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The curved arrows help you in drawing the intermediate. Because you can use curved arrows in only three ways bond to lone pair, bond to bond, and lone pair to bond , you have limited options for drawing intermediates. In the next example, a nucleophile attacks a double bond. In this case, the nucleophile is the hydroxide ion. The process begins with the hydroxide ion attacking the carbon atom at one end of the carbon-carbon bond. This is a lone pair—to—bond step. This is a bond-tobond transfer.

Finally, a bond—to—lone pair transfer takes place. Nucleophilic attack of a double bond. It may help to remember that charge will be conserved. The nucleophile is at the tail of the arrow and the electrophile is at the head of the arrow. Mechanisms Some materials may behave as either a base or a nucleophile. The hydroxide ion is an example. When the nucleophile attacks and removes a hydrogen ion, it is behaving as a base. When the nucleophile is attacking at some other point than a hydrogen atom, it is acting like a nucleophile. For example, both the methoxide ion CH3O— and the t-butoxide ion CH3 3O— are strong bases, but only the methoxide ion is a strong nucleophile.

The t-butoxide ion is too big and bulky to attack efficiently. The effect of the bulky nature of the t-butoxide ion on its reactivity is an example of steric hindrance, which was discussed in your Organic I course and, naturally, in Organic I For Dummies. A molecule with a lone pair of electrons to donate can behave as a nucleophile. The strength of the nucleophile the nucleophilicity is often related to basicity. A strong nucleophile is usually a strong base and vice versa.

The base strength is shown by its equilibrium constant. On the other hand, nucleophilicity refers to the ability of a lone pair of electrons to attack a carbon on an electrophile. When working with nucleophiles, keep a few additional points in mind: Keys to substitution and elimination mechanisms Four types of mechanisms are inherent to Organic Chemistry I. The principles of these four types apply to Organic Chemistry II, and no review would be complete without a few reminders about these processes. The SN refers to a nucleophilic substitution process where some nucleophile attacks an electrophile and substitutes for some part of the electrophile.

The E refers to an elimination process where the nucleophile attacks an electrophile and causes the elimination of something. The 1 and 2 refer to the order of the reaction. A 1 first order means only one molecule determines the rate of the reaction, whereas a 2 second order means that a combination of two molecules determines the rate of the reaction. In many cases, two or more of these mechanisms are competing and more than one product may result. Brushing Up on Important Organic Chemistry I Concepts Increasing the strength of the nucleophile increases the likelihood of a substitution occurring instead of elimination.

Increasing substitution on the electrophile tends to increase the likelihood of a first-order process over a second-order process. All nucleophilic substitution reactions require a good leaving group. The following four lists summarize the main features of each of these mechanisms. Conjugated Unsaturated Systems Figure H Allylic radical resonance H hybrid. The simplest conjugated system is 1,3-butadiene, so we use it here to illustrate the behavior of all conjugated systems.

The structure of 1,3-butadiene is shown in Figure In this compound, the single bond is shorter than the single bond in ethane CH3-CH3. The sp2 hybridized carbon atoms are more electronegative than the sp3 hybridized carbon atoms. They pull the electrons closer to the nuclei, making the atoms smaller and causing the carbon atoms to move closer together. Molecular orbital theory predicts some double-bond character between the middle carbon atoms. At room temperature, the cis and trans forms of 1,3-butadiene are in equilibrium.

The equilibrium favors the trans form, with the distribution 5 percent cis and 95 percent trans. Figure shows the equilibrium and the structures of the two forms. The equilibrium between cis and trans 1,3-butadiene. Organic Chemistry II requires an extension of the basic rules of resonance to other systems.

In addition to constructing reasonable resonance structures, you also need to understand which structures are more stable. In general, resonance makes a species more stable by delocalizing the electrons. Delocalization, among other things, reduces electron-electron repulsion. Following is an expanded set of rules for drawing resonance structures. This list is an expansion of the rules necessary to understand resonance for Organic Chemistry I.

You may want to bookmark this list because these rules apply throughout the remainder of this book. Never, ever move any atoms. All the structures must have reasonable Lewis structures. This includes the same overall charge same number of electrons. For example, carbon should never have five bonds. Second period elements can never exceed an octet of electrons.

Also keep the following rules in mind: Structures doing the opposite are less important less stable. However, such structures are not necessarily impossible. In most cases, this is an octet. No change in the number of unpaired electrons should occur.

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Conjugated Unsaturated Systems 5. An sp3 hybridized carbon atom will not be involved in resonance. The presence of resonance leads to stabilization, which means that the species is more stable than any of the contributing structures. Equivalent resonance structures are equally important to the overall structure. The more stable the resonance structure, the more important it is the more it contributes to the hybrid. In general, the more stable resonance structure is the one with more bonds.

