Ch3NO2 Hybridization

Delocalized Electrons Resonance 1 Delocalization and Conjugation • Lewis structures assign ALL ELECTRONS to specific bonds or as non-bonding pairs localized on specific atoms. The Lewis structure model is very powerful, but this strict localization of electrons to specific bonds and atoms is sometimes just not correct. To properly describe some structures we need to take into account the fact that some of the electrons are DELOCALIZED, i.e., are not "stuck" on specific atoms or in specific bonds. For some these a SINGLE LEWIS STRUCTURE is INSUFFICIENT • At this point we could go to more sophisticated molecular orbital models of bonding, but these are often too detailed for solve simple organic chemistry problems. Instead, we go to the next "level" of bonding model, one that retains the simple Lewis picture, but expands it to take delocalization into account by describing some structures in terms of a MIXTURE of MORE THAN ONE LEWIS STRUCTURE. This is RESONANCE Delocalization Q. Which of the following two Lewis structures is correct? H H resonance brackets C O resonance arrow H H N H C O H H O = N O 50% + 50% two resonance CONTRIBUTORS H C O 1/2 N O 1/2 H "actual" structure resonance hybrid NOT a Lewis structure A. Neither give a proper description of where the electrons are in the molecule, a proper description of CH3NO2 requires a CONTRIBUTION from BOTH LEWIS STRUCTURES, or a MIXTURE OF THE TWO. The "actual" structure (also called the resonance hybrid) is a mixture of the two resonance contributors, a 50:50 mixture in this case • typical N–O bond ~1.3Å and N=O bond ~1.16Å, however, both N–O bonds are same length, ~1.22Å, i.e. "average of the two" • the "actual" structure shows two partial pi bonds and a half negative charge on each oxygen, but the "actual" structure IS NOT a Lewis structure, since Lewis structures can't have "dashed bonds" and electrons that are not specifically associated with a particular atom or a particular bond Conjugation (Origin of the Electron Delocalization Phenomenon) • A CONJUGATED SYSTEM is one in which overlap of p A.O.s occurs "across" what would be considered pure sigma-bonds if we only consider Lewis structures, simple Lewis structures are "deficient" in situations like this, conjugated systems result in electron DELOCALIZATION, as in the example given above • The definition of a conjugated system is one that consist of more that two ADJACENT atoms that have p atomic orbitals that can overlap "sideways" to form an extended pi-system. • We will learn to how to understand bonding from a molecular orbital perspective, and how to draw the molecular orbitals of conjugated systems later in the course, for now we will learn another way to understand the bonding in conjugated systems, which is to use RESONANCE CONTRIBUTORS conjugated system sp2 H H C O N H sp2 O H H C O H N each atom is sp2 hybridized! O sp2 • there is a formal; p A.O. on both of the oxygen and the nitrogen atoms 1 • the three p atomic orbitals form a CONJUGATED system, we should not surprised that they result in a bonding situation that cannot be described by simple Lewis structures • conjugated systems can NOT involve any sp3 hybridized atoms, since they do not have a formal p A.O. The Allyl Cation as an Example • Each carbon of the allyl cation (below) is sp2 hybridized, the three carbon atoms represent a conjugated system since the p atomic orbitals can overlap to form an extended pi-system, resulting in delocalized electrons • Add the p A.O.s that make the pi-bond and the "empty" p A.O. on the carbon with the positive charge • Which 2 p A.O.s make the formal (Lewis) pi bond and which is empty? The molecule isn't sure either!! This is what conjugation and delocalization is all about two resonance CONTRIBUTORS these 2 make the π-bond 50% H H C C C H + H 50% H H C H C H why don't these 2 make a π-bond? they do!! all sp2 - conjugated H C H = H 1/2 H C C C H 1/2 H H "actual" structure resonance hybrid NOT a Lewis structure all carbons sp2 hybridized resonance arrow resonance brackets • The conjugated system consists of all three carbon atoms. If we try to draw the cation according to the rules of Lewis structures then we must draw TWO Lewis "contributors", and understand that neither is a correct representation of the cation on its own, but that the correct, "actual" structure of the cation can be obtained by combining, or mixing the two contributors • The "actual" structure is a RESONANCE MIXTURE or HYBRID of the two RESONANCE CONTRIBUTORS that when MIXED together properly describe the overall conjugated structure. The two resonance contributors are two different ways of describing the SAME THING! • Resonance contributors are contained within resonance brackets, and combined (mixed) using a resonance arrow, each resonance contributor is a correct, but