Alkyl Groups If you imagine removing a hydrogen atom from an alkane, the partial structure that remains is called an alkyl group . Alkyl groups are not stable compounds themselves, they are simply parts of larger compounds and are named by replacing the - ane ending of the parent alkane with an - yl ending. For example, removal of a hydrogen from methane, CH 4 , generates a methyl group, –CH 3 , and removal of a hydrogen from ethane, CH 3 CH 3 , generates an ethyl group, –CH 2 CH 3 . Similarly, removal of a hydrogen atom from the end carbon of any straight-chain alkane gives the series of straight-chain alkyl groups shown in . Combining an alkyl group with any of the functional groups listed earlier makes it possible to generate and name many thousands of compounds. For example: Some Straight-Chain Alkyl Groups Alkane Name Alkyl group Name (abbreviation) CH 4 Methane –CH 3 Methyl (Me) CH 3 CH 3 Ethane –CH 2 CH 3 Ethyl (Et) CH 3 CH 2 CH 3 Propane –CH 2 CH 2 CH 3 Propyl (Pr) CH 3 CH 2 CH 2 CH 3 Butane –CH 2 CH 2 CH 2 CH 3 Butyl (Bu) CH 3 CH 2 CH 2 CH 2 CH 3 Pentane –CH 2 CH 2 CH 2 CH 2 CH 3 Pentyl, or amyl Just as straight-chain alkyl groups are generated by removing a hydrogen from an end carbon, branched alkyl groups are generated by removing a hydrogen atom from an internal carbon. Two 3-carbon alkyl groups and four 4-carbon alkyl groups are possible ( ). Alkyl groups generated from straight-chain alkanes. One further comment about naming alkyl groups: the prefixes sec - (for secondary) and tert - (for tertiary) used for the C 4 alkyl groups in refer to the number of other carbon atoms attached to the branching carbon atom. There are four possibilities: primary (1°), secondary (2°), tertiary (3°), and quaternary (4°). The symbol R is used here and throughout organic chemistry to represent a generalized organic group. The R group can be methyl, ethyl, propyl, or any of a multitude of others. You might think of R as representing the R est of the molecule, which isn’t specified. The terms primary, secondary, tertiary, and quaternary are routinely used in organic chemistry, and their meanings need to become second nature. For example, if we were to say, “Citric acid is a tertiary alcohol,” we would mean that it has an alcohol functional group (–OH) bonded to a carbon atom that is itself bonded to three other carbons. In addition to speaking of carbon atoms as being primary, secondary, or tertiary, we speak of hydrogens in the same way. Primary hydrogen atoms are attached to primary carbons (RCH 3 ), secondary hydrogens are attached to secondary carbons (R 2 CH 2 ), and tertiary hydrogens are attached to tertiary carbons (R 3 CH). There is, however, no such thing as a quaternary hydrogen. (Why not?) Problems

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Naming Alkanes In earlier times, when relatively few pure organic chemicals were known, new compounds were named at the whim of their discoverer. Thus, urea (CH 4 N 2 O) is a crystalline substance isolated from urine; morphine (C 17 H 19 NO 3 ) is an analgesic (painkiller) named after Morpheus, the Greek god of dreams; and acetic acid, the primary organic constituent of vinegar, is named from the Latin word for vinegar, acetum . As the science of organic chemistry slowly grew in the 19th century, so too did the number of known compounds and the need for a systematic method of naming them. The system of naming (nomenclature) we’ll use in this book is that devised by the International Union of Pure and Applied Chemistry (IUPAC, usually spoken as eye -you-pac). A chemical name typically has four parts in the IUPAC system: parent, prefix, locant, and suffix. The parent name identifies the main part of the molecule and tells how many carbon atoms are in that part. Prefixes identify the various substituent groups attached to the parent. Locants give the positions of the attached substituents. And the suffix identifies the primary functional group attached to the parent. As we cover new functional groups in later chapters, the applicable IUPAC rules of nomenclature will be given. In addition, at the back of this book gives an overall view of organic nomenclature and shows how compounds that contain more than one functional group are named. (If preferred, you can study that appendix now.) For the present, let’s see how to name branched-chain alkanes and learn some general rules that are applicable to all compounds. All but the most complex branched-chain alkanes can be named by following four steps. For a very few compounds, a fifth step is needed. STEP 1 Identify the parent hydrocarbon. (a) Find the longest continuous chain of carbon atoms in the molecule, and use the name of that chain as the parent name. The longest chain may not always be apparent from the manner of writing; you may have to “turn corners.” (b) If two different chains of equal length are present, choose the one with the larger number of branch points as the parent. STEP 2 Number the atoms in the longest chain. (a) Beginning at the end nearer the first branch point, number each carbon atom in the parent chain. The first branch occurs at C3 in the proper system of numbering, not at C4. (b) If there is branching an equal distance away from both ends of the parent chain, begin numbering at the end nearer the second branch point. STEP 3 Identify and number the substituents. (a) Assign a number to each substituent to locate its point of attachment to the parent chain. (b) If there are two substituents on the same carbon, give both the same number. There must be as many numbers in the name as there are substituents. STEP 4 Write the name as a single word. Use hyphens to separate the different prefixes, and use commas to separate numbers. If two or more different substituents are present, cite them in alphabetical order. If two or more identical substituents are present on the parent chain, use one of the multiplier prefixes di -, tri -, tetra -, and so forth, but don’t use these prefixes for alphabetizing. Full names for some of the examples we have been using are as follows: STEP 5 Name a branched substituent as though it were itself a compound. In some particularly complex cases, a fifth step is necessary. It occasionally happens that a substituent on the main chain is itself branched. In the following case, for instance, the substituent at C6 is a three-carbon chain with a methyl group. To name the compound fully, the branched substituent must first be named. Number the branched substituent beginning at the point of its attachment to the main chain, and identify it—in this case, a 2-methylpropyl group. The substituent is treated as a whole and is alphabetized according to the first letter of its complete name, including any numerical prefix. It is set off in parentheses when naming the entire molecule. As a further example: For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic, common names, as noted earlier. The common names of these simple alkyl groups are so well entrenched in the chemical literature that IUPAC rules make allowance for them. Thus, the following compound is properly named either 4-(1-methylethyl)heptane or 4-isopropylheptane. There’s no choice but to memorize these common names; fortunately, there are only a few of them. When writing an alkane name, the nonhyphenated prefix iso- is considered part of the alkyl-group name for alphabetizing purposes, but the hyphenated and italicized prefixes sec - and tert - are not. Thus, isopropyl and isobutyl are listed alphabetically under i , but sec -butyl and tert -butyl are listed under b . Naming Alkanes What is the IUPAC name for the following alkane? Strategy Find the longest continuous carbon chain in the molecule, and use that as the parent name. This molecule has a chain of eight carbons—octane—with two methyl substituents. (You have to turn corners to see it.) Numbering from the end nearer the first methyl substituent indicates that the methyls are at C2 and C6. Solution Converting a Chemical Name into a Structure Draw the structure of 3-isopropyl-2-methylhexane. Strategy This is the reverse of and uses a reverse strategy. Look at the parent name (hexane), and draw its carbon structure. C–C–C–C–C–C Hexane Next, find the substituents (3-isopropyl and 2-methyl), and place them on the proper carbons. Finally, add hydrogens to complete the structure. Solution Problems

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Properties of Alkanes Alkanes are sometimes referred to as paraffins , a word derived from the Latin parum affinis , meaning “little affinity.” This term aptly describes their behavior, for alkanes show little chemical affinity for other substances and are chemically inert to most laboratory reagents. They are also relatively inert biologically and are not often involved in the chemistry of living organisms. Alkanes do, however, react with oxygen, halogens, and a few other substances under appropriate conditions. Reaction with oxygen occurs during combustion in an engine or furnace when an alkane is used as a fuel. Carbon dioxide and water are formed as products, and a large amount of heat is released. For example, methane (natural gas) reacts with oxygen according to the equation CH 4 + 2 O 2 ⟶ CO 2 + 2 H 2 O + 890 kJ/mol ( 213 kcal/mol ) The reaction of an alkane with Cl 2 occurs when a mixture of the two is irradiated with ultraviolet light (denoted hυ, where υ is the Greek letter nu ). Depending on the time allowed and the relative amounts of the two reactants, a sequential substitution of the alkane hydrogen atoms by chlorine occurs, leading to a mixture of chlorinated products. Methane, for instance, reacts with Cl 2 to yield a mixture of CH 3 Cl, CH 2 Cl 2 , CHCl 3 , and CCl 4 . We’ll look at this reaction in more detail in Section 6.6 . Alkanes show regular increases in both boiling point and melting point as molecular weight increases ( ), an effect due to the presence of weak dispersion forces between molecules ( ). Only when sufficient energy is applied to overcome these forces does the solid melt or liquid boil. As you might expect, dispersion forces increase as molecular size increases, accounting for the higher melting and boiling points of larger alkanes. A plot of melting and boiling points versus number of carbon atoms for the C1–C14 straight-chain alkanes. There is a regular increase with molecular size. Another effect seen in alkanes is that increased branching lowers an alkane’s boiling point. Thus, pentane has no branches and boils at 36.1 °C, isopentane (2-methylbutane) has one branch and boils at 27.85 °C, and neopentane (2,2-dimethylpropane) has two branches and boils at 9.5 °C. Similarly, octane boils at 125.7 °C, whereas isooctane (2,2,4-trimethylpentane) boils at 99.3 °C. Branched-chain alkanes are lower-boiling because they are more nearly spherical than straight-chain alkanes, have smaller surface areas, and consequently have smaller dispersion forces.

