Some of the following molecular formulas correspond to stable com...

Question

Some of the following molecular formulas correspond to stable compounds. When possible, draw a stable structure for each formula. Propose a general rule for the numbers of hydrogen atoms in stable hydrocarbons.

Accepted Answer

Hey everyone. Let's solve this problem. It says if possible, draw a stable structure for each molecular formula of hydrocarbon shown below. Make a generalization for the numbers of hydrogen atoms in stable hydrocarbons. Okay, well to draw a stable structure or a lewis structure of each of these compounds. We first have to count the total number of valence electrons that we're going to use to connect all of our atoms together. Then we can draw our sigma bond structure, showing the connectivity of the atoms with. And since these are hydrocarbons, we know carbon is always going to be in the center. And lastly we will complete the octet with any loan pairs or maybe even some double bonds. Okay, and it says to make a generalization for the numbers of hydrogen atoms in the stable hydrocarbon. So I'm going to write generalization right here for when we figure it out. Okay, so there's a lot going on here. So let's just start from the beginning and see where we go. So we'll start with C4 H4, C four H 4. And how many valence electrons will we have? Well four times four for carbon is 16 plus four for hydrogen, which gives me 20 valence electrons. So let's draw four carbons 1, 2, 3, 4 and four hydrogen. Okay, sometimes it might take some practice to figure out which exact carbons are going to go on. I'll start at the ends and I'll add some here. Okay, so that uses 14 valence electrons. 2, 4, 6, 8, 10, 12, 14. So we need six more valence electrons. So that will be in the form of lone pairs so we can add some to those carbons that are lacking some electrons. This is not our final form of our structure. Right. These carbons don't have octet, This carbon doesn't have octet. So we need to make some double bonds. So these electrons, these lone pairs will donate towards each other to create a triple bond and this carbon will donate towards that other carbon that has does not have a complete octet to form the structure to save space and time. I'm just going to erase and redraw. So we'll have a triple bond here and our double bond here. And that is our final structure for C four H 4. Okay, that worked out that's a stable structure. So let's move on to the next one. C four H five. I'm going to draw a line here. C four H 5. Okay, how many valence electrons is this? 16 plus five, which will be 21 valence electrons. Okay, interesting. We don't usually see an odd number of electrons. Right, so let's see what this looks like if we draw it out. So still four carbons and let's just add some hydrogen is to it. 1, 2, 3, 45. So that uses 2468, 10, 12, 14, 16 electrons. So that means we need five more valence electrons. Well, four valence electrons would be two lone pairs. But then we would just have one electron. So if we add those lone pairs and then made them into double bonds to satisfy the octet of these carbons, we would still have one more electron. That would be a radical on this carbon. Is that very stable or radical stable? No, they are not stable. So this is not a stable structure for a hydrocarbon. And why is that? We said because this electron is by itself. So can we make a generalization about the number of hydrogen atoms in stable hydrocarbons? Well, we need an even number of valence electrons. So that will happen when we have an even number of hydrogen is right. C four H four was stable because we had an even number of electrons. C four h five has an odd number of hydrogen. So it is not stable with that odd number of hydrogen. So our generalization is stable. Hydrocarbons must have an even number of hydra jen's in the molecular formula. Okay, stable hydrocarbons must have even number of hydrogen in the molecular formula. So let's see if this holds true when we continue to do our other structures. So I'm going to cross out all of the odd number of hydrogen is because we know those will not be stable. So we won't waste our time drawing those and let's just continue with the even number of hydrogen hydrocarbons, hydrocarbons with even number of hydrogen rather. So next will be C four H six which will have valence electrons And we will have our four carbons and six hydrogen 123, 456. Now let's add lone pairs because that is 2468, 10, 12, 14, 16, 18. So I have four more electrons. So I can add two lone pairs. Let's say to these two carbons. Now we see that these outer carbons have a positive charge, they are missing some electrons. So these carbons with the negative formal charge will donate their lone pairs down to share with those carbons forming a double bond. And now all of our carbons have octet. So this will be your structure for C4 H6. Okay, C four H 8. So we did see four H six. Now we're on a C four H