Predict the products of the following reactions.
(a) sec-butyl is...

Question

Predict the products of the following reactions.
(a) sec-butyl isopropyl ether + concd. HBr, heat
(b) 2-ethoxy-2-methylpentane + concd. HBr, heat
(c) di-n-butyl ether + hot concd. NaOH

Accepted Answer

Everyone in today's video, we are going to show the major products expected from the reactions given below a two meth oxy to method propane reacting with an excess amount of hydroponic asset in the presence of heat. Be to each oxy propane reacting with hydroponic asset. In the presence of heat. An excess amount of hydrochloric acid is used and see Toxie ethane reacting with sodium hydroxide concentrated solution in the presence of heat. Let's start with a because we have a multi part problem. So we are going to draw a propane because that's our parent. Let's draw our three member chain. So that would be 123 carbon atoms. Position number two has two substations. It has a meth oxy substitution. So we are going to add oxygen that has a metal group, bonnet it and then it has a metal substitution. So that's our starting material. We understand that we are treating it with an excess amount of hydrodynamic asset in the presence of heat. Now you may recall that hydroponic asset is a strong asset. So it's a perfect proton donor and oxygen has a lone pair. So we are expecting to see the protein nation step in which the lone pair on oxygen attacks the acidic proton, followed by the loss of the leaving group. In this case the bromide and iron is formed. So we may show that in this step we are forming our protein ated intermediate, there will be formal positive charge and oxygen because now it forms three bonds. And now you may notice that the presence of heat, we can essentially promote that ether cleavage. Right? Essentially, if we get rid of our leaving group because the auction has a formal positive charge, we will get a stable tertiary car book at iron because there are three metal groups surrounding that carbon. This is really favorable for us. So we understand that we can think of an sn one reaction. This would correspond to our S. N. One reaction. And in this s. n. one reaction, we're essentially getting rid of our leaving group in forming an alcohol with our car broke Orion. First of all, we control the result of methanol CH 30. H. And then afterwards we can include our tertiary carbo carrion. Now, first of all, let's think about the carbo cast iron. Recall that in the first step we have formed the bromine and iron and the brunette and I can now participate in the reaction. It can act as a nuclear file and it can attack the carbo cast iron to form our first product. Notice that we are getting our al killed bromide. So that would be our first product. Right? We are going to have brimming over here, but now we have to be extra careful, should be a really include methanol among our products. Now, the answer is not really if you notice we're using an excess amount of hydroponic asset. So you may notice that the G. Two, that oxygen over here, we can have another protein nation stop where we are forming the brahma tannin as well. And in that stop. First of all we are getting our protein ated intermediate with a formal positive charge and oxygen. And then afterwards the broadway and I. And that previously left will attack the electra Felix center. So now this is where we have our essence to reaction type. Right? So this one would be and sn two reaction because we have a primary substrate. Right? This my apologies. That would be our metal substrate. It doesn't even have any neighboring carbon atoms so that the bromide and ion would perform the new club like a tech at this center followed by the loss of the leaving group. So to illustrate that properly, we want to make sure that we draw out our bonds. Let's see that that would be our metal group bonnet C O H C with a formal positive charge. And then when the bomb went and an attack that carbon we have the loss of the leaving group which is water. And this is how we are producing our final product because there was an excess amount of hydrochloric acid. So we can include this is our final product for the overall reaction. Okay, well done. So we got our final products. Now let's move on to be to Edith oxy propane reacting with an excess amount of hydroponic acid in the presence of heat. Similar conditions are used here. We just want to make sure that we get our substrate correctly. We may draw to Edith oxy propane. So that means we have a three member chain that would be propane. And because it's us to eat oxy we're going to add that oxygen atom with an ether oxy group. Right? My apologies. Any Toxie group overall because we have oxygen bonded and ethyl groups. So this is our propane two. Youth. Oxy essential implies that we have oxygen bonnets and ethyl group. So that's our starting material. Now, as we already know the first step in this reaction because we are using hydroponic asset and an excess amount of it in the presence of heat. We are going to observe that protein nation stops. So the lone parent oxygen will easily get protein ated, followed by the loss of the leaving group, which is the brunette and iron. And this is where we are getting our protein ated intermediate. So let's draw it. That would be oxygen bonnets, a hydrogen with a formal positive charge. And now this is where we are forming our cleavage of ether. Right. Notice that the carbon atom over here is a secondary carbon because there are two neighbors. So the reaction might be either SN one or SN C. Let's suppose that it's sN one for simplicity because it would be consistent with what we have seen previously. It doesn't really matter which pathway we choose here because we're using heat. It would actually promote the formation of the carbo cast iron. So we can definitely think about N. S. N. One reaction. So now the application of heat here let's suppose that this is SN one heat promotes the lost believing group. So we are essentially going to cleave this carbon auction bond over here. Now the reason why we're doing so is because that would be a more stable secondary carbon cast iron. Then the primary carbon carbon that we we could have got here. If we had broken this carbon auction bond. Right? So we are going to think about this bond instead and therefore if you think about the products formed one of them as ethanol. Right? So that would be our first product plus our secondary carb Okura informed And previously we saw how we got one of our products we said that the bronze and iron must attack that carbo Karen based on an S. N. One reaction pathway. We have that nuclear attack and this is how we obtain Our first product which is an alcohol bromide. So we may essentially joy three member chain and then we are adding our bombing substitution. Now for the second product we are again not going to include ethanol but instead we're going to remember that ethanol will undergo an essence a reaction. We can illustrate the steps once again but you may already remember that we're essentially getting an elk kill bromide as well because we are performing the protein nation stop right If we add HBR first of all, we are protein ating it. And when we get our protein it'd and remediate OH two plus, then there is this nuclear for like a text stop the bromide and I. And attacks in an SNC manner followed by the loss of the leaving group. So you may notice that we're getting our ethyl bromide, which we can include as our second final product. That's once again because we're using an excess amount of hydroponic asset. So, this will be our second product now for c ether oxy ethane. So that would be the ethyl ether. That's exactly the same structure. Right? We are going to draw to ethyl groups bonded oxygen according to the problem. The reaction takes place with the concentrated solution of sodium hydroxide in the presence of heat. This reaction is not possible. We can say no reaction. And the reason why we're doing this is simply because the reaction is not possible as ethers are only cleaved under acidic conditions. So we can say that ethers are only cleaved in acidic conditions. This really makes sense, right? Because we cannot protein eight oxygen conditions are basic. So there's nothing that can actually donate a proton to this oxygen. And because we cannot donate a proton, we cannot perform the bond cleavage because oxygen oxygen would be a bad leaving group, we actually want to make sure that it has a formal positive charge before we break that carbon oxygen bond. And since there are no civic products that would promote this, we cannot really do anything. And we can say that there would be no reaction since conditions are basic. So now, seeing all of this, we can go back to our answers, starting with A because we are asked to show the products. We see that for the first reaction, our products are to alcohol bromides. For me, we got two different L killed bromides, and then for C, we said that no reaction takes place because ethers are only cleaved in acidic conditions. So now, looking at our answers, we can conclude that choice D. Is the correct answer to this multiple choice question. However, if you feel that this concept makes you confused or you don't understand something, Feel free to drop down your message in the comments section below. Thank you for watching.

Predict the products of the following reactions. (a) sec-butyl isopropyl ether + concd. HBr, heat (b) 2-ethoxy-2-methylpentane + concd. HBr, heat (c) di-n-butyl ether + hot concd. NaOH

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