Propose structures and draw condensed formulas for molecules that meet the following descriptions.

(a) A ketone with the formula C₅H₁₀

Video transcript

Welcome back everyone. What is the condensed formula for the ketone with the molecular formula? C six H 12 0. Before we begin, since we're given the molecular formula, we are going to calculate the hydrogen deficiency index, which is one half multiplied by two, multiplied by the number of carbon atoms, which is six plus two minus the number of hygen atoms, which is 12. What do we get? Well, 12 plus two minus 12 gives us two and two divided by two is one, which means that we either have one rank or one pi bond in the structure. And we're going to say that we have one pi bond because it is a ketone that has a car bono group, Ceo, right. So the general formula of the ketone is carbonel bonded to two alkyl groups and we're given six carbon atoms. So let's try to visualize what sort of a ketone we can have, we can begin with the Kanal group. And what we're going to do is essentially bond the shortest possible alkyl group on the left side of the car bal group, which is the methyl group. So that would be CH three and this leaves us with four additional carbons. So essentially CH two, because each carbon must have another siege to one more siege to. And finally, ch three at the end of the chain, that's one of the possible key zones. Now, from here, we can essentially make sure that we still have our carbel. But instead of choosing the shortest possible alky group on the left, we're going to make it longer. What if we go with the ethyl group? C two C three, this leaves us with three more carbons on the right, because we have three total. So we just want to add CH two, ch, two, ch three. And that's our second possibility. Now, what if we change the position of group? Well, if we move, move our car bono to the next carbon, we would have a total of three carbons as one of our LK groups and two more carbons on the other side. So a prop and an ethyl group which we already have in the second structure. And if we move our carbonel even further, we would just end up with the first structure. So we only have two possibilities to get two different alkyl groups. One of them will be a methyl group and a beetle group for the first. And for the second possibility, we would have two alkyl groups, one of them will be an ethyl group and the other one will be a prop group. So we only have two possibilities. And we want to turn these into condensed structures. For condensed structures, we are using a text formula, meaning we're not indicating any bonds. So if we start with the methyl group C three, now we want to simply illustrate the presence of carbonel witches given by co. So we're not indicating any bonds. And then we simply want to show how many CH two units we have, we have a total of three. So we're going to say CH 23 times CH three. So we're using parent C to indicate that we have a total of three units of CH two. And now for the second structure, we have CH three followed by CH two CEO and now we have two units of CH two. So we can just say CH two, two times CH three. Well, then we have our final answers. Let's label them. And thank you for watching.