation pattern are shown in Figures 9-1 and 9-2. The method is extraordinarily useful for detection of minute amounts of impurities provided these are separated from the main peak. Glc also can be used effectively to purify materials as well as to detect impurities. To do this, the sample size and the size of the apparatus may be increased, or an automatic system may be used wherein the products from many small-scale runs are combined.
Liquid-Solid Chromatography
Liquid-solid chromatography originally was developed for the separation of colored substances, hence the name chromatography, which stems from the Greek word chroma meaning color. In a typical examination, a colored substance suspected of containing colored impurities is dissolved in a suitable solvent and the solution allowed to percolate down through a column packed with a solid adsorbent, such as alumina or silica, as shown in Figure 9-3. The "chromatogram" then is "developed" by passing through a suitable solvent that washes the adsorbate down through the column. What one hopes for, but may not always find, is that the components of the mixture will be adsorbed unequally by the solid phase so distinct bands or zones of color appear. The bands at the top of the column contain the most strongly adsorbed components and the bands at the bottom the least strongly held components. The zones may be separated mechanically, or sufficient solvent can be added to wash, or elute, the zones of adsorbed materials sequentially from the column for further analysis.
Liquid-solid chromatography in the form just described was developed first by the Russian biochemist M. S. Tswett, about 1906. In recent years, many variations have been developed that provide greater convenience, better separating power, and wider applicability. In thin-layer chromatography, which is especially useful for rapid analyses, a solid adsorbent containing a suitable binder is spread evenly on a glass plate, a drop of solution to be analyzed is placed near one edge and the plate is placed in a container with the edge of the plate below the spot, dipping into an eluting solvent. The solvent ascends the plate and the materials in the spot move upward at different rates, as on a Tswett column. Various detecting means are used - simple visual observation for colored compounds, differential fluorescence under ultraviolet light, and spraying of the plate with substances that will give colored materials with the compounds present. In favorable cases, this form of liquid-solid chromatography can be carried out with submicrogram quantities of materials.
An extremely important improvement on the Tswett procedure is high-pressure solid-liquid chromatography (HPLC). Increasing the input pressure on the system to $20$-$70 \: \text{atm}$ improves the speed of separations by permitting the use of much smaller solid particles (with more surface area) than would be practical for gravity-flow Tswett columns. Automatic monitoring of the column effluent by ultraviolet spectroscopy (Section 9-9) or by changes in the refractive index usually provides an effective means of determining how the separation is proceeding. With such techniques chromatograms similar to Figure 9-2 are obtained. High-pressure liquid chromatography (hplc) has great advantages for analysis and separation of high-molecular-weight heat-sensitive compounds that are unsuitable for glc.
An ingenious variation of solid-liquid chromatography is to use a solid support to which material is attached that has a specific affinity for a particular substance to be separated. The technique is especially useful for separating enzymes, and the immobile phase can be constructed from compounds known to react with, or be complexed by, the enzyme. Some other forms of chromatography are discussed in Sections 25-4B and 25-7E.
Observation of a single peak in a given chromatographic procedure is evidence, albeit not definitive evidence, for purity. Contaminants with nearly the same properties may be very difficult to separate and, if knowing the degree of purity is highly important, one can run chromatograms with a variety of different adsorbents to see if each gives the same result. If they do, the presumption of purity improves, although it is desirable to determine whether the spectroscopic techniques to be described in the following section permit the same conclusion. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.02%3A_How_do_we_know_when_an_Organic_Compounds_is_Pure.txt |
The most straightforward way to determine the structures of molecules would be to "see" how the nuclei are arranged and how the electrons are distributed. This is not possible with visible light, because the wavelengths of visible light are very much longer than the usual molecular dimensions. A beam of electrons can have the requisite short wavelengths, but small organic molecules are destroyed rapidly by irradiation with electrons of the proper wavelengths. Nonetheless, electron microscopy is a valuable technique for the study of large molecules, such as DNA, which can be stained with heavy-metal atoms before viewing, or are themselves reasonably stable to an electron beam (Figures 9-4 and 9-5).
Virtually all parts of the spectrum of electromagnetic radiation, from x rays lo radio waves, have some practical application for the study of organic molecules. The use of x-ray diffraction for determination of the structures of molecules in crystals is of particular value, and in the past ten years this technique has become almost routine. Figure 9-6 shows the detailed arrangement of the carbons, hydrogens, and bromines in 1,8-bis(bromomethyl)naphthalene, $1$, as determined by x-ray diffraction. The apparatus and techniques used are highly complex and are not available yet to very many organic laboratories.$^3$
Other diffraction methods include electron diffraction, which may be used to’determine the structures of gases or of volatile liquid substances that cannot be obtained as crystals suitable for x-ray diffraction, and neutron diffraction, which has special application for crystals in which the exact location of hydrogens is desired. Hydrogen does not have sufficient scattering power for x rays to be located precisely by x-ray diffraction.
The diffraction methods can be used to determine complete structures of organic molecules, but they are not sufficiently routine to be utilized generally in practical organic laboratory work. For this reason, in the remainder of this chapter we will emphasize those forms of spectroscopy that are generally available for routine laboratory use. As will be seen, these methods are used by organic chemists in more or less empirical ways. In general, spectroscopic methods depend on some form of excitation of molecules by absorption of electromagnetic radiation and, as we have said, virtually all parts of the electromagnetic spectrum have utility in this regard. The commonly used span of the electromagnetic spectrum is shown in Figure 9-7 along with a comparison of the various units that are employed to express energy or wavelength.
The major kinds of spectroscopy used for structural analysis of organic compounds are listed in Table 9-1. The range of frequencies of the absorbed radiation are shown, as well as the effect produced by the radiation and specific kind of information that is utilized in structural analysis. After a brief account of the principles of spectroscopy, we will describe the methods that are of greatest utility to practical laboratory work. Nonetheless, it is very important to be aware of the other, less routine, methods that can be used to solve special problems, and some of these are discussed in this chapter and in Chapters 19 and 27.
You may have problems with the relationships among the variety of wavelength and frequency units commonly used in spectroscopy. The relationship between wavelength, frequency, and velocity should become clear to you by considering yourself standing on a pier watching ocean waves going by. Assuming the waves are uniformly spaced, there will be a uniform distance between the crests, which is $\lambda$, the wavelength. The wave crests will pass by at a certain number per minute, which is $\nu$, the frequency. The velocity, $c$, at which the crests move by you is related to $\lambda$ and $\nu$ by the relationship $c = \lambda \nu$.
This is not really very complicated and it applies equally well to water waves or electromagnetic radiation. What is almost needlessly complicated is the variety of units commonly used to express $\lambda$ and $\nu$ for electromagnetic radiation. One problem is tradition, the other is the desire to avoid very large or very small numbers. Thus, as Figure 9-7 shows, we may be interested in electromagnetic wavelengths that differ by as much as a factor of $10^{16}$. Because the velocity of electromagnetic radiation in a vacuum is constant at $3 \times 10^8 \: \text{m sec}^{-1}$, the frequencies will differ by the same factor.
Units commonly used for wavelength are meters ($\text{m}$), centimeters ($\text{cm}$), nanometers ($\text{nm}$), and microns ($\mu$). In the past, angstroms ($Å$) and millimicrons ($\text{m} \mu m$) also were used rather widely.
$1 \: \text{m} = 10^2 \: \text{cm} = 10^9 \: \text{nm} = 10^6 \: \mu m$
$10^{-2} \: \text{m} = 1 \: \text{cm} = 10^7 \: \text{nm} = 10^4 \: \mu m$
$10^{-6} \: \text{m} = 10^{-4} \: \text{cm} = 10^3 \: \text{nm} = 1 \: \mu m$
$10^{-9} \: \text{m} = 10^{-7} \: \text{cm} = 1 \: \text{nm} = 10^{-3} \: \mu m= 1 \: \text{m} \mu m= 10 \: \text{Å}$
Frequency units are in cycles per second (cps) or hertz ($\text{Hz}$), which are equivalent (radians per second are used widely by physicists).
Table 9-1: Principal Spectroscopic Techniques Currently in Use for Analysis of Molecular Structure
$1 \: \text{Hz} = 10^{-6} \: \text{MHz} \cong 3.3 \times 10^{-11} \: \text{cm}^{-1}$
$10^6 \: \text{Hz} = 1 \: \text{MHz} \cong 3.3 \times 10^{-5} \: \text{cm}^{-1}$
$3 \times 10^{10} \: \text{Hz} = 3 \times 10^4 \: \text{MHz} \cong 1 \: \text{cm}^{-1}$
$^3$A useful description of how molecular structures can be determined by "x-ray vision" is given in Chapter XI of Organic Molecules in Action by M. Goodman and F. Morehouse, Gordon and Breach, New York, 1973. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.04%3A_Why_Can%27t_We_See_Molecules.txt |
The energies of the hydrogenlike orbitals of various atoms were mentioned in Chapter 6 and, in particular, we showed a diagram of the most stable state $\left( 1s \right)^2 \left( 2s \right)^2 \left( 2p \right)^2$ of a carbon atom (Figure 6-4). Transfer of one of the $2p$ electrons to the $3s$ orbital requires excitation of the atom to a higher energy state and this can be achieved by absorption of electromagnetic radiation of the proper wavelength. The usual way that such excitation occurs is by absorption of a single quantum of radiant energy, and we can say that the absorption of this amount of energy $\Delta{E_{12}}$, corresponds to excitation of the atom from the ground state with energy $E_1$ to an excited state of configuration $\left( 1s \right)^2 \left( 2s \right)^2 \left( 2p \right)^1 \left( 3s \right)^1$ and energy $E_2$:
The difference in energy, $\Delta{E_{12}}$, is related directly to the frequency ($\nu$, $\text{sec}^{-1}$) or wavelength ($\lambda$, $\text{nm}$)$^4$ of the absorbed quantum of radiation by the equation
$\Delta{E_{12}} = h \nu = \dfrac{hc}{\lambda} \tag{9.1}$
in which $h$ is Planck's constant and $c$ is the velocity of light. The relationship $\Delta{E} = h\nu$ often is called the Bohr frequency condition.
For chemical reactions, we usually express energy changes in $\text{kcal mol}^{-1}$. For absorption of one quantum of radiation by each atom (or each molecule) in one mole, the energy change is related to $\lambda$ by
$\Delta{E_{12}} = \dfrac{28,600}{\lambda(nm)} \; kcal/mol \tag{9.2}$
As defined, $\Delta{E_{12}}$ corresponds to one einstein of radiation.
What we have developed here is the idea of a spectroscopic change being related to a change in energy associated with the absorption of a quantum of energy. Spectra are the result of searches for such absorptions over a range of wavelengths (or frequencies). If one determines and plots the degree of absorption by a monoatomic gas such as sodium vapor as a function of wavelength, a series of very sharp absorption bands or lines are observed, hence the name line spectra. The lines are sharp because they correspond to specific changes in electronic configuration without complication from other possible energy changes.
$^4$See Section 9-3 for discussion of the units of frequency and wavelength.
9.06: Energy States of Molecules
The energy states and spectra of molecules are much more complex than those of isolated atoms. In addition to the energies associated with molecular electronic states, there is kinetic energy associated with vibrational and rotational motions. The total energy, \(E\), of a molecule (apart from its translational\(^5\) and nuclear energy) can be expressed as the sum of three terms:
\[E = E_{electronic} + E_{vibrational} + E_{rotational}\]
Absorption of electromagnetic radiation by molecules occurs not only by electronic excitation of the type described for atoms, but also by changes in the vibrational and rotational energies.
Both rotations and vibrations of molecules are quantized. This means that only particular values of rotational angular momentum or vibrational energy are possible. We speak of these permitted values of the energies as the vibrational and rotational energy levels.
\(^5\)Translational energy is not very important in connection with spectroscopy and will not be considered here.
9.07: Microwave (Rotational) Spectra
Rotational energy levels normally are very closely spaced so low-energy radiation, such as is produced by radio transmitters operating in the microwave region, suffices to change molecular rotational energies. Because electronic and vibrational energy levels are spaced much more widely, and because changes between them, are induced only by higher-energy radiation, microwave absorptions by gaseous substances can be characterized as essentially pure “rotational spectra.” It is possible to obtain rotational moments of inertia from microwave spectra, and from these moments to obtain bond angles and bond distances for simple molecules.
An example of the use of microwave spectra is provided by Figure 9-8, which shows separate rotational absorptions observed for trans and gauche conformations of propyl iodide (cf. Section 5-2).
Although microwave spectroscopy, being confined to gases, is not a routine method in the organic laboratory, it is important to us here in setting the stage for the consideration of more complex absorptions that occur with infrared radiation. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.05%3A_Atomic_Energy_States_and_Line_Spectra.txt |
At the turn of the nineteenth century Sir William Herschel discovered invisible radiation beyond the red end of the visible region of the electromagnetic spectrum. This radiation appropriately is called infrared, meaning “beneath the red,” and it encompasses the wavelength region from $10^3 \: \text{nm}$ to $10^6 \: \text{nm}$. You probably are familiar with the common applications of infrared to radiant heating and photography. In addition to these uses, infrared spectroscopy has become the most widely used spectroscopic technique for investigating organic structures.
Infrared spectroscopy was the province of physicists and physical chemists until about 1940. At that time, the potential of infrared spectroscopy as an analytical tool began to be recognized by organic chemists. The change was due largely to the production of small, quite rugged infrared spectrophotometers and instruments of this kind now are virtually indispensable for chemical analysis. A brief description of the principles and practice of this spectroscopic method is the topic of this section.
General Considerations of Infrared Spectroscopy
Absorption of infrared radiation causes transitions between vibrational energy states of a molecule. A simple diatomic molecule, such as $\ce{H-Cl}$, has only one vibrational mode available to it, a stretching vibration somewhat like balls on the ends of a spring:
Molecules with three or more atoms can vibrate by stretching and also by bending of the chemical bonds, as indicated below for carbon dioxide:
The absorption frequencies in the infrared spectra of molecules correspond to changes in the stretching or bending vibrations or both. In general, a polyatomic molecule with $n$ atoms will have $3n - 6$ modes of vibration of which $n -1$ are stretching vibrations and $2n - 5$ are bending vibrations. There are circumstances, however, where fewer vibrational modes are possible. If the molecule is linear, like $CO_2$, then there are $3n-5$ possible vibrations, and some of these vibrations may be equivalent (degenerate vibrations in the language of spectroscopists). For example, $CO_2$ should have $3n - 5$ or $4$ vibrational modes, two of which are stretching and two of which are bending modes. However, the two bending modes are equivalent because the direction in which the molecule bends is immaterial; in-plane or out-of-plane bending are the same:
Diatomic molecules such as $HCl$ have one vibrational mode, but it is important to note that symmetrical diatomic molecules, such as $O_2$, $N_2$, $Cl_2$, $F_2$, and $H_2$, do not absorb in the infrared region of the spectrum. This is because absorption cannot occur if the vibration is electrically symmetrical. Fortunately, then, the infrared spectra can be recorded in air because the main components of air, $N_2$ and $O_2$, do not interfere.
In practice, infrared spectra can be obtained with gaseous, liquid, or solid samples. The sample containers (cells) and the optical parts of the instrument are made of rock salt ($NaCl$) or similar material that transmits infrared radiation (glass is opaque).
Typical infrared spectra are shown in Figure 9-9 for 2-propanone (acetone), $\ce{CH_3-CO-CH_3}$, and 2-butanone (methyl ethyl ketone), $\ce{CH_3-CO-CH_2-CH_3}$. In accord with current practice, the position of absorption (horizontal scale) is recorded in units of wave numbers ($\bar{\nu}$, $\text{cm}^{-1}$; see Section 9-3). The vertical scale measures the intensity of radiation transmitted through the sample. Zero transmission means complete absorption of radiation by the sample as at $1740 \: \text{cm}^{-1}$ in Figure 9-9. The other absorption bands in Figure 9-9 that correspond to excitation of stretching or bending vibrations are not as intense as the absorption at $1740 \: \text{cm}^{-1}$.
Characteristic Stretching Vibrations of Infrared Spectroscopy
What information can we derive about molecular structure from the vibrational bands of infrared spectra? Absorption of radiation in the range of $5000$-$1250 \: \text{cm}^{-1}$ is characteristic of the types of bonds present in the molecule, and corresponds for the most part to stretching vibrations. For example, we know that the $\ce{C-H}$ bonds of alkanes and alkyl groups have characteristic absorption bands around $2900 \: \text{cm}^{-1}$; an unidentified compound that shows absorption in this region will very likely have alkane-type $\ce{C-H}$ bonds.
More explicitly, the band observed for 2-propanone (Figure 9-9a) at $3050 \: \text{cm}^{-1}$ arises from absorption of infrared radiation, which causes transitions between the ground vibrational state (or lowest vibrational energy level) of a $\ce{C-H}$ bond and the first excited vibrational energy level for stretching of that $\ce{C-H}$ bond. The band at $1740 \: \text{cm}^{-1}$ corresponds to the infrared absorption that causes transitions between vibrational energy levels of the $\ce{C=O}$ bond. The reason that these are transitions from the vibrational ground state is because, at room temperature, by far the largest portion of the molecules are in this state (cf. Exercise 9-9).
Stretching frequencies characteristic of the most important types of bonds found in organic molecules are given in Table 9-2. You will notice that the absorption band for each bond type is described by its position within a more or less broad frequency range and by its shape (broad, sharp) and intensity (strong, medium, weak).
A qualitative discussion of the factors that determine infrared band position and band intensities follows. To a first approximation, a chemical bond resembles a mechanical spring that vibrates with a stretching frequency $\bar{\nu}$ ($\text{cm}^{-1}$),
$\bar{\nu} = \dfrac{1}{2\pi c} \sqrt{\dfrac{k}{\dfrac{m_1m_2}{m_1+m_2}}} \tag{9.3}$
in which $k$ is the force constant, and $m_1$ and $m_2$ are the masses of the individual atoms at each end of the bond. The force constant $k$ is a measure of the stiffness of the bond and usually is related to the bond strength. From Equation 9-3, we can see that the heavier the bonded atoms, the smaller will be the vibrational frequency of the bond provided $k$ remains essentially constant.$^6$ Thus if we increase $m_2$ while holding $k$ and $m_1$ constant we expect the frequency to decrease. This is just what occurs when we change the $C-H$ bond to a $C-D$ bond. We also see that the frequency decreases in the order $\ce{C-H}$ > $\ce{C-C}$ > $\ce{C-N}$ > $\ce{C-O}$, which also is in the order of increasing $m_2$, but here matters are more complicated because $k$ also changes.
Other things being equal, it requires more energy to stretch a bond than to bend it. Therefore the infrared bands arising from changes in the stretching vibrations are found at higher frequencies than are those arising from changes in the bending vibrations.
Another consequence of Equation 9-3 is that if $m_1$ and $m_2$ remain the same, the larger the value of $k$, the higher will be the vibrational frequency. Because $k$ is expected to run more or less parallel to the bond strength, and because multiple bonds are stronger than single bonds, the absorption frequencies of multiple bonds are higher than for single bonds. Examples are the absorption of $C=C$ at $2100 \: \text{cm}^{-1}$, $C=C$ at $1650 \: \text{cm}^{-1}$, and $C-C$ at $1000 \: \text{cm}^{-1}$.
Other effects besides mass and bond strength also affect infrared absorption frequencies. The structural environment of a bond is particularly important. Thus the absorption frequency of a $\ce{C-H}$ bond depends on whether it is an alkyl, alkenyl, alkynyl, or aryl $\ce{C-H}$ bond (see Table 9-2).
The intensity of an infrared absorption band arising from changes in the vibrational energy is related to the electrical symmetry of the bond. More symmetrical, less polarized bonds give weaker absorptions. In fact, if the bond is completely symmetrical, there is no infrared absorption. In contrast, unsymmetrical molecules in which the bonds are quite polarized, such as $C=O$ bonds, show strong infrared absorptions.
Notice in Figure 9-9 that infrared spectra of organic molecules do not show very sharp absorption lines. This is because changes in rotational energies can occur together with the vibrational changes. The reason can be seen more clearly in Figure 9-10, in which each vibrational level, such as $E_1$ and $E_2$, of a molecule has associated with it closely spaced rotational levels. Transitions between $E_1$ and $E_2$ also may involve changes in rotational levels. This gives a “band” of closely spaced lines for any given vibrational change. For complex molecules, particularly in the liquid state, the “rotational fine structure” of a given vibrational band usually cannot be resolved.
Table 9-2: Some Characteristic Infrared Absorption Frequencies
The Fingerprint Region
Infrared absorption bands between $1250 \: \text{cm}^{-1}$ and $675 \: \text{cm}^{-1}$ generally are associated with complex vibrational and rotational energy changes of the molecule as a whole and are quite characteristic of particular molecules. This part of the spectrum is often called the "fingerprint" region and is extremely useful for determining whether the samples are chemically identical. The spectra of 2-propanone and 2-butanone are seen to be very similar in the region $4000 \: \text{cm}^{-1}$ to $1250 \: \text{cm}^{-1}$ but quite different from $1250 \: \text{cm}^{-1}$ to $675 \: \text{cm}^{-1}$. The fingerprint region of the spectrum is individual enough so that if the infrared spectra of two samples are indistinguishable in the range of frequencies from $3600 \: \text{cm}^{-1}$ to $675 \: \text{cm}^{-1}$, it is highly probable that the two samples are of the same compound (or the same mixture of compounds).
Characteristic stretching and bending frequencies occur in the fingerprint region, but they are less useful for identifying functional groups, because they frequently overlap with other bands. This region is sufficiently complex that a complete analysis of the spectrum is seldom possible.
Alkanes and Cycloalkanes
The infrared spectra of the alkanes show clearly absorptions corresponding to the $C-H$ stretching frequencies at $2850 \: \text{cm}^{-1}$ to $3000 \: \text{cm}^{-1}$. The $C-C$ stretching absorptions have variable frequencies and are usually weak. Methyl ($CH_3-$) and methylene ($-CH_2-$) groups normally have characteristic $C-H$ bending vibrations at $1400 \: \text{cm}^{-1}$ to $1470 \: \text{cm}^{-1}$. Methyl groups also show a weaker band near $1380 \: \text{cm}^{-1}$. Two sample infrared spectra that illustrate these features are given in Figure 9-11.
The infrared spectra of the cycloalkanes are similar to those of the alkanes, except that when there are no alkyl substituents the characteristic
bending frequencies of methyl groups at $1380 \: \text{cm}^{-1}$ are absent. A moderately strong $CH_2$ "scissoring" frequency is observed between $1440 \: \text{cm}^{-1}$ and $1470 \: \text{cm}^{-1}$, the position depending somewhat on the size of the ring. These features of the infrared spectra of cycloalkanes are illustrated in Figure 9-12 using cyclooctane and methylcyclohexane as examples.
Applications of Infrared Spectroscopy to Structure Determination
Infrared spectra are very useful both for identification of specific organic compounds, and for determining types of compounds. For example, Figure 9-13 shows the infrared spectrum of a substance, $C_4H_6O_2$, for which we wish to determine the compound type and, if possible, the specific structure. The most informative infrared absorptions for determining the compound type are between $1500 \: \text{cm}^{-1}$ and $3600 \: \text{cm}^{-1}$. Two groups of bands in this region can be seen at about $1700 \: \text{cm}^{-1} \left( s \right)$ and $3000 \: \text{cm}^{-1} \left( s \right)$, where $\left( s \right)$ means strong; if we used $\left( m \right)$ it would mean medium, and $\left( w \right)$ would mean weak. From Table 9-2 we can see that these bands are indicative of $C=O$ ($1700 \: \text{cm}^{-1}$) and hydrogen-bonded $OH$ of carboxylic acids ($3000 \: \text{cm}^{-1}$). The presumption is that there is a $-CO_2H$ group in the molecule, and we can derive some reassurance from the fact that the molecular formula $C_4H_6O_2$ has enough oxygens to allow for this possibility.
Table 9-2 also shows that a $-CO_2H$ group should have a $C-O$ absorption band between $1350 \: \text{cm}^{-1}$ and $1400 \: \text{cm}^{-1}$ and $O-H$ absorption (bending frequency) between $1000 \: \text{cm}^{-1}$ and $1410 \: \text{cm}^{-1}$, and there is indeed a band of medium intensity at $1350 \: \text{cm}^{-1}$ and a strong band at $1240 \: \text{cm}^{-1}$. These absorptions, being in the fingerprint region, do not prove that the compound is a carboxylic acid; but if there were no absorptions in the $1000 \: \text{cm}^{-1}$ to $1400 \: \text{cm}^{-1}$ range, the presence of a $-CO_2H$ group would be highly questionable.
Tentatively, then, we may write a partial structure for $C_4H_6O_2$ as
Final identification may be possible by comparison with an authentic spectrum of cyclopropanecarboxylic acid, if it is available in one of the several standard compendia of infrared spectra. A total of about 150,000 infrared spectra are available for comparison purposes. You should check with the reference section of your library to see what atlases of spectral data are available to you.
The foregoing example illustrates the way structures can be determined from infrared spectral data. For many purposes, the infrared frequencies given in Table 9-2 are both approximate and incomplete. However, you could be easily frustrated in interpreting spectral data by being burdened with a very detailed table in which the unimportant is mixed with the important. The ability to use extensive tables effectively comes with experience. You should remember that tabulated infrared frequencies indicate only the range in which a given vibrational transition will fall. The exact value for a particular compound usually is meaningless because it will change depending on whether the spectrum is taken of the solid, liquid, or gaseous states, the solvent used, the concentration, and the temperature.
$^6$Remember that lower frequency means longer wavelengths and lower energy. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.08%3A_Infrared_%28Rovibrational%29_Spectroscopy.txt |
Raman spectroscopy often is a highly useful adjunct to infrared spectroscopy. The experimental arrangement for Raman spectra is quite simple in principle. Monochromatic light, such as from an argon-gas laser, is passed through a sample, and the light scattered at right angles to the incident beam is analyzed by an optical spectrometer.
Raman spectra arise as a result of light photons being “captured” momentarily by molecules in the sample and giving up (or gaining) small increments of energy through changes in the molecular vibrational and rotational energies before being emitted as scattered light. The changes in the vibrational and rotational energies result in changes in wavelength of the incident light. These changes are detected as lines falling both above and below the wavelength of the incident light. The line positions in Raman spectra always are reported in wave numbers. Highly efficient laser Raman spectrometers are commercially available.
Although changes in wavelength in Raman scattering correspond to absorption or emission of infrared radiation, infrared and Raman spectra are not always identical. Indeed, valuable information about molecular symmetry may be obtained by comparison of infrared and Raman spectra. When a bond is electrically symmetrical it does not absorb infrared radiation and, for this reason, symmetrical diatomic molecules such as $H_2$ and $O_2$, which are always electrically symmetrical, do not give infrared absorption spectra. However, excitation of symmetrical vibrations does occur in Raman scattering.$^7$ In a molecule such as ethene, $CH_2=CH_2$, the double-bond stretching vibration is symmetrical, because both ends of the molecule are the same. As a result, the double-bond stretching absorption is not observable in the infrared spectrum of ethene and is weak in all nearly symmetrically substituted ethenes. Nonetheless, this vibration appears strongly in the Raman spectrum of ethene and provides evidence for a symmetrical structure for ethene.
Absorption due to the stretching vibration of the double bond in tetrachloroethene ($1570 \: \text{cm}^{-1}$ is strong in the Raman and absent in the infrared, whereas that arising from the less symmetrical double bond of cyclohexene ($1658 \: \text{cm}^{-1}$) is weak in the infrared and slightly stronger in the Raman.
$^7$This is in accord with the spectroscopic "selection rules," derived from theoretical arguments, that predict which transitions between rotational and vibrational energy levels are "allowed" and which are "forbidden."
9.10: Electronic Spectra of Organic Molecules
General Characteristics
A year after Herschel discovered infrared radiation, Johann Ritter discovered radiation beyond the violet end of the visible spectrum. This radiation came to be known as ultraviolet and soon was recognized as being especially effective in causing chemical reactions. Absorption of light in the ultraviolet and visible regions produces changes in the electronic energies of molecules associated with excitation of an electron from a stable to an unstable orbital. Because the energy required to excite the valence-shell electrons of molecules is comparable to the strengths of chemical bonds, absorption may lead to chemical reactions. We discussed this briefly in Chapter 4 in connection with photochemical halogenation of alkanes; a more detailed account of photochemistry is given in Chapter 28.
The transition of an electron from the ground state, $E_1$, to an excited electronic state, $E_2$, is accompanied by vibrational and rotational changes in the molecule, as shown in Figure 9-17. In condense phase samples, it usually is not possible to resolve the resulting absorption bands well enough to see the fine structure due to vibration-rotation transitions. Consequently, absorptions due to electronic excitation are relatively broad.
The ultraviolet spectrum of 2-propanone (acetone) is shown in Figure 9-18. The weak absorption, which peaks (i.e., has $\lambda_\text{max}$ at $280 \: \text{nm}$, is the result of excitation of one of the unshared electrons on oxygen to a higher energy level. This is called an $n \rightarrow \pi^*$ (often $N \rightarrow A$) transition, in which $n$ denotes that the excited electron is one of the unshared $n$ electrons on oxygen and $\pi^*$ (pi star) denotes that the excited electron goes to a high-energy antibonding orbital of the carbon-oxygen double bond (cf. Sections 6-2 and 6-4C). The same kind of $n \rightarrow \pi^*$ transition occurs at about the same wavelength and intensity for many simple compounds of the type $\ce{R_2C=O}$ and $\ce{RCH=O}$, in which $R$ is an alkyl group. In a very schematic way, we can write
Effects of Conjugation on Electronic Spectra
The $\pi \rightarrow \pi^*$ transition for ethene has $\lambda_\text{max} = 175 \: \text{nm}$ and $\epsilon_\text{max} = 10,000$. It would be expected that an alkadiene would give an absorption spectrum similar to that of ethene but with a larger $\epsilon$, because there are more double bonds per mole to absorb radiation. This expectation is more or less realized for compounds such as 1,5-hexadiene and 1,3-dimethylenecyclobutane, which have isolated double bonds, but not for 1,3-butadiene or ethenylbenzene, which have conjugated double bonds (Section 3-3):
In general, conjugated systems of double bonds absorb radiation of longer wavelengths and with greater intensity than corresponding systems of isolated double bonds. This means that the difference in energy between the normal and excited states of conjugated systems is less than for isolated systems of double bonds. For 1,3-butadiene and 1,5-hexadiene we can calculate from Equation 9-2
that the transition energy is about $23 \: \text{kcal}$ less for the conjugated system. The ground state of 1,3-butadiene is stabilized by perhaps $3 \: \text{kcal}$ relative to a nonconjugated system of double bonds, which means that the excited state must be much more stabilized than this if the transition energy is to be $23 \: \text{kcal}$ less than for 1,5-hexadiene.
Why is the excited state of a conjugated system of double bonds stabilized more, relative to the ground state, than for a nonconjugated system? Resonance theory provides an explanation (see Section 6-5). Of the several conventional valence-bond structures that can be written for 1,3-butadiene, four of which are shown here, $2a$-$2d$, only structure $2a$ has a low enough energy to be dominant for the ground state of 1,3-butadiene:
Now, when the molecule is excited to the extent of $132 \: \text{kcal mol}^{-1}$ by $217 \: \text{nm}$ ultraviolet light, its energy is so large that pairing schemes such as $2b$, $2c$, and $2d$, which are too unfavorable to contribute very much to the ground state, can be very important for the excited state. Thus the stabilization energy of the excited state, which has a multiplicity of nearly equal-energy pairing schemes, is expected to be greater than that of the ground state with one dominant pairing scheme.
The more double bonds in the conjugated system, the smaller the energy difference between the normal and excited states. The diphenylpolyenes of formula $C_6H_5-(CH=CH)_n-C_6H_5$ absorb radiation at progressively longer wavelengths as $n$ is increased. This is apparent from the colors of the compounds, which range from colorless with $n = 1$, to orange with $n = 2-7$, to red with $n = 8$, as $\lambda_\text{max}$ goes from the ultraviolet into the visible region of the electromagnetic spectrum.