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Conformations of Ethane Up until now, we’ve viewed molecules primarily in a two-dimensional way and have given little thought to any consequences that might arise from the spatial arrangement of atoms in molecules. Now it’s time to add a third dimension to our study. Stereochemistry is the branch of chemistry concerned with the three-dimensional aspects of molecules. We’ll see on many occasions in future chapters that the exact three-dimensional structure of a molecule is often crucial to determining its properties and biological behavior. We know from that σ bonds are cylindrically symmetrical. In other words, the intersection of a plane cutting through a carbon–carbon single-bond orbital looks like a circle. Because of this cylindrical symmetry, rotation is possible around carbon–carbon bonds in open-chain molecules. In ethane, for instance, rotation around the C–C bond occurs freely, constantly changing the spatial relationships between the hydrogens on one carbon and those on the other ( ). Rotation occurs around the carbon–carbon single bond in ethane because of σ bond cylindrical symmetry. The different arrangements of atoms that result from bond rotation are called conformations , and molecules that have different arrangements are called conformational isomers , or conformers . Unlike constitutional isomers, however, different conformers often can’t be isolated because they interconvert too rapidly. Conformational isomers are represented in two ways, as shown in . A sawhorse representation views the carbon–carbon bond from an oblique angle and indicates spatial orientation by showing all C–H bonds. A Newman projection views the carbon–carbon bond directly end-on and represents the two carbon atoms by a circle. Bonds attached to the front carbon are represented by lines to the center of the circle, and bonds attached to the rear carbon are represented by lines to the edge of the circle. A sawhorse representation and a Newman projection of ethane. The sawhorse representation views the molecule from an oblique angle, while the Newman projection views the molecule end-on. Note that the molecular model of the Newman projection appears at first to have six atoms attached to a single carbon. Actually, the front carbon, with three attached green atoms , is directly in front of the rear carbon, with three attached red atoms . Despite what we’ve just said, we actually don’t observe perfectly free rotation in ethane. Experiments show that there is a small (12 kJ/mol; 2.9 kcal/mol) barrier to rotation and that some conformations are more stable than others. The lowest-energy, most stable conformation is the one in which all six C–H bonds are as far away from one another as possible— staggered when viewed end-on in a Newman projection. The highest-energy, least stable conformation is the one in which the six C–H bonds are as close as possible— eclipsed in a Newman projection. At any given instant, about 99% of ethane molecules have an approximately staggered conformation and only about 1% are near the eclipsed conformation . The extra 12 kJ/mol of energy present in the eclipsed conformation of ethane is called torsional strain . Its cause has been the subject of controversy, but the major factor is an interaction between C–H bonding orbitals on one carbon and antibonding orbitals on the adjacent carbon, which stabilizes the staggered conformation relative to the eclipsed one. Because a total strain of 12 kJ/mol arises from three equal hydrogen–hydrogen eclipsing interactions, we can assign a value of approximately 4.0 kJ/mol (1.0 kcal/mol) to each single interaction. The barrier to rotation that results can be represented on a graph of potential energy versus degree of rotation, in which the angle between C–H bonds on the front and back carbons as viewed end-on (the dihedral angle ) goes full circle from 0 to 360°. Energy minima occur at staggered conformations, and energy maxima occur at eclipsed conformations, as shown in . A graph of potential energy versus bond rotation in ethane. The staggered conformations are 12 kJ/mol lower in energy than the eclipsed conformations.

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Conformations of Other Alkanes Propane, the next-higher member in the alkane series, also has a torsional barrier that results in hindered rotation around the carbon–carbon bonds. The barrier is slightly higher in propane than in ethane—a total of 14 kJ/mol (3.4 kcal/mol) versus 12 kJ/mol. The eclipsed conformation of propane has three interactions—two ethane-type hydrogen–hydrogen interactions and one additional hydrogen–methyl interaction. Since each eclipsing H ⟷ H interaction is the same as that in ethane and thus has an energy “cost” of 4.0 kJ/mol, we can assign a value of 14 – (2 × 4.0) = 6.0 kJ/mol (1.4 kcal/mol) to the eclipsing H ⟷ CH 3 interaction ( ). Newman projections of propane showing staggered and eclipsed conformations. The staggered conformer is lower in energy by 14 kJ/mol. The conformational situation becomes more complex for larger alkanes because not all staggered conformations have the same energy and not all eclipsed conformations have the same energy. In butane, for instance, the lowest-energy arrangement, called the anti conformation , is the one in which the two methyl groups are as far apart as possible—180° away from each other. As rotation around the C2–C3 bond occurs, an eclipsed conformation is reached where there are two CH 3 ⟷ H interactions and one H ⟷ H interaction. Using the energy values derived previously from ethane and propane, this eclipsed conformation is more strained than the anti conformation by 2 × 6.0 kJ/mol + 4.0 kJ/mol (two CH 3 ⟷ H interactions plus one H ⟷ H interaction), for a total of 16 kJ/mol (3.8 kcal/mol). As bond rotation continues, an energy minimum is reached at the staggered conformation where the methyl groups are 60° apart. Called the gauche conformation , it lies 3.8 kJ/mol (0.9 kcal/mol) higher in energy than the anti conformation even though it has no eclipsing interactions. This energy difference occurs because the hydrogen atoms of the methyl groups are near one another in the gauche conformation, resulting in what is called steric strain. Steric strain is the repulsive interaction that occurs when atoms are forced closer together than their atomic radii allow. It’s the result of trying to force two atoms to occupy the same space. As the dihedral angle between the methyl groups approaches zero, an energy maximum is reached at a second eclipsed conformation. Because the methyl groups are forced even closer together than in the gauche conformation, both torsional strain and steric strain are present. A total strain energy of 19 kJ/mol (4.5 kcal/mol) has been estimated for this conformation, making it possible to calculate a value of 11 kJ/mol (2.6 kcal/mol) for the CH 3 ⟷ CH 3 eclipsing interaction: total strain of 19 kJ/mol minus the strain of two H ⟷ H eclipsing interactions (2 × 4.0 kcal/mol) equals 11 kJ/mol. After 0°, the rotation becomes a mirror image of what we’ve already seen: another gauche conformation is reached, another eclipsed conformation, and finally a return to the anti conformation. A plot of potential energy versus rotation about the C2–C3 bond is shown in . A plot of potential energy versus rotation for the C2−C3 bond in butane. The energy maximum occurs when the two methyl groups eclipse each other, and the energy minimum occurs when the two methyl groups are 180° apart (anti). The notion of assigning definite energy values to specific interactions within a molecule is very useful, and we’ll return to it in the next chapter. A summary of what we’ve seen thus far is given in . The same principles just developed for butane apply to pentane, hexane, and all higher alkanes. The most favorable conformation for any alkane has the carbon–carbon bonds in staggered arrangements, with large substituents arranged anti to one another. A generalized alkane structure is shown in . Energy Costs for Interactions in Alkane Conformers Interaction Cause Energy cost (kJ/mol) (kcal/mol) H ⟷ H eclipsed Torsional strain 4.0 1.0 H ⟷ CH 3 eclipsed Mostly torsional strain 6.0 1.4 CH 3 ⟷ CH 3 eclipsed Torsional and steric strain 11.0 2.6 CH 3 ⟷ CH 3 gauche Steric strain 3.8 0.9 The most stable alkane conformation is the one in which all substituents are staggered and the carbon–carbon bonds are arranged anti , as shown in this model of decane. One final point: saying that one particular conformer is “more stable” than another doesn’t mean the molecule adopts and maintains only the more stable conformation. At room temperature, rotations around σ bonds occur so rapidly that all conformers are in equilibrium. At any given instant, however, a larger percentage of molecules will be found in a more stable conformation than in a less stable one. Newman Projections Sight along the C1–C2 bond of 1-chloropropane, and draw Newman projections of the most stable and least stable conformations. Strategy The most stable conformation of a substituted alkane is generally a staggered one in which large groups have an anti relationship. The least stable conformation is generally an eclipsed one in which large groups are as close as possible. Solution Problems Gasoline British Foreign Minister Ernest Bevin once said that “The Kingdom of Heaven runs on righteousness, but the Kingdom of Earth runs on alkanes.” (Actually, he said “runs on oil” not “runs on alkanes,” but they’re essentially the same.) By far, the major sources of alkanes are the world’s natural gas and petroleum deposits. Laid down eons ago, these deposits are thought to be derived primarily from the decomposition of tiny single-celled marine organisms called foraminifera. Natural gas consists chiefly of methane but also contains ethane, propane, and butane. Petroleum is a complex mixture of hydrocarbons that must be separated into fractions and then further refined before it can be used. Gasoline is a finite resource. It won’t be around forever. (credit: “The first oil well” by Unknown/Library of Congress) The petroleum era began in August 1859, when the world’s first oil well was drilled by Edwin Drake near Titusville, Pennsylvania. The petroleum was distilled into fractions according to boiling point, but it was high-boiling kerosene, or lamp oil, rather than gasoline that was primarily sought. Literacy was becoming widespread at the time, and people wanted better light for reading than was available from candles. Gasoline was too volatile for use in lamps and was initially considered a waste by-product. The world has changed greatly since those early days, however, and it is now gasoline rather than lamp oil that is prized. Petroleum refining begins by fractional distillation of crude oil into three principal cuts according to boiling point (bp): straight-run gasoline (bp 30–200 °C), kerosene (bp 175–300 °C), and heating oil, or diesel fuel (bp 275–400 °C). Further distillation under reduced pressure then yields lubricating oils and waxes and leaves a tarry residue of asphalt. The distillation of crude oil is only the first step in gasoline production, however. Straight-run gasoline turns out to be a poor fuel in automobiles because of engine knock, an uncontrolled combustion that can occur in a hot engine causing potentially serious damage. The octane number of a fuel is the measure by which its antiknock properties are judged. It was recognized long ago that straight-chain hydrocarbons are far more prone to inducing engine knock than highly branched compounds. Heptane, a particularly bad fuel, is assigned a base value of 0 octane number, and 2,2,4-trimethylpentane, commonly known as isooctane, has a rating of 100. Because straight-run gasoline burns so poorly in engines, petroleum chemists have devised numerous methods for producing higher-quality fuels. One of these methods, catalytic cracking , involves taking the high-boiling kerosene cut (C 11 –C 14 ) and “cracking” it into smaller branched molecules suitable for use in gasoline. Another process, called reforming , is used to convert C 6 –C 8 alkanes to aromatic compounds such as benzene and toluene, which have substantially higher octane numbers than alkanes. The final product that goes in your tank has an approximate composition of 15% C 4 –C 8 straight-chain alkanes, 25% to 40% C 4 –C 10 branched-chain alkanes, 10% cyclic alkanes, 10% straight-chain and cyclic alkenes, and 25% arenes (aromatics). alcohol aldehyde aliphatic alkane alkene alkyl group alkyl halide alkyne amide amine anti conformation arene branched-chain alkane carbonyl group carboxylic acid conformation conformational isomer conformer constitutional isomer eclipsed conformation ester ether functional group gauche conformation hydrocarbon isomer ketone Newman projection nitrile R group saturated sawhorse representation staggered conformation stereochemistry steric strain straight-chain alkane substituent sulfide thiol torsional strain Summary Alkanes are relatively unreactive and rarely involved in chemical reactions, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we’ve used alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules. A functional group is a group of atoms within a larger molecule that has a characteristic chemical reactivity. Because functional groups behave in approximately the same way in all molecules where they occur, the chemical reactions of an organic molecule are largely determined by its functional groups. Alkanes are a class of saturated hydrocarbons with the general formula C n H 2 n +2 . They contain no functional groups, are relatively inert, and can be either straight-chain ( normal ) or branched . Alkanes are named by a series of IUPAC rules of nomenclature. Compounds that have the same chemical formula but different structures are called isomers . More specifically, compounds such as butane and isobutane, which differ in their connections between atoms, are called constitutional isomers . Carbon–carbon single bonds in alkanes are formed by σ overlap of carbon sp 3 hybrid orbitals. Rotation is possible around σ bonds because of their cylindrical symmetry, and alkanes therefore exist in a large number of rapidly interconverting conformations . Newman projections make it possible to visualize the spatial consequences of bond rotation by sighting directly along a carbon–carbon bond axis. Not all alkane conformations are equally stable. The staggered conformation of ethane is 12 kJ/mol (2.9 kcal/mol) more stable than the eclipsed conformation because of torsional strain . In general, any alkane is most stable when all its bonds are staggered. Additional Problems Visualizing Chemistry Functional Groups Isomers Naming Compounds Conformations General Problems