Stability of conjugated unsaturated systems You can estimate the stability of a conjugated versus nonconjugated system by comparing the energy changes. Reactions of Conjugated Unsaturated Systems Many students make the mistake of reacting conjugated systems as extensions of the reactions they learned for simple alkenes and alkynes.

While in some cases this is acceptable, the unique nature of these systems makes their chemistry different. You investigate some of these features in the next few pages. Put in the second string: Substitution reactions The reaction of chlorine with propene illustrates one difference caused by conjugation. The products of the reaction depend upon the reaction conditions, as illustrated in Figure Two reactions of propene with chlorine. These conditions minimize the possibility of forming chlorine atoms chlorine free radicals , and the presence of oxygen traps the few that do form.

However, when the conditions promote the formation of chlorine atoms, a substitution occurs to produce 3-chloropropene. The mechanism Allylic halogenation is a substitution reaction involving a free-radical mechanism. The general mechanism is in Figure CH2 CH A secondary free radical. The structure of NBS. However, conjugated dienes may behave differently.

An example is the reaction of HBr with 1,3-butadiene as illustrated in Figure In this case, no stereochemistry is implied. The distribution of the products depends on the reaction conditions shown in Table The information in the last column of the table indicates the process is reversible and an equilibrium results upon heating.

The equilibrium leads to the production of the more stable product. The reaction begins with the protonation of one of the carbon-carbon double bonds see Figure by the hydrogen ion from the HBr. A primary or a secondary carbocation can be formed by this reaction. As seen in Organic Chemistry I, a secondary carbocation is more stable than a primary carbocation. The protonation of CH2 CH 1,3-butadiene. Attack on the second carbon gives 1,2-addition, while attack on the fourth carbon gives 1,4-addition. Bromide CH3 CH attack on the allylic carbocation.

Figure uses a reaction diagram to illustrate this situation. The two-step process requires a diagram with two hills. The first step is the same for both products, so the second step is the one that makes the difference. At low temperatures, fewer molecules have sufficient kinetic energy to get over the higher barrier.

Therefore, the 1,2-addition product lower barrier is likely to form. The reaction diagram for 1,2-addition and 1,4-addition. This formation is especially noticeable at low temperatures because they always favor the kinetic product, which is the reaction product with the lower activation energy barrier. In addition, few molecules have sufficient energy to surmount the barrier in the reverse direction to allow the establishment of an equilibrium.

This reaction forms the thermodynamic product.

Organic Chemistry 2 Final Exam Test Review - Reagents & Reaction Mechanisms

At higher temperatures, more molecules have sufficient energy to cross the second barrier in the reverse direction and establish an equilibrium. The equilibrium allows the less stable 1,2-addition product to convert to the more stable 1,4-addition product. Low temperatures favor the kinetic product and high temperatures favor the thermodynamic product. More than a tree: Diels-Alder reactions In the Diels-Alder reaction, a diene, such as 1,3-butadiene, reacts with a dienophile, such as ethylene, to form a product with a six-membered ring.

This is an important reaction, not only to students trying to pass Organic Chemistry, but also in organic synthesis. Any of a number of dienes react as long as a conjugated system is present. Substituents attached to the conjugated system alter the reactivity of the diene. The dienophile is typically a substituted alkene; however, as you see later in this chapter, other species may react. The substituents around the double bond also alter the reactivity of the dienophile.

Figure illustrates the general reaction. In this figure, the arrows are for apparent, not actual, electron movement as a means of keeping track of the process. The final double bond between carbons two and three is between the positions of the double bonds in the original diene. In this example, the aldehyde is an electron-withdrawing group the electronegative oxygen pulls electron density away from the double bond.

The polarity arrow illustrates this electron shift. This shift of electron density speeds up the reaction a lower temperature is necessary. Conditions As long as a diene and a dienophile are present, a Diels-Alder reaction occurs. However, the yield of the reaction can be improved by adjusting the reactivity of the reactants. Conjugated Unsaturated Systems 1 2 Figure Diels-Alder 3 reaction of a substituted dienophile.

Examples of electron-donating groups are alcohols, ethers, and amines. The dienophile has the opposite requirement; that is, an electron-withdrawing group EWG facilitates the reaction. Examples of electronwithdrawing groups are carbonyl groups, cyano groups, and nitro groups. In order for a reaction to occur, the diene must be in the cis configuration and not the trans configuration refer to Figure Normally, both forms are in equilibrium. However, the diene can be locked into the cis conformation and facilitate the reaction.