Similar effects are found with conjugated $\ce{C=O}$ and $\ce{C=N}$ double bonds. For example, the electronic spectra of 2-butanone and 3-buten-2-one are shown in Figure 9-20.
Conjugation also can influence infrared spectra. Transitions arising from $C=C$ and $C=O$ stretching vibrations generally are more intense and are shifted to slightly lower frequencies (longer wavelengths) for conjugated compounds relative to nonconjugated compounds. Thus the $C=C$ stretching of 1-butene occurs at $1650 \: \text{cm}^{-1}$, whereas that of 1,3-butadiene is observed at $1597 \: \text{cm}^{-1}$.
Alkanes and cycloalkanes have no low-energy electronic transitions comparable to conjugated systems or molecules with nonbonding electrons. Therefore alkanes and cycloalkanes show no absorption above $200 \: \text{nm}$ and are good solvents to use for electronic spectroscopy.
Applications of Electronic Spectroscopy
How do we use electronic spectroscopy in chemical analysis? The two principal applications are structure determinations and quantitative analysis. The position and intensity of an electronic absorption band provides information as to chemical structure. Such absorptions normally are not as useful as infrared absorptions because they do not give as detailed information. For our purposes here, the main points to remember are:
1. A weak absorption ($\epsilon = 10-100$) suggests an $n \rightarrow \pi^*$ transition of an isolated carbonyl group. If this absorption is found in the region $270$-$350 \: \text{nm}$ an aldehyde or ketone is probable.
2. Somewhat stronger absorptions ($\epsilon = 100-4,000$) between $200 \: \text{nm}$ and $260 \: \text{nm}$ may correspond to $\pi \rightarrow \sigma^*$ transitions.
3. Strong absorptions ($\epsilon = 10,000-20,000$) usually are characteristic of $\pi \rightarrow \pi^*$ transitions. If absorption occurs above $200 \: \text{nm}$, a conjugated system of multiple bonds is indicated. Each additional carbon-carbon double bond shifts $\lambda_\text{max}$ about $30 \: \text{nm}$ to longer wavelengths and enhances the intensity of absorption. Conjugation also shifts $\lambda_\text{max}$ of $n \rightarrow \pi^*$ transitions to longer wavelengths.
If we are dealing with compounds for which the wavelengths and the molar intensities of the absorption bands are known, then we can use the degree of absorption for quantitative analysis with the aid of the Beer-Lambert law (see Table 9-3 for definitions):
$A = \epsilon cl$
By measuring the absorbance $A$ of a sample of known $\epsilon$ in a cell of known path length $I$, the concentration $c$ may be determined. Because changes in absorbance reflect changes in concentration, it is possible to use absorbance measurements to follow rates of chemical reactions, to determine equilibrium constants (such as the dissociation constants of acids and bases), and to follow conformational changes in bio-organic molecules such as proteins and nucleic acids.
The capacity of electronic spectroscopy for performing qualitative and quantitative of the elements in chemical compounds has been done and applications have been made to elements of low atomic number, such as carbon and oxygen. Electronic spectra has been developed in the study of this elements and it’s compounds. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.09%3A_Raman_Spectroscopy.txt |
Nuclear magnetic resonance (NMR) spectroscopy is extremely useful for identification and analysis of organic compounds. The principle on which this form of spectroscopy is based is simple. The nuclei of many kinds of atoms act like tiny magnets and tend to become aligned in a magnetic field. In NMR spectroscopy, we measure the energy required to change the alignment of magnetic nuclei in a magnetic field. To illustrate the procedure with a simple example, consider the behavior of a proton $\left( ^1H \right)$ in a magnetic field. There are two possible alignments of this magnetic nucleus with respect to the direction of the applied field, as shown in Figure 9-21. The nuclear magnets can be aligned either with the field direction, or opposed to it. The two orientations are not equivalent, and energy is required to change the more stable alignment to the less stable alignment.
A schematic diagram of an NMR instrument is shown in Figure 9-22. When a substance such as ethanol, $CH_3-CH_2-OH$, the hydrogens of which have nuclei (protons) that are magnetic, is placed in the transmitter coil and the magnetic field is increased gradually, at certain field strengths radio-frequency energy is absorbed by the sample and the ammeter indicates an increase in the flow of current in the coil. The overall result is a spectrum such as the one shown in Figure 9-23. This spectrum is detailed enough to serve as a useful "fingerprint" for ethanol, and also is simple enough that we will be able to account for the origin of each line. It is the purpose of this section to explain how the complexities of spectra such as that of Figure 9-23 can be interpreted in terms of chemical structure.
For what kinds of substances can we expect nuclear magnetic resonance absorption to occur? Magnetic properties always are found with nuclei of odd-numbered masses, $^1H$, $^{13}C$, $^{15}N$, $^{17}O$, $^{19}F$, $^{31}P$, and so on, as well as for nuclei of even mass but odd atomic number, $^2H$, $^{10}B$, $^{14}N$, and so on.$^8$ Nuclei such as $^{12}C$, $^{16}O$, and $^{32}S$, which have even mass and atomic numbers, have no magnetic properties and do not give nuclear magnetic resonance signals. For various reasons, routine use of NMR spectra in organic chemistry is confined to $^1H$, $^{19}F$, $^{13}C$, and $^{31}P$. We shall be concerned in this chapter only with NMR spectra of hydrogen ($^1H$) and of carbon ($^{13}C$).
The kind of NMR spectroscopy we shall discuss here is limited in its applications because it can be carried out only with liquids or solutions. Fortunately, the allowable range of solvents is large, from hydrocarbons to concentrated sulfuric acid, and for most compounds it is possible to find a suitable solvent.
Nuclear magnetic resonance spectra may be so simple as to have only a single absorption peak, but they also can be much more complex than the spectrum of Figure 9-23. However, it is important to recognize that no matter how complex an NMR spectrum appears to be, in involves just three parameters: chemical shifts, spin-spin splittings, and kinetic (reaction-rate) processes. We shall have more to say about each of these later. First, let us try to establish the relationship of NMR spectroscopy to some of the other forms of spectroscopy we have already discussed in this chapter.
The Relation of NMR to Other Kinds of Spectroscopy
Nuclear magnetic resonance$^9$ spectroscopy involves transitions between possible energy levels of magnetic nuclei in an applied magnetic field (see Figure 9-21). The transition energies are related to the frequency of the absorbed radiation by the familiar equation $\Delta E - h \nu$. An important difference between nmr and other forms of spectroscopy is that $\Delta E$ is influenced by the strength of the applied field. This should not be surprising, because if we are to measure the energy of changing the direction of alignment of a magnetic nucleus in a magnetic field, then the stronger the field the more energy will be invloved.
Nuclear spin (symbolized as $I$) is a quantized property that correlates with nuclear magnetism such that when $I$ is zero the nucleus has no spin and no magnetic properties. Examples are $^{12}C$ and $^{16}O$. Several nuclei of particular interest to organic chemists - $^1H$, $^{13}C$, $^{15}$, $^{19}$, and $^{31}P$ - have spin of $\frac{1}{2}$. With $I = \frac{1}{2}$ there are only two magnetic energy states of the nucleus in a magnetic field. These states are designated with the spin quantum numbers $+ \frac{1}{2}$ and $- \frac{1}{2}$. The difference in energy between these states, $\Delta E$, is given by
$\Delta E = \gamma h H = h \nu$
or
$\nu = \gamma H$
in which $h$ is Planck's constant, $\nu$ is in hertz, $\gamma$ is a nuclear magnetic constant called the gyromagnetic ratio,$^{10}$, and $H$ is the magnetic field strength at the nucleus. In general, $H$ will not be exactly equal to $H_\text{o}$, the applied magnetic field and, as we will see, this difference leads to important chemical information. Each kind of nucleus ($^1H$, $^{13}C$, $^{15}N$, etc.) has its own $\gamma$ value and, consequently, will undergo transitions at different frequencies at any particular value of $H$. This should become clearer by study of Figure 9-24.
There are several modes of operation of an nmr spectrometer. First and most common, we hold $\nu$ constant and vary (or "sweep") $H_\text{o}$. Close to $\nu = \gamma H$, energy is absorbed by the nuclei and the current flow from the transmitter increases until $\nu$ is exactly equal to $\gamma H$. Further increase of $H_\text{o}$ makes $\nu < \gamma H_\text{o}$ and the current flow decreases. The form of the energy-absorption curve as a function of $H_\text{o}$ when $H_\text{o}$ is changed very slowly is shown in Figure 9-25a. The peak is centered on the point where $\nu = \gamma H$. When $H_\text{o}$ is changed more rapidly, transient effects are observed on the peak, which are a consequence of the fact that the nuclei do not revert instantly from the $- \frac{1}{2}$ to $+ \frac{1}{2}$ state. The resulting
phenomenon is called "ringing" and is shown in Figures 9-25b and 9-25c. Evidence of ringing also will be seen on peaks of Figure 9-23.
An alternative method of running an nmr spectrometer is to hold the magnetic field constant and to sweep the transmitter frequency through the resonances. This mode of operation is more like other forms of spectroscopy and gives the same line shapes as sweeping the field (Figure 9-25).
What energy is associated with a $^1H$ nmr transition? The magnitude of this energy may be calculated from the relationship between energy and wavelength (frequency) of the absorbed radiation (Section 9-4). That is,
$\Delta E = \frac{28,600}{\lambda} \text{kcal mol}^{-1}$ and $\lambda = \frac{c}{\nu}$
The frequency $\nu$ is the operating frequency of the spectrometer, which we will take as $60 \: \text{MHz}$ or $6 \times 10^7 \: \text{Hz}$ (cycles $\text{sec}^{-1}$), and the velocity of light is $3 \times 10^8 \: \text{m sec}^{-1}$. Hence
$\lambda = \frac{3 \times 10^8 \times 10^9 \left( \text{nm sec}^{-1} \right)}{6 \times 10^7 \left( \text{Hz} \right)} = 5 \times 10^9 \: \text{nm}$
and
$\Delta E = \frac{28,600}{5 \times 10^9} = 5.7 \times 10^{-6} \: \text{kcal mol}^{-1}$
This is a very small energy difference, which means that only very few more of the nuclei are in the more stable $+ \frac{1}{2}$ state than in the less stable $- \frac{1}{2}$ state. The equilibrium constant $K$ for $- \frac{1}{2} \rightleftharpoons + \frac{1}{2}$ calculated from Equation 4-2 for $25^\text{o}$ (298 \: \text{K}\)) and neglecting possible entropy effects is 1.000010!
The Chemical Shift
The plot of signal against magnetic field strength for ethanol in Figure 9-23 shows three principal groups of lines corresponding to the three varieties of hydrogen present: methyl ($CH_3$), methylene ($CH_3$), and hydroxyl ($OH$). Differences in the field strengths at which signals are obtained for nuclei of the same kind, such as protons, but located in different molecular environments, are called chemical shifts.
Another very important point to notice about Figure 9-23 is that the intensities of the three principal absorptions are in the ratio of 1:2:3, corresponding to the ratio of the number of each kind of proton ($OH$, $CH_2$, $CH_3$) producing the signal. In general, areas under the peaks of a spectrum such as in Figure 9-23 are proportional to the number of nuclei in the sample that give those peaks. The areas can be measured by electronic integration and the integral often is displayed on the chart, as it is in Figure 9-23, as a stepped line increasing from left to right. The height of each step corresponds to the relative number of nuclei of a particular kind. Unless special precautions are taken, integrals usually should not be considered accurate to better than about $5\%$.
Why do protons in different molecular environments absorb at different field strengths? The field strength $H$ at a particular nucleus is less than the strength of the external magnetic field $H_\text{o}$. This is because the valence electrons around a particular nucleus and around neighboring nuclei respond to the applied magnetic field so as to shield the nucleus from the applied field. The way this shielding occurs is as follows.
First, when an atom is placed in a magnetic field, its electrons are forced to undergo a rotation about the field axis, as shown in Figure 9-26. Second,
rotation of the electrons around the nucleus is a circulation of charge, and this creates a small magnetic field at the nucleus opposite in the direction to $H_\text{o}$. Third, the magnitude of this diamagnetic$^{11}$ effect is directly proportional to $H_\text{o}$ and can be quantified as $\sigma H_\text{o}$, in which $\sigma$ is the proportionality constant. It is important to recognize that $\sigma$ is not a nuclear property but depends on the chemical environment of the atom. Each chemically different proton will have a different value of $\sigma$ and hence a different chemical shift.
The actual field $H$ at the nucleus will be $H_\text{o} - \sigma H_\text{o}$. Because $\sigma$ acts to reduce the strength of the applied field at the nucleus, it is called the magnetic shielding parameter. The more shielding there is, the stronger the applied field must be to satisfy the resonance condition,
Common usage is: upfield, more shielding; downfield, less shielding; and you should remember that field-sweep spectra always are recorded with the field increasing from left to right.
Chemical Shift and Stereochemistry
The value of nmr spectroscopy in structure determination lies in the fact that chemically different nuclei absorb at different field strengths. In later sections we will be concerned with correlating the chemical shifts with structural features. However, before proceeding furher it is extremely important that you be able to identify the number and kind of nonequivalent protons in a given structure, and therefore the number of chemical shifts to expect. This number is not always self-evident, especially when subtle factors of stereochemistry intervene. For this reason, we suggest that you inspect structures $3$-$5$ to convince yourself that the protons labeled with different letter subscripts in any one molecule are indeed chemically different.
One way of checking whether two protons are in equivalent environments is to imagine that each is separately replaced with a different atom or group. If the product of replacing $H_\text{A}$ is identical with that obtained by replacing $H_\text{B}$, then $H_\text{A}$ and $H_\text{B}$ are chemically equivalent. If the two products are nonidentical, then $H_\text{A}$ and $H_\text{B}$ are nonequivalent. For example, replacement of $H_\text{A}$ or $H_\text{B}$ in $3$, $4$, and $5$ by an atom $X$ would give different products. Therefore, $H_\text{A}$ and $H_\text{B}$ are nonequivalent in $3$, $4$, and $5$.
Matters become more complicated with substances such as $6$ and $7$:
Notice that $6$ represents a chiral molecule and if $H_\text{A}$ and $H_\text{B}$ each are replaced with $X$ we get $8$ and $9$, which are diastereomers (see Section 5-5). You can verify this with molecular models if necessary. Diastereomers have different chemical and physical properties; therefore $H_\text{A}$ and $H_\text{B}$ in $6$ are nonequivalent. They often are called diastereotopic hydrogens.
What of the two methylene protons in ethanol, $7$, which we have labeled as $H_\text{A}$ $H_\text{A'}$? Are they identical? In a sense they are not identical because, if each were replaced by $X$, we would have a pair of enantiomers. Therefore, $H_\text{A}$ and $H_\text{A'}$ sometimes are called enantiotopic hydrogens.
But, you will recall that enantiomers are chemically indistinguishable unless they are in a chiral environment. Therefore we expect shifts of enantiotopic hydrogens to be identical, unless they are in a chiral environment. To summarize, enantiotopic protons normally will have the same chemical shifts, whereas diastereotopic protons normally will have different chemical shifts.
We so far have ignored the relationship of chemical shifts to conformational equilibria. Consider a specific example, 1,2-dibromoethane, for which there are three staggered conformations $10a$, $10b$, and $10c$:
Each of these conformations is expected to have its own nmr spectrum. The two gauche forms, $10a$ and $10b$, are enantiomers and their spectra should be identical. The hydrogens $H_\text{A}$ in $10a$ each are trans to the bromine on the adjacent carbon, while the $H_\text{B}$ hydrogens are cis to the same bromines (see Section 5-5A). Consequently the $H_\text{A}$ and $H_\text{B}$ hydrogens are nonequivalent and would be expected to have different chemical shifts. In contrast, all of the hydrogens of the anti conformer, $10c$, are equivalent and would have the same chemical shift. Therefore we would expect to observe three chemical shifts arising from $H_\text{A}$, $H_\text{B}$, and $H_\text{C}$ for a mixture of $10a$, $10b$, and $10c$. However, the actual spectrum of 1,2-dibromoethane shows only one sharp proton signal under ordinary conditions. The reason is that the magnetic nuclei can absorb the exciting radiation. The result is that we observe an average chemical shift, which reflects the relative shifts and populations of the three conformers present. If we can go to a sufficiently low temperature to make interconversion of the conformations slow (on the order of 10 times per second), then we will expect to see the three different chemical shifts $H_\text{A}$, $H_\text{B}$, and $H_\text{C}$ with intensities corresponding to the actual populations of the conformations at the sample temperature. This is one example of the effect of rate processes on nmr spectra. Other examples and a more detailed account of how to relate the appearance of the signal to the rates of the exchange processes are given in Section 27-2.
Chemical-Shift Standards and Units
Chemical shifts always are measured with reference to a standard. For protons or $^{13}C$ in organic molecules, the customary standard is a tetramethylsilane, $\left( CH_3 \right)_4 Si$, which gives strong, sharp nmr signals in regions where only a very few other kinds of protons or carbon nuclei absorb. Chemical shifts often are expressed in $\text{Hz}$ (cycles per second) relative to tetramethylsilane (TMS). These may seem odd units for magnetic field strength but because resonance occurs at $\nu = \gamma H$, either frequency units ($\text{Hz}$, radians $\text{sec}^{-1}$) or magnetic field units (gauss) are appropriate.
Ten years ago, most nmr spectrometers operated for protons with radio-frequency (rf) transmitters set at $60 \: \text{MHz}$ ($6 \times 10^7$ cycles per second) but there has been a proliferation of different proton-operating frequencies and now $30$, $60$, $90$, $100$, $220$, $270$, $300$ and $360 \: \text{MHz}$ machines are commercially available. The cost of these machines is roughly proportional to the square of the frequency, and one well may wonder why there is such an exotic variety available and what this has to do with the chemical shift. High operating frequencies are desirable because chemical shifts increase with spectrometer frequency, and this makes the spectra simpler to interpret. A 12-fold increase in operating frequency (as from $30 \: \text{MHz}$ to $360 \: \text{MHz}$) means a 12-fold increase in $H_\text{o}$ at the point of resonance (remember $\nu = \gamma H$) and this means also a 12-fold increase in $\sigma H_\text{o}$. Thus resonances that differ because they correspond to different $\sigma$ values will be twelve times farther apart at $360 \: \text{MHz}$ than at $30 \: \text{MHz}$. This can produce a dramatic simplification of spectra, as can be seen from Figure 9-27, which shows the effect of almost a factor of four in $\nu$ on the proton nmr spectrum of 2-methyl-2-butanol.$^{12}$
To reiterate, chemical shifts are strictly proportional to spectrometer frequency, thus lines $100 \: \text{Hz}$ apart at $60 \: \text{MHz}$ will be $167 \: \text{Hz}$ apart at $100 \: \text{MHz}$. This might seem to make comparisons of nmr spectra on different spectrometers hopelessly complex but, because of the proportionality of shifts to frequency (or field), if we divide the measured shifts in $\text{Hz}$ (relative to the same standard) for any spectrometer by the transmitter frequency in $\text{MHz}$, we get a set of frequency-independent shifts in parts per million ($\text{ppm}$, which are useful for all nmr spectrometers. Nmr shifts reported in $\text{ppm}$ relative to TMS as zero, as shown in Figure 9-23, are called $\delta$ (delta) values:
$\delta = \frac{\left( \text{chemical shift downfield in Hz relative to TMS} \right) \times 10^6}{\text{spectrometer frequency in Hz}}$
Thus, if at $60 \: \text{MHz}$ a proton signal comes $100 \: \text{Hz}$ downfield relative to tetramethylsilane, it can be designated as being $\left( +100 \: \text{Hz} \times 10^6 \right)/ 60 \times 10^6 \: \text{Hz} = +1.67 : \text{ppm}$ relative to tetramethylsilane. At $100 \: \text{MHz}$, the line then will be $\left( 1.67 \times 10^{-6} \right) \left(100 \times 10^6 \right) = 167 \: \text{Hz}$ downfield from tetramethylsilane. Typical proton chemical shifts relative to TMS are given in Table 9-4.$^{13}$ The values quoted for each type of proton may, in practice, show variations of $0.1$-$0.3 \: \text{ppm}$. This is not unreasonable, because the chemical shift of a given proton is expected to depend somewhat on the nature of the particular molecule involved, and also on the solvent, temperature, and concentration.
A positive $\delta$ value means a shift to lower field (or lower frequency) with respect to TMS, whereas a negative $\delta$ signifies a shift to higher field (or higher frequency). The $\delta$ convention is accepted widely, but you often find in the literature proton shifts with reference to TMS reported as "$\tau$ values." The $\tau$ scale has the TMS reference at $+10 \: \text{ppm}$, so most proton signals fall in the range of $\tau = 0$ to $\tau = +10$. A $\tau$ value can be converted to the appropriate $\delta$ value by subtracting it from 10. Life with nmr spectra would be simpler if the $\tau$ scale would just go away.
Correlations Between Structure and Chemical Shifts
Protonc chemical shifts are very valuable for the determination of structures, but to use the shifts in this way we must know something about the correlations that exist between chemical shift and structural environment of protons in organic compounds. The most important effects arise from differences in electronegativity, types of carbon bonding, hydrogen bonding, and chemical exchange.
Electronegativity
Consider first the chemical shifts of protons attached to an $sp^3$ carbon, .
The degree of shielding of the proton by the carbon valence electrons depends on the character of the substituent atoms and groups present, and particularly on their electron-attracting power, or electronegativity. For a grouping of the type , the shielding will be less as $\ce{X}$ is more electron withdrawing relative to hydrogen:
If $\ce{X}$ is electron-withdrawing, the proton is deshielded.
For example, the proton chemical shifts of the methyl halides (Table 9-4) show decreasing shielding, hence progressively low-field chemical shifts with increasing halogen electronegativity $\left( \ce{F} > \ce{Cl} > \ce{Br} > \ce{I} \right)$:
Table 9-4: Typical Proton Chemical-Shift Values $\left( \delta \right)$ in Dilute $\ce{CHCl_3}$ Solutions
When two electronegative groups, $\ce{X}$ and $\ce{Y}$, are bonded to the same carbon, as in $\ce{XCH_2Y}$, the protons are expected to be less shielded and come into resonance downfield of the methylenes of $\ce{XCH_2CH_2Y}$. There is an approximate relationship (see below) between the shifts of the $\ce{XCH_2Y}$ protons and the effective shielding constants $\left( \sigma \right)$ of $\ce{X}$ and $\ce{Y}$ known as Shoolery's rule.
$\delta = 0.23 + \sigma_x + \sigma_y \tag{9-4}$
Appropriate values of $\sigma$ for use with this equation are given in Table 9-4.
Effects of Carbon Bond Type
The shifts of the protons of alkanes and cycloalkanes fall in the range of $0.9$-$1.5 \: \text{ppm}$ with $\ce{C-H}$ protons coming at the low-field end of this range and $\ce{-CH_3}$ protons coming at the high-field end (see Table 9-4).
Alkenic hydrogens (vinyl hydrogens, ) normally are observed between $4.6$-$6.3 \: \text{ppm}$ toward lower fields than the shifts of protons in alkanes and cycloalkanes. This means that alkenic hydrogens in an organic compound can be easily distinguished from alkane hydrogens.
Aromatic protons, such as those in benzene, have shifts at still lower fields and commonly are observed at $7$-$8 \: \text{ppm}$. In contrast, alkynic protons of the type $\ce{-C \equiv CH}$ give resonances that are upfield of alkenic or aromatic protons and come at $2$-$3 \: \text{ppm}$. Another effect associated with multiple bonds is the large difference in shift between a $\ce{-CH(OCH_3)_2}$ proton, which normally comes at about $5.5 \: \text{ppm}$, and aldehyde protons, $\ce{-CH=O}$, which are much farter downfield at $9$-$11 \: \text{ppm}$.
Clearly, the shifts of a proton depend on whether the carbon forms single, double, or triple bonds. In a magnetic field, the circulation of electrons in the $\pi$ orbitals of multiple bonds induced by the field (Figure 9-26) generates diamagnetic shielding effects in some regions of the multiple bond and paramagnetic deshielding effects in other regions. Apparently, protons attached to double-bonded carbons are in the deshielding zones and thus are downfield while protons attached to triple-bonded carbons are in the shielding zones and are observed at rather high field.
Hydrogen Bonding
When a proton is directly bonded to a strongly electronegative atom such as oxygen or nitrogen its chemical shift is critically dependent on the nature of the solvent, temperature, concentration, and whether acidic or basic impurities are present. The usual variations in chemical shift for such protons are so large (up to $5 \: \text{ppm}$ for alcohols) that no very useful correlations exist.
Hydrogen bonding is the major reason for the variable chemical shifts of $\ce{OH}$ and $\ce{NH}$ protons. In general, hydrogen bonding results in deshielding, which causes the resonances to move downfield. The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in $\ce{CCl_4}$ (Figure 9-29). The hydroxyl resonance will be seen to move upfield by hydrogen bonding through equilibria such as
Section 9-10I. There is no significant change in the relative shifts of the $\ce{CH_2}$ and $\ce{CH_3}$ lines as the concentration is changed.
Chemical Exchange
Many $\ce{OH}$ and $\ce{NH}$ compounds are weak acids and weak bases and can undergo autoprotolysis, which means that a proton can be transferred from one molecule to another. Suppose we have a compound such as 2-aminoethanol, $\ce{H_2NCH_2CH_2OH}$. This substance normally would be expected to have an $\ce{NH_2}$ proton resonance at about $1 \: \text{ppm}$ and an $\ce{OH}$ proton resonance at about $3 \: \text{ppm}$. Autoprotolysis equilibria can exchange the protons between the molecules and also from one end to the other as shown below, even if the equilibria are not very favorable.
\begin{align} \ce{NH_2CH_2CH_2OH} &\overset{\rightarrow}{\longleftarrow} ^\oplus \ce{NH_3CH_2CH_2O}^\ominus \tag{9-5} \ 2 \ce{NH_2CH_2CH_2OH} &\overset{\rightarrow}{\longleftarrow} ^\oplus \ce{NH_3CH_2CH_2OH} + \ce{N_2CH_2CH_2O}^\ominus \tag{9-6} \end{align}
Such equilibria can be established very rapidly, especially if traces of a strong acid or a strong base are present. In such circumstances, a single average $\left( \ce{-NH_2}, \: \ce{-OH} \right)$ proton signal is observed, because the excitation of a given proton from its lower-energy magnetic state to its higher-energy magnetic state occurs while it is partly on oxygen and partly on nitrogen. This is the same kind of chemical shift averaging that occurs for rapidly equilibrating conformations (see Section 9-10C).
Application of Chemical Shifts to Structure Determination
To see how nmr and infrared spectra can be used together for structure determination we shall work through a representative example.
The objective is to assign a structure to the compound $\ce{C_4H_8O_3}$ whose nmr spectrum is shown in Figure 9-30 and whose infrared spectrum shows prominent bands at $2900 \: \text{cm}^{-1}$, $1750 \: \text{cm}^{-1}$, $1000 \: \text{cm}^{-1}$, and $1100 \: \text{cm}^{-1}$.
The infrared spectrum indicates $\left( 1750 \: \text{cm}^{-1} \right)$, $\ce{C-H} \: \left( 2900 \: \text{cm}^{-1} \right)$, and $\ce{C-O} \: \left( 1000 \: \text{cm}^{-1}, \: 1100 \: \text{cm}^{-1} \right)$. The position of the carbonyl band suggests that it is probably an ester, . A carboxylic acid is ruled out because there is no sign of an $\ce{O-H}$ stretch.
The nmr spectrum shows three kinds of signals corresponding to three kinds of protons. The integral shows these are in the ratio of 2:3:3. From this, we can conclude that they are two different kinds of $\ce{CH_3-}$ groups and a $\ce{-CH_2-}$ group.
The chemical shifts of the presumed $\ce{CH_3}$ groups are at $3.70 \: \text{ppm}$ and $3.35 \: \text{ppm}$. Because the compound contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$, the data of Table 9-4 suggest that these resonances arise from $\ce{OCH_3}$ groups. The low-field resonance is likely to be (we know from the infrared that there probably is an ester function), while the higher-field resonance is possibly an ether function, $\ce{-OCH_3}$. If you put all of this information together, you find that $\ce{CH_3OCH_2CO_2CH_3}$ is the only possible structure. To check whether the $\ce{CH_2}$ resonance at $3.9 \: \text{ppm}$ is consistent with the assigned structure we can calculate a shift value from Equation 9-4:
\begin{align} &\delta = 0.23 + \sigma_{OCH_3} + \sigma_{O=COCH_3} \ &\delta = 0.23 + 2.36 + 1.55 = 4.14 \: \text{ppm} \end{align}
The agreement between the calculated and observed shifts is not perfect, but is within the usual range of variation for Equation 9-4. We can be satisfied that the assigned structure is correct.
Spin-Spin Splitting - What We Observe
If you look at the nmr spectrum of ethanol, $\ce{CH_3CH_2OH}$, in Figure 9-23, you will see that the $\ce{CH_2}$ resonance is actually a group of four lines and the $\ce{CH_3}$ resonance is a group of three lines. This three-four line pattern for the grouping $\ce{CH_3CH_2X} \: \left( \ce{X} \neq \ce{H} \right)$ also is evident in the $220 \: \text{MHz}$ spectrum of 2-methyl-2-butanol (Figure 9-27) and in the $60 \: \text{MHz}$ spectrum of ethyl iodide (Figure 9-32).
Why do certain proton resonances appear as groups of equally spaced lines rather than single resonances? The facts are that nonequivalent protons on contiguous carbons , such as ethyl derivatives $\ce{CH_3CH_2X}$, interact magnetically to "split" each other's resonances. This multiplicity of lines produced by the mutual interaction of magnetic nuclei is called "spin-spin splitting", and while it complicates nmr spectra, it also provides valuable structural information, as we shall see.
An example of a complex proton spectrum is that of ethyl iodide (Figure 9-32). To a first approximation, the two main groups of lines appear as equally spaced sets of three and four lines, arising from what are called "first-order spin-spin interactions". Matters are further complicated by additional splitting of the "three-four" pattern of ethyl iodide, as also can be seen in Figure 9-32. This additional splitting is called "second-order" splitting.
When there are so many lines present, how do we know what we are dealing with? From where to we measure the chemical shift in a complex group of lines?
First, the chemical shift normally is at the center of the group of lines corresponding to first-order splitting. In ethyl iodide, the chemical shift of the methyl protons is in the center of the quartet:
Second, the chemical shift can be recognized by the fact that it is directly proportional to the transmitter frequency, $\nu$. If we double $\nu$, the chemical shifts double. In contrast, the first-order spin-spin splittings remain the same. By this we mean that the magnitude (in $\text{Hz}$) of the spacing between the lines of a split resonance is independent of the transmitter frequency, $\nu$. This spacing corresponds to what is called the spin-spin coupling constant, or simply the coupling constant, and is symbolized by $J$.
Third, the second-order splitting tends to disappear with increasing transmitter frequency. For ethyl iodide (Figure 9-32), the second-order splitting at $60 \: \text{MHz}$ is barely discernible at $100 \: \text{MHz}$ and disappears at $200 \: \text{MHz}$. This also can be seen to occur for the three-four splitting pattern of 2-methyl-2-butanol as a function of $\nu$ (Figure 9-27).
The next question is how can we understand and predict what spin-spin splitting patterns will be observed? And how do they give us structural information? The important point is that the multiplicity of lines for protons of a given chemical shift often is seen to be $\left( n + 1 \right)$, in which $n$ is the number of protons on the contiguous carbons. For example, the $\ce{CH_2}$ resonance of the ethyl group of ethyl iodide is a quartet of lines because of the spin-spin interaction with the neighboring three protons $\left( n = 3 \right)$ of the methyl group. Likewise, the $\ce{CH_3}$ group is a triplet of lines because of spin-spin interactions with the two protons $\left( n = 2 \right)$ of the methylene group.
