Alkenes and Alkynes

Hey, guys, In this new video, we're gonna take a look at Halabja Nation. So here we're going to say that al keens and all kinds undergo addition reactions or adding to them, in which elements add across their pie bonds to create new Sigma bonds. Remember, Pi Bon is a double bond or triple bond. Sigma bonds are just single bonds now under Halogen Nation to Hala Jin's will add toe one pie bond. So we need one mole off X two for one double bond for one pie bond. Okay, so we need to halogen. So that's X two. So one mole of X two for every pi bon and here X can either be C, l or B are those tend to be the best intelligence because again, flooring is too reactive. Aydin is too slow. So here determine the major product from the following halogen nation reaction. So here goes. Are double bond it shared by this carbon here in this carbon here now, in halogen nation, what happens here is that double bond is broken so that each carbon can receive a halogen. So doing that, this is what we would get ch three this ch would gain one of those chlorine because there's two of them. The double bond is broken, but there's still a single bomb there. And then that ch two would also receive Ah, chlorine. So we would make here is a die Khallad di Howard meeting to Hala Jin's and not just die Halid, but something called a Fastenal di Hallett Vis any ALS refers to vicinity, meaning that the two halogen are on neighboring carbons. So again, the scene you'll die. How to refers to the same vicinity. The two carbons with the halogen, our neighboring carbons that's called the Fastenal Die Hallett. So we've seen this first one. Let's see if you guys could attempt to do this next one. Now here we have a triple bond, remember, we have two pi bonds. Remember, It takes one mole of X to to remove one pi bon. And since we have two pi bonds in the triple bond, that's why we require two moles of this halogen. So do what we've done before. Get a final answer, come back and take a look at my explanation and see if your answer matches mine. Good luck, guys.

All right guys, let's attempt to the example that was left on the bottom of the page. So here it says, determine the major product for the following halogen nation reaction. So again, we have two pie bonds. That's why we use two moles of B R two. So this carbon and this carbon is triple bonded carbons. Each one gets two halogens apiece. Because why we have two moles of BR 2, which means we have a total of for brom means at our disposal. So two of them will go to one triple bonded carbon. The other two will go to the other triple bonded carbon. And remember when we're adding bees, these will remove the pi bonds. So we have this C H three carbon here connected to this carbon which was once triple bonded, but no longer. It has now 2 BRS on it. It's connected to the other carbon that was triple bonded at one point but no longer. And it has 2 BRS on it and then we're connected to the rest of the chain. So that's what we'd get there. We'd get four halogens being added and this would be called a tetra meaning four halid. So this would be a tetra halid. So just remember one mole of B R two or C L two helps to add two halogens onto a double bond. It removes the double bond to add two halogens. But here we have two pi bonds because it's a triple bond. So we need two moles of B R two to get our final answer.

Hey, guys, let's take a look at the following new video dealing with hydrogenation. So here hydrogenation can be seen as a reduction reaction in which two hydrogen are added to one pi bon. So here the ratio is you need one mole of BR to not be are two of H two to reduce one pi bon. Now here. Let's talk about this word. Reduce and reduction in Gen. Cam and General Chemistry, we said reduction refers to gaining electrons, but in organic chemistry, we don't really say reduction means that specifically what we say in organic chemistry. We say reduction involves gaining off hydrogen ins. And the reason we say gaming of hydrogen instead is that the hydrogen is that we're gaining. Add electrons to something. Okay, so that's what we're saying. They don't physically add riel electrons. In a sense, what they're doing here is they're less electoral negative than the thing they're connected to. So they're gonna be sharing their electron energy to that new compound. So on or go, we say reduction means adding hydrogen and jen can we say reduction means adding or gaining electrons. So knowing the difference between these two is key to getting the answer. Correct. So in this first example, what you guys to take a look just realize here, when it comes to hydrogenation, all we're gonna do is that hydrogen is to the double bond to remove the double bond. And remember what happens when we have Ah, hydrogen on a carbon and we have, ah, skeletal formula. Do we show them or not? Where did those hydrogen is go And are they visible? Knowing that is he key to getting this answer correct? So attempt to do this on your own and then come back and see a video of me explaining what the answer is.

Alright, guys. So let's attempt to do that hydrogenation question that was left. So here it says, determine the major product for the following hydrogenation reaction. So we're gonna basically add Ah, hydrogen to this double bonded carbon in this double bonded carbon and adding hydrogen gets rid of that double bond. So what you're gonna get here at the end would be this structure. So you literally just take off the double bond. And remember, we don't show the hydrogen is because in a skeletal formula, we're going to say here that hydrogen is connected to carbon are invisible, so the carbons are invisible and the hydrogen is connected to them are invisible, so that will be our answer there. Now, look at the trend. We used one mole to get rid of one double bond. So knowing that, try to answer the following question where I give you a structure and I'm asking how many moles of H two would be necessary to completely reduce the structure. So just recall what one mole of h two conduce you. How many pi bonds total do we have in this molecule? That is the answer. That's the key To answer this question correctly. Good luck, guys.

