Show how you would synthesize the following compounds from the ap...

Question

Show how you would synthesize the following compounds from the appropriate carboxylic acids or acid derivatives. (a)

(b)

(c)

Accepted Answer

Hey, everyone and welcome back, propose a synthetic approach to produce each of the given compounds using suitable car book Celik assets or their derivatives. Starting with a, we want to produce an alcohol because we have our carbon, a wage group. So we are essentially going to draw that product. Okay. So that would be our five member ring and we essentially have two carbon atoms. So let's draw two carbon atoms, 12. And then we noticed that this position, we have our alcohol group. So that's our product. We want to form that product out of an asset or and acid derivative. So essentially what we're going to do here, we're going to notice that we're getting our primary alcohol and because it's a primary alcohol, we can use our strong reducing agent whenever it comes to assets, right, we can immediately think of an asset. And the reason why we're going to turn this structure into an asset is because we know a very basic way to turn our assets into corresponding alcohols that will be number one using a strong reducing agent, which is lithium aluminum hydride, followed by reaction number too acidic work up to protein, ate your elk oxide into an alcohol. So now what we did here, we essentially understood that if we have two carbon atoms, we drew two carbon atoms and we turned the one that has an alcohol group into a car book silic acid. And we had, we simply have to remember that whenever you reduce your acid using lithium aluminum high, try to get your elk oxide and you have to protein ate it and you get a corresponding primary alcohol. That would be our answer to A. We are ready to move on to part B and then part B, we have a compound which has two double bonds and an al to hide group. So this might look a bit more complicated. Simply let's redraw our five member ring, let's add our pi bonds and simply let's draw the alga hide group. That's the compound that we want to get. And here we have to be extra careful if we have to choose an acid or a corresponding acid derivative because we have our al to hide to form. It essentially implies another reduction reaction. But remember that lithium aluminum hydrate essentially reduces assets into alcohol. So we see a complete reduction. It's a strong reducing agent. We want to make sure that we make it milder. And there are two things here to consider. First of all, whenever we want to perform a mild reduction, we cannot use an acid, we want to use an acid chloride instead because acid chlorides are more reactive. So we are going to start with a corresponding acid chloride, right. We essentially replace this hydrogen with chlorine and we get a corresponding acid chloride. And now the second thing to keep in mind is that you definitely don't want to use lithium aluminum hydrate because it would reduce your acid chloride into an alcohol. You want to stop at an al to hide and to do that, we actually need only one reaction. Now, first of all, we are going to keep in mind that there are four hydrogen atoms bonded to aluminum and lithium aluminum hydride. So we're going to use a lithium aluminum. First of all, we're going to leave only one hydrogen, right? So basically, only one hydrogen will be reactive. And that's exactly what we want to have here. So let's leave one hydrogen and for the three remaining hydrants, we are going to substitute them with three third beetle groups who are going to say that we have O C bu, right? These are our three third beetle groups. In other words, we're just replacing three hydrogen atoms with three hysterically hinder third beetle groups. Now, whenever we are increasing hysteric hindrance and removing our hydrogen from here, we're essentially making our reducing agent milder. So it allows us to stop our reduction at an al to hide and not reduce it further into an alcohol. So this is the region that we use here, that's sufficient for us to perform the conversion. We don't need to use the stick work up because we're not producing an alcohol. So there's no Alcock side formed and simply the reaction here is over, that's our conversion. There's one more thing to keep in mind, we have two double bonds. The pipe bonds by lithium aluminum hydride are not affected as well as its derivatives, right? A more hysterically hinder reform of it. It has an affinity to carb oxalic acid groups or carbon compounds. In general, pi bonds are selectively affected by sodium Borough high trite. So that would be the priority for sodium carbohydrate to reduce them. But for lithium aluminum hydride, it essentially is selective to carbonell groups. That's why we are not worried about the pi bonds. And now we're going to the final one, that's our c the product that we want to form has two functional groups. One of them is our carbonell group here. We have our key tone and then here we have our alcohol. Now we are definitely not worried about the key tone itself. The reason why we're not worried is because we cannot really turn it into an asset, right? Notice how this carbon is bonded to two carbon atoms and the only way to form our curb oxalic acid group is over here. So for this one, we're going to choose an asset and the reason why we're going to choose an asset will be explained just in a bit. First of all, let's take our structure and let's change our alcohol group into an asset. So we are going to essentially just add that Carbonell group and we got it right. That's essentially our code book, sarcastic group instead of and alcohol. So now the only thing for us here is to think about the region if you want to use, we definitely cannot use lithium aluminum hydrate because it will reduce key tones and assets completely. Here, we would have a secondary alcohol as well. If we use lithium aluminum hydrate, we only want to reduce our asset into an alcohol. And you may recall that for this purpose, you may choose die boring be to age six, essentially die boring is selective towards assets. Whenever you have several different, several different carbonell groups die boring will selectively reduce your asset without affecting the other carbonell groups nearby. So that boring allows you to reduce your asset into an alcohol. But once again, because you are forming an alcohol, we need to use the second reaction acidic work up to protein eight, the result on the Alcock side and I and into an alcohol. So this will be our answer for C. Let's keep in mind that for C once again, we have two carbonell groups, we essentially want to reduce our acid only for this purpose where not using lithium aluminum high try that reduces everything but we're using die barbarian instead to only reduce our chromebook silic acid into an alcohol. And that would be it based on our answer. We may conclude that the correct answer to the multiple choice question is a Thank you for watching.

Show how you would synthesize the following compounds from the appropriate carboxylic acids or acid derivatives. (a) (b) (c)

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