Show how you would synthesize octan-2-one from each compound. ...

Question

Show how you would synthesize octan-2-one from each compound. You may use any necessary reagents.

(a) heptanal

Accepted Answer

Everyone and welcome back and today we are going to show how pence in L. Can be converted to hex ends he own and provide any necessary regions. Let's throw the reaction scheme, the starting material is pence in L. So we have our Aldo hide that be our five member chain 12345. And carbon number five will essentially contain an alga hide groups. So we are going to add carbonell group with hydrogen. So that's 12345 pence in L. Right. And we are going to convert our starting material into hex and T. Own. So let's draw a six carbon atoms at position number two. Or specifically, we can think of this position. We have our carbon Neil group because it's a key tone. Now you may notice that the main difference between them is that this starting material has five carbon atoms and this one has six. You may simply notice how we are essentially changing this hydrogen C metal group and that's it. But this conversion is not really easy. However, let's keep in mind that instead of this hydrogen, we actually have a methyl group. Right? And we need to think about a possible conversion. Unfortunately, there's no easy way to essentially eliminate that hye jin at the metal group somehow, but we already know that carbonell groups are good for us. Whenever we were thinking about the green yard reactions. So why don't we use this is our starting material and essentially add our metal group to it to form our alcohol first because this will allow us to add a metal group easily. Right? We already understand that. We need a metal group and to add a metal group to our Aldo hide, we want to use the graveyard reaction. We call that the world First step is to make sure that we're using a metal group, the one that we desire to add bonded to magnesium bromide or iodide for example. So for the purposes of this problem, let's choose our green yard region to be method magnesium iodide. Okay, now, what happens in this reaction is that you have a strongly polar bond towards carbon because magnesium is a metal right? There is a high electron density which can essentially attacked the electrolytic position and then we're breaking our pi bond. And this is how we get our elk oxide intermediate. So we can draw our five carbon atoms and I'll notice how we are going to get our elk oxide. So instead of having a pi bond we are going to break it and we will only have our sigma bond, carbon bonded to oxygen with a formal negative charge. We have our implicit hygiene. And then we simply are going to add our metal group. Right? Because that's what we added over here. Now Wilcox I'd intermediate must be protein ated because we want to get our alcohol for a complete granular reaction. And then the next step we're going to use acidic conditions or acidic work up right to make sure that we protein it. Our world cock side. The protein nation step will essentially allow us to get our 123456 carbon atoms. Right? 123456. And now at position number two we have our alcohol and that looks better. Right? Because these two structures are very similar. The main difference between them is that the alcohol group now is converted into a carbon ill group. So this is our secondary alcohol, the C key zone. And you may recall that we can easily convert secondary alcohols into key tones if we oxidize them. So we essentially want to make sure that we oxidize the secondary alcohol and key tone. And here we have so many options. Right? You can use the majority of oxidation agents because you're only oxidizing at one step, there's only one possible oxidation. So you can use P CCs. Where an oxidation D. M. P. Right? You can essentially use PCC as your option. A option B You can use swear an oxidation which essentially implies two reactions that B. D. M S O ox a little chloride followed by try FL I mean at low temperature now see you can essentially use sodium hypochlorite in the presence of acetic acid and you can use any strong oxidizing agent for example jones regent. Right? So just chronic acid and so on. For the purposes of this problem, you can actually choose any of them. We are going to choose sodium hypochlorite, that B N. A O. C. L. In the presence of acetic acid. And that's how we oxidize it and that would be it. We understand how the reaction works. Let's just throw a complete sequence so that we don't need to look at our intermediates because the problem didn't ask us to do that. So we can essentially start by drawing our al to hide. Okay, so that's our starting material. Now, based on our analysis, we said that in the first reaction we're going to perform the Green Yard reaction in which we are going to add a method group and form an alcohol. So that corresponds to two steps addition of method magnesium. I died in the presence of ether. Let's not forget the servant. Right? So when we go back to our analysis, of course, recall that greener reactions require us to use solvents which are ethers. So we want to make sure that we indicate the presence of our ether. And the second reaction that be acidic work up. This allowed us to form our secondary alcohol is already drawn, right? 1, 2, 3, 456 carbon atoms in our parents. We understand that we have our alcohol. And when we get our alcohol, we eat just want to oxidize it into our final product. So let's draw our final product from this stuff. Let's indicate that in this reaction step we are going to use so the hipaa chloride in the presence of acetic asset. And now from here we're just oxidizing it into corresponding ketone and that will be it. That's our sentences scheme. Let's label the scheme as our final answer. And based on the outcome, we can conclude that the correct answer to the multiple choice question is a If you have any questions, don't ask this. Leave a comment down below. Thank you for watching.

Show how you would synthesize octan-2-one from each compound. You may use any necessary reagents.
(a) heptanal

Key Concepts

Aldol Condensation

Reduction Reactions

Rearrangement Reactions

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