Propose a mechanism for the entire reaction of pent-1-yne with...

Question

Propose a mechanism for the entire reaction of pent-1-yne with 2 moles of HBr. Show why Markovnikov's rule should be observed in both the first and second additions of HBr.

Accepted Answer

Hey everyone and welcome back in today's today. We are going to draw a mechanism for the reaction of one mole of one pro pine with two most of hydroponic acid. The mechanism should explain the Markovic of regional selectivity of the reaction. If we start with the given material, that's as one pro fine. So we can just draw a triple bond. These are two carbon atoms, we need one more because that's one pro pine. The triple bond starts at carbon number one. So we are good and this is our starting material and we are ready to draw the first equivalent of hydroponic acid because that's what the problem suggests. Right, the first equivalent essentially reacts with the pipe bond and this is how it happens. It's the fact that hydrogen is more electoral positive and grooming is more electro negative or basically, since blooming is more electro negative, it has a partially negative charge and hygiene has a partially positive charge. The bond is polarized towards a more electro negative blooming. The high electron density in the pi bond can essentially attack that hydrogen. And it can essentially perform that addition to one of the carbon atoms followed by the loss of the leaving group. So we are getting rid of bromine, it would form the borough might and ion. And this first step is actually reversible. We have to keep in mind that the first step is reversible because we are going to form a carpool Karen intermediate that can easily restore the original al kind since it would have no charge during an elimination of the hydrogen atom. Okay. So keep that in mind. Now, when we're adding that hydrogen, we are keeping in mind that we want to follow them or cosmic of regio selectivity. That's what it says. So we have to recall that hydrogen must be added to a less substituted position according to the more comical school. Because if we add hydrogen over here, we will get a positive charge for our car broke down at position number two, which would be more stable secondary Carbo Koran as opposed to the primary cardboard Karen. So that's how we can easily explain the more cosmic of regional selectivity based on the fact that a more stable secondary carbo Karen is formed. Otherwise, if we don't follow the rule, we would form a primary carbo Karen at this position, which would be less stable, right? So we have explained more comic of regions selectivity here, meaning we can just proceed with our mechanism. Now, in the next step, we can simply say that the brown white and I that we have previously formed will act as a nuclear file, right. So the brunette and I will essentially attack the carbo can iron and this would be a non reversible stop. We are going to form our first product. Essentially, we are forming that carbon booming bond ride because the broom and and in transfers its electrons into a Covalin bond. The reason why this stuff is not reversible is basically because it's not released its break that carbon booming bond and go backwards to form that carbon Karen. Of course, basically because it's a secondary carbon Karen. And and it's also sp two hybridized, right? If you know that there's a double bond at this position. So it's not something that you expect to see in an S N one reaction where we have a stable tertiary carbon Karen. And then you will see irreversible step. But this step is not reversible simply because it's not favorable to produce that secondary carbon Karen, which is SB two in nature, right? So we are good and we can now simply think about our next step. We have seen the first step with the first equivalent, we are ready to look at the second equivalent top hydrochloric acid, the double bond would once again attack this hygiene fold by the loss of roaming. And as we said before, we are going to use more cosmic of regional selectivity concept to understand that hydrogen is added to a less substituted position and we will form a positive charge of this position as a result because that would be more stable Carbo Karen for us. So you can just draw that result on carbo can iron blooming positive charge or here, right. So once again, this is more cosmic of regional selectivity because we have added another hydrant at a less substituted position, giving us another more stable Carbo kiran than this one, which would be primary, right? So we got our car bull can on. And once again, as we said before, the bronze and iron formed in the previous step would transfer its electrons into the formation of that carbon booming bond and give us our final product. So we can now draw our found product two comma to dye bromo propane, right? So this is our final product. And in addition to that, keeping in mind what we said before, one of the arrows over here must also indicate the qua Librium that would be more accurate to say this is we previously concluded. These two are reversible stops. In particular, you may notice that we are forming carbo can iron and it's very easy to eliminate this hygiene to restore a neutral molecule which is quite favorable. So indeed, there is a competition here. We cannot say that one step is negligible with respect to the other just as we previously explained for the first step, whenever we have that Carbo Karen and the competition between the Carbo Karen and it's neutral form. If we can simply remove that hydrant, to restore the neutral form, we will definitely use equilibrium arrows. So we got our final product, we see all of the steps of the mechanism. And just to keep in mind, the main question in this problem, we observed more cosmic of regional selectivity twice during the formation of two carbo Karen's at this step. And this step in each case, we said that we were adding hydrants, a less substituted position of the triple bond and the double bond respectively, right, to take into account the formation of a more stable secondary carbo Karen as opposed to a primary one. And that's that we can simply show that this is our final answer, our complete mechanism and the correct answer choice to the small cultures question is a. However, if you have any questions, don't have states, leave your comment down below in the comments section and thank you for watching.

Propose a mechanism for the entire reaction of pent-1-yne with 2 moles of HBr. Show why Markovnikov's rule should be observed in both the first and second additions of HBr.

Key Concepts

Alkyne Reactivity

Markovnikov's Rule

Reaction Mechanism

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