3,4-Dimethylpent-1-ene has the formula CH₂=CH—CH(CH...

Question

3,4-Dimethylpent-1-ene has the formula CH₂=CH—CH(CH₃)—CH(CH₃)₂. When pure (R)-3,4-dimethylpent-1-ene is treated with hydrogen over a platinum catalyst, the product is (S)-2,3-dimethylpentane.

a. Draw the equation for this reaction. Show the stereochemistry of the reactant and the product.

b. Has the chiral center retained its configuration during this hydrogenation, or has it been inverted?

Accepted Answer

hi everyone. The structure of three bromo. Four methods and one in is given below. When the sn and humor reacts with hydrogen gas in the presence of a platinum catalyst, then the art form is produced a draw. The reaction equation showing the zero chemistry of the reactant and the product be determined whether the configuration of the carroll center was retained or inverted during hydrogenation. Starting with part A. We are told that there is an aspect of stereo chemistry involved here. And what that means is we have to understand which our carl center is. A carl center is a specific carbon atom that has for unique substations. That said, if I'm told that there's an ask form, there must be one carbon atom with four different substitutions. And if I look at my structure, the only carbon that has four different substations is this carbon over here. Specifically, I noticed that there are four different substitutions which are hydrogen bro mean vinyl group as well as proto. Now these are my four different substitutions. And to correctly draw the stereo chemistry of the reaction, I have to recall um con Ingle pre log rules according to the rules, I have to assign the priorities to my substitutions. Looking at the atoms bonded to my carl center and do that. I have to determine the atomic numbers of each substitution. Specifically. If I start with booming, it has an atomic number of 35 hydrogen has an atomic number of one. Each carbon has an atomic number of sex. Now I'm going to choose another color. To assign the actual priorities. The highest priority or priority number one goes to the atom with the greatest atomic number and the greatest atomic number actually goes to blooming. So that would be my priority one. The atom with the lowest priority. My apologies. The atom with the lowest atomic number has the lowest priority. So hydrogen has an atomic number of one. That'd be my lowest priority number four. So that said, the only thing I want to understand now is simply which of the two carbon atoms would correspond to a lower priority. And to do that, I need to look at the following atoms bonded to each of these two carbon atoms. Now because there is a double bonded carbon to the left, I can actually show I can actually expand this double bond into two single bonds and add exactly same CH two substitue into each side. And there's still this hygiene over here. So I have these three substitue ints wanted to it and here specifically, let's suppose that x is our call center to their wide. The isopropyl group has C With age and two methyl groups. Now, what we want to do, we want to isolate um we first of all went to isolate exactly the same groups for comparison. So those would be hydrogen atoms. They're exactly the same. Two carbon atoms that are exactly the same and again, two carbon atoms are exactly the same. So we can conclude that there that we don't see any atom hiring priority. And here here is where it gets a bit tricky in this case in particular, which is not so common comparing the two carbon atoms. This substitution would be higher in priority because it has higher saturation, it has a double bond between two carbon atoms instead of two single bonds and a double bond implies that such a substitution or in this case by Niall would be hiring priority. So what I can say is now that um final would be priority number two and I so appropriate would be priority number three. Having assigned the priorities. Now, I can think about it this way commingle pre log rules say that the rotation from 1 to 2 and then two or three would indicate my configuration on that center. As long as protecting number four or hydrogen um points down. Or you can basically see it on a dash simply speaking. If we try to assume that hygiene is pointing down and it has a wedge Annette and if we assign the priorities or in 123 order we see that the rotation is clockwise and clockwise rotation corresponds to our configuration. However, our original substrate has an S configuration. So what I have to do, I actually want to change or flip the flip the the original dash of hydrogen into a wedge, flipping a dash into your wedge would essentially give me the s configuration even though you see the rotation clockwise having priority number four pointing up actually implies that you're reversing your original assignment and it would actually correspond to us that said, we now clearly see how the original substrate looks like and now we are ready to draw it out. Let's use bond line structures here and we can notice that we have a carbon chain so that we original chain, we have a muscle group over here, um we also have a double bond specifically here and this carbon atom has hydrogen pointing up to make it as and bro mean would be pointing down, we have successfully managed to draw the original substrate and the problem says that the process here is the process of hydrogenation. So we're using hydrogen gas age too and the catalyst that we're using is the platinum catalyst. Now we are ready to draw the product and because this is hydrogenation, we identify that okay, there is a double bond over here and we will essentially add one hydrogen bond to each end of that double bond to essentially produce an Elk ile substitution. Now notice that in this reaction we were only adding hydrogen towards the double bond and our Cairo center was not even affected. It did not participate in the hydrogenation process. It was simply the double bond, it was reduced and the Caro center should not be affected. We would still expect Haijun be pointing up just as originally and bro mean pointing down just as originally. So that is our final equation. We have our product. And the question for part A was um simply to draw that equation which we managed to do and part B asks us determine whether the configuration of the carroll center was retained or inverted during hydrogenation. We can notice now that assigning priorities to our um product blooming will be priority number one and hygiene be priority number four. But now um the priorities for the two carbons on top would slightly change. This would actually be our priority number two and this will be our priority number three, indicating that if we assign the rotation directly um 123 goes counterclockwise. But because hydrogen is not or basically our lowest priority is pointing up instead of down, we have to flip it. And instead of having counterclockwise rotation, the actual rotation will be clockwise. So that would be indeed our our form. In addition. Let's just try to understand why the priorities now change. If we look at the right most carbon, that's see age with two methyl groups just as before And the left most now has two hydrogen atoms bond it's here. So that's CHHCH three. If we exclude exactly same substitutions, let's say hygiene from here and here as well as a methyl group from here and here, we noticed that we have a comparison between a methyl group and a hydrogen atom and because carbon atom has a greater atomic number of six, that would make it a higher priority. So we are perfectly consistent with a new art form. Now, the question for B is determined whether the configuration of the carroll center is retained were inverted and we can sell that since hydrogenation did not take place at the carroll center, the actual configuration was retained. And the only reason why we are forming and our form from us is just because we're changing the way that the submissions are now prioritized. Simply speaking, we swapped priorities number two and three in the product based on what we already explained that said, um this is these are our reactions and products. And for part B, we can clearly indicate now that the configuration was retained. The correct choice for the multiple choice question is B. I hope you enjoyed this video. And if you feel that you have any questions, don't hesitate to leave your question down below. Thank you.

Key Concepts

Hydrogenation

Stereochemistry

Chirality and Inversion

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