One way to lock the conformation is to use a ring system. The compound 1,3-cyclopentadiene contains a diene locked in the cis conformation. Figure illustrates the reaction of 1,3-cyclopentadiene with ethylene. The reaction of 1,3-cyclopentadiene with ethylene. CH2 CH2 or Stereochemistry The Diels-Alder reaction is a concerted syn addition meaning the addition is on one side with the stereochemistry of the dienophile preserved in the stereochemistry of the product. If the dienophile is cis then the product is also cis, and if the dienophile is trans, the product is also trans.

See Figure for the attack by a cis dienophile. The reaction in Figure is the reaction of isoprene with maleic anhydride cis. The reac- CH3 tion of a cis dienophile. The reaction of a trans dienophile. The two products are the endo- and the exo-product. The endo-product is the major or only product of the reaction. The process leading to the endo-product is Alder endo-addition.

The endo-form is more stable than the exo-form. O H O H Figure Two possible cis products. The endo Endo and exo forms. In order to gain insight into the reaction itself and to increase your score on your organic exams , in this section we look at some typical questions that you might face. Two general types of Diels-Alder questions commonly appear on organic chemistry exams. These may be simple stand-alone questions or part of a larger question.

One type of question involves identifying the product, and the other type of question involves identifying the reactants.

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Indentifying the product What is the Diels-Alder product from the reaction in Figure ? H 65 66 Part I: The product of the reaction in Figure Identifying the reactants What reactants are necessary to form the product in Figure ? O CH3 Figure The product of a Diels-Alder CH3 reaction. O The reactants necessary to form the product in Figure are in Figure In general, the absorption of energy and many times its subsequent release of energy leads to evidence about the presence of some feature of the molecule. In infrared IR spectroscopy, the absorption of energy in the infrared region of the spectrum gives information about the types of bonds present.

Ultraviolet-visible UV-vis spectroscopy provides info about the molecular orbital arrangement in a molecule. Mass spectroscopy uses energy to ionize, or to break up a molecule into ions. The masses of these ions give information concerning the size and structure of the original molecule. Finally, nuclear magnetic resonance NMR spectroscopy utilizes the absorption of energy in the radio wave portion of the spectrum to give information concerning the environments occupied by certain nuclei, especially 1H and 13C.

In this chapter you find out how these different types of spectroscopy can be used to learn about organic molecules. It has some detailed sections on the theory behind instrumentation and exactly how it works. In principle, you can determine the complete structure of an organic molecule from just the IR, NMR, or mass spectrum of a compound. However, the process of determination can be very tedious. Organic chemists use all the data available when attempting to determine the structure of a compound.

Brushing Up on Important Organic Chemistry I Concepts For example, they gather some information from each of the spectra they have available and combine this data to produce a structure consistent with all the available data. When looking at a compound such as C10H20O, they may look at the IR to determine whether the oxygen atom is part of a carbonyl group, an alcohol group, or some other group. Then they may look at the NMR spectrum to get some idea about the carbon backbone. Later in this book you can see the specific functional groups and explore other features of the spectra that are characteristic of the group.

The methods in this chapter apply to the common types of organic compounds. Unusual compounds have their own characteristic behavior. Infrared Spectroscopy Light with energy in the infrared region of the electromagnetic spectrum has enough energy to cause a covalent bond to vibrate. The vibrations are due to the stretching and bending of the bonds.

If the vibration causes a change in the dipole moment of the molecule, energy will be absorbed. Most organic compounds have several absorptions in the 4,— cm—1 region of the spectrum. All of the absorptions give information about the structure of the molecule, but some absorptions are more useful than others are. The region below 1, cm—1 is the fingerprint region of the IR spectrum. The fingerprint region gets its name because this region tends to be unique for every compound, no matter how similar it may be to other molecules.

Instead, look primarily in the important places between 1, and 2, cm—1 and above 3, cm—1. The following sections offer a summary of some of the more important IR absorptions that occur in typical organic compounds. Double bonds The carbon-oxygen double bond and, to a lesser degree, the carbon-carbon double bond are found in many organic compounds. The carbonyl carbonoxygen double bond has a very sharp and intense band around 1, cm—1. Spectroscopy Revisited In many cases, this band is the most prominent feature of the IR spectrum of compounds containing a carbonyl group.

The peak due to a carbon-carbon double bond is characteristic of alkenes. Triple bonds Though not as common as double bonds, both carbon-carbon triple bonds alkynes and carbon-nitrogen triple bonds nitriles are important. Both occur in the 2,—2, cm—1 region of the spectrum. They are usually very sharp. The carbon-nitrogen triple bond tends to give a more intense peak than the alkyne peak.

O-H and N-H stretches Compounds containing a hydroxyl group OH have a strong very broad absorption in the 3,—2, cm—1 region of the spectrum. The most common examples are the alcohols and the carboxylic acids. The combination of a broad 3,—2, cm—1 band with a 1, cm—1 peak often indicates a carboxylic acid or amide — keep reading.