alright, guys. So let's try to attempt to answer this question. I was left on the bottom dealing with hydrogenation. So remember, we need one mole off H two to remove one pie bond. And remember, a double bond has one pie bond and one Sigma bond. We don't care about the Sigma because we can't get rid of Sigma's. A triple bond has two PI bonds and one Sigma bond. So if we look, what do we have? We have one double bond, so that's one pie. We have another double bond, so that's a second. And then we have a triple bond, which adds another two pi bonds. So we have a total off four pi bonds. And remember, for every pi bon, we need one mole of H two. Since we have four pi bonds, that means we need four moles of H two to completely reduced this compound. So our answer here would be four moles of H two and using formals of H two on this. It would give us this as our answer. We get rid of that double bond in the ring, these two carbons, each of them would get another hydrogen the double bomb would disappear. You can draw the hydrogen anyway you want, and then here there's a carbon there. But there's also a carbon here that's not shown. So they gained two. Hydrogen is there, and that would be your answer at the end, where we've completely reduced it by adding four moles of H two.

Hey, guys, In this new video, we're gonna take a look at Hydra Halogen Nation. Now, under Hydrology Nation, a hydrogen and a halogen are added across a pi bon. So remember, hydro does not mean water. Hydro means hydrogen. So we're adding Ah, hydrogen in a halogen. Now, hear your book doesn't explicitly say this, but what we're doing here is we're following something called Markov Knockoffs rule when it comes to organic Cam, Russia and Germany and all these European countries have contributed a vast amount of knowledge in a lot of these reactions. One of the guys to help with this type of reaction was Markov Nankov. So they named it the Markov Markov's rule. Now, under my Kafelnikov's rule, you're gonna focus on the two double bonded carbons. You're going to say the hydrogen goes to the double bonded carbon with mawr. Hydrogen is on it, and then the halogen goes to the double bonded carbon with less hydrogen. So here, the re agent that we use in this reaction is HCL and HBR normally, But we could also use H F and H I HCL in hbr just happened to be the major forms that you'll see now again, I've said this multiple times. We get to organic chemistry. You gonna learn that hydrogenation is not as simple as this. And there are things called rearrangements that to occur here. Luckily, because you guys are learning the beginnings of organic chemistry, you don't have to worry about that. So just simply follow what I said here. This is just Markov. No cost rule Markov Makov Rule tells us h goes tome or hydrogen halogen goes toe less so knowing that we can use that Marco Viticultural to answer these questions. So attempt to do this one on your own Focus on the double bonded carbons which one has mawr hydrogen is directly connected to it will gain the hydrogen. The one that has less hydrogen is connected to it will gain the halogen. And when we're adding these two groups, what happens to our double bond? Knowing all of these things will give you the correct answer so attempted on your own and then come back and see how I approach the question

Alright, guys. So we're gonna attempt to do example one under hydrogenation. So here, if we take a look at those two double bonded carbons, this one here on the left has one h directly connected to it. This one on the right has no hydrogen is directly connected to it. Remember, H goes to more hydrogen. So this here is gonna get a hydrogen halogen goes to the one with less, so the halogen will go here. And what happens when we add things when we add things that double bond disappears? So what you get as your answer at the end would be this So you'd have everything else being exactly the same and actually here, this should be ch three ch two. So the program I use sometimes it doesn't want to do that. So it should be ch three ch two. So the double bond disappears, so this carbon here is gonna have one h and then another h added to it. They were gonna have ch three then the CL because the halogen in question now a c l. And then that carbon still connected the ch two ch two ch three. So that's will be. That would be your answer. Now, here. You could if you wanted to combine those two hydrogen with the carbon right now and just make it ch two. If you wanted, that would also be acceptable. The halogen would still be a branching group. And now we are answer. So again, we're following Markov. No cost rule Focus on the double bonded carbons. Now, hear this one. We have a triple bond, so we have two pi bonds, which means that we can use two moles of H p. R. So it still follows Mark Avnet, Cultural focus. Now on the triple bonded carbons, the one with more hydrogen still gets the h and in this case to hydrogen because we're using two moles and the one with less gets to Hala Jin's. So take a look attempted on your own and then come back and see how your answer compares to my answer. Good luck, guys.

All right guys. So we're gonna attempt to do that last hydrogenation question that was left on the bottom of the page. So here it says, determine the major product. So this one here has one hydrogen directly connected to it. This one has none connected to it. The one with more will gain hydrogens. How many will it gain? We're using two moles of H pr. So it's gonna gain two hydrogens. This one here is gonna gain two halogens. Now we drew this CH 3 this way because it's the carbon that's connected to that carbon, but you could still draw it as CH 3 if you want. OK. So now that triple bonded carbon is no longer triple bonded because we're adding things. We're removing pi bonds. It's gonna gain two halogens because we're using two moles and then the other triple bonded carbon is no longer triple bonded, either it had one H to begin with and now it's gaining two more to become CH 3. So what we make here at the end is a dye ha we get two halogens added on there. So it's a dial and they are on the same carbon when they're on the same carbon, we call it a Gemini geno dial. So think of uh astrology, Geminis twins and stuff like that. So Gemini means on the same carbon, what's on the same carbon two halogens? So remember we saw the dials earlier where two halogens are on neighboring carbons. And now Gemini, they're both on the same carbon. But here again, we're using our Carboni costal and again, we have a triple bond now, so that means we can add more than one mole of H pr. In this case, we chose to use two moles to get it all the way down to single bonds.