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If our hypothetical single crystal of H-D molecules were allowed to melt, the restraints between the molecules would diminish and rapid molecular tumbling would begin. Tumbling molecules present all possible values of the angle $\theta$ between the internuclear lines and the magnetic field axis. Integration of (3cos2$\theta$-1) over all possible values of $\theta$ shows that the time-average direct dipole-dipole interaction between the bonded H-D nuclei is zero. Hence, we would expect that there would then be no observable spin-spin coupling in nuclear resonance spectra of liquids or gases. Nonetheless, small couplings persist even when tumbling is rapid, although they are usually on the order of about 10-2 gauss, roughly 1/1000 of the values expected for direct dipole-dipole interactions. The residual couplings are not merely due to partial averaging of the dipole-dipole interaction through tumbling, since they are temperature independent except in special cases. It has been shown that the residual couplings are the result of magnetic interactions transmitted between nuclei by the bonding electrons in such a way as not to be averaged to zero by tumbling.
Except that the lines are much narrower and very much more closely spaced, the appearance of the NMR spectrum given by tumbling H-D molecules is qualitatively the same as predicted for the single crystal, i.e., a doublet deuteron resonance and a triplet proton resonance. The argument for expecting this pattern is unchanged from that given earlier for the crystal except that now the influence of the magnetic orientation of one nucleus on the precession frequency of the other nucleus is considered to be transmitted by the bonding electrons instead of by direct dipole-dipole interaction.
Gutowsky, McCall, and Slichter1 discuss the relation between the magnitudes of spin-spin interactions among nuclei as a function of various atomic and molecular parameters. Customarily, spin-spin coupling constants are found to decrease monotonically with the number of chemical bonds between the nuclei involved. Several spectacular failures of this generalization have been observed involving fluorine-fluorine and hydrogen-fluorine interactions. For example, it has been found that the coupling between hydrogen and fluorine atoms connected to the 1 and 3 carbons of certain cyclobutene derivatives, and thus four chemical bonds apart, are much larger than the corresponding interactions between the same groups attached to the 1 and 4 carbons, which are only three bonds apart.2 Similarly, the hydrogens on the 1 and 3 carbons of bromoallene (four bonds apart) are coupled more strongly even than the CH2 and CH3 hydrogens of an ethyl group (three bonds apart).
1 H. S. Gutowsky, D. W. McCall, and C. P. Slichter, I. Chem. Phys., 21, 279 (1953).
2 C. M. Sharts and J. D. Roberts, I. Am. Chem. Soc., 79, 1008 (1957).
3.04: Spin-Spin Splitting in the Ethyl Group
The typical three-four line resonance pattern of the ethyl group arises because of spin-spin coupling between the methyl and methylene protons. It turns out that the coupling constant J, which represents the line spacings, is very nearly constant over a wide range of ethyl derivatives and amounts to about 1.6 milligauss or 7 cps.
Reference to Fig. 3-5 shows that the two methylene protons may have any one of four possible combinations of their magnetic quantum numbers. Thus, both magnetic quantum numbers may be the same with either a + or - sign for the total or they may be different two possible ways) and cancel each other's magnetic effect. The ethyl groups from molecule to molecule can then be classified into those in which the total spin of the methylene protons adds to +1, those in which they cancel each other, and those in which they add to -1. The methyl protons in ethyl groups whose methylene protons have a total spin of +1 will come into resonance when the magnetic field is increasing sooner than the others. When the methylene protons have a net spin of zero, they will have no effect on the resonance line position of the methyl, while if the net spin is -1, the line will come late by the same amount as the +1 combination came early. Consequently, there should be a total of three resonance lines for the methyl group because of the adjacent methylene protons. Since there are a total of four equally probable combinations of the methylene magnetic quantum numbers, one way to give +1, two ways to give zero, and one way to give -1, it will be expected that the three resonance lines will have signal-strength ratios of 1:2:1. Similar reasoning applied to the resonance absorption of the methylene protons in the presence of the methyl protons leads to prediction of a four-line spectrum with the signal strengths in the ratio 1:3:3:1.In the above analysis, the spins of a group of equivalent protons were taken to be independent (each with two possible orientations in the external field) and distributed statistically among the possible states as shown in Fig. 3-5. This procedure gives satisfactory predictions for many simple spin-spin splitting problems but fails in others. Chemically equivalent protons in a group like a methyl are actually coupled to one another to give, in effect, a larger magnet which in turn is coupled to the external field. Consequently, such protons are not to be expected to behave independently of one another in all situations. This point will be discussed further in Sec. 3-7. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/03%3A_Spin-Spin_Splitting/3.03%3A_Spin-Spin_Splitting_in_Liquids.txt |
a. Styrene Oxide
In general, we shall anticipate that the resonance of a given nucleus will be split into (n + 1) lines by n equivalently placed magnetic nuclei with I = 1/2. The situation is more complex if more than one J value is involved. Styrene oxide offers a nice example of three nonequivalent protons each coupled to the others with different J's. The spectrum of styrene oxide is shown in Fig. 3-6; the ring protons are not coupled to the side-chain protons, so only the latter will be considered. None of the side-chain protons are in equivalent chemical locations since one ($\alpha$) is adjacent to the phenyl ring and the other two are either cis($\beta$) or trans($\beta$') to the phenyl ring. In the absence of spin-spin coupling, we would then expect a simple three-line proton spectrum as shown at the top of Fig. 3-6. The a proton will be coupled unequally to the $\beta$ protons, and if we assume J$\alpha\beta$ (trans) to be larger than J$\alpha\beta$'(cis), then four lines of roughly equal height will be observed for the a proton. The $\beta$ protons are coupled to each other and in the first-order treatment would be expected to split each other's resonance so as to have four lines of equal strength. As will be shown later, the fact that the chemical shift difference between the $\beta$ protons ($\delta\beta\beta$,H) is not a great deal larger than the coupling constant (J$\beta\beta$,) means that the lines are not all of equal height and the spacings between the centers of the groups are not $\delta\beta\beta$,H but $\sqrt{J\beta\beta^2 + (\delta\beta\beta,H)^2}$. This, however, is a second-order effect which does not change the basic argument. Each line for the P protons will then be split by the coupling between the a and (3 protons with different J values, so that the $\beta$ protons will give eight lines in all, as is clearly evident from the observed spectrum.
b. Aliphatic Alcohols
Nuclear spin-spin coupling and chemical-shift data are extremely useful for structural investigations. As an example, we way consider high resolution NMR spectra of a number of simple aliphatic alcohols (Fig. 3-7). The signal strengths, line positions, and degrees of spin-spin splitting permit each alcohol to be identified unambiguously. Thus, isopropyl alcohol shows an 0-H proton resonance, an $\alpha$-proton resonance split into seven, and the resonance of six $\beta$ protons split into a doublet. In the examples shown, the $\alpha$ hydrogens do not appear to be coupled to the hydroxyl protons, although there are only three bonds (H-C-O-H) separating them. The lack of this expected spin-spin splitting is a consequence of rapid intermolecular chemical exchange of the hydroxyl protons, as will be described later.
c. Methyl Photosantonate
The nuclear resonance spectrum of methyl photosantonate has been shown by van Tamelen3 to be decisive in distinguishing among three rather closely related proposed structures for photosantonic acid. An extraordinarily clear spectrum of the methyl ester at 60 Mc is shown in Fig. 3-8. The group of lines centered on 55 cps is in the vinyl hydrogen region and provides a telling argument in favor of structure I over structures II and III which have no vinyl hydrogens. Furthermore, the vinyl hydrogen resonance is split into three principal lines as would suggest the grouping -CH,-CH=C<. This assignment is corroborated by the two principal lines centered on 216 cps which are in about the right place for a methylene group attached on one side to a carbomethoxy and on the other to a double bond. The spacing of these lines is identical with that observed for the vinyl hydrogen.
The slight doubling of the anticipated principal three-two pattern of the vinyl and methylene protons is invariant with the applied magnetic field and hence is the result of spin-spin coupling. Apparently, the side-chain -CH2- and -CH= protons are coupled to the CH-O- proton with just sufficiently different J values to lead to only barely perceptible fine structure for the components of the 148-cps doublet.
3 E. E. van Tamelen, S. H. Levin, G. Brenner, J. Wolinsky, and P. Aldrich, J. Am. Chem. Soc., 80, 501 (1958), and private communication. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/03%3A_Spin-Spin_Splitting/3.05%3A_More_Complex_Spin-Spin_Splittings.txt |
In saturated noncyclic compounds, spin-spin coupling constants for protons located on adjacent carbons (three bonds apart) are more or less constant from molecule to molecule at about 7 cps. Generally, J becomes immeasurably small with four bonds intervening, except when one or two of these is a double bond. Thus, as mentioned earlier, rather large couplings ( ~10 cps) are observed across the double bonds of an allenic system, and we note now that J is much smaller (1 to 2 cps) across one single and one double C-C bond, as in isobutylene.
An interesting variation of J with bond type has been noted by Shoolery4 for couplings through the double bonds of a quinone and a derivative of the corresponding hydroquinone with a fully aromatic ring. Thus, the benzenoid thymol system shows no resolvable coupling between the ring protons and those on the adjacent alkyl groups whereas, with thymoquinone, one of the ring-proton resonances is split into a 1:3:3:1 quartet through spin-spin coupling involving the neighboring CH3 group, while the other ring-proton resonance is split into a doublet by the $\alpha$ proton of the isopropyl group.
Proton spin-spin couplings are powerfully influenced by molecular geometry. Trans conformations of protons in saturated compounds appear to lead to substantially larger couplings than corresponding gauche or eclipsed arrangements. Couplings between trans protons attached to double bonds are similarly larger than between cis protons.
In some compounds such as cyclobutene, spin-spin splitting is unexpectedly small or completely absent. Consequently, any structure interpretation which suggests that groups of nonequivalent protons are not contiguous because of failure to observe spin-spin couplings may be seriously in error.
4 J. N. Shoolery, Varian Associates Tech. Bull., 2, 8 (1957).
3.07: Coupling between Equivalent and Nearly Equivalent Protons
In connection with spin-spin coupling in substances containing ethyl or similar groups, it is natural to wonder why there is no apparent splitting of the resonance of a proton belonging to an equivalent group of protons by the other members of the group, which are much closer than a group of protons attached to an adjacent carbon. Thus, spin-spin splittings might be expected as the result of couplings between the individual protons at the methylene position of an ethyl group, as well as between the methylene and methyl protons. In practice, except for some special situations to be explained later, magnetic nuclei in equivalent chemical environments do not show spin-spin coupling. This is because equivalent protons do not absorb rf energy independently of one another. In spectroscopic terms, we can say that any transition which would lead to a splitting of the resonance of one proton of an equivalent group by the magnetic moment of another proton in the group would be "forbidden," since it would be a "singlet-triplet" transition. "Allowed" transitions involve absorption of energy by protons as a group, and no splittings result therefrom. With nuclei which are not completely equivalent, the type of transition forbidden for equivalent nuclei is allowed but reduced in probability-the closer the degree of equivalence, the lower the transition probability.5
It is instructive to consider the effect of coupling between two like nuclei as a function of the chemical shift between their resonances. When the chemical shift is large compared with J, the simple first-order treatment holds as described earlier and each of the resonances is split by the interaction between the nuclei into two lines of equal intensity, as shown in Fig. 3-9. The distance between the centers of the doublets is the chemical shift $\delta$H. Now, as the chemical shift decreases at constant J, a more complicated relation obtains between $\delta$H, J, the signal strengths, and the distances between the resonance lines. The splitting of the multiplets is still J, but the distance between the multiplet centers is no longer $\delta$H but $\sqrt{J^2 + \delta^2 H^2}$. The line intensities no longer remain equal-the center peaks get larger and the outer peaks smaller, so
$\frac{Intensity of inner lines}{Intensity of outer lines}$ = $(\frac{1+Q}{1-Q})^2$ where Q = $\frac{J}{ \delta H + \sqrt{\delta^2 H^2 + J^2}}$
As $\delta$H continues to decrease, the outer lines get weaker and weaker, and, finally, when $\delta$H is smaller than I, the spectrum may appear to a casual observer as a closely spaced doublet. At very small $\delta$H values, the center lines coalesce and only one intense line is observed. The limiting case with $\delta$H = 0 corresponds to nuclei in equivalent chemical locations and no splitting is observed.
It should always be kept in mind that spectra involving spin-spin splittings will be distorted from the predictions of the simple first-order treatment and extra lines may even appear whenever the chemical shifts are small. Thus, the second-order spin-spin splitting displayed by ethanol at 40 Mc (Fig. 3-1) is much accentuated at lower frequencies. An excellent example of the striking effect of extreme changes of oscillator frequency on the appearance of NMR spectra has been provided by Muetterties and Phillips6 for the 19F resonances of chlorine trifluoride.
The chemical shifts between nuclei of different kinds, such as between hydrogen and fluorine, hydrogen and nitrogen, etc., are always very large, so that the above-mentioned complications do not arise except at very low field strengths.
5 Cf. H. S. Gutowsky, Ann. N.Y. Acad. Sci., 70, 786 (1958); H. J. Bernstein, I. A. Pople, and W. G. Schneider, Can. J. Chern., 35, 65 (1957).
6 E. L. Muetterties and W. D. Phillips, J. Am. Chem. Soc., 79, 322 (1957). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/03%3A_Spin-Spin_Splitting/3.06%3A_Magnitudes_of_Coupling_Constants.txt |
Difficulties in interpretation of spin-spin couplings are very often encountered in more or less rigid complex molecules such as cyclic compounds. There are several reasons for expecting difficulties. First, as shown for styrene oxide, the protons of adjacent methylenes in a rigid system will generally have at least two spin-spin coupling constants which correspond to interactions between the various combinations of cis and trans protons. Second, J for protons on adjacent carbons shows a strong angular dependence and, as a result, the couplings in more or less rigid systems vary between zero and the values normally encountered for open-chain substances. Furthermore, it should be remembered that whenever J is large compared with the chemical shift, no splitting or chemical shift may be apparent in the spectrum. An example is cyclopentanone, which shows only a single proton resonance line.
A further and serious complication is introduced in rigid systems when chemically equivalent nuclei which are close enough to each other to be strongly coupled are also strongly coupled to other nuclei.7 An excellent illustration is afforded by 1,l-difluoroethylene. Here, the first-order treatment predicts four resonance lines of equal intensity because each proton should be coupled to the two fluorines with different coupling constants, one fluorine being cis to a given proton and the other being trans. However, the observed spectrum shows ten lines (Fig. 3-10), and in the detailed mathematical treatment of spin-spin coupling for this molecule, both the fluorine-fluorine and proton-proton coupling make important contributions to the splitting. Generally, one will expect complications from couplings between formally equivalent nuclei wherever such nuclei are so located as to have different intramolecular coupling constants to other magnetic nuclei.
7 H. M. McConnell, A. D. McLean, and C. A. Reilly, J. Chem. Phys., 23, 1152 (1955).
3.09: Spin-Spin Splitting and Molecular Asymmetry
An interesting and unusual splitting is sometimes observed when a -CH2-Y group is attached to another carbon atom, carrying three different groups.8 An excellent example is methyl 2,3-dibromo-2- methylpropionate, the proton NMR spectrum of which is shown in Fig.3-11. Besides the large three-proton resonance peaks of the ester and the $\alpha$ methyls there are four small peaks which fit the typical pattern of two nonequivalent protons, splitting each other with a coupling constant of about the same magnitude as the chemical shift. All chemical experience indicates that the rate of rotation of the groups with respect to one another around the 2,3-carbon-carbon single bond in this molecule should be reasonably rapid. However, there is good evidence that substituted ethanes of this type are more stable in the staggered conformations, as shown in Fig. 3-12.
In the present case, inspection of the various staggered conformations shows that the $\beta$-methylene hydrogens are not equivalently located in any one of them with respect to the groups attached to the carbon. Therefore, there must be at least some degree of nonequivalence of the $\beta$ protons, and spin-spin splitting is consequently possible. At present, it is not known to what extent the relative residence times of the molecules in particular conformations are important in determining the degree of nonequivalence. If all the conformations had equal residence times and rapid rotation occurred, then each hydrogen would experience nearly the same average magnetic field. On the other hand, if the residence times were quite different, then the average magnetic field would be much like that of the most favored conformation and a larger degree of nonequivalence would be expected.
Ethanes with -CH2-Y attached to -CX2Y are not expected to show nonequivalent protons if rotation is rapid around the C-C bond. Thus, inspection of the possible staggered conformations for 1,2-dibromo-2- methylpropane (see Fig. 3-13) shows that the conformation with the bromines trans to one another has a plane of symmetry and the methylene hydrogens are thereby automatically equivalent. The other two conformations have nonequivalent protons, but since they are mirror images of each other, the protons will spend equal time in each.
Therefore, if all the forms are interconverted rapidly, the two methylene protons will reside in identical average environments (cf. spectrum in Fig. 3-1 1). As will be seen later, slow rotation about the C-C bond in substances of this type will lead to more complex spectra because then the magnetic environments of the individual methylene protons are not identical in all the conformations.
The type of spin-spin splitting observed for methyl 2,3-bromo-2- methylpropionate appears to offer a new approach to detecting asymmetric centers in organic molecules without recourse to resolution into optical antipodes. Strictly, the method provides a test for a carbon with three different groups attached rather than a test for molecules which might be resolvable into optical antipodes. Thus, 2-nitro-2-methyl-1,3- propanediol is not capable of being resolved into optical antipodes but shows nonequivalent methylene protons because each methylene is attached to an asymmetric group (see Fig. 3-14). A comprehensive theoretical treatment of various types of asymmetric molecules has been made by Pople.9
8 P. M. Nair and J. D. Roberts, J. Am. Chem. Soc., 79, 4565 (1957).
9 J. A. Pople, Molecular Physics, 1, 3 (1958). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/03%3A_Spin-Spin_Splitting/3.08%3A_Spin-Spin_Couplings_in_Rigid_Systems.txt |
One of the most important applications of NMR spectroscopy is to the determination of kinetics of very fast reactions. Obviously, reaction kinetics could be followed by measuring rates of increase or decrease of resonance signals attributable to products or reactants. However, there is a very simple and much less conventional NMR method for determination of reaction rates which covers a range of rate constants and reaction types exceedingly difficult to measure in any other way.
As an introduction, we consider the NMR spectra of solutions of acetic acid and water. The spectra of the two separate substances are shown in Fig. 4-1, the carboxyl hydrogen of the acetic acid appearing at a very low field compared with water and the methyl group at a considerably higher field. When the substances are mixed, the spectrum might be expected to show three lines-one for the methyl group and two others corresponding to the two varieties of hydroxyl hydrogen. This prediction is buttressed by the knowledge that a variety of physical methods show that acetic acid in water is almost completely undissociated. Nonetheless, the simple expectation is not realized. Mixtures of acetic acid and water show only two lines-a methyl resonance and a hydroxyl resonance, the latter of which is intermediate between the normal positions of the carboxyl and water proton resonances. The position of this hydroxyl resonance depends on the concentrations, and in fact, a linear relationship has been demonstrated between the line position and the fraction of the hydroxyl protons in the form of acetic acid, as shown in Fig. 4-2.1 This result is of considerable practical interest in that it provides a means of analysis by virtue of measurements of line positions rather than of line areas. This type of NMR analysis is very advantageous where applicable because line positions are far more easily and accurately measured than the integrated resonance intensities.
Apart from applications to analysis, the occurrence of an average hydroxyl resonance line position from two chemically different species of hydroxyl protons is of considerable theoretical significance. The single hydroxyl line is a consequence of rapid chemical exchange between the hydroxyl hydrogens of acetic acid and water. It represents the hydroxyl protons in a time-averaged environment. Such a phenomenon is almost unthinkable to anyone brought up exclusively on visible and ultraviolet spectroscopy. Indeed, the Franck-Condon principle states that the transition time for an electronic excitation resulting from the absorption of ultraviolet or visible electromagnetic radiation is short even compared with the rate at which the atoms vibrate in chemical bonds. However, in NMR spectroscopy, the transition times for excitation of the nuclei to higher energy magnetic states not only are long with respect to the rates of vibration of atoms in bonds but are also long with respect to rotations around single bonds and are even long with respect to rapid chemical reactions. In effect, the NMR spectrometer acts as a camera with a slow shutter speed and so records molecules in an average state taken over the relatively long periods of time required for the transitions of nuclei from one magnetic quantum state to another. The long transition times are associated with the fact that these magnetic transitions are induced by very low frequency radiations compared with the frequencies of ultraviolet and visible light.
The exchange of acetic acid and water hydroxyl protons appears always to be too fast to give anything but an average hydroxyl resonance line. With other substances, this is not always the case, and as will be seen later, separate hydroxyl absorptions can be obtained with mixtures of some hydroxyl compounds.
1 H. S. Gutowsky and A. Saika, J. Chem. Phys., 211, 1688 (1954).
4.02: Relationship between Resonance Line Shapes and Exchange Rates
In general, there will be a gradual change in the appearance of a resonance due to a magnetic nucleus in a particular chemical environment as the mean lifetime in that environment decreases. The changes which take place for the simple case where one type of nucleus undergoes exchange between environments A and B, in which on the average it spends equal lengths of time, are shown in Fig. 4-3. When the mean lifetime before exchange ($\tau$A or $\tau$B) is long compared with the transition time between the magnetic energy states, two separate sharp resonances $\delta$ABH apart are observed. When $\tau$A is short compared with the transition times, a single sharp resonance line at 1/2 ($\delta$AH + $\delta$BH) is observed. An intermediate broad line is obtained when the mean lifetime before exchange is comparable with the transition time.
Gutowsky2 has shown that the point at which the separate lines just coalesce corresponds to $\tau$A equal to $\sqrt 2 (\pi\delta$ABH)-1 sec when ($\delta$ABH) is expressed in cycles per second or 2$\sqrt 2 (\pi\delta$ABH)-1 when ($\delta$ABH) is expressed in radians per second. The rate constant in either direction for A $\rightleftharpoons$ B will then be l/$\tau$A. Expressions have also been derived for determining rate constants from changes in resonance line shape. In some instances, rate constants accurate to a few per cent have been reported.
Since the easily measurable proton chemical shifts range from 5 to several hundred cycles per second, NMR spectroscopy is most useful for study of reactions which have mean lifetimes of the reacting species ranging between 0.1 and 0.0005 sec. We shall illustrate some of the potentialities of this method of determining reaction rates in a number of different kinds of kinetic processes.
2 H. S. Gutowsky and C. H. Holm, I. Chem. Phys., 25, 1228 (1957). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/04%3A_Nuclear_Magnetic_Resonance_and_Reaction_Kinetics/4.01%3A_Introduction._Proton_Exchange_in_Water-Acetic_Acid_Mixtures.txt |
The work of Weinberg and Zimmerman3 on the hydroxyl resonance in mixtures of ethanol and water is particularly interesting because it rounds out the picture of proton-exchange phenomena discussed earlier in connection with mixtures of acetic acid and water. We shall consider first mixtures of ethanol and water with relatively high water content.
The NMR spectrum (Fig. 4-4) of such a mixture with 30 per cent water shows three principal groups of resonance lines. The two groups at high fields arise from the ethyl protons as evidenced by the customary four-three pattern of spin-spin splitting. The large peak at low fields is the hydroxyl resonance and represents an average ethanol-water hydroxyl resulting from rapid exchange. As will be seen from Fig. 4-5, the position of this line is relatively insensitive to the composition of the mixture, down to about 25 per cent water. Now, the spectrum of ethanol containing only little water (and no acid or base or other substance which might catalyze exchange between the hydroxyls) is seen in Fig. 4-4 to have a quite different appearance from ethanol containing water in which exchange is rapid. At low fields, two groups of peaks appear instead of one. These represent the separate resonances of the water and the hydroxyl protons of ethanol in the mixture and can be easily distinguished by the fact that the alcohol hydroxyl resonance is split into three because of spin-spin coupling between the hydroxyl and methylene protons. The protons of the individual water molecules are, of course, equivalent, and their respective resonance is not split. In nonexchanging ethanol, the methylene resonance is seen to be substantially more complex than for exchanging ethanol. This is because the protons of the methylene group are coupled to both the methyl and the hydroxyl protons. If the coupling constants were equal, the methylene resonance would be expected to be split into five lines. This is approximately what is observed although complications are introduced because the respective J's are not exactly equal.
When rapid exchange sets in, the splitting caused by coupling between the hydroxyl and the methylene protons disappears somewhat sooner than the water and alcohol hydroxyl resonances merge to a single line. The splitting disappears with rapid exchange because as a given hydroxyl proton moves from alcohol to water to alcohol, etc., it experiences local fields produced by different combinations of the two spins of the methylene protons of different alcohol molecules. It does not stay on any one molecule long enough to give a resonance line corresponding to its particular spin combination. Instead, the exchanging proton acts as though it were experiencing an average of the possible CH2 magnetic quantum numbers which is zero and, of course, would produce no splitting.
The reason that the spin-spin splittings between the hydroxyl and methylene protons disappear somewhat before the two hydroxyl lines merge into an average is because a splitting of about 5 cps will be averaged to zero by a process having a mean proton lifetime in the various states of less than 0.07 sec. On the other hand, two hydroxyl lines some 30 cps apart will be averaged by exchange only when the mean lifetime is less than about 0.015 sec. This type of situation where the same reaction process causes differently spaced lines to be averaged separately is particularly advantageous, since it allows determination of more than one rate constant as a function of temperature or concentration. Thus, two values of the rate of hydrogen exchange between water and alcohol as a function of concentration can be obtained by increasing the amount of water in the alcohol so as to wash out successively the hydroxyl-methylene spin-spin couplings and the separate hydroxyl resonances. Similarly, with a given alcohol-water composition, the rate of exchange could be ascertained at two different temperatures by determining the temperatures for separate averaging of the spin-spin couplings and the chemical shift, respectively.
The data presented in Fig. 4-5 show that the mean lifetime of a proton before exchange on the left of the transition point is greater than 0.015 sec, less than 0.015 sec on the right of the transition point, and about 0.015 sec in the neighborhood of the transition point. Acids and bases are powerful catalysts for the hydroxyl exchange, since alcohol samples which show the three-line hydroxyl resonance immediately give a single hydroxyl resonance when minute amounts of a strong acid or base are added. The kinetic order of acid or base in causing this exchange has not as yet been obtained accurately.
3 I. Weinberg and J. R. Zimmerman, J. Chem. Phys., 23, 748 (1955). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/04%3A_Nuclear_Magnetic_Resonance_and_Reaction_Kinetics/4.03%3A_Proton_Exchange_in_Ethanol-Water_Mixtures.txt |
Ethyl acetoacetate provides an interesting example of the use of NMR spectra for both a structural and kinetic analysis. The spectrum of the pure ester is shown in Fig. 4-6. The ethoxy group of the ester is easily identified by the typical four-three pattern of resonance lines. The other prominent resonances are due to the $\alpha$ and $\gamma$ protons. These are approximately in the theoretical ratio 2:3 and appear in the anticipated order with respect to field strength. Thus, the protons on the $\alpha$-carbon atom are adjacent to two electron-attracting carbonyl groups and give a resonance line at a considerably lower field than the protons on the $\gamma$-carbon atom, which are adjacent to only one carbonyl group. At room temperature, ethyl acetoacetate contains about 10 per cent of the corresponding enol form. The presence of this material is shown by the NMR spectrum, there being a small band in the vinyl hydrogen region and a hydrogen-bonded hydroxyl proton resonance at very low fields. The strength of these bands agrees with the composition as established by the Kurt Meyer titration. Apparently, the $\gamma$-methyl and ethoxy resonances of the enol form are not separated enough from those of the keto form to make them easily distinguishable in the appropriate regions.
The detection of the separate resonances of the keto and enol forms shows that the enol and keto forms are not interconverted rapidly at room temperature, and this is in agreement with the observation that the enol and keto forms can be separated by "aseptic distillation" and separately preserved at low temperatures. The NMR spectrum of the equilibrium mixture of the ethyl acetoacetate tautomers at room temperature is markedly altered by the addition of a small amount of the sodium enolate (from dissolution of a small piece of sodium in the liquid), as shown in Fig. 4-6. The $\alpha$-proton resonance of the keto form and vinyl and 0-H resonances of the enol form disappear, and a new, rather broad band appears underneath the resonances of the $\alpha$ hydrogens of the ethyl group. In the particular circumstances, exchange is occurring at an intermediate rate among the $\alpha$-keto, vinyl- , and hydroxylenol hydrogens. Cooling the mixture slows down the rate of exchange and the a-hydrogen line reemerges, although somewhat broadened. On heating, the exchange rate is increased and a new rather sharp average line of the exchanging protons is produced.
Separate experiments have shown that the chemical shifts of the resonance lines of the keto form of ethyl acetoacetate are not substantially altered by raising the temperature. With this information, it is possible to calculate that about 10 per cent of the enol is present in the rapidly exchanging mixture at 110 degrees, by virtue of the relative position of the average line with respect to the a-proton line of the keto form. The average consists of a small contribution of enol hydroxyl with a resonance line at a very low field, an equal-sized vinyl resonance at a much higher field, and a large contribution of $\alpha$-keto hydrogens at a still higher field. Thus, the equilibrium composition of the tautomers of ethyl acetoacetate probably does not change markedly with temperature. A comparable analysis with acetylacetone indicates that with this liquid there is a greater temperature dependence for the position of its keto-enol equilibrium. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/04%3A_Nuclear_Magnetic_Resonance_and_Reaction_Kinetics/4.04%3A_Ethyl_Acetoacetate_and_Its_Enol_Form.txt |
Of extraordinary interest and importance is the use of NMR methods to determine rates of rotation around single bonds.4 The first observations of this kind were made with dimethylformamide. The spectrum of dimethylformamide at room temperature is shown in Fig. 4-7. At low fields, there is a one-proton resonance which arises from the aldehydelike hydrogen of the formyl group. The N-methyl proton resonances occur as two lines spaced about 10 cps apart at 40 Mc. The doublet methyl resonance might conceivably arise either from chemically different methyl groups or spin-spin coupling with the formyl proton. It will be noted from Fig. 4-7 that the formyl proton resonance is not split, so that the latter explanation is untenable. A chemical-shift difference has been established by the fact that at 30 Mc the two methyl peaks are only 7 cps apart. The knowledge that the separate methyl resonances arise from methyl groups in different chemical and magnetic environments forces the conclusion that there is restricted rotation around the -CO-N: bond of dimethylformamide. This is quite reasonable because resonance interaction between the carbonyl group and the unshared electron pair on nitrogen will tend to make all the atoms of the molecule lie in one plane, except for the protons on the methyl group. If the amide group is planar, one methyl group must be cis and the other trans to the carbonyl oxygen, and if rotation is slow about the -CO-N: bond, then the protons of each methyl will give a separate resonance line. On heating, the rate of rotation around the -CO-N: bond increases and finally the separate methyl resonances coalesce to a single line, as shown in Fig. 4-7. Considerable work has been done to determine the activation energy for rotation in dimethylformamide, and although the precision obtained by a given investigator is satisfactory, there are considerable discrepancies among the reported values.2,6 This may possibly be due to effects of impurities.
Phillips7 has made elegant use of the slowness of rotation around the -CO-N: bond of dimethylformamide to learn how the molecule accepts a proton in forming the conjugate acid. As shown in the following equation, the proton could add to either the oxygen or the nitrogen.
One could well argue that nitrogen is intrinsically more basic than oxygen, so that the nitrogen might be the favored position, even though addition of the proton to nitrogen would mean loss of resonance energy associated with the interaction of the nitrogen unshared electron pair with the carbonyl group. On the other hand, it could be argued that the amide oxygen would be more prone to accept a proton than usual for carbonyl oxygen because the resulting conjugate acid would be substantially stabilized by resonance involving the unshared electron pair on nitrogen.
Phillips recognized that if the proton adds to nitrogen, this would destroy the double-bond character of the -CO-N: bond and reduce the barrier to rotation to the very low value expected for a C-N single bond. On the other hand, if the proton adds to oxygen, then the amount of double-bond character in the -CO-N: bond would be increased and rotation would be more difficult. Experimentally, it is observed that the rate of rotation around the single bond in dimethylformamide in strong acid solution is not increased compared with neutral solution, and therefore the added proton is preferentially attached to oxygen.
A number of interesting studies of restricted rotation in nitrites, N-nitrosamines, and oximes have been reviewed by Phillips.4
4 W. D. Phillips, Ann. N.Y. Acad. Sci., 70, 817 (1958).
5 W. D. Phillips, J. Chem. Phys., 23, 1363 (1955).
6 Research by J. N. Shoolery reported in Varian Associates Tech. Bull., 2, 7 (1957).
7 Private communication from W. D. Phillips. Similar studies have been made by G. Fraenkel and C. Niemann, Proc. Nut. Acad. Sci., 44, 688 (1958). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/04%3A_Nuclear_Magnetic_Resonance_and_Reaction_Kinetics/4.05%3A_Rates_of_Rotation_Around_-CO-N-_Bonds_of_Amides.txt |
In favorable circumstances, the rates of rotation around C-C bonds in ethane derivatives can be shown by NMR methods to be slow.4,8 An excellent example is afforded by the behavior of 1,1-difluoro-1,2-dibromo-2,2-dichloroethane.8 The 19F nuclear resonance spectrum of this substance at room temperature shows a single sharp line. Inspection of the possible rotational conformations (Fig. 4-8) shows that a single 19F line is to be expected either as a result of averaging of the fluorines by rapid rotation or else, if rotation is slow, because the substance is "locked into" the "meso" conformation (I) with the 1,2-bromines trans to one another and the fluorines equivalently located. The other possible staggered conformations are seen to be a "d,l" pair (II and III) and to have nonequivalent fluorines. Rapid equilibrium of these two rotational isomers would average the fluorine spectrum to a single line because, as one isomer goes to the other, the fluorines exchange magnetic environments. Therefore, the rotational isomers must be rapidly equilibrated or else the substance must exist predominantly in a symmetrical form in order that only a single resonance line be obtained.
The 19F spectrum of 1,1-difluoro-1,2-dibromo-2,2-dichloroethane is quite temperature dependent, as is shown in Fig. 4-9. At -30°, the single line is seen to broaden substantially; at -60°, new lines commence to appear; and at -80°, five sharp lines are observed. These lines can be easily identified by their spin-spin coupling patterns. Four of the lines represent the spectra of the d,l pair II and III, which have nonequivalent fluorines, so that the fluorine resonances are shifted from one another and the concomitant spin-spin interactions give a quartet of lines with the outer lines being weakened by virtue of J being comparable with $\delta$H.
Since the rotational isomers II and III are mirror images, they will give identical spectra, and the symmetrical four-line pattern in Fig. 4-9 represents the contribution of this pair. The large single sharp resonance is due to the symmetrical "meso" rotational isomer with equivalent fluorines. At -80°, the rate of interconversion of these isomers must be much less than 200 times per second. On a purely statistical basis, the ratio of I to the isomers II and III should be 1:2. However, a rough analysis of the mixture at -80° by measuring the ratios of the signal intensities indicates that the composition is actually close to 1.4:1. This ratio is quite reasonable, since the conformation with two bromines trans to each other is expected to be sterically favored.
It is possible that valuable information as to relative steric sizes of groups will be obtainable through NMR studies of rotational isomer populations. An excellent example is afforded by 1,1-difluoro-1,2,2- tribromo-2-phenylethane. The 19F spectrum of this substance at room temperature shows a single line, while at -80° there are five lines as expected from a "freezing out" of the rotational isomers. The "meso" isomer of 1,1 -difluoro-1,2,2-tribromo-2-phenylethane has phenyl trans to bromine, and this would be expected to be the favorable conformation if phenyl were sterically larger than bromine as far as influencing the populations of the various configurations goes (see Fig. 4-10). However, the NMR spectrum (Fig. 4-11) shows that the meso isomer is actually present in far smaller concentration than the d,l pair, and in this case, we conclude that bromine acts as though it were larger than phenyl. Extensions of this type of experiment can clearly provide much information with regard to rotation about single bonds which is not easily available in any other way.
8 P. M. Nair and J. D. Roberts, J. Am. Chem. Soc., 79, 4565 (1957). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/04%3A_Nuclear_Magnetic_Resonance_and_Reaction_Kinetics/4.06%3A_Restricted_Rotation_in_Ethane_Derivatives.txt |
Another interesting process which can be studied by NMR spectroscopy is the rate of inversion of nonplanar nitrogen carrying substituent groups. Ammonia has long been known from its rf spectrum to undergo inversion extremely rapidly. The corresponding inversion frequencies of various alkyl- and aryl-substituted amines have not been determined, but a variety of experiments have failed to provide optically active isomers of trisubstituted amines that are asymmetric solely because of the configuration at the nitrogen atom, presumably because inversion is too rapid. In 1940, Kincaid and Henriques9 published calculations which indicated that the rates of inversion of substituted nitrogen compounds would be much smaller if two of the substituents on nitrogen were connected together so as to form a three-membered N-C-C ring. However, prior and subsequent attempts to resolve such substituted ethylenimine derivatives into optically active forms were unsuccessful.
Nonetheless, the predictions of Kincaid and Henriques have been confirmed through the NMR spectra of cyclic imines.10
The proton resonance spectrum of N-ethylethylenimine at various temperatures is shown in Fig. 4-12. At room temperature, the typical three-four line pattern of the N-ethyl group is easily recognized and there remain two groups of lines to be assigned which arise from the ring protons. Models of N-ethylethylenimine show that, if the nitrogen is not planar, one would expect two kinds of ring protons-ones which are cis to the ethyl group and ones which are trans to the ethyl group.
These will be in different magnetic environments and can account for the two observed peaks about 30 cps apart. Since nitrogen inversion results in exchange of the environments of the ring protons, the mean lifetime of an ethylenimine molecule before inversion occurs must be much longer than 0.015 sec if the cis- and trans-ring protons are to give separate sharp resonances. At higher temperatures, the inversion rate increases, and, finally, at 120°, the two ring-hydrogen resonances coalesce to a single line. The intermediate temperature is about 110°, and at this point the mean lifetime of the molecule before inversion is about 0.015 sec.
Effects of substituents on the inversion rates of cyclic imines have been studied by the NMR method, and it has been found that bulky groups either on the ring or on the nitrogen tend to increase the inversion rates, presumably by destabilizing the separate nonplanar configurations relative to the transition state. Unsaturated groups connected to either the ring or the nitrogen of ethylenimine cause the inversion rates to increase markedly. Presumably, with such substituents, the planar transition state is stabilized relative to the nonplanar ground state by resonance interaction between the unshared pair of electrons on nitrogen and the unsaturated centers. N-Ethylallenimine provides an excellent example of this type of behavior and the temperature dependence of its NMR spectrum is shown in Fig. 4-13.
Solvents like water or alcohol which can form hydrogen bonds to the unshared electron pairs of ethylenimines substantially reduce the imine inversion frequencies. Presumably, the hydrogen bonds stabilize the nonplanar configuration by tending to anchor the unshared electron pairs on nitrogen.
9 J. F. Kincaid and F. C. Henriques, Jr., J. Am. Chem. Snc., 62, 1474 (1940).
10 A. T. Bottini and J. D. Roberts, J. Am. Chem. Soc., 78, 5126 (1956), and 80, 5203 (1958).
4.08: Proton Exchange in Ammonia and Ammonium Ions
A number of interesting exchange experiments of other kinds have been done with nitrogen compounds. Only one sharp proton resonance has been observed lipX2 for ammonia-water mixtures, as would be anticipated for a rapidly exchanging system. Extremely anhydrous liquid ammonia or gaseous ammonia gives a triplet resonance pattern with all of the peaks being the same height (see Fig. 4-14).11,12 This pattern arises from spin-spin coupling between the protons and the 14N nucleus, which has a spin of 1 and hence the magnetic quantum numbers +1, 0, -1, which are equally probable. The line at the lowest field represents the proton resonance absorption of those ammonia molecules having nitrogen with a magnetic quantum number +1, while the center and high-field lines correspond to the nitrogen nuclei with magnetic quantum numbers 0 and -1, respectively.
In the presence of water or minute amounts of amide ion, exchange of hydrogens between ammonia molecules is extremely rapid, so that the spin-spin splitting is washed out when the mean lifetime of a proton on any given nitrogen is substantially less than 0.007 sec. In aqueous solution, exchange is so rapid between ammonia and water molecules that separate 0-H and N-H resonances are not obtained and only a single average proton resonance is obtained. Similar results are obtained with ammonium nitrate solutions in water containing enough additional ammonia to make the solution nearly neutral. Under these circumstances, ammonium ions, ammonia, and water exchange protons with one another sufficiently rapidly to give only a single line. A startling change takes place on the acidification of such solutions with nitric acid.13 As is seen in Fig. 4-15, a triplet pattern appears which corresponds to the three possible spin orientations of the 14N nuclei and there is an additional large single proton resonance, that of water. This result proves that water is singularly ineffective in removing protons from ammonium ions while ammonia is extremely effective. Similar observations14 have been made for methylammonium ion, the spectrum of which in acidic solution is shown in Fig. 4-16. The broad N-H resonances which contrast to the sharp lines observed for ammonium ion result from quadrupole effects as will be discussed later and not from intermediate proton exchange rates, because further addition of acid does not cause the lines to sharpen. It is possible to obtain the exchange rate constants by observations of line-shape changes with temperature or pH. The kinetics and mechanisms of such processes have been studied in detail by Grunwald and coworkers.15
11 R. A. Ogg, Jr., J. Chem. Phys., 22, 560 (1954).
12 H. S. Gutowsky and S. Fujiwara, J. Chem. Phys., 22, 1782 (1954).
13 R. A. Ogg, Jr., Discussions Faraday Soc., 215 (1954).
14 J. D. Roberts, J. Am. Chem. Soc., 78, 4495 (1956).
15 E. Grunwald, A. Loewenstein, and S. Meiboom, J. Chem. Phys., 27, 630 (1957). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/04%3A_Nuclear_Magnetic_Resonance_and_Reaction_Kinetics/4.07%3A_Nitrogen_Inversion_Frequencies_of_Cyclic_Imines.txt |
The broad N-H lines observed for the proton resonance spectrum (Fig. 4-16) of methylammonium ion in acid solution are worthy of special mention. Similar N-H resonances have been noted for quite a variety of substances provided the protons are not undergoing rapid exchange.l Pyrrolidine hydrochloride affords a particularly striking example of this behavior. The methylene protons of pyrrolidine hydrochloride in a solution containing a slight excess of pyrrolidine (Fig. 5-1) show reasonably normal spin-spin couplings. Under these circumstances, the N-H proton and the water proton resonances are averaged. Acidification of the pyrrolidine hydrochloride solution effectively stops exchange but results in broad N-H resonances which, as mentioned before in connection with methylammonium ion, could conceivably be due to intermediate proton exchange rates with the water. However, this interpretation is ruled out by virtue of the fact that Fig. 5-1 shows the N-H protons to be coupled to the a-methylene protons. If there were intermediate exchange rates, then one would expect that the N-H:C-H splittings would be washed out before the N-H resonances themselves were appreciably broadened.
Besides being broad, the N-H resonances of substituted ammonium ions behave anomalously with temperature. Thus, lowering the temperature causes the three-line pattern to disappear and be replaced by a very broad single N-H resonance.l On the other hand, raising the temperature tends to sharpen the triplet pattern, although, in any case, the lines are much broader than for the resonances of the protons of ammonium ion itself.
1 J. D. Roberts, I. Am. Chem. Soc., 78, 4495 (1956).
5.02: Nuclear Quadrupoles and Quadrupole-induced Relaxation
The above observations are the result of changes in the magnetic quantum numbers of the 14N nuclei in substituted ammonium ions through interaction of the 14N electric quadrupoles with surrounding asymmetrical electric fields. The mechanism for this is not so formidable as it might sound. In the first place, we noted earlier (pages 6 and 7) that nuclei with spins I of 0 or 1/2 act as though their charges were distributed over a spherical surface. With those nuclei like 14N having I>1/2, however, the nuclear charge appears to be distributed over an ellipsoidal surface, and such nuclei act like electric quadrupoles. The symmetry axis of a nuclear quadrupole is collinear with the magnetic and angular momentum vectors of the nucleus. Now, consider a 14N nuclear quadrupole such as the nitrogen of pyrrolidinium ion surrounded by a cloud of valence electrons which is not spherically symmetrical. The rapid tumbling motions of the ion in solution will cause a timevariable electric torque to be exerted on the quadrupole which will tend to shift the quadrupole orientation. Changes in the orientation of the quadrupole also result in changes in the direction of the magnetic vector of the nucleus. A relaxation mechanism is thereby provided for the 14N nuclear magnet (see Fig. 5-2). Now, if the nitrogen nucleus of an 14N-H compound is caused to flip back and forth among its several possible magnetic quantum states, the attached proton will "see" more or less of an average of three orientations of the nitrogen nucleus depending on the rate of Ripping.
In solutions of pyrrolidinium ion at room temperature, relaxation of the nitrogen nucleus takes place at a rate such as to render the hydrogens attached to the nitrogen somewhat confused as to the 14N spin orientation. This results in broadened N-H proton resonance lines. At room temperature, the N-H line is intermediate between a singlet and triplet absorption, so that the mean lifetime of the nitrogen with a given magnetic quantum number is on the order of the reciprocal of zG/2 times the triplet line separation, or about 0.009 sec. When the temperature is raised, the increased rate of tumbling of the molecules apparently results in less efficient nitrogen relaxation and the proton resonance goes to a broad triplet. When the temperature is lowered, the decreased tumbling rate allows for rather more efficient nitrogen relaxation and the triplet proton resonance lines coalesce to a broad singlet. Only a sharp triplet absorption is observed for nonexchanging ammonium ions because in these ions the electric field around the nucleus is spherically symmetrical and quadrupole-induced relaxation is not effective. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/05%3A_Nuclear_Quadrupole_Relaxation_Effects_and_Double_Resonance/5.01%3A_Proton_Resonance_Line_Broadening_by_14N.txt |
Broadening of N-H proton resonance lines by quadrupole-induced relaxation has been observed with a number of types of nitrogen compounds. A particularly striking example is afforded by pyrrole. The NMR spectrum of pyrrole at room temperature shows no N-H line on casual inspection (see Fig. 5-3). The principal observed peaks are those of the C-H protons. However, very careful inspection of the spectrum shows a very broad peak to the low-field side of the C-H resonances, which rises only slightly above the base line. Apparently, the 14N nucleus in pyrrole is undergoing relaxation at just the right rate to cause the protons attached to it to give the broadest possible line intermediate between the singlet and triplet patterns.
Any process which causes the N-H protons to "see" either a more slowly or more rapidly moving sequence of nitrogen magnetic quantum states will sharpen the resonance lines to give the triplet or singlet lines, respectively. As is seen in Fig. 5-3, this may be achieved by raising or lowering the temperature so as to change the rates of tumbling of the molecules and thus influence the effectiveness of the quadrupole relaxation of the nitrogen nucleus. As with pyrrolidine hydrochloride, lowering the temperature gives a sharpened single resonance, while raising the temperature causes a triplet pattern to appear. A given N-H proton of pyrrole can be exposed to a more rapid sequence of nitrogen magnetic quantum states by addition of potassium pyrrolate, which induces intermolecular proton exchange. As expected, exchange causes the N-H resonance line to sharpen.
By far the most elegant procedure for eliminating the magnetic relaxation effects of the 14N nucleus is the "double-resonance" or "spindecoupling" technique. As applied by Shoolery2 to pyrrole, this involves observing the proton spectrum in the normal way while subjecting the sample to a powerful rf field at the 14N resonance frequency (2.9 Mc at 9400 gauss). The proton signal is detected with a narrow-bandwidth receiver, so that there is no pickup of the second oscillator frequency. The rf input at the 14N frequency causes the nitrogen nuclei to change their magnetic quantum numbers rather more rapidly than is possible for quadrupole-induced relaxation alone. As a result, each N-H proton sees its nitrogen nucleus with the magnetic quantum numbers effectively averaged to zero and thus gives a sharpened N-H resonance (Fig. 5-4). Several other applications of double-resonance technique will be described in Sec. 5-5.
2 Private communication from J. N. Shoolery.
5.04: Quadrupole-induced Relaxation with Other Nuclei
All nuclei which have a spin of greater than \$\$ act like electric quadrupoles and undergo more or less efficient relaxation through interactions with surrounding dissymmetric electric fields produced by their valence electrons. Most of the halogen nuclei (chlorine, bromine, and iodine, but not fluorine) are relaxed very rapidly by the quadrupole electric field interactions. As a result, even though these nuclei have considerable magnetic moments, they do not normally cause spin-spin splitting as might be expected for proton-halogen coupling in compounds like methyl chloride, bromide, or iodide, because rapid relaxation averages the magnetic effects of the halogens to zero. 14N is notorious for its tendency to relax at intermediate rates when unsymmetrically substituted and this causes broadening of N-H proton resonance signals. However, with a number of amino compounds, such as ethylamine, sharp proton resonances are observed for the N-H protons. This is surely a consequence of rapid intermolecular exchange between the N-H protons rather than rapid quadrupole-electric field relaxation.
5.05: Applications of Double Resonance
The double-resonance technique described in Sec. 5-3 has important uses in analysis of complex spin-spin coupling patterns when two or more varieties of nuclei with different precession frequencies are involved. Valuable information for the analysis of the NMR spectra of boron hydrides has been obtained by Shoolery3 by collapsing of spinspin multiplets due to 10B and 11B while observing proton resonances.
Double resonance has also been utilized to aid in direct determinations of the H-F spin couplings in fluorobenzene.4 With ordinary fluorobenzene, spin-spin splitting makes the spectrum complex and difficult to analyze. In principle, the problem of determining coupling constants for interactions between the fluorine and the hydrogens at the o, rn, and p positions can be greatly simplified through study of splittings in various deuterium-substituted fluorobenzenes. Thus, to obtain the coupling between the fluorine and meta hydrogens, 2,4,6-trideuterofluorobenzene can be employed. However, the proton spectrum (Fig. 5-5a) of this molecule is rendered more complex than might otherwise be expected because of F-D and H-D couplings which are at least as difficult to unravel as the spectrum of ordinary fluorobenzene. However, the double-resonance technique permits averaging of the deuterium spins to zero, so that the residual clean doublet observed in the proton spectrum (Fig. 5-5b) is due only to 1,3-proton-fluorine coupling with J equal to 5.8 cps. A similar study of 2,3,5,6-tetradeuterofluorobenzene shows that coupling between fluorine and para protons is negligible.
Double resonance would be extremely valuable in the analysis of spin-spin couplings in compounds with interacting nuclei of the same type. For example, one might use a suitable oscillator to "stir up" the methylene protons of ethanol while observing the methyl protons at another precession frequency. In such circumstances, the methyl resonance would be averaged to a single line and it would then be known that the splitting ordinarily observed arises from the various possible combinations of magnetic quantum numbers of the methylene protons. Unfortunately, this type of double resonance is not experimentally easy because it is necessary to use a considerably higher rf power level to stir up one variety of proton than is necessary or desirable for the observation of the resonance of another variety of proton. As a result, one needs a receiving arrangement with high discriminating power to distinguish between the decoupling frequency and the observing frequency which may differ only by 100 cps at 40 Mc. Successful experiments of this type have been reported by Anderson,5 but the method cannot yet be regarded as being applicable to routine work.
3 J. N. Shoolery, Discussions Faraday Soc., 215 (1955); R. Schaeffer, J. N. Shoolery, and R. Jones, J. Am. Chem. Soc., 79,4606 (1957).
4 B. Bak, J. N. Shoolery, and G. A. Williams, 111, J. Mol. Spectroscopy, 2, 525 51958).
5 W. A. Anderson, Phys. Rev., 102, 151 (1956). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/05%3A_Nuclear_Quadrupole_Relaxation_Effects_and_Double_Resonance/5.03%3A_Proton_N-H_Resonance_of_Pyrrole._Double_Resonance.txt |
As we will see, organic reactions can be classified using a small set of reaction types—the largest and most all-encompassing of which are those involving acid–base reactions. Understanding acid–base reactions, therefore, provides a broadly useful conceptual framework within which to consider a wide range of organic reactions. Although it is likely that you have already been introduced to acid–base reactions (especially if you used the CLUE general chemistry curriculum[1]), we are going to review this class of reactions in order to emphasize their general features. Our goal is that you learn how to recognize their role in a range of reaction mechanisms; understanding how and why acid–base reactions occur will give you to a set of tools to understand phenomena as diverse as why most drugs are usually administered as in their salt form (a conjugate acid or base), why biological systems are buffered to specific pH levels (and why different pH levels are found in different cellular and organismic compartments), and why molecular oxygen (O2) transport systems require a metal ion complex (within the proteins involved, e.g. myoglobin, hemoglobin, cytochromes). As we will see, acid–base reactions are by far the most common types of reactions in biological systems.
01: AcidBase Reactions
There are a number of ways to discuss acid–base reactions, depending on what aspects of the reaction we want to highlight. They range from the extremely simplified (and not useful) Arrhenius model, to the Brønsted–Lowry model that we use only for reactions in which protons are transferred, and finally to the Lewis model, which can encompass any type of acid–base reaction.
Arrhenius:
The Arrhenius acid–base model is probably the first acid–base model that you were introduced to in the course of your education. In this model, when an acid dissolves in water it dissociates to release a hydrogen ion (H+); when a base dissolves it releases a hydroxide ion (–OH).
Acid: $\mathrm{HCl}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$ (sometimes written as or $\mathrm{HCl}(\mathrm{aq})$)
Base: $\mathrm{NaOH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+{ }^{-} \mathrm{OH}(\mathrm{aq})$
Acid–Base Reaction: $\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \text { (l) }$[2]
Although simple, the Arrhenius model is not particularly useful when it comes to understanding the reactions considered in organic chemistry. This of course raises the obvious question: so why are we mentioning it? The answer is two fold: i) because you might well vaguely remember it as a description of acid–base behaviors and ii) so that we can consider why it is not useful and why you should not use it. The Arrhenius acid–base model applies only when water is the solvent—as we will see many organic reactions do not occur in water. The Arrhenius model also, falsely implies that there are free protons ($\mathrm{H}^{+}$) roaming around in water and it restricts bases to those substances that release a hydroxide ion. Finally, it implies that an acid can exist independent of a base—and vice versa, which doesn’t make a great deal of sense.
Brønsted–Lowry:
The Brønsted–Lowry model is a much more useful and flexible model for considering acid base reactions. In this model an acid is a proton ($\mathrm{H}^{+}$) donor and a base is a proton acceptor. In the Brønsted–Lowry model you cannot have an acid without a base, and vice versa; the acid has to donate its $\mathrm{H}^{+}$ to something (the base), and similarly the base has to accept it. The $\mathrm{H}^{+}$ doesn’t just “drop off”—it is transferred.[3] In the case of reactions that occur within aqueous solution, the $\mathrm{H}^{+}$ is transferred to a water molecule to form $\mathrm{H}_{3} \mathrm{O}^{+}$. Consider, as an example, $\mathrm{HCl}$; in aqueous solution $\mathrm{HCl}$ transfer a $\mathrm{H}^{+}$ group to a water molecule. The products are $\mathrm{H}_{3} \mathrm{O}^{+}$ (the conjugate acid of water) and $\mathrm{Cl}^{-}$, the conjugate base of $\mathrm{HCl}$.
HCl(g) +
H2O(l)
H3O+ (aq)
+
Cl(aq)
acid
base
conjugate acid
conjugate base
The key point here is that the $\mathrm{H}^{+}$ is transferred from one molecule to the other—it doesn’t drop off and then reattach.
The flexibility of the Brønsted–Lowry model lies in the fact that the base does not necessarily have to be water. For example, if we look at the reaction of hydrogen chloride and ammonia ($\mathrm{NH}_{3}$), we see that the proton transfer from acid to base is analogous to the reaction in water.
HCl
+
NH3
⇄ NH4+ +
Cl
acid
base
conjugate acid
conjugate base
In the Brønsted–Lowry model, as for all chemical reactions considered at the molecular level, there is the possibility for the reaction to reverse, which is denoted by the use of equilibrium arrows ($\rightleftarrows$).
At the macroscopic level the extent to which the reaction proceeds (from reactants on the left to products on the right) is determined by a number of factors. That is, we need more information to predict (or calculate) the concentrations of reactants and projects at equilibrium. This is information that also enables us to predict whether the reaction will proceed in the forward direction (to the right) or not and how the reaction might change if we add or remove reactants (or products).
We can identify a potentially acidic $\mathrm{H}^{+}$ because it will be bonded to a more electronegative atom; the result is that the electron density in the bond will lie mainly with the more electronegative atom (e.g. $\mathrm{O}$, $\mathrm{N}$, or $\mathrm{Cl}$). The outcome is that, for example, an $\mathrm{H-O}$ bond will be weakened (require less energy to break); the $\mathrm{H}$ will have a large partial positive charge on it, and will be strongly attracted to basic centers (as described in the next section). Similarly, simple bases can be identified by the presence of an atom (within the molecule) that has a partial negative charge; this partial negative charge arises because the atom (the basic center) is bonded to less electronegative atoms. Now we add one further consideration, this base center atom also needs to be able to accept the incoming $\mathrm{H}^{+}$. In practice, this means that a basic molecule will contain an atom that has a lone (non-bonding) pair of electrons that can form a bond to the $\mathrm{H}^{+}$.
The Brønsted–Lowry model is useful for acid–base reactions that involve proton transfer, but even so, it is limited to proton transfer reactions. We also note here that the solvent in which the proton transfer takes place will have an effect on the reaction, and we will return to this idea later in the course. If we extend the Brønsted model to other reactions where a base uses its lone electron pair to form a new bond with an electropositive center, we can expand the class of acid–base reactions even further. Which brings us to the next model of acid base chemistry: the Lewis model.
Lewis:
The Lewis model allows us to describe exactly the same set of reactions as does the Brønsted–Lowry model, but from a different perspective, and it also allows us to expand on the model. In the Lewis model a base has the ability to donate an electron pair to form a new bond with the acid that accepts this new bond, often but not always with the concomitant breaking of a bond within the acid molecule. We use same rationale for why the reaction occurs between two oppositely charged centers, but from the perspective of the electrons, rather than the $\mathrm{H}^{+}$. A base must therefore have a lone pair of electrons that can take part in a bond while an acid must have an atom that can accept that lone pair of electrons. Using the reaction of $\mathrm{HCl}$ and water as an example, we use the curved arrow notation to denote how the electrons move between base and acid. Recall[4] that we use this curved arrow notation to indicate the movement of electron pair
from a source of electrons to a sink. Here the source is the lone pair on the oxygen, and the sink is the hydrogen (which has a $\delta^{+}$ due to its bonding to a $\mathrm{Cl}$) The second arrow moves from the source (the bond between $\mathrm{H}$ and $\mathrm{Cl}$, to the sink—the electronegative $\mathrm{Cl}$ which ends up with the negative charge, while the $\mathrm{O}$ that donated the original electron pair ends up with a positive charge).
The Lewis model encompasses the Brønsted–Lowry model, that is, all Brønsted–Lowry acid–base reactions that can be described using the Lewis model. However, the Lewis model extends the range of reaction types that can be considered as acid–base reactions. Take for example the reaction of ammonia ($\mathrm{NH}_{3}$) and boron trifluoride ($\mathrm{BF}_{3}$). This reaction is classified as a Lewis acid–base reaction, but it is not a Brønsted acid–base reaction.
Why use different models of acid–base chemistry? While at first the idea of using different models to explain acid–base chemistry may be a little confusing. Why not use the all-encompassing Lewis model for everything? It turns out that both the Brønsted–Lowry and Lewis models are particularly useful depending on the system under consideration. The trick is to recognize which is the most useful when describing, predicting, and explaining a particular type of chemical reaction.[5]
In our explorations in organic chemistry we will be using both Brønsted–Lowry (proton transfer) and Lewis (electron pair donation) models to describe acid–base chemistry, depending on the type of reaction. In practice, the Brønsted–Lowry model is simple and useful; it tells you what is happening (proton transferred from acid to base) but nothing about the mechanism by which the $\mathrm{H}^{+}$ moves. For that we must turn to the Lewis model, which tells us how the electrons rearrange during the reaction. It is also important to keep in mind why these reactions happen—they are caused by an electrostatic interaction between two oppositely charged parts of molecules: $\delta^{-}$ is attracted to $\delta^{+}$.
One further note, all reactions are initiated by random collisions of molecules, but only collisions that allow the electrostatic interaction of the acid and base to occur are productive (that is, collisions that involve two similarly charged parts of molecules will not give rise to a reaction. Again we will have more to say about this later. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/01%3A_AcidBase_Reactions/1.01%3A_A_Quick_Review_of_the_Models_of_Acid-Base_Reactions..txt |
Acid base reactions begin because of electrostatic interactions, but the extent to which the reaction proceeds depends on the relative Gibbs free energy of the reactants and products, that is, the overall Gibbs free energy change ($\Delta \mathrm{G}$) for the reaction. This is a subtle but important point: the reaction does not occur because the products are more stable, it occurs because there is an attractive force between two reactants that have polar structures, As we will see, we can predict the relative amounts of reactants and products in a mixture (at equilibrium), based both on an understanding of molecular structures and by comparing their $\mathrm{pK}_{\mathrm{a}}$.
Acid Strength (using the Brønsted–Lowry model):
The strength of an acid, that is the degree to which it donates $\mathrm{H}^{+}$ to (or accepts electron pairs from) other molecules, depends on a number of factors including, obviously, the strength of the base (that is the degree to which the base donates electron pairs to other molecules) it reacts with. Acid and base strengths are usually reported using water as the solvent (i.e. as the base or acid respectively), so that acid strengths can be compared directly. Since biological reactions take place in aqueous solution we will be able to extend our understanding of simple acid base reactions to much more complex ones as we move forward.
The reaction for any acid HA is: $\mathrm{HA}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{A}^{-}$
We can estimate the extent of the reaction (i.e., how far the reaction goes, that is the concentrations of reactants and products when the reaction reaches equilibrium) by determining the equilibrium constant $\mathrm{K}_{\mathrm{a}}$. $\mathrm{K}_{\mathrm{a}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right] /[\mathrm{HA}]$
In contrast to strong inorganic acids (such as $\mathrm{HCl}$, or $\mathrm{HNO}_{3}$), the equilibrium constants for many organic acids are small (ranging from 10–1 to 10-55) and it is more common to report $\mathrm{pK}_{\mathrm{a}}$ – which, as you will remember, is $=-\log \mathbf{K}_{\mathbf{a}}$. A strong acid such as $\mathrm{HCl}$ has a large $\mathrm{K}_{\mathrm{a}}$ (in fact it is so large as to be meaningless) and therefore a very small (negative) $\mathrm{pK}_{\mathrm{a}}$.
Some representative Ka and pKa values.
Acid
Ka
pKa
HCl (hydrochloric acid)
~107
–7
CF3COOH (Trifluoroacetic acid)
3.2 x 10–1
0.5
HF (hydrofluoric acid)
7.2 x 10–4
3.14
CH3COOH (acetic acid)
1.8 x 10–5
4.8
H2O
10-14
14
CH3CH2OH (acetic acid)
10-16
16
NH4+ (ammonia in NH4Cl) 5.6 x 10–10 9.25
CH4 (methane) ~10–55 55
It helps to be able to interpret these numbers in terms of the extent of the associated reaction. For example, water (which acts as both an acid and a base) dissociates to a very small extent. In a liter of pure water, which contains ~54 moles of water molecules (or $\sim 54 \times 6.02 \times 10^{23}$ molecules or $\sim 3.25 \times 10^{25}$ molecules), $\sim 10^{-7}$ moles (or $\sim 10^{-7} \times 54 \times 6.02 \times 10^{23}$ molecules or $\sim 3.25 \times 10^{16} \mathrm{H}_{3} \mathrm{O}^{+}$ ions). The weaker the acid the higher the $\mathrm{pK}_{\mathrm{a}}$ (can you explain why that is the case and what it means in terms of the relative concentrations of species at equilibrium?).
It will help you greatly if you memorize a few important approximate $\mathrm{pK}_{\mathrm{a}}$ values for common acids, for example alcohols tend to have a $\mathrm{pK}_{\mathrm{a}}$ of $\sim 15$, while amines have a $\mathrm{pK}_{\mathrm{a}} \sim 33$. As we will see the $\mathrm{pK}_{\mathrm{a}}$ of various carbon species is very dependent on the environment of the $\mathrm{C-H}$ bond, but remembering that $\mathrm{sp}^{3}$ carbon-hydrogen bonds ($\mathrm{pK}_{\mathrm{a}} \sim 55$) are not likely to ionize under any circumstances is helpful. However, it is even more important to understand the factors that affect acid strength, and be able to use them to predict and explain the outcomes of reactions.
Another important idea to remember is that the extent of a reaction (as measured by its equilibrium constant $K$) is related to the change in Gibbs free energy ($\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$) associated with that reaction. That is when we think about the extent of a reaction (the concentration of reactants and products when the reaction reaches equilibrium) in terms of the relative stabilities of the reactants and products we need to take into account both the enthalpy change ($\mathrm{H}^{\circ}$), which reflects the changes in bonding and intermolecular interactions involving both reactants and products, and the entropy change ($\Delta \mathrm{S}^{\circ}$) associated with the reaction system. Recall that $\Delta \mathrm{S}^{\circ}$ reflects change in the number of possible energy states and positions in the reaction system. For most organic (weak) acids, it turns out that the $\mathrm{H}^{\circ}$ of the dissociation reaction in water is approximately zero, because the types of bonding and interactions that are broken and formed during the reaction are similar. Differences in $\Delta \mathrm{G}$ for the reaction (and therefor $\mathrm{K}_{\mathrm{a}}$ and $\mathrm{pK}_{\mathrm{a}}$) are typically due to differences in $\Delta \mathrm{S}$.
Questions to Answer
• Explain why acids and bases are always (as pairs) found together in a system.
• What is meant by the terms conjugate acid or conjugate base?
• In the Lewis model for the $\mathrm{HCl}$ + water reaction, explain why you draw the arrow pointing from $\mathrm{O}$ to $\mathrm{H}$.
• Complete these acid base reactions and predict the relative amounts of reactants and products when the reaction reaches equilibrium for each reaction. Explain your predictions using your knowledge of atomic and molecular structures and electronegativity. \begin{aligned} &\mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{HCL} \rightleftarrows \ &\mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \ &\mathrm{CH}_{3} \mathrm{NH}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \ &\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \end{aligned}
Organic Acids and Bases
Having reviewed acids and bases using rather simple molecules ($\mathrm{HCl}$ and $\mathrm{NH}_{3}$), let us move on to the more complex world of organic acids and bases, how to identify them, how to determine relative strengths, and how to predict what will happen in any given mixture. We begin by comparing the $\mathrm{pK}_{\mathrm{a}}$‘s of some organic acids. Let us begin with ethanol ($\mathrm{pK}_{\mathrm{a}} \sim 16$), a molecule that we typically do not consider to be an acid, and acetic acid ($\mathrm{pK}_{\mathrm{a }} 4.8$). There is clearly a huge difference between the $\mathrm{pK}_{\mathrm{a}}$’s of these two molecules, the question is can we understand why this is the case?
If we draw out their structures we see that both have (as expected) the acidic hydrogen bonded to the electronegative oxygen. (Make sure you remember why the hydrogens bonded to carbons are not as acidic as those bonded to oxygen). So why the huge difference in $\mathrm{pK}_{\mathrm{a}}$‘s? To answer this question we have to remember that the extent of the reaction depends on the relative thermodynamic stability of the products—that is, the system containing the conjugate base of the acid and the hydronium ion. The reactions and conjugate bases of the two are shown here ($\downarrow$). Based on their $\mathrm{pK}_{\mathrm{a}}$ values, we would predict that the ethanol dissociation reaction is rare (few ethoxide ions form) while the acetic acid dissociation reaction is more frequent. However note that even in the case of the acetic acid only about 3% of the acid molecules are dissociated in a 1M solution.
The first step in both reactions appears to be more or less the same, an electron pair from the oxygen in water forms a bond to the electron deficient hydrogen while the $\mathrm{O-H}$ bond of the acid breaks and the electrons originally associated with it the move back to the oxygen. The difference between the two reactions lies mainly in the way that the negatively charged conjugate bases (ethoxide and acetate) behave, and the way that they are solvated by the solvent (water). For ethoxide (ethanol’s conjugate base), the extra negative charge is localized onto the oxygen, which leads to a concentration of charge. Water molecules are strongly attracted to the ethoxide anion, an interaction that limits the mobility of the Resonance Structures Resonance Hybrid water molecules and results in a decrease in entropy ($\Delta \mathrm{S}$ is negative). In contrast, in acetate (acetic acid’s conjugate base), the negative charge is delocalized onto both oxygens (even though it is often drawn as if it was associated with one but not the other). We can illustrate this in two ways (or more!) by drawing arrows to indicate how the extra electron pair can move from one oxygen to the other; it looks like this ($\rightarrow$).
The actual structure has a partial negative charge on both oxygens. This pair of structures is often referred to as a resonance structure and the process is termed resonance but the name is misleading. In fact the actual structure, the resonance hybrid, does NOT involve the electrons (and the double bond) moving back and forth between the two oxygen atoms. By a biological (and not completely sensical) analogy we might say it is a mule or a hinny—the offspring of a cross between a horse and a donkey.[6] Just as a mule (or a hinny) is not bouncing back and forth between being a horse and being a donkey, so the resonance hybrid actually exists as a new species[7], with an actual structure that is partway between the two (drawn) resonance structures. In this case, we are using two bonding models (a valence bond and a delocalized molecular orbital model) to describe the structure of acetate anion. The localized valence bond model involves a sigma single bond framework that connects the atoms and provides the molecular shape. The delocalized molecular orbital model describes a pi bond that connects both $\mathrm{O}$s to the $\mathrm{C}$. We can visualize the anion as a planar $\mathrm{sp}^{2}$ hybridized carbon connected to a methyl group and two oxygens by sigma bonds together with a 3 atom two electron pi bond that extends over the $\mathrm{O-C-O}$ framework ($\rightarrow$). The result is that in the acetate ion the negative charge is delocalized over two oxygens, rather than being concentrated on only one atom as it is in the ethoxide ion. The result is that the interactions of the acetate with solvent water molecules is not as strong, so that the water molecules are not as ordered, meaning that the water is not as ordered around the molecule and the entropy change is not as negative. The effects of delocalizing charge over more than one atom play a major role in predicting the outcomes of a wide range of reactions. We note that $\Delta \mathrm{S}$ is still negative since the creation of a charged species still leads to increased ordering of solvent molecules.
One way to predict whether charge can be delocalized is to determine whether resonance structures can be drawn for the charged species. For example: try convincing yourself that you cannot draw resonance structures for ethanol.
Resonance is not the only way to stabilize charge. Typically, resonance occurs through aconjugated pi bond system, such as occurs within the $-\mathrm{CO}_{2}^{-}$ part of an organic acid, but how do we account for the difference in the acidities of acetic acid ($\mathrm{pK}_{\mathrm{a }} 4.8$) and trifluoroacetic acid ($\rightarrow) (\(\mathrm{pK}_{\mathrm{a }} 0.5$), even though they both have the carboxylate functional group? The difference between the two lies in the fact that the charge on the trifluoroacetate anion is delocalized by two distinct mechanisms. As in acetate, the negative charge is delocalized by resonance through the pi bonding system; in addition it is also delocalized onto the fluorines by the fact that the highly electronegative fluorine atoms (more electronegative than $\mathrm{O}$) withdraw electrons from the methyl carbon through the sigma bonds, which in turn withdraws electrons from the next carbon, and in turn from the two oxygens (a process known as “induction”). The result is that the negative charge is “smeared out” over even more atoms, making the anion even less likely to cause a solvent molecule ordering (reducing the effect on $\Delta \mathrm{S}$). As you might expect, the inductive effect is distance dependent (perhaps you can predict the effect of adding more $\mathrm{CH}_{2}$ groups between the $\mathrm{CF}_{3}$ and $\mathrm{CO}_{2}$ groups).
Questions to Answer:
• Using resonance structures predict which is more acidic: $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$?
• Draw structures to show how sodium ethoxide and sodium acetate are solvated in water, and use them to show why the negative entropy change for the formation of sodium acetate is smaller than that of sodium ethoxide.
• Consider the $\mathrm{pK}_{\mathrm{a}}$‘s of the three chlorobutanoic acids: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}\left(\mathrm{pK}_{\mathrm{a}}\right. \text { 2.86) }$, $\mathrm{CH}_{3} \mathrm{CHClCH}_{2} \mathrm{COOH}\left(\mathrm{pK}_{\mathrm{a}}\right. \text { 4.05) }$, and $\mathrm{CH}_{2} \mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\left(\mathrm{pK}_{\mathrm{a}}\right. \text { 4.53) }$. Draw structures and use them to explain why these carboxylic acids have different $\mathrm{pK}_{\mathrm{a}}$‘s.
Organic Bases
As noted previously, there are no acids without bases, and vice versa. Even if we are only discussing $\mathrm{H}^{+}$ (proton) transfer, it is (arguably) easier to think about the base using a Lewis model. That is, a base has an electron pair available for donation into a bond with the acid. Recall that almost everything that has a pair of non-bonding electrons (sometimes called a lone pair) can act as a base. The most common types of organic bases often have a nitrogen atom somewhere in their structure. If we compare the basicity of $\mathrm{N}$, $\mathrm{O}$ and $\mathrm{F}$, each of which have lone pairs that are could potentially be donated, nitrogen is the least electronegative and therefore the best able to donate its electrons into a bond, since its lone pair is least attracted by the nucleus. Fluorine, the most electronegative element, holds its electrons very close to the nucleus, and under normal circumstances would not be considered as a base.
Oxygen, since it is more electronegative than nitrogen is not as strong a base, therefore when ammonia and water are mixed, the only reaction that occurs (and that to a relatively small extent) is a proton transfer from water to ammonia. $\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \leftrightarrows \mathrm{NH}_{4}^{+}+{ }^{-} \mathrm{OH}$
The equilibrium constant for this reaction is $1.8 \times 10^{-5}$ (most of the species in the mixture at equilibrium are reactants)
*insert image here*
Here are some organic bases ($\rightarrow$). Note that they are components of a wide range of biologically active molecules, including DNA, hormones and pharmaceuticals. As we will see the basic nitrogen provides an important way to understand the reactivity of a particular species.
For now, however, let us start with a simpler base such as methylamine ($\mathrm{CH}_{3} \mathrm{NH}_{2}$) the simplest nitrogenous organic base. Methylamine reacts with acids ($\downarrow$) in much the same way that ammonia does; it will react with a strong acid like $\mathrm{HCl}(\mathrm{aq})$ to produce methylammonium chloride.
Recall that the position of equilibrium can be predicted by comparing the strength ($\mathrm{pK}_{\mathrm{a}}$’s) of the two acids. $\mathrm{HCl}\left(\mathrm{pK}_{\mathrm{a}}-7\right)$ is a much stronger acid than $\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\left(\mathrm{pK}_{\mathrm{a}} \sim 10\right)$ and therefore we predict that the equilibrium of the methylamine + $\mathrm{HCl}$ reaction will lie well to the right. Now consider the reaction in which methylamine reacts with acetic acid ($\downarrow$).
Again we can predict the position of equilibrium by comparing $\mathrm{pK}_{\mathrm{a}}$’s of the conjugate acids (acetic acid $4.8$ and $\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+} \sim 10$). Notice that you can predict the structure of the products simply by following the flow of electrons. We could change the $\mathrm{CH}_{3}$ (methyl) groups on either methylamine and acetic acid to a wide range of different groups and still be able to predict the product easily, as long as you recognize that the reaction that takes place is a (simple) proton transfer (acid–base). For example, look at the structure of cocaine (above): can you predict what will happen if it were reacted with acetic acid? What would be the structure of the product?
Molecules that contain both an acid and a base:
The most common example of a molecule that act as both an acid and a base is of course water because it has both a potentially acidic hydroged, and a lone pair that can accept the proton. However, since this is organic chemistry, where water is not as common a solvent, let us consider the class of molecules that have both acidic and basic domains simultaneously. The most biologically important such molecules are the amino acids, which have both an amino group and a carboxylic acid. A subset of the possible amino acids are those used in biological systems to assemble polypeptides. Amino acids (or rather the $\alpha$-amino acids) contain both a carboxylic acid and an amino group attached to a central carbon (the $\alpha$-carbon). The generic structure is given here ($\rightarrow$) where R stands for a wide range of side chains.[8]At $\mathrm{pH } 7$ the amino acid exists in what is know as a zwitterionic form, in which the carboxylic acid group is negatively charged while the amino group is positively charged. At no time would an amino acid (dissolved in water) exist in an un-ionized form. We can predict what form would be present at different $\mathrm{pH}$’s by considering the $\mathrm{pK}_{\mathrm{a}}$‘s of the species involved. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/01%3A_AcidBase_Reactions/1.02%3A_Acid-Base_Reaction_Direction_and_Position_of_Equilibrium.txt |
So far, we have discussed situations when the acid or base is added to solution of pure water. Pure water has a $\mathrm{pH}$ of 7, and $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[{ }^{-} \mathrm{OH}\right]$. Now let us consider what happens when we change the $\mathrm{pH}$ of the solution. For example consider a situation in which we dissolve a simple organic acid ($\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}$, acetic acid) in a solution that has a $\mathrm{pH}>7$, that is where the $\left[{ }^{-} \mathrm{OH}\right]>\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$; under these conditions the extent of the acetic acid’s ionization is increased. Recall that in 1M acetic acid only $\sim 3%$ of the acid is ionized at $\mathrm{pH } 7$. If we change the solution to make it basic by adding $\mathrm{NaOH}$, the excess of strong base (${ }^{-} \mathrm{OH}$) will completely deprotonate the acid. At equilibrium, the reaction will now favor products over reactants (i.e. it will move to the right). What we have done here is drive the acetic acid $\leftrightarrows$ acetate reaction to the right, increasing the concentration of acetate, which is an application of Le Chatelier’s principle). Note that $\mathrm{Na}^{+}$, derived from the addition of the $\mathrm{NaOH}$ used to adjust the $\mathrm{pH}$, is present but does not take part in the reaction – for this reason it referred to as a “spectator ion”. Another, perhaps simpler, way to predict the outcome of this reaction is to use the $\mathrm{pK}_{\mathrm{a}}$ values of the two acids ($\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, 4.8$ and $\mathrm{H}_{2} \mathrm{O}, 14$), clearly acetic acid is a much stronger acid than water, and therefore the equilibrium position for this reaction will lie over to the right in favor of the weakest acid and the weakest base. What we have done here is change the acetic acid, which is a polar organic molecule, into acetate, an ionic species.