eight. So that will have 24 electrons. Right? Because we're just adding two hydrogen from the previous structure and that will look like this carbons. I'm gonna give each end carbon three hydrogen ins so we have two metals and then one hydrogen each to those inner carbons. And now we only have two more electrons left over. So I'll add them to one of these inner hydrogen or carbons. But this carbon is still not happy. So instead of just this carbon keeping those electrons, they will share them. So those that lone pair will come down here and they will share them as a double bond. So now all of their octaves are happy with four electrons touching them. Cool, Let's go on to C four H 10. Oh, I didn't follow my color scheme. C4 each turn So again two more electrons. So that will be 26 valence electrons. So we have four carbons and 10 hydrogen ins. So do we have any electrons left over? No. So this is a fully substituted hydrocarbon, fully substituted. Right. There's no double bonds, no long pairs. Just carbons with full octet. It's so this shows that the two N plus two rule is satisfied right, when we have the number of hydrogen zones. So and being the number of carbons, right, that two N plus two is are a number of hydrogen. Right, so two times four is eight plus two is 10. So this is the formula for a fully substituted carbon. Cool, let's keep on going. So now we are going up to five carbons five Carbon C five H 4. So we have 24 valence electrons here. 1, 2, 3, 4, 5 carbons only four hydrogen ins. I'm gonna add them to the ends because they can share electrons with double bonds in the center carbons. But the end carbons are harder to do that with. So they need those hydrogen. So we have eight more electrons. We can adhere well let's add a lone pair to each carbon. Why not? Okay, so these electrons will come down in between carbon is one and two and three and four and they will form two triple bonds because there's not a lot of electrons to go around. So they're going to have to do a lot of sharing. So you'll notice the less hydrogen there are the more shared bonds in the form of double bonds and triple bonds. Right, So this is our structure for C5 age four. Now go on to C Age six And we're adding two more electrons. So that's 26 valence electrons. And we will have 1, 2, 3, 45 six hydrogen ins one, 234, 56. So that gives us six more electrons which would be three lone pairs. So let's start with these empty ones. And at these ones here, since this carbon is going to need some electrons as well. So these two carbons need to share that lone pair. And then these carbons have well this carbon has no hydrogen and this carbon only has one hydrogen. So they are going to share those lone pairs together to form a triple bond. And these will have a double bond between them. Okay, Now we have C5 H Which is 28 valence electrons. 12345 carbons, 12 hydrogen ions, 12 or sorry, 5345 hydrogen. And if I add one to each of the remaining carbons, that's eight. So that is 28 valence electrons. And we need that uses 24. We need four more which would be two lone pairs. So let's add them here. Since we can't add a lone pair to each of these four carbons, they're going to want to share them. Right? Because they all need more electrons to satisfy their octet. So there will be two double bonds between each of those sets of carbons to satisfy their octet. Okay, we're almost there, you guys. Two more? We have C5H 10 Which is two more electrons gives us 30 valence electrons 1, 2, 3, 4, 5 carbons. And almost all of them have carbons. Or sorry, hydrogen 89, 10. And that gives us two more leftover electrons. But this carbon also needs to satisfy its octet needs more electrons. So this carbon will share those electrons with it forming a double bond and that will be for that one. So you'll see the patterns as you draw all of these out. So the closer you get to being fully saturated, the less double bonds and triple bonds you will have. And lastly C five H Will give us 32 valence electrons 1, 345. And you'll see that there are no double bonds, no triple bonds here. This hydrocarbon is fully saturated. And does our number of hydrogen satisfy the two N plus two rule. Yes, it does. Right. So we have five carbons. So two times five is 10 plus two is 12. So since this is satisfied this hydrocarbon is fully saturated. Okay, So that is all the structures for these hydrocarbons. And remember our generalization is that if we have we have to have an even number of hydrogen in the molecular formula of a hydrocarbon in order for it to be stable. If there's an odd number, it will result in a radical, which is definitely not stable. Okay, I hope this video is helpful, and thanks for watching.

Some of the following molecular formulas correspond to stable compounds. When possible, draw a stable structure for each formula. Propose a general rule for the numbers of hydrogen atoms in stable hydrocarbons.

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