The ratios of the line intensities in the spin-spin splitting patterns of Figure 9-33 usually follow simple rules. A doublet appears as two lines of equal intensity; a triplet as three lines in the ratio 1:2:1; a quartet as four lines in the ratio 1:3:3:1; a quintet as 1:4:6:4:1, and so on. The intensities follow the binomial coefficients for $\left( x + y \right)^n$, where $n$ is the number of protons in the splitting group. Thus when $n = 4$, we have $x^4 + 4 x^3y + 6 x^2 y^2 + 4 x y^3 + y^4$, or 1:4:6:4:1.
The spectrum of $\ce{(CH_3O)_2CHCH_3}$ (Figure 9-34) provides an excellent example of how nmr shows the presence of contiguous protons. The symmetrical doublet and 1:3:3:1 quartet are typical of the interaction between a single proton and an adjacent group of three, that is, . The methyl protons of the $\ce{(CH_3O)}$ groups are too far from the others to give demonstrable spin-spin splitting; thus they appear as a single six-proton resonance.
In general, the magnitude of the spin-spin splitting effect of one proton on another proton (or group of equivalent protons) depends on the number and kind of intervening chemical bonds and on the spatial relationships between the groups. For simple systems without double bonds and with normal bond angles, we usually find for nonequivalent protons (i.e., having different chemical shifts):
Where restricted rotation or double- and triple-bonded groups are involved, widely divergent splittings are observed. For double bonds, the two-bond couplings between two nonequivalent hydrogens located on one end are characteristically small, while the three-bond couplings in $\ce{-HC=CH}-$ are larger, especially for the trans configuration:
Coupling through four or more bonds is significant for compounds with double or triple bonds. Examples of these so-called long-range couplings and some other useful splitting values follow:
Finally, chemically equivalent protons do not split each other's resonances.
Proton-Proton Splittings and Conformational Analysis
A very important characteristic of three-bond proton-proton couplings, $\ce{H-C-C-H}$, is the way that they depend on the conformation at the $\ce{C-C}$ bond. Typical values for several particular conformations are
For protons in groups such as ethyl groups, in which rotation is rapid and the favored conformations are staggered (but none of the staggered conformations is preferred over the others), average proton-proton splittings are observed. The average for $\ce{CH_3CH_2}-$ splittings is about $7 \: \text{Hz}$, which corresponds to $\left[ 2 \times 4.5 \: \text{(gauche)} \: \div 12 \: \text{(trans)} \right]/3$.
Proton-Proton Splittings and Chemical Exchange
You may have wondered why the hydroxyl proton of ethanol produces a single resonance in the spectrum of Figure 9-23. It is quite reasonable to expect that the hydroxyl proton would be split by the neighboring methylene protons because they are only three bonds apart, however, this coupling will not be observed if the hydroxyl protons are exchanging rapidly between the ethanol molecules (Section 9-10E). When proton exchange is rapid, the spin interactions between the $\ce{-CH_2}-$ and $\ce{-OH}$ protons average to zero. At intermediate exchange rates, the coupling manifests itself through line broadening or by actually giving multiple lines. If you look at the several spectra of ethanol in Figure 9-29, you will notice how the shape of the $\ce{OH}$ resonance varies from a broad singlet to a distinct triplet.
Rapid chemical exchange of magnetic nuclei is not the only way that spin-coupling interactions can be averaged to zero. The same effect can be achieved by a technique known as double resonance. To understand how this is done, consider two coupled protons $\ce{H}_\text{A}$ and $\ce{H}_\text{B}$ having different chemical shifts. Suppose that $\ce{H}_\text{A}$ is selectively irradiated at its resonance frequency $\nu_\text{A}$ while at the same time we observe the resonance signal of $\ce{H}_\text{B}$. The coupling between $\ce{H}_\text{A}$ and $\ce{H}_\text{B}$ disappears, and $\ce{H}_\text{B}$ shows a single resonance. Why is this so? By irradiation of $\ce{H}_\text{A}$, the $\ce{H}_\text{A}$ nuclei are changed from the +1/2 state to -1/2 and back again sufficiently rapidly that the neighboring nucleus $\ce{H}_\text{B}$ effectively “sees” neither one state nor the other. The magnetic interaction between the states therefore averages to zero. This decoupling of magnetic nuclei by double resonance techniques is especially important in $\ce{^{13}C}$ NMR spectroscopy (Section 9-10L) but also is used to simplify proton spectra by selectively removing particular couplings.
Use of Nuclear Magnetic Resonance Spectroscopy in Organic Structural Analysis
The solution of a typical structural analysis problem by nmr methods utilizes at least four kinds of information obtained directly from the spectrum. They are: chemical shifts $\left( \delta \right)$, line intensities (signal areas), spin-spin splitting patterns (line mulitplicities), and coupling constants $\left( J \right)$. We already have shown how chemical shifts are used in the absence of spin-spin splitting. We now will illustrate how more complex spectra may be analyzed.
Figure 9-35 shows the proton nmr spectrum for a compound of formula $\ce{C_3H_6O}$. There are three principal groups of lines at $9.8$, $2.4$, and $1.0 \: \text{ppm}$. Look at the multiplicity of these groups before reading further.
There are several ways to approach a problem such as this, but probably the easiest is to start with the integral. The relative heights of the stepped integral for the principal groups of lines can be obtained by a pair of dividers, with a ruler, or with horizontal lines as in Figure 9-35. The integral suggests that one hydrogen is responsible for the resonance at $9.8 \: \text{ppm}$, two hydrogens at $2.4 \: \text{ppm}$, and three at $1.0 \: \text{ppm}$. Three hydrogens in a single group suggest a $\ce{CH_3}-$ group, and because there is a three-four splitting pattern, it is reasonable to postulate $\ce{CH_3-CH_2}-$. Subtracting $\ce{C_2H_5}$ from the given formula $\ce{C_3H_6O}$ leaves $\ce{CHO}$, which, with normal valences, has to be $\ce{-CH=O}$. The spectrum thus appears to be consistent with the structure $\ce{CH_3CH_2CH=O}$ (propanal) as judged from the molecular formula and the spin-spin splitting pattern, which indicates the $\ce{CH_3CH_2}-$ grouping. To be sure of the structure, we should check it against all of the available information. First, from the shifts (Table 9-4) we see that the single proton at $9.8 \: \text{ppm}$ fits almost perfectly for $\ce{RCHO}$, the two-proton $\ce{-CH_2C=O}$ resonance at $2.4 \: \text{ppm}$ is consistent with that reported for $\ce{-CH_2COR}$, while the three-proton line at $1.0 \: \text{ppm}$ checks with $0.9 \: \text{ppm}$ for $\ce{CH_3R}$.
What about the couplings? The three-four pattern has a spacing of slightly over $7 \: \text{Hz}$, which is just right for an ethyl group (compare Figures 9-23 and 9-32). The doubling up (almost obscured by second-order splitting) of the $\ce{-CH_2}-$ resonance and the splitting of the $\ce{-CH=O}$ resonance into a 1:2:1 triplet indicate about a $2$-$\text{Hz}$ coupling for the $\ce{-CH_2-CH=O}$ group. Three-bond couplings between $\ce{-CHO}$ and adjacent $\ce{-CH_2}-$ protons appear to be generally smaller than $\ce{-CH_2-CH_3}$ couplings.
We usually would not rely on nmr alone in a structure-analysis problem of this kind, but would seek clues or corroboration from the infrared, electronic, or other spectra, as well as chemical tests. In later chapters we will have many problems that will be facilitated by the use of both nmr and infrared spectra. A further worked example will illustrate the approach.
A compound has the composition $\ce{C_3H_3Br}$ and gives the infrared and nuclear magnetic resonance spectra shown in Figure 9-36. The problem is how to use this information to deduce the structure of the compound. The molecular formula tells us the number and kind of atoms and the number of multiple bonds or rings. The formulas of the corresponding $\ce{C_3}$ hydrocarbon without the bromine would be $\ce{C_3H_4}$, or four hydrogens less than the saturated alkane $\ce{C_3H_8}$. This means there must be two double bonds or the equivalent - one triple bond or one ring and one double bond.$^{14}$ Because from the formula we suspect unsaturation, we should check this out with the infrared spectrum. There is a band at $2120 \: \text{cm}^{-1}$, which is indicative of an unsymmetrically substituted $\ce{-C \equiv C}-$ group (Table 9-2). The strong, sharp band at $3300 \: \text{cm}^{-1}$ further tells us that the substance is a 1-alkyne $\ce{-C \equiv C-H}$.
The proton nmr spectrum shows that there are only two principal groups of lines - a two-proton doublet at $3.85 \: \text{ppm}$ and a one-proton triplet at $2.45 \: \text{ppm}$. The two-three splitting pattern combined with the 2:1 proton ratio suggests a $\ce{CH_2}$ group coupled with a $\ce{CH}$ group. The structure must be a 3-bromo-propyne, $\ce{BrCH_2C \equiv CH}$. To confirm the assignment, the chemical shifts should be checked (Table 9-4). The $\ce{\equiv C-H}$ at $2.45 \: \text{ppm}$ agrees well with the tabulated value of $2.5 \: \text{ppm}$. There is no tabulated data for $\ce{-C \equiv C-CH_2Br}$ but the observed shift at $3.85 \: \text{ppm}$ is at slightly lower fields than the tabulated $3.33 \: \text{ppm}$ for $\ce{-CH_2Br}$. This is expected because of the triple bond. The correlation of Equation 9-4 predicts a value of $4.0 \: \text{ppm}$.
Very often, a proton will be spin-coupled to two or more different protons, and the couplings are not necessarily the same. When this happens, the resulting spectrum can be quite complex, as our next example shows.
A compound $\ce{C_9H_{10}}$ gives the nmr spectrum of Figure 9-37. There are clearly four kinds of protons in the molecule at $\delta = 7.28 \: \text{ppm}$, $5.35 \: \text{ppm}$, $5.11 \: \text{ppm}$, and $1.81 \: \text{ppm}$. Although the integral is not shown, the main groups of lines have intensities from the low-field to high-field in the ratio of 5:1:1:3.
The five-proton signal at $7.28 \: \text{ppm}$ is typical of a phenyl group, $\ce{C_6H_5}$, and the one-proton signals at $5.35$ and $5.11 \: \text{ppm}$ are in the region for alkenic protons, . The three-proton signal at $1.81 \: \text{ppm}$ is typical of a methyl group on a carbon-carbon double bond, .
There are only three ways to put together a phenyl ring, , and two $\ce{HC=}$ protons such that they add up to $\ce{C_9H_{10}}$. They are
Coupling between A and B (designated by the constant $J_\text{AB}$) should give four lines, two for A and two for B, as shown in Figure 9-38. Because A and B also are coupled to the three hydrogens of the methyl group (C), each of the four lines corresponding to $J_\text{AB}$ will be further split (into 1:3:3:1 quartets). If $J_\text{AC} \neq J_\text{BC}$, then the spacing of the lines in the two sets of quartets will not be the same.
According to the foregoing analysis, the maximum number of lines observable for the A and B resonances is sixteen (8 for A and 8 for B). In fact, only eleven are visible (6 for A and 5 for B), which means that some of the sixteen possible lines must overlap. Without examining all possibilities, we can see that the actual situation can be reproduced if $J_\text{AB} \cong J_\text{BC} = 2J_\text{AC}$. The only structure that is consistent with $J_\text{AB} = 1.5 \: \text{Hz}$ is $13$, or 2-phenylpropene; the other possibilities are excluded because $J_\text{AB}$ should be about $10 \: \text{Hz}$ for $12$ and $16 \: \text{Hz}$ for $11$.
Chemical-Shift Effects on Spin-Spin Splitting
The simple $n + 1$ rule for predicting the multiplicity of spin-coupled proton signals often breaks down whenever the chemical-shift difference between the protons in different groups becomes comparable to coupling constants for magnetic interaction between the groups. Under these circumstances, you may expect to see more lines, or lines in different positions with different intensities, than predicted from the simple first-order treatment. One example is the effect of changing chemical shift on a two-proton spectrum with $J = 10 \: \text{Hz}$ (Figure 9-44).
We see in Figure 9-44 that even when the shift is 7.5 times larger than the coupling, the outside lines are weaker than the inside lines. This general kind of asymmetry of line intensities also is apparent in the spectrum of ethyl iodide (Figure 9-32), in which the lines of each group are more like 0.7:2.5:3.5:1.3 and 1.2:2.0:0.8, rather than the 1:3:3:1 and 1:2:1 ratios predicted from the first-order treatment. The asymmetry is such that two groups of lines that are connected by spin-spin splitting in effect "point" to one another - the lines on the "inside" of the pattern are stronger than predicted from the first-order treatment, whereas those on the "outside" are weaker. The effect can be put to practical use, as illustrated in the following exercise.
To explain the effect of chemical shifts on second-order splitting is beyond the scope of this book. In fact, we haven't really explained first-order splitting, although more on this topic will be found in Section 27-3. But regardless of how many lines appear in a complex nmr spectrum, they can be rationalized in terms of the chemical shifts, coupling constants, and exchange effects. Furthermore, the overall signal intensities remain proportional to the number of protons giving rise to the signals.
When there are many hydrogens and small chemical-shift differences, as in alkanes, the proton nmr spectra may have so many closely spaced resonance lines that they merge together to give a series of smooth, more-or-less featureless peaks. The proton spectrum of octane (Figure 9-46a) is an excellent example of this type of spectrum. Useful information often can be obtained from such spectra as to the ratio of $\ce{CH_3}$ : $\ce{CH_2}$ : $\ce{CH}$ by investigation of the integrals over the range of alkane proton absorptions. Figure 9-46 illustrates how this can be done for octane and 2,2,4-trimethylpentane.
Carbon-13 Nuclear Magnetic Resonance Spectroscopy
In recent years $\ce{^{13}C}$ nmr spectroscopy using $\ce{^{13}C}$ of natural abundance $\left( 1.1 \% \right)$ has become an important tool for organic structural analysis. That this did not happen sooner is because $\ce{^{13}C}$ has a much smaller magnetic moment than $\ce{^1H}$ and the small moment combined with the small natural abundance means that $\ce{^{13}C}$ is harder to detect in the nmr than $\ce{^1H}$ by a factor of 5700. This is a large difference and can be put in the proper context in the following way. Suppose two people are talking in a noisy room and one is trying to hear the other. The common request is "talk louder". If this is not possible then the request is "say it again" or "talk more slowly". Either of the latter requests amounts to an integration of signal versus noise and takes time. Improvement in signal-to-noise for a given communication is achieved as the square root of the time of communication. On the crucial time basis, $\ce{^{13}C}$ nmr signals require $\left( 5700 \right)^2 \cong 30,000,000$ times more time to get the same signal-to-noise ratio as in $\ce{^1H}$ nmr for the same number of nuclei per unit volume. This is a problem.
Electronic improvements and use of communication theory, with emphasis on the "say-it-again" technique, have provided the means for obtaining routine $\ce{^{13}C}$ spectra for even fairly dilute solutions of quite complex molecules.
Some of the same kinds of structural effects are important for $\ce{^{13}C}$ chemical shifts as for proton chemical shifts (Section 9-10E). For example, there is a similar parallel between $\ce{^{13}C}$ shift differences in compounds of the type $\ce{CH_3-CH_2-X}$ and electronegativity (Figure 9-47) as between the corresponding proton shifts and electronegativity (Figure 9-28). It is important to notice that $\ce{^{13}C}$ shifts in $\text{ppm}$ units are much larger than those of protons. This is because carbon uses $p$ orbitals in forming bonds, whereas hydrogen uses $s$ orbitals. We therefore will expect to find the the nuclei of other elements that use $p$ orbitals in bonding, such as $\ce{^{15}N}$, $\ce{^{19}F}$, and $\ce{^{31}P}$, also will have larger shifts than for protons, as indeed they do.
A structural application of $\ce{^{13}C}$ nmr, which shows its power in an area where $\ce{^1H}$ nmr is indecisive, is shown in Figure 9-48. Here, we see the high-field $\ce{^{13}C}$ resonances of a substance known variously as Coumadin, or the sodium salt of warfarin, $14$, which is used widely as a blood anticoagulant in the treatment of diseases such as phlebitis. It also has substantial utility as a rat poison because of its anticoagulant action.
There is no indication of any abnormality in the chemical shifts of carbons 11, 12, and 14 shown in Figure 9-48a. Furthermore, there is a downfield resonance $216.5 \: \text{ppm}$ from the carbons of TMS (not shown in Figure 9-48a) which is typical of a $\ce{C=O}$ carbon corresponding to C13. When $14$ is treated with acid, we expect the product (warfarin) of structure $15$ to be formed, which should have a $\ce{^{13}C}$ spectrum much like that shown in Figure 9-48a.
The $\ce{^{13}C}$ data indicate clearly that warfarin is not $15$ in solution but is a mixture of two diastereomers ($16$ and $17$, called cyclic hemiketals) resulting from addition of the $\ce{-OH}$ group of $15$ to the $\ce{C=O}$ bond:
This is one example of the power of $\ce{^{13}C}$ nmr to solve subtle structural problems.
$^8$Although the principal isotopes of $Cl$, $Br$, and $I$ have magnetic properties, because of the special character of all of these isotopes, they act in organic compounds as though they were nonmagnetic.
$^9$Resonance in the sense used here means that the radio-frequency absorption takes place at specified "resonance" frequencies. However, you will see that almost all of the forms of spectroscopy we discuss in this book involve "resonance" absorption in the same sense.
$^{10}$Here, $\gamma$ is in $\text{Hz}$ per gauss; physicists usually define $\gamma$ in radians per second per gauss.
$^{11}$From the Greek prefix dia meaning through, across. The opposite of diamagnetic is paramagnetic; para meaning alongside. We shall use this term later.
$^{12}$In addition to giving better separation of the lines and clearer spectra, going to higher fields also has the beneficial effect of increasing the proportions of the nuclei in the $+ \frac{1}{2}$ state, thereby giving more intense, easier-to-detect resonances.
$^{13}$Many other proton-shift values are available in NMR Spectra Catalog, Volume 1 and 2, Varian Associates, Palo Alto, Calif., 1962, 1963.
$^{14}$If two rings were present, this also would give four hydrogens less than the alkane. However, two rings are not possible with only three carbons. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.11%3A_Nuclear_Magnetic_Resonance_Spectroscopy.txt |
The usual application of mass spectroscopy to organic molecules involves bombardment with a beam of medium-energy electrons ($50$-$100 \: \text{eV}$ or $1150$-$2300 \: \text{kcal mol}^{-1}$) in high vacuum, and analysis of the charged particles and fragments so produced. Most mass spectrometers are set up to analyze positively charged fragments, although negative-ion mass spectrometry also is possible. The elements of a mass spectrometer are shown in Figure 9-51. The positive ions produced by electron impact are accelerated by the negatively charged accelerating plates and sweep down to the curve of the analyzer tube where they are sorted as to their mass-to-charge ($\left( m/e \right)$ ratio by the analyzing magnet. With good resolution, only the ions of a single mass number will pass through the slit and impinge on the collector, even when the mass numbers are in the neighborhood of several thousand. The populations of the whole range of mass numbers of interest can be determined by plotting the rate of ion collection as a function of the magnetic field of the analyzing magnet.
Mass spectra of 2-propanone, 2-butanone, and propanal are shown in Figure 9-52. Each peak represents ions of particular masses formed as the result of fragmentation of the molecule produced by electron impact into $\ce{CH_3^+}$, $\ce{CH_3CH_2^+}$, $\ce{CH_3CO}^\oplus$, and so on. The "cracking patterns" are, of course, functions of the energy of the bombarding electrons and serve as an extraordinarily individual fingerprint of the particular molecules. For instance, 2-propanone and propanal are isomers, yet their cracking patterns are strikingly different.
The peak that is highest in mass number is of considerable importance because it corresponds to the parent molecule $M$ minus one electron (designated as $M^+$) and provides a highly accurate method for measuring molecular weights. Incorrect molecular weights will be obtained if the positive ion, $M^+$, becomes fragmented before it reaches the collector, or if two fragments combine to give a fragment heavier than $M^+$. The peak of $M^+$ is especially weak with alcohols and branched-chain hydrocarbons, which readily undergo fragmentation by loss of water or side-chain groups. With such compounds the peak corresponding to $M^+$ may be $0.1\%$ or less of the highest peak in the spectrum, which is called the base peak and usually is assigned an arbitrary intensity of 100.
The pressure of the sample in the ion source of a mass spectrometer is usually about $10^{-5} \: \text{mm}$, and, under these conditions, buildup of fragments to give significant peaks with $m/e$ greater than $M^+$ is rare. One exception to this is the formation of $\left( M + 1 \right)^+$ peaks resulting from transfer of a hydrogen atom from $M$ to $M^+$. The relative intensities of such $\left( M + 1 \right)^+$ peaks are usually sensitive to the sample pressure and may be identified in this way.
With the molecular weight available from the $M^+$ peak with reasonable certainty, the next step is to determine the molecular formula. If the resolution of the instrument is sufficiently high, quite exact masses can be measured, which means that ions with $m/e$ values differing by one part in 50,000 can be distinguished. At this resolution it is possible to determine the elemental composition of each ion from its exact $m/e$ value.
Many mass spectrometers in routine use are incapable of resolving ions with $m/e$ values that differ by less than one mass unit. In this event, the determination of elemental composition can be determined by the method of isotope abundance. We will illustrate this with the following simple example.
The highest peaks corresponding to $M^+$ in the mass spectrum of an unknown sample have $m/e$ equal to 64 and 66 with relative intensities of 3:1. What is the elemental composition? The 3:1 abundance ratio is uniquely characteristic of the chlorine isotopes, $\ce{^{35}Cl}$:$\ce{^{37}Cl} =$ 3:1. The mass peaks at 64 and 66 are therefore both molecular ions; the 64 peak is of an ion containing $\ce{^{35}Cl}$ and the 66 peak is of an ion containing $\ce{^{37}Cl}$. The remaining atoms in the molecule must add up to $\left( 64 - 35 \right) = 29$, or $\left( 66 - 37 \right) = 29$ mass units. There are several possible combinations of $\ce{C}$, $\ce{H}$, $\ce{N}$, and $\ce{O}$ that give mass 29; they are $\ce{N_2H}$, $\ce{CHO}$, $\ce{CH_3N}$, and $\ce{C_2H_5}$.$^{15}$ Of these, the combination with $\ce{Cl}$ that makes the most chemical sense is $\ce{C_2H_5}$, and the formula of the molecule therefore is $\ce{C_2H_5Cl}$, chloroethane.
This example illustrates how $m/e$ values of ions that differ only in isotopic composition can be used to determine elemental compositions. The important isotopes for this purpose in addition to those of chlorine are the stable isotopes of natural abundance, $\ce{^{13}C}$ $\left( 1.1\% \right)$, $\ce{^{15}N}$ $\left( 0.37\% \right)$, $\ce{^{17}O}$ $\left( 0.04\% \right)$, and $\ce{^{18}O}$ $\left( 0.20\% \right)$. As a further example, suppose that we have isolated a hydrocarbon and have determine from its mass spectrum that $M^+ =$ 86 mass units. In the absence of any combination reactions there will be an $\left( M + 1 \right)^+$ ion corresponding to the same molecular ion but with one $\ce{^{13}C}$ ion in place of $\ce{^{12}C}$. The intensity ratio $\left( M + 1 \right)^+/M^+$ will depend on the number of carbon atoms present, because the more carbons there are the greater the probability will be that one of them is $\ce{^{13}C}$. The greater the probability, the larger the $\left( M + 1 \right)^+/M^+$ ratio. For $n$ carbons, we expect
$\frac{\text{abundance of} \left( M + 1 \right)^+}{\text{abundance of} \: M^+} = n \times \% \ce{^{13}C} \: \text{abundance}/100$
If the measured $\left( M + 1 \right)^+/M^+$ ratio is 6.6:100, then
\begin{align} \frac{6.6}{100} &= n \times 1.1/100 \ n &= 6 \end{align}
The only hydrocarbon formula with $M^+ =$ 86 and $n=6$ is $\ce{C_6H_{14}}$.
Nitrogen $as \(\ce{^{15}N}$) and oxygen (as $\ce{^{17}O}$) also contribute to $\left( M + 1 \right)^+$, if present, while $\ce{^{18}O}$ and two $\ce{^{13}C}$'s contribute to $\left( M + 2 \right)^+$. The calculated intensities of $\left( M + 1 \right)^+$ and $\left( M + 2 \right)^+$ relative to $M^+$ (as 100) are tabulated in Table 9-5 for elemental composition of ions up to $\ce{C_{20}}$. The table applies to fragment ions as well as molecular ions, but the intensity data from fragment ions very often is complicated by overlapping peaks.
Table 9-5: Isotopic Contributions for Carbon and other Elements to Intensities of $\left( M + 1 \right)^+$ and $\left( M + 2 \right)^+$ relative to $M^+$ (100)
The next step in the analysis of a mass spectrum is to see what clues as to structure can be obtained from the fragment ions. It would be a serious error to imagine that in mass spectra nothing is observed but simple nonspecific fragmentation of organic molecules on electron impact. Actually, even though electron impact produces highly unstable molecular ions, there is a strong tendency for breakdown to occur by reasonable chemical processes, and this may involve straightforward fragmentation or rearrangement of atoms from one part of the molecule to another.
In general, fragmentation occurs at the weakest bonds, and the most abundant fragments also are the most stable ones. For instance, hydrocarbons fragment preferentially at branch points, partly because the $\ce{C-C}$ bonds are weaker here than elsewhere along the chain, and partly because the ionic fragments are more stable. As an example, consider 2,2-dimethylbutane. There is no molecular ion evident in its mass spectrum because it cleaves so readily at the quaternary carbon to give the $m/e$ 57 peak corresponding to the most abundant fragment ion. This ion is presumably the tert-butyl cation and the alternate cleavage to the less stable ethyl cation with $m/e =$ 29 is much less significant.
An excellent example of a rearrangement with fragmentation is provided by the $M^+$ ion of ethyl butanoate, which breaks down to give ethene and the $M^+$ ion of an isomer of ethyl ethanoate called its "enol form".
An interesting and complex rearrangement occurs on electron impact with methylbenzene (toluene). An intense peak is observed having $m/e$ for $\ce{C_7H_7^+}$, but the ion involved appears to be a symmetrical $\ce{C_7H_7^+}$ ion, rather than a phenylmethyl cation. The evidence for this is that the fragmentation patterns found in the mass spectrometry of the ion itself are the same, no matter which of the monodeuteriomethylbenzenes is used as starting material. This rearrangement occurs because of the high delocalization energy of the symmetrical $ce{C_7H_7}^\oplus$ ion (usually called "tropylium cation") and because its charge is spread out more evenly over the carbons than would be the charge for the phenylmethyl cation (see Section 8-7B).
$^{15}$Tabulations of elemental compositions of $\ce{C}$, $\ce{H}$, $\ce{N}$ and $\ce{O}$ for mass values up to 250 are listed in many texts on mass spectrometry. Consult these tables to see all possible alternatives. See also J. H. Beynon, Mass Spectrometry and its Applications to Organic Chemistry, Elsevier Publishing Co., Amsterdam, 1960. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.12%3A_Mass_Spectroscopy.txt |
With few exceptions, the multitude of reactions discussed in introductory texts are classified as ionic reactions. By this we mean that nucleophilic and electrophilic sites in reacting molecules bond to each other. Furthermore, charged species such as carbocations, carbanions, conjugate acids and conjugate bases are often intermediates on the reaction path, the overall transformation taking place in two or more discrete steps. Ionic reactions normally occur in solution, and changes in solvents may have dramatic consequences. Free-radical addition is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non-radical, or between two radicals.
• 10.1: Prelude to Alkenes and Alkynes
Carbon-carbon double and triple bonds undergo a wide variety of addition reactions in which one of the multiple bonds is broken and two new bonds to carbon are formed. The importance of such reactions to synthetic organic chemistry is paramount. It is our intention in this and the following chapter to show the great diversity, utility, and specificity of addition reactions of alkenes and alkynes.
• 10.2: Physical and Spectroscopic Properties of Alkenes and Alkynes
In general, the physical properties of alkenes are similar to those of alkanes. Like the continuous-chain alkanes, the 1-alkenes form a homologous series of compounds that show regular changes in physical properties with increasing chain length. The boiling points, melting points, and densities of the simple alkynes are somewhat higher than those of the corresponding alkanes or alkenes, and these properties also show regular changes as the chain length is increased.
• 10.3: The Reactivity of Multiple Carbon-Carbon Bonds
In the early days of organic chemistry, alkenes were described as "unsaturated" because, in contrast to the "saturated" alkanes, they were found to react readily with substances such as halogens, hydrogen halides, oxidizing agents, and so on. Therefore, the "chemical affinity" of alkenes was regarded as unsatisfied or "unsaturated". One reason alkenes and alkynes react more readily than alkanes is because the carbon-carbon bonds of a multiple bond are individually weaker than normal carbon-carb
• 10.4: Electrophilic Additions to Alkenes
The reactions of alkanes discussed previously are homolytic processes, which means that the bonds are made and broken through radical or atomic intermediates. In contrast, the reactions of alkyl halides involve heterolytic bond cleavage and ionic reagents or products. Many important reactions of alkenes are heterolytic reactions because the electrons in the alkene double bonds are more exposed and accessible than the electrons in an alkane C−C bond
• 10.5: Orientation in Addition to Alkenes
Addition of an unsymmetrical substance such as HX to an unsymmetrical alkene can give two products. If the ratio of the products is determined by the ratio of their equilibrium constants, the reaction is under "equilibrium (or thermodynamic) control". and the reaction is reversible. When a reaction is carried out under conditions in which it is not reversible, the ratio is determined by the relative rates of formation of the various products and the reactions are under "kinetic control."
• 10.6: Electrophilic Addition Reactions of Alkynes
The alkynes behave in many ways as if they were doubly unsaturated alkenes. However, alkynes are substantially less reactive than corresponding alkenes toward many electrophiles. A simple but reasonable explanation is that the carbocation formed from the alkyne is less stable than that from the alkene because it cannot achieve the sp2 hybrid-orbital configuration expected to be the most stable arrangement for a carbocation.
• 10.7: Nucleophilic Addition Reactions
When a stepwise ionic addition reaction involves nucleophilic attack at carbon as a first step, it is described as a nucleophilic addition. Reactions of this type often are catalyzed by bases, which generate the required nucleophile.
• 10.8: Radical-Chain Addition Reactions to Alkenes
The addition of hydrogen bromide to unsymmetrical alkenes involves two reaction mechanisms, each giving a different product. These are the ionic addition and the radical-chain addition mechanisms.
• 10.9: Polymerization of Alkenes
One of the most important technical reactions of alkenes is their conversion to higher-molecular-weight compounds or polymers. A polymer is defined as a long-chain molecule with recurring structural units.
• 10.10: Alkylation of Alkenes
Addition of a saturated hydrocarbon to an alkene to yield a saturated hydrocarbon of higher molecular weight is known as alkylation. Such reactions are used by the petroleum industry to produce medium-molecular-weight hydrocarbons from smaller molecules. The key to the mechanism of hydrocarbon alkylation was provided by the discovery by P. D. Bartlett, in 1940, that a carbocation can react rapidly with a hydrocarbon having a tertiary hydrogen to yield a new carbocation and a new hydrocarbon
• 10.E: Alkenes and Alkynes I (Exercises)
These are the homework exercises to accompany Chapter 10 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."
10: Alkenes and Alkynes I - Ionic and Radical Addition Reactions
Carbon-carbon double and triple bonds undergo a wide variety of addition reactions in which one of the multiple bonds is broken and two new bonds to carbon are formed:
The importance of such reactions to synthetic organic chemistry is paramount. It is our intention in this and the following chapter to show the great diversity, utility, and specificity of addition reactions of alkenes and alkynes.
We will begin with a brief discussion of the physical and spectroscopic properties of alkenes and alkynes. But the major emphasis in the chapter is on two main types of reactions, ionic addition and radical-chain addition. For ionic additions we will make extensive use of the classification of reagents as electrophiles and nucleophiles, as described in Chapter 8.
