RESONANCE IN CHEMISTRY, High school final essays of Chemistry

Download RESONANCE IN CHEMISTRY and more High school final essays Chemistry in PDF only on Docsity! 68 CHAPTER 2 Molecular Representations Methadone, developed in Germany during World War II, is used to treat heroin addicts suffering from withdrawal symptoms. Methadone binds to the same receptor as heroin, but it has a longer retention time in the body, thereby enabling the body to cope with the decreasing levels of drug that normally cause withdrawal symptoms. Etorphine is over 3000 times more potent | than morphine and is used exclusively in veterinary medicine to immobilize elephants and other large mammals. Scientists are constantly searching for new lead com- pounds. In 1992, researchers at NIH (National Institutes of Health) in Bethesda, Maryland, isolated epibatidine from the skin of the Ecuadorian frog, Epipedobates tricolor. Epibatidine was found to be an analgesic that is 200 times more potent than morphine. Further studies indi- cated that epibatidine and mor- phine bind to different receptors. This discovery is very exciting, because it means that epibatidine can serve as a new lead compound. Although this compound is too toxic for clinical use, a significant number of researchers are currently working to identify the pharmacophore of epibati- dine and to develop nontoxic derivatives. This area of research indeed looks promising. Epibatidine @®® CONCEPTUAL CHECKPOINT 2.20 Troglitazone, rosiglitazone, and pioglitazone, all anti- diabetic drugs introduced to the market in the late 1990s, are believed to act on the same receptor: HO O Troglitazone Rosiglitazone Oo i! ° SS N—-H Ss 0 x oO Pioglitazone (a) Based on these structures, try to identify the likely pharmacophore that is responsible for the anti- diabetic activity of these drugs. (b) Consider the structure of rivoglitazone (below). This compound is currently being studied for potential anti- diabetic activity. Based on your analysis of the likely pharma- cophore, do you believe that rivoglitazone will exhibit antidia- betic properties? or Rivoglitazone 2.7 Introduction to Resonance The Inadequacy of Bond-Line Structures We have seen that bond-line structures are generally the most efficient and preferred way to draw the structure of an organic compound. Nevertheless, bond-line structures suffer from one major defect. Specifically, a pair of bonding electrons is always represented as a line that is drawn between two atoms, which implies that the bonding electrons are confined to a region of space directly in between two atoms. In some cases, this assertion is acceptable, as in the following structure: = FIGURE 2.3 The overlapping p orbitals of an allyl carbocation. Antibonding MO Nonbonding MO Bonding MO FIGURE 2.4 The molecular orbitals associated with the = electrons of an allylic system. 2.7 Introduction to Resonance y * 69 In this case, the 1 electrons are in fact located where they are drawn, in between the two central car- bon atoms, But in other cases, the electron density is spread out over a larger region of the molecule. For example, consider the following ion, called an allyl carbocation: > Ie might seem from the drawing above that there are two & electrons on the left side and a positive charge on the right side. But this is not the entire picture, and the drawing above is inadequate. Let’s take a closer look and first analyze the hybridization states. Each of the three carbon atoms above is sp” hybridized. Why? The two carbon atoms on the left side are each sp” hybridized because each of those carbon atoms is utilizing a p orbital co form the x bond (Section 1.9). The third carbon atom, bearing the positive charge, is also sp” hybridized because it has an empty p orbital. Figure 2.3 shows the three p orbitals associated with an allyl carbocation. This image focuses our attention on the continuous system of p orbitals, which functions as a “conduit,” allowing the two 7 electrons to be associated with all three carbon atoms. Valence bond theory is inadequate for analysis of this system because it treats the electrons as if they were con- fined between only two atoms. A more appropriate analysis of the allyl cation requires the use of molecular orbital (MO) theory (Section 1.8), in which electrons are associated with the molecule as a whole, rather than individual atoms. Specifically, in MO theory, the entire molecule is treated as one entity, and all of the electrons in the entire molecule occupy regions of space called molecu- lar orbitals. Two electrons are placed in each orbital, starting with che lowest energy orbital, until all electrons occupy orbitals. According to MO theory, the