Show the resonance forms for the enolate ions that result
when t...

Question

Show the resonance forms for the enolate ions that result 
when the following compounds are treated with a strong base.
(a) ethyl acetoacetate 
(b) pentane-2,4-dione

Accepted Answer

Everyone and welcome back in this video, we are going to draw the resonance structures of the final eight ions produced by the deep rotation of the below given compounds through a base a cyclops, benzene 13 Dione and be methanol acetone acetate. If we start with a it's just sufficient to start by drawing the structure because it says cycle pence in 13 die on. We understand that the structure has a five member drain which represents cycle pencil and on positions one and three was supposed to this position number one and this is position number three. Right? If we go clockwise, they contain two carbonell groups because it says one comma three D. I own. So we are going to add two carbonell groups. Okay, so this is our structure. Now, what we want to understand is that we are going to produce an in a late iron and stew that we first of all want to identify our most acidic hydrogen. We are always interested in the alpha position. So if you look at this carbon group, it has two alpha positions, this one and that one right now looking at the other carbon group, it also has two alpha positions. This one and that one notice how each carbon group shares that exactly the same alpha position, which increases the acidity of that hygiene. So this hydrogen would be more acidic than any of the two below. So we are going to remove that hydrogen using a base. Right as a problem suggests we need to use a base and a commonly used based on this reaction whenever we're producing in a late would be hydroxide. So what we're going to do, we're going to produce our insulate and we are going to form just one possible residence form for now. And as we know, we are essentially attacking our stick proton. We are transferring these carbon hydrogen electrons into a pi bond. Rights were forming a double bond at this position and then we are breaking the adjacent pi bond over here, transferring our electrons onto oxygen in the form of a lone pair. And now you may notice that we are getting our first residence form. Indeed, That would be our five member growing. This oxygen is still double bonded, but now we have a pipe on a disposition and a single bonded oxygen with a negative charge. Okay, so this is our first resident structure. We got our in a late, we just want to think about the possible electron shift. So what we want to do, we want to make sure that we first of all add brackets which indicate that we are drawing residence forms. And we now understand that the negative charge and oxygen can essentially be used in the restoration of that pi bond, right, We're using that lone parent oxygen and we are essentially de localizing it forming our pi bond. But if we form a pi bond at this position, we have to bring the adjacent pi bon and we are going to transfer these pi electrons in the form of a lone pair on carbon, which will allow us to get our second residence form. So we are using our residence arrows. Now we are going to draw our five member grain. We understand that now we have our double bond for that left side oxygen. Now because we have broken that pi bond over here we will have a lone pair, it's just sufficient for you to indicate a negative charge, which essentially implies that there is a lone pair. Okay, and now for the other end we have our carbonell group. So this is our second residence form. And finally we can through our third residence form, let's indicate the presence of residence arrows. And now we understand that the lone parent, this carbon can also be de localized into our system because if we form a pi bond disposition and we have that adjacent pi bond between carbon and oxygen. If we break that pi bond and transfer the pi electrons onto oxygen in the form of a lone pair, that would give us our final residence structure because we don't have any other double bonds adjacent. So that would be our oxygen from carbon double bonded. Now here we are forming our pi bond from these electrons and here we have a single bond carbon oxygen bond that will be negative charge on oxygen. And that would be it because we don't have any other adjacent pi bonds. There's no other possible residence form. We can simply say that these three residents structures would correspond to our insulate ions right now, this will be our final answer for the first part. We can immediately label this. And let's move on to part B. Now, for part B, we want to draw method aceto acetate. So let's start by drawing that structure, methanol, acetone acid, it would essentially have a metal groupe carbonell group. And then we would have a siege to group that allow that allows us to connect these two carbonell groups. And because we have a metal group that will be oxygen and a metal substitution, right? Which comes from an ester derivative. Now this is methyl acetate acetate and we once again want to think about that acidic proton. When we think about our alpha positions. This carbon, your group only has one alpha position. This carbon your group has two on the left and on the right, so they both share the same alpha position, which makes this hydrogen the most acidic one. Because this is the most acidic hydrogen. We are going to use our base Once again, it should be hydroxide because this is the general base whenever we are forming in a late ions. And we understand that we are observing that the proto nation stop in which we control our first in a late form, we are going to transfer these electrons in the form of a pi bon. And therefore we have to break the adjacent pi bon over here on carbonell by transferring pi electrons in the form of a lone pair and oxygen. So now we are ready to draw our first insulate resonance form. Right? Because the welfare structure that we get is included in resonance and we may see that first of all we have oxygen which is single bonded to carbon. Now here we are forming our double bond and we can draw the remaining fragment of the structure. Okay, perfect. So we got our first residence form and now we need to think about the possible electron shifts. What if we take this lone parent oxygen, restore the double bond and as a result we have to break the adjacent pi bond. So the way we can see that is by transferring these pi electrons in the form of a lone pair onto this carbon. So this would give us our next residence form and we may simply draw it. So that would be our double bonded oxygen. Now, all right, now we meet some play troll the other carbon steel group oxygen metal and I'll notice that what we're getting essentially is a lone pair on that carbon and again, it's just sufficient to indicate the presence of a negative charge. Okay, it's not so important to indicate the presence of that lone pair because the negative charge already represents the presence of that lone pair. And now you may notice that there is still an adjacent pi bond which allows the D localization of this lone pair. So what we're going to do, we're going to essentially take this lone pair form a double bond by breaking the adjacent pi bond. And this would give us our final residence form because there are no more adjacent pi bonds. Now let's draw the resultant and the final residence form here, that would be our carbonell group on the left. We may notice that we are forming a double bond at this position and now the oxygen atom on the right would essentially have a negative charge because we have transferred that that pair of pi electrons in the form of Olympia and oxygen. And then on the right side we have oxygen bonded to the metal group and that would correspond to our final structure. So once again we got three residents forms for be and we may simply conclude that each part has three residents forms. Now, if you look at the answer choices, we may simply conclude that the correct answer choice to the small cultures question is d However, if you have any questions, don't hesitate to leave a comment down below and thank you for watching

Show the resonance forms for the enolate ions that result when the following compounds are treated with a strong base. (a) ethyl acetoacetate (b) pentane-2,4-dione

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