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2.2: Resonancia

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2.2A: The meaning of resonance contributors: benzene and its derivatives

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.

These two drawings are an example of what is referred to in organic chemistry as 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. A few chapters from now when we begin to study organic reactions - a process in which electron density shifts and covalent bonds between atoms break and form - this ‘curved arrow notation’ will become extremely important in depicting electron movement. 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.

Usually, benzene and substituted derivatives of benzene are depicted with only one resonance contributor, and it is assumed that the reader understands that resonance delocalization is implied. This is the convention that will be used for the most part in this book.  In other books or articles, you may sometimes see benzene or a phenyl group drawn with a circle inside the hexagon, either solid or dashed, as a way of depicting the resonance delocalization.

2.2B: Resonance contributors for the carboxylate group

The convention of drawing two or more resonance contributors to approximate a single structure may seem a bit clumsy to you at this point, but as you gain experience you will see that the practice is actually very useful when discussing the manner in which many functional groups react. Let’s next consider the carboxylate ion (the conjugate base of a carboxylic acid). As our example, we will use formate, the simplest possible carboxylate-containing molecule. The conjugate acid of formate is formic acid, which causes the painful sting you felt if you have ever been bitten by an ant.

Usually, you will see carboxylate groups drawn with one carbon-oxygen double bond and one carbon-oxygen single bond, with a negative formal charge located on the single-bonded oxygen.  In actuality, however, the two carbon-oxygen bonds are the same length, and although there is indeed an overall negative formal charge on the group, it is shared equally between the two oxygens. Therefore, the carboxylate 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 π bond and the negative charge located in between the two oxygens.

Let’s see if we can correlate these drawing conventions to a valence bond theory 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 sbonds, then, are formed from the overlap of carbon sp2 orbitals and oxygen sp2 orbitals.

In addition, the carbon and both oxygens each have an unhybridized 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.

Resonance contributor A shows oxygen #1 sharing a pair of electrons with carbon in a pi bond, and oxygen #2 holding a lone pair of electrons in its 2pz orbital.  Resonance contributor B, on the other hand, shows oxygen #2 participating in the pi bond with carbon, and oxygen #1 holding a lone pair in its 2pz orbital.  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. This is the kind of 3D picture that resonance contributors are used to approximate, and once you get some practice you should be able to quickly visualize overlapping 2pz orbitals and delocalized $$pi$$ electrons whenever you see resonance structures being used. In this text, carboxylate groups will usually be drawn showing only one resonance contributor for the sake of simplicity, but you should always keep in mind that the two C-O bonds are equal, and that the negative charge is delocalized to both oxygens.

2.2C: Rules for drawing resonance structures

As you work on learning how to draw and interpret resonance structures, there are a few basic rules that you should keep in mind in order to avoid drawing nonsensical structures. All of these rules make perfect sense as long as you keep in mind that resonance contributors are merely a human-invented convention for depicting the delocalization of electrons in overlapping p orbitals.

1. When you see two different resonance contributors, you are not seeing a chemical reaction! Rather, you are seeing the exact same molecule or ion depicted in two different ways.
2. Resonance contributors involve the ‘imaginary movement’ of pi-bonded electrons and  of lone pairs 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 sbond forming or breaking, you are showing a chemical reaction taking place (see rule #1).
3. All resonance contributors for a molecule or ion must have the same net charge – if not, re-check your formal charge calculations.
4. Be very careful in your electron accounting so as not to break the octet rule. Never show "electron movement' arrows that would lead to a situation where a second-row element has more than eight electrons. Sometimes, however, resonance contributors can be drawn in which a second-row atom has only six electrons (see resonance rule #5 below).

To expand a bit on rule #4, there are really only three things we can do with curved arrows when drawing resonance structures. First, we can take a lone pair on an atom and put those two electrons into a pi bond on the same atom (arrow ‘a’ below). Second, we can take the two electrons in a pi bond and shift them up into a lone pair (arrow ‘b’ below).

Notice that the electron movement shown by arrow ‘b’ allows us to avoid placing ten electrons around the carbon.

Third,  we can take the two electrons in a pi bond and shift them over by one bond. Arrow ‘c’ below illustrates this third allowable possibility.

Notice that we do not exceed the octet rule on any carbons when we move the pi electrons, because the positively charged carbon only has six electrons around it to start off with.

When drawing resonance structures, it will help you to keep track of electrons and bonds if you carefully draw arrows to show how you are moving electrons to go from one resonance contributor to another. Always be very careful that your arrows do only the three types of electron movement described above, and that you never exceed the octet rule on a second-row element.

