For each of the following reactions, identify the bonds that a...

Question

For each of the following reactions, identify the bonds that are broken and formed. Be sure to indicate whether the bond that is broken is a σ bond or a π bond.

(e)

Accepted Answer

Hey, everyone. And welcome back to another video indicate which bonds are broken and which bonds are formed during the following reaction also show whether each of the bonds that are broken or formed is a sigma bond or a pi bond. We're given an acid al formation reaction in which propan Toone reacts with ethane 12 diol in the presence of acidic conditions to produce 22 dimethyl, 13 dioxane and water. So to solve this problem, we really need to think about the mechanism of the reaction. Let's recall that first of all, we're starting with our carbonyl compound in this case, ketone. And we understand that we have acidic conditions. So we're going to proton our oxygen from the carbonyl group to make a better electrolytic center or increase the electro Felicity of the carbon atom. For that purpose, we're going to illustrate the protein capture by the oxygen atom of the carbonyl group. And we will learn that this increases the electro Felicity of the carbon atom within the carbonyl group because now we have a positively charged oxygen, it has a formal positive charge. And we understand that the given structure is less than stabilized because we can simply shift our pi bond electrons in the form of a lone pair onto oxygen. And this will allow us to get the second resonance form where we now have an alcohol group and a positively charged carbon, which really explains how the electro Felicity of that carbon is increased, right? Just by looking at the resonance form well done. Now, in the next step, we're going to use our dial. So we have two alcohol groups and because it's symmetrical, we can just use any of these two groups. We are going to use the lone pair and oxygen which will perform the nucleic attack on that electro Pyle carbon atom. So in this step, we are going to form an intermediate product, but we want to de proin it, right. Let's see that oxygen gets bonded to carbon, but it has that extra proton which will be removed by water from the acid because we have a solution of an acid, there's plenty of water which will allow us to achieve that deep protonation stop. So let's go ahead and draw the result on intermediate. We first will have our original alcohol group and that we're burning oxygen. We still have two carbon atoms in our chain and there's one more alcohol group at the end of that chain. Now, from here, we're going to proton one of our alcohol groups once again and we are going to proton the alcohol group that is bonded to the original carbon atom that belonged to the carbon group because there's an adjacent oxygen that is electron we joining. So that makes the O H bond, in this case, more polarized than the other O H bond, which is farther away from the oxygen that polarization increases the acidity of the carbon hydrogen. So we're going to proin that if we think about the mechanism, we can just throw another arrow capturing a proton. And from here, we can just draw the protein at intermediate, specifically. What we're going to do is just take the whole structure and change the o age group into O H two, right? Because now we have two hydrogen atoms bonded to the oxygen atom. There is no necessity to indicate the lone pair anymore. We can just add the formal positive charge on oxygen. And of course, this makes a very good living group. So we can remove the water molecule from the structure and we will get the following. So we have a central carbon atom with two metal groups bonded to it. And then we also have our ether, right. So we're going to add oxygen then and Ethyl substi actually does be S H two S H two, right? And at the end of the chain, we have an alcohol group and the carbon atom is now positively charged. Of course, we can have the resonance form, there's a resonance stabilization because we have the parent oxygen that is capable of forming a pi bond and getting that formal positive charge. But it's sufficient to look at that resonance form that we already have. There's no necessity to draw an additional resonance form for the mechanism. And now what's happening is that the remaining alcohol group or specific at the lone payer can attack the electro pic carbo Canion. And now we are going to form a, a ring. So let's go ahead and see that for simplicity, we can just label our carbon atoms or more precisely, we can label all of the atoms that will belong to the ring. So if we start with oxygen as number one, we have carbon number two carbon, number three, oxygen, number five and carbon number oxygen number four, right? And carbon number five. So once again, oxygen number one carbons two and three, oxygen number four and carbon number five, right. So we will have the five member grain, we can start drawing carbons two and three. Then on the left, we will have our oxygen on the right, we will have auction number four, that oxygen number four is bonded to carbon number five. And now the oxygen from the alcohol group will also be bonded to that