In addition, primary amines usually have two bands. The observation of a 3,—3, cm—1 band with a 1, cm—1 peak is often indicative of an amide — or carboxylic acid. C-H stretches Nearly every organic compound has one or more carbon-hydrogen bonds. Some guidelines are helpful: IR absorption spectra of various functional groups. Ultraviolet and Visible Spectroscopy Ultraviolet and visible spectroscopy UV-vis is an analytical technique useful in the investigation of some organic molecules.

Absorption of energy in Chapter 5: Spectroscopy Revisited this region of the electromagnetic spectrum can excite an electron from the ground state to an excited state — usually from the HOMO highest occupied molecular orbital to the LUMO lowest unoccupied molecular orbital. This technique is particularly useful for compounds containing multiple bonds. The useful spectral range is usually found between and nm. The region below nm is the vacuum ultraviolet, which requires special instrumentation and, for this reason, is less important.

The spectrum is the result of measuring the absorption of light versus wavelength. UV-vis spectra tend to be much simpler than IR spectra. Most UV-vis spectra contain only a few peaks; in many cases, only one or two. Organic compounds with no double bonds or only one double bond typically show absorption only in the vacuum ultraviolet portion of the spectrum.

The main use of this method is for molecules that have conjugated double bonds. The more double bonds in a conjugated system, the longer the wavelength is where the molecule absorbs light. The presence of eight or more conjugated double bonds can shift the absorption maximum into the visible portion of the spectrum, which occurs in many organic dyes. Compounds containing carbon-oxygen double bonds also exhibit absorptions in the UV-vis region.

In general, these absorptions occur at longer wavelengths than absorptions that are due to carbon-carbon double bonds. Conjugation shifts the absorption maximum to still longer wavelengths. In most cases, information from the UV-vis spectrum of a molecule is useful to corroborate data and information from other sources. This method is also useful in the quantitative determination of the concentration of an absorbing molecule. Figure shows a typical UV-vis spectrum. A typical UV-vis spectrum. Mass Spectroscopy In mass spectroscopy an organic molecule is vaporized and is injected into a mass spectrometer where it undergoes bombardment by electrons.

Other molecules fragment to produce pieces that may or may not be ions. The fragments with a positive charge are important. The mass spectrometer then takes the positive ions and sorts them according to their masses. The result is a mass spectrum. The molecular ion The molecular ion represents the molecular mass of the compound. Specifically, the molecular mass is the sum of the masses of the most abundant isotope of each element present.

You can derive an approximate formula for a compound based on its molecular mass, especially if only some or all of the following elements are present: If the compound contains or may contain nitrogen, the nitrogen rule is applicable. According to this rule, any molecule with an odd number of nitrogen atoms has an odd mass. The primary source of this ion is the presence of carbon This isotope is about 1. If more than one chlorine or bromine atom is present, the pattern is more complicated, but it includes a group of peaks separated by two mass units.

Fragmentation In addition to the molecular ion, the molecule generates a number of fragments. In general, the fragments result by breaking the weakest bonds. Different types of compounds often have characteristic fragmentation patterns. The loss of 15 mass units from a molecular ion generally indicates the loss of a methyl CH3 group.

The loss of 29 mass units often indicates the loss of an ethyl CH2CH3 group. Cleavage the breaking of a bond next to a heteroatom any atom other than carbon or hydrogen is also relatively common. This is particularly important if fragmentation involves the loss of a very stable molecule such as H2O or CO2 since this would indicate the presences of very particular functional groups and thus would give clues to the structure of the molecule.

NMR spectroscopy may be used on nuclei that behave as small magnets. Organic chemists usually rely on 1H proton and 13C as the most important isotopes because most organic compounds contain hydrogen and all organic compounds contain carbon. Nuclei that behave like small magnets have a magnetic moment are subject to the influence of other magnets. In an NMR spectrometer, the sample resides in a large external magnetic field.

This external field forces the nuclei to align themselves either with the field or against the field. Nuclei with one alignment 73 74 Part I: Brushing Up on Important Organic Chemistry I Concepts can absorb energy and switch to the other alignment and vice versa in a process called spin flipping.

When spin flipping occurs, the nucleus is in resonance. The energy required to induce this transition is in the radio frequency region of the electromagnetic spectrum. In addition to the external magnetic field, other magnetic fields influence the nuclei. The electrons in the molecule also have their own magnetic field. The field due to the electrons tends to oppose the external magnetic field, which results in electron shielding. The amount of shielding depends on the number of electrons. The more electron-shielding taking place, the lower the energy requirement for resonance resulting in a downfield shift.