Acetic acid is a small organic molecule; since it is polar it can interact with water (though intermolecular forces), therefore acetic acid is very soluble in water (indeed it is miscible with water (it has unlimited solubility.)[9] But now, let us consider the effect of increasing the length of the hydrocarbon group of the organic acid on its molecular properties. Acetic acid has a methyl ($\mathrm{CH}_{3}-$) group, the smallest possible hydrocarbon. In contrast dodecanoic (lauric) acid has a 12-carbon hydrocarbon chain ($\mathrm{CH}_{3}\left[\mathrm{CH}_{2}\right] 11 -$) and has a solubility in water of $0.063 \mathrm{g} / \mathrm{L}(\sim 30 \mathrm{mM})$ at $25^{\circ} \mathrm{C}$, which is much less that of acetic acid.[10] As the hydrocarbon (non-polar) part of the molecule increases in length, solubility in water decreases: the $\Delta \mathrm{G}$ of the process of dissolving the organic acid in water becomes more positive. This decrease in solubility is primarily due to a negative entropy change ($\Delta \mathrm{S}$) caused by the self-organization of water molecules around the hydrocarbon “tail” of the molecule. Now let us consider the behavior of ionized sodium dodecanoate (the sodium salt of dodecanoic acid); it, like many ionic species, it is soluble in water. Although as you may remember, this is a different form solubility – the soluble species is not isolated molecules but rather molecular complexes known as micelles ($\rightarrow$).[11] The upshot of this is we can “solubilize” organic acids in water by deprotonating them, but if we then lower the pH, the organic acid will separate from solution again.
Organic bases can be solubilized in a similar way, except that now the solution must be made acidic. For example, a nitrogenous base with a large non-polar group such as dodecylamine ($\mathrm{C}_{12} \mathrm{H}_{27} \mathrm{~N}$) has a solubility of about $3.5 \mathrm{~g} / \mathrm{L}(\sim 20 \mathrm{mM})$, but at acidic $\mathrm{pH}$s it is completely soluble. Contrast the solubility of dodecyl amine with cadaverine ($\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$), the compound that smells like its name, which is completely miscible with water because it has two polar amino groups. It turns out that we can predict the pH at which a particular acid or base protonates or deprotonates. You may recall from general chemistry that the pH of weak acids and their conjugate bases (like most organic species) can be described using the Henderson Hasselbalch equation ($rightarrow$).
One way to work with this equation is to note that $[\text { acid }]=\left[\text { conjugate } p H=p K_{a}+\log \frac{[\text { base }]}{[\text { acid }]} \text { base }\right]$ when the $\mathrm{pH}$ of the system is equal to $\mathrm{pK}_{\mathrm{a}}$ of the acid. At $\mathrm{pH}$’s below $\mathrm{pK}_{\mathrm{a}}$, $\text { [acid] }>\text { [conjugate base] }$, there is more acid than base, and vice versa for $\mathrm{pH}>\mathrm{pK}_{\mathrm{a}}$. Therefore, by adjusting the $\mathrm{pH}$ we can change the concentrations of conjugate acid and base to suit our purposes, or we can predict the relative concentrations at any $\mathrm{pH}$. For example, acetic acid with a $\mathrm{pK}_{\mathrm{a}}$ of $4.8$ would have $50 \% \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, and $50 \% \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \mathrm{Na}^{+}$, at a $\mathrm{pH}$ of $4.8$. If the $\mathrm{pH}$ falls below $4.8$ the concentration of protonated acid will increase, and if it rises the concentration of acetate ion will increase.
This ability to transform an organic substance from an insoluble (in water) molecule to a soluble ionic species can be very useful. One common example stems from the fact that many pharmaceutical drugs are organic substances that are insoluble in aqueous solutions (like cytoplasm or blood). If these substances were introduced into the body in their non-ionized form they would not dissolve, and therefore be inactive. If you check the labels on many prescription bottles you will see that the drug is administered as a salt. Consider norepinephrine ($rightarrow), a hormone that is often administered intravenously to counteract the effects of allergic reactions. It is administered as a salt of tartaric acid to ensure that it is soluble in the blood stream. You may come across another example of this phenomenon (that acids are soluble in basic solution, and bases are soluble in acid solution) if you take the organic chemistry laboratory course. If your product has an acidic or basic moiety in its structure, you can extract the substance into aqueous (acid or basic) solution, washing away all the organic by-products with an organic solvent, and then regenerating the acidic or basic substance. This is an important purification method for many substances, because it allows the compound of interest to be separated into aqueous solution and then regenerated simply by adding or subtracting a proton. When we consider biomolecules (that is, organic molecules found in organisms) the situation is not so clear cut; most biomolecules have a variety of acidic and basic groups as part of their structure. Even the simplest amino acid, glycine (\(rightarrow$) exist in a variety of protonated and deprotonated forms depending on the $\mathrm{pH}$.
One thing that becomes clear is that individual amino acids are always charged regardless of the $\mathrm{pH}$, so they are water-soluble. But the extent of the protonation/deprotonation reactions is $\mathrm{pH}$ dependent. As we will see this has a number of ramifications for a wide range of biological molecules, because they will behave very differently in different pH solutions. This is one important reason why most biological systems are buffered so that they remain at a fairly constant $\mathrm{pH}$.
Questions to Answer:
• If you have a mixture of benzoic acid $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{pK}_{\mathrm{a}}\right. \text { 4.2) }$, toluene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$ and aniline hydrochloride ($\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+} 4.6$). Which substance will be soluble in aqueous acidic solution, which will be soluble in aqueous basic solution, which will not be soluble in water?
• Outline a scheme for separating these three substances by using their differing solubilities in organic and aqueous solutions of different $\mathrm{pH}$s. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/01%3A_AcidBase_Reactions/1.03%3A_Effect_of_pH_on_Acid_Base_Reactions.txt |
As we have seen, any reaction in which a proton ($\mathrm{H}^{+}$) is transferred from one molecule to another can be considered as a Lewis acid–base reaction, but now it is time to broaden the scope of Lewis acid–base reactions. The structural requirement for a Lewis base is essentially the same as those we discussed for a Brønsted base. That is, a Lewis base must have an accessible lone pair of electrons that can be donated into a bond with a Lewis acid. For example, many (but not all) nitrogen and oxygen containing molecules have such available lone electron pairs and so can be considered as Lewis bases.[12] It is the Lewis acid that can take a number of different forms (and so, can be harder to recognize). A Lewis acid must be able to accept a pair of electrons. In practice this means a variety of substances (besides $\mathrm{H}^{+}$) can act as Lewis acids: for example, any species with empty orbitals that are energetically accessible can be a Lewis acid. Common examples of this situation are compounds of Group III elements (specifically $\mathrm{B}$ and $\mathrm{Al}$); these have only three valence electrons. Examples include $\mathrm{BF}_{3}$ ($\uparrow$) and $\mathrm{AlCl}_{3}$,[13] both of which have a partial positive charge on the central atom and an empty orbital that can accept electrons.
Other examples of Lewis acids are metal ions such as $\mathrm{Fe}^{2+}$, $\mathrm{Fe}^{3+}$, $\mathrm{Cu}^{2+}$, and $\mathrm{Mg}^{2+}$; these, by definition, have empty orbitals. The same situation holds true for many transition metal salts, for example $\mathrm{TiCl}_{4}$ and $\mathrm{NiCl}_{2}$.[14] In biological systems, examples of Lewis acid–base complexes include the active site of the oxygen transport complex in hemoglobin (and myoglobin), which consists of an iron ion complexed with 4 nitrogens, which are part of a porphyrin ring. A similar iron-porphyrin complex is found associated with the cytochrome proteins that participate in the ATP synthesis reaction associated with in the electron transport chain of the mitochondria ($\rightarrow$). Chlorophyll, the green pigment that is part of the light capture system in algae and plants has a similar structure, except that the Lewis acid at the center of the complex is $\mathrm{Mg}^{2+}$ rather than $\mathrm{Fe}^{2+}$. This has the interesting effect of making chlorophyll species appear to be green, rather than the red observed in blood. This is caused by the difference energy gaps between the molecular orbitals in an $\mathrm{Fe}$ complex as compared to a $\mathrm{Mg}$ complex with a porphyrin ring. We will discuss this effect in more detail later. As we will also see later, Lewis acids are important class of reagents in organic chemistry because they can interact with a wide range of bases.
Electrophiles and Nucleophiles
The next logical step in expanding our ideas about Lewis acids and bases is to consider reactions that involve carbon. We will first consider reactions in which carbon acts like the Lewis acid, that is, it accepts a pair of electrons to form a new bond with a Lewis base. So, what situations would we make a carbon act in this way? We can rule out (for now) carbon compounds with an empty orbital (akin to boron). Why? Because all stable carbon compounds form 4 bonds and there are no low-lying empty orbitals that can be used to accept electrons.
But let us first look at the proton ($\mathrm{H}^{+}$) transfer reaction as a model ($\rightarrow$). In this case the bond with the Lewis base ($\mathrm{OH}^{-}$) is formed at the same time as the bond to the conjugate base (of the acid) is broken. We see that the $\delta^{+}$ on the $\mathrm{H}$ means that the bond to the $\mathrm{H}$ is partially ionized. The $\mathrm{H}$ is “on the way” to becoming $\mathrm{H}^{+}$—a species that does have an empty and accessible orbital. The $\delta^{+}$ on the $\mathrm{H}$ attracts the negative (or $\delta^{-}$) charge on the base, and the reaction is initiated, forming a new bond between the $\mathrm{O}$ and the $\mathrm{H}$, and at the same time breaking the old $\mathrm{O-H}$ bond.
We can imagine that a carbon compound with a $\delta^{+}$ on the $\mathrm{C}$ might behave in a very similar way. In this molecule ($\mathrm{H}_{3} \mathrm{CBr}$) the $\mathrm{C-Br}$ bond is polarized so that there is a small positive charge on the $\mathrm{C}$, which attracts the negatively charged hydroxide ($\rightarrow$). Formation of the $\mathrm{O-C}$ bond occurs with the simultaneous breaking of the $\mathrm{C-Br}$ bond.
Consider the analogies between these two reactions – the mechanisms of how and why the electrons move are similar. The only real difference between the two reactions is that in the first an $\mathrm{H}^{+}$ is transferred from an $\mathrm{C}$ (on the carboxylic acid) to the $\mathrm{OH}^{-}$, while in the second, a methyl group is transferred to the $\mathrm{OH}^{-}$. Now for a change in nomenclature: when such a reaction involves a $\mathrm{C}$ atom (a carbon center) rather than call the electron deficient carbon a Lewis acid, we call it an electrophile (electron or negative charge loving). Similarly, the hydroxide ion (which acts as a Lewis base) is now called a nucleophile (positive charge loving). This change in terminology is not just to confuse students! In fact, there are subtle differences between Lewis acids and bases and electrophiles and nucleophiles that make the distinction between the two useful. In particular, while all Lewis bases are nucleophiles, as we will see, not all nucleophiles are bases.
So now we have to ask ourselves, what factors make a particular $\mathrm{C}$ within a molecule an electrophile? How can we recognize a nucleophile? What criteria do we use to estimate the strength of a particular electrophile or an nucleophile? Can we ever get carbon to act as a nucleophile? If we can answer these questions, we can predict the outcome of a wide array of reactions.
What makes a particular carbon an electrophile?
The simplest of organic compounds are hydrocarbons, and the simplest of hydrocarbons are known as alkanes. Alkanes typically have the formula $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}$ (or if $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}$ if there is one ring of carbons, subtract $2 \mathrm{H}$ for every extra ring). All of the bonds within an alkane are sigma (single) bonds; they do not contain pi (double) bonds.[15] In an alkane, each carbon is fully saturated, it makes four single bonds and (as noted above) there are no double or triple bonds. $\mathrm{C-C}$ bonds are of course, completely non-polar since the electrons are equally distributed between two identical atoms, however $\mathrm{C-H}$ bonds are also relatively non-polar since the electronegativities of $\mathrm{C}$ and $\mathrm{H}$ are quite similar. In practice this means that alkanes are limited in their reactivity. The most common reactions that an alkane can take part in are reactions with oxygen to produce $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$. This reaction is highly exothermic, although there is a significant activation energy, so it requires an initial input of energy (typically a spark, a burning match) to start the reaction, but then the energy from the formation of the strong $\mathrm{C=O}$ and $\mathrm{O-H}$ bonds (which is why the reaction is exothermic) can be used to initiate more reaction. The actual reaction mechanism is complex; it proceeds via a series of highly reactive (unstable) free radicals (species with unpaired electrons)[16]. While this reaction is obviously highly important—this is still how we generate much of the energy to run our cars and electrical power stations, from an organic chemistry perspective it is not very interesting in large part because it is more or less uncontrollable. That is, if you have enough oxygen once started the reaction generates $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$, regardless of which hydrocarbon you begin with. [17]
All this is to say that alkanes are not good candidates for the kinds of reactions we are considering, they have neither nucleophile nor electrophilic carbons. So, let us turn our attention to carbon compounds with elements other than $\mathrm{C}$ and $\mathrm{H}$ and both sigma and pi bonds (this is, of course, the rest of organic chemistry). Here we find a very different situation: the range of reactions and the types of products can seem almost unlimited. While it is impossible (and certainly undesirable) to memorize every reaction and every potential product, it is possible to organize your understanding of chemical systems so that you can make plausible predictions as to which reactions may occur. By knowing reaction mechanisms, and when they are relevant, you can also predict which reactions will occur and therefore what products will form. As you might recognize, this is the same strategy we have used to consider acid–base reactions, which can be understood much more broadly than simple proton ($\mathrm{H}^{+}$) transfer reactions. Thinking in an electrophile-nucleophile context provides an entrée into much of organic chemistry.
For reactions (other than reactions involving free radicals, like combustion) to occur, there is generally a “handle” within the substrate: a place where the electron density is not evenly distributed, a site at which reactants of opposite charge interact (and react). In the example we used previously, the electrophilic carbon has a $\delta^{+}$ on it; this partial charge arose because the $\mathrm{C}$ was bonded to a more electronegative element. Such a partially positively charged $\mathrm{C}$ is attractive to any species with a negative (or partial negative) charge. Note that, for now, we are going to restrict the type of carbon atom that we are considering to either a primary (that is a carbon with only one alkyl group (denoted by $\mathrm{R}$) and 2 hydrogens, $\mathrm{CH}_{2} \mathrm{R-}$) or a methyl carbon ($\mathrm{CH}_{3}-$). As we will see things become more complicated when we start to add more alkyl groups around the site of attack—so we will come back to that later.
To identify such a partially positively charged $\mathrm{C}$ one would look for $\mathrm{C}$’s bonded to groups (atoms) that are more electronegative, that is, that will act to withdraw electrons from the carbon (denoted by $\mathrm{L}$ below). But since carbon cannot form more than four bonds as the nucleophile comes in and forms a bond, another bond must break. The electronegative atom ($\mathrm{L}$) (or group of atoms), is known as the “leaving group” (oh, how dull) needs to be stable when it leaves with the extra pair of electrons. We can, in fact, predict the characteristics of a good leaving group. For example, the bond to the leaving group should be polarized, and since the leaving group takes the electron pair with it, the group should be stable with this extra pair of electrons on it ($\mathrm{L-}$). Another way of saying this is that the leaving group should be electronegative and breaking the $\mathrm{C-L}$ bond should produce a weak base. Halide ions are examples of good leaving groups, and their order of reactivity is $\mathrm{I-}>\mathrm{Br-}>\mathrm{Cl-}>\mathrm{F-}$. This ranking mirrors their acid strength rankings—that is, $\mathrm{HI}$ is the strongest acid and $\mathrm{HF}$ is the weakest—which means $\mathrm{F-}$ is the strongest base (and therefore least likely to leave)
So, what about oxygen, in the form of an alcohol $\mathrm{O-H}$ group, as a leaving group? ($\rightarrow$) It certainly fulfills the requirement that the $\mathrm{C-O}$ bond be polarized, but if you follow the reaction through it would mean that the leaving group would be a hydroxide ion ($\mathrm{-OH}$), a very strong base. Therefore, alcohols ($\mathrm{ROH}$) are not likely to be attacked by a nucleophile.
There are ways, however, ways to make an alcohol reactive. For example, if we can carry out the reaction in an acidic solution, the alcohol will be protonated (at least some of the time), and therefore the leaving group will be a water molecule, a stable entity ($\rightarrow$).
What makes a good nucleophile?
As we have noted, a Lewis base is also a nucleophile, so the trends you have learned about the strengths of Lewis bases also hold for nucleophiles. So, for example, nucleophilicity decreases across a row in the periodic table so $\mathrm{NH}_{3}>\mathrm{H}_{2} \mathrm{O}>\mathrm{HF}$ in the same way as base strength does (recall this is because the lone pair is more available on the $\mathrm{N}$ than on $\mathrm{F}$). But since this is organic chemistry, we should have some organic groups dangling off the nucleophiles. So for example, instead of a hydroxide nucleophile, we could use an alkoxide nucleophile (for example, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O-Na}^{+}$ sodium ethoxide), or we could use amine nucleophiles like serotonin (the nitrogen in the $\mathrm{NH}_{2}$ group here is more nucleophilic than the $\mathrm{OH}$ group, and the $\mathrm{N}$ in the ring). In addition, if we compare nucleophiles with the same nucleophilic atom, a negatively charged species is more nucleophilic than the uncharged form, so $\mathrm{OH-}>\mathrm{H}_{2} \mathrm{O}$, and $\mathrm{NH}_{2}\mathrm{-}>\mathrm{NH}_{3}$ (and by analogy any organic derivatives behave the same way).
Besides the nucleophiles that are easily recognizable because they are bases, there is another class of nucleophiles that are somewhat different; they have a lone pair of electrons, but they are not particularly basic. The most common examples are the halide ions, which are weak bases and good leaving groups. So, the question arises: why are halide ions such good nucleophiles? The reason for this has to do with their polarizability (that is, the extent to which an electron cloud can get distorted) of the nucleophile. A very large anion-like iodide has a very polarizable electron cloud because the electrons extend much further out from the nucleus than, for example, the electron cloud in fluoride. This means that the electron cloud for iodide can begin partial bond formation to the carbon much earlier than the one for fluoride, and therefore iodide reacts much faster than fluoride.[18] This logic allows us to explain why the nucleophilicity of halide ions increases as you go down a group: $\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{F}^{-}$.
Although we will return to this reaction in greater detail later, let us take a look at the range of possible reactions that this generic scheme enables us to predict – with the caveat that we are considering simple carbon substrates. Reactions like this are called nucleophilic substitutions, because the species that attacks the carbon is a nucleophile, and the overall effect of the reaction is that we have substituted the nucleophile for the leaving group. This particular example is called an $\mathbf{S}_{\mathrm{N}}\mathbf{2}$ reaction which stands for Substitution, Nucleophilic, 2nd Order, and we will come back to discuss the reaction in much more detail later.
Another type of carbon nucleophile
The $\mathrm{S}_{\mathrm{N}}\mathrm{2}$ reaction is a mainstay of organic chemistry, by varying the substrate (carbon electrophile) the leaving group, and the nucleophile we can construct a huge array of different compounds. Another very important type of compound that has an electrophilic carbon (i.e. a carbon that is subject to nucleophilic attack) is one which contains a carbonyl group ($\mathrm{C=O}$). The carbonyl group is highly polarized, with a large $\delta^{+}$ on the carbon. This can be rationalized by the idea that there are two bonds to the electronegative oxygen and therefore the oxygen has even more tendency to pull electrons away from the carbon than a single bonded oxygen would. One way to visualize this is to draw resonance structures for the carbonyl group as shown, where the electrons from the double bond are now located on the $\mathrm{O}$. We will come back to how to draw resonance forms in much more detail later.
Once we understand how compounds with carbonyl groups are polarized, we can predict (at least for the first step) how these compounds will react. For example, if we have a reasonably good nucleophile (here shown as $\mathrm{Nu-}$)we might predict that it would attack at the carbonyl carbon. The difference in this reaction and an $\mathrm{S}_{\mathrm{N}}\mathrm{2}$ reaction is that there is no leaving group. Instead the electrons from one of the $\mathrm{C-O}$ bonds move onto the oxygen as shown.
There are a number of ways that this reaction can continue, the most obvious of which is that if the reaction is in contact with a solvent that has acidic protons (e.g. water or an alcohol), the $\mathrm{O}^{-}$ can simply protonate in an acid base reaction. As we will see later, the course of the reaction also depends on what the nucleophile is. Here we will give the simplest example which is the reaction of a ketone (acetone) with a carbon nucleophile ($\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Li}$, ethyl lithium). For now, we will not worry about how to make ethyl lithium, but rest assured it is possible! When the negatively charged carbon electrophile adds to the carbonyl we make a new carbon-carbon bond. This is followed by addition of water to protonate the oxygen, to produce an alcohol. The overall reaction is a nucleophilic addition.
Questions to Answer
Try your hand at predicting the outcomes for these reactions by drawing arrow pushing mechanisms.
• $\mathrm{CH}_{3} \mathrm{CH} 2 \mathrm{I}+\mathrm{NaOH} \rightarrow$
• $\mathrm{CH}_{3} \mathrm{Br}+\mathrm{NaN} 3 \rightarrow$
• $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{NH}_{2} \mathrm{CH}_{3} \rightarrow$
• $\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+}\rightarrow$
What nucleophile and electrophile would you react together to form these products?
• $\mathrm{CH}_{3} \mathrm{OH}+\mathrm{Br}^{-}$
• $\mathrm{CH}_{3} \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{Cl}^{-}$
• Construct a generalizable model for the $\mathrm{S}_{\mathrm{N}}\mathrm{2}$ reaction and explain the role of the substrate (the carbon electrophile), the leaving group, and the nucleophile.
Construct a generalizable model for the nucleophilic addition reaction and explain the role of the substrate (the carbon electrophile), and the nucleophile. What functional groups would undergo a nucleophilic addition?
• What would make a carbon in a compound a nucleophile? How could you go about making a particular carbon nucleophilic? | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/01%3A_AcidBase_Reactions/1.04%3A_Lewis_Acids_and_Bases_Electrophiles_and_Nucleophiles.txt |
1. If you did not take the CLUE general chemistry curriculum, we recommend that you take a look at the materials on the web here.
2. Here the nomenclature can be confusing, since when in solution $\mathrm{HCl}$, $\mathrm{NaCl}$, and $\mathrm{NaOH}$ exist primarily in their dissociated forms, i.e. $\mathrm{H}^{+}$ and $\mathrm{Cl}^{-}$, $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$, and $\mathrm{Na}^{+}$ and $\mathrm{OH}^{-}$.
3. In fact, later on we will use abbreviated notation to indicate the transfer of $\mathrm{H}^{+}$ from one molecule to another simply because there are so many reactions in which multiple proton transfers take place. However, you should always be aware that protons are always transferred (not simply ionized or dropped off).
4. We will have MUCH more to say about arrow pushing as we proceed, but it is good to get a head start on this skill.
5. The situation is similar to when we use different atomic and molecular structure models depending on our purpose. The most sophisticated of these models are based on quantum mechanical calculations, but in many cases, producing a calculated model (which considers each and every atom in a system) may be impossible or not even particularly useful; when a simpler model is adequate it makes sense to use it. As an example, when considering bonding within molecules we typically use the valence bond (VB) model in which we think of each bond as composed of two electrons that are attracted to both of the atomic nuclei involved in the bond. Alternatively we could use molecular orbital (MO) model, in which we consider bonding (and anti-bonding) orbitals that involve the entire molecule. Each MO contains a maximum of two electrons and the bonding interactions are considered over the whole molecule. While useful in certain contexts, the MO model is usually unnecessarily complex for everyday use. As we will see shortly, there are situations when we will choose to use one or another, or both models of bonding, depending on the circumstances.
6. Whether a mule or a hinny is produced from such a cross depends upon whether the horse in the mother or father and is due primarily to parental imprinting of genes expression within the placenta: http://www.ncbi.nlm.nih.gov/pubmed/23754418
7. Please excuse this analogy, since hinnys and mules are typically sterile and so do not represent species: https://en.Wikipedia.org/wiki/Mule#Fertility
8. There are 20 naturally occurring amino acids that are encoded for by the genetic code (actually there are 22 but selenocysteine and pyrrolysine are used only in a very few microbes).
9. The implication is that we can take pure acetic acid (known as glacial acetic acid) which is 17.4 Moles/L and add any amount of water and have a solution of acetic acid in water (or perhaps water in acetic acid, depending on the relative amounts of two compounds).
10. For fuller explanation of this phenomenon, you should review the chapter in CLUE on solubility (chapter 6).
11. In a micelle, the non-polar groups are oriented towards the center of the structure while the ionic “heads” are in contact with the water solvent.
12. Nitrogenous compounds with non-available lone pairs include ammonium salts (where the $\mathrm{N}$ is bonded to four other atoms), and aromatic heterocyclic compounds, where the lone pair is part of the aromatic pi system.
13. In fact $\mathrm{AlCl}_{3}$(l) exists as a dimers $\mathrm{Al}_{2} \mathrm{Cl}_{6}$, but this structure decomposes and then reacts as $\mathrm{AlCl}_{3}$ in the presence of a Lewis base. Diborane $\mathrm{B}_{2} \mathrm{H}_{6}$ is also a dimer of $\mathrm{BH}_{3}$, that will break apart to react as $\mathrm{BH}_{3}$.
14. In general transition metal complexes and ions have empty d orbitals that are energetically accessible.
15. If there are pi bonds present the substance is not an alkane – we will get to that presently
16. There is an excellent (but simple) video on the mechanism of reactions that produce flames such as this one. https://vimeo.com/40271657
17. In fact there is a great deal of interest in controlling this reaction—that is, finding ways to take hydrocarbons and selectively introducing a functional group—rather than fully oxidizing the hydrocarbon. http://pubs.acs.org/doi/full/10.1021...entsci.6b00139
18. Another very important factor in understanding nucleophile strength is the structure of the solvent. We will return to this idea later. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/01%3A_AcidBase_Reactions/1.05%3A_In-Text_References.txt |
Spectroscopy is the study of how energy (particularly electromagnetic radiation) and matter interact. By analyzing these interactions, we can infer a great deal of evidence about the structure of matter. In organic chemistry, spectroscopy allows us to determine the structure of products and reactants (and in some cases we can also get information about intermediates of reactions). By using spectroscopy of different kinds, we can gather evidence about reaction rates and from this information we can infer mechanisms of reactions. In this chapter, we will briefly review the background that you (probably) learned in general chemistry, and then we will move on to the various type of spectroscopy and discuss what each type can and cannot tell us about the structure of organic compounds.
From your earlier studies, you will recall that at the atomic/molecular level, the energies of atoms and molecules are quantized: that is, there are discrete, separate energy states with nothing in between. This applies not only to the energies of electrons, but also to the energies of vibration of bonds and rotation around bonds. Nuclear energies are also quantized. It is possible to switch from one energy state to another by absorbing or emitting an amount of electromagnetic energy (a photon) that corresponds to the energy difference between the two states. Photons with the energy that does not correspond to differences between two energy states are not absorbed or emitted, which means that we can use the energies of the photons absorbed or emitted to tell us about the energy differences between quantum states in atoms and molecules. The differences between these quantized energy levels is highly dependent on the identity and environment of the atoms and molecules, and therefore we can the photons emitted to identify particular species. For example, in isolated atoms the energy differences between levels often correspond to electromagnetic energy in the ultraviolet or visible. This results in the atomic absorption or emission spectra that allow us to determine what elements are present in (for example) stars, and interstellar space.
02: Spectroscopy- how we know what we know about the structure of matter
Now we will extend this idea from atoms to molecules. Just as electrons occupy atomic orbitals in atoms, the electrons in molecules occupy molecular orbitals. $\left. A+hv \rightarrow A^{*}(\text { excited state }) \rightarrow A \text { (ground state }\right)+hv$
As with atomic orbitals, electrons in molecular orbitals can absorb or release photons of a specific energy as they move from one molecular orbital to another. However, there is a significant difference between the absorption/emission process in isolated atoms (or ions) versus that of molecules. When an electron is promoted to a higher energy level in an atom, the product is an atom in an excited state—generally the excited atom (or ion) will decay back to the ground state by emitting a photon ($\downarrow$). However, when an electron within a molecule is excited it moves (or is “promoted”) from its original molecular orbital to another. Now there are a number of different consequences that can occur. For example, if the electron absorbs a photon and is promoted from a bonding molecular orbital to an antibonding orbital, the result will be that the bond will break, since there is now no overall stabilizing interaction. Consider $\mathrm{H-H}$, which is the simplest possible molecule. The set of molecular orbitals for hydrogen includes a $\sigma$ (sigma) bonding and a $\sigma^{\star}$ antibonding orbital. In the ground (or lowest-energy) state, molecular hydrogen has a $\sigma$ bonding orbital containing both of the molecule’s electrons. If one of the bonding electrons absorbs a photon that has just the right amount of energy (the energy difference between the bonding and antibonding orbital) it will be promoted and move into the destabilized antibonding orbital—causing the bond between the atoms to break because there is now no overall bonding interaction. As you might imagine, if chemical bonds were susceptible to breaking merely by being exposed to low-energy electromagnetic radiation (such as that of visible light) the world would be a different (and rather boring) place. For example, life would not be possible since it depends upon the stability of molecules. In fact, the energy of the photons required to bring about bond-breaking is quite large. For example, the energy required to break an $\mathrm{H-H}$ bond (the bond energy) is $436 \mathrm{kJ} / \mathrm{mol} \text {. }$. If you calculate the wavelength of a photon that could deliver this amount of energy, the amount of energy required to break one $\mathrm{H-H}$ bond would be in the far UV section of the electromagnetic spectrum (~280nm). The typically strong covalent $\sigma$ (or single) bond requires quite high-energy photons to break them.
In fact, the Earth’s atmosphere blocks out most ($>98 \%$) high-energy (ultraviolet) photons and most biologically important molecules cannot absorb visible light, so that leaves us with the question of why is there a need for sunscreen, which filters out the UV A (400 - 315 nm) and UV B (315 - 280 nm) photons. The answer is that a number of biological molecules contain more than simple sigma bonds. For example, most complex biological molecules also contain $\pi$ (pi) bonds and non-bonding electrons in addition to $\sigma$ bonds; transitions between these orbitals may be observed. The energy gaps between these different orbitals are quite are smaller than the $\sigma-\sigma$* energy gap. Photons with enough energy to cause these electron transitions are present in sunlight. For example, a double bond typically involves both a $\sigma$ and a $\pi$ bond. Absorption of a photon that would promote an electron from a $\pi$ bonding orbital to a $\pi^{\star}$ anti-bonding orbital would have the effect of breaking the original pi bond. One way to represent this is shown here ($\rightarrow$). In this case, one of electrons that was in the pi bond is now in the high-energy pi* antibonding orbital and is far more reactive. Another way to think about it is that the electrons are now unpaired, and are much more likely to react to form a more stable entity.[1] An obvious way to regain stability is for the electron in the $\pi$ antibonding orbital to drop back down to the bonding energy level and emit a photon of the same energy, and in most cases this is what happens—ultimately causing no damage. (As we will see later, since double-bonds are rotationally constrained, another possible occurrence is that there can be rotation around the single [$\sigma$] bond) and then reformation of the $\pi$ bond, leading to an isomer of the original alkene). However, in some instances, if there is another potentially reactive species in proximity, reactions between molecules (or in the case of biological macromolecules, between distinct regions of these molecules) can occur and cause problems. For example: most of us are aware that exposure to the sun causes skin damage that can lead to skin cancer. A major mechanism occurs in DNA where two thymidine bases are adjacent to one another. A UV photon can be absorbed by a pi bond in one thymine base. This broken pi bond (and resulting unpaired electron) is very reactive. It can react with a pi bond in an adjacent thymine leading to a new bond, a reaction that produces a four-membered carbon ring, known as a thymine dimer. The DNA replication machinery cannot accurately replicate a sequence containing a thymine dimer, resulting in a change in DNA sequence—a mutation. Mutations of this type are a common early step in the generation of cancerous skin cells (carcinomas) and pigment cells (melanomas).[2]
A more benign example of photon absorption in biological systems underlies the mechanism by which we (and other organisms) detect light—that is how we can see things! While it was originally thought (at least by some) that vision involved rays emitted from eyes,[3] we now understand that to see we need to detect photons that are reflected or emitted by the objects around us. The process begins when the photons of light fall on cells known as photoreceptors. In our eyes, these cells are located within the retina: a sheet of cells that line the interior surface of the eye. Within a subset of retinal cells are a number of different types of molecules that contain pi bonds. These molecules are proteins known generically as opsins. An opsin is composed of a polypeptide (or apoprotein) that is covalently bound to another molecule, 11-cis-retinal.[4]
This molecule is derived from vitamin A (all-trans-retinal). The complex of apoprotein and retinal is the functional opsin protein. There are a number of different opsin components that influence the wavelength of the photons absorbed by the functional opsin protein. When a photon is absorbed, it promotes an electron from one of the retinal’s pi bonds to an antibonding orbital. Instead of reacting with another molecule, like thymine, there is a rotation around the remaining single (sigma) bond, and then the formation of a new pi bond, this leads to the isomerization of the original 11-cis form into the trans-isomer. This change in retinal shape influences the shape of the opsin protein which initiates a cascade of electrochemical events that carry signals to the rest of the brain (the retina is considered an extension of the brain) that are eventually recognized as visual input. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.01%3A_Interactions_of_Electromagnetic_Radia.txt |
One common recommendation from doctors is that we eat plenty of highly colored fruits and vegetables. The compounds that give these foods their strong color have a number of commonalities. For example, the compound that gives carrots and sweet potatoes their distinctive orange color is beta-carotene. You might well notice its similarity to retinal. The compound that contributes to the red color of tomatoes is lycopene. Molecules of this type are known generically as pigments.
The wavelengths at which a compound absorbs light depends on the energy gap between the orbitals that are involved in the transition. This energy gap is determined by the structure of the molecule. A molecule with only single bonds absorbs light at shorter wavelengths (in the high-energy UV), while more complex bonding patterns are associated with the absorption of visible light. For example, the presence of multiple pi bonds and their interactions within the molecule can affect the energy gap between the molecular orbitals. Recall our discussion of graphite. Rather than thinking of graphite as sheets of fused six-membered rings with alternating single and double bonds, we can think of each bond as a localized sigma bond and a delocalized pi bond. There are huge numbers of pi molecular orbitals spread over the whole sheet of carbon atoms. The more pi MO’s there are, the more the energy gap between these orbital decreases; that is, the less energy (longer wavelength light) is needed to move an electron from a pi to pi* orbital . In the case of network substances like graphite and metals, the energy gap between the orbitals becomes negligible, and we think of the bonding model as a band of molecular orbitals. In these cases, many wavelengths of light can be absorbed and then re-emitted which gives graphite and metals their characteristic shininess. In substances like lycopene or $\beta$-carotene we also find this pattern of alternating single and double bonds. We say that compounds with this pattern of alternating single and double bonds (e.g. $\mathrm{-C=C-C=C-}$) are conjugated, and we can model the bonding in the same way as graphite. There are pi MO’s that can extend over the region of the molecule, and the more orbitals there are, the closer together in energy they get.
For an isolated $\mathrm{C=C}$ double bond, the energy required to promote an electron from the pi to the pi* orbital corresponds to the light in the UV region (around 170 nm), but as the number of double-bonds that are conjugated (separated by single bonds) increases, the energy gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) decreases. Eventually, the wavelength of light needed to promote an electron from the HOMO to the LUMO moves into the visible region, and the substance becomes colored. (Note that it does not become the color of the light that is absorbed, but rather the remaining light that is transmitted or reflected). These conjugated regions of molecules are called chromophores.[5] The longer the conjugated section of the molecule, the longer the wavelength of light that it absorbs. You will notice that both lycopene and $\beta$-carotene contain large chromophore regions.
Samples of UV-VIS absorption spectra are shown here. Note that in contrast to the atomic absorption spectra we saw earlier (which consisted of sharp lines corresponding to the wavelength of light absorbed by atoms) these spectra are broad and ill-defined. In addition, you can see that the longer (larger) the chromophore, the longer the wavelength that is absorbed—and each of these compounds appears to be a different color. The fact that the peaks in these spectra are not sharp means that UV-VIS spectroscopy is typically not used for identification of compounds (see below for IR and NMR spectroscopy which can be used for this purpose). However, the amount of light absorbed is proportional to the concentration of the substance and therefore UV-VIS spectroscopy can be used to determine the concentration of samples.
Questions to Answer:
• Construct a representation that can help you explain why compounds with longer chromophores absorb lower energy photons than those with shorter chromophores
• Which compound do you think absorbs photons of the lowest energy light? Explain your reasoning, using a molecular orbital diagram to illustrate. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.02%3A_UV-Vis_Spectroscopy_and_Chormophores_.txt |
Up to now we have concentrated on the absorption (and emission) of energy associated with transitions of electrons between quantized energy levels. However, as we discussed earlier, electron energies are not the only quantized energies at the atomic/molecular level. In molecules, both vibrations and rotations are quantized, but the energies involved are much lower than those needed to break bonds. Let us begin with the simplest of molecular systems, consisting of two atoms bonded together. In such a system, the atoms can move back and forth relative to each other along the bond axis (vibrations). As they vibrate and change rotational speeds (and directions), the potential energy of the system changes (Why is that? What factors influence these changes?). There are also motions associated with rotations around bonds. But (weirdly, and quantum-mechanically) rather than being able to assume any value, the energies of these vibrations (and rotations) are also quantized. The energy gaps between the vibrational energy states tend to be in the range of infrared radiation. If three or more atoms are bonded together the molecule can also bend, changing the bond angle or the shape of the molecule.[6]
Infrared radiation is of lower energy than visible light (longer wavelength, lower frequency). You are probably familiar with IR heat lamps that are used for warming and night vision goggles that allow the wearer to “see” at night.[7] Recall that objects tend to emit radiation (the phenomenon is called black-body radiation) as the kinetic energy of the atoms and molecules in the object is converted to electromagnetic radiation. Around 300K (room temperature or body temperature) the radiation emitted is in the IR region of the spectrum.[8] Conversely, when IR radiation falls on our skin, we feel that as a warming sensation, mainly because it is causing the molecules in our skin to vibrate and rotate—increasing the kinetic energy and thus the temperature.
When we investigate the light absorbed or emitted as molecules undergo vibrational energy changes it is known as infrared spectroscopy.[9] Why, you might ask, are we interested in the vibrations of molecules? The vibrations, rotations, and bending movements of molecules are influenced by the structure of the molecule as a whole (as well as its environment). The result is that that many molecules and fragments of molecules have very distinctive IR absorption patterns that can be used to identify them. Infrared spectroscopy allows us to identify substances based on patterns in the lab and, for example, in interstellar dust clouds. The presence of quite complex molecules in space (hundreds of millions of light years away from earth) has been detected by the use of IR spectroscopy.
The types of changes that can be detected using IR spectroscopy are associated both with stretching (vibrations of) one particular bond, and with movements associated with three or more atoms such as bending or twisting. Consider methane for example: Each $\mathrm{C-H}$ bond can vibrate separately, but we can also imagine that they could vibrate “in phase” (at the same time) so that the $\mathrm{C-H}$ bonds lengthen and shorten at the same time. This movement is called the symmetric stretch. Conversely, we can imagine that one might lengthen as the other shortens: this is called the asymmetric stretch. Furthermore, the molecule can bend and twist in a number of ways so that there are quite a number of possible “vibrational modes.”[10]
Vibrational Modes of CH2
Here we see the the vibrational modes for part of an organic molecule with a $\mathrm{CH}_{2}$ group. Not all these modes can be observed through infrared absorption and emission: vibrations that do not change the dipole movement (or charge distribution) for the molecules (for example, the symmetrical stretch) do not result in absorption of IR radiation, but since there are plenty of other vibrational modes[11] we can always detect the presence of symmetrical molecules like methane (and most other molecules) by IR spectroscopy. In addition, the more the charge distribution changes as the bond stretches, the greater the intensity of the peak. Therefore, as we will see, polar molecules tend to have stronger absorptions than non-polar molecules.
For historical reasons, IR spectra are typically plotted as transmittance (that is the amount of light that is allowed through the sample) versus wavenumber ($\mathrm{cm}^{-1}$).[12] That is, the peaks are inverted, so that at the top of the spectrum 100% of the light is transmitted and at the bottom it is all absorbed.
The position of the absorption in an IR spectrum depends three main factors:
• Bondstrength: it makes sense that the energy needed to stretch a bond (i.e. make it vibrate) depends on the strength of the bond. Therefore, multiple bonds appear at higher frequency (wavenumber) than single bonds.
• Whether the vibrations involved involve bond stretching or bending: it is easier to bend a molecule than to stretch a bond.
• The massesoftheatoms in a particular bond or group of atoms: bonds to very light atoms (particularly appear at higher frequency).[13]
The figure below shows each of the general areas of the IR spectrum and the types of bonds that give rise to absorptions in each area of the spectrum. In general $\mathrm{C-H}$, $\mathrm{O-H}$, and $\mathrm{N-H}$ bond stretches appear above $3000 \mathrm{ cm}^{-1}$. Between $2500$ and $2000$ is typically where triple-bond stretches appear. Between $2000$ and $1500 \mathrm{ cm}^{-1}$ is the region where double-bond stretches appear, and the region below $1600$ is called the fingerprint region. Typically, there are many peaks in this fingerprint region which may correspond to $\mathrm{C-C}$, $\mathrm{C-O}$ and $\mathrm{C-N}$ stretches and many of the bending modes. In fact, this region is usually so complex that it is not possible to assign all the peaks, but rather the pattern of peaks may be compared to a database of compounds for identification purposes.
Vibrational Frequencies for Common Bond Types or Functional Groups
The figure below ($\downarrow$) shows a spectrum of acetone ($\rightarrow$) in which you can see a number of absorptions, the strongest of which appears around $1710 \mathrm{ cm}^{-1}$. Note also that there is a discontinuity in scale above and below $2000 \mathrm{ cm}^{-1}$. The strong peak around $1710 \mathrm{ cm}^{-1}$ can be ascribed to one particular part of the acetone molecule: the $\mathrm{C=O}$ (or carbonyl) group. It turns out that carbonyl groups can be identified by the presence of a strong peak in this region although the wavenumber of absorption may change a little depending on the chemical environment of the carbonyl. The presence of a strong peak between around $1700 \mathrm{ cm}^{-1}$ almost always signifies the presence of a $\mathrm{C=O}$ group within the molecule, while its shift from $1700 \mathrm{ cm}^{-1}$ is influenced by the structure of the rest of the molecule. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.03%3A_Infrared_%28IR%29_Spectroscopy_-_Look.txt |
The idea that a carbonyl group can be recognized, regardless of the structure of the rest of the molecule, is evidence for a major organizing idea of organic chemistry—that of functional groups. Functional groups are recognizable clusters of atoms within a larger molecule that have predictable properties allowing us to organize and make sense of organic chemistry. Rather than having to look at very large molecules as whole entities, (for example biomolecules that have many thousands to millions of carbon atoms) we can identify functional groups, and if we know how they behave we can predict how the molecule will behave.[14] A list of common functional groups is shown in the table ($\rightarrow$) and it is important that over time you learn to recognize each of them. Later, we will consider each of these groups in more detail and discuss their chemistry. For the moment, it is enough to be able to recognize these groups and understand how to use the evidence for the existence of functional groups that can be obtained from IR spectroscopy. Molecules with functional groups always give rise to particular patterns of peaks that correspond to the vibrations associated with that group.
IR spectra can provide evidence for the presence of many of these functional groups, although in practice the most easily detected groups are double bonds—especially the carbonyl ($1600-1800 \mathrm{~cm}^{-1}$), and $\mathrm{OH}$ and $\mathrm{N-H}$ groups (above $3000 \mathrm{~cm}^{-1}$). For example, the spectrum of acetic acid (below) shows two very strong absorptions (above the fingerprint region): a very broad absorption around $3000 \mathrm{~cm}^{-1}$, and a stronger, narrower one around $1720 \mathrm{~cm}^{-1}$.
IR spectrum of Acetic Acid
The broad peak corresponds to the $\mathrm{O-H}$ group in the carboxylic acid. It is so broad because this group participates in hydrogen bonding of various types and strengths, meaning that the energy required to promote this bond stretch varies with the degree of hydrogen bonding. The peak at $1720$ corresponds to the carbonyl group. It should be noted that there are also $\mathrm{C-H}$ stretches around $3000$ that are hidden by the strong $\mathrm{OH}$ stretches. The take-home message here is that IR spectroscopy is good for identifying clusters of atoms (functional groups) within a larger molecule, but it does not provide us with information about how the atoms in the molecule are actually arranged and connected. For that we need to turn to another kind of spectroscopy: nuclear magnetic resonance.
Questions to Answer
• What factors predict the strength of the IR absorption?
• What factors predict the position of the IR absorption?
• Draw a structure for a carboxylic acid in solution in methanol and use it to explain why the $\mathrm{O-H}$ bond absorption is so broad.
Nuclear Magnetic Resonance (NMR) Spectroscopy
is a form of spectroscopy based on the fact that atomic nuclei behave like tiny spinning charges that generate a magnetic field. Just as with electrons, nuclei have a characteristic that we call “spin” that is quantized.[15] Nuclei can have spins that are integers ($(1,2,3 \ldots)$) or non-integral spins ($\frac{1}{2}, \frac{3}{2}, \frac{5}{2} \ldots)$), and some have no spin. Those nuclei with a non-integral-spin quantum number are said to be NMR active, and of these the two most relevant to NMR analyses of organic molecules are $\mathrm{H-} 1$ and $\mathrm{C-} 13$, which have spin quantum numbers of $\frac{1}{2}$. When materials that contain carbon or hydrogen are placed in a magnetic field their nuclei can assume either of two possible orientations with respect to the field: a low energy orientation in which the spin- associated nuclear magnet is aligned parallel to the field and a high energy orientation in which the nuclear magnet is aligned anti-parallel to (against) the field. A parallel nucleus can be flipped to the anti-parallel orientation by absorbing the appropriate amount of electromagnetic energy—which happens to be in the radio-wave range. From the pattern of energy absorbed, we can deduce the structure of the molecule being analyzed (see below)
The energies of electromagnetic radiation that cause the nuclei to spin flip are referred to as resonance frequencies, (which is why the method is called nuclear magnetic resonance (NMR) spectroscopy). The same phenomena is involved in magnetic resonance imaging (MRI)—although the “nuclear” part of the name was removed because of concerns that patients might be worried (needlessly) about exposure to damaging radiation. Both NMR and MRI involve placing the specimen, either a chemical sample or a human body, into a very strong magnetic field, but the energy used to “flip” the spin of the nuclei within the sample is relatively low, that is in the radiofrequency range. We are constantly bombarded by low energy radio-waves but their energy is not enough to be absorbed by most molecules, and so they have little effect.[16] It is only when very high magnetic fields are applied to a sample (or body) that the spin states of the nuclei are split (into orientations that are parallel and anti-parallel to the field) and we observe the absorption (or emission) of these low energy radio-waves.
An NMR spectrum (or an MRI image for that matter) requires a sample to be placed into a strong magnetic field. In older NMR spectrometers, the field strength is changed, and the resonance frequencies are recorded. Newer (now almost all) NMR machines operate with a fixed magnetic field strength; the sample is irradiated with radio frequency (RF) radiation at all relevant frequencies which excites all NMR active nuclei into the more energetic spin state. As the nuclei relax, the machine records the energies of the emissions, which are then resolved into distinct wavelength using a Fourier-transform-based deconvolution.
The separation of the spin states (and therefore the amount of energy required to flip them) depends upon the strength of the external magnetic field. For example, for a relatively small magnet of 1.4 Tesla the RF used is $60 \mathrm{ MHz}$; at 7 Tesla the frequency is around $300 \mathrm{ MHz}$.[17] The most modern instruments use very strong magnets, operating in the $1000 \mathrm{ MHz}$ range. In practice stronger magnets provide better, more detailed spectra. Since different NMR instruments use different magnetic field strengths, they will also produce different RF absorption/emission peaks for same molecule. In order to make measurements from different NMR instruments comparable, the scale is set to show signals in terms of parts per million ($\mathrm{ppm}$) from a reference signal (see following section on chemical shift). | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.04%3A_Functional_Groups.txt |
We begin by considering the use of $\mathrm{C-} 13$ NMR spectroscopy because it can provide the simplest type of NMR spectrum. $\mathrm{C-} 13$ is a minor isotope of carbon, usually $\sim 1 \%$ of carbon nuclei present in sample are $\mathrm{C-} 13$ (the majority are $\mathrm{C-} 12$ and a very small percentage (less than 1 in a million are $\mathrm{C-} 14$). In a $\mathrm{C-} 13$ spectrum, each chemically different carbon atom will give rise to a signal or peak in the spectrum. By chemically different, we mean that the carbons are in different environments; these environments impact the local magnetic field that a particular nucleus will experience. So, for example, while ethane $\mathrm{CH}_{3} \mathrm{CH}_{3}$ (obviously) has two carbons, because of the symmetry of molecule, both carbon nuclei experience the same chemical (and local magnetic) environment; we would expect ethane to show a single $\mathrm{C-} 13$ NMR peak.
Now consider a more complex hydrocarbon such as 2-methylbutane ($\beta$), which has a total of five carbons; of these the $\mathrm{C}1$ and $\mathrm{C}2$ methyl carbons are in identical chemical environments. The molecule as a whole has four distinct environments and therefore there are 4 peaks in its $\mathrm{C-} 13$ NMR spectrum.
Shielding and Deshielding
We can answer the question of why NMR signals appear at different places by first remembering that we are dealing not with isolated nuclei, but with molecules that consist of nuclear cores surrounded by electrons that can be described as occupying various molecular orbitals. The electron density around the nucleus has a marked effect on the local magnetic field—that is, the magnetic field that is “felt” by each nucleus. The electrons also “feel” the effect of the magnetic field and begin to circulate around the nuclei to induce a new magnetic field that opposes the original one. This overall impact of this is to reduce the effective nuclear field as shown in the figure below. It now takes a stronger external field to bring the nucleus to resonance at the same frequency (the flip the nuclear spin).
Nuclei that are surrounded by a larger amount of electron density are said to be shielded; they require a larger magnetic field to bring to spin flip than nuclei that are de-shielded (that is, surrounded by lower electron density). Absorption by shielded nuclei tend to be higher (“upfield”) in the spectrum and de-shielded nuclei appear downfield. Consider for example, the $\mathrm{C-} 13$ NMR spectrum of 1-chloropropane.
As we might expect, there are 3 peaks in the spectrum, but now we can figure out which is which, because we can predict the relative charge densities on each carbon. The carbon attached to the chlorine ($\mathrm{C-} 1$) is most de-shielded by the inductive effect, and therefore should appear at lowest field. Indeed the signal at around $47 \mathrm{ ppm}$ belongs to the $\mathrm{C-} 1$. We can also see how the inductive effect dissipates with distance from the electron withdrawing group to the $\mathrm{C-} 3$ signal at around $11 \mathrm{ ppm}$. This is direct evidence for the inductive electron-withdrawing effect.
Now let us compare this spectrum to that of acetone. As we might predict, this spectrum has only two peaks in it because acetone only has two types of carbon, but what is even more interesting is that the $\mathrm{C=O}$ carbon appears at such low field, around $205 \mathrm{ ppm}$. This means that the $\mathrm{C=O}$ carbon must be very electron deficient—much more so than a carbon single-bonded to an oxygen ($\mathrm{C-O}$) in an alcohol, which appears around $64 \mathrm{ ppm}$ (as shown in the spectrum of 1-propanol) . This is direct evidence for a phenomenon that we will see time and again, which is that the $\mathrm{C=O}$ carbon is highly electron deficient and is very susceptible to nucleophilic attack. Note that the spectrum of acetone also shows this highly downfield shifted peak corresponding to the $\mathrm{C=O}$ carbon. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.05%3A_Carbon-13_NMR_Spectroscopy.txt |
While $\mathrm{C-} 13$ NMR provides information about how many chemically distinct carbons there are in a compound, and something about their electronic environments, it does not tell us anything about how the molecules are organized or which carbons are connected together.[18] For that information, we often turn to NMR spectroscopy using the $\mathrm{H-} 1$ isotope. There are a number of advantages to $\mathrm{H-} 1$ NMR: $\mathrm{H-} 1$ is the naturally-occurring, most-abundant isotope of hydrogen, and is much more abundant than $\mathrm{C-} 13$. The result is that the concentration of $\mathrm{H-} 1$ in a sample is much greater than the concentration of $\mathrm{C-} 13$. It is therefore easier to obtain a spectrum and instruments with a lower field can be used. As we will see, $\mathrm{H-} 1$ NMR can provide information about which atoms are connected to each other. On the other hand, the resulting $\mathrm{H-} 1$ NMR spectra are more complex.
The basic theory of both $\mathrm{H-} 1$ and $\mathrm{C-} 13$ NMR are the same and, in fact, both spectra can be recorded on the same instrument. The sample is placed in a magnetic field that splits the spin states of the $\mathrm{H-} 1$ nuclei; the energy required to flip the nuclei from one spin state to another is detected and is transformed into a spectrum such as the one below.
There are similarities and differences in the appearance of $\mathrm{H-} 1$ and $\mathrm{C-} 13$ NMR spectra. As can be seen in the spectrum of 1,2-dichloro-2-methylpropane ($\leftarrow$), the scales over which the spectra are recorded are different. Both involve a chemical shift from a reference peak at 0, (TMS [tetramethylsilane] is used in both), but the shift range (often referred to as $\delta$, as well as $\mathrm{ppm}$, in proton NMR) is smaller for $\mathrm{H-} 1$ compared to $\mathrm{C-} 13$ NMR. Typically an $\mathrm{H-} 1$ NMR spectrum is recorded between $0-10 \mathrm{ ppm}$ (as opposed to $0-200$ for $\mathrm{C-} 13$), although in this case it is truncated because there are no peaks between $4–10 \mathrm{ ppm}$. That said, the same trends in chemical shifts are observed: the more de-shielded the atoms (in this case protons) are, the further downfield they appear. In the 1,2-dichloro-2-methylpropane spectrum above there are two peaks corresponding to the two types of chemically distinct hydrogens (that is, the two identical $-\mathrm{CH}_{3}$ and the $\mathrm{-CH}_{2}$ group). The peak’s lowest field, around $3.6 \mathrm{ ppm}$ can be assigned to the $\mathrm{CH}_{2}$ (methylene) group, which is directly attached to the electron withdrawing chlorine atom, while the second peak, corresponding to the six equivalent hydrogens of the two methyl groups, is further upfield, around $1.6 \mathrm{ ppm}$.
In a proton NMR spectrum, the area under the peak is proportional to the number of protons that give rise to the signal. By integrating the area under the peak, we generate an estimate of the relative numbers of hydrogens giving rise to each signal. In the spectrum of 1,2-dichloro-2-methylpropane the ratio of the two peak areas is 1:3, thus supporting our assignment of the downfield peak to the $\mathrm{CH}_{2}$ and the upfield peak to the two equivalent $-\mathrm{CH}_{3}$ (six) methyl hydrogens. As a technical note, $\mathrm{C-} 13$ NMR signals are not reliably proportional to the number of equivalent carbons involved: they are dependent on the number of $\mathrm{H}$’s attached to them and so integration cannot provide a reliable estimate of the proportions of the types of carbon in a compound.[19]
Spin-Spin Splitting
Another way that $\mathrm{H-} 1$ NMR differs from a $\mathrm{C-} 13$ NMR is shown in the spectrum of 1,1-dichloroethane $\rightarrow$, which has two types of equivalent hydrogens. Each type gives rise to a distinct signal but each of those signals is split into multiple peaks. The upfield signal around $2 \mathrm{ ppm}$, from the 3 equivalent $\mathrm{H}$'s of the methyl group, appears as a doublet (two separate peaks) because each of the hydrogens in the methyl group is affected by the magnetic field generated by the neighboring hydrogen’s possible spin state, thereby altering (adding to or subtracting from) the atoms’ local field. The downfield signal around $5.5 \mathrm{ ppm}$ is due to the single proton on $\mathrm{C-} 1$, de-shielded because of the proximity of the electronegative chlorine, but now it appears as a quartet.
For example, consider a molecule that has two non-equivalent protons ($\mathrm{Ha}$ and $\mathrm{Hb}$) on adjacent carbons ($\rightarrow$). $\mathrm{Ha}$ will experience the magnetic fields generated by $\mathrm{Hb}$ which has two spin states, which will produce two possible electronic environments for $\mathrm{Ha}$, resulting in two signals for $\mathrm{Ha}$.
A single proton produces a doublet in the adjacent signal
The splitting effect is small and does not extend (in a significant way) beyond the protons on adjacent carbons. The local magnetic field also depends on the number of protons that do the splitting, for example if there are two protons on the adjacent carbon, the signal is split into a triplet, and for three adjacent protons the signal is a quartet. The general rule is that for $n$ adjacent protons, the signal is split into $n+1$ peaks.
Three adjacent protons produce a quartet in the adjacent signal
The width, in $\mathrm{Hz}$, of the splitting (that is, the distance between the split peaks) is called the coupling constant $j$. The value of $j$ between the protons on adjacent carbons is the same. The number of split peaks therefore allows us to determine how many hydrogens there are on adjacent carbons, which in turn allows us to determine the how the atoms are connected together in the structure. Signals that are split by more than one kind of hydrogen can get quite messy as shown in the spectrum of D-glucopyranose ($\rightarrow$).
Fortunately, there are ways to simplify such spectrum by selectively irradiating particular frequencies but that is beyond the scope of this course (but something you can look forward to in the future!).
That said, this raises another question: why do $\mathrm{C-} 13$ NMR spectra appear as single lines for each carbon? Shouldn’t the carbon signals also be split both by adjacent carbons, and by the hydrogens attached to them? There are two reasons that this does not happen. First, the abundance of $\mathrm{C-} 13$ is so low that the probability of finding more than one $\mathrm{C-} 13$ in a molecule is very low; in the absence of a second (or third) $\mathrm{C-} 13$, no splitting by adjacent carbon nuclei will occur, (recall $\mathrm{C-} 12$ does not generate a spin-based magnetic field). Second, when we originally introduced $\mathrm{C-} 13$ NMR spectroscopy we presented a simplified model. In fact, the carbon signals are split by the adjacent protons, but this leads to a complex spectrum that is often hard to interpret. Therefore, for most purposes, the sample is irradiated with a radiofrequency that promotes all the hydrogens to the higher-energy spin state. The carbon nuclei do not experience two (or more) magnetic field environments, and therefore only one signal is produced. This technique is called broadband decoupling, and these types of spectra are the most common examples of $\mathrm{C-} 13$ NMR.
Solvents and Acidic hydrogens in the NMR
Most NMR spectra, whether $\mathrm{C-} 13$ or $\mathrm{H-} 1$, are recorded in solution (although it is possible to obtain spectra on solids—as evidenced by the fact that we can obtain MRI data on people, that say, people [biological systems] are mostly [$>70 \%$] water). However, since the solvent is usually present in much greater concentration than the actual sample, as long as we use a solvent that does not generate a strong signal the effects of the solvent can be minimized; typically this involves solvents in which the hydrogens present are replaced with deuterium (D), an isotope of hydrogen that is not NMR active. Common solvents are $\mathrm{CDCl}_{3}$, (deuterated chloroform) and dimethylsulfoxide $\mathrm{d-} 6$ (DMSO) ($\rightarrow$).
Another instance in which deuterated solvents are used is to detect the presence of potentially acidic hydrogens. Proton transfer is a rapid and reversible process, and any hydrogen attached to an electronegative element is potentially available for exchange by reacting with even very weak base such as water.In fact, by adding a drop of $\mathrm{D}_{2} \mathrm{O}$ (“heavy” water) to the NMR sample, the signal from any $\mathrm{O-H}$ or $\mathrm{N-H}$ disappears as the $\mathrm{H}$ is replaced by $\mathrm{D}$.
As shown here ($\rightarrow$), the signal from the $\mathrm{OH}$ (shaded in yellow) disappears when the sample is shaken with $\mathrm{D}_{2} \mathrm{O}$.
Questions to Answer:
Type your exercises here.
• Draw a diagram of an atom and use it to explain why the electron density surrounding the nucleus affects the strength of the external field required to bring the nucleus to resonance (to flip the spin state).
• Using the table below, use your diagram from above and explain a) why the signals for $\mathrm{C-} 1$ in the three compounds are different, and b) why the signals for $\mathrm{C-} 1$ and $\mathrm{C-} 2$ are different (for any single compound).
Compound
Signal for $\mathrm{C-} 1$
Signal for $\mathrm{C-} 1 (\mathrm{ppm})$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{F}$
79.1
15.4
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$
40.0
18.9
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$
27.5
19.3 | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.06%3A_H-1_%28proton%29_NMR.txt |
1. Species with unpaired electrons are called radicals or free radicals—they are typically highly reactive and are thought to be implicated in many processes involving cellular damage and aging.
2. Fortunately there exist cellular mechanisms that can detect and repair (most) these kinds of radiation-induced mutations.
3. http://nivea.psycho.univ paris5.fr/FeelingSupplements/AncientVisions.htm
4. Other related molecules are found throughout the biological world, for more examples see http://www.ncbi.nlm.nih.gov/pubmed/3416013
5. http://phototroph.blogspot.com/2006/...n-spectra.html
6. A video of various vibrations is here https://www.youtube.com/watch?v=iy-8rguvGnM
7. Snakes can see in the infrared which allows them to hunt mammals. In contrast, an ability to see in the infrared would not generally help a mouse see a snake. Can you explain why?
8. As the temperature of the object gets higher the energy of the electromagnetic radiation also increases, until it moves into the visible region. This is why very hot objects glow red and then white as they get even hotter.
9. In addition to vibrational transitions, we can also investigate transitions from one rotational energy level to another which are associated with microwave radiation, leading to microwave spectroscopy.
10. For animated versions of these modes see https://en.Wikipedia.org/wiki/Molecular_vibration
11. In fact, for linear molecules the number of vibrational modes is $3 \mathrm{N}-5$, and for non-linear molecules it is $3 \mathrm{N}-6$, where $\mathrm{N} =$ the number of atoms in the molecule. Therefore, $\mathrm{CH}4$ has $5 \times 3-6=9$ possible vibrational modes, although they are not all IR active.
12. Spectra are plotted as the % of light transmitted v wavenumber. The units of wavenumber are $1 / \mathrm{cm}(\mathrm{cm}-1)$, which, if you think about it, are directly related to the frequency of the light (remember that frequency $\times$ wavelength = constant (the speed of light). So units of 1/wavelength are in fact directly related to the frequency.
13. The vibrations of a molecule can be modeled by thinking of the bonds as springs, and using Hooke’s Law which states that the vibrational frequency is proportional to the bond strength, and inversely proportional to the masses of the atoms in the bond. Therefore, the smaller the masses, the higher the vibrational frequency
14. Although it is true that all functional group behavior is affected by the local chemical environment, it is usually possible to predict how this will affect the properties of the functional group.
15. This spin arises from the sum of the spins of their component parts, protons and neutrons, which in turn arise from the sum of their component quarks.
16. In contrast, more energetic electromagnetic radiation, such as microwaves, are absorbed by water; when absorbed they are converted into kinetic energy, and heat the sample—something that would quickly damage (and kill) living tissue.
17. The Earth’s magnetic field ranges from $25-60 \times 10^{-6}$ Tesla
18. Although some types of $\mathrm{C-} 13$ NMR spectroscopy can tell us about connectivity it is too complex to discuss here.
19. Some of the $\mathrm{C-} 13$ nuclei take much longer to relax back to the lower spin state than others (and the energy that is emitted is what is recorded), so it is not really feasible to integrate the peaks. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter/2.07%3A_In-Text_References.txt |
Now it is time to turn our attention to the ways in which the three-dimensional structure of organic molecules affects their stability, reactivity, and the ways in which they interact with one another and with solvent molecules. To begin with, we will consider compounds that are composed of $\mathrm{sp}^{3}$ hybridized (i.e. tetrahedral) carbons which are inherently three-dimensional since each of the four bonds points toward the vertex of a tetrahedron (a three-dimensional as opposed to a planar or two-dimensional organization). It is important to remember and understand this aspect of organic chemistry, since we tend to represent organic compounds in ever more simplified (and abstract) two-dimensional diagrams. The Lewis structure of a molecule is a $2\mathrm{-D}$ (that is, flat) cartoon that contains a great deal of information (if you know how to look for it). As we move to even more abstracted representations of molecular structures it is easy to forget about the wealth of information encoded within a structural diagram. This includes both how many carbons and hydrogens are present and the three-dimensional relationships between the parts of the molecule. While there are ways to show the three-dimensional nature of $\mathrm{sp}^{3}$ hybridized carbon (such as through the use of wedge-dash structures à), they can be quite cumbersome when it comes time represent complex molecules and we don’t often use them unless we want to specifically address the three-dimensional arrangement of atoms within the molecule. The other idea that gets lost in a static two-dimensional representation is that the bonds between $\mathrm{sp}^{3}$ hybridized carbon atoms are sigma ($\sigma$) bonds; the two parts of the molecule linked by a $\sigma$ bond can rotate relative to each other, without disrupting the overlap between the bonding orbitals.[1] At room temperatures the carbons in most σ bonds are rotating freely and rapidly with respect to each other, although we have to portray them in a fixed orientation when we draw them.
03: Conformations and Configurations - the consequences of the three-dimensional nature of carbon compounds
While most $\mathrm{C-C}$ single bonds do allow for free rotation, there are energy costs that are associated with such rotations which arise from the fact that as the carbons rotate around the bond axis, the distance between groups attached to each carbon (even if they are hydrogens) changes. To understand the implications of this changing distance between these groups, recall our previous discussions of London dispersion forces and van der Waals interactions.[2] As two uncharged atoms, molecules, or groups within a molecule approach each other there is initially an attraction between them due to the instantaneous and induced dipoles in their electron clouds (London dispersion forces). If the molecules (or parts of a molecule) also have a permanent dipole (i.e. the molecule is polar), then there is an additional attractive (or repulsive) interaction in the form of dipole-dipole interactions and hydrogen bonding. Attractive interactions lead to a decrease in the potential energy of the system and repulsive interactions lead to an increase. Even when the interaction is attractive, as the interacting molecules (or regions of molecules) get closer, the repulsive interactions between their electron clouds increase and the potential energy of the system increases. This gives rise to the familiar (or at least it should be familiar) potential energy v distance curve that we have encountered many times in CLUE.
While we have previously discussed this change in potential energy for separate molecules (and atoms), the same principles apply as parts of molecules approach each other, for example, because of rotations around $\mathrm{C-C}$ sigma bonds. The different rotational structures of a molecule are called conformations, and one of the best ways to represent different possible conformations is called the Newman projection. In a Newman representation, we look down the $\mathrm{C-C}$ bond of interest; this has the effect of making the groups (atoms, etc.) attached to each carbon (known as substituents) appear to be at angles of $120^{\circ}$ to each other (this is due to the perspective from which we are viewing the molecule; the bond angles are still $109^{\circ}$). If we imagine the carbons rotating with respect to one another, we can see how the relative distances between the groups on each carbon change with the “dihedral angle” between the substituents on the front and back carbons. The two conformations shown for ethane are the fully eclipsed conformation, where the dihedral angle is equal to $0^{\circ}$, and the staggered conformation, where the dihedral angle is $60^{\circ}$. As the rotation around the $\mathrm{C-C}$ bond continues, these two conformations repeat.
Based on NMR studies at low temperature, these two conformations are not equally stable, the staggered conformation is about $13.8 \mathrm{~kJ} / \mathrm{mol}$ more stable than the eclipsed conformation. At room temperature, this energy difference is negligible, that is, a typical molecular collision supplies much more energy, and the two conformations rapidly interconvert between one another. As the sample is cooled down, however, the energy available from collisions is reduced (because the molecules are moving more slowly) and we can actually find, based on NMR studies, that we can see different conformations present in different proportions—in part because in the eclipsed conformation the hydrogens on the two carbons come closer to their van der Waals radii and so begin to repel one another, raising the potential energy of the molecule.
Potential Energy v Dihedral Angle for Butane
If we look at butane ($\rightarrow$) down the $\mathrm{C}2\mathrm{-C} 3$ bond axis we can see a more exaggerated version of this repulsion due to the presence of the bulkier methyl groups. There are two types of eclipsed conformations, and two types of staggered conformations—the anti conformation where the methyl groups are $180^{\circ}$ from each other, and the gauche conformation where they are $60^{\circ}$ apart. If we plot their relative energies vs dihedral angle we can see how the potential energy changes as the larger centers of electron density (the methyl groups) get closer to one another. The anti conformation is the most stable and the eclipsed methyl group conformation is the least stable. Again, these energy differences are not all that large. The difference between the highest-energy eclipsed conformation and the lowest-energy anti conformation is about $21 \mathrm{~kJ} / \mathrm{mol}$, which is not a big enough barrier to prevent rotation at room temperature.
By applying these ideas, we can predict conformations of long-chain alkanes as well. We usually draw long-chained alkanes in the most stable (all-“anti”) conformation, but there are occasions when the gauche conformation is also present—the result of this conformation is a bend or kink in the chain. This has implications how long-chained alkyl groups interact with one another, and influences the molecules’ physical properties, such as melting point. Many biologically-significant lipid molecules contain long-chain alkane groups. For example phospholipids form the structural basis of biological membranes. These lipids are a major component of the cell membrane that is the permeable barrier that surrounds the cell. The long-chain fatty acids that make up the membrane are typically found in the all-anti conformation, which is not only the lowest-energy conformation, but it also makes for the largest surface area. If the long-chains were in the gauche conformation, they would be more clumped up and have a lower surface area. Having a large surface area means that the chains are more likely to attract each other by London dispersion forces, which also stabilizes the structure of the membrane.
Questions to Answer
• Construct an explanation for why the potential energy of ethane changes as you rotate around the $\mathrm{C-C}$ bond. What is the interaction that is raising the potential energy? Where does the energy come from—so that the rotation around the $\mathrm{C-C}$ bond can occur?
• Construct a potential energy v dihedral angle diagram for propane, as you look down the $\mathrm{C}1\mathrm{-C} 2$ bond. Label the maxima and minima with the corresponding Newman projections.
• Draw a schematic picture of a lipid bilayer made up of phospholipids (no need to draw every bond and atom). What would be the effect if the $\mathrm{C-C}$ bonds in phospholipids took up the less stable gauche conformation?
Conformations of Cyclic Compounds
As we will see, cyclic compounds show some of the same kinds of energy changes with rotations around $\mathrm{C-C}$ bonds; conformations change from eclipsed to staggered. However, when the molecule is cyclic two other factors come into play and they complicate things a little. The first is that, by their very nature, ring compounds are more constrained; it is not possible to do a full $360^{\circ}$ rotation around the $\mathrm{C-C}$ bonds without breaking a covalent bond (which requires more energy than is available through thermal collisions). There is, however, a range of rotations that most ring compounds can move through. The other factor involves rings that would have bond angles that are inconsistent with the $\sim 109^{\circ}$ bond angle that is usually found for $\mathrm{sp}^{3}$ hybridized carbons.
For example, consider the $\mathrm{C-}3$ to $\mathrm{C-}6$ cycloalkanes ($\rightarrow$). We can calculate the bond angles for these regular polyhedrons if they were flat, that is, two-dimensional: they would range from $60^{\circ}$ for cyclopropane to $120^{\circ}$ for cyclohexane, both far outside the range of $\mathrm{sp}^{3}$ bond angles. Not surprisingly, cycloalkanes behave in a range of different ways to minimize the strain imposed by both bond angles and the constraints on bond rotation. This ring strain can be experimentally determined by measuring the heats of combustion for cycloalkanes: the energy released by this reaction is a proxy for the stability of the particular cycloalkane. To compare among cycloalkanes, we need to calculate the heat of combustion per $\mathrm{CH}_{2}$ group, and as we will see shortly, six-membered rings do not, in fact, have any ring strain so we can determine the stabilities of differently sized rings compared to cyclohexane.
Cycloalkane
Heat of combustion($\mathrm{kJ} / \mathrm{mol}$)
Heat of combustion per $\mathrm{CH}_{2}$ ($\mathrm{kJ} / \mathrm{mol}$)
Ring Strain Compared to $\mathrm{C}_{6} \mathrm{H}_{12}$ ($\mathrm{kJ} / \mathrm{mol}$)
Total Ring Strain
$= \text { Ring strain } \times \# \text { of } \mathrm{CH}_{2}$ ($\mathrm{kJ} / \mathrm{mol}$)
C3H6
-499.8
-166.3
9.2
27.6
C4H8
-655.9
-164
6.6
26.4
C5H10
-793.5
-158.7
1.3
6.5
C6H12
-944.5
-157.4
0
0
Heats of Combustion for the Cycloalkanes
Cyclopropane is actually the worst-case scenario: it has the highest strain because it is forced to exist as a flat, $2\mathrm{-D}$ ring (since three points define a plane). If we look at a Newman projection of cyclohexane we can see that all the $\mathrm{C-H}$ and $\mathrm{C-C}$ bonds are eclipsed, which raises the potential energy of the molecule.
In an attempt to relieve some of the strain imposed by having a $60^{\circ}$ bond angle, cyclopropane has bent bonds (sometimes called banana bonds). The electron density bulges out rather than being located between the two carbons, thus making the bond angle a little larger—however, there is still considerable strain. The combination of angle and eclipsing strain (also called torsional strain) explains why cyclopropane is a much more reactive species than its straight-chain analog, propane. There are a few naturally occurring three-membered ring compounds, but in general they are quite chemically reactive since the ring structure is quite unstable. Four-membered rings are more stable—they have less angle strain because the ring is bigger and because the larger ring can bend a little, which has the effect of relieving some of the torsional strain. At room temperature, the cyclobutane ring is constantly bending so that each carbon can be relieved of a little of the strain. There are a number of important four-membered ring compounds that you may have run across: the antibiotics penicillin (1) and cephalosporin (2) have a four-membered amide ring as part of their structure (à). Normally (as we will see), amide groups are relatively stable (chemically nonreactive), but because this amide in the ring structure is constrained in a high-energy, four-membered ring, it is much more reactive than normal. Antibiotics act by targeting some part of the biochemistry of bacteria that will not affect the host organism (us). These antibiotics interfere with the synthesis and replication of the microbial cell wall, which is a type structure that most animals do not have (we have cell membranes).
Ring stability improves with cyclopentane. If the molecule were flat, it would have almost exactly the $\mathrm{sp}^{3}$ bond angles between the ring carbons; this would mean, however, that all the bonds would be in the eclipsed conformation and the resulting torsional energy would be high. To relieve torsional strain the molecule bends into a sequence of “envelope” type shapes—again in dynamic and rapid motion so that the energy associated with the eclipsed conformation is reduced.
This brings us to cyclohexane, the “goldilocks” of cycloalkanes. The bending of the cyclohexane molecule can completely relieve all angle strain and torsional strain. In fact, cyclohexane is as stable as hexane—there is no ring strain associated with a six-membered ring. The most stable conformation of cyclohexane is known as the chair conformation, because one might imagine it as a chair with a footrest, seat, and back. In the chair conformation all the bonds are staggered, and all bond angles are $109^{\circ}$.
The groups attached to the cyclohexane ring can assume two types of positions: the positions that seem to point straight up or down are called axial, and the ones that are more in the plane of the ring are called equatorial. On each carbon, there is one axial and one equatorial position as shown ($\uparrow$).
Chair conformations can “ring-flip” as shown below. Here the structure on the left has all the axial positions in blue and the equatorial in red. Just as with the other rings, the ring is in constant motion and the bonds are rotating (as far as they can within the confines of the ring) so that the ring will “flip” back and forth between chair conformations, by bringing the chair back down and the footrest up. When this happens, axial substituents are converted to equatorial and vice versa. It’s worth noting that axial components can be “above” or “below” the ring plane; and the equatorial positions are also above or below the plane—but on average they lie in the plane of the ring. The other thing to notice is that while groups can flip between axial and equatorial, they do not flip from above to below. That is an axial substituent above the plane becomes and equatorial substituent above the plane.
There are a number of conformations between one chair conformation and another, for example the boat conformation ($\rightarrow$). This is a much more high-energy species since many of the bonds are eclipsed and there is a strong repulsion of the H’s at the 1 and 4 position.
Mono-substituted cyclohexanes:
Now consider methylcyclohexane, the methyl group can be either axial or equatorial. When we look at Newman projections we see that the equatorial methyl group is in a more stable position, it is anti to the $\mathrm{C-C}$ bond in the ring, whereas an axial group is gauche to the ring $\mathrm{C-C}$ bond.
Equatorial methylcyclohexane
In addition, an axial methyl group is also close to the axial hydrogens on carbons 3 and 5 in what is called a 1.3 diaxial interaction—which introduces another source of strain. Therefore, the more stable conformation is always equatorial.
Axial methylcyclohexane
For a methyl group the difference in energy is still quite small, and both axial and equatorial are present at room temperature, but as bulkier groups are introduced onto the ring, the preference for equatorial becomes even stronger. For example, cyclohexane rings with isopropyl ($-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$) or t-butyl ($-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$) groups are locked into conformations where these groups are equatorial.
Disubstitutedcyclohexanes:
When we get to cyclohexane rings that have two (or more) substituents, we need to consider whether the substituents are on the same (cis) or opposite (trans) sides of the ring. This is easiest to see if we first draw the ring flat and use a wedge dash representation to show relative positions of the groups. In these representations we don’t know whether the methyl groups oriented in axial or equatorial configurations (but we can figure it out), but what we do know is that these are different compounds, they cannot be interconverted by ring-flips like axial and equatorial—which are conformations of the same compound. To interconvert we would have to break the $\mathrm{C-C}$ bonds in the ring. These two compounds are called geometric isomers—they have the same molecular formula and connectivity but their atoms have a different (permanent) relationship in space (a different geometry).
