Show the mechanism for tle following conversion (5 pwitts)H;c CH, ~OHyc El H;c" ~CHHBrH;c...

Show the mechanism for alkylation of benzene by an alkene.

We start with a molecule. Combat looks like this, and our product replaces hydroxide with brahmi. For this to happen, our molecule must react with hydro bro MK acid. And then our product will be that hydrogen plus the age from the original molecule, which, of course, is just water.

We start with a molecule. Combat looks like this, and our product replaces hydroxide with brahmi. For this to happen, our molecule must react with hydro bro MK acid. And then our product will be that hydrogen plus the age from the original molecule, which, of course, is just water.

So this problem from organic chemistry Ask us to give the method for converting three Mitchell Beauty in one all into three Mitchell Beauty one in and also to give the mechanism for the reaction So this reaction can be done by using mental asset like sulfuric acid, which act as a catalyst as well as a dehydrating agent were los off water molecule from the reaction takes place to form the Elkin. The mechanism for this reaction can be given as below. First. The reactive molecule digs up proton from the sulfuric acid molecule and forms an intermediate. In the next step, the intermediate loses a water more off molecule, along with the loss off Proton to form the Elkin. Yeah, here. The catalyst sulfuric, I said, has regained at the end off the reaction.

This is the answer to Chapter 21. Problem number 74 from the Smith Organic Chemistry textbook. In this problem asks us to draw a stepwise mechanism for the conversion of this Dikky tune into a fury. Um, and, uh, we told that this takes place under city conditions. Uh, yeah. Okay, So the problem tells us that the only region that we're using is so fair gases. Um and so the first step is going to be pro nation of one of these key tones. So the lone pair on the oxygen will grab a proton. Um, like that. And so our first intermediate is going to be the pro nated key tone there. I'm in. So remember, um, for carbon compounds. Um, when they are pro donated like this, uh, the rationale behind prone eating them is it makes them more susceptible to nuclear filic attack at the carbon or carbon. And so in this, in this instance, susceptible enough that the oxygen of the other carbon eel can act as nuclear file and add in to the carbon or carbon of the carbon. You know that we pro nated in the first step. So that's going to go ahead and form out ring eyes. Um, that's going to look like this. So, uh, there's the oxygen. Um, the oxygen has three bonds now, so of course it will have. Ah, positive charge on it. I wish Nuffer at these metal groups. Um, and then also this carbon eyes going to have the alcohol that is the residue of the carbon. You know, that we permitted in the first step, that was then add it into s. Oh, there we go. There's that, um the Mexican A draw, uh, hydrogen there, and it will become a parent y in moment. So there is, as I said, a plus charge on that oxygen. It has three bonds. Um, And so, uh, in order to relieve that plus charge on the oxygen, um, and to form the first the double bonds in our furin, we can use the conjugated base of the sofa, a gas that we used in the first step. So, um, it has a negative charge. This oxygen has three lone pairs. Um, and one of those lone pairs is going to grab that hydrogen. And those electrons will, uh, go to form new double bond their knees, electrons will refer to the furin oxygen. And so, uh, after those three arrows, we're going to be at this intermediate. It's now this oxygen is uncharged. We have a double bond that's formed there. Um, still have our metal groups here and here, of course. Um and then our alcohol there on DSO. The next thing that's going to happen in this mechanism is we're going to have to get rid of that alcohol. And the way to do that is again Thio use pro nation by sulfuric acid because, of course, there is so sulfuric acid floating around. So that's going to look like this, um, one of the lone pairs of this alcohol oxygen. We'll grab this proton. And so this alcohol has never been pro nated. So obviously there's a plus charge on the oxygen. Um, and one thing that we should definitely know by now is that that is going to make this into a good leaving group, So alcohol is not a good leaving group. Um, water is a great a great leaving. Um, okay. So that bond will break, that water will leave, and of course, that's going to result in a carbo cat eye on being formed at the carbon to which that alcohol or water was attached. And I do apologize. I omitted a double bond there, So there it is. Okay, um and so at this point, we're here, and I am just going, I'm gonna draw hydrogen there and again. The reason why will become apparent momentarily. Um, so we have a carbo cat eye on. They're at that Corbin. Um, And then again, we can use the contra get base of sulfuric acid. Um, this oxygen has negative charge and three lone pairs. One of those lone pairs will grab that hydrogen. Those electrons will kick down to form a new double bond between the cargo cat eye on and that carbon. Oh, um, right. And so that is going thio, uh, results in the formation of our final product. So there we go. There's our fury in the Monica What we're trying to form. And so this is ah, pretty straightforward mechanism, uh, pro nation of one of the key tones so that the other one can nuclear physically attack it. Um, removal of a likely hydrogen by the Khan ticket base of sulfuric acid. in order to relieve the positive charge on the fury and oxygen and to form the first double bond and then just pro nation of the alcohol. That's a residue of our first key tone, Um, leaving off that leaving group and the removal of a second hydrogen in order to form the second double bond on That's the mechanism. And that's the answer to Chapter 21. Problem number 74.