Draw the condensed structural formulas and give the IUPAC names for all the aldehydes and ketones that have the formula C₅H₁₀O. (12.3)

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Welcome everyone. Let's take a look at the next problem here. It says, sketch, the condensed structural formulas for all the ketones and aldehydes with the formula C 4h 80 provide the Iupac name for each structure. So this one sounds really daunting, but we need to look at the fact we only have four carmens there. So there's only going to be so many arrangements possible. So don't worry, it's not going to get too out of control here. So let's first look at our base structure. We have four carbons. So let's start by the most basic arrangement of four of them in a straight chain, all in a row. So we'll put four carbons and leaving a space after each one to make room to fill in those hydrogens. And it condensed structural form. Manila, we write in the hydrogens though, we write them as groups of letters with the carbons, we show our bonds with lines and make sure all other atoms besides hydrogen are shown on here explicitly. So we've got our four Carmens in a row. So what are the possible aldehydes that can be made with this compound? Well, for an aldehyde of course, we'll only have one, our group attached to our carbon. So we'll have our group attached to a carbonel group. So it's attached to that carbon, carbon C double VDO. And there'll just be a hydrogen on the other side. So an aldehyde would need to be on either of the terminal carbons here. However, because of the nature of this molecule, if I put it on carbon one versus carbon four, that will be an identical molecule. So there's only one possible structure. And with this al one possible way of making an aldehyde from this four carbon chain. So I'll put this just right here on the end, put a double bond to an oxygen on that last carbon, it will have one hydrogen there and let's fill in the other hydrogens. That first carbon on the left will have three hydrogens as its terminal with no substituents. The next two internal carbons with no substituents will have two hydrogens. We'll double check that the formula matches four carbons, one oxygen and three plus two is five plus two, is seven plus one on the aldehyde is eight hydrogens. So there's one aldehyde possibility. We would say that the aldehyde is on carbon one, you always number your chain so that your most important functional group has the lowest number possible. So this will be one and our four carbon chain has that name butane. And since it's an aldehyde, we add the ending A L one Butan now is the Iupac name for this possible structure. So there's one aldehyde, that's the only aldehyde possible with four Carmens in a straight chain. Now, what happens if we make a branch chain? Well, the only way to do that with four Carmens is we have a three carbon chain, but three carbons and if we attach a carbon to the middle, a carbon on either end of a three carbon chain just makes it a four carbon chain. So only one possibility for branching here. Now, we have two terminal carbons where we could add an aldehyde on the end on the propane chain. But our branch here, our methyl group here could also be made into an aldehyde. It's a terminal carbon. But what do we notice about the structure of this molecule? An aldehyde on the end of our branch here, if we look at it in another way, it's an aldehyde on the end of a three carbon chain with a branch on carbon two. So those are identical possibilities. So, out of this whole structure, there's only one possible aldehyde structure. Everything else is just it, the identical molecule. If you rearranged it, if you rotated it, they would match. So just pick in a terminal carbon, I'll do the one on the right, we add a double bond to an oxygen and one hydrogen and then fill on our other hydrogens ch three on each of the two terminal carbons. Our middle carbon here already has three bonds. So it just gets one hydrogen check our molecular formula again, four carbons, one oxygen and three plus one is four plus one, is five plus three, is eight. So that matches to name this, we name the main chain. So that would be propane. It's an aldehyde. So we have that a L ending propel, but we have a substituent here. So we're going to call the carbon one as the carbon with the aldehyde group on it. And then we'll number our chain 123. We don't have to number that propel. It's assumed that our aldehyde being the most important group will be on carbon one, but we do need to number the branch. So in front of the propel, we'll have two metal to indicate a substitute group. So we have a two hyphen methyl. So we've got this condensed structure here that three carbon chain with an aldehyde on the end and a methyl group and that's two methyl proin now, so that's another aldehyde possibility. So we've kind of exhausted our aldehyde possibilities here. What are our possible ketones? Let's recall that a ketone as an R group attached to a carbon group. So that's C double Bondo and that carbon has a second alkyl group on the other side. So it's an internal carbonel group. So we'll start again with that straight four carbon chain, draw four carbons, the two terminal carbons are out those are aldehydes. We can put a carbel on one of the internal carbons. So let's try the second one from the left, add a double bond to an O note that if we put that double bond on the carbon three, as we would see it here, we generate the identical molecule just numbered from the other side. So just one possibility for ketone from this arrangement of carbons. So let's fill in our hydrogens. Our first carbon will have ch three, it's a terminal carbon. Then our carbon carbon doesn't have any hydrogens. It has four bonds already one on either side to the alkyl groups and two to the oxygen. Third carbon has two hydrogens being an internal carbon. And the last one is terminal with three hydrogens, double check our molecular formula, four carbons, one oxygen and three plus two is five plus three is eight hydrogens. So now we just need our OPEC name four carbon chain. So we've got butane as our root, it's a ketone. So we add the ending one, Butan and where is the carbonel group located? Well, again, we number our chain so that our functional group has the lowest possible number. So we would start from the left carbon one and then carbon two will have the carbon group. So our full name would be two hyphen Butan know with this molecule. So there's our structural formula that carbon you group on carbon two. And our name being too bean know, finally, we just want to check, can we do a ketone if we have branch structure as we start with our aldehyde? So just briefly look at if we had three carbons and then a methyl branch, well, we only have one internal carbon here, there's three terminal carbons. Could we add a double bond to an oxygen there? No, we can't because this would give carbon five bonds since it has, it's a tertiary carbon has three carbon bonds. Put a big red X over that. That is not a possibility for ketone. So all together we have a pretty modest three possibilities. Here. Two aldehydes are one butanol or two methyl propanol and one ketone. Two butan know, see you in the next video.

Draw the condensed structural formulas and give the IUPAC names for all the aldehydes and ketones that have the formula C₅H₁₀O. (12.3)

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Draw the condensed structural formulas and give the IUPAC names for all the aldehydes and ketones that have the formula C5H10O. (12.3)

Verified video answer for a similar problem:

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Draw the condensed structural formulas and give the IUPAC names for all the aldehydes and ketones that have the formula C₅H₁₀O. (12.3)