For each molecule shown below,
2. draw the important re...

Question

For each molecule shown below,

2. draw the important resonance contributors of the anion that results from removal of the most acidic hydrogen.

(a)

(b)

(c)

Accepted Answer

Welcome back everyone to another video, draw the important resonance structures of the anion formed from the removal of the most acidic hydrogen in each molecule given below, we're given three molecules and we are going to start with the first one. What we notice is that we have a carboxylic acid group. And of course, the hydrogen atom from the carboxylic acid will be the most acidic. So let's simply remove it. We're going to draw a five membered ring to begin with. We're going to add that pi bond and we are going to expand our carboxylic acid group which is coo h but we have removed that acidic hydrogen. So we got our carboxylate and this is our first resonance for now to get the second resonance form, we are simply going to shift our pi electrons right, because we have an adjacent pi bond for that de localization. So we are forming our P bond breaking the adjacent pi bond. And this is how we get our second resonance contributor. So let's go ahead and try it our single bonded oxygen, our double bonded oxygen. And now let's add the result and negative charge. Well done. So we got our resonance contributors. For the first structure. There is no necessity to think about the Python because it's too far away to participate in conjugation. Now, for the second structure, we have a di ketone and we have to remember that alpha hydrogens are the most acidic, but the metal one will be the most acidic of all the hydrogens present in the structure because it's adjacent to two carbon groups are basically it is alpha relative to each carbon group and carbon groups are electron withdrawing. So the acidity of the given hygen is significantly high. Now, the removal of this hydrogen will essentially produce a carbon ion. So we can draw that carbon ion. But we know that the negative charge on carbon is not that stable because carbon is not that electron negative. So what we can do is just shift our electrons form a pi bond and break the adjacent pi bond to get our second resonance form. So let's draw our second residence form. We can see that by redrawing our chain, now oxygen will be single bonded with a negative charge. We're going to get our pi bond and the adjacent carbon group. So now this resonance form is actually much more stable because the negative charge is present on a more electron oxygen atom. Compared to the less electron carbon. For the for the third resonance structure, we can actually delocalized our pi electrons change the position of the pi bond and break the pi bond from the carbonyl group. And this will produce our third resonance form and that would be it. So the two final resin structures are more stable than the first one. Meaning the first one is not that significant because it has a negative charge on carbon. And since oxygen is much more electron, we are only going to leave the final two retinal structures because they are significant well done. Now, moving on to structure number three, we have our ketone which also has a nitro group. So let's recall once again that the alpha positions will have the most acidic hydrogens. But the alpha position on the right will have a more acidic hydrogen because the nitro group is electron withdrawing. So the removal of that hydrogen once again will give us a carbon ion. So let's go ahead and throw that carbon on. We will have our six membered crane as well as our nitro group. What we can do is just expand our nitro group, there is one double bonded oxygen and one single bonded oxygen. Let's add that negative charge on carbon. So that could be our first re structure. And now let's consider the other ones. So what we can do is just shift our pi electrons into the formation of a pi bond and break the pi bond from the carny. This will be a more stable resonance structure because the negative charge would appear on oxygen which is once again, more electron negative than carbon. So let's get rid of the negative charge and place it on oxygen instead, which is now single bonded. So we get our E Nate well done and now that we got our EO eight, we can consider the resonance within the nitro group itself. So what we can do is just shift our biel electrons from the negatively charged oxygen right, and break the pipe bond between nitrogen and oxygen. So that would give us an additional resonance form. Let's go ahead and add it. So now what happens is that we are changing the position of the pi bond, we are choosing another oxygen. Let's get rid of the negative charge and place it on the other oxygen. Well done. And finally, what else can we see? Well, we can essentially use the lump pair on oxygen from the original carbonyl group. We formed the carbonyl shift the pi bond between carbon and nitrogen by breaking the pi bond between nitrogen and oxygen. And this would give us an additional resonance form with two negatively charged oxygens, meaning the first resonance form was actually insignificant because once again, there was a negative charge on carbon. So what we can do is just leave our first one, the second one. And now let's toward the final one, which will be the third resonance structure. And to do that, we can simply do what the arrow suggests. It essentially suggests us adding that pi bond. So we get our carbonyl and drawing a double bond between carbon and nitrogen as well as breaking the pipe bond between nitrogen and oxygen. So that oxygen gets a negative charge. And that's how we got all of the significant resin structures that would be it. Thank you for watching.

For each molecule shown below,
2. draw the important resonance contributors of the anion that results from removal of the most acidic hydrogen.
(a)
(b)
(c)

Key Concepts

Resonance Structures

Acidity and pKa

Anions and Charge Distribution

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