6.2. resonance | Organic Chemistry 1: An Open Book (2023)

First, watch this video:6.2. resonance | Organic Chemistry 1: An Open Book (1)

If we were to draw the structure of an aromatic molecule like 1,2-dimethylbenzene, there are two ways to draw the double bonds:

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Which path is correct? There are two simple answers to this question: "both" and "neither". Both ways of drawing the molecule are equally good approximations of the binding picture of the molecule, but neither is by itself an accurate picture of the delocalized pi bonds. The two alternative designs, on the other handwhen they see each other, give a much more accurate picture than each separately. This is because together they imply that the carbon-carbon bonds are not double bonds, nor are they single bonds, but somewhere in between.

These two drawings are an example of what is called in organic chemistryresonance contributors: two or more distinct Lewis structures representing the same molecule or ion that, when viewed together, approximate the delocalized pi bond better than either structure individually. By convention, resonance contributors are connected by a double arrow and sometimes enclosed in square brackets:

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Small curved arrows are often used to make the difference between two resonance contributions more visible. Each of these arrows shows the "movement" of two pi electrons. In later chapters, as we begin to study organic reactions, a process in which electron density changes and covalent bonds between atoms are broken and formed, this "curved arrow notation" will become extremely important for representing the process of motion. of electrons. However, when we draw resonance contributions, this "movement" of electrons only occurs in our minds when we try to visualize delocalized pi bonds. However, the use of curved arrow notation is an essential skill to develop when designing resonance contributors.

The representation of benzene with the two resonance contributions A and B in the figure above does the job.Noimplies that the molecule looks like structure A one moment and then changes to look like structure B the next moment. Rather, the molecule at any point in time is a combination orresonance hybridambos a e B.

Caution🇧🇷 It is very important to make it clear that when drawing two (or more) resonance contributions, we are not drawing two different molecules: they are simpledifferent representations of exactly the same molecule🇧🇷 Also, the double-headed resonance arrow does NOT mean that a chemical reaction has occurred.

Conventionally, benzene derivatives (and phenyl groups when the benzene ring is incorporated into a larger organic structure) are presented with a single resonance contributor, and it is assumed that the reader understands that resonance hybridization is implied. This is the convention used for most of this book. In other books or articles, you sometimes see benzene or a phenyl group drawn with a circle inside the hexagon, solid or discontinuous to draw a resonance hybrid.

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Resonance contributions to the carboxylate group.

The convention of drawing two or more resonance contributions to approximate a single structure may seem a little clunky to you at the moment, but as you gain experience you will find that the practice is actually very useful when analyzing how many functional groups react. Next. , let's look at carboxylation (the conjugate base of a carboxylic acid). For our example, we're going to use formate, the simplest carboxylate-containing molecule possible. The conjugate acid of formate is formic acid, which causes the painful sting you get if you've ever been stung by an ant.

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Typically, you will see carboxylate groups drawn with a carbon-oxygen double bond and a carbon-oxygen single bond, with a formal negative charge on the single-bonded oxygen. In reality, however, the two carbon-oxygen bonds are the same length, and although the group does indeed have an overall formal negative charge, it is shared equally between the two oxygen atoms. Therefore, the carboxylate can be represented more specifically by aparof resonators. Alternatively, a single structure can be used, with a dashed line representing the pi bond delocalized by resonance and the negative charge placed between the two oxygen atoms.

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Let's see if we can correlate these drawing conventions with a picture of the valence bond theory of bonding in a carboxylate group. We know that carbon must be sp2-hybridized, (the bond angles are close to 120˚ and the molecule is flat), and we will treat both oxygen atoms as sp2- also hybridized. Both sigma carbon-oxygen bonds are formed from overlapping sp carbon2orbitals and sp oxygen2Orbital.

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Furthermore, carbon and oxygen atoms each have an unhybridized 2.pzOrbital perpendicular to the plane of sigma bonds. these three 2pzThe orbitals are parallel to each other and can overlap side by side to form a delocalized pi bond.