We can determine the actual conformations that each isomer can take up by drawing the chair representation, and we can ring-flip to find the most stable conformation. Remember that equatorial is always favored, so when we do this we can see that the trans isomer has a conformation in which both methyl groups are equatorial; whereas the cis conformation has both conformations where one is axial and one is equatorial. Therefore, we can conclude that the trans isomer is more stable than the cis. Note that the groups do not change their cis or trans relationship when the ring-flips.
It is possible to do this kind of analysis for 1,3- and 1,4-disubstituted cyclohexanes, and even for more complex substances. For example, most sugars actually exist as ring structures in which the many alcohol groups (which are bigger than hydrogen) take up the equatorial position. For example, glucose exists as cyclic six-membered rings (one of the atoms is an oxygen but the principle is the same).
Note how almost all of the $\mathrm{OH}$ groups take up the equatorial position, except for the $\mathrm{OH}$ group on $\mathrm{C-}1$ in $\alpha$-D-glucose. These two forms of glucose are actually separate compounds. It is not possible to convert $\alpha$-D-glucose to $\beta$-D-Glucose simply by a ring-flip (convince yourself that this is true by making model). We will see much more of these cyclic sugar molecules later in the course.
Questions to Answer
• Use excel to plot total ring strain vs ring size (number of $\mathrm{CH}_{2}$ units) as shown in Table 1. In addition use these total ring strain values for $\mathrm{C}_{7} \mathrm{H}_{14}$ ($6.3 \mathrm{~kJ} / \mathrm{mol})$), and $\mathrm{C}_{8} \mathrm{H}_{16}$ ($9.6 \mathrm{~kJ} / \mathrm{mol})$), $\mathrm{C}_{9} \mathrm{H}_{18}$ ($12.6 \mathrm{~kJ} / \mathrm{mol})$). What trends do you see? How do you account for them?
• Draw two chair forms of cyclohexanol, which one is most stable? Is this a geometric isomer or a conformational isomer.
• Draw chair forms of cis and trans-1,3-dimethylcyclohexane and predict which geometric isomer is the most stable. Do the same for cis and trans-1,4-dimethylcyclohexane.
• Does 1,5-dimethylcyclohexane exist? | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/03%3A_Conformations_and_Configurations_-_the_consequences_of_the_three-dimensional_nature_of_carbon_compounds/3.01%3A_Co.txt |
In the previous section we looked at conformational and geometric stereoisomerism, by which we mean that molecules that have the same molecular formula and connectivity of the atoms, but that have different arrangements in space. Conformational isomers can be interconverted by rotations around $\mathrm{C-C}$ bonds while geometric (cis/trans) isomers have permanently distinct geometries. Now we introduce another type of stereoisomer, known as optical isomers. Optical isomers have a property called chirality—they are mirror images of one another. While they have exactly the same overall shape, they are not superimposable on one another. One obvious example of macroscopic objects that are chiral are your hands.
Your hands are mirror images of each other, but they are not superimposable on one another placing one hand on top of the other shows you that they are different as does the fact that a left hand glove does not fit your right hand. There are many objects in the world that are chiral. At the molecular level, many molecules are chiral—and it is often said that they exist in left- and right-handed forms.
The simplest examples of chiral molecules are those that have a single carbon atom with four different substituents attached. While these compounds have exactly the same structure, connectivity and relationship between the atoms in space, they are different in the same way that a right and left hand are different: they are known as enantiomers—non-superimposable mirror images of each other. An equal mixture of enantiomers is called a racemate (racemic). Under normal laboratory conditions it is difficult to differentiate between a pair of enantiomers. If you were able to separate the left- from the right-handed forms, you would find that they have the same structure, the same physical and chemical properties, that is, the same melting point, boiling point, solubility, and chemical reactivity. Nevertheless, they are not identical—they can be distinguished by the way they interact with certain types of light.
Electromagnetic radiation can be considered as being composed of electrical and magnetic (e-m) waves that oscillate perpendicular to each other and to the direction of wave transmission.
But a beam of light is not a single e-m wave; it consists of many such waves. The orientation of the various e-m waves within the beam are independent on one another and oriented randomly with respect to the direction of the beam. We can simplify an e-m wave by passing the light beam through what is known as a polarizer. A polarizer is a transparent material (it allows light to pass through it) composed of oriented crystalline that block all e-m oscillations that do not align with the crystal structure. The result is that the light that passes through the polarizer oscillates in only one plane; known as plane-polarized light. You can now make a prediction; if you place a second polarizing filter in the same orientation as the first, in the path of a plane-polarized light beam, the beam will pass through, but if you rotate the second filter by $90^{\circ}$ to the first, the beam will be absorbed (blocked). This is an experiment you can do if you have polarizing sunglasses, which reduce glare by allowing light vibrating in only one direction to pass through. If the plane-polarized light interacts with an asymmetric electric field (such as the one generated by a chiral molecule), the plane of vibration will be rotated. We can measure the angle of rotation using another polarizing lens and in this way, we can detect the presence of chiral molecules. On the other hand, passing plane-polarized light through a racemic mixture (with 50% of each enantiomers present), or through a solution containing a non-chiral solute, will not affect the direction of the polarization plane.
In a sample that rotates the plane of a plane-polarized light beam to the right, the enantiomer is designated as the (+ or dextrorotatory ) isomer; if it rotates the plane-polarized light beam to the left it is referred to as the (– or levorotatory ) isomer. This phenomenon is direct evidence of the existence of chiral molecules. The amount of rotation is dependent on the concentration of the substance in the solution, the path length of the sample tube, the wavelength of the incident light, and also the nature of the substance itself. We often report these rotations as a specific rotation, where all of these variables are defined, so that compounds can be identified by this property. Because of this phenomenon, substances that are chiral are said to be optically active. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/03%3A_Conformations_and_Configurations_-_the_consequences_of_the_three-dimensional_nature_of_carbon_compounds/3.02%3A_Op.txt |
Since we cannot distinguish between enantiomers using the naming conventions that we have considered so far, we have to invoke a new convention to unambiguously specify the arrangement of bonds around a chiral center which is known as the configuration (not to be confused with conformation). The most common naming strategy used is referred to as the Cahn-Ingold-Prelog Convention: a set of rules that assigns a configuration (known as R or S) to a specific chiral (or stereogenic) center. An important fact to remember is that the configuration assigned to a molecule in this way has nothing to do with the molecule’s observed optical rotation.
Cahn-Ingold-Prelog Convention
1. Look at each atom directly connected to a chiral carbon and rank by atomic number (Z); highest first (e.g. $\mathrm{O}>\mathrm{N}>\mathrm{C}$). For isotopes the higher atomic mass receives a higher priority (e.g. $\mathrm{D}>\mathrm{H}$).
2. If you can’t make a decision here, go out to the next atom in each chain—and so on until you get to the first point of difference. So for example, $-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$ takes priority over $-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$.
3. Multiple bonds are equivalent to the same number of single-bonded atoms. For example, $\mathrm{CH}_{2}\mathrm{=CH}_{2}$ takes priority over $-\mathrm{CH}_{2} \mathrm{CH}_{3}$ because it is counted as if it were $\mathrm{CH}_{2}\left(\mathrm{CH}_{3}\right)_{2}$
4. Now place your eye so that you are looking down the bond from carbon to the lowest ranking group.
• If the sequence, high$\rightarrow$low, is clockwise: R (Rectus).
• If the sequence, high$\rightarrow$low, is counterclockwise: S (Sinister).
The skill of assigning configurations is one that takes practice and it is necessary to understand what the rules mean. Some common errors that people make are:
1. Looking at the substituent as a whole rather than looking only at the atom (or the first atom that is different). So for example, $\mathrm{O}>\mathrm{N}>\mathrm{C}$, but also $-\mathrm{OH}>-\mathrm{CO}_{2} \mathrm{H}$ even though $\mathrm{CO}_{2} \mathrm{H}$ is overall larger, it is attached by a $\mathrm{C}$ which is a lower priority than an $\mathrm{O}$. Consider the example of this stereoisomer of the amino acid serine (It’s the L isomer but we will get to that in a while). The highest ranking group is the $\mathrm{N}$ (since its atomic number is higher than $\mathrm{C}$), next is the $\mathrm{CO}_{2} \mathrm{H}$ because there are an equivalent of three $\mathrm{C-O}$ bonds on that carbon and only one $\mathrm{C-O}$ bond on the $\mathrm{CH}_{2} \mathrm{OH}$ group. Since the $\mathrm{H}$ is pointing back, we can now directly see that $1 \rightarrow 2 \rightarrow 3$ is counterclockwise and the configuration is S.
2. Forgetting to look down the bond from the $\mathrm{C}$ to the lowest ranking group (often, but not always, an $\mathrm{H}$). If you look down any other bond (or from the low ranking atom to the carbon) then you will almost certainly get the wrong assignment.
If the molecule is not drawn with the lowest ranking group pointing away, there are a number of options:
1. Particularly when you are first getting started, you should MAKE A MODEL. There is research evidence that shows using an actual, physical model is the best way to learn this particular skill. Be careful to construct the model so that it looks exactly like the drawing. Then you can physically rotate it and identify the configuration.
or
2. You can switch two groups—either redrawing or in your imagination (we strongly suggest you redraw it), so that the lowest group is now pointing back out of the plane. Then the configuration you get will be the opposite of the actual configuration.
or
3. You could imagine yourself moving in space to look down the bond from C to the lowest ranking group. This is quite difficult for some people.
or
4. You could rotate the molecule in your head so that you are looking down the bond from C to the lowest ranking group. This is also quite difficult for some people.
You should always use at least two methods – and make sure you get the same answer both times!
After a while, this skill will come more easily to you.
R and S and D and L isomers:
As you probably already know, most biological molecules are chiral. For example, all of the naturally-occurring amino acids (aside from glycine) have a chiral carbon (see above), sugars have several chiral centers, and so the large molecules made up from these smaller monomers (such as nucleic acids, carbohydrates, lipids, polypeptides, and a range of smaller molecules) contain one or more (sometimes many) chiral carbons. For historical reasons, however, most simple biological molecules (that is the monomers from which polymers are constructed) are referred to as either D or L isomers, rather than R or S. The convention for naming substances as D or L dates back to the early 1900s and has to do with a compound’s similarity to glyceraldehyde which exists as a pair of enantiomers. The enantiomer that rotated light to the right ($+$) was identified as D, and the other one ($–$) as L. For example, the naturally-occurring amino acids are the L-isomer, and many of the simple sugars are the D isomer. It should be noted that a compound such as glucose can be identified as D-glucose, even though it contains 5 chiral centers. The D/L nomenclature is applied to the molecule as a whole, whereas the R/S nomenclature applies to a specific chiral carbon. The D/L nomenclature does not, in fact, predict the direction of rotation of polarized light in most larger molecules—but it is a much simpler way of naming compounds with more than one chiral center.
Speaking of which…
Molecules with two or more chiral (stereogenic) centers
Molecules that have more than one chiral center bring in another level of complexity. It can be quite difficult to draw molecules accurately that have more than one chiral center, and therefore we often turn to a representation that is more stylized than the wedge-dash; this is known as the Fischer projection, named after Emil Fischer (1852-1919) who elucidated the structures of sugars (and who also introduced the D/L nomenclature). Fischer projections are written as vertical and horizontal lines. The carbon backbone is the vertical line, and the other substituents are horizontal. We assume that the horizontal bonds are coming out towards you ($\downarrow$). This convention makes it easier to draw and to assign configuration to the chiral centers (this structure is S).
Now if we look at Fischer projections for a four carbon sugar that has two chiral centers, we see that there can be several possibilities.
Numbering from the aldehyde carbon as carbon 1, both $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are chiral. We also see that there are two pairs of enantiomers: A and B are mirror images of each other and so are C and D. In general, for a molecule with n chiral centers, there are a possibility of $2^{n}$ stereoisomers. When we look at the relationship between isomer A and isomer C (or D) we note that they are not mirror images of each other (the chiral centers are the same at one carbon and different at the other). These compounds are known as diastereomers, that is, stereoisomers that are not mirror images of one another; they do not have identical distances between all of their atoms. Because they have different arrangements in space, diastereomers have different properties, both physical and chemical, and can be separated. In fact, the two compounds have different names: A and B are called erythrose while C and D are called threose. Diastereomers actually belong to the same class of stereoisomers as cis/trans ring compounds (and alkenes), they have the same molecular formula and connectivity, but have different arrangements in space and cannot be interconverted by bond rotations. In contrast, enantiomers have the same atomic arrangements in space, but cannot be superimposed on one another.
So now you might well ask yourself, given that stereoisomers have the same physical properties: how could you possibility separate enantiomers? The existence of enantiomers was first observed by Louis Pasteur (1882-1895), who was actually able to identify different crystalline forms of tartaric acid[3] (actually, the potassium salt), which turned out to be the two enantiomers (but not the meso isomer), which appeared as tiny crystals that were mirror images of each other. In fact, this selective recrystallization is highly unusual, and it is normally not possible to selectively recrystallize isomers.
So why does the stereo-asymmetry of a molecule matter? The answer is that organisms (living systems) are—apparently due to an accident of their origin and subsequent evolution—asymmetric at the molecular level. For example, polypeptides and proteins are polymers of amino acids. While organisms can make and use both D- and L-form amino acids, all of the amino acids used in polypeptide/protein synthesis (with the exception of glycine, which is achiral) exist in L- and D- forms. Only L-form amino acids are used in polypeptides and proteins. The use of only L-form amino acids means that each polypeptide/protein has a distinct three-dimensional shape that influence how other molecules bind to it. For example, a particular drug can be designed to inhibit a particular enzyme (protein-based catalyst).
If the drug is itself chiral, then it is extremely likely that one or the other of its chiral forms will be an effective inhibitor while the others will not. In fact, it is common for drug companies to first patent the racemic mixture of a drug, and then later the purified enantiomers. For example, Fluoxetine (Prozac)($\rightarrow$) is a racemic mixture of the two enantiomers (can you find the chiral center?). In the human body, the two forms of fluoxetine are metabolized into corresponding forms of norfluoxetine, one of which is significantly more active than the other.[4] Another example of the use of a racemic drug is the terrible story of thalidomide, which was marketed in the 1960’s in Europe and Canada as a drug for morning sickness.
The original drug was administered as the racemic mixture but we know that it is the R isomer that is the sedative. The S isomer is teratogenic. That is, it causes birth defects in this case associated with the development of limbs. Many children were born without limbs before Thalidomide was removed from the market. Now we know that even the pure R enantiomer racemizes at physiological pH. In fact, Thalidomide was re-introduced to the market because it is one of the few drugs that can be used to treat leprosy.
Non-chiral species with chiral centers
Just because a compound has a chiral (stereogenic) center, doesn’t mean the actual compound is optically active. Remember that the requirement for a chiral substance is that there is no symmetry element. Consider the set of compounds below.
There appear to be four stereoisomers here, but, in fact, there are only three. Isomers A and B are identical due to the a mirror plane through the center of the molecule (bisecting the $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond), the top half of the molecule is identical to the bottom half. Compounds like this are called meso isomers. The meso isomer is a diastereomer of the pair of enantiomers C and D.
Questions to Answer
• Without looking at the diagram above, construct your own representation and explanation for the various types of isomers that we have encountered. Give an example of each type of isomerism (using two such isomers) and explain how and why they differ from each other and from other types of isomers.
• How is it possible that compounds with chiral carbons are not themselves chiral?
• Do you think it is possible to have a chiral compound that does not have chiral carbons in it? What structural features would you look for? Why?
Questions to Ponder
• Most biomolecules are chiral, how do you think they got that way?
• Most biomolecules are chiral, what are the implications for the synthesis of pharmacologically active compounds?
3.04: In
1. Recall that this is not the case for $\mathrm{sp}^{2}$ or $\mathrm{sp}$ hybridized carbon, which also form pi bonds, where the orbital overlap is destroyed by rotation. Therefore, as we will see later, a much higher input of energy is required for compounds with pi bonds to rotate the bonding carbons relative to each other.
2. See Chapter $1$ and Chapter $4$ in CLUE text.
3. https://en.Wikipedia.org/wiki/Tartaric_acid
4. Wikipedia on fluoxetine: https://en.Wikipedia.org/wiki/Fluoxetine | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/03%3A_Conformations_and_Configurations_-_the_consequences_of_the_three-dimensional_nature_of_carbon_compounds/3.03%3A_Co.txt |
In Chapter $1$, we learned about one of the most fundamental reactions in organic chemistry: nucleophilic substitution. Before we move on, it is important to make sure that you have a good understanding of what the terms nucleophile, electrophile, and leaving group mean and that you are able to predict the products for a range of substrate molecules (electrophiles) with different leaving groups and nucleophiles. In this section, we move forward and look at nucleophilic substitution reactions in more detail by examining the evidence that leads us to understand how the mechanisms of nucleophilic substitutions were determined.
04: Nucleophilic Substitution Part II
One of the most powerful sources of evidence for how a particular reaction proceeds comes from the study of reaction rates or reaction kinetics. In such studies, how the rate of a reaction changes[1] is measured as a function of the concentration of each reactant. One of the most common ways of measuring this change is by using a spectroscopic technique. For example, if the compound absorbs in the UV-VIS region of the spectrum, the absorbance is proportional to the concentration. Therefore, if the concentration of the substance changes, it can be measured by changes in the absorbance. The reaction is carried out several times with all but one of the reactants set as constant; then a different concentration of the remaining reactant is added and the rate of the reaction measured. This is repeated for each reactant over several concentrations. The end result of such a study is to produce what is known as the rate equation; for a generic reaction $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$ the rate equation takes the form: $\text { Rate }=k[\mathrm{A}]^{x}[\mathrm{B}]^{y}$
In this equation, $k$ is the rate constant, and the exponents x and y tell us about how the concentration of each reactant influences the rate. The sum of the exponents (that is, $x+y+\ldots$) is the order of the reaction. For example, if $x = 1$, then the rate is directly related to the concentration of $\mathrm{A}$. If both $x$ and $y = 1$, then the rate is directly proportional to both $[\mathrm{A}]$ and $[\mathrm{B}]$, and the overall reaction order is = 2. If an exponent = 0 then the rate is not dependent on that reactant concentration, and that reactant can be removed from the rate law equation (since $[n]^{0}=1$ no matter what the value of concentration of n is).
The most important idea to remember is that the rate equation only contains the reactants that are involved in the rate-determining step (that is, the slowest step) of the reaction. If the reaction proceeds by a number of steps, then the step with the highest activation energy will be rate-determining, and only those reactants that participate in this step will be present in the rate law.[2] Since the rate law is determined empirically, the rate law provides us with evidence about the mechanism of the reaction.
Evidence for the $\mathbf{S}_{\mathbf{N}} 2$ Mechanism:
The reaction we discussed earlier in the course is known as an $\mathrm{S}_{\mathrm{N}} 2$ reaction that is shorthand for Substitution, Nucleophilic, Second order. We proposed a mechanism for this reaction without providing any empirical evidence, but now let us use some of what you have learned to consider more carefully the evidence for this mechanism.
The reaction is second order:
the first piece of evidence comes from the kinetic rate law. The rate of reaction depends on both the concentration of the substrate and the nucleophile: $\text { rate }=k[\mathrm{RX}][\mathrm{Nu}]$. This means that both must be present in the rate-determining step. The simplest explanation that is consistent with this finding is the one we have already proposed: the nucleophile attacks the electrophilic carbon at the same time as the leaving group leaves. That is, the reaction takes place in one continuous step. A reaction energy diagram in which we plot Energy v reaction progress looks like this ($\rightarrow$). In this reaction, there is only one energy barrier, only one maximum in the reaction pathway. The energy of this barrier is known as the activation energy $\Delta \mathrm{E}_{+}^{+}$. Looking at the reaction diagram, we also note that the reaction is exothermic (or exergonic if we plot Gibbs energy), since the $\Delta \mathrm{E}$ of the overall reaction is negative). The species at the peak of the activation energy barrier is known as the transition state, and its structure and associated energy determines the rate of the reaction.
The structure of the substrate affects the rate:
Perhaps you have noticed in our earlier discussions of nucleophilic substitution that the organic substrate was always either a methyl or primary carbon attached to a good leaving group. The reason was that in the reactions we have considered, both the rate and mechanism of reaction is highly dependent on the structure of the substrate. As the number of methyl groups attached to the primary carbon increases (from 0 for a methyl group itself to three [tertiary]) the reaction rate slows, as shown. The rate of an $\mathrm{S}_{\mathrm{N}} 2$ reaction for a tertiary substrate is negligible.
So two questions arise: first, why are these reaction rates different? and second, why is this change in reaction rates evidence for the $\mathrm{S}_{\mathrm{N}} 2$ mechanism? Both can be answered by taking a closer look at the reaction from a molecular perspective. Remember, all of the reactants are dissolved in a solvent; thermal motion leads to their colliding with one another and with solvent molecules. For the nucleophile and the substrate to react with each other, they first have to collide with one another. For a reaction to occur, that collision has to transfer enough energy so that the complex (substrate + nucleophile) can form the transition state—moreover to form the transition-state molecule, the molecules must collide with one another in the correct orientation. Once formed, the transition state can decay to form the products of the reaction.
Recall that our proposed structure for the transition state for this reaction has the central carbon connected to five groups: the incoming nucleophile, the leaving group, and the three other substituents that do not change during the reaction (they are not part of the reaction). As the bond forms between the nucleophile and the substrate carbon, and the bond breaks between the carbon and leaving group, the carbon changes its hybridization state. What does that mean? In the substrate molecule, the reacting carbon is attached to surrounding groups ($\mathrm{H-}$ or $\left.\mathrm{CH}_{3}-\right)$) with bonds formed from $\mathrm{sp}^{3}$ orbitals. In the transition state, this carbon is still attached to those groups that will remain in the product molecule, but now with bonds formed from $\mathrm{sp}^{2}$ orbitals. Additionally, it is still attached to both the leaving group and the incoming nucleophile using a p orbital to form these partial bonds. You can think of this process as electron density being funneled from the nucleophile through the carbon and out the other side to the leaving group. However, for this to occur the nucleophile can only begin to bond when it approaches from the back of the bond to the leaving group $\rightarrow$.
At this point, you might well find yourself asking: what does all this have to do with the structure of the substrate? For a reaction to occur, the only productive collisions are those where the nucleophile begins to form a bond with the back part of the $\mathrm{sp}^{3}$ hybrid orbital; but the structure of the substrate influences the probability of such an event. In the tertiary substrates (for example $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}$) the approach to the substrate is hindered by the bulky alkyl groups such that the probability of the nucleophile interacting with the reactive center is low. This phenomenon is called steric hindrance and provides an explanation for the order of reaction of $\mathrm{S}_{\mathrm{N}} 2$ reactions.
$\mathrm{S}_{\mathrm{N}} 2$ reactions at a chiral center:
Another piece of evidence for the $\mathrm{S}_{\mathrm{N}} 2$ mechanism is what happens when an $\mathrm{S}_{\mathrm{N}} 2$ reaction takes place at a chiral center (within a molecule). It turns out that the configuration at that center is changed; the carbon inverts (like an umbrella blowing inside out in the wind) so that an S enantiomer is converted to an R enantiomer. In fact, it is possible to follow the progress of an SN2 reaction involving a chiral center using a polarimeter (the instrument used to measure optical activity); as the reaction proceeds to completion, the optical rotation of the solution changes over time. For each particular substrate, the direction and magnitude of the rotation for the product will be different. This phenomenon is called the Walden inversion and provides another piece of evidence to support the proposed reaction mechanism.
The role of solvent in an $\mathrm{S}_{\mathrm{N}} 2$ reaction:
$\mathrm{S}_{\mathrm{N}} 2$ reactions are generally carried out in a solvent (why is that?). Empirical studies reveal that such reactions proceed more rapidly when carried out in what is known as a polar aprotic solvent. So what is a polar aprotic solvent? The term means that the solvent is polar but without acidic protons. Examples of polar aprotic solvents are acetone, dimethyl formamide (DMF), and dimethylsulfoxide (DMSO): each is polar, but lacks a potentially acidic proton such as the H that is bonded to the $\mathrm{O}$ in ethanol $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ or in water $\mathrm{H-O-H}$. Water (and methanol and ethanol) is a polar protic solvent. In a polar aprotic solvent, the negative end of the $\mathrm{C=O}$ or $\mathrm{S=O}$ dipole is localized to the $\mathrm{O}$, while the positive end is diffuse and delocalized. For example, in acetone, the oxygen has a $\delta-$ charge on the oxygen while the positive charge of the dipole is delocalized over both the $\mathrm{C}$ and the methyl groups as shown in the electrostatic potential map ($\rightarrow$). In practice, polar aprotic solvents can solvate cations well through interactions with the localized negative end of the dipole, but they cannot solvate anions very well.
Recall that solvation is an interaction that lowers the energy of the system, making it more stable (less reactive). Therefore, a solvent that leaves the nucleophile (the anion) unsolvated will make it more reactive. In contrast, a polar protic solvent (such as water or ethanol) can solvate the nucleophile, through interactions with the positive end of a dipole that is localized on an acidic $\mathrm{H}$, stabilizing the nucleophile and making it less reactive. In summary, $\mathrm{S}_{\mathrm{N}} 2$ reactions occur in one step with inversion at a chiral center. Such reactions are generally faster for unhindered substrates and are accelerated when carried out in polar aprotic solvents. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/04%3A_Nucleophilic_Substitution_Part_II/4.01%3A_Kinetics_and_Mechanisms-.txt |
If you have already taken a laboratory course in chemistry, you have no doubt observed that you do not get a 100% yield for a particular reaction and often more than one product is generated. This is not (generally) a case of faulty experimental technique, but rather reflects the complexity of reaction systems. Given our experience using evidence to support proposed reaction mechanisms, let us take a look at another set of conditions for nucleophilic substitutions. Consider the reaction:
Relative Rates of $\mathrm{S}_{\mathrm{N}} 1$ Reactions
In this scenario, water is both the nucleophile and the solvent. Water is not a very strong nucleophile and it is a protic solvent. Under these conditions, a nucleophilic substitution takes place, but this reaction differs in several empirically observable ways from the $\mathrm{S}_{\mathrm{N}} 2$ reactions discussed earlier.
1. The rate of reaction depends only on the substrate. The reactivity of the nucleophile is irrelevant. The rate equation for these reactions is: $\text { rate }=k[\mathrm{RBr}]$. The reaction is first order, and is therefore named as an $\mathrm{S}_{\mathrm{N}} 1$ reaction (Substitution, Nucleophilic, First Order).
2. The relative rates of reaction for substrates are reversed from the $\mathrm{S}_{\mathrm{N}} 2$ reaction. That is, reaction rate is higher for the tertiary and lowest for the primary form.
3. When a chiral center undergoes an $\mathrm{S}_{\mathrm{N}} 1$ reaction, the product contains a mixture of both possible enantiomers, rather than the inversion of configuration found with $\mathrm{S}_{\mathrm{N}} 2$ reactions..
4. The reaction is accelerated by polar protic solvents (which slow $\mathrm{S}_{\mathrm{N}} 2$ reactions).
How can we explain these behaviors? Our assumption is that a different reaction mechanism is involved; compared to the $\mathrm{S}_{\mathrm{N}} 2$ reactions we have been considering. What might that mechanism be?
Given that the observed reaction rate is dependent only on the substrate concentration (1) we can assume that only the substrate molecule is involved in the rate-determining step. So how does the reaction begin? We must assume that the reaction involves bond-breaking (since there is nothing else that can happen if there is only one molecule in the rate-determining step. Since bond breaking requires energy, thermal collisions with solvent molecules must drive this bond breaking event. However, we also know that the the solvent molecule does not take part in this step of the reaction (because it is not in the rate law). One possibility is that the bond to the leaving group breaks, resulting in a positively charged carbon (a carbocation) and the leaving group anion $\rightarrow$. This is the first instance we have seen of a carbocation. The formation of this carbocation requires energy (since the bond is breaking), and we can infer that it has a high activation energy. Carbocations themselves are high-energy species that are very reactive. However, they are different from transition states in that it is possible to generate and detect the a carbocation—it usually has a short lifetime—but we can detect its presence by spectroscopic methods. This is different from a transition state, which is the highest-energy species on the reaction energy profile. Transition states exist for only one molecular vibration and have lifetimes on the order femtoseconds.
We might expect that such a carbocation would rapidly react with any potential nucleophile present, which in this case is the solvent water molecule.
Although water is a poor nucleophile, it will react with the highly reactive carbocation to give the intermediate protonated form. This is followed by a proton transfer to another water (solvent) molecule to form the final product.
Such a mechanism satisfies our experimental observations. It has a slow (rate limiting, high activation, energy requiring) first step, and a faster (lower activation energy) second step. Because of the differences in the activation energies of the two steps, only the first step is involved in determining the overall reaction rate. To figure out which is the rate determining step we can see that $\Delta \mathrm{G}^{\dagger}$ for step 1 is larger than $\Delta \mathrm{G}^{\dagger}$ for step 2.
Why does the structure of the substrate matter?
The formation of intermediate carbocation is the rate-determining step in such a reaction, and therefore it follows that the structure of the carbocation has an impact on the activation energy of the reaction, and therefore its rate. The more stable the carbocation, the lower the energy of the transition state leading to the carbocation.[3] Factors that stabilize the carbocation will also stabilize the transition state and lower the activation energy.
There are two mechanisms that can stabilize carbocations and both predict that, as the number of alkyl groups attached to the $\mathrm{C+}$ increase, so will its stability. One mechanism we have already encountered is induction. The alkyl groups attached to the central carbon, are polarizable and their electron density is attracted towards the positive charge of the central carbon thus delocalizing the positive charge over the alkyl groups. The more alkyl groups attached to the central carbon, the more pronounced this stabilization is. The second mechanism, known as hyperconjugation, also delocalizes the positive charge. In hyperconjugation, the electron density from any adjacent $\mathrm{C-H}$ or $\mathrm{C-C}$ bond can overlap with the empty p orbital on the $\mathrm{sp}^{2}$ hybridized carbocation, forming a sort of pi bond and, again, delocalizing the positive charge over the rest of the molecule. The more alkyl groups present (attached to the $\mathrm{C+}$), the more pronounced this effect will be.
Together, both induction and hyperconjugation explain why an $\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds faster with tertiary substrates. The tertiary carbocation is more stable (relative to secondary and primary carbocations) so that the reaction has a lower activation energy.
Why do chiral centers racemize?
The answer to this question lies in the structure of the carbocation. It is a planar, $\mathrm{sp}^{2}$ hybridized, symmetrical structure. Once formed, it can be attacked from either side by a nucleophile; in simple compounds which side the carbocation will be attacked on involves a random collision event, giving a mixture of enantiomers.
Why are $\mathrm{S}_{\mathrm{N}} 1$ reactions accelerated by polar protic solvents?
Remember a polar protic solvent (such water or ethanol) contains a dipole: a partially positive and partially negative domain.
Attack at a chiral center gives a racemic mixture of products.
This solvent molecule dipole serves two functions: it can solvate the leaving group, in effect helping to remove it from the carbocation through interactions with the positive end of the solvent dipole and it can solvate the carbocation through interactions with the solvent dipole’s negative domain. In essence, the solvent assists in the ionization of the leaving group, and it lowers the energy of the intermediate carbocation.
$\mathrm{S}_{\mathrm{N}} 1$ reactions in resonance-stabilized systems:
As we have seen, $\mathrm{S}_{\mathrm{N}} 1$ reactions tend to occur when a stabilized carbocation intermediate can be formed. In addition to those tertiary cations discussed earlier, which are stabilized by induction and hyperconjugation, there are also occasions when even primary carbocations can be formed if there is a way to stabilize them. For example, any primary carbon that can be conjugated with a pi system can be stabilized by resonance. For example, a benzylic (that is any carbon attached to a benzene ring) carbocation can be stabilized by delocalizing the positive charge into the benzene ring. This lowers the activation energy for the reaction and makes an $\mathrm{S}_{\mathrm{N}} 1$ reaction possible.
A similar phenomenon can happen in substrates with leaving groups in the allylic position: that is, on the carbon next to a double bond, where the resulting carbocation can be resonance stabilized by delocalization onto the pi system of the double bond. We will discuss this phenomenon in more detail later (in Chapter $7$).
$\mathrm{S}_{\mathrm{N}} 1$ or $\mathrm{S}_{\mathrm{N}} 2$?
Before we move on, let us review what we know about $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions. While it may seem a little confusing, there are a number of factors that can help us predict what the potential mechanism for a reaction might be, and also to predict the product. As we move forward to more complex reaction systems, it will be important to remember that there is typically more than one pathway a reaction can take, but by understanding how reactions occur we can adjust conditions so that the product we want is the major product. The table below summarizes what we have discussed so far with regards to $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions.
$\mathrm{S}_{\mathrm{N}} 1$ $\mathrm{S}_{\mathrm{N}} 2$
Substrate Tertiary > secondary > benzylic~allylic
> primary > methyl
Methyl > primary > secondary> tertiary
Leaving Group Good leaving groups increase rate by lowering the energy of the TS Good leaving groups increase rate by lowering the energy of the TS Accelerated by strong nucleophiles
Nucleophile Non-(Bronsted) basic nucleophiles, weak nucleophiles (strong bases promote elimination—see next section)
Solvent Polar (solvates carbocation) protic (helps pull off the leaving group and solvates it. Polar aprotic—solvates the cation leaving the nucleophile unsolvated and more reactive.
Stereochemistry Racemization results from attack at both sides of the planar carbocation Inversion of configuration if reaction takes place at the chiral center
In practice, tertiary substrates only undergo $\mathrm{S}_{\mathrm{N}} 1$ and methyl and primary only undergo $\mathrm{S}_{\mathrm{N}} 2$, it is the secondary substrates where the ambiguity lies, and for that we have to consider the other factors such as solvent and nucleophile strength. However, reactions that generate carbocations may also undergo other reaction pathways besides substitution. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/04%3A_Nucleophilic_Substitution_Part_II/4.02%3A_The_%28mathrmS_mathrmN_1%29_Reaction.txt |
Reactions that generate carbocations can undergo reaction pathways besides substitution.
Consider this reaction:
The substrate is a secondary alkyl halide, and since the solvent is methanol, a polar protic solvent, and with a weak nucleophile, we would expect an $\mathrm{S}_{\mathrm{N} 1$ reaction to occur, via a carbocation at the secondary carbon. The first product (A) is exactly that, but how did the second product (B) form? To understand this, let us look at the mechanism of the reaction. the first step is the ionization of $\mathrm{Br}^{-}$ to leave behind the secondary carbocation (which produces product A). Product B must have been formed by nucleophilic attack of the $\mathrm{O}$ in methanol onto a different carbocation. The precursor to B is a tertiary carbocation, which is more stable than the secondary, and it is formed by what is known as a hydride shift in which the hydrogen shifts with its pair of electrons.
Formation of carbocations is often accompanied by a skeletal rearrangement: Here a 1,2 hydride shift occurs
This rearrangement produces a more stable intermediate that then undergoes reaction; it can also involve the shift of an alkyl group with its electrons from one carbon to the next.
4.04: Eliminations
Skeletal rearrangements are a drawback of exposing substrates to conditions in which the leaving group ionizes. Unfortunately, they are not the only complication—there is also the possibility that another type of reaction may occur—an elimination to produce an alkene.
In this case, the reaction proceeds through the same carbocation intermediate, and then a proton is eliminated from a carbon next to the carbocation (a $\beta$ carbon).
An E1 elimination reaction
This is called an elimination reaction, and it is first-order since the rate-determining step is the formation of the carbocation, and so it is an $\mathrm{E} 1$ reaction. In fact, $\mathrm{S}_{\mathrm{N} 1$ reactions are often accompanied by $\mathrm{E} 1$ reactions (and vice versa). If there is a possibility of forming more than one alkene (because there are different $\beta$ carbons), usually the most substituted alkene is the major product.[4] For example, alcohols undergo $\mathrm{E} 1$ eliminations when treated with concentrated sulfuric acid. In this case, there is no substitution product because the sulfate anion is not a good nucleophile (it is highly stabilized by resonance).
Here, the major product has three alkyl groups on the double bond, while the minor product only has two. In fact, this acid catalyzed dehydration of alcohols is quite a synthetically useful reaction, but if the substrate has the potential for rearrangements (i.e. the resulting carbocation can be stabilized by a hydride or alkyl shift), then there is the potential for the formation of even more products. For example:
Elimination and Rearrangements
Obviously, these kind of rearrangements and eliminations are not synthetically useful on substrates that are prone to skeletal rearrangements. However, there is an elimination reaction that typically provides us with much more control.
The $\mathrm{E} 2$ Reaction.
As we will see shortly, the synthesis of alkenes by elimination of $\mathrm{H–L}$ (where L is a leaving group) is an important reaction, but we are much more likely to have control over the products if the reaction does not go through a carbocation. That is, if we can simultaneously eliminate both the $\mathrm{H}^{+}$ and the leaving group there is less chance of side reactions. This reaction is an $\mathrm{E} 2$ reaction (elimination second order), and is promoted by the presence of a strong base. For example, the reaction of t-butyl bromide with hydroxide (or any strong base), shown above. In this case, there is little substitution product, and instead the base simultaneously removes a proton from the $\beta$ carbon as shown.
An E2 Elimination
The rate therefore depends on both the substrate and the base:$\text { Rate }=k[\mathrm{RL}][\text{base}]$—that is, a second order reaction. But wait—didn’t we see that strong bases are good nucleophiles? From the beginning, we have shown that methyl and primary substrates with strong nucleophiles undergo $\mathrm{S}_{\mathrm{N} 2$ reactions. How can we bring about an elimination in this case? Well, just as a sterically hindered substrate will not undergo an $\mathrm{S}_{\mathrm{N} 2$ reaction, we can use a sterically hindered base to avoid such reactions. If the base is too bulky around its reactive site, then it cannot approach the substrate at the electrophilic center, and will instead pick of a proton from one of the $\beta$ carbons. One such base is the salt of t-butanol, potassium t-butoxide ($\mathrm{tBuOK}$), which is used to bring about $\mathrm{E} 2$ eliminations for primary and secondary substrates.
Just as with $\mathrm{E} 1$ reactions, the most substituted double bond is the major product:
Another factor that must be accounted for in $\mathrm{E} 2$ reactions is that for such a reaction to occur, the leaving group and the proton that is eliminated must be in an orientation that allows the rehybridizing orbitals to overlap in the transition state.
This orientation is called antiperiplanar, and this need for a specific arrangement for the reaction to occur is called the stereoelectronic requirement.
Antiperiplanar (or trans diaxial) stereoelectronic requirement for $\mathrm{E} 2$ eliminations
In systems where free rotation is possible, this lining up of the groups is not usually a problem, but if the elimination is to take place in a ring system, then the $H$ and the leaving group must be trans and diaxial, otherwise the stereoelectronic requirement cannot be met.
So, for example, 1-bromo-2-methylcylohexane produces 3-methylcyclohexene, not 1- methylcyclohexene. This is because the hydrogen on the same carbon as the methyl group must be equatorial (if the methyl group is trans). Therefore the axial hydrogen on the other beta carbon is eliminated instead.
$\mathrm{E} 2$ elimination requires a trans diaxial (antiperiplanar) conformation
Nucleophile/base strength Methyl Primary Secondary Tertiary
Strong/strong e.g. ${}^{-} \mathrm{OCH}_{3}$ $\mathrm{S}_{\mathrm{N}} 2$ $\mathrm{S}_{\mathrm{N}} 2$ $\mathrm{E} 2$ $\mathrm{E} 2$
Strong/weak e.g. $\mathrm{RSH}$, halide ions $\mathrm{S}_{\mathrm{N}} 2$ $\mathrm{S}_{\mathrm{N}} 2$ $\mathrm{S}_{\mathrm{N}} 2$ $\mathrm{NR}$
Weak/strong e.g. ${}^{-} \mathrm{OBu}^{\mathrm{t}}$, NaH $\mathrm{NR}$ $\mathrm{E} 2$ $\mathrm{E} 2$ $\mathrm{E} 2$
Weak/weak e.g. $\mathrm{H}_{2} \mathrm{O}$, $\mathrm{CH}_{3} \mathrm{OH}$ $\mathrm{NR}$ $\mathrm{NR}$ $\mathrm{S}_{\mathrm{N}} 1 / \mathrm{E} 1$ $\mathrm{S}_{\mathrm{N}} 1 / \mathrm{E} 1$
One thing is certain: the interplay between substrate and solvent can be very confusing. It is impossible to memorize all the possible outcomes from a given set of reaction conditions, and although some generalizations can be made, the best way to manage all of this is to try to work through the reaction by writing a plausible mechanism. That being said, the following table summarizes some of the potential outcomes by type of substrate and strength of the nucleophile/base.
4.05: In-Text References
1. i.e. the change in concentration of reactant divided by the time $d[\mathrm{A}] / dt$ (where $\mathrm{A}$ is one of the reactants). For more information, see CLUE Chapter $8$.
2. See Chemistry, Life, the Universe and Everything: Chapter $8$ for a longer discussion.
3. This hypothesis is known as Hammond’s postulate. It states that for an endothermic reaction, the transition state is closer in structure to the product (in this case the carbocation), and for an exothermic reaction the transition state is closer in structure to the reactant. See https://en.Wikipedia.org/wiki/Hammond%27s_postulate
4. As we will discuss shortly, this is because the most substituted alkene is the most stable. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/04%3A_Nucleophilic_Substitution_Part_II/4.03%3A_Rearrangements-_A_Consequence_of_Generating_Unstable_Carbocations.txt |
When a carbon is bonded to one or more electronegative atoms, it takes on a partial positive charge and it is electrophilic. Such electrophilic carbons can undergo nucleophilicsubstitution or elimination reactions, or both, depending upon the structures of the reacting molecules, the strength of the nucleophile, and the type of solvent in which the reaction occurs. Now, we turn to reactions that electron-rich carbon species can undergo.
Alkenes and alkynes.
Both alkenes and alkynes are “unsaturated,” which means that they contain double or triple carbon-carbon bonds. The term unsaturated comes from the fact that more $\mathrm{H}$ atoms can be added to these molecules across the double or triple bonds. A simple alkene contains a pair of carbons linked by a double bond; this double bond consists of a sigma bond and a pi bond. The sigma bond is formed by end-to-end overlap of $\mathrm{sp}^{2}$ hybrid orbitals, and the pi bond by side-to-side overlap of the p orbitals. A pi bond has two lobes of electron density above and below the plane of the molecule. There are a number of consequences to this arrangement:
1. the resulting region of the molecule is planar (the molecule is said to have trigonal planar geometry),
2. the electron density between the two carbons is high because there are four electrons in this region instead of two, and
3. rotation around a double bond is constrained (in contrast to rotation around a single bond).
Rotation around a double bond requires breaking the overlap of the pi bond and its subsequent reformation. As with all bond-breaking phenomena, the bond-breaking step requires energy; in fact, significantly more energy than is required to bring about rotation around a single bond where no bond-breaking occurs. As we will see, these three factors have a marked effect on the behavior of alkenes.
Alkynes are compounds that contain triple bonds. The triple bond consists of one sigma bond formed from end-to-end overlap of sp-hybrid orbitals and two pi bonds formed from side to side overlap. The carbons are $\mathrm{sp}$-hybridized and the molecule is linear in the region of the triple bond; again rotation around a triple bond is constrained—two pi bonds must be broken for it to occur (which requires an input of energy). This bonding arrangement results in a very electron rich $\mathrm{C-C}$ region with the sigma bond inside what looks like a cylinder of pi electron density.
Naming Alkenes
Since alkenes have restricted rotation around the $\mathrm{C=C}$ group, they can exist as stereoisomers. For example, in 2-butene there is a methyl and an $\mathrm{H}$ bonded to each of the double-bonded carbons (carbons 2 and 3 of the molecule). Because the $\mathrm{C=C}$ group is planar, the $\mathrm{CH}_{3}$ groups can be on either the same (“cis”) or opposite (“trans”) sides of the double bond ($\rightarrow$); this cis/trans nomenclature is similar to that we used with cyclohexane rings. As the groups attached to each carbon get more complex, such nomenclature quickly becomes confusing. To cope, we turn to another established naming scheme; in this case, the Cahn-Ingold-Prelog convention we previously used with chiral centers. This involves ranking the groups linked to each double-bond carbon. If the high groups are together (same side), the name is prefixed by Z (from the German word for together: zusammen). If they are on opposite sides, they are labeled E (entgegen; away). E and Z isomers are diastereoisomers: they have the same connectivity but neither can be superimposed on its mirror image. In E-3-bromo-2-pentene, the $\mathrm{CH}_{3}$ and $\mathrm{CH}_{2} \mathrm{CH}_{3}$ groups are closer to one another than they are in Z-3-bromo-2-pentene; the result is that they have different physical and chemical properties. These differences make it possible to separate E and Z isomers (and cis/trans since they are just a special case of E/Z) from one another.
Stability of alkenes:
Elimination reactions that produce alkenes tend to favor the most substituted alkene as the major product. The relative stabilities of various alkenes can be determined by reacting the alkene with hydrogen and determining the enthalpy change ($\Delta \mathrm{H}$).
For example, shown ($\rightarrow$), the three different alkenes produce the same product, and therefore the differences in the energy released must arise from the fact that the initial alkenes have different energies. The more alkyl groups attached to the double bond, the more stable (less reactive) the alkene is, and therefore a lower amount of energy is released. Molecular stability in alkenes is attributed to the same causes as the relative stabilities of carbocations; alkyl groups stabilize the pi bond by hyperconjugation and induction.
05: Alkenes and Alkynes
The double-bonded carbons of an alkene are electron-rich, that is, the electron density is high in the region of the double bond. Therefore, the “signature” reaction of alkenes involves initial attack on an electrophile. Instead of a substitution, alkenes undergo electrophilic addition, a reaction in which a two-component reactant adds across the double bond. The reaction begins with an electrophilic attack by the double bond onto the reactant which produces a carbocation that then undergoes nucleophilic attack. In the case of unsymmetrical alkenes (where the groups attached to the double-bonded carbons are not exactly the same), the most stable carbocation is produced. This reaction is regioselective, that is, we can predict the orientation of reactant addition across the double bond. If we designate the reagent as $\mathrm{E}$ (for electrophile) or $\mathrm{N}$ (for nucleophile), the reaction would proceed as outlined below.
Reagent Electrophile Nucleophile Typical Conditions
$\mathrm{HBr}$ $\mathrm{H}^{+}$ $\mathrm{Br}^{-}$ Low Temp
$\mathrm{H}_{2}\mathrm{O}$ $\mathrm{H}^{+}$ $\mathrm{H}_{2}\mathrm{O}$ (with loss of $\mathrm{H}^{+}$ after addition) Aqueous acid/Low Temp
$\mathrm{ROH}$ $\mathrm{H}^{+}$ $\mathrm{ROH}$ $\mathrm{ROH} / \mathrm{H}_{3} \mathrm{O}^{+}$
$\mathrm{Br}_{2}$ $\mathrm{Br}^{+}$ $\mathrm{Br}^{-}$ $\mathrm{Br}_{2} / \mathrm{CCl}_{4}$
$\mathrm{BrOH}$ $\mathrm{Br}^{-}$ ${}^{-} \mathrm{OH}$ $\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}$
$\mathrm{BrOR}$ $\mathrm{Br}^{+}$ ${}^{-} \mathrm{OR}$ $\mathrm{Br}_{2} / \mathrm{ROH}$
The intermediate carbocation is the tertiary carbocation, (rather than the primary carbocation that would be produced by addition to the $\mathrm{=CH}_{2}$ end of the double bond). This pattern of reaction is referred to as Markovinkov addition, after the person[1] who first discovered that $\mathrm{HBr}$ adds in this way to a double bond. We can classify many reagents as combinations of electrophile and nucleophile and, in this way, predict how they will add across the double bond. Examples of such reagents are shown ($\uparrow$). Rather than memorizing the product of every type of addition across a double bond, it is much more productive to write a mechanism by determining which part is the electrophile, adding it to give the most stable carbocation, followed by the nucleophile.
Additions to alkenes are reversible:
Let us now take a closer look at the addition of water across a double bond. Such a reaction can be accomplished by reacting the alkene with dilute sulfuric acid at low temperatures. The first step is addition of a proton to produce the most stable carbocation—which is then attacked by water (the nucleophile). The final product is the alcohol that forms after a proton is transferred to water. In this case, we can consider the proton (or more accurately $\mathrm{H}_{3}\mathrm{O}^{+}$) as a catalyst since it is regenerated at the end of the reaction sequence.
Acid-catalyzed addition of water across a double bond
At this point you might be asking yourself: well didn’t we just talk about the reverse reaction—that is, the elimination of $\mathrm{H}_{2}\mathrm{O}$ from alcohols to give alkenes? Indeed we did! Many organic reactions are reversible[2], it is just a matter of manipulating the conditions. The exact reaction conditions will determine which reaction is favored.
As is the case with most addition reactions, the addition of water across an alkene is exothermic, that is, $\Delta \mathrm{H}$ is negative because stronger (sigma) bonds are formed during the reaction and energy is released into the environment. A typical energy diagram is shown below.
Reaction energy diagram for addition/elimination across a double bond.
This means that $\Delta \mathrm{H}$ for the elimination reaction must be positive (i.e. going from right to left on the diagram above). The question then is: why does an elimination reaction ever occur? To answer that, we have to recall that the thermodynamic criterion for a reaction to proceed is not simply a negative enthalpy change, but rather a negative change in the Gibbs change ($\Delta \mathrm{G}$). Recall that $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$. The change in entropy also influences the thermodynamic favorability of a reaction. In the elimination reaction, two molecules (alkene and water) are produced one alcohol molecule –the entropy change will be (Recall that entropy is associated with the number of possible arrangements of the system. Since two molecules will have more possible arrangements than one, this reaction will always be accompanied by an increase in entropy of the system). To encourage the equilibrium to shift to the right (the addition reaction) we need increase the temperature, which will increase the magnitude of the $-\mathrm{T} \Delta \mathrm{S}$\) term, making $\Delta \mathrm{G}$ more negative (assuming that $\Delta \mathrm{S}$ is positive). In general, additions to double bonds are carried out at lower temperatures, while elimination reactions involve heating the reaction solution. Another way to influence the equilibrium state is to change the relative concentrations of reactants or products. Because water is a reactant, increasing the concentration of water shifts the equilibrium position towards the addition product while lowering the water concentration favors the elimination reaction.
Specific reagents for additions across a double bond that reduce the carbocation problem
The problem with many of these simple addition reactions to a double bond is that they generate carbocations, which as we have seen already can lead to further reactions, resulting in skeletal rearrangements and the production of racemic mixtures (rather than a single stereoisomer). To address this issue, a number of reagents have been developed that minimize this problem. For example, a reagent that involves mercuric acetate ($\mathrm{Hg}(\mathrm{OAc})_{2}$) and sodium borohydride ($\mathrm{NaBH}_{4}$) as an intermediate can be used to add $\mathrm{H}_{2}\mathrm{O}$, (or alcohol) across a double bond ($\downarrow$).
The reaction involves a mercury-stabilized cation ($\rightarrow$) that prevents unwanted rearrangements. The product is still a Markovnikov product (see above) but is often formed more cleanly, that is, without unwanted alternatives.
Another set of reactions that can be used to constrain molecular rearrangements and lead to stereospecific products are those that begin with the addition of bromine across the double bond. The simplest of these co-reactions is addition of $\mathrm{Br}_{2}$ itself; since $\mathrm{Br}$ is a large polarizable atom, the bromine molecule can become polarized and interact with the double bond as shown ($\downarrow$) to form a bromonium ion (rather than a carbocation).
The bromonium ion can now undergo nucleophilic attack at either carbon (since in this example they are the same, that is, they are attached to identical groups), to produce the trans-dibromo addition product. The trans product is formed because the second step is an $\mathrm{S}_{\mathrm{N}} 2$ reaction with the bromide nucleophile attacking the carbon from the back-side.
Addition of $\mathrm{Br}_{2}$ is accomplished by using a reaction solvent such as carbon tetrachloride that does not interfere with the reaction. If water or an alcohol is used as the solvent, then attack on the bromonium ion comes from the solvent acting as the nucleophile in the second step.
Again, the addition is trans, but now an incoming nucleophile ($\mathrm{H}_{2}\mathrm{O}$) will attack the carbon that is most carbocation-like, that is it is the most stabilized, as shown here $\rightarrow$. The reaction is both regiospecific and stereospecific. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/05%3A_Alkenes_and_Alkynes/5.01%3A_Reactions_of_Alkenes-_Electrophilic_Addition.txt |
While the heading for this section is called “anti-Markovnikov” addition, this does not mean that the reaction mechanism is actually different. In the two examples we will discuss here, the difference is merely that the first addition to the double bond is not the H, which as we will see makes it appear that we have added a particular reagent the opposite way to the normal addition. For example, if we want to add water across the double bond in to give the anti-Markovnikov product a different set of reagents is used: a Lewis acid-base complex of $\mathrm{BH}_{3}$ and the ether tetrahydrofuran ($\mathrm{THF}$), followed by a solution of hydrogen peroxide in base. This reagent adds across the double bond in the direction that you would expect, that is the electrophile (Lewis acid) boron adds to the least substituted carbon, but atthesametime, a hydrogen adds to the most substituted carbon from the same side of the molecule.
Mechanism of syn addition of $\mathrm{BH}_{3}$ across the double bond
This process happens twice more, and then the boron species is replaced by reaction with hydrogen peroxide and sodium hydroxide.
Mechanism of removal of boron moiety from the double bond
The overall reaction appears to have added the elements of water in an anti-Markovinkov direction. This reaction is not only regiospecific, but it is also stereospecific. The $\mathrm{H}$ and $\mathrm{OH}$ are added on the same (cis) side of the double bond and it is termed a syn addition.
Anti-Markovnikov addition of $\mathrm{HBr}$ across a double bond.
Another reaction which appears to violate what we have learned about the regiochemistry of addition across double bonds is the reaction of an alkene with $\mathrm{HBr}$ in the presence of light or peroxides. In contrast to the reaction we discussed previously, under conditions of light and in the presence of peroxides, the $\mathrm{HBr}$ adds in the reverse direction. Clearly something different is happening here: the reaction is proceeding by another $\mathrm{Br}$ mechanism. The clue is the presence of peroxides, which almost always signify that a reaction is proceeding via a radical mechanism rather than a polar mechanism.
Radicals are species with unpaired electrons, and, as such, are very reactive. The reaction begins with an initiation step in which the peroxide (which contains a weak $\mathrm{O–O}$ bond) is broken homolytically to give two oxygen radicals. These react with $\mathrm{HBr}$ by abstracting a hydrogen, and leaving a bromine radical. Note that the oxy radical abstracts $\mathrm{H}$ and not $\mathrm{Br}$, because $\mathrm{Br}$ is a more stable radical than $\mathrm{H}$. Bromine radical is a large polarizable species and which can help stabilize the unpaired electron. A hydrogen radical is actually a hydrogen atom, it is highly unstable and reactive.
Note: when a mechanism involves single electrons moving (as in a homolytic bond cleavage, or any reaction of a radical species) we use what is called a fishhook arrow—with only one head, rather than the typical arrow that denotes movement of two electrons.
The resulting bromine radical now reacts with the alkene double bond to produce the most stable intermediate, which is (just as in the carbocations) the tertiary. Carbon radicals show the same trends in stability as carbocations for the reason that they are also electron deficient and can be stabilized by the same mechanisms as carbocations (induction and hyperconjugation). The resulting carbon radical now abstracts an H from another molecule of HBr, to produce the anti-Markovnikov addition product, plus another bromine radical that can begin the cycle again. This is called a radical chain reaction—because it produces another reactive species that can continue the chain reaction.
Note: Even though this reaction produces a different addition product than the typical addition of $\mathrm{HBr}$ across the double bond, the principles guiding the reaction are the same. The first addition produces the most stable intermediate; the difference is that bromine adds first. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/05%3A_Alkenes_and_Alkynes/5.02%3A_Anti-Markovnikov_Addition_across_Double_Bonds.txt |
The historical meaning of “reduction” involved reactions with hydrogen ($\mathrm{H}_{2}$), and conversely, oxidation meant reaction with oxygen ($\mathrm{O}_{2}$). This makes sense from the perspective that carbon is slightly more electronegative than hydrogen, so that a $\mathrm{C-H}$ bond is polarized as $\mathrm{C}^{\partial-}$ and $\mathrm{H}^{\partial+}$. Therefore, adding hydrogen to a $\mathrm{C=H}$ will increase (slightly) the negative charge on the carbon. (Similarly, a $\mathrm{C-O}$ bond is polarized $\mathrm{C}^{\partial+}$ and $\mathrm{O}^{\partial-}$, so that adding more oxygens to a carbon increases the amount of positive charge on the carbon.) Even today we refer to adding hydrogen across pi bonds as a reduction. However, alkenes do not normally react with hydrogen; typically a catalyst (usually a transition metal) is necessary for the reaction to occur. In general, the catalyst is supplied as a finely divided powder adsorbed onto an inert substance such as charcoal. The, most common catalysts are platinum or palladium on charcoal ($\mathrm{Pt} / \mathrm{C}$ or $\mathrm{Pt} / \mathrm{C}$). Typically, the substance to be reduced is dissolved in a solvent, the catalyst is added, and then hydrogen is bubbled through the mixture. The catalyst adsorbs both $\mathrm{H}_{2}$ and the alkene onto its surface and this interaction weakens both the $\mathrm{H}_{2}$ bond and the pi bond. The hydrogen then migrates to the adsorbed alkene and adds across the double bond. The reaction is stereospecific in that both H’s add from the same side—a syn addition. This can be seen more clearly if we use deuterium instead of hydrogen—both the $\mathrm{D}$’s add from the same side.
5.04: Oxidation of Alkenes
There are a variety of reagents that can result in the oxidation (i.e. the addition of oxygen to both carbons) of an alkene.[3] These reactions are synthetically useful because they enable us to place functional groups on adjacent carbons and these groups can subsequently be modified. The reagents used in these transformation reactions are highly reactive, and most include species in a high oxidation states, such as permanganate ($\mathrm{MnO}_{4}{ }^{-}$) and or Osmium tetroxide ($\mathrm{OsO}_{4}$), or contain unstable oxygen-oxygen bonds (e.g. Ozone $\mathrm{O}_{3}$) or a peroxy-acid (see below). The common factor in these reagents is that they are able to add oxygen in various ways to the $\mathrm{C=C}$ bond. Many of resulting reactions are quite complex, and we will not delve into their mechanistic details except where necessary: for example, to explain why a particular stereochemistry is produced.
Epoxidation:
Epoxides (also known as oxiranes) ($\rightarrow$) are three-membered ring ethers, and can be formed by the reaction of an alkene with a per-acid, that is, a carboxylic acid with an extra oxygen ($\leftarrow$). The reaction occurs via a concerted (coordinated) movement of electrons. The result is that both of the carbons in the original double bond end up linked to the same $\mathrm{O}$ atom.
Recall that earlier we looked at relative stabilities of rings, and found that their stability depends on the ring size and the torsional (eclipsing) strain. A three membered carbon ring is highly strained because the bond angles are distorted away from the $109^{\circ}$ angle that $\mathrm{sp}^{3}$ hybridization calls for; moreover, all of the bonds are eclipsed. The result is that epoxides are susceptible to nucleophilic attack at a ring carbon ($\rightarrow$). An $\mathrm{S}_{\mathrm{N}} 2$ reaction that proceeds via attack from the back side of the ring, leading to the production of the trans product. Such ring opening reactions can be accomplished by a range of nucleophiles, including water. The reaction with water results in a trans diol. In general, under $\mathrm{S}_{\mathrm{N}} 2$ conditions the ring opening is also stereospecific—that is the nucleophile will attack the least hindered carbon ($\downarrow$).
Epoxides tend to be reactive and for this reason can be useful as synthetic intermediates. Within biological systems, their reactivity can lead to chemical modification of DNA, leading to mutations (for that reason, many are known as genoxic or toxic to the genome). As a defense against such epoxides, organisms encode enzymes known as epoxide hydrolyzes.[4]
Cis-diols:
Alkenes can be oxidized to produce cis-diols using a different type of reagent that adds atoms across the double bond via a cyclic intermediate. For example permanganate ($\mathrm{MnO}_{4}{ }^{-}$) and osmium tetroxide ($\mathrm{OsO}_{4}$), both of which contain transition metals in high-oxidation states, can accomplish this transformation ($\rightarrow$). It is worth noting that by controlling the reaction conditions, we can choose to produce either cis or trans diols. As we move into more complex organic chemistry we will see that the ability to choose and predict outcomes is a major component of organic chemistry.
Ozonlysis:
Another type of alkene double-bond oxidation involves a reaction with ozone ($\mathrm{O}_{3}$), the highly reactive allotrope of oxygen.[5] The mechanism is quite complex as shown below (no need to memorize it!).
Typically, ozone cleaves the double bond and the reaction is treated with a mild reducing agent such as tin ($\mathrm{Sn}$)[6], leading to the production of the corresponding aldehydes or ketones ($\downarrow$).
As we will see later, the ozonolysis reaction can be useful in identifying the position of a double bond within a molecule, as well as in the synthesis of aldehydes and ketones. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/05%3A_Alkenes_and_Alkynes/5.03%3A_Reduction_of_Alkenes-.txt |
As you might predict, alkynes often behave in a similar way to alkenes. The triple-bonded carbons are an electron-rich region of the molecule and we would expect them to undergo electrophilic addition, in a similar manner to alkenes. So, for example, we see Markovikov addition across the triple bond with $\mathrm{HBr}$ ($\rightarrow$), the only difference being that if excess $\mathrm{HBr}$ is present, two—rather than one—bromine atom will be added; one to each of the originally triple-bonded carbons. Other reagents behave in a similar manner. For example $\mathrm{Br}_{2}$ will also add across the triple bond to give first the dibromo, and then the tetrabromo compound.
In contrast, when water is added across the triple bond we find a somewhat different outcome. While the initial steps are the same: the electrophile ($\mathrm{H}^{+}$) adds to the least-substituted carbon, and the nucleophile ($\mathrm{H}_{2}\mathrm{O}$) adds to the carbocation that is produced. This produces a new functionality called an enol (A combination of alkene and alcohol). The enol now undergoes what is known as a tautomerism: the proton from the alcohol moiety is removed (by water as a base), and another proton is picked up on the alkene $\mathrm{CH}_{2}$ carbon ($\rightarrow$). As we have seen many times before this type protonation/deprotonation reaction occurs readily on either oxygen or nitrogen, but this is the first time we have seen it on a carbon; keto-enol tautomerism is an important part of the reactions of carbonyl groups.
The keto and enol forms appear to be different compounds and we might be tempted to classify them as structural isomers—but they are not. The keto- and enol- forms always exist in an equilibrium with one another, and even though we usually write the structure with the carbonyl group (the keto form), there is always a small amount of the enol form present. The transition between keto- and enol- forms of the nucleotide bases initially confused Watson and Crick in their modeling of DNA structure.[7]
Reduction of alkynes:
Addition of hydrogen ($\mathrm{H}_{2}$) to alkynes can be accomplished in several ways. It is possible to completely reduce the alkyne to the corresponding fully-saturated alkane through the addition of two $\mathrm{H}_{2}$ molecules. In fact, when using catalysts such as $\mathrm{Pd}$ (palladium) or $\mathrm{Pt}$ (platinum) the reaction cannot be stopped at the intermediate alkene stage. There are, however, specialized catalysts that allow for partial hydrogenation to the alkene. One of these is known as Lindlar’s catalyst, which is less efficient (poisoned) catalyst. As one might expect (by analogy with alkene reduction), the cis hydrogenated product is formed ($\downarrow$).
It is also possible to reduce an alkyne to the trans product, but to do this we have to use a different method; a method that involves the stepwise addition of the components of the reduction. The conditions for this reduction require a source of protons and a separate source of electrons. For example, a solution of sodium metal (a source of electrons) in liquid ammonia (a source of protons) at low temperature (since ammonia boils at $-33^{\circ} \mathrm{C}$), can be used to reduce an alkyne, but since the reaction proceeds via a stepwise (not concerted) addition, the product formed is the trans alkene: the most stable product ($\rightarrow$).
Acidity of Terminal Alkynes:
One alkyne-specific reaction involves the acidity of protons attached to sp hybridized carbons. The $\mathrm{pK}_{\mathrm{a}$ of such protons is around $25$, which is much lower than that of alkanes ($> 55$) or alkenes ($\sim 45$). In fact, “terminal” alkyne protons can be removed by strong bases such as $\mathrm{NH}_{2}-$ (the amide ion), since the $\mathrm{pK}_{\mathrm{a}$ of $\mathrm{NH}_{3}$ (ammonia) is $33$ ($\downarrow$).
The acidity of terminal alkyne protons can be explained by the idea that the negative charge (the lone pair on the resulting anion) is located in an sp hybrid orbital. The more “s” character in the hybrid orbital, the closer to the nucleus. Since the sp orbital is more “s” than either $\mathrm{sp}^{2}$ or $\mathrm{sp}^{3}$ orbitals, then the electron of the carbon anion is held closer to the nucleus and is therefore more stable than if the carbon were $\mathrm{sp}^{2}$ or $\mathrm{sp}^{3}$ hybridized. The effect is similar to the effective nuclear charge explanation for the trends in electronegativity across the periodic table (i.e. why fluorine is more electronegative than oxygen). The alkyne anion is very useful because it is now a carbon nucleophile, and will attack electrophilic carbon species in an $\mathrm{S}_{\mathrm{N} 2$ reaction. This is the first example we have seen of carbon carbon bond formation (although we will see many more).
5.06: In-Text References
1. https://en.Wikipedia.org/wiki/Vladimir_Markovnikov
2. In fact ALL reactions are reversible in theory (this is called the principle of microscopic reversibility, https://en.Wikipedia.org/wiki/Micros..._reversibility . However, in practice it is extremely difficult to reverse some reactions in the laboratory. For example, combustion of hydrocarbons is not something you would try to reverse in the lab, since the products are gases and will be very difficult to bring back together, and the reaction is highly exergonic. However, plants can do the reverse reaction quite well using energy from sunlight.
3. While we have seen that alkenes can add water (as $\mathrm{H}^{+}$ and ${}^{–}\mathrm{OH}$) across a double bond, this is not classified as an oxidation. There is no change in oxidation state of the $\mathrm{O}$ or $\mathrm{H}$ that add to the double bonded carbons.
4. epoxide hydrolases: http://www.annualreviews.org/doi/pdf....120403.095920
5. Ozone is generated during the reaction by using a special generator because it is too reactive to store. It is generated in the same way that lightning generates ozone—by passing a spark of electric current through oxygen.
6. The reducing agent is present to stop “over oxidation” to the carboxylic acid.
7. Tautomers: evil twins of the bases!: http://blc.arizona.edu/courses/181La...Tautomers.html | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/05%3A_Alkenes_and_Alkynes/5.05%3A_Reactions_of_Alkynes.txt |
In this chapter, we are going to take a closer look at the families of compounds that have carbon linked through a single covalent bond to an $\mathrm{O}$, $\mathrm{N}$, or $\mathrm{S}$. These are known as alcohols ($\mathrm{R-OH}$), amines ($\mathrm{R-NH}_{2}$, $\mathrm{RR}^{\prime}\mathrm{-NH}$, $\mathrm{RR}^{\prime}\mathrm{R}^{\prime \prime}\mathrm{-N}$), thiols ($\mathrm{R-SH}$), ethers ($\mathrm{R-OR}^{\prime}$), and sulfides ($\mathrm{R-SR}^{\prime}$). We group these compounds together based on the predictable similarities and differences in their chemical and physical properties, specifically the fact that each of these functional groups has a relatively electronegative element ($\mathrm{O}$, $\mathrm{N}$ or $\mathrm{S}$) attached by a single bond to carbon and each has available lone electron pairs that can be donated to H+ or other electrophiles. The result is that alcohols, thiols, and amines (primary and secondary) all have relatively acidic hydrogens, which influences their chemical reactivities, and all show nucleophilic properties.
Table $6.0.1$ Examples of Functional groups, their names and approximate $\mathrm{pK}_{\mathrm{a}}$‘s
Functional Group Example Name $\mathrm{pK}_{\mathrm{a}}$
Alcohol Remove -ane, add -ol.
4-methylpentan-2-ol
(approximate)
$\sim 15-16$
Alcohol Alcohols take precedence over
alkenes, But-3-en-2-ol
Thiol Longest chain, add -thiol
Propane-1-thiol
$\sim 10$
Primary amine Longest chain, remove e, add
-amine
Propanamine or Propyl amine
$\sim 33$
Secondary amine N-methylethanamine $\sim 33$
Tertiary amine N-ethyl-N-methylpropanamine N/A
Ether Methoxyethane
Ethyl methyl ether
N/A
Sulfide Dimethylsulfane
Dimethyl sulfide
N/A
We will concentrate our discussion on oxygenated compounds, but we will note reactivities across the various groups to illustrate their similarities (and differences).
06: Alcohols and an introduction to thiols amines ethers and sulfides
Recall that there are several factors that can influence Brønsted acidity. These include: the strength of the bond between the $\mathrm{R}$ ($\mathrm{C}$) and the $\mathrm{O}$, $\mathrm{S}$, or $\mathrm{N}$ (denoted by “$\mathrm{Y}$” below) and $\mathrm{H}$; the polarity of the bond; and the stability of the resulting anion (“$\mathrm{Y}^{-}$“). $\mathrm{R}-\mathrm{Y}-\mathrm{H}+\mathrm{B}^{-} \rightarrow \mathrm{BH}+\mathrm{R}-\mathrm{Y}^{-}$
Simple alcohols have acidities that are about the same as water, with a $\mathrm{pK}_{\mathrm{a}}$ of around $15-16$ (remember the $\mathrm{pK}_{\mathrm{a}}$ of water is $14$). In comparison, amines are much less acidic, with $\mathrm{pK}_{\mathrm{a}}$‘s around $33-36$. We understand this difference based on the fact that $\mathrm{O}$ is more electronegative than $\mathrm{N}$, therefore the $\mathrm{O-H}$ bond is more polarized than the $\mathrm{N-H}$ bond—making the partial positive charge on the $\mathrm{H}$ larger in $\mathrm{O-H}$ than in $\mathrm{N-H}$ bonds. In polar solvents, this means that the $\mathrm{H}$ bonded to an $\mathrm{O}$ is better solvated and the proton is easier to remove than an $\mathrm{H}$ bonded to an $\mathrm{N}$. Similarly, the resulting negative charge on the anion is more stable on $\mathrm{O}$ than on $\mathrm{N}$ because the effective nuclear charge on $\mathrm{O}$ is greater (which is the cause of its greater electronegativity, compared with $\mathrm{N}$). In contrast, thiols are more acidic than alcohols because the $\mathrm{S-H}$ bond is weaker—the size of $\mathrm{S}$ and $\mathrm{H}$ orbitals results in smaller overlap and therefore weaker bonds (just like $\mathrm{HBr}$ is more acidic than $\mathrm{HCl}$) and the resulting anion is more stable because the larger size of $\mathrm{S}$ results in the negative charge being distributed over a larger volume.
To deprotonate a simple alcohol, with a $\mathrm{pK}_{\mathrm{a}}$ of around $15$, requires a base that forms a bond with $\mathrm{H}$ that is than the bond that would be formed with the resultant alkoxide ($\mathrm{RO}^{-}$)—otherwise the acid-base reaction would reverse. For this reason, we cannot use a base like sodium hydroxide, because its conjugate acid $\mathrm{H}_{2} \mathrm{O}$ has a $\mathrm{pK}_{\mathrm{a}}$ of around $14$. Therefore, although some alcohol deprotonation would occur in water, the equilibrium position would be somewhere in the middle, so that both ${}^{-}\mathrm{OH}$ and $\mathrm{RO}^{-}$ would be present in the reaction mixture. To completely deprotonate an alcohol with a $\mathrm{pK}_{\mathrm{a}}$ of around $15$, we would need to use a stronger base, such as sodium hydride[1] ($\mathrm{NaH}$), or sodium amide[2] ($\mathrm{NaNH}_{2}$).
Note that the reaction must take place in an aprotic solvent; in this case diethyl ether is used (otherwise the $\mathrm{NaH}$ would react with the solvent as well as the alcohol). Another common method to deprotonate alcohols is through a redox reaction, using a group one metal—usually sodium.
Some alcohols are much more acidic; for example $-\mathrm{OH}$ groups attached to an aromatic ring (which are called phenols) typically have $\mathrm{pK}_{\mathrm{a}}$‘s around $10$. This difference in $\mathrm{pK}_{\mathrm{a}}$ must lie with the nature of the conjugate base (the anion), since the same $\mathrm{O-H}$ bond breaks during the proton transfer. In this case, the conjugate base is stabilized by delocalizing the electrons to the aromatic ring by resonance. Recall that the more delocalized a charge (in this case the electrons), the more stable the resultant ion is.
Phenols are more acidic than typical alcohols because the conjugate base is stabilized by resonance
Phenols can be deprotonated by $\mathrm{NaOH}$ because the phenolate anion is more stable than hydroxide. Therefore, phenols are soluble in aqueous solutions of sodium hydroxide. This also provides a way of separating phenols from other (non-acidic) organic substances, since the phenol can be regenerated simply by adding acid. One practical way that this phenomenon can be used is to remove the highly allergenic substances secreted by poison ivy (or oak or sumac) plants. The major allergen belongs to a family of di-phenols called urushiol. The $\mathrm{R}$ can by any of a number of long chained hydrocarbons. Washing the affected part with a basic solution (soap for example) will help solubilize the urushiol and remove it from the skin.
Alcohol acidity can also be increased by inductive electron withdrawal (due to the presence of electronegative atoms linked through sigma bonds) just as we discussed earlier in the case of carboxylic acids: for example $\mathrm{CF}_{3} \mathrm{OH}$ is more acidic than $\mathrm{CH}_{3} \mathrm{OH}$.
We might also predict the effects relative to acidities of amines and thiols in terms of resonance and inductive stabilization, but, in fact, most of their chemistry is not associated with acidity and we will not dwell on this idea here. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/06%3A_Alcohols_and_an_introduction_to_thiols_amines_ethers_and_sulfides/6.01%3A_%28Brnsted%29_Acidity_of_Alcohols_Thiols.txt |
Earlier (Chapters $1$ and $4$), we discussed (at great length) that all three functional groups (${}^{-}\mathrm{OH}$, ${}^{-}\mathrm{NH}$, and ${}^{-}\mathrm{SH}$) are nucleophilic: that is, they will react at the carbon center that is electron-deficient. For functional groups that contain nucleophilic centers from the same row of the periodic table, the trends in nucleophilicity parallel Bronsted basicity: amines are more nucleophilic (and basic) than alcohols. However, in functional groups that contain nucleophilic centers from the same group of the periodic table (nucleophilicity increases down the group, while basicity decreases), thiols are more nucleophilic than alcohols. Both amines and thiols are very nucleophilic. All three groups participate in nucleophilic substitutions as discussed in Chapters $1$ and $4$.
Examples of these kinds of nucleophilic substitutions are the reactions of alcohols, thiols, and amines with alkyl halides to give the corresponding ethers, sulfides, and (secondary, tertiary or quaternary) amines. Alcohols are not as nucleophilic as thiols and amines, and therefore typically the corresponding alkoxide must be used (because it is more reactive), for the synthesis of ethers.
In the case of amines, the nitrogen can react several times with the electrophile (alkyl halide), and in practice it is difficult to stop the reaction at any intermediate step (in the laboratory).
Amines typically react with electrophiles to give poly-alkylated amines
6.03: (mathrmO) (mathrmS) and (mathr
Recall that a good leaving group should be able to accept (in a stable form) the pair of electrons from the bond that breaks. Typically, good leaving groups are weak bases. For this reason, hydroxide (${}^{-}\mathrm{OH}$) and amide (${}^{-}\mathrm{NH}_{2}$) are unlikely to be produced during a nucleophilic substitution reaction. However, as noted earlier, alcohols can be converted into good leaving groups by protonation, which results in $\mathrm{H}_{2}\mathrm{O}$ as the leaving group.
Alcohols can also be modified (or derivatized) to produce better leaving groups. This is particularly useful when we need to carry out a reaction that is sensitive to acidic conditions when the method we have used earlier (protonation of the $\mathrm{OH}$) cannot be used. The most common derivative used to make the $\mathrm{OH}$ group into a good leaving group is the Tosyl group (para-toluenesulphonate). It can be formed by reacting an alcohol with p-toluenesulfonylchloride ($\mathrm{TosCl}$) in the presence of a base (such as pyridine) that acts to remove the $\mathrm{HCl}$ that is produced).
We can consider the derivatization reaction as mechanistically similar to other nucleophilic substitutions we have considered, except that it takes place at an $\mathrm{S}$ instead of a $\mathrm{C}$.
The resulting $\mathrm{OTos}$ group is a very good leaving group, making the molecule reactive to nucleophilic substitution reactions. In effect, we have changed the leaving group from ${}^{-}\mathrm{OH}$, which is a relatively strong base, to ${}^{-}\mathrm{OTos}$ which is a very weak base—it is the organic equivalent of sulfate, the conjugate base of sulfuric acid. The negative charge on ${}^{-}\mathrm{OTos}$ becomes delocalized to the other oxygens bound to the $\mathrm{S}$, thereby stabilizing the base.
In a similar manner, sulfides can be transformed into leaving groups, most commonly through the methylation of the sulfide, which produces a powerful reagent that can be used to methylate other species.
In biological systems, a common methylating agent, S-adenosylmethionine ($\mathrm{SAM} \rightarrow$), uses this mechanism | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/06%3A_Alcohols_and_an_introduction_to_thiols_amines_ethers_and_sulfides/6.02%3A_Nucleophilicity_of_%28mathrmROH%29_%28ma.txt |
Before we discuss oxidation of alcohols, it should be clear what we mean by the “oxidation” and “reduction” of carbon compounds. Recall that in earlier discussions we used the term reduction to mean the addition of hydrogen and oxidation to mean the addition of oxygen, rather than calculating changes in oxidation numbers (decrease for reduction, increase for oxidation). The reason is because oxidation numbers in organic compounds can be hard to calculate and apply[3]. In this section, we consider how alcohols can be oxidized to give aldehydes, ketones, or carboxylic acids. In general, we consider a carbon compound to be oxidized when the number of bonds between the $\mathrm{C}$ and electronegative atoms (often, but not always, $\mathrm{O}$) is increased.