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Why This Chapter? The musk gland of the male Himalayan musk deer secretes a substance once used in perfumery that contains cycloalkanes of 14 to 18 carbons . (credit: modification of work “Siberian musk deer in the tiaga” by ErikAdamsson/Wikimedia Commons, CC0 1.0) We’ll see numerous instances in future chapters where the chemistry of a given functional group is affected by being in a ring rather than an open chain. Because cyclic molecules are encountered in most pharmaceuticals and in all classes of biomolecules, including proteins, lipids, carbohydrates, and nucleic acids, it’s important to understand the behavior of cyclic structures. Although we’ve only discussed open-chain compounds up to now, most organic compounds contain rings of carbon atoms. Chrysanthemic acid, for instance, whose esters occur naturally as the active insecticidal constituents of chrysanthemum flowers, contains a three-membered (cyclopropane) ring. Prostaglandins , potent hormones that control an extraordinary variety of physiological functions in humans, contain a five-membered (cyclopentane) ring. Steroids , such as cortisone, contain four rings joined together—three six-membered (cyclohexane) and one five-membered. We’ll discuss steroids and their properties in more detail in Sections 27.6 and 27.7 .

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Naming Cycloalkanes Saturated cyclic hydrocarbons are called cycloalkanes , or alicyclic compounds ( ali phatic cyclic ). Because cycloalkanes consist of rings of −CH 2 −units, they have the general formula (CH 2 ) n , or C n H 2 n , and can be represented by polygons in skeletal drawings. Substituted cycloalkanes are named by rules similar to those we saw in ( ) for open-chain alkanes. For most compounds, there are only two steps. STEP 1 Find the parent. Count the number of carbon atoms in the ring and the number in the largest substituent. If the number of carbon atoms in the ring is equal to or greater than the number in the substituent, the compound is named as an alkyl-substituted cycloalkane. If the number of carbon atoms in the largest substituent is greater than the number in the ring, the compound is named as a cycloalkyl-substituted alkane. For example: STEP 2 Number the substituents, and write the name. For an alkyl- or halo-substituted cycloalkane, choose a point of attachment as carbon 1 and number the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found. (a) When two or more different alkyl groups are present that could potentially take the same numbers, number them by alphabetical priority, ignoring numerical prefixes such as di- and tri-. (b) If halogens are present, treat them just like alkyl groups. Some additional examples follow: Problems

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Cis–Trans Isomerism in Cycloalkanes In many respects, the chemistry of cycloalkanes is like that of open-chain alkanes: both are nonpolar and fairly inert. There are, however, some important differences. One difference is that cycloalkanes are less flexible than open-chain alkanes. In contrast with the relatively free rotation around single bonds in open-chain alkanes ( and ), there is much less freedom in cycloalkanes. Cyclopropane, for example, must be a rigid, planar molecule because three points (the carbon atoms) define a plane. No bond rotation can take place around a cyclopropane carbon–carbon bond without breaking open the ring ( ). Bond rotation in ethane and cyclopropane. (a) Rotation occurs around the carbon–carbon bond in ethane, but (b) no rotation is possible around the carbon–carbon bonds in cyclopropane without breaking open the ring. Larger cycloalkanes have increasing rotational freedom, and very large rings (C 25 and up) are so floppy that they are nearly indistinguishable from open-chain alkanes. The common ring sizes (C 3 –C 7 ), however, are severely restricted in their molecular motions. Because of their cyclic structures, cycloalkanes have two faces when viewed edge-on, a “top” face and a “bottom” face. As a result, isomerism is possible in substituted cycloalkanes. For example, there are two different 1,2-dimethylcyclopropane isomers, one with the two methyl groups on the same face of the ring and one with the methyl groups on opposite faces ( ). Both isomers are stable compounds, and neither can be converted into the other without breaking and reforming chemical bonds. There are two different 1,2-dimethylcyclopropane isomers , one with the methyl groups on the same face of the ring (cis) and the other with the methyl groups on opposite faces of the ring (trans). The two isomers do not interconvert. Unlike the constitutional isomers butane and isobutane, which have their atoms connected in a different order ( ), the two 1,2-dimethylcyclopropanes have the same order of connections but differ in the spatial orientation of the atoms. Such compounds, with atoms connected in the same order but differing in three-dimensional orientation, are called stereochemical isomers, or stereoisomers . As we saw in , the term stereochemistry is used generally to refer to the three-dimensional aspects of structure and reactivity. The 1,2-dimethylcyclopropanes are members of a subclass of stereoisomers called cis–trans isomers . The prefixes cis - (Latin “on the same side”) and trans - (Latin “across”) are used to distinguish between them. Cis–trans isomerism is a common occurrence in substituted cycloalkanes and in many cyclic biological molecules. Naming Cycloalkanes Name the following substances, including the cis - or trans - prefix: Strategy In these views, the ring is roughly in the plane of the page, a wedged bond protrudes out of the page, and a dashed bond recedes into the page. Two substituents are cis if they are both out of or both into the page, and they are trans if one is out of and one is into the page. Solution (a) trans -1,3-Dimethylcyclopentane (b) cis -1,2-Dichlorocyclohexane Problems

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Stability of Cycloalkanes: Ring Strain Chemists in the late 1800s knew that cyclic molecules existed, but the limitations on ring size were unclear. Although numerous compounds containing five-membered and six-membered rings were known, smaller and larger ring sizes had not been prepared, despite many attempts. A theoretical interpretation of this observation was proposed in 1885 by Adolf von Baeyer, who suggested that small and large rings might be unstable due to angle strain —the strain induced in a molecule when bond angles are forced to deviate from the ideal 109° tetrahedral value. Baeyer based his suggestion on the simple geometric notion that a three-membered ring (cyclopropane) should be an equilateral triangle with bond angles of 60° rather than 109°, a four-membered ring (cyclobutane) should be a square with bond angles of 90°, a five-membered ring should be a regular pentagon with bond angles of 108°, and so on. Continuing this argument, large rings should be strained by having bond angles that are much greater than 109°. What are the facts? To measure the amount of strain in a compound, we have to measure the total energy of the compound and then subtract the energy of a strain-free reference compound. The difference between the two values should represent the amount of extra energy in the molecule due to strain. The simplest experimental way to do this for a cycloalkane is to measure its heat of combustion, the amount of heat released when the compound burns completely with oxygen. The more energy (strain) the compound contains, the more energy (heat) is released by combustion. ( CH 2 ) n + 3 n / 2 O 2 ⟶ n CO 2 + n H 2 O + Heat (add) Because the heat of combustion of a cycloalkane depends on size, we need to look at heats of combustion per CH 2 unit. Subtracting a reference value derived from a strain-free acyclic alkane and then multiplying by the number of CH 2 units in the ring gives the overall strain energy. shows the results. Cycloalkane strain energies, as calculated by taking the difference between cycloalkane heat of combustion per CH 2 and acyclic alkane heat of combustion per CH 2 , and multiplying by the number of CH 2 units in a ring. Small and medium rings are strained, but cyclohexane rings and very large rings are strain-free. The data in show that Baeyer’s theory is only partially correct. Cyclopropane and cyclobutane are indeed strained, just as predicted, but cyclopentane is more strained than predicted, and cyclohexane is strain-free. Cycloalkanes of intermediate size have only modest strain, and rings of 14 carbons or more are strain-free. Why is Baeyer’s theory wrong? Baeyer’s theory is wrong for the simple reason that he assumed all cycloalkanes to be flat. In fact, as we’ll see in the next section, most cycloalkanes are not flat; instead, they adopt puckered three-dimensional conformations that allow bond angles to be nearly tetrahedral. As a result, angle strain occurs only in three- and four-membered rings, which have little flexibility. For most ring sizes, particularly the medium-ring (C 7 –C 11 ) cycloalkanes, torsional strain caused by H ⟷ H eclipsing interactions at adjacent carbons ( ) and steric strain caused by the repulsion between nonbonded atoms that approach too closely ( ) are the most important factors. Thus, three kinds of strain contribute to the overall energy of a cycloalkane. Angle strain —the strain due to expansion or compression of bond angles Torsional strain —the strain due to eclipsing of bonds between neighboring atoms Steric strain —the strain due to repulsive interactions when atoms approach each other too closely Problems

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Conformations of Cycloalkanes Cyclopropane Cyclopropane is the most strained of all rings, primarily due to the angle strain caused by its 60° C−C−C bond angles. In addition, cyclopropane has considerable torsional strain because the C−H bonds on neighboring carbon atoms are eclipsed ( ). The structure of cyclopropane, showing the eclipsing of neighboring C−H bonds that gives rise to torsional strain. Part (b) is a Newman projection along a C−C bond. How can the hybrid-orbital model of bonding account for the large distortion of bond angles from the normal 109° tetrahedral value to 60° in cyclopropane? The answer is that cyclopropane has bent bonds. In an unstrained alkane, maximum bonding is achieved when two atoms have their overlapping orbitals pointing directly toward each other. In cyclopropane, though, the orbitals can’t point directly toward each other; instead, they overlap at a slight angle. The result is that cyclopropane bonds are weaker and more reactive than typical alkane bonds—255 kJ/mol (61 kcal/mol) for a C−C bond in cyclopropane versus 370 kJ/mol (88 kcal/mol) for a C−C bond in open-chain propane. Cyclobutane Cyclobutane has less angle strain than cyclopropane but has more torsional strain because of its larger number of ring hydrogens. As a result, the total strain for the two compounds is nearly the same—110 kJ/mol (26.4 kcal/mol) for cyclobutane versus 115 kJ/mol (27.5 kcal/mol) for cyclopropane. Cyclobutane is not quite flat but is slightly bent so that one carbon atom lies about 25° above the plane of the other three ( ). The effect of this slight bend is to increase angle strain but to decrease torsional strain, until a minimum-energy balance between the two opposing effects is achieved. The conformation of cyclobutane. Part (c) is a Newman projection along a C−C bond, showing that neighboring C−H bonds are not quite eclipsed. Cyclopentane Cyclopentane was predicted by Baeyer to be nearly strain-free, but it actually has a total strain energy of 26 kJ/mol (6.2 kcal/mol). Although planar cyclopentane has practically no angle strain, it has a large torsional strain. Cyclopentane therefore twists to adopt a puckered, nonplanar conformation that strikes a balance between increased angle strain and decreased torsional strain. Four of the cyclopentane carbon atoms are in approximately the same plane, with the fifth carbon atom bent out of the plane. Most of the hydrogens are nearly staggered with respect to their neighbors ( ). The conformation of cyclopentane. Carbons 1, 2, 3, and 4 are nearly coplanar, but carbon 5 is out of the plane. Part (c) is a Newman projection along the C1–C2 bond, showing that neighboring C−H bonds are nearly staggered. Problems

