a. Draw the condensed structures and give the systematic names...

Question

a. Draw the condensed structures and give the systematic names for all the alkenes with molecular formula C₆H₁₂, ignoring stereoisomers. (Hint: There are 13.)

b. Which of the alkenes have E and Z isomers?

c. Which of the alkenes is the most stable?

d. Which of the alkenes is the least stable?

Accepted Answer

Hey everyone. And welcome back in today's video, we have a three part problem. And in the first part of the problem, we want to draw the condensed formula of all Calkins with molecular formula C five age son and assign systematic names to each ignore stereo is mars. In part B, we will identify the Elkins which can show E N C I summarize. Um And in part C, we are going to identify the most stable and the least stable elkin among them. Let's begin solving our problem with part A. We want to draw condensed formulas. And essentially that means we are only going to show our double bond. First of all, we have C five H son because the number of hydrogen atoms is twice as large as compared to the number of carbon atoms. This indeed corresponds to an elk in which has one double bond or a cyclo al caine. But that's irrelevant because we're only concentrating on al canes. So we are going to draw condensed formulas, meaning the only bond that will be shown will be our double bond. Let's start with part A, we have five carbon atoms. So first of all, we can choose our parents to be five carbon long and we are going to change the position of our double bond. Let's suppose that first of all, we are going to think about the structure which has the double bonds starting at carbon. Number one, then we have 12, three for five carbon atoms. Notice that we're not yet drawing the condensed formula. We just want to look at the pattern of bonding. So we can have our five member chain, then we can have our five member chain with a different position of the double bonds. We can also say that we can have this scenario where the double bond starts as carbon number two. Okay. Now, that's it. If we switch our double bond to carbon number three, that would correspond to carbon number two, we will just change the numbering system. So we only have two possibilities here. And now we simply want to shorten our chain and we will start again with the double bond at position one. So we can have four carbon atoms in the longest chain. And then we have several possibilities. We can assign our remaining carbon atom at this position or that position, right? So we have two possibilities here. Okay. So we got it. And finally, we are going to think about a four member parent where the double one actually starts at carbon number two. And notice that specifically in this case, we can only have one possibility to bond our carbon either to this carbon or that carbon, but they are equivalent because we have that line of symmetry, right? Otherwise, if we bonded here, we will just end up with our five member chains. So we don't want to have that. So we see our bonding patterns, we will have five different structures. We are just ready to draw them out. Now, for the first one, let's essentially indicate our carbon carbon double bond and let's add the remaining carbon and hydrogen. So if this has two bonds, we need two hydrogen atoms, this one will be bonded to another carbon. So we need one hydrogen to add, then we have our third carbon now because it's single bonded, it will have to C C bonds and therefore, it must have two hydrogen, our fourth one exactly the same, right? And the final one only has one bond. So it needs three hydrogen. That's the first structure that we got. Now, moving on to the next one, we get cc single bond. So that would be three hydrogen atoms bonded to that carbon, then we will have this carbon double bond it. So it must have only one hydrogen. So we're going to say H C then clearly indicate that double bond between the two. And because that carbon is bonded to another, we need one hydrogen to bond to it. So we are essentially going to say that we have ch Then our 4th one will have two hydrants because we see two bonds. So that BC is two and finally see three just as we had it before. That's our second structure. Let's move on. See the 3rd 1. So we start with CC double bond, meaning we will have exactly the same thing as before. We will have ceased to you on the left that destroyed here H two C double bond, right? And then we will have our carbon, which has one carbon atom bonded. So that means we must have a metal group. But because we're dealing with a condensed formula, we want to show inferences that this metal group is bonded to that carbon, right? We don't want to show that carbon carbon bond. And then after we through our metal group, we see that there is one carbon atom with two bonds. So it must be a siege to group. And finally, we have our C3 group, another metal group moving on to the next one, we once again have cc double bonds. So we will end up with a church to see every carbon must have four bonds. Then this one has three bonds. So it needs one hydrogen atom, then the next one has a metal group, right? As we said, we want to have that carbon atom over here specifically, let's add them to clearly indicate that these are actually carbon atoms that we have here. Okay. So if we're looking at the structure, we have cc double bond. Then this is CH because it needs one hygiene. And this C will first of all have one hydrant because we see three bonds around that rights were going to say CH ch once again. But inferences we will indicate that this carbon has a metal group C three. And then finally, we have our external carbon which will correspond, which will correspond to another metal group. So we can just say that there are two metal groups bonded to that carbon, right? You can clearly see them as labeled and read these two will be our metal groups. So we are using a subscript to indicate the multiplicity. And now moving on to the final one, We noticed that we have CC single bonds. So it must be C3 to begin with. Okay. Now we have our carbon carbon double bond and that essentially means that this carbon needs one hydrogen, it has three bonds total. So we're going to say H C, we're going to show that carbon carbon double bond. Okay. And now this carbon once again has two metal groups, right? Two carbon atoms which are external and it needs one hydrogen because it has three bonds were going to say ch And then C 3, 2 times. Okay. Well done. So we got our structures, we can essentially eliminate our analysis at the very top and we can just replace these models where the actual structures that we have drawn. So let's see that let's just Adam here and let's name them. So now the first one is our pension because the longest chain has five carbon atoms. But because it has a double bond that we pen