For example, a primary alcohol can be oxidized (which we will denote by $\mathrm{O}$ for the time being) to an aldehyde; depending upon the reagent used, the reaction can proceed through a second step to produce the corresponding carboxylic acid. At each step, the oxidation level of the carbon is increasing.
Starting with a secondary alcohol, the product of an oxidation reaction is the corresponding ketone, but tertiary alcohols do not give useful products and may simply lead to degradation ($\mathrm{C-C}$ bond breaking). Generally, it is not possible to oxidize a secondary carbon beyond the ketone level without breaking carbon-carbon bonds, and similarly, tertiary alcohols cannot be oxidized under normal circumstances.
Typical oxidizing reagents include transition metals in high-oxidation states (that is able to accept [bond to] $\mathrm{O}$ atom). For example, chromium (VI) in the form of chromium trioxide ($\mathrm{CrO}_{3}$) or sodium dichromate ($\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$), when in concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$, are both powerful oxidizing agents and both will oxidize a primary alcohol through both steps, that is, all the way to the carboxylic acid form. Pyridinium chlorochromate ($\mathrm{PCC} \rightarrow$) is a milder oxidizing agent that will oxidize primary alcohols, only through the first step, to produce an aldehyde.
The general mechanism of oxidation is shown below, note electrons leave the alcohol and end up on the $\mathrm{Cr}$, reducing its oxidation state from $6$ to $4$, and the alcohol carbon ends up oxidized.
Primary alcohols can be selectively oxidized to aldehydes with PCC
One problematic aspect of such oxidizing reagents is that they contain highly toxic and carcinogenic $\mathrm{Cr}$ (VI) in one form or another. Such materials oxidize a range of biomolecules such as vitamin C (ascorbic acid) and some thiols (such as the amino acid cysteine). Reduced chromium also reacts with nucleic acids and can lead to mutations, which can lead to cell death and/or cancer. To avoid using such toxic chemicals, there has been increasing in what has come to be known as green chemistry[4]. One of the tenets of green chemistry is to minimize the use of toxic reagents (such as chromium compounds).[5]
6.05: Oxidation of Thiols
In alcohols, oxidation generally occurs at the carbon bonded to oxygen. In contrast, with thiols the oxidation site is often at the sulfur. For example, many oxidizing agents (even molecular oxygen in air) oxidize thiols to disulfides. The reverse reaction is also readily accomplished using a reducing agent such as $\mathrm{Zn} / \mathrm{HCl}$. The disulfide bond is relatively weak, that is, requires less energy to break (about half the strength of a typical $\mathrm{C-C}$ or $\mathrm{C-H}$ bond).[6]
In fact, the amino acids cysteine and diamino acid cystine are readily interconverted in biological systems (usually through the $\mathrm{NADH} / \mathrm{NAD}$ oxidation/reduction system; see below). These disulfide crosslinks between cysteine moieties in polypeptides and proteins often serve to stabilize the 3D structure of proteins.
Sulfides ($\mathrm{R-S-R}$) are also susceptible to oxidation, which can lead to the formation of a sulfoxide, which can be further oxidized to form a sulfone. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/06%3A_Alcohols_and_an_introduction_to_thiols_amines_ethers_and_sulfides/6.04%3A_Oxidation_of_Alcohols.txt |
We have already seen several methods by which alcohols can be produced, mostly in Chapter $5$. For example, the addition of water across a double bond, either through acid catalysis (Markovnikov addition) or by hydroboration/oxidation (Anti-Markovnikov addition), produces alcohols. We have also seen, under certain conditions, that alcohols can be produced by nucleophilic substitution. Both $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions can produce alcohols, and now would be a good time to review all of these reactions (covered in Chapters $1$, $3$, $4$ and $5$).
A reaction that we have not yet encountered is the reduction of carbonyl compounds. For example, a ketone such as acetone can be reduced through a reaction with sodium borohydride ($\mathrm{NaBH}_{4}$) or lithium aluminum hydride ($\mathrm{LiAlH}_{4}$)[7]; both of these molecules can deliver hydride ($\mathrm{H}^{-}$) to the partially positive carbon of the carbonyl.
Sodium borohydride ($\mathrm{NaBH}_{4}$) is generally the reagent of choice as it is less reactive and the reaction can be carried in an open flask, whereas $\mathrm{LiAlH}_{4}$ typically must be used with solvents that do not contain water and under a dry atmosphere. The intermediate $\mathrm{R-O-BH}_{3}$ complex is destroyed by adding aqueous acid to give the final alcohol product.
Reactions where hydride is delivered to a carbonyl are similar to a reaction found in biological systems. $\mathrm{NADH}$ (Nicotinamide Adenine Dinucleotide Hydride) is an unstable intermediate generated through a number of metabolic processes (such as fermentation), while not as reactive as $\mathrm{NaBH}_{4}$, and (like, essentially, all biological reactions) requires a catalyst (an enzyme) to bring about the reduction of carbonyls; but the mechanism is similar. The reaction (like all reactions) is also reversible, so that the oxidized version of $\mathrm{NADH}$, $\mathrm{NAD}^{+}$, can accept a hydride from an alcohol to produce a ketone. In the mechanism (with only the nicotinamide part of $\mathrm{NADH}$ shown) the “$\mathrm{R}$” group attached to the $\mathrm{N}$ in the ring is actually a complex molecule consisting of an adenine moiety (a base found in nucleic acids and nucleotides), two sugar units (ribose), and two phosphate linkages. For now, let us focus on the similarities between the reduction reactions discussed above and those that take place in biological systems.
Reduction of a carbonyl by $\mathrm{NADH}$ by delivery of $\mathrm{H}^{-}$ to the carbonyl carbon
The conversion of pyruvic acid to lactic acid during glycolysis is just such an example. By looking at simpler systems, we can understand (and model) the types of reactions that occur in organisms.
Alcohols can also be produced by direct reduction with $\mathrm{H}_{2}$(g) using a transition metal catalyst, in a way similar to the reduction of $\mathrm{C=C}$, except that the hydrogens add across the $\mathrm{C=O}$.
The choice of reducing agent depends on presence of other functional groups within the molecule. For example, if we wanted to reduce a carbonyl group in a molecule that also had a carbon-carbon double bond, we would not use $\mathrm{H}_{2} / \mathrm{Pd}$ as the reagent/catalyst, since it would also reduce the double bond as shown here ($\rightarrow$).
Preparation of alcohols with Grignard reagents:
Just as we can add hydride ion by a nucleophilic attack at a carbonyl, we can also add an alkyl group, which formally contains a carbanion (a negatively charged carbon). The most common way to do this is via a Grignard[8] reagent, produced by reacting an alkyl halide with magnesium metal in a dry atmosphere with a non-protic solvent such as diethyl ether ($\mathrm{Et}_{2}\mathrm{O}$). The resulting Grignard reagent, $\mathrm{RMgBr}$, is now polarized with a large partial negative charge on the carbon.
For our purposes, we can treat the Grignard as if it were a carbanion, which will react with a carbonyl group in much the same way as a hydride ion.[9]
Reaction of a Grignard reagent with a ketone to give an alcohol
This reaction is applicable to a wide range of compounds that contain carbonyl groups including aldehydes and ketones, and (as we will see later) to esters and acid chlorides, but not carboxylic acids (why do you think that is?).
6.07: In-Text References
1. $\mathrm{NaH}$ is synthesized by the reaction of sodium with hydrogen—a redox reaction. It is an ionic compound consisting of $\mathrm{Na}^{+}$ and $\mathrm{H}^{-}$ (hydride) ions; hydride cannot be produced by deprotonating $\mathrm{NaH}_{2}$.
2. The $\mathrm{pK}_{a}$ of $\mathrm{NaH}_{3}$ (the conjugate acid of $\mathrm{NaH}_{2} {}^{-}$) is $33$
3. To use the oxidation number method, we must remember that $\mathrm{H}$ is less electronegative than $\mathrm{C}$; so in $\mathrm{CH}_{4}$, the ON ofcarbon is –4 and each $\mathrm{H}$ is +1. (This is confusing since we usually consider $\mathrm{C-H}$ bonds as non-polar). In $\mathrm{CO}_{2}$, each $\mathrm{O}$ is –2 and the $\mathrm{C}$ is +4. Therefore, in $\mathrm{CO}_{2}$ the carbon is in a higher oxidation state than in CH4
4. For more information about green chemistry see: https://www.epa.gov/greenchemistry
5. It is not necessary here to provide a long list of such reagents since many of them are complex, but it is important to know that there are alternatives should you ever need to oxidize an alcohol.
6. The formation of the analogous peroxide $\mathrm{O-O}$ bond (Bond Dissociation Energy $140 \mathrm{~kJ} / \mathrm{mol}$) is even less likely, this bond is even weaker than $\mathrm{S-S}$ (BDE $230 \mathrm{~kJ} / \mathrm{mol}$).
7. $\mathrm{NaBH}_{4}$ and $\mathrm{LiAlH}_{4}$ both contain a group III element ($\mathrm{B}, \(\mathrm{Al}$) here found in the form of the Lewis acid-base complex $\mathrm{BH}_{4}$ or $\mathrm{AlH}_{4}$. They are sources of Hydride ion, as shown above. $\mathrm{LiAlH}_{4}$ is more reactive than $\mathrm{NaBH}_{4}$.
8. Victor Grignard won a Nobel prize for this discovery: https://en.Wikipedia.org/wiki/Victor_Grignard.
9. The reaction mechanism is a little more complex than this—actually occurring via one electron transfer—but the result is the same. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/06%3A_Alcohols_and_an_introduction_to_thiols_amines_ethers_and_sulfides/6.06%3A_Preparation_of_Alcohols.txt |
There is a set of organic compounds that incorporates the carbonyl group ($\mathrm{C=O}$) which includes aldehyde ketones, carboxylic acids, and carboxylic acid derivatives such as: esters, amides, acid anhydrides, and acid chlorides (as shown in Table $7.0.1$).
Table $7.0.1$: Functional groups that contain a carbonyl group
Functional Group Example Name
Aldehyde Longest chain with aldehyde:
remove -ane, add -al
Ethanal (IUPAC)[1]
Acetaldehyde
Ketone Longest chain: remove -e, add -one
2-propanone (IUPAC)
Acetone
Carboxylic Acid Longest chain with $\mathrm{CO}_{2}\mathrm{H}$: remove -e, add -oic acid
Ethanoic acid (IUPAC)
Acetic acid
Ester Named as a derivative of the $\mathrm{CO}_{2}\mathrm{H}$, remove
-icacid and -ateEthyl ethanateEthyl acetate
Amide Named as a derivate of the $\mathrm{CO}_{2}\mathrm{H}$, remove
-icacid, add -amideEthanamideAcetamide
Acid chloride Named as a derivate of the $\mathrm{CO}_{2}\mathrm{H}$, remove
-icacid, add -yl chloride (or other halide)Ethanoyl chlorideAcetyl chloride
Acid anhydride Named as a derivative of the $\mathrm{CO}_{2}\mathrm{H}$, remove “acid” and add “anhydride”
Ethanoic anhydride
Acetic anhydride
These functional groups have a number of similarities and some notable differences in their properties, which can be predicted on the basis of our understanding of structure and function relationships. These carbonyl compounds can be classified into two broad groups:
1. ketones and aldehydes and
2. carboxylic acids and derivatives (amide, chloride, ester, and anhydride).
These two, broad groups differ in the oxidation level of the carbonyl carbon: aldehydes and ketones have two bonds to the electronegative oxygen; acids and derivatives have three bonds to electronegative atoms ($\mathrm{O}$, $\mathrm{N}$ or $\mathrm{Cl}$); and, of course, alcohols only have one bond. Interconverting between alcohols, ketones, and carboxylic acids involves some kind of redox reaction. We have already discussed the alcohol to ketone (or aldehyde) transformation and, later, we will discuss further oxidation to the acid level.
As we discussed in Chapter $6$, the carbonyl carbon is highly polarized; the large $\sigma^{+}$ on the carbon makes it susceptible to nucleophilic attack. There are a large number of reactions that begin by the attack of a nucleophile on a carbonyl group. To make understanding these reactions more manageable (intelligible), we will consider these reactions in a sequence of increasing complexity, beginning with reactions of aldehydes and ketones.[2] We will then cycle back around and visit similar reactions involving acids and their derivatives.
07: Nucleophilic attack at the carbonyl carbon-
As we have just seen, esters can be made from carboxylic acids and vice versa: we can control the outcome of the reaction by using Le Chatelier’s principle. This is because the tetrahedral intermediate contains only oxygen leaving groups and the reactants and products are of similar stability. However, if we react a derivative such as an acid chloride with an oxygen nucleophile, we see that the resulting reaction tends not to be reversible. It is possible to go from the acid chloride to the carboxylic acid (with $\mathrm{H}_{2}\mathrm{O}$ nucleophile), to the ester (with an alcohol nucleophile), or to the amide (with an amine nucleophile). The reverse reaction is not feasible because 1) the tetrahedral intermediate, for either the forward or the reverse reaction, now has different leaving groups as shown below and 2) the acid chloride is highly reactive (it is destabilized by inductive withdrawal) and the reaction is unlikely to reverse under typical reaction conditions.
The tetrahedral intermediate reacts to produce the carbonyl by expelling chloride, which is the best leaving group. It is, therefore, difficult to get this reaction to reverse. In fact, the acid chloride can be used to produce all the other acid derivatives, and carboxylic acids. Reaction of acid chlorides with amines will produce amides and, with a carboxylate anion, will produce acid anhydrides as shown. In each case chloride is the best leaving group when the tetrahedral intermediate collapses, and the more stable product is formed.
You may now be asking yourself if acid chlorides are such good reactants, how can we make them in the first place? We cannot use chloride ion to do a nucleophilic addition/elimination on any other acid derivative, so how can we get around that problem? The answer is to introduce an even better leaving group into the molecule. One example is the reaction of carboxylic acids with thionyl chloride ($\mathrm{SOCl}_{2}$). The sulfur in thionyl chloride is highly susceptible to nucleophilic attack—much more so than the carbonyl—because of all the electronegative groups attached to it. The first step is attack by the carboxylic acid oxygen on the $\mathrm{SOCl}_{2}$, as shown below.
The intermediate that is formed has an excellent leaving group (in the shadowed box). In effect, we have activated the carbonyl and made it more reactive. Attack of chloride ion can now proceed and the reaction will move forward because we have a better leaving group than chloride. The tetrahedral intermediate now collapses and, at the same time, the leaving group decomposes to give $\mathrm{SO}_{2}$ and hydrogen chloride[10], both of which are gases that are expelled from the reaction mixture which drives the reaction towards products (Le Chatelier’s principle).
In this reaction, we have seen a very important and powerful idea: we have been able to drive a reaction forward to produce a thermodynamically unfavorable product by producing a highly reactive intermediate. We encountered simpler examples earlier: for example, protonating an alcohol or preparing a tosyl derivative to transform $\mathrm{OH}$ into a good leaving group. In this case, the reaction is messy and toxic (it’s no fun at all to do this reaction in a laboratory), but biological systems use this strategy to bring about thermodynamically unfavorable reactions. As we will discuss later in the course, substrate reactants can be activated by this same strategy of making an $\mathrm{OH}$ into a good leaving group that then leads to the formation of a product that could not be produced under normal circumstances. For example, the formation of sucrose from glucose and fructose proceeds using such a (enzyme-catalyzed) strategy. In this case, activation involves the activation of an $\mathrm{OH}$ group by coupling with $\mathrm{ATP}$. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.01%3A_Interconversion_of_Acids_and_Derivatives-_Predicting_Outcomes.txt |
All the derivatives of carboxylic acids can be produced from the acid form, although it may occur via the acid chloride. We have also seen several reactions in which carboxylic acids can be produced by alternative mechanisms: for example, via the oxidation of primary alcohols or aldehydes. It turns out that we can also produce acids by ozonolysis (reaction with ozone $\mathrm{O}_{3}$) if one or both of the alkene carbons are bonded to a hydrogen. When we looked at such reactions previously, they were accompanied by a reductive workup (either $\mathrm{Zn}$, or dimethyl sulfide) and arrested at the aldehyde level. If instead we use an oxidative workup with hydrogen peroxide, the oxidation goes all the way through to the carboxylic acid.
Another reaction that is often used to produce carboxylic acids is the hydrolysis of nitriles ($\mathrm{RCN}$). The nitrile carbon is at the same oxidation state as a carboxylic acid, and when treated with aqueous acid (or base), it undergoes a hydrolysis reaction in just the same way as we have seen on numerous occasions (try it—you will see!).
Finally, a different approach to producing carboxylic acids occurs when we react a Grignard reagent (or alkyl lithium) with carbon dioxide. You may recall from general chemistry that $\mathrm{CO}_{2}$ is a linear molecule and has no overall molecular dipole. However, each $\mathrm{C=O}$ bond in the molecule is polarized, leading to a partial positive charge on the central carbon that makes it susceptible to nucleophilic attack. Therefore, when reacted with a Grignard reagent, the end result is the formation of a new $\mathrm{C-C}$ bond with the $\mathrm{CO}_{2}$ becoming a carboxyl group.
7.03: The Wittig Reaction
This reaction allows us to synthesize alkenes by adding the two carbons of the alkene double bond together. This is a very powerful synthetic technique that allows us to construct larger molecules from smaller ones. It is much more efficient to add two molecules together, than to try and synthesize a large molecule one carbon-carbon bond at a time.
The Wittig reaction starts by preparation of a reagent that involves addition of an alkyl halide to a phosphine (the phosphorus analog of an amine), such as triphenylphosphine ($\mathrm{Ph}_{3}\mathrm{P}$). This reaction is directly analogous to the alkylation of amines that we have seen previously to produce the phosphonium salt as shown. However, the reaction can go further in the presence of a base to produce the stabilized carbanion by deprotonation of the carbon next to the phosphorus as shown[11].[12]
The Wittig reagent can then react with an aldehyde or ketone as shown. The first step, as we might imagine, involves nucleophilic attack at the carbonyl, but since the carbanion is also attached to the phosphorus, another avenue of reaction is now available. The tetrahedral intermediate undergoes further reaction in which the oxygen bonds to the phosphorus and then, via a cyclic rearrangement of electrons, the four-membered ring rearranges to form the alkene and triphenylphosphine oxide. In fact, it is the formation of the very strong $\mathrm{P=O}$ double bond that makes this reaction essentially irreversible, and drives the reaction towards products.
The Wittig reaction can be used to prepare many different alkenes. In general, if there is a possibility of E/Z isomerism, the most stable alkene will be produced (for example a trans isomer rather than a cis isomer). But, just as we have seen many times, it is possible to manipulate the reactants and reaction conditions to produce the desired product—a subject for a more advanced treatment of organic chemistry[13]. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.02%3A_Preparations_of_Carboxylic_Acids.txt |
Up to now, we have focused mostly on the reactivity of organic molecules, particularly from the standpoint of being able to predict reactivity from the structure of a particular compound. We have also discussed how spectroscopic techniques allow us to determine molecular structure. However, we have not really discussed another of the major areas of organic chemistry, that is: the design and synthesis of molecules, particularly those that may be expected to have biological activity (drugs of various sorts). Now that we have a fairly large repertoire of reactions to choose from, let us take a closer look at the kinds of decision-making that goes into designing molecular structures.
Typically, molecular synthesis involves a target structure, which may be an actual molecule, or it could be a substance that has particular properties: for example, the active site of an enzyme or a regulatory domain of a protein. To achieve specificity, the molecule to be synthesized must fit into the surface of the protein and influence its structure or catalytic activity. The better the fit, the more specific (higher affinity, fewer non-target interactions) interactions the drug will make. We will begin by thinking about how to go about the synthesis of a given molecule. Molecular synthesis is both an art and a science: it requires that you have at your fingertips a good collection of reactions you have organized in such a way as to make them accessible to you, but it also requires creativity and imagination. There is always more than one way to design a synthesis and, in reality, there are many setbacks and path changes since reactions may not go as planned, so alternative routes have to be considered. In this section, we will look at strategies that you might employ to design a synthesis of a target molecule.
Retrosynthetic analysis
Retrosynthetic analysis is exactly what it sounds like: you begin with the target and move backwards one step at a time to identify what reactants and reagents could have produced the products. This is often the situation when you are dealing with a natural product that was originally isolated based on its biological activity. A classic example is the molecule Paclitaxel, which was isolated from the Pacific yew tree, Taxus bervifolia, based on its anti-cancer properties.[14]
The process of molecular synthesis can be broken down into a number of steps, but as we will see, depending on the nature of the task, we do not always approach synthesis in a linear manner.
Step 1: Identify the number of carbons in the target molecule and determine whether you will need (at some point in the synthesis) to make new carbon-carbon bonds.
Step 2: Identify what functional groups are present in the molecule. Functional groups are where the reactivity of any molecule lies and they give you a place to begin because you now know ways to produce that functional group.
Step 3: Identify which bonds are going to be made during the reaction in which the product is produced.
Step 4: This will allow you to work backwards to identify what the precursor is and (using your knowledge of functional group transformations) decide what reaction will product the product.
Step 5: Repeat until you reach a recognizable target material.
Let us approach the process with a (significantly simpler) target.[15] Let us design a synthesis of the lactone (cyclic ester) from cyclopentene. We begin by noting that both the starting material and the product have five carbons, which means we do not need to consider any carbon-carbon bond formation reactions. However, the route from cyclopentene to the lactone is not obvious, so let us go back one step at a time as outlined below.
So, we have the immediate precursor to the lactone, but how do we get that from cyclopentene? While you could continue moving backwards through the steps, the knowledge that you don’t have to construct the carbon skeleton makes a difference in the way that you might approach the problem. For example, if you look at the starting material, it is clear that there is no five-membered ring in the product and no alkene. Not only that, but at each end of the carbon chain is at least one oxygen. Is there a reaction that will allow us to open up that ring and, at the same time, introduce oxygens at each carbon? Yes! Recall that alkenes can be cleaved by ozonolysis (under oxidizing conditions), to give the dicarboxylic acid. If one of those carboxylate groups can be reduced to the alcohol, something that might well be be difficult to do in practice, the ring would cyclize spontaneously to give the lactone as shown below.
Note that, in this synthesis, we mixed up both retro and forward synthetic analysis to produce an overall synthetic pathway. This is where some of the art and imagination come into play. When you are trying to design a synthesis, it is a good idea to start from the product and work backwards—but at some point you may have to also work forward. Every synthesis is different and, as we noted before, there are often many synthetic routes that can be designed for any single target molecule. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.04%3A_Synthesis.txt |
1. IUPAC is the International Union of Pure and Applied Chemists. This body is responsible (among other things) for setting the rules about systematic nomenclature of chemical substances.
2. The preparation of aldehydes and ketones has been discussed earlier, including reactions in which alkenes are cleaved (broken apart) by oxidation with ozone (ozonolysis) by addition of water across triple bonds (Chapter $5$) and the oxidation of alcohols (Chapter $6$).
3. Why do you think that pyruvate and lactate are present in the form of their conjugate bases?
4. Sometimes this grouping is called a ketal (when the starting $\mathrm{C=O}$ is a ketone), but general “acetal” and “hemiacetal” can refer to either an aldehyde or a ketone.
5. Remember, Le Chateliers Principle is just a rule of thumb—it tells us what happens but not why. Adding more reactants increases the rate of the forward reaction, removing products decreases the rate of the reverse reaction.
6. However, as we will see later, carbonyl compounds are often acidic; the alpha carbon can be deprotonated; more on that later.
7. See chapter $9$ of CLUE general chemistry text for more information on the Henderson Hasselbalch equation and its uses.
8. The $\mathrm{pK}_{\mathrm{a}}$ of protonated amines ($\mathrm{RNH}_{3} {}^{+}$) is about 10. Using the Henderson Hasselbalch equation, we see that the ratio of $\left[\mathrm{RNH}_{2}\right] /\left[\mathrm{RNH}_{3}{ }^{+}\right]$ is about $0.001$ —that is, there is 1000 times more protonated than unprotonated amine.
9. For a more in-depth discussion of this phenomenon, including the entropic and enthalpic contributions to micelle formation, see the CLUE Chapter $6$.
10. Both $\mathrm{HCl}$ and $\mathrm{SO}_{2}$ are highly toxic, requiring special precautions—another one not to try at home.
11. Note that this stabilization uses d orbitals on the phosphorus; this reaction could not happen with an amine.
12. Species such as this carbanion are called ylides, because they can be written as containing both negative and positive charges on adjacent atoms (in contrast to Zwitterions: forms of amino acids at different $\mathrm{pH}$’s that also have both positive and negative charges on them—just not on adjacent atoms).
13. See http://www.organic-chemistry.org/nam...-reaction.shtm for example.
14. It functions (In at least one way, by binding to the protein tubulin: the structural basis of the cytoplasmic microtubules found in eukaryotic cells. When bound, Paclitaxel acts to make microtubules more stable (i.e. less likely to depolymerize). Since microtubule function depends on dynamic assembly and disassembly, this has effects on cell behavior, specifically microtubule-based cell division
15. The total synthesis of taxol is described here: http://www.nature.com/nature/journal.../367630a0.html
7.06: Aldehydes and Ketones
Is there a difference in reactivity between aldehydes and ketones? Not really; both types of compounds undergo nucleophilic attack, although, in general, aldehydes react faster than ketones for two reasons:
1. Aldehydes are less hindered at the carbonyl carbon than ketones (since there is always at least one $\mathrm{H}$ attached whereas ketones always have two bulkier alkyl groups).
2. Alkyl groups are electron-donating so the partial positive charge on the carbon is partially offset by induction from the alkyl groups.
In practice, however, there is generally little difference between the two and so unless we point out a difference you can assume that they undergo similar reactions.
The similarities between aldehydes and ketones are supported by an examination of their spectra. For example, both aldehydes and ketones display a strong carbonyl absorption around $1700 \mathrm{cm}^{-1}$. As we will see shortly, not all carbonyls absorb in this area—their $\mathrm{C=O}$ stretching frequencies are dependent on their electronic environment. In $\mathrm{C-13}$ NMR, both aldehydes and ketones show a low field peak for the $\mathrm{C=O}$ around $200 \mathrm{ppm}$ indicating that the electronic environment of the carbonyls are similar. The major difference in the spectra is the aldehyde proton resonance in the HNMR that appears downfield. This peak often appears as a singlet, even though it may be adjacent to a $\mathrm{C-H}$ group, because the coupling constant is often small and, depending on the sensitivity of the instrument used, splitting may not be detectable.
7.07: Nucleophilic Attack by Hydride or Carbanions
As we discussed in Chapter $6$, aldehydes and ketones react with reagents that are able to deliver hydride (for example from sodium borohydride) or a carbanion (in the form of a Grignard reagent) to the carbonyl group.
In addition to Grignard reagents, carbanions can also be generated by treating alkyl halides with lithium under dry conditions. These alkyl lithium reagents ($\mathrm{RLi}$) behave in a very similar way to Grignard reagents, although they are somewhat more reactive.
These two reactions: addition of hydride (reduction) or a carbanion (resulting in carbon-carbon bond formation) to a carbonyl are analogous. They also have something else in common in that, under normal laboratory conditions, they are not reversible because reversing the reaction would require that the hydride ion or carbanion be expelled from the central carbon. These species are very unstable, very strong bases. The reagents that produced them ($\mathrm{NaBH}_{4}$, $\mathrm{RMgX}$, $\mathrm{RLi}$) are specialized reagents that do not contain “naked” hydride ions or carbanions. If we wanted to reverse them, we would have to use completely different reaction conditions that avoided the expulsion of the high-energy hydride or carbanion. For example, to accomplish the reverse of the reduction reaction, we would have to use an oxidizing agent (such as $\mathrm{Cr}$(VI)) under completely different conditions. The reduction of a ketone group is central to the reactions of glycolysis, in which pyruvate (the conjugate base of pyruvic acid[3]) is reduced to lactate (the conjugate base of lactic acid) by $\mathrm{NADH}$ (see Chapter $6$). This reaction can be also reversed under different conditions (in this case, the presence of the enzyme lactate dehydrogenase). Glycolysis will be discussed in more detail in Chapter $9$.
Reaction with other carbanions:
There are a number of other ways to generate carbanions: for example, terminal alkynes are quite acidic ($\mathrm{pK}_{\mathrm{a}} 22$) and can be deprotonated by sodium amide (Chapter $5$). The resulting carbanion adds to the carbonyl just as we might expect. Similarly, cyanide ion ($\mathrm{CN}^{-}$) is another source of a negatively charged carbon. It is a good nucleophile, and just as one might expect, it adds to carbonyl groups, and after reaction with a dilute acid, the resulting cyanohydrin is formed. There are two items to note here:
1. Sodium cyanide $\mathrm{NaCN}$ (the usual form of cyanide ion) is highly toxic, so don’t try this at home.
2. The oxidation state of the carbon in the cyano group is the same as a carboxylic acid. As we will see later, this reaction will come in very useful. All of these reactions are the result of a nucleophilic addition to the carbonyl group, during the course of which the carbonyl carbon rehybridizes from $\mathrm{sp}^{2}$ to $\mathrm{sp}^{3}$. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.05%3A_In-Text_References.txt |
In contrast to the addition of hydrogen or carbon nucleophiles, the addition of oxygen and nitrogen nucleophiles is reversible under the conditions in which the reaction occurs. This is because (as we will see) the addition of an oxygen or nitrogen nucleophile results in a tetrahedral intermediate that can regenerate the carbonyl by expelling a leaving group. In contrast to the cases with carbon or hydrogen nucleophiles, oxygen and nitrogen nucleophiles can be good leaving groups. Typically, the reaction is catalyzed either by acid or base as discussed below. For example: in aqueous solution, most aldehydes and ketones will react with water to produce a hydrate.
In acid, the first step is protonation of the carbonyl oxygen, the resultant positive charge is partially delocalized onto the carbon $\rightarrow$, which makes the carbonyl more susceptible to nucleophilic attack even by a relatively poor nucleophile such as water.
This reaction is completely reversible in aqueous solution, and if any aldehyde or ketone is dissolved in water there is always some equilibrium concentration of the hydrate. In fact, formaldehyde ($\mathrm{H}_{2} \mathrm{C=O}$) exists almost exclusively as the hydrate in aqueous solution, whereas most other aldehydes and ketones exist mainly in the carbonyl form. However, it is important to keep in mind that both forms are typically present, and therefore further reactions can proceed from either the hydrate or the carbonyl form.
Base catalysis differs in that the first step is attack by the hydroxide (rather than water) on the carbonyl. Since hydroxide is more reactive than water, the carbonyl does not need to be activated by protonation. What the two mechanisms have in common is the rapid protonation/deprotonation reactions that take place in the intermediate steps. We have already seen this many times and with the reactions of aldehydes and ketones, it becomes more important to appreciate just how ubiquitous protonation and deprotonation are. By controlling the pH or the amounts of reactants or products we will see that it is possible to direct such reactions so that the product we desire is produced.
The hydrate is an example of a structure that will play a major part in our discussion of all carbonyl compounds—which we will refer to as the “tetrahedral intermediate.” In this tetrahedral form, the carbon is in the same oxidation state as the ketone (two bonds to oxygen) but in a different hybridization state, $\mathrm{sp}^{3}$ a for the hydrate and $\mathrm{sp}^{2}$ for the carbonyl. In most cases, the $\mathrm{C=O}$ bond (in the $\mathrm{sp}^{2}$ hybridized form) is stronger ($745 \mathrm{~kJ} / \mathrm{mol}$ in a typical ketone) than two single $\mathrm{C-O}$ bonds ($2 \times 358 \mathrm{~kJ} / \mathrm{mol}$) (in the $\mathrm{sp}^{3}$ hybridized form), which explains why (when there is a low-energy stable leaving group) the tetrahedral intermediate that is formed by attack on the $\mathrm{sp}^{2}$ carbon usually collapses back down to a $\mathrm{C=O}$, expelling a leaving group at the same time. As we move forward, we will see the move from tetrahedral to $\mathrm{C=O}$ many times, the difference in many of the reactions is which group will leave during this process.
If we change the solvent to an alcohol, we see that the same type of reaction occurs. The alcohol oxygen attacks the carbonyl carbon, but we find that the reaction proceeds further. The first product, formed by addition of one alcohol to the carbonyl is called a hemiacetal but then the reaction continues. Each step is reversible (with low activation energy), each protonation and deprotonation is reversible. All of the oxygens in the molecule can be protonated and deprotonated. When the $\mathrm{OH}$ group of the hemiacetal is protonated, it is turned into a good leaving group ($\mathrm{H}_{2} \mathrm{O}$) and the carbon undergoes another attack by an alcohol molecule. The end result is an acetal[4] and water.
Since these reactions are reversible, you may be asking how we can control the reaction. In this case we can use Le Chatelier’s principle: we can either add a lot of starting material or we can remove one of the products as it is formed. In this case, acetal formation is usually done using the alcohol as solvent, and the water that is formed is removed, so that the position of equilibrium is shifted over to produce the acetal. However, as we will see shortly, sometimes we want to regenerate the carbonyl compound, and this can be done by adding water (and acid catalyst) so that the equilibrium shifts back.
If we use a diol such as ethylene glycol ($\mathrm{OHCH}_{2} \mathrm{CH}_{2} \mathrm{OH}$), the resulting cyclic acetal is formed. This is frequently used to protect carbonyl groups in more complex molecules, for example, if we wanted to do a reaction in another part of the molecule. Again, the carbonyl group is easily regenerated. The mechanism of hydrolysis is simply the reverse of the acetal formation, beginning with protonation, attack by water, and so on as shown below.
While this mechanism may look (a bit) complicated, in fact, each step is simple and we have seen similar things many times. The issue with these kinds of reactions is that all the oxygens are being protonated and deprotonated all the time.
Note that the ways that tetrahedral intermediates behave depends on which oxygen is protonated, and which nucleophile (water or alcohol) attacks the protonated acetal or ketone.
One common example of hemiacetal formation is the intramolecular cyclization of D-glucose to form a six-membered ring which contains a hemiacetal group (among many others). This case is actually a rare example of the tetrahedral form (hemiacetal) being more stable than the carbonyl form. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.08%3A_Reactions_of_Aldehydes_and_Ketones_with_Oxygen_Nucleophiles.txt |
There is a similarity between the reactions of oxygen nucleophiles and nitrogen nucleophiles. For example, let us look at the reaction of a primary amine with a ketone. The mechanism begins in the same way with the nucleophile(N) attacking the carbonyl to form a tetrahedral intermediate, which can undergo various reversible protonation/deprotonation reactions until an intermediate is formed that can collapse down to a new product with a $\mathrm{C=N}$ function—which is called an imine. It is the nitrogen analog of the ketone and behaves in much the same way.
There are many other nitrogenous nucleophiles that can react with aldehydes and ketones, for example hydroxylamine ($\mathrm{NH}_{2} \mathrm{OH}$), or hydrazine ($\left(\mathrm{NH}_{2} \mathrm{NH}_{2}\right)$) or a whole range of substituted hydrazines, all react with aldehydes and ketones to produce the corresponding imine.
Generally, we do not see the nitrogen analog of an acetal, the intermediate is unstable and reacts to form the imine. However, in the case of a secondary amine, the reaction proceeds in exactly the same way until the imine stage—with the one difference that the nitrogen is now in the quaternary state (an iminium ion). Instead of addition of another amine, a proton is removed to produce a new functionality, the enamine.
A note on how these reactions proceed:
While these mechanisms may seem complex with many steps, those individual steps are very similar. Students often ask how they can know which way the reaction goes, and the way we write out mechanisms does tend to give the idea that each step is marching purposefully along like clockwork—from one intermediate to another as if the molecules had an “end goal” in sight. Nothing could be further from the truth: each step in the reaction, each protonation and deprotonation, and each nucleophilic attack and loss of a leaving group is occurring all at the same time in a stochastic and chaotic fashion. However, we can control the reaction as discussed earlier by using Le Chatelier’s principle[5]: adding reactants or removing products can shift the position of equilibrium to produce the product that we are after.
7.10: Carboxylic Acids and Derivatives
Now that we have a fairly solid understanding of the reactions of aldehydes and ketones, we are going to move up one oxidation state to look at the behavior of carboxylic acids and their derivatives (Table $7.0.1$), a group of compounds that includes the acids, esters, amides, acid chlorides, and acid anhydrides. Just as we did with aldehydes and ketones, we will highlight and discuss the reasons for both the similarities and differences observed. The most obvious difference between this group of compounds is that the carboxylic acids are acidic; the other derivatives lack an acidic hydrogen bonded to an $\mathrm{O}$ and, therefore, do not participate in simple acid-base reactions[6]. Since we have discussed the reasons for the acidity of carboxylic acids earlier, we will not go over that here at great length, but be sure to check Chapter $1$ if you need a refresher. However, we do want to remind you that many organic compounds are acidic (or basic) and can exist as their conjugate base (or acid) in aqueous solutions, and that the relative amounts of conjugate acid or bases change as $\mathrm{pH}$ changes. The degree to which a molecule exists in an acidic or basic form (in water) is particularly important for biological systems that (in humans) are buffered at around $7.3–7.4$. Recall that we can relate the $\mathrm{pH}$ of a buffered solution to the $\mathrm{pK}_{\mathrm{a}}$ of any acid that is participating in the solution using the Henderson-Hasselbalch equation[7]: $\mathrm{pH}=\mathrm{p} \mathrm{K}_{\mathrm{a}}+\log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}$
where $\mathrm{HA}$ and $\mathrm{A}^{-}$ are the concentrations of the acid and its conjugate base, respectively. We can rewrite this equation (by taking the anti-log of the terms) so that: $\frac{[\mathrm{A}^{-}]} {[\mathrm{HA}]}$
which allows us to estimate the relative amounts of acid and base.
For example, a typical carboxylic acid has a $\mathrm{pK}_{\mathrm{a}}$ of around $4$. At physiological $\mathrm{pH}$ ($\sim 7$), the ratio of the conjugate base to conjugate acid is $\sim 10^{(7-4)}=10^{3}$, that is, there is about 1000 times more of the conjugate base than the conjugate acid for most common carboxylic acids in biological systems.
There are many naturally occurring (that is, biologically relevant) carboxylic acids and most of them exist as the conjugate base at physiological $\mathrm{pH}$. One consequence is that these species are soluble in water because of the favorable ion-dipole interactions that can be formed. Biological molecules that do not contain such polar (ionized) groups ($-\mathrm{COO}^{-}$ or $-\mathrm{NH}_{3} {}^{+}$) are typically insoluble in water: indeed, the presence of this type of polar (ionic) side chain explains the water solubility of many large biological molecules. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.09%3A_Reactions_with_Nitrogen_Nucleophiles.txt |
Carboxylic acids and their derivatives have a number of features in common, the most obvious being that the carbonyl carbon has three bonds to electronegative elements—the carbonyl carbon is, therefore, in a higher oxidation state than aldehydes and ketones. However, the nature of the electronegative atom bonded to the carbonyl does impact the properties of the molecule as a whole. For example, if we examine the carbonyl absorptions of these derivatives, we find quite a wide range of frequencies. Recall that most aldehydes and ketones absorb around $1710-30 \mathrm{cm}^{-1}$, similar to the value displayed by carboxylic acids. Derivatives of carboxylic acids, however, range from around $1670 \mathrm{cm}^{-1}$ for amides to 1810-30 cm–1 for acid chlorides and anhydrides. How can we explain the difference and use it to predict (and explain) differences in the properties of these derivatives?
Recall that the IR absorption frequency of carbonyls depends on the energy required to stretch the bond (which is determined by the bond energy). Since the amide absorption frequency is much lower than the absorption frequency of the acid chloride we can conclude that the carbonyl group in the amide requires less energy to stretch than the acid chloride (the bond is weaker). If we look at the structures of these two functional groups we see that both involve an electronegative element bonded to the carbonyl carbon, leading to electron withdrawal by induction through the sigma bond to the carbon. The difference between the two (amide and acid chloride) arises because the amide nitrogen can also donate its electron pair to the carbonyl oxygen. The result is that in amides there is significant overlap between the lone pair of the amide nitrogen and the carbonyl pi bond system. The $\mathrm{C-O}$ bond now has less double bond character and therefore it takes less energy to stretch. In contrast chlorine (or any halogen) is not basic and so does not participate in this kind of resonance through the pi system. In acid chlorides, the chlorine is removing electron density from the carbon by induction through the pi system; there is more $\mathrm{C=O}$ double bond character than in an amide and it takes more energy to stretch the carbonyl bond (leading to a higher IR absorption).
Electron donation from the nitrogen means that the lone pair on the amide nitrogen is not as available for donation to acids. This means that amides are not basic, or rather, nowhere near as basic as amines in which the lone pair is freely available for donation to an acid. This has important ramifications in biological systems. In polypeptides, which are amino acids linked by amide functional groups (circled below), amines are protonated at physiological $\mathrm{pH}$ (they exist as $\mathrm{RNH}_{3}{}^{+}$)[8]. It is an interesting thought experiment to predict how proteins and peptides would behave if amide nitrogens were more basic.
While there are relatively few naturally occurring acidic or basic side chains in polypeptides (from amino acids such as glutamic acid or lysine), every peptide is linked by innumerable amide bonds between the individual amino acids. If all these amide nitrogens were protonated, the peptides and proteins would take up very different structures (since they would have a large positive charge which would, in the absence of counter ions, repel other parts of the molecule). | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.11%3A_Infra-red_Spectra_as_Evidence_of_Carboxylic_Acid_Derivative_St.txt |
The evidence from IR spectroscopy can help us predict the relative reactivities of carbonyls. For example: IR spectroscopy evidence tells us that the amide functional group is stabilized and the nitrogen lone pair is conjugated to the carbonyl group, whereas the partially positive charge at the acid chloride carbonyl carbon is increased (because of induction) compared to the amide. Therefore, a plausible prediction is that the acid chloride is more reactive than the amide and, as we shall see shortly, this is true. In general, the order of reactivity parallels the absorption frequency of the carbonyl group, acid chlorides are more reactive than anhydrides, esters and carboxylic acids are fairly similar in their reactivity (except with bases), and amides are the least reactive. In general, aldehydes and ketones are more reactive than all carboxylic acid derivatives except acid chlorides.
7.13: Reactions at the Carbonyl Group of Acid Derivatives with Irrev
Just as we saw with aldehydes and ketones, we can reduce a carbonyl group by the addition of a hydride ion. Typically, Lithium Aluminum Hydride (rather than sodium borohydride) is used in such a reaction because acid derivatives (amides, esters, anhydrides, chlorides) are usually not reactive enough to react with sodium borohydride. With $\mathrm{LiAlH}_{4}$, the acid or acid derivative is reduced all the way down to the primary alcohol or amine. With acids or esters, the reaction does not stop at the aldehyde step since aldehydes are generally more reactive and the reducing reagent will preferentially reduce any aldehyde as it is formed.
The situation is a different when acid derivatives are reacted with highly reactive carbanions such as Grignard reagents or alkyl lithium reagents. In this case, any derivative that has acidic protons (the carboxylic acid itself or most amides) will simply deprotonate and the reaction will go no further. With esters or acid chlorides, however, two equivalents of the carbanion add to the carbonyl. The reaction goes through the intermediate step of forming a ketone, but just like the $\mathrm{LiAlH}_{4}$ reduction, since ketones are more reactive than acid derivatives, the ketone will undergo nucleophilic attack as they are formed, resulting in the alcohol.
7.14: Nucleophilic Addition and Elimination Reactions of Acids and D
Just as with aldehydes and ketones, the reaction of acids and derivatives with oxygen and nitrogen nucleophiles is somewhat more complex: at each step, there is the potential for reversal. The formation and decomposition of the tetrahedral intermediate again plays a central role in the outcome of the reaction, and it is possible to use the knowledge of structure and properties to predict how the reaction will proceed. Furthermore, with a knowledge of how concentrations of reactants and products affect equilibrium positions, we can control the outcome of reactions using Le Chatelier’s Principle.
First, we will consider the reaction of a carboxylic acid with an alcohol leading to the formation of an ester. The reaction is typically performed with an acid catalyst and given that information, you should be able to write the mechanism. The first step is protonation; while there are two potentially basic oxygens, protonation tends to occur preferentially on the carbonyl oxygen (not the $\mathrm{OH}$) because the resulting cation can be resonance-stabilized. As with aldehydes and ketones, protonation activates the carbonyl and the next step is attack by the nucleophile—in this case an alcohol. In the reaction scheme below, ethanol reacts with acetic acid to give ethyl acetate.
The crucial part of this mechanism is the series of tetrahedral intermediates that are interconverted by protonating and deprotonating the three different oxygens. Since all these groups are similar ($\mathrm{OH}$ or $\mathrm{OR}$) the probability of each of these groups leaving is more or less the same once they are protonated. Just as with aldehydes and ketones, the system will become more stable (with stronger bonds) if the carbonyl group reforms by the elimination of one of the groups attached to the carbon. Again, we can shift the equilibrium for this reaction by manipulating the reaction conditions. Typically, esterifications are carried out using the alcohol as solvent (so it is in large excess), and the water produced is removed as it is formed.
As you might expect, this reaction is entirely reversible, and the reverse reaction is typically carried out in aqueous solution with either acid or base catalysis. In fact, this reaction is the basis of saponification (soap making); in which long-chained fatty acid esters of glycerol (triglycerides) are hydrolyzed in an aqueous solution with a base catalyst.
The triglyceride (fat or oil) is insoluble in water, while the sodium salt of the long-chain fatty acid (soap) is soluble. The soap molecules aggregate to form spherical micelles in which the polar head groups lie on the outside and the non-polar tails are inside[9]. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/07%3A_Nucleophilic_attack_at_the_carbonyl_carbon-/7.12%3A_Relative_Reactivities_of_Carboxylic_Acids_and_Derivatives.txt |
Conjugation: Conjugation is the term we use to describe an arrangement of alternating single and double bonds. To explain how conjugated systems behave differently from non-conjugated systems we will compare 1,3-pentadiene (which is conjugated) and 1,4 pentadiene (which is not conjugated). To recognize the differences between the two, let us look at the orbitals that are involved in the $\pi$ system bonding. Recall that $\pi$ bonds can be considered as the side-to-side overlap of p orbitals, so that the electron density lies above and below the plane of the rest of the molecule. Consider the orbitals in 1,3-pentadiene: there is overlap of $\mathrm{p}$ orbitals that result in a continuous
π orbital system over carbons 1-4. The consequence of this is that there is some partial double-bond character between carbons 2 and 3. In 1,4-pentadiene there is no possibility of overlap between the two separate $\pi$ bonds. Note that in the non-conjugated system, there is an $\mathrm{sp}^{3}$ hybridized carbon between the two sets of $\mathrm{sp}^{2}$ hybridized carbon-carbon double bonds which prevents any overlap of ($\mathrm{p}$) orbitals on carbons 2 and 4.
One way to indicate and predict where this partial double bond character can occur is to use resonance structures. We can write resonance structures for the conjugated system which have double bond character between $\mathrm{C}-2$ and $\mathrm{C}-3$. Note that since resonance contributors A and B are equivalent, there is no actual charge separation in this molecule. It is not possible (without breaking a sigma bond) to write resonance structures like this for 1,4-pentadiene (try and convince yourself that this is true ).
MolecularOrbitalTheory: Another model that can be used to describe the bonding is to consider the π system in terms of molecular orbital theory. Molecular orbital theory considers the bonding orbitals as extending over the whole molecule. The number of molecular orbitals is equal to the sum of all the atomic orbitals. In practice, this approach is far too complex for even the smallest of organic molecules, which is why we usually use the simpler valence bond model in which we consider bonds as being located between two atoms. In the case of conjugated systems, it is often helpful to use the valence bond approach for the sigma (single bonds) framework and then consider the conjugated π system using molecular orbital theory. In this case, if we have a conjugated system of two π bonds, then four p atomic orbitals are involved in forming the four molecular orbitals ($\mathrm{MO}$s). Recall that when molecular orbitals are formed from atomic orbitals ($\mathrm{AO}$s), if the (quantum mechanical) wave functions add in phase, the resulting energy of the $\mathrm{MO}$ is lower—that is, the interaction is stabilizing and the $\mathrm{MO}$s are bonding $\mathrm{MO}$s. If the $\mathrm{AO}$s add out of phase, the interaction is destabilizing and the result is antibonding $\mathrm{MO}$s. Note that, in the diagram ($\downarrow$), only the lowest energy $\mathrm{MO}$ has electron density between carbons 2 and 3. All the other $\mathrm{MO}$s have a node (no electron density) at this position. Overall, we see that there is more $\pi$ electron density between $\mathrm{C}-1$ and $\mathrm{C}-2$, and between $\mathrm{C}-3$ and $\mathrm{C}-4$.
Since there are only 4 $\pi$ electrons, only the two bonding $\mathrm{MO}$s are occupied while the two antibonding $\mathrm{MO}$s are unoccupied. As we will see, if we consider reactions using $\mathrm{MO}$ theory, the Highest Occupied $\mathrm{MO}$ ($\mathrm{HOMO}$) and the Lowest Unoccupied $\mathrm{MO}$ ($\mathrm{LUMO}$) are the orbitals that participate in new bonding interactions. In general, we use the simplest model that allows us to predict and explain the outcome of reactions, which is usually valence bond theory, but we will call on $\mathrm{MO}$ theory when necessary—and discussions of conjugated systems sometimes require $\mathrm{MO}$ theory to explain phenomena.
Stability of Conjugated Systems: As we have seen before, one way to identify the thermodynamic stability of alkenes is to reduce them to the corresponding alkane by adding $\mathrm{H}_{2}$ (hydrogens) across the double bond and determine the enthalpy change.[1] In the case of 1,3- and 1,4-pentadiene, we can compare their heats of hydrogenation to produce the corresponding pentane; we find that the conjugated diene is about $25 \mathrm{kJ/mol}$ more stable. This is a general finding: the more conjugated a system, the more stable it is and the less reactive it is.
08: Conjugated compounds and aromaticity
The position next to the benzene ring is special because reactive species such as carbanions, carbocations, or radicals at that site can be conjugated (and therefore stabilized) with the benzene pi system. For example, we have already seen that phenol ($\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{OH}$, $\mathrm{pK}_{\mathrm{a}} \sim 10$) is much more acidic than a typical alcohol ($\mathrm{pK}_{\mathrm{a}} \sim 16$) because the negative charge can be stabilized in the aromatic ring. Similarly, benzyl halides ( e.e. $\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}_{2}\mathrm{Br}$) undergo nucleophilic substitution under $\mathrm{S}_{\mathrm{N}} 1$ conditions, even though these compounds appear to be primary halides, a carbocation next to the ring can be stabilized.
The position next to the ring is called the benzylic position, and is particularly reactive because of the ability to stabilize any intermediate in the aromatic ring. For example, it is possible to selectively introduce a bromine at the benzylic position via a reaction in which radicals are generated. We have spent little time on the reactions of organic compounds with radicals[9], because, generally, these reactions are very difficult to control. For example, an alkane will react with halogens in the presence of light or peroxides (which initiate the reaction by forming a radical), but the reaction is not synthetically useful and typically the halogen can end up in all the possible positions.
However, we can selectively introduce a bromine atom at a benzylic position, because the intermediate benzylic radical is most stable, and will have a lower activation energy to formation. The reaction begins by producing a bromine radical from $\mathrm{Br}_{2}$ by breaking the bond with light to give two $\mathrm{Br}$ radicals. The $\mathrm{Br}$ radical abstracts (removes) an $\mathrm{H}$ from the benzylic position to give the resonance-stabilized benzylic radical which then abstracts a $\mathrm{Br}$ from bromine ($\mathrm{Br}_{2}$).
In practice, we use a source of bromine radicals that is easier to handle than elemental bromine, N-bromosuccinimide (NBS), which has a weak $\mathrm{N-Br}$ bond that will break homolytically (i.e. to give two radicals) in the presence of light or peroxides. NBS will react with alkyl benzenes to introduce a bromine specifically at the benzylic position.
Another unique reaction of benzylic positions is that they can be oxidized by reagents such as $\mathrm{KMnO}_{4}$ to give the corresponding carboxylic acid; any other carbons in the side chain are removed in the process. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.01%3A_Reactions_of_Substituted_Benzenes-_Reaction_at_the_Benzylic_Position.txt |
$An image of Lewis acid base electrophile-nucleophile reactions.$
Diels-Alder Reactions:
The best-known of these reactions is a cycloaddition reaction known as the Diels-Alder reaction. It typically occurs between a conjugated diene and another alkene known as the dienophile. This reaction is very useful in synthesis because it forms two new $\mathrm{C-C}$ bonds at the same time. It is also stereospecific and regiospecific. The simplest diene that can participate in a Diels-Alder is butadiene. Typically, butadiene tends to exist in the more stable s-trans[10] conformation—however, there is an equilibrium concentration of the s-cis conformation through which the reaction can occur. In fact, cyclopentadiene, in which the diene is permanently fused in the s-trans conformation, reacts rapidly with itself (it can be both diene and dienophile) in a reversible reaction.
The “dienophile” can be a simple alkene, but the presence of an electron-withdrawing group (such as a carbonyl group) on the double bond improves yields. We can envision the reaction as taking place in a concerted fashion. Again, the reaction is reversible, and the reverse reaction is known as a retro Diels-Alder. We can use our knowledge of thermodynamics to predict the most appropriate conditions for the reaction. Recall that the extent of any reaction can be predicted from the Gibbs free-energy change $\Delta \mathrm{G} = \Delta \mathrm{H} – T \Delta \mathrm{S}$. Since the reaction produces two new $\mathrm{C-C}$ single bonds and one new $\mathrm{C-C}$ pi bond while breaking two $\mathrm{C-C}$ pi bonds, we can assume that the enthalpy change for this reaction is negative (bond formation releases energy and bond-breaking uses energy). Therefore, the $\Delta \mathrm{H}$ term is always favorable. We can also predict the sign of the entropy change for the system; since we are producing one molecule from two, we would expect $\Delta \mathrm{S}$ to be negative also. The entropy term is unfavorable. From this analysis, we can see that the temperature at which the reaction is carried out is crucial. High temperatures would favor the reverse reaction. Therefore Diels-Alder reactions are typically run at fairly moderate temperatures that are between room temperature and $150^{\circ} \mathrm{C}$).
The possibilities for Diels-Alder reactions are quite extensive and since this is a concerted reaction, any stereochemistry in the starting materials is conserved in the products. For example, if the dienophile has cis or trans stereochemistry, this is conserved in the product. The cis dienophile gives the cis product (and vice versa for the trans).
Some examples of Diels-Alder reactions are given here.
While we can draw mechanistic arrows for pericyclic reactions such as Diels-Alder, they are best understood by using molecular orbital theory. In this treatment, we can consider the reaction as taking place between the highest occupied molecular orbital (HOMO) of the diene and the lowest unoccupied molecular orbital (LUMO) of the dienophile.
For our purposes, this treatment is too complex, but if you go on to further studies of pericyclic reactions, MO theory will be the approach that will allow you to predict the outcome of many different reactions.
If a fused ring is formed during the reaction, there are two possibilities for the orientation of the substituents on the diene as shown below: exo and endo.
Because of the possibility of a stabilizing interaction in the endo position, (the pi system of the carboxylic acid can interact with the pi system of the diene), the endo product is usually produced.
As noted earlier, there are other types of pericyclic reactions. All of the common reactions that occur simply by heating up the starting materials involve the cyclic movement of six electrons. The transition states for these reactions all involve molecular orbitals that extend throughout the system; these orbitals have considerably lower energy than one might expect. In some ways, this is analogous to the 6 pi electrons of aromatic systems, which are also stabilized in lower energy molecular orbitals.
Interestingly, these reactions do not occur in the same way if they are initiated by electromagnetic radiation (that is, if we shine light on them rather than heat them up). In this case the electrons that participate in the reaction are actually in higher energy orbitals (the electron absorbs a photon and is promoted to a higher energy level). Absorbing light leads to a completely different set of reactions and outcomes, something that is explored in subsequent organic chemistry courses. There is, however, one particularly interesting (and biologically relevant) photochemically induced reaction, namely the reaction of adjacent thymine bases within a DNA molecule. Upon the absorption of a UV photon, such adjacent thymines can undergo an electrocyclic reaction that results in dimer formation.
The presence of a thymine dimer results in conformational changes in the DNA that, unrepaired, can lead to mutations. Thymine dimers are recognized and repaired via two distinct cellular-repair mechanisms. Unrepaired damage can lead to skin cancer (a range of carcinomas and melanomas).
This is one reason to limit skin exposure to the sun.
8.03: In-Text References
1. Recall that this is how relative stabilities of alkenes were determined (Chapter 4).
2. A longer version of this brief overview is found in Chapter 2.
3. http://phototroph.blogspot.com/2006/...n-spectra.html
4. Recall we saw this same tautomerism when water adds across a triple bond.
5. https://en.Wikipedia.org/wiki/Gilman_reagent
6. The explanation for this phenomenon goes beyond the scope of this course and is best explained using the theory of hard and soft acids and bases. For more information see https://en.Wikipedia.org/wiki/HSAB_theory
7. This is in contrast to simple hydrocarbons which do not smell. In fact, methane thiol ($\mathrm{CH}_{3}\mathrm{SH}$) must be added to methane and propane which are used for heating so that they can be detected in the event of a gas leak.
8. TNT is an explosive compound, as are many nitrated organic compounds (for example nitroglycerin). The nitro group is relatively unstable ($\mathrm{NO}$ bonds are weak) and these compounds can decompose explosively to produce more stable nitrogen oxides and $\mathrm{CO}_{2}$, releasing a great amount of energy at the same time. https://en.Wikipedia.org/wiki/Trinitrotoluene
9. Except for the radical-induced addition of $\mathrm{HBr}$ in an anti-Markovinkov manner across a double bond.
10. Recall that there is some double bond character between carbons 2 and 3 and, therefore, rotation is somewhat restricted around this bond unlike a normal $\mathrm{C-C}$ single bond. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.02%3A_Pericyclic_reactions.txt |
As you probably remember, the more atomic orbitals that combine to produce molecular orbitals, the smaller the energy gap between the $\mathrm{MO}$s becomes. For example, an electron in an isolated pi bond absorbs energy in the far UV ($\sim 170 \mathrm{~nm}$), a rather high energy. As we increase the number of conjugated double bonds, the energy gap between the orbitals (in fact the gap between the HOMO and LUMO) gets smaller and smaller, so that lower energy photons can bring about this transition. Eventually, the wavelength of light needed to promote an electron from highest occupied to the lowest unoccupied $\mathrm{MO}$ ($\mathrm{HOMO} \rightarrow \mathrm{LUMO}$) moves into the visible region, and the substance becomes colored. Note that its color does not represent the the light that is absorbed, but rather the light that is transmitted or reflected. These conjugated regions of molecules are called chromophores.[3] The longer the conjugated section of the molecule, the longer the wavelength that is absorbed. The compounds responsible for highly colored fruits and vegetables—such as lycopene and B-carotene, as well as your ability to see visible light (retinals) contain large chromophore regions.
Samples of UV-VIS absorption spectra are shown here ($\rightarrow$). Note that, in contrast with most other spectroscopic techniques (which usually produce sharp lines), the peaks in these spectra are more broad; the longer the conjugated section of the chromophore is, the longer the wavelength (and lower energy) that it absorbs. This means that each of these compounds has a different color. Moreover, the fact that the peaks in these spectra are not sharp means that UV-VIS spectroscopy is typically not used for identification of compounds. However, the amount of light absorbed is proportional to the concentration of the substance, so UV-VIS spectroscopy can be used to determine the concentration of samples. The visible spectrum runs from about $300 \mathrm{~nm}$ to $750 \mathrm{~nm}$.
Reactions of Conjugated Systems: Kinetic and Thermodynamic Control of Reactions:
Now let us consider the reaction of 1,3-butadiene with a reagent such as $\mathrm{HCl}$. Just as we saw with isolated alkenes, the first step in the reaction is the addition of the electrophile $\mathrm{H}^{+}$ to produce the most stable carbocation. In this reaction, the proton adds to C-1; the resulting carbocation is resonance-stabilized with positive charge at both C-2 and C-4. The question, then, is: where does the nucleophile add?
In fact, the position of equilibrium depends upon the conditions under which the reaction occurs. First, let us consider the two potential sites of attack. Attack at C-2 would mean that the reagent has added across the C-1/C-2 pi bond—this is called 1,2 addition. Attack at C-4 is called 1,4-addition and the two products are clearly different.
The partial positive charge at C-2 is located on a secondary carbon, whereas that on the C-4 is on a primary carbon. This means that the intermediate carbocation has more partial positive charge on C-2 than on C-4 and the transition state for the reaction to give the 1,2-addition product will have a lower activation energy than the transition state for the 1,4 product because it is more stabilized (by induction and hyper-conjugation). Therefore, the attack of the nucleophile (chloride) would occur faster at C-2; that is, 1,2 addition is the kinetically favored product.
However, if we consider the 1,4-addition product, the alkene product itself is more substituted (there are two alkyl groups on the double bond) and, therefore, it is more stable than the product of 1,2 addition. Even though the 1,4-addition product is formed more slowly through a less-stabilized transition state, the product itself is more stable; it is the thermodynamically favored product.
In fact, by controlling the reaction conditions, it is possible to produce either the kinetic or the thermodynamic product. If the reaction is run at relatively low temperature, there will not be enough energy to overcome the activation energy barrier associated with the 1,4-addition product reaction and the kinetic (1,2) product will be produced. However, at higher temperatures, there is enough energy; the 1,4 activation energy barrier will be reached more often, so that even if the kinetic product is formed, the fact that the reaction is reversible will, over time, lead to the most stable product accumulating at equilibrium. It is important to note that the most stable product is not always formed—it depends upon reaction conditions.
Conjugated carbonyl compounds:
Carbonyl compounds can also be conjugated with carbon-carbon double bonds. These compounds are often referred to as α,β unsaturated carbonyls (the carbon next to the carbonyl carbon is termed the alpha carbon). By drawing resonance forms for this system, we see that there is a partial positive charge on the carbonyl carbon and on the $\beta$ carbon. This means that $\alpha , \beta$ unsaturated carbonyl compounds are susceptible to nucleophilic attack at both the $\mathrm{C=O}$ and at the $\beta$ carbon. Which is analogous to the 1,2- and 1,4-additions to conjugated dienes.
Just as the 1,3-conjugated diene case, attack at the carbonyl carbon is the kinetically preferred product since there is more positive charge there and the reaction is faster. Reagents that attack carbonyls irreversibly, such as grignards, alkyl lithiums, or reducing agents such as $\mathrm{LiAlH}_{4}$ will, therefore, tend to produce the product of attack at the carbonyl. This is the 1,2-addition product.
If we use a reversible nucleophile ($\mathrm{ROH}$, $\mathrm{H}_{2}\mathrm{O}$, $\mathrm{RNH}_{2}$), then the most thermodynamically stable product will predominate, that is, the product of 1,4 addition. However, this product undergoes a tautomerism[4] that regenerates the carbonyl (which is the source of the stability—recall $\mathrm{C=O}$ bonds are very strong).
Another reagent that can produce a product resulting from attack at the beta carbon is a reagent that we have not seen yet, an organocuprate, otherwise known as the Gilman[5] reagent, which consists of a complex of alkyl groups, copper, and lithium and has the generic formula $\mathrm{R}_{2}\mathrm{CuLi}$. The general reaction is shown here. The organocuprate has the two alkyl groups bonded to the copper species (formally $\mathrm{Cu}^{2+}$), but the carbon-copper bond is more covalent than ionic (the charge on the carbon is lower than on the equivalent Grignard or alkyl lithium reagent) and this makes attack at the beta carbon more likely.[6]
Gilman reagents are also very useful because they can accomplish nucleophilic attack on alkyl halides, which does not occur with Grignard or alkyl lithium reagents. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.04%3A_UV-VIS_Spectroscopy_and_Conjugated_Systems-_Review.txt |
After considering some of the properties of conjugated systems, we now move on to what might be the ultimate in conjugated systems, as exemplified by benzene $\mathrm{C}_{6}\mathrm{H}_{6}$. While many of the properties of benzene and its derivatives are similar in some ways to those of open-chained conjugated systems, there are important differences. Benzene has the property known as aromaticity and we say that benzene is aromatic. In everyday language, the term aromatic implies that something smells; usually in a good way. While benzene does have a fairly strong smell[7], in chemistry, aromatic has come to mean a particular set properties that emerge from the molecular structure of some molecules. Benzene is the simplest and most common example of an aromatic compound. The structure of benzene was something of a puzzle for quite a long time; it eventually came to be written in the form of what are now called Kekulé structures in which double and single bonds appear to alternate around the ring. We can write two equivalent resonance structures which contribute equally to the overall structure of the molecule. While these models can serve us well when trying to figure out what the electrons are doing during reactions, neither is not adequate to represent the actual structure of benzene. Sometimes, you will see benzene written with a circle in the middle ($\beta$) to indicate that, in reality, there are no single or double bonds present; rather, there is the same electron density (1.5 bonds) between all the carbons.
Benzene has some rather remarkable properties which led chemists to classify it as a member of a completely different type of functionality. For example, benzene is much more stable than one might imagine—even for a conjugated system. The heat of hydrogenation of benzene is $-208 \mathrm{~kJ/mole}$, while the $\Delta \mathrm{H}$ of hydrogenation of the isolated double bond in cyclohexene is $130 \mathrm{~kJ/mol}$. We can see the effects of conjugation in 1,3-cyclo-hexadiene, which is $\mathrm{232 \mathrm{~kJ/mol}$ (not 260 as might be expected if it were conjugated). Similarly, if benzene were not conjugated, we would expect a heat of hydrogenation to be $3 \times 130 = 390 \mathrm{~kJ/mol}$. Therefore, the difference between the expected and the actual hydrogenation energy must be due to the stabilization conferred by resonance.
This stability is called the resonance energy and, in aromatic compounds like benzene, it has a significant effect on the properties of the substance and types of reactions that a molecule participates in. For example, benzene does not react with electrophiles in the same way as isolated alkenes or even open-chained conjugated alkenes. Recall that the most common reactivity of simple alkenes is the electrophilic addition of E-Nu, where E is an electrophile and Nu is the nucleophile, across the double bond. In contrast, benzene undergoes electrophilic substitution; typically the reaction conditions require a catalyst and extensive heating. We will discuss the mechanism of this reaction shortly, but for now the important thing to note is that it is difficult to disrupt the aromatic ring of electrons and, when that does happen, the aromatic ring regenerates.
8.06: What is aromaticity and how do we recognize aromatic systems
To recap, benzene is uniquely stable and it is relatively difficult to get it to react, despite the high electron density in the ring. It is significantly more stable than cyclohexadiene which is conjugated, but only contains two bonds. There is clearly something special about a conjugated ring system above and beyond simple conjugation properties.
If we examine other systems, we can begin to find the parameters that govern the property of aromaticity. We can begin by looking at some other cyclic, conjugated systems. Many fused six-membered rings (for example: naphthalene and anthracene) are aromatic. However, cyclobutadiene and cyclooctatetraene are not aromatic and have quite different properties.
By investigating many cyclic conjugated systems, it has been possible to identify the factors that lead to aromaticity.
These are known as Huckel’sRule: Aromatic compounds are planar, cyclic, conjugated, and have $4 n + 2 \(\pi$ electrons in the $\pi$ electron cloud. Benzene is the archetypal aromatic compound: it is planar (all carbons $\mathrm{sp}^{2}$ hybridized), cyclic (obviously) conjugated (apparent alternating single and double bonds), and it has $6 \pi$ electrons ($n=1$). If we look at the bonding within benzene, we see that overlapping $\mathrm{p}$ orbitals form a ring of pi electron density above and below the sigma $C-C$ framework. More generally, aromatic compounds must be planar, cyclic, and conjugated so that the p orbitals overlap to form this continuous ring of electron density. So why do aromatic compounds have 4n+2 π electrons? The answer involves molecular orbital theory.
As previously noted, a consideration of the number and type of atomic orbitals that contribute to the bonding system enables $\mathrm{MO}$ theory to predict the molecular orbitals that span the whole molecule. In benzene, we have six $\mathrm{p}$ atomic orbitals and, therefore, expect that they combine to give six molecular orbitals as shown here. Note that there are three bonding and three antibonding $\mathrm{MO}$s and, since there are only six electrons in the system, we get a total of three bonds.
This type of analysis can be done for any cyclic conjugated system. While it is too complex here to go into the mathematical underpinnings, there is a relatively simple way to determine the relative energies of the $\mathrm{MO}$s. The approach requires that you inscribe the cyclic system into a circle, with one corner of the ring at the bottom. The places where the corners meet the ring are represent the relative energies of the $\mathrm{MO}$s, as shown below. This arrangement is the origin of the $4 n + 2$ rule as we will see.
So, for example, if we consider cyclobutadiene (which has $4 n \pi$ electrons), we see that there are 4 $\mathrm{MO}$s and 4 electrons. However, two of those electrons are in non-stabilized orbitals AND are uncoupled because the orbitals are of the same energy. Remember Hund’s rule: electrons occupy orbitals singly until they have to start doubling up.
This means that if cyclobutadiene were aromatic, it would be highly unstable because it would contain two unpaired electrons—it would exist as a diradical. The presence of unpaired electrons generally makes compounds very reactive. In fact, cyclobutadiene is not aromatic and instead exists as two isolated double bonds by lengthening the single bonds. Even so, it is still unstable and does not exist above $-78^{\circ}\mathrm{C}$.
Cyclooctatetraene also has $4 n \pi$ electrons (where $n=2$), and again we see the problem is that there are a pair of degenerate (same energy) $\mathrm{MO}$s, where the last two π electrons are located. Again, this is highly destabilizing and, to avoid this electron configuration, cyclooctatetraene actually bends so the double bonds are not conjugated with each other. The origin of the $n+2$ rule, then, has to do with the arrangement of $\mathrm{MO}$s. Any cyclic conjugated compound with $4 n \pi$ electrons will not be able to take up a stable conjugated arrangement because it will involve the highly unstable diradical. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.05%3A_Aromaticity.txt |
The stability conveyed by aromaticity can be a powerful driving force. For example, cyclopentadiene is quite acidic with a $\mathrm{pK}_{a}$ of 15 (compared to $> 50$ for alkanes). It is the $\mathrm{sp}^{3}$ hybridized carbon that is deprotonated and it rehybridizes to $\mathrm{sp}^{2}$ so that conjugation around the ring is possible. The product cyclopentadienyl anion is aromatic since it now has $6 \pi$ electrons; two from the lone pair resulting from the deprotonation and four from the original pi system. It is the drive towards aromaticity that makes cyclopentadiene so much more acidic than a normal alkene.
Similarly, cycloheptatriene can be induced to become a positively charged aromatic ion by treating the corresponding alcohol with acid which will allow the $\mathrm{OH}$ to leave as $\mathrm{H}_{2}\mathrm{O}$, leaving behind a positively charged delocalized aromatic ion (the tropylium cation), which has six electrons delocalized over seven carbon atoms. In fact, there are quite a few ways to generate such aromatic anions, and they are all more stable than might be predicted if you didn’t know about aromaticity.
8.08: Heterocyclic Aromatic Compounds
While the aromatic ions are interesting curiosities, there is a class of aromatic compounds that are of practical importance from a biological perspective: the heterocyclic aromatic compounds. These compounds typically have one or more carbons replaced by $\mathrm{N}$, $\mathrm{O}$, or $\mathrm{S}$. For example, pyridine can be considered as the nitrogen analog of benzene. The nitrogen is $\mathrm{sp}^{2}$ hybridized and contributes an electron to the aromatic system. Pyridine is an important component of many biological molecules; for example, nicotine, a component of the $\mathrm{NAD}$ and $\mathrm{NADH}$ oxidation/reduction system discussed earlier. The lone pair on the pyridine nitrogen sits in an $\mathrm{sp}^{2}$ hybrid orbital at right angles to the pi system (like the $\mathrm{C-H}$ bonds) and, therefore, pyridine is basic because the lone pair is accessible. In fact, pyridine is often used as a base to react with any by-product acid such as $\mathrm{HCl}$ that might be produced in a reaction.
Indole is another heterocyclic aromatic system, but, in contrast to pyridine, indole uses the nitrogen lone pair to contribute the aromatic pi system. Indole has $10 \pi$ electrons ($n=2$), of which two are from the nitrogen lone pair.
Consequently, indole is not basic. The ring system of indole also appears in many biologically important molecules, including the neurotransmitter serotonin, amino acid tryptophan, and hallucinogen lysergic acid diethylamide (LSD).
There are many biologically important nitrogenous aromatic heterocycles, for example the bases in $\mathrm{DNA}$ and $\mathrm{RNA}$ all contain heterocyclic aromatic rings. For example, cytosine, which is usually written in the keto form (structure A) – which may not at first sight seem to be aromatic. However, all the atoms in the ring are planar $\mathrm{sp}^{2}$ hybridized and there are six electrons in the ring. The result is that cytosine can exist in its tautomeric enol form (B), but the keto form is actually more stable (because of the $\mathrm{C=O}$). Both enol and keto forms are aromatic.
8.09: Spectroscopy of Aromatic Compounds-
As you might expect, both the $\mathrm{C}-13$ and the ${}^{1}\mathrm{H}$ NMR spectrum of benzene show only one peak (H-NMR $7.3 \mathrm{~ppm}$ and C-NMR $128 \mathrm{~ppm}$) meaning that there is only one type of carbon and one type of hydrogen present in the molecule. However, these single peaks appear at lower field strengths relative to those found in alkenes or even conjugated alkenes (which normally appear between $5$ and $6 \mathrm{~ppm}$ in the $\mathrm{H}$-NMR). This low field absorption is caused by a phenomenon that occurs when the cyclic electron cloud in the ring is placed into an external magnetic field. The ring of electron density begins to cycle, producing a ring current and an induced magnetic field; the resulting intrinsic field reinforces the external field. The result is that the external field does not need to be very high to bring the carbon or hydrogen nuclei to resonance. In effect, the aromatic carbons and hydrogens are deshielded and appear at low field.
Interestingly, the induced field opposes the external field in the center of the ring, and, in fact, there are cyclic aromatic polyenes where (because of structural constraints) some of the hydrogens do point to the center of the ring. There is a marked difference between the inner and outer hydrogen resonances in these compounds because the hydrogens are in different areas of the induced magnetic field. This effect is called diamagnetic anisotropy.
The IR spectra of aromatic compounds typically show a $\mathrm{C-H}$ stretch above $3,000 \mathrm{~cm}^{-1}$, and a $\mathrm{C-C}$ bend around $1600 \mathrm{~cm}^{-1}$, but these signals are often mixed in with others and, in general, IR is not very useful for identifying aromatic compounds. On the other hand, UV-VIS spectroscopy is often used to identify the aromatic chromophore. Benzene itself has a broad absorption around $254 \mathrm{~nm}$ and, as we will see, this absorption changes depending on the electron withdrawing and donating properties of groups on the ring itself. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.07%3A_Aromatic_Ions.txt |
As we have seen, aromatic compounds are considerably more stable than one might predict. Consequently, it takes more energy to make aromatic compounds undergo reactions since, in order to react the aromatic overlap of orbitals in the ring, it must be destroyed at some point during the reaction. Since the aromatic ring is so electron-rich we might predict that it would undergo electrophilic attack but, rather than undergoing electrophilic addition like alkenes and conjugated alkenes do, aromatic compounds typically undergo electrophilic substitution. For example, benzene will react with bromine in the presence of a catalyst such as iron (III) bromide (FeBr3) to give bromobenzene.
Let us now take a closer look at the steps involved in this reaction to see how this substitution ($\mathrm{Br}$ for $\mathrm{H}$ivc) is accomplished. Since benzene is so stable, a more reactive electrophile is needed to react with the ring. This is accomplished by adding a Lewis acid catalyst $\mathrm{FeBr}_{3}$, which forms a complex with the bromine to produce $\mathrm{Br}^{+}$ (stabilized by the $\mathrm{FeBr}_{4} {}^{-}$). The $\mathrm{Br}^{+}$ electrophile now reacts with the electron-rich benzene ring to produce a resonance-stabilized intermediate called a sigma complex.
Now, instead of a nucleophile attacking, the aromatic ring is regenerated by loss of a proton.
The resulting bromobenzene is much more stable than the corresponding addition product.
c
This electrophilic substitution reaction is the primary mechanism by which most aromatic compounds react. A very reactive electrophile must be generated; it then adds to one of the ring carbons followed by loss of a proton from the same carbon.
There are a number of substituents that can be introduced onto the ring in this way, including nitro, alkyl, acyl, and sulfonyl groups. Each proceeds via a similar mechanism (via a reactive electrophile that is generated in the reaction) either by using a catalyst or by using very reactive reagents.
Alkylation and acylation can be accomplished by treatment of an alkyl halide or acyl halide with a Lewis acid catalyst such as aluminum trichloride ($\mathrm{AlCl}_{3}$). In this case, the reactive electrophile is generated when the alkyl (or acyl) halide interacts with the catalyst to produce an intermediate that is carbocation-like. It is this very reactive species that reacts with the benzene ring to produce the substituted benzene. This reaction is called a Friedel-Crafts alkylation (or acylation). The acylation reaction is usually preferred because it is difficult to stop an alkylation reaction at just one substitution since the product is more reactive than the starting material (see below).
Nitration is accomplished by treating benzene with a mixture of concentrated nitric and sulfuric acids. This mixture generates a nitronium ion ($\mathrm{NO}_{2} {}^{+}$), which is the reactive electrophile.
Sulfonation is accomplished by using fuming sulfuric acid, which actually contains sulfur trioxide ($\mathrm{SO}_{3}$) dissolved in the sulfuric acid. It is the actually the $\mathrm{SO}_{3}$ that is the electrophile in this case.
Some groups cannot be introduced directly onto the ring: for example, groups that we might normally consider as nucleophiles (such as $\mathrm{NH}_{2}$ or $\mathrm{OH}$) have to be introduced indirectly. For example, aniline (aminobenzene) can be produced from nitrobenzene by reduction.