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)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.01%3A_Prelude_to_Alkenes_and_Alkynes.txt |
Physical Properties
In general, the physical properties of alkenes are similar to those of alkanes. The data of Table 10-1 allow comparison of the boiling points, melting points, and densities of several alkenes with the corresponding alkanes that have the same carbon skeleton. Like the continuous-chain alkanes, the 1-alkenes form a homologous series of compounds that show regular changes in physical properties with increasing chain length. The boiling points, melting points, and densities of the simple alkynes (also included in Table 10-1) are somewhat higher than those of the corresponding alkanes or alkenes, and these properties also show regular changes as the chain length is increased.
Table 10-1: Comparison of Physical Properties of Alkanes, Alkenes, and Alkynes
Spectroscopic Properties of Alkenes
The infrared spectra of alkenes are sufficiently different from those of alkanes in most instances to make it possible to recognize when a double bond is present. For example, in the infrared spectrum of 1-butene (Figure 10-1) the absorption band near $1650 \: \text{cm}^{-1}$ is characteristic of the stretching vibration of the double bond. In general, the intensity and position of this band depends on the structure of the alkene; it varies with the degree of branching at the double bond, with the presence of a second unsaturated group in conjugation with the first (i.e., or ), and with the symmetry of the substitution of the double bond (see Section 9-7B). However, in many cases the absorption bands caused by the various modes of vibration of the alkenic $\ce{C-H}$ bonds frequently are more useful for detecting a double bond and identifying its type than is the absorption band caused by $\ce{C=C}$ stretch. With 1-butene, absorptions arising from the $\ce{C-H}$ vibrations of the terminal $\ce{=CH_2}$ group occur near $3100 \: \text{cm}^{-1}$, $1420 \: \text{cm}^{-1}$, and $915 \: \text{cm}^{-1}$, and those of the $\ce{-CH=}$ grouping near $3020 \: \text{cm}^{-1}$, $1420 \: \text{cm}^{-1}$, and $1000 \: \text{cm}^{-1}$. In general, absorption bands at these frequencies are from the grouping $\ce{-CH=CH_2}$. The bands near $1420 \: \text{cm}^{-1}$ are due to in-plane bending, whereas those at $915 \: \text{cm}^{-1}$ to $1000 \: \text{cm}^{-1}$ arise from out-of-plane bending. The other intense absorptions, near $1460 \: \text{cm}^{-1}$ and $3000 \: \text{cm}^{-1}$, are due to $\ce{C-H}$ vibrations of the $\ce{CH_3CH_2}-$ group (see Section 9-7D). These illustrate a further point - namely, the positions of the infrared absorptions of alkyl $\ce{C-H}$ bonds are significantly different from those of alkenic $\ce{C-H}$ bonds.
The double bonds of an alkene with no alkenic hydrogens are difficult to detect by infrared spectroscopy and in such cases Raman spectroscopy is helpful (see Section 9-8).
The infrared absorption of 1-butene that occurs at $1830 \: \text{cm}^{-1}$ (Figure 10-1) falls in the region where stretching vibrations of alkene bonds usually are not observed. However, this band actually arises from an overtone (harmonic) of the $\ce{=CH_2}$ out-of-plane bending at $915 \: \text{cm}^{-1}$. Such overtone absorptions come at exactly twice the frequency of the fundamental frequency, and whenever an absorption like this is observed that does not seem to fit with the normal fundamental vibrations, the possibility of its being an overtone should be checked.
With regard to electronic spectra, a $\pi$ electron of a simple alkene can be excited to a higher energy $\left( \pi^* \right)$ state by light of wavelength $180 \: \text{nm}$ to $100 \: \text{nm}$. However, many other substances absorb in this region of the spectrum, including air, the quartz sample cell, and most solvents that might be used to dissolve the sample, and as a result the spectra of simple alkenes are not obtained easily with the usual ultraviolet spectrometers. When the double bond is conjugated as in or , then the wavelengths of maximum absorption shift to longer wavelengths and such absorptions are determined more easily and accurately (also see Section 9-9B).
In proton nmr spectra, the chemical shifts of alkenic hydrogens are toward lower fields than those of alkane hydrogens and normally fall in the range of $4.6$-$5.3 \: \text{ppm}$ relative to TMS (see section 9-10E and Table 9-4). Spin-spin couplings of alkenic hydrogens are discussed in Section 9-10G and 9-10J.
Spectroscopic Properties of Alkynes
The infrared spectrum of a monosubstituted alkyne such as ethynylbenzene, $\ce{C_6H_5C \equiv CH}$ (Figure 10-4), has a strong band near $3300 \: \text{cm}^{-1}$, which is characteristic of the carbon-hydrogen stretching vibration in the grouping $\ce{\equiv C-H}$. At a lower frequency (longer wavelength) around $2100 \: \text{cm}^{-1}$, there is a band associated with the stretching vibration of the triple bond (also see Figure 9-36). Therefore the presence of the grouping $\ce{-C \equiv CH}$ in a molecule may be detected readily by infrared spectroscopy. However, the triple bond of a disubstituted alkyne, $\ce{R-C \equiv C-R}$, is detected less easily because there is no $\ce{\equiv C-H}$ absorption near $3300 \: \text{cm}^{-1}$, and furthermore the $\ce{C \equiv C}$ absorption sometimes is of such low intensity that it may be indiscernible. Raman spectroscopy (Section 9-8) or chemical methods must then be used to confirm the presence of a triple bond.
Alkynes, like alkenes, undergo electronic absorption strongly only at wavelengths in the relatively inaccessible region below $200 \: \text{nm}$. However, when a triple bond is conjugated with one or more unsaturated groups, radiation of longer wavelength is absorbed. To illustrate, ethyne absorbs at $150 \: \text{nm}$ and $173 \: \text{nm}$, whereas 1-buten-3-yne $\left( \ce{CH_2=CH-C \equiv CH} \right)$ absorbs at $219 \: \text{nm}$ and $227.5 \: \text{nm}$. The effects of such conjugation on spectra is discussed in more detail in Section 9-9B.
The proton nuclear magnetic resonance spectrum of ethynylbenzene is shown in Figure 10-5. The peaks near $435 \: \text{Hz}$ and $185 \: \text{Hz}$ correspond to resonances of the phenyl and $\ce{\equiv C-H}$ protons, respectively. The difference in chemical shift between the two types of protons is considerably larger than between alkenic and aromatic protons come into resonance at higher magnetic fields (i.e., they are subject to more diamagnetic shielding, Section 9-10E) than alkenic or aromatic protons. In fact, the $\ce{\equiv C-H}$ protons of alkynes have chemical shifts approaching those of alkyl protons. (Also see Figure 9-36.)
The mass spectra of alkenes and alkynes usually give distinct molecular ions; however, the fragmentation is often complex and not easily interpreted.
10.03: The Reactivity of Multiple Carbon-Carbon Bonds
In the early days of organic chemistry, alkenes were described as "unsaturated" because, in contrast to the "saturated" alkanes, they were found to react readily with substances such as halogens, hydrogen halides, oxidizing agents, and so on. Therefore, the "chemical affinity" of alkenes was regarded as unsatisfied or "unsaturated". (Also see Section 1-1I.)
One reason alkenes and alkynes react more readily than alkanes is because the carbon-carbon bonds of a multiple bond are individually weaker than normal carbon-carbon single bonds. Consider the bond energies involved. According to Table 4-3, the strengths of carbon-carbon single, double, and triple bonds are $83$, $146$, and $200 \: \text{kcal}$, respectively. From these values we can calculate that cleavage of one-half of a carbon-carbon double bond should require $63 \: \text{kcal}$ and cleavage of one-third of a carbon-carbon triple bond should require $54 \: \text{kcal}$:
As a result, addition reactions to multiple bonds are expected to be about $20$-$30 \: \text{kcal}$ more exothermic than the corresponding cleavage reactions of carbon-carbon single bonds, as estimated here for reaction with bromine:
The substantial difference in the heats of reaction of ethane, ethene, and ethyne with bromine is reflected in a very important practical consideration in handling ethyne (acetylene), namely its thermodynamic stability relative to solid carbon and hydrogen gas. Unlike ethane, both ethene and ethyne can be shown from bond energies to be unstable with respect to formation of solid carbon and gaseous hydrogen:
Although this does not seem to offer particular problems with ethene, an explosive decomposition of ethyne to carbon and hydrogen may occur if the gas is compressed to $10$-$20 \: \text{kg cm}^{-2}$. Even liquid ethyne (bp \sim 83^\text{o}\)) must be handled with care. Ethyne is not used commercially under pressure unless it is mixed with an inert gas and handled in rugged equipment. Ethyne burns with pure oxygen to give a very hot flame that is widely used for welding. For this purpose, the gas is dissolved under about $15 \: \text{kg cm}^{-2}$ in 2-propanone (acetone, , bp $56.5^\text{o}$) and contained in cylinders packed with diatomaceous earth.
Why is ethyne so much less stable than ethene or ethane? First, $\ce{C-C}$ bonds are not as strong as $\ce{C-H}$ bonds. Therefore a gain in stability usually is to be expected when $\ce{C-H}$ bonds are made at the expense of $\ce{C-C}$ bonds; ethene and ethane each have more $\ce{C-H}$ bonds than ethyne has. Second, ethyne has six electrons held between the two carbons and these electrons experience considerable mutual interelectronic repulsion. This accounts for the fact that the average $\ce{C-C}$ bond strength for the triple bond of an alkyne is $200/3 = 67 \: \text{kcal}$, compared to $146/2 = 73 \: \text{kcal}$ for the double bond of an alkene and $83 \: \text{kcal}$ for a normal single bond of an alkane. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.02%3A_Physical_and_Spectroscopic_Properties_of_Alkenes_and_Alkynes.txt |
The reactions of alkanes discussed in Chapter 4 are homolytic processes, which means that the bonds are made and broken through radical or atomic intermediates. In contrast, the $S_\text{N}$ and $E$ reactions of alkyl halides, considered in Chapter 8, involve heterolytic bond cleavage and ionic reagents or products. An especially important factor contributing to the differences between the reactions of the alkanes and alkyl halides is the slight ionic character of $\ce{C-H}$ compared to $\ce{C}$-halide bonds (see Section 1-3). The alkenes are like the alkanes in being nonpolar compounds (Section 4-1) and it may come as a surprise that many important reactions of alkenes are heterolytic reactions. Why should this be so? No doubt because the electrons in the alkene double bonds are more exposed and accessible than the electrons in an alkane $\ce{C-C}$ bond.
This is evident from the atomic-orbital models of ethene described in Section 6-4C. The electrons of the double bond are pushed outward by their mutual repulsions, and their average positions are considerably farther from the bond axis than the electron positions of a single bond (Figure 10-6). In such circumstances, electrophilic reagents, which act to acquire electrons in chemical reactions (Section 8-1), are expected to be particularly reactive. This is actually the case. Furthermore, reagents that are primarily nucleophilic (electron-donating) are notoriously poor for initiating reactions at carbon-carbon double bonds. Exceptions occur when the double bonds carry substituents with a sufficiently high degree of electron-attracting power to reduce the electron density in the double bond enough to permit attack by a nucleophilic agent.
Example of electrophilic reagents that normally add to carbon-carbon double bonds of alkenes to give saturated compounds include halogens ($\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$), hydrogen halides ($\ce{HCl}$ and $\ce{HBr}$), hypohalous acids ($\ce{HOCl}$ and $\ce{HOBr}$), water, and sulfuric acid:
The mechanisms of these reactions have much in common and have been studied extensively from this point of view. They also have very considerable synthetic utility. The addition of water to alkenes (hydration) is particularly important for the preparation of a number of commercially important alcohols. Thus ethanol and 2-propanol (isopropyl alcohol) are made on a very large scale by the hydration of the corresponding alkenes (ethene and propene) using sulfuric or phosphoric acids as catalysts. The nature of this type of reaction will be described later.
The Stepwise Ionic Mechanism, Halogen Addition
We shall give particular attention here to the addition of bromine to alkenes because this reaction is carried out very conveniently in the laboratory and illustrates a number of important points about electrophilic addition reactions. Much of what follows applies to addition of the other halogens, except fluorine.
A significant observation concerning bromine addition is that it and many of the other reactions listed above proceed in the dark and are not influenced by radical inhibitors. This is evidence against a radical-chain mechanism of the type involved in the halogenation of alkanes (Section 4-4D). However, it does not preclude the operation of radical-addition reactions under other conditions, and, as we shall see later in this chapter, bromine, chlorine, and many other reagents that commonly add to alkenes by ionic mechanisms also can add by radical mechanisms.
One alternative to a radical-chain reaction for bromine addition to an alkene would be the simple four-center, one-step process shown in Figure 10-7.
The mechanism of Figure 10-7 cannot be correct for bromine addition to alkenes in solution for two important reasons. First, notice that this mechanism requires that the two $\ce{C-Br}$ bonds be formed on the same side of the double bond, and hence produce suprafacial addition. However, there is much evidence to show that bromine and many other reagents add to alkenes to form antarafacial addition products (Figure 10-8).
Cyclohexene adds bromine to give trans-1,2-dibromocyclohexane:
The cis isomer is not formed at all. To give the trans isomer, the two new $\ce{C-Br}$ bonds have to be formed on opposite sides of the double bond by antarafacial addition. But this is impossible by a one-step mechanism because the $\ce{Br-Br}$ bond would have to stretch too far to permit the formation of both $\ce{C-Br}$ bonds at the same time.
The second piece of evidence against the mechanism of Figure 10-7 is that bromine addition reactions carried out in the presence of more than one nucleophilic reagent usually give mixtures of products. Thus the addition of bromine to an alkene in methanol solution containing lithium chloride leads not only to the expected dibromoalkane, but also to products resulting from attack by chloride ions and by the solvent:
The intervention of extraneous nucleophiles suggests a stepwise mechanism in which the nucleophiles compete for a reactive intermediate formed in one of the steps.
A somewhat oversimplified two-step mechanism that accounts for most of the foregoing facts is illustrated for the addition of bromine to ethene. [In the formation shown below, the curved arrows are not considered to have real mechanistic significance, but are used primarily to show which atoms can be regarded as nucleophilic (donate electrons) and which as electrophilic (accept electrons). The arrowheads always should be drawn to point to the atoms that are formulated as accepting a pair of electrons.]
The first step (which involves electrophilic attack by bromine on the double bond) produces a bromide ion and a carbocation, as shown in Equation 10-1.$^1$
As we know from our study of $S_\text{N}1$ reactions (Section 8-4), carbocations react readily with nucleophilic reagents. Therefore in the second step of the bromine-addition mechanism, shown in Equation 10-2, the bromoethyl cation is expected to combine rapidly with bromide ion to give the dibromo compound. However, if other nucleophiles, such as $\ce{Cl}^\ominus$ or $\ce{CH_3OH}$, are present in solution, they should be able to compete with bromide ion for the cation, as in Equations 10-3 and 10-4, and mixtures of products will result:
To account for the observation that all of these reactions result in antarafacial addition, we must conclude that the first and second steps take place from opposite sides of the double bond.
Why Antarafacial Addition?
The simple carbocation intermediate of Equation 10-1 does not account for formation of the antarafacial-addition product. The results with $S_\text{N}1$ reactions (Section 8-6) and the atomic-orbital representation (see Section 6-4E) predict that the bonds to the positively charged carbon atom of a carbocation should lie in a plane. Therefore, in the second step of addition of bromine to cycloalkenes, bromide ion could attack either side of the planar positive carbon to give a mixture of cis- and trans-1,2-dibromocyclohexanes. Nonetheless, antarafacial addition occurs exclusively:
To account for the stereospecificity of bromine addition to alkenes, it has been suggested that in the initial electrophilic attack of bromine a cyclic intermediate is formed that has bromine bonded to both carbons of the double bond. Such a "bridged" ion is called a bromonium ion because the bromine formally carries the positive charge:
An $S_\text{N}2$-type of attack of bromide ion, or other nucleophile, at carbon on the side opposite to the bridging group then results in formation of the antarafacial-addition product:
We may seem to have contradicted ourselves because Equation 10-1 shows a carbocation to be formed in bromine addition, but Equation 10-5 suggests a bromonium ion. Actually, the formulation of intermediates in alkene addition reactions as "open" ions or as cyclic ions is a controversial matter, even after many years of study. Unfortunately, it is not possible to determine the structure of the intermediate ions by any direct physical method because, under the conditions of the reaction, the ions are so reactive that they form products more rapidly than they can be observed. However, it is possible to generate stable bromonium ions, as well as the corresponding chloronium and iodonium ions. The technique is to use low temperatures in the absence of any strong nucleophiles and to start with a 1,2-dihaloalkane and antimony pentafluoride in liquid sulfur dioxide:
The $\ce{C_2H_4Br}^\oplus$ ions produced in this way are relatively stable and have been shown by nmr to have the cyclic halonium ion structure.
Complexes of Electrophilic Agents with Double Bonds
There is a further aspect of polar additions to alkenes that we should consider, namely, that electrophilic reagents form loose complexes with the $\pi$ electrons of the double bonds of alkenes prior to reaction by addition. Complexes of this type are called charge-transfer complexes (or $\pi$ complexes). Formation of a complex between iodine and cyclohexene is demonstrated by the fact that iodine dissolves in cyclohexene to give a brown solution, whereas its solutions in cyclohexane are violet. The brown solution of iodine in cyclohexene slowly fades as addition occurs to give colorless trans-1,2-diiodocyclohexane.
Precise Lewis structures cannot be written for charge-transfer complexes, but they commonly are represented as
with the arrow denoting that electrons of the double bond are associated with the electrophile. These complexes probably represent the first stage in the formation of addition products by a sequence such as the following for bromine addition:
Addition of Proton Acids
We have seen that electrophiles can react with alkenes to form carbon-halogen bonds by donating positive halogen, $\ce{Br}^\oplus$, $\ce{Cl}^\oplus$, or $\ce{I}^\oplus$. Likewise, carbon-hydrogen bonds can be formed by appropriately strong proton donors, which, of course, are typically strong proton acids. These acids are more effective in the absence of large amounts of water because water can compete with the alkene as a proton acceptor (also see Section 10-3E). Hydrogen chloride addition to ethene occurs by way of a proton-transfer step to give the ethyl cation and a chloride ion (Equation 10-6) followed by a step in which the nucleophilic chloride ion combines with the ethyl cation (Equation 10-7):
All of the hydrogen halides $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$) will add to alkenes. Addition of hydrogen fluoride, while facile, is easily reversible. However, a solution of $70\%$ anhydrous hydrogen fluoride and $30\%$ of the weak organic base, pyridine, which is about 1/10,000 times as strong as ammonia, works better, and with cyclohexene gives fluorocyclohexane. With hydrogen iodide, care must be taken to prevent $\ce{I_2}$ addition products resulting from iodine formed by oxidation reactions such as
$4 \ce{HI} + \ce{O_2} \rightarrow 2 \ce{I_2} + 2 \ce{H_2O}$
With hydrogen bromide, radical-chain addition may intervene unless the reaction conditions are controlled carefully (this will be discussed in Section 10-7).
The stereochemistry of addition depends largely on the structure of the alkene, but for simple alkenes and cycloalkenes, addition occurs predominantly in an antarafacial manner. For example, hydrogen bromide reacts with 1,2-dimethylcyclohexene to give the antarafacial addition product:
Hydration
We mentioned previously that the hydration of alkenes required a strong acid as a catalyst, because water itself is too weak an acid to initiate the proton-transfer step. However, if a small amount of a strong acid such as sulfuric acid is present, hydronium ions, $\ce{H_3O}^\oplus$, are formed in sufficient amount to protonate reasonably reactive alkenes, although by no means as effectively as does concentrated sulfuric acid. The carbocation formed then is attacked rapidly by a nucleophilic water molecule to give the alcohol as its conjugate acid,$^2$ which regenerates hydronium ion by transferring a proton to water. The reaction sequence follows for 2-methylpropene:
In this sequence, the acid acts as a catalyst because the hydronium ion used in the proton addition step is regenerated in the final step.
Sulfuric acid (or phosphoric acid) is preferred as an acid catalyst for addition of water to alkenes because the conjugate base, $\ce{HSO_4-}$ (or $\ce{H_2PO_4-}$), is a poor nucleophile and does not interfere in the reaction. However, if the water concentration is kept low by using concentrated acid, addition occurs to give sulfate (or phosphate) esters. The esters formed with sulfuric acid are either alkyl acid sulfates $\ce{R-OSO_3H}$ or dialkyl sulfates $\ce{(RO)_2SO_2}$. In fact, this is one of the major routes used in the commercial production of ethanol and 2-propanol. Ethen and sulfuric acid give ethyl hydrogen sulfate, which reacts readily with water in a second step to give ethanol:
Aqueous versus Nonaqueous Acids. Acid Strengths
One of the more confusing features of organic chemistry is the multitude of conditions that are used to carry out a given kind of reaction, such as the electrophilic addition of proton acids to different alkenes. Strong acids, weak acids, water, no water - Why can't there be a standard procedure? The problem is that alkenes have very different tendencies to accept protons. In the vapor phase, $\Delta H^0$ for addition of a proton to ethene is about $35 \: \text{kcal}$ more positive than for 2-methylpropene, and although the difference should be smaller in solution, it still would be large. Therefore we can anticipate (and we find) that a much more powerful proton donor is needed to initiate addition of an acid to ethene than to 2-methylpropene. But why not use in all cases a strong enough acid to protonate any alkene one might want to have a proton acid add to? Two reasons: First, strong acids can induce undesirable side reactions, so that one usually will try not to use a stronger acid than necessary; second, very strong acid may even prevent the desired reaction from occurring!
In elementary chemistry, we usually deal with acids in more or less dilute aqueous solution and we think of sulfuric, hydrochloric, and nitric acids as being similarly strong because each is essentially completely disassociated in dilute water solution:
$\ce{HCl} + \ce{H_2O} \overset{\longrightarrow}{\leftarrow} \ce{H_3O}^\oplus + \ce{Cl}^\ominus$
This does not mean they actually are equally strong acids. It means only that each of the acids is sufficiently strong to donate all of its protons to water. We can say that water has a "leveling effect" on acid strengths because as long as an acid can donate its protons to water, the solution has but one acid "strength" that is determined by the $\ce{H_3O}^\oplus$ concentration, because $\ce{H_3O}^\oplus$ is where the protons are.
Now, if we use poorer proton acceptors as solvent we find the proton-donating powers of various "strong" acids begin to spread out immensely. Furthermore, new things begin to happen. For example, ethene is not hydrated appreciably by dilute aqueous acid; it just is too hard to transfer a proton from hydronium ion to ethene. So we use concentrated sulfuric acid, which is strong enough to add a proton to ethene. But now we don't get hydration, because any water that is present in concentrated sulfuric acid is virtually all converted to $\ce{H_3O}^\oplus$, which is non-nucleophilic!
$\ce{H_2SO_4} + \ce{H_2O} \rightarrow \ce{H_3O}^\oplus + \ce{HSO_4-}$
However, formation of $\ce{H_3O}^\oplus$ leads to formation of $\ce{HSO_4-}$, which has enough nucleophilic character to react with the $\ce{CH_3CH_2+}$ to give ethyl hydrogen sulfate and this is formed instead of the conjugate acid of ethanol (Section 10-3E). The epitome of the use of stronger acid and weaker nucleophile is with liquid $\ce{SO_2}$ (bp $\sim 10^\text{o}$) as the solvent and $\ce{HBF_6}$ as the acid. This solvent is a very poor proton acceptor (which means that its conjugate acid is a very good proton donor) and $\ce{SbF_6-}$ is an extremely poor nucleophile. If we add ethene to such a solution, a stable solution of $\ce{CH_3CH_2+} \ce{SbF_6-}$ is formed. The reason is that there is no better proton acceptor present than $\ce{CH_2=CH_2}$ and no nucleophile good enough to combine with the cation.
A Biological Hydration Reaction
The conversion of fumaric acid to malic acid is an important biological hydration reaction. It is one of a cycle of reactions (Krebs citric acid cycle) involved in the metabolic combustion of fuels (amino acids and carbohydrates) to $\ce{CO_2}$ and $\ce{H_2O}$ in a living cell.
$^1$An alternative to Equation 10-1 would be to have $\ce{Br_2}$ ionize to $\ce{Br}^\oplus$ and $\ce{Br}^\ominus$, with a subsequent attack of $\ce{Br}^\oplus$ on the double bond to produce the carbocation. The fact is that energy required for such an ionization of $\ce{Br_2}$ is prohibitively large even in water solution $\left( \Delta H^0 \geq 80 \: \text{kcal} \right)$. One might well wonder why Equation 10-1 could possibly be more favorable. The calculated $\Delta H^0$ for $\ce{CH_2=CH_2} + \ce{Br_2} \rightarrow \cdot \ce{CH_2-CH_2Br} + \ce{Br} \cdot$ is $+41 \: \text{kcal}$, which is only slightly more favorable than the $\Delta H^0$ for $\ce{Br_2} \rightarrow 2 \ce{Br} \cdot$ of $46.4 \: \text{kcal}$. However, available thermochemical data suggest that the ease of transferring an electron from $\cdot \ce{CH_2CH_2Br}$ to $\ce{Br} \cdot$ to give $^\oplus \ce{CH_2CH_2Br} + \ce{Br}^\ominus$ is about $80 \: \text{kcal}$ more favorable than $2 \ce{Br} \cdot \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus$. Thus the overall $\Delta H^0$ of Equation 10-1 is likely to be about $85 \: \text{kcal}$ more favorable than $\ce{Br_2} \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus$.
$^2$The terms conjugate acid and conjugate base are very convenient to designate substances that are difficult to name simply as acids, bases, or salts. The conjugate acid of a compound $\ce{X}$ is $\ce{XH}^\oplus$ and the conjugate base of $\ce{HY}$ is $\ce{Y}^\ominus$. Thus $\ce{H_3O}^\oplus$ is the conjugate acid of water, while $\ce{OH}^\ominus$ is its conjugate base. Water itself is then both the conjugate base of $\ce{H_3O}^\oplus$ and the conjugate acid of $\ce{OH}^\ominus$ | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.04%3A_Electrophilic_Additions_to_Alkenes.txt |
Addition of $\ce{HX}$
Addition of an unsymmetrical substance such as $\ce{HX}$ to an unsymmetrical alkene theoretically can give two products:
and
Both products are seldom formed in equal amounts; in fact, one isomer usually is formed to the exclusion of the other. For example, the hydration of propene gives 2-propanol (not 1-propanol), and hydrogen chloride adds to 2-methylpropene to give tert-butyl chloride (not isobutyl chloride):
To understand the reason for the pronounced selectivity in the orientation of addition of electrophiles, it will help to consider one example, hydrogen bromide addition to 2-methylpropene. Two different carbocation intermediates could be formed by attachment of a proton to one or the other of the double bond carbons:
Subsequent reactions of the cations with bromide ion give tert-butyl bromide and isobutyl bromide. In the usual way of running these additions, the product is very pure tert-butyl bromide.
How could be have predicted which product would be favored? The first step is to decide whether the prediction is to be based on (1) which of the two products is the more stable, or (2) which of the two products if formed more rapidly. If we make a decision on the basis of product stabilities, we take into account $\Delta H^0$ values, entropy effects, and so on, to estimate the equilibrium constants $K_\text{eq}$ for the reactants and each product. When the ratio of the products is determined by the ratio of their equilibrium constants, we say the overall reaction is subject to equilibrium (or thermodynamic) control. Equilibrium control requires that the reaction be reversible.
When a reaction is carried out under conditions in which it is not reversible, the ratio of the products is determined by the relative rates of formation of the various products. Such reactions are said to be under kinetic control.
The products obtained in a reaction subject to kinetic control are not necessarily the same as those obtained under equilibrium control. Indeed, the equilibrium constant for interconversion of tert-butyl bromide and isobutyl bromide at $25^\text{o}$ is 4.5, and if the addition of hydrogen bromide to 2-methylpropene were under equilibrium control, the products would be formed in this ratio:
$K_\text{eq} = \dfrac{\left[ \text{tert-butyl bromide} \right]}{\left[ \text{isobutyl bromide} \right]} = 4.5$
But the addition product is $99+\%$ tert-butyl bromide so the reaction clearly is kinetically controlled, tert-butyl being formed considerably faster than isobutyl bromide. The slow, or rate-determining, step in this reaction is the formation of the intermediate cation rather than the reaction of the cation with bromide ion. So to account for the formation of tert-butyl bromide we have to consider why the tert-butyl cation is formed more rapidly than the isobutyl cation:
As we have seen in Section 8-7B, alkyl groups are more electron donating than hydrogen. This means that the more alkyl groups there are on the positive carbon of the cation, the more stable and the more easily formed the cation will be. The reason is that electron-donating groups can partially compensate for the electron deficiency of the positive carbon. As a result, we can predict that the tert-butyl cation with three alkyl groups attached to the positive center will be formed more readily than the primary isobutyl cation with one alkyl group attached to the positive center.
Thus the problem of predicting which of the two possible products will be favored in the addition of unsymmetrical reagents to alkenes under kinetic control reduces to predicting which of two possible carbocation intermediates will be formed most readily. With simple alkenes, we shall expect the preference of formation of the carbocations to be in the order:
tertiary > secondary > primary.
The reaction scheme can be represented conveniently in the form of an energy diagram (Figure 10-10). The activation energy, $\Delta H^1_\text{tert}$ for the formation of the tert-butyl cation is less than $\Delta H^1_\text{prim}$ for the formation of the isobutyl cation because the tertiary ion is much more stable (relative to the reactants) than the primary ion, and therefore is formed at the faster rate. The second step, to form the product from the intermediate cation, is very rapid and requires little activation energy. Provided that the reaction is irreversible, it will take the lowest-energy path and form exclusively tert-butyl bromide. However, if the reaction mixture is allowed to stand for a long time, isobutyl bromide begins to form. Over a long period, the products equilibrate and, at equilibrium, the product distribution reflects the relative stabilities of the products rather than the stability of the transition states for formation of the intermediates.
A rather simple rule, formulated in 1870 and known as Markownikoff's rule, correlates the direction of kinetically controlled additions of $\ce{HX}$ to unsymmetrical alkenes. This rule, an important early generalization of organic reactions, may be stated as follows: In addition of $\ce{HX}$ to an unsymmetrical carbon-carbon double bond, the hydrogen of $\ce{HX}$ goes to that carbon of the double bond that carries the greater number of hydrogens. It should be clear that Markownikoff's rule predicts that addition of hydrogen bromide to 2-methylpropene will give tert-butyl bromide.
Addition of Other Reagents to Unsymmetrical Alkenes. The Electronegativity Chart
We can extend Markownikoff's rule to cover additions of substances of the general type $\ce{X-Y}$ to unsymmetrically substituted alkenes when a clear-cut decision is possible as to whether $\ce{X}$ or $\ce{Y}$ is the more electrophilic atom of $\ce{X-Y}$. If the polarization of the $\ce{X-Y}$ bond is such that $\ce{X}$ is positive, $^{\delta \oplus} \ce{X-Y} ^{\delta \ominus}$, then $\ce{X}$ will be expected to add as $\ce{X}^\oplus$ to the alkene to form the more stable carbocation. This step will determine the direction of addition. For example, if we know that the $\ce{O-Br}$ bond of $\ce{HOBr}$ is polarized as $\overset{\delta \ominus}{\ce{HO}} - \overset{\delta \oplus}{\ce{Br}}$, then we can predict that addition of $\ce{HOBr}$ to 2-methylpropene will give 1-bromo-2-methyl-2-propanol:
Pauling's value for the electronegativity of carbon makes it slightly more electron-attracting than hydrogen. However, we expect that the electron-attracting power of a carbon atom (or of other elements) will depend also on the electronegativities of the groups to which it is attached. In fact, many experimental observations indicate that carbon in methyl or other alkyl groups is significantly less electron-attracting than hydrogen. Conversely, the $\ce{CF_3}-$ group is, as expected, far more electron-attracting than hydrogen.