incomplete, Lewis Structure • resonance contributors are NOT isomers, they represent the REQUIRED different ways of indicating where delocalized electrons are within the "confines" of the Lewis structure "language", they are different ways of drawing the SAME THING, together they CONTRIBUTE to the actual conjugated structure • The "actual" structure has all of the electrons that can be involved in bonding, in partial (dashed) bonds. The Allyl Anion as an Example these 2 make the πbond H EACH CONTRIBUTES 50% H H C H C C H H H C C C H H these 2 MUST also make a π-bond 2 sp2 H = 1/2 H C H C sp2 C H H 1/2 H H "actual" structure resonance hybrid all sp2 hybridized C H H compare –CH3 sp3 hybridized (VSEPR) • ALL THREE carbons must be sp hybridized to allow formation of both pi-bonds, this is a three carbon conjugated system • "actual" hybrid (mixed) structure has all of the electrons that can be involved in bonding in partial (dashed) bonds, so no non-bonding electrons are included in this "actual" structure. 2 Q. Why is the carbon with the formal negative charge sp2 hybridized, when the methyl anion is sp3 hybridized? A. Because, by being sp2 hybridized it can provide a p A.O. to the conjugated system, and in doing to it gets involved in extended pi-bonding, which means that the electrons on that carbon are not LOCALIZED on one carbon, they are DELOCALIZED. Delocalization allows ALL of the electrons to "see" more nuclei, which lowers their energy! The hybridization is not rapidly oscillating back and forth between the resonance contributors, the nuclei are NOT IN DIFFRENT POSITIONS in the resonance contributors, since each is simply a different way of describing the SAME THING. A Mule an as Example • Image that there was no word in English for a mule. A mule is a mixture of a horse and a donkey, and so we could make up a word for a mule by combining the words horse and donkey (German speakers would have NO PROBLEM doing this!), so…… MULE = Horse-donkey • This is exactly the situation we have when trying to draw the bonding and electron distribution in conugated systems, there is no way of properly describing the bonding in conjugated systems using the Lewis structure "language", which is why we use resonance contributors • And sooooo, we can describe a mule using resonance contributors as follows…. ~50% ~50% [ HORSE DONKEY ] = MULE "actual" resonance hybrid resonance contributors Benzene as an example 2 • Each carbon in the 6-membered ring is sp hybridized, each has 1 p A.O., we have a CONJUGATED SYSTEM • draw the pi-bonds in benzene sp2 H C sp2 C H sp2 H sp2 sp2 H C C draw the sp2 π-bonds C C H H H H C C H C C H H C H C C H H C C C H H C C H H H EACH CONTRIBUTES 50% = C H C C H C H C H "actual" hybrid structure all carbons sp2 hybridized H C • 50% of the class will draw the structure on the left, 50% the structure on the right, BOTH are required resonance contributors to the actual structure of benzene • resonance does not involve only charged structures • delocalization of the electrons in benzene in this way lowers their energy, increases stability, lowers reactivity • "actual" structure has all electrons that can be involved in bonding in partial (dashed) bonds. • benzene is properly described as a 50:50 MIXTURE of the two Lewis contributors above 2 Curved Arrow "Pushing" • although writing resonance contributors seems more complex than simply drawing one "actual" structure, we will see that this is an essential way to write and understand most chemical processes involving delocalized electrons • it is also important to be able to describe how the different resonance contributors are related using the conventional "curved arrow pushing" language of organic chemistry 3 Allyl anion again • The resonance contributors we looked at previously are not the same thing just "flipped over", we can show this using deuterium "labeling". Recall, D is just an isotope of H deuterium "labels" these two H atoms H H D C C C D H D C C C D H H = 1/2 D sp2 C Hsp2 2 C sp H C 1/2 D H H "actual" resonance hybrid structure all carbons sp2 hybridized "actual" = 50 : 50 hybrid mixture of resonance CONTRIBUTORS • the allyl cation consists of a 50:50 mixture of the two resonance contributors shown • the formal negative charge can either be on the carbon that had the hydrogen atoms, or on the carbon that has the deuterium atoms, the deuterium labeling show that the second contributor is not the first one "flipped over" • the "actual" hybrid structure has all electrons that can be involved in bonding in partial (dashed) bonds, so no non-bonding electrons are included in this "actual" structure. Example Problem : Draw the resonance contributors for the following radical δ means "part" of an electron H H H H δ δ δ H H C C C H C H H C C C C H = C C C C C C C C C C C C H H H H H H H H H H H H "actual" resonance