As we will see later, other nucleophilic groups have to be introduced by a different approach. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.10%3A_Reactions_of_Aromatic_Compounds-_Introduction_of_one_group_onto_the_r.txt |
$An image of disubstituted benzenes.$
As we have seen, nitrobenzene can be produced by treating benzene with a mixture of nitric and sulfuric acids. If this reaction is carried out at room temperature, nitrobenzene is produced. If the reaction mixture is heated, eventually another product is produced. While there are three possible products (1,2- or 1,3- or 1,4-dinitrobenzene) only 1,3-dinitrobenzene (or meta-nitrobenzene) is formed. In addition, it is formed much more slowly and it is quite easy to isolate only the mono-nitro product.
To understand why only one further product is formed and why the rate of the second substitution is slower (so that it requires more energy to surmount the activation energy barrier), let’s take a look at nitrobenzene itself. Notice that the positively charged nitrogen is adjacent to the ring. We might expect that this would remove some electron density from the ring and indeed it does.We can see this effect both in the $\mathrm{C-}$13 and ${}^{1}\mathrm{H}$ NMR spectra, where the peaks are shifted to the lower field. Recall that benzene $\mathrm{H}$ signals all appear at $7.3$ while in nitrobenzene we see that not only are there different signals, but that they are all at a lower field than benzene. There are chemically distinguishable hydrogens in nitrobenzene (2 ortho, 2 meta and 1 para).
The lowest field ${}^{2}\mathrm{H}$ doublet at $8.2 \mathrm{~ppm}$ corresponds to the ortho $\mathrm{H}$s, the ${}^{2}\mathrm{H}$ signal at $7.6 \mathrm{~ppm}$ is the two meta $\mathrm{H}$s, and the signal at $7.5$ is the para $\mathrm{H}$. In the $\mathrm{C}-13$ NMR spectrum, all the carbons in benzene appear at $128 \mathrm{~ppm}$. In nitrobenzene, $\mathrm{C}-1$, the carbon bonded to the nitro group is, as we might expect, shifted downfield to $148 \mathrm{~ppm}$ because of its proximity to the positively charged nitrogen.
The evidence shows us that nitrobenzene is less electron-rich than benzene—in fact, it reacts with electrophiles 10,000 times more slowly than benzene—but why does it only produce one product on dinitration? The answer to this lies in the structure of the intermediate sigma complex. If you draw out the resonance forms for reaction at the ortho, meta, and para positions, what becomes clear is that the meta position is the only one that doesnot (cannot) place the negative charge on the original nitro-substituted carbon. Both ortho and para substitution are highly disfavored because these intermediates are of such high energy (because of the adjacent positive charges), and, therefore, only the meta product is formed.
Nitro groups belong to the class of substituents that are deactivating meta directors. Other substitutents that belong in this group are sulfonic acids ($\mathrm{SO}_{3}\mathrm{H}$) and any group with a carbonyl next to the ring (aldehydes, ketones, carboxylic acids, and derivatives). All of these groups are destabilized by having a positive charge on the adjacent carbon.
It makes sense then that there is another group of substituents that are activating ortho, para directors. For example, all alkyl groups fall into this classification because they are electron donating by induction. Toluene (methyl benzene) is the simplest example, and indeed when we nitrate toluene, we see a mixture of ortho and para products because now these are the intermediate sigma complexes that can be directly stabilized by induction from the methyl group.
In this case, the ortho and para substitution products are stabilized and produced preferentially. The ratios of products are approximately 2:1 ortho to para (because there are two possible ortho positions). Indeed toluene can be nitrated further until all the ortho and para positions are substituted to give 2,4,6-trinitrotoluene, otherwise known as TNT.[8]
Another group of ortho, para directors are substituents that are attached to the ring by an $\mathrm{O}$ or $\mathrm{N}$ (that is, phenol) alkoxybenzenes or aniline and its derivatives. While it might seem, at first glance, that these electronegative atoms would be electron-withdrawing by induction, we know that they can also donate electrons through resonance with the ring. We can see both modes in operation if we look at the evidence from the $\mathrm{C}-13$ NMR. $\mathrm{C}-1$ appears at a much lower field ($158 \mathrm{~ppm}$) than benzene ($128 \mathrm{~ppm}$), presumably because it is directly attached to the electron-withdrawing $\mathrm{O}$. But the ortho and para carbons appear at a slightly higher field than benzene because these positions are enriched in electron density by resonance.
In fact, phenol is more reactive than benzene and it is difficult to stop the reaction at the mono-nitration step. Again, we observe a mixture of ortho and para products, because the transition state on the way to the intermediate sigma complex can be stabilized by resonance.
Generally, any benzene derivatives with a lone pair that is available for stabilizing the intermediate will be an ortho, para director including aniline (and any N-alkylated analogs) and anisole (methoxybenzene). This also includes compounds that have a lone pair that must be shared between the ring and a carbonyl group: for example, phenyl esters or amides.
There is one more classification for directing groups: the halogens are anomalous in that they are deactivating ortho, para directors. Halogens are electronegative and withdraw electrons from the ring by induction, thus reducing the reactivity. However, halogens also have lone pairs that can be used to stabilize positive charge by resonance. The resonance effect is not as large as it is for $\mathrm{N}-$ or $\mathrm{O}-$substituted rings, but it can still operate. Therefore halo-substituted benzene rings tend to react more slowly than benzene, but they do produce ortho and para products.
Multiple substituents: While it is relatively easy to predict the position of substitution on a monosubstituted ring (one you know how to do), predicting the outcome when there are two or more often becomes problematic. Sometimes, both substituents will “point” to the same positions. For example, in p-nitrophenol, both the $\mathrm{OH}$ and $\mathrm{NO}_{2}$ direct the substituent to the same position, but in m-nitrophenol, they direct to different positions as shown.
It is, therefore, sometimes possible to predict the position of the third substituent, but, often, it is not. Regardless of this problem, most aromatic substitution reactions have the potential for producing multiple products, either because both ortho and para products are formed or because the product may be more reactive than the reactant. For example, any electrophilic aromatic substitution that adds an activating group to the ring may be difficult to stop at one substitution. Friedel Crafts alkylation is difficult to stop at only one alkylation. In fact, rather than alkylate the ring, it is often preferable to acylate the ring, followed by a reduction of the acyl group. There are a number of reducing reagents that can take the aldehyde or ketone right down to an alkyl group: for example, as shown here, the Clemmensen reduction involves a zinc-mercury amalgam in $\mathrm{HCl}$. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.11%3A_Multiple_Substituents-_Directing_Effects.txt |
All of the reactions on the aromatic ring so far have proceeded by one mechanism: electrophilic aromatic substitution. Up to now, we have not solved the problem of introducing substituents that usually react as nucleophiles onto an aromatic ring. There are a number of ways to accomplish this and we will consider three of these mechanisms.
The first, Nucleophilic Aromatic Substitution ($\mathrm{SNAr}$), is somewhat analogous to an $\mathrm{S}_{\mathrm{N}} 2$ reaction. It will occur if two conditions are met: the first is there must be a leaving group present on the ring, typically a halide; the second is that that the electron density of the ring must be reduced. This is typically accomplished by having electron-withdrawing groups such as nitro groups on the ring. We can observe the effect of adding electron-withdrawing groups by looking at the $\mathrm{NMR}$ spectra of 2-chloro-1,3,5-trinitrobenzene. The only peak in the $\mathrm{H-NMR}$ spectrum is a ${}^{2}\mathrm{H}$ singlet at $9.1 \mathrm{~ppm}$. This is considerably downfield of the $7.3 \mathrm{~ppm}$ of benzene because the of the effect of the electron withdrawing nitro groups. This effect explains why the ring is susceptible to nucleophilic attack.
The reaction proceeds by an initial attack by a nucleophile such as $\mathrm{OH}$, $\mathrm{OR}$, followed by loss of the leaving group which takes the pair of electrons with it.
The intermediate anion can be directly stabilized by the nitro group, but only if that group is in the ortho or para position. If the nitro group is meta to the leaving group the reaction will not occur, because such stabilization is not possible (try to draw resonance forms for it).
A second mechanism involves elimination of $\mathrm{HX}$ from the ring, followed by a rapid addition of $\mathrm{HY}$; it can be considered analogous to elimination and addition reactions of alkenes and alkynes. The first step is the elimination of $\mathrm{HX}$ from benzene, which produces a highly reactive, strained species which is called a benzyne. This intermediate undergoes very rapid reaction to add $\mathrm{H}^{+}$ and $\mathrm{Y}^{–}$ across the double bond.
Evidence for this mechanism, rather than $\mathrm{SNAr}$, comes from isotope labeling studies. If the original site is isotopically labeled (e.g. with $\mathrm{C}-13$, the final product has only 50% of the nucleophile at the labeled site and 50% at the adjacent carbon. If this were a straight nucleophilic substitution, all of the substitution would take place at the labeled carbon.
Diazonium ions:
Another approach that allows access to multiple products involves the reaction of aniline ($\mathrm{PhNH}_{2}$) with the nitronium ion (produced in the reaction mixture) to what is known as a diazonium ion.
The diazonium ion is highly unstable: it must be prepared at temperatures around $0^{\circ}\mathrm{C}$; if the solution is warmed above this temperature, it will decompose by losing a molecule of nitrogen ($\mathrm{N}_{2}$) to produce a carbocation. It is this decomposition that can be captured by a range of nucleophiles. In some ways this reaction is akin to an $\mathrm{S}_{\mathrm{N}} 1$ reaction in which a carbocation is produced, which then undergoes rapid nucleophilic attack.
The intermediate carbocation can be captured in the presence of a nucleophile: for example water, alcohol, halide, or cyanide ions. Note that all these reactions typically require a raised temperature or the presence of a metal ion catalyst. This is to enhance the rate of decomposition of the diazonium ion, so that the resulting carbocation can be captured by the nucleophile. It is important to note that this carbocation is NOT resonance-stabilized (you cannot draw resonance forms to stabilize—try it).
Diazo coupling reactions:
Besides being a way to introduce some nucleophiles into aromatic rings, diazonium salts also undergo what is known as a coupling reaction. The diazonium ion itself is susceptible to nucleophilic attack if it is not decomposed too rapidly—usually by another aromatic system. This reaction has been used to produce a wide range of dyes and indicators.
These azo compounds have a long-conjugated chromophore that typically results in absorption of visible light, making the compounds highly colored (as a result of the light that is reflected and not absorbed). Manipulation of groups on either benzene ring can extend the conjugation even further. About 50% of dyes belong to this family of compounds. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity/8.12%3A_Nucleophilic_Substitutions_on_Aromatic_Systems-_Expanding_the_range_o.txt |
As we have seen carbonyl compounds undergo both acid and base catalyzed reactions involving nucleophilic attack at the carbonyl carbon (or at the beta carbon of conjugated carbonyls). This reaction, in its many guises, can produce an impressive range of products from the formation of an ester from an acid to the production of alcohols from carbonyls. While these reactions may seem superficially different, if you understand the underlying mechanism involved, it is possible to make plausible predictions for the outcome for literally thousands of reactions. By this time, you should have come to understand such processes. Now, it is time to reconsider carbonyl groups in light of the fact that there is a completely different set of reactions that involve the reactivity of the alpha carbon of the carbonyl groups. Carbonyl compounds typically exist in two tautomeric forms: the keto and enol forms. The keto form is usually the major tautomer and there is always some enol present as well.
The structure of the enol form can provide clues about its different reactivity, which is distinct from that of the keto form. The enol form consists of an alcohol directly attached to a $\mathrm{C}$ that is involved in a double bond. As we know, alkenes are electron-rich and tend to undergo electrophilic attack; the presence of an attached $\mathrm{-OH}$ group makes such an electrophilic attack even more likely. Just like an $\mathrm{-OH}$ group on an aromatic ring, the $\mathrm{OH}$ can donate electrons through resonance with the $\mathrm{-C=C-}$ and make the enol more reactive. Carbonyl groups can react, through the small percentage of the enol form present, to undergo electrophilic attack at the alpha carbon.
We have already seen that aldehydes and ketones exist as keto-enol tautomers, but, in fact, carboxylic acids, esters, and other acid derivatives also have the potential to exist in the corresponding enol form.
Another implication of the alcohol-nature of an enol is that we expect it to be acidic—and indeed it is. The conjugate base of the enol is called the enolate ion and it is resonance-stabilized so that the negative charge is delocalized on both the oxygen and on the alpha carbon.
The $\mathrm{pK}_{\mathrm{a}}$ of acetone is $19$—somewhat higher than a typical alcohol ($\mathrm{pK}_{\mathrm{a}} \sim 15$). In the enolate form, the majority of the charge sits on the more electronegative oxygen, but a significant proportion of the negative charge is associated with the alpha carbon: the enolate ion is a stabilized carbanion. The enolate anion is often written in its carbanion form because this is the form that produces most of the interesting chemistry. Treatment of a carbonyl compound with a base, such as an alkoxide, results in the reversible formation of a small amount of the enolate ion (although the equilibrium lies on the side of the unprotonated form). Similarly, many carbonyl compounds can be deprotonated to give the corresponding enolate anion. For example, esters can be treated with a base to give the corresponding enolate anion.
Ethyl acetate has a $\mathrm{pK}_{\mathrm{a}}$ of around $25$ (less acidic than acetone: $\mathrm{pK}_{\mathrm{a }} 19$), but still well within reach of many of the strong bases. For example, sodium amide ($\mathrm{NaNH}_{2}$), the conjugate base of ammonia ($\mathrm{pK}_{\mathrm{a }} 33$), is strong enough to deprotonate the ester. In fact, we typically use what is known as a hindered base, such as lithium di-isopropylamide ($\mathrm{LDA}$), which is similar to sodium amide but the nitrogen has two bulky isopropyl groups attached to it.[1]
Since $\mathrm{LDA}$ is such a strong base, treatment of most carbonyl compounds with $\mathrm{LDA}$ produces essentially 100% of the corresponding enolate anion. However, there are exceptions. Any carbonyl compound that has a more acidic proton than the $\mathrm{H}$ associated with the alpha carbon will not undergo this reaction. For example, treatment of carboxylic acids with $\mathrm{LDA}$ will merely result in the loss of the acidic proton from the carboxylic acid OH group.
Most carbonyl compounds have $\mathrm{pK}_{\mathrm{a}}$‘s between $19$ and $25$. Compounds that have carbonyl groups that are beta to each other (that is, separated by an intervening carbon), have significantly lower $\mathrm{pK}_{\mathrm{a}}$‘s (around $9$), because the resulting anion can be stabilized on both carbonyl oxygens.
They can be easily deprotonated by bases such as sodium ethoxide or sodium hydroxide.
09: A return to the carbonyl
The keto and enol forms of carbonyl compounds can undergo completely different reactions. The carbonyl (keto) form undergoes nucleophilic attack at the carbonyl carbon and the enol/enolate form undergoes electrophilic attack, usually at the alpha carbon (although the $\mathrm{O}$ is also reactive). For example, aldehydes and ketones can be halogenated at the alpha carbon just by treatment with a solution of the halogen, either with acid or base catalysis. The first step is enolization, which produces the very electron-rich alkene that attacks the bromine (just like the first step of addition to a normal alkene). This intermediate then loses a proton to give to the halogenated compound and $\mathrm{HBr}$.
The reaction can also be done in a base via the enolate, but in this case the reaction is difficult to stop after one halogen has added and, typically, all alpha positions will end up brominated. Such a reaction is analogous to the first step of addition of halogens to an alkene, but the second step involves the regeneration of the carbonyl. Just as reversible nucleophilic addition to the carbonyl typically produces the $\mathrm{sp}^{2}$ hybridized product, these enol/enolate forms also end up as substitutions rather than additions.
A reaction of an alpha carbon that has no analogy in alkene chemistry involves their acting as a nucleophile in an $\mathrm{S}_{\mathrm{N}} 2$ reaction. The reaction occurs via the enolate anion, which then attacks any appropriate alkyl halide via an $\mathrm{S}_{\mathrm{N}} 2$ reaction.
If the ketone undergoing such a reaction has the possibility of forming two different enolates, and therefore producing two different alpha alkylation products, the enolate that has the most substituted double bond is the most stable and is thermodynamically favored. Typically, the enolate formed from the least-hindered carbon is formed fastest (it has the lowest activation energy). It is therefore possible to control the product of such a reaction by carefully controlling the reaction conditions. At very low temperatures, the kinetic product is formed, while at higher temperatures thermodynamic product is formed. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/09%3A_A_return_to_the_carbonyl/9.01%3A_Reactions_of_Enols_and_Enolates.txt |
As we have previously discussed, the carbonyl group has a kind of split personality. The carbonyl group is susceptible to nucleophilic attack at the carbonyl carbon and carbonyl compounds can be nucleophiles at the alpha carbon. Therefore, we should not be too surprised to learn that carbonyl compounds can (and do) react with themselves. For example, when acetaldehyde, the simplest carbonyl compound that is capable of forming an enol, is treated with a reversible base such as $\mathrm{NaOH}$, it will form a small amount of the enolate anion, which can then react with the carbonyl of another molecule as shown. This reaction is known as the aldol reaction. As the enolate is used up in the reaction, more is formed and more aldol reaction occurs. The product is a beta-hydroxy aldehyde.
Typically, aldehydes undergo this reaction readily and the aldol product is formed in good yield. The reaction is reversible, however, and ketones often do not give good yields of the aldol product. The reverse reaction is called a “retro aldol” and occurs via deprotonation of the alcohol and loss of the enolate anion as shown.
In fact, the aldol reaction rarely ends at the simple addition of one carbonyl and its corresponding enolate. Usually, the reaction is heated and, under these conditions, the alcohol that is formed undergoes an elimination to form the alpha, beta unsaturated carbonyl. When this happens, the reaction is called an aldol condensation (the term “condensation” is usually used for reactions in which water is lost).
Aldol condensation is often used to synthesize rings via an intramolecular aldol condensation. In these cases, although there may be the potential to form different ring sizes, typically only the most stable rings are produced: that is, five- or six-membered rings.
9.03: The Claisen Condensation
As we have noted, all carbonyl compounds are capable of forming enols and enolate anions, and just as aldehydes and ketones undergo condensation reactions with each other, so then to esters. The ester version of the aldol is called a Claisen condensation, but the essential details are very similar in terms of the initial mechanistic sequence.
The enolate anion of the ester attacks the carbonyl of another molecule and the resulting tetrahedral intermediate collapses back to the carbonyl by regenerating the ethoxide anion that was used to initiate enolate formation. The difference between the Claisen and the aldol reactions is that the Claisen product is a $\beta$-ketoester, which can be a useful species in its own right. Claisen condensations can also form rings via intramolecular condensations (which are known as Dieckmann cyclizations).
Both Aldol and Claisen condensations can be carried out between two different carbonyl compounds: however, if both are capable of forming enolates, there is the possibility of forming four different products. Consider two carbonyl compounds A and B, if enol A reacts with carbonyl B, we get product AB, but if enol B reacts with carbonyl A we get product BA (which would have a different structure). A can also condense with another A to form product AA, and similarly we could get BB. Therefore, it is important to control the reaction conditions carefully: for example, by using an irreversible base such as LDA to form the enolate first. This precludes the possibility of the enolate reacting with itself. Then, the other component can be added slowly to the reaction mixture and the condensation can be carried out.
9.04: -Ketoacids Decarboxylate
All of these Claisen condensation reactions produce a difunctional compound in which a carbonyl group is located on the beta position of an ester. There is a useful reaction that can be carried out if the ester is hydrolyzed to the corresponding acid. If the resulting $\beta$-keto acid is heated, it decarboxylates (loses $\mathrm{CO}_{2}$) in a pericyclic reaction that involves the cyclic rearrangement of six electrons as shown below.
The $\beta$-ketoester here is known as acetoacetic ester. The $\mathrm{CH}_{2}$ group between the two carbonyls is easily deprotonated, and the resulting anion can do a nucleophilic attack on any susceptible substrate: for example, an alkyl halide. A subsequent hydrolysis and decarboxylation results in a compound that has three more carbons in it than the original alkyl halide, as shown below.
The “acetoacetic ester” synthesis is a powerful way of adding a 3-carbon unit. A similar reaction involves malonic ester (below), which can be used to add a 2-carbon unit.
In the decarboxylation step, only one of the carboxylic acids decarboxylates and the alkyl group is extended by two carbon atoms. Interestingly, fatty acids (long chain carboxylic acids) are synthesized by a mechanism that is analogous to this malonic ester synthesis. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/09%3A_A_return_to_the_carbonyl/9.02%3A_The_Aldol_Reaction.txt |
As we move forward, we will discuss some biosynthetic pathways. It is not our intention that you reproduce these pathways, complete with enzymes and co-factors, but that you understand the underlying organic chemistry behind them. As you will see, most biochemical reactions are quite simple: it is their surroundings (the other parts of the molecules, enzymes, and co-factors) that make it possible for these reactions to occur (generally around room temperature) in a crowded aqueous environment; these pathways appear complex.
Fatty acids are built two carbons at a time by the following mechanism: the two-carbon unit is provided by a malonyl unit that is formed from malonyl-CoA, (a thioester) which is attached to a carrier protein for recognition purposes. It is formed from acetyl CoA (co-enzyme A)—which is a product of the breakdown of glucose (glycolysis) by reaction with bicarbonate. The result of adding two carbon units is that most biological fatty acids have an even number of carbons. They are synthesized by a sequence of reactions that is highly analogous to the malonic ester synthesis. The syntheses of fatty acids is one of the mechanisms that the body uses to “store” energy and to generate various membranes. Fatty acids often occur as esters of glycerol and are therefore called triglycerides.
Acetyl CoA is transferred to the acyl carrier protein (ACP) and is then attacked by the malonyl anion with loss of the S-ACP group (this is analogous to the $\mathrm{S}_{\mathrm{N}} 2$ reaction on an alkyl halide). Decarboxylation produces a new beta-keto thioester, extended by two carbons.
The next step is reduction of the carbonyl (using NADPH—which is analogous to NADH—and delivers hydride ion), elimination, and reduction to the fully saturated chain. The system then cycles around to add two more carbons from malonyl ACP.
All of these reactions are very similar to those we have learned throughout the course. They seem more complex because of the biological “bits” that control the direction of reaction and activation of the functional groups, but once you understand organic chemistry, biochemistry makes much more sense!
9.06: Michael Reactions
Recall that aldol condensations result in $\alpha ,\beta$-unsaturated carbonyl compounds, a functionality that we have already discussed at some length.
These conjugated carbonyl groups can undergo nucleophilic attack at either the carbonyl carbon or at the $\beta$ carbon, depending on the nature of the nucleophile. For example: highly reactive (or “hard”) nucleophiles like Grignards or alkyl lithiums tend to react at the carbonyl carbon, while less reactive (soft) nucleophiles like dialkyl cuprates or reversible nucleophiles like amines or alcohols tend react at the $\beta$ carbon.
Anions formed from $\beta$-diketones are relatively unreactive (they are stabilized) and, therefore, we might predict that they will attack the conjugated carbonyl at the $\beta$ position – and we would be correct! This reaction is called the Michael reaction.
In fact, we can condense the same $\beta$-diketone with a different conjugated ketone (not formed from an aldol condensation) to produce an intermediate that can then undergo an intramolecular aldol condensation as shown below. This two-step procedure is called the Robinson Annulation.
Unfortunately, this reaction only works well with beta-diketones; a simple ketone does not attack the conjugated system. However, there is a relatively simple solution to this, which is to modify the ketone to form an enamine, which will then react as shown below. This variation is known as the Stork enamine synthesis. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/09%3A_A_return_to_the_carbonyl/9.05%3A_Biosynthesis_of_Fatty_Acids-.txt |
Glycolysis is the name we give to the group of reactions that result in the splitting of a $\mathrm{C-} 6$ glucose molecule into two $\mathrm{C-} 3$ units, which is accompanied by the overall production of $\mathrm{ATP}$, the molecule that can be used to provide energy to drive unfavorable chemical reactions such as building up biopolymers like peptides, nucleic acid polymers ($\mathrm{DNA}$ and $\mathrm{RNA}$), and production of fats. This process is usually depicted schematically, particularly in biology texts, but every step of the process is a relatively simple organic reaction that can be understood in terms of the principles that we have learned over the past year. The overall schematic for glycolysis is given below[2], but our intent here is not that you memorize each step so that it can be regurgitated; it’s to allow you to understand how and why these reactions occur the way that they do (or at least the way they do in biological systems).
We will be treating these reactions from an organic chemistry perspective, but it is important to note that in the body all of these reactions are mediated by enzymes and co-factors that lower the activation energy for each reaction. The first part of the glycolysis pathway involves the conversion of the sugar glucose, to a different sugar, fructose. Therefore we will begin by looking briefly at the structure and properties of sugars.
Glucose:
While glucose looks complex, we have already investigated the functional groups present and all the reactions that are important here. Glucose belongs to the family of compounds called sugars part of a larger group, known as carbohydrates, denoted by the suffix -ose. Since it has six carbons, glucose is known as hexose (similarly, a five-carbon sugar would be known as a pentose). Glucose can exist in several forms; both open and closed chain. We consider the open-chained form first. The easiest way to represent sugars is by using a Fischer projection (in fact these representations were invented for just this purpose). Remember that in a Fischer projection, the horizontal bonds are pointing out of the plane and are all eclipsed. The naturally occurring form of glucose is D-Glucose $\rightarrow$.
Note that there are four chiral centers in glucose, and therefore there are $16$ ($2^{4}$) possible stereoisomers, many of which do occur naturally. The D designation has to do with the stereochemistry at position 5 and does not refer to the direction of the rotation of plane-polarized light. (While it is possible to designate R or S for each chiral center[3], it is not possible to designate R or S for the molecule as a whole). Note that glucose also has an aldehyde group (at position 1) and, therefore, also belongs to the class of sugars called aldoses.
In its open chain form, glucose has an aldehyde group and five hydroxyl groups and, as you might expect, there is great potential here for reactions, both inter- and intra-molecular. In a solution, glucose commonly exists in the hemiacetal form, in which the $\mathrm{OH}$ group on $\mathrm{C-} 5$ has attacked the carbonyl group to form a six-membered ring which is referred to as the pyranose form (pyran is a six-membered heterocyclic ring with an oxygen atom in it). The pyranose form is usually drawn in the chair form as shown below.
Hemiacetal formation produces two possible configurations, but rather than calling them R and S, we label them alpha and beta. In alpha form, the $\mathrm{OH}$ on carbon 1 is on the opposite side of the ring from the $\mathrm{CH}_{2}\mathrm{OH}$ ($\mathrm{C-} 6$ of the original chain), the beta form has the $\mathrm{OH}$ group on the same side as the $\mathrm{CH}_{2}\mathrm{OH}$. These two forms are stereoisomers because they have the opposite configuration at $\mathrm{C-} 1$, but unlike typical stereoisomers, they can be interconverted by ring-opening of the hemiacetal and reclosure of the ring. They are called anomeric forms and $\mathrm{C-} 1$ is referred to as the anomeric carbon. Such carbons can be identified by the fact that they have two oxygens attached to them. In an aqueous equilibrium solution of glucose, less than 1% is present in the open chain form, but since there is always an equilibrium concentration present, the alpha and beta forms can and will interconvert via this open-chain form. As might be expected, the beta form is more stable (because the $\mathrm{OH}$ is equatorial) and at equilibrium there is about 64% beta. For sugars such as glucose, there is also the possibility that a five-membered ring can form by the reaction of the alcohol at $\mathrm{C-} 4$ with the carbonyl. Again, two forms (alpha and beta) are possible, which can be interconverted via ring-opening. This five-membered ring form is called the furanose form (furan is a five-membered ring with one oxygen).
Fructose:
Fructose is another six-carbon sugar. It differs from glucose in that it that has a ketone rather than an aldehyde at $\mathrm{C-} 2$; for this reason, it is called a ketose (rather than an aldose). Fructose usually exists in a five-membered hemiacetal ring formed by reaction of the $\mathrm{OH}$ at $\mathrm{C-} 5$ with the ketone carbonyl at $\mathrm{C-} 2$. The resulting five-membered ring is called the furanose form (furan is a five-membered ring with one oxygen) and typically furanose rings are depicted using yet another structural representation: the Haworth projection, shown here. In this structure, the ring is drawn as a plane (although it isn’t, of course), and the substituents are either above or below the plane of the ring. In the same way as glucose, fructose can exist in either the alpha or beta forms. | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/09%3A_A_return_to_the_carbonyl/9.07%3A_Glycolysis.txt |
The first steps of the biological cycle known as glycolysis involve the conversion of glucose to fructose. As we will see, this sets up the reverse aldol reaction that results in the splitting of glucose into two 3-carbon fragments—but we will get to that shortly. The first step in glycolysis is the phosphorylation of the $\mathrm{OH}$ group at $\mathrm{C-} 6$. This reaction is analogous to the formation of a carboxylate ester, the only difference being that the attack occurs at a phosphorus rather than a carbon. The source of phosphate is adenosine triphosphate ($\mathrm{ATP}$) and, in this case, it is the terminal phosphate that is transferred to the glucose and the leaving group is $\mathrm{ADP}$ (adenosine diphosphate). The reaction is mediated by an enzyme (a kinase): this reaction is highly thermodynamically favorable ($\Delta \mathrm{G}$ is negative).
The consequences of glucose phosphorylation reaction are twofold: First, it serves to keep the concentration of glucose in the cell low, so that glucose can continuously diffuse into the cell (through membrane channel proteins). Glucose-6-phosphate is highly charged (the oxygens on the phosphate group are almost entirely ionized at physiological $\mathrm{pH}$) and cannot diffuse back out of the cell (through the channel). The second consequence is that the hydroxyl group on $\mathrm{C-} 6$ has been activated. The phosphate is an excellent leaving group.
The next step is the transformation of glucose into fructose. In organic chemistry terms, this is a simple tautomerization as fructose is a structural isomer of glucose. The enzyme that regulates this conversion is known as an isomerase.
The glucose aldehyde undergoes enolization as shown, followed by another tautomerization to produce the fructose ketone. All of these reactions are entirely analogous to the keto-enol interconversions that we have seen in simpler systems.
The next step is another phosphorylation reaction at $\mathrm{C-} 1$ to produce fructose-1,6-diphosphate; it involves the same reaction mechanism that produced phosphorylation at $\mathrm{C-} 6$ using $\mathrm{ATP}$ as the source of phosphate. You might be asking, how is it that these reactions occur in this particular sequence and why don’t the other oxygens undergo phosphorylation? The answer to this lies in the fact that these reactions are mediated by enzymes that guide the substrates into the correct (pre-established, evolutionarily) orientations. Reactions in solution depend entirely on random collisions with enough energy to surmount the activation energy barrier and in the correct orientation. In biological systems, the substrate first collides with, and its orientation is constrained through interactions with, the enzyme; limiting subsequent aspects (steps) of the reaction. While there is potential for other hydroxyls to be phosphorylated (and it would certainly be difficult to control the site of attack if the molecules were free in solution, the reactant-enzyme complex favors certain ones dramatically over others. Remember the enzyme itself is result of evolutionary mechanisms. Given the ubiquity of this process, it was likely established early, and present in the last common ancestor of life.
The Reverse Aldol:
The system is now set up for the cleavage of the six-carbon sugar into two three-carbon species via a reverse aldol reaction as shown.
The result is the production of two molecules of glyceraldehyde-3-phosphate from one molecule of glucose. Up to this stage, glycolysis has involved the use of two $\mathrm{ATP}$ molecules associated with the two phosphorylation reactions.
In the next stage of glycolysis, another phosphate is added to each glyceraldehyde-3-phosphate molecule; this phosphate is not derived from $\mathrm{ATP}$, but from inorganic phosphate. The mechanism involves an addition of a phosphate unit to the aldehyde carbonyl, followed by oxidation as the tetrahedral intermediate collapses, with loss of the hydride ion that adds to $\mathrm{NAD}^{+}$, to form $\mathrm{NADH}$.
The resultant species now contains two phosphate groups, but they are different chemically: the phosphate attached to the newly oxidized carbon is much more reactive than the other. Attack by a nucleophile at the phosphorus preferentially expels a carboxylate anion from the intermediate, rather than an alkoxide. In the next step of the reaction, the highly reactive phosphate is transferred to $\mathrm{ADP}$ through an attack by the $\mathrm{O}^{-}$ on the terminal phosphate of $\mathrm{ADP}$ onto the $\mathrm{P}$ of the 1,3-diphosophoglycerate, with subsequent loss of 3-phosphoglycerate (because the carboxylate is a good leaving group).
This reaction produces two $\mathrm{ATP}$ molecules (and 2 reduced $\mathrm{NADH}$ molecules) from one original glucose (because there are now two three-carbon units), so, at this stage, the net production of $\mathrm{ATP } = 0$. The transfer of the other phosphate unit from $\mathrm{C-} 3$ to another $\mathrm{ADP}$ does not occur because the leaving group (an alkoxide) is not thermodynamically favorable. Instead, the phosphate group is transferred from $\mathrm{C-} 3$ to $\mathrm{C-} 2$ by an intermolecular nucleophilic attack that forms a five-membered ring intermediate and subsequent elimination of water.
The product of this elimination reaction is called phosphoenol pyruvate (PEP). In essence, it is an enol that has been trapped by esterification with the phosphate. The enol is an excellent leaving group (since it is really a carbonyl group) and, therefore, PEP will also undergo attack by $\mathrm{ADP}$ to produce $\mathrm{ATP}$ and pyruvate, resulting in a in a net $+ 2 \mathrm{ ATP}$ molecules for the overall glucose to pyruvate reaction (plus the two $\mathrm{NADH}$ molecules).
Once again, our intent here is not to have you memorize this long sequence of reactions, but rather to recognize that even in systems that appear highly complex, when each reaction is viewed at the molecular level, it is recognizable as the same reactions that we have been studying. In fact, the mechanisms of a large majority of reactions in biological systems can be understood in relatively simple terms. Many of the reactions are the same: attack at carbonyl (or phosphate) groups, aldol and retro aldol condensations, and keto-enol tautomerizations.
9.09: In-Text References
1. Recall that we have used hindered bases before. In that case, it was to prevent nucleophilic attack when we wanted to bring about an $\mathrm{E} 2$ elimination reaction and tertiary butoxide ($\mathrm{KOC}\left(\mathrm{CH}_{3}\right)_{3}$), as our hindered base.
2. Diagram from Wikipedia https://en.Wikipedia.org/wiki/File:G...tated.svg#file licensed under a creative commons license.
3. In fact the name of the most common form of glucose is $(2 \mathrm{R},3 \mathrm{R},4 \mathrm{S},5 \mathrm{S},6 \mathrm{R})-6(\text { hydroxymethyl }) \text { tetrahydro-} 2 \mathrm{H} \text{-pyran-2,3,4,5-tetraol}$. There is no need to remember this! | textbooks/chem/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/09%3A_A_return_to_the_carbonyl/9.08%3A_Glycolysis-_From_Glucose_to_Fructose.txt |
Chapter Objectives
This chapter provides a review of material covered in a standard freshman general-chemistry course through a discussion of the following topics:
• the differences between organic and inorganic chemistry.
• the shapes and significance of atomic orbitals.
• electron configurations.
• ionic and covalent bonding.
• molecular orbital theory.
• hybridization.
• the structure and geometry of the compounds methane, ethane, ethylene and acetylene.
• 1.0: Introduction to Organic Chemistry
Organic compounds contain carbon atoms bonded hydrogen and other carbon atoms. Organic chemistry studies the properties and reactions of organic compounds.
• 1.1: Atomic Structure - The Nucleus
Atoms are comprised of protons, neutrons and electrons. Protons and neutrons are found in the nucleus of the atom, while electrons are found in the electron cloud around the nucleus. The relative electrical charge of a proton is +1, a neutron has no charge, and an electron’s relative charge is -1. The number of protons in an atom’s nucleus is called the atomic number, Z. The mass number, A, is the sum of the number of protons and the number of neutrons in a nucleus.
• 1.2: Atomic Structure - Orbitals
An atomic orbital is the probability description of where an electron can be found. The four basic types of orbitals are designated as s, p, d, and f.
• 1.3: Atomic Structure - Electron Configurations
The order in which electrons are placed in atomic orbitals is called the electron configuration and is governed by the aufbau principle. Electrons in the outermost shell of an atom are called valence electrons. The number of valence electrons in any atom is related to its position in the periodic table. Elements in the same periodic group have the same number of valence electrons.
• 1.4: Development of Chemical Bonding Theory
Lewis Dot Symbols are a way of indicating the number of valence electrons in an atom. They are useful for predicting the number and types of covalent bonds within organic molecules. The molecular shape of molecules is predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. The shapes of common organic molecules are based on tetrahedral, trigonal planar or linear arrangements of electron groups.
• 1.5: Describing Chemical Bonds - Valence Bond Theory
Covalent bonds form as valence electrons are shared between two atoms. Lewis Structures and structural formulas are common ways of showing the covalent bonding in organic molecules. Formal charge describes the changes in the number of valence electrons as an atom becomes bonded into a molecule. If the atom has a net loss of valence electrons it will have a positive formal charge. If the atom has a net gain of valence electrons it will have a negative formal charge.
• 1.6: sp³ Hybrid Orbitals and the Structure of Methane
The four identical C-H single bonds in methane form as the result of sigma bond overlap between the sp3 hybrid orbitals of carbon and the s orbital of each hydrogen.
• 1.7: sp³ Hybrid Orbitals and the Structure of Ethane
The C-C bond in ethane forms as the result of sigma bond overlap between a sp³ hybrid orbital on each carbon. and the s orbital of each hydrogen. The six identical C-H single bonds in form as the result of sigma bond overlap between the sp³ hybrid orbitals of carbon and the s orbital of each hydrogen.
• 1.8: sp² Hybrid Orbitals and the Structure of Ethylene
The C=C bond in ethylene forms as the result of both a sigma bond overlap between a sp2 hybrid orbital on each carbon and a pi bond overlap of a p orbital on each carbon
• 1.9: sp Hybrid Orbitals and the Structure of Acetylene
The carbon-carbon triple bond in acetylene forms as the result of one sigma bond overlap between a sp hybrid orbital on each carbon and two pi bond overlaps of p orbitals on each carbon.
• 1.10: Hybridization of Nitrogen, Oxygen, Phosphorus and Sulfur
The atomic orbitals of nitrogen, oxygen, phosphorus and sulfur can hybridize in the same way as those of carbon.
• 1.11: Describing Chemical Bonds - Molecular Orbital Theory
Molecular Orbital theory (MO) is a more advanced bonding model than Valence Bond Theory, in which two atomic orbitals overlap to form two molecular orbitals – a bonding MO and an anti-bonding MO.
• 1.12: Drawing Chemical Structures
Kekulé Formulas or structural formulas display the atoms of the molecule in the order they are bonded. Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space. Skeleton formulas or Shorthand formulas or line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. Isomers have the same molecular formula, but different structural formulas
01: Structure and Bonding
Objectives
After completing this section, you should be able to
1. Define organic chemistry as the study of carbon-containing compounds.
2. Explain why the results of the experiments carried out by Chevreul and Wöhler contributed to the demise of the “vital force” theory.
Key Terms
Make certain that you can define, and use in context, the key term below.
• organic chemistry
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds.
Jöns Jacob Berzelius, a physician by trade, first coined the term “organic chemistry” in 1806 for the study of compounds derived from biological sources. Up through the early 19th century, naturalists and scientists observed critical differences between compounds that were derived from living things and those that were not.
In 1828, Friedrich Wöhler (widely regarded as a pioneer in organic chemistry) successfully completed an organic synthesis by heating ammonium cyanate and synthesizing of the biological compound urea (a component of urine in many animals) in what is now called “the Wöhler synthesis.” Until this discovery it was widely believed by chemists that organic substances could only be formed under the influence of the “vital force” in the bodies of animals and plants. Wöhler’s synthesis dramatically proved that view to be false.
Urea synthesis was a critical discovery for biochemists because it showed that a compound known to be produced in nature only by biological organisms could be produced in a laboratory under controlled conditions from inanimate matter. This “in vitro” synthesis of organic matter disproved the common theory (vitalism) about the vis vitalis, a transcendent “life force” needed for producing organic compounds.
The ability to manipulate organic compounds includes fermentation to create wine and the making of soap, both of which have been a part of society so long their discovery has been lost in antiquity. Evidence has shown the Babylonians, as early as 2800 BC, were creating soap by mixing animal fat with wood ashes. It wasn't until the 19th century that the chemical nature of the creation of soap was discovered by Eugène Chevreul. In a reaction now call saponification, fats are heated in the presence of a strong base (KOH or NaOH) to produce fatty acid salts and glycerol. The fatty acid salts are the soap which improve water's ability to dissolve grease.
Although originally defined as the chemistry of biological molecules, organic chemistry has since been redefined to refer specifically to carbon compounds — even those with non-biological origin. Some carbon molecules are not considered organic, with carbon dioxide being the most well known and most common inorganic carbon compound, but such molecules are the exception and not the rule. Organic chemistry focuses on carbon compounds and following movement of the electrons in carbon chains and rings, and also how electrons are shared with other carbon atoms and heteroatoms. Organic chemistry is primarily concerned with the properties of covalent bonds and non-metallic elements, though ions and metals do play critical roles in some reactions.
Why is carbon so special? The answer to this question involves carbon's special ability to bond with itself, which will be discussed in this chapter. Carbon is unique in its ability to form a wide variety of compounds from simple to complex. There are literally millions of organic compounds known to science from methane, which contains one carbon atom, to DNA which contains millions of carbons. More importantly, organic chemistry gives us the ability to make and alter the structure of organic compounds, which is the main topic in this book. The applications of organic chemistry are myriad, and include all sorts of plastics, dyes, flavorings, scents, detergents, explosives, fuels and many, many other products. Read the ingredient list for almost any kind of food that you eat — or even your shampoo bottle — and you will see the handiwork of organic chemists listed there.
The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings. The field of organic chemistry is probably the most active and important field of chemistry at the moment, due to its extreme applicability to both biochemistry (especially in the pharmaceutical industry) and petrochemistry (especially in the energy industry). Organic chemistry has a relatively recent history, but it will have an enormously important future, affecting the lives of everyone around the world for many, many years to come | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.00%3A_Introduction_to_Organic_Chemistry.txt |
Objective
After completing this section, you should be able to describe the basic structure of the atom.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• atomic number
• atomic weight
• electron
• mass number
• neutron
• proton
The Nuclear Atom
The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that:
• Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge whose value identifies each element and is known as the atomic number of the element.
• Almost all of the volume of an atom consists of empty space in which electrons, the fundamental carriers of negative electric charge, reside. The extremely small mass of the electron (1/1840th the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud. The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm.
The nucleus is itself composed of two kinds of particles. Protons are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the nuclear charge) is balanced by the same number of electrons outside the nucleus. The other nuclear particle is the neutron. As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom.
Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an ion.
Atomic Number (Z)
What single parameter uniquely characterizes the atom of a given element? It is not the atom's relative mass, as we will see in the section on isotopes below. It is, rather, the number of protons in the nucleus, which we call the atomic number and denote by the symbol Z. Each proton carries an electric charge of +1, so the atomic number also specifies the electric charge of the nucleus. In the neutral atom, the Z protons within the nucleus are balanced by Z electrons outside it.
Atomic numbers were first worked out in 1913 by Henry Moseley, a young member of Rutherford's research group in Manchester.
Moseley searched for a measurable property of each element that increases linearly with atomic number. He found this in a class of X-rays emitted by an element when it is bombarded with electrons. The frequencies of these X-rays are unique to each element, and they increase uniformly in successive elements. Moseley found that the square roots of these frequencies give a straight line when plotted against Z; this enabled him to sort the elements in order of increasing atomic number.
You can think of the atomic number as a kind of serial number of an element, commencing at 1 for hydrogen and increasing by one for each successive element. The chemical name of the element and its symbol are uniquely tied to the atomic number; thus the symbol "Sr" stands for strontium, whose atoms all have Z = 38.
Mass Number (A)
The mass number equals the sum of the numbers of protons and the number of neutrons in the nucleus. It is sometimes represented by the symbol $A$, so
$A = Z + N \nonumber$A=Z+N
in which $Z$ is the atomic number and $N$ is the neutron number.
Elements
To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table $1$, which illustrates three important points:
1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively.
2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
Table $1$: Properties of Subatomic Particles*
Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to indicate their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table $2$.
Table $2$: Element Symbols Based on Names No Longer in Use
Element Symbol Derivation Meaning
antimony Sb stibium Latin for “mark”
copper Cu cuprum from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold Au aurum Latin for “gold”
iron Fe ferrum Latin for “iron”
lead Pb plumbum Latin for “heavy”
mercury Hg hydrargyrum Latin for “liquid silver”
potassium K kalium from the Arabic al-qili, “alkali”
silver Ag argentum Latin for “silver”
sodium Na natrium Latin for “sodium”
tin Sn stannum Latin for “tin”
tungsten W wolfram German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin
Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.
Carbon Isotopes
The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $^A_Z X$, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $_6^{12} C$. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, $_6^{12} C$ is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.
In addition to $^{12}C$, a typical sample of carbon contains 1.11% $_6^{13} C$ (13C), with 7 neutrons and 6 protons, and a trace of $_6^{14} C$ (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archaeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table $3$.
Table $3$: Properties of Selected Isotopes
Element Symbol Atomic Mass (amu) Isotope Mass Number Isotope Masses (amu) Percent Abundances (%)
hydrogen H 1.0079 1 1.007825 99.9855
2 2.014102 0.0115
boron B 10.81 10 10.012937 19.91
11 11.009305 80.09
carbon C 12.011 12 12 (defined) 99.89
13 13.003355 1.11
oxygen O 15.9994 16 15.994915 99.757
17 16.999132 0.0378
18 17.999161 0.205
iron Fe 55.845 54 53.939611 5.82
56 55.934938 91.66
57 56.935394 2.19
58 57.933276 0.33
uranium U 238.03 234 234.040952 0.0054
235 235.043930 0.7204
238 238.050788 99.274
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
Example $1$
An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.
Given: number of protons and neutrons
Asked for: element and atomic symbol
Strategy:
1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.
Solution:
A The element with 82 protons (atomic number of 82) is lead: Pb.
B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are $^{206}_{82}Pb$, $^{207}_{82}Pb$, and $^{208}_{82}Pb$, which are usually abbreviated as $^{206}Pb$, $^{207}Pb$, and $^{208}Pb$.
Exercise $1$
Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.
Answer
$\ce{^{79}_{35}Br}$ and $\ce{^{81}_{35}Br}$ or, more commonly, $\ce{^{79}Br}$ and $\ce{^{81}Br}$.
Summary
The atom consists of discrete particles that govern its chemical and physical behavior. Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.01%3A_Atomic_Structure_-_The_Nucleus.txt |
Objectives
After completing this section, you should be able to
1. describe the physical significance of an orbital.
2. list the atomic orbitals from 1s to 3d in order of increasing energy.
3. sketch the shapes of s and p orbitals.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• nodal plane
• node
• orbital
• quantum mechanics
• wave function
For a refresher on quantum numbers view this text
Atomic Orbitals
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
One way of representing electron probability distributions was illustrated in Figure 6.5.2 for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at r = 0 (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx − 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (part (a) in Figure \(1\)), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (part (b) in Figure \(1\)), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure \(1\). In contrast, the surface area of each spherical shell is equal to 4πr2, which increases very rapidly with increasing r (part (c) in Figure \(1\)). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (part (d) in Figure \(1\)). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (part (d) in Figure \(1\)).
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the quantum mechanical Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure \(2\) compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in Figure \(2\)). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability.
s Orbitals
Three things happen to s orbitals as n increases (Figure \(2\)):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in Figure \(2\). Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure \(3\). Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a 2pz orbital. As shown in Figure \(4\), the other two 2p orbitals have identical shapes, but they lie along the x axis (2px) and y axis (2py), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative.
Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure \(4\).
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.02%3A_Atomic_Structure_-_Orbitals.txt |
Objective
After completing this section, you should be able to write the ground-state electron configuration for each of the elements up to and including atomic number 36.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• ground-state electronic configuration
• Hund's rule
• Pauli exclusion principle
• aufbau principle
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.
Electron Configurations
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.
Before assigning the electrons of an atom into orbitals, one must become familiar with the basic concepts of electron configurations. Every element on the periodic table consists of atoms, which are composed of protons, neutrons, and electrons. Electrons exhibit a negative charge and are found around the nucleus of the atom in electron orbitals, defined as the volume of space in which the electron can be found within 95% probability. The four different types of orbitals (s,p,d, and f) have different shapes, and one orbital can hold a maximum of two electrons. The p, d, and f orbitals have different sublevels, thus can hold more electrons.
As stated, the electron configuration of each element is unique to its position on the periodic table. The energy level is determined by the period and the number of electrons is given by the atomic number of the element. Orbitals on different energy levels are similar to each other, but they occupy different areas in space. The 1s orbital and 2s orbital both have the characteristics of an s orbital (radial nodes, spherical volume probabilities, can only hold two electrons, etc.) but, as they are found in different energy levels, they occupy different spaces around the nucleus. Each orbital can be represented by specific blocks on the periodic table. The s-block is the region of the alkali metals including helium (Groups 1 & 2), the d-block are the transition metals (Groups 3 to 12), the p-block are the main group elements from Groups 13 to 18, and the f-block are the lanthanides and actinides series.
Using the periodic table to determine the electron configurations of atoms is key, but also keep in mind that there are certain rules to follow when assigning electrons to different orbitals. The periodic table is an incredibly helpful tool in writing electron configurations. For more information on how electron configurations and the periodic table are linked, visit the Connecting Electrons to the Periodic Table module.
Rules for Assigning Electron Orbitals
Pauli Exclusion Principle
The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. The first three (n, l, and ml) may be the same, but the fourth quantum number must be different. A single orbital can hold a maximum of two electrons, which must have opposing spins; otherwise they would have the same four quantum numbers, which is forbidden. One electron is spin up (ms = +1/2) and the other would spin down (ms = -1/2). This tells us that each subshell has double the electrons per orbital. The s subshell has 1 orbital that can hold up to 2 electrons, the p subshell has 3 orbitals that can hold up to 6 electrons, the d subshell has 5 orbitals that hold up to 10 electrons, and the f subshell has 7 orbitals with 14 electrons.
Example 1: Hydrogen and Helium
The first three quantum numbers of an electron are n=1, l=0, ml=0. Only two electrons can correspond to these, which would be either ms = -1/2 or ms = +1/2. As we already know from our studies of quantum numbers and electron orbitals, we can conclude that these four quantum numbers refer to the 1s subshell. If only one of the ms values are given then we would have 1s1 (denoting hydrogen) if both are given we would have 1s2 (denoting helium). Visually, this is be represented as:
As shown, the 1s subshell can hold only two electrons and, when filled, the electrons have opposite spins.
Hund's Rule
When assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy (also referred to as degenerate) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. When visualizing this processes, think about how electrons are exhibiting the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals they first try to get as far as possible from each other before having to pair up.
Example 2: Oxygen and Nitrogen
If we look at the correct electron configuration of the Nitrogen (Z = 7) atom, a very important element in the biology of plants: 1s2 2s2 2p3
We can clearly see that p orbitals are half-filled as there are three electrons and three p orbitals. This is because Hund's Rule states that the three electrons in the 2p subshell will fill all the empty orbitals first before filling orbitals with electrons in them. If we look at the element after nitrogen in the same period, oxygen (Z = 8) its electron configuration is: 1s2 2s2 2p4 (for an atom).
Oxygen has one more electron than nitrogen and as the orbitals are all half filled the electron must pair up.
Occupation of Orbitals
Electrons fill orbitals in a way to minimize the energy of the atom. Therefore, the electrons in an atom fill the principal energy levels in order of increasing energy (the electrons are getting farther from the nucleus). The relative energy of the orbitals is shown in Figure \(2\)
The order of levels filled is then:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p
The general order in which orbitals are filled is depicted in Figure \(3\). Subshells corresponding to each value of n are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4s orbital is filled prior to the 3d orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4s1, and the configuration of calcium is [Ar]4s2. Five 3d orbitals are filled by the next 10 elements, the transition metals, followed by three 4p orbitals. Notice that the last member of this row is the noble gas krypton (Z = 36), [Ar]4s23d104p6 = [Kr], which has filled 4s, 3d, and 4p orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5s, 4d, and 5p orbitals are filled sequentially.
The Aufbau Process
Aufbau comes from the German word "aufbauen" meaning "to build." When writing electron configurations, orbitals are built up from atom to atom. When writing the electron configuration for an atom, orbitals are filled in order of increasing atomic number. However, there are some exceptions to this rule.
Example 3: 3rd row elements
Following the pattern across a period from B (Z=5) to Ne (Z=10), the number of electrons increases and the subshells are filled. This example focuses on the p subshell, which fills from boron to neon.
• B (Z=5) configuration: 1s2 2s2 2p1
• C (Z=6) configuration:1s2 2s2 2p2
• N (Z=7) configuration:1s2 2s2 2p3
• O (Z=8) configuration:1s2 2s2 2p4
• F (Z=9) configuration:1s2 2s2 2p5
• Ne (Z=10) configuration:1s2 2s2 2p6
The Number of Valence Electrons
The number of valence electrons of an element can be determined by the periodic table group (vertical column) in which the element is categorized. With the exception of groups 3–12 (the transition metals), the units digit of the group number identifies how many valence electrons are associated with a neutral atom of an element listed under that particular column. For example in group 16, the units digit is 6 and elements in this group have 6 valence electrons.
Table \(1\) Valence electrons derived from periodic table group
Periodic table group Valence electrons
Group 1: alkali metals 1
Group 2: alkaline earth metals 2
Groups 3-12: transition metals 2* (The 4s shell is complete and cannot hold any more electrons)
Group 13: boron group 3
Group 14: carbon group 4
Group 15: pnictogens 5
Group 16: chalcogens 6
Group 17: halogens 7
Group 18: noble gases 8**
* The general method for counting valence electrons is generally not useful for transition metals. Instead the modified d electron count method is used.
** Except for helium, which has only two valence electrons.
The electron configuration of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry.
Example 1.3.1
Draw an orbital diagram and use it to derive the electron configuration of phosphorus, Z = 15. What is its valence electron configuration?
Given: atomic number
Asked for: orbital diagram and valence electron configuration for phosphorus
Strategy:
1. Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
2. Referring to Figure 1.3.1, draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
3. Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.
Solution:
A Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus.
B The additional five electrons are placed in the next available orbitals, which Figure 1.2.5 tells us are the 3s and 3p orbitals:
Because the 3s orbital is lower in energy than the 3p orbitals, we fill it first:
Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3p orbitals separately but with their spins aligned:
The electron configuration is [Ne]3s23p3.
C We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3s23p3.
Exercise 1.3.1
Draw an orbital diagram and use it to derive the electron configuration of chlorine, Z = 17. What is its valence electron configuration?
Answer
[Ne]3s23p5; 3s23p5
The sixth row of the periodic table will be different from the preceding two because the 4f orbitals, which can hold 14 electrons, are filled between the 6s and the 5d orbitals. The elements that contain 4f orbitals in their valence shell are the lanthanides. When the 6p orbitals are finally filled, we have reached the next (and last known) noble gas, radon (Z = 86), [Xe]6s24f145d106p6 = [Rn]. In the last row, the 5f orbitals are filled between the 7s and the 6d orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.
Example 1.3.2
Write the electron configuration of mercury (Z = 80), showing all the inner orbitals.
Given: atomic number
Asked for: complete electron configuration
Strategy:
Using the orbital diagram in Figure 1.3.1 and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.
Solution:
By placing the electrons in orbitals following the order shown in Figure 1.3.1 and using the periodic table as a guide, we obtain
1s2 row 1 2 electrons
2s22p6 row 2 8 electrons
3s23p6 row 3 8 electrons
4s23d104p6 row 4 18 electrons
5s24d105p6 row 5 18 electrons
row 1–5 54 electrons
After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to Figure 1.3.1, we need to fill the 6s (2 electrons), 4f (14 electrons), and 5d (10 electrons) orbitals. The result is mercury’s electron configuration:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 = Hg = [Xe]6s24f145d10
with a filled 5d subshell, a 6s24f145d10 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.)
Summary
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the valence electrons. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.03%3A_Atomic_Structure_-_Electron_Configurations.txt |
Objectives
After completing this section, you should be able to
1. draw Lewis Dot Symbols for main group elements and ions.
2. describe the three-dimensional nature of molecules.
3. sketch a tetrahedral molecule, CX4, using the “wedge-and-broken-line” method of representation.
4. make a ball-and-stick model of a simple tetrahedral molecule such as methane, CH4.
5. draw Lewis Dot Structures for 2 electron group molecules.
6. draw Lewis Dot Structures for 3 electron group molecules.
7. draw Lewis Dot Structures for 4 electron group molecules.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bond strength
• covalent bond
• ionic bond
• Lewis structure
• lone-pair electron
• non-bonding electron
Study Notes
To draw Lewis structures successfully, you need to know the number of valence electrons present in each of the atoms involved. Memorize the number of valence electrons possessed by each of the elements commonly encountered in organic chemistry: C, H, O, N, S, P and the halogens.
When drawing any organic structure, you must remember that a neutral carbon atom will almost always have four bonds. Similarly, hydrogen always has one bond; neutral oxygen atoms have two bonds; and neutral nitrogen atoms have three bonds. By committing these simple rules to memory, you can avoid making unnecessary mistakes later in the course.
The “wedge-and-broken-line” type of representation, which helps to convey the three-dimensional nature of organic compounds, will be used throughout the course.
Bonding Overview
Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond and the general properties found in typical substances in which the bond type occurs will be discussed.
1. Ionic bonds results from electrostatic forces that exist between ions of opposite charge. These bonds typically involve a metal with a nonmetal
2. Covalent bonds result from the sharing of electrons between two atoms. The bonds typically involve one nonmetallic element with another
3. Metallic bonds are found in solid metals (copper, iron, aluminum) with each metal atom bonded to several neighboring metal atoms and the bonding electrons are free to move throughout the 3-dimensional structure.
Each bond classification is discussed in detail in subsequent sections of the chapter. Let's look at the preferred arrangements of electrons in atoms when they form chemical compounds.
Lewis Symbols
At the beginning of the 20th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols, often shortened to Lewis dot symbols—that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons.
Lewis Dot symbols:
• provide a convenient representation of valence electrons
• allows you to keep track of valence electrons during bond formation
• consists of the chemical symbol for the element plus a dot for each valence electron
To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. Fluorine, for example, with the electron configuration [He]2s22p5, has seven valence electrons, so its Lewis dot symbol is constructed as follows:
Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure $2$. The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds. Lewis symbols are a tool to help draw structures. We will see why bonding in molecular compounds follow Lewis' theory in the next section.
Elements in the same group have the same number of valence electrons and similar Lewis symbols. For example, the electron configuration for atomic sulfur is [Ne]3s23p4, thus there are six valence electrons. Its Lewis symbol would therefore be similar to oxygen and look like:
The Octet Rule
Lewis’s major contribution to bonding theory was to recognize that atoms tend to lose, gain, or share electrons to reach a total of eight valence electrons, called an octet. This so-called octet rule explains the stoichiometry of most compounds in the s and p blocks of the periodic table. We now know from quantum mechanics that the number eight corresponds to one ns and three np valence orbitals, which together can accommodate a total of eight electrons. Remarkably, though, Lewis’s insight was made nearly a decade before Rutherford proposed the nuclear model of the atom. Common exceptions to the octet rule are helium, whose 1s2 electron configuration gives it a full n = 1 shell, and hydrogen, which tends to gain or share its one electron to achieve the electron configuration of helium.
Lewis's idea of an octet explains why noble gases rarely form compounds. They have the stable s2p6 configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration. This explains why atom combine together to form compounds. By forming bond the makes the atoms more stable and lower in energy. Making bonds releases energy and represents a driving force for the formation of compounds.
Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table.
Lewis Structures
Lewis structures represent how Lewis symbols gain, lose, or share electrons to obtain an octet by forming compounds.
Lewis Structures of Ionic Compounds
Whenever there is a metal present in the structure of an organic compound, there is a high likelihood that at least one ionic bond is present. Ionic bonds are represented differently in Lewis structures than covalent bonds. Great care should be taken whenever drawing the Lewis structure of an organic compounds which contains and ionic bond. Ionic bonds typically are formed when a metal and a nonmetal are part of a compound. Some atoms achieve an octet by fully gaining or losing electrons to form ions. Ionic bonds form through the eletrostatic attraction of the created ions. The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium (1s22s22p63s1) has one valence electron, so giving it up would result in the same electron configuration as neon (1s22s22p6). Chlorine (1s22s22p63s23p7)has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon (1s22s22p63s23p8) when it gains an electron.
The Lewis structure of an ionic compound show the movement of electrons. For NaCl, sodium is in group 1 and has one valence electron and chlorine is in group 17 and has seven valence electrons. Sodium loses its sole valence electron thereby becomes positively charged. Chlorine gains this electron, gaining a full octet, and a negative charge. After the gain/loss of an electron the new Lewis structures of Na+ and Cl- are written next to each other representing the ionic bond in NaCl.
Examples of Lewis Structures of Ionic Compounds
Covalent Bonds and the Lewis Structures of Molecular Compounds
While alkali metals (such as sodium and potassium), alkaline earth metals (such as magnesium and calcium), and halogens (such as fluorine and chlorine) often form ions in order to achieve a full octet, the principle elements of organic chemistry - carbon, hydrogen, nitrogen, and oxygen - instead tend to fill their octet by sharing electrons with other atoms, forming covalent bonds. Consider the simplest case of hydrogen gas. An isolated hydrogen atom has only one electron, located in the 1s orbital. If two hydrogen atoms come close enough so that their respective 1s orbitals overlap, the two electrons can be shared between the two nuclei, and a covalently bonded H2 molecule is formed. In the Lewis structure of H2, each pair of electrons that is shared between two atoms is drawn as a single line, designating a single covalent bond.
Hydrogen represents a special case, because a hydrogen atom cannot fulfill the octet rule; it needs only two electrons to have a full shell. This is often called the ‘doublet rule’ for hydrogen.
One of the simplest organic molecules is methane with the molecular formula CH4. Methane is the ‘natural gas’ burned in home furnaces and hot water heaters, as well as in electrical power generating plants. To illustrate the covalent bonding in methane using a Lewis structure, we first must recognize that, although a carbon atom has a total of six electrons it's Lewis symbol has four unpaired electrons. Following Lewis' theory the carbon atom wants to form four covalent bonds to fill its octet. In a methane molecule, the central carbon atom shares its four valence electrons with four hydrogen atoms, thus forming four bonds and fulfilling the octet rule (for the carbon) and the ‘doublet rule’ (for each of the hydrogens).
The next relatively simple organic molecule to consider is ethane, which has the molecular formula C2H6. If we draw each atom's Lewis symbol separately, we can see that the octet/doublet rule can be fulfilled for all of them by forming one carbon-carbon bond and six carbon-hydrogen bonds.
The same approach can be used for molecules in which there is no carbon atom. In a water molecule, the Lewis symbol of the oxygen atom has two unpaired electrons. These are paired with the single electron in the Lewis symbols of the hydrogens' two O-H covalent bonds. The remaining four non-bonding electrons on oxygen called ‘lone pairs’.
Since the lone pair electrons are often NOT shown in chemical structures, it is important to see mentally add the lone pairs. In the beginning, it can be helpful to physically add the lone pair electrons.
methylamine ethanol chloromethane
Exercise
For the following structure, please fill in all of the missing lone pair electrons.
Answer
When two or more electrons are shared between atoms a multiple covalent bond is formed. The molecular formula for ethene (also known as ethylene, a compound found in fruits, such as apples, that signals them to ripen) is C2H4. Arranging Lewis symbols of the atoms, you can see that the octet/doublet rule can be fulfilled for all atoms only if the two carbons share two pairs of electrons between them. Ethene contains a carbon/carbon double bond.
Following this pattern, the triple bond in ethyne molecular formula C2H2, (also known as acetylene, the fuel used in welding torches), is formed when the two carbon atoms share three pairs of electrons between them.
Exercise $1$
Draw the Lewis structure for ammonia, NH3.
Answer
Molecular Shape
A stick and wedge drawing of methane shows the tetrahedral angles...(The wedge is coming out of the paper and the dashed line is going behind the paper. The solid lines are in the plane of the paper.)
The following examples make use of this notation, and also illustrate the importance of including non-bonding valence shell electron pairs when viewing such configurations.
Methane Ammonia Water
Bonding configurations are readily predicted by valence-shell electron-pair repulsion theory, commonly referred to as VSEPR in most introductory chemistry texts. This simple model is based on the fact that electrons repel each other, and that it is reasonable to expect that the bonds and non-bonding valence electron pairs associated with a given atom will prefer to be as far apart as possible. The bonding configurations of carbon are easy to remember, since there are only three categories.
Configuration Bonding Partners Bond Angles Example
Tetrahedral 4 109.5º
Trigonal Planar 3 120º
Linear 2 180º
Figure 1.4.3
In the three examples shown above, the central atom (carbon) does not have any non-bonding valence electrons; consequently the configuration may be estimated from the number of bonding partners alone. However, for molecules of water and ammonia, the non-bonding electrons must be included in the calculation. In each case there are four regions of electron density associated with the valence shell so that a tetrahedral bond angle is expected. The measured bond angles of these compounds (H2O 104.5º & NH3 107.3º) show that they are closer to being tetrahedral than trigonal planar or linear. Of course, it is the configuration of atoms (not electrons) that defines the the shape of a molecule, and in this sense ammonia is said to be pyramidal (not tetrahedral). The compound boron trifluoride, BF3, does not have non-bonding valence electrons and the configuration of its atoms is trigonal. Nice treatments of VSEPR theory have been provided by Oxford and Purdue. The best way to study the three-dimensional shapes of molecules is by using molecular models. Many kinds of model kits are available to students and professional chemists.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, BeH2.
AX2: BeH2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
2. There are two electron groups around the central atom. We see from Figure 1.4.3 that the arrangement that minimizes repulsions places the groups 180° apart.
3. Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.
4. From Figure 1.4.3 we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.
AX2: CO2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure 1.4.3).
Three Electron Groups
AX3: BCl3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure 1.4.3).
3. All electron groups are bonding pairs (BP), so the structure is designated as AX3.
4. From Figure 1.4.3 we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar.
AX3: CO32−
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
2. The structure of CO32 is a resonance hybrid. It has three identical bonds, each with a bond order of 4/3.We minimize repulsions by placing the three groups 120° apart (Figure 1.4.3).
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.
4. We see from Figure 1.4.3 that the molecular geometry of CO32 is trigonal planar.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
• AX2E: SO2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure 1.4.3).
3. There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure 1.4.4). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures 1.4.2.1 and 1.4.3).
As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX4: CH4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure 1.4.2, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX4.
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure 1.4.3).
AX3E: NH3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure 1.4.3). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions.
AX2E2: H2O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure 1.4.3.
3. With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one.. This molecular shape is essentially a tetrahedron with two missing vertices. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.04%3A_Development_of_Chemical_Bonding_Theory.txt |
Objectives
After completing this section, you should be able to
1. explain how covalent bonds are formed as a result of the ability of atoms to share electrons.
2. describe the formation of covalent bonds in terms of the overlapping of atomic orbitals.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bond strength
• covalent bond
• bond length
• sigma (σ) bond
• pi (π) bond
• valence bond theory
Valence Bond Theory
As we have been discussing how to use Lewis structures to depict the bonding in organic compounds, we have been very vague so far in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms – but how does this happen, and how does it lead to the formation of a bond holding the two atoms together? Two main models have been developed to described how covalent bonds are formed: valence bond theory and molecularly orbital theory.
Valence bond theory is most often used to describe bonding in organic molecules. In this model, covalent bonds are considered to form from the overlap of two atomic orbitals on different atoms, each orbital containing a single electron. The electrons become paired in the orbital overlap bonding the atoms together.
The simplest example valence bond theory can be demonstrated by the H2 molecule. We can see from the periodic table that each hydrogen atom has a single valence electron. If 2 hydrogen atoms come together to form a bond, then each hydrogen atom effectively has a share in both electrons and thus each resembles the noble gas helium and is more stable. The 2 electrons shared in the orbital overlap are represented by a single dash between the atoms.
Valence bond theory describes a chemical bond as the overlap of atomic orbitals. In the case of the hydrogen molecule, the 1s orbital of one hydrogen atom overlaps with the 1s orbital of the second hydrogen atom to form a molecular orbital called a sigma bond which contains two electrons of opposite spin. The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap.
ne more characteristic of the covalent bond in H2 is important to consider at this point. The two overlapping 1s orbitals can be visualized as two spherical balloons being pressed together. This means that the bond has cylindrical symmetry: if we were to take a cross-sectional plane of the bond at any point, it would form a circle. This type of bond is referred to as a σ(sigma) bond.
The energy of the system depends on how much the orbitals overlap. The energy diagram below illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is called the the bond lengths between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration.
This optimal internuclear distance is the bond length. For the H2 molecule, the distance is 74 pm (picometers, 10-12 meters). Likewise, the difference in potential energy between the lowest energy state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the bond dissociation energy, or, more simply, bond strength. For the hydrogen molecule, the H-H bond strength is equal to about 435 kJ/mol. This means it would take 435 kJ to break one mole of H-H bonds.
Every covalent bond in a given molecule has a characteristic length and strength. In general, the length of a typical carbon-carbon single bond in an organic molecule is about 150 pm, while carbon-carbon double bonds are about 130 pm, carbon-oxygen double bonds are about 120 pm, and carbon-hydrogen bonds are in the range of 100 to 110 pm. The strength of covalent bonds in organic molecules ranges from about 234 kJ/mol for a carbon-iodine bond (in thyroid hormone, for example), about 410 kJ/mole for a typical carbon-hydrogen bond, and up to over 800 kJ/mole for a carbon-carbon triple bond.
Representative Bond Energies and Lengths
Bond Length (pm) Energy (kJ/mol) Bond Length (pm) Energy (kJ/mol)
H-H 74 436 C-O 140.1 358
H-C 106.8 413 C=O 119.7 745
H-N 101.5 391 C≡O 113.7 1072
H-O 97.5 467 H-Cl 127.5 431
C-C 150.6 347 H-Br 141.4 366
C=C 133.5 614 H-I 160.9 298
C≡C 120.8 839 O-O 148 146
C-N 142.1 305 O=O 120.8 498
C=N 130.0 615 F-F 141.2 159
C≡N 116.1 891 Cl-Cl 198.8 243
Exercises
1) For the following energy diagram for energy vs. intermolecular distance is for a fluorine molecule (F2). Please describe the importance for points A, B, & C on the graph.
Solutions
1)
A - Repulsive Forces are present, nuclei are too close to one another.
B - Optimal distance between the two orbitals to have a bond (the bond length)
C - Cannot form a bond, the orbitals are too far apart. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.05%3A_Describing_Chemical_Bonds_-_Valence_Bond_Theory.txt |
Objective
After completing this section, you should be able to describe the structure of methane in terms of the sp3 hybridization of the central carbon atom.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bond angle
• hybridization
• sp3 hybrid
Study Notes
The tetrahedral shape is a very important one in organic chemistry, as it is the basic shape of all compounds in which a carbon atom is bonded to four other atoms. Note that the tetrahedral bond angle of \(\ce{\sf{H−C−H}}\) is 109.5°.
Valence Bond Theory
Valence bond theory's use of overlapping atomic orbitals to explain how chemical bonds form works well in simple diatomic molecules such as H2. However, when molecules with more than two atoms form stable bonds, we require a more detailed model. A good example is methane (CH4). According to valence bond theory, the structure of a covalent species can be depicted using a Lewis structure.
Experimentally, it has been shown that the four carbon-hydrogen bonds in the methane molecule are identical, meaning they have the same bond energy and the same bond length. Also, VSEPR theory suggests that the geometry at the carbon atom in the methane molecule is tetrahedral (2), and there exists a large body of both theoretical and experimental evidence supporting this prediction.
According to valence bond theory, to form a covalent bond forms when an unpaired electron in one atom overlaps with an unpaired electron in a different atom. Now, consider the the electron configuration of the four valence electrons in carbon.
There is a serious mismatch between the electron configuration of carbon (1s22s22p2) and the predicted structure of methane. The modern structure shows that there are only 2 unpaired electrons to share with hydrogens, instead of the 4 needed to create methane. Also, the px and py orbitals are at 90o to each other. They would form perpendicular bonds instead of the tetrahedral 109.5o bond angle predicted by VSEPR and experimental data. Lastly, there are two different orbitals, 2s and 2p, which would create different types of C-H bonds. As noted earlier, experimentally, the four carbon-hydrogen bonds in the methane molecule are identical.
Hybrid Orbitals
An answer to the problems posed above was offered in 1931 by Linus Pauling. He showed mathematically that an s orbital and three p orbitals on an atom can combine to form four equivalent hybrid atomic orbitals.
Important Ideals in Understanding Hybridization
1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.
2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.
4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.
5. The type of hybrid orbitals formed in a bonded atom create the molecular geometry as predicted by the VSEPR theory.
6. Hybrid orbitals overlap to form σ bonds.
7. Lone pair electrons are often contained in hybrid orbitals
sp3 Hybridization in Methane
In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by equations) to form four equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital creating four unpaired electrons.
The shape of an sp3 hybridized orbital is a combination of s and p atomic orbitals.
Each sp3-hybridized orbital bears an electron, and electrons repel each other. To minimize the repulsion between electrons, the four sp3-hybridized orbitals arrange themselves around the carbon nucleus so that they are as far away as possible from each other, resulting in the tetrahedral arrangement predicted by VSPER. The carbon atom in methane is called an “sp3-hybridized carbon atom.” The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5o.
Bonding in Methane
Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in four hydrogen atoms and the larger lobe of one of the four half-filled sp3 hybrid orbitals form a four equivalent sigma (σ) bond. This orbital overlap is often described using the notation: sp3(C)-1s(H). The formation of sp3 hybrid orbitals successfully explains the tetrahedral structure of methane and the equivalency of the the four C-H bonds.
What remains is an explanation of why the sp3 hybrid orbitals form. When the s and 3 p orbitals in carbon hybridize the resulting sp3 hybrid orbital is unsymmetrical with one lobe larger than the other. This means the larger lobe can overlap more effectively with orbitals from other bonds making them stronger. Hybridizing allows for the carbon to form stronger bonds than it would with unhybridized s or p orbitals.
The four carbon-hydrogen bonds in methane are equivalent and all have a bond length of 109 pm (1.09 x 10-10 m), bond strength of of 429 kJ/mol. All of the H-C-H bond angles are 109.5o.
Looking Closer: Linus Pauling
Arguably the most influential chemist of the 20th century, Linus Pauling (1901–1994) is the only person to have won two individual (that is, unshared) Nobel Prizes. In the 1930s, Pauling used new mathematical theories to enunciate some fundamental principles of the chemical bond. His 1939 book The Nature of the Chemical Bond is one of the most significant books ever published in chemistry.
Pauling's big contribution to chemistry was valence bond theory, which combined his knowledge of quantum mechanical theory with his knowledge of basic chemical facts, like bond lengths and and bond strengths and shapes of molecules. Valence bond theory, like Lewis's bonding theory, provides a simple model that is useful for predicting and understanding the structures of molecules, especially for organic chemistry. .
By 1935, Pauling’s interest turned to biological molecules, and he was awarded the 1954 Nobel Prize in Chemistry for his work on protein structure. (He was very close to discovering the double helix structure of DNA when James Watson and James Crick announced their own discovery of its structure in 1953.) He was later awarded the 1962 Nobel Peace Prize for his efforts to ban the testing of nuclear weapons.
Linus Pauling was one of the most influential chemists of the 20th century.
In his later years, Pauling became convinced that large doses of vitamin C would prevent disease, including the common cold. Most clinical research failed to show a connection, but Pauling continued to take large doses daily. He died in 1994, having spent a lifetime establishing a scientific legacy that few will ever equal | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.06%3A_sp_Hybrid_Orbitals_and_the_Structure_of_Methane.txt |
Objective
After completing this section, you should be able to describe the structure of ethane in terms of the sp3 hybridization of the two carbon atoms present in the molecule ethane.
Bonding in Ethane
Representations of Ethane
The simplest molecule with a carbon-carbon bond is ethane, C2H6. In ethane (CH3CH3), both carbons are sp3-hybridized, meaning that both have four bonds with tetrahedral geometry. An sp3 orbital of one carbon atom overlaps end to end with an sp3 orbital of the second carbon atom to form a carbon-carbon σ bond. This orbital overlap is often described using the notation: sp3(C)-sp3(C). Each of the remaining sp3 hybrid orbitals overlaps with the s orbital of a hydrogen atom to form carbon–hydrogen σ bonds.
The σ carbon-carbon bond has a bond length of 154 pm, and a bond strength of 377 kJ/mol. The carbon-hydrogen σ bonds are slightly weaker, 421 kJ/mol, than those of methane. The C-C-H bond angles in ethane are 111.2o which is close to the what is expected for tetrahedral molecules.
The orientation of the two CH3 groups is not fixed relative to each other. Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate. This means, in the case of ethane molecule, that the two methyl (CH3) groups can be pictured as two wheels on a hub, each one able to rotate freely with respect to the other. In Section 3.7 we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the ‘conformation’ of organic molecules
How did they know?
The tetrahedral geometry of carbon was predicted as far back as 1874. But how did they know? A question came up when looking at ethane with a bromine substituent (C2H5Br). When looking at the possible structures of the compound C2H5Br there are several possible structural formulas.here was a serious problem as to whether these formulas represent the same or different compounds. All that was known in the early days was that every purified sample of C2H5Br, no matter how prepared, had a boiling point of 38 oC and density of 1.460 gml−1. Furthermore, all looked the same, all smelled the same, and all underwent the same chemical reactions. There was no evidence that C2H5Br was a mixture or that more than one compound of this formula could be prepared. One might conclude, therefore, that all of the structural formulas above represent a single substance but how? A brilliant solution to the problem came when J. H. van 't Hoff proposed that all four bonds of carbon are equivalent and directed to the corners of a regular tetrahedron. If we redraw the structures for C2H5Br with both carbons having tetrahedral geometry, we see that there is only one possible arrangement. This theory hints at the idea of free rotation around sigma bonds which will be discussed later.