The direction of polarization of bonds between various elements may be predicted from Figure 10-11. For example, an $\ce{O-Cl}$ bond should be polarized so the oxygen is negative; a $\ce{C-N}$ bond should be polarized so the nitrogen is negative:
$\overset{\delta \ominus}{\ce{O}}---\overset{\delta \oplus}{\ce{Cl}} \: \: \: \: \: \overset{\delta \oplus}{\ce{C}}---\overset{\delta \ominus}{\ce{N}}$
We then can predict that, in the addition of $\ce{HOCl}$ to an alkene, the chlorine will add preferentially to form the more stable of two possible carbon cations. Generally, this means that chlorine will bond to the carbon carrying the greater number of hydrogens:
A number of reagents that are useful sources of electrophilic halogen are included in Table 10-2. Some of these reagents, notably those with $\ce{O}-$halogen or $\ce{N}-$halogen bonds, actually are sources of hypohalous acids, $\ce{HOX}$, and function to introduce halogen and hydroxyl groups at carbon. There are very few good fluorinating agents whereby the fluorine is added as $\ce{F}^\oplus$.
Table 10-2: Reagents that add to Alkenes by Electrophilic Attack:
Additions to Substituted Alkenes
For alkenes that have halogen or similar substituents at the doubly bonded carbons, the same principles apply as with the simple alkenes. That is, under kinetic control the preferred product will be the one derived from the more stable of the two possible intermediate carbon cations. Consider a compound of the type $\ce{Y-CH=CH_2}$. If $\ce{Y}$ is more electron-attracting than hydrogen, then hydrogen halide should add in such a way as to put the proton of $\ce{HX}$ on the $\ce{YCH=}$ end and $\ce{X}$ on the $\ce{=CH_2}$ end. The reason is that the positive carbon is expected to be more favorably located if it is not attached directly to an electron-attracting substituent:
The addition goes as predicted, provided that the atom directly attached to the carbon of the double bond carries no unshared (nonbonding) electron pairs. For example,
$\ce{CF_3-CH=CH_2} + \ce{HCl} \rightarrow \ce{CF_3-CH_2-H_2-Cl}$
Such substituents are relatively uncommon, and most of the reported $\ce{H-X}$ additions have been carried out with $\ce{Y}$ groups having unshared electron pairs on an atom connected directly to a carbon of the double bond:
These substituents usually are strongly electronegative relative to hydrogen, and this often causes diminished reactivity of the double bond toward electrophiles. Nonetheless, the preferred orientation of $\ce{HX}$ additions situates the positive charge of the intermediate carbocation next to the substituent:
The electron-attracting power of the substituent is more than counterbalanced by stabilization of the intermediate cation by the ability of the substituents to delocalize their unshared electrons to the adjacent positive carbon (see Section 6-6). | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.05%3A_Orientation_in_Addition_to_Alkenes.txt |
The alkynes behave in many ways as if they were doubly unsaturated alkenes. For example, bromine adds to ethyne in two stages - first to give trans-1,2-dibromoethene by antarafacial addition, and finally to give 1,1,2,2-tetrabromoethane:
Likewise, anhydrous hydrogen fluoride adds first to give fluoroethene and ultimately to give 1,1-difluoroethane:
However, there is an interesting contrast in reactivity. Alkynes are substantially less reactive than corresponding alkenes toward many electrophiles. This is perhaps surprising because the electrons of a triple bond, like those of a double bond, are highly exposed, which suggests that the reactivity (nucleophilicity) of a triple bond should be high. Evidently this is not the case. A simple but reasonable explanation is that the carbocation formed from the alkyne is less stable than that from the alkene because it cannot achieve the \(sp^2\) hybrid-orbital configuration expected to be the most stable arrangement for a carbocation (see Section 6-4E):
Hydration of Alkynes
Alkynes, unlike alkenes, are not hydrated readily in aqueous acid unless a mercuric salt is present as a catalyst. Also, the products that are isolated are either aldehydes or ketones instead of alcohols. Even though the addition of one molecule of water to ethyne probably gives ethenol (vinyl alcohol) initially, this compound is unstable relative to its structural isomer (ethanal) and rapidly rearranges:
Similarly, addition of water to propyne leads to 2-propanone by way of its unstable isomer, 2-propenol:
In general, the position of equilibrium for interconversion of a carbonyl compound with the corresponding alkenol (or enol), having the hydroxyl group attached to the double bond, lies far on the side of the carbonyl compound:
Because mercuric salts catalyze the hydration of alkynes, they probably are acting as electrophiles. Mercuric salts are known to add to both alkenes and alkynes, and if the reaction mixture is acidic, the carbon-mercury bond is cleaved to form a carbon-hydrogen bond. The overall sequence in propyne hydration may be written as follows:
10.07: Nucleophilic Addition Reactions
When a stepwise ionic addition reaction involves nucleophilic attack at carbon as a first step, it is described as a nucleophilic addition. Reactions of this type often are catalyzed by bases, which generate the required nucleophile. For example, consider the addition of some weakly acidic reagent $\ce{HX}$ to an alkene. In the presence of a strong base $\left( ^\ominus \ce{OH} \right)$, $\ce{HX}$ could give up its proton to form the conjugate base $\ce{X}^\ominus$, which is expected to be a much better nucleophile than $\ce{HX}$:
$\ce{H}:\ce{X} + ^\ominus \ce{OH} \rightleftharpoons \ce{H_2O} + :\ce{X}^\ominus$
What can follow with an alkene is an ionic chain reaction with the following two propagating steps. First, the nucleophile attacks at carbon to form a carbon anion (carbanion) intermediate (Equation 10-8). Second, electrophilic transfer of a proton from $\ce{HX}$ to the carbanion forms the adduct and regenerates the nucleophile (Equation 10-9). The overall reaction is the addition of $\ce{HX}$ to the double bond:
Net reaction:
The $\ce{HX}$ reagent can be water, an alcohol $\left( \ce{ROH} \right)$, a thiol $\left( \ce{RSH} \right)$, an amine $\left( \ce{RNH_2} \right)$, or hydrogen cyanide $\left( \ce{HCN} \right)$ or other carbon acids (i.e., compounds with acidic $\ce{C-H}$ bonds). However, nucleophilic addition of these reagents to simple alkenes rarely is encountered. To have nucleophilic addition the double bond must be substituted with strongly electron-withdrawing groups such as carbonyl-containing groups, $\ce{NO_2}$, $\ce{C \equiv N}$, or positively charged ammonium or sulfonium groups. However, alkynes generally are more reactive towards nucleophiles than they are toward electrophiles. For example, with a base catalyst, 2-hexen-4-yne adds methanol across the triple bond, leaving the double bond untouched:
(Nonetheless, the double bond seems to be necessary because a corresponding addition is not observed for 2-butyne, $\ce{CH_3C \equiv CCH_3}$.)
Many nucleophilic addition reactions have considerable synthetic value, particularly those involving addition of carbon acids, such as $\ce{HCN}$, because they provide ways of forming carbon-carbon bonds. More of their utility will be discussed in Chapters 14, 17, and 18. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.06%3A_Electrophilic_Addition_Reactions_of_Alkynes.txt |
The early literature concerning the addition of hydrogen bromide to unsymmetrical alkenes at best is confused. Sometimes the same alkene was reported to give addition both according to, and in opposition to, the principles discussed for electrophilic ionic addition (Section 10-4). Much of the uncertainty on the addition of hydrogen bromide was removed by the classical researches of M. S. Kharasch and F. R. Mayo (1933) who showed that there must be two reaction mechanisms, each giving a different product. Kharasch and Mayo found, in the presence of radical inhibitors, hydrogen bromide adds to propene in a rather slow reaction to give pure 2-bromopropane:
With light, peroxides, radical initiators, and in the absence of radical inhibitors a rapid radical-chain addition of hydrogen bromide occurs to yield $80\%$ or more of 1-bromopropane:
Similar effects have been noted occasionally with hydrogen chloride, but never with hydrogen iodide or hydrogen fluoride. A few substances apparently add to alkenes only by radical mechanisms, and always add in the opposite way to that expected for electrophilic ionic addition.
The ionic addition of hydrogen bromide was discussed in Section 10-4 and will not be considered further at this point. Two questions with regard to the so-called abnormal addition will be given special attention. Why does the radical mechanism give a product of different structure than the ionic addition? Why does the radical addition occur readily with hydrogen bromide but rarely with the other hydrogen halides?
The abnormal addition of hydrogen bromide is catalyzed strongly by peroxides, which have the structure $\ce{R-O-O-R}$ and decompose thermally to give $\ce{RO} \cdot$ radicals (see Section 4-5B):
The $\ce{RO} \cdot$ radicals can react with hydrogen bromide in two ways, to abstract either hydrogen atoms or bromine atoms:
Clearly, the formation of $\ce{ROH}$ and a bromine atom is energetically more favorable. The overall process of decomposition of peroxide and attack on hydrogen bromide, which results in the formation of a bromine atom, can initiate a radical-chain addition of hydrogen bromide to an alkene.
The two chain-propagating steps, taken together, are exothermic by $16 \: \text{kcal}$ and have a fairly reasonable energy balance between the separate steps. The reaction chains apparently are rather long, because the addition is strongly inhibited by radical traps and only traces of peroxide catalyst are needed.
Orientation of Addition
The direction of addition of hydrogen bromide to propene clearly depends on which end of the double bond the bromine atom attacks. The important question is which of the two possible carbon radicals that may be formed is the more stable, the 1-bromo-2-propyl radical, $5$, or the 2-bromo-1-propyl radical, $6$:
From $\ce{C-H}$ bond-dissociation energies of alkanes (see Table 5-6), the ease of formation and stabilities of the carbon radicals is seen to follow the sequence tertiary $>$ secondary $>$ primary. By analogy, the secondary 1-bromo-2-propyl radical, $5$, is expected to be more stable and more easily formed than the primary 2-bromo-1-propyl radical, $6$. The product of radical addition should be, and indeed is, 1-bromopropane:
Other reagents, such as the halogens, also can add to alkenes and alkynes by both radical-chain and ionic mechanisms. Radical addition usually is initiated by light, whereas ionic addition is favored by low temperatures and no light. Nevertheless, it often is difficult to keep both mechanisms from operating at the same time. This is important even when the alkene is symmetrical because, although the adduct will then have the same structural formula regardless of mechanism, the stereochemical configurations may differ. Electrophilic addition of halogens generally is a stereospecific antarafacial addition, but radical-chain additions are less stereospecific.
There are many reagents that add to alkenes only by radical-chain mechanisms. A number of these are listed in Table 10-3. They have in common a relatively weak bond, $\ce{X-Y}$, that can be cleaved homolytically either by light or by chemical initiators such as peroxides. In the propagation steps, the radical that attacks the double bond does so to produce the more stable carbon radical. For addition to simple alkenes and alkynes, the more stable carbon radical is the one with the fewest hydrogens or the most alkyl groups at the radical center.
Table 10-3: Reagents that add to Alkenes by Radical-Chain Mechanisms
The principles of radical addition reactions of alkenes appear to apply equally to alkynes, although there are fewer documented examples of radical additions to triple bonds. Two molecules of hydrogen bromide can add to propyne first to give cis-1-bromopropene (by antarafacial addition) and then 1,2-dibromopropane: | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.08%3A_Radical-Chain_Addition_Reactions_to_Alkenes.txt |
One of the most important technical reactions of alkenes is their conversion to higher-molecular-weight compounds or polymers (Table 10-4). A polymer is defined as a long-chain molecule with recurring structural units. Thus polymerization of propene gives a long-chain hydrocarbon with recurring units:
Table 10-4: Alkene Monomers and Their Polymers
Most technically important polymerizations of alkenes occur by chain mechanisms and may be classed as anion, cation, or radical reactions, depending upon the character of the chain-carrying species. In each case, the key steps involve successive additions to molecules of the alkene, the differences being in the number of electrons that are supplied by the attacking agent for formation of the new carbon-carbon bond. For simplicity, these steps will be illustrated by using ethene, even though it does not polymerize very easily by any of them:
Anionic Polymerization
Initiation of alkene polymerization by the anion-chain mechanism may be formulated as involving an attack by a nucleophilic reagent $\ce{Y}^\ominus$ on one end of the double bond and formation of a carbanion:
Attack by the carbanion on another alkene molecule would give a four-carbon carbanion, and subsequent additions to further alkene molecules would lead to a high-molecular-weight anion:
The growing chain can be terminated by any reaction (such as the addition of a proton) that would destroy the carbanion on the end of the chain:
Anionic polymerization of alkenes is quite difficult to achieve because few anions (or nucleophiles) are able to add readily to alkene double bonds (see Section 10-6). Anionic polymerization occurs readily only with alkenes substituted with sufficiently powerful electron-attracting groups to expedite nucleophilic attack. By this reasoning, alkynes should polymerize more readily than alkenes under anionic conditions, but there appear to be no technically important alkyne polymerizations in operation by this or any other mechanism. Perhaps this is because the resultant polymer would be highly conjugated, and therefore highly reactive, and may not survive the experimental conditions:
Cationic Polymerization
Polymerization of an alkene by acidic reagents can be formulated by a mechanism similar to the addition of hydrogen halides to alkene linkages. First, a proton from a suitable acid adds to an alkene to yield a carbocation. Then, in the absence of any other reasonably strong nucleophilic reagent, another alkene molecule donates an electron pair and forms a longer-chain cation. Continuation of this process can lead to a high-molecular-weight cation. Termination can occur by loss of a proton. The following equations represent the overall reaction sequence:
Ethene does not polymerize by the cationic mechanism because it does not have sufficiently electron-donating groups to permit easy formation of the intermediate growing-chain cation. 2-Methylpropene has electron-donating alkyl groups and polymerizes much more easily than ethene by this type of mechanism. The usual catalysts for cationic polymerization of 2-methylpropene are sulfuric acid, hydrogen fluoride, or a complex of boron trifluoride and water. Under nearly anhydrous conditions a very long chain polymer called polyisobutylene is formed.
Polyisobutylene fractions of particular molecular weights are very tacky and are used as adhesives for pressure-sealing tapes.
In the presence of $60\%$ sulfuric acid, 2-methylpropene is not converted to a long-chain polymer, but to a mixture of eight-carbon alkenes. The mechanism is like that of the polymerization of 2-methylpropene under nearly anhydrous conditions, except that chain termination occurs after only one 2-methylpropene molecule has been added:
The short chain length is due to the high water concentration; the intermediate carbocation loses a proton to water before it can react with another alkene molecule.
The proton can be lost in two different ways, and a mixture of alkene isomers is obtained. The alkene mixture is known as "diisobutylene" and has a number of commercial uses. Hydrogenation yields 2,2,4-trimethylpentane (often erroneously called "isooctane"), which is used as the standard "100 antiknock rating" fuel for internal-combustion gasoline engines:
Radical Polymerization
Ethene can be polymerized with peroxide catalysts under high pressure ($1000 \: \text{atm}$ or more, literally in a cannon barrel) at temperatures in excess of $100^\text{o}$. The initiation step involves formation of radicals, and chain propagation entails stepwise addition of radicals to ethene molecules.
Chain termination can occur by any reaction resulting in combination or disproportionation of free radicals.
The polyethene produced in this way has from 100 to 1000 ethene units in the hydrocarbon chain. The polymer possesses a number of desirable properties as a plastic and is used widely for electrical insulation, packaging films, piping, and a variety of molded articles. Propene and 2-methylpropene do not polymerize satisfactorily by radical mechanisms.
Coordination Polymerization
A relatively low-pressure, low-temperature ethene polymerization has been achieved with an aluminum-molybdenum oxide catalyst, which requires occasional activation with hydrogen (Phillips Petroleum process). Ethene also polymerizes quite rapidly at atmospheric pressure and room temperature in an alkane solvent containing a suspension of the insoluble reaction product from triethylaluminum and titanium tetrachloride (Ziegler process). Both the Phillips and Ziegler processes produce very high-molecular-weight polyethene with exceptional physical properties. The unusual characteristics of these reactions indicate that no simple anion, cation, or radical mechanism can be involved. It is believed that the catalysts act by coordinating with the alkene molecules in somewhat the same way that hydrogenation catalysts combine with alkenes (Section 11-2A).
Polymerization of propene by the Ziegler process gives a very useful plastic material. It can be made into durable fibers or molded into a variety of shapes. Copolymers (polymers with more than one kind of monomer unit in the polymer chains) of ethene and propene made by the Ziegler process have highly desirable rubberlike properties and are potentially the cheapest useful elastomers (elastic polymers). A Nobel Prize was shared in 1963 by K. Ziegler and G. Natta for their work on alkene polymerization.
The properties and uses of polymers are discussed in greater detail in Chapters 13 and 29. The most important alkene monomers used in addition polymerizations are listed in Table 10-4 along with some names and uses of the corresponding polymers.
10.10: Alkylation of Alkenes
Addition of a saturated hydrocarbon $\left( \ce{R-H} \right)$ to an alkene to yield a saturated hydrocarbon of higher molecular weight is known as alkylation:
Such reactions are used by the petroleum industry to produce medium-molecular-weight hydrocarbons from smaller molecules. A particularly important example is afforded by the addition of 2-methylpropane to 2-methylpropene in the presence of sulfuric acid or anhydrous hydrogen fluoride to yield 2,2,4-trimethylpentane:
The overall reaction appears to be different from any so far discussed, because it involves addition of a nonpolar reagent $\left( \ce{RH} \right)$ to an alkene bond.
The key to the mechanism of hydrocarbon alkylation was provided by the discovery by P. D. Bartlett, in 1940, that a carbocation can react rapidly with a hydrocarbon having a tertiary hydrogen to yield a new carbocation and a new hydrocarbon. Some of these "hydrogen-transfer" reactions are extraordinarily fast and may be complete in seconds or less. The hydrogen is transferred with both bonding electrons $\left( \ce{H}^\ominus \right)$. For example,
With the knowledge that the hydrogen transfer is fast, the alkylation of 2-methylpropene with 2-methylpropane can be formulated as involving first polymerization of two 2-methylpropene molecules under the influence of the sulfuric acid catalyst to give the same octyl cation as was postulated for the dimerization of 2-methylpropene:
The octyl cation then can undergo a hydrogen-transfer reaction with 2-methylpropane to furnish 2,2,4-trimethylpentane and a tert-butyl cation:
Attack by the tert-butyl cation on another molecule of 2-methylpropene produces an eight-carbon tertiary cation, which them proceeds to another molecule of "alkylate":
This is an important example of a cationic chain reaction. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.09%3A_Polymerization_of_Alkenes.txt |
Further chemistry of alkenes and alkynes is described in this chapter, with emphasis on addition reactions that lead to reduction and oxidation of carbon- carbon multiple bonds. First we explain what is meant by the terms reduction and oxidation as applied to carbon compounds. Then we emphasize hydrogenation, which is reduction through addition of hydrogen, and oxidative addition reactions with reagents such as ozone, peroxides, permanganate, and osmium tetroxide. We conclude with a section on the special nature of 1-alkynes, their acidic behavior, and how the conjugate bases of alkynes can be used in synthesis to form carbon-carbon bonds.
• 11.1: Oxidation-Reduction of Organic Compounds
An organic compound commonly is said to be "reduced" if reaction leads to an increase in its hydrogen content or a decrease in its oxygen content. The compound would be "oxidized" if the reverse change took place.
• 11.2: Hydrogenation with Heterogeneous Catalysts
Addition of hydrogen to a multiple bond is hydrogenation. It is applicable to almost all types of multiple bonds and is of great importance in synthetic chemistry, particularly in the chemical industry.
• 11.3: Heats of Hydrogenation
Catalytic hydrogenation is useful for analytical and thermochemical purposes. The analysis of a substance for the number of carbon-carbon double bonds it contains is carried out by measuring the uptake of hydrogen for a known amount of sample. Measurement of the heat evolved in the hydrogenation of alkenes gives information as to the relative stabilities of alkenes.
• 11.4: Hydrogenation with Homogeneous Catalysts
Hydrogen addition to multiple bonds is catalyzed by certain complex metal salts in solution. This may be described as homogeneous catalysis and, compared to heterogeneous catalysis, is a relatively new development in the area of hydrogenation reactions. Rhodium and ruthenium salts appear to be generally useful catalyst.
• 11.5: Hydrogenation with Diimide
There are alternative ways to add hydrogen to a multiple bond besides the catalytic methods described in the previous sections. For example, hydrogen can be added to a multiple bond via homogeneous reactions utilizing diimide and diborane.
• 11.6: Addition of Boron Hydrides to Alkenes. Organoboranes
Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes.
• 11.7: Oxidation Reactions
Most alkenes react readily with ozone, even at low temperatures, to yield cyclic peroxidic derivatives known as ozonides. Ozonization of alkenes has been studied extensively for many years, but there is still disagreement about the mechanism (or mechanisms) involved because some alkenes react with ozone to give oxidation products other than ozonides. Several oxidizing reagents react with alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide.
• 11.8: Terminal Alkynes as Acids
A characteristic and synthetically important reaction of ethyne and terminal alkynes is salt (“acetylide”) formation with very strong bases. In such reactions the alkynes behaves as an acids in the sense that they give up protons to suitably strong bases.
• 11.E: Alkenes and Alkynes II (Exercises)
These are the homework exercises to accompany Chapter 11 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."
11: Alkenes and Alkynes II - Oxidation and Reduction Reactions. Acidity of Alkynes
An organic compound commonly is said to be "reduced" if reaction leads to an increase in its hydrogen content or a decrease in its oxygen content. The compound would be "oxidized" if the reverse change took place:
This is a very unsatisfactory definition because many oxidation-reduction or redox reactions do not involve changes in hydrogen or oxygen content, as the following example illustrates:
Redox reactions are better defined in terms of the concept of electron transfer. Thus an atom is said to be oxidized if, as the result of a reaction, it experiences a net loss of electrons; and is reduced if it experiences a net gain of electrons. This simple definition can be used to identify oxidation or reduction processes at carbon in terms of a scale of oxidation states for carbon based on the electronegativities of the atoms attached to carbon. The idea is to find out whether in a given reaction carbon becomes more, or less, electron-rich. We will use the following somewhat arbitrary rules:
1. Elementary carbon is assigned the zero oxidation state.
2. The oxidation state of any chemically bonded carbon may be assigned by adding $-1$ for each more electropositive atom and $+1$ for each more electronegative atom, and $0$ for each carbon atom bonded directly to the carbon of interest (see Figure 10-11) for the Pauling electronegativity scale). That is,
$-1$ for electropositive atoms, $\ce{H}$, $\ce{B}$, $\ce{Na}$, $\ce{Li}$, $\ce{Mg}$,
$+1$ for electronegative atoms, halogens $\ce{O}$, $\ce{N}$, $\ce{S}$, and
$0$ for carbon.
The rationale for this mode of operation can be seen if we look more closely at the example of $\ce{CH_3Cl} + \ce{Mg} \rightarrow \ce{CH_3-Mg-Cl}$. Chlorine is more electronegative than either carbon or magnesium. Carbon is more electronegative than magnesium. Thus $\ce{CH_3Cl}$ is written properly with a polar bond as $\overset{\delta \oplus}{\ce{CH_3}} --- \overset{\delta \ominus}{\ce{Cl}}$, whereas the $\ce{C-Mg}$ bond is oppositely polarized, $\overset{\delta \ominus}{\ce{CH_3}} --- \overset{\delta \oplus}{\ce{Mg}}$. If all of the bonds were ionized completely, we could write
$\ce{CH_3^+} + \ce{Cl}^\ominus + \ce{Mg}^0 \rightarrow \ce{CH_3} :^\ominus + \ce{Mg}^{2 \oplus} + \ce{Cl}^\ominus$
and it would be completely clear that carbon gains two electrons (is reduced), while magnesium loses two electrons (is oxidized). But because covalent, or at most polar, bonds actually are involved, it is much more difficult to determine whether oxidation or reduction occurs.
3. In compounds with multiple bonds (, ), the attached heteroatom is counted twice or three times, depending on whether the bond is double or triple.
4. A formal positive charge on carbon changes the oxidation state by $+1$, and a formal negative charge by $-1$; an odd electron on carbon leaves the oxidation state unchanged.
To illustrate, the oxidation state of carbon in four representative examples is determined as follows:
Using this approach, we can construct a carbon oxidation scale, as in Table 11-1. Any reaction that increases the degree of oxidation of carbon corresponds to a loss of electrons (oxidation), and a reaction that decreases the oxidation level corresponds to a gain of electrons (reduction). Two examples follow:
Table 11-1: Carbon Oxidation States of Representative Organic Compounds ($\ce{R} =$ alkyl)
We recommend this scheme of oxidation states only as an aid to identify and balance redox reactions. Also, the terminology "redox" should not be confused with the mechanism of a reaction, as there is no connection between them. A moment's reflection also will show that virtually all reactions theoretically can be regarded as redox reactions, because in almost every reaction the reacting atoms experience some change in their electronic environments. Traditionally, however, reactions are described as redox reactions of carbon only when there is a net change in the oxidation state of the carbon atoms involved. An indication of just how arbitrary this is can be seen by the example of addition of water to ethene. This reaction usually is not regarded as an oxidation-reduction reaction because there is no net change in the oxidation state of the ethene carbons, despite the fact that, by our rules, one carbon is oxidized and the other reduced:
$\overset{-2}{\ce{CH_2}} = \overset{-2}{\ce{CH_2}} + \ce{H_2O} \rightarrow \overset{-3}{\ce{CH_3}} \overset{-1}{\ce{CH_2}} \ce{OH}$
Apart from indicating when oxidation or reduction occurs, the oxidation scales is useful in balancing redox equations. For example, consider the following oxidation of ethenylbenzene (styrene) with potassium permanganate:
To determine how many moles of permanganate ion are required to oxidize one mole of styrene in this reaction, first determine the net change in oxidation state of the reacting carbons:
Second, determine the net change in oxidation state of manganese for $\ce{MnO_4^-} \rightarrow \ce{MnO_2}$:
Therefore we need three moles of styrene for every eight moles of permanganate:
To get the overall atom and electrical balance for Equation 11-1, the requisite amounts of $\ce{H_2O}$ must be added, but the 3:8 ratio will remain unchanged:
Because $\ce{KOH}$ reacts in a nonoxidative way with carboxylic acids to form carboxylate salts \(\left( \ce{RCO_2H} + \ce{KOH} \rightarrow \ce{RCO_2K} + \ce{H_2O} \right), the final equation is
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)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.01%3A_Oxidation-Reduction_of_Organic_Compounds.txt |
Addition of hydrogen to a multiple bond is hydrogenation. It is applicable to almost all types of multiple bonds and is of great importance in synthetic chemistry, particularly in the chemical industry. Probably the most important technical example is production of ammonia by the hydrogenation of nitrogen:
This may appear to be a simple process, but in fact it is difficult to carry out because the equilibrium is not very favorable. High pressures ($150$-$200 \: \text{atm}$) are required to get a reasonable conversion, and high temperatures ($430$-$510^\text{o}$) are necessary to get reasonable reaction rates. A catalyst, usually iron oxide, also is required. The reaction is very important because ammonia is used in ever-increasing amounts as a fertilizer either directly or through conversion to urea or ammonium salts.
Production of ammonia requires large quantities of hydrogen, most of which comes from the partial oxidation of hydrocarbons with water or oxygen. A simple and important example is the so-called "methane-steam gas" reaction, which is favorable only at very high temperatures because of the entropy effect in the formation of $\ce{H_2}$ (see Section 4-4B):
Therefore the fertilizer industry is allied closely with the natural gas and petroleum industries, and for obvious reasons ammonia and hydrogen often are produced at the same locations.,
Alkenes and alkynes add hydrogen much more readily than does nitrogen. For example, ethene reacts rapidly and completely with hydrogen at ordinary pressures and temperatures in the presence of metal catalysts such as nickel, platinum, palladium, copper, and chromium:
These reactions are unlike any we have encountered so far. They are heterogeneous reactions, which means that the reacting system consists of two or more phases. Usually, the metal catalyst is present as a finely divided solid suspension in the liquid or solution to be reduced. Alternatively, the metal is deposited on an inert solid support such as carbon, barium sulfate, alumina $\left( \ce{Al_2O_3} \right)$, or calcium carbonate. Then the mixture of the liquid substrate and solid catalyst is shaken or stirred in a hydrogen atmosphere. However, then actual reaction takes place at the surface of the metal catalyst and is an example of heterogeneous or surface catalysis.
Mechanism of Hydrogenation
The exact mechanisms of heterogeneous reactions are difficult to determine, but much interesting and helpful information has been obtained for catalytic hydrogenation. The metal catalyst is believed to act by binding the reactants at the surface of a crystal lattice. As an example, consider the surface of a nickel crystal (Figure 11-1). The nickel atoms at the surface have fewer neighbors (lower covalency) than the atoms in the interior of the crystal. The surface atoms therefore have residual bonding capacity and might be expected to combine with a variety of substances.
It has been shown experimentally that ethene combines exothermically $\left( \Delta H^0 = -60 \: \text{kcal/mol} \right)$ and reversibly with a metal surface. Although the precise structure of the ethene-nickel complex is unknown, the bonding to nickel must involve the electrons of the double bond because saturated hydrocarbons, such as ethane, combine only weakly with the nickel surface. A possible structure with carbon-nickel $\sigma$ bonds is shown in Figure 11-1.
Hydrogen gas combines with nickel quite readily with dissociation of the $\ce{H-H}$ bonds and formation of $\ce{Ni-H}$ bonds (nickel hydride bonds). The overall hydrogenation process is viewed as a series of reversible and sequential steps, as summarized in Figure 11-2. First the reactants, hydrogen and ethene, are adsorbed on the surface of the metal catalyst. The energies of the metal-hydrogen and metal-carbon bonds are such that, in a second step, a hydrogen is transferred to carbon to give an ethyl attached to nickel. This is the halfway point. In the next step, the nickel-carbon bond is broken and the second carbon-hydrogen bond is formed. Hydrogenation is now complete and the product is desorbed from the catalyst surface.
Ethane has a low affinity for the metal surface and, when desorbed, creates a vacant space for the adsorption of new ethene and hydrogen molecules. The cycle continues until one of the reagents is consumed or some material is adsorbed that "poisons" the surface and makes it incapable of further catalytic activity. Because the reaction occurs only on the surface, small amounts of a catalyst poison can completely stop the reaction.
As might be expected for the postulated mechanism, the spacings of the metal atoms in the crystal lattice are quite important in determining the hydrogenation rates. The mechanism also accounts for the observation that hydrogen usually adds to an alkene in the suprafacial manner. To illustrate, 1,2-dimethylcyclohexene is reduced to cis-1,2-dimethylcyclohexane:
Catalyst Activity and Selectivity
For maximum catalytic activity, the metal usually is prepared in a finely divided state. This is achieved for platinum and palladium by reducing the metal oxides with hydrogen prior to hydrogenation of the alkene. A specially active form of nickel ("Raney nickel") is prepared from a nickel-aluminum alloy. Sodium hydroxide is added to the alloy to dissolve the aluminum. The nickel remains as a black powder which is pyrophoric (burns in air) if not kept moist:
$2 \ce{Ni-Al} + 2 \ce{OH}^\ominus + 2 \ce{H_2O} \rightarrow 2 \ce{Ni} + 2 \ce{AlO_2^-} + 3 \ce{H_2}$
Highly active platinum, palladium, and nickel catalysts also can be obtained by reduction of metal salts with sodium borohydride $\left( \ce{NaBH_4} \right)$.