three p orbitals shown in Figure 2.3 no longer exist. Instead, they have been replaced by three MOs, illustrated in Figure 2.4 in order of increasing energy. Notice that the lowest energy MO, called the bonding molecular orbital, has no vertical nodes. The next higher energy MO, called the nonbonding molecular orbital, has one vertical node. The highest energy MO, called the antibonding molecular orbital, has two vertical nodes. The ™ electrons of the allyl system will fill these MOs, starting with the lowest energy MO. How many 7 electrons will occupy these MOs? The allyl carbocation has only two & electrons, rather than three, because one of the carbon atoms bears a positive formal charge indicating that one electron is missing. The two x electrons of the allyl system will occupy the lowest energy MO (the bonding MO). If the missing electron were to return, it would occupy the next higher energy MO, which is the nonbonding MO. Focus your attention on the nonbonding MO. The nonbonding molecular orbital (from Figure 2.4) associated with the = electrons of an allylic system. There should be an electron occupying this nonbonding MO, but the electron is missing. Therefore, the colored lobes are empty and represent regions of space that are electron deficient. In conclusion, MO theory suggests that the positive charge of the allyl carbocation is associated with the two ends of the system, rather than just one end. Ina situation like this, any single bond-line structure that we draw will be inadequate. How can we draw a positive charge that is spread out over two locations, and how can we draw two 7 electrons that are associated with three carbon atoms? Resonance The approach that chemists use to deal with the inadequacy of bond-line structures is called resonance. According to this approach, we draw more than one bond-line structure and then Zr a ZS mentally meld them together: ® 72° CHAPTER 2 Molecular Representations SKILLBUILDER ) LEARN the skill Inspect the arrow drawn on the following structure and determine whether it violates either of the two rules for drawing curved arrows: XK o @ sovution i In order to determine if either rule has been broken, we must look carefully at the tail and i the head of the curved arrow. The tail is placed on a double bond, and therefore, this curved arrow does not break a single bond. So the first rule is not violated. : STEP 1 Next, we look at the head of the arrow: Has the octet rule been violated? Is there a fifth t Make sure that the bond being formed here? Remember that a carbocation (C*) only has three bonds, not four. tail of the curved Two of the bonds are shown, which means that the C* has only one bond to a hydrogen atom: t arrow is not located : ona single bond. & ! STEP 2 } Make sure that the : head of the curved Therefore, the curved arrow will give the carbon atom a fourth bond, which does not violate } arrow does not the octet rule. violate the octet rule The curved arrow is valid, because the two rules were not violated. Both the tail and head of the arrow are acceptable. PRACTICE the skill 2.21 For each of the problems below, determine whether each curved arrow violates either of the two rules and describe the violation, if any. (Don't forget to count all hydrogen atoms and all lone pairs.) ZONS o @ 4 ” (b) AR O Q o (e) a“ ( @) (hy Ae Bs : 1S AS y Hse N= N: en }) APPLY the skill 2.22 Drawing the resonance structure of the following compound requires one curved } arrow. The head of this curved arrow is placed on the oxygen atom, and the tail of the curved arrow can only be placed in one location without violating the rules for drawing curved arrows. Draw this curved arrow. | or --> need more PRACTICE? Try Problem 2.51 2 ‘Whenever more than one curved arrow is used, all curved arrows must be taken into account in order to determine if any of the rules have been violated. For example, the following arrow violates the octet rule: This carbon atom cannot form a fifth bond WATCH OUT The electrons are not really moving. We are just treating them as if they were. 2.9 Formal Charges in Resonance Structures y * 73 However, by adding another curved arrow, we remove the violation: | 0 | 6 “NS ye > UN aX The second curved arrow removes the violation of the first curved arrow. In this example, both arrows are acceptable, because taken together, they do not violate our rules. Arrow pushing is much like bike riding. The skill of bike riding cannot be learned by watching someone else ride, Learning to ride a bike requires practice. Falling occasionally is a necessary part of the learning process. The same is true with arrow pushing. The only way to learn is with practice. This chapter is designed to provide ample opportunity for practicing and mastering resonance structures. 