There was a serious problem as to whether these formulas represent the same or different compounds. All that was known in the early days was that every purified sample of C2H5Br, no matter how prepared, had a boiling point of 38 oC and density of 1.460 gml−1. Furthermore, all looked the same, all smelled the same, and all underwent the same chemical reactions. There was no evidence that C2H5Br was a mixture or that more than one compound of this formula could be prepared. One might conclude, therefore, that all of the structural formulas above represent a single substance but how? A brilliant solution to the problem came when J. H. van 't Hoff proposed that all four bonds of carbon are equivalent and directed to the corners of a regular tetrahedron. If we redraw the structures for C2H5Br with both carbons having tetrahedral geometry, we see that there is only one possible arrangement. This theory hints at the idea of free rotation around sigma bonds which will be discussed later. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.07%3A_sp_Hybrid_Orbitals_and_the_Structure_of_Ethane.txt |
Objectives
After completing this section, you should be able to
1. account for the formation of carbon-carbon double bonds using the concept of sp2 hybridization.
2. describe a carbon-carbon double bond as consisting of one σ bond and one π bond.
3. explain the difference between a σ bond and a π bond in terms of the way in which p orbitals overlap.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• pi (π) bond
• sp2 hybrid
Bonding in Ethylene
Thus far valence bond theory has been able to describe the bonding in molecules containing only single bonds. However, when molecules contain double or triple bonds the model requires more details. Ethylene (commonly knows as ethene), CH2CH2, is the simplest molecule which contains a carbon carbon double bond. The Lewis structure of ethylene indicates that there are one carbon-carbon double bond and four carbon-hydrogen single bonds. Experimentally, the four carbon-hydrogen bonds in the ethylene molecule have been shown to be identical. Because each carbon is surrounded by three electron groups, VSEPR theory says the molecule should have a trigonal planar geometry. Although each carbon has fulfilled its tetravalent requirement, one bond appears different. Clearly, a different type of orbital overlap is involved.
The sigma bonds formed in ethene is by the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized. Three of the four valence electrons on each carbon are distributed to the three sp2 hybrid orbitals, while the remaining electron goes into the unhybridized pz orbital. Each carbon in ethene is said to be a “sp2-hybridized carbon.” The electron configuration of the sp2 hybridized carbon shows that there are four unpaired electrons to form bonds. However, the unpaired electrons are contained in two different types of orbitals so it is to be expected that two different types of bonds will form.
The shape of the sp2-hybridized orbital has be mathematically shown to to be roughly the same as that of the sp3-hybridized orbital. To minimize the repulsion between electrons, the three sp2-hybridized orbitals are arranged with a trigonal planar geometry. Each orbital lobe is pointing to the three corners of an equilateral triangle, with angles of 120° between them. Again, geometry and hybrization can be tied together. Atoms surrounded by three electron groups can be said to have a trigonal planar geometry and sp2 hybridization.
The unhybridized 2pz orbital is perpendicular to the plane of the trigonal planar sp2 hybrid orbtals.
In the ethylene molecule, each carbon atom is bonded to two hydrogen atoms. Thus, overlap two sp2-hybridized orbitals with the 1s orbitals of two hydrogen atoms for the C-H sigma bonds in ethylene (sp2(C)-1s(H). Consequently, consistent with the observations, the four carbon-hydrogen bonds in ethylene are identical.
The C-C sigma bond in ethylene is formed by the overlap of an sp2 hybrid orbital from each carbon.
The overlap of hybrid orbitals or a hybrid orbital and a 1s orbtial from hydrogen creates the sigma bond framework of the ethylene molecule. However the unhybridized pz orbital on each carbon remains.
The unhybridized pz orbitals on each carbon overlap to a π bond (pi). The orbital overlap is commonly written as pz(C)-1pz(C). In general multiple bonds in molecular compound are formed by the overlap of unhybridized p orbitals. It should be noted that the carbon-carbon double bond in ethlene is made up of two different types of bond, a sigma and a pi.
Overall, ethylene is said to contain five sigma bonds and one pi bond. Pi bonds tend to be weaker than sigma bonds because the side-by-side overlap the p orbitals give a less effective orbital overlap when compared to the end-to-end orbital overlap of a sigma bond. This makes the pi much easier to break which is one of the most important ideas in organic chemistry reactions as we will see in Chapter 7 and subsequent chapters.
An ethylene molecule is said to be made up of five sigma bonds and one pi bond. The three sp2 hybrid orbitals on each carbon orient to create the basic trigonal planer geometry. The H-C-C bond angle in ethylene is 121.3o which is very close to the 120o predicted by VSEPR. The four C-H sigma bonds in ethylene . The carbon-carbon double bond in ethylene is both shorter (133.9 pm) and almost twice as strong (728 kJ/mol) than the carbon- carbon single bond in ethylene (154 pm & 377 kJ/mol). Each of the four carbon-hydrogen bond in ethylene are equivalent has have a length of 108.7 pm
Rigidity in Ethene
Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. If free rotation were to occur the p-orbitals would have to go through a phase where they are 90° from each other, which would break the pi bond because there would be no overlap. Since the pi bond is essential to the structure of ethene it must not break, so there can be not free rotation about the carbon-carbon sigma bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane.
Exercise
1) Consider the following molecule:
At each atom, what is the hybridization and the bond angle and the bond angle predicted by VSPER?
2) Please identify the types of orbitals shown in the following diagram:
3)
a: Describe the orbitals which overlap to the carbon-nitrogen sigma bond and pie bond in the molecule below:
b: What kind of orbital holds the nitrogen lone pair?
4) For the following molecule please indicate with atoms are being held in the same plane by the carbon-carbon double bond:
Solutions
1)
A - sp2, 120°
B - sp3, 109°
C - sp2, 120° (with the lone pairs present)
D - sp3, 109°
2)
3)
a) The carbon and nitrogen atoms are both sp2 hybridized. The carbon-nitrogen double bond is composed of a sigma bond formed from two sp2 orbitals, and a pi bond formed from the side-by-side overlap of two unhybridized 2p orbitals.
b) As shown in the figure above, the nitrogen lone pair electrons occupy one of the three sp2 hybrid orbitals.
4) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.08%3A_sp_Hybrid_Orbitals_and_the_Structure_of_Ethylene.txt |
Objectives
After completing this section, you should be able to
1. use the concept of sp hybridization to account for the formation of carbon-carbon triple bonds, and describe a carbon-carbon triple bond as consisting of one σ bond and two π bonds.
2. list the approximate bond lengths associated with typical carbon-carbon single bonds, double bonds and triple bonds. [You may need to review Sections 1.7 and 1.8.]
3. list the approximate bond angles associated with sp3-, sp2- and sp‑hybridized carbon atoms and predict the bond angles to be expected in given organic compounds. [If necessary, review Sections 1.6, 1.7 and 1.8.]
4. account for the differences in bond length, bond strength and bond angles found in compounds containing sp3-, sp2- and sp‑hybridized carbon atoms, such as ethane, ethylene and acetylene.
Key Terms
Make certain that you can define, and use in context, the key term below.
• sp hybrid orbital
Study Notes
The bond angles associated with sp3-, sp2- and sp‑hybridized carbon atoms are approximately 109.5°, 120° and 180°, respectively.
Bonding in acetylene
Finally, the hybrid orbital concept applies well to triple-bonded groups, such as alkynes and nitriles. Consider, for example, the structure of ethyne (another common name is acetylene), the simplest alkyne.
ethyne
(acetylene)
This molecule is linear: all four atoms lie in a straight line. The carbon-carbon triple bond is only 1.20Å long. In the hybrid orbital picture of acetylene, both carbons are sp-hybridized. In an sp-hybridized carbon, the 2s orbital combines with the 2px orbital to form two sp hybrid orbitals that are oriented at an angle of 180°with respect to each other (eg. along the x axis). The 2py and 2pz orbitals remain non-hybridized, and are oriented perpendicularly along the y and z axes, respectively.
The C-C sigma bond is formed by the overlap of one sp orbital from each of the carbons, while the two C-H sigma bonds are formed by the overlap of the second sp orbital on each carbon with a 1s orbital on a hydrogen. Each carbon atom still has two half-filled 2py and 2pz orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).
sigma bonding in ethylene pi bonding in ethylene
Acetylene is said to have three sigma bonds and two pi bonds. The carbon-carbon triple bond in acetylene is the shortest (120 pm) and the strongest (965 kJ/mol) of the carbon-carbon bond types. Because each carbon in acetylene has two electron groups, VSEPR predicts a linear geometry and and H-C-C bond angle of 180o.
Comparison of C-C bonds Ethane, Ethylene, and Acetylene
Molecule Bond Bond Strength (kJ/mol) Bond Length (pm)
Ethane, CH3CH3 (sp3) C-C (sp3) 376 154
Ethylene, H2C=CH2 (sp2) C=C (sp2) 728 134
Acetylene, HC≡CH (sp) C≡C (sp) 965 120
Notice that as the bond order increases the bond length decreases and the bond strength increases.
The hybrid orbital concept nicely explains another experimental observation: single bonds adjacent to double and triple bonds are progressively shorter and stronger than ‘normal’ single bonds, such as the one in a simple alkane. The carbon-carbon bond in ethane (structure A below) results from the overlap of two sp3 orbitals.
A B C
In propene (B), however, the carbon-carbon single bond is the result of overlap between an sp2 orbital and an sp3 orbital, while in propyne (C) the carbon-carbon single bond is the result of overlap between an sp orbital and an sp3 orbital. These are all single bonds, but the single bond in molecule C is shorter and stronger than the one in B, which is in turn shorter and stronger than the one in A.
The explanation here is relatively straightforward. An sp orbital is composed of one s orbital and one p orbital, and thus it has 50% s character and 50% p character. sp2 orbitals, by comparison, have 33% s character and 67% p character, while sp3 orbitals have 25% s character and 75% p character. Because of their spherical shape, 2s orbitals are smaller, and hold electrons closer and ‘tighter’ to the nucleus, compared to 2p orbitals. Consequently, bonds involving sp + sp3 overlap (as in alkyne C) are shorter and stronger than bonds involving sp2 + sp3 overlap (as in alkene B). Bonds involving sp3-sp3overlap (as in alkane A) are the longest and weakest of the group, because of the 75% ‘p’ character of the hybrids.
Hybridization Summary
• A single bond is a sigma bond.
• A double bond is made up of a sigma bond and a pi bond.
• A triple bond is made up of a sigma bond and two pi bonds.
• Sigma bonds are made by the overlap of two hybrid orbitals or the overlap of a hybrid orbital and a s orbital from hydrogen.
• Pi bonds are made by the overlap of two unhybridized p orbitals.
• Lone pair electrons are usually contained in hybrid orbitals.
The hybrid orbitals used (and hence the hybridization) depends on how many electron groups are around the atom in question. An electron group can mean either a bonded atom or a lone pair. Molecular geometry is also decided by the number of electron groups so it is directly linked to hybridization.
# of Electron Groups Hybrid Orbital Used Example Basic Geometry Basic Bond Angle
2 sp Linear 180o
3 sp2 Trigonal Planar 120o
4 sp3 Tetrahedral 109.5o
Exercises
1) For the molecule acetonitrile:
a) How many sigma and pi bonds does it have?
b) What orbitals overlap to form the C-H sigma bonds?
c) What orbitals overlap to form the C-C sigma bond?
d) What orbitals overlap to form the C-N sigma bond?
e) What orbitals overlap to the form the C-N pi bonds?
f) What orbital contains the lone pair electrons on nitrogen?
Solutions
1)
a) 5 sigma and 2 pi
b) An sp3 hybrid orbital from carbon and an a s orbital from hydrogen.
c) An sp3 hybrid orbital from one carbon and an a sp3 orbital from the other carbon.
d) An sp hybrid orbital from carbon and an a sp orbital from nitrogen.
e) An py and pz orbital from carbon and an py and pz orbital from nitrogen.
f) An sp hybrid orbital. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.09%3A_sp_Hybrid_Orbitals_and_the_Structure_of_Acetylene.txt |
Objective
After completing this section, you should be able to apply the concept of hybridization of atoms such as N, O, P and S to explain the structures of simple species containing these atoms.
Key Terms
Make certain that you can define, and use in context, the key term below.
• lone pair electrons
Study Notes
Nitrogen is frequently found in organic compounds. As with carbon atoms, nitrogen atoms can be sp3-, sp2- or sp‑hybridized.
Note that, in this course, the term “lone pair” is used to describe an unshared pair of electrons.
The valence-bond concept of orbital hybridization can be extrapolated to other atoms including nitrogen, oxygen, phosphorus, and sulfur. In other compounds, covalent bonds that are formed can be described using hybrid orbitals.
Nitrogen
Bonding in NH3
The nitrogen in NH3 has five valence electrons. After hybridization these five electrons are placed in the four equivalent sp3 hybrid orbitals. The electron configuration of nitrogen now has one sp3 hybrid orbital completely filled with two electrons and three sp3 hybrid orbitals with one unpaired electron each. The two electrons in the filled sp3 hybrid orbital are considered non-bonding because they are already paired. These electrons will be represented as a lone pair on the structure of NH3. The three unpaired electrons in the hybrid orbitals are considered bonding and will overlap with the s orbitals in hydrogen to form N-H sigma bonds. Note! This bonding configuration was predicted by the Lewis structure of NH3.
The four sp3 hybrid orbitals of nitrogen orientate themselves to form a tetrahedral geometry. The three N-H sigma bonds of NH3 are formed by sp3(N)-1s(H) orbital overlap. The fourth sp3 hybrid orbital contains the two electrons of the lone pair and is not directly involved in bonding.
Methyl amine
The nitrogen is sp3 hybridized which means that it has four sp3 hybrid orbitals. Two of the sp3 hybridized orbitals overlap with s orbitals from hydrogens to form the two N-H sigma bonds. One of the sp3 hybridized orbitals overlap with an sp3 hybridized orbital from carbon to form the C-N sigma bond. The lone pair electrons on the nitrogen are contained in the last sp3 hybridized orbital. Due to the sp3 hybridization the nitrogen has a tetrahedral geometry. However, the H-N-H and H-N-C bonds angles are less than the typical 109.5o due to compression by the lone pair electrons.
Oxygen
Bonding in H2O
The oxygen in H2O has six valence electrons. After hybridization these six electrons are placed in the four equivalent sp3 hybrid orbitals. The electron configuration of oxygen now has two sp3 hybrid orbitals completely filled with two electrons and two sp3 hybrid orbitals with one unpaired electron each. The filled sp3 hybrid orbitals are considered non-bonding because they are already paired. These electrons will be represented as a two sets of lone pair on the structure of H2O . The two unpaired electrons in the hybrid orbitals are considered bonding and will overlap with the s orbitals in hydrogen to form O-H sigma bonds. Note! This bonding configuration was predicted by the Lewis structure of H2O.
The four sp3 hybrid orbitals of oxygen orientate themselves to form a tetrahedral geometry. The two O-H sigma bonds of H2O are formed by sp3(O)-1s(H) orbital overlap. The two remaining sp3 hybrid orbitals each contain two electrons in the form of a lone pair.
Methanol
The oxygen is sp3 hybridized which means that it has four sp3 hybrid orbitals. One of the sp3 hybridized orbitals overlap with s orbitals from a hydrogen to form the O-H sigma bonds. One of the sp3 hybridized orbitals overlap with an sp3 hybridized orbital from carbon to form the C-O sigma bond. Both the sets of lone pair electrons on the oxygen are contained in the remaining sp3 hybridized orbital. Due to the sp3 hybridization the oxygen has a tetrahedral geometry. However, the H-O-C bond angles are less than the typical 109.5o due to compression by the lone pair electrons.
Phosphorus
Methyl phosphate
The bond pattern of phosphorus is analogous to nitrogen because they are both in period 15. However, phosphorus can have have expanded octets because it is in the n = 3 row. Typically, phosphorus forms five covalent bonds. In biological molecules, phosphorus is usually found in organophosphates. Organophosphates are made up of a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to a carbon. In methyl phosphate, the phosphorus is sp3 hybridized and the O-P-O bond angle varies from 110° to 112o.
Sulfur
Methanethiol & Dimethyl Sulfide
Sulfur has a bonding pattern similar to oxygen because they are both in period 16 of the periodic table. Because sulfur is positioned in the third row of the periodic table it has the ability to form an expanded octet and the ability to form more than the typical number of covalent bonds. In biological system, sulfur is typically found in molecules called thiols or sulfides. In a thiol, the sulfur atom is bonded to one hydrogen and one carbon and is analogous to an alcohol O-H bond. In a sulfide, the sulfur is bonded to two carbons. The simplest example of a thiol is methane thiol (CH3SH) and the simplest example of a sulfide is dimethyl sulfide [(CH3)3S]. In both cases the sulfur is sp3 hybridized, however the sulfur bond angles are much less than the typical tetrahedral 109.5o being 96.6o and 99.1o respectively.
methanethiol
dimethyl sulfide
Exercises
1) Insert the missing lone pairs of electrons in the following molecules, and tell what hybridization you expect for each of the indicated atoms.
a) The oxygen is dimethyl ether:
b) The nitrogen in dimethyl amine:
c) The phosphorus in phosphine:
d) The sulfur in hydrogen sulfide:
Solutions
1)
a) sp3 hybridization
b) sp3 hybridization
c) sp3 hybridization
d) sp3 hybridization | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.10%3A_Hybridization_of_Nitrogen_Oxygen_Phosphorus_and_Sulfur.txt |
Objectives
After completing this section, you should be able to
1. describe the formation of covalent bonds in terms of molecular orbitals.
2. account for differences in bond length and strength in terms of the efficiency with which atomic orbitals overlap.
3. draw simple molecular orbital diagrams (e.g., for the H2 molecule) showing the formation of bonding and anti-bonding orbitals.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti-bonding molecular orbital
• bonding molecular orbital
• molecular orbital (MO) theory
As we have seen, valence bond theory does a remarkably good job of explaining the bonding geometry and properties of many organic compounds. There are some areas, however, where the valence bond theory falls short. It fails to adequately account, for example, for some interesting properties of compounds that contain alternating double and single bonds. In order to understand these properties, we need to think about chemical bonding in a new way, using the ideas of molecular orbital (MO) theory.
Another look at the H2 molecule: bonding and anti-bonding sigma molecular orbitals
Let’s consider again the simplest possible covalent bond: the one in molecular hydrogen (H2). When we described the hydrogen molecule using valence bond theory, we said that the two 1s orbitals from each atom overlap, allowing the two electrons to be shared and thus forming a covalent bond. In molecular orbital theory, we make a further statement: we say that the two atomic 1s orbitals don’t just overlap, they actually combine to form two completely new orbitals. These two new orbitals, instead of describing the likely location of an electron around a single nucleus, describe the location of an electron pair around two or more nuclei. The bonding in H2, then, is due to the formation of a new molecular orbital (MO), in which a pair of electrons is delocalized around two hydrogen nuclei.
An important principle of quantum mechanical theory is that when orbitals combine, the number of orbitals before the combination takes place must equal the number of new orbitals that result – orbitals don’t just disappear! We saw this previously when we discussed hybrid orbitals: one s and three p orbitals make four sp3 hybrids. When two atomic 1s orbitals combine in the formation of H2, the result is two molecular orbitals called sigma (σ) orbitals. According to MO theory, the first sigma orbital is lower in energy than either of the two isolated atomic 1s orbitals – thus this sigma orbital is referred to as a bonding molecular orbital. The second, sigma-star (σ*) orbital is higher in energy than the two atomic 1s orbitals, and is referred to as an anti-bonding molecular orbital. In MO theory, a star (*) sign always indicates an anti-bonding orbital.
Following the aufbau ('building up') principle, we place the two electrons in the H2 molecule in the lowest energy molecular orbital, which is the (bonding) sigma orbital.
The bonding sigma orbital, which holds both electrons in the ground state of the molecule, is egg-shaped, encompassing the two nuclei, and with the highest likelihood of electrons being in the area between the two nuclei. The high-energy, anti-bonding sigma-star orbital can be visualized as a pair of droplets, with areas of higher electron density near each nucleus and a ‘node’, (area of zero electron density) midway between the two nuclei.
Remember that we are thinking here about electron behavior as wave behavior. When two separate waves combine, they can do so with what is called constructive interference, where the two amplitudes reinforce one another, or destructive interference, where the two amplitudes cancel one another out. Bonding MO’s are the consequence of constructive interference between two atomic orbitals which results in an attractive interaction and an increase in electron density between the nuclei. Anti-bonding MO’s are the consequence of destructive interference which results in a repulsive interaction and a ‘canceling out’ of electron density between the nuclei (in other words, a node).
MO theory and pi bonds
In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals. If two atoms are located along the x-axis in a Cartesian coordinate system, the two px orbitals overlap end to end and form σpx (bonding) and σ∗px (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with s-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.
The advantage of MO theory becomes more apparent when we think about pi bonds, especially in those situations where two or more pi bonds are able to interact with one another. Let’s first consider the pi bond in ethene from an MO theory standpoint (in this example we will be disregarding the various sigma bonds, and thinking only about the pi bond). According to MO theory, the two atomic 2pz orbitals combine to form two pi (π) molecular orbitals, one a low-energy π bonding orbital and one a high-energy π-star (π*) anti-bonding molecular orbital. These are sometimes denoted, in MO diagrams like the one below, with the Greek letter psi (Ψ) instead of π.
In the bonding Ψ1 orbital, the two shaded lobes of the 2pz orbitals interact constructively with each other, as do the two unshaded lobes (remember, the shading choice represents mathematical (+) and (-) signs for the wavefunction). Therefore, there is increased electron density between the nuclei in the molecular orbital – this is why it is a bonding orbital.
In the higher-energy anti-bonding Ψ2* orbital, the shaded lobe of one 2pz orbital interacts destructively with the unshaded lobe of the second 2pz orbital, leading to a node between the two nuclei and overall repulsion.
By the aufbau principle, the two electrons from the two atomic orbitals will be paired in the lower-energy Ψ1 orbital when the molecule is in the ground state.
Example \(1\)
Draw a simple molecular orbital diagram for each of the following molecules
1. nitrogen, N2.
2. oxygen, O2.
Answers | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.11%3A_Describing_Chemical_Bonds_-_Molecular_Orbital_Theory.txt |
Objectives
After completing this section, you should be able to
1. propose one or more acceptable Kekulé structures (structural formulas) for any given molecular formula
2. write the molecular formula of a compound, given its Kekulé structure.
3. draw the shorthand structure of a compound, given its Kekulé structure.
4. interpret shorthand structures and convert them to Kekulé structures.
5. write the molecular formula of a compound, given its shorthand structure.
Study Notes
When drawing the structure of a neutral organic compound, you will find it helpful to remember that
• each carbon atom has four bonds.
• each nitrogen atom has three bonds.
• each oxygen atom has two bonds.
• each hydrogen atom has one bond.
Through general chemistry, you may have already experienced looking at molecular structures using Lewis structures. Because organic chemistry can involve large molecules it would be beneficial if Lewis structures could be abbreviated. The three different ways to draw organic molecules include Keku Formulas, Condensed Formulas, and Skeletal structures (also called line-bond structures or line formulas). During this course, you will view molecules written in all three forms. It will be more helpful if you become comfortable going from one style of drawing to another, and look at drawings and understanding what they represent. Developing the ability to convert between different types of formulas requires practice, and in most cases the aid of molecular models. Many kinds of model kits are available to students and professional chemists, and the beginning student is encouraged to obtain one.
Simplification of structural formulas may be achieved without any loss of the information they convey. Kekule formulas is just organic chemistry's term for Lewis structures you have previously encountered. In condensed structural formulas, the bonds to each carbon are omitted, but each distinct structural unit (group) is written with subscript numbers designating multiple substituents, including the hydrogens. Line formulas omit the symbols for carbon and hydrogen entirely (unless the hydrogen is bonded to an atom other than carbon). Each straight line segment represents a bond, the ends and intersections of the lines are carbon atoms, and the correct number of hydrogens is calculated from the tetravalency of carbon. Non-bonding valence shell electrons are omitted in these formulas.
Kekulé (a.k.a. Lewis Structures)
A Kekulé Formula or structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. All atom labels are shown and all lone pairs are shown.
A B C
Condensed Formula
A condensed formula is made up of the elemental symbols. Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. The order of the atoms suggests the connectivity in the molecule. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together. Also, if more than one of the same substituent is attached to a given atom, it is show with a subscript number. An example is CH4, which represents four hydrogens attached to the same carbon. Condensed formulas can be read from either direction and H3C is the same as CH3, although the latter is more common.
Look at the examples below and match them with their identical molecule under the Kekulé structures and the line formulas.
CH3CH2OH ClCH2CH2CH(OCH3)CH3 CH3NHCH2COOH
A B C
Let's look closely at example B. As you go through a condensed formula, you want to focus on the carbons and other elements that aren't hydrogen. The hydrogen's are important, but are usually there to complete octets. Also, notice the -OCH3 is in written in parentheses which tell you that it not part of the main chain of carbons. As you read through a a condensed formula, if you reach an atom that doesn't have a complete octet by the time you reach the next hydrogen, then it's possible that there are double or triple bonds. In example C, the carbon is double bonded to oxygen and single bonded to another oxygen. Notice how COOH means C(=O)-O-H instead of CH3-C-O-O-H because carbon does not have a complete octet and oxygens.
Line Formula
Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letter "C" with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are omitted but are assumed to be present to complete each of carbon's four bonds. Hydrogens that are attached to elements other than carbon are shown. Atom labels for all other elements are shown. Lone pair electrons are usually omitted. They are assumed to be present to complete the octet of non-carbon atoms. Line formulas help show the structure and order of the atoms in a compound.
A B C
These molecules correspond to the exact same molecules depicted for Kekulé structures and condensed formulas. Notice how the carbons are no longer drawn in and are replaced by the ends and bends of a lines. In addition, the hydrogens have been omitted, but could be easily drawn in (see practice problems). Although we do not usually draw in the H's that are bonded to carbon, we do draw them in if they are connected to other atoms besides carbon (example is the OH group above in example A) . This is done because it is not always clear if the non-carbon atom is surrounded by lone pairs or hydrogens. Also in example A, notice how the OH is drawn with a bond to the second carbon, but it does not mean that there is a third carbon at the end of that bond/ line.
Kekulé Formula Condensed Formula Line Formula
CH3(CH2)3OH
CH3CH2CH(OH)CH3
(CH3)2CHCH2OH
(CH3)3COH
CH3CH2OCH2CH3
Table \(1\): Structural Formulas for C4H10O isomers
Example: Converting between Structural Formulas
It is helpful to convert compounds into different structural formulas (Kekule, Line, and Condensed) depending on the type of question that is asked. Standardized exams frequently include a high percentage of condensed formulas because it is easier and cheaper to type letters and numbers than to import figures. Initially, it can be difficult writing a Line structure directly from a condensed formula. First, write the Kekule structure from the condensed formula and then draw the Line structure from the Kekule.
a) The condensed formula for propanal is CH3CH2CHO. Draw the Kekule structure.
The Kekule structure for propanal is shown below. Remember that every carbon will have four bonds and oxygens octet is filled with lone pairs.
The bond-line structure for propanal is shown below. First, remove hydrogens. The hydrogen attached to the aldehyde group remains because it is part of a functional group. The remove the "C" labels from the structure and keep the lines in place. Lastly, remove any lone pairs.
All three structures represent the same compound, propanal.
b) The following is the line structure of the molecule trimethyl amine.
To convert it to a Kekule structure first identify the carbons in the molecule. The will be at the corners and ends of line without an atom label. Trimethyl amine has three carbons. Next, add hydrogens to the carbons until four bonds are present. Each carbon in trimethyl amine is singly bonded to nitrogen. This means each carbon will need three additional C-H bonds to create its octet. Lastly, add lone pairs to other elements to fill their octets. The nitrogen in trimethyl amine is bonded to three carbons. This means it will require one of lone pair electrons to complete its octet.
Exercises
1. How many carbons are in the following drawing? How many hydrogens?
2. How many carbons are in the following drawing? How many hydrogens?
3. How many carbons are in the following drawing? How many hydrogens?
4. Look at the following molecule of vitamin A and draw in the hidden hydrogens and electron pairs.
(hint: Do all of the carbons have 4 bonds? Do all the oxygens have a full octet?)
5. Draw ClCH2CH2CH(OCH3)CH3 in Kekulé and line form.
6. Write down the molecular formula for each of the compounds shown here.
Answers:
1. Remember the octet rule and how many times carbons and hydrogens are able to bond to other atoms.
2. Electron pairs drawn in blue and hydrogens draw in red.
6.
1. C7H7N
2. C5H10
3. C5H4O
4. C5H6Br2 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.12%3A_Drawing_Chemical_Structures.txt |
Concepts & Vocabulary
1.0: Prelude to Structure and Bonding
• Organic compounds contain carbon atoms bonded hydrogen and other carbon atoms.
• Organic chemistry studies the properties and reactions of organic compounds.
1.1: Atomic Structure: The Nucleus
• Atoms are comprised of protons, neutrons and electrons. Protons and neutrons are found in the nucleus of the atom, while electrons are found in the electron cloud around the nucleus. The relative electrical charge of a proton is +1, a neutron has no charge, and an electron’s relative charge is -1.
• The number of protons in an atom’s nucleus is called the atomic number, Z.
• The mass number, A, is the sum of the number of protons and the number of neutrons in a nucleus.
• The type of element an atom represents is defined by the atomic number, Z in the atom. All atoms of one specific element have the same number of protons (Z).
• Atoms that have the same atomic number (Z), but different mass numbers (A) are called isotopes.
1.2: Atomic Structure: Orbitals
• An atomic orbital is the probability description of where an electron can be found. The four basic types of orbitals are designated as s, p, d, and f.
1.3: Atomic Structure: Electron Configurations
• The order in which electrons are placed in atomic orbitals is called the electron configuration and is governed by the aufbau principle.
• Electrons in the outermost shell of an atom are called valence electrons. The number of valence electrons in any atom is related to its position in the periodic table. Elements in the same periodic group have the same number of valence electrons.
1.4: Development of Chemical Bonding Theory
• Lewis Dot Symbols are a way of indicating the number of valence electrons in an atom. They are useful for predicting the number and types of covalent bonds within organic molecules.
• The molecular shape of molecules is predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. The shapes of common organic molecules are based on tetrahedral, trigonal planar or linear arrangements of electron groups.
1.5: The Nature of Chemical Bonds: Valence Bond Theory
• Covalent bonds form as valence electrons are shared between two atoms.
• Lewis Structures and structural formulas are common ways of showing the covalent bonding in organic molecules.
• Formal charge describes the changes in the number of valence electrons as an atom becomes bonded into a molecule. If the atom has a net loss of valence electrons it will have a positive formal charge. If the atom has a net gain of valence electrons it will have a negative formal charge.
• Atomic orbitals often change as they overlap to form molecular orbitals. This process is known as orbital hybridization. The common types of hybrid orbitals in organic molecules are sp3, sp2, and sp.
1.6: \(sp^3\) Hybrid Orbitals and the Structure of Methane
• The four identical C-H single bonds in CH4 form as the result of sigma bond overlap between the sp3 hybrid orbitals of carbon and the s orbital of each hydrogen.
1.7: \(sp^3\) Hybrid Orbitals and the Structure of Ethane
• The C-C bond in C2H6 forms as the result of sigma bond overlap between a sp3 hybrid orbital on each carbon. and the s orbital of each hydrogen. The six identical C-H single bonds in form as the result of sigma bond overlap between the sp3 hybrid orbitals of carbon and the s orbital of each hydrogen.
1.8: \(sp^2\) Hybrid Orbitals and the Structure of Ethylene
• The C=C bond in C2H4 forms as the result of both a sigma bond overlap between a sp2 hybrid orbital on each carbon and a pi bond overlap of a p orbital on each carbon
1.9 \(sp\) Hybrid Orbitals and the Structure of Acetylene
• The carbon-carbon triple bond in C2H4 forms as the result of one sigma bond overlap between a sp hybrid orbital on each carbon and two pi bond overlaps of p orbitals on each carbon.
1.10: Hybridization of Nitrogen, Oxygen, Phosphorus and Sulfur
• The atomic orbitals of nitrogen, oxygen, phosphorus and sulfur can hybridize in the same way as those of carbon.
1.11: The Nature of Chemical Bonds: Molecular Orbital Theory
• Molecular Orbital theory (MO) is a more advanced bonding model than Valence Bond Theory, in which two atomic orbitals overlap to form two molecular orbitals – a bonding MO and an anti-bonding MO.
1.12: Drawing Chemical Structures
• Kekulé Formulas or structural formulas display the atoms of the molecule in the order they are bonded.
• Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space.
• Skeleton formulas or Shorthand formulas or line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines.
• Isomers have the same molecular formula, but different structural formulas
Summary Problems
Exercise \(1\)
The following molecule is highly reactive and contains a functional group called a ketene. For each numbered atom, list the geometry, bond angle, hybridization, orbitals present, and orbital function. Notes: Ignore the geometry and bond angle for atom #1. For orbitals present, list all of the orbitals at that atom. For orbital function, describe what each orbital is doing (e.g., participating in a sigma bond, containing lone pair electrons, etc.).
Answer
#1 - sp2 hybridization; 3 sp2 orbitals and 1 p orbital present; p orbital participates in a pi bond, 1 sp2 orbital participates in a sigma bond, 2 sp2 orbitals contain lone pair electrons
#2 - linear; 180 degrees; sp hybridization; 2 sp orbitals and 2 p orbitals present; p orbitals participate in two pi bonds, s sp2 orbitals participate in two sigma bonds
#3 - trigonal planar; 120 degrees; sp2 hybridization; 3 sp2 orbitals and 1 p orbital present; p orbital participates in a pi bond, 3 sp2 orbitals participate in three sigma bonds
#4 - tetrahedral; 109.5 degrees; sp3 hybridization; 4 sp3 orbitals present; 4 sp3 orbitals participate in four sigma bonds
Exercise \(2\)
First, add lone pair electrons and formal charges to the molecule shown below; its net charge is plus one. (The only formal charge on a carbon atom has already been added for you.) Second, for each numbered atom, list the geometry, bond angle, hybridization, orbitals present, and orbital function. Notes: For orbitals present, list all of the orbitals at that atom. For orbital function, describe what each orbital is doing (e.g., participating in a sigma bond, containing lone pair electrons, etc.).
Answer
#1 - linear; 180 degrees; sp hybridization; 2 sp orbitals and 2 p orbitals present; p orbitals participate in two pi bonds, s sp2 orbitals participate in two sigma bonds
#2 - trigonal planar; 120 degrees; sp2 hybridization; 3 sp2 orbitals and 1 p orbital present; p orbital is empty, 3 sp2 orbitals participate in three sigma bonds
#3 - bent; <109.5 degrees; sp3 hybridization; 4 sp3 orbitals present; 2 sp3 orbitals participate in two sigma bonds, 2 sp3 orbitals contain two lone pairs
#4 - tetrahedral; 109.5 degrees; sp3 hybridization; 4 sp3 orbitals present; 4 sp3 orbitals participate in four sigma bonds
Exercise \(3\)
First, add lone pair electrons and formal charges to the molecule shown below. Second, for each numbered atom, list the geometry, bond angle, hybridization, orbitals present, and orbital function. Third, clearly identify on the structure below sigma bonds formed by overlap of the following orbitals: sp-sp2 (label as “a”), sp-sp3 (label as “b”), sp2-sp2 (label as “c”), sp2-sp3 (label as “d”), and sp3-sp3 (label as “e”). Notes: For orbitals present, list all of the orbitals at that atom. For orbital function, describe what each orbital is doing (e.g., participating in a sigma bond, containing lone pair electrons, etc.). When identifying particular types of sigma bonds, there will be more than one correct answer for some options.
Answer
All atoms are neutral in this molecule.
#1 - bent; <109.5 degrees; sp3 hybridization; 4 sp3 orbitals present; 2 sp3 orbitals participate in two sigma bonds, 2 sp3 orbitals contain two lone pairs
#2 - linear; 180 degrees; sp hybridization; 2 sp orbitals and 2 p orbitals present; p orbitals participate in two pi bonds, s sp2 orbitals participate in two sigma bonds
#3 - trigonal planar; 120 degrees; sp2 hybridization; 3 sp2 orbitals and 1 p orbital present; p orbital participates in a pi bond, 3 sp2 orbitals participate in three sigma bonds
For a and b, these are the only bonds that are correct. For c, there are several correct options. For d, there is one other option (from the benzene ring to the O in the five-membered ring. For e, there are several correct options.
Skills to Master
Skill 1.1 Determine the number of protons, neutrons, and electrons in a nuclide.
Skill 1.2 Write the electron configuration and orbital diagram for an atom.
Skill 1.3 Determine the number of valence electrons in an atom.
Skill 1.4 Draw the molecular formula, Lewis Dot Structure, structural formula, condensed structural formula, shorthand formula and wedge-dash structure of simple organic molecules.
Skill 1.5 Use Lewis Dot structures to predict molecular shape, bond angle, hybridization.
Skill 1.6 Calculate formal charge on an atom in a molecule.
Skill 1.7 Determine the number of sigma and pi bonds in organic molecules.
Skill 1.8 Determine relative bond energy and bond length based on atoms involved in the bond and bond type.
Skill 1.9 Describe and draw the orbital overlap and types of bonding in simple organic molecules like methane, ethane, ethylene and acetylene.
Skill 1.10 Describe the bonding in organic molecules using both the Valence Bond Theory and Molecular Orbital Theory.
Memorization Tasks (MT)
MT 1.1 Memorize the number of valence electrons in the atoms - C, H, N, O, and the halides.
MT 1.2 Memorize the number of bonds and lone pairs to atoms of carbon, hydrogen, oxygen and nitrogen that result in formal charges of zero. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.S%3A_Structure_and_Bonding_%28Summary%29.txt |
Chapter Objectives
This chapter provides a review of the more advanced material covered in a standard introductory chemistry course through a discussion of the following topics:
• the use of electronegativity to determine bond polarity, and the application of this knowledge to determine whether a given molecule possesses a dipole moment.
• the drawing and interpretation of organic chemical structures.
• the concept and determination of formal charge.
• resonance and drawing of resonance forms
• the Brønsted-Lowry and Lewis definitions of acids and bases, acidity constants and acid-base reactions.
• intermolecular forces
• 2.1: Polar Covalent Bonds - Electronegativity
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity, defined as the relative ability of an atom to attract electrons to itself in a chemical compound.
• 2.2: Polar Covalent Bonds - Dipole Moments
Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment.
• 2.3: Formal Charges
A formal charge is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.
• 2.4: Resonance
Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structure are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds.
• 2.5: Rules for Resonance Forms
The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite formal charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on one terminal oxygen can be delocalized by resonance through the other terminal oxygen.
• 2.6: Drawing Resonance Forms
Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer.
• 2.7: Acids and Bases - The Brønsted-Lowry Definition
In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur.
• 2.8: Acid and Base Strength
The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the dissociation constant, abbreviated Ka. The common base chosen for comparison is water.
• 2.9: Predicting Acid-Base Reactions from pKa Values
pKa values can be used to predict the equilibrium of an acid-base reaction. The equilibrium will favor the side with the weaker acid.
• 2.10: Organic Acids and Organic Bases
In the absence of pKa values, the relative strength of an organic acid can be predicted based on the stability of the conjugate base that it forms. The acid that forms the more stable conjugate base will be the stronger acid. The common factors that affect the conjugate base's stability are 1) the size and electronegativity of the the atom that has lost the proton, 2) resonance effects, 3) inductive effects, and 4) solvation effects.
• 2.11: Acids and Bases - The Lewis Definition
A broader definition is provided by the Lewis theory of acids and bases, in which a Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. This definition covers Brønsted-Lowry proton transfer reactions, but also includes reactions in which no proton transfer is involved.
• 2.12: Noncovalent Interactions Between Molecules
In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. The most common intermolecular forces in organic chemistry are from strongest to weakest are hydrogen bonds, dipole-dipole interactions, and London Dispersion (van der Waals) forces.
• 2.MM: Molecular Models
• 2.S: Polar Covalent Bonds; Acids and Bases (Summary)
02: Polar Covalent Bonds Acids and Bases
Objectives
After completing this section, you should be able to
1. describe how differences in electronegativity give rise to bond polarity.
2. arrange a given series of the elements most often encountered in organic chemistry (C, H, O, N, S, P and the halogens) in order of increasing or decreasing electronegativity, without referring to a table of electronegativities.
3. predict the partial positive and partial negative ends of a given bond formed between any two of the elements listed in Objective 2, above, without the use of a table of electronegativities or a periodic table.
4. predict the partial positive and partial negative ends of a given bond formed between any two elements not listed in Objective 2, above, using a periodic table.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electronegativity inductive effect
• polar covalent bond
Study Notes
Students often wonder why it is important to be able to tell whether a given bond is polar or not, and why they need to know which atoms carry a partial positive charge and which a partial negative charge. Consider the chloromethane (CH3Cl) molecule. The carbon atom is shown as carrying a partial positive charge. Now, recall that opposite charges attract. Thus, it seems reasonable that the slightly positive carbon atom in chloromethane should be susceptible to attack by a negatively charged species, such as the hydroxide ion, OH. This theory is borne out in practice: hydroxide ions react with chloromethane by attacking the slightly positive carbon atom in the latter. It is often possible to rationalize chemical reactions in this manner, and you will find the knowledge of bond polarity indispensible when you start to write reaction mechanisms.
Note: Because of the small difference in electronegativity between carbon and hydrogen, the C-H bond is normally assumed to be nonpolar.
Electronegativity
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity (represented by the Greek letter chi, χ, pronounced “ky” as in “sky”), which is defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by the neighboring atoms in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. Figure \(1\) shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling. In this scale a value of 4.0 is arbitrarily given to the most electronegative element, fluorine, and the other electronegativities are scaled relative to this value. In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0 as previously noted). It is important to notice that the elements most important to organic chemistry, carbon, nitrogen, and oxygen have some of the highest electronegativites in the periodic table (EN = 2.5, 3.0, 3.5 respectively). Metals, on the left, tend to be less electronegative elements, with cesium having the lowest (EN = 0.7). Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell.
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The larger the electronegativity value, the greater the attraction.
Electronegativity and Bond Type
The two idealized extremes of chemical bonding: (1) ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and (2) covalent bonding, in which electrons are shared equally between two atoms. Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Electronegativity determines how the shared electrons are distributed between the two atoms in a polar covalent bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms. Recall that a lowercase Greek delta ( δ ) is used to indicate that a bonded atom possesses a partial positive charge, indicated by δ+ , or a partial negative charge, indicated by δ , and a bond between two atoms that possess partial charges is a polar bond.
Whether a bond is ionic, nonpolar covalent, or polar covalent can be estimated by by calculating the absolute value of the difference in electronegativity (ΔEN) of two bonded atoms. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds H–H, H–Cl, and Na–Cl are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 7.2.4 shows the relationship between electronegativity difference and bond type. This table is just a general guide, however, with many exceptions. The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.
Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as OH, NO3, and NH4+, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, KNO3, contains the K+ cation and the polyatomic NO3 anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions K+ and NO3, as well as covalent between the nitrogen and oxygen atoms in NO3.
Example \(1\): Electronegativity and Bond Polarity
Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Table A2, arrange the following covalent bonds—all commonly found in amino acids—in order of increasing polarity. Then designate the positive and negative atoms using the symbols δ+ and δ–:
C–H, C–N, C–O, N–H, O–H, S–H
Solution
The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the δ– designation is the more electronegative of the two. Table \(1\) shows these bonds in order of increasing polarity.
Table \(1\): Bond Polarity and Electronegativity Difference
Bond ΔEN Polarity
C–H 0.4 \(\overset{δ−}{\ce C}−\overset{δ+}{\ce H}\)
S–H 0.4 \(\overset{δ−}{\ce S}−\overset{δ+}{\ce H}\)
C–N 0.5 \(\overset{δ+}{\ce C}−\overset{δ−}{\ce N}\)
N–H 0.9 \(\overset{δ−}{\ce N}−\overset{δ+}{\ce H}\)
C–O 1.0 \(\overset{δ+}{\ce C}−\overset{δ−}{\ce O}\)
O–H 1.4 \(\overset{δ−}{\ce O}−\overset{δ+}{\ce H}\)
Visualizing Bonding
Calculated charge distributions in molecules can easily be visualized by using electrostatic potential maps. The color red is used to indicate electron-rich regions of a molecule while the color blue is used to indicated electron-poor regions. An easier method for visually representing electron displacement in a molecule uses a crossed arrow. By convention the arrow point in the direction of the electron-rich region of a molecule and away from the electron-poor. An example is shown in the molecule fluoromethane. The C-F bond is polarized drawing the bonding electrons toward the more electronegative fluorine giving it a partial negative charge. Consequently, the bonding electrons are drawn away from the less electronegative carbon giving it a partial positive charge. The the electron-rich fluorine is shown as red in the electrostatic potential map and while the electron-poor carbon is shown as blue. The crossed arrow points in the direction of the electron-rich fluorine.
Electrostatic Potential Map and Dipole Moment of Fluoromethane
Chemists often use the term, inductive effect, to describe the shifting of electrons in a sigma by the electronegativity of atoms. Relatively electronegative atoms, such as fluorine, tend to inductively draw electrons towards themselves and away from nearby atoms. The inductive effect will be used to explain chemical reactivity in many situations in organic chemistry. An excellent example of the inductive effect is seen when comparing the O-H bond polarities of water (H2O) and hypochlorous acid (ClOH). Replacing the less electronegative hydrogen (EN = 2.1) in water with the more electronegative chlorine (EN = 3.0) in hypochlorous acid creates a greater bond polarity. The chlorine draws electrons away giving the hydrogen a greater partial positive charge. This is shown in the electrostatic potential map as an increase in the blue color around hydrogen.
water hypochlorous acid
A "spectrum" of bonds
There is no clear-cut division between covalent and ionic bonds. In a pure non-polar covalent bond, the electrons are held on average exactly half way between the atoms. In a polar bond, the electrons have been dragged slightly towards one end. How far does this dragging have to go before the bond counts as ionic? There is no real answer to that. Sodium chloride is typically considered an ionic solid, but even here the sodium has not completely lost control of its electron. Because of the properties of sodium chloride, however, we tend to count it as if it were purely ionic. Lithium iodide, on the other hand, would be described as being "ionic with some covalent character". In this case, the pair of electrons has not moved entirely over to the iodine end of the bond. Lithium iodide, for example, dissolves in organic solvents like ethanol - not something which ionic substances normally do. Many bonds between metals and non-metal atoms, are considered ionic, however some of these bonds cannot be simply identified as one type of bond. Examples of this are the lithium - carbon bond in methyllithium which is usually considered as polar covalent (somewhat between covalent and ionic) and the potassium - oxygen bond in potassium tert-butoxide which is considered more ionic than covalent.
methyllithium potassium tert-butoxide
Summary
Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.
• No electronegativity difference between two atoms leads to a non-polar covalent bond.
• A small electronegativity difference leads to a polar covalent bond.
• A large electronegativity difference leads to an ionic bond.
Exercises
1. Identify the positive and negative ends of each of the bonds shown below.
2. Which of the following elements is the more electronegative?
a) Br or C
b) C or H
c) Cl or I
d) C or Li
3. Which of the following molecules would you expect to have the more polarized O-H bond?
4. Predict the direction of polarizing C-O bond in methanol by looking at its electrostatic potential map.
Solutions
a) Br
b) C
c) Cl
d) C
3.
The molecule on the right would have the more polorized O-H bond. The presence of the highly electronegative fluorines would draw electrons away by the inductive effect.
4. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.01%3A_Polar_Covalent_Bonds_-_Electronegativity.txt |
Learning Objectives
After completing this section, you should be able to
1. explain how dipole moments depend on both molecular shape and bond polarity.
2. predict whether a molecule will possess a dipole moment from the molecular formula or structure.
3. use the presence or absence of a dipole moment as an aid to deducing the structure of a given compound.
Key Terms
Make certain that you can define, and use in context, the key term below.
• dipole moment
Study Notes
You must be able to combine your knowledge of molecular shapes and bond polarities to determine whether or not a given compound will have a dipole moment. Conversely, the presence or absence of a dipole moment may also give an important clue to a compound’s structure. BCl3, for example, has no dipole moment, while NH3 does. This suggests that in BCl3 the chlorines around boron are in a trigonal planar arrangement, while the hydrogens around nitrogen in NH3 have a less symmetrical arrangement - trigonal pyramidal.
Remember that the C-H bond is assumed to be non-polar.
Molecular Dipole Moments
In molecules containing more than one polar bond, the molecular dipole moment is just the vector combination of what can be regarded as individual "bond dipole moments". Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (Figure $\PageIndex{1a}$). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (Figure $\PageIndex{1b}$); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules.
The following is a simplified equation for a simple separated two-charge system that is present in diatomic molecules or when considering a bond dipole within a molecule.
$\mu_{diatomic} = Q \times r \label{1a}$
This bond dipole, µ (Greek mu) is interpreted as the dipole from a charge separation over a distance $r$ between the partial charges $Q^+$ and $Q^-$ (or the more commonly used terms $δ^+$ - $δ^-$); the orientation of the dipole is along the axis of the bond. The units on dipole moments are typically debyes (D) where one debye is equal to 3.336 x 1030 coulomb meters (C · m) in SI units. Consider a simple system of a single electron and proton separated by a fix distance. The unit charge on an electron is 1.60 X 1019 C and the proton & electron are 100 pm apart (about the length of a typical covalent bond), the dipole moment is calculated as:
\begin{align} \mu &= Qr \nonumber \[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align}
\begin{align} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \[4pt] &= 4.80\; D \label{3} \end{align}
$4.80\; D$ is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. However, if the charge separation were increased then the dipole moment increases (linearly):
• If the proton and electron were separated by 120 pm:
$\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \nonumber$
• If the proton and electron were separated by 150 pm:
$\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \nonumber$
• If the proton and electron were separated by 200 pm:
$\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \nonumber$
Example $1$: Water
The water molecule in Figure $1$ can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).
The bond moment of O-H bond =1.5 D, so the net dipole moment is
$\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber$
Dipoles with Differing Atoms
In more complex molecules with more than one polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether the molecule has a net dipole moment. The manner in which the individual bonds contribute to the dipole moment of the molecule is nicely illustrated by the series of chloromethanes shown below. The electronegative chlorine draws electrons towards itself.
Consider $CCl_4$, (left panel in figure below), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right. In contrast, $CHCl_3$ is a polar molecule (right panel in figure above). However, although a molecule like CHCl3 has a tetrahedral geometry, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments do not cancel one another, and the result is a molecule which has a dipole moment. The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
Electron densities in a molecule (and the dipole moments that unbalanced electron distributions can produce) are easily visualized with electrostatic potential maps. In this example, $CHCl_3$, the blue area centered on carbon represents a electron-deficient positive area, of the molecule, the red area centered on the chlorine, represents a electron-abundant negative area. This charge separation creates a net dipole moment of 1.87 D which points in the direction of the chlorine. In, $CCl_4$ the evenly spaced red areas represent that there is no separation of charge in the molecule. $CCl_4$ has a net dipole moment of zero which makes it a nonpolar molecule.
chloromethane tetrachloromethane
(μ = 1.87 D) (μ = 0 D)
Figure $4$: Electrostatic potential maps and dipole moments for chloromethane and tetrachloromethane (carbon tetrachloride).
Molecules with asymmetrical charge distributions have a net dipole moment
Other examples of molecules with polar bonds are shown in Figure $2$. In molecules like BCl3 and CCl4, that have only one type of bond and a molecular geometries that are highly symmetrical (trigonal planar and tetrahedral), the individual bond dipole moments completely cancel, and there is no net dipole moment. However, although a molecule like CHCl3 has a tetrahedral geometry, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments do not cancel one another, and the result is a molecule which has a dipole moment.
net dipole no net dipole net dipole net dipole no net dipole net dipole no net dipole no net dipole
HCl BCl3 CH2O NH3 CCl4 CHCl3 PF5 SF6
Figure $5$: Molecules with Polar Bonds. Individual bond dipole moments are indicated in black. Due to their different three-dimensional structures, some molecules with polar bonds have a net dipole moment (HCl, CH2O, NH3, and CHCl3), indicated in red, whereas others do not because the bond dipole moments cancel (BCl3, CCl4, PF5, and SF6).
Table $1$: Dipole Moments of Some Compounds
Compound Dipole Moment (Debyes)
NaCl 9.0 (measured in the gas phase)
CH2O 2.33
CH3Cl 1.87
H2O 1.85
CH3OH 1.70
NH3 1.47
CH3NH2 1.31
CO2 0
CCl4 0
CH4 0
CH3CH3 0
The table above give the dipole moment of some common substances. Sodium chloride has the largest dipole listed (9.00 D) because it is an ionic compounds. Even small organic compounds such as formaldehyde (CH2O, 2.33 D) and methanol (CH3OH, 1.70 D) have significant dipole moments. Both of these molecules contain the strongly electronegative oxygen atom lone pair electrons which gives rise to considerable dipole moments.
methanol formaldehyde
(μ = 1.70 D) (μ = 2.33 D)
In contrast many organic molecule have a zero dipole moment despite the fact that they are made up of polar covalent bonds. In, structures with highly symmetrical molecular geometries, the polar bonds and the lone pair electrons can can exactly cancel leaving no overall charge separation.
carbon dioxide carbon tetrachloride ethane
(μ = 0 D) (μ = 0 D) (μ = 0 D)
Exercise $1$
Which of the molecules below have molecular dipole moments?
Answer
Only molecule (b) does not have a molecular dipole, due to its symmetry (bond dipoles are equal and in opposite directions). Add texts here.
Exercise $2$
Draw out the line structure of the molecule with a molecular formula of C2Cl4. Indicate all of the individual bond polarities and predict if the molecule is polar or nonpolar.
Answer
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl2C=CCl2" id="MathJax-Element-47-Frame" role="presentation" style="position:relative;" tabindex="0">does not have a net dipole moment.Cl2C=CCl2
Exercise $3$
The following molecule has no net dipole moment, explain.
Answer
The hydroxyl groups are oriented opposite of one another and therefore the dipole moments would “cancel” one another out. Therefore having a zero net-dipole.
Exercise $4$
Within reactions with carbonyls, such as a hydride reduction reaction, the carbonyl is attacked from the carbon side and not the oxygen side. Using knowledge of electronegativity explain why this happens.
Answer
The oxygen is more electronegative than the carbon and therefore creates a dipole along the bond. This leads to having a partial positive charge on the carbon and the reduction can take place.
Exercise $1$
Which molecule(s) has a net dipole moment?
1. H2S
2. NHF2
3. BF3
Answer
Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of H2S is bent. The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF2 cannot cancel one another. This means that NHF2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
• The molecular geometry of BF3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF3 has a net dipole moment of zero:
Additional Exercises
1. Determine whether each of the compounds listed below possesses a dipole moment. For the polar compounds, indicate the direction of the dipole moment.
1. $\ce{O=C=O}$
2. $ICl$
3. $SO_2$
4. $\ce{CH3-O-CH3}$
5. $\ce{CH3C(=O)CH3}$
Answers: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.02%3A_Polar_Covalent_Bonds_-_Dipole_Moments.txt |
Objectives
After completing this section, you should be able to
1. calculate the formal charge of an atom in an organic molecule or ion.
2. identify and recognize the bonding patterns for atoms of carbon, hydrogen, oxygen, nitrogen and the halogens that have a formal charge of zero.
Key Terms
Make certain that you can define, and use in context, the key term below.
• valence electrons
• bonding and non-bonding electrons
• formal charge
• carbocations
Study Notes
It is more important that students learn to easily identify atoms that have formal charges of zero, than it is to actually calculate the formal charge of every atom in an organic compound. Students will benefit by memorizing the "normal" number of bonds and non-bonding electrons around atoms whose formal charge is equal to zero.
Determining the Formal Charge on an Atom
A formal charge ($FC$) compares the number of electrons around a "neutral atom" (an atom not in a molecule) versus the number of electrons around an atom in a molecule. Formal charge is assigned to an atom in a molecule by assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
• Non-bonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the two bonded atoms, so one electron from each bond goes to each atom.
The formal charge of each atom in a molecule can be calculated using the following equation:
$FC = \text{(# of valence electrons in free atom)} − \text{(# of lone-pair electrons)} −\dfrac{1}{2} (\text{# of bonding electrons}) \label{2.3.1}$
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia ($\ce{NH3}$) whose Lewis structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From the Lewis structure, the nitrogen atom in ammonia has one lone pair and three bonds with hydrogen atoms. Substituting into Equation \ref{2.3.1}, we obtain
\begin{align*} FC (N) &= (\text{5 valence electrons}) − (\text{2 lone pair electrons}) − \dfrac{1}{2} (\text{6 bonding electrons}) \[4pt] &= 0 \end{align*}
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule has no non-bonding electrons and one bond. Using Equation \ref{2.3.1} to calculate the formal charge on hydrogen, we obtain
\begin{align*} FC (H) &= (\text{1 valence electrons}) − (\text{0 lone pair electrons}) − \dfrac{1}{2} (\text{2 bonding electrons}) \[4pt] &= 0 \end{align*}
The sum of the formal charges of each atom must be equal to the overall charge of the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall neutral charge of the $\ce{NH3}$ molecule.
Typically, the structure with the most formal charges of zero on atoms is the more stable Lewis structure. In cases where there MUST be positive or negative formal charges on various atoms, the most stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges for polyatomic ions.
Example 2.3.1
Calculate the formal charges on each atom in the $\ce{NH4^{+}}$ ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the $\ce{NH4^{+}}$ ion. Use the Lewis electron structure of $\ce{NH4^{+}}$ to identify the number of bonding and non-bonding electrons associated with each atom and then use Equation \ref{2.3.1} to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the $\ce{NH4^{+}}$ ion is as follows:
The nitrogen atom in ammonium has zero non-bonding electrons and 4 bonds. Using Equation \ref{2.3.1}, the formal charge on the nitrogen atom is therefore
\begin{align*} FC (N) &= (\text{5 valence electrons}) − (\text{0 lone pair electrons}) − \dfrac{1}{2} (\text{8 bonding electrons}) \[4pt] &= +1 \end{align*}
Each hydrogen atom in has one bond and zero non-bonding electrons. The formal charge on each hydrogen atom is therefore
\begin{align*} FC (H) &= (\text{1 valence electrons}) − (\text{0 lone pair electrons}) − \dfrac{1}{2} (\text{2 bonding electrons}) \[4pt] &= 0 \end{align*}
The formal charges on the atoms in the $\ce{NH4^{+}}$ ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = 1+, which is the same as the total charge of the ammonium polyatomic ion.
Exercise $1$
Write the formal charges on all atoms in $\ce{BH4^{−}}$.
Answer
.
Determining the Charge of Atoms in Organic Structures
The calculation method reviewed above for determining formal charges on atoms is an essential starting point for a novice organic chemist, and works well when dealing with small structures. But this method becomes unreasonably time-consuming when dealing with larger structures. It would be exceptionally tedious to determine the formal charges on each atom in 2'-deoxycytidine (one of the four nucleoside building blocks that make up DNA) using Equation \ref{2.3.1}. As you get more experience with organic structures, you will be able to quickly look at this type of complicated structure and determine charges on each atom.
2'-deoxycytidine
You need to develop the ability to quickly and efficiently draw large structures and determine formal charges. Fortunately, this only requires some practice with recognizing common bonding patterns.
Organic chemistry only deals with a small part of the periodic table, so much so that it becomes convenient to be able to recognize the bonding forms of these atoms. The figure below contains the most important bonding forms. These will be discussed in detail below. An important idea to note is most atoms in a molecule are neutral. Pay close attention to the neutral forms of the elements below because that is how they will appear most of the time.
Figure $1$: Structures of common organic atoms and ions.
Atom Positive Neutral Negative
C
N
O
Cl (halogens)
Carbon
Carbon, the most important element for organic chemists. In the structures of methane, methanol, ethane, ethene, and ethyne, there are four bonds to the carbon atom. And each carbon atom has a formal charge of zero. In other words, carbon is tetravalent, meaning that it commonly forms four bonds.
Carbon usually makes four bonds
Carbon is tetravalent in most organic molecules, but there are exceptions. Later in this chapter and throughout this book are examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom has a positive or negative formal charge, respectively. Carbocations occur when a carbon has only three bonds and no lone pairs of electrons. Carbocations have only 3 valence electrons and a formal charge of 1+. Carbanions occur when the carbon atom has three bonds plus one lone pair of electrons. Carbanions have 5 valence electrons and a formal charge of 1−.
carbocation: 3 bonds & no lone pair carbanion: 3 bonds & one lone pair carbon radical: 3 bonds & one unpaired electron
Two other possibilities are carbon radicals and carbenes, both of which have a formal charge of zero. A carbon radical has three bonds and a single, unpaired electron. Carbon radicals have 4 valence electrons and a formal charge of zero. Carbenes are a highly reactive species, in which a carbon atom has two bonds and one lone pair of electrons, giving it a formal charge of zero. Though carbenes are rare, you will encounter them in section 8.10 Addition of Carbenes to Alkenes.
You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify carbons that bear positive and negative formal charges by a quick inspection.
Hydrogen
The common bonding pattern for hydrogen is easy: hydrogen atoms in organic molecules typically have only one bond, no unpaired electrons and a formal charge of zero. The exceptions to this rule are the proton, H+, the hydride ion, H-, and the hydrogen radical, H.. The proton is a hydrogen with no bonds and no lone pairs and a formal charge of 1+. The hydride ion is a is a hydrogen with no bonds, a pair of electrons, and a formal charge of 1−. The hydrogen radical is a hydrogen atom with no bonds, a single unpaired electron and a formal charge of 0. Because this book concentrates on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge.
Neutral Hydrogen: one bond, no lone pair
Oxygen
The common arrangement of oxygen that has a formal charge of zero is when the oxygen atom has 2 bonds and 2 lone pairs. Other arrangements are oxygen with 1 bond and 3 lone pairs, that has a 1− formal charge, and oxygen with 3 bonds and 1 lone pair that has a formal charge of 1+. All three patterns of oxygen fulfill the octet rule.
neutral oxygen: 2 bonds & 2 lone pairs negative oxygen: 1 bond & 3 lone pairs positive oxygen: 3 bonds & one lone pair
If it has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of 1−. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of 1+.
Oxygen can also exist as a radical, such as where an oxygen atom has one bond, two lone pairs, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for most oxygen containing molecules you will encounter in organic chemistry.
Nitrogen
Nitrogen has two major bonding patterns, both of which fulfill the octet rule:
neutral nitrogen: 3 bonds & 1 lone pair positive nitrogen: 4 bonds negative nitrogen: 2 bonds & 2 lone pairs
If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of 1+. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs.
Phosphorus and Sulfur
Two third row elements are commonly found in biological organic molecules: phosphorus and sulfur. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO43), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that elements in the third row of the periodic table have d orbitals in their valence shell as well as s and p orbitals, and thus are not bound by the octet rule.
Halogens
The halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of 1+.
Common Neutral Bonding Patterns for Halogens
Common Positive Bonding Pattern for Halogens
These rules, if learned and internalized so that you don’t even need to think about them, will allow you to draw large organic structures, complete with formal charges, quite quickly.
Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear.
Exercise $2$
Please identify an atom with a non-neutral charge in the following atom:
Answer
Drawing the Lewis Structure of Ionic Molecular Compounds
The hydroxide ion, OH-, is drawn simply by showing the oxygen atom with its six valence electrons, then adding one more electron to account for the negative charge. By changing the number of valence electrons the bonding characteristic of oxygen are now changed. Now the oxygen has three non-bonding lone pairs, and can only form one bond to a hydrogen.
To draw a Lewis structure of the hydronium ion, H3O+, you again start with the oxygen atom with its six valence electrons, then take one away to account for the positive charge to give oxygen five valence electrons. The oxygen has one non-bonding lone pair and three unpaired electrons which can be used to form bonds to three hydrogen atoms.
Using Formal Charges to Distinguish between Lewis Structures
As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO2. Both structures conform to the rules for Lewis electron structures.
CO2
1. C is less electronegative than O, so it is the central atom.
2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.
3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.
4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:
This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.
5. No electrons are left for the central atom.
6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:
Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO2 are as follows:
Both Lewis structures have a net formal charge of zero, but the structure on the right has a 1+ charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.
Note
The Lewis structure with the set of formal charges closest to zero is usually the most stable.
Example 2.3.2: Thiocyanate Ion
The thiocyanate ion ($\ce{SCN^{−}}$), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.
Given: chemical species
Asked for: Lewis electron structures, formal charges, and preferred arrangement
Strategy:
A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN.
B Calculate the formal charge on each atom using Equation \ref{2.3.1}.
C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.
Solution:
A Possible Lewis structures for the SCN ion are as follows:
B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore 6 - (6 + 2/2) = 1−. In (b), the sulfur atom has a formal charge of 0. In (c), the sulfur atom has a formal charge of 1+. Continuing with the nitrogen, we observe that in (a) the nitrogen atom shares three bonding pairs and has one lone pair and has a total of 5 valence electrons. The formal charge on the nitrogen atom is therefore 5 - (2 + 6/2) = 0. In (b), the nitrogen atom has a formal charge of 1−. In (c), the nitrogen atom has a formal charge of 2−.
C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, 1− versus 1+, 2−.
Exercise $2$: Fulminate Ion
Salts containing the fulminate ion ($\ce{CNO^{−}}$) are used in explosive detonators. Draw three Lewis electron structures for $\ce{CNO^{−}}$ and use formal charges to predict which is more stable. (Note: $\ce{N}$ is the central atom.)
Answer
The second structure is predicted to be the most stable.
Exercises
1. Draw the Lewis structure of each of these molecules: $\ce{CH3^{+}}$, $\ce{NH2^{-}}$, $\ce{CH3^{-}}$, $\ce{NH4^{+}}$, $\ce{BF4^{-}}$. In each case, use the method of calculating formal charge described to satisfy yourself that the structures you have drawn do in fact carry the charges shown.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.03%3A_Formal_Charges.txt |
Objective
After completing this section, you should be able to
• draw resonance forms for molecules and ions.
Key Terms
Make certain that you can define, and use in context, the key term below.
• resonance form
• delocalization
Resonance Delocalization
Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance contributors or canonical forms). Resonance contributors involve the ‘imaginary movement’ of pi-bonded electrons or of lone-pair electrons that are adjacent to pi bonds. Note, sigma bonds cannot be broken during resonance – if you show a sigma bond forming or breaking, you are showing a chemical reaction taking place. Likewise, the positions of atoms in the molecule cannot change between resonance contributors.
When looking at the structure of the molecule, formate, we see that there are two equivalent structures possible. Which one is correct? There are two simple answers to this question: 'both' and 'neither one'. Both ways of drawing the molecule are equally acceptable approximations of the bonding picture for the molecule, but neither one, by itself, is an accurate picture of the delocalized pi bonds. The two alternative drawings, however, when considered together, give a much more accurate picture than either one on its own. This is because they imply, together, that the carbon-carbon bonds are not double bonds, not single bonds, but about halfway in between.
Formate Ion Structures are Equivalent in Energy
When it is possible to draw more than one valid structure for a compound or ion, we have identified resonance contributors: two or more different Lewis structures depicting the same molecule or ion that, when considered together, do a better job of approximating delocalized pi-bonding than any single structure. By convention, resonance contributors are linked by a double-headed arrow, and are sometimes enclosed by brackets:
The depiction of formate using the two resonance contributors A and B in the figure above does not imply that the molecule at one moment looks like structure A, then at the next moment shifts to look like structure B. Rather, at all moments, the molecule is a combination, or resonance hybrid of both A and B. Each individual resonance contributor of the formate ion is drawn with one carbon-oxygen double bond (120 pm) and one carbon-oxygen single bond (135 pm), with a negative formal charge located on the single-bonded oxygen. However, the two carbon-oxygen bonds in formate are actually the same length (127 pm) which implies that neither resonance contributor is correct. Although there is an overall negative formal charge on the formate ion, it is shared equally between the two oxygens. Therefore, the formate ion can be more accurately depicted by a pair of resonance contributors. Alternatively, a single structure can be used, with a dashed line depicting the resonance-delocalized pi bond and the negative charge located in between the two oxygens.
The electrostatic potential map of formate shows that there is an equal amount of electron density (shown in red) around each oxygen.
Valence bond theory can be used to develop a picture of the bonding in a carboxylate group. We know that the carbon must be sp2-hybridized, (the bond angles are close to 120˚, and the molecule is planar), and we will treat both oxygens as being sp2-hybridized as well. Both carbon-oxygen sigma bonds, then, are formed from the overlap of carbon sp2 orbitals and oxygen sp2 orbitals.
In addition, the carbon and both oxygens each have an un-hybridized 2pz orbital situated perpendicular to the plane of the sigma bonds. These three 2pz orbitals are parallel to each other, and can overlap in a side-by-side fashion to form a delocalized pi bond.
Overall, the situation is one of three parallel, overlapping 2pz orbitals sharing four delocalized pi electrons. Because there is one more electron than there are 2pz orbitals, the system has an overall charge of 1–. Resonance contributors are used to approximate overlapping 2pz orbitals and delocalized pi electrons. Molecules with resonance are usually drawn showing only one resonance contributor for the sake of simplicity. However, identifying molecules with resonance is an important skill in organic chemistry.
This example shows an important exception to the general rules for determining the hybridization of an atom. The oxygen with the negative charge appears to be sp3 hybridized because it is surrounded by four electron groups. However, this representation of the oxygen atom is not correct because it is actually part of a resonance hybrid. A pair of lone pair of electrons on the negatively charged oxygen are not localized in an sp3 orbital, rather, they are delocalized as part of a conjugated pi system. The stability gained though resonance is enough to cause the expected sp3 to become sp2. The sp2 hybridization gives the oxygen a p orbital allowing it to participate in conjugation. As a general rule sp3 hybridized atoms with lone pair electrons tend to become sp2 hybridized when adjacent to a conjugated system.
Example: Carbonate (CO32−)
Like formate, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the Lewis structures describing CO32 has three equivalent representations.
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the 2− charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the 2− charge:
5. There are no electrons left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with formate, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
As the case for formate, the actual structure involves the formation of a molecular orbital from pz orbitals centered on each atom and sitting above and below the plane of the CO32 ion.
Example 2.4.1
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (C6H6) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
A Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
B Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
C Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of 1+, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pair-wise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of 1−, while three carbon atoms have only 6 electrons and a formal charge of 1+. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
This combination of p orbitals for benzene can be visualized as a ring with a node in the plane of the carbon atoms. As can be seen in an electrostatic potential map of benzene, the electrons are distributed symmetrically around the ring.
Exercise \(1\)
The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO2).
Answer
Extra example - O3
A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical forms). Such is the case for ozone (O3), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. If the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures . The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The resonance structure of ozone involves a molecular orbital extending all three oxygen atoms.In ozone, a molecular orbital extending over all three oxygen atoms is formed from three atom centered pz orbitals. Similar molecular orbitals are found in every resonance structure. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.04%3A_Resonance.txt |
Objectives
After completing this section, you should be able to
• use the concept of resonance to explain structural features of molecules and ions.
• understand the relationship between resonance and relative stability of molecules and ions.
Rules for Drawing and Working with Resonance Contributors
Recognizing, drawing, and evaluating the relative stability of resonance contributors is essential to understanding organic reaction mechanisms. When learning to draw and interpret resonance structures, there are a few basic guidelines to help. .
1) There is ONLY ONE REAL STRUCTURE for each molecule or ion. This real structure (the resonance hybrid) takes its character from the average of all the individual resonance contributors. When looking at a resonance contributors, we are seeing the exact same molecule or ion depicted in different ways. Resonance hybrids are really a single, unchanging structure.
Major resonance contributors of the formate ion
Representations of the formate resonance hybrid
2) The resonance hybrid is more stable than any individual resonance structures. Often, resonance structures represent the movement of a charge between two or more atoms. The charge is spread out amongst these atoms and therefore more stabilized. When looking at the picture above the resonance contributors represent the negative charge as being on one oxygen or the other. The resonance hybrid shows the negative charge being shared equally between two oxygens. In the resonance hybrid, the negative charge is spread out over a larger part of the molecule and is therefore more stable.
3) Resonance contributors do not have to be equivalent. Because of this, resonance structures do necessarily contribute equally to the resonance hybrid. The two resonance structures shown below are not equivalent because one show the negative charge on an oxygen while the other shows it on a carbon. Later, we will show that the contributor with the negative charge on the oxygen is the more stable of the two. Also, this means that the resonance hybrid will not be an exact mixture of the two structures.
4) All resonance contributors must be correct Lewis structures. Each atom should have a complete valence shell and be shown with correct formal charges. A carbocation (carbon with only 6 valence electrons) is the only allowed exception to the valence shell rules. The structure below is an invalid resonance structure even though it only shows the movement of a pi bond. The resulting structure contains a carbon with ten electrons, which violates the octet rule, making it invalid.
5) All resonance contributors must have the same molecular formula, the same number of electrons, and same net charge. The molecules in the figure below are not resonance structures of the same molecule because then have different molecular formulas (C2H5NO Vs. C2H6NO). Also, the two structures have different net charges (neutral Vs. positive).
6) Resonance contributors only differ by the positions of pi bond and lone pair electrons. Sigma bonds are never broken or made, because of this atoms must maintain their same position. The molecules in the figure below are not resonance structures of the same molecule even though they have the same molecular formula (C3H6O). These molecules are considered structural isomers because their difference involves the breaking of a sigma bond and moving a hydrogen atom.
Major and Minor Resonance Contributors
As previously state the true structure of a resonance hybrid is the combination of all the possible resonance structures. If the resonance structures are equal in stability they the contribute equally to the structure of the hybrid. However, if the resonance structures have different stabilities they contribute to the hybrid's structure in proportions related to their relative stabilities. It can be said the the resonance hybrid's structure resembles the most stable resonance structure. Because of this it is important to be able to compare the stabilities of resonance structures. In the example below, structure B is much less important in terms of its contribution to the hybrid because it contains the violated octet of a carbocation. The relative stabilities of the two structures are so vastly different that molecules which contain a C=O bond are almost exclusively written in a form like structure A. However, as will learn in chapter 19, the positively charged carbon created by structure B will explain how the C=O bond will react with electron rich species.
Rules for Estimating Stability of Resonance Structures
1. The resonance structures in which all atoms have complete valence shells is more stable. This means most atoms have a full octet. In the example below structure A has a carbon atom with a positive charge and therefore an incomplete octet. Based on this criterion, structure A is less stable and is a more minor contributor to the resonance hybrid than structure B.
2. The structures with the least number of formal charges is more stable. Based on this, structure B is less stable because is has two atoms with formal charges while structure A has none. Structure A would be the major resonance contributor.
3. The structures with a negative charge on the more electronegative atom will be more stable. The difference between the two resonance structures is the placement of a negative charge. Structure B is the more stable and the major resonance contributor, because it places the negative charge on the more electronegative oxygen.
4. The structures with a positive charges on the least electronegative atom (most electropositive) is more stable.
5. The structures with the least separation of formal charges is more stable. The only difference between the two structures below are the relative positions of the positive and negative charges. In structure A the charges are closer together making it more stable.
6. Resonance forms that are equivalent have no difference in stability. When looking at the two structures below no difference can be made using the rules listed above. This means the two structures are equivalent in stability and would make equal structural contributions to the resonance hybrid.
Example 1:
Example 2:
Example 3:
Carboxylate example
In the case of carboxylates, contributors A and B below are equivalent in terms of their relative contribution to the hybrid structure. However, there is also a third resonance contributor C, in which the carbon bears a positive formal charge (a carbocation) and both oxygens are single-bonded and bear negative charges.
Structure C makes a less important contribution to the overall bonding picture of the group relative to A and B. How do we know that structure C is the ‘minor’ contributor? Apply the rules below
• The carbon in contributor C does not have an octet. In general, resonance contributors in which a carbon does not fulfill the octet rule are relatively less important. (rule #1)
• In structure C, there are only three bonds, compared to four in A and B. In general, a resonance structure with a lower number of total bonds is relatively less important. (rule #2)
• Structure C also has more formal charges than are present in A or B. In general, resonance contributors in which there is more/greater separation of charge are relatively less important. (rule #3)
• Structures A and B are equivalent and will be equal contributors to the resonance hybrid. (rule #5).
• The resonance contributor in which a negative formal charge is located on a more electronegative atom, usually oxygen or nitrogen, is more stable than one in which the negative charge is located on a less electronegative atom such as carbon. An example is in the upper left expression in the next figure. (rule #4)
Example with 3 resonance structures
Molecules with more than 2 resonance structures can also be considered using the rules listed above. Of the resonance structures listed below, structure A would be the most stable because all the non-hydrogen atoms have a full octet and the negative charge is on the more electronegative atom (oxygen). Structure C would be next in stability because all of the non-hydrogen atoms have full octets, though now the negative charge is on carbon rather than oxygen. Structure B would be the least stable of the three because it has the carbocation does not have an octet. Of the three, structure A would be the major resonance structure and would most resemble the structure of the true resonance hybrid. Structure B is considered a minor resonance contributor and would have very little effect on the structure of the resonance hybrid.
Example \(1\)
Draw the major resonance contributor of the structure below. Include in your figure the appropriate curved arrows showing how you got from the given structure to your structure. Explain why your contributor is the major one. In what kind of orbitals are the two lone pairs on the oxygen?
Solution
In the structure above, the carbon with the positive formal charge does not have a complete octet of valence electrons. Using the curved arrow convention, a lone pair on the oxygen can be moved to the adjacent bond to the left, and the electrons in the double bond shifted over to the left (see the rules for drawing resonance contributors to convince yourself that these are 'legal' moves).
The resulting resonance contributor, in which the oxygen bears the formal charge, is the major one because all atoms have a complete octet, and there is one additional bond drawn (resonance rules #1 and #2 both apply). This system can be thought of as four parallel 2p orbitals (one each on C2, C3, and C4, plus one on oxygen) sharing four pi electrons. One lone pair on the oxygen is in an unhybridized 2p orbital and is part of the conjugated pi system, and the other is located in an sp2 orbital.