same carbon during the ring formation. Now we still have one of the oxygen atoms proton. So it will get the formal positive charge. And now we simply want to add two methyl groups to carbon. Number five, of course, we can now be used another de protonation step, water comes in and captures the proton. So let's just use a shorthand notation minus H plus. So that's how we are forming our final product. Well done. We got our mechanism and we just want to understand which bonds are broken and which bonds are formed. Starting with the carbon. We noticed that there is one pi bond that we are breaking. So let's label the bonds that we are breaking in blue, we are breaking it based on the resonance form. So first of all, we're going to say that this would be our pi bond that we break. Because within the double bond, we have one pi bond and one sigma bond. Then we noticed that in the reaction mechanism, the O H group leaves. So we're breaking the sigma bond from the carbonyl group as well later in the reaction sequence. So we're breaking these two bonds within the carbon group. One of them is a pi bond. The other one is a sigma bond. Later on in the mechanism, we also noticed that we are breaking our 20 H bonds from the dial, right? Specifically, let's take a look at this de protonation step. So the water molecule comes in and accepts the proton by breaking the O H bond. Specifically, when we think about the mechanism, the water molecule with a lone pair will capture that proton by breaking the O H bond So we're breaking 20 H bonds in the sequence of that reaction by two D protonation stop at the very end of the mechanism, we're producing water because once again, we're repeating that step or the remaining O H for the proton at O H, so we are going to break T O H bonds. OK. So let's, we will label them in, in the final, in the original scheme actually. So we have identified all of the bonds that we have broken. Those would be two sigma bonds, 10 wage and the error wage bond will also be a Sigma bond because it's a single bond, right? Well done. And now we have to think about the bonds formed. Let's label them in green. So first of all, in the mechanism, we see how oxygen attacks the carbo. So we are creating our carbon oxygen bond, let's label it in green. OK? So that's the first one. And then during the intra molecular attack, we have another O H bond form. So within the ace, we have two carbon A, actually, we're, we're forming another carbon oxygen bond, right? To be more accurate. So we're forming two carbon oxygen bonds within the ace all green. And I was thinking about the water molecule, it's important for us to understand that we are introducing 20 H sigma bonds, right? Because we have the loss of the leaving group and this O H two has been formed from the original auction within the carbon group. First of all, we added one oxygen hydrogen bond. OK. Now, after residence, we just have a wage which is later on proton. Once again, we are forming 20 wage bonds total. So that means we, if we expand the structure of H2O, that'd be Hyde bonded to oxygen, oxygen is bonded to another Hyde and we are forming these 20 H bonds. Well done. So now let's go back to the original reaction scheme and uh label all of the bonds that we're forming and breaking. So choosing a blue color, we can say that we are breaking the bonds in blue. Those would be 11 of them is a sigma bond. The other one is a pi bond. So we're breaking the carbon. Now, we're also breaking just as we said, our old age bonds. So a better idea is just to expand them. And for that purpose, we are just going to introduce in O H bond and we are going to label I'm in blue. So we're breaking the 20 H bonds from the dial. OK? They're in blue and they're both sigma bonds because there are single bonds well done. And I was thinking about the product in green, we are forming two C O sigma bonds because they are single bonds. And now for the water molecule because we need to expand that H O H, we're going to label T O H bonds formed, they are both sigma bonds because they are single bonds, well done. So to make it more, to make it easier for us to stand, we just going to get rid of our mechanism for now. Ok. And a retro it as follows, right? So we have our final answer. The bonds broken are given in blue. So first of all, we have bonds broken within the carbon in group. One of them is a Sigma bond. Another one is a pi bond because that's what a double bond is all the time. Then we have 20 H bonds broken from the dial. There are Sigma bonds because they are single bonds. The bonds formed correspond to two C O bonds within the ace. They are both Sigma bonds because they're single bonds and C O H bonds within the water molecule. They're both Sigma bonds because they are single bonds. That will be our final answer. And thank you for watching and I hope to see you in the next video.

For each of the following reactions, identify the bonds that are broken and formed. Be sure to indicate whether the bond that is broken is a σ bond or a π bond.
(e)

Key Concepts

Types of Chemical Bonds

Bond Breaking and Formation

Reaction Mechanisms

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