Show how each of the following compounds can be synthesized fr...

Question

Show how each of the following compounds can be synthesized from benzene:

a. o-bromopropylbenzene

Accepted Answer

All right. Hello everyone. So this question is asking us to propose a synthesis route for one bromo two iso beetle benzene starting from benzene. All right. So based on the names of both the starting material and the product, we can see that the idea here is to add substituents onto a benzene ring. So up on the screen, I have a benzene ring and let's go ahead and draw out the final product first. So on position number one, we have bromine attaching to benzene and on carbon number two, we have an iso beetle substituent. So that's a three carbon chain with a methyl group branching off of the second carbon of the substituent making a total of four carbons. All right. So first and foremost, given the fact that our substituents are on adjacent carbons, we can see that they are Ortho to each other. So we have an ocho group and a halogen specifically bromine attaching to benzene recall that electro pilic aromatic substitution or eas reactions are common ways to install certain substituents onto aromatic rings. But because we're dealing with multiple substituents and therefore a series of reactions we have to consider the effect of these substituents on subsequent EAG reactions. If you recall the positioning of one substituent on an aromatic ring is going to affect the positioning of a new substituent. So let's go ahead and talk about the directing effects of the substituents that we have in this problem. First, we have bromine, which is of course a halogen. Now, bromine being a halogen is actually a deactivated. It's going to make the benzene ring less reactive towards eas. However, unlike most deactivating groups, halogens are actually Ortho pares, which mean that they would direct existing substituents onto either the Ortho or the paras position. And we also have an O kill group like the Isobel group. Here, ocho groups are activating groups, meaning they donate electrons to the benzene ring, making it more reactive and they are Ortho pair directors. So here, even though both substituents are Ortho pares, we should go ahead and proceed with adding the keel group first, considering it's an activator, right? It's going to make subsequent eas reactions easier. And so with that, let's go ahead and start by drawing out benzene, which is our starting material at the bottom of the screen here to start off our synthesis. Now, here it's worth mentioning that there are two ways to install an O group onto a benzene ring. The first is Freedom Crafts Oculus, which would directly install an al Q group on two benzene. But the second approach is known as Freedom Crafts isolation. Now, the difference between the two is that photo crafts isolation would install an ay group with a car bal to achieve the Isobel group that we see in the product, we would have to reduce that car bono in a separate step. So to understand which of these approaches is better, it's worth talking about the main limitation of freedom. Craft's Oculus compared to isolation because recall that fried craft's Oculus is formed or is characterized by the formation of a carbo cion. This means that it is susceptible to rearrangement. So the group that you add may not be the oche group that ends up installed on benzene because of rearrangement. The Isobel group is very vulnerable to rearrangement considering it does have a tertiary carbon that could make an excellent carbo cion. So instead to avoid rearrangement entirely, we'll proceed with freedom craft's isolation followed by reduction of the carbo. All right. So let's start with our fruto crafts isolation, which of course requires an ayl chloride. So here, because we're installing an Iso beyl group, it's best to start with iso butyl chloride as our other reagent. So here's iso butyl chloride. And for the isolation step, we're going to need a Lewis acid catalyst such as for example, ACL three. Now isolation is unique in that it's actually a two step sequence because the second step requires the addition of alcohol. So I'm just going to write that on the same arrow here. All right. So after this, we end up with our isolated benzene. And so our next step is to go ahead and reduce the car bal of the AO group to give us an O group instead. Now to do this, we can go ahead and proceed with the clemenson reduction. And in the clemenson reduction, our reagents are zinc amalgam and HCL. So for my second arrow, I have zinc and mercury in parentheses with HCL beneath the arrow, this is going to reduce the car bono and give us an ore group. OK. So from here, we officially have our all Q group, excuse me, all Q group attached, which is an activating group and an Ortho para director. So from here, we would proceed with the halogen reaction that would install bromine onto our ring. But there's one more thing to consider because notice how the ocho substituent and actually Bromine or halogen are Ortho and per directors. This means that there's two possible products, one in the Ortho position and the other in the para position. Recall it, the paras position tends to be the preferred product. This is because if the substituents are on opposite sides of the ring, they experience less stare kindreds. So here because our target product is Ortho, we would want the Ortho product to be preferred. We want that to be our majority product. So what that means is that we actually need an additional step because we have to protect the paras position and make sure that the para product is not the major product. So we proceed with sulfur because recall that a sulfonic acid group or so three H can function as a protecting group on a certain position. This is because it can be removed very easily in the presence of some acid and water. So before we can proceed with brominated, we have to block the para opposition from being substituted and we can do this with sulfur nation. So that requires so three and Sulphuric acid that's H two. So four. So at this moment, because the pairs position is available, the majority product is going to be the pair product, which means that RSO three H would be para to the group. And because these substituents are going to experience less STAIC hindrance, this is going to be actually the major product with a yield of about 90%. So now that the preposition is protected, we can proceed with our halogen nation. So because we're adding bromine, we would need BR two in the presence of a Lewis acid catalyst like Febr three, the he or your group, our protecting group. And now because the pairs position has been blocked, bromine is going to add to the Ortho position relative to our all kill group. So finally, we can finish off the synthesis with our de protection step to remove the sulfonic acid group and actually give us our final product. So to do that. As mentioned before, all we need is some acid in the presence of water and some heat and there you have it, our synthesis is now complete. And so with that being said, if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Show how each of the following compounds can be synthesized from benzene:
a. o-bromopropylbenzene

Key Concepts

Electrophilic Aromatic Substitution

Bromination of Aromatic Compounds

Alkylation of Aromatic Compounds

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