Also note that one additional contributor can be drawn, but it is also minor because it has a carbon with an incomplete octet:
Exercises
1) For the following resonance structures please rank them in order of stability. Indicate which would be the major contributor to the resonance hybrid.
2) Draw four additional resonance contributors for the molecule below. Label each one as major or minor (the structure below is of a major contributor).
3) Draw three resonance contributors of methyl acetate (an ester with the structure CH3COOCH3), and order them according to their relative importance to the bonding picture of the molecule. Explain your reasoning.
4) Below is a minor resonance contributor of a species known as an ‘enamine’, which we will study more in Section 19.8 (formation of enamines) Section 23.12 (reactions of enamines). Draw the major resonance contributor for the enamine, and explain why your contributor is the major one.
5) Draw the major resonance contributor for each of the anions below:
Solutions
1) Structure I would be the most stable because all the non-hydrogen atoms have a full octet and the negative charge is on the more electronegative nitrogen. Structure III would be the next in stability because all of the non-hydrogen atoms have full octets. Structrure II would be the least stable because it has the violated octet of a carbocation.
2)
3)
The contributor on the left is the most stable: there are no formal charges.
The contributor on the right is least stable: there are formal charges, and a carbon has an incomplete octet.
The contributor in the middle is intermediate stability: there are formal charges, but all atoms have a complete octet.
4) This contributor is major because there are no formal charges.
5)
Recognizing Resonance
Resonance contributors involve the ‘imaginary movement’ of pi-bonded electrons or of lone-pair electrons that are adjacent to (i.e. conjugated to) pi bonds. You can never shift the location of electrons in sigma bonds – if you show a sigma bond forming or breaking, you are showing a chemical reaction taking place. Likewise, the positions of atoms in the molecule cannot change between two resonance contributors.
Because benzene will appear throughout this course, it is important to recognize the stability gained through the resonance delocalization of the six pi electrons throughout the six carbon atoms. Benzene also illustrates one way to recognize resonance - when it is possible to draw two or more equivalent Lewis structures. If we were to draw the structure of an aromatic molecule such as 1,2-dimethylbenzene, there are two ways that we could draw the double bonds:
Which way is correct? There are two simple answers to this question: 'both' and 'neither one'. Both ways of drawing the molecule are equally acceptable approximations of the bonding picture for the molecule, but neither one, by itself, is an accurate picture of the delocalized pi bonds. The two alternative drawings, however, when considered together, give a much more accurate picture than either one on its own. This is because they imply, together, that the carbon-carbon bonds are not double bonds, not single bonds, but about halfway in between.
When it is possible to draw more than one valid structure for a compound or ion, we have identified resonance contributors: two or more different Lewis structures depicting the same molecule or ion that, when considered together, do a better job of approximating delocalized pi-bonding than any single structure. By convention, resonance contributors are linked by a double-headed arrow, and are sometimes enclosed by brackets:
In order to make it easier to visualize the difference between two resonance contributors, small, curved arrows are often used. Each of these arrows depicts the ‘movement’ of two pi electrons. In the drawing of resonance contributors, however, this electron ‘movement’ occurs only in our minds, as we try to visualize delocalized pi bonds. Nevertheless, use of the curved arrow notation is an essential skill that you will need to develop in drawing resonance contributors.
The depiction of benzene using the two resonance contributors A and B in the figure above does not imply that the molecule at one moment looks like structure A, then at the next moment shifts to look like structure B. Rather, at all moments, the molecule is a combination, or resonance hybrid of both A and B.
Caution! It is very important to be clear that in drawing two (or more) resonance contributors, we are not drawing two different molecules: they are simply different depictions of the exact same molecule. Furthermore, the double-headed resonance arrow does NOT mean that a chemical reaction has taken place.
Benzene is often drawn as only one of the two possible resonance contributors (it is assumed that the reader understands that resonance hybridization is implied). However, sometimes benzene will be drawn with a circle inside the hexagon, either solid or dashed, as a way of drawing a resonance hybrid.
Examples of Resonance
Molecules with a Single Resonance Configuration
Example 1:
Example 2:
Example 3:
Example 4:
The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite formal charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge one terminal oxygen can be delocalized by resonance through the other terminal oxygen.
Benzene is an extremely stable molecule due to its geometry and molecular orbital interactions, but most importantly, due to its resonance structures. The delocalized electrons in the benzene ring make the molecule very stable and with its characteristics of a nucleophile, it will react with a strong electrophile only and after the first reactivity, the substituted benzene will depend on its resonance to direct the next position for the reaction to add a second substituent.
Example 2.5.1: Multiple Resonance of other Molecules
Molecules and ions with more than one resonance form:
Some structural resonance conformations are the major contributor or the dominant forms that the molecule exists. For example, if we look at the above rules for estimating the stability of a molecule, we see that for the third molecule the first and second forms are the major contributors for the overall stability of the molecule. The nitrogen is more electronegative than carbon so, it can handle the negative charge more than carbon. A carbon with a negative charge is the least favorable conformation for the molecule to exist, so the last resonance form contributes very little for the stability of the Ion.
Hybrid Resonance
The different resonance forms of the molecule help predict the reactivity of the molecule at specific sites.
The Hybrid Resonance forms show the different Lewis structures with the electron been delocalized. This is very important for the reactivity of chloro-benzene because in the presence of an electrophile it will react and the formation of another bond will be directed and determine by resonance. The lone pair of electrons delocalized in the aromatic substituted ring is where it can potentially form a new bond with an electrophile, as it is shown there are three possible places that reactivity can take place, the first to react will take place at the para position with respect to the chloro- substituent and then to either ortho- position. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.05%3A_Rules_for_Resonance_Forms.txt |
Objectives
After completing this section, you should be able to
• draw the resonance structures of molecules or ions that exhibit delocalization.
• determine the relative stability of resonance structures using a set of rules.
• use the concept of resonance to explain structural features of molecules and ions.
key words
• resonance structure
• resonance hybrid
Curved Arrows Communicate Electron Flow (movement)
Organic chemistry has developed a system to show how electrons move between resonance structures. This system will also be used to help describe how electrons from in reactions. Curved double barbed arrows indicates the flow of two electrons. The base of the curved arrow is placed at the source of the electrons that are moving. The head of the arrow is placed at the destination of the electrons.
It is also important to consciously use the correct type of arrow. There are four primary types of arrows used by chemists to communicate one of the following: completion reaction, equilibrium reaction, electron movement, and resonance forms. The three other types of arrows are shown below to build discernment between them. Note, the electron movement arrows are the only ones that are curved.
Reaction Arrows
Resonance Arrows
Using Curved Arrows to Show Electron Movement
Because the double barbed arrow represents the movement of two electrons, they usually involve lone pair electrons or pi bonds. There are only three types of electron "motion" in resonance. They are:
1. A lone pair forms a pi bond to an adjacent atom
2. A pi bond forms a new pi bond to an adjacent atom
3. A pi bond forms a lone pair on adjacent atom
Let's look at the resonance within acrylic acid to demonstrate these three types of resonance.
The curved arrow in structure A represents the type 3 resonance "motion" - the pi bond between the carbon and oxygen breaks to form another lone pair on the oxygen. The curved arrow in structure B represents type 2 resonance "motion" - the pi bond breaks to form a new pi bond to the carbocation carbon. In structure C, there are two curved arrows. The curved arrow from the oxygen lone pair is type 1 resonance motion - the lone pairs forms a new pi bond between the oxygen and carbon. The other arrow in structure C moves the pi bond to the end of the chain and represents resonance type 2. By combining these three basic types of electron movement we can describe virtually any type of resonance.
Example:
Below are a few more examples of ‘legal’ resonance expressions. Confirm for yourself that the octet rule is not exceeded for any atoms, that formal charges are correct, and identify which type of electron movement is being represented by each arrow.
Exercise \(1\)
Draw the resonance contributors that correspond to the curved, two-electron movement arrows in the resonance expressions below. Then identify the type of resonance motion in each structure below.
Answer
Exercise \(2\)
In each resonance expression, identify the type of resonance motion. Then draw curved arrows on the left-side contributor that shows how we get to the right-side contributor.
Answer
Recognizing Common Patterns of Resonance
If you examine a large number of resonance examples, you will begin to notice that they nearly always match common patterns, of which there are only three. It is important to be able to identify atoms that participate in resonance. In complex resonance cases, multiple types of resonance may occur simultaneously.
Type I - Neutral Species
The electrons of a pi bond move to become a set of lone pair electrons on a electronegative atom. The resonance structure made has a carbon with a violated octet which make it a minor contributor. This type of resonance is commonly used to the polarity in certain double bonds.
Neutral Species
Note: Y is an electronegative atom, usually N, O, or S.
Type II - Charged Species
Type II resonance is only seen with a + charge, and usually involves a positive charge on oxygen or nitrogen being shared onto a carbon; the carbocation form has only six valence electrons on the carbon, so it is a less stable form than the major form (which has complete octets).
Charged Species
Note: Y is an electronegative atom, usually N, O, S, sometimes halogen
Type III - Allylic Resonance
Type III resonance is very common and important because is serves to stabilize positive charges, negative charges, or lone pairs. It is sometimes referred to as “allylic” resonance, especially in cases with all carbon. This type of resonance can be identified by a three-atom group of atoms each with sp2 hybridization and a p orbital.
* represents a charge or an unpaired electron
Type III Example
Atoms with lone pair electrons next to a pi bond can be sp2 hybridized and have the lone pair of electrons in a p orbital despite the fact that they are surrounded by four electron groups. The lone pair electrons contained in the p orbital cause the ion to be stabilized due to resonance.
Similarly, carbocations are sp2-hybridized, with an empty 2p orbital oriented perpendicular to the plane formed by three sigma bonds. If a carbocation is adjacent to a double bond, then three 2p orbitals can overlap and share the two pi electrons - another kind of conjugated pi system in which the positive charge is shared over two carbons.
charge is delocalized over two carbons
Exercise \(3\)
Each of the 'illegal' resonance expressions below contains one or more mistakes. Explain what is incorrect in each.
Answer
The first pair are not resonance structures since there is an additional hydrogen on the second structure oxygen. The second pair pushed electrons toward nitrogen which already has a lone pair and would exceed its octet. The third pair includes a structure with 5 bonds to carbon. The fourth pair requires moving carbon-hydrogen bonds, therefore is not resonance. The fifth pair show electrons moving toward the negatively charged oxygen which would exceed an octet. The fifth pair shows a sigma bond breaking on the ring, rather than pi bond.
Exercise \(4\)
a) Draw three additional resonance contributors for the carbocation below. Include in your figure the appropriate curved arrows showing how one contributor is converted to the next.
b) Fill in the blanks: the conjugated pi system in this carbocation is composed of ______ 2p orbitals sharing ________ delocalized pi electrons.
Answer
a)
b) The conjugated pi system in this carbocation is composed of seven p orbitals containing six delocalized pi electrons.
Example \(1\)
Draw the major resonance contributor of the structure below. Include in your figure the appropriate curved arrows showing how you got from the given structure to your structure. Explain why your contributor is the major one. In what kind of orbitals are the two lone pairs on the oxygen?
Solution
In the structure above, the carbon with the positive formal charge does not have a complete octet of valence electrons. Using the curved arrow convention, a lone pair on the oxygen can be moved to the adjacent bond to the left, and the electrons in the double bond shifted over to the left (see the rules for drawing resonance contributors to convince yourself that these are 'legal' moves).
The resulting resonance contributor, in which the oxygen bears the formal charge, is the major one because all atoms have a complete octet, and there is one additional bond drawn (resonance rules #1 and #2 both apply). This system can be thought of as four parallel 2p orbitals (one each on C2, C3, and C4, plus one on oxygen) sharing four pi electrons. One lone pair on the oxygen is in an unhybridized 2p orbital and is part of the conjugated pi system, and the other is located in an sp2 orbital.
Also note that one additional contributor can be drawn, but it is also minor because it has a carbon with an incomplete octet:
Exercise \(5\)
The figure below shows how the negative formal charge on an oxygen (of an enol) can be delocalized to the carbon indicated by an arrow. More resonance contributors can be drawn in which negative charge is delocalized to three other atoms on the molecule.
a) Circle these atoms that can also have a resonance structure with a negative charge.
b) Draw the two most important resonance contributors for the enolate ion.
Answer
The two major contributors are those in which the negative formal charge is located on an oxygen rather than on a carbon.
Exercise 2.6.5:
a) Draw three additional resonance contributors for the carbocation below. Include in your figure the appropriate curved arrows showing how one contributor is converted to the next.
b) Fill in the blanks: the conjugated pi system in this carbocation is composed of ______ 2p orbitals sharing ________ delocalized pi electrons.
Exercise 2.6.6: Draw the major resonance contributor for each of the anions below.
c) Fill in the blanks: the conjugated pi system in part (a) is composed of ______ 2p orbitals containing ________ delocalized pi electrons.
Exercise 2.6.7: The figure below shows how the negative formal charge on the oxygen can be delocalized to the carbon indicated by an arrow. More resonance contributors can be drawn in which negative charge is delocalized to three other atoms on the molecule.
a) Circle these atoms.
b) Draw the two most important resonance contributors for the molecule.
A word of advice
Becoming adept at drawing resonance contributors, using the curved arrow notation to show how one contributor can be converted to another, and understanding the concepts of conjugation and resonance delocalization are some of the most challenging but also most important jobs that you will have as a beginning student of organic chemistry. If you work hard now to gain a firm grasp of these ideas, you will have come a long way toward understanding much of what follows in your organic chemistry course. Conversely, if you fail to come to grips with these concepts now, a lot of what you see later in the course will seem like a bunch of mysterious and incomprehensible lines, dots, and arrows, and it will be difficult to be successful in organic chemistry.
References
1. Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications. New Jersey: Pearson Prentice Hall, 2007. Print.
2. Ahmad, Wan-Yaacob and Zakaria, Mat B. "Drawing Lewis Structures from Lewis Symbols: A Direct Electron Pairing Approach." Journal of Chemical Education: Journal 77.3: n. pag. Web. March 2000. Link to this journal: pkukmweb.ukm.my/~mbz/c_penerb...83%29/p329.pdf
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.06%3A_Drawing_Resonance_Forms.txt |
Objectives
After completing this section, you should be able to
1. state the Brønsted-Lowry definition of an acid and a base.
2. identify the Brønsted-Lowry acid and base in a given acid-base reaction.
3. identify the conjugate base of an acid and identify the conjugate acid of a base.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Brønsted-Lowry acid
• Brønsted-Lowry base
• conjugate acid
• conjugate base
Study Notes
You should already be familiar with the Brønsted-Lowry concept of acidity and the differences between strong and weak acids. You may wish to review this topic before proceeding.
In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur.
Brønsted-Lowry Definition
J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid-base reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by whether the molecule produces hydrogen ion or hydroxide ion when dissolved in water was too limiting, because not all chemical reactions, especially organic reactions, occur in water. The Brønsted-Lowry Theory defines an acid a proton donor, while a base is a proton acceptor. This is illustrated in the following reactions:
$HCl + HOH \rightarrow H_3O^+ + Cl^- \nonumber$
$HOH + NH_3 \rightarrow NH_4^+ + OH^- \nonumber$
Acid Base
Donates hydrogen ions Accepts hydrogen ions.
HCl HOH → H3O+ + Cl
HOH NH3 NH4+ + OH
The determination of a substance as a Brønsted-Lowry acid or base can only be done by examining the reaction, since many chemicals can be either an acid or a base. For example, HOH is a base in the first reaction and an acid in the second reaction.
Bronsted-Lowry Acids and Bases
To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the number of hydrogens has decreased, then that substance is the acid (donates hydrogen ions). If the number of hydrogens has increased, then that substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is viewed in reverse a new acid and base can be identified. The substances on the right side of the equation are called the conjugate acid and conjugate base compared to those on the left. Also note that an acid turns into a conjugate base, and the base turns into the conjugate acid after the reaction is over.
Conjugate Acid–Base Pairs
In aqueous solutions, acids and bases can be defined in terms of the transfer of a proton from an acid to a base. Thus for every acidic species in an aqueous solution, there exists a species derived from the acid by the loss of a proton. These two species that differ by only a proton constitute a conjugate acid–base pair. For example, in the reaction of HCl with water shown below, HCl , the parent acid, donates a proton to a water molecule, the parent base, thereby forming Cl-. Thus HCl and Cl- constitute a conjugate acid–base pair. By convention, we always write a conjugate acid–base pair as the acid followed by its conjugate base. In the reverse reaction, the Cl- ion in solution acts as a base to accept a proton from H3O+, forming H2O and HCl. Thus H3O+ and H2O constitute a second conjugate acid–base pair. In general, any acid–base reaction must contain two conjugate acid–base pairs, which in this case are HCl/Cl- and H3O+/H2O.
All acid–base reactions contain two conjugate acid–base pairs.
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, H3O+ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (CH3CO2H/CH3CO2-) and the parent base and its conjugate acid (H3O+/H2O).
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are NH4+/NH3 and H2O/OH-.
Example $1$
Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.
Solution
C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows:
$\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber$
Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.
Example $1$
Identify the conjugate acid-base pairs in this equilibrium.
$(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base.
The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Some common conjugate acid–base pairs are shown in Figure 2.7.1. The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.
Figure 2.7.1 The Relative Strengths of Some Common Conjugate Acid–Base Pairs
Exercises
1. Identify the Brønsted-Lowry acids and bases in the reactions given below.
1. $CH3CH2O^- + H2O <=> CH3CH2OH + OH^-$
2. $CH3CH2OH + H2SO4 <=> CH3CH2OH2+ + HSO4-$
2. Show the structures of species X and Y in the following acid-base reactions:
Answer:
1. a)
b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.07%3A_Acids_and_Bases_-_The_Brnsted-Lowry_Definition.txt |
Objectives
After completing this section, you should be able to
• write the expression for the Ka of a weak acid.
• convert a given Ka value into a pKa value, and vice versa.
• arrange a series of acids in order of increasing or decreasing strength, given their Ka or pKa values.
• arrange a series of bases in order of increasing or decreasing strength, given the Ka or pKa values of their conjugate acids.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• acidity constant, Ka
• equilibrium constant, Keq
Study Notes
Calculations and expressions involving Ka and pKa were covered in detail in your first-year general chemistry course. Note that acidity constant is also known as the acid dissociation constant.
You are no doubt aware that some acids are stronger than others. Sulfuric acid is strong enough to be used as a drain cleaner, as it will rapidly dissolve clogs of hair and other organic material.
Not surprisingly, concentrated sulfuric acid will also cause painful burns if it touches your skin, and permanent damage if it gets in your eyes (there’s a good reason for those safety goggles you wear in chemistry lab!). Acetic acid (vinegar), will also burn your skin and eyes, but is not nearly strong enough to make an effective drain cleaner. Water, which we know can act as a proton donor, is obviously not a very strong acid. Even hydroxide ion could theoretically act as an acid – it has, after all, a proton to donate – but this is not a reaction that we would normally consider to be relevant in anything but the most extreme conditions.
The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the acid dissociation constant, abbreviated Ka. The common base chosen for comparison is water.
We will consider acetic acid as our first example. When a small amount of acetic acid is added to water, a proton-transfer event (acid-base reaction) occurs to some extent.
Notice the phrase ‘to some extent’ – this reaction does not run to completion, with all of the acetic acid converted to acetate, its conjugate base. Rather, a dynamic equilibrium is reached, with proton transfer going in both directions (thus the two-way arrows) and finite concentrations of all four species in play. The nature of this equilibrium situation, as you recall from General Chemistry, is expressed by an equilibrium constant, K.
The equilibrium constant is actually a ratio of activities (represented by the symbol $a$), but activities are rarely used in courses other than analytical or physical chemistry. To simplify the discussion for general chemistry and organic chemistry courses, the activities of all of the solutes are replaced with molarities, and the activity of the solvent (usually water) is defined as having the value of 1.
In our example, we added a small amount of acetic acid to a large amount of water: water is the solvent for this reaction. Therefore, in the course of the reaction, the concentration of water changes very little, and the water can be treated as a pure solvent, which is always assigned an activity of 1. The acetic acid, acetate ion and hydronium ion are all solutes, and so their activities are approximated with molarities. The acid dissociation constant, or Ka, for acetic acid is therefore defined as:
$K_{eq} = \dfrac{a_{CH_3COO^-}·a_{H_3O^+}}{a_{CH_3COOH}·a_{H_2O}} ≈ \dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH][1]} \nonumber$
Because dividing by 1 does not change the value of the constant, the "1" is usually not written, and Ka is written as:
$K_{eq} = K_{a} = \dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]} = 1.75 \times 10^{-5} \nonumber$
In more general terms, the dissociation constant for a given acid is expressed as:
$K_a = \dfrac{[A^-][H_3O^+]}{[HA]} \label{First}$
or
$K_a = \dfrac{[A][H_3O^+]}{[HA^+]} \label{Second}$
Equation $\ref{First}$ applies to a neutral acid such as like HCl or acetic acid, while Equation $\ref{Second}$ applies to a cationic acid like ammonium (NH4+).
The value of Ka = 1.75 x 10-5 for acetic acid is very small - this means that very little dissociation actually takes place, and there is much more acetic acid in solution at equilibrium than there is acetate ion. Acetic acid is a relatively weak acid, at least when compared to sulfuric acid (Ka = 109) or hydrochloric acid (Ka = 107), both of which undergo essentially complete dissociation in water.
A number like 1.75 x 10- 5 is not very easy either to say or to remember. Chemists often use pKa values as a more convenient term to express relative acidity. pKa is related to Ka by the following equation
$pK_a = -\log K_a \nonumber$
Doing the math, we find that the pKa of acetic acid is 4.8. The use of pKa values allows us to express the acidity of common compounds and functional groups on a numerical scale of about –10 (very strong acid) to 50 (not acidic at all). Table $1$ at the end of the text lists exact or approximate pKa values for different types of protons that you are likely to encounter in your study of organic and biological chemistry. Looking at Table $1$, you see that the pKa of carboxylic acids are in the 4-5 range, the pKa of sulfuric acid is –10, and the pKa of water is 14. Alkenes and alkanes, which are not acidic at all, have pKa values above 30. The lower the pKa value, the stronger the acid.
Table $1$: Representative acid constants
sulfuric acid
pKa −10
hydrochloric acid
pKa −7
hydronium
pKa 0.00
protonated ketone
pKa ~ −7
protonated alcohol
pKa ~ −3
phosphate monoester
pKa ~ 1
phosphate diester
pKa ~ 1.5
phosphoric acid
pKa 2.2
protonated aniline
pKa ~ 4.6
carboxylic acid
pKa ~ 4-5
pyridinium
pKa 5.3
carbonic acid
pKa 6.4
hydrogen cyanide
pKa ~ 9.2
ammonium
pKa 9.2
phenol
pKa 9.9
thiol
pKa ~ 10-11
water
pKa 14.00
amide
pKa ~ 17
alcohol
pKa ~ 16-18
alpha-proton
pKa ~ 18-20
terminal alkyne
pKa ~ 25
terminal alkene
pKa ~ 35
ammonia
pKa ~ 35
It is important to realize that pKa is not the same thing as pH: pKa is an inherent property of a compound or functional group, while pH is the measure of the hydronium ion concentration in a particular aqueous solution:
$pH = -\log [H_3O^+] \nonumber$
Any particular acid will always have the same pKa (assuming that we are talking about an aqueous solution at room temperature) but different aqueous solutions of the acid could have different pH values, depending on how much acid is added to how much water.
Our table of pKa values will also allow us to compare the strengths of different bases by comparing the pKa values of their conjugate acids. The key idea to remember is this: the stronger the conjugate acid, the weaker the conjugate base. Sulfuric acid is the strongest acid on our list with a pKa value of –10, so HSO4- is the weakest conjugate base. You can see that hydroxide ion is a stronger base than ammonia (NH3), because ammonium (NH4+, pKa = 9.2) is a stronger acid than water (pKa = 14.00).
The stronger the conjugate acid, the weaker the conjugate base.
While Table $1$ provides the pKa values of only a limited number of compounds, it can be very useful as a starting point for estimating the acidity or basicity of just about any organic molecule. Here is where your familiarity with organic functional groups will come in very handy. What, for example, is the pKa of cyclohexanol? It is not on the table, but as it is an alcohol it is probably somewhere near that of ethanol (pKa = 16). Likewise, we can use Table $1$ to predict that para-hydroxyphenyl acetaldehyde, an intermediate compound in the biosynthesis of morphine, has a pKa in the neighborhood of 10, close to that of our reference compound, phenol.
Notice in this example that we need to evaluate the potential acidity at four different locations on the molecule.
pKa Ha ~ 10
pKa Hb = not on table (not acidic)
pKa Hc ~ 19
pKa Hd = not on table (not acidic)
Aldehyde and aromatic protons are not at all acidic (pKa values are above 40 – not on our table). The two protons on the carbon next to the carbonyl are slightly acidic, with pKa values around 19-20 according to the table. The most acidic proton is on the phenol group, so if the compound were to be reacted with a single molar equivalent of strong base, this is the proton that would be donated first.
As you continue your study of organic chemistry, it will be a very good idea to commit to memory the approximate pKa ranges of some important functional groups, including water, alcohols, phenols, ammonium, thiols, phosphates, carboxylic acids and carbons next to carbonyl groups (so-called a-carbons). These are the groups that you are most likely to see acting as acids or bases in biological organic reactions.
A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
Example $1$: Acidic Groups
Using the pKa table, estimate pKa values for the most acidic group on the compounds below, and draw the structure of the conjugate base that results when this group donates a proton. Use the pKa table above and/or from the Reference Tables.
Answer
a. The most acidic group is the protonated amine, pKa ~ 5-9
b. Alpha proton by the C=O group, pKa ~ 18-20
c. Thiol, pKa ~ 10
d. Carboxylic acid, pKa ~ 5
e. Carboxylic acid, pKa ~ 5
Example $2$
Acetic acid (CH3COOH) is known to have a pKa of 4.76. Please determine the Ka for acetic acid.
Solution
Solving for Ka algebraically you get the following:
pKa = -Log(Ka)
-pKa = Log(Ka)
10-pKa = Ka
Using a calculator first enter in the value for the pKa (4.76). The make the number negative (-4.76). Next, use the inverse log function. All calculators are slightly different so this function may appear as: ANTILOG, INV LOG, or 10X. Often it is the second function of the LOG button.
Ka for acetic acid = 10-pKa = 1.74 x 10-5
Exercises
1. Write down an expression for the acidity constant of acetic acid, CH3COOH.
2. The pKa of acetic acid is 4.72; calculate its Ka.
3. The Ka of benzoic acid is 6.5 × 10−5; determine its pKa.
4. From your answers to the questions above, determine whether acetic acid or benzoic acid is stronger
Answers
1. $K_a = \dfrac{[CH_3CO_2^-][H^+]}{[CH_3CO_2H]}$ or $K_a = \dfrac{[CH_3CO_2^-][H_3O^+]}{[CH_3CO_2H]}$
2. $pK_a =− \log_{10} K_a =4.74$ $\text{Thus,}{\mathrm{log}}_{10}{\text{K}}_{\text{a}}\text{=}-4.72{\text{and K}}_{\text{a}}\text{= anti-log}\left(-4.72\right)\text{= 1}\text{.9}×{\text{10}}^{-5}$
3. $pK_a =−\log_{10} K_a =− \log_{10} 6.5 \times 10^{−5} =−(−4.19) =4.19$
4. Benzoic acid is stronger than acetic acid. [Benzoic acid has a higher Ka and a lower pKa.] | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.08%3A_Acid_and_Base_Strength.txt |
Objective
After completing this section, you should be able to
• use pKa values to calculate Keq
• use pKa values to predict the equilibrium direction of an acid-base reaction.
Key Terms
Make certain that you can define, and use in context, the key term below.
• pKa
Using pKa values to predict reaction Equilibria
By definition, the pKa value tells us the extent to which an acid will react with water as the base, but by extension, we can also calculate the equilibrium constant for a reaction between any acid-base pair. Mathematically, it can be shown that:
Keq (for the acid base reaction in question) = 10ΔpKa
where $ΔpK_a$ is the $pK_a$ of product acid minus $pK_a$ of reactant acid
Consider a reaction between methylamine and acetic acid:
First, we need to identify the acid species on either side of the equation. On the left side, the acid is of course acetic acid, while on the right side the acid is methyl ammonium. The specific pKa values for these acids are not on our very generalized pKa table, but are given in the figure above. Without performing any calculations, you should be able to see that this equilibrium lies far to the right-hand side: acetic acid has a lower pKa, is a stronger acid, and thus it wants to give up its proton more than methyl ammonium does. Doing the math, we see that
$K_{eq} = 10^{ΔpK_a} = 10^{10.6 – 4.8} = 10^{5.8} = 6.3 \times 10^5 \nonumber$
So $K_{eq}$ is a very large number (much greater than 1) and the equilibrium lies far to the right-hand side of the equation, just as we had predicted.
If you had just wanted to approximate an answer without bothering to look for a calculator, you could have noted that the difference in pKa values is approximately 6, so the equilibrium constant should be somewhere in the order of 106, or one million. Using the pKa table in this way, and making functional group-based pKa approximations for molecules for which we don’t have exact values, we can easily estimate the extent to which a given acid-base reaction will proceed.
Example $1$
Show the products of the following acid-base reactions, and estimate the value of Keq. Use the pKa table from Section 2.8 and/or from the Reference Tables.
Answer
The pKa of water is 14.0. Thus the Keq for reaction d) is ~ 104.
Exercises
Exercise $1$
Use the pKa table from Section 2.8 and/or from the Reference Tables to determine if the following reactions would be expected to occur:
a)
b)
c)
Answer
a) Yes - alkenes have pKa values of ~35 while alkynes have pKa values of ~25. This means that alkynes are more acidic and more likely to donate a proton.
b) Yes - alkenes have pKa values of ~35 while alcohols have pKa values of ~16-18. This means that alcohols are more acidic and more likely to donate a proton.
c) No - carboxylic acids have pKa values of ~4-5 while alcohols have pKa values of ~16-18. This means that carboxylic acids are more acidic and more likely to donate a proton (so the reverse reaction would be expected to occur). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.09%3A_Predicting_Acid-Base_Reactions_from_pKa_Values.txt |
Objective
After completing this section, you should be able to
• predict the relative acidity of two organic molecules from their structures.
• predict the relative basicity of two organic molecules from their structures.
This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths.
Organic acids as weak acids
For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water.
An acid in solution sets up this equilibrium:
$AH_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
A hydronium ion is formed together with the anion (negative ion) from the acid. This equilibrium is sometimes simplified by leaving out the water to emphasize the ionization of the acid.
$AH_{(aq)} \rightleftharpoons A^-_{(aq)} + H^+_{(aq)} \nonumber$
If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions.
The organic acids are weak in the sense that this ionization is very incomplete. At any one time, most of the acid will be present in the solution as un-ionized molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionized. The position of equilibrium therefore lies well to the left.
Weak acid polarization
Organic acids can usually can be characterized in electrostatic potential maps by the presence of of a positively polarized hydrogen atom shown in blue. When looking at the maps below, methanol has a slightly polarized O-H bond and is considered very weakly acidic. The O-H bond in methyl amine is less polarized, as shown by the lighter blue color around the hydrogen, making it less acidic than methanol. However, the C-H bond in ethane lack virtually any polarity, as shown by the lack of a blue color, making it non-acidic. The following discussion will explain the difference in acidity of these and other organic molecules molecules.
methanol
methylamine
ethane
Comparing the strengths of weak acids
Acid strength is strongly correlated to stability of the conjugate base that will form by removing a proton. In order to analyze how acidic a molecule is likely to be, then you need to estimate the stability of its conjugate base.
Stabilization of the Conjugate Base - Four Main Considerations:
1. Size and electronegativity of the atom holding the charge
2. Can the charge be delocalized by resonance?
3. Are there any inductive effects?
4. Hybridization of orbital holding the charge
These considerations are listed in order of importance and are explained individually, but must be looked at collectively.
1. Size and Electronegativity Effects in Acidity
When comparing elements, it depends on the positional relationship of the elements on the periodic table. When moving a period (aka across a row) of the main group elements, the valence electrons all occupy orbitals in the same shell. These electrons have comparable energy, so this factor does not help us discern differences relative stability. Differences in electronegativity are now the dominant factor. This trend is shown when comparing the pKa values of ethane, methyl amine, and methanol which reflects the relative electronegativities of the C < N < O. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable the conjugate base, the stronger the acid. In general, the more electronegative an atom, the better it is able to bear a negative charge. In the ethyl anion, the negative charge is borne by carbon, in the methylamine anion by nitrogen, and in the methoxide anion by an oxygen. Remember the periodic trend in electronegativity: it also increases as we move from left to right along a row, meaning that oxygen is the most electronegative of three elements being considered. This makes the negative charge on the methoxide anion the most stable of the three conjugate bases and methanol the strongest of the three acids. Likewise, carbon is the least electronegative making ethane the weakest of the three acids.
Within a Group (aka down a column) As we move down the periodic table, the electrons are occupying higher energy subshells creating a larger atomic size and volume. As the volume of an element increases, any negative charge present tends to become more spread out which decreases electron density and increases stability. The figure below shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius.
This relationship of atomic size and electron density is illustrated when we compare the relative acidities of methanol, CH3OH, with methanethiol, CH3SH. The lower pKa value of 10.4 for methanethiol indicates that it is a stronger acid than methanol with a pKa value of 15.5. It is important to remember that neither compound is considered an acid. These relationships become useful when trying to deprotonate compounds to increase their chemical reactivity in non-aqueous reaction conditions.
The difference in size can easily be seen when looking at the electrostatic potential maps for methanol (Left) and methanethiol (Right). The sulfur atom methanethiol is larger than the oxygen atom in methanol. The larger size of sulfur will be better able to delocalize and stabilize the negative charge in its conjugate base metanethiolate.
This section will focus on how the resonance structures of different organic groups contributes to their relative acidity even though the same element acts as the proton donor. When evaluating conjugate bases for the presence of resonance contributors, remember to look for movable electrons (lone pairs and pi bonding electrons). Delocalizing electrons over two or more atoms spreads out the electron density, increasing the stability of the conjugate base, and increasing the acidity of the corresponding acid. A classic example compares the relative acidity of ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups. Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are very different.
In both species, the negative charge on the conjugate base is held by an oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: a resonance contributor can be drawn in which the negative charge is drawn on the second oxygen of the group. The two resonance forms for the conjugate base are equal in energy, according to our ‘rules of resonance’ (Section 2.5). What this means is that the negative charge on the acetate ion is not located on one oxygen or the other: rather it is shared between the two. Chemists use the term ‘delocalization of charge’ to describe this situation. In the ethoxide ion, by contrast, the negative charge is ‘locked’ on the single oxygen. This stabilization leads to a markedly increased acidity.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of nearly 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a difference of over 1012 between the acidity constants for the two molecules). The acetate ion is much more stable than the ethoxide ion, due to the effects of resonance delocalization.
The effects of conjugation can be seen when comparing the electrostatic potential maps of ethanol and acetic acid. Conjugation creates a greater polarization in the O-H bond in acetic acid as shown by its darker blue color.
ethanol
acetic acid
Why is Phenol Acidic?
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base. For example, in aqueous solution:
Since phenol is a very weak acid, the position of equilibrium lies well to the left. However, phenol can lose a hydrogen ion because the phenoxide ion (or phenolate ion - the two terms can be used interchangeably) formed is stabilized due to resonance. The negative charge on the oxygen atom is delocalized around the ring since one of the lone pairs on the oxygen atom can be in a p orbital and overlap with the pi electrons on the benzene ring.
This overlap leads to a delocalization which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localized on the oxygen, but is spread out around the whole ion. Spreading the charge around makes the ion more stable than it would be if all the charge remained on the oxygen. However, oxygen is the most electronegative element in the ion and the delocalized electrons will be drawn towards it. That means that there will still be a lot of charge around the oxygen which will tend to attract the hydrogen ion back again. That is why phenol is only a very weak acid.
This explains why phenol is a much stronger acid than cyclohexanol. As can be seen in the following energy diagram, resonance stabilization is increased for the conjugate base of phenol vs. cyclohexanol after removal of a proton.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of contributors to the resonance hybrid. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Acidity of hydrogen α (alpha) to carbonyl
Alkyl hydrogen atoms bonded to a carbon atom in a α (alpha) position (directly adjacent) relative to a C=O group display unusual acidity. While the pKa values for alkyl C-H bonds in is typically on the order of 40-50, pKa values for these alpha hydrogens is more on the order of 19-20. This is almost exclusively due to the resonance stabilization of the product carbanion, called an enolate, as illustrated in the diagram below. The effect of the the stabilizing C=O is seen when comparing the pKa for the α hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25).
1. Inductive Effects
The inductive effect is an experimentally observed effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond. For example, in a carboxylic acid group the presence of chlorine on adjacent carbons increases the acidity of the carboxylic acid group. A chlorine atom is more electronegative than hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. This further spreads out the electron density of the conjugate base, which has a stabilizing effect. In this context, the chlorine substituent is called an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
negative charge is delocalized by being pulled out onto chlorine atom
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The inductive effects of chlorine can be clearly seen when looking at the electrostatic potential maps of acetic acid (Left) and trichloroacetic acid (Right). The O-H bond in trichloroacetic acid is highly polarized as shown by the dark blue color. This illustrates that tricholoracetic acid is a much stronger acid than acetic acid.
acetic acid
trichloroacetic acid
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowering effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away. 2-chloropropanoic acid has a pKa of 2.8 while for 3-chloropropanoic acid, the pKa is 4.0.
Alkyl groups (hydrocarbons) are weak inductive electron donators. In this case the inductive effect pushes electron density onto the conjugate base, causing the electron density to become more concentrated and producing a destabilizing effect.
more negative charge pushed towards already negative end
The inductive effects of alkyl groups causes a significant variation in the acidities of different carboxylic acids. Notice that the inductive effect drops off after the alkyl chain is about three carbons long.
pKa
HCOOH (Methanoic Acid or Formic Acid) 3.75
CH3COOH (Ethanoic Acid or Acetic Acid) 4.76
CH3CH2COOH (Propanoic Acid) 4.87
CH3CH2CH2COOH (Butanoic Acid) 4.82
1. Orbital Hybridization
The hybridization of an orbital affects its electronegativity. Within a shell, the s orbitals occupy the region closer to the nucleus than the p orbitals. Therefore, the spherical s orbitals are more electronegative than the lobed p orbitals. The relative electronegativity of hybridized orbitals is sp > sp2 > sp3 since the percentage of s-character is decreasing as more p-orbitals are added to the hybrids. This trend indicates the sp hybridized orbitals are more stable with a negative charge than sp3 hybridized orbitals. The table below shows how orbital hybridization influences relative acidity.
compound conjugate base hybridization s character pKa
sp3 25% 50 weakest acid
sp2 33% 44
36
sp 50% 25
16 strongest acid
Comparing the Strengths of Weak Bases
Technically, organic bases are characterized by the presence of an atom with lone pair electrons. These lone pairs contain a high electron density, which is shown red in the electrostatic potential maps, and can bond to H+. Below are the maps of methanol, methyl amine, and acetone. All three compounds can be protonated with a sufficiently strong acid. Note, that all three of these compounds also have the ability to donate a proton when reacted with a strong enough base. Whether these compounds act as a acid or base depends on the conditions.
methylamine
methanol
acetone
It is common to compare basicity's quantitatively by using the pKa's of their conjugate acids rather than their pKb's. Since pKa + pKb = 14, the higher the pKa the stronger the base, in contrast to the usual inverse relationship of pKa with acidity. Recall that ammonia (NH3) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. The conjugate acid of most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their water solutions are basic (have a pH of 11 to 12, depending on concentration). This can be illustrated by the reaction below where an amine removes a proton from water to form substituted ammonium (e.g. NH4+) ions and hydroxide (OH) ions:
Amines are one of the only neutral functional groups which are considered basic. This is a direct consequence of the presence of the unshared electron pair on the nitrogen. The unshared electron pair is less tightly held by the nitrogen of an amine than the corresponding oxygen of an alcohol, which makes it more available to act as a base. As a specific example, methylamine reacts with water to form the methylammonium ion and the OH ion.
Example: Ammonia
All of the have similarities to ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water.
Ammonia in solution sets up this equilibrium:
$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\tag{2.10.1}$
An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left.
The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water.
When looking at the table below, it is clear that the basicity of nitrogen containing compounds is greatly influenced by their structures. The variance in the basicity of these compounds can mostly be explained by the effects of electron delocalization discussed above.
Table $1$: pKa of conjugate acids of a series of amines.
Compound
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.0
Inductive Effects in Nitrogen Basicity
Alkylamines are more basic than ammonia since alkyl groups donate electrons to the more electronegative nitrogen. This inductive effect makes the electron density on the alkylamine nitrogen greater than the nitrogen of ammonium. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
methyl group pushes electron density toward the nitrogen, making it more basic
Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion:
In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there is not any way of spreading the charge.
To summarize:
• The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily.
• The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again.
Taken together, these mean that methylamine is a stronger base than ammonia.
Compound pKa
NH3 9.3
CH3NH2 10.66
(CH3)2NH 10.74
(CH3)3N 9.81
Resonance Effects in Nitrogen Basicity
The resonance effect also explains why a nitrogen atom is basic when it is in an amine, but not significantly basic when it is part of an amide group. While the lone pair of electrons in an amine nitrogen is localized in one place, the lone pair on an amide nitrogen is delocalized by resonance. The lone pair is stabilized by resonance delocalization. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too stable being part of the delocalized pi-bonding system. The electrostatic potential map show the effect of resonance on the basicity of an amide. The map shows that the electron density, shown in red, is almost completely shifted towards the oxygen. This greatly decreases the basicity of the lone pair electrons on the nitrogen in an amide.
amine
pKa of conjugate acid ~ 11
amide
pKa of conjugate acid ~ −1
Aniline, the amine analog of phenol, is substantially less basic than an amine (as evidenced by the pKa of the conjugate acids).
cyclohexylamine
pKa of conjugate acid ~ 10
aniline
pKa of conjugate acid ~ 5
We can use the same reasoning that we used when comparing the acidity of a phenol to that of an alcohol. In aniline, the lone pair on the nitrogen atom is stabilized by resonance with the aromatic pi system, making it less available for bonding and thus less basic.
lone pair is stabilized through resonance
In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydronium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case.
Exercises
Exercise $1$
Select the more basic from each of the following pairs of compounds.
(a)
(b)
Answer
a) The lone pair of electrons on the amide nitrogen are less available to react with a proton.
(b) NaOH -- The hydroxide has a negative charge with three lone pairs of electrons that can react with a proton.
Exercise $2$
The 4-methylbenzylammonium ion has a pKa of 9.51, and the butylammonium ion has a pKa of 10.59. Which is more basic? What's the pKb for each compound?
Answer
The butylammonium is more basic. Remember that pKa+pKb = 14. The pKb for butylammonium is 3.41, the pKb for 4-methylbenzylammonium is 4.49. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.10%3A_Organic_Acids_and_Organic_Bases.txt |
Objectives
After completing this section, you should be able to
1. state the Lewis definition of an acid and a base.
2. identify a given compound as being a Lewis acid or Lewis base, given its Lewis structure or its Kekulé structure.
3. identify an organic molecule or ion in a reaction as either an electrophile or nucleophile.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Lewis acid
• Lewis base
• Electrophile
• Nucleophile
Study Notes
The Lewis concept of acidity and basicity will be of great use to you when you study reaction mechanisms. The realization that an ion such as
is electron deficient, and is therefore a Lewis acid, should help you understand why this ion reacts with substances which are Lewis bases (e.g., H2O).
Introduction to Lewis Acid-Base Theory
The Brønsted acid-base theory has been used throughout the history of acid and base chemistry. However, this theory is very restrictive and focuses primarily on acids and bases acting as proton donors and acceptors. Sometimes conditions arise where the theory doesn't necessarily fit, such as in solids and gases. In 1923, G.N. Lewis from UC Berkeley proposed an alternate theory to describe acids and bases. His theory gave a generalized explanation of acids and bases based on structure and bonding. Through the use of the Lewis definition of acids and bases, chemists are now able to predict a wider variety of acid-base reactions. Lewis' theory used electrons instead of proton transfer and specifically stated that an acid is a species that accepts an electron pair while a base donates an electron pair.
Example of Lewis base (oxygen atom from carbonyl) reacting with Lewis Acid (Mg2+ ion).
The reaction of a Lewis acid and a Lewis base will produce a coordinate covalent bond. A coordinate covalent bond is just a type of covalent bond in which one reactant donates both electrons to form the bond. In this case the Lewis base donates its electrons to form a bond to the Lewis acid. The resulting product is called an addition compound, or more commonly a complex. The electron-pair flow from Lewis base to Lewis acid is shown using curved arrows much like those used for resonance structures in Section 2.5. Curved arrows always mean that an electron pair moves from the atom at the tail of the arrow to the atom at the head of the arrow. In this case the lone pair on the Lewis base attacks the Lewis acid forming a bond. This new type of electron pair movement, along with those described in Section 2.5 will be used throughout this text to describe electron flow during reactions.
• Lewis Acid: a species that accepts an electron pair and will typically either have vacant orbitals or a polar bond involving hydrogen such that it can donate H+ (which has an empty 1s orbital)
• Lewis Base: a species that donates an electron pair and will have lone-pair electrons of pi bonding electrons.
Lewis Acids
Neutral compounds of boron, aluminum, and the other Group 13 elements, (BF3, AlCl3), which possess only six valence electrons, have a very strong tendency to gain an additional electron pair. Because these compounds are only surrounded by three electron groups, they are sp2 hybridized, contained a vacant p orbital, and are potent Lewis acids. Trimethylamine's lone pair elections are contained in an sp3 hybrid orbital making it a Lewis base. These two orbitals overlap, creating a covalent bond in a boron trifluoride-trimethylamine complex. The movement of electrons during this interaction is show by by an arrow.
Positive ions are often Lewis acids because they have an electrostatic attraction for electron donors. Examples include alkali and alkaline earth metals in the group IA and IIA columns. K+, Mg2+ and Ca2+ are sometimes seen as Lewis acidic sites in biology, for example. These ions are very stable forms of these elements because of their low electron ionization potentials. However, their positive charges do attract electron donors.The interaction between a magnesium cation (Mg+2) and a carbonyl oxygen is a common example of a Lewis acid-base reaction. The carbonyl oxygen (the Lewis base) donates a pair of electrons to the magnesium cation (the Lewis acid). As we will see in Chapter 19 when we begin the study of reactions involving carbonyl groups, this interaction has the very important effect of increasing the polarity of the carbon-oxygen double bond.
The eight-electron rule does not hold throughout the periodic table. In order to obtain noble gas configurations, some atoms may need eighteen electrons in their valence shell. Transition metals such as titanium, iron and nickel may have up to eighteen electrons and can frequently accept electron pairs from Lewis bases. Transition metals are often Lewis acids. For example, titanium has four valence electrons and can form four bonds in compounds such as titanium tetrakis (isopropoxide), below, or titanium tetrachloride, TiCl4. However, the titanium atom in that compound has only eight valence electrons, not eighteen. It can easily accept electrons from donors.
For example, when THF and TiCl4 are combined, a Lewis acid-base complex is formed, TiCl4(THF)2. TiCl4(THF)2 is a yellow solid at room temperature.
The Proton as a Common Lewis Acid
Perhaps the most common example of a Lewis acid is also the simplest. It is the hydrogen cation (H+) or proton. It is called a proton because, in most hydrogen atoms, the only particle in the nucleus is a proton. If an electron is removed to make a cation, a proton is all that is left. A proton is a Lewis acid for a number of reasons. It has a positive charge, and so it will attract electrons, which are negative. Also, it lacks the electron configuration of its noble gas neighbor, helium. Helium has two electrons. If a Lewis base or nucleophile donates a pair of electrons to a proton, the proton will obtain a helium noble gas configuration.
There is something about hydrogen cations that is not so simple, however. They are actually not so common. Instead, protons are generally always bound to a Lewis base. Hydrogen is almost always covalently (or coordinately) bonded to another atom. Many of the other elements commonly found in compounds with hydrogen are more electronegative than hydrogen. As a result, hydrogen often has a partial positive charge making is still act as a Lewis Acid.
A acid-base reaction involving protons might better be expressed as:
The Lewis acid-base interactions we have looked at so far are slightly different here. Instead of two compounds coming together and forming a bond, we have one Lewis base replacing another at a proton. Two specific movements of electrons are shown in the reaction both of which are show by arrows. Lone pair electrons on oxygen attack the hydrogen to form an O-H bond in the product. Also, the electrons of the H-Cl bond move to become a lone pair on chlorine as the H-Cl bond breaks. These two arrows together are said to represent the mechanism of this acid-base reaction.
Lewis Bases
What makes a molecule (or an atom or ion) a Lewis base? It must have a pair of electrons available to share with another atom to form a bond. The most readily available electrons are those that are not already in bonds. Bonding electrons are low in energy. Non-bonding electrons are higher in energy and may be stabilized when they are delocalized in a new bond. Lewis bases usually have non-bonding electrons or lone pairs this makes oxygen and nitrogen compounds common Lewis bases. Lewis bases may be anionic or neutral. The basic requirement is that they have a pair of electrons to donate.
Note 1: Ammonia
Ammonia, NH3, has a lone pair and is a Lewis base. It can donate to compounds that will accept electrons.
Ammonia donating to an electron acceptor or Lewis acid.
Not all compounds can act as a Lewis base. For example, methane, CH4, has all of its valence electrons in bonding pairs. These bonding pairs are too stable to donate under normal conditions therefore methane is not a Lewis base. Neutral boron compounds also have all electrons in bonding pairs. For example, borane, BH3 has no lone pairs; all its valence electrons are in bonds. Boron compounds are not typically Lewis bases.
Exercise \(1\)
Which of the following compounds would you expect to be Lewis bases?
a) SiH4 b) AlH3 c) PH3 d) SH2 e) -SH
Answer
a) No, silicon has 4 valence electrons (like carbon) and all 4 are involved in sigma bonds
b) No, aluminum has 3 valence electrons and all 3 are involved in sigma bonds
c) Yes, phosphorus has 5 valence electrons, so there is one lone pair available
d) Yes, sulfur has 6 valence electrons, so there are two lone pairs available
e) Yes, this ion has 3 lone pairs available
Lewis Acid-Base Complexes
What happens when a Lewis base donates a pair of electrons to a Lewis acid? The electron pushing formalism (arrows) we have been using to illustrate the behavior of Lewis acids and Lewis bases is meant to show the direction of electron movement from the donor to the acceptor. However, given that a bond can be thought of as a pair of electrons that are shared between two atoms (in this case, between the donor and the acceptor), these arrows also show where bonds are forming.
The electrons donated from a Lewis base to a Lewis acid form a new bond. A new, larger compound is formed from the smaller Lewis acid and Lewis base. This compound is called a Lewis acid-base complex. A simple example of Lewis acid-base complexation involves ammonia and boron trifluoride. The nitrogen atom has a lone pair and is an electron donor. The boron has no octet and is an electron acceptor. The two compounds can form a Lewis acid-base complex or a coordination complex together.
When the nitrogen donates a pair of electrons to share with the boron, the bond that forms is sometimes called a coordinate bond. A coordinate bond is any covalent bond that arose because one atom brought a pair of its electrons and donated them to another.
In organic chemistry terminology, the electron donor is called a nucleophile and the electron acceptor is called an electrophile. Ammonia is a nucleophile and boron trifluoride is an electrophile.
• Because Lewis bases are attracted to electron-deficient atoms, and because positive charge is generally associated with the nucleus of an atom, Lewis bases are sometimes refered to as "nucleophiles". Nucleophile means nucleus-loving.
• Because Lewis acids attract electron pairs, Lewis acids are sometimes called "electrophiles". Electrophile means electron-loving.
Exercise \(2\)
For the following reaction, add curved arrows (electron pushing formalism) to indicate the electron flow.
Answer
Exercise \(3\)
A Lewis acid-base complex is formed between THF (tetrahyrofuran) and borane, BH3.
a) Which compound is the Lewis acid? Which one is the Lewis base?
b) Which atom in the Lewis acid is the acidic site? Why?
c) Which atom in the Lewis base is the basic site? Why?
d) How many donors would be needed to satisfy the acidic site?
e) Show, using arrow notation, the reaction to form a Lewis acid-base complex.
f) Borane is highly pyrophoric; it reacts violently with air, bursting into flames. Show, using arrow notation, what might be happening when borane contacts the air.
g) Borane-THF complex is much less pyrophoric than borane. Why do you suppose that is so?
Add exercises text here.
Answer
a) Borane is the Lewis acid. THF has lone pair electrons so it is the Lewis base.
b) The Boron atom has an unfilled octet so it has an empty p orbital that can accept electrons.
c) The oxygen atom in THF has lone pair electrons contained in a sp3 hybridized orbital.
d) The boron in borane has six electrons around it so it would only need one lone pair donor to reach an octet.
e) Show, using arrow notation, the reaction to form a Lewis acid-base complex.
f) The Borane initially reacts with water in air.
g) After the Borane-THF complex is formed, the boron atom has a complete octet making it less reactive. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.11%3A_Acids_and_Bases_-_The_Lewis_Definition.txt |
Objectives
After completing this section, you should be able to
1. identify the various intermolecular forces that may be at play in a given organic compound.
2. describe how intermolecular forces influence the physical properties, 3‑dimensional shape and structure of compounds.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• dipole-dipole forces
• London dispersion forces
• hydrogen bond
• intermolecular forces
• noncovalent interaction
• van der Waals forces
Study Notes
Much of the material in this section should be familiar to you from your pre-requisite general chemistry course. Nonetheless, this section is important, as it covers some of the fundamental factors that influence many physical and chemical properties.
Introduction
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
Note
The properties of liquids are intermediate between those of gases and solids but are more similar to solids.
Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law.
In this section, we explicitly consider three kinds of intermolecular interactions: dipole-dipole forces, dispersion forces, and hydrogen bonds. These intermolecular interactions are also called van der Waals forces or noncovalent interactions. There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. These are less common with organic molecules, so will not be described further in this section.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure 2.12.1 parts (a and b). Arrangements in which two positive or two negative ends are adjacent (parts (c and d) in Figure 2.12.1) are higher energy as the similar charges would repel one another. Hence dipole–dipole interactions, such as those in parts (a and b) in Figure 2.12.1, are attractive intermolecular interactions, whereas those in part (c and d) are repulsive intermolecular interactions.
Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure 2.12.2. On average, however, the attractive interactions dominate.
Chloromethane is an example of a polar molecule. An electrostatic potential map shows a high electron density (seen in red) around the electronegative chlorine giving it a partial negative charge. The other end of the molecule has electron density pulled away from it giving it a partial positive charge seen in blue. The positive and negative ends of different chloromethane molecules are attracted to one another through this electrostatic interaction.
chloromethane
(μ = 1.87 D)
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r6, so doubling the distance between the dipoles decreases the strength of the interaction by 26, or 64-fold. Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas NaCl, which is held together by ionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table 2.12.1. Using what we learned about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules.
Table 2.12.1: Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
Note
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Example 2.12.1
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativity values. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone. This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise $1$
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both.
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table 2.12.2: Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure 2.12.3, the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure 2.12.3, tends to become more pronounced as atomic and molecular masses increase (Table 2.12.3). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
Note
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure 2.12.4). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure 2.12.4 shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Example 2.12.2
Arrange n-butane, propane, 2-methylpropane [isobutane], and n-pentane in order of increasing boiling points.
Strategy:
Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise $2$
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer
GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure 2.12.5. Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure 2.12.6.
A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cage-like structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Note
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example 2.12.3
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise 2.12.3
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer: CH3CO2H and NH3;
Example 2.12.4
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise 2.12.4
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer: KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Summary
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/r6, where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/r6. Larger atoms tend to be more polarizable than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong OHhydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water.
Key Takeaway
• Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds.
Problems
1. Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers.
2. Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions?
3. Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases?
4. In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why?
5. Identify the most important intermolecular interaction in each of the following.
a) SO2
b) HF
c) CO2
d) CCl4
e) CH2Cl2
6. Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds.
7. Do you expect the boiling point of H2S to be higher or lower than that of H2O? Justify your answer.
1. Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input?
Solutions
1. Dipole-Diple interactions are stronger because molecules polarity is permanent. Molecules involved in London dispersion forces are only temporarily polar
2. Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule.
3. In solids and liquids the molecules are in direct contact. In gases the molecules only contact during collisions. Intermolecules forces become inmportant when gases are cooled to the point where they form a liquid. 4. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure. 5. a) The V-shaped SO2 molecule has a large dipole moment due to the polar S=O bonds, so dipole–dipole interactions will be most important.
b) The H–F bond is highly polar, and the fluorine atom has three lone pairs of electrons to act as hydrogen bond acceptors; hydrogen bonding will be most important.
c) Although the C=O bonds are polar, this linear molecule has no net dipole moment; hence, London dispersion forces are most important.
d) This is a symmetrical molecule that has no net dipole moment, and the Cl atoms are relatively polarizable; thus, London dispersion forces will dominate.
e) This molecule has a small dipole moment, as well as polarizable Cl atoms. In such a case, dipole–dipole interactions and London dispersion forces are often comparable in magnitude.
6) Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of four hydrogen bonds per H2O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form two hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water.
7) H2O would have the higher boiling point because it can form the stronger hydrogen bonding intermolecular force. 8) Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.12%3A_Noncovalent_Interactions_Between_Molecules.txt |
Objective
After completing this section, you should be able to
• use ball-and-stick molecular models to make models of simple organic compounds (e.g., ethane, ethylene, acetylene, ethanol, formaldehyde, acetone, acetic acid), given their Kekulé structures or molecular formulas.
Study Notes
You will have noticed that we have given two names for most of the compounds discussed up to this point. In general we shall be using systematic (i.e., IUPAC—International Union of Pure and Applied Chemistry) names throughout the course. However, simple compounds are often known by their common names, which may be more familiar than their IUPAC counterparts. We shall address the subject of nomenclature (naming) in Chapter 3.
ethanol
formaldehyde (methanal),
acetone (propanone),
acetic (ethanoic) acid,
If your instructor is having you work with molecular models in class, they may use this section for you to practice creating specific structures.
Exercises
1. Construct a molecular model of each of the compounds listed below.
1. \$\ce{\sf{CH3-CN}}\$
2. \$\ce{\sf{CH3-N=C=O}}\$
3. \$\ce{\sf{CH3-CH2-O-CH3}}\$
Hint: Use the curved sticks to form the multiple bonds and the straight sticks for single bonds.
2.S: Polar Covalent Bonds Acids and Bases (Summary)
Concepts & Vocabulary
2.1 Polar Covalent Bonds: Electronegativity
• The difference in electronegativity values of two atoms determines whether the bond between those atoms is classified as either ionic, polar covalent, or non-polar covalent.
• Ionic bonds result from large differences in electronegativity values, such as that between a metal and non-metal atom (Na and Cl).
• Covalent bonding generally results when both atoms are non-metals, like C, H, O, N and the halides.
• When both atoms the same and/or have the same electronegativity value, then the bonding electrons are shared equally and the bond is classified as non-polar covalent.
• Polar covalent bonds occur when the difference in electronegativity values is small, and the bonding electrons are not shared equally.
2.2 Polar Covalent Bonds: Dipole Moments
• The molecular dipole moment is the sum of all the bond dipoles within a molecule and depends on both the molecular geometry and the bond polarity.
• Molecules that contain no polar bonds, like CH4, and/or completely symmetrical molecules, like CO2, generally have no net dipole moment.
• Asymmetrical molecules that contain bonds of different polarities or non-bonding lone pairs typically have a molecular dipole moment.
2.3 Formal Charges
• Formal Charge compares how many valence electrons surround a free atom versus how many surround that same type of atom bonded with a molecule or ion.
• Formal Charge = (# of valence electrons in free atom) - (# of lone-pair electrons) - (1/2 # of bond pair electrons) Eqn. 2.3.1
• Formal charges of zero generally represent the most stable structures.
• These bonding patterns for the atoms commonly found in organic molecules result in a formal charge of zero
• Carbon - 4 bonds, no lone pairs
• Hydrogen - 1 bond, no lone pairs
• Nitrogen - 3 bonds, 1 lone pair
• Oxygen - 2 bonds, 2 lone pairs
• Halogens - 1 bond, 3 lone pairs.
2.4 Resonance
• Resonance Theory is often used when the observed chemical and physical properties of a molecule or ion cannot be adequately described by a single Lewis Structure. A classic example is the benzene molecule, C6H6. The Lewis Structure of benzene could be drawn in two different ways. Both structures have alternating double bond and single bonds between the carbons. The only difference is the location of the pi bonds.
If these structures are correct, then the benzene molecule should have two different C-C bond lengths and bond energies, corresponding to a C-C single bond and to a C=C double bond. However, analysis shows that benzene contains only one type of carbon-carbon bond and it's bond length and energy are half between those of a single bond and double bond. Resonance theory states that benzene exists as the "average" of the two structures called a resonance hybrid, in which the six pi electrons delocalized over all six carbon atoms. Each C-C bond in benzene would be the average of a single bond and double bond or a "bond and a half". Dashed lines are often used to show type of "partial" bonding in a resonance hybrid of benzene
2.5 Rules for Resonance Forms
• The rules for estimating stability of resonance structures are
• The resonance form in which all atoms have complete valence shells is more stable.
• The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
• The structure with the least number of formal charges is more stable
• The structure with the least separation of formal charges is more stable
• A structure with a negative charge on the more electronegative atom will be more stable
• Positive charges on the least electronegative atom (most electropositive) is more stable
• Resonance forms that are equivalent have no difference in stability and contribute equally. (eg. benzene)
• If these rules are applied to the two Lewis Structures of benzene, the result would be that both structures will have the same relative stability and will both contribute equally to the character of the resonance hybrid.
2.6 Drawing Resonance Forms
• In resonance structures, the electrons are able to move to help stabilize the molecule. This movement of the electrons is called delocalization.
• The rules for drawing resonance structures are:
• Resonance structures should have the same number of electrons, do not add or subtract any electrons. (You can check the number of electrons by counting them)
• All resonance structures must follow the rules of writing Lewis Structures.
• The hybridization of the structure must stay the same.
• The skeleton of the structure can not be changed (only the electrons move).
• Resonance structures must also have the same amount of lone pairs.
• A Brønsted-Lowry acid is a proton (H+) donor and a Brønsted-Lowry base is a proton acceptor.
• The strength of Brønsted-Lowry acids is measured indicated by its pKa value. The lower the pKa - the stronger the acid.
• A strong acid will have a weak conjugate base. A strong base will have a weak conjugate acid.
2.9 Predicting Acid-Base Reactions from pKa Values
• The equilibrium of an acid-base reaction favors the formation of weaker acids from stronger acids. To predict the direction of the equilibrium, identify Brønsted-Lowry acid on each side of the reaction. Assign/look up pKa values for each acid. The equilibrium will favor the side that has the weakest acid (the highest pKa).
2.10 Organic Acids and Organic Bases
• Organic acids are stronger when the conjugate base that is formed upon loss of a proton is more stable.
• Some factors that effect the stability of the conjugate base (often an anion) are the anionic atom's size and electronegativity, resonance effects, inductive effects, and solvation.
2.11: Acids and Bases - The Lewis Definition
• A Lewis acid is a lone pair acceptor and a Lewis base is a lone pair donor.
2.12: Non-covalent Interactions between Molecules
• Non-covalent Interactions, also known as Intermolecular Forces, significantly effect the physical properties of organic molecules. Hydrogen bonding is the most important of these interactions, but others include ion-dipole, dipole-dipole, and London Dispersion Forces.
2.MM: Molecular Models
Summary Problems
Exercise \(1\)
Draw all possible resonance structures to demonstrate delocalization of the positive charge in the following molecule. Circle the most stable resonance structure and explain your answer. Also, draw the resonance hybrid.
Answer
The circled structure is the most stable because it is the only structure with a full octet on all atoms. (Remember that carbocations must have an incomplete octet, and an oxygen with a positive charge and a full octet is more stable than a carbocation.)
As a reminder, the resonance hybrid structure is a combination of all of your resonance structure showing partial pi bonds and partial formal charges.
Exercise \(2\)
Draw all possible resonance structures to demonstrate delocalization of the negative charge in the following molecule. Also, draw the resonance hybrid and circle the most stable resonance contributor among all of your resonance structures.
Answer
The circled structure is the most stable because the negative charge is on oxygen, the most electronegative element sharing the negative charge. As a reminder, in resonance structures with negative charges, all of the atoms have octets, so you only need to focus on electronegativity differences to find the most stable structure.
Exercise \(3\)
For the equilibrium shown below, answer the following questions: a) Draw curved arrows to illustrate bond breakage and formation in the reaction. b) At equilibrium, are the products or reactants favored? Explain. c) What percent reactants and percent products are present at equilibrium? d) Use resonance structures to explain why N is the most basic atom in the conjugate base. e) Use resonance structures to explain why the hydrogen that is removed from the conjugate acid is the most acidic proton.
Answer
a)
b) The lower pKa is the stronger acid, and the equilibrium will always favor the weak acid (higher pKa), so this reaction favors the reactants. Note: It's normal to assume that the neutral products would be favored, but this is a good example that shows our chemical intuition isn't always right. To make sure we understand the equilibrium, we need to have pKa values to compare.
c) Keq = 1012.5-17.5 = 10-5 = 0.00001 So, the ratio of reactants to products is 1:0.00001. The percent reactants is 99.999% and the percent products is 0.001%.
d) There are two atoms with lone pairs in the conjugate base, O and N, so we should evaluate both to show that N is the most basic atom. Protonating O yields a positive charge localized on O. Protonating N yields a charge delocalized over 3 Cs, 1 N, and 1 O. The N is most basic because the charge in the acid is delocalized and there are two structures with full octets (positive charge on N and O).
e) To find the most acidic proton, we should focus on Hs on atoms next to (not on) double bonds. This will yield a delocalized negative charge when the H is removed. In the conjugate acid, we have two options. Option 1 yields a compound with a negative charge delocalized over 2 atoms, O and C. Option 2 yields the base from the original reaction because the charge is delocalized over 4 atoms, C and 3 Os. This is very stable because it is spread over 3 electronegative oxygen atoms.
Exercise \(4\)
This question focuses on the reaction of the two molecules shown below. a) Draw an acid-base reaction of these molecules. Clearly label the acid, the base, the conjugate acid, and the conjugate base. Draw curved arrows to clearly indicate electron movement involved in bond formation and bond cleavage. For the base and the conjugate base, label the hybridization of the charged atoms. b) If you drew the reaction correctly, the pKa of the conjugate acid is 9. Use the table in Section 2.8 to determine the pKa of the acid. Are the reactants or products favored at equilibrium? What is the approximate ratio of reactants to products? What is the approximate percentage of reactants and products? c) Draw the structures of the two charged molecules in the reaction. Draw resonance structures to illustrate charge delocalization in these molecules. Based on your analysis in part b, which charged molecule is more stable? Briefly explain this result.
Answer
a)
Don't forget that the charged atoms in the base and conjugate base are sp2 hybridized so that the lone pair can be in a p orbital and then delocalized by resonance. For negative charges to be delocalized, the lone pair must be in a p orbital so that electrons can flow to adjacent pi bonds by resonance.
b)
The acid has the larger pKa, so it is the weaker acid and the reactants are favored. The approximate difference in pKa values is 1. So, Keq is 10-1 and for every 1 product molecule there are 10 reactant molecules. (Remember, since the reactants are favored, there are more reactants and the Keq must be less than 1.) The approximate percentage of reactants is 90% and products is 10%. (10/(10+1) is approximately 90% and 1/(10+1) is approximately 10%)
c)
We know that the equilibrium favors the reactants. We also know that in acid base reactions, the stronger acid reacts with the stronger base to yield the weaker base and the weaker acid. In this reaction, the products are the stronger acid-base pair. We also know that "stronger" means more reactive and less stable, while "weaker" means less reactive and more stable. So, since the reactants are favored, that means the base (as labeled above) is more stable than the conjugate base. This seems strange since the charge is more delocalized in the conjugate base. This comparison highlights the importance of quality of resonance structures versus quantity of resonance structures. The charge is more stable in the base since in two of the resonance structures, the negative charge is on O. In the conjugate base, only one of the structures has the negative charge on O.
Exercise \(5\)
Determine the position of the most acidic proton in the following molecule. You should draw resonance structures and a resonance hybrid to justify your answer.
Answer
Donation of the most acidic proton (H+) yields the most stable conjugate base; the one that has the charge on the most electronegative element and/or the most delocalized charge. You should focus on protons attached to electronegative elements (e.g., O or N) and protons on atoms next to pi bonds, but not attached to atoms with pi bonds. In this molecule, all Hs are attached to carbon. So, we should focus on the Hs next to pi bonds. These are Hs on sp3 hybridized carbons next to sp2 hybridized carbons. Loss of these protons will yield a negative charge that can be delocalized by resonance.
Option #1 yields a conjugate base with the charge delocalized over 1 carbon and 1 oxygen. Option #2 yields a conjugate base with a charge delocalized over 3 carbons and 2 oxygens. This is the most stable conjugate base, so the most acidic proton is the red proton (removed in option #2).
Exercise \(6\)
Determine the position of the most basic atom in the following molecule. You should draw resonance structures and a resonance hybrid to justify your answer.
Answer
There are four basic atoms (atoms with lone pair electrons) in this molecule. The strongest base will yield the most stable conjugate acid (the one with the most delocalized positive charge and/or the charge on the least electronegative atom).
Options #1 and #2 yield conjugate acids with localized charges. Options #3 and #4 yield conjugate acids with delocalized charges.This illustrates a key point: For atoms that can't have an expanded octet, atoms with a lone pair and no pi bond will accept a proton and yield a localized charge in the conjugate acid while atoms with both a lone pair and a pi bond will accept a proton and yield a delocalized charge in the conjugate acid. So, we should focus our energy on the latter atoms. In this problem, that means Options #3 and #4. In option #3, the charge is delocalized over 3 atoms, 2 that have full octets on all atoms (positive charge on N and O). In option #4, the charge is delocalized over 4 atoms, 3 that have full octets on all atoms (positive charge on N and 2 Os). So, Option #4 has more total resonance structures and more structures that have a full octet. This is the most stable conjugate acid which means the circled O is the most basic atom.
Skills to Master
• Skill 2.1 Predict whether a bond is ionic, polar covalent, or non-polar covalent based on the position of the atoms in the periodic table.
• Skill 2.2 Identify the partial positive and partial negative atoms of a polar covalent bond based on relative electronegativity.
• Skill 2.3 Determine the dipole moment of a molecule based on molecular geometry and bond polarity.
• Skill 2.4 Identify the chemicals in a reaction as Brønsted-Lowry acids or bases, and conjugate acids and bases.
• Skill 2.5 Predict the products of an acid-base reaction.
• Skill 2.6 Use pKa values to predict the equilibrium direction of an acid-base reaction.
• Skill 2.7 Predict the relative strength of an organic acid by examining the stability of the conjugate base.
• Skill 2.8 Use molecular structure and analysis of intermolecular forces to rank a series of organic molecules with respect to physical properties like melting point and boiling point.
• Skill 2.9 Identify the chemicals in a reaction as Lewis acids or bases.
Memorization Tasks (MT)
• MT 2.1 Memorize that the C-H bond is considered to be non-polar.
• MT 2.2 Memorize the common bonding patterns for C, H, N, O and the halogens that have a zero formal charge.
• MT 2.3 Memorize the factors that affect the relativity stability of conjugate bases. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.MM%3A_Molecular_Models.txt |
Learning Objectives
After you have completed Chapter 3, you should be able to
1. fulfill the detailed objectives listed under each section.
2. identify some of the commonest functional groups.
3. write the structures and names of the first ten straight-chain alkanes.
4. recognize and name the simple alkyl substituents, and give the systematic names for branched-chain alkanes.
5. briefly describe some of the processes used during the refining of petroleum.
6. briefly describe the physical properties of alkanes.
7. draw a number of possible conformations of some simple alkanes and alkane-like compounds, and represent the energies of such conformations on energy versus rotation diagrams.
8. define, and use in context, the key terms introduced in this chapter.
This chapter begins with an introduction to the concept of the functional group, a concept that facilitates the systematic study of organic chemistry. Next, we introduce the fundamentals of organic nomenclature (i.e., the naming of organic chemicals) through examination of the alkane family of compounds. We then discuss, briefly, the occurrence and properties of alkanes, and end with a description of cis-trans isomerism in cycloalkanes.
• 3.1: Functional Groups
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups
• 3.2: Alkanes and Alkane Isomers
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula CnH2n+2 and can be subdivided into the following three groups: the linear straight-chain alkanes, branched alkanes, and cycloalkanes. Alkanes are also saturated hydrocarbons. Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring.
• 3.3: Alkyl Groups
The IUPAC system requires first that we have names for simple unbranched chains, as noted above, and second that we have names for simple alkyl groups that may be attached to the chains. Examples of some common alkyl groups are given in the following table. Note that the "ane" suffix is replaced by "yl" in naming groups. The symbol R is used to designate a generic (unspecified) alkyl group.
• 3.4: Naming Alkanes
There are too many organic molecules to memorize a name for each one. The IUPAC nomenclature system provides an unique name for each different molecule based on functional groups, the longest carbon chain and other attached substituents.
• 3.5: Properties of Alkanes
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.
• 3.6: Conformations of Ethane
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers, are really just two of the exact same molecule that differ only in terms of the angle about one or more sigma bonds.
• 3.7: Conformations of Other Alkanes
Ethane has only two conformers of note - staggered and eclipsed. Alkanes that are more complex than ethane, like propane and butane have a greater variety in possible conformers and their relative energies.
• 3.8: Gasoline - A Deeper Look
The petroleum that is pumped out of the ground at locations around the world is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundances of the components vary depending on the source.
• 3.S: Organic Compounds- Alkanes and Their Stereochemistry (Summary)
03: Organic Compounds- Alkanes and Their Stereochemistry
Objectives
After completing this section, you should be able to
1. explain why the properties of a given organic compound are largely dependent on the functional group or groups present in the compound.
2. identify the functional groups present in each of the following compound types: alkenes, alkynes, arenes, (alkyl and aryl) halides, alcohols, ethers, aldehydes, ketones, esters, carboxylic acids, (carboxylic) acid chlorides, amides, amines, nitriles, nitro compounds, sulfides and sulfoxides.
3. identify the functional groups present in an organic compound, given its structure.
4. Given the structure of an organic compound containing a single functional group, identify which of the compound types listed under Objective 2, above, it belongs to.
5. draw the structure of a simple example of each of the compound types listed in Objective 2.
Key Terms
Make certain that you can define, and use in context, the key term below.
• functional group
Study Notes
The concept of functional groups is a very important one. We expect that you will need to refer back to tables at the end of Section 3.1 quite frequently at first, as it is not really feasible to learn the names and structures of all the functional groups and compound types at one sitting. Gradually they will become familiar, and eventually you will recognize them automatically.
Functional groups are small groups of atoms that exhibit a characteristic reactivity. A particular functional group will almost always display its distinctive chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have specific names that often carry over in the naming of individual compounds incorporating the groups.
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups tend to behave in organic reactions.
Drawing abbreviated organic structures
Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest:
The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule.
Common Functional Groups
In the following sections, many of the common functional groups found in organic chemistry will be described. Tables of these functional groups can be found at the bottom of the page.
Hydrocarbons
The simplest functional group in organic chemistry (which is often ignored when listing functional groups) is called an alkane, characterized by single bonds between two carbons and between carbon and hydrogen. Some examples of alkanes include methane, CH4, is the natural gas you may burn in your furnace or on a stove. Octane, C8H18, is a component of gasoline.
Alkanes
Alkenes (sometimes called olefins) have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds. Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening. (If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas). Ethyne, commonly called acetylene, is used as a fuel in welding blow torches.
Alkenes and alkynes
Alkenes have trigonal planar electron geometry (due to sp2 hybrid orbitals at the alkene carbons) while alkynes have linear geometry (due to sp hybrid orbitals at the alkyne carbons). Furthermore, many alkenes can take two geometric forms: cis or trans (or Z and E which will be explained in detail in Chapter 7). The cis and trans forms of a given alkene are different molecules with different physical properties there is a very high energy barrier to rotation about a double bond. In the example below, the difference between cis and trans alkenes is readily apparent.
Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens - in other words, they are saturated with hydrogen atoms. The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. As we will see in Chapter 7, hydrogen can be added to double and triple bonds, in a type of reaction called 'hydrogenation'.
The aromatic group is exemplified by benzene (which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic), and naphthalene, a compound with a distinctive 'mothball' smell. Aromatic groups are planar (flat) ring structures, and are widespread in nature. We will learn more about the structure and reactions of aromatic groups in Chapter 15.
Functional Groups with Carbon Single Bonds to other Atoms
Halides
When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or haloalkane. The presence of a halogen atom (F, Cl, Br, or I), is often represented by X due to the similar chemistry of halogens. Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules.
Alcohols and Thiols
In the alcohol functional group, a carbon is single-bonded to an OH group (the OH group, by itself, is referred to as a hydroxyl). Except for methanol, all alcohols can be classified as primary, secondary, or tertiary. In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon. In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol.
The sulfur analog of an alcohol is called a thiol (the prefix thio, derived from the Greek, refers to sulfur).
Ethers and sulfides
In an ether functional group, a central oxygen is bonded to two carbons. Below are the line and Lewis structures of diethyl ether, a common laboratory solvent and also one of the first medical anaesthesia agents.
In sulfides, the oxygen atom of an ether has been replaced by a sulfur atom.
Amines
Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations. In the case where a nitrogen has four bonds to carbon (which is somewhat unusual in biomolecules), it is called a quaternary ammonium ion.
Caution
Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols and amines - the definitions are different. In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.
Carbonyl Containing Functional Groups
Aldehydes and Ketones
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
Carboxylic acids and acid derivatives
If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a heteroatom (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl (OH) group.
As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a carboxylate (much more about carboxylic acids in Chapter 20).
In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine.
In esters, the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
In an acid anhydride, there are two carbonyl carbons with an oxygen in between. An acid anhydride is formed from combination of two carboxylic acids with the loss of water (anhydride).
