Predict the major products of the following reactions, and pro...

Question

Predict the major products of the following reactions, and propose mechanisms to support your predictions.

a. pent-1-ene + HCl

Accepted Answer

welcome back everyone. In today's video we are analyzing the problem which states determine the products for the following reaction where one metal cycle Bentsen reacts with hydrochloric acid and we are also asked to draw a mechanism that shows the formation of the predicted products. We will start this problem by drawing the substrate. We are given one metal cycle benzene and that means we have a five member ring with a double bond. Because that's what cyclops Benzene is. We may place our double bond at any position position. Number one has a metal substitution. So you may also add a muscle substitution supposed to this position. The reaction is hydroponic assad and we may recognize this reaction as hydro hallucination because we have hydrogen and halogen and also we have a double bond. So this is an al keen and we have an electra filic edition reaction in which hydrogen and bromine are both added along this double bond. We call that according to the more cosmic of school, we are adding the hydrogen atom to a less subsided carbon. So we may say that hygiene is bonded to a less substituted carbon while acts in general any halogen in this case bro mean is added to a more substituted carbon. That said if we look at our original elkin, we wish to identify which of the two carbon atoms within the double bond is more substituted. Notice that this carbon atom in particular has three substitutions. In other words it has three carbon atoms around it. While this one only has two and therefore this is a more substituted carbon atom. What we said before is that acts or blooming in this case is added to a more substituted carbon atoms. So we are going to add blooming to this position because it's more substituted. And to this position we will be adding hydrogen because it's less substituted. And as a result because we have an addition reaction. The original double bond will be reduced essentially will not have any double bond. We will have a single bond because we're forming two additional bonds across it. Now our product as a result would be cycle pen sign. We would have that still that metal group over there and we would bond blooming at this position as well as hydrogen in that position because the product has a plane of symmetry. We don't worry about stereo chemistry at all. And we are not asked to show any stereo chemistry in this product. And therefore we may simply draw our product without any hydrogen shown. We may just think about them as implicit hydrogen atoms and all that. We have Those are a method group and april no suspicion. This is our final product. Well done. So we have our product and now we wish to draw the mechanism for this reaction. So let's see how the mechanism looks like and why more cognitive school works here. If we start with the given Elkin, we will just again redraw it for a complete clarity. This is our starting Elkin. We have a double bond, a methyl group and then we are adding a molecule of H B. R. We have electra Filic edition. So you may 1st of all expand the hbr bond. This stop is the electrolytic edition stop in which we are forming a carbo. Karen now notice that the double bond essentially attacks hydrogen and simultaneously we have a loss of bronze and iron. And now the question is, where should I add my hydrogen? Should I add a disposition or that position? I have to think about the stability of carbo Karen's form. If I add hygiene at this position, that would form a tertiary carbon Karen because here I would have three substations. On the other hand, if I had hygiene at this position, I would form a secondary carbo Karen that would be less stable. And that's why more cognitive school works for us because we wish to form a more stable carbo Karelian. In this case we wish to form a third sherry carbo Karen. And therefore I'm going to add ha jin a disposition making this the position of the car broke Orion, the sea tertiary carbo carry on. And now the bronze and iron that we have previously formed would attack the car broke down in the step, we would have a nuclear attack step notice that if this was a Cairo carbon atom because it's actually not. We already know that there's an internal line of symmetry. We would form a systemic mixture of R and S. In and humor. But in this case we are forming a single product because this carbon atom is not Cairo, meaning it's just sufficient to show that the bromine and iodine, which acts as a nuclear file, simply attacks that carbo Ryan. And we are forming our final product in this step, the product is formed. So we are trying the ring, the methyl group and the most efficient attached attached it notice that this is indeed our product. We got our mechanism and our product. That's all that we had to do for this problem. To correct multiple choice answer is B. And if you have any questions, feel free to live down, down below. Thank you for watching.

Predict the major products of the following reactions, and propose mechanisms to support your predictions.
a. pent-1-ene + HCl

Key Concepts

Electrophilic Addition

Carbocation Stability

Markovnikov's Rule

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