As mentioned previously, multiple bonds are not hydrogenated with equal facility. This fact can be used to advantage in carrying out selective reactions. For instance, hydrogenation of a carbon-carbon double bond can be achieved without simultaneously reducing a carbonyl bond in the same molecule. For example the carbon-carbon double bond of the following aldehyde can be reduced selectively:
Alkynes are hydrogenated more easily than alkenes mainly because alkynes are adsorbed more readily on the catalyst surface. Hydrogenation proceeds in stages, first to the cis-alkene and then to the alkane. For example,
Normally, it is not possible to stop the hydrogenation of an alkyne at the alkene stage, but if the catalyst is suitably deactivated, addition to the triple bond can be achieved without further addition occurring to the resulting double bond. The preferred catalyst for selective hydrogenation of alkynes is palladium partially "poisoned" with a lead salt (Lindlar catalyst). This catalyst shows little affinity for adsorbing alkenes and hence is ineffective in bringing about hydrogenation to the alkane stage:
Aromatic hydrocarbons are hydrogenated with considerable difficulty, requiring higher temperatures, higher pressures, and longer reaction times than for alkenes or alkynes:
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)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.02%3A_Hydrogenation_with_Heterogeneous_Catalysts.txt |
In addition to having synthetic applications, catalytic hydrogenation is useful for analytical and thermochemical purposes. The analysis of a substance for the number of carbon-carbon double bonds it contains is carried out by measuring the uptake of hydrogen for a known amount of sample. Measurement of the heat evolved in the hydrogenation of alkenes gives information as to the relative stabilities of alkenes, provided that the differences in $\Delta S^0$ values are small.
The experimental values of $\Delta H^0$ for hydrogenation of a number of alkenes and alkynes are listed in Table 11-2. The $\Delta H^0$ calculated from average bond energies is $-30 \: \text{kcal/mol}$ for a double bond and $-69 \: \text{kcal/mol}$ for a triple bond. The divergences from these values reflect the influence of structure on the strengths of multiple bonds. Some important generalizations can be made:
1. The more alkyl groups or other substituents there are on the multiple bond, the less heat is evolved on hydrogenation. Because less heat evolved signifies a stronger, more stable bond, it appears that alkyl substitution increases the stability (strength) of the multiple bond.
2. Trans isomers of 1,2-dialkyl-substituted ethenes evolve less heat (are more stable) than the corresponding cis isomers. This is the result of molecular overcrowding in the cis isomers from nonbonded interactions between two alkyl groups on the same side of the double bond. The effect amounts to almost $10 \: \text{kcal/mol}$ with two cis-tert-butyl groups. This effect is another manifestation of steric hindrance and can be seen most clearly with space-filling models (Figure 11-3).
3. Conjugated dienes are more stable than isolated dienes (compare 1,3- and 1,4-pentadiene).
4. Cumulated dienes appear to be less stable than conjugated or isolated dienes (see 1,2-propadiene).
Table 11-2: Heats of Hydrogenation of Gaseous Alkenes and Alkynes $\left( \text{kcal/mol}, \: 1 \: \text{atm}, \: 25^\text{o} \right)$
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."
11.04: Hydrogenation with Homogeneous Catalysts
Hydrogen addition to multiple bonds is catalyzed by certain complex metal salts in solution. This may be described as homogeneous catalysis and, compared to heterogeneous catalysis, is a relatively new development in the area of hydrogenation reactions. Rhodium and ruthenium salts appear to be generally useful catalysts:
At present, homogeneous catalysis for routine hydrogenation reactions offers little advantage over the convenience and simplicity of heterogeneous catalysis. Suprafacial addition of hydrogen is observed with both types of catalytic systems. However, greater selectivity can be achieved with homogeneous catalysts because they appear to be more sensitive to steric hindrance and are less likely to cause rearrangement, dissociation, and hydrogenation of other bonds (e.g., $\ce{-NO_2}$ and ).
The most thoroughly investigated homogeneous hydrogenation catalyst is the four-coordinated rhodium complex $\ce{Rh} \left[ \ce{(C_6H_5)_3P} \right]_3 \ce{Cl}$. This catalyst is called Wilkinson's catalyst after its discoverer, G. Wilkinson. In 1973, the Nobel Prize in chemistry was awarded jointly to Wilkinson and E. O. H. Fischer for their respective contributions to the field of organometallic chemistry. As you will see in this and later chapters, compounds with carbon-metal bonds (organometallic compounds) are extremely useful reagents, reactive intermediates, or catalysts in organic reactions. To a very large extent, the work of Fischer and Wilkinson created the current interest and developments in the field of transition-metal organic chemistry, which will be discussed in Chapter 31.
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."
11.05: Hydrogenation with Diimide
There are alternative ways to add hydrogen to a multiple bond besides the catalytic methods described in the previous sections. The most useful of these are homogeneous reactions utilizing diimide, $\ce{HN=NH}$, and diborane, $\ce{B_2H_6}$.
The behavior and reactivity of diimide can be understood best by considering the thermochemistry of hydrogenation of nitrogen:
The first step is strongly endothermic and is the main hurdle to overcome in the hydrogenation of nitrogen to ammonia. Conversely, the reverse reaction, which is the dehydrogenation of diimide, is strongly exothermic. Therefore we may expect that diimide will have a pronounced tendency to revert to molecular nitrogen. This is in fact so and, at normal temperatures, diimide exists only as a transient intermediate that cannot be isolated. It is extremely reactive and readily transfers hydrogen to carbon-carbon multiple bonds:
In practice, diimide is generated as it is needed in the presence of the compound to be hydrogenated. There are various ways to do this, but one of the simplest methods is dehydrogenation of hydrazine with oxidizing agents such as atmospheric oxygen or hydrogen peroxide:
$\ce{H_2N-N_2} \underset{\text{or} \: \ce{O_2}}{\overset{\ce{H_2O_2}}{\longrightarrow}} \ce{HN=NH} + 2 \ce{H}^\oplus + 2 \ce{e}^\ominus$
Hydrazine actually has been used as a hydrogenating agent for over sixty years, but it was not until the 1960's that the diimide intermediate in such reactions was recognized.
The hydrogenation step is stereospecific and transfers hydrogen in the suprafacial manner. For example, alkynes are converted to cis-alkenes:
There are no detectable intermediate stages or rearrangements in diimide hydrogenation. The reaction is visualized as a six-center (pericyclic) process in which the bonds are broken and made in a concerted fashion:
An important difference between diimide hydrogenation and catalytic hydrogenation is that diimide will react only with symmetrical or nonpolar bonds $\left( \ce{C=C}, \: \ce{C \equiv C}, \: \ce{N=N} \right)$, whereas hydrogen can add, albeit reluctantly, to polar bonds $\left( \ce{C=O}, \: \ce{C=N} \right)$. Diimide does not attack the stronger polar bonds probably because it does not survive long enough to do so. It selfdestructs in the absence of a reactive substrate to give nitrogen and hydrazine:
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)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.03%3A_Heats_of_Hydrogenation.txt |
An especially valuable group of intermediates can be prepared by addition of an compound to carbon-carbon double or triple bonds:
The reaction is called hydroboration and is a versatile synthesis of organoboron compounds. One example is the addition of diborane, $\ce{B_2H_6}$, to ethene. Diborane behaves as though it is in equilibrium with $\ce{BH_3}$ $\left( \ce{B_2H_6} \rightleftharpoons 2 \ce{BH_3} \right)$, and addition proceeds in three stages:
The monoalkylborane, $\ce{RBH_2}$, and the dialkylborane, $\ce{R_2BH}$, seldom are isolated because they rapidly add to the alkene. These additions amount to reduction of both carbons of the double bond:
Organoboranes can be considered to be organometallic compounds. Elemental boron does not have the properties of a metal, and boron-carbon bonds are more covalent than ionic. However, boron is more electropositive than either carbon or hydrogen and when bonded to carbon behaves like most metals in the sense that bonds are polarized with $\ce{R}$ negative and boron positive:
Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes.
Hydroboration
The simplest borane, $\ce{BH_3}$, exists as the dimer, $\ce{B_2H_6}$, or in complexed form with certain ethers or sulfides:
Any of these $\ce{BH_3}$ compounds adds readily to most alkenes at room temperature or lower temperatures. The reactions usually are carried out in ether solvents, although hydrocarbon solvents can be used with the borane-dimethyl sulfide complex. When diborane is the reagent, it can be generated either in situ or externally through the reaction of boron trifluoride with sodium borohydride:
$3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{H_4} + 4 \ce{BF_3} \rightarrow 2 \ce{B_2H_6} + 3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{F_4}$
Hydroborations have to be carried out with some care, because diborane and many alkylboranes are highly reactive and toxic substances; many are spontaneously flammable in air.
With unsymmetrical alkenes, hydroboration occurs so that boron becomes attached to the less-substituted end of the double bond:
These additions are suprafacial additions:
Furthermore, when there is a choice, addition occurs preferentially from the less crowded side of the double bond:
If the alkene is a bulky molecule, borane may add only one or two alkene molecules to give either mono- or dialkylborane, $\ce{RBH_2}$ or $\ce{R_2BH}$, respectively, as the following reactions show:
These bulky boranes still possess $\ce{B-H}$ bonds and can add further to a multiple bond, but they are highly selective reagents and add only if the alkene or alkyne is unhindered. This selectivity can be useful, particularly to 1-alkynes, which are difficult to stop at the alkenylborane stage when using diborane:
With a bulky dialkylborane, such as di-(1,2-dimethylpropyl)borane, further addition to the alkenylborane does not occur.
An especially selective hydroborating reagent is prepared from 1,5-cyclooctadiene and borane. The product is a bicyclic compound of structure $1$ (often abbreviated as 9-BBN), in which the residual $\ce{B-H}$ bond adds to unhindered alkenes with much greater selectivity than is observed with other hydroborating reagents. It is also one of the few boranes that reacts sufficiently slowly with oxygen that it can be manipulated in air.
An example of the difference in selectivity in the hydroboration of cis-4-methyl-2-pentene with $\ce{B_2H_6}$ and $1$ follows:
Mechanism of Hydroboration
According to the electronegativity chart (Figure 10-11), the boron-hydrogen bond is polarized in the sense $\overset{\delta \oplus}{\ce{B}} --- \overset{\delta \ominus}{\ce{H}}$. Therefore the direction of addition of $\ce{B_2H_6}$ to propene is that expected of a polar mechanism whereby the electrophilic boron atom becomes bonded to the less-substituted carbon of the double bond.
Stepwise mechanism
However, there is no firm evidence to suggest that a carbocation intermediate is formed through a stepwise electrophilic addition reaction. For this reason, the reaction often is considered to be a four-center concerted addition.
Concerted mechanism
The stepwise formulation explains why boron becomes attached to the less-substituted carbon, but does not account for the fact that the reactions show no other characteristics of carbocation reactions. This could be because of an expected, extraordinarily fast rate of hydride-ion transfer to the carbocation. A more serious objection to the stepwise mechanism is that alkynes react more rapidly than alkenes, something which normally is not observed for stepwise electrophilic additions (cf. Section 10-5).
Isomerization of Alkylboranes
Some alkylboranes rearrange at elevated temperatures $\left( 160^\text{o} \right)$ to form more stable isomers. For example, the alkylborane $2$, produced by hydroboration of 3-ethyl-2-pentene, rearranges to $3$ on heating:
In general, the boron in alkylboranes prefers to be at the end of a hydrocarbon chain so it is bonded to a primary carbon where steric crowding around boron is least severe. Thus rearrangement tends to proceed in the direction
Rearrangement is associated with the fact that hydroboration is reversible at elevated temperatures. This makes possible a sequence of elimination-addition reactions in which boron becomes attached to different carbons and ultimately leads to the most stable product that has boron bonded to the carbon at the end of the chain:
Rearrangement of alkylboranes can be used to transform alkenes with double bonds in the middle of the chain into less stable 1-alkenes; for example, $\ce{RCH=CHCH_3} \rightarrow \ce{RCH_2-CH=CH_2}$. The procedure involves hydroboration of the starting alkene in the usual manner; the borane then is isomerized by heating. An excess of 1-decene (bp $170^\text{o}$) then is added to the rearranged borane and the mixture is reheated. Heating causes the alkylborane to dissociate into 1-alkene and $\ce{HBR_2}$; the 1-decene "scavenges" the $\ce{HBR_2}$ as it forms, thereby allowing a more volatile 1-alkene (bp $<170^\text{o}$) to be removed by simple distillation. Thus, for the rearrangement of 3-ethyl-2-pentene to 3-ethyl-1-pentene,
Synthetic Reactions of Organoboranes
Alkylboranes formed in the hydroboration of alkenes and alkynes seldom are isolated; for the most part they are used as reactive intermediates for the synthesis of other substances. In the reactions of alkylboranes, the $\ce{B-C}$ bond is cleaved in the sense $\ce{B}^\oplus - \ce{C}^\ominus$ so that carbon is transferred to other atoms, such as $\ce{H}$, $\ce{O}$, $\ce{N}$, and $\ce{C}$, with its bonding electron pair:
In the first of these reactions (Equation 11-2), a hydrocarbon is produced by the cleavage of a borane, $\ce{R_3B}$, with aqueous acid, or better, with anhydrous propanoic acid, $\ce{CH_3CH_2CO_2H}$. The overall sequence of hydroboration-acid hydrolysis achieves the reduction of a carbon-carbon multiple bond without using hydrogen and a metal catalyst or diimide (Table 11-3):
Table 11-3: Some Methods of Hydrogenation of Carbon-Carbon Multiple Bonds
The second reaction (Equation 11-3) achieves the synthesis of a primary alcohol by the oxidation of the alkylborane with hydrogen peroxide in basic solution. Starting with a 1-alkene, one can prepare a primary alcohol in two steps:
This sequence complements the direct hydration of 1-alkenes, which gives secondary alcohols:
Hydroboration of an alkene and subsequent reactions of the product trialkylborane, either with hydrogen peroxide or with acid, appear to be highly stereospecific. For example, 1-methylcyclopentene gives exclusively trans-2-methylcyclopentanol on hydroboration followed by reaction with alkaline hydrogen peroxide. This indicates that, overall, the reactions result in suprafacial addition of water to the double bond:
Hydroboration of an alkyne followed by treatment of the alkenylborane with basic peroxide provides a method of synthesis of aldehydes and ketones. Thus hydroboration of 1-hexyne and oxidation of the 1-hexenylborane, $4$, with hydrogen peroxide gives hexanal by way of the enol:
If $4$ is treated with deuteriopropanoic acid, replacement of $\ce{-BR_2}$ by deuterium occurs with retention of configuration, forming trans-hexene-1-$\ce{D_1}$:
Mechanism of Oxidation of Alkylboranes
The stereospecific oxidation of alkylboranes occurs with hydrogen peroxide by an interesting and important general type of rearrangement which, for these reactions, involves migration of an organic group from boron to oxygen. The first step in the oxidation depends on the fact that tricoordinate boron has only six electrons in its valence shell and therefore behaves as if it were electron-deficient. The first step is bond formation at boron by the strongly nucleophilic peroxide anion (from $\ce{H_2O_2} + \ce{OH}^\ominus \rightleftharpoons ^\ominus \ce{OOH} + \ce{H_2O}$) to give a tetracovalent boron intermediate:
In the second step, an alkyl group moves with its bonding electron pair from boron to the neighboring oxygen and, in so doing, displaces hydroxide ion. The stereochemical configuration of the migrating $\ce{R}$ group is retained:
Reaction is completed by hydrolysis of the $\ce{B-O}$ bond:
All three groups on boron are replaced in this manner.
The rearrangement step (Equation 11-5) is an example of many related rearrangements in which a group, $\ce{R}$, migrates with its bonding electrons from one atom to an adjacent atom. We already have encountered an example in the rearrangement of carbocations (Section 8-9B):
The difference between the carbocation rearrangement and the rearrangement of Equation 11-5 is that $\ce{R}$ migrates from boron to oxygen as $\ce{HO}^\ominus$ departs in what might be considered an internal $S_\text{N}2$ reaction. We can generalize this kind of reaction of boron with a substance, $\ce{X-Y}$, as in Equation 11-6:
An example of the use of an $\ce{X-Y}$ reagent is conversion of alkylboranes to primary amines with hydroxylaminesulfonic acid, $\ce{H_2NOSO_3H}$ (Equation 11-4). The key steps are attack of the nucleophilic nitrogen at boron,
followed by rearrangement,
and hydrolysis,
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)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.06%3A_Addition_of_Boron_Hydrides_to_Alkenes._Organoboran.txt |
Ozonization
Most alkenes react readily with ozone $\left( \ce{O_3} \right)$, even at low temperatures, to yield cyclic peroxidic derivatives known as ozonides. For example,
These substances, like most compounds with peroxide $\left( \ce{O-O} \right)$ bonds, may explode violently and unpredictably. Therefore ozonizations must be carried out with appropriate caution. The general importance of these reactions derives not from the ozonides, which usually are not isolated, but from their subsequent products. The ozonides can be converted by hydrolysis with water and reduction, with hydrogen (palladium catalyst) or with zinc and acid, to carbonyl compounds that can be isolated and identified. For example, 2-butene gives ethanal on ozonization, provided the ozonide is destroyed with water and a reducing agent which is effective for hydrogen peroxide:
An alternative procedure for decomposing ozonides from di- or trisubstituted alkenes is to treat them with methanol $\left( \ce{CH_3OH} \right)$. The use of this reagent results in the formation of an aldehyde or ketone and a carboxylic acid:
The overall ozonization reaction sequence provides an excellent means for locating the positions of double bonds in alkenes. The potentialities of the method may be illustrated by the difference in reaction products from the 1- and 2-butenes:
Mechanism of Ozonization
Ozonization of alkenes has been studied extensively for many years, but there is still disagreement about the mechanism (or mechanisms) involved because some alkenes react with ozone to give oxidation products other than ozonides. It is clear that the ozonide is not formed directly, but by way of an unstable intermediate called a molozonide. the molozonide then either isomerizes to the "normal" ozonide or participates in other oxidation reactions. Although the structure of normal ozonides has been established beyond question, that of the molozonide, which is very unstable even at $-100^\text{o}$, is much less certain.
The simplest and most widely accepted mechanism involves formation of a molozonide by a direct cycloaddition of ozone to the double bond.$^1$
Isomerization of the molozonide appears to occur by a fragmentation-recombination reaction, as shown in Equations 11-7 and 11-8:
Hydroxylation of Alkenes
Several oxidizing reagents react with alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide as $\ce{HO-OH}$. Of particular importance are alkaline permanganate $\left( \ce{MnO_4^-} \right)$ and osmium tetroxide $\left( \ce{OsO_4} \right)$, both of which react in an initial step by a suprafacial cycloaddition mechanism like that postulated for ozone.
Each of these reagents produces cis-1,2-dihydroxy compounds (diols) with cycloalkenes:
Osmium tetroxide is superior to permanganate in giving good yields of diol, but its use is restricted because it is a very costly and very toxic reagent.
Oxidation with Peroxidic Compounds. Oxacyclopropane (Oxirane) Formation
Alkenes can be oxidized with peroxycarboxylic acids, $\ce{RCO_3H}$, to give oxacyclopropanes (oxiranes, epoxides), which are three-membered cyclic ethers:
The reaction, known as epoxidation, is valuable because the oxacyclopropane ring is cleaved easily, thereby providing a route to the introduction of many kinds of functional groups. In fact, oxidation of alkenes with peroxymethanoic acid (peroxyformic acid), prepared by mixing methanoic acid and hydrogen peroxide, usually does not stop at the oxacyclopropane stage, but leads to ring-opening and the subsequent formation of a diol:
This is an alternative scheme for the hydroxylation of alkenes (see Section 11-7C). However, the overall stereochemistry is opposite to that in permanganate hydroxylation. For instance, cyclopentene gives trans-1,2-cylcopentanediol. First the oxirane forms by suprafacial addition and then undergoes ring opening to give the trans product:
The ring opening is a type of $S_\text{N}2$ reaction. Methanoic acid is sufficiently acidic to protonate the ring oxygen, which makes it a better leaving group, thus facilitating nucleophilic attack by water. The nucleophile always attacks from the side remote from the leaving group:
The peroxyacids that are used in the formation of oxacyclopropanes include peroxyethanoic $\left( \ce{CH_3CO_3H} \right)$, peroxybenzoic $\left( \ce{C_6H_5CO_3H} \right)$, and trifluoroperoxyethanoic $\left( \ce{CF_3CO_3H} \right)$ acids. A particularly useful peroxyacid is 3-chloroperoxybenzoic acid, because it is relatively stable and is handled easily as the crystalline solid. The most reactive reagent is trifluoroperoxyethanoic acid, which suggests that the peroxyacid behaves as an electrophile (the electronegativity of fluorine makes the $\ce{CF_3}$ group strongly electron-attracting). The overall reaction can be viewed as a cycloaddition, in which the proton on oxygen is transferred to the neighboring carbonyl oxygen more or less simultaneously with formation of the three-membered ring:
A reaction of immense industrial importance is the formation of oxacyclopropane itself (most often called ethylene oxide) by oxidation fo ethene with oxygen over a silver oxide catalyst at $300^\text{o}$:
Oxacyclopropane is used for many purposes, but probably the most important reaction is ring opening with water to give 1,2-ethanediol (ethylene glycol, bp $197^\text{o}$). This diol, mixed with water, is employed widely in automotive cooling systems to provide both a higher boiling and lower freezing coolant than water alone:
Propene and higher alkenes are not efficiently epoxidized by oxygen and $\ce{Ag_2O}$ in the same way as ethene because of competing attack at other than the double-bond carbons.
$^1$The ozone structure shown here with single electrons having paired spins on the terminal oxygens accords both with the best available quantum mechanical calculations and the low dipole moment of ozone, which is not consonant with the conventional $\ce{O=} \overset{\oplus}{\ce{O}} - \overset{\ominus}{\ce{O}}$ structure. See W. A. Goddard III, T. H. Dunning, Jr., W. J. Hunt, and P. J. Hay, Accounts of Chemical Research 6, 368 (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)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.07%3A_Oxidation_Reactions.txt |
A characteristic and synthetically important reaction of ethyne and 1-alkynes is salt (“acetylide”) formation with very strong bases. In such reactions the alkynes behaves as an acids in the sense that they give up protons to suitably strong bases:
Water is too weak a base to accept protons from alkynes; consequently no measurable concentration of $\ce{H_3O}^\oplus$ is expected from the ionization of alkynes in dilute aqueous solutions. Therefore we have no quantitative measure of 1-alkyne acidity in aqueous solution other than that it probably is about $10^{10}$ times less acidic than water, as judged from measurements in other solvents to be discussed shortly. In the gas phase, however, the situation is reversed, and ethyne is a stronger acid than water:
This reversal is of little practical value because organic reactions involving ions normally are not carried out in the gas phase. However, it should alert us to the tremendous role that solvents play in determining acidities by their abilities (some much more than others) to stabilize ions by the property known as solvation. (Section 11-8A.)
Liquid ammonia is a more useful solvent than water for the preparation of 1-alkyne salts. A substantial amount of the alkyne can be converted to the conjugate base by amide anions (potassium or sodium amide) because a 1-alkyne is a stronger acid than ammonia.
The acidity of the terminal hydrogen in 1-alkynes provides a simple and useful test for 1-alkynes. With silver-ammonia solution ($\ce{AgNO_3}$ in aqueous ammonia), 1-alkynes give insoluble silver salts, whereas disubstituted alkynes do not:
The silver “acetylides” appear to have substantially covalent carbon-metal bonds and are less ionic than sodium and potassium alkynides. Silver-ammonia solution may be used to precipitate 1-alkynes from mixtures with other hydrocarbons. The 1-alkynes are regenerated easily from the silver precipitates by treatment with strong inorganic acids. It should be noted, however, that silver alkynides may be shock sensitive and can decompose explosively, especially when dry.
Thermodynamics of Solvation of Ions and Its Importance
Some idea of the importance of solvation can be gained from the calculated $\Delta H$ for the following process:
$\ce{Na^+ (g) + Cl^- (g) \rightarrow Na^+ (aq) + Cl^- (aq)}$
with $Delta H^o = -187\,kcal$.
The solvation energies of ions are so large that relatively small differences for different ions can have a very large effect on equilibrium constants. Thus, the ratio between the relative acidities of ethyne and water in the gas phase and in water of $10^{12}$ corresponds at $25^\text{o}$ to an overall $\Delta G^0$ difference in solvation energies of approximately $16 \: \text{kcal}$, which is less than $10\%$ of the total solvation energies of the ions. Further difficulties arise because of differences between solvation energies and interactions between the ions in different solvents. Thus the acidities of 1-alkynes relative to other acids have been found to change by a factor of $10^{11}$ in different solvents. For this reason, we must be particularly careful in comparing the rates and equilibrium constants of ionic reactions to take proper account of solvation and ion interaction effects.
An excellent rule of thumb is that, other things being equal, large ions are more stable than small ions in the gas phase, with the opposite being true in polar solvents, where small ions are more strongly solvated (thus more stable) than large ions. For comparison,
From (1) minus (3), the solvation energy of gaseous $\ce{Li}^\oplus$ is $47 \: \text{kcal mol}^{-1}$ greater than $\ce{K}^\oplus$; and from (1) minus (2), that of $\ce{F}^\ominus$ is $35 \: \text{kcal mol}^{-1}$ greater than that of $\ce{I}^\ominus$. Such differences in solvation energies can have considerable effects on reactivity, and you may remember from Section 8-7E that $\ce{F}^\ominus$ is a weaker nucleophile than $\ce{I}^\ominus$, largely because of its greater solvation energy.
Why is Ethyne a Stronger Acid than Ethane or Ethene?
If we compare acid strengths of the simple hydrocarbons, we find that ethyne is substantially more acidic than ethene or ethane in the gas phase or in solution. Why is this? The simplest explanation is that there is a direct connection between $\ce{C-H}$ acidity and the amount of $s$ character associated with the $\sigma$-bonding carbon orbital. Other things being equal, acidity increases with increasing $s$ character in the carbon orbital.
Alkynide Anions as Nucleophiles
1-Alkynes are very weak acids, hence their conjugate bases, $\ce{RC \equiv C}^\ominus$, are quite strong bases. These anions also are reactive carbon nucleophiles, and it is this property that makes them useful for organic synthesis. Recall from Chapter 8 that one of the most generally useful organic reactions is a displacement reaction in which an anionic nucleophile, $\ce{Nu}^\ominus$, attacks an alkyl derivative, $\ce{RX}$, to displace $\ce{X}^\ominus$ and form a new bond between carbon and the nucleophile:
The displaced group $\ce{X}$ often is a halide ion (chloride, bromide, or iodide), and if the entering nucleophile $\ce{Nu}^\ominus$ is an alkynide anion, the reaction leads to formation of a carbon-carbon bond:
With those $\ce{RX}$ derivatives that undergo nucleophilic displacement readily, this is a general method of forming a $\ce{C-C}$ bond, thereby leading to substituted alkynes. The alkynide salts generally used are those of lithium, sodium, potassium, or magnesium.
Coupling Reactions of Alkynes
Another reaction of 1-alkynes that extends the carbon chain is a coupling reaction in which the alkyne dimerizes under the influence of a cuprous salt, usually cuprous ammonium chloride:
This addition of one molecule of alkyne to another is formally analogous to the dimerization of alkenes under the influence of sulfuric acid (see Section 10-9), but the mechanisms are quite different.
If the reaction is carried out in the presence of an oxidizing agent, such as $\ce{O_2}$ or a cupric salt dissolved in pyridine (a weak base), a different product is obtained. Under these conditions, oxidative coupling occurs to give a conjugated diyne:
Although the details of the mechanisms of these alkyne reactions are not known, it is likely that the ability of 1-alkynes to form carbon-metal bonds with metals such as copper is a key factor. Other oxidative coupling reactions occur with transition metals, and this will be discussed in detail in Chapter 31.
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)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.08%3A_Terminal_Alkynes_as_Acids.txt |
Many important hydrocarbons, known as cycloalkanes, contain rings of carbon atoms linked together by single bonds. The simple cycloalkanes of formula \((CH_2)_n\), make up a particularly important homologous series in which the chemical properties change in a much more dramatic way with increasing n than do those of the acyclic hydrocarbons \(\ce{CH_3(CH_2)}_{n-1} \ce{H}\). The cycloalkanes with small rings (\(n = 3\)-\(6\)) are of special interest in exhibiting chemical properties intermediate between those of alkanes and alkenes. In this chapter we will show how this behavior can be explained in terms of angle strain and steric hindrance, concepts that have been introduced previously and will be used with increasing frequency as we proceed further. We also discuss the conformations of cycloalkanes, especially cyclohexane, in detail because of their importance to the chemistry of many kinds of naturally occurring organic compounds. Some attention also will be paid to polycyclic compounds, substances with more than one ring, and to cycloalkenes and cycloalkynes.
• 12.1: Nomenclature and Physical Properties of Cycloalkanes
The nomenclature and physical properties of cycloalkanes covers the the IUPAC systems of naming cycloalkanes. The physical properties of cycloalkanes can explain each cycloalkane molecular structure and the relative size from simple propane to multiple carbon containing cycloakane like cyclononane.
• 12.2: Spectroscopic Properties of Cyclohexanes
The spectroscopic properties of cycloalkanes are considerably similar to those of alkanes. We mentioned previously the main features of their infrared spectra, and that their lack of ultraviolet absorption at wavelengths down to 200 nm makes them useful solvents for the determination of ultraviolet spectra of other substances. Some spectoscopic properties of cycloalkanes has been notable in the proton NMR spectra.
• 12.3: Conformations of Cycloalkanes
The equilibria (relative stabilities) and equilibration (rate of interconversion) of the rotational conformations of ethane and butane were discussed previously. If you review this material, it will be clear that forming a ring from a hydrocarbon chain will greatly reduce the number of possible staggered and eclipsed conformations. We will begin our discussion with cyclohexane because of its special importance, proceed to smaller rings, then give a brief exposition of the conformations of the l
• 12.4: Strain in Cycloalkane Rings
Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes.
• 12.5: Chemical Properties
Strain in small-ring cycloalkanes has a profound influence on their heats of combustion. We reasonably expect that other chemical properties also will be affected. Indeed, like alkenes, cyclopropane and cyclobutane undergo C−C bond cleavage reactions that are not observed for cyclopentane and cyclohexane, or for saturated, open-chain hydrocarbons.
• 12.6: The Larger Cycloalkanes and their Conformations
The Baeyer strain theory suggested that the larger cycloalkanes ring are difficult to synthesize because of angle strain associated with planar rings. We now know that, except for cyclopropane, none of the cycloalkanes have planar carbon rings and that the higher cycloalkanes have normal or nearly normal bond angles.
• 12.7: Cycloalkenes and Cycloalkanes
The C−C=C angle in alkenes normally is about 22°, which is 10° larger than the normal C−C−C angle in cycloalkanes. This means that we would expect about 20° more angle strain in small-ring cycloalkenes than in the cycloalkanes with the same numbers of carbons in the ring. Comparison of the data for cycloalkenes and for cycloalkanes reveals that this expectation is realized for cyclopropene, but is less conspicuous for cyclobutene and cyclopentene.
• 12.8: Nomenclature of Polycycloalkanes
There are many hydrocarbons and hydrocarbon derivatives with two or more rings having common carbon atoms. Compounds of this type usually are named by attaching the prefix bicyclo to the name of the open-chain hydrocarbon with the same total number of carbon atoms as in the rings.
• 12.9: Conformations of Decalin
The six-membered rings of decalin, like those of cyclohexane, are expected to be most stable in the chair form. However, there are two possible ways in which two chairs can be joined. The ring-junction hydrogens may be either on the same side of the molecule (cis-decalin) or on opposite sides (trans-decalin). When the two rings are joined through two equatorial-type bonds, trans-decalin results, whereas an axial-equatorial union gives cis-decalin.
• 12.10: Strain in Polycyclic Molecules
Knowing the importance of angle and eclipsing strain in the small-ring cycloalkanes, we should expect that these strains would become still more important in reactions.
• 12.E: Cycloalkanes, Cycloalkenes, and Cycloalkynes (Exercises)
These are the homework exercises to accompany Chapter 12 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."
12: Cycloalkanes Cycloalkenes and Cycloalkynes
The nomenclature and physical properties of cycloalkanes covers the the IUPAC systems of naming cycloalkanes. The physical properties of cycloalkanes can explain each cycloalkane molecular structure and the relative size from simple propane to multiple carbon containing cycloakane like cyclononane.
