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.

j. NaNH₂

k. H₂SO₄/HgSO₄, H₂O

l. Sia₂BH, then H₂O₂, ^–OH

Accepted Answer

Alright. Hey, everyone. So for this question, we must determine the identity of products obtained in the reaction of beauty in with each of the re agents shown below. And so let's get started with our part one here now, but one line is a four carbon chain with a triple bond in between carbons, one and two. So we have a four carbon chain With a triple bond in between positions one and 2 like so, so we are reacting our beauty one line with sodium and mind. So we have sodium and we have negatively charged N H two minus over here. Now notice how our beauty one line is what we refer to as a terminal, all kind, right? It's a terminal all kind because it begins on position one, which is at the very end of the chain. So a terminal all kind is going to have a hydrogen on position one which is relatively acidic. It's kind of acidic. And this of mind this NH two minus is a base. So what's going to happen is simply an acid base reaction in which our N H two minus is going to go ahead and take that terminal hydrogen, Therefore, um bouncing this pair of electrons towards carbon Wanna Beaute one i. So in our final product, we're simply going to have a deep patinated butte one eye and we're going to have an an eye on a carbon ion. So this is going to be our carbon ion, right? We now have a negative charge on carbon one. And now um be sure to always take into account your spectator ions, right? So recall that sodium is here in this case to serve as a counter a counter balance, you could say to the negative charge of our mind. So now that carbon one of but one man has a negative charge, sodium is going to go ahead and counterbalance it instead And we're going to end up with neutral Ammonia NH three. So notice how because NH three is now neutral sodium is going to go ahead and counterbalance the negative charge of our carbon ion. So that's gonna be it for part two or for part one. Sorry. Now we can proceed to part two. So for part two, I'm gonna redraw our starting material because that didn't change. And now what's interesting to me is that we have a sulfuric acid which is a strong acid in the presence of water. So recall that when we're talking about all Keen's H two S 04 in water are the re agents for acid catalyzed hydration, which is a Markov Markov edition of an O H group. Markov, I am 100% not spelling that correctly. There we go. Sorry about that. But anyways, like we said before, this is a pretty standard market competition of um of an O H group when we're talking about all kings and the same is true for all kinds. So what we're gonna do is we're going to add an O H group on the more substituted carbon of our all kind here, which in this case happens to be carbon too. So if I draw an arrow with our re agents, with our re agents in the presence of water, we're going to end up with a double bond in between carbons, one and two and an O H. So we're going to have A double bond like so and we're going to have an OH group on position two. Now recall that when we're talking about all kind reactions, this is never the end of the story, right? Because what I want to bring to your attention really quickly is the fact that we have here an Enel, we have an Enel because we have an alcohol and O H group and an all keen in the same molecule. So we have an all keen and an alcohol in the same molecule. It's an eternal. Now, recall once again that when you have an Enel, you're going to have an equilibrium with its alternative form, right? We have Keto taught memorization that is going to happen in this case where R O H group becomes a carbonnel instead. So we would end up with a ketone as our Keto four. And the keto form is actually going to be our final product in this case, because keto forms are more stable than in all forms in most cases. And it's going to be true in this case. So the Keto form is actually going to be our product for part two. And now we can move on to part three. So I'm gonna scroll down a little bit to give myself some space. And for part three, let me just redraw our starting material really quickly is our starting material. And now what I want to bring to your attention in our list of free agents is the fact that we have a starting material that contains boron and a separate step with hydrogen peroxide and O H minus. Now, a compound containing boron followed by hydrogen peroxide and O H minus sounds like hydro operation oxidation, which recall is an anti Markov Markov and time mark called Nick off addition of an O H group. So if it's an time are commonly called, then here in Butte one line, we're going to have an O H group on the less substituted position, right? It's going to be on carbon one, whereas in part two, it was on carbon too. So I'm just gonna write our re agents on the arrow. This part one And part two is hydrogen peroxide with OH -. So here for part three, we're going to end up with an all keen But now we're going to have an OH group on carbon one instead. Now notice once again that this is not the end of the story, right? It's not that simple because this is also an N, all right, we have an all keen in between carbons one and two and we have an O H group which makes it an alcohol. So once again, we are going to see keto memorization where this primary alcohol here will become an alga hide. Oh, I'm not sure you guys can see that. Let me just move this over. There we go. So this is what you're going to end up with because like we mentioned before, the keto form is generally more stable than the final form. So just to recap um part one was in simple acid based reaction. So we're going to end up with a carbon ion from our beautiful nine. Part two is going to yield a ketone as the final product because of Keto enoteca memorization And part three is going to yield an alga hide once again because of keto memorization. Now, with that being said, thanks so much for watching and I hope you found this helpful

Predict the products of reaction of pent-1-yne with the following reagents.
j. NaNH₂
k. H₂SO₄/HgSO₄, H₂O
l. Sia₂BH, then H₂O₂, ^–OH

Key Concepts

Alkyne Reactivity

Hydroboration-Oxidation

Acid-Catalyzed Hydration

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