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The A resonance contribution shows that oxygen #1 shares an electron pair with carbon in a pi bond and oxygen #2 retains a lone pair of electrons in its 2 bond.pzorbital. The resonance contribution B, on the other hand, shows oxygen #2 involved in pi bonding to carbon and oxygen #1 containing a lone pair on its 2.pzorbital. In general, the situation isthree parallel, overlapping 2pzOrbitals shared by four delocalized pi electrons🇧🇷 Why is there one more electron than 2pzorbitals, the system has a total charge of -1. This is the approximate type of 3D image by resonance contributors, and after practicing, you can quickly visualize 2 superimposed.pzOrbitals and pi electrons delocalized wherever you see resonant structures used. In this text, carboxylate groups are usually drawn with a single resonance contribution for simplicity, but you should always keep in mind that the two C-O bonds are the same and the negative charge is delocalized on both oxygen atoms.


There is a third resonant contributor to the format (which we'll soon discover is considered a "minor" contributor). Draw this resonance contribution.

Here's another example, this time with a carbocation. Remember ofSection 2.1🇧🇷 what are carbocationssp2-hybridized, with a void2pOrbital oriented perpendicular to the plane formed by three sigma bonds. If there is a carbocation next to a double bond, then three2pThe orbitals can overlap and share the two pi electrons, another type of conjugated pi system where the positive charge is shared between two carbons.

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1. Draw the resonance contributions that correspond to the curved arrows of motion of two electrons in the following resonance expressions.

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2. In each resonance expression, draw curved arrows of motion of two electrons in the left contributors, showing how we got to the right contributor. Be sure to include any formal fees.

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(Video) Chem 109 Zoom of Class Thurs Oct 1

Some common resonance patterns

Se você examinar um grande número de exemplos de ressonância, descobrirá que eles quase sempre concordam com certos padrões comuns, dos quais existem apenas quatro. Destas, duas correspondem a estruturas sem carga e duas a estruturas carregadas, como se pode observar na figura:

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They are listed as Types I-IV, but those terms are not used outside of this book! The most important are Types III and IV, where resonance is used to share or "offset" the load around the structure to stabilize it. In more complex cases, both types can occur simultaneously.

type III resonanceis only seen with a + charge and usually involves a positive charge of oxygen or nitrogen shared on a carbon; The carbocation form has only six valence electrons on carbon, making it less stable than the original form (which has full octets).

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type IV resonanceis very common and is observed for both positive and negative charges, although the cationic (+) forms always (in this course) involve only carbon (where X, Y, and Z are all carbon). It is sometimes referred to as an "allylic" resonance, particularly in all-carbon cases, because it is modeled after the allyl (all-carbon) group. In the type IV resonance of anions, the primary mode is to place the negative charge on the atom that can best handle the charge, often the more electronegative oxygen or nitrogen.

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Rules for drawing resonance structures.

When learning to draw and interpret resonance structures, there are some basic rules to remember to avoid drawing meaningless structures. These rules all make perfect sense as long as you consider that resonance contributions are simply a man-made convention for representing delocalization of pi electrons in conjugated systems.

Rules for drawing resonance structures:

1)If you see two different echo messages, it's you.Nosee a chemical reaction! Instead, you see the exact same molecule or ion represented in two different ways. This means you cannot add, remove or move atoms - not even Hs!

2)Resonance contributions include the "imaginary motion" of pi-bonded electrons or lone electron pairs adjacent to (dh🇧🇷 conjugate a) pi bonds. You mayOh noChanging the position of electrons in sigma bonds: When you show that a sigma bond is being formed or broken, you are showing that a chemical reaction is taking place (see rule #1). Likewise, the positions ofAtomin the molecule cannot switch between two resonance contributions.

3)All resonance contributions to a molecule or ion must be equal.The netcharge.

4)All resonance contributions must be drawn as correct Lewis structures with correct formal charges. Never show "curved arrows of electron movement" which would lead to a situation where an element in the second row (eg carbon, nitrogen or oxygen) has more than eight electrons: this would violate the "octet rule". However, we sometimes draw resonance contributions where a carbon atom has only six electrons (i.e. a carbocation). In general, all oxygen and nitrogen atoms must have a full octet of valence electrons.