For a review, cycloalkanes contain rings of of carbon atoms linked together by a single bonds. The simple cycloalkanes of formula \(\ce{(CH_2)}_n\) make up a particularly homologous series in which the chemical properties change in a much more dramatic way with increasing \(n\) than do those of the acyclic hydrocarbons \(\ce{CH_3(CH_2)}_{n-1}\ce{H}\). The cycloalkanes with small rings (\(n=3\)-\(6\)) are of special interest in exhibiting chemical properties intermediate between those of alkanes and alkenes. In this chapter we will show how this behavior can be explained in terms of angle strain and steric hindrance, concepts that have been introduced previously and will be used with increasing frequency as we proceed further.
The IUPAC system for naming cycloalkanes and cycloalkenes was presented in some detail in Sections 3-2 and 3-3, and you may wish to review that material before proceeding further. Additional procedures are required for naming polycyclic compounds, which have rings with common carbons, and these will be discussed later in this chapter. Furthermore, given below are the table of physical properties of cycloalkanes from propane to cyclononane.
Table 12-1: Physical Properties of Alkanes and Cycloalkanes
Compounds Boiling Point \(^oC\) Melting Point \(^oC\) Density (g/ml)
propane -42 -187 0.580
cyclopropane -33 -127 0.689
butane -0.5 -135 0.579
cyclobutane 13 -90 0.689
pentane 36 -130 0.626
cyclopentane 49 -94 0.746
hexane 69 -95 0.659
cyclohexane 81 7 0.778
heptane 98 -91 0.684
cycloheptane 119 -8 0.810
octane 126 -57 0.703
cyclooctane 151 15 0.830
nonane 151 -54 0.718
cyclononane 178 11 0.845
The melting and boiling points of cycloalkanes (Table 12-1) are somewhat higher than those of the corresponding alkanes. In contrast to the more rigid cyclic compounds, the general “floppiness” of open-chain hydrocarbons makes them harder to fit into a crystal lattice (hence their lower melting points) and less hospitable toward neighboring molecules of the same type (hence their lower boiling points). The nomenclature and physical properties of cycloalkanes also includes conformation of cycloalkanes, especially cyclohexane in later discussion because of their importance to the chemistry of many kinds of naturally occurring organic compounds. The nomenclature and physical properties of cycloalkanes will also cover polycyclic compounds, substances with more than one ring, and to the cycloalkenes and cycloalkynes.
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."
12.02: Spectroscopic Properties of Cyclohexanes
The spectroscopic properties of cycloalkanes are considerably similar to those of alkanes. We mentioned previously the main features of their infrared spectra (Section 9-7D), and that their lack of ultraviolet absorption at wavelengths down to 200 nm makes them useful solvents for the determination of ultraviolet spectra of other substances (Section 9-9B).
Some spectoscopic properties of cycloalkanes has been notable in the proton NMR spectra. The proton NMR spectra of alkanes and cycloalkanes are characteristic but difficult to interpret because the chemical shifts between the various kinds of protons are usually small. Although proton spectra of simple cycloalkanes, $\ce{(CH_2)}_n$, show one sharp line at room temperature, when alkyl substituents are present, small differences in chemical shifts between the ring hydrogens occur and, with spin-spin splitting, provide more closely spaced lines than normally can be resolved. The complexity so introduced can be seen by comparing the proton spectra of cyclooctane and methylcyclohexane shown in Figure 12-1. For methyl-substituted cycloalkanes the methyl resonances generally stand out as high-field signals centered on $0.9 \: \text{ppm}$, and the area of these signals relative to the other $\ce{C-H}$ signals may be useful in indicating how many methyl groups there are (see Section 9-10K, especially Figure 9-46). However, in cyclopropanes the ring protons have abnormally small chemical shifts ($\delta = 0.22$ for cyclopropane), which often overlap with the shifts of methyl groups $\left( \delta \cong 0.9 \: \text{ppm} \right)$.
Although spectroscopic properties of cycloalkanes in proton spectra are not very useful for identification purposes, $\ce{^{13}C}$ NMR spectra are very useful. Chain-branching and ring-substitution normally cause quite large chemical-shift changes, and it is not uncommon to observe $\ce{^{13}C}$ shifts in cycloalkanes spanning $35 \: \text{ppm}$. Some special features of application of $\ce{^{13}C}$ NMR spectra to conformational analysis of cycloalkanes are described in Section 12-3D.
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)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.01%3A_Nomenclature_and_Physical_Properties_of_Cycloalkanes.txt |
The equilibria (relative stabilities) and equilibration (rate of interconversion) of the rotational conformations of ethane and butane were discussed in Section 5-2. If you review this material, it will be clear that forming a ring from a hydrocarbon chain will greatly reduce the number of possible staggered and eclipsed conformations. We will begin our discussion with cyclohexane because of its special importance, proceed to smaller rings, then give a brief exposition of the conformations of the larger rings.
Cyclohexane Conformations
If the carbons of a cyclohexane ring were placed at the corners of a regular planar hexagon, all the $\ce{C-C-C}$ bond angles would have to be $120^\text{o}$. Because the expected normal $\ce{C-C-C}$ bond angle should be near the tetrahedral value of $109.5^\text{o}$, the suggested planar configuration of cyclohexane would have angle strain at each of the carbons, and would correspond to less stable cyclohexane molecules than those with more normal bond angles. The actual normal value for the $\ce{C-C-C}$ bond angle of an open-chain $\ce{-CH_2-CH_2-CH_2}-$ unit appears to be about $112.5^\text{o}$, which is $3^\text{o}$ greater than the tetrahedral value. From this we can conclude that the angle strain at each carbon of a planar cyclohexane would be $\left( 120^\text{o} - 112.5^\text{o} \right) = 7.5^\text{o}$. Angle strain is not the whole story with regard to the instability of the planar form, because in addition to having $\ce{C-C-C}$ bond angles different from their normal values, the planar structure also has its carbons and hydrogens in the unfavorable eclipsed arrangement, as shown in Figure 12-2.
If the carbon valence angles are kept near the tetrahedral value, you will find that you can construct ball-and-stick models of the cyclohexane six-carbon ring with two quite different conformations. These are known as the "chair" and "boat" conformations (Figure 12-3). It has not been possible to separate cyclohexane at room temperature into pure isomeric forms that correspond to these conformations, and actually the two forms appear to be rapidly interconverted. The chair conformation is considerably more stable and comprises more than $99.9\%$ of the equilibrium mixture at room temperature.$^1$
Why is the boar form less stable than the chair form, if both have normal $\ce{C-C-C}$ bond angles? The answer is that the boat form has unfavorable nonbonded interactions between the hydrogen atoms around the ring. If we make all of the bond angles normal and orient the carbons to give the "extreme boat" conformation of Figure 12-4, a pair of 1,4 hydrogens (the so-called "flagpole" hydrogens) have to be very close together $\left( 1.83 \: \text{Å} \right)$. Hydrogens this close together would be on the rising part of a repulsion potential energy curve, such as Figure 4-6, for hydrogen-hydrogen nonbonded interactions. This steric hindrance at an $\ce{H-H}$ distance of $1.83 \: \text{Å}$ corresponds to a repulsion energy of about $3 \: \text{kcal mol}^{-1}$. There is still another factor that makes the extreme boat unfavorable; namely, that the eight hydrogens along the "sides" of the boat are eclipsed, which brings them substantially closer together than they would be in a staggered arrangement (about $2.27 \: \text{Å}$ compared with $2.50 \: \text{Å}$). This is in striking contrast with the chair form (Figure 12-5), for which adjacent hydrogens are in staggered positions with respect to one another all around the ring. Therefore the chair form is expected to be more stable than the boat form because it has less repulsion between the hydrogens.
You should make and inspect models such as those in Figure 12-3 to see the rather striking difference between the chair and boat conformations that is not obvious from the diagrams. You will find that the chair structure is quite rigid, and rotation does not occur around the $\ce{C-C}$ bonds with interconversion to the boat structure. In contrast, the boat form is quite flexible. Rotation about the $\ce{C-C}$ bonds permits the ring to twist one way or the other from the extreme boat conformation to considerably more stable, equal-energy conformations, in which the flagpole hydrogens move farther apart and the eight hydrogens along the sides become largely but not completely staggered. These arrangements are called the twist-boat (sometimes skew-boat) conformations (see Figure 12-6) and are believed to be about $5 \: \text{kcal mol}^{-1}$ less stable than the chair form.
It is possible to measure the spectral properties of the twist-boat form by a very elegant technique employed by F. A. L. Anet. Because the equilibrium constant for conversion of chair to boat increases with temperature, a considerable proportion of the molecules exist as the twist-boat form in the vapor at $800^\text{o}$. If such vapor is allowed to impinge on a surface cooled to $20 \: \text{K}$, the film condensate contains about $25\%$ of the twist-boat form. At this low temperature, the twist-boat form is converted to the more stable chair form at a very slow rate. Infrared spectra can be taken of the boat-chair mixture at $10 \: \text{K}$. If the mixture is allowed to warm to $75 \: \text{K}$, the normal equilibrium favoring the chair form is established in a short time.
Dreiding Models
The spatial arrangement (stereochemistry) of cyclohexane and other organic compounds are studied conveniently with the aid of Dreiding models, which are made with standard bond angles and scaled bond distances. The bonds have stainless-steel rods that make a snap-fit into stainless-steel sleeves. Rotation is smooth about the bonds and there is sufficient flexibility to accommodate some angle strain. Dreding models of the conformations of cyclohexane are shown in Figure 12-7. Notice that these models correspond closely to the sawhorse representations in Figures 12-4, 12-5, and 12-6.
Conformational Equilibria and Equilibration for Cyclohexane Derivatives
Figure 12-5 shows that there are two distinct kinds of hydrogen in the chair form of cyclohexane - six that are close to the "average" plane of the ring (called equatorial hydrogens) and three above and three below this average plane (called axial hydrogens). This raises interesting questions in connection with substituted cyclohexanes: For example, is the methyl group in methylcyclohexane equatorial or axial? Since only one methylcyclohexane is known, the methyl group must be exclusively equatorial $\left( e \right)$, exclusively axial $\left( a \right)$, or the two forms must be interconverted so rapidly that they cannot be separated into isomeric forms. It appears that the latter circumstance prevails, with the ring changing rapidly from one chair form to another by flipping one end of the chair up and the other end down:
Such a change would cause a substituent in an axial position to go to an equatorial position and vice versa. This process is called ring inversion and its rate often is called the inversion frequency. With cyclohexane, inversion is so fast at room temperature that, on average, the molecules, flip about 100,000 times per second, over an energy barrier of about $11 \: \text{kcal mol}^{-1}$.
You will understand this flipping process if you make a model of a cyclohexane ring carrying a single substituent. By manipulating the model you can discover some of the different ways the process can occur. The simplest route is simply to flip up one corner of the ring to convert the chair into a boat and then flip down the opposite carbon:
Because of the flexibility of the boat conformation, it is possible to transform it to other boat conformations whereby carbons other than the one indicated flip down and complete the interconversion.
At room temperature the conformation of methylcyclohexane with the methyl equatorial is more stable than the one with the methyl axial by $1.7 \: \text{kcal mol}^{-1}$. The same is true of all monosubstituted cyclohexanes to a greater or lesser degree. Reasons for this can be seen from space-filling models (Figure 12-8), which show that a substituent group has more room when the substituent is equatorial than when it is axial. In the axial position the substituent is considerably closer to the two axial hydrogens on the same side of the ring than to other hydrognes, even hydrogens on adjacent carbons when the substituent is in the equatorial position (Figure 12-8). For example, when the substituent is bromine, which has a $\ce{C-Br}$ bond length of $1.94 \: \text{Å}$, the distance from axial bromine to the axial hydrogen at $\ce{C_3}$ or $\ce{C_5}$ on the same side of the ring is about $2.7 \: \text{Å}$. In contrast, the distance from equatorial bromine to any of the hydrogens on the adjacent carbons is about $3.1 \: \text{Å}$.
There is a very important general aspect of the difference between these two nonbonded $\ce{H} \cdot \cdot \cdot \ce{Br}$ interactions at $2.7 \: \text{Å}$ and $3.1 \: \text{Å}$. Whenever two nonbonded atoms are brought close together, and before the massive repulsion sets in (which is so evident in Figure 4-6), there is a slight dip in the energy curve corresponding to attraction.$^2$ For nonbonded $\ce{H} \cdot \cdot \cdot \ce{Br}$ interactions the bottom of the dip comes at about $3.1 \: \text{Å}$ (Figure 12-9), and the resulting attraction between the atoms will provide some stabilization of the equatorial conformation relative to the axial conformation.
Weak attractive forces between nonbonded atoms are called van der Waals attractive forces, London$^3$ forces, or dispersion forces, and are of great importance in determining the properties of liquids. They also can be expected to play a role in determining conformation equilibria whenever the distances between the atoms in the conformations correspond to the so-called van der Walls minima.
Table 12-2 shows the contribution made by various substituents to the free-energy change from the axial to the equatorial orientations of the substituent. Thus, for bromine, the free-energy change, $\Delta G^0$, is $-0.5 \: \text{kcal mol}^{-1}$, which means that at $25^\text{o}$, the equilibrium constant, $K$, for the axial $\rightleftharpoons$ equatorial equilibrium is about 2.3 (from $-2.303 RT \: \text{log} \: K = \Delta G^0$; see Section 4-4A).
From many studies it is known that the interconversion of conformations with the substituent in the equatorial and the axial positions occurs about 100,000 times per second, which corresponds to a transition-state energy (activation energy) of about $11 \: \text{kcal mol}^{-1}$ above the ground-state energy. The rate decreases as the temperature is lowered. If one cools chlorocyclohexane to its melting point $\left( -44^\text{o} \right)$, the substance crystallizes to give the pure equatorial isomer. The crystals then can be cooled to $-150^\text{o}$ and dissolved at this temperature in a suitable solvent. At $-150^\text{o}$ it would take about 130 days for half of the equatorial form to be converted to the axial form. However, when the solution is warmed to $-60^\text{o}$ the equatorial conformation is converted to the equilibrium mixture in a few tenths of a second.
Table 12-2: A Selection of $\Delta G^0$ for the Change from Axial to Equatorial Orientation of Substituents for Monosubstituted Cyclohexanes
Cis-Trans Isomerism and Conformational Equilibria for Cyclohexane Derivatives
The cis-trans isomerism of cyclohexane derivatives (Section 5-1A) is complicated by conformational isomerism. For example, 4-tert-butylcyclohexyl chloride theoretically could exist in four stereoisomeric chair forms, $1$, $2$, $3$, and $4$.
trans
cis
Use of $\ce{^{13}C}$ nmr spectroscopy to determine whether a substituent is in an axial or equatorial position is well illustrated with cis- and trans-4-tert-butylcyclohexanols, $5$ and $6$:
Cyclopentane
The five $\ce{-CH_2}-$ groups of cyclopentane theoretically could form a regular planar pentagon (internal angles of $108^\text{o}$) with only a little bending of the normal $\ce{C-C-C}$ bond angles. Actually, cyclopentane molecules are not flat. The planar structure has completely eclipsed hydrogens, which makes it less stable by about $10 \: \text{kcal mol}^{-1}$ than if there were no eclipsed hydrogens. The result is that each molecule assumes a puckered conformation that is the best compromise between distortion of bond angles and eclipsing of hydrogens. The best compromise conformations have the ring twisted with one or two of the $\ce{-CH_2}-$ groups bent substantially out of a plane passed through the other carbons (Figure 12-14). The flexibility of the ring is such that these deformations move rapidly around the ring.
Cyclobutane
Formation of a four-membered ring of carbon atoms can be achieved only with substantial distortion of the normal valence angles of carbon, regardless of whether the ring is planar or nonplanar. In cyclobutane, for example, if the valence bonds are assumed to lie along straight lines drawn between the carbon nuclei, each $\ce{C-C-C}$ bond angle will be $19.5^\text{o}$ smaller than the $109.5^\text{o}$ tetrahedral value:
Cyclopropane
The three carbon atoms of the cyclopropane ring lie in a plane. Therefore the angle strain is expected to be considerable because each $\ce{C-C-C}$ valence angle must be deformed $49.5^\text{o}$ from the tetrahedral value. It is likely that some relief from the strain associated with the eclipsing of the hydrogens of cyclopropane is achieved by distortion of the $\ce{H-C-H}$ and $\ce{H-C-C}$ bond angles:
"Cycloethane" (Ethene)
If one is willing to consider a carbon-carbon double bond as a two-membered ring, then ethene, $\ce{C_2H_4}$, is the simplest possible cycloalkane ("cycloethane"). As such, $\ce{C_2H_4}$ has $\ce{C-C-C}$ valence angles of $0^\text{o}$ and therefore an angle strain of $109.5^\text{o}$ at each $\ce{CH_2}$ group compared to the tetrahedral value:
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."
$^1$Pioneering work on the conformations of cyclohexane and its derivatives was carried out by O. Hassel (Norway) and D. H. R. Barton (United Kingdom) for which they shared a Nobel Prize in 1969.
$^2$The vertical scale of Figure 4-6 does not permit seeing the dip in the curve resulting from attractive forces between neon atoms. It is deepest when $r$ is about $3.12 \: Å$ and amounts to $0.070 \: \text{kcal mol}^{-1}$.
$^3$After F. London, who developed a quantum-mechanical theory of the origin of these forces and also pioneered many quantum calculations of great consequence to chemistry, including bonding in $\ce{H_2}$, which will be discussed in Section 21-1. | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.03%3A_Conformations_of_Cycloalkanes.txt |
The Baeyer Theory
Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes. In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane derivatives are different from cyclopentane and cyclohexane, because their $\ce{C-C-C}$ angles cannot have the tetrahedral value of $109.5^\text{o}$. At the same time, Baeyer hypothesized that the difficulties encountered in synthesizing cycloalkane rings from $\ce{C_7}$ upward was the result of the angle strain that would be expected if the large rings were regular planar polygons (see Table 12-3). Baeyer also believed that cyclohexane had a planar structure like that shown in Figure 12-2, which would mean that the bond angles would have to deviate $10.5^\text{o}$ from the tetrahedral value. However, in 1895, the then unknown chemist H. Sachse suggested that cyclohexane exists in the strain-free chair and boat forms discussed in Section 12-3. This suggestion was not accepted at the time because it led to the prediction of several possible isomers for compounds such as chlorocyclohexane. The idea that such isomers might act as a single substance, as the result of rapid equilibration, seemed like a needless complication, and it was not until 1918 that E. Mohr proposed a definitive way to distinguish between the Baeyer and Sachse cyclohexanes. As will be discussed in Section 12-9, the result, now known as the Sachse-Mohr theory, was complete confirmation of the idea of nonplanar large rings.
Table 12-3: Strain and Heats of Combustion of Cycloalkanes
Because cyclopentane and cyclobutane (Sections 12-3E and 12-3F) also have nonplanar carbon rings, it is clear that the Baeyer postulate of planar rings is not correct. Nonetheless, the idea of angle strain in small rings is important. There is much evidence to show that such strain produces thermodynamic instability and usually, but not always, enhanced chemical reactivity.
Heats of Combustion of Cycloalkanes and Strain Energies
The strain in ring compounds can be evaluated quantitatively by comparing the heats of combustion per $\ce{CH_2}$ group, as in Table 12-3. The data indicate that cyclohexane is virtually strain-free, because the heat of combustion per $\ce{CH_2}$ is the same as for alkanes $\left( 157.4 \: \text{kcal mol}^{-1} \right)$. The increase for the smaller rings clearly reflects increasing angle strain and, to some extent, unfavorable interactions between nonbonded atoms. For rings from $\ce{C_7}$ to $\ce{C_{12}}$ there appears to be a residual strain for each additional $\ce{CH_2}$ of $1$ to $1.5 \: \text{kcal mol}^{-1}$. These rings can be puckered into flexible conformations with normal $\ce{C-C-C}$ angles, but as will be shown in Section 12-6, from $\ce{C_7}$ to $\ce{C_{13}}$ such arrangements all have pairs of partially eclipsed or interfering hydrogens. The larger cycloalkanes such as cyclopentadecane appear to be essentially strain-free.
We expect that the total strain in cycloalkanes of the type $\ce{(CH_2)}_n$ should decrease rapidly in the order $n = 2 > n = 3 > n = 4$. However, the data of Table 12-3 show that the order actually is $3 \cong 4 > 2$. This difference in order often is disguised by dividing the heats of combustion by the numbers of $\ce{CH_2}$ groups and showing that the heats of combustion per $\ce{CH_2}$ are at least in the order expected from bond-angle strain. This stratagem does not really solve the problem.
It is important to recognize that when we evaluate strain from the heats of combustion per $\ce{CH_2}$ group, we are assuming that the $\ce{C-H}$ bonds have the same strength, independent of $n$. However, the bond-dissociation energies of each of the $\ce{C-H}$ bonds of ethene and cyclopropane are greater than of the $\ce{C_2-H}$ bonds of propane (Table 4-6). Any amount that these bonds are stronger than normal will make the strain energies judged from heats of combustion appear to be less. If we take the $\ce{C-H}$ bonds to be on average $2 \: \text{kcal mol}^{-1}$ stronger in cyclobutane, $6 \: \text{kcal mol}^{-1}$ stronger in cyclopropane, and $13 \: \text{kcal mol}^{-1}$ in ethene, we can correct the carbon-carbon strain energies accordingly. For cyclobutane the corrected strain then is $8 \times 2$ (for the eight $\ce{C-H}$ bonds) $+ \: 26.3$ (total strain from Table 12-3) $= 42.3 \: \text{kcal mol}^{-1}$. The corresponding figures for cyclopropane are $6 \times 6 + 27.6 = 63.6 \: \text{kcal mol}^{-1}$, and for ethene, $4 \times 13 + 22.4 = 74.4 \: \text{kcal mol}^{-1}$. The results support the intuitive expectations by giving larger differences in the right direction for the strain energies of cyclobutane, cyclopropane, and ethene. Whether this analysis is quantitatively correct or not, it does give some indication of why strain energy is not a very precise concept - unless we can reliably estimate the net effects of a strain.
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."
12.05: Chemical Properties
Strain in small-ring cycloalkanes has a profound influence on their heats of combustion (Table 12-3). We reasonably expect that other chemical properties also will be affected. Indeed, like alkenes, cyclopropane and cyclobutane undergo \(\ce{C-C}\) bond cleavage reactions that are not observed for cyclopentane and cyclohexane, or for saturated, open-chain hydrocarbons. A summary of these reactions is presented in Table 12-4. It will be seen that the reactions result in cleavage of a \(\ce{C-C}\) bond to give an open-chain compound with normal bond angles. Relief of angle strain is an important contributing factor to the driving force for these reactions. Therefore, ethene is highly reactive, whereas cyclopropane and cyclobutane are somewhat less reactive. The \(\ce{C-C}\) bonds of the larger, relatively strain-free cycloalkanes are inert, so these substances resemble the alkanes in their chemical behavior. Substitution reactions, such as chlorination of cyclopentane and higher cycloalkanes, generally are less complex than those of the corresponding alkanes because there are fewer possible isomeric substitution products. Thus cyclohexane gives only one monochloro product, whereas hexane gives three isomeric monochloro hexanes.
Conformation has a major influence on the chemical reactivity of cycloalkanes. To understand its effect in any one reaction, we first need to know what the conformation is of the transition state, and this requires a knowledge of the reaction mechanism. Next, we have to decide what amount of energy is required for the reactants to achieve transition-state conformations. For example, consider the \(E_2\) elimination discussed in Section 8-8D. The preferred transition state requires the leaving groups to be antarafacial and coplanar:
For cyclohexane derivatives to react in this way, the transition-state conformation must have both leaving groups axial:
Table 12-4: Ring-Cleavage Reactions of Cycloalkanes
For this reason, compounds such as cis-4-tert-butylchlorocyclohexane eliminate \(\ce{HCl}\) much more readily by the \(E_2\) mechanism than do the corresponding trans isomers.
To have the antarafacial coplanar mechanism of cycloalkanes operate with the trans isomer, the transition state would have to have the tert-butyl group in the highly unfavorable axial position.
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)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.04%3A_Strain_in_Cycloalkane_Rings.txt |
The Baeyer strain theory suggested that the larger cycloalkanes ring are difficult to synthesize because of angle strain associated with planar rings, as calculated in Table 12-3. We now know that, except for cyclopropane, none of the cycloalkanes have planar carbon rings and that the higher cycloalkanes have normal or nearly normal bond angles. The reason that the higher cycloalkanes are generally difficult to synthesize from open-chain compounds is not so much angle strain, as Baeyer hypothesized, but the low probability of having reactive groups on the two fairly remote ends of a long hydrocarbon chain come together to effect cyclization. Usually, coupling of reactive groups on the ends of different molecules occurs in preference to cyclization, unless the reactions are carried out in very dilute solutions. This is called the high-dilution technique for achieving ring formation when the ring-forming reaction has to compete with rapid inter-molecular reactions.
With regard to conformations of the larger cycloalkanes, we first note that the chair form of cyclohexane is a “perfect” conformation for a cycloalkane. The $\ce{C-C-C}$ bond angles are close to their normal values, all the adjacent hydrogens are staggered with respect to one another, and the 1,3-axial hydrogens are not close enough together to experience nonbonded repulsions. About the only qualification one could put on the ideality of the chair form is that the trans conformation of butane is somewhat more stable than the gauche conformation (Section 5-2), and that all of the $\ce{C-C-C-C}$ segments of cyclohexane have the gauche arrangement. Arguing from this, J. Dale$^6$ has suggested that large cycloalkane rings would tend to have trans $\ce{C-C-C-C}$ segments to the degree possible and, indeed, cyclotetradecane seems to be most stable in a rectangular conformation with trans $\ce{C-C-C-C}$ bond segments (Figure 12-16). This conformation has a number of possible substituent positions, but because only single isomers of monosubstituted cyclotetradecanes have been isolated, rapid equilibration of the various conformational isomers must occur. Other evidence indicates that the barrier to interconversion of these conformations is about $7 \: \text{kcal mol}^{-1}$.
With the cycloalkanes having 7 to 10 carbons, there are problems in trying to make either trans or gauche $\ce{C-C-C-C}$ segments, because the sizes of these rings do not allow the proper bond angles or torsional angles, or else there are more or less serious nonbonded repulsions. Consequently each of these rings assumes a compromise conformation with some eclipsing, some nonbonded repulsions, and some angle distortions. Brief comments on some of these conformations follow. It will be useful to use molecular models to see the interactions involved.
Cycloheptane
Possible conformations for cycloheptane include the “comfortable” appearing chair form, $7$. However, this form has eclipsed hydrogens at $\ce{C_4}$ and $\ce{C_5}$ as well as nonbonded interactions between the axial-like hydrogens on $\ce{C_3}$ and $\ce{C_6}$. The best compromise conformation is achieved by a $30^\text{o}$-$40^\text{o}$ rotation around the $\ce{C_4-C_5}$ bond to relieve the eclipsing of the hydrogens. This spreads the interfering hydrogens at $\ce{C_3}$ and $\ce{C_6}$ and results in a somewhat less strained conformation called the twist chair. The twist chair, $8$, is very flexible and probably only about $3 \: \text{kcal mol}^{-1}$ of activation is required to interconvert the various possible monosubstituted cycloheptane conformations.
Cyclooctane
There are several more or less reasonable looking cyclooctane conformations. After much research it now is clear that the favored conformation is the boat-chair, $9$, which is in equilibrium with a few tenths percent of the crown conformation, $10$ :
The activation energy for interconversion of these two forms is about $10 \: \text{kcal mol}^{-1}$. The boat-chair conformation $9$ is quite flexible and movement of its $\ce{CH_2}$ groups between the various possible positions occurs with an activation energy of only about $5 \: \text{kcal mol}^{-1}$.
Cyclononane
Several more or less reasonable conformations of cyclononane also can be developed, but the most favorable one is called the twist-boat-chair, which has three-fold symmetry (Figure 12-17). The activation energy for inversion of the ring is about $6 \: \text{kcal mol}^{-1}$.
Cyclodecane
The stable conformation of cyclodecane (Figure 12-18) is similar to that of cyclotetradecane (Figure 12-16). However, there are relatively short $\ce{H} \cdot \cdot \cdot \cdot \ce{H}$ distances and the $\ce{C-C-C}$ bond angles are somewhat distorted because of cross-ring hydrogen-hydrogen repulsions. The most stable position for a substituent on the cyclodecane ring is the one indicated in Figure 12-18. The least stable positions are those in which a substituent replaces any of the six hydrogens shown, because nonbonded interactions are particularly strong at these positions. The activation energy for interconversion of substituent positions is about $6 \: \text{kcal mol}^{-1}$.
Most stable conformation of cyclodecane; Dale and sawhorse representations. The shaded area in the sawhorse convention indicates substantial nonbonded $\ce{H} \cdot \cdot \cdot \cdot \ce{H}$ interactions.
$^6$Pronounced Dalluh.
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)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.06%3A_The_Larger_Cycloalkanes_and_their_Conformations.txt |
The $\ce{C-C=C}$ angle in alkenes normally is about $122^\text{o}$, which is $10^\text{o}$ larger than the normal $\ce{C-C-C}$ angle in cycloalkanes. This means that we would expect about $20^\text{o}$ more angle strain in small-ring cycloalkenes than in the cycloalkanes with the same numbers of carbons in the ring. Comparison of the data for cycloalkenes in Table 12-5 and for cycloalkanes in Table 12-3 reveals that this expectation is realized for cyclopropene, but is less conspicuous for cyclobutene and cyclopentene. The reason for this is not clear, but may be connected in part with the $\ce{C-H}$ bond strengths (see Section 12-4B).
Cyclopropene has rather exceptional properties compared to the other cycloalkenes. It is quite unstable and the liquid polymerizes spontaneously although slowly, even at $-80^\text{o}$. This substance, unlike other alkenes, reacts rapidly with iodine and behaves like an alkyne in that one of its double-bond hydrogens is replaced in silver-ammonia solution to yield an alkynide-like silver complex.
Table 12-5: Properties of Some Cycloalkenes and Cycloalkynes
The $\ce{C-C=C}$ bond angles in alkynes normally are $180^\text{o}$ and the angle strain involved in making a small-ring cycloalkyne, such as cyclopropyne, apparently is prohibitive. The smallest reasonably stable member of the series is cyclooctyne, and its properties, along with those of some higher homologs, are shown in Table 12-5. Strong evidence has been adduced for the existence of cyclopentyne, cyclohexyne, and cycloheptyne as unstable reaction intermediates.
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."
12.08: Nomenclature of Polycycloalkanes
There are many hydrocarbons and hydrocarbon derivatives with two or more rings having common carbon atoms. Such a substance is decalin, which has ten carbons arranged in two six-membered rings:
Compounds of this type usually are named by attaching the prefix bicyclo to the name of the open-chain hydrocarbon with the same total number of carbon atoms as in the rings. Thus decalin, which has ten carbons in the ring system, is a bicyclodecane. Next, we have to have a way to specify the sizes of the rings, which is done by counting the number of carbon atoms in each of the chains connecting the two atoms that constitute the ring junctions or bridgeheads. Decalin has four carbons in each of two chains and none in the third. Therefore, decalin is bicyclo[4.4.0]decane. Notice that the numbers are enclosed in square brackets after the prefix “bicyclo” and before the name of the hydrocarbon. The numbers are listed in order of decreasing magnitude and are properly separated by periods, not commas. Some other examples follow:
To name substituted polycycloalkanes, a numbering system is required. In the IUPAC system the main ring is the one containing the largest number of carbon atoms. Two of the carbons in the main ring serve as junctions for the main bridge, which is chosen to be as large as possible, consistent with the choice of the main ring. Additional rules are required for more complex cases, but these are not of interest to us here.
In numbering bicyclic ring systems that have two ring junctions, one of them is chosen as $\ce{C_1}$. The numbering proceeds along the longest chain of carbons to the next junction, then continues along the next longest chain, and finally is completed along the shortest chain. For example,
Here, the main ring has seven carbons ($\ce{C_1}$ to $\ce{C_7}$) and there is a one-carbon bridge $\left( \ce{C_8} \right)$.
