Alkynes undergo hydrogenation to give alkanes, just as alkenes...

Question

Alkynes undergo hydrogenation to give alkanes, just as alkenes do. Draw and name the products that would result from hydrogenation of the alkynes shown in Problem 13.25.

Accepted Answer

All right. Hi, everyone. So this question says similar to all Kines, all kinds can also undergo hydrogenation to yield all kine draw the line structures and provide the Iupac names of the products of the hydrogenation of the molecules shown below. And here we have two different models which we will discuss in a little bit. Now, just a quick note about the reaction that's being discussed here, right? Recall that the hydrogenation of an all kind involves two moles of hydrogen gas in the presence of a catalyst because what essentially happens is that the triple bond between two carbons is replaced by a single bond and both carbons receive two hydrogen atoms each. Now, the nice thing about this question is that in this case, we are dealing with only line structures. Now, recall it in a line structure, carbons are depicted as either corners or open ends and hydrogens are simply implied, meaning that we do not actually have to draw them. So the first thing I'm going to do here is scroll up to consider my models and then draw the line structure of the reagent and then proceed to draw the product. Now, sometimes the trickiest part about these models is finding where the triple bond actually is. So recall that all kinds have linear geometry. So any region in a given model that appears to be a straight line is most likely going to be your all kind, right. So if we consider the first model on the left, then the left side of the structure appears more linear and therefore is our triple want. So there is our triple wand. But now the question is how long the parent chain happens to be? So starting off with my all kind carbons, I can continue numbering as far as I can. And I noticed that I end up with a six carbon parent chain. So I have a six carbon chain but a triple bond in between carbons, one and two. And in addition to this, I have 21 carbon substituents attached to the carbon that I previously labeled number five. So that's two methyl groups on that one carpet, right? So in this case, then because we're dealing with a line structure, in order to draw the product, we are simply going to go ahead and remove the triple wand. Since hydrogens do not actually have to be shown. Now, in addition to simply removing the triple wand for the purposes of our line structure, recall, the geometry also has to be changed just a little bit because in going from an all kind to an all cane, the geometry changes from linear to tetrahedral. So instead of drawing a straight line or more linear geometry between carbons, one and two, I'm going to go ahead and draw them as a simple zigzag line in my final product because that reflects the change in geometry when I go from an all kind to an all K. So that gives me just to make sure I have the right number of carbons. Yes, right. I have six carbons in total and I have two methyl substituents on carbon five. So this is the product of part what? So now we can proceed with drawing out the structure of part two. And in part two, if I consider the model, the area of the model itself, that appears to be the most linear is towards the center of the compound itself. And that is where our triple bond is going to be. So in this case, to find the parent chain, I'm going to go ahead and see how many carbons I can list in a straight chain that incorporates my triple bond as well. So starting from the left and going towards the right, I can see that the longest continuous carbon chain is eight carbons long. So I have an eight carbon chain with a triple bond in between carbons four and five. So I'm going to go ahead and draw that first just to be careful about the geometry. So it's three carbons, four and five, six, seven and eight Now, I'm going to go ahead and make some space for myself on the bottom so that I can improve that drawing just a little bit. OK. That looks better. So now I'm going to scroll up one more time to go ahead and consider my substituents. Now, in this case, it looks like I have 41 carbon substituents branching off of the perry chain. I have one on carbon two, one on 31 on six and one on seven. So if I scroll down here, I've got one on two, another on three, another on six and one more on seven. So now here is the all kind represented by the second model and just like we did for the first part, right? In this step, I am going to simply remove my triple bond and proceed with a single bond in between carbons four and five. So that results in an eight carbon chain, I can do that better there, right? This is the same eight carbon chain, but I'm keeping the the orientation of it similar to what I had in the beginning. And now I have a substituent on two on carbon three on carbon six and on carbon seven. So here is the product of part two. So now let's proceed with our Iupac naming right. The nice thing in solving this question is that we already have quite a bit of information regarding the parent chain and any substituents on here. So the first product was a five or sorry, six carbon chain. So that gives it the name Hexane and we have two methyl substituents. Now, the direction of numbering should make sure that substituents have the smallest number possible. So because my methyl groups are closest to the right side of the molecule, I'm going to start numbering from right to left. And that means that both methyl groups are on carbon too. Now because I have multiple copies of the same substituent, I have to add a prefix to indicate their quantity. And because I have two, that prefix is going to be die as in dimethyl. So, so far I have dimethyl hexane and now I have to add position numbers for each substituent that I have. So because I have two are both substituents on Carvi two. I'm going to write the number two twice. So that's two comma two dash dimethyl hexane and notice how I use commas to separate numbers from each other and dashes to separate numbers from letters. So now we can do the same thing with the product for part two. But to do that, I'm going to go ahead and move our example reaction down one more time. Yeah. So now here for the second product, our parent chain was eight carbons long. So that's octane and with respect to numbering, notice how our substituents are a pretty even distance apart from both ends of the molecule. This means that no matter how or in what direction I decide to number my parent chain, the substituents are going to be on the same numbers right, on the same positions. So in this case, I have four copies of the same substituent because I have four metal groups in total. So instead of saying dimethyl, like in the first part, my prefix becomes tetra. So that's tetra methyl octane. And now I need numbers to indicate their positions on the parent chain. So that's 236 and seven. So that's two comma three comma six comma seven dash tetra methyl octane. Again, keep in mind that comma separate numbers and dashes, separate numbers from letters and there you have it right. So here are not only the hydrogenation products for both parts but also their respective iupac names. And so with that being said, if you stuck around to the end of this video, thank you so very much for watching. And I hope you found this helpful.

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Alkynes undergo hydrogenation to give alkanes, just as alkenes do. Draw and name the products that would result from hydrogenation of the alkynes shown in Problem 13.25.

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