Identify each organic compound as an alkane, alkene, alkyne,

aromatic hydrocarbon, alcohol, ether, aldehyde, ketone, carboxylic acid, ester, or amine, and provide a name for the compound.

c.

Video transcript

Welcome back, everyone, provide the systematic name of the following compound and identify its organic functional group. Choose from alkane alkin alky aromatic hydrocarbon alcohol, ether aldehyde ketone carboxylic acid ester. Or I mean, what we can immediately notice in our structure is that we have a hydrocarbon chain. And at the end of the chain, this is where we encounter oxygens. And specifically, we can consider all of the chain on the left and alkyl group and coo H at the end of the chain as a fragment, which represents a carboxylic acid. So this structure has a general form of RC oh which is a carboxylic acid. And that's the first part or the very first step of the problem that we want to accomplish. We want to identify the class of the compound before we name it. Now that we know that this is a carboxylic acid. We want to identify the longest continuous carbon chain. We want to also understand that carbon number one is the carbon atom that is part of the carboxyl group. And now we can essentially move to the left, we will have a total of 123, 45. That's one route right, to have a total of five carbon atoms. And the second route is to go to the left, we will have 12345, no matter what route we take, we always have a total of five carbon atoms in the longest continuous carbon chain. What we want to recall is that whenever we name a carboxylic acid, we're essentially using a prefix that corresponds to the longest continuous carbon chain. So the prefix would be dancer because we have five total, right? And a prefix that corresponds to five is Panza. Now, whenever we name carboxylic acids, we are essentially naming them as alcaic acid in a general form using a suffix O IC. So in this case, if we have five carbon atoms, the prefix is Pansa, we're adding that. And so our parent, would it be 10 an aic acid? Well, then we have our parent and now we simply want to identify our substituents. Now, which route should we take? Well, essentially, it doesn't matter because if we look at carbon number three, it contains CH two ch three going to the left and CH two ch three going down. So those are equivalent routes because carbon number three contains two ethyl groups, it's essentially irrelevant, right. So let's choose the route from right to left. And this tells us that we have our first substitute bonded to carbon number two, it's called methyl because it's CH three. And now if we look at carbon number three, it contains CH two C three, which we call an ethyl substituent. So that would be two dash methyl, indicating that we have a methyl group bonded to carbon number two and three ethyl, indicating that we have an ethyl subscription bonded to carbon number three. And that we want to combine our subscriptions with the parent. Let's recall that we want to arrange our subscriptions alphabetically. So we are going to begin with three sl because we're, we're comparing E against M, that would be three el followed by T MAT. And this is where we are simply incorporating the name of the parent Panino acid. And that's our final answer. Let's label it and conclude the video. Thank you for watching.