There are four amide isomers with the molecular formula C₃H₇NO. Draw their condensed structural formulas and write the IUPAC name for each.

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Welcome back everyone. The molecular formula C 4h 9, no has many amide isomers propose plausible structures of the isomers and determine the IP name of each. So essentially, as we know, we have a four carbon chain, what we can do is simply begin by drawing a general structure of an aide which consists of car bono and the amino group. We're going to begin with a primary aide. Now, what does that mean? Well, we have NH two directly bonded to carbo and this means that the remaining four carbons belong to the parent, we have 12, we and four carbons. Now we simply want to check if the molecular formula is consistent or carbon atoms, one nitrogen, one oxygen. What about hydrogens? We have two plus two plus two and then finally plus three. So this gives us nine, which means that we don't have any additional pi bonds. So we have our first structure and now we can just proceed, right. So if we want to consider another possible primary aide, we can just begin by drawing carba with NH two. And we can simply start branching the parent instead of having a total of four carbon atoms, we're going to draw three and we will just branch the chain, right. So that's another primary, am I possible and we cannot branch it any further instead. However, we can just try to draw the structure of a secondary amide, which means that we have carbona with nitrogen bonded to another carbon. And this leaves us with a 12, we carbon atoms on the left including carbonel, right. So 123 and four. And essentially, let's suppose that we add the muscle group nitro must have a total of three bonds. So we're adding hydrogen and that's how we get the third structure. Now for the four structure. Well, we can essentially think about a tertiary A mi which means that nitrogen is going to be bonded to three carbons. If we add two metal groups, we have 123 carbons. And this leaves us with an additional metal group bonded to carbo. From here, we can once again consider a potential secondary aide. Let's go ahead and throw carbo with nitrogen bonded to carbo. And instead of adding carbon atoms directly bonded to carbona, we will just add hydrogen because that's also a possibility. And this leaves us with a total of three carbons, right. So what can we do? Well, we can essentially draw a three member chain. Let's start with a branched one and then we can just draw a linear one. So first of all, we have a branched one which can be followed by a linear chain. So instead we can just draw 1 to 3 carbon atoms, right? And those would be the possibilities. In addition to that, we haven't considered a tertiary aide where the parents does not have any carbon atoms directly bonded to Carboni. So instead we can add hydrogen, which gives us we carbons bondage to nitrogen, right. So we can essentially add a one two. And of course, we want to get an ethyl group with the third carbon because nitro can only have we bonds, right, we cannot attach wi muscle groups. And that would be the final structure that we can get. You can see that there are seven possible structures we just want to name them. So the first one is beats anam, the parent has four carbon atoms, right? So it simply beats an amide. Now, what about the second structure? It is propan amide because the parent has we carbons and there's a metal group bonded to nitrogen. So we're going to state N methyl, which means that the methyl group is bonded to nitrogen, followed by the parent's name. So we're going to state pro tan aite and let's make sure that we get everything on the same line in one word. So N methyl propan amide, let's consider the next structure. In this case, there are or there is an isopropyl subit bonded to nitrogen rights. We're going to state N isopropyl, right? Because we have an isopropyl substi bonded to nitrogen. Let's carefully look at this specific structure. We can notice that nitrogen lacks hydrogen. So let's add the missing hydrogen, right, because nitrogen always must have three bonds. This is also applicable to the structure below. So now for this specific structure, we have n isopropyl and the parent is methan amide because we only have one carbon moving on, we have another structure. And in this case, if we count the number of carbon atoms in the longest continuous chain, that would be propan amide and carbon two is bonded to the metal group. So overall, that would be two metal prop pan amide well done. Now, let's focus on the next structure. Let's begin with the substituents bonded to nitrogen and there are two of them. So that'd be NN. And because there are two muscle substi that be dime and the parent has two carbons, right? So that would be the muscle Ethan amide. Now, let's take a look at the other structure. We can clearly see that there's a three member chain bonded too nitrogen. So what we're going to do is simply state that we have an propel bonnets tonight and we have a pro the chant followed by methane amide because there's only one carbon in the parent. And now let's look at the final structure once again, it is methan amide. But in this case, we have two different substitutions bonded to nitrogen. Let's alphabetically arrange them, we have ethyl and methyl. So we want to begin with the ethyl substituent. That'd be N ethyl M methyl and the parent is methan amide. Well, then we have got all of the structures needed to this problem. Thank you for watching.

There are four amide isomers with the molecular formula C₃H₇NO. Draw their condensed structural formulas and write the IUPAC name for each.

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