When the hydrocarbon rings have only one carbon in common, they are called spiranes and are given systematic names in accord with the following examples:
Notice that for spiranes the numbering starts next to the junction point in the smaller ring.
The naming of tricycloalkanes follows the same general system.$^7$ The largest ring and its main linkage form a bicyclic system, and the location of the fourth or secondary linkage is shown by superscripts. The systematic name of the interesting hydrocarbon adamantane is given below as an example; its conformation also is shown. The largest ring in adamantane is eight-membered and the carbons that constitute it could be selected in several different ways. The carbon chosen as $\ce{C_9}$ lies between $\ce{C_1}$ and $\ce{C_5}$, not between the higher-numbered $\ce{C_3}$ and $\ce{C_7}$:
To generate a structure from a name such as 8-chlorobicyclo[3.2.1]octane, $11$, start with a pair of junction atoms, connect them as prescribed, then number the initial skeleton, make the final connections, and locate the substituents. The steps follow:
A further and more complicated example is 1,4-dichloropentacyclo[4.2.0.0$^{2.5}$.0$^{3.8}$.0$^{4.7}$]octane:
The most difficult part of the whole procedure may be generating the final structure in appropriate perspective. The task of doing this can be simplified greatly by the use of molecular models.
$^7$To determine whether a given bridged polycylic ring system should be bicyclo-, tricyclo-, and so on, use the rule that the number of rings is equal to the minimum number of bond cleavages to convert the ring system into an acyclic hydrocarbon having the same number of carbons.
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)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.07%3A_Cycloalkenes_and_Cycloalkanes.txt |
The six-membered rings of decalin, like those of cyclohexane, are expected to be most stable in the chair form. However, there are two possible ways in which two chairs can be joined (Figure 12-21). The ring-junction hydrogens may be either on the same side of the molecule (cis-decalin) or on opposite sides (trans-decalin). When the two rings are joined through two equatorial-type bonds, trans-decalin results, whereas an axial-equatorial union gives cis-decalin. Both isomers are known, and the trans isomer is about $2 \: \text{kcal mol}^{-1}$ more stable than the cis isomer, largely because of relatively unfavorable nonbonded interactions within the concave area of cis-decalin (see Figure 12-22).
It is of historical interest to note that the Baeyer strain theory with its planar rings predicts only one form of decalin with the ring-junction hydrogens on the same side of the molecule (Figure 12-23). The Sachse-Mohr concept of puckered strain-free rings allows for two isomers. In fact, Mohr predicted that the two isomers of decalin should exist before W. Hückel (1925) succeeded in preparing them. Both isomers occur in petroleum.
At this point, it probably will be helpful to construct models of cis- and trans-decalins to appreciate the following: (a) The two compounds cannot interconvert unless $\ce{C-C}$ or $\ce{C-H}$ bonds first are broken. (b) trans-Decalin is a relatively rigid system and, unlike cyclohexane, the two rings cannot flip from one chair form to another. Accordingly, the orientation of the substituent is fixed in the chair-chair conformation of trans-decalin. (c) The chair-chair forms of cis-decalin are relatively flexible, and inversion of both rings at once occurs fairly easily (the barrier to inversion is about $14 \: \text{kcal mol}^{-1}$). A substituent therefore can interconvert between axial and equatorial conformations (Figure 12-24).
The ramifications of conformational analysis of flexible and rigid ring systems are of considerable importance to the understanding of stability and reactivity in polycyclic systems. This will become increasingly evident in later discussions.
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."
12.10: Strain in Polycyclic Molecules
Knowing the importance of angle and eclipsing strain in the small-ring cycloalkanes, we should expect that these strains would become still more important in going from cyclobutane to bicyclo[1.1.0]butane or from cyclooctane to pentacyclo[4.2.0.0$^{2.5}$.0$^{3.8}$.0$^{4.7}$]octane (cubane). This expectation is borne out by the data in Table 12-6, which gives the properties of several illustrative small-ring polycyclic molecules that have been synthesized only in recent years.
The extraordinary strain energy of cubane $\left( \sim 142 \: \text{kcal mol}^{-1} \right)$ is worthy of special note. It is roughly equal to six times the strain energy of a single cyclobutane ring $\left( \sim 26 \: \text{kcal mol}^{-1} \right)$ as befits a molecule made up of six cyclobutane rings as faces. Despite this, cubane is surprisingly stable to spontaneous decomposition processes, although it will rearrange under the influence of metal or acid catalysts.
Another extraordinarily strained polycyclic hydrocarbon that has been prepared is prismane (the Ladenburg structure for benzene).
This substance is a liquid that decomposes explosively when heated. In dilute solution at $100^\text{o}$, it is converted slowly to benzene.
One of the most interesting types of polycyclic carbon compounds prepared in recent years is the group of tricyclic substances known as “propellanes.” A typical example is tricyclo [3.2.2.0$^{1.5}$]nonane, which sometimes is called [3.2.2]propellane, $12$. The physical properties of several of these are included in Table 12-6. A quick look at formula $12$ probably does not suggest any great structural difference from the bicyclic compounds we have discussed previously. However, if one tries to construct a ball-and-stick model of $12$, one soon concludes that the propellanes are truly extraordinary substances in that all four carbon bonds at the bridgehead carbons extend, not to the comers of a tetrahedron, or even a distorted tetrahedron as for a cyclopropane ring, but away from the carbon on the same side of a plane through the carbon as in $13$:
Table 12-6: Properties of Some Small-Ring Polycyclic Hydrocarbons
Angle strain is severe. Accordingly, [3.2.1]propellane reacts rapidly with bromine at $-60^\text{o}$ and with hydrogen over palladium at room temperature:
Still another way to torture the valence angles of carbon is to twist a double bond by connecting it to the bridgehead carbon of a bicyclic system with reasonably small rings, as in bicyclo[3.3.1]-1-nonene, $14$:
As with $12$, it might appear that there is nothing unusual about $14$. But a ball-and-stick model of $14$ reveals that the carbon-carbon double bond is in a strained configuration like $15$. Some of the properties of $14$ are given in Table 12-6. That compounds with a double bond to a bridgehead carbon, such as $14$, should be highly strained is known as “Bredt’s Rule.” The most spectacular example of this form of molecular distortion reported so far is bicyclo[2.2.1]-1-heptene, $16$, for which evidence has been adduced that it is an unstable reaction intermediate:
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)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.09%3A_Conformations_of_Decalin.txt |
Organic compounds of natural origin rarely have simple structures. Most have more than one functional group in each molecule. Usually the chemical behavior of a functional group is influenced significantly by the presence of another functional group, especially when the groups are in close proximity. Indeed, the complexities that are associated with polyfunctionality are of central importance in biochemical reactions and in the design of organic syntheses. For this reason, you will need to gain experience in judging how and when functional groups in the same molecule interact with one another. We will start by considering the chemistry of alkadienes, which are hydrocarbons with two carbon-carbon double bonds.
• 13.1: General Comments on Alkadienes
The molecular properties of alkadienes depend on the relationship between the double bonds, that is, whether they are cumulated, conjugated, or isolated. The emphasis here will be on the effects of conjugation on chemical properties. The reactions of greatest interest are addition reactions, and this chapter will include various types of addition reactions: electrophilic, radical, cycloaddition, and polymerization.
• 13.2: 1,3- or Conjugated Dienes. Electrophilic and Radical Addition
The reactions of 1,3-butadiene are reasonably typical of conjugated dienes. The compound undergoes the usual reactions of alkenes, such as catalytic hydrogenation or radical and polar additions, but it does so more readily than most alkenes or dienes that have isolated double bonds. Furthermore, the products frequently are those of 1,2 and 1,4 addition.
• 13.3: Cycloaddition Reactions
There are a variety of reactions whereby rings are formed through addition to double or triple bonds. Diels-Alder reaction, which has proved so valuable in synthesis, is the most famous and can be called a [4 + 2] cycloaddition and results in the formation of a six-membered ring.
• 13.4: Polymerization Reactions of Conjugated Dienes
The general character of alkene polymerization by radical and ionic mechanisms was discussed previosuly. The same principles apply to the polymerization of alkadienes, with the added feature that there are additional ways of linking the monomer units. The polymer chain may grow by either 1,2 or 1,4 addition to the monomer.
• 13.5: Cumulated Alkadienes
The 1,2-dienes, which have cumulated double bonds, commonly are called allenes. Allenes of the type may be chiral molecules and can exist in two stereoisomeric forms, i.e., enantiomers. Verification of the chirality of such allenes (originally proposed by van’t Hoff in 1875) was slow in coming and was preceded by many unsuccessful attempts to resolve suitably substituted allenes into their enantiomers.
• 13.6: Approaches to Planning Practical Organic Syntheses
Chemical synthesis is not a science that can be taught or learned by any well-defined set of rules. Some classify it as more art than science because, as with all really creative endeavors, to be very successful requires great imagination conditioned by a wealth of background knowledge and experience. The problems of synthesis basically are problems in design and planning. There always is a variety of ways that the objective can be achieved either from the same or different starting materials.
• 13.7: Building the Carbon Skeleton
According to the suggested approach to planning a synthesis, the primary consideration is how to construct the target carbon skeleton starting with smaller molecules (or, alternatively, to reconstruct an existing skeleton). Construction of a skeleton from smaller molecules almost always will involve formation of carbon-carbon bonds. Up to this point we have discussed only a few reactions in which carbon-carbon bonds are formed.
• 13.8: Introducing Functionality
It may be easy to construct the carbon skeleton of the target compound of a synthesis, but with a reactive functional group at the wrong carbon. Therefore it is important also to have practice at shifting reactive entry points around to achieve the final desired product. We shall illustrate this form of molecular chess with reactions from previous post.
• 13.9: Construction of Ring Systems by Cycloaddition
Another example of a synthesis problem makes use of the cycloaddition reactions discussed here. Whenever a ring has to be constructed, you should consider the possibility of cycloaddition reactions, especially [4 + 2] cycloaddition by the Diels-Alder reaction.
• 13.10: Protecting Groups in Organic Synthesis
Functional groups usually are the most reactive sites in the molecule, and it may be difficult or even impossible to insulate one functional group from a reaction occurring at another. Therefore any proposed synthesis must be evaluated at each step for possible side reactions that may degrade or otherwise modify the structure in an undesired way. To do this will require an understanding of how variations in structure affect chemical reactivity.
• 13.11: Building the Carbon Skeleton
According to the suggested approach to planning a synthesis, the primary consideration is how to construct the target carbon skeleton starting with smaller molecules (or, alternatively, to reconstruct an existing skeleton). Construction of a skeleton from smaller molecules almost always will involve formation of carbon-carbon bonds. Up to this point we have discussed only a few reactions in which carbon-carbon bonds are formed.
• 13.E: Polyfunctional Compounds, Alkadienes, and Approaches to Organic Synthesis (Exercises)
These are the homework exercises to accompany Chapter 13 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."
13: Polyfunctional Compounds Alkadienes and Approaches to Organic Synthesis
The molecular properties of alkadienes depend on the relationship between the double bonds, that is, whether they are cumulated, conjugated, or isolated:
• 2,3-pentadiene (cumulated)
\[\ce{CH_3CH=C=CHCH_3}\]
• 1,3-pentadiene (conjugated)
\[\ce{CH_2=CH-CH=CHCH_3}\]
• 1,4-pentadiene (isolated)
\[\ce{CH_2=CH-CH_2-CH=CH_2}\]
The properties of a compound with isolated double bonds, such as 1,4-pentadiene, generally are similar to those of simple alkenes because the double bonds are essentially isolated from one another by the intervening \(\ce{CH_2}\) group. However, with a conjugated alkadiene, such as 1,3-pentadiene, or a cumulated alkadiene, such as 2,3-pentadiene, the properties are sufficiently different from those of simple alkenes (and from each other) to warrant separate discussion. Some aspects of the effects of conjugation already have been mentioneed, such as the influence on spectroscopic properties (Section 9-9B). The emphasis here will be on the effects of conjugation on chemical properties. The reactions of greatest interest are addition reactions, and this chapter will include various types of addition reactions: electrophilic, radical, cycloaddition, and polymerization.
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."
13.02: 13- or Conjugated Dienes. Electrophilic and Radical Addit
The reactions of 1,3-butadiene are reasonably typical of conjugated dienes. The compound undergoes the usual reactions of alkenes, such as catalytic hydrogenation or radical and polar additions, but it does so more readily than most alkenes or dienes that have isolated double bonds. Furthermore, the products frequently are those of 1,2 and 1,4 addition:
Formation of both 1,2- and 1,4-addition products occurs not only with halogens, but also with other electrophiles such as the hydrogen halides. The mechanistic course of the reaction of 1,3-butadiene with hydrogen chloride is shown in Equation 13-1. The first step, as with alkenes, is formation of a carbocation. However, with 1,3-butadiene, if the proton is added to $\ce{C_1}$ (but not $\ce{C_2}$), the resulting cation has a substantial delocalization energy, with the charge distributed over two carbons (review Sections 6-5 and 6-5C if this is not clear to you). Attack of $\ce{Cl}^\ominus$ as a nucleophile at one or the other of the positive carbons yields the 1,2- or the 1,4- addition product:
An important feature of reactions in which 1,2 and 1,4 additions occur in competition with one another is that the ratio of the products can depend on the temperature, the solvent, and also on the total time of reaction. The reason for the dependence on the reaction time is that the formation of the carbocation is reversible, and the ratio of products at equilibrium need not be the same as the ratio of the rates of attack of $\ce{Cl}^\ominus$ at $\ce{C_1}$ and $\ce{C_3}$ of the carbocation. This is another example of a difference in product ratios resulting from kinetic control versus equilibrium control.
The fact is that at low temperatures the 1,2 product predominates because it is formed more rapidly, and the back reactions, corresponding to $k_{-1}$ or $k_{-3}$, are slow (Equation 13-2). However, at equilibrium$^1$ the 1,4 product is favored because it is more stable, not because it is formed more rapidly.
Conjugated dienes also undergo addition reactions by radical-chain mechanisms. Here, the addition product almost always is the 1,4 adduct. Thus radical addition of hydrogen bromide to 1,3-butadiene gives l-bromo-2-butene, presumably by the following mechanism:
$^1$The equilibrium ratio is obtained as follows. At equilibrium $k_1/k_{-1} = \left[ \ce{CH_3CH=CHCH_2Cl} \right]/\left[ \ce{R}^\oplus \right] \left[ \ce{Cl}^\ominus \right]$ and $k_{-3}/k_3 = \left[ \ce{R}^\oplus \right] \left[ \ce{Cl}^\ominus \right]/\left[ \ce{CH_3CHClCH=CH_2} \right]$, in which $\ce{R}^\oplus$ is the concentration of delocalized carbocation. Multiplication of these equations gives $k_1k_{-3}/k_{-1}k_3 = \left[ \ce{CH_3CH=CHCH_2Cl} \right]/\left[ \ce{CH_3CHClCH=CH_2} \right] = K_\text{eq}$.
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)/13%3A_Polyfunctional_Compounds_Alkadienes_and_Approaches_to_Organic_Synthesis/13.01%3A_General_Comments_on_Alkadienes.txt |
There are a variety of reactions whereby rings are formed through addition to double or triple bonds. An especially simple example is the addition of ethene to 1,3-butadiene to give cyclohexene:
This is the prototype Diels-Alder reaction, which has proved so valuable in synthesis that it won its discoverers, O. Diels and K. Alder, the Nobel Prize in chemistry in 1950.
The Diels-Alder reaction is both a 1,4 addition or ethene to 1,3-butadiene and a 1,2 addition of butadiene to ethene. It can be called a [4 + 2] cycloaddition and as such results in the formation of a six-membered ring. Many other cycloadditions are known, such as [2 + 2], other types of [4 + 2], and [2 + 2 + 2], which give different size of rings. Some specific examples follow:
The synthetic importance of these reactions is very great and, because many of them often involve dienes, we will discuss their general characteristics in this chapter. The most valuable cycloaddition reaction almost certainly is the [4 + 2], or Diels-Alder, reaction and will be discussed in detail.
[4 + 2] Cycloadditions
There is one very important point you should remember about the Diels-Alder reaction: The reaction usually occurs well only when the [2] component is substituted with electron-attracting groups and the [4] component is substituted with electron-donating groups, or the reverse. The most common arrangement is to have the alkene (usually referred to as the dienophile) substituted with electron-attracting groups such as $\ce{-CO_2H}$, $\ce{-COR}$, or $\ce{-C \equiv N}$. For example,
A list of the more reactive dienophiles carrying electron-attracting groups is given in Table 13-1. Ethene and other simple alkenes generally are poor dienophiles and react with 1,3-butadiene only under rather extreme conditions and in low yield.
However, when the diene is substituted with several electron-attracting groups such as chlorine or bromine, electron-donating groups on the dienophile facilitate the reaction. Many substances, such as 2-methylpropene, that act as dienophiles with hexachlorocyclopentadiene simply will not undergo [4 + 2] addition with cyclopentadiene itself:
Table 13-1: Reactive Dienophiles with 1,3-Butadiene and Similar Dienes
The Diels-Alder reaction is highly stereospecific. The diene reacts in an unfavorable conformation in which its double bonds lie in a plane on the same side (cis) of the single bond connecting them. This s-cis (or cisoid) conformation is required to give a stable product with a cis double bond. Addition of ethene to the alternate and more stable (transoid) conformation would give an impossibly strained trans-cyclohexene ring. Possible transition states for reaction in each conformation follow, and it will be seen that enormous molecular distortion would have to take place to allow addition of ethene to the transoid conformation:
Cyclic dienes usually react more readily than open-chain dienes, probably because they have their double bonds fixed in the proper conformation for [4 + 2] cycloaddition, consequently the price in energy of achieving the s-cis configuration already has been paid:
Further evidence of stereospecificity in [4 + 2] additions is that the configurations of the diene and the dienophile are retained in the adduct. This means that the reactants (or addends) come together to give suprafacial addition. Two examples follow, which are drawn to emphasize how suprafacial addition occurs. In the first example, dimethyl cis-butadioate adds to 1,3-butadiene to give a cis-substituted cyclohexene:
In the second example, suprafacial approach of a dienophile to the 2,5 carbons of trans,trans-2,4-hexadiene is seen to lead to a product with two methyl groups on the same side of the cyclohexene ring:
(The use of models will help you visualize these reactions and their stereochemistry.)
There is a further feature of the Diels-Alder reaction that concerns the stereochemical orientation of the addends. In the addition of cis-butanedioic anhydride (maleic anhydride) to cyclopentadiene there are two possible ways that the diene and the dienophile could come together to produce different products. These are shown in Equations 13-3 and 13-4:
In practice, the adduct with the endo$^2$ configuration usually is the major product. As a general rule, Diels-Alder additions tend to proceed to favor that orientation that corresponds to having the diene double bonds and the unsaturated substituents of the dienophile closest to one another. This means that addition by Equation 13-3 is more favorable than by Equation 13-4, but the degree of endo-exo stereospecificity is not as high as the degree of stereospecificity of suprafacial addition to the diene and dienophile.
There are exceptions to favored endo stereochemistry of Diels-Alder additions. Some of these exceptions arise because the addition reaction is reversible, dissociation being particularly important at high temperature. The exo configuration is generally more stable than the endo and, given time to reach equilibrium (cf. Section 10-4A), the exo isomer may be the major adduct. Thus endo stereospecificity can be expected only when the additions are subject to kinetic control.
The reactivities of dienes in the Diels-Alder reaction depend on the number and kind of substituents they possess. The larger the substituents are, or the more of them, at the ends of the conjugated system, the slower the reaction is likely to be. There also is a marked difference in reactivity with diene configuration. Thus trans-1,3-pentadiene is substantially less reactive toward a given dienophile (such as maleic anhydride) than is cis-1,3-pentadiene. In fact, a mixture of the cis and trans isomers can be separated by taking advantage of the difference in their reactivities on cycloaddition:
Mechanism of the Diels-Alder Reaction
There is little evidence to support simple radical or polar mechanisms (such as we have discussed previously) for the Diels-Alder reaction. As the result of many studies the reaction seems best formulated as a process in which the bonds between the diene and the dienophile are formed essentially simultaneously:
We already have discussed a few addition reactions that appear to occur in a concerted manner. These include the addition of diimide, ozone, and boron hydrides to alkenes (Sections 11-5, 11-7A, and 11-6B). Concerted reactions that have cyclic transition states often are called pericyclic reactions. Other examples will be considered in later chapters.
A [4 + 2] Cycloaddition
We indicated previously that sulfur dioxide $\left( \ce{SO_2} \right)$ and 1,3-butadiene form a [4 + 1] cycloaddition product:
This reaction is more readily reversible than most Diels-Alder reactions, and the product largely dissociates to the starting materials on heating to $120^\text{o}$. The cycloadduct is an unsaturated cyclic sulfone, which can be hydrogenated to give the saturated cyclic sulfone known as "sulfolane":
This compound is used extensively in the petrochemical industry as a selective solvent.
The reversibility of the diene-$\ce{SO_2}$ cycloaddition makes it useful in the purification of reactive dienes. 2-Methyl-1,3-butadiene (isoprene) is purified commercially in this manner prior to being polymerized to rubber (Section 13-4):
Neither 1,3-cyclopentadiene nor 1,3-cyclohexadiene react with sulfur dioxide, probably because the adducts would be too highly strained:
Some [2 + 2] Cycloadditions
Many naturally occurring organic compounds contain six-membered carbon rings, but there are relatively few with four-membered carbon rings. After encountering the considerable ease with which six-membered rings are formed by [4 + 2] cycloaddition, we might expect that the simpler [2 + 2] cycloadditions to give four-membered rings also should go well, provided that strain is not too severe in the products. In fact, the dimerization of ethene is thermodynamically favorable:
Nonetheless, this and many other [2 + 2] cycloaddition reactions do not occur on simple heating.
However, there are a few exceptions. One is the dimerization of tetrafluoroethene, which perhaps is not surprising, considering the favorable thermodynamic parameters:
What is surprising is that addition of $\ce{CF_2=CF_2}$ to 1,3-butadiene gives a cyclobutane and not a cyclohexane, although the [2 + 2] product probably is about $25 \: \text{kcal mol}^{-1}$ less stable than the [4 + 2] product:
Such [2 + 2] thermal additions generally are limited to polyhaloethenes and a few substances with cumulated double bonds, such as 1,2-propadiene $\left( \ce{CH_2=C=CH_2} \right)$ and ketenes $\left( \ce{R_2C=C=O} \right)$. Some examples follow:
Many [2 + 2] cycloadditions that do not occur by simply heating the possible reactants can be achieved by irradiation with ultraviolet light. The following example, [2 + 2] addition of 2-cyclopentenone to cyclopentene, occurs photochemically but not thermally:
In all such photochemical cycloadditions the energy required to achieve a cycloaddition transition state, which can amount to $100 \: \text{kcal mol}^{-1}$ or more, is acquired by absorption of light.
Thermodynamically unfavorable cycloaddition products can be formed photochemically. A striking example is the photochemical conversion of norbornadiene to quadricyclene. The reverse of this reaction can occur with almost explosive violence in the presence of appropriate metal catalysts or on simple heating:
Why do some [2 + 2] cycloadditions occur thermally and others photochemically? What is special about fluoroalkenes and cumulated dienes? The answers are complex, but it appears that most thermal [2 + 2] cycloadditions, unlike the Diels-Alder [4 + 2] cycloadditions, go by stepwise routes (see Section 21-11). Why the two types of thermal cycloaddition have different mechanisms will be discussed in Sections 21-10A and B.
$^2$In general, the designation endo or exo refers to configuration in bridged or polycyclic ring systems such as those shown in Equations 13-3 and 13-4. With reference to the bridge atoms, a substituent is exo if it is on the same side as the bridge, or endo if it is on the opposite side. Further examples are
In drawing endo and exo isomers, it is best to represent the actual spatial relationships of the atoms as closely as possible. The cyclohexane ring is shown here in the boat form (Section 12-3A) because it is held in this configuration by the methylene group that bridges the 1,4 positions. If you do not see this, we strongly advise that you construct models.
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."
• Layne A. Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/13%3A_Polyfunctional_Compounds_Alkadienes_and_Approaches_to_Organic_Synthesis/13.03%3A_Cycloaddition_Reactions.txt |
The general character of alkene polymerization by radical and ionic mechanisms was discussed briefly in Section 10-8. The same principles apply to the polymerization of alkadienes, with the added feature that there are additional ways of linking the monomer units. The polymer chain may grow by either 1,2 or 1,4 addition to the monomer. With 1,3-butadiene, for example,
In 1,2 addition, a chiral carbon (marked with \(^*\)) is created as each molecule of the monomer adds to the growing chain radical. The physical properties of the polymer greatly depend on whether these carbons have the same or different configurations, as we will show in greater detail in Chapter 29. However, in polymer nomenclature, an isotactic polymer is one with essentially all chiral carbons having the same configuration, whereas an atactic polymer has a random ordering of the chiral carbons with different configurations.
For polymerization of 1,3-butadiene by 1,4 addition, there are no chiral carbons, but there is the possibility of cis-trans isomerism:
A polymer made of identical repeating units is called a homopolymer. If the units are nonidentical, as when different monomers are polymerized, the product is called a copolymer.
Many of the polymers formed from conjugated dienes are elastic and are used to manufacture synthetic rubbers. The raw polymers usually are tacky and of little direct use, except as adhesives and cements. They are transformed into materials with greater elasticity and strength by vulcanization, in which the polymer is heated with sulfur and various other substances called accelerators, with the result that the polymer chains become cross-linked to one another by carbon-sulfur and carbon-carbon bonds. Some of the cross-linking appears to occur by addition to the double bonds, but the amount of sulfur added generally is insufficient to saturate the polymer. With large proportions of sulfur, hard rubber is formed such as is used in storage-battery cases.
Because of the many double bonds present, diene rubbers are sensitive to air oxidation unless antioxidants are added to inhibit oxidation.
The more important dienes for the manufacture of synthetic rubbers are 1,3-butadiene, 2-chloro-1,3-butadiene (chloroprene), and 2-methyl-1,3-butadiene (isoprene):
Several rubbers that have desirable properties of elasticity, flexibility, abrasive resistance, and resistance to chemicals are listed in Table 13-2. The homogeneity of these polymers depends greatly on the way in which they are prepared, particularly on the polymerization catalyst employed. A synthetic rubber that is virtually identical to natural Hevea rubber is made from 2-methyl-1,3-butadiene (isoprene) using finely divided lithium metal or transition-metal catalysts; the product is formed almost exclusively by cis 1,4 addition\(^3\):
Table 13-2: Synthetic Rubbers
Polymerization of 2-methylpropene in the presence of small amounts of 2-methyl-1,3-butadiene (isoprene) gives a copolymer with enough double bonds to permit cross-linking of the polymer chains through vulcanization. The product is a hard-wearing, chemically resistant rubber called “butyl rubber.” It is highly impermeable to air and is used widely for inner tubes for tires.
\(^3\)Synthetic rubber has provided severe competition for natural rubber and, for many years, it seemed as though rubber plantations eventually would become extinct. However, rising petroleum prices and higher 2-methyl-1,3-butadiene costs coupled with methods developed for greatly increasing the output of rubber latex per tree, and the fact that natural rubber has superior properties in radial automobile tires, have reversed the trend and rubber plantations currently are being expanded.
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)/13%3A_Polyfunctional_Compounds_Alkadienes_and_Approaches_to_Organic_Synthesis/13.04%3A_Polymerization_Reactions_of_Conjugated_Dienes.txt |
Structure and Stereoisomerism
The 1,2-dienes, which have cumulated double bonds, commonly are called allenes. The simplest example is 1,2-propadiene,
Allenes of the type $\ce{RR'C=C=CRR'}$ are chiral molecules and can exist in two stereoisomeric forms, one being the mirror image of the other and neither being superimposable on the other (i.e., enantiomers, Figure 13-5).
Verification of the chirality of such allenes (originally proposed by van’t Hoff in 1875) was slow in coming and was preceded by many unsuccessful attempts to resolve suitably substituted allenes into their enantiomers. The first successful resolutions were achieved in 1935 for the enantiomers of two compounds $1$ and $2$. This was a classic achievement because it dispelled the suspicion prevalent at the time that rotation about the bonds of the cumulated diene system was free enough to preclude the isolation of configurationally stable enantiomers.
The chirality observed in this kind of substituted allene is a consequence of dissymmetry resulting from restricted rotation about the double bonds, not because of a tetrahedral atom carrying four different groups. Restricted rotation occurs in many other kinds of compounds and a few examples are shown in Table 13-3, which includes trans-cycloalkenes (Section 12-7), cycloalkylidenes, spiranes, and ortho-substituted biphenyl compounds. To have enantiomers, the structure must not have a plane or center of symmetry (Section 5-5).
Table 13-3: Examples of Chiral Substances Resulting from Restricted Rotation About Double or Single Bonds$^a$
The chirality of biphenyls results from restricted rotation about a single bond imposed by the bulky nature of ortho substituents. Models will help you visualize the degree of difficulty of having the substituents pass by one another. If $\ce{X} = \ce{H}$ and $\ce{Y} = \ce{F}$ (Table 13-3), the enantiomers are not stable at room temperature; if $\ce{X} = \ce{H}$ and $\ce{Y} = \ce{Br}$, they are marginally stable; if $\ce{X} = \ce{H}$ and $\ce{Y} = \ce{I}$, the rate of loss of optical activity is about 700 times slower than with $\ce{Y} = \ce{Br}$. This is in keeping with the fact that the atomic size of the halogens increases in the order $\ce{F} < \ce{Br} < \ce{I}$.
Cis-Trans Isomerism
In a cumulated triene, or any cumulated polyene with an odd number of double bonds, the atoms connected to the terminal carbons lie in the same plane, just as they do in an ordinary alkene. Van’t Hoff pointed out that suitably substituted cumulated polyenes of this type should then have cis and trans forms:
Like the resolution of allenes, the separate existence of cis and trans isomers of cumulated trienes was not verified until many years after van’t Hoff’s original predictions, but a separation finally was achieved, in 1954, by R. Kuhn and K. Scholler for compounds $3$ and $4$:
There are relatively few cis-trans forms of 1,2,3-alkatrienes known. They appear to interconvert readily on mild heating, which suggests that one of the double bonds has a lower rotational barrier than is normal for an alkene double bond.
Chemistry of Allenes
The properties of allenes are similar to those of alkenes, although the pure compounds often are difficult to prepare and are not indefinitely stable. Allenes undergo many of the usual double-bond reactions, being readily hydrogenated, adding bromine, and being oxidized with potassium permanganate solution. The hydration of allenes resembles the hydration of alkynes in giving initially an unstable enol that rapidly rearranges to a ketone:
Allenes are not as stable as dienes with conjugated or isolated double bonds. The heats of hydrogenation (Table 11-2) indicate that the order of stability is conjugated dienes $>$ isolated dienes $>$ cumulated dienes. The relative instability of allenes probably reflects extra strain as the result of one carbon atom forming two double bonds. 1,2-Propadiene is slightly more strained than that of propyne. It is not surprising then that 1,2-propadiene isomerizes to propyne. This isomerization occurs under the influence of strongly basic substances such as sodium amide in liquid ammonia or potassium hydroxide in ethyl alcohol:
Indeed, one of the difficulties associated with syntheses of allenes and alkynes (which often are carried out in the presence of strong bases) is the concurrent formation of isomerization products.
The basic catalyst in the isomerization of 1,2-butadiene to butynes acts by removing an alkenic proton from the hydrocarbon. Two different anions can be formed, each of which is stabilized by electron delocalization involving the adjacent multiple bond. Either anion can react with the solvent by proton transfer to form the starting material or an alkyne. At equilibrium the most stable product, which is 2-butyne, predominates [1-butyne $\left( g \right) \rightleftharpoons$ 2-butyne $\left( g \right)$, $\Delta G^0 = -4.0 \: \text{kcal mol}^{-1}$]:
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)/13%3A_Polyfunctional_Compounds_Alkadienes_and_Approaches_to_Organic_Synthesis/13.05%3A_Cumulated_Alkadienes.txt |
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