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Conformations of Cyclohexane Substituted cyclohexanes are the most common cycloalkanes and occur widely in nature. A large number of compounds, including steroids and many pharmaceutical agents, have cyclohexane rings. The flavoring agent menthol, for instance, has three substituents on a six-membered ring. Cyclohexane adopts a strain-free, three-dimensional shape that is called a chair conformation because of its similarity to a lounge chair, with a back, seat, and footrest ( ). Chair cyclohexane has neither angle strain nor torsional strain—all C−C−C bond angles are near the 109° tetrahedral value, and all neighboring C−H bonds are staggered. The strain-free chair conformation of cyclohexane. All C−C−C bond angles are 111.5°, close to the ideal 109° tetrahedral angle, and all neighboring C−H bonds are staggered. The easiest way to visualize chair cyclohexane is to build a molecular model if you have access to a model kit, or alternatively to explore with one of the many computer-based modeling programs you may have access to. The chair conformation of cyclohexane can be drawn in three steps. STEP 1 Draw two parallel lines, slanted downward and slightly offset from each other. This means that four of the cyclohexane carbons lie in a plane. STEP 2 Place the topmost carbon atom above and to the right of the plane of the other four, and connect the bonds. STEP 3 Place the bottommost carbon atom below and to the left of the plane of the middle four, and connect the bonds. Note that the bonds to the bottommost carbon atom are parallel to the bonds to the topmost carbon. When viewing cyclohexane, it’s helpful to remember that the lower bond is in front and the upper bond is in back. If this convention isn’t defined, it can appear that the reverse is true. For clarity, all cyclohexane rings drawn in this book will have the front (lower) bond heavily shaded to indicate nearness to the viewer. In addition to the chair conformation of cyclohexane, there is an alternative conformation of cyclohexane that bears a slight resemblance to a twisted boat. Called the twist-boat conformation , it is nearly free of angle strain. It does, however, have both steric strain and torsional strain and is about 23 kJ/mol (5.5 kcal/mol) higher in energy than the chair conformation. As a result, molecules adopt the twist-boat geometry only rarely.

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Axial and Equatorial Bonds in Cyclohexane The chair conformation of cyclohexane has many consequences. We’ll see in Section 11.9 , for instance, that the chemical behavior of many substituted cyclohexanes is influenced by their conformation. In addition, we’ll see in Section 25.5 that simple carbohydrates, such as glucose, adopt a conformation based on the cyclohexane chair and that their chemistry is directly affected as a result. Another trait of the chair conformation is that there are two kinds of positions for substituents on the cyclohexane ring: axial positions and equatorial positions (as shown in ). The six axial positions are parallel to the ring axis , while the six equatorial positions are in the rough plane of the ring, around the ring equator . Axial and equatorial positions in chair cyclohexane. The six axial hydrogens are parallel to the ring axis, and the six equatorial hydrogens are in a band around the ring equator. As shown in , each carbon atom in chair cyclohexane has one axial and one equatorial hydrogen. Furthermore, each side of the ring has three axial and three equatorial hydrogens in an alternating arrangement. For example, if the top side of the ring has axial hydrogens on carbons 1, 3, and 5, then it has equatorial hydrogens on carbons 2, 4, and 6. The reverse is true for the bottom side: carbons 1, 3, and 5 have equatorial hydrogens, but carbons 2, 4, and 6 have axial hydrogens ( ). Alternating axial and equatorial positions in chair cyclohexane, looking directly down the ring axis. Each carbon atom has one axial and one equatorial position, and each face has alternating axial and equatorial positions. Note that we haven’t used the words cis and trans in this discussion of cyclohexane conformation. Two hydrogens on the same side of the ring are always cis, regardless of whether they’re axial or equatorial and regardless of whether they’re adjacent. Similarly, two hydrogens on opposite sides of the ring are always trans. Axial and equatorial bonds can be drawn following the procedure shown in . If possible, look at a molecular model as you practice. A procedure for drawing axial and equatorial bonds in chair cyclohexane. Because chair cyclohexane has two kinds of positions—axial and equatorial—we might expect to find two isomeric forms of a monosubstituted cyclohexane. In fact, we don’t. There is only one methylcyclohexane, one bromocyclohexane, one cyclohexanol (hydroxycyclohexane), and so on, because cyclohexane rings are conformationally mobile at room temperature. Different chair conformations readily interconvert, exchanging axial and equatorial positions. This interconversion, called a ring-flip , is shown in . A ring-flip in chair cyclohexane interconverts axial and equatorial positions. What is axial in the starting structure becomes equatorial in the ring-flipped structure, and what is equatorial in the starting structure is axial after ring-flip. As shown in , a chair cyclohexane can be ring-flipped by keeping the middle four carbon atoms in place while folding the two end carbons in opposite directions. In so doing, an axial substituent in one chair form becomes an equatorial substituent in the ring-flipped chair form and vice versa. For example, axial bromocyclohexane becomes equatorial bromocyclohexane after a ring-flip. Since the energy barrier to chair–chair interconversion is only about 45 kJ/mol (10.8 kcal/mol), the process is rapid at room temperature and we see what appears to be a single structure rather than distinct axial and equatorial isomers. Drawing the Chair Conformation of a Substituted Cyclohexane Draw 1,1-dimethylcyclohexane in a chair conformation, indicating which methyl group in your drawing is axial and which is equatorial. Strategy Draw a chair cyclohexane ring using the procedure in , and then put two methyl groups on the same carbon. The methyl group in the rough plane of the ring is equatorial, and the one directly above or below the ring is axial. Solution Problems

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Conformations of Monosubstituted Cyclohexanes Even though cyclohexane rings flip rapidly between chair conformations at room temperature, the two conformations of a monosubstituted cyclohexane aren’t equally stable. In methylcyclohexane, for instance, the equatorial conformation is more stable than the axial conformation by 7.6 kJ/mol (1.8 kcal/mol). The same is true of other monosubstituted cyclohexanes: a substituent is almost always more stable in an equatorial position than in an axial position. You might recall from your general chemistry course that it’s possible to calculate the percentages of two isomers at equilibrium using the equation Δ E = – RT ln K, where Δ E is the energy difference between isomers, R is the gas constant [8.315 J/(K·mol)], T is the Kelvin temperature, and K is the equilibrium constant between isomers. For example, an energy difference of 7.6 kJ/mol means that about 95% of methylcyclohexane molecules have an equatorial methyl group at any given instant while only 5% have an axial methyl group. plots the relationship between energy and isomer percentages. A plot of the percentages of two isomers at equilibrium versus the energy difference between them. The curves are calculated using the equation Δ E = − RT ln K . The energy difference between axial and equatorial conformations is due to steric strain caused by 1,3-diaxial interactions . The axial methyl group on C1 is too close to the axial hydrogens three carbons away on C3 and C5, resulting in 7.6 kJ/mol of steric strain ( ). Interconversion of axial and equatorial methylcyclohexane, represented in several formats. The equatorial conformation is more stable than the axial conformation by 7.6 kJ/mol. The 1,3-diaxial steric strain in substituted methylcyclohexane is already familiar—we saw it previously as the steric strain between methyl groups in gauche in . Gauche butane is less stable than anti butane by 3.8 kJ/mol (0.9 kcal/mol) because of steric interference between hydrogen atoms on the two methyl groups. Comparing a four-carbon fragment of axial methylcyclohexane with gauche butane shows that the steric interaction is the same in both ( ). Because axial methylcyclohexane has two such interactions, it has 2 × 3.8 = 7.6 kJ/mol of steric strain. Equatorial methylcyclohexane has no such interactions and is therefore more stable. The origin of 1,3-diaxial interactions in methylcyclohexane. The steric strain between an axial methyl group and an axial hydrogen atom three carbons away is identical to the steric strain in gauche butane. (To display clearly the diaxial interactions in methylcyclohexane, two of the equatorial hydrogens are not shown.) The exact amount of 1,3-diaxial steric strain in a given substituted cyclohexane depends on the nature and size of the substituent, as indicated in . Not surprisingly, the amount of steric strain increases through the series H 3 C− < CH 3 CH 2 − < (CH 3 ) 2 CH–<< (CH 3 ) 3 C−, paralleling the increasing size of the alkyl groups. Note that the values given in refer to 1,3-diaxial interactions of the substituent with a single hydrogen atom. These values must be doubled to arrive at the amount of strain in a monosubstituted cyclohexane. Steric Strain in Monosubstituted Cyclohexanes 1,3-Diaxial strain Y (kJ/mol) (kcal/mol) F 0.5 0.12 Cl, Br 1.0 0.25 OH 2.1 0.5 CH 3 3.8 0.9 CH 2 CH 3 4.0 0.95 CH(CH 3 ) 2 4.6 1.1 C(CH 3 ) 3 11.4 2.7 C 6 H 5 6.3 1.5 CO 2 H 2.9 0.7 CN 0.4 0.1 Problems