The position of the absorption is referred to as the chemical shift. For 13C NMR, the chemical shift is normally in the 0— ppm region. Chemically equivalent nuclei absorb at the same energy level. Consider, for example, the structure of ethanol see Figure Three distinct types of hydrogen atoms appear in this structure. Therefore, the proton NMR spectrum of ethanol begins with three signals. The strucH ture of ethanol. Coupling results in the splitting of some of the absorptions.

Proton Proton NMR spectra follow the generalizations expressed in the previous sections, but this section discusses some additional factors. A proton adjacent to an atom that has a high electronegativity has a lower electron density than a proton adjacent to an atom with a low electronegativity. Therefore, a proton adjacent to an oxygen atom, for example, comes into resonance at a higher frequency than a proton adjacent to, for example, a carbon atom lower electronegativity.

Often, drawing all the hydrogens on the molecule can help you see the different chemical environments different interaction due to nearby hydrogen atoms each hydrogen experiences. Integration One important additional aspect of proton NMR spectra is the ability to integrate the spectra. Integration provides a means to determine how many hydrogen atoms are in each equivalent set, and it involves determining the area under each of the peaks in the spectrum.

The area is proportional to the number of hydrogen atoms contributing to that particular peak. For example, in the proton NMR of ethanol, the integration of the peak due to the alcoholic hydrogen would represent one hydrogen atom. The integration of the peak due to the CH2 group would be twice this value twice the area , because twice as many hydrogen atoms contribute to the intensity.

Finally, the integration of the CH3 absorption would be three times the OH absorption. The total of all the integrations must equal the total number of hydrogen atoms in the compound. In organic chemistry, coupling is another feature of proton NMR spectra. Coupling is the result of the interaction of the magnetic field of some hydrogen nuclei with other hydrogen nuclei. Hydrogen atoms on adjacent carbon atoms or separated by a double bond do couple. The amount of splitting is expressed by the coupling constant J.

The peak due to the hydrogen atoms in the CH3 group couples with the two hydrogen atoms of the adjacent CH2 group. The different types doublet, triplet, and so on exhibit a characteristic ratio of intensities. Doublets are equally intense. Triplets have a more intense central peak flanked by two equal peaks of lesser intensity.

A quartet has two equally intense central peaks with two smaller outer peaks that are equal to each other in intensity. See Figure for the NMR spectrum of ethanol. NMR spectrum of ethanol showing the splitting and intensities. However, in most organic compounds the hydrogen atoms on one carbon couple with hydrogen atoms on two or three adjacent carbon atoms, resulting in a multiplet of indistinguishable peaks appearing in the spectrum. In this case, the CH3 hydrogen atoms split the CH2 hydrogen atoms into a quartet and the remaining hydrogen atom on the CHClBr splits each of the members of the quartet into a doublet to give a total of eight peaks.

Instead of integration, the height of each absorption peak is approximately proportional to the number of carbon atoms contributing to the peak. This lack of coupling greatly simplifies the spectra. You can see this simplification in the carbon spectra of butyric acid in Figure O OH Butyric acid Figure Carbon spectrum of butyric acid.

Next you study benzene derivatives and heterocyclic aromatic compounds, and then we address the spectroscopy of these aromatic compounds. And in Chapters 7 and 8 we introduce you to aromatic substitution by both electrophiles and nucleophiles, and you get to see a lot of reactions and a lot of examples.

In this part you also start working with many more named reactions. One division deals with aliphatic fatty compounds, the first compounds you encountered in Organic Chemistry I.

Methane is a typical example of this type of compound. The second division includes the aromatic fragrant compounds, of which benzene is a typical example. Compounds in the two groups differ in a number of ways. The two differ chemically in that the aliphatic undergo free-radical substitution reactions and the aromatic undergo ionic substitution reactions.

In this chapter you examine the basics of both aromatic and heterocyclic aromatic compounds, concentrating on benzene and related compounds. Where It All Starts Benzene is the fundamental aromatic compound. An understanding of the behavior of many other aromatic compounds is much easier if you first gain an understanding of benzene. For this reason, you may find it useful to examine a few key characteristics of benzene, which we discuss in the following sections.

Later, chemists determined that it had the molecular formula C6H6. Further investigation of its chemical behavior showed that benzene was unlike other hydrocarbons in both structure and reactivity. Some proposed structures of benzene. This reaction produced only one monobromo product and three distinct dibromo products ortho, meta, and para.

These two forms are in Figure Br Br Br Br The ring structure with the rapidly moving double bonds explained many of the facts known about benzene at the time. However, as more information became available and as chemistry advanced, it became obvious that more was going on in this system than just rapidly interconverting structures. Chemists determined that only one benzene structure existed, not an equilibrium between two related structures. The term delocalized bonding refers to a situation in which one or more bonding pairs of electrons are spread out over a number of atoms.