radical hybrid structure • resonance describes delocalization of single electrons in addition to electron pairs • note the use of "single-headed" fishhook arrows for single electrons • note the use of a Greek delta to indicate a "partial electron" in the actual structure, we use this terminology when there isn't an obvious simple fractional electron population on the atom • ALL carbon atoms involved in resonance must "contribute" a p atomic orbital therefore all must be sp2 or sp hybridized, all of the carbons are part of the CONJUGATED SYSTEM H H H H H H 3 Major and Minor Resonance Contributors The Enolate anion as an Example Conjugated System H H C H O sp2 sp2 H C C H A < 50% O >1/2 O H H C H C H C sp2 H B > 50% = sp2 C H C sp2 C <1/2 H H H "actual" resonance hybrid structure H sp2 • oxygen is more electronegative than carbon, the electrons would be lower in energy in contributor B compared to A • the two do not contribute equally to the "actual" structure, B is the major resonance contributor • there is MORE OF B in the MIXTURE • The unequal resonance contributors tell us that in this case there is an unequal distribution of charge between (in this case) the carbon and the oxygen atoms, i.e. there is greater than half a negative charge on the oxygen and less than half a negative charge on the carbon atom • the "actual" hybrid structure has all electrons that can be involved in bonding in partial (dashed) bonds. 4 Example Problem: Which is the major resonance contributor? Look at the formal electron distribution in the various contributors, the one(s) with the lowest energy formal electron energy is(are) the major contributor(s) sp3 not part of the conjugated system conjugated >1/2 C N = H3C CH "actual" resonance hybrid structure N H3C CH C sp sp sp major N CH C sp sp sp minor H3C <1/2 • the negative charge is on the more electronegative nitrogen in the major contributor • the second contributor has a formal negative charge on oxygen, the formal electron energy in the second contributor is lower, which is why it is the major contributor • the second (major) contributes more to the mixture, therefore, there is greater than 1/2 a negative charge on nitrogen in the actual structure • the "actual" hybrid structure has all of the electrons that can be involved in bonding in partial (dashed) bonds. Example Problem: Draw other resonance contributors, indicate the major and minor and draw "actual" structure conjugated system NH2 H3C CH2 C sp3 NH2 C H3C CH2 NH2 SP2 δ NH2 NH2 NH2 H3C CH2 C NH2 equal major (more bonds) minor SP2 NH2 H3C CH2 C δ δ "actual" resonance hybrid structure • the most important factor in determining the major contributor is usually the number of electrons that are in bonds, the more electrons in bonds, the more important the resonance contributor • resonance delocalization stabilizes charged ions, the more resonance contributors the more stable the ion 4 Resonance and Hybridization (more) • EACH ATOM that is involved in resonance MUST be sp2 or sp hybridized, each atom must be capable of providing a p A.O., resonance requires each atom to "provide" a p A.O., i.e. as we have now said many times, resonance occurs in CONJUGATED SYSTEMS Example resonance/conjugated system sp3 not involved not conjugated H H H H C C H C C H sp2 2 H sp H C C H C C H H H H H conjugated Example Problem: What is the hybridization of the oxygens and the central carbon in the following structure? all sp2 conjugated O O H 3C H3C C O 5 conjugated = C O H3 C O 1/2 C O 1/2 • both oxygens MUST be sp2 hybridized AND the central carbon must also be sp2 hybridized • any atom to which you can draw a double bond to in ANY resonance contributor must be sp or sp2 hybridized • hybridization must be the same in all resonance contributors, because the contributors are only different ways of drawing the SAME THING, all atoms involved in resonance must be either sp2 or sp hybridized (or they could not be conjugated) Example Problem: What is the hybridization of the circled nitrogen? sp2 X sp3 H3C N H3C N N δ N N = H3 C N N H3 C N conjugated N δ N N N 2 although possible to draw, this is "unreasonable", see below • it must be sp2 hybridized since we can draw another resonance contributor with a double bond to that nitrogen • the hybridization of all atoms must be the same in all resonance contributors, because the contributors are only different ways of drawing the SAME THIN G • EACH atom involved in resonance is conjugated, all are sp2 or sp hybridized, each must be capable of contributing a p A.O. (which is impossible; if they are sp2 hybridized) Example Problem: What is the hybridization of the nitrogen? H3C H3C C N H major δ O O O H C = N H3C H H this has "unnecessary charges" and is thus a very minor contributor resonance hybrid C N δ H H • here we introduce another factor to consider when deciding on major and minor contributors. The minor one has the same number of pi-bonds as the major, but is minor because it has unnecessarily separated charges and violates the normal rules of valence unnecessarily. 