Which reactions will produce the desired product in good yield...

Question

Which reactions will produce the desired product in good yield? You may assume that aluminum chloride is added as a catalyst in each case. For the reactions that will not give a good yield of the desired product, predict the major products.
(d) Reagents: benzamide (PhCONH₂) + CH₃CH₂Cl
Desired Product: p-ethylbenzamide

(d) Reagents: benzamide (PhCONH₂) + CH₃CH₂Cl

Desired Product: p-ethylbenzamide

Accepted Answer

Hey everyone and welcome back. Consider the following reaction. Assume that iron three chloride is added as a catalyst. Will this reaction produce the indicated product in good yield. If not predict which product will form instead. And look at this reaction, we have to recall that this reaction corresponds to Friedel Crafts calculation. The reason why this is a Friedel Crafts calculation is because we are starting with our benzene derivative. We are using an al kill chloride and we are using our loose acid iron three chloride in general. As you know, Friedel Crafts calculation takes place by the generation of an electro house, we're going to remove that chloral group. We understand that the carbo Caroline would be tertiary carbo Karelian. So the electro file indeed corresponds to our third beetle group. There will be no rearrangement. We understand that the substitution takes place with the required electro file. So that's good. This fragment is our electro file. However, we have to think about the Ortho para directors and the substitution on our benzene ring. This is try floral methane group, right. C. F three. So try floral methane. And because this is a trifle or methane group, we have to keep in mind that it is highly deactivating group because this is a highly deactivating group. Try floral methane is a very common. The activating group because it contains three florin atoms. Each fluorine atom has a high electro negativity. So the electron density is essentially decreased on the benzene ring because flooring is so electro negative. Right? So because the electron density within the ring decreases the rate of an electrical aromatic substitution decreases as well. Right? We want to make sure that our ranges electron rich and this is not the case whenever we have a highly deactivating substitution. However, so far it's not so important. We will keep that in mind. Highly deactivating substitutions are meta directors. Right? So we want to make sure that we identify our meta positions. And if we look at our substitution, this is worth the position. This is So we have two meta positions over here. So we expect our substitution to take place at any of these two positions. They're basically equivalent because the compound is symmetrical with respect to the vertical. So there's actually one unique position, let's suppose here. Now, on the other hand, the product is formed at the power position. So, this is the first inconsistency. We can say that immediately. No, this reaction will not produce the indicated product. The reason for that is because our trifle oral methane group is deactivating and meta Director. Well, the products shown is para substituted. Okay, we understand that it's far substituted. This product be extremely negligible. Right? We actually expect the product to be formed at the meta position. So because the problem is asking us to predict what product will form instead, we are going to draw the expected product that product would correspond to our benzene ring. Try Flora methane group. Right now, we are going to add torpedo group add the meta position because try floral methane is method director. So this is the expected product theoretically, we can keep that in mind. But in practice the reaction will fail. The reason why the reaction fails in practice is that simply trifle methane group is highly deactivating Friedel Crafts calculation requires activated ranks. So this is one of the requirements keep in mind that Friedel Crafts calculation requires activated rings to proceed if the ring is highly deactivated due to the fact that there is a highly deactivating trifle or methane group. Even though this is a theoretically expected product, the substitution takes place at the matter position. In practice the reaction feels, and even this product that we theoretically predicted is formed in a world very low yield. However, based on the nature of the question, if we want to predict which product will form, instead ignoring the actual yield, this would be the most likely product. Because we know that the substitution will only take place at the meta position, predominantly. The part substituted product is not expected, right? Because it's Method trifle methane group is meta directing and therefore this is the theoretically expected product. But in practice, if we want to supply this contact concept practically, we can say that the reaction feels since Friedel Crafts calculation requires activated ranks. Now based on the theoretically expected product and is bond line structure, we can say that the correct answer to the multiple choice question is C. No. This reaction will not produce the indicated product and good yield, and the actual predicted product is the product which produces substitution at the meta position. That would be it? Hope to see you in the next video and thank you for watching.

Key Concepts

Friedel-Crafts Alkylation

Regioselectivity

Carbocation Stability

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