The presence of resonance explains why the carboncarbon bonds in benzene are of equal length and strength. As contributing structures, they have no independent existence. The only form present is the hybrid. Notice the use of the double-headed arrow in Figure To review the resonance arrow and others, see Chapter 2. The proposed resonance structures of benzene. A representation of the resonance hybrid of benzene. The presence of the double bonds in the resonance structures typically implies that benzene should react like an alkene in terms of addition reactions and similar reactions.

However, as Figure shows, benzene does not react like alkenes. Examples of reactions where benzene does not behave like an alkene. This contributes to the observed stability of benzene. The stability of benzene One way to investigate the stability of benzene is to compare the amount of heat produced by the reactions of benzene to similar compounds that are not aromatic. For example, a simple comparison of the heat of hydrogenation for a series of related compounds allows us to see the difference.

Figure shows the hydrogenation of cyclohexane, 1,3-cyclohexadiene, and benzene, which make a suitable set because all three yield cyclohexane. The hydrogenation of cyclohexane, 1,3-cyclohexadiene, and benzene. If the reaction of one double bond releases this amount of energy, then the reaction of two double bonds 1,3-cyclohexadiene should release about twice this amount of energy. The classical, three double-bond benzene should Chapter 6: Introducing Aromatics release three times as much energy.

The small value for benzene indicates that it is significantly more stable than a triene. This difference is a direct measure of the resonance energy. Physical properties of benzene Benzene is a typical nonpolar compound that, like other nonpolar compounds, has a low solubility in water.

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It has a characteristic odor which most people find unpleasant. It was widely used in academic labs as a solvent, but that use has been largely discontinued since it was found that benzene may be carcinogenic. Table compares the melting and boiling points of benzene to those of cyclohexane, which indicates some differences found in aromatic compounds. The values for benzene are higher than the corresponding values for cyclohexane. In most other situations, when dealing with compounds with similar structures, the lower molecular-weight compound has the lower melting and boiling point; however, when comparing benzene to cyclohexane, you see that the reverse is true.

This is because of the delocalization of the electron pairs in benzene. We know that benzene is more stable than expected. The increased stability of benzene is due to resonance having a stabilizing influence. Following are the requirements for a species to be aromatic: This means that an sp3 atom cannot be part of an aromatic system. This last requirement is an important characteristic of all aromatic systems.

Next, solve for n, and if n is an integer a whole number , the system is aromatic. Two additional aromatic systems. Naphthalene Anthracene Common examples of systems often mistaken as being aromatic because of their alternating double and single bonds are cyclobutadiene and cyclooctatetraene shown in Figure In these two compounds, n is not an integer, so these systems are anti-aromatic nonaromatic. Introducing Aromatics Figure Cyclobutadiene Cyclooctatetraene Other aromatics A number of other molecules in addition to those shown in Figures and are aromatic.

The first five possible values of n are 0, 1, 2, 3, and 4. Pyridine refer to Figure illustrates the fact that aromatic compounds are not necessarily hydrocarbons. However, the replacement of the nitrogen in pyridine with oxygen places an sp3 hybridized atom in the ring, so the system is no longer aromatic. Two species in Figure are not molecules but ions. Aromaticity is not restricted to molecules. The Aromatic Family Benzene is an important and common aromatic compound. However, many other aromatic compounds are based on benzene: In this section we take a look at the properties and reactions of some of these compounds.

Nomenclature of the aromatic family The nomenclature naming of aromatic compounds begins with numbering the ring positions. This numbering is similar to the numbering of cycloalkane systems. The key group is at the number one position see Figure Older usage uses ortho o- in place of 1,2-numbers, meta m- in place of 1,3-numbers, and para p- in place of 1,4-numbers. Cl Two examples demonstrating different ways of naming Cl aromatic compounds.

The methyl groups may be at the one and two positions orthoxylene , the one and three positions meta-xylene , or the one and four positions para-xylene. Alternate names are o-xylene, m-xylene, and p-xylene. In the other cases, one group is attached at the number one position. All numbering begins at this position. Introducing Aromatics Branches of aromatic groups Aromatic groups may be branches on other systems. Be careful to notice that the third example, the benzyl group, is not truly aromatic.

The benzyl group consists of an aromatic ring and an antiaromatic -CH2— group. For a branch to be truly aromatic, the connection must be directly to the ring. Some aromatic group names plus the benzyl group. Heterocyclic Aromatic Compounds Up to this point, this chapter has discussed aromatic systems composed exclusively of rings of carbon atoms.