2.9 Formal Charges in Resonance Structures In Section 1.4, we learned how to calculate formal charges. Resonance structures very often contain formal charges, and it is absolutely critical to draw them properly. Consider the following example: a? In this example, there are two curved arrows. The first arrow pushes one of the lone pairs to form a bond, and the second arrow pushes the 7 bond to form a lone pair on a carbon atom. When both arrows are pushed at the same time, neither of the rules is violated. So, let’s focus on how to draw the resonance structure by following the instructions provided by the curved arrows. We delete one lone pair from oxygen and place a 7 bond between carbon and oxygen. Then we must delete the C—C 7 bond and place a lone pair on carbon: oor _. oO So ‘The arrows are really a language, and they tell us what to do. However, the structure is not complete without drawing formal charges. If we apply the rules of assigning formal charges, oxygen acquires a positive charge and carbon acquires a negative charge: 6® 7 “ Bo AL q Another way to assign formal charges is to think about what the arrows are indicating, In this case, the curved arrows indicate that the oxygen atom is losing a lone pair and gaining a bond. In other words, it is losing two electrons and only gaining one back. The net result is the loss of one electron, indicating that oxygen must incur a positive charge in the resonance structure. A similar analysis for the carbon atom on the bottom right shows that it must incur a negative charge. Notice that the overall net charge is the same in each resonance structure. Let’s practice assigning formal charges in resonance structures. 74 CHAPTER 2 Molecular Representations « SKILLBUILDER ae y 2.7 ASSIGNING FORMAL CHARGES IN RESONANCE STRUCTURES ¥<)) LEARN the skill Draw the resonance structure below. Be sure to include formal charges. my ay — ? @ sovution i The arrows indicate that one of the lone pairs on oxygen is coming down to form a bond, : and the C=C double bond is being pushed to form a lone pair on a carbon atom. This is very similar to the previous example. The arrows indicate that we must delete one lone pair on oxygen, place a double bond between carbon and oxygen, delete the carbon-carbon double bond, and place a lone pair on carbon: STEP 1 7 fo) Carefully read what — the curved arrows Ay A indicate. Finally, we must assign formal charges. In this case, oxygen started with a negative charge, and this charge has now been pushed down (as the arrows indicate) onto a carbon atom. Therefore, the carbon atom must now bear the negative charge: STEP 2 Assign formal charges Earlier in this chapter, we said that it is not necessary to draw lone pairs, because they are implied by bond+-line structures. In the example above, the lone pairs are shown for clarity. This raises an obvious question. Look at the first curved arrow above: The tail is drawn on a lone pair. If the lone pairs had not been drawn, how would the curved arrow be drawn? In situations like this, organic chemists will sometimes draw the curved arrow coming from the negative charge: Fo AY is the same as _— Nevertheless, you should avoid this practice, because it can easily lead to mistakes in certain situations. It is highly preferable to draw the lone pairs and then place the tail of the curved arrow ona lone pair, rather than placing it on a negative charge. After drawing a resonance structure and assigning formal charges, it is always a good idea to count the total charge on the resonance structure. This total charge MUST be the same as on the original structure (conservation of charge). If the first structure had a negative charge, then the resonance structure must also have a net negative charge. If it doesn’t, then the resonance structure cannot possibly be correct. The total charge must be the same for all resonance structures, and there are no exceptions to this rule. Ae 7 2.10 Drawing Resonance Structures via Pattern Recognition a Recognizing this pattern (a lone pair next to a 7 bond) will save time in calculating formal charges and determining if the octet rule is being violated. @ CONCEPTUAL CHECKPOINT 2.25 For each of the compounds below, locate the pattern we just learned (lone pair next to a 7 bond) and draw the appropriate reso- nance structure: O° 8 S 0 oO ° Ore NC. Oo ..0 O we (a) (b) (Co) (d) (e) AN Q O Be who @) NS @) po (h) ° Acetylcholine 5-Amino-4-oxopentanoic acid (a neurotransmitter) (used in therapy and diagnosis of hepatic tumors) 2. Anallylic positive charge. Again we are focusing on allylic positions, but this time, we are