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Conformations of Disubstituted Cyclohexanes Monosubstituted cyclohexanes are always more stable with their substituent in an equatorial position, but the situation with disubstituted cyclohexanes is more complex because the steric effects of both substituents must be taken into account. All steric interactions for both possible chair conformations must be analyzed before deciding which conformation is favored. Let’s look at 1,2-dimethylcyclohexane as an example. There are two isomers, cis -1,2-dimethylcyclohexane and trans -1,2-dimethylcyclohexane, which must be considered separately. In the cis isomer, both methyl groups are on the same face of the ring and the compound can exist in either of the two chair conformations shown in . (It may be easier for you to see whether a compound is cis- or trans-disubstituted by first drawing the ring as a flat representation and then converting it to a chair conformation.) Conformations of cis -1,2-dimethylcyclohexane. The two chair conformations are equal in energy because each has one axial methyl group and one equatorial methyl group. Both chair conformations of cis -1,2-dimethylcyclohexane have one axial methyl group and one equatorial methyl group. The top conformation in has an axial methyl group at C2, which has 1,3-diaxial interactions with hydrogens on C4 and C6. The ring-flipped conformation has an axial methyl group at C1, which has 1,3-diaxial interactions with hydrogens on C3 and C5. In addition, both conformations have gauche butane interactions between the two methyl groups. The two conformations are equal in energy, with a total steric strain of 3 × 3.8 kJ/mol = 11.4 kJ/mol (2.7 kcal/mol). In trans -1,2-dimethylcyclohexane, the two methyl groups are on opposite sides of the ring and the compound can exist in either of the two chair conformations shown in . The situation here is quite different from that of the cis isomer. The top conformation in has both methyl groups equatorial with only a gauche butane interaction between them (3.8 kJ/mol) but no 1,3-diaxial interactions. The ring-flipped conformation, however, has both methyl groups axial. The axial methyl group at C1 interacts with axial hydrogens at C3 and C5, and the axial methyl group at C2 interacts with axial hydrogens at C4 and C6. These four 1,3-diaxial interactions produce a steric strain of 4 × 3.8 kJ/mol = 15.2 kJ/mol and make the diaxial conformation 15.2 − 3.8 = 11.4 kJ/mol less favorable than the diequatorial conformation. We therefore predict that trans -1,2-dimethylcyclohexane will exist almost exclusively in the diequatorial conformation. Conformations of trans -1,2-dimethylcyclohexane. The conformation with both methyl groups equatorial (top) is favored by 11.4 kJ/mol (2.7 kcal/mol) over the conformation with both methyl groups axial (bottom). The same kind of conformational analysis just carried out for cis - and trans -1,2-dimethylcyclohexane can be done for any substituted cyclohexane, such as cis -1- tert -butyl-4-chlorocyclohexane (see ). As you might imagine, though, the situation becomes more complex as the number of substituents increases. For instance, compare glucose with mannose, a carbohydrate present in seaweed. Which do you think is more strained? In glucose, all substituents on the six-membered ring are equatorial, while in mannose, one of the −OH groups is axial, making it more strained. A summary of the various axial and equatorial relationships among substituent groups in the different possible cis and trans substitution patterns for disubstituted cyclohexanes is given in . Axial and Equatorial Relationships in Cis- and Trans-Disubstituted Cyclohexanes Cis/trans substitution pattern Axial/equatorial relationships 1,2-Cis disubstituted a,e or e,a 1,2-Trans disubstituted a,a or e,e 1,3-Cis disubstituted a,a or e,e 1,3-Trans disubstituted a,e or e,a 1,4-Cis disubstituted a,e or e,a 1,4-Trans disubstituted a,a or e,e Drawing the Most Stable Conformation of a Substituted Cyclohexane Draw the more stable chair conformation of cis -1- tert -butyl-4-chlorocyclohexane. By how much is it favored? Strategy Draw the two possible chair conformations, and calculate the strain energy in each. Remember that equatorial substituents cause less strain than axial substituents. Solution First draw the two chair conformations of the molecule: In the conformation on the left, the tert -butyl group is equatorial and the chlorine is axial. In the conformation on the right, the tert -butyl group is axial and the chlorine is equatorial. These conformations aren’t of equal energy because an axial tert -butyl substituent and an axial chloro substituent produce different amounts of steric strain. shows that the 1,3-diaxial interaction between a hydrogen and a tert -butyl group costs 11.4 kJ/mol (2.7 kcal/mol), whereas the interaction between a hydrogen and a chlorine costs only 1.0 kJ/mol (0.25 kcal/mol). An axial tert -butyl group therefore produces (2 × 11.4 kJ/mol) − (2 × 1.0 kJ/mol) = 20.8 kJ/mol (4.9 kcal/mol) more steric strain than an axial chlorine, and the compound preferentially adopts the conformation with the chlorine axial and the tert -butyl equatorial. Problems

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Conformations of Polycyclic Molecules The final point we’ll consider about cycloalkane stereochemistry is to see what happens when two or more cycloalkane rings are fused together along a common bond to construct a polycyclic molecule —for example, decalin. Decalin consists of two cyclohexane rings joined to share two carbon atoms (the bridgehead carbons, C1 and C6) and a common bond. Decalin can exist in either of two isomeric forms, depending on whether the rings are trans fused or cis fused. In cis -decalin, the hydrogen atoms at the bridgehead carbons are on the same side of the rings; in trans -decalin, the bridgehead hydrogens are on opposite sides. shows how both compounds can be represented using chair cyclohexane conformations. Note that the two decalin isomers are not interconvertible by ring-flips or other rotations. They are cis–trans stereoisomers and have the same relationship to each other that cis - and trans -1,2-dimethylcyclohexane have. Representations of cis - and trans -decalin. Hydrogen atoms at the bridgehead carbons are on the same face of the rings in the cis isomer but on opposite faces in the trans isomer. Polycyclic compounds are common in nature, and many valuable substances have fused-ring structures. For example, steroids, such as testosterone, the primary sex hormone in males, have three six-membered rings and one five-membered ring fused together. Although steroids look complicated compared with cyclohexane or decalin, the same principles that apply to the conformational analysis of simple cyclohexane rings apply equally well (and often better) to steroids. Another common ring system is the norbornane, or bicyclo[2.2.1]heptane, structure. Like decalin, norbornane is a bicycloalkane, so called because two rings would have to be broken open to generate an acyclic structure. Its systematic name, bicyclo[2.2.1]heptane, reflects the fact that the molecule has seven carbons, is bicyclic, and has three “bridges” of 2, 2, and 1 carbon atoms connecting the two bridgehead carbons. Norbornane has a conformationally locked boat cyclohexane ring ( ) in which carbons 1 and 4 are joined by an additional CH 2 group. In drawing this structure, a break in the rear bond indicates that the vertical bond crosses in front of it. Making a molecular model is particularly helpful when trying to see the three-dimensionality of norbornane. Substituted norbornanes, such as camphor, are found widely in nature, and many have been important historically in developing organic structural theories. Problems Molecular Mechanics All the structural models in this book are computer-drawn. To make sure they accurately represent bond angles, bond lengths, torsional interactions, and steric interactions, the most stable geometry of each molecule has been calculated on a desktop computer using a commercially available molecular mechanics program based on work by Norman Allinger at the University of Georgia. The idea behind molecular mechanics is to begin with a rough geometry for a molecule and then calculate a total strain energy for that starting geometry, using mathematical equations that assign values to specific kinds of molecular interactions. Bond angles that are too large or too small cause angle strain; bond lengths that are too short or too long cause stretching or compressing strain; unfavorable eclipsing interactions around single bonds cause torsional strain; and nonbonded atoms that approach each other too closely cause steric, or van der Waals, strain. E total = E bond stretching + E angle strain + E torsional strain + E van der Waals Computer programs make it possible to accurately represent molecular geometry. (credit: “Molecular geometry” by Jane Whitney/Flickr, CC BY 4.0) After calculating a total strain energy for the starting geometry, the program automatically changes the geometry slightly in an attempt to lower strain—perhaps by lengthening a bond that is too short or decreasing an angle that is too large. Strain is recalculated for the new geometry, more changes are made, and more calculations are done. After dozens or hundreds of iterations, the calculation ultimately converges on a minimum energy that corresponds to the most favorable, least strained conformation of the molecule. Similar calculations have proven to be particularly useful in pharmaceutical research, where a complementary fit between a drug molecule and a receptor molecule in the body is often the key to designing new pharmaceutical agents ( ). The structure of Tamiflu (oseltamivir), an antiviral agent active against type A influenza , along with a molecular model of its minimum-energy conformation as calculated by molecular mechanics. alicyclic angle strain axial position (cyclohexane) boat cyclohexane Bridgehead atom chair conformation cis–trans isomers conformational analysis cycloalkanes 1,3-diaxial interaction Equatorial position (cyclohexane) polycyclic molecule ring-flip (cyclohexane) stereoisomer steric strain torsional strain twist-boat conformation Summary Cyclic molecules are so commonly encountered throughout organic and biological chemistry that it’s important to understand the consequences of their cyclic structures. Thus, we’ve taken a close look at cyclic structures in this chapter. Cycloalkanes are saturated cyclic hydrocarbons with the general formula C n H 2 n . In contrast to open-chain alkanes, where nearly free rotation occurs around C−C bonds, rotation is greatly reduced in cycloalkanes. Disubstituted cycloalkanes can therefore exist as cis–trans isomers . The cis isomer has both substituents on the same side of the ring; the trans isomer has substituents on opposite sides. Cis–trans isomers are just one kind of stereoisomer —compounds that have the same connections between atoms but different three-dimensional arrangements. Not all cycloalkanes are equally stable. Three kinds of strain contribute to the overall energy of a cycloalkane: (1) angle strain is the resistance of a bond angle to compression or expansion from the normal 109° tetrahedral value, (2) torsional strain is the energy cost of having neighboring C−H bonds eclipsed rather than staggered, and (3) steric strain is the repulsive interaction that arises when two groups attempt to occupy the same space. Cyclopropane (115 kJ/mol strain) and cyclobutane (110.4 kJ/mol strain) have both angle strain and torsional strain. Cyclopentane is free of angle strain but has a substantial torsional strain due to its large number of eclipsing interactions. Both cyclobutane and cyclopentane pucker slightly away from planarity to relieve torsional strain. Cyclohexane is strain-free because it adopts a puckered chair conformation , in which all bond angles are near 109° and all neighboring C–H bonds are staggered. Chair cyclohexane has two kinds of positions: axial and equatorial . Axial positions are oriented up and down, parallel to the ring axis, while equatorial positions lie in a belt around the equator of the ring. Each carbon atom has one axial and one equatorial position. Chair cyclohexanes are conformationally mobile and can undergo a ring-flip , which interconverts axial and equatorial positions. Substituents on the ring are more stable in the equatorial position because axial substituents cause 1,3-diaxial interactions . The amount of 1,3-diaxial steric strain caused by an axial substituent depends on its size. Additional Problems Visualizing Chemistry Cycloalkane Isomers Cycloalkane Conformation and Stability Cyclohexane Conformational Analysis General Problems

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Why This Chapter? Like the mountain whose image is reflected in a lake, many organic molecules also have mirror-image counterparts . (credit: modification of work “Crystal Lake sunrise reflection” by Sandy Horvath-Dori/Wikimedia Commons, CC BY 2.0) Understanding the causes and consequences of molecular handedness is crucial to understanding organic and biological chemistry. The subject can be a bit complex at first, but the material covered in this chapter nevertheless forms the basis for much of the remainder of the book. Are you right-handed or left-handed? You may not spend much time thinking about it, but handedness plays a surprisingly large role in your daily activities. Many musical instruments, such as oboes and clarinets, have a handedness to them; the last available softball glove always fits the wrong hand. The reason for these difficulties is that our hands aren’t identical; rather, they’re mirror images . When you hold a left hand up to a mirror, the image you see looks like a right hand. Try it. Handedness is also important in organic and biological chemistry, where it arises primarily as a consequence of the tetrahedral stereochemistry of sp 3 -hybridized carbon atoms. Many drugs and almost all the molecules in our bodies—amino acids, carbohydrates, nucleic acids, and many more—have a handedness. Furthermore, molecular handedness enables the precise interactions between enzymes and their substrates that are involved in the hundreds of thousands of chemical reactions on which life is based.