To extend rule #4 a bit, there are only three things we can do with curved arrows when drawing resonant structures. First, we can take the two electrons in a pi bond and exchange them so that they become a lone pair on a neighboring atom (arrow "a" below).

Second, we can take a lone pair of electrons in an atom and place those two electrons in a pi bond on the same atom (arrow "b"). Third, we can move a pi link one position (arrow c).

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Resonance arrows can also be combined; then we show arrows a and b together:

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Note that when we move the electrons with the a, b, and c arrows, we are not breaking the octet rule for any atom. The following resonance image shows "illegal" movement of electrons, as this would result in a carbon with five bonds or 10 valence electrons (this would violate the octet rule):

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Always be very careful when drawing resonant structures your arrows create.onlythe three modes of electron motion described above, and that youOh noexceed the octet rule for a second row element. It is often useful (but optional) to include all oxygen and nitrogen pairs in the drawing to keep track of valence electrons, avoid breaking the octet rule, and identify whether atoms have a formal negative or positive charge. Proper "electron accounting" is a big part of working with resonance donors.

Below are some more examples of "cool" resonating expressions. Convince yourself that the octet rule is not exceeded for any atom and that the formal charges are correct.

(Video) LTHS CHEM PPT 6.2 Stoich and Enthalpy 17-18 HARBIN

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Each of the following "illegal" resonance expressions contains one or more errors. Explain what is wrong in each case.

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Key Contributions of Resonance vs. secondary

The different resonance contributors do not always make the same contribution to the overall structure of the hybrid; in many cases, one contributor is closer to the true image of the title than another. In the case of carboxylates, the following contributors A and B are equivalent in their relative contribution to the hybrid structure. But there is a third 'C resonance contribution, in which the carbon carries a formal positive charge and both oxygen atoms are single bonded and carry negative charges.

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Compared to A and B, structure C makes a less important contribution to the overall picture of group unity. How do we know that the C structure is the "minor" contribution? There are four basic rules you need to learn to assess the relative importance of different contributions to resonance. We'll number them 5 through 8 so they can be added to the "Resonance Rules" list earlier on this page.

Rules for determining principal and spurious resonance contributions:

5)The carbon in contributor C has no octet; in general, resonance contributors where a carbon does not satisfy the octet rule are relatively less important.

6)Charge separation was introduced in structure C, which is not present in A or B. In general, resonance contributions where there is greater charge separation are relatively less important.

7)There are only three bonds in the C structure compared to four in A and B. In general, a resonance structure with fewer total bonds is relatively less important.

8)The resonance contribution where a formal negative charge is on a more electronegative atom, usually oxygen or nitrogen, is more stable than where the negative charge is on a less electronegative atom, such as carbon.An example is in print in the upper left corner of the following figure.

Here are some additional examples of primary and secondary resonance contributions:

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Why bother with a resonance contribution when it's the smallest? We will see later that often a small contribution to our understanding of how a molecule reacts can be extremely important.


1. a) Draw a small resonance structure for acetone (propan-2-one). Explain why the contribution is small.

b) Do acetone and 2-propanol mutually participate in resonance? Explain.

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2. Draw four additional resonance contributions for the molecule below. Label each as a major or minor contributor (the structure below is for a major contributor).

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(Video) Section 6.2 example 1

3. Draw three resonance contributions from methyl acetate (an ester with the structure CH3KOCH3) and rank them according to their relative importance in the binding picture of the molecule. Explain your reasoning.

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Resonance and peptide bonds

What is the hybridization state of the nitrogen atom in an amide? At first glance, it seems logical to say that it issp3-hybridized because the Lewis structure, like nitrogen in an amine, has three single bonds and a lone pair of electrons. The picture is quite different if we consider another resonance contribution where the nitrogen has a double bond with the carbonyl carbon: in this case we would have to say that hybridization is applicable.sp2, and the bond geometry is trigonal planar rather than tetrahedral.

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In fact, the last picture is more accurate: the lone pair of electrons in a nitrogen amide is not located in asp3Orbital, in contrast, they are delocalized as part of a conjugate pi system, and the bond geometry around nitrogen is trigonal planar, as expected.sp2hybridization. This is a good illustration of an important point: conjugation and the corresponding delocalization of electron density are leveling out, so if conjugation can occur, it probably will.