2 • the nitrogen atom must be (partially at least) sp hybridized (because we can draw a resonance contributor that has a double bond to this nitrogen), however, this question becomes more and more difficult to answer as the resonance contributors involving double bonds become less and less important (actually we start to talk about fractional hybridizations, but we won't get into this in a General organic chemistry course….) 5 Minor and Unreasonable Resonance Contributors • We will often be able to draw many resonance contributors that are minor, in some cases these can be important and in other cases they will not be important, they will be unreasonable, whether they are useful or unreasonable it will depend upon the context Example 1: Only some of the carbon atoms in this cation will be reactive, those that carry a partial positive charge. Which are they? reactive δ δ X = reactive 6 "legitimate" but very minor and unreasonable in this context • the unreasonable contributor is not "wrong", it represents a legitimate way to distribute the electrons in this cation, it is very minor because compared to the other two the this electron distribution would be high in energy, it has one less formal bond and also separates charges, which costs energy according to the Coulomb law * the other two contributors tell us what we need to know about this cation, which is how are the electrons and charges delocalized? * The unreasonable one does not help to answer this question, in this context we are not concerned with minor contributors Example: Many stable neutral organic molecules have resonance contributors that although are very minor (and would be unreasonable in the context above), are still useful in predicting reactivity, for example, acetone, shown below H H C O C H C H H major >> 50% H H H C O C O C H = H H H minor << 50% H H C C δ δ C H H H H "actual" resonance hybrid structure • The second resonance contributor is clearly VERY minor (it has one less bond and separated charges), and in the context of the first example above would have been considered "unreasonable", so why is it considered here • In this context, however, the minor contributor tells us how the electrons are unevenly distributed in the molecule, specifically in the C=O double bond, even if they are only a little uneven, as in this case • "Actual" acetone is a MIXTURE of the two resonance contributors, thus, there must be at least a partial negative charge on the oxygen, with more negative charge the more the second on contributes to the mixture • In this context, the minor structure tells us that there is a small partial charge on oxygen, and a small partial charge on the carbon of the C-O bond • note, this model gives the SAME information as the inductive effect/dipole moment model, the electrons in the C=O bond are polarized towards the oxygen O δ H + Cδ C C H H H H H • "delta" symbol often used to indicate "some" charge, especially when the actual charge is very small (as in this case), or it is not obvious how much charge there is Example: Phenol OH OH OH OH OH OH δ =δ phenol minor resonance contributors δ δ • We need to consider very minor, otherwise "unreasonable" resonance contributors in this context in order to figure out which specific carbons in the benzene ring carry small partial negative charges • This is not something that we can figure out using dipole moments in this case, the only way we can get the information we need is via minor resonance contributors, so in this context they are not unreasonable, but they are very informative specifically because we are interested in minor redistributions of electrons • we will find that consideration of minor resonance contributors such as these help us to understand reactivity of many organic molecules 7 So…. Whether a minor resonance contributor is "unreasonable" or is informative depends upon the context • when we are asking about how electrons are distributed generally in a structure (usually in anions, cations radicals etc), then minor contributors are not useful and are "unreasonable" • when we are asking 6 "Guidelines" for Resonance Some "guidelines" when considering resonance contributors • When considering major and minor contributors • Consider the total electron energy for each contributor individually, the lower the electron energy associated with that particular electron distribution, the more important (likely to be major) that contributor is • Non-bonding electrons are usually higher in energy than bonding electrons, electrons in bonds are usually the most important factor to consider for electron energy • Electrons associated with