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Enantiomers and the Tetrahedral Carbon What causes molecular handedness? Look at generalized molecules of the type CH 3 X, CH 2 XY, and CHXYZ shown in . On the left are three molecules, and on the right are their images reflected in a mirror. The CH 3 X and CH 2 XY molecules are identical to their mirror images and thus are not handed. If you make a molecular model of each molecule and its mirror image, you find that you can superimpose one on the other so that all atoms coincide. The CHXYZ molecule, by contrast, is not identical to its mirror image. You can’t superimpose a model of this molecule on a model of its mirror image for the same reason that you can’t superimpose a left hand on a right hand: they simply aren’t the same. Tetrahedral carbon atoms and their mirror images. Molecules of the type CH 3 X and CH 2 XY are identical to their mirror images, but a molecule of the type CHXYZ is not. A CHXYZ molecule is related to its mirror image in the same way a right hand is related to a left hand. A molecule that is not identical to its mirror image is a kind of stereoisomer ( ) called an enantiomer (e- nan -tee-oh-mer, from the Greek enantio, meaning “opposite”). Enantiomers are related to each other as a right hand is related to a left hand and result whenever a tetrahedral carbon is bonded to four different substituents (one need not be H). For example, lactic acid (2-hydroxypropanoic acid) exists as a pair of enantiomers because there are four different groups (−H, −OH, −CH 3 , −CO 2 H) bonded to the central carbon atom. The enantiomers are called (+)-lactic acid and (−)-lactic acid. Both are found in sour milk, but only the (+) enantiomer occurs in muscle tissue. No matter how hard you try, you can’t superimpose a molecule of (+)-lactic acid on a molecule of (−)-lactic acid. If any two groups match up, say −H and −CO 2 H, the remaining two groups don’t match ( ). Attempts at superimposing the mirror-image forms of lactic acid. (a) When the −H and −OH substituents match up, the −CO 2 H and −CH 3 substituents don’t; (b) when −CO 2 H and −CH 3 match up, −H and −OH don’t. Regardless of how the molecules are oriented, they aren’t identical.

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The Reason for Handedness in Molecules: Chirality A molecule that is not identical to its mirror image is said to be chiral ( ky -ral, from the Greek cheir, meaning “hand”). You can’t take a chiral molecule and its enantiomer and place one on the other so that all atoms coincide. How can you predict whether a given molecule is or is not chiral? A molecule is not chiral if it has a plane of symmetry. A plane of symmetry is a plane that cuts through the middle of a molecule (or any object) in such a way that one half of the molecule or object is a mirror image of the other half. A coffee mug, for example, has a plane of symmetry. If you were to cut the coffee mug in half from top to bottom, one half would be a mirror image of the other half. A hand, however, does not have a plane of symmetry. One “half” of a hand is not a mirror image of the other half ( ). The meaning of symmetry plane. (a) An object like the coffee mug has a symmetry plane cutting through it so that right and left halves are mirror images. (b) An object like a hand has no symmetry plane; the right “half” of a hand is not a mirror image of the left half. A molecule that has a plane of symmetry in any conformation must be identical to its mirror image and must be nonchiral, or achiral . Thus, propanoic acid, CH 3 CH 2 CO 2 H, has a plane of symmetry when lined up as shown in and is achiral, while lactic acid, CH 3 CH(OH)CO 2 H, has no plane of symmetry in any conformation and is chiral. The achiral propanoic acid molecule versus the chiral lactic acid molecule. Propanoic acid has a plane of symmetry that makes one side of the molecule a mirror image of the other. Lactic acid has no such symmetry plane. The most common, although not the only, cause of chirality in organic molecules is the presence of a tetrahedral carbon atom bonded to four different groups—for example, the central carbon atom in lactic acid. Such carbons are referred to as chirality centers , although other terms such as stereocenter, asymmetric center, and stereogenic center have also been used. Note that chirality is a property of the entire molecule, whereas a chirality center is the cause of chirality. Detecting a chirality center in a complex molecule takes practice because it’s not always immediately apparent whether four different groups are bonded to a given carbon. The differences don’t necessarily appear right next to the chirality center. For example, 5-bromodecane is a chiral molecule because four different groups are bonded to C5, the chirality center (marked with an asterisk). A butyl substituent is similar to a pentyl substituent, but it isn’t identical. The difference isn’t apparent until looking four carbon atoms away from the chirality center, but there’s still a difference. As other possible examples, look at methylcyclohexane and 2-methylcyclohexanone. Methylcyclohexane is achiral because no carbon atom in the molecule is bonded to four different groups. You can immediately eliminate all −CH 2 − carbons and the −CH 3 carbon from consideration, but what about C1 on the ring? The C1 carbon atom is bonded to a −CH 3 group, to an −H atom, and to C2 and C6 of the ring. Carbons 2 and 6 are equivalent, however, as are carbons 3 and 5. Thus, the C6–C5–C4 “substituent” is equivalent to the C2–C3–C4 substituent, and methylcyclohexane is achiral. Another way of reaching the same conclusion is to realize that methylcyclohexane has a symmetry plane, which passes through the methyl group and through C1 and C4 of the ring. The situation is different for 2-methylcyclohexanone. 2-Methylcyclohexanone has no symmetry plane and is chiral because its C2 is bonded to four different groups: a −CH 3 group, an −H atom, a −COCH 2 − ring bond (C1), and a −CH 2 CH 2 − ring bond (C3). Several more examples of chiral molecules are shown below. Check for yourself that the labeled carbons are chirality centers. You might note that carbons in −CH 2 −, −CH 3 , C═O , C═C , and C≡C groups can’t be chirality centers. (Why not?) Drawing the Three-Dimensional Structure of a Chiral Molecule Draw the structure of a chiral alcohol. Strategy An alcohol is a compound that contains the −OH functional group. To make an alcohol chiral, we need to have four different groups bonded to a single carbon atom, say −H, −OH, −CH 3 , and −CH 2 CH 3 . Solution Problems

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Optical Activity The study of chirality originated in the early 19th century during investigations by the French physicist Jean-Baptiste Biot into the nature of plane-polarized light. A beam of ordinary light consists of electromagnetic waves that oscillate in an infinite number of planes at right angles to its direction of travel. When a beam of ordinary light passes through a device called a polarizer , however, only the light waves oscillating in a single plane pass through and the light is said to be plane-polarized. Light waves in all other planes are blocked out. Biot made the remarkable observation that when a beam of plane-polarized light passes through a solution of certain organic molecules, such as sugar or camphor, the plane of polarization is rotated through an angle, α . Not all organic substances exhibit this property, but those that do are said to be optically active . The angle of rotation can be measured with an instrument called a polarimeter, represented in . A solution of optically active organic molecules is placed in a sample tube, plane-polarized light is passed through the tube, and rotation of the polarization plane occurs. The light then goes through a second polarizer called the analyzer. By rotating the analyzer until the light passes through it , we can find the new plane of polarization and can tell to what extent rotation has occurred. Schematic representation of a polarimeter. Plane-polarized light passes through a solution of optically active molecules, which rotate the plane of polarization. In addition to determining the extent of rotation, we can also find the direction. From the vantage point of the observer looking directly at the analyzer, some optically active molecules rotate polarized light to the left (counterclockwise) and are said to be levorotatory , whereas others rotate polarized light to the right (clockwise) and are said to be dextrorotatory . By convention, rotation to the left is given a minus sign (−) and rotation to the right is given a plus sign (+). (−)-Morphine, for example, is levorotatory, and (+)-sucrose is dextrorotatory. The extent of rotation observed in a polarimetry experiment depends on the number of optically active molecules encountered by the light beam. This number, in turn, depends on sample concentration and sample pathlength. If the concentration of the sample is doubled, the observed rotation doubles. If the concentration is kept constant but the length of the sample tube is doubled, the observed rotation doubles. It also happens that the angle of rotation depends on the wavelength of the light used. To express optical rotations in a meaningful way so that comparisons can be made, we have to choose standard conditions. The specific rotation, [ α ] D , of a compound is defined as the observed rotation when light of 589.6 nanometer (nm; 1 nm = 10 −9 m) wavelength is used with a sample pathlength l of 1 decimeter (dm; 1 dm = 10 cm) and a sample concentration c of 1 g/cm 3 . (Light of 589.6 nm, the so-called sodium D line, is the yellow light emitted from common sodium street lamps.) [ α ] D = Observed rotation (degrees) Pathlength, l (dm) × Concentration, c ( g/cm 3 ) = α l × c When optical rotations are expressed in this standard way, the specific rotation, [ α ] D , is a physical constant characteristic of a given optically active compound. For example, (+)-lactic acid has [ α ] D = +3.82, and (−)-lactic acid has [ α ] D = −3.82. That is, the two enantiomers rotate plane-polarized light to exactly the same extent but in opposite directions. Note that the units of specific rotation are [(deg · cm 2 )/g] but the values are usually expressed without units. Some additional examples are listed in . Specific Rotations of Some Organic Molecules Compound [ α ] D Compound [ α ] D Penicillin V +233 Cholesterol −31.5 Sucrose +66.47 Morphine −132 Camphor +44.26 Cocaine −16 Chloroform 0 Acetic acid 0 Calculating an Optical Rotation A 1.20 g sample of cocaine, [ α ] D = −16, was dissolved in 7.50 mL of chloroform and placed in a sample tube having a pathlength of 5.00 cm. What was the observed rotation? Strategy Since [ α ] D = α l × c Then α = l × c × [ α ] D where [ α ] D = −16; l = 5.00 cm = 0.500 dm; c = 1.20 g/7.50 cm 3 = 0.160 g/cm 3 Solution α = (−16) (0.500) (0.160) = −1.3°. Problems

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Pasteur’s Discovery of Enantiomers Little was done to build on Biot’s discovery of optical activity until 1848, when Louis Pasteur began work on a study of crystalline tartaric acid salts derived from wine. On crystallizing a concentrated solution of sodium ammonium tartrate below 28 °C, Pasteur made the surprising observation that two distinct kinds of crystals were obtained. Furthermore, the two kinds of crystals were nonsuperimposable mirror images and were related in the same way that a right hand is related to a left hand. Working carefully with tweezers, Pasteur was able to separate the crystals into two piles, one of “right-handed” crystals and one of “left-handed” crystals, like those shown in . Although the original sample, a 50 : 50 mixture of right and left, was optically inactive, solutions of the crystals from each of the sorted piles were optically active and their specific rotations were equal in magnitude but opposite in sign. Drawings of sodium ammonium tartrate crystals taken from Pasteur’s original sketches. One of the crystals is dextrorotatory in solution, and the other is levorotatory. Pasteur was far ahead of his time. Although the structural theory of Kekulé had not yet been proposed, Pasteur explained his results by speaking of the molecules themselves, saying, “There is no doubt that [in the dextro tartaric acid] there exists an asymmetric arrangement having a nonsuperimposable image. It is no less certain that the atoms of the levo acid have precisely the inverse asymmetric arrangement.” Pasteur’s vision was extraordinary, for it was not until 25 years later that his ideas regarding asymmetric carbon atoms were confirmed. Today, we would describe Pasteur’s work by saying that he had discovered enantiomers. Enantiomers , also called optical isomers , have identical physical properties, such as melting point and boiling point, but differ in the direction in which their solutions rotate plane-polarized light.

