Show how you would synthesize 2-phenylhex-3-yn-2-ol, starting ...

Question

Show how you would synthesize 2-phenylhex-3-yn-2-ol, starting with acetophenone (PhCOCH₃) and any other reagents you need. ("2-ol" means there is an OH group on C2.)

Accepted Answer

Hey everyone and welcome back to this video. We are going to answer the following question, How would you synthesize one final Beauty to I and one Oh, Starting from Bansal to hide and any other required regions. If we think about this reaction is the problem suggests our starting material is Ben Zelda hide, meaning we're going to draw our Benzene ring and we are going to add an al high group, C H. O bond. It's it. So that would be our buns out hide. Right? And according to the problem, we wish to synthesize a compound that has quite a complicated name, but don't worry, we will just try to break it down. So it says one final Beauty to iron one. Oh, so the main part here is Beauty to iron. Right? We have B two, iron as our parents. We can think about four carbon atoms having a triple bond at position number one. So supposed to simply withdraw your first carbon atom in that position, so that B one and then position number two essentially has a triple bond. Okay, so what that means says you're going to draw a triple bond at position number two, then you have position number three over here and then that would be your position number four. So that's beauty to iron. But we have to keep in mind that carbon number one has a final final substitution. So we are going to add final at position number one and then position number one also has an alcohol group, it says one. Oh, so let's draw that O. H. Over here, that's our final product, right? And if we look at the two structures, we understand that the starting materials in al to hide. But much more important is to understand that this is a carbon compound, right? It has a carbon steel group that is reactive towards nuclear filic edition. If we look at the final product, this is an alcohol. And what is more important is that this alcohol is actually secondary alcohol. Recall that there's one important reaction which we can use to form secondary alcohols out of an alga hides. We can either use a green yard reaction or in nuclear filic edition. So if you think about the structure there is that implicit hygiene. If you compare the two structures, there's this hygiene, this final ring right bonded to that carbon atoms. So this is exactly what we see within our al to hide. But instead of seeing our carbon group, we're seeing an alcohol group because there is one more substitution and edits it. And that substitution is a substitution. That has how many carbon items. So that would be 123. Right? Meaning we want to perform a nuclear attack. And what kind of nuclear file can we use if we simply break it apart? We notice that we get our Aldo hide after the removal of this acidic hydrogen and restoring the double bond. So this could be our nuclear file. And if we perform that bond cleavage over here, we noticed that we get our satellite eye on our triple bond with a negative charge could be our nuclear file over here. Also there is a method group. Right? So we get our satellite and essentially, if we form that a satellite, it will attack our Aldo hide, break the pipe on and we will get our L cox it and on. So we can look at the intermediate form in this stuff. We can just redraw our benzene ring. There will be this hydrogen Alcock side and uninformed. And then we are essentially adding our triple bond in this position for by the metal group. And this looks exactly the same as our final alcohol. Actually. Not exactly the same yet, because we have Alcock side. Right. So that's not really a problem. We can always protein. Eight hour Alcock said. And we can say if we use acidic electrolysis were basically acidic work up, right, We are going to protein ate it and we are going to form our final alcohol. So there's no necessity to redraw the structure. Because we're just thinking ahead, we can just say that we are going to use acidic workups to produce our final alcohol. The product is already shown. Right? So, this is our idea for the synthesis. Let's just put everything together. The only thing is to think about how we can produce that a satellite and iron. So we can actually start with beat to iron itself. Keep in mind that if you start with B two iron. But in this case, let's see if we need to be to iron, that'd be 123 carbon atoms. So we want to be protein ate it at this position. So that would be actually pro Pine itself. Right? It has acidic hydrogen here. So we're using pro Pine. And because it has this acidic hydrogen at position number one, this acidic hydrogen can be removed by a strong base. And recall that the strong base that we're using is sodium A might N A NH two because um it has a negative charge can easily deeper donate this hydrogen and form our satellite required. So we understand how to proceed now. Right, step number one, let's label this. This will be step number one treating Pro Pine with sodium amy to produce a satellite. Now in step number two, A satellite attacks our benzel to hide, forming Alcock side and then our L cox. It has to be protein ated to form an alcohol. So that makes sense. Right. We can simply label our steps as follows. So let's clearly draw all of our scheme. You understand everything that we had so far. So we are going to perform our first step. We are going to take pro Pine and we are going to treat it with sodium Amytal which is a strong base and it will depart in eight that acidic hydrogen to form a satellite. A satellite is a strong base strong nuclear file. So we can essentially use it as our nuclear file in that nuclear attack reaction on Carbonell. And we want to treat a satellite with our Ben's Aldo hide which is our starting material. So we are going to draw bends out hide which has that electra Philip position that gets attacked. And in this step we are producing our cock side. So we control our Alcock sign. That would be oxygen with a negative charge. Now we are adding our triple bonded out position. So we can say there's a triple bond, there's a metal group and then this position still has hydrogen and finally, we need to acidified. Right? We want to make sure that we produce our alcohol. So after a stick work up, we're using a source of protons. So we in the general case you can say hydro ni um there's hydro new max is a proton donor and it's essentially protein. It's that negatively charged oxygen. So that's how we are producing our final alcohol. And we may now draw our final product as benzene ring with our alcohol group, our Triple one over here. Right. And there's this hygiene. So this is how we form our final product. Notice that it's not important to show this hygiene. It can be implicit right? But we have shown it for complete clarity that it's still there. It comes from the allied itself. However, it's not so important. We generally don't show these implicit hodgins, right. But this is our general reaction scheme for the synthesis of the required compound. We started with propane. Use sodium AM I to produce a satellite reacted, reacted it with Ben Saleh High to produce Alcock Side and Alcock Side got protein ated to produce our final alcohol. And therefore looking at these steps, we can conclude that the correct answer to this multiple choice question is the However, if you have any questions, don't have States, leave a comment down below in the comments section and thank you for watching.

Show how you would synthesize 2-phenylhex-3-yn-2-ol, starting with acetophenone (PhCOCH₃) and any other reagents you need. ("2-ol" means there is an OH group on C2.)

Key Concepts

Alkyne Synthesis

Nucleophilic Addition to Carbonyls

Functional Group Interconversion

Watch next

Show how you would synthesize 2-phenylhex-3-yn-2-ol, starting with acetophenone (PhCOCH3) and any other reagents you need. ("2-ol" means there is an OH group on C2.)

Key Concepts

Alkyne Synthesis

Nucleophilic Addition to Carbonyls

Functional Group Interconversion

Watch next

Show how you would synthesize 2-phenylhex-3-yn-2-ol, starting with acetophenone (PhCOCH₃) and any other reagents you need. ("2-ol" means there is an OH group on C2.)