2.2D: Major vs minor resonance contributors: four more rules to follow

Different resonance contributors do not always make the same contribution to the overall structure of the hybrid.  If one resonance structure is more stable (lower in energy) than another, then the first will come closer to depicting the ‘real’ (hybrid) structure than the second.  In the case of carboxylates, contributors A and B are equivalent to each other in terms of their relative stability and therefore their relative contribution to the hybrid structure.  However, there is a third resonance contributor ‘C’ that we have not considered yet, in which the carbon bears a positive formal charge and both oxygens are single-bonded and bear negative charges.

Structure C is relatively less stable, and therefore makes a less important contribution to the overall structure of the hybrid relative to A and B.

How do we know that structure C is the less stable, and thus the ‘minor’ contributor? There are four basic rules which you need to learn in order to evaluate the relative stability of different resonance contributors. We will number them 5-8 so that they may be added to in the 'rules for resonance' list from section 2.2C.

1. 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.
2. In structure C, a separation of charge has been introduced that is not present in A or B. In general, resonance contributors in which there is a greater separation of charge are relatively less important.
3. In structure C, there are no double bonds.  In general, a resonance structure with a lower number of multiple bonds is relatively less important.

There is one more important rule that does not apply to this particular example, but we will list it here in the interest of completeness (we will soon see an actual example).

1. The resonance contributor in which a negative formal charge is located on a more electronegative atom (such as oxygen or nitrogen) is more stable than one in which the negative charge is located on a less electronegative atom (such as carbon).

When discussing other examples of resonance contributors, we will often see cases where one form is less stable – but these minor resonance contributors can still be very relevant in explaining properties of structure or reactivity, and should not be disregarded.

Exercise 2.3

Draw a second resonance structure for acetone (IUPAC name 2-propanone).  Explain why this is a minor (less stable) contributor.

Exercise 2.4

Draw four additional resonance contributors for the molecule below.  Label each one as major or minor.

Solution

2.2E: More examples of resonance contributors: peptide bonds, enolates, and carbocations

A consideration of resonance contributors is crucial to any discussion of the amide functional group. One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.

Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond.

This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly  suggests that the nitrogen is sp2-hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen.

Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond.

Exercise 2.5

Draw two pictures showing the unhybridized p orbitals and the location of pi electrons in an amide group. One picture should represent the major resonance contributor A, the other the minor contributor B (from the figure above).

Solution

‘Enolate ions’ are extremely important species in organic chemistry – if you look ahead, you will see that two whole chapters are devoted to reactions in which enolates play a role! In an enolate, a negatively-charged carbon (carbanion) is adjacent to a carbonyl group (this could be on an aldehyde, a ketone, or a carboxylic acid derivative).

A second resonance contributor B puts the negative charge on the oxygen, and shows a carbon-carbon double bond.

Which of the two resonance contributors is more important?  Consider now rule #8 in the list presented earlier. We have not yet discussed electronegativity (that is the subject of section 2.3), but you probably remember from your general chemistry class that oxygen is more electronegative than carbon.  Therefore, contributor B is the more stable of the two. However, as we will see in chapters 13 and 14, enolate ions usually react in ways that suggest that the carbon bears a substantial degree of negative charge,  so contributor A is still extremely relevant.

Exercise 2.6

Draw the structure of a more stable resonance contributor for the molecule shown below.

Solution

Carbocations such as the one depicted below by two resonance contributors are important in the synthesis of cholesterol and many other related biological molecules.

In this case, three parallel 2pz orbitals are sharing two pi electrons: because there is one fewer pi electron than there are 2pz orbitals, there is a positive charge overall.

For reasons that will be discussed in chapter 7, contributor B is the more stable of the two. The use of these two resonance contributors expresses the idea that both carbon #1 and carbon #3 bear some degree of positive charge, and could react accordingly.

Exercise 2.8

a) Draw three more resonance contributors for the carbocation shown.

b) Fill in the blanks:  the conjugated pi system in this carbocation is composed of  ______ p orbitals containing ________ delocalized $$\pi$$ electrons.

Exercise 2.8

Draw the most stable resonance contributor for each of the anions below.

c) Fill in the blanks:  the conjugated pi system in structure a) is composed of  ______ p orbitals containing ________ delocalized pi electrons.

Solutions

A final word of advice: becoming adept at drawing resonance contributors - and understanding the overall concept of resonance delocalization - is one of the 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 mysterious and incomprehensible, and you will be in for a rough ride, to say the least.  Keep working problems, keep asking questions, keep pounding at it until you get it!

This page titled 2.2: Resonancia is shared under a not declared license and was authored, remixed, and/or curated by Tim Soderberg.