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Sequence Rules for Specifying Configuration Structural drawings provide a visual representation of stereochemistry, but a written method for indicating the three-dimensional arrangement, or configuration , of substituents at a chirality center is also needed. The method used a set of sequence rules to rank the four groups attached to the chirality center and then looks at the handedness with which those groups are attached. Called the Cahn–Ingold–Prelog rules after the chemists who proposed them, the sequence rules are as follows: RULE 1 Look at the four atoms directly attached to the chirality center, and rank them according to atomic number. The atom with the highest atomic number has the highest ranking (first), and the atom with the lowest atomic number (usually hydrogen) has the lowest ranking (fourth). When different isotopes of the same element are compared, such as deuterium ( 2 H) and protium ( 1 H), the heavier isotope ranks higher than the lighter isotope. Thus, atoms commonly found in organic compounds have the following order. RULE 2 If a decision can’t be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the chirality center until the first difference is found. A −CH 2 CH 3 substituent and a −CH 3 substituent are equivalent by rule 1 because both have carbon as the first atom. By rule 2, however, ethyl ranks higher than methyl because ethyl has a carbon as its highest second atom, while methyl has only hydrogen as its second atom. Look at the following pairs of examples to see how the rule works: RULE 3 Multiple-bonded atoms are equivalent to the same number of single-bonded atoms. For example, an aldehyde substituent ( –CH═O ), which has a carbon atom doubly bonded to one oxygen, is equivalent to a substituent having a carbon atom singly bonded to two oxygens: As further examples, the following pairs are equivalent: Having ranked the four groups attached to a chiral carbon, we describe the stereochemical configuration around the carbon by orienting the molecule so that the group with the lowest ranking (4) points directly away from us. We then look at the three remaining substituents, which now appear to radiate toward us like the spokes on a steering wheel ( ). If a curved arrow drawn from the highest to second-highest to third-highest ranked substituent (1 → 2 → 3) is clockwise, we say that the chirality center has the R configuration (Latin rectus, meaning “right”). If an arrow from 1 → 2 → 3 is counterclockwise, the chirality center has the S configuration (Latin sinister , meaning “left”). To remember these assignments, think of a car’s steering wheel when making a R ight (clockwise) turn. Assigning R and S configurations to chirality centers. When the molecule is oriented so that the lowest-ranked group ( 4 ) is toward the rear, the remaining three groups radiate toward the viewer like the spokes of a steering wheel. If the direction of travel 1 → 2 → 3 is clockwise (right turn), the center has the R configuration. If the direction of travel 1 → 2 → 3 is counterclockwise (left turn), the center is S . Look at (−)-lactic acid in for an example of how to assign configuration. Sequence rule 1 says that −OH is ranked 1 and −H is ranked 4, but it doesn’t allow us to distinguish between −CH 3 and −CO 2 H because both groups have carbon as their first atom. Sequence rule 2, however, says that −CO 2 H ranks higher than −CH 3 because O (the highest second atom in −CO 2 H) outranks H (the highest second atom in −CH 3 ). Now, turn the molecule so that the fourth-ranked group (−H) is oriented toward the rear, away from the observer. Since a curved arrow from 1 (−OH) to 2 (−CO 2 H) to 3 (−CH 3 ) is clockwise (right turn of the steering wheel), (−)-lactic acid has the R configuration. Applying the same procedure to (+)-lactic acid leads to the opposite assignment. Assigning configuration to (a) ( R )-(−)-lactic acid and (b) ( S )-(+)-lactic acid. Further examples are provided by naturally occurring (−)-glyceraldehyde and (+)-alanine, which both have the S configuration as shown in . Note that the sign of optical rotation, (+) or (−), is not related to the R,S designation. ( S )-Glyceraldehyde happens to be levorotatory (−), and ( S )-alanine happens to be dextrorotatory (+). There is no simple correlation between R,S configuration and direction or magnitude of optical rotation. Assigning configuration to (a) (−)-glyceraldehyde. (b) (+)-alanine. Both happen to have the S configuration, although one is levorotatory and the other is dextrorotatory. One additional point needs to be mentioned—the matter of absolute configuration . How do we know that the assignments of R and S configuration are correct in an absolute sense , rather than a relative, sense? Since there is no correlation between the R,S configuration and the direction or magnitude of optical rotation, how do we know that the R configuration belongs to the levorotatory enantiomer of lactic acid? This difficult question was finally solved in 1951, when an X-ray diffraction method was found for determining the absolute spatial arrangement of atoms in a molecule. Based on those results, we can say with certainty that the R , S conventions are correct. Assigning Configuration to Chirality Centers Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign R or S configuration to each: Strategy It takes practice to be able to visualize and orient a chirality center in three dimensions. You might start by indicating where the observer must be located—180° opposite the lowest-ranked group. Then imagine yourself in the position of the observer, and redraw what you would see. Solution In (a) , you would be located in front of the page toward the top right of the molecule, and you would see group 2 to your left, group 3 to your right, and group 1 below you. This corresponds to an R configuration. In (b) , you would be located behind the page toward the top left of the molecule from your point of view, and you would see group 3 to your left, group 1 to your right, and group 2 below you. This also corresponds to an R configuration. Drawing the Three-Dimensional Structure of a Specific Enantiomer Draw a tetrahedral representation of ( R )-2-chlorobutane. Strategy Begin by ranking the four substituents bonded to the chirality center: (1) −Cl, (2) −CH 2 CH 3 , (3) −CH 3 , (4) −H. To draw a tetrahedral representation of the molecule, orient the lowest-ranked group (−H) away from you and imagine that the other three groups are coming out of the page toward you. Then, place the remaining three substituents such that the direction of travel 1 → 2 → 3 is clockwise (right turn), and tilt the molecule toward you to bring the rear hydrogen into view. Using molecular models is a real help in working problems of this sort. Solution Problems

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Diastereomers Molecules like lactic acid, alanine, and glyceraldehyde are relatively simple because each has only one chirality center and thus only two stereoisomers. The situation becomes more complex, however, with molecules that have more than one chirality center. As a general rule, a molecule with n chirality centers can have up to 2 n stereoisomers (although it may have fewer, as we’ll see below). Take the amino acid threonine (2-amino-3-hydroxybutanoic acid), for example. Since threonine has two chirality centers (C2 and C3), there are four possible stereoisomers, as shown in . Check for yourself that the R , S configurations of all stereoisomers are correct. The four stereoisomers of 2-amino-3-hydroxybutanoic acid. The four stereoisomers of 2-amino-3-hydroxybutanoic acid can be grouped into two pairs of enantiomers. The 2 R ,3 R stereoisomer is the mirror image of 2 S ,3 S , and the 2 R ,3 S stereoisomer is the mirror image of 2 S ,3 R . But what is the relationship between any two molecules that are not mirror images? What, for instance, is the relationship between the 2 R ,3 R isomer and the 2 R ,3 S isomer? They are stereoisomers, yet they aren’t enantiomers. To describe such a relationship, we need a new term— diastereomer . Diastereomers (dia- stair -e-oh-mers) are stereoisomers that are not mirror images. Since we used the right-hand/left-hand analogy to describe the relationship between two enantiomers, we might extend the analogy by saying that the relationship between diastereomers is like that of hands from different people. Your hand and your friend’s hand look similar , but they aren’t identical and they aren’t mirror images. The same is true of diastereomers: they’re similar, but they aren’t identical and they aren’t mirror images. Note carefully the difference between enantiomers and diastereomers: enantiomers have opposite configurations at all chirality centers, whereas diastereomers have opposite configurations at some (one or more) chirality centers but the same configuration at others. A full description of the four stereoisomers of threonine is given in . Of the four, only the 2 S ,3 R isomer, [ α ] D = −28.3, occurs naturally in plants and animals and is an essential nutrient for humans. This result is typical: most biological molecules are chiral, and usually only one stereoisomer is found in nature. Relationships among the Four Stereoisomers of Threonine Stereoisomer Enantiomer Diastereomer 2 R ,3 R 2 S ,3 S 2 R ,3 S and 2 S ,3 R 2 S ,3 S 2 R ,3 R 2 R ,3 S and 2 S ,3 R 2 R ,3 S 2 S ,3 R 2 R ,3 R and 2 S ,3 S 2 S ,3 R 2 R ,3 S 2 R ,3 R and 2 S ,3 S In the special case where two diastereomers differ at only one chirality center but are the same at all others, we say that the compounds are epimers . Cholestanol and coprostanol, for instance, are both found in human feces, and both have nine chirality centers. Eight of the nine are identical, but the one at C5 is different. Thus, cholestanol and coprostanol are epimeric at C5. Problems