But aromatic systems can contain heteroatoms, which in this case means any atoms in the ring other than carbon. Single-bonded heteroatoms can donate a single lone-pair to the pi system but not two, because one lone-pair must be in an unhybridized p orbital orthogonal at 90 degrees to the sp2 ring plane.

Discovering Aromatic And Not So Aromatic Compounds Aromatic nitrogen compounds In addition to pyridine refer to Figure , other aromatic nitrogen systems exist and show up on organic chemistry tests. Two examples of aromatic nitrogen systems are shown in Figure Two nitrogencontaining aromatic systems. In both compounds, the heteroatom has two lone pairs. However, only one of the pairs is in a p-orbital perpendicular to the plane of the ring. The other electron pair is in the plane of the ring.

A sulfurand an oxygencontaining aromatic compound. Aromatic compounds, because of the delocalization of the electrons, have unique features in their spectra. In fact, spectral evidence can indicate what atoms or functional groups are attached to the aromatic ring or whether the ring itself contains an atom other than carbon. Aromatic compounds have a characteristic C-H peak near 3, cm—1. NMR The nuclear magnetic resonance spectrum of aromatic compounds commonly contain the following features: The downfield shift is due to the aromatic system.

The aromatic ring has a ring current, which gives rise to an induced field. The following tips help you interpret the mass spectral data of aromatic compounds: Chapter 7 Aromatic Substitution Part I: Nevertheless, a number of reactions involving aromatic systems can be carried out. However, with the exception of combustion, the conditions required by the anti-aromatic systems for reactions that you studied in your first semester organic course are different than the conditions for aromatic systems. In this chapter we focus on the substitution attack on an aromatic molecule by an electrophile.

Throughout the chapter we show you how these electrophilic substitutions occur, first by using benzene as an example. These mechanisms are closely related and fit together quite nicely. Look for the relationships; concentrate on understanding how and why the reactions occur as they do and avoid simple memorization. Like the substitution reactions you learned in your first semester of Organic Chemistry, this process involves the substitution of something for a hydrogen atom. In this reaction, a nucleophile the aromatic system attacks the electrophile.

The stability of the aromatic system makes it a poor nucleophile, though, so a very strong electrophile is needed to force the reaction to occur. Loss of a hydrogen ion from the intermediate to a base completes the reaction. The general mechanism is shown in Figure This mechanism is the key to all electrophilic substitution reactions. You need to grasp this basic mechanism and be able to recognize it in each of the mechanisms in this chapter and the next. For this loss to occur, a base must be present to pull it away.

In this particular case a wide variety of bases would work. Structure III in Figure represents an arenium ion, more commonly called a sigma complex. Figure shows the energy changes occurring during the reaction. The general mechanism for an electrophilic substitution reaction. E Energy changes during an electrophilic substitution reaction. I IV Progress of reaction Reactions of Benzene In this section we use benzene as a typical aromatic compound to study three basic reactions: In each case, part of the mechanism involves the generation of the electrophile. Let your understanding of one mechanism reinforce your understanding of other mechanisms.

Halogenation of benzene In a halogenation reaction, a catalyst is necessary to generate the electrophile. For chlorination, X is Cl, and for bromination, X is Br. Adding iodine I requires slightly different reaction conditions.

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Figure illustrates the general reaction for the chlorination of benzene. Figure shows a partial mechanism for the reaction. The chlorination of benzene. The partial mechanism for the chlorination of benzene. This step would be the same if iron III chloride were the catalyst. The presence of resonance in this intermediate stabilizes it and helps the reaction along. These resonance forms and similar forms are important to all the electrophilic substitution mechanisms in this chapter. However, as shown in the mechanism, the tetrachloroaluminate ion AlCl4— is a sufficiently strong base.

Nitration of benzene The generation of the electrophile for a nitration reaction begins with the reaction of nitric acid with sulfuric acid. The remainder of the reaction is shown in Figure Attack of the Electrophiles The nitrogen atom in all structures in Figure has four bonds. Nitrogen has a zero formal charge when it has three bonds and a lone pair and a negative formal charge —1 when it has two bonds and two lone pairs.

The formation of the H nitronium ion. The nitration of benzene. Unlike the base that causes the loss of hydrogen ion in the halogenation reaction, the base that removes the hydrogen ion in this mechanism is the hydrogen sulfate ion HSO4—. Nitroaromatic compounds are useful in synthesis because converting the nitro -NO2 group to an amino -NH2 group is relatively easy. For example, the reaction of nitrobenzene with acidic tin II chloride SnCl2 converts nitrobenzene to aniline, an important industrial chemical used in the production of medicines, plastics, and dyes, to name but a few.

