a. Draw resonance contributors for the following species. Do n...

Question

a. Draw resonance contributors for the following species. Do not include structures that are so unstable that their contributions to the resonance hybrid would be negligible. Indicate which are major contributors and which are minor contributors to the resonance hybrid.
b. Do any of the species have resonance contributors that all contribute equally to the resonance hybrid?
14.

b. Do any of the species have resonance contributors that all contribute equally to the resonance hybrid?

14. Structural representation of a molecule with resonance contributors, showing major and minor contributors.

Accepted Answer

Welcome back everyone to another video, draw the resonance structures for the given I and below. Excluding extremely unstable structures with negligible impact on resonance structures, determine the major and minor contributors do. The resonance contributors contribute equally to the resonance hybrid. And we're given the structure of E for oxo to E in one I, we want to think about the possible resonance structures. And what we want to remember is that resonance can simply be explained by the de localization off high electrons in a conjugate system. Specifically electrons flow from higher density to lower density. And what we notice is that we have that lone pair of electrons on carbon and the adjacent pi bond which will allow us to de localize that lone pair into the conjugate system. So what we're going to do to begin with is just redraw the original structure and use our arrows to illustrate that de localization of electrons starting with the long pair of electrons, we're going to use it in the formation of a pi bond. But if we do so we have to break the adjacent pi bond. So what we get is our second resonance form because the first one is already considered to be a resonance form. So what do we get based on the arrows? Well, basically we're going to get red of that long pair of electrons, we're going to get rid of the pi bond. And now the last arrow is telling us that we end up with a lone pair on the other carbon atom due to the presence of that long pair of electrons, we can add a negative charge on carbon. And now the negative charge on the first carbon should not be there anymore because the the loan pair of electrons is now de localized right carbon now has a total of four bonds. These two group has a double bond and two high gens bonded to that carbon. So we don't have any negative charge because it now appears on the adjacent carbon atom. Now, it is also not necessary for us to indicate that long pair of electrons because it is implicit if we simply state that negative charge. So we can just get rid of our long pair of electrons and illustrate it by the presence of a negative charge. And we can do the same for the first structure even though we can represent both the negative charge and the long pair of electrons. Now, we can also draw another resonance form. And the reason why we can do so is because of the adjacent pi bond. So what we're going to do is just de localize that long pair of electrons since the formation of a pi bond and break the pi bond from the carbon group. So what, what do we get in this case? Well, what we're going to do is just essentially break our pyon from the carbonyl group. That's very important for us to do, right? Because that's what the last arrow suggest we're going to add the remaining pi bonds. So first of all, notice how we have that pi bond for the siege two group, nothing changes. But now we are forming a pi bond between two carbons, one carbon from the previous resins form and the other from the carbonyl group. And now the last arrow tells us that there must be a negative charge on oxygen because we're placing a low pair of electrons. And from here, we cannot relate to anything. We can just add brackets to illustrate that those are resins forms. So let's go ahead and add those brackets. And essentially here we can stop drawing out resonance forms because we don't have any additional ones. Simply speaking, there are no adjacent pi bonds that we can use for new resonance forms. And now our goal is to consider minor and major contributors notice how the first and the second rein forms they have a negative charge on carbon atoms. And the third one has a negative charge on oxygen. Recall that the electron negativity of oxygen is greater than the electron negativity of carbon. And we once remember that the negative charge is expected to be present on a more electron negative atom that's more favorable. So this means that the third resonance form is the major contributor because there's a negative charge on a more electron negative oxygen atom. And the first ones are minor contributors because they contain a negative charge on a less electron negative carbon atom. And this finalizes our problem, we can state that the reins contributors. So the resonance contributors do not equally contribute to the reins hybrid or to the hybrid. Now, the hybrid is just the average of them. And because one of the resonance forms is the major one, while the other ones are minor ones, we can state that there is no equal contribution and they do not equally contribute to the resonance hybrid. That would be our final answer and our we possible resonance forms. Thank you for watching.

Key Concepts

Resonance Structures

Major and Minor Contributors

Resonance Hybrid

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