scene, and we want to indicate the position of the double bond so that we pent one in because the double bond starts at carbon one, this one is again pence seen, but the double bond starts as carbon C. So we can call it meant to in for the third one, the longest chain has 1234, we don't count that carbon atom because it's pointing up, right? So the longest chain has four carbon atoms and therefore it's be thine. But now this is your carbon one, this is your carbon too, right? So the double bond starts as carbon one, but carbon number two has a metal group. So we're going to say to metal, Right? Because the metal group is at carbon # two. And then because its beauty scene, we are going to say beauty, the double bond starts at carbon number one. So that's one in moving on to the next structure that's once again, be seen because we have four carbon atoms in the longest chain and one metal group pointing up right away from the straight line that we see in which our carbons are distributed. And this condensed formula. So we can first of all understand that this is beating the double bond once again starts at carbon one, but the metal group is bonded to carbon number three. So the name would be three metal be it. And the double bond starts at carbon number one. So that's one in and for this one, the whole difference between this one and the one above it is the fact that the metal group is still bonded at carbon number two, right? We are essentially going from right to left now. So we have C three siege that has seized three, right? So the metal group is bonded at that carbon, which is carbon number two. And the double bond starts at carbon number two as well. So that would be to metal beat to end. Okay. So we got our answers for a party. Let's label these structures and the corresponding names. And now we are ready to move on to part B. We want to identify which of the five can show an E or C I summaries. Um In particular, whenever you're dealing with E N C S Armory's, um you want to take a look at your double bond, identify your carbon atoms and see if you have two different substitutions on one of your carbons. So this one has two hydrants, we can eliminate this one. For this one. We have one hydrogen and one metal group right on the other one, we have one hydrogen and one ethyl group bonded to that carbon. So it indeed can show E N Z A summer's. Um If we clearly draw it out with the bonds shown, you will have your CC double bond right. On the left carbon, you have hydrogen And then your metal group. So that's C3 And on the other one, you have your hygiene and your ethyl group sees C3. So this is our cynicism. All right, because you have two hydrants on the same side of the double bond. And then you can have a trans one, you can make a trans out of it if you essentially swap these two substitutions. So that the two hydrogen czar pointing in opposite directions with respect to the double bond. Okay. So this is your trans. So we can definitely see that this is the one that can exhibit our cis trans is tamar is um this one cannot because there are exactly the same two hydrogen atoms bonded to that carbon, same thing here to exactly the same hydrants bonded to that carbon. And we cannot have cis trans isomerization whenever two identical substitutes are bonded to the same carbon. And looking at this one, we essentially understand that first of all, we had that extra hygiene that we want to eliminate, right? Because um I'm sorry, I did a mistake, right? We had our double bonded carbon and then see method groups bonded sit. So we don't need that hydrogen because carbon already has two metal substitutions bonnets it. So that's 1234 bonds. Okay. So we had to make this quick fix and are based on our fixed. We can clearly see that there are two exactly same metal groups bonded to that carbon. So it cannot exhibit cis trans isomerization. Meaning for part B, we can just say that only and two in can exhibit cis trans isomerization. Okay. That's the only structure. And I'm moving on to part C, we want to identify the most and the least stable al king. So first of all, if we identify the most stable and then the least stable, we have to look at how substituted our double bonded carbons are. So starting with the first one, you have two high jeans wanted to this carbon on this carbon, you have one hygiene and one L kill group. So it's mono substitute that we can say mono or basically one substitution for this one. You have one hygiene, one metal group. So that's one L kill substitution and then one hygiene, one ethyl substitution. So that's why I substituted for this one to hide Jen's. And for this one, for this side, basically we have one metal group and one ethnic group. So it's di substituted to hide Jen's no substitutions, no L kill substitutions, right? And then see each one hydrogen, one isopropyl substitution. So that's essentially mono substituted. And for this 11 hygienic, one methyl group. So that's one alcohol group. And then here we have two alcohol groups. So it's tri substituted. Okay. So the most stable will be two metal beauty in because this Elkin is substituted the most among all of them. Remember that the more substituted and Elkin is the more stable it is. So I can say to metal beat two in it's the most stable Al king. Now, the least stable Al king corresponds to an alc in which is least substituted. And actually, We had to Elkins that were mono substituted in particular structures, one and 4.1 in and three metal beat one in. So let's write them down because they're both mono substituted, they correspond to two least stable Elkins. So our answer for the least stable Elkins would be Antoine in and three methanol be it one. In so these are our answers to part C the most stable, one is two metal beach scene and the least stable actually are right? Because we have two structures. Those would be pent one in And three muscle butte one in. Okay. So we got our answers once again, don't forget that we fixed part A, we had this missing, we had that extra hygiene. We should have had four bonds total. So we don't need that hydrogen over here. We already see these two bonds. And therefore, because there are two bonds already over here, there are two method groups. We don't need that hygiene. So my apologies for that mistake, but based on our answers, we may conclude that the correct answer to the multiple choice question is a thank you for watching.

a. Draw the condensed structures and give the systematic names for all the alkenes with molecular formula C₆H₁₂, ignoring stereoisomers. (Hint: There are 13.)
b. Which of the alkenes have E and Z isomers?
c. Which of the alkenes is the most stable?
d. Which of the alkenes is the least stable?

Key Concepts

Alkene Structure and Nomenclature

E/Z Isomerism

Stability of Alkenes

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