Sulfonation of benzene The generation of the electrophile for the sulfonation generally begins with adding fuming sulfuric acid a mixture of sulfuric acid and sulfur trioxide to benzene. The electrophile can then attack the benzene ring in a manner analogous to the attack by the nitronium ion. The general reaction is shown in Figure and the mechanism is in Figure Notice the similarity between the mechanism in Figures and Keep this reaction in mind when dealing with any synthesis problem involving an aromatic system.

Sulfonation is easily reversible. Simply diluting the fuming sulfuric acid leads to the removal of the -SO3H. This is an important synthetic technique for protecting certain sites from reaction. Sulfonation can act as a placeholder while other reactions are performed, and then the easy removal of the sulfonic acid group makes the site available for reaction in a later step in a series of reaction steps. The sulfonation of benzene. The mechanism for the sulfonation of benzene. Attack of the Electrophiles Friedel-Crafts Reactions Friedel-Crafts reactions are electrophilic substitution reactions in which the electrophile is a carbocation or an acylium ion.

The removal of a halide ion from an alkyl halide is the means of generating the carbocation. An acylium ion is created by removing a chloride ion from an acid chloride R-CO-Cl. Both of these processes require a Lewis acid as a catalyst. The most commonly used Lewis acid is aluminum chloride. Alkylation Figure illustrates a typical Friedel-Crafts alkylation. Once formed, the carbocation is a very strong electrophile. A complication that may occur is the rearrangement of the carbocation to a more stable carbocation, as seen in SN1 mechanisms of alkyl halides.

These rearrangements may involve a hydride or other shift. Tertiary cations are more stable than either secondary, primary, or methyl cations. Methyl and primary cations are, in fact, the least stable. A typical Friedel-Crafts alkylation. However, AlCl3 and FeCl3 are such good Lewis acids that even these elusive primary carbocations can form. You should still avoid primary carbocations in SN1 mechanisms. The reaction begins with the formation of a complex with the catalyst.

Figure illustrates the generation of the electrophile the acylium ion from an acid chloride. The presence of resonance stabilizes the acylium ion, and that reduces the possibility of rearrangement. The generation of an acylium ion. The reaction produces an aryl ketone, which is useful in synthesis because it makes it relatively easy to convert the ketone RCOR group to an alkyl R group.

The attack leads to the formation of the resonance-stabilized sigma complex, followed by the loss of a hydrogen ion to a base. A FriedelCrafts acylation reaction. Attack of the Electrophiles The ketone can be reduced with a hot mixture of HCl and zinc amalgam zinc metal dissolved in mercury. Why Do an Alkylation?

Alkylated aromatics are useful in organic synthesis. If the alkyl group has a hydrogen atom benzylic hydrogen on the carbon adjacent to the ring, the alkyl group is susceptible to oxidation. A powerful oxidizing agent, such as acidic potassium permanganate KMnO4 or acidic dichromate Cr2O72— , converts the alkyl group to a carboxylic acid group with the elimination of all carbon atoms except the one connected to the ring. Figure illustrates this reaction. Any R group with a benzylic hydrogen gives the same product. If the ring is disubstituted, this can produce a number of isomers. Xylenes give phthalic acid, isophthalic acid, and terephthalic acid.

The oxidation of an alkylated benzene. Modifying the Reactivity of an Aromatic An electrophilic substitution reaction can take place on a substituted benzene. You can replace one of the hydrogen atoms on the substituted benzene with an electrophile. Based on those ratios, substitution of a monosubstituted benzene should be 40 percent ortho, 40 percent meta, and 20 percent para.

However, you never see this distribution, so something else must be happening. The group already present on the aromatic ring must be influencing further substitution on the aromatic ring. The ortho, ortho meta, and para positions of a monosubmeta stituted benzene. All substituents influence the resonance in any aromatic system, and they may do this in a number of ways: The presence of a meta-director means the major or only product of the reaction will be a metadisubstituted benzene.

In the next section, you see why ortho- and para-directors are a single group. Directing A group G attached to a ring makes its presence known in one of the two following ways: Groups that increase the electron density of the ring are activating. Groups that withdraw electron density from the ring are deactivating, which also helps direct an attacking group to certain positions on the ring. Figure illustrates activation and deactivation. How a group G influences the reactivity of an aromatic system. G G In order to understand any chemical process, you need to remember that stability is the key.

In the case of aromatic systems, resonance is important to stability. To find out more about stability and resonance, we begin by examining the resonance in each of the different sigma complexes that may form, listed below and shown in Figure The entering group can attack in one of three relative positions: Figure shows that each attack yields three equivalent resonance structures.

This implies that the simple electrophilic attack is not the key to what product will form. It must be the identity of the original substituent G. All attacks will give three stable resonance structures. The sigma complexes arising from ortho, meta, and para attack. The answer is that if the substituent can create another resonance structure, the sigma complex is further stabilized.