look- ing for a positive charge located in an allylic position: ee Allylic positive charge When there is an allylic positive charge, only one curved arrow will be required; this arrow goes from the 7 bond to form a new x bond: ADR eo a ® Notice what happens to the formal charge in the process. The positive charge is moved to the other end of the system. In the previous example, the positive charge was next to one 7 bond. The following example contains two 7 bonds, which are said to be conjugated, because they are separated from each other by exactly one o bond (we will explore conjugated x systems in more detail in Chapter 17). oY e In this situation, we push each of the double bonds over, one at a time: ADT TO OBE OE BY ® ® ® It is not necessary to waste time recalculating formal charges for each resonance structure, because the arrows indicate what is happening. Think of a positive charge as a hole of electron density—a place that is missing an electron. When we push 7 electrons to plug up the hole, a new hole is created nearby. In this way, the hole is simply moved from one location to another. Notice that in the above structures the tails of the curved arrows are placed on the 7 bonds, not on the positive charge. Never place the tail of a curved arrow on a positive charge (that is a common mistake). 78 CHAPTER 2 Molecular Representations @ CONCEPTUAL CHECKPOINT 2.26 Draw the resonance structure(s) for each of the compounds below: ® ® () “\F ©) © ) 3. A lone pair adjacent to a positive charge. Consider the following example: The oxygen atom exhibits three lone pairs, all of which are adjacent to the positive charge. This pattern requires only one curved arrow. The tail of the curved arrow is placed on a lone pair, and the head of the arrow is placed to form a x bond between the lone pair and the posi- tive charge: Bom A Notice what happens with the formal charges above. The atom with the lone pair has a negative charge in this case, and therefore the charges end up canceling each other. Let’s consider what happens with formal charges when the atom with the lone pair does not bear a negative charge. For example, consider the following: 2 | Aas — Ack | Once again, there is a lone pair adjacent to a positive charge. Therefore, we draw only one curved arrow: The tail goes on the lone pair, and the head is placed to form a x bond. In this case, the oxygen atom did not start out with a negative charge. Therefore, it will incur a positive charge in the resonance structure (remember conservation of charge). @ CONCEPTUAL CHECKPOINT 2.27 For each of the compounds below, locate the lone pair adjacent to a positive charge and draw the resonance structure: | Qe Oe NNN or @ ® 0) 4% i) 2.10 Drawing Resonance Structures via Pattern Recognition a * 79 In one of the previous problems, a negative charge and a positive charge are seen canceling each other to become a double bond. However, there is one situation where it is not possible to combine charges to form a double bond—this occurs with the nitro group. The structure of the nitro group looks like this: In this case, there is a lone pair adjacent to a positive charge, yet we cannot draw a single curved arrow to cancel out the charges: Nota valid resonance structure Why not? The curved arrow shown above violates the octet rule, because it would give the nitrogen atom five bonds. Remember that second-row elements can never have more than four bonds. There is only one way to draw the curved arrow above without violating the octet rule— we must draw a second curved arrow, like thi Look carefully. These two curved arrows are simply our first pattern (a lone pair next to a 7 bond). Notice that the charges have not been canceled. Rather, the location of the negative charge has moved from one oxygen atom to the other. The two resonance structures above are the only two valid resonance structures for a nitro group. In other words, the nitro group must be drawn with charge separation, even though the nitro group is overall neutral. The structure of the nitro group cannot be drawn without the charges. 4, Ax bond between two atoms of differing electronegativity. Recall that electronegativity measures the ability of an atom to attract electrons. A chart of electronegativity values can be found in Section 1.11. For purposes of recognizing this pattern, we will focus on C=O and C=N double bonds a A In these situations, we move the 1 bond up onto the electronegative atom to become alone pair: Notice what happens with the formal charges. A double bond is being separated into a positive and negative charge (this is the opposite of our third pattern, where the charges came together to form a double bond).