Predict the products of reaction of pent-1-yne with the follow...

Question

Predict the products of reaction of pent-1-yne with the following reagents.

d. H₂, Pd/BaSO₄, quinoline

e. 1 equivalent of Br₂

f. 2 equivalents of Br₂

Accepted Answer

Everyone and welcome back in this video, we are going to tackle the following problem which states determine the identity of products obtained in the reaction of butte 1 with each of the regions shown below in particular. We have three different regions but in each case the starting material is exactly the same. So why don't we just label these three reactions first And through the starting material which is beat one iron because the ending is fine. We understand that this is an al Qaeda and it has a triple bond In particular. The triple bond starts at position number one and because the parent is be time, that means we need two additional carbon atoms since we already see two and there's a total of four carbon atoms. So we may draw two more. That's the structure of butte one eye and we understand that we have three different reactions. So let's label those reactions as one. Then we have a second one. So that would be our second one over here. And then we have the third reaction. Now for these three reactions, we are going to look at each of them separately to break the problem down as much as possible and try to understand what's happening were the first reaction we have hydrogen gas and then it says palladium over barium sulfate catalyst kino line. Now recall that whenever you're seeing palladium over barium sulfate or calcium carbonate catalyst in kino line. This is often written as landlords catalyst, which might be easier for you to understand. So we may also indicate Lynn blurs catalyst recall that landlords catalyst is a very famous region that we're using when we're dealing with al kinds because it allows us to perform a partial reduction of a triple bond simply speaking, we will form an AL keen using one of the pi bonds in our hydrogenation reactions. We will be adding two hydrogen atoms across one of the pi bonds and we will stop at the elkin. Also keep in mind that the arrangement of these to hye jin's will be says this is a selective reduction which exclusively produces a seismic but this is not so important here because we will not have any possibility for assist Transit Tamar is um so we are ready to draw first product going back to what we previously said, we will form an L. Keen which will be our final product. There will now be two hydrants in this position. So there's no trans is amateurism and we still have that remaining ethyl group. So we made troll and a group well done. That's our final product. We have to understand that lenders catalyst stops at the elk in part when we are thinking about our product and we are not reducing it further. Moving to reaction # two We have bromine one equivalent. So we made right on blooming one equivalent and because the third reactions were similar, we will also look at reaction number we at the same time to have some sort of comparison. Now if you look at these two reactions, we are applying bromine, which is a halogen. So this is a hallucination reaction which follows an electra Filic edition mechanism because we have a triple bond, we actually may add up to two equivalents of halogen. In particular, if we are only using one equivalent, we will only be using one pi bond in that carbon bromine bond formation. So to summarize, we're adding roaming atom across one of the pi bonds, we may now say that the product would still have one of the pi bonds remaining. So if we're using the upper one, the bottom one stays and there are two blooming atoms that are added in anti arrangement, recall that due to the backside attack in the mechanism that there is an anti arrangement for the two booming atoms. So one of them will be pointing up, the other one will be pointing down, We still have that Essel group had this position. So that would be our final product. Now, for comparison, if we add two equivalents, we may simply look at this structure and we may understand that we are adding an additional equivalent to that structure. And if we do that, we're essentially using that double bond, the remaining pi bond will form two additional carbon bromine bonds. So we're adding one broom in a disposition and another one at that position. Another electra filic edition. So knowing this, we may now draw our final product if we draw it in the same perspective, there was this initial blooming. Now we're adding one more across the double bond to make it a single bond. So there would be another blooming adam attached. And similarly here we will now have three different groups, one of them will be blooming, another one will be the original ethyl group and the one that we have just added is just another blooming. That's our final product for number three. Great job. So now we may summarize that for the first reaction we used hydrogen and lenders catalyst to form an AL keen assist elkin, but in this case is not CIS or trans. Just an AL keen in general from an AL kind for reaction number two, we used an electra Filic edition mechanism with one equivalent of bromine to form an AL keen and for number three we use two equivalents of roaming to perform a complete electra filic edition to not have any pi bonds remaining and to form an AL kill hey light, we will circle our final answer. These are our reactions and in particular, we are mainly interested in these three products. If you look at them, you may conclude that the correct answer to this multiple choice question is C However, if you have any questions don't have states, leave them down below in the comments section and thank you for watching this video

Predict the products of reaction of pent-1-yne with the following reagents.
d. H₂, Pd/BaSO₄, quinoline
e. 1 equivalent of Br₂
f. 2 equivalents of Br₂

Key Concepts

Hydrogenation

Electrophilic Addition of Bromine

Regioselectivity and Stereochemistry

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