One of the most important examples of amide groups in nature is the "peptide bond" that connects amino acids to form polypeptides and proteins.

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Crucial to the structure of proteins is the fact that, although traditionally designed as a single bond,The C-N bond in a peptide bond has a significant barrier to rotation, indicating that there is some degree of C-N-Pi overlap; in other words, there is some double bond character and nitrogen is presentsp2hybridize as trigonal planar geometry.

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The rotational barrier at peptide bonds is an integral part of protein structure and provides more rigidity to the protein backbone. If there were no barrier to rotation in a peptide bond, proteins would be much "lazier" and three-dimensional folding would be quite different.


1. Take two photos showing the non-hybridized2pOrbitals and position of pi electrons in a simple amide. One image should represent the major resonance contribution, the other the minor contribution. how many overlap2pHow many pi-bonded electrons share orbitals?

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2. Make two drawings showing the non-hybridized2pOrbitals and the position of the pi electrons in the "enolate" anion shown below. One image should represent the major resonance contribution, the other the minor contribution. how many overlap 2pHow many pi-bonded electrons share orbitals?

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3. Below is a spurious contribution of a kind known as 'enamine', which we will examine in more detail in Chapter 12. Draw the main resonance contribution to enamine and explain why its contribution is the main contribution (see resonance rules) . 5-8 of this section).

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Resolved example:Draw the main resonance contribution of the following structure. Include in your figure the appropriate curved arrows that show how you got from the given structure to your structure. Explain why your partner is the most important. What are the orbitals of the two lone pairs of electrons in oxygen?

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solution:In the above structure, the carbon with the formal positive charge does not have a full octet of valence electrons. Using the curved arrow convention, a lone pair on oxygen can be shifted to the left in the adjacent bond, and the electrons in the double bond are shifted to the left (see rules for drawing resonance contributions to ensure this is 'legal motions'). ').

(Video) EDEXCEL Topic 6 Organic Chemistry I (Part 1 of 2) REVISION

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The resulting resonance contribution, where oxygen carries the formal charge, is the main contribution, as all atoms have a full octet and an additional bond is drawn (resonance rules #5 and #7 apply). This system can be considered as four in parallel2pOrbitals (one each in C2, C3, y C4, plus one on oxygen) sharing four pi electrons. A lone pair on oxygen is in an unhybridized state.2porbital and is part of the conjugate pi system, and the other is in asp2orbital.

Also note that an additional contributor can be drawn, but it is also smaller as it has a carbon with an incomplete octet:

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1. a) Draw three more resonance contributions to the carbocation below. In your figure, include the appropriate curved arrows that show how one contributor transforms into the next.

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b) Fill in the blanks: The conjugated pi system in this carbocation consists of ______2pOrbitals that share ________ delocalized pi electrons.

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2. Draw the main resonance contribution for each of the anions below.

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c) Fill in the blanks: The conjugate pi system in part (a) consists of ______2pOrbitals with ________ delocalized pi electrons.

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3. The figure below shows how the formal negative charge of oxygen can be delocalized to the carbon indicated by the arrow. Further resonance contributions can be extracted where the negative charge is delocalized to three other atoms in the molecule.

a) Circle these atoms.

b) Draw the two most important resonance contributions to the molecule.

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An advice

Drawing resonance contributors, using curved arrow notation to show how one contributor can transform into another, and understanding the concepts of conjugation and resonance displacement are some of the most challenging but also important tasks for a beginner. organic chemistry. Now, if you work hard to firmly grasp these ideas, you will understand much of what follows in your organic chemistry course. On the other hand, if you don't delve into these concepts now, much of what you'll see later in the course will look like a bunch of mysterious, incomprehensible lines, dots, and arrows, and you'll be on a rocky road, to say the least. More than many other topics in organic chemistry,Understanding attachment, conjugation, and resonance is something most students really need to work on "in person" with a teacher or tutor.,preferably with a molecular modeling kit🇧🇷 Keep working on the problems, keep asking questions, and persist until it all makes sense!

Khan Academy video tutorials

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