When diethyl ether (CH3CH2OCH2CH3)
is treated with concentrated H...

Question

When diethyl ether (CH3CH2OCH2CH3) is treated with concentrated HBr, the initial products are CH3CH2Br and CH3CH2OH. Propose a mechanism to account for this reaction.

Accepted Answer

welcome everyone in this problem. We are considering the fulling compound and when treated with hydrochloric acid, this compound undergoes cleavage to form five bromo five metal hex and one. Oh we wish to write a plausible mechanism for this reaction. Before we begin. Let's just complete the overall reaction equation. We're treating the original substrate with hydrochloric acid and we are now going to convert the aipac name into an actual structure. We can draw a six member chain of carbon atoms because its hexagonal, we can label it from left to right Carbon No 1, 2, 3, 4, 5 and six at position number one we are going to add an O. H. Group because it's hexagonal In position number five will contain a metal submission as well as a bromo substation. We can clearly now see what's happening here. We indeed have a bond cleavage because this oxygen is not a part of the ring system anymore. We can now concentrate on the actual mechanism if we start with our substrate and recall that hydrochloric acid is a strong asset from general chemistry as well as the fact that assets are perfect proton donors in this case, the only thing to to realize is that there's something that gets protein ated and because oxygen contains lone pairs, those lone pairs could be used in the protein nation step so the long pair of oxygen attacks Haijun and simultaneously we have a loss of roaming. This is called the proto nation stop during the protein ation step. We are forming a protein ated intermediate. So now the oxygen atom Has three bonds on it, meaning it has a formal positive charge. Now we have to think about it carefully. Can we actually open that ring immediately or do we need to perform a nuclear attack? We have two possibilities to get rid of our formal balls of charge and oxygen by either transferring these bonding electrons on set or the electrons from the bottom. But because breaking this bond specifically will form a more stable tertiary carbon. Karen intermediate. This is really much more favorable for us. So what we're going to do because a stable tertiary carbon Karen will be formed, we will actually immediately open the ring without a nuclear attack because we now recognize that this is an sn one reaction of role. So in this step we have a loss of the leaving group inside of the molecule itself. That's it. We can draw our next intermediate. So we can also identify this step as um the laws of the leaving group and forming a tertiary carbon cannon. To draw it once again we can just label our carbon atoms, Draw a six remembered um chain of carbon atoms And a metal substitution at position number five at position number one, we will add O. H. With no more charge because now auction only contains two bonds but the charge now actually belongs to this carbon atom because we have formed a stable tertiary carbon Karen. Now having formed this carbo cast iron, we can think about it this way. What if we just compare the final product and our intermediate? The main difference between them is that the final product has blooming attached to this carbon while our intermediate has a positive charge. So what that means is the brunette and are informed in the first step will now act as a nuclear file. And the step that we will be drawing now is the nuclear filic attack. Stop bromide will act as a nuclear file and it will attack the carbon can are informed to produce the final product. Now let's compare our product with the project that we drew previously. They're exactly the same. That said. Our mechanism makes sense. And this is our final answer. The correct multiple choice answer is C. And in case you have any questions, feel free to leave them down below. I will see you in the next video.

When diethyl ether (CH3CH2OCH2CH3) is treated with concentrated HBr, the initial products are CH3CH2Br and CH3CH2OH. Propose a mechanism to account for this reaction.

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