LESS electronegative atoms are HIGHER in energy than those on more electronegative atoms • Electrons on atoms with formal negative charges are usually higher in energy • Separation of charges costs energy according to the Coulomb law • Usefulness and "unreasonableness" for resonance contributors depends upon the context, see above • When moving electrons • for delocalized structures with non-bonding electrons that are involved in resonance, "get" the high energy electrons into a bond first, these are usually the non-bonding electrons • because of the requirement for conjugation, you can only put electrons into a bond to an adjacent atom • for delocalized structures with no non-bonding electrons, usually cations, the best place to start is usually to move electrons towards the site of the positive charge • ONLY atoms that are part of the conjugated system, i.e. that can provide a p A.O. and thus must be sp2 or sp hybridized, can be involved in generating resonance contributors • look for the "smallest" movement of the electrons (arrows) to avoid missing an important contributor Example Problem Which carbons on the following cation carry partial positive charges? Draw all reasonable resonance contributors. X δ = unreasonable in THIS CONTEXT H δ H sp3, not involved in resonance Example Problem: Draw all reasonable resonance contributors for the following cation • start by moving pi-bonding electrons towards the positively charged carbon • note that drawing 2 arrows at once, as above, is not wrong, it just "misses" the central resonance contributor, drawing the arrows as shown below does not miss any of the contributors 8 δ Example Problem. Draw all reasonable resonance contributors for the following anion to determine which atoms will have partial negative charge, and assign the major and minor contributors N H N H N H O O O major minor • start by moving the non-bonding electrons into a bond with an adjacent atom, and move the pi-electrons in the C=O pi-bond to avoid having 5 bonds to the carbon atom of the C=O bond • it would have been equally correct and useful to move the pi-electrons towards the carbon of the C=N pi-bond also • in this case it is necessary to draw 2 arrows at once this time to generate the third contributor from the second one 7 Resonance and Stability Consider the following two cations A and B……… H H H C C C H H H H C C H H C H H H H A resonance stabilized, less reactive C H C H C H more reactive H B Q. Which has the delocalized electrons? A. A Q. Does delocalization of electrons mean that the electrons "see" more nuclei are thus lower in energy? A. YES! Q. Are species with lower energy electrons generally less reactive? A. YES! Q. Which is the more reactive cation, A or B? A. B! • Thermodynamically (energetically), the resonance stabilized cation A is less reactive • The resonance stabilized cation A is also less reactive kinetically, because the positive charge is "diluted" onto more than one carbon atom X Consider the following two anions, A or B, which is more reactive? • remember, higher energy electrons are more reactive • delocalization lowers the energies of electrons since it allows them to "see" more nuclei • resonance means delocalization H H A B can NOT move electrons to sp3 hybridized atom 3 resonance contributors more stable LESS reactive anion 2 resonance contributors, less stable 9 • the non-bonding electrons in BOTH are delocalized, in neither are the electrons localized on the carbon atoms indicated in the structures provided • the non-bonded electrons in A are delocalized onto 2 carbon atoms, 2 resonance contributors can be drawn • the non-bonded electrons in B are MORE delocalized, 3 resonance contributors can be drawn • the electrons on B are more stable, anion B is less chemically reactive Consider the following two radicals A and B……… H H H C C C H H H C C H H H H H C H H C H C H A resonance stabilized, less reactive C H more reactive H B Q. Which is more reactive? A. B is not resonance stabilized, is more reactive Q. Which of the indicated C-H bonds below has the lower BDE? H3C H2C CH H3C H H CH2 H 2C H2C CH CH B.D.E. kcal/mol 104 H H CH2 ~87 Length Å 1.09 ~1.07 CH2 • the resonance stabilized radical is easier to form, the non-bonding electron is lower in energy (it is stabilized be delocalization, resonance), thus the energy required to break the bond is lower, the B.D.E. is lower H3C• + •H Relative Energy R–H costs this much energy to break these bonds H2C CH higher energy electron higher BDE lower energy electron smaller BDE CH + H H2C CH2 CH2 energies normalized here rR–H •IMPORTANT: it is resonance stabilization in the RADICAL that is formed upon bond fragmentation, not resonance in the bond (there is none!) that lowers the B.D.E. We will use these concepts that relate resonance and stability extensively in this course! 10