In an acyl phosphate, the carbonyl carbon is bonded to the oxygen of a phosphate, and in an acid chloride, the carbonyl carbon is bonded to a chlorine.
Nitriles and Imines
In a nitrile group, a carbon is triple-bonded to a nitrogen. Nitriles are also often referred to as cyano groups.
Molecules with carbon-nitrogen double bonds are called imines, or Schiff bases.
Phosphates
Phosphorus is a very important element in biological organic chemistry, and is found as the central atom in the phosphate group. Many biological organic molecules contain phosphate, diphosphate, and triphosphate groups, which are linked to a carbon atom by the phosphate ester functionality.
Because phosphates are so abundant in biological organic chemistry, it is convenient to depict them with the abbreviation 'P'. Notice that this 'P' abbreviation includes the oxygen atoms and negative charges associated with the phosphate groups.
Molecules with Multiple Functional Groups
A single compound may contain several different functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups.
The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table found below provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
Exercise \(1\)
Identify the functional groups in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Answer
a) carboxylate, sulfide, aromatic, two amide groups (one of which is cyclic)
b) tertiary alcohol, thioester
c) carboxylate, ketone
d) ether, primary amine, alkene
2: Draw one example each (there are many possible correct answers) of compounds fitting the descriptions below, using line structures. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule).
a) a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
b) an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
c) A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group.
Functional Group Tables
Exclusively Carbon Functional Groups
Group Formula Class Name Specific Example IUPAC Name Common Name
alkene H2C=CH2 ethene ethylene
alkyne HC≡CH ethyne acetylene
arene C6H6 benzene benzene
Functional Groups with Single Bonds to Heteroatoms
Group Formula Class Name Specific Example IUPAC Name Common Name
halide H3C-I iodomethane methyl iodide
alcohol CH3CH2OH ethanol ethyl alcohol
ether CH3CH2OCH2CH3 diethyl ether ether
amine H3C-NH2 aminomethane methylamine
nitro compound H3C-NO2 nitromethane
thiol H3C-SH methanethiol methyl mercaptan
sulfide H3C-S-CH3 dimethyl sulfide
Functional Groups with Multiple Bonds to Heteroatoms
Group Formula Class Name Specific Example IUPAC Name Common Name
nitrile H3C-CN ethanenitrile acetonitrile
aldehyde H3CCHO ethanal acetaldehyde
ketone H3CCOCH3 propanone acetone
carboxylic acid H3CCO2H ethanoic Acid acetic acid
ester H3CCO2CH2CH3 ethyl ethanoate ethyl acetate
acid halide H3CCOCl ethanoyl chloride acetyl chloride
amide H3CCON(CH3)2 N,N-dimethylethanamide N,N-dimethylacetamide
acid Anhydride (H3CCO)2O ethanoic anhydride acetic anhydride | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.01%3A_Functional_Groups.txt |
Objectives
After completing this section, you should be able to
1. draw the Kekulé structure, condensed structure and shorthand structure of each of the first ten straight-chain alkanes.
2. name each of the first ten straight-chain alkanes, given its molecular formula, Kekulé structure, condensed structure or shorthand structure.
3. explain the difference in structure between a straight- and a branched-chain alkane, and illustrate the difference using a suitable example.
4. explain why the number of possible isomers for a given molecular formula increases as the number of carbon atoms increases.
5. draw all the possible isomers that correspond to a given molecular formula of the type Cn H2n+2, where n is ≤ 7.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• branched-chain alkane
• constitutional or structural isomer
• homologous series
• isomer
• saturated hydrocarbon
• straight-chain alkane (or normal alkane)
Study Notes
A series of compounds in which successive members differ from one another by a CH2 unit is called a homologous series. Thus, the series CH4, C2H6, C3H8 . . . CnH2n+2, is an example of a homologous series.
It is important that you commit to memory the names of the first 10 straight-chain alkanes (i.e., from CH4 to C10H22). You will use these names repeatedly when you begin to learn how to derive the systematic names of a large variety of organic compounds. You need not remember the number of isomers possible for alkanes containing more than seven carbon atoms. Such information is available in reference books when it is needed. When drawing isomers, be careful not to deceive yourself into thinking that you can draw more isomers than you are supposed to be able to. Remember that it is possible to draw each isomer in several different ways and you may inadvertently count the same isomer more than once.
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes are often called saturated hydrocarbons because they have the maximum possible number of hydrogens per carbon. In Section 1.7, thealkane molecule, ethane, was shown to contain a C-C sigma bond. By adding more C-C sigma bond larger and more complexed alkanes can be formed. Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner.
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. Consider the series in Figure 25.3.3. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner.
Figure 25.3.2: Members of a Homologous Series. Each succeeding formula incorporates one carbon atom and two hydrogen atoms more than the previous formula.
The homologous series allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
Molecular Formulas
Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only. Alkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The first six alkanes are as follows:
Table $1$: Molecular formulas for small alkanes
methane CH4
ethane C2H6
propane C3H8
butane C4H10
pentane C5H12
hexane C6H14
You can work out the formula of any of the alkanes using the general formula CnH2n+2
Isomerism
All of the alkanes containing 4 or more carbon atoms show structural isomerism, meaning that there are two or more different structural formulas that you can draw for each molecular formula. Isomers (from the Greek isos + meros, meaning "made of the same parts") are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. Alkanes with 1-3 carbons, methane (CH4), ethane (C2H6), and propane (C3H8,) do not exist in isomeric forms because there is only one way to arrange the atoms in each formula so that each carbon atom has four bonds. However, C4H10, has more than possible structure. The four carbons can be drawn in a row to form butane or the can branch to form isobutane. The two compounds have different properties—for example, butane boils at −0.5°C, while isobutane boils at −11.7°C.
Likewise the molecular formula: C5H12 has three possible isomer. The compound at the far left is pentane because it has all five carbon atoms in a continuous chain. The compound in the middle is isopentane; like isobutane, it has a one CH3 branch off the second carbon atom of the continuous chain. The compound at the far right, discovered after the other two, was named neopentane (from the Greek neos, meaning “new”). Although all three have the same molecular formula, they have different properties, including boiling points: pentane, 36.1°C; isopentane, 27.7°C; and neopentane, 9.5°C.
Of the structures show above, butane and pentane are called normal alkanes or straight-chain alkanes, indicating that all contain a single continuous chain of carbon atoms and can be represented by a projection formula whose carbon atoms are in a straight line. The other structures, isobutane, isopentane, and neopentane are called called branched-chain alkanes. As the number of carbons in an akane increases the number of possible isomers also increases as shown in the table below.
Table $2$: Number of isomers for hydrocarbons
Molecular Formula Number of Structural Isomers
CH4 1
C2H6 1
C3H8 1
C4H10 2
C5H12 3
C6H14 5
C7H16 9
C8H18 18
C9H20 35
C10H22 75
C14H30 1858
C18H38 60,523
C30H62 4,111,846,763
Akanes can be represented in many different ways. The figure below shows some of the different ways straight-chain butane can be represented. Most often chemists refer to butane by the condensed structure CH3CH2CH2CH3 or n-C4H10 where n denotes a normal straight alkane.
Note that many of these structures only imply bonding connections and do not indicate any particular geometry. The bottom two structures, referred to as "ball and stick" and "space filling" do show 3D geometry for butane. Because the four-carbon chain in butane may be bent in various ways the groups can rotate freely about the C–C bonds. However, this rotation does not change the identity of the compound. It is important to realize that bending a chain does not change the identity of the compound; all of the following represent the same compound, butane:
The nomenclature of straight alkanes is based on the number of carbon atoms they contain. The number of carbons are indicated by a prefix and the suffix -ane is added to indicate the molecules is an alkane. The prefix for three carbons is prop so adding -ane, the IUPAC name for C3H8 is propane. Likewise, the prefix for six is hex so the name for the straight chain isomer of C6H14 is called hexane. The first ten prefixes should be memorized, because these alkane names from the basis for naming many other organic compounds.
9.2.1" role="presentation" style="position:relative;" tabindex="0">9.2.1
Table $3$: The First 10 Straight-Chain Alkanes
Molecular Formula Prefix Condensed Structural Formula Name
CH4 Meth CH4 methane
C2H6 Eth CH3CH3 ethane
C3H8 Prop CH3CH2CH3 propane
C4H10 But CH3CH2CH2CH3 butane
C5H12 Pent CH3CH2CH2CH2CH3 pentane
C6H14 Hex CH3(CH2)4CH3 hexane
C7H16 Hept CH3(CH2)5CH3 heptane
C8H18 Oct CH3(CH2)6CH3 octane
C9H20 Non CH3(CH2)7CH3 nonane
C10H22 Dec CH3(CH2)8CH3 decane
Example $1$: Chain Isomers in Pentane
Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models.
Exercises
Exercise $1$
Draw all of the isomers for C6H14O that contain a 6 carbon chain and an alcohol (-OH) functional group.
Answer
Exercise $2$
Draw all possible isomers for C6H14 (There are five total).
Answer
The top structure is when it is a 6 carbon chain. The middle row contains the 5 carbon chained isomers with branching at the 2nd and 3rd carbon. The bottom row contains the two 4 carbon chain isomers that can be drawn.
Exercise $3$
Draw all possible isomers for C3H8.
Answer
The first structure is when an alcohol comes off the first carbon. The second structure is when the alcohol is coming off the central carbon. The third structure is the only possible ether form of C3H8.
Exercise $4$
Draw all possible isomers for C4H8O2 that contain a carboxylic acid.
Answer
There are only 2 possibilities.
Exercise $5$
Draw all possible isomers for C3H9N and indicate whether each amine is primary, secondary, or tertiary.
Answer
The first and second structures are primary amines. The third structure is a secondary amine. The last structure is a tertiary amine.
Exercise $6$
Indicate whether each of the following sets are constitutional isomers, the same compound, or different compounds.
Answer
a) Both structures have formulas of C10H22 and have different connectivity which makes these constitutional isomers.
b) Both structures have formulas of C7H16 and have the same connectivity which makes these the same compound.
c) Both structure have formulas of C7H16 and have different connectivity which makes these constitutional isomers.
d) The structure on the left has a formula of C9H20 and the structure on the right has a formula of C10H22 so these are different compounds.
Exercise $7$
Draw the 5 constitutional isomers of C7H16 (of the 9 total isomers possible) that have 5 carbons as the longest carbon chain length.
Answer
The 5 constitutional isomers with a 5 carbon chain length are shown above. Since there needs to be 7 carbons total, the 2 extra carbons are added as substituents. From left to right, the methyl group substitution pattern is 2,2, 2,3, 2,4, and 3,3, and the last one (on right) has a 3-ethyl substituent.
The other 4 possible constitutional isomers (with different length carbon chains) are shown below. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.02%3A_Alkanes_and_Alkane_Isomers.txt |
Objectives
After completing this section, you should be able to
1. recognize and name any alkyl group that can be considered to have been formed by the removal of a terminal hydrogen atom from a straight-chain alkane containing ten or fewer carbon atoms.
2. explain what is meant by a primary, secondary, tertiary or quaternary carbon atom.
3. represent the various types of organic compounds using the symbol “R” to represent any alkyl group.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alkyl group
• methyl group
• isopropyl group
• sec-butyl group
• isobutyl group
• tert-butyl group
• primary carbon
• secondary carbon
• tertiary carbon
• quaternary carbon
Study Notes
The differences among primary, secondary, tertiary and quaternary carbon atoms are explained in the following discussion. A convenient way of memorizing this classification scheme is to remember that a primary carbon atom is attached directly to only one other carbon atom, a secondary carbon atom is attached directly to two carbon atoms, and so on.
The IUPAC system requires first that we have names for simple unbranched chains and second, that we have names for simple alkyl groups that may be attached to the chains. An alkyl group is formed by removing one hydrogen from the alkane chain. The removal of this hydrogen results in a stem change from -ane to -yl to indicate an alkyl group. The removal of a hydrogen from methane, CH4, creates a methyl group -CH3. Likewise, the removal of a hydrogen from ethane, CH3CH3, creates an ethyl group -CH2CH3. The nomenclature pattern can continue to provide a series of straight-chain alkyl groups from straight chain alkanes with a hydrogen removed from the end. Note, the letter R is used to designate a generic (unspecified) alkyl group.
Table \(2\): Straight chain alkane and alkyl group names
Alkane Name Alkyl Group Name (Abbreviation)
CH4 Methane -CH3 Methyl (Me)
CH3CH3 Ethane -CH2CH3 Ethyl (Et)
CH3CH2CH3 Propane -CH2CH2CH3 Propyl (Pr)
CH3CH2CH2CH3 Butane -CH2CH2CH2CH3 Butyl (Bu)
CH3CH2CH2CH2CH3 Pentane -CH2CH2CH2CH2CH3 Pentyl
CH3CH2CH2CH2CH2CH3 Hexane -CH2CH2CH2CH2CH2CH3 Hexyl
Prior to the systematic nomenclature developed for organic chemistry, prefixes were used to specify the connection point of straight-chain and branched-chain alkyl groups. Although the modern nomenclature system, discussed in the next section, is preferred these older terms are still often used, especially in solvents and reagents. Thus, an understanding of these prefixes is important to understanding organic chemistry. Notice that the total number of carbons in the alkyl subsistent is still indicated with the prefix + yl. For methyl and ethyl alkyl groups there is only one possible connection point so connection prefixes are not necessary. Starting with a three carbon alkyl group (propyl) the possibility of multiple connection points necessitates connection prefixes. These prefixes are often abbreviated with a letter which is italicized.
Normal (n)
The prefix "n" is used to indicate a connection at the end of a straight-chain alkane. This prefix is not commonly used to just indicate alkyl subsistent as discussed above. However it is sometime used to indicate the connection of a functional group onto a straight alkane.
Iso (i)
Starting with propyl alkyl groups there is the possibility of a connection other than the very end. The prefix "iso" implies that the connection ends with a (CH3)2CH- group.
Secondary (Sec)
With butyl straight-chain alkyl groups there is the possibility of a connection on the second carbon from the end of the chain. These alkyl groups are given the prefix "Sec." This is not used for pentyl or hexyl groups because there is more than one structure that are not identical that could be named as sec-pentyl or sec-hexyl.
Tertiary (tert or t)
Starting with four carbon alkyl groups, there is an isomer which can have a connection to a tertiary carbon. These alkyl groups get the prefix "t."
Example:
The naming system described above is often used to describe halogens which contain only a few carbons. The halogen is shown as bonded to the connection point of the alkyl group.
Classification of carbon atoms
Carbons have a special terminology to describe how many other carbons they are attached to. This allows for an easy description of branching in alkanes. Also, we will find that the number of carbons attached to a given atom will have subtle effects on its chemistry.
• Primary carbons (1o) attached to one other C atom.
• Secondary carbons (2o) are attached to two other C’s.
• Tertiary carbons (3o) are attached to three other C’s.
• Quaternary carbons (4o) are attached to four C's.
The figure below use the group "R" to represent an alkyl group of unspecified length. R typically used to represent alkyl groups but an also represent a part of a molecule which is either unspecified or not germane to the discussion.
This terminology will be used repeatedly in organic chemistry to describe the number of carbons attached to a specific atom, however, the atom will not always a carbon.
Example \(1\)
Please indicate the the number of 1o, 2o, 3o, and 4o carbons in the following molecule:
• The molecule has six Primary carbons (1o).
• The molecule has one Secondary carbon (2o).
• The molecule has one Tertiary carbon (3o).
• The molecule has one Quaternary carbon (4o)
Hydrogen atoms are also classified in this manner. A hydrogen atom attached to a primary carbon atom is called a primary hydrogen ect.
• Primary hydrogens (1o) are attached to carbons bonded to one other C atom
• Secondary hydrogens (2o) are attached to carbons bonded to two other C’s
• Tertiary hydrogens (3o) are attached to carbons bonded to three other C’s
• It is not possible to have a quaternary hydrogen (4o).
Example \(2\)
Please indicate the the number of 1o, 2o, and 3o, hydrogens are in the following molecule:
• The molecule has fifteen Primary (1o) hydrogens.
• The molecule has two Secondary (2o) hydrogens.
• The molecule has one Tertiary (3o) hydrogen.
Exercises
Exercise \(1\)
Determine whether the H’s indicated in the following structure are 1o, 2o, or 3o.
Answer
Exercise \(2\)
Determine whether the H’s indicated in the following structure is 1o, 2o, or 3o.
Answer
Exercise \(3\)
Determine whether the carbons indicated in the structure below are 1o, 2o, 3o, or 4o.
Answer
Exercise \(4\)
Determine whether the carbons indicated in the structure below are 1o, 2o, 3o, or 4o.
Answer
Exercise \(5\)
Please indicate the total number of each type 1o, 2o, 3o, and 4o carbons in the following molecule.
Answer
There are 5 primary (1o) C’s, 2 secondary (2o) C’s, 1 tertiary (3o) C and 1 quaternary (4o) C in the structure (seen color coded below).
Exercise \(6\)
Please indicate the total number of each type 1o, 2o, 3o, and 4o carbons in the following molecule.
Answer
There are 8 primary (1o) C’s, 7 secondary (2o) C’s, 2 tertiary (3o) C’s and 2 quaternary (4o) C’s in the structure (seen color coded below). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.03%3A_Alkyl_Groups.txt |
Objectives
After completing this section, you should be able to
1. provide the correct IUPAC name for any given alkane structure (Kekulé, condensed or shorthand).
2. draw the Kekulé, condensed or shorthand structure of an alkane, given its IUPAC name.
Key Terms
Make certain that you can define, and use in context, the key term below.
• IUPAC system
Study Notes
The IUPAC system of nomenclature aims to ensure
1. that every organic compound has a unique, unambiguous name.
2. that the IUPAC name of any compound conveys the structure of that compound to a person familiar with the system.
One way of checking whether the name you have given to an alkane is reasonable is to count the number of carbon atoms implied by the chosen name. For example, if you named a compound 3‑ethyl-4‑methylheptane, you have indicated that the compound contains a total of 10 carbon atoms—seven carbon atoms in the main chain, two carbon atoms in an ethyl group, and one carbon atom in a methyl group. If you were to check the given structure and find 11 carbon atoms, you would know that you had made a mistake. Perhaps the name you should have written was 3‑ethyl-4,4‑dimethylheptane!
When naming alkanes, a common error of beginning students is a failure to pick out the longest carbon chain. For example, the correct name for the compound shown below is 3‑methylheptane, not 2‑ethylhexane.
Remember that every substituent must have a number, and do not forget the prefixes: di, tri, tetra, etc.
You must use commas to separate numbers, and hyphens to separate numbers and substituents. Notice that 3‑methylhexane is one word.
Hydrocarbons having no double or triple bond functional groups are classified as alkanes or cycloalkanes, depending on whether the carbon atoms of the molecule are arranged only in chains or also in rings. Although these hydrocarbons have no functional groups, they constitute the framework on which functional groups are located in other classes of compounds, and provide an ideal starting point for studying and naming organic compounds. The alkanes and cycloalkanes are also members of a larger class of compounds referred to as aliphatic. Simply put, aliphatic compounds are compounds that do not incorporate any aromatic rings in their molecular structure.
The following table lists the IUPAC names assigned to simple continuous-chain alkanes from C-1 to C-10. A common "ane" suffix identifies these compounds as alkanes. Longer chain alkanes are well known, and their names may be found in many reference and text books. The names methane through decane should be memorized, since they constitute the root of many IUPAC names. Fortunately, common numerical prefixes are used in naming chains of five or more carbon atoms.
Table 3.4.1 : Simple Unbranched Alkanes
Name Molecular
Formula
Structural
Formula
Isomers Name Molecular
Formula
Structural
Formula
Isomers
methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Some important behavior trends and terminologies
1. The formulas and structures of these alkanes increase uniformly by a CH2 increment.
2. A uniform variation of this kind in a series of compounds is called homologous.
3. These formulas all fit the CnH2n+2 rule. This is also the highest possible H/C ratio for a stable hydrocarbon.
4. Since the H/C ratio in these compounds is at a maximum, we call them saturated (with hydrogen).
Beginning with butane (C4H10), and becoming more numerous with larger alkanes, we note the existence of alkane isomers. For example, there are five C6H14 isomers, shown below as abbreviated line formulas (A through E):
Although these distinct compounds all have the same molecular formula, only one (A) can be called hexane. How then are we to name the others?
The IUPAC system requires first that we have names for simple unbranched chains, as noted above, and second that we have names for simple alkyl groups that may be attached to the chains. Examples of some common alkyl groups are given in the following table. Note that the "ane" suffix is replaced by "yl" in naming groups. The symbol R is used to designate a generic (unspecified) alkyl group.
Table 3.4.2 : Alkyl Groups Names
Group CH3 C2H5 CH3CH2CH2 (CH3)2CH– CH3CH2CH2CH2 (CH3)2CHCH2 CH3CH2CH(CH3)– (CH3)3C– R–
Name Methyl Ethyl Propyl Isopropyl Butyl Isobutyl sec-Butyl tert-Butyl Alkyl
IUPAC Rules for Alkane Nomenclature
1. Find and name the longest continuous carbon chain.
2. Identify and name groups attached to this chain.
3. Number the chain consecutively, starting at the end nearest a substituent group.
4. Designate the location of each substituent group by an appropriate number and name.
5. Assemble the name, listing groups in alphabetical order.
6. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.
Example 3.4.1 : Alkanes
The IUPAC names of the isomers of hexane are: A hexane B 2-methylpentane C 3-methylpentane D 2,2-dimethylbutane E 2,3-dimethylbutane
Halogen Groups
Halogen substituents are easily accommodated, using the names: fluoro (F-), chloro (Cl-), bromo (Br-) and iodo (I-).
Example 3.4.2 : Halogen Substitution
For example, (CH3)2CHCH2CH2Br would be named 1-bromo-3-methylbutane. If the halogen is bonded to a simple alkyl group an alternative "alkyl halide" name may be used. Thus, C2H5Cl may be named chloroethane (no locator number is needed for a two carbon chain) or ethyl chloride.
Alkyl Groups
Alkanes can be described by the general formula CnH2n+2. An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same concept can be applied to any of the straight chain alkane names provided in the table below.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Three Rules for Naming Alkanes
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substituent present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substituents and/or functional groups are written in alphabetical order.
Example 3.4.3
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in blue and consists of eight carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
Rule 3: In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the following examples.
The name of this molecule is 2,5-dimethyloctane
Example 3.4.4
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in blue and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substituent present must have the lowest possible number. In this example, numbering the chain from either the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
The name of this molecule is: 2-bromo-6-chloroheptane
Example 3.4.5
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in blue and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Parent chain: heptane Substitutents: 2-chloro 3-methyl 6-bromo
The name of this molecule is: 6-bromo-2-chloro-3-methylheptane
Exercises
Exercise 3.4.1
Give the proper IUPAC names of the following compounds.
Answer
a) Since this structure is an unbranched alkane (all single bonds) with a 5 carbon chain length, its name would be pentane.
b) This alkane has a 7 carbon longest continuous chain length that we number from right to left to get the first methyl substituent we encounter to have the lowest possible number (3 versus being 4 if numbering from left to right). This causes it to have 2 methyl substituents at positions 3 & 4 so we would name it indicating those numbers and the prefix dimethyl which gives a proper IUPAC name of 3,4-dimethylheptane.
c) This alkane has a 5 carbon longest continuous chain length (which could be numbered from left to right or right to left due to the symmetry at C-3). It has two methyl substituents off of C-3 so the proper IUPAC name is 3,3-dimethylpentane.
Exercise 3.4.2
Give the proper IUPAC names of the following compounds.
Answer
a) This alkane has a 9 carbon longest continuous chain length that we number from left to right (structure on left numbered in blue) to make the ethyl substituent be number 4. For the structure on the right (numbered in red) going from right to left, the methyl substituent is number 4. Since ethyl is higher in the alphabetical order, you want to make it have the lower number so the structure on the left (blue numbering) takes priority and the name is 4-ethyl-6-methylnonane.
b) This alkane has a 6 carbon longest continuous chain length that we number from right to left to make the first methyl be C-2 (versus the opposite direction which would make the first methyl C-3). Since there are 3 methyl substituents at positions 2,3, & 4, this compound would have the name 2,3,4-trimethylhexane.
c) This 6 carbon alkane can be numbered along different chains (see below) as well as in the opposite directions. This shows the two different chains that can be drawn (making the first substituent in that chain the lowest number). The structure on the left (numbered in blue) is the correct choice since it causes more substituents to be on the longest continuous chain (3 vs 2 in the structure on the right). This would make the IUPAC name of the structure 3-ethyl-2,4-dimethylhexane. (Notice how ethyl takes priority over methyl and the di- is not considered for alphabetizing.)
Exercise 3.4.3
All of the following names represent a compound that has been named improperly. Draw out the structure from the name and give the proper IUPAC name for the compounds.
1. 1,3-dimethylbutane
2. 4-ethylpentane
3. 2-ethyl-3-methylpentane
Answer
a) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), the correct name should be 2-methylpentane.
b) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is 6 C’s and we start the numbering on the end to the right to make the methyl substituent come off at C-3 (instead of beong at C-4 if we numbered it the opposite direction) the correct name should be 3-methyl hexane.
c) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is 6 C’s and since this molecule is symmetrical (between carbon 3 & 4), you can start the numbers from either end. In this case, we have methyl substituents coming off of carbons 3 & 4 so the proper name is 3,4-dimethyl hexane.
Exercise 3.4.4
All of the following names represent a compound that has been named improperly. Draw out the structure from the name and give the proper IUPAC name for the compounds.
1. 2,2-diethylheptane
2. 2-propylpentane
3. 4,4-diethylbutane
Answer
a) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is now 8 C’s and you have an ethyl substituent at C-3 and a methyl substituent also at C-3 so the proper name is 3- ethyl-3-methyloctane.
b) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is now 7 C’s and since this molecule is symmetrical (at carbon 4), you can start the numbers from either end. There is a methyl substituent at C-4 so the proper name is 4-methylheptane.
c) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right) going from right to left (to make the ethyl substituent have the lowest number possible), the correct name is 3-ethylhexane | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.04%3A_Naming_Alkanes.txt |
Objectives
After completing this section, you should be able to
1. arrange a number of given straight-chain alkanes in order of increasing or decreasing boiling point or melting point.
2. arrange a series of isomeric alkanes in order of increasing or decreasing boiling point.
3. explain the difference in boiling points between a given number of alkanes.
Key Terms
Make certain that you can define, and use in context, the key term below.
• van der Waals force (also known as London Dispersion force)
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless non-polar compounds. The relative weak London dispersion forces of alkanes result in gaseous substances for short carbon chains, volatile liquids with densities around 0.7 g/mL for moderate carbon chains, and solids for long carbon chains. For molecules with the same functional groups, there is a direct relationship between the size and shape of molecules and the strength of the intermolecular forces (IMFs) causing the differences in the physical states.
Boiling Points
Table 3.5.1 describes some of the properties of some straight-chain alkanes. There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.
For isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.
The boiling points shown are for the "straight chain" isomers of which there is more than one. The first four alkanes are gases at room temperature, and solids do not begin to appear until about $C_{17}H_{36}$, but this is imprecise because different isomers typically have different melting and boiling points.3.2.1
Table 3.5.1: Physical Properties of Some Alkanes
Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C)
methane CH4 –182 –164 0.668 g/L gas
ethane C2H6 –183 –89 1.265 g/L gas
propane C3H8 –190 –42 1.867 g/L gas
butane C4H10 –138 –1 2.493 g/L gas
pentane C5H12 –130 36 0.626 g/mL liquid
hexane C6H14 –95 69 0.659 g/mL liquid
octane C8H18 –57 125 0.703 g/mL liquid
decane C10H22 –30 174 0.730 g mL liquid
*Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure.
The boiling points for the "straight chain" isomers and isoalkanes isomers are shown to demonstrate that branching decreases the surfaces area, weakens the IMFs, and lowers the boiling point.
Example 3.5.1: Boiling Points of Alkanes
For example, the boiling points of the three isomers of $C_5H_{12}$ are:
• pentane: 309.2 K
• 2-methylbutane: 301.0 K
• 2,2-dimethylpropane: 282.6 K
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them better able!
Exercise 3.5.1
For each of the following pairs of compounds, select the substance which you expect to have the higher boiling point:
1. octane and nonane.
2. octane and 2,2,3,3-tetramethylbutane.
Solution
1. nonane, since it has more atoms it will have greater IMF
2. octane, since it is not branched, the molecules can pack closer together increasing IMF
Solubility
Alkanes are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.
Solubility in Water
When a molecular substance dissolves in water, the following must occur:
• break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water.
As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water. The alkane does not dissolve.
Note
This is a simplification because entropic effects are important when things dissolve.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Exercise $1$
For each of the following pairs of compounds, select the substance you expect to have the higher boiling point.
1. octane and nonane.
2. octane and 2,2,3,3‑tetramethylbutane.
Answer
Nonane will have a higher boiling point than octane, because it has a longer carbon chain than octane. Octane will have a higher boiling point than 2,2,3,3‑tetramethylbutane, because it branches less than 2,2,3,3‑tetramethylbutane, and therefore has a larger “surface area” and more van der Waals forces.
Note: The actual boiling points are
nonane, 150.8°C
octane, 125.7°C
2,2,3,3‑tetramethylbutane, 106.5°C
Reactions of Alkanes
Alkanes undergo very few reactions. There are two important reactions that are still possible, combustion and halogenation. The halogenation reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Combustion
Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide, water, and a significant amount of heat. Due to the exothermic nature of these combustion reactions, alkanes are commonly used as a fuel source (for example: propane for outdoor grills, butane for lighters). The hydrocarbons become harder to ignite as the molecules get bigger. This is because the larger molecules don't vaporize as easily. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Larger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and become a gas. An example combustion reaction is shown for propane:
$\ce{ C_3H_8 + O_2 -> 3CO_2 + 4H_2O + 2044 kJ/mol} \nonumber$
Halogenation
Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed.
Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following unbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride.
$\ce{CH_4 + Cl_2 \rightarrow CH_3Cl + CH_2Cl + CHCl_3 + CCl_4 + HCl} \nonumber$
Looking Closer: An Alkane Basis for Properties of Other Compounds
An understanding of the physical properties of alkanes is important since petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fat like compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.05%3A_Properties_of_Alkanes.txt |
Objectives
After completing this section, you should be able to
1. explain the concept of free rotation about a carbon-carbon single bond.
2. explain the difference between conformational isomerism and structural isomerism.
3. draw the conformers of ethane using both sawhorse representation and Newman projection.
4. sketch a graph of energy versus bond rotation for ethane, and discuss the graph in terms of torsional strain.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• conformation (conformer, conformational isomer)
• dihedral angle
• eclipsed conformation
• Newman projection
• staggered conformation
• strain energy
• torsional strain (eclipsing strain)
Study Notes
You should be prepared to sketch various conformers using both sawhorse representations and Newman projections. Each method has its own advantages, depending upon the circumstances. Notice that when drawing the Newman projection of the eclipsed conformation of ethane, you cannot clearly draw the rear hydrogens exactly behind the front ones. This is an inherent limitation associated with representing a 3-D structure in two dimensions.
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity of the atoms or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers, are really just two of the exact same molecule that differ only in rotation of one or more sigma bonds.
Ethane Conformations
Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.
In order to better visualize these different conformations, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.
The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons at 120°angles, which is what the actual tetrahedral geometry looks like when viewed from this perspective and flattened into two dimensions.
Interactive Element
Figure 3.6.1: A 3D Model of Staggered Ethane.
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation. In the staggered conformation, all of the C-H bonds on the front carbon are positioned at an angle of 60° relative to the C-H bonds on the back carbon. This angle between a sigma bond on the front carbon compared to a sigma bond on the back carbon is called the dihedral angle. In this conformation, the distance between the bonds (and the electrons in them) is maximized. Maximizing the distance between the electrons decreases the electrostatic repulsion between the electrons and results in a more stable structure.
If we now rotate the front CH3 group 60° clockwise, the molecule is in the highest energy ‘eclipsed' conformation, and the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon.
This is the highest energy conformation because of unfavorable electrostatic repulsion between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol (12 kJ/mol) higher than that of the staggered conformation. Torsional strain (or eclipsing strain) is the name give to the energy difference caused by the increased electrostatic repulsion of eclipsing bonds.
Another 60° rotation returns the molecule to a second eclipsed conformation. This process can be continued all around the 360° circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of variations in between. We will focus on the staggered and eclipsed conformers since they are, respectively, the lowest and highest energy conformers.
Unhindered (Free) Rotations Do Not Exist in Ethane
The carbon-carbon bond is not completely free to rotate – the 3 kcal/mol torsional strain in ethane creates a barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not large enough to prevent rotation except at extremely cold temperatures. So at normal temperatures, the carbon-carbon bond is constantly rotating. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other conformer. The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown in Figure 3.6.2.
Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time. The animation below illustrates the relationship between ethane's potential energy and its dihedral angle
Exercises
1) What is the most stable rotational conformation of ethane and explain why it is preferred over the other conformation?
Solutions
1) Staggered, as there is less repulsion between the hydrogen atoms. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.06%3A_Conformations_of_Ethane.txt |
Objectives
After completing this section, you should be able to
1. depict the staggered and eclipsed conformers of propane (or a similar compound) using sawhorse representations and Newman projections.
2. sketch a graph of energy versus bond rotation for propane (or a similar compound) and discuss the graph in terms of torsional strain.
3. depict the anti, gauche, eclipsed and fully eclipsed conformers of butane (or a similar compound), using sawhorse representations and Newman projections.
4. sketch a graph of energy versus C2-C3 bond rotation for butane (or a similar compound), and discuss it in terms of torsional and steric repulsion.
5. assess which of two (or more) conformers of a given compound is likely to predominate at room temperature from a semi-quantitative knowledge of the energy costs of the interactions involved.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti conformation
• gauche conformation
• eclipsed conformation
• steric repulsion (strain)
In butane, there are three rotating carbon-carbon sigma bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position.
Eclipsed interaction Energy (kcal/mol) Energy (kJ/mol)
H-H 1.0 4.0
H-CH3 1.4 6.0
CH3-CH3 2.6 11.0
The CH3-CH3 groups create the significantly larger eclipsed interaction of 11.0 kJ/mol. There are also two H-H eclipsed interactions at 4.0 kJ/mol each to create a total of 2(4.0 kJ/mol) + 11.0 kJ/mol = 19.0 kJ/mol of strain. This is the highest energy conformation for butane, due to torsional strain caused by the electrostatic repulsion of electrons in the eclipsed bonds, but also because of another type of strain called ‘steric repulsion’, between the two rather bulky methyl groups. Steric strain comes about when two large groups, such as two methyl groups, try to occupy the same space. What results is a repulsive non-covalent interaction caused by their respective electron densities.
If we rotate the front, (blue) carbon by 60°clockwise, the butane molecule is now in a staggered conformation.
This is more specifically referred to as the ‘gauche’ conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be. There is still significant steric repulsion between the two bulky groups.
A further rotation of 60° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
Due to steric repulsion between methyl and hydrogen substituents, this eclipsed conformation B is higher in energy than the gauche conformation. However, because there is no methyl-to-methyl eclipsing, it is lower in energy than eclipsed conformation A. One more 60° rotation produces the ‘anti’ conformation, where the two methyl groups are positioned opposite each other and steric repulsion is minimized.
The anti conformation is the lowest energy conformation for butane. The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
Interactive Element
Figure \(1\): A 3D Structure of the Anti Butane Conformer.
At room temperature, butane is most likely to be in the lowest-energy anti conformation at any given moment in time, although the energy barrier between the anti and eclipsed conformations is not high enough to prevent constant rotation except at very low temperatures. For this reason (and also simply for ease of drawing), it is conventional to draw straight-chain alkanes in a zigzag form, which implies the anti conformation at all carbon-carbon bonds. For example octane is commonly drawn as:
Drawing Newman Projections
Newman projections are a valuable method for viewing the relative positions of groups within molecule. Being able to draw the Newman projection for a given molecule is a valuable skill and will be used repeatedly throughout organic chemistry. Because organic molecules often contain multiple carbon-carbon bonds it is important to precisely know which bond and which direction is being sighted for the Newman projection. The details of the Newman projection change given the molecule but for typical alkanes a full conformational analysis involves a full 360o rotation in 60o increments. This will produce three staggered conformers and three eclipsed conformers. Typically, the staggered conformers are more stable and the eclipsed conformers are less stable. The least stable conformer will have the largest groups eclipsed while the most stable conformer will have the largest groups anti (180o) to each other.
Example
Draw the Newman projection of 2,3 dimethylbutane along the C2-C3 bond. Then determine the least stable conformation.
First draw the molecule and locate the indicated bond:
Because the question asks for the least stable conformation, focus on the three possible eclipsed Newman projections. Draw out three eclipsed Newman projections as a template. Because it is difficult to draw a true staggered Newman projection, it is common to show the bonds slightly askew.
Place the substituents attached to the second carbon (C3) on the back bonds of all three Newman projections. In this example they are 2 CH3s and an H. Place the substituents in the same position on all three Newman projections.
Then place the substituents attached to the first carbon (C2) on the front bonds of the Newman projection. In this example, the substituents are also 2 CH3s and an H. Move the substituents through two 60o rotations to create the remaining two eclipsed Newman projections. Leave the substituents on the back carbon in place. Attempting to rotate the front and back carbons simultaneously is a common mistake and often leads to incorrect Newman projections.
Compare the Newman projections by looking the eclipsed interactions. Remember that the order of torsional strain interactions are CH3-CH3 > CH3-H > H-H. The third structure has two CH3-CH3 torsional interactions which will make it the least stable conformer of 2,3 dimethyl butane.
Example \(1\)
Draw Newman projections of the eclipsed and staggered conformations of propane, as if viewed down the C1-C2 bond.
Answer
Example \(2\)
Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.07%3A_Conformations_of_Other_Alkanes.txt |
Objectives
After completing this section, you should be able to
1. describe the general nature of petroleum deposits, and recognize why petroleum is such an important source of organic compounds.
2. explain, in general terms, the processes involved in the refining of petroleum.
3. define the octane number of a fuel, and relate octane number to chemical structure.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• catalytic cracking
• catalytic reforming
• fractional distillation
• octane number (octane rating)
Study Notes
The refining of petroleum into usable fractions is a very important industrial process. In the laboratory component of this course, you will have the opportunity to compare this industrial process to the distillation procedure as it is performed in the student laboratory.
Petroleum
The petroleum that is pumped out of the ground is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundance of the components vary depending on the source - Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules.
Gasoline
Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall from Chapter 1 that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in Figure 3.8.1 shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F) and becomes a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in Figure 3.8.1 shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs.
The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions are converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking, in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Using metals such as platinum brings about the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation.
Octane Ratings
The quality of a fuel is indicated by its octane rating, which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion (Figure 3.8.2), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level.
The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture:
$0.89(100) + 0.11(0) = 89 \label{3.8.1}$
As shown in Table $1$, many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, anti-knock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyl lead [(C2H5)4Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place. They combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing.
Table $1$: The Octane Ratings of Some Hydrocarbons and Common Additives
Name Condensed Structural Formula Octane Rating Name Condensed Structural Formula Octane Rating
n-heptane CH3CH2CH2CH2CH2CH2CH3 0 o-xylene skeletal structure of o-xylene.cdxml 107
n-hexane CH3CH2CH2CH2CH2CH3 25 ethanol CH3CH2OH 108
n-pentane CH3CH2CH2CH2CH3 62 t-butyl alcohol (CH3)3COH 113
isooctane (CH3)3CCH2CH(CH3)2 100 p-xylene 116
benzene 106 methyl t-butyl ether H3COC(CH3)3 116
methanol CH3OH 107 toluene 118 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.08%3A_Gasoline_-_A_Deeper_Look.txt |
Concepts & Vocabulary
• Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity.
• Functional groups have characteristic names that often carry over into the naming of compounds.
• The most common organic functional groups that will be encountered is this course are: alkanes, alkenes, alkynes, arenes, (alkyl and aryl) halides, alcohols, ethers, aldehydes, ketones, esters, carboxylic acids, acid chlorides, amides, amines, nitriles, nitro compounds, sulfides and sulfoxides.
3.2: Alkanes and Alkane Isomers
• Hydrocarbons are a common class of organic molecules that contain only carbon and hydrogen atoms.
• Alkanes are one type of hydrocarbon that contains only carbon-carbon and carbon hydrogen single bonds.
• Straight chain and branched alkanes have the generic formula CnH2n+2, where n is equal to the number of carbons. Cycloalkanes have the generic formula CnH2n.
• Structural isomers are molecules with the same molecular formula, but different structures.
3.3: Alkyl Groups
• Alkyl groups are small hydrocarbon chains attached to the parent alkane chain. The names of alkyl groups use the same prefixes to indicate the number of carbons (meth-, eth-, etc.), but use "-yl" as the ending, instead of "-ane".
3.4: Naming Alkanes
• The IUPAC System of nomenclature provides a set of rules for assigning every molecule a unique name.
3.5: Properties of Alkanes
• The boiling point of an alkane depends upon molecular weight and number of branches in the chain. Boiling points tend to increase with increasing molecular weight. Boiling points tend to decrease within a set of isomers as the number of branches increases.
• Alkanes and cycloalkanes are generally more soluble in organic solvents than in water.
3.6: Conformations of Ethane
• Rotation about the carbon-carbon sigma bonds in ethane results in different rotational isomers (also known as conformational isomers or conformers). Newman projections are a very common way of depicting conformers.
• The two most common conformers of ethane are called staggered and eclipsed. The staggered conformer is lower in energy (more stable) than the eclipsed conformer, because it has less torsional strain.
3.7: Conformations of Other Alkanes
• Alkanes more complex than ethane, will have a greater variety of possible conformers. The anti and gauche conformers of butane are specific types of staggered conformations.
3.8: Gasoline - A Deeper Look
Summary Problems
Exercise \(1\)
Atorvastatin (Lipitor) is a synthetic pharmaceutical marketed by Pfizer for lowering blood cholesterol and is one of the top selling drugs in the world. Circle and label all functional groups in Lipitor. (Be sure to label any amines or alcohols as primary, secondary, or tertiary.)
Answer
Exercise \(2\)
Vincristine is a natural product that was originally isolated in 1958 and has been studied as an anticancer agent. Answer the following questions related to this molecule. a) Circle and label all functional groups. (Be sure to label any amines or alcohols as primary, secondary, or tertiary.) b) Clearly label one primary carbon, one secondary carbon, one tertiary carbon, and one quaternary carbon.
Answer
Exercise \(3\)
Tamiflu (oseltamivir) is an anti-influenza medication that was discovered by Gilead Sciences and is marketed by the pharmaceutical company Genetech. Circle and label all functional groups in Tamiflu. (Be sure to label any amines or alcohols as primary, secondary, or tertiary.)
Answer
Notes: Alkanes are not functional groups, so there is no reason to circle groups like methyl or ethyl. (They are structural fragments.) Also, this molecule does not contain an aromatic ring. This is a cyclohexene ring. An aromatic ring is benzene and requires three double bonds in a six-membered ring.
Exercise \(4\)
For the following structure, a) Draw a Newman projection looking down the C2-C3 bond. (C2 should be in front.) b) Rotate the C2-C3 bond by 180 degrees (keep 3, 4, and 5 in the same orientation as the original structure). Draw the new skeletal structure and the new Newman projection looking down the C2-C3 bond. c) Label the Newman projections from parts a and b as staggered or eclipsed.
Answer
a)
b)
Skills to Master
• Skill 3.1 Identify the following functional groups that are present in a given organic molecule: alkanes, alkenes, alkynes, arenes, (alkyl and aryl) halides, alcohols, ethers, aldehydes, ketones, esters, carboxylic acids, acid chlorides, amides, amines, nitriles, and nitro compounds.
• Skill 3.2 Name and draw structures of straight chain alkanes up to ten carbons in length.
• Skill 3.3 Name and draw structures for all the structural isomers of a given molecular formula.
• Skill 3.4 Identify methyl, primary, secondary, tertiary, and quaternary carbons in organic structures.
• Skill 3.5 Provide the IUPAC name of any given alkane or cycloalkane structure.
• Skill 3.6 Draw the structure of an alkane or cycloalkane given its IUPAC name.
• Skill 3.7 Arrange a series of alkanes in order of increasing or decreasing boiling point.
• Skill 3.8 Be able to draw Newman Projections of different conformers of alkanes.
• Skill 3.9 be able evaluate a conformer in terms of torsional and steric strain.
• Skill 3.10 Be able to identify the staggered, eclipsed, anti and gauche conformers of alkanes and to order them with respect to relative energy.
Memorization Tasks (MT)
• MT 3.1 Memorize the name and structure of each of the common functional groups listed in Skill 3.1.
• MT 3.2 Memorize the names and be able to draw the first ten straight chain alkanes.
• MT 3.3 Memorize the structures and common names of the alkyl substitutent groups - isopropyl, sec-butyl, isobutyl,, and tert-butyl. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.S%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry_%28Summary%29.txt |
Learning Objectives
After you have completed Chapter 4, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. draw the cis-trans isomers of some simple disubstituted cycloalkanes, and write the IUPAC names of such compounds.
3. define, and use in context, the key terms introduced in this chapter.
This chapter deals with the concept of stereochemistry and conformational analysis in cyclic compounds. The causes of various ring strains and their effects on the overall energy level of a cycloalkane are discussed. We shall stress the stereochemistry of alicyclic compounds.
• 4.1: Naming Cycloalkanes
Cycloalkanes have one or more rings of carbon atoms, and contain only carbon-hydrogen and carbon-carbon single bonds. The naming of cycloalkanes follows a set of rules similar to that used for naming alkanes.
• 4.2: Cis-Trans Isomerism in Cycloalkanes
Stereoisomers are molecules that have the same molecular formula, the same atom connectivity, but they differ in the relative spatial orientation of the atoms. Di-substituted cycloalkanes are one class of molecules that exhibit stereoisomerism. Di-substituted cycloalkane stereoisomers are designated by the nomenclature prefixes cis (Latin, meaning on this side) and trans (Latin, meaning across).
• 4.3: Stability of Cycloalkanes - Ring Strain
Small cycloalkanes, like cyclopropane, are dramatically less stable than larger cycloalkanes due to ring strain. Ring strain is caused by increased torsional strain, steric strain, and angle strain, in the small, nearly planar ring of cyclopropane. Larger rings like cyclohexane, have much lower ring straing because they adopt non-planar conformations.
• 4.4: Conformations of Cycloalkanes
Overall ring strain decreases in cycloalkane rings that are large enough to allow the carbon-carbon bonds to rotate away from planar structures. For this reason, cyclopentane is significantly more stable, than cyclopropane and cyclobutane.
• 4.5: Conformations of Cyclohexane
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations.
• 4.6: Axial and Equatorial Bonds in Cyclohexane
The hydrogens of cyclohexane exist in two distinct locations - axial and equatorial. The two chair conformations of cyclohexane rapidly interconverts through a process called ring flip.
• 4.7: Conformations of Monosubstituted Cyclohexanes
Mono-substituted cyclohexane prefers the ring flip conformer in which the substituent is equatorial. 1,3-diaxial interactions occur when the substituent is axial, instead of equatorial. The larger the substituent, the more pronounced the preference.
• 4.8: Conformations of Disubstituted Cyclohexanes
The most stable configurational isomer of a disubstituted cyclohexane will be the isomer that has the most stable conformational isomer.
• 4.9: Conformations of Polycyclic Molecules
Polycyclic molecules are common and important in nature. Biologically important polycyclic molecules are found in cholesterol, sex hormones, birth control pills, cortisone, and anabolic steroids
• 4.S: Organic Compounds- Cycloalkanes and their Stereochemistry (Summary)
Thumbnail: Ball-and-stick model of cyclobutane. (Public Domain; Ben Mills via Wikipedia)
04: Organic Compounds- Cycloalkanes and their Stereochemistry
Objectives
After completing this section, you should be able to
1. name a substituted or unsubstituted cycloalkane, given its Kekulé structure, shorthand structure or condensed structure.
2. draw the Kekulé, shorthand or condensed structure for a substituted or unsubstituted cycloalkane, given its IUPAC name.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• cycloalkane
Study Notes
Provided that you have mastered the IUPAC system for naming alkanes, you should find that the nomenclature of cycloalkanes does not present any particular difficulties.
Many organic compounds found in nature contain rings of carbon atoms. These compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The simplest examples of this class consist of a single, un-substituted carbon ring, and these form a homologous series similar to the unbranched alkanes.
Like alkanes, cycloalkane molecules are often drawn as skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens. Cyclohexane, one of the most common cycloalkanes is shown below as an example.
Cyclic hydrocarbons have the prefix "cyclo-". The IUPAC names, molecular formulas, and skeleton structures of the cycloalkanes with 3 to 10 carbons are given in Table 4.1.1. Note that the general formula for a cycloalkane composed of n carbons is CnH2n, and not CnH2n+2 as for alkanes. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so are still considered to be saturated with hydrogen.
Table \(1\): Examples of Simple Cycloalkanes
Cycloalkane Molecular Formula Skeleton Structure
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Cyclohexane C6H12
Cycloheptane C7H14
Cyclooctane C8H16
Cyclononane C9H18
Cyclodecane C10H20
IUPAC Rules for Nomenclature
The naming of substituted cycloalkanes follows the same basic steps used in naming alkanes.
1. Determine the parent chain.
2. Number the substituents of the ring beginning at one substituent so that the nearest substituent is numbered the lowest possible. If there are multiple choices that are still the same, go to the next substituent and give it the lowest number possible.
3. Name the substituents and place them in alphabetical order.
More specific rules for naming substituted cycloalkanes with examples are given below.
1. Determine the cycloalkane to use as the parent. If there is an alkyl straight chain that has a greater number of carbons than the cycloalkane, then the alkyl chain must be used as the primary parent chain. Cycloalkanes substituents have an ending "-yl". If there are two cycloalkanes in the molecule, use the cycloalkane with the higher number of carbons as the parent.
Example \(1\)
The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. The parent chain in this molecule is decane and cyclopropane is a substituent. The name of this molecule is 3-cyclopropyl-6-methyldecane.
Example \(2\)
Name the cycloalkane structure.
Solution
There are two different cycloalkanes in this molecule. Because it contains more carbons, the cyclopentane ring will be named as the parent chain. The smaller ring, cyclobutane, is named as a substituent on the parent chain. The name of this molecule is cyclobutylcyclopentane.
2) When there is only one substituent on the ring, the ring carbon attached to the substituent is automatically carbon #1. Indicating the number of the carbon with the substituent in the name is optional.
Example \(3\)
1-chlorocyclobutane or cholorocyclobutane 1-propylcyclohexane or propylcyclohexane
3) If there are multiple substituents on the ring, number the carbons of the cycloalkane so that the carbons with substituents have the lowest possible number. A carbon with multiple substituents should have a lower number than a carbon with only one substituent or functional group. One way to make sure that the lowest number possible is assigned is to number the carbons so that when the numbers corresponding to the substituents are added, their sum is the lowest possible.
4) When naming the cycloalkane, the substituents must be placed in alphabetical order. Remember the prefixes di-, tri-, etc. , are not used for alphabetization.
Example \(4\)
In this example, the ethyl or the methyl subsistent could be attached to carbon one. The ethyl group attachment is assigned carbon 1 because ethyl comes before methyl alphabetically. After assigning carbon 1 the cyclohexane ring can be numbered going clockwise or counterclockwise. When looking at the numbers produced going clockwise produces lower first substituent numbers (1,3) than when numbered counterclockwise (1,5). So the correct name is 1-ethyl-3-methylcyclohexane.
Example \(5\)
Name the following structure using IUPAC rules.
Solution
Remember when dealing with cycloalkanes with more than two substituents, finding the lowest possible 2nd substituent numbering takes precedence. Consider a numbering system with each substituent attachment point as being carbon one. Compare them and whichever produces the lowest first point of difference will be correct.
The first structure would have 1,4 for the relationship between the first two groups. The next structure would have 1,3. The final 2 structures both have 1,2 so those are preferable to the first two. Now we have to determine which is better between the final 2 structures. The 3rd substituent on structure 3 would be at the 5 position leading to 1,2,5 while in the final structure the 3rd methyl group is on carbon 4 leading to 1,2,4. This follows the rules of giving the lowest numbers at the first point of difference.
The correct name for the molecule is 4-Bromo-1,2-dimethylcyclohexane.
Example \(6\)
2-bromo-1-chloro-3-methylcyclopentane
Notice that "b" of bromo alphabetically precedes the "m" of methyl. Also, notice that the chlorine attachment point is assigned carbon 1 because it comes first alphabetically and the overall sum of numbers would be the same if the methyl attachment carbon was assigned as 1 and the chlorine attachment as 3.
Example \(7\)
(2-bromo-1,1-dimethylcyclohexane)
Although "di" alphabetically precedes "f", "di" is not used in determining the alphabetical order.
Example \(8\)
2-fluoro-1,1,-dimethylcyclohexane NOT 1,1-dimethyl-2-fluorocyclohexane
also
2-fluoro-1,1,-dimethylcyclohexane NOT 1-fluoro-2,2-dimethylcyclohexane (as that would give a larger number to the first point of difference)
Although "di" alphabetically precedes "f", "di" is not used in determining the alphabetical order of the substituents. Notice that the attachment point of the two methyl groups is assigned carbon 1 despite the fact that fluorine comes first alphabetically. This is because this assignment allows for a lower overall numbering of substituents, so assigning alphabetical priority is not necessary.
Polycyclic Compounds
Hydrocarbons having more than one ring are common, and are referred to as bicyclic (two rings), tricyclic (three rings) and in general, polycyclic compounds. The molecular formulas of such compounds have H/C ratios that decrease with the number of rings. In general, for a hydrocarbon composed of \(n\) carbon atoms associated with \(m\) rings the formula is: \(\ce{C_nH_{2n + 2 - 2m}}\). The structural relationship of rings in a polycyclic compound can vary. They may be separate and independent, or they may share one or two common atoms. Some examples of these possible arrangements are shown in the following table.
Table \(2\): Examples of Isomeric \(\ce{C_8H_{14}}\) Bicycloalkanes
Isolated Rings Spiro Rings Fused Rings Bridged Rings
No common atoms One common atom One common bond Two common atoms
Polycyclic compounds, like cholesterol shown below, are biologically important and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules, and will not be covered here.
Cholesterol (polycyclic)
Exercise \(1\)
Give the IUPAC names for the following cycloalkane structures.
Answer
a) 1,3-Dimethylcyclohexane
b) 2-Cyclopropylbutane
c) 1-Ethyl-3-methylcyclooctane
d) 1-Bromo-3-methylcyclobutane
e) 1,2,4-Triethylcycloheptane
f) 1-Chloro-2,4-dimethylcyclopentane
Exercise \(2\)
Draw the structures for the IUPAC names below.
1. 2,3-dicyclopropylpentane
2. 1,2,3-triethylcyclopentane
3. 3-cyclobutyl-2-methylhexane
4. 2-bromo-1-chloro-4-methylcyclohexane
5. 1-bromo-5-propylcyclododecane
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.01%3A_Naming_Cycloalkanes.txt |
Objectives
After completing this section, you should be able to
1. draw structural formulas that distinguish between cis and trans disubstituted cycloalkanes.
2. construct models of cis and trans disubstituted cycloalkanes using ball-and-stick molecular models.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• constitutional isomer
• stereoisomer
• cis-trans isomers
Previously, constitutional isomers have been defined as molecules that have the same molecular formula, but different atom connectivity. In this section, a new class of isomers, stereoisomers, will be introduced. Stereoisomers are molecules that have the same molecular formula, the same atom connectivity, but they differ in the relative spatial orientation of the atoms.
Cycloalkanes are similar to open-chain alkanes in many respects. They both tend to be nonpolar and relatively inert. One important difference, is that cycloalkanes have much less freedom of movement than open-chain alkanes. As discussed in Sections 3.6 and 3.7, open-chain alkanes are capable of rotation around their carbon-carbon sigma bonds. The ringed structures of cycloalkanes prevent such free rotation, causing them to be more rigid and somewhat planar.
Di-substituted cycloalkanes are one class of molecules that exhibit stereoisomerism. 1,2-dibromocyclopentane can exist as two different stereoisomers: cis-1,2-dibromocyclopentane and trans-1,2-dibromocyclopentane. The cis-1,2-dibromocyclopentane and trans-1,2-dibromocyclopentane stereoisomers of 1,2-dibromocyclopentane are shown below. Both molecules have the same molecular formula and the same atom connectivity. They differ only in the relative spatial orientation of the two bromines on the ring. In cis-1,2-dibromocyclopentane, both bromine atoms are on the same "face" of the cyclopentane ring, while in trans-1,2-dibromocyclopentane, the two bromines are on opposite faces of the ring. Stereoisomers require an additional nomenclature prefix be added to the IUPAC name in order to indicate their spatial orientation. Di-substituted cycloalkane stereoisomers are designated by the nomenclature prefixes cis (Latin, meaning on this side) and trans (Latin, meaning across).
Interactive Element
The 3D Structure of cis-1,2-dibromocyclopentane
Interactive Element
The 3D Structure of trans-1,2-dibromocyclopentane
Representing 3D Structures
By convention, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the plane of the ring (coming out of the page), a dashed line for bonds to atoms or groups located below the ring (going back into the page), and solid lines for bonds in the plane of the page.
In general, if any two sp3 carbons in a ring have two different substituent groups (not counting other ring atoms) cis/trans stereoisomerism is possible. However, the cis/trans designations are not used if both groups are on the same carbon. For example, the chlorine and the methyl group are on the same carbon in 1-chloro-1-methylcyclohexane and the trans prefix should not be used.
If more than two ring carbons have substituents, the stereochemical notation distinguishing the various isomers becomes more complex and the prefixes cis and trans cannot be used to formally name the molecule. However, the relationship of any two substituents can be informally described using cis or trans. For example, in the tri-substituted cyclohexane below, the methyl group is cis to the ethyl group, and also trans to the chlorine. However, the entire molecule cannot be designated as either a cis or trans isomer. Later sections will describe how to name these more complex molecules (5.5: Sequence Rules for Specifying Configuration)
Example \(1\)
Name the following cycloalkanes:
Solution
These two example represent the two main ways of showing spatial orientation in cycloalkanes.
a) In example "a" the cycloalkane is shown as being flat and in the plane of the page. The positioning of the substituents is shown by using dash-wedge bonds. Cis/trans positioning can be determined by looking at the type of bonds attached to the substituents. If the substituents are both on the same side of the ring (Cis) they would both have either dash bonds or wedge bonds. If the the substituents are on opposite side of the ring (Trans) one substituent would have a dash bond and the other a wedge bond. Because both bromo substituents have a wedge bond they are one the same side of the ring and are cis. The name of this molecule is cis-1,4-Dibromocyclohexane.
b) Example "b" shows the cycloalkane ring roughly perpendicular to the plane of the page. When this is done, the upper and lower face of the ring is defined and each carbon in the ring will have a bond one the upper face and a bond on the lower face. Cis substituents will either both be on the upper face or the lower face. Trans substituents will have one on the upper face and one one the lower face. In example "b", one of the methyl substituents is on the upper face of the ring and one is on the lower face which makes them trans to each other. The name of this molecule is trans-1,2-Dimethylcyclopropane.
Exercises
Exercise \(1\)
Draw the following molecules:
1. trans-1,3-dimethylcyclohexane
2. trans-1,2-dibromocyclopentane
3. cis-1,3-dichlorocyclobutane
Answer
2) Cis/Trans nomenclature can be used to describe the relative positioning of substituents on molecules with more complex ring structures. The molecule below is tesosterone, the primary male sex hormone. Is the OH and the adjacent methyl group cis or trans to each other? What can you deduce about the relative positions of the indicated hydrogens?
3) Name the following compounds:
Solutions
2) Both the OH and the methyl group have wedge bonds. This implies that they are both on the same side of the testosterone ring making them cis. Two of the hydrogens have wedge bonds while one has a wedge. This means two of the hydrogens are on one side of the testosterone ring while one is on the other side.
3)
Cis-1-Bromo-3-Chlorocyclobutane
Trans-1,4-Dimethylcyclooctane
Trans-1-Bromo-3-ethylcyclopentane | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.02%3A_Cis-Trans_Isomerism_in_Cycloalkanes.txt |
Objectives
After completing this section, you should be able to
1. describe the Baeyer strain theory.
2. describe how the measurement of heats of combustion provides information about the amount of strain present in a cycloalkane ring.
3. determine the relative stability of cyclic compounds, by assessing such factors as angle strain, torsional strain and steric strain.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• angle strain
• steric strain
• torsional strain
• ring strain
• heat of combustion
Heat of Combustion as a Measure of Bond Strength
The combustion of carbon compounds, especially hydrocarbons, has been the most important source of heat energy for human civilizations throughout recorded history. The practical importance of this reaction cannot be denied, but the massive and uncontrolled chemical changes that take place in combustion make it difficult to deduce mechanistic paths. Using the combustion of propane as an example, we see from the following equation that every covalent bond in the reactants has been broken and an entirely new set of covalent bonds have formed in the products. No other common reaction involves such a profound and pervasive change, and the mechanism of combustion is so complex that chemists are just beginning to explore and understand some of its elementary features.
$\ce{CH_3CH_2CH_3 + 5O_2 -> 3CO_2 + 4H_2O + heat} \nonumber$
Since all the covalent bonds in the reactant molecules are broken, the quantity of heat evolved in this reaction, and any other combustion reaction, is related to the strength of these bonds (and, of course, the strength of the bonds formed in the products). Precise heat of combustion measurements can provide useful information about the structure of molecules and their relative stability.
For example, heat of combustion is useful in determining the relative stability of isomers. Pentane has a heat of combustion of -782 kcal/mol, while that of its isomer, 2,2-dimethylpropane (neopentane), is –777 kcal/mol. These values indicate that 2,3-dimethylpentane is 5 kcal/mol more stable than pentane, since it has a lower heat of combustion.
Ring Strain
Table $1$ lists the heat of combustion data for some simple cycloalkanes. These cycloalkanes do not have the same molecular formula, so the heat of combustion per each CH2 unit present in each molecule is calculated (the fourth column) to provide a useful comparison. From the data, cyclopropane and cyclobutane have significantly higher heats of combustion per CH2, while cyclohexane has the lowest heat of combustion. This indicates that cyclohexane is more stable than cyclopropane and cyclobutane, and in fact, that cyclohexane has a same relative stability as long chain alkanes that are not cyclic. This difference in stability is seen in nature where six membered rings are by far the most common. What causes the difference in stability or the strain in small cycloalkanes?
Table $1$: Heats of combustion of select hydrocarbons
Cycloalkane
(CH2)n
CH2 Units
n
ΔH25º
kcal/mol
ΔH25º
per CH2 Unit
Ring Strain
kcal/mol
Cyclopropane n = 3 468.7 156.2 27.6
Cyclobutane n = 4 614.3 153.6 26.4
Cyclopentane n = 5 741.5 148.3 6.5
Cyclohexane n = 6 882.1 147.0 0.0
Cycloheptane n = 7 1035.4 147.9 6.3
Cyclooctane n = 8 1186.0 148.2 9.6
Cyclononane n = 9 1335.0 148.3 11.7
Cyclodecane n = 10 1481 148.1 11.0
CH3(CH2)mCH3 m = large 147.0 0.0
The Baeyer Theory on the Strain in Cycloalkane Rings
In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane are less stable than cyclohexane, because the the smaller rings are more "strained". There are many different types of strain that contribute to the overall ring strain in cycloalkanes, including angle strain, torsional strain, and steric strain. Torsional strain and steric strain were previously defined in the discussion of conformations of butane. Angle Strain occurs when the sp3 hybridized carbons in cycloalkanes do not have the expected ideal bond angle of 109.5o, causing an increase in the potential energy. An example of angle strain can be seen in the diagram of cyclopropane below in which the bond angle is 60o between the carbons. The compressed bond angles causes poor overlap of the hybrid orbitals forming the carbon-carbon sigma bonds which in turn creates destabilization.
The C-C-C bond angles in cyclopropane (diagram above) (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o. This bond angle causes cyclopropane and cyclobutane to be less stable than molecules such as cyclohexane and cyclopentane, which have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o. Changes in chemical reactivity as a consequence of angle strain are dramatic in the case of cyclopropane, and are also evident for cyclobutane.
In addition to angle strain, there is also steric (transannular) strain and torsional strain in many cycloalkanes. Transannular strain exists when there is steric repulsion between atoms.
Because cycloalkane lack the ability to freely rotate, torsional (eclipsing) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond. Torsional strain is especially prevalent in small cycloalkanes, such as cyclopropane, whose structures are nearly planar.
The Eclipsed C-H Bonds in Cyclopropane
Larger rings like cyclohexane, deal with torsional strain by forming conformers in which the rings are not planar. A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds, which decreases the angle and torsional strain in the ring. The non-planar structures of cyclohexane are very stable compared to cyclopropane and cyclobutane, and will be discussed in more detail in the next section.
The Types of Strain Which Contribute to Ring Strain in Cycloalkanes
Angle Strain The strain caused by the increase or reduction of bond angles
Torsional Strain The strain caused by eclipsing bonds on adjacent atoms
Steric Strain The strain caused by the repulsive interactions of atoms trying to occupy the same space
Exercise $1$
trans-1,2-Dimethylcyclobutane is more stable than cis-1,2-dimethylcyclobutane. Explain this observation.
Answer
The trans form does not have eclipsing methyl groups, therefore lowering the energy within the molecule. It does however have hydrogen-methyl eclipsing interactions which are not as high in energy as methyl-methyl interactions.
Exercise $2$
Cyclobutane has more torsional stain than cyclopropane. Explain this observation.
Answer
Cyclobutane has 4 CH2 groups while cyclopropane only has 3. More CH2 groups means cyclobutane has more eclipsing H-H interactions and therefore has more torsional strain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.03%3A_Stability_of_Cycloalkanes_-_Ring_Strain.txt |
Objectives
After completing this section, you should be able to
1. describe, and sketch the conformation of cyclopropane, cyclobutane, and cyclopentane.
2. describe the bonding in cyclopropane, and use this to account for the high reactivity of this compound.
3. analyze the stability of cyclobutane, cyclopentane and their substituted derivatives in terms of angular strain, torsional strain and steric interactions.
Study Notes
Notice that in both cyclobutane and cyclopentane, torsional strain is reduced at the cost of increasing angular (angle) strain.
Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different.
cyclopropane cyclobutane cyclopentane cyclohexane cycloheptane cyclooctane
Cyclopropane is necessarily planar (flat), with the carbon atoms at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 40 kJ/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible.
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360º, going through many different conformations. That complete rotation isn't possible in a cyclic system, because the parts that would be trying to twist away from each other would still be connected together. Thus cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation".
Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings. Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them. In nature, three- to six-membered rings are frequently encountered, so we'll focus on those.
Cyclopropane
A three membered ring has no rotational freedom whatsoever. A plane is defined by three points, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane. This lack of flexibility does not allow cyclopropane to form more stable conformers which are non-planar.
The main source of ring strain in cyclopropane is angle strain. All of the carbon atoms in cyclopropane are tetrahedral and would prefer to have a bond angle of 109.5o The angles in an equilateral triangle are actually 60o, about half as large as the optimum angle. The large deviation from the optimal bond angle means that the C-C sigma bonds forming the cyclopropane ring are bent. Maximum bonding occurs when the overlapping orbitals are pointing directly toward each other. The severely strained bond angles in cyclopropane means that the orbitals forming the C-C bonds overlap at a slight angle making them weaker. This strain is partially overcome by using so-called “banana bonds”, where the overlap between orbitals is no longer directly in a line between the two nuclei, as shown here in three representations of the bonding in cyclopropane:
The constrained nature of cyclopropane causes neighboring C-H bonds to all be held in eclipsed conformations. Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a Newman projections of cyclopropane as shown from the side.
Newman Projection of cyclopropane
Cyclopropane isn't large enough to introduce any steric strain. Steric strain does not become a factor until we reach six membered rings. Before that point, rings are not flexible enough to allow for two ring substituents to interact with each other.
Interactive Element
The 3D Structure of Cyclopropane
The combination of torsional and angle strain creates a large amount of ring strain in cyclopropane which weakens the C-C ring bonds (255 kJ/mol) when compared to C-C bonds in open-chain propane (370 kJ/mol).
Cyclobutane
Cyclobutane is a four membered ring. The larger number of ring hydrogens would cause a substantial amount of torsional strain if cyclobutane were planar.
In three dimensions, cyclobutane is flexible enough to buckle into a "puckered" shape which causes the C-H ring hydrogens to slightly deviate away from being completely eclipsed. This conformation relives some of the torsional strain but increases the angle strain because the ring bond angles decreases to 88o.
In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines.
The deviation of cyclobutane's ring C-H bonds away from being fully eclipsed can clearly be seen when viewing a Newman projections signed down one of the C-C bond.
Newman projection of cyclobutane
• With bond angles of 88o rather than 109.5o degrees, cyclobutane has significant amounts of angle strain, but less than in cyclopropane.
• Although torsional strain is still present, the neighboring C-H bonds are not exactly eclipsed in the cyclobutane's puckered conformation.
• Steric strain is very low. Cyclobutane is still not large enough that substituents can reach around to cause crowding.
• Overall the ring strain in cyclobutane (110 kJ/mol) is slightly less than cyclopropane (115 kJ/mol).
Interactive Element
The 3D Structure of Cyclobutane
Cyclopentane
Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common cycloalkane ring in nature, after cyclohexanes. Planar cyclopentane has virtually no angle strain but an immense amount of torsional strain. To reduce torsional strain, cyclopentane addops a non-planar conformation even though it slightly increases angle strain.
The lowest energy conformation of cyclopentane is known as the ‘envelope’, with four of the ring atoms in the same plane and one out of plane (notice that this shape resembles an envelope with the flap open). The out-of-plane carbon is said to be in the endo position (‘endo’ means ‘inside’). The envelope removes torsional strain along the sides and flap of the envelope. However, the neighboring carbons are eclipsed along the "bottom" of the envelope, away from the flap.
3D structure of cyclopentane (notice that the far top right carbon is the endo position).
At room temperature, cyclopentane undergoes a rapid bond rotation process in which each of the five carbons takes turns being in the endo position.
Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four. The envelope removes torsional strain along the sides and flap of the envelope by allowing the bonds to be in an almost completely staggered position. However, the neighboring bonds are eclipsed along the "bottom" of the envelope, away from the flap. Viewing a Newman projections of cyclopentane signed down one of the C-C bond show the staggered C-H bonds.
Newman projection of cyclopentane
• The angle strain in the envelope conformation of cyclopentane is low. The ideal angle in a regular pentagon is about 107o, very close to the preferred 109.5o tetrahedral bond angle.
• There is some torsional strain in cyclopentane. The envelope conformation reduces torsional strain by placing some bonds in nearly staggered positions. However, other bonds are still almost fully eclipsed.
• Cyclopentane is not large enough to allow for steric strain to be created.
• Overall, cyclopentane has very little ring strain (26 kJ/mol) when compared to cyclopropane and cyclobutane.
Interactive Element
The 3D Structure of Cyclopentane
C3-C5 Cycloalkanes in Nature
If one of the carbon-carbon bonds is broken in cyclopropane or cyclobutane, the ring will ‘spring’ open, releasing energy as the bonds reassume their preferred tetrahedral geometry. The effectiveness of two antibiotic drugs, fosfomycin and penicillin, is due in large part to the high reactivity of the three- and four-membered rings in their structures.
One of the most important five-membered rings in nature is a sugar called ribose – DNA and RNA are both constructed upon ‘backbones’ derived from ribose. Pictured below is one thymidine (T) deoxy-nucleotide from a stretch of DNA. Since the ribose has lost one of the OH groups (at carbon 2 of the ribose ring), this is part of a deoxyribonucleic acid (DNA). If the OH at carbon 2 of the ribose ring was present, this would be part of a ribonucleic acid (RNA).
The lowest-energy conformations for ribose are envelope forms in which either C3 or C2 are endo, on the same side as the C5 substituent.
Exercises
1) If cyclobutane were to be planar, how many H-H eclipsing interactions would there be? Assuming 4 kJ/mol per H-H eclipsing interaction what would the strain be on this “planar” molecule?
2) In the two conformations of trans-1,2-Dimethylcyclopentane one is more stable than the other. Explain why this is.
3) In methylcyclopentane, which carbon would most likely be in the endo position?
Solutions
1) There are 8 eclipsing interactions (two per C-C bond). The extra strain on this molecule would be 32 kJ/mol (4 kJ/mol x 8).
2) The first conformation is more stable. Even though the methyl groups are trans in both models, they are anti to one another in the first structure (which is lower energy) while they are gauche in the second structure increasing strain within the molecule.
3) The ring carbon attached to the methyl group would most likely be the endo carbon. The large methyl group would create the most torsional strain if eclipsed. Being in the endo position would place the bonds is a more staggered position which would reduce strain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.04%3A_Conformations_of_Cycloalkanes.txt |
Objectives
After completing this section, you should be able to
1. explain why cyclohexane rings are free of angular strain.
2. draw the structure of a cyclohexane ring in the chair conformation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• chair conformation
• twist-boat conformation
We will find that cyclohexanes tend to have the least angle strain and consequently are the most common cycloalkanes found in nature. A wide variety of compounds including, hormones, pharmaceuticals, and flavoring agents have substituted cyclohexane rings.
testosterone, which contains three cyclohexane rings and one cyclopentane ring
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring). Cyclohexane has the possibility of forming multiple conformations each of which have structural differences which lead to different amounts of ring strain.
planar structure
severe angle strain (120°)
severe eclipsing strain (all bonds)
small steric strain
boat conformation
slight angle strain
eclipsing strain at two bonds
steric crowding of two hydrogens
twist boat conformation
slight angle strain
small eclipsing strain
small steric strain
chair conformation
no angle strain
no eclipsing strain
small steric strain
Conformations of Cyclohexane
A planar structure for cyclohexane is clearly improbable. The bond angles would necessarily be 120º, 10.5º larger than the ideal tetrahedral angle. Also, every carbon-hydrogen bond in such a structure would be eclipsed. The resulting angle and eclipsing strains would severely destabilize this structure. The ring strain of planar cyclohexane is in excess of 84 kJ/mol so it rarely discussed other than in theory.
Cyclohexane in the strained planar configuration showing how the hydrogens become eclipsed.
Chair Conformation of Cyclohexane
The flexibility of cyclohexane allows for a conformation which is almost free of ring strain. If two carbon atoms on opposite sides of the six-membered ring are bent out of the plane of the ring, a shape is formed that resembles a reclining beach chair. This chair conformation is the lowest energy conformation for cyclohexane with an overall ring strain of 0 kJ/mol. In this conformation, the carbon-carbon ring bonds are able to assume bonding angles of ~111o which is very near the optimal tetrahedral 109.5o so angle strain has been eliminated.
Also, the C-H ring bonds are staggered so torsional strain has also been eliminated. This is clearly seen when looking at a Newman projection of chair cyclohexane sighted down the two central C-C bonds.
Newman projection of cyclohexane
Interactive Element
The 3D Structure of Chair Cyclohexane
Boat Conformation of Cyclohexane
The Boat Conformation of cyclohexane is created when two carbon atoms on opposite sides of the six-membered ring are both lifted up out of the plane of the ring creating a shape which slightly resembles a boat. The boat conformation is less stable than the chair form for two major reasons. The boat conformation has unfavorable steric interactions between a pair of 1,4 hydrogens (the so-called "flagpole" hydrogens) that are forced to be very close together (1.83Å). This steric hindrance creates a repulsion energy of about 12 kJ/mol. An additional cause of the higher energy of the boat conformation is that adjacent hydrogen atoms on the 'bottom of the boat' are forced into eclipsed positions. For these reasons, the boat conformation about 30 kJ/mol less stable than the chair conformation.
A boat structure of cyclohexane (the interfering "flagpole" hydrogens are shown in red)
Twist-Boat Conformation of Cyclohexane
The boat form is quite flexible and by twisting it at the bottom created the twist-boat conformer. This conformation reduces the strain which characterized the boat conformer. The flagpole hydrogens move farther apart (the carbons they are attached to are shifted in opposite directions, one forward and one back) and the eight hydrogens along the sides become largely but not completely staggered. Though more stable than the boat conformation, the twist-boat (sometimes skew-boat) conformation is roughly 23 kJ/mol less stable than the chair conformation.
A twist-boat structure of cyclohexane
Half Chair Conformation of Cyclohexane
Cyclohexane can obtain a partially plane conformation called "half chair" but with only with excessive amounts of ring strain. The half chair conformation is formed by taking planar cyclohexane and lifting one carbon out of the plane of the ring. The half chair conformation has much of the same strain effects predicted by the fully planar cyclohexane. In the planar portion of half chair cyclohexane the C-C bond angles are forced to 120o which creates significant amounts of angle strain. Also, the corresponding C-H bonds are fully eclipsed which create torsional strain. The out-of-plane carbon allows for some of the ring's bond angles to reach 109.5o and for some of C-H bonds to not be fully eclipsed. Overall, the half chair conformation is roughly 45 kJ/mol less stable than the chair conformation.
Conformation Changes in Cyclohexane - "Ring Flips"
Cyclohexane is rapidly rotating between the two most stable conformations known as the chair conformations in what is called the "ring flip" shown below. The importance of the ring flip will be discussed in the next section.
"Ring flip" describes the rapid equilibrium of cyclohexane rings between the two chair conformations
It is important to note that one chair does not immediately become the other chair, rather the ring must travel through the higher energy conformations as transitions. At room temperature the energy barrier created by the half chair conformation is easily overcome allowing for equilibration between the two chair conformation on the order of 80,000 times per second. Although cyclohexane is continually converting between these different conformations, the stability of the chair conformation causes it to comprises more than 99.9% of the equilibrium mixture at room temperature.
1" id="MathJax-Element-12-Frame" role="presentation" style="position:relative;" tabindex="0">Image of energy diagram of cyclohexane conformations
1" role="presentation" style="position:relative;" tabindex="0">1
Exercises
1) Consider the conformations of cyclohexane: half chair, chair, boat, twist boat. Order them in increasing ring strain in the molecule.
Solutions
1) Chair < Twist Boat < Boat < half chair (most ring strain) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.05%3A_Conformations_of_Cyclohexane.txt |
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