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Meso Compounds Let’s look at another example ( ) of a compound with more than one chirality center: the tartaric acid used by Pasteur. The four stereoisomers can be drawn as follows: The 2 R ,3 R and 2 S ,3 S structures are nonsuperimposable mirror images and therefore represent a pair of enantiomers. A close look at the 2 R ,3 S and 2 S ,3 R structures, however, shows that they are superimposable, and thus identical, as can be seen by rotating one structure 180°. The 2 R ,3 S and 2 S ,3 R structures are identical because the molecule has a plane of symmetry and is therefore achiral. The symmetry plane cuts through the C2–C3 bond, making one half of the molecule a mirror image of the other half ( ). Because of the plane of symmetry, the molecule is achiral, despite the fact that it has two chirality centers. Such compounds, which are achiral, yet contain chirality centers, are called meso compounds (me-zo) . Thus, tartaric acid exists in three stereoisomeric forms: two enantiomers and one meso form. A symmetry plane through the C2–C3 bond of meso -tartaric acid makes the molecule achiral. Some physical properties of the three stereoisomers are listed in . The (+)- and (−)-tartaric acids have identical melting points, solubilities, and densities, but they differ in the sign of their rotation of plane-polarized light. The meso isomer, by contrast, is diastereomeric with the (+) and (−) forms. It has no mirror-image relationship to (+)- and (−)-tartaric acids, is a different compound altogether, and thus has different physical properties. Some Properties of the Stereoisomers of Tartaric Acid Stereoisomer Melting point ( ° C) [ α ] D Density (g/cm 3 ) Solubility at 20 ° C (g/100 mL H 2 O) (+) 168–170 +12 1.7598 139.0 (−) 168–170 −12 1.7598 139.0 Meso 146–148 0 1.6660 125.0 Some Properties of the Stereoisomers of Tartaric Acid Distinguishing Chiral Compounds from Meso Compounds Does cis -1,2-dimethylcyclobutane have any chirality centers? Is it chiral? Strategy To see whether a chirality center is present, look for a carbon atom bonded to four different groups. To see whether the molecule is chiral, look for the presence or absence of a symmetry plane. Not all molecules with chirality centers are chiral overall—meso compounds are an exception. Solution A look at the structure of cis -1,2-dimethylcyclobutane shows that both methyl-bearing ring carbons (C1 and C2) are chirality centers. Overall, though, the compound is achiral because there is a symmetry plane bisecting the ring between C1 and C2. Thus, the molecule is a meso compound. Problems

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Racemic Mixtures and the Resolution of Enantiomers To end this discussion of stereoisomerism, let’s return for a last look at Pasteur’s pioneering work, described in . Pasteur took an optically inactive tartaric acid salt and found that he could crystallize from it two optically active forms having what we would now call 2 R ,3 R and 2 S ,3 S configurations. But what was the optically inactive form he started with? It couldn’t have been meso -tartaric acid, because meso -tartaric acid is a different chemical compound and can’t interconvert with the two chiral enantiomers without breaking and re-forming chemical bonds. The answer is that Pasteur started with a 50 : 50 mixture of the two chiral tartaric acid enantiomers. Such a mixture is called a racemate ( rass -uh-mate), or racemic mixture, and is denoted by either the symbol (±) or the prefix d,l to indicate an equal mixture of dextrorotatory and levorotatory forms. Racemates show no optical rotation because the (+) rotation from one enantiomer exactly cancels the (−) rotation from the other. Through good luck, Pasteur was able to separate, or resolve, racemic tartaric acid into its (+) and (−) enantiomers. Unfortunately, the fractional crystallization technique he used doesn’t work for most racemates, so other methods are needed. The most common method for resolving the racemate of a chiral carboxylic acid (RCO 2 H) is to carry out an acid-base reaction between the acid and an amine base (RNH 2 ) to yield an ammonium salt: To understand how this method of resolution works, let’s see what happens when a racemic mixture of chiral acids, such as (+)- and (−)-lactic acids, reacts with an achiral amine base, such as methylamine, CH 3 NH 2 . The situation is analogous to what happens when left and right hands (chiral) pick up a ball (achiral). Both left and right hands pick up the ball equally well, and the products—ball in right hand versus ball in left hand—are mirror images. In the same way, both (+)- and (−)-lactic acid react with methylamine equally well, and the product is a racemic mixture of the two enantiomers methylammonium (+)-lactate and methylammonium (−)-lactate ( ). Reaction of racemic lactic acid with achiral methylamine gives a racemic mixture of ammonium salts. Now let’s see what happens when the racemic mixture of (+)- and (−)-lactic acids reacts with a single enantiomer of a chiral amine base, such as ( R )-1-phenylethylamine. The situation is analogous to what happens when left and right hands (chiral) put on a right-handed glove ( also chiral ). Left and right hands don’t put on the right-handed glove in the same way, so the products—right hand in right glove versus left hand in right glove—are not mirror images; they’re similar but different. In the same way, (+)- and (−)-lactic acids react with ( R )-1-phenylethylamine to give two different products ( ). ( R )-Lactic acid reacts with ( R )-1-phenylethylamine to give the R,R salt, and ( S )-lactic acid reacts with the R amine to give the S,R salt. The two salts are diastereomers, not enantiomers. They have different chemical and physical properties, and it may therefore be possible to separate them by crystallization or some other means. Once separated, acidification of the two diastereomeric salts with a strong acid makes it possible to isolate the two pure enantiomers of lactic acid and to recover the chiral amine for reuse. Reaction of racemic lactic acid with ( R )-1-phenylethylamine yields a mixture of diastereomeric ammonium salts, which have different properties and can be separated. Predicting the Chirality of a Reaction Product We’ll see in Section 21.3 that carboxylic acids (RCO 2 H) react with alcohols (R′OH) to form esters (RCO 2 R′). Suppose that (±)-lactic acid reacts with CH 3 OH to form the ester, methyl lactate. What stereochemistry would you expect the product(s) to have? What is the relationship of the products? Solution Reaction of a racemic acid with an achiral alcohol such as methanol yields a racemic mixture of mirror-image (enantiomeric) products. Problems

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A Review of Isomerism As noted on several previous occasions, isomers are compounds with the same chemical formula but different structures. We’ve seen several kinds of isomers in the past few chapters, and it’s a good idea at this point to see how they relate to one another ( ). A summary of the different kinds of isomers. There are two fundamental types of isomers, both of which we’ve now encountered: constitutional isomers and stereoisomers. Constitutional isomers ( ) are compounds whose atoms are connected differently. Among the kinds of constitutional isomers we’ve seen are skeletal, functional, and positional isomers. Stereoisomers ( ) are compounds whose atoms are connected in the same order but with a different spatial arrangement. Among the kinds of stereoisomers we’ve seen are enantiomers, diastereomers, and cis–trans isomers of cycloalkanes. Actually, cis–trans isomers are just a subclass of diastereomers because they are non–mirror-image stereoisomers: Problems

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Chirality at Nitrogen, Phosphorus, and Sulfur As noted previously, the most common cause of chirality in a molecule is the presence of four different substituents bonded to a tetrahedral atom. Although that atom is usually carbon, it doesn’t necessarily have to be. Nitrogen, phosphorus, and sulfur atoms are all commonly encountered in organic molecules, and can all be chirality centers. We know, for instance, that trivalent nitrogen is tetrahedral, with its lone pair of electrons acting as the fourth “substituent” ( ). Is trivalent nitrogen chiral? Does a compound such as ethylmethylamine exist as a pair of enantiomers? The answer is both yes and no. Yes in principle, but no in practice. It turns out that most trivalent nitrogen compounds undergo a rapid umbrella-like inversion that interconverts enantiomers, so we can’t isolate individual enantiomers except in special cases. A similar situation occurs in trivalent phosphorus compounds, called phosphines, but the inversion at phosphorus is substantially slower than inversion at nitrogen, so stable chiral phosphines can be isolated. ( R )- and ( S )-methylpropylphenylphosphine, for instance, are configurationally stable for several hours at 100 °C. We’ll see the importance of phosphine chirality in Section 26.7 in connection with the synthesis of chiral amino acids. Divalent sulfur compounds are achiral, but trivalent sulfur compounds called sulfonium salts (R 3 S + ) can be chiral. Like phosphines, sulfonium salts undergo relatively slow inversion, so chiral sulfonium salts are configurationally stable and can be isolated. Perhaps the best known example is the coenzyme S -adenosylmethionine, the so-called biological methyl donor, which is involved in many metabolic pathways as a source of CH 3 groups. (The “ S ” in the name S -adenosylmethionine stands for sulfur and means that the adenosyl group is attached to the sulfur atom of the amino acid methionine.) The molecule has S stereochemistry at sulfur and is configurationally stable for several days at room temperature. Its R enantiomer is also known but is not biologically active.

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Prochirality Closely related to the concept of chirality, and particularly important in biological chemistry, is the notion of prochirality. A molecule is said to be prochiral if it can be converted from achiral to chiral in a single chemical step. For instance, an unsymmetrical ketone like 2-butanone is prochiral because it can be converted to the chiral alcohol 2-butanol by the addition of hydrogen, as we’ll see in Section 17.4 . Which enantiomer of 2-butanol is produced depends on which face of the planar carbonyl group undergoes reaction. To distinguish between the possibilities, we use the stereochemical descriptors Re and Si. Rank the three groups attached to the trigonal, sp 2 -hybridized carbon, and imagine curved arrows from the highest to second-highest to third-highest ranked substituents. The face on which the arrows curve clockwise is designated the Re face (similar to R ), and the face on which the arrows curve counterclockwise is designated the Si face (similar to S ). In this example, addition of hydrogen from the Re face gives ( S )-2-butane, and addition from the Si face gives ( R )-2-butane. In addition to compounds with planar, sp 2 -hybridized atoms, compounds with tetrahedral, sp 3 -hybridized atoms can also be prochiral. An sp 3 -hybridized atom is said to be a prochirality center if, by changing one of its attached groups, it becomes a chirality center. The −CH 2 OH carbon atom of ethanol, for instance, is a prochirality center because changing one of its attached −H atoms converts it into a chirality center. To distinguish between the two identical atoms (or groups of atoms) on a prochirality center, we imagine a change that will raise the ranking of one atom over the other without affecting its rank with respect to other attached groups. On the −CH 2 OH carbon of ethanol, for instance, we might imagine replacing one of the 1 H atoms (protium) by 2 H (deuterium). The newly introduced 2 H atom ranks higher than the remaining 1 H atom, but it remains lower than other groups attached to the carbon. Of the two identical atoms in the original compound, the atom whose replacement leads to an R chirality center is said to be pro- R and the atom whose replacement leads to an S chirality center is pro- S . A large number of biological reactions involve prochiral compounds. One of the steps in the citric acid cycle by which food is metabolized, for instance, is the addition of H 2 O to fumarate to give malate. Addition of −OH occurs on the Si face of a fumarate carbon and gives ( S )-malate as product. As another example, studies with deuterium-labeled substrates have shown that the reaction of ethanol with the coenzyme nicotinamide adenine dinucleotide (NAD + ), catalyzed by yeast alcohol dehydrogenase, occurs with exclusive removal of the pro- R hydrogen from ethanol and with addition only to the Re face of NAD + . Determining the stereochemistry of reactions at prochirality centers is a powerful method for studying detailed mechanisms in biochemical reactions. As just one example, the conversion of citrate to cis -aconitate in the citric acid cycle has been shown to occur with loss of a pro- R hydrogen, implying that the OH and H groups leave from opposite sides of the molecule. Problems

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